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SOLUTIONS MANUA for Principles of Geotechnical Engineering 10th Edition by Braja M. Das

Contents Chapter 2 .......................................................................................................................................... 1 Chapter 3 ........................................................................................................................................11 Chapter 4 ........................................................................................................................................19 Chapter 5 ........................................................................................................................................23 Chapter 6 ........................................................................................................................................29 Chapter 7 ........................................................................................................................................35 Chapter 8 ........................................................................................................................................45 Chapter 9 ........................................................................................................................................53 Chapter 10 ......................................................................................................................................63 Chapter 11 ......................................................................................................................................77 Chapter 12 ......................................................................................................................................81 Chapter 13 ......................................................................................................................................95 Chapter 14 ....................................................................................................................................105 Chapter 15 ....................................................................................................................................119 Chapter 16 ....................................................................................................................................125 Chapter 17 ....................................................................................................................................139 Chapter 18 ....................................................................................................................................149

Chapter 2 2.1

= Cu

= Cc

2.2

= Cu

Cc = 2.3

D60 0.41 = = 5.13 D10 0.08

( D30 ) 2 (0.22) 2 = = 1.48 ( D10 )( D60 ) (0.08)(0.41) D60 1.81 = = 7.54 D10 0.24

( D30 ) 2 (0.82) 2 = = 1.55 ( D10 )( D60 ) (0.24)(1.81)

Sieve no.

Mass of soil retained Percent retained on each sieve (g) on each sieve Percent finer

0.0

100.0

18.5

4.4

95.6

53.2

12.6

83.0

90.5

21.5

61.5

81.8

19.4

42.1

100

92.2

21.9

20.2

200

58.5

13.9

6.3

Pan

26.5

6.3

Σ 421.2 g The grain-size distribution is shown in the figure.

b. D60 = 0.4 mm; D30 = 0.2 mm; D10 = 0.095 mm Cu c. =

= d. Cc 2.4

D60 0.4 = = 4.21 D10 0.095

( D30 ) 2 (0.2) 2 = = 1.05 ( D10 )( D60 ) (0.4)(0.095)

a. Mass of soil retained Sieve no. on each sieve (g)

Percent retained on each sieve

Percent finer

0.0

100

6.0

94.0

48.7

9.74

84.26

127.3

25.46

58.80

96.8

19.36

39.44

76.6

15.32

24.12

100

55.2

11.04

13.08

200

43.4

8.68

4.40

Pan

4.40

Σ 500 g The grain-size distribution is shown in the figure. 2 © 2022 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

b. D10 = 0.13 mm; D30 = 0.3 mm; D60 = 0.9 mm

2.5

Cu c. =

D60 0.9 = = 6.923 ≈ 6.92 D10 0.13

Cc d.=

D302 0.32 = = 0.769 ≈ 0.77 ( D60 )( D10 ) (0.9)(0.13)

a. Sieve no.

Mass retained (g)

Percent retained on each sieve Percent finer

4.54

95.46

6.81

88.65

7.78

80.87

128

20.75

60.12

221

35.82

24.3

100

13.94

10.36

200

6.48

3.88

Pan

3.88

Σ 617 g The graph for percent finer versus grain size is shown. 3 © 2022 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

b. From the graph, D10 = 0.14 mm, D30 = 0.27 mm, D60 = 0.42 mm Cu c. =

= d. Cc

2.6

D60 0.42 = = 3 D10 0.14

( D30 ) 2 (0.27) 2 = = 1.24 ( D60 )( D10 ) (0.42)(0.14)

a. Mass of soil retained Percent retained Sieve no. on each sieve (g) on each sieve The grain-size distribution is shown in the figure. 4 0 0.0

Percent finer 100

0.0

100

0.0

100

9.1

1.82

98.18

249.4

49.88

48.3

179.8

35.96

12.34

100

22.7

4.54

7.8

200

15.5

3.1

4.7

Pan

23.5

4.7

b. D10 = 0.21 mm; D30 = 0.39 mm; D60 = 0.45 mm Cu c. =

D60 0.45 = = 2.142 ≈ 2.14 D10 0.21

D302 0.392 Cc = = 1.609 ≈ 1.61 d.= ( D60 )( D10 ) (0.45)(0.21) 2.7

a. The grain-size distribution curve is shown in the figure.

b. Percent passing 2 mm = 100

2.8

GRAVEL: 100 − 100 = 0%

Percent passing 0.06 mm = 84 Percent passing 0.002 mm = 11

SAND: 100 − 84 = 16% SILT: 84 − 11 = 73% CLAY: 11 − 0 = 11%

c. Percent passing 2 mm = 100 Percent passing 0.05 mm = 80 Percent passing 0.002 mm = 11

GRAVEL: 100 − 100 = 0% SAND: 100 − 80 = 20% SILT: 80 − 11 = 69% CLAY: 11 − 0 = 11%

d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 90 Percent passing 0.002 mm = 11

GRAVEL: 100 − 100 = 0% SAND: 100 − 90 = 10% SILT: 90 − 11 = 79% CLAY: 11 − 0 = 11%

The grain-size distribution curve is shown in the figure.

a. Percent passing 2 mm = 100 Percent passing 0.05 mm = 94 Percent passing 0.002 mm = 42

GRAVEL: 100 − 100 = 0% SAND: 100 − 94·= 6% SILT: 94 − 42 = 52% CLAY: 42 − 0 = 42%

b. Percent passing 2 mm = 100 Percent passing 0.075 mm = 97 Percent passing 0.002 mm = 42

GRAVEL: 100 − 100 = 0% SAND: 100 − 97 = 3% SILT: 97 − 42 = 55% CLAY: 42 − 0 = 42%

2.9

2.10

a. The grain-size distribution curve is shown below.

b. Percent passing 2 mm = 100 Percent passing 0.06 mm = 84 Percent passing 0.002 mm = 28

GRAVEL: 100 − 100 = 0% SAND: 100 − 84 = 16% SILT: 84 − 28 = 56% CLAY: 28 − 0 = 28%

c. Percent passing 2 mm = 100 Percent passing 0.05 mm = 83 Percent passing 0.002 mm = 28

GRAVEL: 100 − 100 = 0% SAND: 100 − 83 = 17% SILT: 83 − 28 = 55% CLAY: 28 − 0 = 28%

d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 90 Percent passing 0.002 mm = 28

GRAVEL: 100 − 100 = 0% SAND: 100 − 90 = 10% SILT: 90 − 28 = 62% CLAY: 28 − 0 = 28%

Gs = 2.60; temperature = 24°; R = 43; time = 60 min. Referring to Table 2.10, L = 9.2. Eq. (2.6): D (mm) = K

L(cm) t (min)

From Table 2.9 for Gs = 2.60 and temperature = 24°, K = 0.01321. D = 0.01321

9.2 = 0.0052 mm 60

2.11

For Gs = 2.70 and temperature = 23°, K = 0.01297 (Table 2.9); R = 25, L = 12.2 (Table 2.10).

L(cm) 12.2 = D(mm) K= 0.01297 = 0.0041 mm t (min) 120 2.12

24,000 kg a. Total mass in the ternary mix = 8000 × 3 = Percent of each soil in the mix =

8000 ×100 = 33.33% 24, 000

ΣmB = ΣmC = 500 g Mass of each soil used in the sieve analysis, ΣmA = If a sieve analysis is conducted on the ternary mix using the same set of sieves, the percent of mass retained on each sieve, mM (%), can be computed as follows: m  m  m  (%) 0.333  A ×100  + 0.333  B ×100  + 0.333  C ×100  mM=  500   500   500 

The calculated values are shown in the following table.

Sieve size (mm)

Mass retained mA (g) mB (g)

mC (g)

mM (%)

Percent passing for the mixture

25.0

0.0

19.0

6.66

93.34

12.7

130

18.65

74.69

9.5

12.65

62.04

4.75

100

165

23.64

38.4

2.36

9.32

29.08

0.6

11.65

17.43

0.075

105

14.98

2.45

Pan

2.33

≈0

100

b. The grain-size distribution curve for the mixture is drawn below.

From the curve, D10 = 0.21; D30 = 2.5; D60 = 9.0; = Cu

D60 9.0 D302 2.52 = = 42.85 ; = C x = = 3.31 D10 0.21 ( D60 )( D10 ) (9.0)(0.21)

Chapter 3 3.1

(G + e)γ w Gs γ w eγ w  e  = + = γd +  a. γ sat = s γw 1+ e 1+ e 1+ e  1+ e 

Gs γ w = (1 + e)

γd = b.

( Se /w) γ w

eS γ w = (1 + e) (1 + e) w

(G + e)γ w Gs γ w eγ w  e  = + = γd +  c. γ sat = s γw 1+ e 1+ e 1+ e  1+ e 

Rearranging, γ sat (1 + e)= γ d (1 + e) + eγ w Therefore, e =

γ sat − γ d γ d − γ sat + γ w

 1 + wsat   1 + wsat  eγ w (1 + wsat )nγ w = Gs γ w = d. γ sat  =   wsat  1+ e   1 + e  wsat

nγ w Rearranging, wsat (γ sat − nγ w ) = Therefore, wsat =

nγ w γ sat − nγ w

 711.2 − 623.9  = 3.2 a. w =  (100) 14% 623.9  

ρ b. =

M 711.2 = = 1778 kg /m 3 0.4 V

ρd c.=

623.9 = 1559.75 kg /m 3 0.4

d. ρ d =

Gs ρ w 1+ e

e =

Gs ρ w

ρd

= −1

(2.68)(1000) = − 1 0.718 1559.75

0.718 e = = 0.418 1 + e 1 + 0.718

n e.=

a. γ d 3.3 =

Gs γ w

e b.=

3.4

γ 17.8 = = 15.6 kN /m 3 (1 + w) (1 + 0.14)

γd

= −1

(2.69)(9.81) = − 1 0.69 15.6

S c.=

wGs (0.14)(2.69) = = 0.545 = 54.5% e 0.69

γ a. =

W 23 = = 115 lb /ft 3 V 0.2

γd b.=

115 γ = = 103.6 lb /ft 3 1 + w 1 + 0.11

Gs γ w

e c.=

γd

= −1

(2.7)(62.4) = − 1 0.626 103.6

n d.=

e 0.626 = = 0.385 1 + e 1 + 0.626

S e.=

wGs  (0.11)(2.7)  (100) =  = 47.4% e  0.626 

Ws f.=

23 W = = 20.72 lb 1 + w 1 + 0.11

Ww =− 23 20.72 = 2.28 lb V = w

3.5

Ww 2.28 = = 0.0365 ft 3 γ w 62.4

γd a.=

γ sat 19.8 = = 16.91 kN /m 3 1 + w 1 + 0.171

b. γ d =

(Gs )(9.81) Gs γ w ; Gs = 2.44 ; 16.91 = 1 + 0.171Gs 1+ e

e wG = (0.171)(2.44) = 0.417 c.= s 3.6

a. γ =

Gs γ w + wGs γ w Se ; Gs = ; wGs = Se w 1+ e

 Se    γ w + Seγ w w γ=  1+ e

 (0.6)(e)   0.192  (62.4) + (0.6)(e)(62.4) = ; e 0.69 1+ e

b. G= s

= c. γ sat

3.7

γd a.=

e b.=

Se (0.6)(0.69) = = 2.16 w 0.192

γ w (Gs + e)

= 1+ e

(62.4)(2.16 + 0.69) = 105.2 lb /ft 3 1 + 0.69

112 γ = = 101.1 lb /ft 3 1 + w 1 + 0.108

Gs γ w

γd

= −1

(2.67)(62.4) = − 1 0.648 101.1

n c.=

0.648 e = = 0.39 1 + e 1 + 0.648

S d.=

wGs  (0.108)(2.67)  =  = 44.5%  (100) e 0.648 

= 3.8 a. γ

(Gs + Se)γ w [2.67 + (0.8)(0.648)](62.4) = = 120.73 lb/ft 3 1+ e 1 + 0.648

Water to be added: 120.73 - 112 = 8.73 lb/ft3 = b. γ

(Gs + Se)γ w (2.67 + 0.648)(62.4) = = 125.6 lb/ft3 1+ e 1.648

Water to be added: 125.6 - 112 = 13.6 lb/ft3 3.9

ρ 1680 = = 1423.7 kg /m 3 1 + w 1 + 0.18

ρd a.=

Gs ρ w

e b.=

= n

ρd

= −1

(2.73)(1000) = − 1 0.918 1423.7

e 0.918 = = 0.479 1 + e 1 + 0.918

wGs  (0.18)(2.73)  =  (100) = 53.5% e  0.918 

S c.=

= d. ρsat

(Gs + e)γ w (2.73 + 0.918)(1000) = = 1901.98 kg/m3 1+ e 1 + 0.918

Water to be added: rsat – r = 1901.98 – 1680 ≈ 222 kg/m3 Gs γ w (2.68)(9.81) (1780)(9.81) e = −1 = − 1 0.506 = 17.46 kN/m3 ; = γd 17.46 1000

= γd 3.10

= w

e  0.506  =  = 18.88%  (100) Gs  2.68 

a. e 3.11 =

0.35 = 0.538 1 − 0.35

= γ sat

(Gs + e)γ w (2.69 + 0.538)(9.81) = = 20.59 kN/m3 1+ e 1 + 0.538

b. γ =

Gs γ w (1 + w) (2.69)(9.81)(1 + w) ; 17.5 = ; w = 1.99% 1+ e 1 + 0.538

= (0.23)(2.62) = 0.603 e wG 3.12 = s = ρd

Gs ρ w (2.62)(1000) = = 1634.4 kg/m3 1+ e 1 + 0.603

ρsat = ρd (1 + w) = (1634.4)(1 + 0.23) = 2010.3 kg/m3 14 © 2022 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Gs γ w (1 + w) (2.71)(62.4)(1 + 0.215) = = 117.4 lb/ft3 1+ e 1 + 0.75

= a. γ 3.13

γd b.=

117.4 γ = = 96.6 lb/ft3 1 + w 1 + 0.215

S c.=

wGs  (0.215)(2.71)  =  = 77.7%  (100) 0.75 e 

e 3.14 =

= γd

wGs (0.182)(2.67) = = 0.607 S 0.8 Gs γ w (2.67)(62.4) = = 103.7 lb/ft3 1+ e 1.607

γ = γ d (1 + w) = (103.7)(1.182) = 122.6 lb/ft3 3.15

a. γ =

Gs γ w (1 + w) (2.7)(62.4)(1.1) ; 112.32 = ; e = 0.65 1+ e 1+ e

γ w (Gs + e)

(62.4)(2.7 + 0.65) = 126.7 lb/ft3 1.65

γ w (Gs + Se)

(62.4)(Gs + 0.5e) 1+ e

= b. γ sat

3.16

a. γ =

= 1+ e

1+ e

; 105.73 =

Gs 1.694 + 1.194e (a) = 112.67 =

(62.4)(Gs + 0.75e) (b) 1+ e

From Eqs. (a) and (b), 112.67 =

(62.4)(1.694 + 1.194e + 0.75e) ; e = 0.81 1+ e

b. From Eq. (a): 2.66 Gs = 1.694 + (1.194)(0.81) =

γd 3.17 =

Gs γ w (2.66)(62.4) = = 917 lb/ft3 1+ e 1 + 0.81