CE04015_QA_2

MINISTRY OF SCIENCE AND TECHNOLOGY

DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION

CE-04015 GEOTECHNICAL ENGINEERING

Worked Out Examples

B.Tech. (Second Year)

Civil Engineering

PART - I

1

1.(1-06)* Define: (a) Effective size, (b) Uniformity Coefficient, (c) Coefficient of Gradation. Solution

(a) Effective size (D10): This parameter is the diameter in the particle size distribution curve corresponding to 10% finer.

(b) Uniformity coefficient (Cu): This parameter is the ratio of the maximum particle size of the smallest 60% to the effective size.

10

60u D

DC =

(c) Coefficient of gradation (Cc): This parameter is a measure of the shape of the curve between theD60 and D10 grain size.

1060

230

c DDD

C =

2.(2-06)** In its natural state, a moist soil has a volume of 0.33 ft3 and weights 39.93 lb. The over-dried weight of the soil is 34.54 lb. If Gs = 2.67, calculate ( a ) Moisture content ( % ) ( b ) Moist unit weight ( lb/ft3) ( c ) Dry unit weight ( lb/ft3) ( d ) Void ratio ( e ) Porosity ( f ) Degree of saturation ( % ) Solution

( a ) 6.15100x54.34

54.3493.39WW

s

=−

==ω ω %

( b ) 3ft/lb12133.093.39

VW

===γ

( c ) 3sd ft/lb7.104

33.054.34

VW

===γ

( d ) Volume of solids is 3

s

ss ft207.0

4.62x67.254.34

GW

V ==γ

=ω

Thus, Vv = V - Vs = 0.33 – 0.207 = 0.123 ft3

2

Volume of water is 3ft086.04.62

54.3493.39WV =

−=

γ=

ω

ωω

59.0207.0123.0

VV

es

v ===

( e ) 37.033.0

123.0VV

n v ===

( f ) %9.69699.0123.0086.0

VV

Sv

==== ω

3(2-06)*** A sample of soil weighing 30.6 kg had a volume of 0.0183 m3. When dried out in an oven its weight was reduced to 27.2 kg. The specific gravity of the solids was found to be 2.65. Determine the following: ( a ) Bulk density ( b ) Dry density ( c ) Percentage moisture content ( d ) Void ratio ( e ) Porosity ( f ) Degree of saturation ( g ) Critical hydraulic gradient. Solution

( a ) Bulk density 3m/kg16720183.0

6.30==ℑ

( b ) Dry density 3sd m/kg1486

0183.02.27

vM

===ℑ

( c ) Weight of water in sample = 30.6 - 27.2 = 3.4 kg

Moisture content 125.02.27

4.3===

sWW

m ω

( or ) percentage moisture content = 12.5%

( d ) 30034.01000

4.3 mV ==ω

30103.0100065.22.27 mxG

WV

s

ss ===

ωγ

Vv = V – Vs = 0.0183 - 0.0103 = 0.008m3

Void ratio 777.00103.0008.0

===s

v

VV

e

3

( e )Porosity 437.00183.0008.0

===VV

n v

( f ) Degree of saturation 425.0008.0

0034.0===

vr VV

S ω

( g ) Critical hydraulic gradient, 93.0777.01

165.21

1=

+−

=+−

=e

Gi sc

4.(3-06)**Classify the following soils by the AASHTO classification system.

Description Soil

A B C Percent finer than No. 10 sieve Percent finer than No. 40 sieve Percent finer than No. 200 sieve Liquid limit Plasticity index

83 48 100 48 28 82 20 6 38 20 - 42 5 Nonplastic 23

Solution Soil A According to Table 3.1, because 20% of the soil is passing through the No. 200 sieve, it falls under granular material classification – that is,A-1,A-3 or A-2. Proceeding from left to right, that is falls under category A-1-b. The group index for A-1-b is 0. So, the classification is A-1-b (0). Soil B The percentage passing through the No.200 sieves is less than 35, so the soil is a granular material. Proceeding from left to right in Table 3.1, that it is type A-1-a. The group index is 0, so the soil is type A-1-a (0). Soil C The percentage passing through the N0.200 sieve is 38, which is greater than 35%, so the soil is a silty clay material. Proceeding from left to right in Table 3.1, that is falls under category A-7. In this case, PI > LL – 30, so the soil is type A-7-6. From equation GI = ( F200 – 35 ) [ 0.2 + 0.005 ( LL – 40 )] + 0.01 ( F200 – 15 ) ( PI – 10 ) For this soil, F200 = 38, LL = 42 and PI = 23, so GI = ( 38 – 35 ) [ 0.2 + 0.005 ( 42 – 40 )] + 0.01 ( 38 – 15 ) (23 – 10 ) = 3.62∼4 Hence, the soil is type A-7-6(4)

4

5.(4-06)** A homogeneous isotropic embankment dam section is detailed in Figure, the coefficient of permeability in the x and z directions being 8 x 10-8m/s and 2 x 10-8m/s. Construct the flow net and determine the quantity of seepage through the dam.

Solution The scale factor for transformation in the x direction is:

5.082

kk

x

z ==

The equivalent isotropic permeability is: ( )zx

' kkk = = √ ( 8 x 2 ) x 10-8 = 4 x 10-8 m/s The section is drawn to the transformed scale as in Figure. The focus of the basic parabola is at point A. The basic parabola passes through point G such that: GC = 0.3 HC = 0.3 x 25 = 7.5m

The coordinates of G are: x = - 41.5, z = 20

o

2

o x4zxx −=

o

2

o x420x5.41 −=−

Hence xo = 2.28m The coordinates of a number of points on the basic parabola are now calculated: z 0 4.6 10 15 20 x 2.3 0 -8.68 -22.29 -41.5 The quantity of seepage can also be determined from equation, q = 2 k′ xo = 2 x 4 x 10-8 x 2.28 = 1.824 x 10-7 m3/s

5

6.(4-06)*** The section through a dam is shown in Figure. Determine the quantity of seepage under the dam and plot the distribution of uplift pressure on the base of the dam. The coefficient of permeability of the foundation soil is 2.5 x 10-5m/s.

6

Solution Total head at first equipotential = 0 Total head at last equipotential = 4 Total head loss between first and last equipotentials = 4 No of flow channels = 4.7 No of equipotential drops = 15

15

7.400.4105.2NN

hkq 5

d

f ×××== −

= 3.1 x 10-5 m3/s (per m) The pore water pressure is calculated at the points of intersection of the equipotentials with the base of the dam.

Point

h (m)

z (m)

h - z (m)

u = γω (h – z) (kN/m2)

1 2 3 4 5 6 7

7 1/2

0.27 0.53 0.80 1.07 1.33 1.60 1.87 2.00

-1.80 -1.80 -1.80 -2.10 -2.40 -2.40 -2.40 -2.40

2.07 2.33 2.60 3.17 3.73 4.00 4.27 4.40

20.3 22.9 25.5 31.1 36.6 39.2 41.9 43.1

7.(5-06)** A 5m depth of sand overlies a 6m layer of clay, the water table being at the surface; the permeability of the clay is very low. The saturated unit weight of the sand is 19 kN/m3 and that of the clay 20 kN/m3. A 4m depth of fill material 0f unit weight 20 kN/m3 is placed on the surface over an extensive area. Determine the total vertical stress and effective vertical stress at the center of the clay layer and center of the sand layer.( a ) immediately after the fill has been placed, assuming this to take place rapidly, ( b ) many year the fill has been placed. Solution ( a ) Immediately after the fill has been placed. Depth ( m )

σv ( kN/m2)

u ( kN/m2)

σ′v ( kN/m2)

2.5 8

20 x 4 + 19 x 2.5 = 127.5 20 x 4 + 19 x 5 + 20 x 3 = 235

9.81 x 2.5 = 24.5 9.8 x 8 + 20 x 4 = 158.4

103 76.6

( b ) Many year after the fill has been placed.

7

Depth ( m )

σv ( kN/m2)

u ( kN/m2)

σ′v ( kN/m2)

2.5 8

127.5

235

24.5

9.8 x 8 = 78.4

103

156.6 8.(5-06)*** The soil stratum is shown in Figure. The saturated unit weight of the clay and sand are 18.5 kN/m3 and 20.6 kN/m3 respectively. Above the water table the unit weight of the sand is 17.4 kN/m3. ( a ) Plot the total vertical stress and effective vertical stress at against depth. ( b ) If sand a height of 0.9 m above the water table is saturated with capillary water, how are the above stress effected?

Solution (a)

Depth (m)

σv (kN/m2)

u (kN/m2)

σv′

(kN/m2) 0

0.9 2.4 5.4

0 ( 17.4 x 0.9) = 15.66

15.66 + (20.6 x 1.5) = 46.56 46.56 + (18.5 x 3 ) = 102.06

0 0

(9.81 x 1.5) = 14.715 (9.81 x 4.5) = 44.145

0 15.66 31.845 57.915

(b)

Depth (m)

σv (kN/m2)

u (kN/m2)

σv′

(kN/m2) 0

0.9 2.4 5.4

0 20.6 x 0.9 = 18.54

18.54 + (20.6 x 1.5) = 49.44 49.44 + (18.5 x 3 ) = 104.94

0 0

14.715 44.145

0 18.54 34.725 60.795

8

9.(6-06)** For the stressed soil element shown in Figure. Determine the magnitude of the principal stress, normal and shear stress on plane AE. ( a ) By using Mohr’s circle method, ( b ) By using equation.

Solution ( a ) By using Mohr’s circle method On plane AD: Normal stress = 90 lb/in2 Shear stress = -60 lb/in2 On plane AB: Normal stress = 150 lb/in2 Shear stress = 60 lb/in2 The Mohr’s circle is plotted in Figure. From the plot,

Major principal stress = 187.1 lb/in2 Minor principal stress = 52.9 lb/in2 The coordinates of point Q gives the stress on the plane AE.

9

Normal stress = 60 lb/in2

Shear stress = 30 lb/in2 ( b ) By using equation

2xy

2

xyxy31 τ

2σ-σ

2σσ

σ,σ +

±+

=

( )22

602

901502

90150−+

−

±+

=

σ1 = 187.1 lb/in2 σ3 = 52.92 lb/in2

θτ+θσ−σ

+σ+σ

=σ 2sin2cos22

xyxyn

( ) ( ) ( )45x2sin6045x2cos2

901502

90150−+

−+

+=

= 60 lb/in2

θτ−θσ−σ

=τ 2cos2sin2

xyn

( ) ( ) ( )45x2cos6045x2sin2

90150−−

−=

= 30 lb/in2 10.(6-06)** Consider a point load P = 2000lb. Plot the variation of vertical stress increase ∆pz, in the x-z plane ( this is y = 0 ) for ( a ) z = 2ft z = 0, ±0.5, ±1, ±1.5, and ± 2ft ( b ) z = 4ft z = 0, ±0.5, ±1, ±1.5, and ± 2ft ( c ) z = 2ft z = 1,2,3,5, and 10ft Solution

10

a. and b. The following table can be prepared: z

( ft ) x

( ft ) r

( ft )

r/z

I1 ∆pz = ( P/z2)I1

( lb/ft2) 2 0

±0.5 ±1.5 ±1.5 ±2.0

0 ±0.5 ±1.0 ±1.5 ±2.0

0 ±0.25 ±0.50 ±0.75 ±1.00

0.4775 0.4103 0.2733 0.1565 0.0844

238.8 205.2 136.7 78.3 42.2

4 0 ±0.5 ±1.5 ±1.5 ±2.0

0 ±0.5 ±1.0 ±1.5 ±2.0

0 ±0.5 ±1.0 ±1.5 ±2.0

0.4775 0.4103 0.2733 0.1565 0.0844

59.7 51.3 31.2 19.6 10.6

b. As in part (a) and (b), the following table can be prepared.

x = r ( ft )

z ( ft )

r/z

I1

∆pz = ( P/z2)I1 ( lb/ft2)

2 1 2 3 5 10

2 1

0.67 0.4 0.2

0.0085 0.0844 0.1889 0.3294 0.4329

17 42.2 42.0 26.4 8.7

The stress increases are plotted in Figure.

11

PART - II

1.(1-7)*** The following results were obtained at failure in a series of drained triaxial tests on fully saturated clay specimens originally 38mm diameter by 76 mm long. Determine the values of the shear strength parameters c' and φ'. All round pressure (kN/m2) 200 400 800 Axial compression (mm) 7.22 8.36 9.41 Axial load (N) 480 895 1300 Volume change (ml) 5.25 7.40 9.30 Solution The original values of A0 = (π/ 4) x d2 = 1134 mm2 l0 = 76 mm V0 = 86200 mm3

σ3'

(kN/m2) ∆V/V0 ∆l / l0 Area

(mm2) Load (N)

σ1 - σ3

(kN/m2) σ1'

(kN/m2) 200 400 600

0.061 0.086 0.108

0.095 0.110 0.124

1177 1165 1155

480 895 1300

408 768 1126

608 1168 1726

From graph, c' = 15 kN/m2, φ' = 28o