Exercise of Soil Mechainics

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1

Exercise I.1 A moist sample of soil in a bottle had a mass of 25.25g. And the bottle, when empty, had a

mass 14.39g. After dry in an oven for 24 hours, the bottle and soil sample had a mass of 21.63 g. find the water content of the soil.

Solution

Let T mass of a bottle m total mass of soil ms mass of soil particles

We have: m + T = 25.25 g , T = 14.39 g ms + T = 21.63 g m = 25.25 g – T = 25.25 – 14.39 = 10.86 g ms = 21.63 g – T = 21.63 – 14.39 = 7.24 g mw = 10.86 – 7.24 = 3.62 g

3.62

0.5 50%7.24

w⇒ = = = S0 50%w=

Exercise I.2

A dry soil sample has the void ratio e = 0.65 and unit weight of the solid

particles 326 /s kN mγ = . Determine its total unit weight.

Solution

Determine the total unit weight (γ )

dγ γ= (Saturated soil) , e = 0.65 , 326 /s kN mγ =

32615.75 /

1 1 0.65s

d kN me

γγ γ⇒ = = = =+ +

S0 315.75 /kN mγ =

Exercise I.3

The undisturbed soil sample was taken from a soft clay layer, which was under ground water level. Some measurements were done on a part of this sample in laboratory as indicated in the following table:

a. Determine the unit weight γ and the water content w. b. Determine the void ratio e. c. To verify the degree of saturation, we measure the unit weight of solid particle,

γs = 27 kN/m3 , calculate the degree of saturation Sr.

Solution

a. Determine the unit weight γ and the water content w.

By the formula: (I.1) ⇒ P

Vγ = , ( 3 5 30.47 0.47 10 , 3.13 10P N kN V m− −= = × = × )

⇒ 3

35

0.47 1015 /

3.13 10kN mγ

−

−

×= =×

Total Weight Total Volume Weight after dry at 105 oC

0.47 N 3.13 x 10-5 m3 0.258 N

⇒

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By the formula: (I.6) ⇒ 100% 100%w s

s s

P P Pw

P P

−= × = × , ( 0.258sP N= )

⇒ 0.47 0.258

100% 82.17%0.258

w−= × =

b. Determine the void ratio e.

Supposed that the soil is saturated: satγ γ=

By the formula: (I.15) ⇒ 31 18.24 /

1 1 0.8217d kN mw

γ = = =+ +

By the formula: (I.20) ⇒ sat d wnγ γ γ= + × 15 8.24

0.67710

sat d

w

nγ γ

γ− −

⇒ = = =

By the formula: (I.11) ⇒ 0.677

2.11 1 1 0.677

e nn e

e n= ⇒ = = =

+ − −

c. To verify the degree of saturation, we measure the unit weight of solid particle,

γs = 27 kN/m3 , calculate the degree of saturation Sr.

By the formula: (I.14) ⇒ 27

1 1 2.271 8.24

s sd

d

ee

γ γγγ

= ⇒ = − = − =+

By the formula: (I.18) ⇒ 0.8217 27

0.97 97%10 2.27

sr

w

wS

e

γγ

× ×= = = =× ×

Exercise I.4

The water content of a saturated soil w and the unit weight of solid particles γs have known, determine:

a. its dry unit weight (γd) b. its void ratio (e).

Solution

a. Determine its dry unit weight (γd) We have known w & γs and a soil is saturated soil so we get: Va = 0 & Sr = 1

0aV

VA

V⇒ = =

By the formula: (I.18) ⇒ ( )

( )( )

( )1 1 0w s V w s

dw s w s

A

w w

γ γ γ γγ

γ γ γ γ× × − × × −

= =+ × + ×

So: ( )w s

dw sw

γ γγγ γ

×=+ ×

b. Determine its void ratio (e).

Soil is saturated so Sr = 1 s sr

w w

w wS e

e

γ γγ γ

× ×⇒ = ⇒ =

×

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Exercise I.5 A pycnometer having a mass of 620g was used to determine the specific gravity of an oven-dried sample of soil. If the combined mass of the soil sample and the pycnometer was 1600g and the mass of the pycnometer with the sample and filled up with water was 2112g, determine the specific gravity of the soil particles. The mass of the pycnometer when filled with water only was 1495g.

Solution

Determine the specific gravity (Gs) We have: T = 620 g mp = T + mw = 1495 g m’p = T + ms + mw = 2112 g T + ms = 1600 g ⇒ ms = 1600 – 620 = 980 g By the formula (I.22):

⇒ 980

2.70' 1495 980 2112

sS

p s p

mG

m m m= = =

+ + + −

So The specific gravity is 2.70

Exercise I.6 A saturated sample of soil was found t have a water content of 27% and a bulk density of 1.97 t/m3. Determine the dry density and the void ratio of the soil, and the specific gravity of the particles.

Solution

Determine the dry density, void ratio and the specific gravity - Dry density:

We have: 327% , 1.97 /w t mρ= =

Based on the equation (I.15): 31.971.55 /

1 1 0.27d t mw

ρρ = = =+ +

- Void Ratio:

Since the soil is saturated: satρ ρ=

Based on the equation (I.20): 1.97 1.55

0.421

sat dsat d w

w

n nρ ρρ ρ ρ

ρ− −= + × ⇒ = = =

Based on the equation (I.11): 0.42

0.7241 1 0.42

ne

n= = =

− −

- The specific gravity:

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Exercise IV.1 The following data were obtained from a test on a sample of sand using a constant head permeameter, which has 100mm diameter with manometer tapping points 200mm a part.

Calculate the coefficient of permeability (k) of the sample.

Solution

By the formula: Q L

kA h

×=× ∆

, 2

4

dA

π= ( )2

23.14 0.10.00785

4m

×= =

⇒ 6

61

145 10 0.244509.25 10 / min

0.00785 0.083k m

−−× ×= = ×

×

6

62

135 10 0.244668.70 10 / min

0.00785 0.077k m

−−× ×= = ×

×

6

63

163 10 0.246661.42 10 / min

0.00785 0.089k m

−−× ×= = ×

×

6

64

154 10 0.245622.87 10 / min

0.00785 0.089k m

−−× ×= = ×

×

( ) 661 2 3 4

44509.25 44668.70 46661.42 45622.87 1045365.56 10 / min

4 4

k k k kk m−+ + ++ + += = = ×

47.56 10 /m s−= ×

So 47.56 10 /k m s−= ×

Water collected In min (ml)

145 135 163 154

Loss of head between

manometer (mm) 83 77 89 86

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Exercise IV.2 In a constant head permeameter test the following results were obtained:

- Duration of test is 4 min - Quantity of water collected is 300 ml - Head difference in manometer is 50 mm - Distance between manometer tapping is 100 mm - Diameter of test sample is 100 mm.

Determine the coefficient of permeability in m/s.

Solution

Rate of flow : 300

75 / min4min

mlQ ml= = 6 31.25 10 /m s−= × 6 31.25 10 /m s−= ×

50 0.05h mm m∆ = = , 100 0.1L mm m= =

( )2223.14 0.1

0.007854 4

dA m

π ×= = =

6

41.25 10 0.13.18 10 /

0.00785 0.05k m s

−−× ×

⇒ = = ××

So 43.18 10 /k m s−= ×

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Exercise IV.3 A falling head permeameter has a diameter of 75mm and the length of the soil sample is mm. The diameter of the standpipe is mm. During the test, the head decrease from 1300mm to 800mm in 135s. Calculate the coefficient of permeability of the soil in m/s.

Solution

We have: D = 75mm , L = 150mm d = 15mm , h1 = 1300mm t = 135s , h2 = 800mm k = ?

by the formula:

2

1 12

2 2

4ln ln

4

dLh ha L

kDA t h h

t

π

π

× ×= × = × × ×

( )( )

2

2 52

15 150 1300ln 2.15 10 / 2.15 10 /

80075 135mm s m s− −× = × = × = ×

×

So 52.15 10 /k m s−= ×

Exercise IV.4

An undisturbed soil sample was test in a falling head permeameter results were: - Initial head in a standpipe is 1500mm - Final head of water in standpipe is 605mm - Duration of test is 281s - Sample diameter is 100mm - Standpipe diameter is 5mm. - Determine the permeability of the soil in m/s.

Solution

We have: h1 = 1500mm , h2 = 605mm , L = 150mm

t = 281s , D = 100mm , d = 5mm

2

1 12

2 2

4ln ln

4

dLh ha L

kDA t h h

t

π

π

× ×= × = × × ×

( )

23 6

2

5 150 1500ln 1.2 10 / 1.2 10 /

605100 281k mm s m s− −×

⇒ = × = × = × ×

S0 61.2 10 /k m s−= ×

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Exercise IV.5 The result of constant-head permeability test for a fine sand sample having a diameter of 150mm and a length of 300mm are as follow:

- Constant head difference is 500mm - Time of collection water is 5min - Volume of water collected is 350cm3 - Temperature of water during test is 24oC - Determine the permeability of the soil in m/s at 20oC.

Solution

We have: D = 150mm ( )2

23.14 15017662.5

4A mm

×⇒ = =

L = 300mm , H = 500mm

t = 5min = 300s , V = 350cm3 = 3.5 x 105 mm3

52 33.5 10

11.67 10 /300

Q mm s×

⇒ = = ×

22 5

24

11.67 10 3003.96 10 / 3.96 10 /

17662.5 500o C

k mm s m s− −× ×⇒ = = × = ×

×

24

20 2420

o

o o

o

CC C

C

k kηη

= ×

25

20

0.8909

1.0019

o

o

C

C

ηη

=

= 5 0.111 , 4o oC C x⇒ = =

4 0.111

0.08885

x×

⇒ = = 24 20

4 1.0019 0.0888 0.9131o o

o

C CCη η⇒ = − = − =

5 50.9131

3.96 10 3.61 10 /1.0019

k m s− −⇒ = × = ×

S0 53.61 10 /k m s−= ×

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Exercise V.1

Draw the mohr’s circle and determine the normal and shear stresses, σα and τα , on the plane of inclination α = 35ο . ( below figure)

1 3 52 1220 2

2 2R kPa units

σ σ− −= = = =

1 3 52 1232 3.2

2 2C kPa units

σ σ+ += = = =

S0 35

35

3.7 38.8

1.8 18.8

o

o

units kPa

units kPa

στ

= =

= =

σ1 = 52 kPa

σ3 = 12 kPa

35ο Horizontal Plane

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Exercise V.2

Draw the mohr’s circle and determine the normal and shear stresses, σα and τα , on the plane of inclination α = 60ο . ( below figure)

1 3 100 3035 1.75

2 2R kPa units

σ σ− −= = = =

1 3 100 3065 3.25

2 2C kPa units

σ σ+ += = = =

S0 60

60

2.37 47.4

1.52 30.4

o

o

units kPa

units kPa

στ

= =

= =

σ1 = 100 kPa

σ3 = 30 kPa

60ο Horizontal Plane

1.52

2.37

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Exercise V.3

1 3 52 1220 2

2 2R kPa units

σ σ− −= = = =

1 3 52 1232 3.2

2 2C kPa units

σ σ+ += = = =

S0 35

35

3.88 38.8

1.88 18.8

o

o

units kPa

units kPa

στ

= =

= =

35ο

20ο

σ3 = 12 kPa

σ1 = 52 kPa

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Exercise V.4

Based on the data of the figure, find the values and the direction of major and minor principle stresses.

Solution

4 kPa

2 kPa8 kPa

-2 kPa

2 kPa

4 kPa

-2 kPa

8 kPa

45?

y

y

x

x

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Exercise V.5

There is a soil mass; the water level varies during the year from +2.5m to -2.5m (we

considered and the variation of water level is linear). The unit weight of soil is 316 /kN mγ = , draw

the curve representing the variations of the effective stress σ’ during the whole year (σ’ = f(t) ) at a depth of 10m.

Solution

Depth at +2.5m Depth at 0m Depth at -2.5m

σ ’ (kPa) (16-10)x10 = 60 (16-10)x10 = 60 (16x2.5)+(16-10)x7.5

= 85

+2.5

-2.5

-10

0

+2.5

-2.5

-10

0

σ’ 10 σ’ 10

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Exercise V.6

Draw the evolution of total and effective stresses (σ and σ ’) according to the depth z.

Solution

Depth 2m Depth 5m Depth 12m

σ (kPa) 22x2 = 44 (22x2)+(22x3) = 110 (22x2)+(22x3)+(20x7) = 250

σ’ (kPa) 22x2 = 44 (22x2)+(22-10)3 = 80 (22x2)+(22-10)3+(20-10)7 =

150

0

5

12

2 Fine Sand

γ = 22 kN/m3

Silt

γ = 20 kN/m3

Z (m) γw = 10 kN/m3

Evolution Diagram of Total & Effective Stress

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Exercise V.7 A Layer of sand 6m deep overlies a thick bed of clay. Draw diagram indicating the total and effective stresses and pore water pressure on horizontal planes to a depth of 10m below the ground surface: - If the water table is at ground level - If the water table is at 1m below ground level and the sand above remains saturated with capillary moisture. - If the water table is at the top of the clay and the sand above dry.

- Saturated unit weight of sand 20.9 kN/m3 - Dry unit weight of sand 17.4 kN/m3 - Unit weight of clay 17.8 kN/m3 - g = 9.81 m/s2

Solution

- If the water table is at ground level


u (kPa) 0 (9.81x6) = 58.86 (9.81x10) = 98.10

σ (kPa) 0 (20.9x6) = 125.40 (20.9x6)+(17.8x4) =

196.60

σ’ (kPa) 0 (20.9-9.81)x6 = 66.54 (20.9-9.81)x6 +

(17.8-9.81)x4 = 98.50

0

6

10

Sand

Clay

Diagram of Total & Effective Stresses And Pore Water Stresses

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- If the water table is at 1m below ground level and the sand above remains

saturated capillary moisture


u (kPa) 0 (9.81x5) = 49.05 (9.81x9) = 88.29

σ (kPa) (20.9x1) = 20.90 (20.9x1)+(20.9x5) = 125.40 (20.9x1)+(20.9x5)+(17.8x4)

= 196.60

σ’ (kPa) (20.9x1) = 20.90 (20.9x1)+(20.9-9.81)x5 =

76.35 (20.9x1)+(20.9-9.81)x5 +(17.8-9.81)x4 = 108.31

0

6

10

Sand

Clay

1


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- If the water table is at the top of the clay and the sand above dry

Depth 6m Depth 10m

u (kPa) 0 (9.81x4) = 39.24

σ (kPa) (17.4x6) = 104.40 (17.4x6)+(17.8x4) = 175.60

σ’ (kPa) (17.4x6) = 104.40 (17.4x6)+(17.8-9.81)x4 =

136.36

0

6

10

Sand

Clay


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Exercise VI.1 A uniform stress of 120 kPa was applied at ground surface due to a rectangular footing of 3 x 4m.

1. Calculate the vertical incremental stress ( )Zσ∆ beneath a corner of the footing at a depth

of 2m.

2. Calculate the vertical incremental stress ( )Zσ∆ under the centre of the footing at a depth

of 2m.

Solution

1. Calculate the vertical incremental stress ( )Zσ∆ beneath a corner of the footing at a depth

of 2m.

By the formula: (VI.11) Z oq Iσ∆ = ×

And: 120oq kPa=

3 , 4 , 2B m L m Z m= = =

⇒ 3 4

1.5 , 22 2

B Lm n

Z Z= = = = = =

1.5 1.4 1.6m is between m and m= = =

By the figure VI.5: we get

1.4 0.221

1.6 0.226

m I

m I

= → == → =

0.221 0.226

1.5 , 0.22352

m I+

⇒ = = =

2120 0.2235 26.82 /Z kN mσ⇒ ∆ = × =

SO 226.82 /Z kN mσ∆ =

2. Calculate the vertical incremental stress ( )Zσ∆ under the centre of the footing at a depth

of 2m. At the Centre: (Z = 2m)

3 4

' 1.5 , ' 22 2

B LB L

Z Z⇒ = = = = = =

' 1.5 ' 2

' 0.75 , ' 12 2

B Lm n

Z Z⇒ = = = = = =

0.75 0.7 0.8m is between m and m= = =

By the figure VI.5: we get

' 0.7 0.15

' 0.8 0.16

m I

m I

= → == → =

0.15 0.16

0.75 , 0.1552

m I+

⇒ = = =

4 120 0.155 74.4Z kPaσ⇒ ∆ = × × =

SO 74.4Z kPaσ∆ =

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Exercise VI.2 A uniform stress of 120 kPa was applied at ground surface from a circular reservoir of 3.91m diameter.

1. Calculate the vertical incremental stress ( )Zσ∆ at a depth of 2m under the centre of the

reservoir.

2. Determine the vertical incremental stress ( )Zσ∆ at a depth of 2m under the perimeter of

the reservoir. Solution

1. Calculate the vertical incremental stress ( )Zσ∆ at a depth of 2m under the centre of the

reservoir.

100Z o

Iqσ∆ = × , ( ) ( )120 , 2oq kPa Z m= =

d = 3.91 m ⇒ R = 1.955 m By the figure: VI.7

⇒ 2

1.0231.955

Z

R= = ,

00

1.955

r

R= =

By the Interpolation:

2 29

1 63

ZI

RZ

IR

= ⇒ =

= ⇒ =

⇒ 1 = - 34 ⇒ 0.023 = - 0.782 1.023 , 63 0.782 62.22Z

IR

⇒ = = − =

2120 62.2274.66 /

100Z kN mσ ×⇒ ∆ = = So 274.66 /Z kN mσ∆ =

2. Determine the vertical incremental stress ( )Zσ∆ at a depth of 2m under the perimeter of

the reservoir.

At the perimeter:

100Z o

Iqσ∆ = × , ( ) ( )120 , 2oq kPa Z m= =

By the figure: VI.7

⇒ 2

1.0231.955

Z

R= = ,

1.9551

1.955

r

R= =

By the Interpolation:

2 19.5

1 33

ZI

RZ

IR

= ⇒ =

= ⇒ = 1 13.5 0.023 0.311⇒ = − ⇒ = −

1.023 , 33 0.311 32.7Z

IR

⇒ = = − =

2120 32.7039.24 /

100Z kN mσ ×⇒ ∆ = = So 239.24 /Z kN mσ∆ =

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Exercise VI.3 There is an embankment of a dam represented in the figure below. It is supported that the average density of soil is 2t/m3. Determine the additional vertical stress under the center of the embankment at a depth of 5m and 10 m.

Solution Determine the additional vertical stress under the center of the embankment at a depth of 5m.

We have: 20 2 9.81 5 98.10 /q h g h kN mγ ρ= × = × × = × × =

a = 10 m , b = 4 m By the figure: VI.8

102

54

0.85

a

Zb

Z

= =

= = 0.46I⇒ =

( ) 22 98.1 0.46 90.25 /Z kN mσ⇒ ∆ = × =

So 290.25 /Z kN mσ∆ =

Determine the additional vertical stress under the center of the embankment at a depth of

10m. By the figure: VI.8

101

104

0.410

a

Zb

Z

= =

= = 0.358I⇒ =

( ) 22 98.1 0.358 70.24 /Z kN mσ⇒ ∆ = × =

So 270.24 /Z kN mσ∆ =

4 m 4 m

2

1

5 m

10 m

ρ = 2 t/m? 5 m3

5

m

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Exercise VI.4

A normally consolidated soft clay layer is 15m thick with natural moisture content of 45 %. The clay has a saturated unit weight of 17.2 kN/m3, a particle specific gravity of 2.68 and a compression index of 0.495. A foundation load will increase a vertical stress of 10 kN/m2 to the centre of the layer. Determine an approximate value for the settlement of the foundation if the ground water level is at the surface of the clay.

Solution Determine an approximate value for the settlement of the foundation if the ground water

level is at the surface of the clay.

We have: Wn = 45 % , Gs = 2.68

317.2 /sat kN mγ = , 210 /kN mσ∆ =

0.495 , 15CC H m= =

It is a normally consolidated soil so:

(VI.37) 0

'lg

1 'zfC

Czo

CS H

e

σσ

= × × +

(I.18) 0

0.45 2.681.206

1n s n s

ro r

w G w GS e

e S

× × ×= ⇒ = = = ( Sr = 1 soil is saturated)

And ( ) ( ) 215' 17.2 10 7.5 54 /

2sat w kN mσ γ γ= − = − =

Final vertical effective stress 2' ' 10 54 64 /zf kN mσ σ σ⇒ = ∆ + = + =

Initial vertical effective stress 2' ' 54 /zo kN mσ σ⇒ = =

So 0.495 64

15 lg1 1.206 54CS

= × × + = 0.248m = 24.8 cm

15m

Clay , Wn=45 % , GS = 2.8

γsat = 17.2 kN/m3 , Cc = 0.495

∆σ = 10 kN/m2

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21

Exercise VI.5

A soil profile is shown in the next slide. If a uniformly distribution load, ∆σ , is applied at the ground surface, what is the settlement of the clay layer caused by primary consolidation if: 1. The clay is normally consolidated.

2. The preconsolidated pressure is σ’P = 190 kN/m2

3. The preconsolidated pressure is σ’P = 170 kN/m2 Use: Cc = 0.009x (WL-10)

Cr = 6

cC

Solution

1. Determine the settlement if the clay is normally consolidated (Sc)

(VI.37): 0

'lg

1 'zfC

Czo

CS H

e

σσ

= × × + , eo = 0.8

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

2

' 2 4 2

14 2 18 10 4 18 10 2 76 /

d sand sat sand w sat clay w

kN m

σ γ γ γ γ γ= × + − × + − ×

= × + − + − =

And ( ) ( )0.009 10 0.009 40 10 0.27C LC W= × − = × − =

⇒ 2' ' 100 76 176 /zf kN mσ σ σ= ∆ + = + =

2' ' 76 /zo kN mσ σ= =

0.27 1764 lg 0.215 215

1 0.8 76CS m mm = × × = = +

So 0.215 215CS m mm= =

2. Determine the settlement if the preconsolidated pressure is σ’P = 190 kN/m2

' ' ' , 78 178 190zo zf Pσ σ σ⇒ < < < < ( so it is the over-consolidated case I )

By the equation (VI.38):

We get: 0

'lg

1 'zfr

Czo

CS H

e

σσ

= × × + ,

0.270.045

6 6c

r

CC = = =

⇒ 0.045 178

4 lg 0.036 361 0.8 78CS m mm

= × × = = +

So 0.036 36CS m mm= =

2m

4m

4m

∆σ = 100 kN/m2

Sand, γd = 14 kN/m3

Sand, γsat = 18 kN/m3

Clay, eo = 0.8 , WL=40

γsat = 18 kN/m3

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22

3. Determine the settlement if the preconsolidated pressure is σ’P = 170 kN/m2

' ' ' , 78 170 178zo P zfσ σ σ⇒ < < < < ( so it is the over-consolidated case II )

By the equation (VI.39):

We get: 0 0

' 'lg lg

1 ' 1 'p zfCr

Czo p

CCS H H

e e

σ σσ σ

= × × + × × + +

0.045 170 0.27 1784 lg 4 lg

1 0.8 78 1 0.8 170

0.034 0.012 0.046 46

CS

mm mm

= × × + × × + +

= + = =

So 0.046 46CS m mm= =

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23

Exercise VI.6 A 12m of clay layer is drained by two sand layers.

Knowing that a structure is creates an average vertical stress of 100 kPa on the compressible layer (clay) , we ask to:

1. Estimate the consolidation settlement (∆h or Sc) of the clay layer. 2. Estimate the necessary times to obtain a settlement corresponding to 50 % and 90 % of

consolidation.

Solution

1. Estimate the consolidation settlement (∆h or Sc) of the clay layer

By the equation (I.12): 00

11

s sd

d

ee

γ γγγ

= − ⇒ =+

326.616.42 /

1 0.62kN m= =

+

By the equation (I.11): 0

0

0.620.38

1 1 0.62

en

e= = =

+ +

By the equation (I.20): ( )sat d wnγ γ γ= + ×

( ) 316.42 0.38 10 20.22 /kN m= + × =

( ) ( ) ( ) 2' 18 1.5 18 10 3 20.22 10 6 112.32 /kN mσ⇒ = × + − × + − × =

' ' 100 112.32 212.32zf kPaσ σ σ⇒ = ∆ + = + = , ' ' 112.32zo kPaσ σ= =

By the equation (VI.37): 0

'lg

1 'zfC

Czo

CS H

e

σσ

= × × +

0.25 212.32

12 lg 0.512 5121 0.62 112.32

m mm = × × = = +

So 0.512 512CS m mm= =

0

1.5m

4.5m Sand, γ = 18 kN/m3

Clay, eo = 0.62 , Wn=23.2% , Cc = 0.25

γs = 26.6 kN/m3 , Cv = 8.10-8 m2/sec 16.5m

h1

h2

h3

Dense Sand

Sand

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24

2. Estimate the necessary times to obtain a settlement corresponding to 50 % and 90 % of consolidation.

- For 50 %

We have: CV = 8.10-8 m2/sec

It is the open layer: 12

62 2

Hh m⇒ = = =

By the equation (VI.35): 2 2

50 5050

50

V VV

V

T h T hC t

t C

× ×= ⇒ =

On page 106 chapter VI: 50 0.197VT =

( )2

450 8

0.197 68865 10 1026 9

8 10t s days h−

×⇒ = = × =

×

So if SC = 512 mm ⇒ t50 = 1026 days 9 h

- For 90 %

By the equation (VI.36): 2 2

90 9090

90

V VV

V

T h T hC t

t C

× ×= ⇒ =

On page 106 chapter VI: 90 0.848VT =

( )2

590 8

0.848 63816 10 4416 16

8 10t s days h−

×⇒ = = × =

×

So if SC = 512 mm ⇒ t90 = 4416 days 16 h

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25

Exercise VII.1

Determine the cohesion and the angle of friction of the soil, with respect to total stress.

Solution

Determine the cohesion and the angle of friction of the soil, with respect to total stress.

By the upper curve we get: C = 57 , ϕ = 22ο

Undrained shear box tests were carried out on a series of soil samples with the following results:

Test No. Total Normal Stress (kPa) Total Shear Stress ar

failure (kPa)

1 100 98

2 200 139

3 300 180

4 400 222

100 200 300 400

98

139

180

222

τ

σ

(kPa)

(kPa)

C=57

ϕ=22? o

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26

Exercise VII.2

Determine the internal friction angles at peak and residual states from the following direct shear test data. The shear box is 64mm square in plan.

Vertical load (kg) Shear force at peak (N) Shear force at residual (N)

50 399.8 228.7

100 801.9 457.4

150 1214.0 686.1

Solution

Determine the internal friction angles at peak and residual states.

Vertical load (N) Shear force at peak (N) Shear force at residual (N) 490.5 399.8 228.7

981.0 801.9 457.4

1471.5 1214.0 686.1

So The internal friction angles at peak = 39o The internal friction angles at residual = 25o

Peak & Residual shear force (N)

Vertical load (N)

Internal friction angle at peak

39o

25o

Internal friction angle at residual

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27

Exercise VII.3

The table below was recorded the results of an undrained shear box test carried out on a set of undisturbed soil samples. The dimension of shear box in plan is 60mm square. Determine the strength parameter of the soil in terms of total stress.

Solution

Normal load (kN) 0.2 0.4 0.8

Strain (ε) , (%) Shearing force (N)

0 0 0 0 1 21 33 45 2 46 72 101 3 70 110 158 4 89 139 203 5 107 164 248 6 121 180 276 7 131 192 304 8 136 201 330 9 138 210 351 10 137 217 370 11 136 224 391 12 - 230 402 13 - 234 410 14 - 237 414 15 - 236 416 16 - - 417 17 - - 417 18 - - 415

Shear Strain (%)

Shear Force (N)

417 N

237 N

138 N

Normal Force 0.8 (kN)



Determine the strength parameter of the soil in terms of total stress.

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28

Exercise VII.4

The results of an unconfined compression test were recorded and listed in the tables below. Based on this data, plot stress-strain relationship of the soil sample and determine its undrained cohesion.

General Data

Soil Description: CL Mass of Tare: 19.11g Sample Number: 09 Mass of Tare &moist sample: 68.75g Moist mass of specimen: 20.41g Mass of Tare &dry sample: 60.48g Specimen length: 70mm Moisture Content: 20% Specimen diameter: 35mm Sample Area (mm2): 961.625

Test Data

Axial Displacement (mm)

Piston force (N) Axial Displacement

(mm) Piston force (N)

0 0 6.25 229 0.5 50 7 229 1 90.5 7.5 228 1.5 113.5 8 225 2 136 8.5 225 3 181.5 9.25 225 3.5 191 10 225 4.25 206 10.5 225 5 218 - -

Solution

Plot stress-strain relationship of the soil sample and determine its undrained cohesion.

A = 961.625 mm2 = 9.616x10-4 m2

ol L

εε = , 1

o

l

AA

ε=

− ,

P

Aσ =

Piston Force(kN)

Axial

Strain (εl) Cross

Section Area Normal

Stress(kPa) Piston

Force(kN) Axial

Strain (εl) Cross

Section Area Normal

Stress(kPa)

0 0 0.00096163 0 0.229 0.082143 0.00104768 218.577

0.05 0.007143 0.00096854 51.624 0.229 0.089286 0.0010559 216.876

0.0905 0.014286 0.00097556 92.767 0.229 0.1 0.00106847 214.325

0.1135 0.021429 0.00098268 115.500 0.228 0.107143 0.00107702 211.695

0.136 0.028571 0.00098991 137.386 0.225 0.114286 0.00108571 207.238

0.1815 0.042857 0.00100468 180.654 0.225 0.121429 0.00109453 205.567

0.191 0.05 0.00101224 188.691 0.225 0.132143 0.00110805 203.060

0.206 0.060714 0.00102378 201.214 0.225 0.142857 0.00112190 200.553

0.218 0.071429 0.00103560 210.507 0.225 0.15 0.00113132 198.882

2max 218.577109.289 /

2 2uC kN mσ= = =

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29

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30

Exercise VII.5 The following results were obtained from a series of unconsolidated undrained triaxial tests

carried out on undisturbed samples of a compacted soil.

Cell Pressure (kPa) Additional axial load at failure (N)

200 342

400 388

600 465

Each sample, originally 76mm long and 38mm in diameter, experienced a vertical deformation of 5.1mm. Draw the failure envelope and determine the coulomb equation for the shear strength of the soil in terms of total stresses.

Solution

Draw the failure envelope and determine the coulomb equation for the shear strength of the soil in

terms of total stresses.

5.1

0.067176

ol L

εε = = = , ( )5.1 , 76o mm L mmε = =

⇒

( )2

3 2

0.038

4 1.215 101 1 0.0671

o

l

AA m

π

ε−= = = ×

− − , ( )38D mm=

Deviater stress 1 3

N

Aσ σ= − = 1 3

N

Aσ σ⇒ = +

Cell Pressure , 3σ ,(kN/m2) Additional axial (N) Normal Stress , 1σ , (kPa)

200 342 481

400 388 719

600 465 983

200 300 800100 400 500 600 700 900 1000

100

200

300

(τ)

(σ)

7?

By the Mohr Circle: ⇒ 100uC kPa= , 7o

uϕ =

Colomb equation is: tan 100 tan 7ou uCτ σ ϕ σ= + × = +

100 0.123σ= +

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31

Exercise VII.6 A sample of clay was subjected to an unconsolidated undrained triaxial test with a cell

pressure of 100kN/m2 and the additional axial stresses necessary to cause failure was found to be 188 kN/m2, assuming u = 0. Determine the value of additional axial stress that would require causing failure of a further sample of the soil if it was tested undrained with a cell pressure of 200 kN/m2.

Solution

Determine the value of additional axial stress that would require causing failure of a further sample of the soil if it was tested undrained with a cell pressure of 200 kN/m2.

3 100kPaσ =

1 3 1188 288kPa kPaσ σ σ− = ⇒ =

If 3 200kPaσ =

⇒ 1 3 1188 388kPa kPaσ σ σ− = ⇒ =

Additional axial stress = 188 kPa 388 200

942uC kPa−

⇒ = =

200 300100 400 500

100

200

300

(σ)

Cu = 94kPa

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32

Exercise VII.7 A Series of drained triaxial tests were performed on a soil. Each test was continued until

failure and the effective principle stresses for the tests were in table below.

Test No. '3σ (kPa) '

1σ (kPa)

1 200 570

2 300 875

3 400 1162

Plot the relevant Mohr stress circles and hence determine the strength envelope of the soil with respect to effective stress.

Solution

Mohr Circle:

By the Mohr Circle we get the strength envelop: 29oϕ =

200 300 800100 400 500 600 700 900 1000 1100 1200

100

200

300

(τ)

(σ)

29?

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33

Exercise VII.8 A vane, used to test a deposit of soft alluvial clay, required a torque of 67.5 Nm. The dimensions of the vane were: D = 75 mm , H = 150 mm , determine a value for the undrained shear strength of the clay.

Solution

Determine a value for the undrained shear strength of the clay We have: T = 67.5 Nm = 0.0675 kN.m D = 75 mm = 0.075 m H = 150 mm = 0.15 m

By the formula (VII.11): 2

2 6

u

TC

H DDπ

= × × +

⇒

( )2

2

0.067543.67 /

0.15 0.0753.14 0.075

2 6

uC kN m= = × × +

So: 43.67 44uC kPa kPa= ≃

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34

Exercise VII.9

A granular soil was subjected to standard penetration tests (SPT) at depth of 3m. Ground water level occurred at a depth of 1.5m below the surface of the soil which was saturated and had a unit weight of 19kPa. The borehole was 100mm, the Donut hammer was used with the rod length of 3.5m, and the standard sampler was used. The average N count was 15. Determine the normalized corrected SPT blows count, (N1)60.

Solution

Determine the normalized corrected SPT blow count, (N1)60

By the formula: (VII.17) : 1 60 60( ) NN C N= ×

We have: 219 /sat kN mγ = , Pa = 100 kN/m2

Borehole (CB) = 100mm 1.0BC⇒ =

Rod length (CR) = 3.5m 0.75RC⇒ =

Donut hammer 0.5EC⇒ =

Standard Sampler 1.0SC⇒ =

The average N = Nm = 15

By the formula: (VII.16) : 60 m E B S RN N C C C C= × × × ×

15 1 1 1 0.75 11.25= × × × × =

By the formula: (VII.18) : '

n

aN

vo

PC

σ

=

, n = 0.5 (sand)

' (19 1.5) (19 10)1.5 42vo kPaσ = × + − =

⇒

0.5100

42NC = =

1.54

So 1( )60 1.54 11.25 17N = × = blows

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35

Exercise VIII.1 A returning wall has a vertical backfill and is 4m. The upper surface of the backfill soil is

horizontal. The unit weight of the backfill soil is 319 /kN mγ = and its angle of friction is of 35o.

Determine the active force exerted on the wall and its point of application.

Solution Determine the active force exerted on the wall and its point of application

1 sin ' 1 sin 35

0.2711 sin ' 1 sin 35

o

a oK

ϕϕ

− −= = =+ +

, ( )' 35oϕ =

At z = 1.5m ⇒ ' 219 4 76 /v h kN mσ γ= × = × =

⇒ ' ' 276 0.271 20.596 /h v ak kN mσ σ= × = × =

So ( )14 76 41.19 /

2aF m kPa kN m= × =

And its point is: 4

1.333 3

hd m= = =

319 /

' 35o

kN mγ

ϕ

=

=

4m

Diagram of active earth force

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36

Exercise VIII.2 From the given data in the below figure:

- Draw the diagram of lateral active earth pressure. - Determine the total active force Fa ,and its point of application.

Solution � Draw the diagram of lateral active earth pressure

We have: Sand: 319 /kN mγ = , ' 33oϕ =

⇒ 1

1 sin ' 1 sin 330.295

1 sin ' 1 sin 33

o

a ok

ϕϕ

− −= = =+ +

Gravel: 321 /kN mγ = , ' 39oϕ =

⇒ 2

1 sin ' 1 sin 390.228

1 sin ' 1 sin 39

o

a ok

ϕϕ

− −= = =+ +

At z = 0m ⇒ ' 0v kPaσ =

At Interface:

z = 1.5m ⇒ '1 19 1.5 28.5v sand h kPaσ γ= × = × =

⇒1

' ' 28.5 0.295 8.408h v ak kPaσ σ= × = × = (sand)

z = 1.5m ⇒2

' ' 28.5 0.228 6.498h v ak kPaσ σ= × = × = (gravel)

z = 5.5m ⇒ ( ) ( ) ( ) ( )'1 2 19 1.5 21 10 4 72.5v sand gravel wh h kPaσ γ γ γ= × + − = × + − =

⇒2

' ' 72.5 0.228 16.53h v ak kPaσ σ= × = × =

The pore water pressure: 4 10 4 40wu kPaγ= × = × =

Total lateral earth pressure: ' 16.53 40 56.530a h u kPaσ σ= + = + =

h1 = 1.5m

3: 19 /

' 25o

Sand kN mγ

ϕ

=

=

h2 = 4m

3: 21 /

' 39o

Gravel kN mγ

ϕ

=

=

3: 21 /

' 39o

Gravel kN mγ

ϕ

=

=

310 /w kN mγ =

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37

� Determine the total active force Fa ,and its point of application

( )1

11.5 8.408 6.306

2aF kN= × = ⇒ 1

1.54 4.5

2d m= + =

( )2

4 6.498 25.992aF kN= × = ⇒ 2

42

2d m= =

( )3

14 50.032 100.064

2aF kN= × = ⇒ 3

41.33

3d m= =

⇒ 1 2 3

6.306 25.992 100.064 132.362a a a aF F F F kN= + + = + + =

⇒ ( ) ( )1 2 31 2 3 6.306 4.5 25.992 2 100.064 1.33

132.362

a a a

a

F d F d F dd

F

× + × + × × + × + ×= =

1.61m=

So The total active force: 132.362aF kN=

Its point: 1.61d m=

Fa = 132.362 kN

Fa1 = 6.306 kN

Fa3 = 100.064 kN1.61 m

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38

Exercise VIII.3 A retaining wall has a vertical back and is 8m. A backfill consist of a cohesive soil with upper surface is horizontal (figure). Determine the active force and its point of application.

Solution

Determine the active force and its point of application

We have: 18 , ' 10 , ' 26 , 8okPa c kPa h mγ ϕ= = = =

⇒ 1 sin 26

0.391 sin 26

o

a ok

−= =+

⇒ 2 ' 2(10) 0.39 12.49ac k− = − = −

The depth of tensile crack: 2 ' 2 10

1.7818 0.39

c

a

cz m

kγ×= = =

⇒ ( )1

112.49 1.78 11.116

2aF kN= − × = − ⇒ 1

21.78 (8 1.78) 7.407

3d m= + − =

And ' 18 8 144v h kPaσ γ= = × = ⇒ ' ' 144 0.39 56.16h v ak kPaσ σ= × = × =

⇒ ( )2

16.22 56.16 174.658

2aF kN= × = ⇒ 2

6.222.07

3d m= =

So the total active force: 1 2

11.116 174.658 163.542a a aF F F kN= + = − + =

⇒ ( ) ( )1 21 2 11.116 7.407 174.658 2.07

1.707163.542

a a

a

F d F dd m

F

× + × − × + ×= = =

- Diagram

8m

3

2

18 /

' 10 /

' 25o

kN m

c kN m

γ

ϕ

=

=

=

1aF

2aF

aF

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39

Exercise VIII.4 From the given data in the below figure, determine the active force and its point of application.

Solution

Determine the active force and its point of application

We have: 320 /kN mγ = , ' 36 , 20o oϕ β= =

⇒ 2 2 2 2

2 2 2 2

cos cos cos ' cos 20 cos 20 cos 360.325

cos cos cos ' cos 20 cos 20 cos 36

o o o

a o o ok

β β ϕβ β ϕ

− − − −= = =+ − + −

At z = 10m ' cos 20 10 cos 20 187.939ov h kPaσ γ β⇒ = = × × =

( )' ' 187.939 0.325 61.1h v aak kPaσ σ⇒ = × = × =

⇒ ( )161.1 10 305.5

2aF kN= × =

⇒ 10

3.333

d m= =

So 305.5 / , 3.33aF kN m d m= =

3: 20 /

' 35o

Sand kN mγ

ϕ

=

=

10m

20oβ =

Fa = 305.5 kN

3.33 m

β = 20?

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40

Exercise VIII.5

For the retaining wall shown in the below figure, H = 8.22 m , 318 /kN mγ = , ' 20oϕ = , 2' 14 /c kN m= and 10oβ = . Calculate the Rankine active force, Fa per unit length of the wall and the

location of the resultant force after the occurrence of the tensile crack.

Solution

We have: 310 , 18 / , ' 20 , 8.22 , ' 14o okN m h m c kPaβ γ ϕ= = = = =

⇒ 2 2

2 2

cos10 cos 10 cos 200.539

cos10 cos 10 cos 20

o o o

a o o ok

− −= =+ −

And 2 ' 2 14

2.1218 0.539

c

a

cz m

kγ×= = = (Depth of Tensile Crack)

2 ' 2 14 0.539 20.557ac k kPa= × =

⇒ ' cos 18 8.22cos10 145.712ov h kPaσ γ β= = × =

⇒ ' ' 145.712 0.539 78.539h v ak kPaσ σ= × = × =

⇒ Total Active earth pressure: ' '( ) 2 ' 78.539 20.557 57.982h a h c ka kPaσ σ= − = − =

⇒ Total Active Force: ( )157.982 8.22 2.12 176.845

2aF kN= × − =

Its point: 6.1 23d m= =

3

2

18 /

' 14 /

' 20o

kN m

c kN m

γ

ϕ

=

=

=

8.22m

10oβ =

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41

- Diagram of total active force:

aF 6m

2m

20oβ =

( )'h a

σ

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42

Exercise VIII.6 Determine the active force due to a solid mass of non-cohesion soil with upper surface

inclined of 20oβ = to the horizontal. All characteristics are specified in the figure.

Solution

Determine the active force due to a solid mass of non-cohesion soil We have:

- Sand: 220 , 20 / , ' 30o okN mβ γ ϕ= = =

⇒ 2 2 2 2

2 2 2 2

cos cos cos ' cos 20 cos 20 cos 300.441

cos cos cos ' cos 20 cos 20 cos 30

o o o

a o o ok

β β ϕβ β ϕ

− − − −= = =+ − + −

- Gravel: 220 , 23 / , ' 40o okN mβ γ ϕ= = =

⇒ 2 2

2 2

cos 20 cos 20 cos 400.266

cos 20 cos 20 cos 40

o o o

a o o ok

− −= =+ −

At z = 5m ⇒ ' 2cos 20 5 cos 20 93.97 /ov sand h kN mσ γ β= = × × =

⇒ 1

' ' 293.97 0.44 41.44 /h v ak kN mσ σ= × = × =

At z = 5m ⇒ 2

' ' 293.97 0.266 25 /h v ak kN mσ σ= × = × = (Interface)

At z = 10m ⇒ '1 2cos cos 93.97 (23 5 cos 20 )o

v sand gravelh hσ γ β γ β= + = + × ×

2202.035 /kN m=

⇒ ( )2

' ' 2202.035 0.266 53.74 /h v aak kN mσ σ= × = × =

Active force: ( )1

15 41.44 103.60

2aF kN= × = ⇒ 1

55 6.67

3d m= + =

( )2

5 25 125aF kN= × = ⇒ 2

52.5

2d m= =

( )3

15 28.74 71.85

2aF kN= × = ⇒ 3

51.67

3d m= =

⇒ Total force 1 2 3

103.35 125 71.85 300.20a a a aF F F F kN= + + = + + =

⇒ ( ) ( )1 2 31 2 3 103.60 6.67 125 5 71.85 1.67

3.74300.20

a a a

a

F d F d F dd m

F

× + × + × × + × + ×= = =

20oβ =

20oβ =

Sand: 320 /

' 30o

kN mγ

ϕ

=

=

Gravel: 323 /

' 40o

kN mγ

ϕ

=

=

5m

5m

Robot

43

- Diagram of total active force:

1aF

3aF 2aF

aF

20oβ =

10m

3.74m 53.74kPa

25kPa

41.44kPa

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44

Exercise VIII.7 Determine the active force, which is being exerted on the wall illustrated below, where is its point of application?

Solution

We have: 220 , 20 / , ' 30o okN mβ γ ϕ= = =

25 , 100 / , 30 , 20o o oq kN m Hλ δ= = = =

- The coefficient of lateral active earth pressure corresponds to the value of θ giving the maximum of active force:

( )

( ) ( ) ( )( ) ( )

2

2

2

cos '

sin ' sin 'cos .cos 1

cos cos

ackϕ λ

ϕ δ ϕ βλ δ λ

λ δ λ β

−=

− × −+ +

+ × −

( )

( ) ( ) ( )( ) ( )

2

2

2

cos 30 50.4728

sin 30 20 sin 30 20cos 5 .cos 5 20 1

cos 5 20 cos 5 20

o o

o o o o

o o o

o o o o

−= =

− × − + + + × −

- The coefficient of lateral active earth pressure:

( ) ( )cos 0.4728 cos5

0.4876cos cos 20 5

oac

q o o

kk

λβ λ

× ×= = =− −

- The total active force:

2

2

1

2

120 10 0.4728 0.4876 100 10 960.20

2

ac ac qF H k k q H

kN

γ= + × ×

= × × × + × × =

320 /

' 30o

kN mγ

ϕ

=

=

320 /

' 30o

kN mγ

ϕ

=

=

20oβ =

20oδ = H

λ H=10m

5oλ = q=100kPa

Robot

45

( )1

1'( ) , '( )

2

194.56 10 472.6

2

a a a acF h H h H k

kN

σ σ γ= × = ×

= × × =

2

100 10 0.4876 487.60

a qF q H k

kN

= × ×

= × × =

( ) ( )472.6 5 487.6 3.33

4.15960.2

d m× + ×

= =

So 960.4 , 4.15aF kN d m= =

5m

3.33m 4.15m

'( ) 94.56h a acH k kPaσ γ= × =

1aF 2aF

acF

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46

Exercise VIII.8 The cross section of a cantilever retaining wall is shown in the figure. Check the factor of safety with respect to overturning, sliding, bearing capacity.

Solution

Check the factor of safety with respect to overturning, sliding, and bearing capacity � Check the factor of safety with respect to overturning - The safety of coefficient of lateral active earth pressure:

2 2 2 2

2 2 2 2

cos cos cos ' cos10 cos 10 cos 300.355

cos cos cos ' cos10 cos 10 cos 30

o o o

a o o ok

β β ϕβ β ϕ

− − − −= = =+ − + −

- The Rankine active force per unit length of wall:

From the figure: 0.7 6 2.6 tan10 7.16oH m= + + × =

⇒ 2 21 ( )

1 1cos 18 (7.16) cos10 0.355 161.30 /

2 2o

a aF H k kN mβγ β= × × × × = × × × × =

⇒ cos 161.30 cos10 158.85 /oh aF F kN mβ= × = × =

sin 161.30 sin10 28 /ov aF F kN mβ= × = × =

- Table of section to determine the total resisting moment: Section No.

Area (m2) Weight per unit length

(kN/m2) Moment arm from A

Resisting moment (Mr) (kN.m/m)

1 0.5 x 6 = 3 3 x 24 = 72 0.7+0.2+0.5/2 =

1.15 72 x 1.15 = 82.8

2 (0.7-0.5)x6x0.5 = 0.6 0.6 x 24 = 14.4 0.7+(0.2x2)/3 = 0.83 14.4 x 0.83 = 11.952

3 0.7 x (0.7+0.7+2.6) =

2.8 2.8 x 24 = 67.2 (0.7x2+2.6)/2 = 2 67.2 x 2 = 134.4

4 2.6 x 6 = 15.6 15.6 x 18 = 280.8 (0.7x2)+2.6/2 = 2.7 280.8 x 2.7 = 758.16

5 2.6 x 0.46 x 0.5 =

0.598 0.598 x 18 = 10.764 1.4+(2.6x2)/3 = 3.13 10.764 x 3.13 = 33.7

Vertical force: Fv = 28 4 28 x 4 = 112

Total vW F v+ = ∑ = 473.16 1133.07

10o

1

2

3

4

5

31

' 21

'1

3

18 /

0 /

30

24 /

o

concrete

kN m

c kN m

kN m

γ

ϕ

γ

=

=

=

=

6m

0.7m

2.6m

1.5m

0.7m A

0.5m

3 ' 2 '2 2 219 / , 30 / , 28okN m c kN mγ ϕ= = =

Robot

47

- Driving moment can determine by:

7.16158.85 379.122 . /

3 3d h

HM F kN m m= × = × =

- Factor of safety against overturning the following table can now be prepared for

determine the resisting moment.

1133.07

3 1.5379.122

rover

d

MF ok

M= = ≈ >

� Check the factor of safety with respect to sliding

- Factor of safety against sliding can be determined by:

( ) tan '

1.5vs

h

W FF

F

ϕ+ ×= ≥

⇒ 473.16 tan 28

1.58 1.5158.85

o

sF ok×= = >

� Check the factor of safety against bearing capacity failure

- Eccentricity: 4 1133.07 379.122

0.40662 2 433.16

r dx

M MBe m

v

− −= − = − =Σ

⇒ 4

0.6666 6x

Be m< = =

- Therefore: 2max

6 473.16 6 0.40661 1 190.44 /

4 4

v ekN m

B Bδ Σ × = + = + =

2min

6 473.16 6 0.40661 1 46.15 /

4 4

v ekN m

B Bδ Σ × = − = − =

- The ultimate bearing capacity of soil can be determined from equations (V.III.35):

0.5 'u q q q x c c c cq DN d i e N S d i B N S d iγ γ γ γγ γ= + +

D= 1.5 m , ' 2 4 2 0.4066 3.187xB B e m= − = − × =

2190.44 /kN m

246.15 /kN m

Robot

48

• ( )tan ' 2 3.14 tan 28 2'tan 45 2.72 tan 59 14.72

2

oo oqN eπ ϕ ϕ × = × + = × =

• ( ) ( )1 cot ' 14.72 1 cot 28 25.80oc qN N ϕ= − × = − × =

• ( ) ( ) ( ) ( )1 tan 1.4 ' 14.72 1 tan 1.4 28 11.19oqN Nγ ϕ= − × = − × × =

• Since the length of the wall (L) is normally much more the width of footing (B):

⇒ 1q cS S Sγ= = =

• ' 1

1 0.1 tan 45 1 0.1 tan 59 1.0782 ' 3.187

o oq

Dd d

Bγϕ = = + × + × = + × × =

• ( )' 11 0.2 tan 45 1 0.2 tan 58 1.157

2 ' 3.187o o

c

Dd

B

ϕ = + × + × = + × × =

•

2158.85

1 , tan 0.33572 18.55890 47.316

ohq c o

Fi i

v

δ δ = = − = = = = Σ

⇒

218.558

1 0.6390

o

q c oi i

= = − =

• 0.114iγ =

⇒ ( ) ( )

( )

17 1.5 14.72 1.078 0.63 0.4066 25.80 1 1.157 0.63

0.5 17 3.187 11.19 1 1.078 0.114 890.72

uq

kPa

= × × × × + × × × ×

+ × × × × × × =

⇒ ( ) 890.72 18 1.5

287.403

unetall

q Dq kPa

F

γ− × − ×= = =

And, max min3 3 190.44 846.15154.37 287.40 !

4 4netallq kPa ok

δ δ+ × += = < =

Robot

49

Exercise IX.1 Compute the ultimate bearing capacity using the Terzaghi equation for the strip footing with dimension and soil parameter shown in the below figure.

Solution

Compute the ultimate bearing capacity using the Terzaghi equation for the strip footing

We have: 318 / , ' 4kN m c kPaγ = =

For ' 30oϕ = by table IX.2 we get: 19.13 , 22.46 , 37.16q cN N Nγ = = =

So for the strip footing we consider by the equation (IX.1):

2 10.5 'u q cq B N D N c Nγγ γ= × × × + × × + ×

⇒ 0.5 18 3 19.13 18 1.5 22.46 4 37.16 1271.57uq kPa= × × × + × × + × =

So The ultimate bearing capacity ( uq ) is 1271.57 kPa

D=1.5m

B=3m

318 /

' 4

' 30o

kN m

c kPa

γ

ϕ

=

=

=

Robot

50

Exercise IX.2 Compute the ultimate bearing capacity using the Terzaghi equation for the strip footing with dimension and soil parameter shown in the below figure.

Solution


We have: 3 3 317 / , 19 / , 10 /sat wkN m kN m kN mγ γ γ= = =



( ) ( ) ( ) ( )0.5 1 1 'u sat w sat w q cq B N D N c Nγγ γ γ γ γ= × − × × + × + − − × + ×

( ) ( ) ( ) ( )0.5 19 10 2 54.36 17 1 19 10 1.5 1 47.16 0 63.53uq = × − × × + × + − − + ×

1503.18kPa=


D=1.5m

B=2m

1m3

3

3

17 /

19 /

10 /

' 36

sat

w

o

kN m

kN m

kN m

γ

γ

γ

ϕ

=

=

=

=

Robot

51

Exercise IX.3 Compute the ultimate bearing capacity using the Terzaghi equation for the strip footing with

dimension and soil parameter shown in the below figure.

Solution


We have: 3 3 318 / , 20 / , 10 /sat wkN m kN m kN mγ γ γ= = =



2 10.5 'u q cq B N D N c Nγγ γ= × × × + × × + ×

But d<B ⇒ 2 avγ γ→ : ( )'2 2

1av d B d

Bγ γ γ = × × + × −

( ) ( ) 3118 1 20 10 2 1 14 /

2kN m= × × + − × − =

⇒ ( )0.5 14 2 78.61 18 1.5 61.55 0 77.50 2762.39uq kPa= × × × + × × + × =


D=1.5m

B=2m 1m

3

3

3

18 /

20 /

10 /

' 38

sat

w

o

kN m

kN m

kN m

γ

γ

γ

ϕ

=

=

=

=

Robot

52

Exercise IX.4

Compute the ultimate & net allowable bearing capacity ( uq and netallq ) using the Terzaghi

equation of square footing in the below figure.

Solution

Compute the ultimate & net allowable bearing capacity ( uq and netallq )

For the square footing we consider by the equation (IX.2):

2 10.4 1.3 'u q cq B N D N c Nγγ γ= × × × + × × + ×


⇒ 0.4 20 3 3.64 18 2 7.44 1.3 15 17.69 700.155uq kPa= × × × + × × + × × =

So we determine the net allowable bearing capacity by equation (IX.19):

( )1

700.155 15 2223.385

3net uall

S

q Dq kPa

F

γ − ×− ×= = =


The net allowable bearing capacity ( )netallq is 223.385 kPa

D=2m

L=B=3m

31 18 /

Fill Material

kN mγ =

3 ' '2 2 2: 20 / , 15 , 20oNatural soil under footing kN m c kPaγ ϕ= = =

Robot

53

Exercise IX.5

What is the ultimate and net allowable bearing capacity ( uq and netallq ) of the mate

foundation in the below figure.

Solution

Compute the ultimate & net allowable bearing capacity ( uq and netallq )


For the mat foundation we consider the ultimate bearing capacity by the equation (IX.4):

2 11 0.2 1 0.3 '2u q c

B B Bq N D N c N

L Lγγ γ = − × × × + × × + + × × ×

⇒ 20 20 20

1 0.2 16 0 16 4 1 1 0.3 30 5.7 260.6540 2 40uq kPa

= − × × × × + × × + + × × × =

So we determine the net allowable bearing capacity by equation (IX.19):

( )1

260.65 16 465.55

3net uall

S

q Dq kPa

F

γ − ×− ×= = =

The ultimate bearing capacity ( uq ) is 260.65 kPa

The net allowable bearing capacity ( )netallq is 65.55 kPa

B x L = 20 x 40 m?

D=4m

3

:

16 /

30

0

u

ou

Clay

kN m

c kPa

γ

ϕ

=

=

=

So

Robot

54

Exercise IX.6

By supposing the factor of safety 3sF = , determine the dimensions of the below mat

foundation .

Solution

Determine the dimensions of the below mat foundation

We know that: Q = 15000kN

The net allowable bearing capacity: netall

Q Qq

S B B= =

×

And the ultimate bearing capacity by equation (IX.14): 1net

u all Sq q F Dγ= × + ×

So we get:

( ) ( ) ( )

( ) ( )

1

2 2

1.5 3 1.5

15000 450003 18 1.5 20 10 1.5 42

u S sat w

Qq F

S

B B

γ γ γ= × + × + − × −

= × + × + − = +

The ultimate bearing capacity of mate foundation by equation (IX.9):

2 11 0.2 1 0.3 '2u q c

B B Bq N D N c N

L Lγγ γ = − × × × + × × + + × × ×

( ) ( ) ( )1 0.2 18 10 0 18 1.5 20 10 1.5 1 0.3 40 5.72

B B B

B B = − × − × × + × + − + + × ×

⇒ 2

4500042 338.4

B+ = ⇒

4500012.32

338.4 42B m= =

−

So The dimension of footing is B=12.32 m

B x B

D=3m

Q=15000kN 1.5m

31

3

:

18 /

20 /sat

Sand

kN m

kN m

γ

γ

=

=

3: 18 / , 40 , 0ou uClay kN m c kPaγ ϕ= = =

Robot

55

Exercise IX.7 A rectangular footing of 3x1.5m is shown in the below figure. Determine the net allowable load with can be supported by this foundation.

Solution

Determine the net allowable load with can be supported by this foundation

The effective dimension of footing can be determined by:

' 2 1.5 2(0.1) 1.3BB B e m= − = − =

' 2 3 2(0.2) 2.6LL L e m= − = − =

The ultimate bearing capacity is given by (IX.9):

2 1

' ' '1 0.2 1 0.3 '

' 2 'u q c

B B Bq N D N c N

L Lγγ γ = − × × × + × × + + × × ×


⇒ 1.3 1.3 1.3

1 0.2 18 19.13 18 2 22.46 1 0.3 0 37.162.6 2 2.6uq

= − × × × + × × + + × × ×

1010kPa=

The net allowable bearing capacity is: 1 1010 18 2324

3net uall

S

q Dq kPa

F

γ− − ×= = =

0.1m

2m

3m

1.5m

0.1m

Q

0.2m

318 /

' 30 , ' 0o

kN m

c

γ

ϕ

=

= =

Robot

56

Exercise IX.8 Verify the dimension of the footing shown in the below figure, whether it is adequate or not. The factor of safety is taken as 3.

Solution Verify the dimension of the footing

The eccentricities in both directions are:

270

0.151780

BB

Me m

Q= = =

160

0.091780

LL

Me m

Q= = =

Therefore the effective dimension of the footing can be estimated:

' 1.8 2(0.15) 1.5B m= − =

' 1.8 2(0.09) 1.62L m= − =

The net allowable bearing capacity is given by:

( )2 1

' ' '1 0.2 1 1 0.3

' 2 'q cnetall

s

B B BN D N c N

L Lq

F

γγ γ − × × × + × − + + × × × =


⇒

( )1.5 1.5 1.51 0.2 18 19.13 18 1.8 22.46 1 1 0.3 95 37.16

1.62 2 1.623

netallq

− × × × + × − + + × × × =

954kPa=

Subsequently, 954(1.5 1.62) 2318 1780 !!!netallQ kN Q kN ok= × = > =

D=1.8m

1.8m x 1.8m

Q=1780 kN

MB=270 kN.mML=160 kN.m

3

:

18 /

95

30o

Silty Clayed Sand

kN m

c kPa

γ

ϕ

=

=

=

Robot

57

Exercise IX.9 Check the factor of safety in the clay layer of the below figure.

Solution

Check the factor of safety in the clay layer

By the figure we get: 1 2.51.5 1.66 3.5

1.5

h

B< = = <

⇒ ' 1.5 2.5 4B B h m= + = + =

' 3 2.5 5.5L L h m= + = + =

So a repartition load can be determined by: 1800

' 81.81' ' 4 5.5

Q kNq kPa

B L m m= = =

× ×

The net allowable bearing capacity: 1 1net u uall S net

S all

q D q Dq F

F q

γ γ− × − ×= ⇒ =

But ( )1 2 1

' ' '1 0.2 1 1 0.3

' 2 'u q c

B B Bq D N D N c N

L Lγγ γ γ − = − × × × + × − + + × × ×

For ' 0oϕ = (the clay layer): by table IX.2 we get: 0.00 , 1.00 , 5.7q cN N Nγ = = =

⇒ ( )1

4 4 41 0.2 22 0 20 1 1 1 1 0.3 35 5.7

5.5 2 5.5uq Dγ − = − × × × × + × − + + × × ×

243.027kPa=

In brief that ' netallq q≤ ⇒ 1 243.027

2.97 !!!' 81.81

uS

q DF ok

q

γ− ×= = =

S0 2.97 !SF ok=

3

3

:

20 /

22 /

40

sat

o

Sandy gravel

kN m

kN m

γ

γ

ϕ

=

=

=

D=1m

1.5m x 3m

Q=1800 kN

, 35 , 0ou uClay c kPa ϕ= =

2.5m

Robot

58

Exercise IX.10

A strip footing is located near a slope. The site investigation and laboratory test provide the following results:

Slope 20oβ =

Cohesion less soil with the frictional angle ' 40oϕ =

Unit weight of the soil 318 /kN mγ =

What is the ultimate bearing capacity of this footing in regard to its width if: b/B = 0.5 and D/B = 1 b/B = 1 and D/B = 0 In both cases the effect of ground water is neglected.

Solution

The bearing capacity of the soil under the foundation in these cases can be determined from equation (IX.18):

2 2u cq q

Bq c N Nγγ= × + × ×

Since 0c = , the above equation can be written as: 2 2u q

Bq Nγ= × ×

- Determine of bearing capacity factor , qNγ :

. for 0.5 ; 1 ; 20 ' 40o ob Dand

B Bβ ϕ= = = =

The chart of figure IX.13 give 135qNγ =

. for 1 ; 0 ; 20 ' 40o ob Dand

B Bβ ϕ= = = =

The chart of figure IX.13 give 50qNγ =

Therefore: . for 0.5 ; 1b D

B B= =

⇒ 2 18 135 (1215 )2 2u q

B Bq N B kPaγγ= × × = × × = ×

. for 1 ; 0b D

B B= =

⇒ 2 18 50 (450 )2 2u q

B Bq N B kPaγγ= × × = × × = ×

Robot

59

Exercise IX.11 Consider a rectangular foundation that is located on a sand layer extending to a great depth. The necessary data is given in below figure, estimate the elastic or immediate settlement assuming that the foundation is rigid.

Solution

Estimate the elastic or immediate settlement assuming that the foundation is rigid

The immediate settlement for foundation by the equation (IX.23): 21

i ps q B IE

υ−= × × ×

We have: E = 14000 kPa , 0.30υ = , B= 1m , L = 2m , D= 1.5m q = 100 kPa

It is the rigid foundation by table (IX.3): 2

2 1.181 p

LI

B= = ⇒ =

⇒ 21 0.30

100 1 1.18 0.00767 7.6714000is m mm−= × × × = =

So The immediate settlement ( ) 7.67is mm=

D=1.5m

1m x 2m

q = 100 kPa

:

14000

0.30

Sand

E kPa

υ

=

=

Robot

60

Exercise X.1 Estimate the maximum allowable static load on a driven pile, 200mm X 200mm, shown in the below figure. The unit weight of concrete pile is 24 kN/m3.

Solution - The ultimate base bearing capacity in sand layer:

Equation (X.7): ' *'b v qq Nσ= ×

From figure (X.8): for ' 38oϕ = ⇒ 15crD

B=

For the given problem: 1

50.2

D

B= =

Since in this case: 5 0.5 7.5crDD

B B= ≤ × =

So *qN become '

qN

For ' 38oϕ = , the chart of figure X.9 gives * 129qN = and the chart of figure X.10 gives

* 226qN =

Therefore the value of 'qN derive from the two values of *

qN are:

1

* 5129 86

0.5 150.5q q

cr

D

BN ND

B

= × = × =××

1

* 5226 150.67

0.5 150.5q q

cr

D

BN ND

B

= × = × =××

And: ( ) ( ) ( )' 17 4 17.5 10 6 18 10 1 121v kPaσ = × + − + − =

Then substituting equation X.7: 1

121 86 10406bq kPa= × =

2

121 150.67 18231bq kPa= × =

0 m

-4 m

-10 m

-11 m

Loose Sand

clay

Dense sand

31

'1

17 /

28o

kN mγ

ϕ

=

=

32

'2

17.5 /

40

0

u

o

kN m

S kPa

γ

ϕ

=

=

=

33

'3

18 /

38o

kN mγ

ϕ

=

=

Robot

61

And: 1 1

*0.5 tan ' 0.5 100 86 tan 38 3359.53ol a q bq P N kPa qϕ= × × × = × × × = <

2 2

*0.5 tan ' 0.5 100 150.67 tan 38 5885.82ol a q bq P N kPa qϕ= × × × = × × × = <

So the ultimate base bearing capacity is: ( )1 0.2 0.2 3359.53 134.38bQ kN= × × =

( )1 0.2 0.2 5885.82 235.43bQ kN= × × =

- The ultimate base bearing capacity in sand layer:

. In loose sand layer:

The skin friction increase linearly up to a depth D’ and constant there after:

' 15 15 0.2 3D B m= × = × =

Equation (X.11): ' tans vf K σ δ= × ×

Driven pile: ( ) ( )1.4 1.4 1 sin ' 1.4 1 sin 28 0.743ooK K ϕ= × = − = − =

3

'4

δ ϕ= ⇒ 3

tan tan 28 0.3844

oδ = × =

Thus, Z between 0 and 3m: 1 0.743 17 0.384 4.85 ,sf z z kPa= × × × =

Z at or below 3m: 2 0.743 17 3 0.384 14.55sf kPa= × × × =

. In clay layer:

Equation (X.13): 2s uf Sα= ×

'

400.442 1

617 4 (17.5 10)

2

u

vo

S

σ= = <

× + −

Equation (X.16):

0.5 0.5

'

400.5 0.5 0.752

90.50u

vo

Sασ

− − = × = × =

So 2 0.752 40 30.08sf kPa= × =

. In dense sand layer: The skin friction also increases from linearly up to a depth D’ and constant there after: ' 15 15 0.2 3D B m= × = × =

Equation (X.11): ' tans vf K σ δ= × ×

Driven pile: ( ) ( )1.4 1.4 1 sin ' 1.4 1 sin 38 0.538ooK K ϕ= × = − = − =

3

'4

δ ϕ= ⇒ 3

tan tan 38 0.5434

oδ = × =

At the top of dense sand layer:

( ) ( )2 0.538 17 4 17.5 10 6 0.543 33sf kPa= × × + − × =

At a depth of 1m from the top of the dense sand layer:

( ) ( ) ( )2 0.538 17 4 17.5 10 6 18 10 1 0.543 35.35sf kPa= × × + − + − × =

Robot

62

Since the pile perimeter is constant throughout the depth the total skin friction force can be computed by multiplying the skin friction distribution shown in the above figure by the perimeter of 0.8m therefore,

( )0.8 0.5 3 14.55 1 14.55 1 1 30.08 6 1 0.5 33 35.35 1SQ = × × × + × × + × × + + ×

200.68kN=

Therefore, the allowable load can be obtained by equation X.2:

1

134.38 200.68106.31

3 3b s

allb s

Q QQ kN

F F= + = + =

2

235.43 200.68(24 0.04 11) 136.44

3 3b s

all pb s

Q QQ w kN

F F= + − = + − × × =

14.55 kPa

30 kPa

33 kPa

35.35 kPa

Loose sand

Clay

Dense sand

0 m

-4 m

-10 m

-11 m

-3 m

Robot

63

Exercise X.2 Determine the maximum load that can be carried by a driven pile in the below figure.

Solution

- Determine the maximum allowable load that can be carried by a driven pile:

b sall p

s s

Q QQ W

F F= + − , (X.2)

- We supposed the unit weight of concrete is 324 /kN m :

( )22 24 0.25 10

(24 10) 11.7754 4p

BW kN

ππ × ×= × × = =

- Determine the point load or base bearing capacity of pile: *b c uq N S= × , (X.9)

The bearing capacity factor * 6.5cN = at 25uS kPa=

⇒ 6.5 25 162.5bq kPa= × =

⇒ ( )20.25

162.5 7.9734b b bQ q A kN

π= × = × =

- Determine the shaft capacity of pile by equation (X.10): s s sQ f A= ×

At the sand layer (0m to 7m): The critical depth: ' 15 15 0.25 3.75D B m= = × =

Unit skin friction resistance by equation (X.11): ' tans vf K σ δ= × ×

The coefficient of lateral earth pressure: ( ) ( )1.4 1 sin ' 1.4 1 sin 22 0.876oK ϕ= − = − =

-2 m

-7 m

-10 m

Sand

clay

3

( ) 16.5 /d sand kN mγ =

3( )

'2

17.5 /

22

sat sand

o

kN mγ

ϕ

=

=

3( ) 17.5 /

25

sat clay

u

kN m

S kPa

γ =

=

???allQ

25B cm=

Circular pile 0 m

Robot

64

The frictional angle (soil &pile): 3 3

tan tan ' tan 22 0.2964 4

oδ ϕ = = =

Z between 0 and 2m: ( )0.876 16.5 0.296 4.278 ,sf z z kPa= × × × =

Z at 2m: ( )0.876 16.5 2 0.296 8.557sf kPa= × × × =

Z at 3.75m or greater than D’=3.75m:

( ) ( )0.876 16.5 2 17.5 10 1.75 0.296 11.960sf kPa= × × + − × =

At the clay layer (7m to 10m):

Unit skin friction resistance by equation (X.13): s uf Sα= ×

The effective strees: ( ) ( ) ( )' 316.5 2 17.5 10 5 17.5 10 81.75

2v kPaσ = × + − + − =

⇒ Equation (X.16):

0.5 0.5

'

250.5 0.5 0.904

81.75u

vo

Sασ

− − = × = × =

⇒ 0.904 25 22.604sf kPa= × =

⇒ The skin friction load of pile

( )0.785 0.5 8.557 2 1 0.5 8.557 11.960 1.75 11.960 3.25 1 22.604 3 1sQ = × × × + + + × × + × ×

104.55kN=

Therefore, the allowable load can be obtained by equation X.2:

7.973 104.55

11.775 25.733 3

b sall p

b s

Q QQ W kN

F F= + − = + − =

Sand

Clay

8.557 kPa

11.960 kPa

22.604 kPa

-3.75 m

-7 m

-10 m

-2 m

Robot

65

Exercise X.3 Based on the given in the below figure, determine the total drag force (total force of negative skin friction).

Solution

Clay fill over granular soil:

Determine total drag force by equation (X.27): ' 2 tan

2o f f

n

p K HQ

γ δ× × × ×=

And we have: 3.14 0.30 0.942p d mπ= = × =

1 sin ' 1 sin 30 0.5ooK ϕ= − = − =

' 316 /f kN mγ =

' 3fH m=

( ) ( )tan tan 0.6 ' tan 0.6 30 0.325oδ ϕ= = × =

⇒ 20.942 0.5 16 3 0.325

11.022nQ kN

× × × ×= =

Dense Sand

0.30d m=

Circular pile Fill materials:

316 /

' 30 , 0.6 '

f

o

kN mγ

ϕ δ ϕ

=

= = 3fH m=

D

Robot

66

Exercise X.4 Based on the given in the below figure, determine the total drag force (total force of

negative skin friction). Assumed the acceleration of gravity 210 / .g m s=

Solution

We have: 3.14 0.30 0.942p d mπ= = × =

1 sin ' 1 sin 25 0.577ooK ϕ= − = − =

' 317.5 /f kN mγ = , ' 3fH m= , 20D m=

( ) ( )tan tan 0.6 ' tan 0.6 25 0.268oδ ϕ= = × =

Determine the neutral depth by equation (X.28):

( ) ' '2

2 ' 'f f f f f

NSF fNSF

D H D H HH H

H

γ γγ γ

− − ×= × + − ×

( )20 3 20 3 17.5 3 2 17.5

32 7 7NSFH

− − × × = × + − ×

17 17 52.5 105

2 7 7NSFH = × + −

272

15NSFH

= − ⇒ 2 15 272 0NSF NSFH H+ − =

2 24 15 4 ( 272) 1 1313b ac∆ = − = − × − × =

⇒ 15 1313

10.6182 2 1NSF

bH m

a

− + ∆ − += = =×

Clay:

317 / ,

' 25 , 0.6 '

sat

o

kN mγ

ϕ δ ϕ

=

= =

0.30d m=

Circular pile

Fill materials: 317.5 /f kN mγ = 3fH m=

20D m= z NSFH

Neutral plane

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67

Granular soils fill over clay: Determine total drag force by equation (X.30):

( ) ( )' 21tan ' tan

2n o f f NSF NSF oQ p K H H H p Kγ δ γ δ= × × × × × + × × × ×

⇒ ( ) ( )210.942 0.577 17.5 3 0.268 10.618 10.618 0.942 0.577 7 0.268

2= × × × × × + × × × ×

81.201 57.480 138.681kN= + =

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68

Exercise X.5 A fully embedded precast, prestressed concrete pile is 12m long and driven into a homogenous layer of sand (c’ = 0). The pile is square in cross section, with sides measuring 305mm.

The dry unit weight of sand ( )dγ is 316 /kN m , and the average effective soil friction angle is

' 35oϕ = . The allowable working load is 338kN. If 240kN is contributed by the frictional resistance

and 98kN is from the point load, determine the elastic settlement of the pile. Use 6 221 10 /pE kN m= × , 230000 /sE kN m= and 0.3.sυ =

Solution

Determine the elastic settlement of the pile

- Elastic settlement of the pile (X.32): ( )

(1)wb ws

ep p

Q QS L

A E

ξ+= ×

×

The point load: 98wbQ kN= , The frictional resistance: 240wsQ kN=

Area of cross section 20.305 0.305 0.093025pA m= × = , (B=30.5 cm)

Modulus of elasticity of pile 6 221 10 /pE kN m= × , 0.6ξ =

⇒ ( )

(1) 6

98 0.6 24012 0.001487 1.487

0.093025 21 10eS m mm+ ×

= × = =× ×

- Settlement of the pile caused by point load (X.33): ( )2(2) 1wb

e s wbs

q BS I

Eυ×= × − ×

0.3sυ = , 0.85wbI =

Modulus of elasticity of soil below the pile tip 30000sE kPa=

98

1053.480.093025

wbwb

p

Qq kPa

A= = =

⇒ ( )2(2)

1053.48 0.3051 0.3 0.85 0.00828 8.28

30000eS m mm×= × − × = =

- Settlement of the pile caused by the shaft around the pile (X.35):

( )2(3) 1ws

e s wss

Q BS I

p D Eυ

= × × − ×

And: 12

2 0.35 2 0.35 4.1950.305ws

DI

B= + × = + × = , (X.36)

The perimeter of pile = 0.305 4 1.22m× =

⇒ ( )2(3)

240 0.3051 0.3 4.195 0.000636 0.636

1.22 12 30000eS m mm = × × − = = ×

⇒ Total settlement 1.487 8.28 0.636 10.40mm= + + =

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69

Exercise X.6 Using a factor of safety equal to 3, determine the allowable bearing capacity of pile group in the below figure.

Solution - Option 1 From equation (X.39) and (X.41):

( ) ( )*( ) ( )g u u b c u b uQ Q m n A N S S p Dη η α = ×Σ = × × × × × + Σ × × × ∆

In which: 20.30 0.30 0.09 , 4 0.30 1.20bA m p m= × = = × =

11D D m∆ = = and 2( ) 80 /u b u uS S C kN m= = =

From equation (X.9): 2 *50 / , 8u cS kN m N= = and 2100 /uS kN m≥ * 9cN =

by interpolation these value we get * 8.60cN = and 280 /uS kN m=

The average values of the effective overburden stress is 'voσ :

' 21119 104.504 /

2vo kN mσ = × =

It follow that: '

800.766 1

104.50u

vo

S

σ= = < , so

0.580

0.5 0.57104.50

α−

= × =

Therefore,

( ) ( )4 3 0.09 8.60 80 0.57 80 1.2 11 7966uQ kNΣ = × × × × + × × × =

Bg

Lg

Soil data: Homogenous saturated clay:

2

3

80 / ,

19 /

uC kN m

kN mγ

=

=

Groundwater table is located at a depth 15 m below the ground surface.

Given:

4, 3, 300

1200 , 11

m n B mm

S mm D m

= = =

= =

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70

Group efficiency (Equation X.40): ( ) ( )1 1

190

n m m n

m nη θ

− × + − ×= −

× ×

( ) ( )3 1 4 4 1 30.3

1 arctan 0.781.2 90 4 3

− × + − ×= − × =

× ×

Hence, ( ) 0.78 7966 6213.48g u uQ Q kNη= ×Σ = × =

And ( )

( )

6213.482071

3g u

g alls

QQ kN

F= = =

- Option 2

From option 1 , 7966uQ kNΣ =

Again from equation (X.42), the ultimate black capacity is:

( ) ( )*( ) 2u g g u b c g g uQ L B S N L B S DΣ = × × × + Σ + × × ∆

In which: ( ) ( ) 0.31 2 4 1 1.2 2 3.9

2 2g

BL m S m= − × + × = − × + × =

( ) ( ) 0.31 2 3 1 1.2 2 2.7

2 2g

BB n S m= − × + × = − × + × =

So, 3.9

1.442.7

g

g

L

B= = and

114.07

2.7g

D

B= =

From figure (X.17): * 8.60cN =

Therefore, the black capacity:

( ) ( )13.9 2.7 80 8.6 2 3.9 2.7 80 11 18860.64uQ kNΣ = × × × + + × × =

Hence: ( ) 7966 18860.64g uQ kN kN= <

And: ( )

( )

79662655

3g u

g us

QQ kN

Fα = = =

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71

Exercise X.7 Determine the consolidation settlement of the piles in the below figure. All clays are normally consolidated.

Solution

Determine the consolidation settlement of the piles

The consolidation settlement (X.49): '

'lg

1zfc

eo zo

CS H

e

σσ

= × × +

- At clay 1 layer:

( ) ( ) ( )' 2717 2.5 18 10 7 18 10 126.5 /

2zo kN mσ = × + − + − =

The effective stress at the middle of layer: ( )( )'g i g i

Q

B Z L Zσ∆ =

+ + ,

73.5

2iZ m = =

( ) ( )22000

43.592 /2.7 3.5 3.9 3.5

kN m= =+ +

⇒ 1

0.3 126.5 43.5927 lg 0.015 150

1 0.8 126.5eS m mm+ = × × = = +

Lg = 3.9 m

z

Clay 2:

318.5 /

0.7 , 0.2

sat

o c

kN m

e C

γ =

= =

Clay 3:

319 /

0.75 , 0.25

sat

o c

kN m

e C

γ =

= =

Clay 1:

318 /

0.8 , 0.3

sat

o c

kN m

e C

γ =

= =

Sand: 317 /sat kN mγ =

GWT

Rock

2 m

3 m

14 m

2.5 m

1 m

7 m

7 m

Block width, 2.7gB m=

2000gQ kN=

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72

- At clay 2 layer:

( ) ( ) ( )' 2317 2.5 18 10 14 18.5 10 167.25 /

2zo kN mσ = × + − + − =


Q

B Z L Zσ∆ =

+ + ,

37 8.5

2iZ m = + =

( )( )22000

14.40 /2.7 8.5 3.9 8.5

kN m= =+ +

⇒ 2

0.2 167.25 14.403 lg 0.01266 12.66

1 0.7 167.25eS m mm+ = × × = = +

- At clay 3 layer:

( ) ( ) ( ) ( )' 2217 2.5 18 10 14 18.5 10 3 19 10 189 /

2zo kN mσ = × + − + − + − =


Q

B Z L Zσ∆ =

+ + ,

210 11

2iZ m = + =

( ) ( )22000

9.80 /2.7 11 3.9 11

kN m= =+ +

⇒ 3

0.25 189 9.802 lg 0.00627 6.27

1 0.75 189eS m mm+ = × × = = +

So the total consolidation settlement of clay 150 12.66 6.27 168.9mm= + + =

Exercise of Soil Mechainics

Documents

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40o ob dandb

internal friction

negative skin

vertical incremental

vertical incremental

rankine active

additional