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

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)

a.

Sieve no.

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

4

0.0

0.0

100.0

10

18.5

4.4

95.6

20

53.2

12.6

83.0

40

90.5

21.5

61.5

60

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

0

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

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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

4

0.0

0.0

100

6

30

6.0

94.0

10

48.7

9.74

84.26

20

127.3

25.46

58.80

40

96.8

19.36

39.44

60

76.6

15.32

24.12

100

55.2

11.04

13.08

200

43.4

8.68

4.40

Pan

22

4.40

0

Σ 500 g The grain-size distribution is shown in the figure.

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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

28

4.54

95.46

10

42

6.81

88.65

20

48

7.78

80.87

40

128

20.75

60.12

60

221

35.82

24.3

100

86

13.94

10.36

200

40

6.48

3.88

Pan

24

3.88

0

Σ 617 g The graph for percent finer versus grain size is shown.

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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

6

0

0.0

100

10

0

0.0

100

20

9.1

1.82

98.18

40

249.4

49.88

48.3

60

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

0

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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.

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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%

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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

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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

0

0

0.0

19.0

60

10

30

6.66

93.34

12.7

130

75

75

18.65

74.69

9.5

65

80

45

12.65

62.04

4.75

100

165

90

23.64

38.4

2.36

50

25

65

9.32

29.08

0.6

40

60

75

11.65

17.43

0.075

50

70

105

14.98

2.45

Pan

5

15

15

2.33

≈0

100

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

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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)

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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

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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

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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

95

 (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

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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

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γ w (Gs + e)

= γ sat

= 1+ e

(62.4)(2.66 + 0.81) = 119.6 lb/ft3 1 + 0.81

− γ d ) (2.5)(119.6 −= 91.7) 69.75 lb Water = (2.5) (γ sat= 3.18

For the borrow pit, γ d −= borrow

Gs γ w Gs γ w = = 0.476Gs γ w 1 + eborrow 1 + 1.1

For the compacted backfill, γ d −= backfill

Gs γ w Gs γ w = = 0.555Gs γ w 1 + ebackfill 1 + 0.8

 γ d − backfill  0.555Gs γ w (1) 1.16 m3 Volume of borrow soil needed =  =  (V ) = γ 0.476 G γ s w  d − borrow 

a. γ borrow 3.19 =

S b.=

wGs (0.11)(2.7) = = 0.27 = 27% eborrow 1.1

= c. γ backfill

3.20

Gs γ w (2.67)(9.81) = = 16.87 kN/m3 1+ e 1 + 0.553

e =emax − Dr (emax − emin ) =0.75 − (0.78)(0.75 − 0.46) =0.524

= γ

3.22

(1 + w)Gs γ w (1 + 0.11)(2.7)(9.81) = = 16.3 kN/m3 1 + ebackfill 1 + 0.8

e =emax − Dr (emax − emin ) =0.78 − (0.65)(0.78 − 0.43) =0.553

γ = 3.21

(1 + w)Gs γ w (1 + 0.11)(2.7)(9.81) = = 14 kN/m3 1 + eborrow 1 + 1.1

Gs γ w (1 + w) (2.68)(9.81)(1.09) = = 18.8 kN/m3 1+ e 1.524

γ d (min) =

Gs γ w (2.65)(62.4) ; 92 = ; emax = 0.8 1 + emax 1 + emax

γ d (max) =

Gs γ w (2.65)(62.4) ; emin = 0.532 ; 108 = 1 + emin 1 + emin

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e =emax − Dr (emax − emin ) =0.8 − (0.6)(0.8 − 0.532) =0.639 = γ

3.23

Gs γ w (1 + w) (2.65)(62.4)(1.08) = = 109 lb/ft3 1+ e 1.639

a. γ d =

= S

Gs γ w (2.65)(62.4) ; 100 = ; e = 0.654 1+ e 1+ e

wGs (0.184)(2.65) = = 74.6% 0.654 e

b. This is when S = 100%, so e = wGs = (0.184)(2.65) = 0.488 = γd

3.24

Gs γ w (2.65)(62.4) = = 111.1 lb/ft3 1+ e 1 + 0.488

a. Dr =

emax − e1 emax − emin

e1 =emax − Dr (emax − emin ) =0.9 − (0.4)(0.9 − 0.46) =0.724

γd =

Gs γ w (2.65)(62.4) = = 95.9 lb/ft3 1 + e1 1 + 0.724

b. e2 =emax − Dr (emax − emin ) =0.9 − (0.75)(0.9 − 0.46) =0.57 e − e ∆H 0.724 − 0.57 ∆H ∆e = = 1 2 ;= H 1 + e1 1 + e1 6 1 + 0.724

∆H = 0.536 ft = 6.43 in.

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Chapter 4 4.1

a. Refer to the plot of w versus N. LL = 29.0 b. PI = LL – PL = 29.0 – 13.4 = 15.6

= LI 4.2

4.3

w − PL 32 − 13.4 = = 1.19 LL − PL 15.6

a. From the plot, LL = 39.7

b. PI = LL – PL = 39.7 – 18.7 = 21 19 © 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.

= LI 4.4

4.5

w − PL 26 − 18.7 = = 0.348 LL − PL 39.7 − 18.7

LL = 28.5; PL = 12.2 PI = 28.5 – 12.2 = 16.3 Eq. (4.14):

 LL + 43.5  = SL 46.4   − 43.5  PI + 46.4   28.5 + 43.5  = 46.4  =  − 43.5 9.78  16.3 + 46.4   LL + 43.5  = 4.6 SL 46.4   − 43.5  PI + 46.4   39.7 + 43.5  = 46.4  =  − 43.5 13.78  21 + 46.4 

4.7

From the plot, LL ≈ 33.8 I FC =

=

w2 (%) − w1 (%) log d 2 − log d1

42.6 − 26.3 ≈ 43.2 log(31) − log(13)

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4.8

= LL Eq. (4.5):

w 28.5 = = 30.08 0.77 log d (0.77) log(17)

= Eq. (4.6): LL

w 28.5 = = 30.08 0.65 + 0.0175d 0.65 + (0.0175)(17) 0.33

0.33

 20   20  Eq. (4.7): LL w= = (28.5)= 30.07      d   17 

4.9

a. The liquidity index is given by: LI =

w − PL LL − PL

The range of the liquidity index corresponding to the range in natural water content is calculated and listed in the table.

Soil

% clay (<0.002 mm in size)

wn(%)

LL

PL

LI

1

34

59-67

49

26

1.43-1.78

2

44

18-36

37

21

−0.18-0.94

3

54

51-56

61

26

0.71-0.86

4

81

61-70

58

24

1.08-1.35

5

28

441-600

511

192

0.78-1.28

6

67

98-111

132

49

0.59-0.75

7

72

51-65

89

31

0.34-0.57

b. Soils 1 and 4: Since the LI range is greater than 1.0, the water content is greater than the liquid limit.

From Figure 4.1, the soil behaves like a viscous fluid with practically no shearing resistance.

Soil 2: At a water content of 18%, the LI < 0, and the soil behaves like a brittle solid with high shear resistance. At a water content of 36%, the soil is in a plastic state (0 < LI < 1) showing moderate shearing resistance and ductile behavior.

Soils 3, 6, and 7: Since 0 < LI < 1, the water content is less than the liquid limit. From Figure 4.1, the soil is in the plastic state showing moderate shearing resistance and ductile behavior.

Soil 5: At a water content of 441%, the soil is in the plastic state (0 < LI < 1) with moderate shear resistance. At a water content of 600%, the LI > 1, and the soil becomes a viscous fluid with practically no shearing resistance.

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4.10

a. Eq. (4.10): PL = 0.04(LL) + 0.26(CF) + 10 = (0.04)(61) + (0.26)(54) + 10 = 26.48 b. Eq. (4.11): PI = 0.96(LL) − 0.26(CF) − 10 = 0.96(61) − 0.26(54) − 10 = 34.52

c. Eq. (4.19):

0.96( LL) − 0.26(CF ) − 10 PI = CF CF 34.52 = ≈ 0.64 54 A

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Chapter 5 5.1

SOIL A: From Table 5.1, the soil is A-2-4. The GI for A-2-4 is zero. So, the classification is A-2-4(0).

SOIL B: Table 5.1. Soil is A-3. GI = 0. Classification: A-3 (0) SOIL C: Table 5.1. Soil is A-2-6. Equation (5.2): GI = 0.01(F200 − 15)(PI − 10) = 0.01(12 − 15)(13 − 10) = −0.09 ≈ 0 Classification: A-2-6(0) SOIL D: Table 5.1. Soil is A-2-7. Equation (5.2): GI = 0.01(F200 − 15)(PI − 10) = 0.01(30 − 15)(18 − 10) = 1.2 ≈ 1 Classification: A-2-7(1)

5.2

SOIL E: Table 5.1. Soil is A-1-b. The group index for A-1-b is 0.

Classification is A-1-b(0) SOIL A: Table 5.1. Soil is A-7-5.

Note: PI = 21, which is less than LL − 30 = 52 − 30 = 22 GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (72 − 35)[0.2 + 0.005(52 − 40)] + 0.01(72 − 15)(21 − 10) = 15.89 ≈ 16 Classification: A-7-5(16) SOIL B: Table 5.1. Soil is A-6.

GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (58 − 35)[0.2 + 0.005(38 − 40)] + 0.01(58 − 15)(12 − 10) = 5.23 ≈ 5 Classification: A-6(5) SOIL C: Table 5.1. Soil is A-7-6.

Note: PI = 14 is greater than LL − 30 = 11 GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (64 − 35)[0.2 + 0.005(41 − 40)] + 0.01(64 − 15)(14 − 10) = 7.905 ≈ 8 Classification: A-7-6(8) SOIL D: Table 5.1. Soil is A-6.

GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (82 − 35)[0.2 + 0.005(32 − 40)] + 0.01(82 − 15)(12 − 10) = 8.86 ≈ 9 Classification: A-6(9) SOIL E: Table 5.1. Soil is A-6.

GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (48 − 35)[0.2 + 0.005(30 − 40)] + 0.01(48 − 15)(11 − 10) = 2.28 ≈ 2 23 © 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.

Classification: A-6(2) 5.3

Soil 1: F200 = 50; PI = 38 − 29 = 9. From Table 5.1, soil is A-4. Eq. (5.1): GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (50 − 35)[0.2 + 0.005(38 − 40)] + 0.01(50 − 15)(9 − 10) = 2.8 ≈ 3 Classification: A-4(3) Soil 2: F200 = 80; PI = 56 − 23 = 33. Soil is A-7-6. Eq. (5.1): GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (80 − 35)[0.2 + 0.005(56 − 40)] + 0.01(80 − 15)(33 − 10) = 27.55 ≈ 28 Classification: A-7-6(28) Soil 3: F200 = 65. LL = 37; PI = 37 − 22 = 15. From Table 5.1, soil is A-6.

GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (65 − 35)[0.2 + 0.005(37 − 40)] + 0.01(65 − 15)(15 − 10) = 8.05 ≈ 8 Classification: A-6(8) Soil 4: F200 = 45. LL = 28; PI = 28 − 20 = 8. From Table 5.1, soil is A-4.

GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (45 − 35)[0.2 + 0.005(28 − 40)] + 0.01(45 − 15)(8 − 10) = 0.8 ≈ 1 Classification: A-4(1) Soil 5: F200 = 62. LL = 43; PI = 43 − 28 = 15. From Table 5.1, soil is A-7-6.

GI = (F200 − 35)[0.2 + 0.005(LL − 40)] + 0.01(F200 − 15)(PI − 10) = (62 − 35)[0.2 + 0.005(43 − 40)] + 0.01(62 − 15)(15 − 10) = 8.155 ≈ 8 Classification: A-7-6(8) Soil 6: F200 = 6. From Table 5.1, the soil is A-1-a.

GI = 0 Classification: A-1-a(0) Soil 7: F200 = 30. LL = 32; PI = 32 − 24 = 8. From Table 5.1, soil is A-2-4.

GI = 0 Classification: A-2-4(0) Soil 8: F200 = 34. LL = 37; PI = 37 − 25 = 12. From Table 5.1, soil is A-2-6.

Eq. (5.2): 24 © 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.

GI = 0.01(F200 − 15)(PI − 10) = 0.01(34·− 15)(12 − 10) = 0.38 ≈ 0 Classification: A-2-6(0) Soil 9: F200 = 8. From Table 5.1, the soil is A-3.

GI = 0 Classification: A-3(0) Soil 10: F200 = 32. LL = 44; PI = 44 − 35 = 9. From Table 5.1, the soil is A-2-5.

GI = 0 Classification: A-2-5(0) 5.4

SOIL 1: Fine fraction = % passing No. 200 sieve = 30% Coarse fraction = 100 − 30 = 70% Gravel fraction = 100 − 70 = 30% Sand fraction: 70 − 30 = 40% More than 50% of coarse fraction passing No. 4 sieve, so sandy soil.

Table 5.2 and Figure 5.2: SC Figure 5.3; more than 15% gravel. Clayey sand with gravel SOIL 2: Coarse fraction = 200 − 20 = 80% Gravel fraction = 100 − 48 = 52% Sand fraction = 80 − 52 = 28% Table 5.2 Gravelly soil.

Table 5.2 and Figure 5.2: GC Figure 5.3, ≥15% sand, so clayey gravel with sand SOIL 3: Coarse fraction = 100 − 70 = 30% Gravel fraction = 100 − 95 = 5% Sand fraction = 95 − 70 = 25% LL = 52; PI = 28. From Table 5.2, it is a fine-grained soil.

Table 5.2 and Figure 5.2: CH

From Figure 5.4, ≥30% plus 200, % sand > % gravel, <15% gravel, so sandy fat clay SOIL 4: Coarse fraction = 100 − 82 = 18% Gravel fraction = 100 − 100 = 0% Sand fraction = 18 − 0 = 18% LL = 30; PI = 19. From Table 5.2 and Figure 5.2: CL Figure 5.4: lean clay with sand SOIL 5: Coarse fraction = 100 − 74 = 26% Gravel fraction = 100 − 100 = 0% 25 © 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.

Sand fraction = 26% Fine-grained soil. LL = 35; PI = 21 From Table 5.2 and Figure 5.2: CL Figure 5.4: lean clay with sand SOIL 6: Coarse fraction = 100 − 26 = 74% Gravel fraction = 100 − 87 = 13% Sand fraction = 74 − 13 = 61% Table 5.2: coarse-grained soil; LL = 38; PI = 18 Table 5.2 and Figure 5.2: SC Figure 5.3: <15% gravel: clayey sand SOIL 7: Coarse fraction = 100 − 78 = 22% Gravel fraction = 100 − 88 = 12% Sand fraction = 22 − 12 = 10% Table 5.2: fine-grained soil; LL = 69; PI = 38 Table 5.2 and Figure 5.2: CH Figure 5.4: <30% plus 200; % sand < % gravel: fat clay with gravel SOIL 8: Coarse fraction = 100 − 57 = 43% Gravel fraction = 100 − 99 = 1% Sand fraction = 43 − 1 = 42% LL = 54; PI = 26. Table 5.2 and Figure 5.2: CH Figure 5.4: ≥30% plus 200; % sand > % gravel: sandy fat clay SOIL 9: Coarse fraction = 100 − 11 = 89% Gravel fraction = 100 − 71 = 29% Sand fraction = 89 − 29 = 70% LL = 32; PI = 16; Cu = 4.8; Cc = 2.9. Table 5.2 and Figure 5.2: SP-SC Figure 5.3: poorly graded sand with clay and gravel SOIL 10: Coarse fraction = 100 − 2 = 98% Gravel fraction = 100 − 100 = 0% Sand fraction = 98 − 0 = 98% Cu = 7.2; Cc = 2.2. Table 5.2: SW Figure 5.3: <15% gravel: well graded sand SOIL 11: Coarse fraction = 100 − 65 = 35% Gravel fraction = 100 − 89 = 11% Sand fraction = 35 − 11 = 24% LL = 44; PI = 21. Table 5.2 and Figure 5.2: CL 26 © 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.

Figure 5.4: sandy lean clay SOIL 12: Coarse fraction = 100 − 8 = 92% Gravel fraction = 100 − 90 = 10% Sand fraction = 92 − 10 = 82% LL = 39; PI = 31; Cu = 3.9; Cc = 2.1 Table 5.2 and Figure 5.2: SP-SC Figure 5.3: poorly graded sand with clay Soil 1: 3% passing No. 200 sieve; 90% passing No. 4 sieve. So, it is sandy soil.

From Table 5.2, it is SW or SP. Cu = 4.48 < 6, so the symbol is SP.

5.5

Gravel fraction = 100 − 94 = 6%. From Figure 5.4, group name is poorly graded sand.

Soil 2: 77% passing No. 200 sieve, so it is fine grained soil. LL = 63; PI = 25.

From Table 5.2 and Figure 5.2, the group symbol is MH. Plus No. 200 sieve is 100 − 77 = 23% < 30%; % sand = 23%; % gravel = 0%.

From Figure 5.4, the group name is elastic silt with sand.

Soil 3: 86% passing No. 200 sieve, so it is fine grained soil. LL = 55; PI = 28.

From Table 5.2 and Figure 5.2, the group symbol is CH. Plus No. 200 sieve is 100 − 86 = 14%; gravel fraction = 0%; sand fraction = 14%.

From Figure 5.4, the group name is fat clay.

Soil 4: 45% passing No. 200 sieve, so it is coarse grained soil. 100% passing No. 4 sieve, so it is sandy soil. From Table 5.2, it is SM or SC. LL = 36; PI = 22. From Figure 5.3, the group symbol is SC. From Figure 5.3, the group name is clayey sand.

Soil 5: 48% passing No. 200 sieve and 92% passing No. 4 sieve, so it is sandy soil (SC or SM).

LL = 30; PI = 8.

From Table 5.2 and Figure 5.2, the group name is SC. Gravel fraction is 100 − 92 = 8%.

From Figure 5.3, the group name is clayey sand.

Soil 6: 40% passing No. 200 sieve and 60% passing No. 4 sieve. So, Gravel fraction = 100 − 60 = 40%; coarse fraction = 100 − 40 = 60%; Sand fraction = 60 − 40 = 20%. LL = 26; PI = 4.

From Table 5.2 and Figure 5.2, the group symbol is GM-GC. From Figure 5.3, the group name is silty clayey gravel with sand.

Soil 7: 76% passing No. 200 sieve, so it is fine grained soil. LL = 60; PI = 32.

From Table 5.2 and Figure 5.2, the group symbol is CH. 27 © 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.

5.6

From Figure 5.4, the group name is fat clay with sand.

Coarse fraction = 100 − 11 = 89% Gravel fraction = 9% Sand fraction = 91 − 11 = 80% Table 5.2: coarse-grained soil. More than 50% of the coarse fraction passing No. 4, so sandy soil. Since fine fraction is between 5 and 12%, dual symbols are required.

LL = 32; PI = 32 − 27 = 5; Figure 5.2: plots below the A-line.

From the given data, D60 = 1.9 mm; D30 = 0.8 mm; D10 = 0.1 mm C = u

2 D60 1.9 D30 0.82 = == 19; Cc = = 3.36; poorly graded D10 0.1 ( D60 )( D10 ) (1.9)(0.1)

Table 5.2: Soil is SP-SM Figure 5.3: <15% gravel, so poorly graded sand with silt.

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Chapter 6 6.1

γw 62.4 62.4 γ zav = = = 1 1 w + 0.3636 Eq. (6.4): w+ w+ Gs 2.75 The table can now be prepared.

6.2

w(%)

γzav(lb/ft3)

5

150.9

8

140.7

10

134.6

12

129.0

15

121.5

ρd =

Gs ρ w G w [from Eq. (6.3)] 1+ s S

ρd@S(kg/m3) Gs 2.67

ρw (kg/m3) w(%) 80% 1000

90% 100%

10

2002 2059

2107

20

1601 1676

1741

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6.3

γd =

w(%) 100 3 (lb/ft )

Weight of soil γ = W (lb/ft3) mass, W (lb) w(%) V

Volume (ft3)

γ

1+

1 30

3.26

97.8

8.4

90.2

1 30

4.15

124.5

10.2

113.0

1 30

4.67

140.1

12.3

124.8

1 30

4.02

120.6

14.3

105.5

1 30

3.63

108.9

16.8

93.2

The plot of γd versus w is shown. γd ≈ 125 lb/ft3 @ w = 12.4%

6.4

Gs γ w (2.72)(62.4) ; 125 ; e ≈ 0.358 γ= d = 1+ e

S =

1+ e

wGs (0.124)(2.72) = = 0.94 = 94% e 0.358

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6.5 M (kg/m3) V

w(%)

ρd (kg/m3)

1.68

1781.0

9.9

1620.6

943.3

1.71

1812.8

10.6

1639.1

943.3

1.77

1876.4

12.1

1673.9

943.3

1.83

1940.0

13.8

1704.7

943.3

1.86

1971.8

15.1

1713.1

943.3

1.88

1993.0

17.4

1697.6

943.3

1.87

1982.4

19.4

1660.3

943.3

1.85

1961.2

21.2

1618.2

Volume of mold V (cm3)

Mass of wet soil M (kg)

943.3

ρ=

The plot of ρd versus w is shown.

From the graph, ρd(max) = 1716 kg/m3 and wopt = 15.5%.

6.6

= ρd

= R

ρ 1705 = 1 + w 1 + 10.5 = 1543 kg/m3 100

ρd

=

ρd (max)

1543 = 89.9% 1716

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6.7

3

γ (in situ ) = 105 lb/ft ;

γ= d (in situ )

105 = 88.98 lb/ft 3 ; γ d (compacted) = 103.5 lb/ft 3 18 1+ 100

 103.5  3 = 000)  Volume of soil = to be excavated (10,  11, 632 yd  88.98 

= (11, 632 × 27)(105) lb Weight of moist soil to be transported = Number of truck loads 6.8

(11, 632)(27)(105) = 824.4 ≈ 825 (20)(2000)

Dry unit weight of solids required

 Gs × 9.81  = 8000 = γ d kN (8000)  =  46,164.7Gs kN  1.7 

Volume to be excavated from borrow pit = Cost/m3 [46,164.7 Gs/γd(borrow pit)] ($)

Borrow pit

γd at borrow pit (kN/m3)

A

Gs × 9.81 = 5.39Gs 1 + 0.82

8564.9 m3

11

94,213.90

B

Gs × 9.81 = 4.67Gs 1 + 1.1

9885.4 m3

8

79,083.32

C

Gs × 9.81 = 5.16Gs 1 + 0.9

8946.6 m3

9

80,519.40

D

Gs × 9.81 = 5.51Gs 1 + 0.78

8378.3 m3

10

83,783.00

Total cost ($)

Borrow Pit B

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6.9

From Eq. (6.15), w= = = 14.3 0.65( PL ) 0.65(38 − 16) opt

From Eq. (6.17),

γ d (max) 2.312e −0.012( PL ) 2.312 == = e −0.012(38−16) 1.78 γw 6.10

From Eq. (6.22),

 LL  wopt (%) = −0.86( LL) + 3.04   + 2.2  Gs   38  = −0.86(38) + 3.04  12.62%  + 2.2 =  2.68 

From Eq. (6.24),

γ d (max) γw

(

)

−0.295 = 4.11 w opt ( PI 0.032 ) − 0.2446

0.032 = 4.11(12.62−0.295 )(16) = − 0.2446 1.88

6.11

 γ d (field) − γ d (min)   γ d (max)  From Eq. (6.26): Dr =     γ d (max) − γ d (min)   γ d (field)   γ d (field) − 93   104  3 0.78 =  ; γ d (field) 101.35 lb/ft   104 − 93   γ d (field) 

= R(%)

6.12

Dr

γ d (field)  101.35  = (100)  =  (100) 97.5% γ d (max)  104 

 ρ d (field) − ρ d (max)   ρ d (max)   ρ d (field) − 1510   1682  =   ; 0.7     1682 − 1510   ρ d (field)   ρ d (max) − ρ d (min)   ρ d (field) 

ρ d (field) = 1626.4 kg /m 3 = R(%)

ρd (field)  1626.4  ×100 =  = 96.7%  (100) ρd (max)  1682  33

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6.13

= R 0.9 a. =

γ d (field) γ d (field) = = 97.2 lb /ft 3 ; γ d (field) γ d (max) 108

 γ d (field) − γ d (min)   γ d (max)   97.2 − 93   108  = = = 31.1% D= b. D=     r r   0.311  γ d (max) − γ d (min)   γ d (field)   108 − 93   97.2  c. γ = γ d (1 + w) = (97.2)(1 + 0.12) = 108.86 lb /ft 3 6.14

In the field: Sand used to fill the hole and cone: 5.99 kg − 2.81 kg = 3.18 kg Sand used to fill the hole: 3.18.kg − 0.117 kg = 3.063 kg Volume of the hole:

Moist density of compacted soil:

γ =

γd =

3.063 kg = 0.001837 m3 1667 kg/m3

3.331 = 1813.2 kg/m3 0.001837

(1813.2)(9.81) = 17.79 kN/m3 1000

γ 17.79 = = 15.94 kN /m 3 w(%) 11.6 1+ 1+ 100 100

S N (1.7) 6.15 =

3 1 1 3 1 1 + + = (1.7) + + = 18 2 2 2 2 2 ( D50 ) ( D20 ) ( D10 ) (1.3) (0.19) (0.11) 2

Rating: GOOD

S N (1.7) 6.16 =

3 1 1 + + = 20.65 2 2 (0.61) (0.25) (0.09) 2

Rating: FAIR

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Chapter 7 k 7.1 =

7.2

QL (350)(45) = = 0.054 cm/s Aht (23)(70.0)(180 s)

k a.=

(620 cm 2 )(30 cm) QL = = 1.18×10−2 cm /s 2 Aht (175 cm )(50 cm)(180 s)

 1+ e  = b. vs v=   ; v ki  e   1+ e   50 cm   1 + 0.58  = vs ki=   (0.0118)   =  0.0536 cm /s  e   30 cm   0.58 

7.3

k=

(7300)(30.5) QL ; 0.015 = (95)(h)(60 × 60) Aht

h = 43.03 cm  43.03  v= ki= (0.015)  =  0.0212 cm /s  30.5 

 h1   (1.3)(50.0)   aL   76.0  = = log10  7.4 a. k 2.303   log10   (2.303)     At   30.0   (26)(10)   h2 

= 0.2324 cm /min h   aL  b. k = 2.303   log10  1   At   h2   76   76.0   (1.3)(50.0)  0.2324 (2.303) =  (26)(5)  log10  h  1.15log  h     2  2 

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 76.0   76.0  = = 0.2021 log  ; 1.59   10   h2   h2 

h2 = 47.8 cm = 478 mm 7.5

h   aL   a × 38   650  k = 2.303   log10  1  ; 0.175 = (2.303)   log10    6.5 × 8   300   At   h2 

a = 0.31 cm 2 7.6

h   aL  a. k = 2.303   log 10  1   At   h2   1.05 × 70   800  = (2.303)   log10   = 0.216 cm/min = 0.000036 m/s  20 × 8   500 

= K

kη (0.000036 m/sec)(1.005 × 10−3 N ⋅ s/m 2 ) = = 3.7 × 10−12 m 2 3 3 γw 9.789 × 10 N/m

h   aL  b. k = 2.303   log10  1   At   h2   800   1.05 × 70  0.216 cm/min = (2.303)    log10   20 × 6   h2  h2 = 562.3 mm 7.7

q = kiA

= i

k

160 − 145 = 0.0833 180 2.8 2.8 m/day = 0.00194 m/min = (24)(60)

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q = (0.00194)(0.0833)(1280) = 0.207 m3/min

7.8

i = From Figure 7.9:

head loss S tan α = = sin α length  S     cos α 

= q kiA = (k )(sin α )(3.8cos α )(1) 5.2 ×10−4 cm/s = 5.2 ×10−6 m/s k= q= (5.2 ×10−6 )(sin12°)(3.8cos12°) (3600) = 0.01447 m3 /hr/m  to change to m/hr

≈ 14.47 ×10−3 m 3 /hr /m 7.9

i=

h  L     cos α 

   h cos α   0.05   3.1cos 5  q = kiA = k   ( H1 cos α × 1) =  2    (2.8cos 5)  L     60   10  m/s 

(0.0005)(0.0515)(2.79) = 7.18×10−5 m 3 /s /m

7.10

e13 k 1+ e 0.022 Eq. (7.31): 1 = 3 1 , or= k2 e2 k2 1 + e2

 0.53    1 + 0.5  0.0833 =  0.73  0.2018    1 + 0.7 

k2 = 0.053 cm/s n1 0.31; = e1 7.11 =

= n2 0.4; = e2

n1 0.31 = = 0.449 1 − n1 1 − 0.31

0.4 = 0.667 1 − 0.4

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3

 e23   1 + e1   1 + e1   e2   1.449   0.667  Eq. (7.31): k2 k1 = =   3  k=    (6.1)  1    1.667   0.449   1 + e2   e1   1 + e2   e1 

3

= 17.38 cm/min 7.12

e =emax − (emax − emin ) Dr e =0.7 − (0.7 − 0.46)(0.6) =0.556  e3  Eq. (7.32): k (cm/sec) = 2.4622  D102  1+ e  

0.7825

0.7825

3   2  0.556 = (2.4622) (0.2) 0.0354 cm /s    1 + 0.556   

7.13 Sieve no. Opening (cm)

Percent passing

Fraction between two consecutive sieves (%)

30

0.06

100

{– – – – 20

40

0.0425

80

{– – – – 12

60

0.02

68

{– – – – 40

100

0.015

28

{– – – – 28

200

0.0075

0

For fraction between  fi 20 = = 407.98  0.404 0.595 0.404 sieve Nos. 30 and 40  Dli × Dsi × 0.04250.595 0.06 For fraction between  fi 12 441.03 = =  0.404 0.595 0.404 sieve Nos. 40 and 60  Dli × Dsi 0.0425 × 0.020.595 For fraction between  fi 40 2362.8 = =  0.404 0.595 0.404 sieve Nos. 60 and 100  Dli × Dsi 0.02 × 0.0150.595

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For fraction between  fi 28 2812.2 = =  0.404 0.595 0.404 sieve Nos. 100 and 200  Dli × Dsi 0.015 × 0.00750.595 100 Σ100% = Σ 0.404 = 0.0166 0.595 404.98 + 441.03 + 2362.8 + 2812.2 Dli × Dsi

2

3  1   0.5  k= 0.0108 cm /s (1.99 ×10 )(0.0166)  =    6.5   1 + 0.5  4

2

k1  e1n   1 + e2   1 + e2   e1  = 7.14 =      k2  1 + e1   e2n   1 + e1   e2 

n

n

1.2 ×10−6  2.4   0.8  n = =   ; 0.25 0.571 −6 3.6 × 10  1.8   1.4 

n =

log 0.25 −0.602 = = 2.477 log 0.571 −0.243

 e1n  (1.2 ×10−6 )(1 + 0.8) k1 = C  = = 3.754 × 10−6 cm/s ; C 2.477 1 + e 0.8  1   0.92.477  = = ×10−6 ) 1.52×10−6 cm /s k3   (3.754 1.9  

7.15

e −e  2.1 − 1.1  −6 = log(0.91 ×10−6 ) −  log k = log k0 − 0  = 0.1×10 cm/s × 0.5e0 0.5 2.1  

7.16

k H (eq)=

= k H (eq)

1 (k1 H1 + k2 H 2 + ....)

H 1 [(10−5 )(1.5) + (300 ×10−5 )(2.5) + (3.5 ×10−5 )(3)] 7 39

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10−5 (1.5 + 750 + 10.5)= 108.86 ×10−5 cm/s = 7

kV (eq) =

H 7 = H1 H 2 1.5 2.5 3 + + + + .... −5 −5 10 300 ×10 3.5 ×10−5 k1 k2

= 2.96 ×10−5 cm/s k H (eq) 108.86 ×10−5 = = 36.8 2.96 ×10−5 kV (eq) H 5 kV (eq) = = 1.5 1 1.5 1 H1 H 2 H 3 H 4 7.17 + + −4 + + + + −4 −4 20 ×10 2 ×10 10 3 ×10−4 k1 k2 k3 k4

= 2.079 ×10−4 cm/s = k H (eq)

=

1 ( H1k1 + H 2 k2 + H 3 k3 + H 4 k4 ) H

1 [(1.5)(20 × 10−4 ) + (1)(2 × 10−4 ) + (1.5)(10−4 ) + (1)(3 × 10−4 )] 5

= 7.3 × 10–4 cm/s

k H (eq) 7.3 × 10−4 = = 3.51 kV (eq) 2.079 × 10−4 60 H = kv (eq) = 20 20 20 H1 H 2 H 3 7.18 a. + + + + −3 −2 5 ×10 4.2 ×10 3.9 ×10−4 k1 k2 k3

= 1.076 × 10–3 cm/s  47   π  2 = (0.001076)    = (15 ) 0.149 cm3 /s = 536.4 cm 3 /hr q k= v (eq) iA  60   4 

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b. x = 0 mm Z = –220 mm u

γw h=

= 470 + 220 = 690 mm

u

γw

+ Z = 690 − 220 = 470 mm

x = 200 mm Z = –220 mm kv (eq)i = k1i1

 47   ∆h  (0.001076)   = (0.005)    60   20 

∆h = 3.371 cm = 33.71 mm Therefore, h =470 − 33.71 =436.29 mm u = 436.29 − (−= 220) 656.29 mm

γw

x = 400 mm Z = –220 mm kv (eq)i = k2i2

 47   ∆h  (0.001076)   = (0.042)    60   20 

∆h =0.4 cm = 4 mm Therefore,= h 436.29= − 4 432.29 mm u = 432.29 − (−= 220) 652.29 mm

γw

x = 600 mm Z = –220 mm kv (eq)i = k3i3

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 47   ∆h  (0.001076)   = (0.00039)    60   20 

∆h =43.224 cm = 432.24 mm Therefore, h = 432.29 − 432.24 ≈ 0 mm u

γw

= 0 − (−220) = 220 mm

c. The variation of heads with distance is shown in the figure.

v  47  = (0.001076) = = vs d. v k= v (eq) i  0.000843 cm/s. n  60 

Soil I: v= s

v 0.000843 = = 0.00168 cm /s n 0.5 42

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Soil II: v= s

v 0.000843 = = 0.0014 cm /s n 0.6

= Soil III: vs

0.000843 = 0.00255 cm /s 0.33

These values are plotted in the above figure (in Part c). e. Height of water column is equal to the piezometric or pressure head at a point; therefore, Height of water in A = pressure head at x = 200 mm = 656.29 mm Height of water in B = pressure head at x = 400 mm = 652.29 mm

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Chapter 8 8.1

The flow net is shown.

k= 4 ×10−4 cm/s

H = H1 − H 2 = 6.0 − 1.5 = 4.5 m

 4 ×10−4   4.5 × 4  q=   2  10   8  = 9 × 10−6 m3/m/s = 77.76 × 10−2 m3/m/day

8.2

The flow net is shown. = N f 3;= Nd 5

 Nf  q = kH    Nd   4 ×10−4  3 =  m/s  (3 − 0.5)   2 5  10  = 6 × 10−6 m3/m/s = 0.518 m3/m/day 45 © 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.

8.3

Based on the notations in Figure 8.12, H = (4 − 1.5) m = 2.5 m; S = D = 3.6 m; T′ = D1 = 6 m; S/T′ = 3.6/6 = 0.6 From the figure,

q ≈ 0.44 kH

 4 ×10−4  = q (0.44)(2.5)  × 60 × 60 × 24 m/day  = 0.38 m 3 /m /day 2  10  8.4

The flow net is shown.

 N f   0.002  5 3 = × 60 × 60 × 24 m/day  (10)  = q kH  =    7.2 m /m /day 2   12   N d   10 8.5

Refer to the flow net given in Problem 8.4 and the figure below. The flow net has 12 potential drops.

Also, H = 10 m. So the head loss for each drop = (10/12) m. Thus,

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Pressure head at D = (10 + 3.34) − (2)(10/12) = 11.67 m Pressure head at E = (10 + 3.34) − (3)(10/12) = 10.84 m

Pressure head at F = (10 + 1.67) − (3.5)(10/12) = 8.75 m

Pressure head at G = (10 + 1.67) − (8.5)(10/12) = 4.586 m Pressure head at H = (10 + 3.34) − (9)(10/12) = 5.84 m Pressure head at I = (10 + 3.34) − (10)(10/12) = 5 m The pressure heads calculated are shown in the figure. The hydraulic uplift force per unit length of the structure can now be calculated to be = γw(area of the pressure head diagram)(1)

 11.67 + 10.84    10.84 + 8.754   (1.67) +   (1.67)   2 2      =   8.75 + 4.586    4.586 + 5.84   5.84 + 5  +   (18.32) +   (1.67) +   (1.67)  2 2     2    

= (9.81)(18.8 + 16.36 + 122.16 + 8.71 + 9.05) = 1717.5 kN/m

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8.6

The flow net is shown. Nf = 3; Nd = 5.

 N f   10−3   4  4 −5 q = kH   =  2  (10 − 1.5)   = (10 )(8.5)    14   14   N d   10  = 2.429 × 10−5 m3/m/s ≈ 2.1 m3/m/day 8.7

b B= /2 4 m.

For this case, T ′ = 8 m; S = 4 m; H = H1 − H 2 = 6 m; B = 8 m;=

a.

b 4 S 4 x 3 − 0.75; = = 0.5 = = 0.5; x = b − x′ = 4 − 1 = 3 m; = b 4 T′ 8 T′ 8

From Figure 8.13, q/kH = 0.37.  0.001  = q (0.37)  × 60 × 60 × 24  (6) ≈ 1.92 m 3 /m /day 2  10 

b.

S b x 2 = 0.5; = 0.5; x = b − x′ = 4 − 2 = 2 m; = = 0.5. So q/kH = 0.4.

T′ T′ b 4

 0.001  = q (0.4)  × 60 × 60 × 24  (6) ≈ 2.07 m 3 /m /day 2  10 

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8.8

m; ∆ 7 = cot 35 10 m; 0.3∆ =3 m 40°; H = 7= 35°; α= α= 1 2

= d H1 cot α2 + L1 + ( H1 − H ) cot α1 + 0.3∆ = (10)(cot 40) + 5 + (10 − 7) cot = 35 + 3 24.2 m 2

2

d d2 H2 24.2  24.2   7  L= − − = −   −  =1.94 m 2 2 cos α2 cos α2 sin α2 cos 40  cos 40   sin 40 

 3 ×10−4   = q kL = tan α2 sin α2   (1.94)  (tan 40)(sin 40) 2  10  

= 3.139 ×10−6 m3 /m/s ≈ 0.271 m 3 /m /day 8.9

From Problem 8.8, d = 24.2 m; H = 7 m; a2 = 40° d 24.2 = = 3.46; m ≈ 0.25 (Figure 8.17) 7 H

= L

(0.25)(7) mH = = 2.72 m sin α2 sin 40

 3 ×10−4  2 = q kL = sin 2 α2   (2.72)(sin 40) 2  10 

3.37 ×10−6 m3 /m/s ≈ 0.291 m 3 /m /day = 49 © 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.

8.10

For this case, β = α = tan −1 ×

1 = 26.57° 2

Eq. (8.30): q = kL tan α sin α

 d2 d H2  − Eq. (8.32): L = −   2 2 cos α  cos α sin α 

d = (2)(30) + 5 + (2)(5) + 0.3[(2)(25)] = 60 + 5 + 10 + 15 = 90 m 2

90 90 25     L= −   −  cos 26.57  cos 26.57   sin 26.57 

2

= 100.63 − (100.63) 2 − (55.89) 2 = 16.95 m

11.37 ×10−4 m 3 /m /min q= (3 ×10−4 )(16.95)(tan 26.57)(sin 26.57) = 8.11

d 90 β 26.57° From Problem 8.10, d = 90 m, H = 25 m, = = 3.6 m,= H 25 = α 26.57° and From Figure 8.17, for

= L

d = 3.6, the value of m ≈ 0.34 H

mH (0.34)(25) = = 19 m sin α sin 26.57

Eq. (8.34):

(3 ×10−4 )(19)(sin 26.57) 2 = q= kL sin 2 α = 11.4×10−4 m 3 /m /min 8.12

Eq. (8.45): 2

h2=

B  B  + Hd −  + H d  − h12 (a) cot α  cot α 

Eq. (8.46): Hd H − h1 h = 2 (b) ln cot β H d − h1 cot α

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B = 7 m; α = 26.57°; H d = 35 m From Eq. (a): 2

7 7   h2 = + 35 −  + 35  − h12 = 49 − 2401 − h12 (c) cot 26.57  cot 26.57  From Eq. (b):

32 − h1  35  h2 ln  = 2  35 − h1  2 or

 35  = h2 (32 − h1 ) ln   (d)  35 − h1 

h1 (m)

h2 [Eq. (c)] (m)

h2 [Eq. (d)] (m)

h1 (m)

h2 [Eq. (c)] (m)

h2 [Eq. (d)] (m)

6

0.37

4.89

20

4.27

10.1

8

0.66

6.22

25

6.85

8.77

10

1.03

7.4

26

7.47

8.15

15

2.35

5.7

27

8.11

7.35

18

3.42

10.01

26.5

7.78

7.78

k= 4 ×10−5 mm/s = (4 ×10−5 )(60)(

1 )= 2.4 ×10−6 m/min 103

Eq. (8.44): q =

kh2 (2.4 ×10−6 )(7.78) = = 37.33×10−6 m 3 /m /min cot α cot 26.57

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Chapter 9 9.1

Point A: Point B:

σ = 0; u = 0; σ′ = 0 σ = (4)(16.2) = 64.8 kN/m2 u=0 σ′ = 64.8 − 0 = 64.8 kN/m2

Point C:

Point D:

σ = 64.8 + (1.5)(18.4) = 92.4 kN/m2 u = (1.5)(9.81) = 14.72 kN/m2 σ′ = 92.4 − 14.72 = 77.68 kN/m2 σ = 92.4 + (9)(19.81) = 270.69 kN/m2 u = 14.72 + (9)(9.81) = 103.01 kN/m2 σ′ = 270.69 − 103.01 = 167.68 kN/m2

9.2

γ d= (sand)

Gs γ w (2.65)(9.81) = = 17.33 kN/m3 1+ e 1 + 0.5

= γ sat (sand)

(Gs + e)γ w (2.68 + 0.65)(9.81) = = 19.8 kN/m3 1+ e 1 + 0.65

γ sat (clay) =

(Gs + e)γ w (2.71 + 0.81)(9.81) = = 19.08 kN/m3 1+ e 1 + 0.81

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kN/m2 Point

σ

A

0

B

u

(4)(17.33) = 69.32

σ′

0

0

0

69.32

C

69.31 + (1.5)(19.8) = 99.02

(9.81)(1.5) = 14.72

84.3

D

99.02 + (3)(19.08) = 156.26

(9.81)(1.5 + 3) = 44.15

112.11

9.3

γ d= (sand)

γ sat (sand) =

γ sat (clay) =

Gs γ w (2.65)(9.81) = = 17.1 kN/m3 1+ e 1 + 0.52 Gs γ w + eγ w (2.65 + 0.52)(9.81) = = 20.46 kN/m3 1+ e 1 + 0.52 Gs γ w + eγ w 1+ e

wG= e; G= s s

= γ sat (clay)

e 1.22 = = 2.71 w 0.45

(2.71 + 1.22)(9.81) = 17.37 kN/m3 2.22

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kN/m2 Point

u

σ

A

σ′

0 (17.1)(3) = 51.3

B

0

0

0

51.3

C

51.3 + (20.46)(2.5) = 102.45

(2.5)(9.81) = 24.53

77.92

D

102.45 + (17.37)(2) = 137.19

(2.5 + 2)(9.81) = 44.15

93.04

9.4

a. γ d= (sand)

= γ sat (clay)

Gs γ w (2.68)(62.4) = = 112.2 lb/ft 3 1+ e 1 + 0.49 (Gs + e)γ w (2.75 + 0.9)(62.4) = = 119.9 lb/ft 3 1+ e 1 + 0.9

lb/ft2 Depth (ft) 0 15 27

σ′

u

σ 0 (112.2)(15) = 1683 1683 + (119.9)(12) = 3121.8

0

0

0

1683

(62.4)(12) = 748.8

2373

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= γ sat (sand) b.

(Gs + e)γ w (2.68 + 0.49)(62.4) = = 132.8 lb/ft 3 1+ e 1 + 0.49

σ = (15)(132.8) + (12)(119.9) = 3430.8 lb/ft2 u = (27)(62.4) = 1684.8 lb/ft2 σ′ = 3430.8 − 1684.8 = 1746 lb/ft2 Change of σ′: 2373 − 1746 = 627 lb/ft2 c. Let the height of rise be h. So, at any time, at the bottom of the clay layer, σ = (15 − h)(112.2) + (h)(l32.8) + (12)(119.9) u = (12 + h)(62.4) Change of σ′: 300 = 2373 − [(1683 − 112.2h + 132.8h + 1438.8) − (748.8 + 62.4h)] h = 7.18 ft 9.5

Eq. (9.9):

icr =

γ ′ Gs − 1 2.68 − 1 1.68 = = = γ w 1+ e 1+ e 1+ e

e

icr

0.4

1.2

0.5

1.12

0.6

1.05

0.7

0.99

γ sat(clay) 9.6=

(1 + w)Gs γ w (1 + 0.4)(2.7)(62.4) = = 113.4 lb/ft 3 1 + wGs 1 + (0.4)(2.7)

Let the depth of the excavation be H. So, (25 − H)(113.4) − (18)(62.4) = 0 = σ′ H ≈ 15.1 ft 9.7

Let the depth of excavation be H, and height of water be h. Given: H = 18 ft. So, (25 − H)(l13.4) + (h)(62.4) − (18)(62.4) = 0 h

(18)(62.4) − (25 − 18)(113.4) = 5.28 ft 62.4

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i 9.8 =

h 0.4 = = 0.4 1 H2

 0.12  q= kiA = × 60 m/min  (0.4)(0.45) = 0.013 m 3 /min   100 

9.9

h 1.2 = = 0.6 H2 2

i a.=

300 cm 3 /s q= kiA = (0.1)(0.6)(0.5 ×1002 cm 2 ) = γ ′ Gs − 1 2.68 − 1 = = = 1.08 γ w 1 + e 1 + 0.55

icr b. =

Since i < icr, no boiling

i icr= c. =

h h ; 1.08= H2 2

h = 2.16 m 9.10

Equation (9.34): h(mm) =

C eD10

C == 10 mm2; h

10 = 85.5 mm (0.65)(0.18)

C == 50 mm2; h

50 = 427.4 mm (0.65)(0.18)

Range: 85.5 mm to 427.4 mm

= γ d (sand) 9.11

γ clay

(2.66)(9.81) = 17.4 kN/m3 1 + 0.5

(9.81)[2.71 + (0.6)(0.75)] = 17.71 kN/m3 1 + 0.75

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= γ sat(clay)

(9.81)(2.72 + 0.95) = 18.46 kN/m3 1 + 0.95

kN/m2 Depth (m) 0

u

σ 0 (17.4)(4) = 69.6

4 4 + 2.5 = 6.5

69.6 + (17.71)(2.5) = 113.88

6.5 + 3.5 = 10

113.88 + (18.46)(3.5) = 178.49

9.12

σ′ 0

0

0

69.6

(−0.6)(9.81)(2.5) = −14.72 0 (3.5)(9.81) = 34.34

84.32 113.88 144.15

From Problem 9.11, γd(sand) = 17.4 kN/m3; γsat (clay) = 18.46 kN/m3 γ (clay)

(9.81)[2.71 + (0.55)(0.75)] = 17.5 kN/m3 1 + 0.75

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kN/m2 Depth (m) 0

σ′

u

σ 0

3

(17.4)(3) = 52.2

4.8

52.2 + (1.8)(17.5) = 83.7

8.8

83.7 + (4)(18.46) = 157.54

0

0

0

52.2

(−0.55)(9.81)(1.8) = −9.71 0 (4)(9.81) = 39.24

61.91 83.7 118.3

The plot is given.

9.13

From Eq. (9.22), FS =

Dγ ′ Coγ w ( H1 − H 2 )

D = 4.5 m; γ′ = 17 − 9.81 = 7.19 kN/m3; H1 − H2 = 7 − 3 = 4 m; D/T = 4.5/12 = 0.375. From Table 9.1, Co = 0.354 (by linear interpolation).

= FS

(4.5)(7.19) = 2.33 (0.354)(9.81)(4)

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9.14

a. The flow nets for both cases are given below:

b. Determination of q/k: From Eq. (8.21): q = kH

Nf Nd

Case 1: Nf = 4, Nd = 11, H = 3.5 − 1 = 2.5 m

q 4 = (2.5) =   0.909 m k  11  60 © 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.

Case 2: Nf = 3.5, Nd = 13, H = 2.5 m

q  3.5  = (2.5) =   0.673 m k  13  c. FS =

icr [Eq. (9.23)] iexit

Gs − 1 2.66 − 1 = = 1.071 1+ e 1.55

icr =

Case 1: Refer to the flow net and Eq. (9.25):

iexit =

= FS

H 2.5 = = 0.643 N d l (11)(0.353) 1.071 ≈ 1.67 0.643

Case 2: Refer to the flow net and Eq. (9.25):

= iexit

= FS

H 2.5 = = 0.545 N d l (13)(0.353) 1.071 ≈ 1.97 0.545

d. From Eq. (9.18), seepage force per unit volume is γwi Case 1: Refer to the flow net. At A, = i

∆H (2.5/11) = = 0.143 l 1.59

Seepage force, γwi = (9.81)(0.143) = 1.4 kN/m3 Case 2: Refer to the flow net. At A, = i

∆H (2.5/13) = = 0.136 1.41 l

Seepage force, γwi = (9.81)(0.136) = 1.33 kN/m3 Installation of the sheet pile cut-off wall reduced the exit gradient and increased the factor of safety against heaving. Accordingly, at any point A, the seepage force also decreased due to a drop in the hydraulic gradient.

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Chapter 10 2

10.1

σ  σy +σx  σy −σx  2 a. 1  = ±   + τ xy σ3  2 2   σx = 60 kN/m2; σy = 100 kN/m2; τxy = +45 kN/m2

σ1  100 + 60  100 − 60 2 ±  + (45) 2 =  σ3  2 2   σ1 = 129.24 kN/m2; σ3 = 30.76 kN/m2

σ +σ 2

y x b. σ n = +

100 + 60 100 − 60 + cos[(2)(150)] + 45sin[(2)(150)] = 51.03 kN/m2 2 2

= σn

= τn

=

10.2

σy −σx cos 2θ + τ xy sin 2θ ; θ = 150° 2

σy −σx sin 2θ − τ xy cos 2θ 2

100 − 60 sin[(2)(150)] − 45 cos[(2)(150)] = 39.82 kN/m2 2 2

2

2

a. σ x = 36 kN/m ; σ y = 19 kN/m ; τ xy = −14 kN/m ; θ= 45°

σ1  19 + 36  19 − 36 2 = ±  + (−14) 2   σ3  2  2  σ1 = 43.88 kN/m2; σ3 = 11.12 kN/m2 19 + 36 19 − 36 + cos 90 − 14 sin 90 =13.5 kN /m 2 2 2

b. σ n =

19 − 36 sin 90 − (−14) cos 90 =−8.5 kN /m 2 2

τn =

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10.3

The Mohr’s circle is shown.

= a. OO1

150 + 100 = 125 kN/m 2 ; 2

= O1 B

 150 − 100  + (50) 2 55.9 kN/m 2   = 2  

O1O2 = 150 − 125 = 25 kN/m 2

2

σ 3 =OS =125 − 55.9 =69.1 kN/m 2 (+) σ1 =ON =125 + 55.9 =180.9 kN/m 2 (+) −1

 50   =63.43°  25 

b. ∠BO1O2 =tan 

σn = OO1 + O1 D cos 36.57 = 125 + 55.9 cos 36.57 = 169.89 kN/m 2 (+) = τ n O1= D sin 36.57 33.3 kN/m 2 (−)

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10.4

The Mohr’s circle is shown.

= OO1 a.

14.4 + 6 = 10.2 kN/m 2 ; 2

O1 B =

(4.2) 2 + (2.6) 2 = 4.94 kN/m 2

O1O2= 10.2 − 6= 4.2 kN/m 2

σ1 =ON =10.2 + 4.94 =15.14 kN /m 2 σ 3 =OS =10.2 − 4.94 =5.26 kN /m 2 −1

 2.6   =31.76°  4.2 

b. ∠CO1O2 =tan 

σ n = OO1 − O1D cos(31.76 + 40) = 10.2 − 4.94 cos 71.76 = 8.65 kN /m 2

= τ n 4.94sin = 71.76 4.69 kN /m 2 10.5

a. The Mohr’s circle is shown on the following page.

= σ1 ON = 95 kN /m 2 ; σ= OS = 30 kN /m 2 3 b. σn and τn are coordinates of D. So

σ n ≈ 94.2 kN /m 2 ; τ n ≈ 7.1 kN /m 2 (−) 65 © 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.

10.6

a. The Mohr’s circle is shown.

OS = 1.25 kN /m 2 = σ1 ON = 5.25 kN /m 2 ; σ= 3

b. σn and τn are coordinates of D. So

σ n ≈ 1.39 kN /m 2 ; τ n ≈ 0.74 kN /m 2

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10.7 r z

P ∆σ z = I1 z2

Load @

P (kN)

A

8.9

(32 + 1.52)0.5 = 3.35

3

1.12

0.0626

0.0619

B

17.8

(32 + 1.52)0.5 = 3.35

3

1.12

0.0626

0.1238

C

26.7

3

0.5

0.2733

0.8108

r (m)

z (m)

1.5

I1 (Table 10.1)

(kN/m2)

∆σ Z = Σ0.9965 kN /m 2 10.8

Eq. (10.18):

2q1 z 3 2q2 z 3 (2)(75)(2)3 (2)(300)(2)3 = ∆σ z + = + π[( x1 + x2 ) 2 + z 2 ]2 π[ x22 + z 2 ]2 π[(5) 2 + (2) 2 ]2 π[32 + 22 ]2 = 9.49 kN/m2 = 10.9 ∆σ z

2q1 z 3 2q2 z 3 + π[( x1 + x2 ) 2 + z 2 ]2 π[ x22 + z 2 ]2

(2)(300)(3)3 (2)(260)(3)3 + = 15.32 kN /m 2 2 2 2 2 2 2 π[(4 + 3) + (3) ] π[3 + 3 ]

=

= 10.10 ∆σ z

2q1 z 3 2q2 z 3 + π[( x1 + x2 ) 2 + z 2 ]2 π[ x22 + z 2 ]2

2q2 (0.9)3 (2)(10.9)(0.9)3 1.7 = + = 0.0248 + 0.0879q2 π[3.67 2 + 0.92 ]2 π[1.222 + 0.92 ]2 q2 = 19.06 kN/m 10.11

2q1 z 3 ∆σz at A due to q1 = π[ x 2 + z 2 ]2 or

= (∆σ z )1

(2)(250)(2)3 = 19.89 kN/m 2 2 2 2 π[(2) + (2) ]

Vertical component of q2 = q2 sin 45 67 © 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.

= (∆σ z ) 2

2q2 (sin 45) z 3 = ;(∆σ z ) 2 0.0043q2 π[(5) 2 + (2) 2 ]2

Horizontal component of q2 = q2 cos 45 From Eq. (10.20): (∆= σ z )3

2q2 xz 2 2q2 (cos 45)(5)(2) 2 = = 0.0107 q2 π ( x12 + x22 ) 2 π[52 + 22 ]2

Total vertical stress:

∆σ z = 30 kN/m 2 = (∆σ z )1 + (∆σ z ) 2 + (∆σ z )3

30 = 19.89 + 0.0043q2 + 0.0107q2 = q2

10.12

30 − 19.89 = 674 kN /m 0.015

B = 3.7 m; q = 16.8 kN/m2; x = 2.7 m; z = 1.5 m

∆σ z 2 x (2)(2.7) 2 z (2)(1.5) ≈ 0.19 = = 1.46; = = 0.82. From Table 10.4, B 3.7 B 3.7 q

= ∆σ z (0.19)(16.8) = 3.19 kN /m 2 2x B

10.13 =

∆σ z (2)(1.5) 2 z (2)(3) = 0.409 = 1;= = 2. From Table 10.4, B 3 3 q

= ∆σ z (60)(0.409) = 24.54 kN /m 2 10.14

Refer to the figure.

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For the left side (with the notations given in Figure 10.27): B1 0 B 12 = = 0; 2= = 2. From Figure 10.28, I2(L) = 0.37 6 z 6 z

For the right side: B1 9 B 12 = = 1.5; 2= = 2. From Figure 10.28, I2(R) = 0.485 z z 6 6

∆σ= q[ I 2( L ) + I 2( R= (132)(0.37 + 0.485) = 112.86 kN /m 2 z )] 10.15

At A:

For the left side:

For the right side:

B1 3 = = 0.5 6 z

B1 = 0.5 z

B2 24 = = 4 z 6

B2 =4 z

I2 = 0.468

I2 = 0.468

= ∆σ z (12)(18.87)(0.468 + 0.468) ≈ 211.95 kN /m 2

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At B:

For the left side:

For the right side:

B1 0 = = 0 6 z

B1 6 = = 1 z 6

B2 24 = = 4 6 z

B2 24 = = 4 z 6

I2 = 0.42

I2 = 0.48

= ∆σ z (12)(18.87)(0.42 + 0.48) ≈ 203.8 kN /m 2 At C:

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For the left side:

For the right side:

B1 =0 z

B1 30 = = 5 z 6

B2 24 = = 4 6 z

B2 24 = = 4 z 6

I2 = 0.42

I2 = 0.5

= ∆σ z (12)(18.87)(0.5 − 0.42) ≈ 18.81 kN /m 2 10.16

Eq. (10.31) and Table 10.8: q = 200 kN/m2

R (m) z (m)

z R

∆σ z q

∆σZ (kN/m2)

4

1.5

0.375

0.9567

191.34

4

3

0.75

0.784

156.8

4

6

1.5

0.4240

84.8

4

9

2.25

0.2369

47.38

4

12

3.0

0.1436

28.72

10.17

Eq. (10.32) and Tables 10.9 and 10.10: q = 96 kN/m2

z (m) r (m) R (m)

z R

r R

A′

B′

∆σZ (kN/m2)

1.5

0

3

0.5

0

0.55279

0.35777

87.4

1.5

0.6

3

0.5

0.2

0.54403

0.35752

86.5

1.5

1.2

3

0.5

0.4

0.51622

0.35323

83.5

1.5

2.4

3

0.5

0.8

0.38390

0.26236

62.04

1.5

3.6

3

0.5

1.2

0.18556

0.02165

19.9

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10.18

Refer to the Newmark’s chart.

The plan is drawn to scale.

= AB 4 m. M ≈ 65.

∆σ z = ( IV )qM = (0.005)(300)(65) = 97.5 kN/m2

10.19

a. Eqs. (10.40) and (10.39): n=

= Eq. (10.37): ∆σ z

L 4 B 2 = = 2; m= = = 1 z 2 z 2

qI 3 ; = I 3 0.1999

= ∆σ z (100)(0.1999) = 19.99 kN/m 2 ≈ 20 kN /m 2 b. Refer to the figure below:

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m For rectangle 1: =

1.2 2.4 n = 1.2; I3 = 0.1431 = 0.6; = 2 2

m For rectangle 2: =

0.8 2.4 = 0.4; = n = 1.2; I3 = 0.1063 2 2

m For rectangle 3: =

1.2 1.6 = 0.6; = n = 0.8; I3 = 0.1247 2 2

m For rectangle 4: =

0.8 1.6 = 0.4; = n = 0.8; I3 = 0.0931 2 2

∆σ= q[ I 3(1) + I 3(2) + I 3(3) + I 3(4)= ] (100)(0.1431 + 0.1063 + 0.1247 + 0.0931) z = 46.72 kN /m 2 c. Refer to the figure.

 stress at C due   stress at C due      = ∆σ z  to rectangular  −  to rectangular   area 5.2 m × 2 m   area 2 m × 1.2 m     

For rectangular area 5.2 m × 2 m: m=

5.2 2 n = 2.6; I3 = 0.202 = 1; = 2 2

m For rectangular area 1.2 m × 2 m: =

1.2 2 = 0.6; n= = 1; I3 = 0.1361 2 2

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∆σ z = q(0.202 − 0.1361) = (100)(0.202 − 0.1361) = 6.59 kN /m 2 10.20

Eqs. (10.44), (10.45), and (10.46): b=

B 2 = = 1m 2 2

m1=

L 4 = = 2 B 2

n1=

z 3.5 = = 3.5 b 1

From Table 10.12, I4 ≈ 0.242

∆σ z = qI 4 = (100)(0.242) = 24.2 kN /m 2 10.21

Eq. (10.31) and Table 10.8: q = (4)(9.81) = 39.24 kN/m2

R (m)

z (m)

z R

∆σ z q

∆σZ (kN/m2)

2.5

0

0

1

39.24

2.5

2

0.8

0.7562

29.53

2.5

4

1.6

0.3961

15.54

2.5

8

3.2

0.1322

5.18

2.5

10

4

0.0869

3.41

10.22

Eq. (10.32) and Tables 10.9 and 10.10: q = (5)(9.81) = 49.05 kN/m2

z (m)

r (m) R (m)

z R

r R

A′

B′

∆σZ (kN/m2)

2

0

4

0.5

0

0.55279

0.35777

44.66

2

2

4

0.5

0.5

0.49075

0.34214

40.85

2

4

4

0.5

1.0

0.28156

0.13591

20.48

2

6

4

0.5

1.5

0.09499

−0.03455

2.964

2

8

4

0.5

2.0

0.03701

−0.02651

0.51

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10.23

Refer to the figure below:

2 = α1 tan −1 =  18.43° 6 2 = α2 tan −1 =   63.43° 1

α= α2 − α1= 63.43° − 18.43°= 45° 1 δ = tan −1   = 26.56°; 2δ = 53.13° 2 From Eq. (10.24), ∆σ z 1  2x   1  2 × 6   π  = = 2δ  × 45  − sin 53.13 = °  0.1726  α − sin    q 2π  B  2π  5   180  

= ∆σ z (400)(0.1726) = 69.04 kN /m 2 75 © 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.

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Chapter 11 11.1

1 − µ s2 Se (flexible, center) = ∆σ (α B′) Is I f Es

∆= σ

600 = 300 kN/m 2 (1)(2) 1 = 0.5; µ s = 0.3 2

Given: α= 4; B=′ Es = 20,600 kN/m2

I= F1 + s

1 − 2µ s F2 1 − µs H

5.75

2

2

= n′ = = 11.5 L 2 m=′ = = 2; B 1 B 1    

From Table 11.1, F1 ≈ 0.65; from Table 11.2, F2 = 0.03.

1 − (2)(0.3) Is = 0.65 + (0.03) = 0.667 1 − 0.3

D

f Also, with =

B

L 0.75 = 0.75 and = 2, Table 11.3 gives If = 0.75. Hence, B 1

 1 − 0.32  Se (flexible, = = 0.0133 = (300)[(4)(0.5)] m 13.3 mm   (0.667)(0.75) center)  20, 600 

= Se (rigid) (0.93)(13.3) ≈ 12.4 mm 11.2

Se (rigid) = 0.93∆σ (α B′)

1 − µ s2 I s I f (Note: H′ = 21 ft; H = 21 + 4 = 25 ft) Es 77

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= Es

ΣEs (i ) ∆z (2200)(5) + (980)(8) + (9100)(8) = 4364 lb/in2 = 628,416 lb/ft2 = 21 z

Given: B = L = 8 ft; µs = 0.4; α = 4

= ∆σ

87,000 ≈ 1360 lb/ft 2 (8)(8)

B=′

B 8 = = 4 ft 2 2

I= F1 + s

1 − 2µ s F2 1 − µs

H = H ′ + D f = 21 + 4 = 25 ft ′ m=

H 25 L n′ = = 6.25 = 1;= B ( B /2) (8/2)

From Table 11.1, F1 = 0.461; from Table 11.2, F2 = 0.025.

1 − (2)(0.4) 0.461 + (0.025) = 0.469 Is = 1 − 0.4

Also,

Df 4 = = 0.5. From Table 11.3, If = 0.82. So, B 8

 1 − 0.42  (0.93)(1360)(4 × 4)  0.0104 ft ≈ 0.13 in.

Se (rigid) =  (0.469)(0.82) =  628, 416 

11.3

Eq.= (11.20): Se

qo Be I G I F I E 1 − µ s2 Eo

(

)

qo = 150 kN/m2 = Be

4B2 =

π

(4)(3) 2 = 3.385 m

π

µs = 0.3; Eo = 16,000 kN/m2 78 © 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.

Eo 16,000 = = 11.82 kBe (400)(3.385)

= β

20 H′ = = 5.91 Be 3.385 From Figure 11.7, IG ≈ 0.89. From Eq. (11.21),

π

I F=

4

+

1

  3    2t  Ef 4.6 + 10  B    Eo + e k   Be  2   1 π 0.815 = + = 4      (2)(0.25) 3 15 ×106  4.6 + 10   16,000 +  3.385  (400)   3.385       2 

From Eq. (11.22),

IE = 1−

1

B  3.5 exp (1.22µ s − 0.4)  e + 1.6   Df    1 1− 0.923 = =  3.385  3.5 exp[(1.22)(0.3) − 0.4]  + 1.6   1.5 

 1 − 0.32  Se = (150)(3.385)(0.89)(0.815)(0.923)   = 0.0193 m = 19.3 mm  16,000 

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Chapter 12 Hs 12.1=

Ws 91 = = 1.056 cm AGs γ w (π )(3.175) 2 (2.72)(1) e=

Hv Hs

Final height, H (cm)

Hs (cm)

Hv = H – Hs (cm)

0

1.9

1.056

0.843

0.799

0.25

1.763

1.056

0.707

0.670

0.5

1.742

1.056

0.686

0.650

1.0

1.702

1.056

0.646

0.612

2.0

1.656

1.056

0.6

0.568

4.0

1.604

1.056

0.548

0.519

8.0

1.550

1.056

0.494

0.468

σ′ (kg/cm2)

The e-log σ′ graph is plotted.

From the graph, σ c′ = 0.7 kg/cm2

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= Cc

12.2

e1 − e2 0.568 − 0.519 = = 0.163  σ ′2  4 log   log   2  σ1′ 

a. The plot of e vs. σ ′ is shown. b. σ c′ = 3.1 ton /ft 2

= Cc

c.

12.3

e1 − e2 0.79 − 0.63 =  σ′   16  log   log  2  8  σ1′  = 0.53

a. The plot is shown. b. σ c′ = 96 kN /m 2

Cc = c.

e1 − e2  σ′  log  2   σ1′ 

1.06 − 0.925  500  log    200  = 0.339

=

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12.4

Hs = a. Height of solids:

σ′ (lb/ft2) H (in.) Hs (in.)

Ws 95.2 g = = 1.12 cm = 0.441 in.

AGs γ w (4.91)(2.54) 2 (2.68)(1) e=

Hv = H – Hs (in.)

500

0.6947

0.441

0.2537

0.575

1000

0.6850

0.441

0.244

0.553

2000

0.6705

0.441

0.2295

0.52

4000

0.6520

0.441

0.211

0.478

8000

0.6358

0.441

0.1948

0.442

16,000

0.6252

0.441

0.1842

0.418

Hv Hs

The e-log σ′ graph is plotted.

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b. From the graph, σ′c = 940 lb/ft2

e1 − e2 0.52 − 0.478 = = 0.133  σ ′2   4000  log   log    2000   σ1′ 

= Cc c.

12.5

a. Sc =

 σ ′ + ∆σ ′  Cc H log  o  1 + eo  σ o′ 

Cc = 0.009(LL − 10) = (0.009)(50 − 10) = 0.36

σ o′ = γ d (sand) H1 + [γ sat(sand) − 62.4]H 2 + [γ sat(clay) − 62.4]

H3 2

 17  2 = (110)(8) + (115 − 62.4)(15) + (120 − 62.4)  =  2158.6 lb/ft  2

Sc

(0.36)(17 ×12)  2158 + 1000  log  =  6.39 in.

1 + 0.9  2158.6 

 σ′  C H  σ ′ + ∆σ ′  Cs H log  c  + c log  o  1 + eo  σ o′  1 + eo  σ c′ 

= Sc

b.

(17)(12)  0.36  2600   3158.6   log   + 0.36 log  =   3.79 in.  1 + 0.9  6  2158.6   2600  

=

γ d (sand) = 12.6

(2.65)(9.81) = 15.85 kN/m3 1 + 0.64

′ = γ sand

(2.65 − 1)(9.81) = 9.87 kN/m3 1 + 0.64

′ = γ clay

(2.75 − 1)(9.81) = 9.04 kN/m3 1 + 0.9

= Cc 0.009( LL −= 10) (0.009)(45) = 0.405

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3  

σ o′ = (2.5)(15.85) + (2.5)(9.87) +   (9.04) = 77.86 kN/m 2 2

= Sc

12.7

(0.405)(3)  77.86 + 100  log  = = m 229 mm  0.229 1 + 0.9  77.86 

γ d= (sand)

Gs γ w (2.65)(62.4) = = 104.66 lb/ft 3 1+ e 1 + 0.58

′ γ sand =

(Gs − 1)γ w (2.65 − 1)(62.4) = = 65.16 lb/ft 3 1+ e 1.58

′ γ clay =

(Gs − 1)γ w (2.72 − 1)(62.4) = = 51.11 lb/ft 3 1+ e 1 + 1.1 6  

σ o′ = (5)(104.66) + (7)(65.16) +   (51.11) = 1132.75 lb/ft 2 2 = Cc 0.009( LL −= 10) (0.009)(45 −= 10) 0.315 = Sc

 σ′  C H  σ ′ + ∆σ ′  Cc H log  c  + c log  o  1 + eo  σ o′  1 + eo  σ c′ 

 0.315    (6)  5  log  3500  + (0.315)(6) log  1132.75 + 3000      2.1 2.1 3500  1132.75    = 0.153 ft ≈ 1.84 in.

Cc = 12.8

Cc =

e1 − e2 0.82 − 0.7 = = 0.588  σ ′2   4000  log   log    2500   σ1′  e1 − e3  σ′  log  3   σ1′ 

 σ′   6000  e3 = e1 − Cc log  3  = 0.82 − 0.588log  0.596 =  2500   σ1′ 

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= Cc 12.9

1.1 − 0.9 = 0.419 3 log   1

 3.5  1.1 − 0.419 log  e3 = 0.872 =  1 

12.10

T90 =

cv t ; U = 60%; Tv = 0.286 (Table 12.3) H dr2

0.286 =

12.11

(2.8 ×10−6 m 2 / min)(t ) 3   m 2 

2

;

t = 229,821 min = 159.6 days

The plot of e-log σ′ is shown.

σ o′ = 0.7 ton/ft2; e1 = 0.96; σ o′ + ∆σ ′ = 2.0 ton/ft2; e = 0.83 ∆= e 0.96 − 0.83 = 0.13

Sc =

H ∆e (4.5 ×12)(0.13) = 1 + eo 1 + 0.96

= 3.58 in.

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12.12

 ∆e    av ∆σ ′  a.  = mv = 1 + eav 1 + eav ∆e = e1 − e2 = 1.7 − 1.48 = 0.22 ∆σ ′ = σ ′2 − σ1′ = 400 − 150 = 250 kN/m 2

= eav

1.7 + 1.48 = 1.59 2

 0.22    av  250  0.00034 m 2 /kN = mv = = 1 + eav 1 + 1.59 cv = b.

k = 0.002 cm 2 /s mv γ w

0.002 cm 2 /s =

k  9.81  (0.00034 ×1002 cm 2 /kN)  kN/cm3 3   100 

k = 6.67 × 10−8 cm/s 12.13

In the laboratory: T50 =

cv t50 H dr2

0.197 =

(cv )(140 s) 1   in.  2 

2

;

cv = 35.17 × 10−5 in2/s

U = 30%; T30 = 0.0707 (Table 12.3) In the field: T30 =

cv t30 H dr2

0.0707 =

(35.17 ×10−5 in 2 /s)(t30 ) (8 ×12 in.) 2

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t30 = 1,852,634 s = 21.44 days

12.14

 ∆e    av a.  ∆σ ′  mv = = 1 + eav 1 + eav ∆= e 1.21 − 0.96 = 0.25

∆σ ′ = 4 − 2 = 2 ton/ft 2 = eav

1.21 + 0.96 = 1.085 2

 0.25    2  = mv = 0.06 ft 2 /ton 1 + 1.085 1.8 ×10−4 = 0.0962 ft 2 /day  62.4  (0.06 ft 2 /ton)  ton/ft 3   2000 

= cv

k = mv γ w

t60 =

Tv H dr2 (0.286)(9) 2 = = 240.8 days cv 0.0962

= Cc b.

Sc =

e1 − e2 1.21 − 0.96 = = 0.83  σ ′2  4 log   log   2  σ1′ 

 σ ′ + ∆σ ′  (0.83)(9) Cc H 4 = log  o log   1.018 ft  = 1 + eo 2  σ o′  1 + 1.21

Sc at 60% = (0.6)(1.018) = 0.611 ft ≈ 7.33 in.  ∆e   av σ ′  ∆ mv = = 12.15 = 1 + eav 1 + eav

 0.73 − 0.61     400 − 200  = 0.3592 ×10−3 m 2 /kN + 0.73 0.61   1+   2   2

 0.025  (0.197)  2  Tv H dr  2 =  1.09 ×10−5 m 2 /min cv = = t50 2.8 k = cvmvgw = (0.3592 × 10−3 m2/kN)(1.09 × 10−5 m2/min)(9.81 kN/m3) = 3.84 × 10−8 m/min 88 © 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.

T50 12.16 =

cv tL cv tF = 2 2 H (dr)L H dr(F)

tL tF tF 225 s = = ; 2 2 H (dr)L H dr(F)  0.025  (2 m) 2 m   2 

= tF 5,760,000 = s 66.7 days

12.17

cv t90 0.848 = a. T90 = 2 ; H dr

cv (100 × 24 × 60 × 60) 3   ×100 cm  2 

2

cv = 2.21×10-3 cm 2 /s

b. T80 =

cv t80 H dr2

From Table 12.3, U = 80%; T80 = 0.567

2

 2.5  (0.567)  cm  2 T80 H dr  2 =  t80 = 400.9 s = −3 cv 2.21 ×10  25  a. U (%) = 12.18 =  (100) 31.25%  80 

b. Tv =

cv t ; U = 50%; Tv = 0.197 H dr2

0.197 =

(0.002)(t ) (300 cm) 2

t50 = 8,865,000 s = 102.6 days 89 © 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.

= = Tv 0.197 c.

(0.002)(t )  300     2 

2

t50 = 2,216,250 s = 25.65 days 12.19

∆σ ′ + 4∆σ ′m + σ b′ ′ =t Eq. (12.58): ∆σ av 6 Eq. (10.42): ∆σ′ = qI4

m1=

= q

B L 3 z b = 0.5 m; n1 = = = 3; = b 2 B 1 110 kN = 36.7 kN/m 2 (3)(1)

z b

m1

B z (m) b = 2 (m)

n1 =

3

1.5

0.5

3

36.7

0.348

12.77

3

2.75

0.5

5.5

36.7

0.15

5.51

3

4.0

0.5

8

36.7

0.079

2.9

′ = ∆σ av

γ sat(clay) =

q (kN/m2) I4 (Table 10.12)

∆σ′ = qI4 (kN/m2)

12.77 + (4 × 5.51) + 2.9 = 6.29 kN/m 2 6 Gs γ w + wGs γ w (2.7)(9.81)(1 + 0.35) = = 18.38 kN/m3 1 + wGs 1 + (0.35)(2.7)

σ o′ = (1.5 ×15) + (1.5)(18 − 9.81) + (1.25)(18.38 − 9.81) = 45.5 kN/m 2 = Cc 0.009( LL −= 10) (0.009)(38 −= 10) 0.252 = Sc

(0.252)(2.5)  45.5 + 6.29  log  m 18.2 mm = =  0.0812 1 + 0.945 45.5  

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12.20

a. For the clay layer:

∆σ ′ + 4∆σ ′m + ∆σ b′ ′ =t Eq. (12.58): ∆σ av 6 qI 4 Eq. (10.42): ∆σ ′ =

m1=

L 8 B z b = 4 m; n1 = = = 1; = 2 b B 8

= q (1.75)(20.1) = 35.17 kN/m 2

B 2 (m)

m1 z (m)

b=

n1 =

z ∆σ′ = qI4 b q (kN/m2) I4 (Table 10.12) (kN/m2)

1

3

4

0.75

35.17

0.823

28.94

1

5

4

1.25

35.17

0.599

21.06

1

7

4

1.75

35.17

0.403

14.17

′ = ∆σ av

28.94 + (4 × 21.06) + 14.17 = 21.22 kN/m 2 6

4 = = 57.56 kN/m 2 σ o′ (1.5)(17) + (1.5)(19.2) +   (18.8) − (1.5 + 2)(9.81) 2

Sc-clay

(0.31)(4)  57.56 + 21.22  log  =  0.081 m 1 + 1.08 57.56  

For the peat layer:

m1=

L 8 z B = = 1; = b = 4 m; n1 = 2 B 8 b

= q (1.75)(20.1) = 35.17 kN/m 2

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B (m) 2

n1 =

7

4

1

7.9

1

8.8

m1

z (m)

1

b=

′ = ∆σ av

z b

q (kN/m2)

I4 (Table 10.12)

∆σ′ = qI4 (kN/m2)

1.75

35.17

0.403

14.17

4

1.975

35.17

0.342

12.03

4

2.2

35.17

0.304

10.69

14.17 + (4 ×12.03) + 10.69 = 12.16 kN/m 2 6

σ o′ (1.5)(17) + (1.5)(19.2) + (4)(18.8) + (0.9)(15) − (5.5 + 0.9)(9.81) = = 80.21 kN/m 2 (7.2)(1.8)  80.21 + 12.16  log  =  0.107 m 1 + 6.4 80.21  

Sc-peat

Total consolidation settlement, Sc = 0.081 + 0.107 = 0.188 m b. For the clay layer, a double drainage condition is assumed since the bottom peat layer has high void ratio and is considered permeable. = t99− clay

( H dr2 )(T99 ) (2002 )(1.781) = = 11, 873,333.33 s ≈ 138 days 0.006 cv

For the peat layer, a single drainage is assumed since the top clay layer is considered to have relatively low permeability. = t99− peat

( H dr2 )(T99 ) (1802 )(1.781) = = 1, 989, 806.9 s ≈ 23 days cv 0.029

c. Secondary compression in clay:

 σ o′ + ∆σ ′   57.56 + 21.22  = ∆eprimary Cc log = =  0.31log   0.042 57.56    σ o′  e p= eo − ∆eprimary= 1.08 − 0.042= 1.038

= Cα′

Cα 0.048 = = 0.0235 1 + e p 1 + 1.038 92

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 t2   (18)(30)  ′ = Ss − clay C=  (0.0235)(4) log= α H log    0.055 m  138   t1  Secondary compression in peat:

 σ o′ + ∆σ ′   80.21 + 12.16  = ∆eprimary Cc log = =  7.2 log   0.441 ′ σ 80.21   o   e p= eo − ∆eprimary= 6.4 − 0.441= 5.959

= Cα′

Cα 0.273 = = 0.039 1 + e p 1 + 5.959

 t2   (18)(30)  ′ = Ss − peat C=  (0.039)(1.8) log= α H log    0.096 m  23   t1  d. Total settlement: Sc + Ss−clay + Ss−peat = 0.188 + 0.055 + 0.096 = 0.339 m e. Time factor in 2 months: Tv =

cv t (0.006)(2)(30)(24)(3600) = = 0.77 H dr2 2002

Determine the degree of consolidation Uz from Figure 12.24.

3.2 z = = 1.6; U z ≈ 0.9 2 H dr

Eq. 12.47: U z = 1 −

uz uo

Initial excess pore water pressure: uo ≈ ∆σ′avg = 21.22 kN/m2 Remaining excess pore water pressure at point A after 2 months: uz = (1 − Uz)uo = (1 − 0.9)(21.22) = 2.122 kN/m2 f. The increase in effective stress after 2 months = 21.22 − 2.122 = 19.09 kN/m2

σ o′ − A = (1.5)(17) + (1.5)(19.2) + (3.2)(18.8) − (1.5 + 3.2)(9.81) = 68.35 kN/m2 Therefore, the final effective stress after 2 months is 68.35 + 19.09 = 87.44 kN/m2 93 © 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.

g. The piezometer reading will be the total pore water pressure at any given time. Therefore, after 2 months, the piezometer reading should be upiezometer = (1.5 + 3.2)(9.81) + 2.122 = 48.23 kN/m2

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Chapter 13 13.1

a. c′ = 0. From Eq. (13.3): τ f = σ ′ tan φ ′

τf =

300 = 75 kN/m 2 (1000)(0.063) 2

So, 75 = 105 tan φ ′ −1  75  = = φ ′ tan   35.5°  105 

b. For σ′ = 180 kN/m2, τf = 180 tan 35.5° = 128.39 kN/m2 Shear force, S = (128.39)(1000)(0.063)2 = 509.5 N 13.2

The point O (180, 128.4) represents the failure stress conditions on the Mohr-Coulomb failure envelope. The perpendicular line OC to the failure envelope determines the center, C of the Mohr’s circle.

With the center at C, and the radius as OC, the Mohr’s circle is drawn by trial and error such that the circle is tangent to the failure envelope at O. From the graph,

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a. σ 3′ ≈ 115 kN /m 2 ; σ1′ ≈ 420 kN /m 2 b. The horizontal line OP drawn from O determines the pole P. Therefore, the orientation or the major principal plane with the horizontal is given by the angle a ≈ 65°.

13.3

For σ' = 28 lb/in2, τf = 28 tan 33° = 18.18 lb/in2 Shear force, S = (18.18)(2.5)2 = 113.65 lb

13.4

2

 π  63.5  2 a. Area of specimen = A =   0.00316 m  4  1000 

Normal Test No. force, N (N)

N A (kN/m2)

σ′ =

S φ ′ = tan −1  τ f   ′ τf = σ  Shear force, A 2 S (N) (kN/m ) (deg)

1

84

26.58

28.9

9.14

18.97

2

168

53.16

59.6

18.86

19.53

3

254

80.38

89.1

28.19

19.33

4

360

113.92

125.3

39.65

19.19

A graph of τf vs. σ′ will yield an average φ′ ≈ 19.2°.

γd b. For the silty sand:= γ sat =

Gs γ w (2.69)(9.81) = = 15.34 kN/m3 1+ e 1 + 0.72

(Gs + e)γ w (2.69 + 0.72)(9.81) = = 19.45 kN/m3 (1 + e) (1 + 0.72)

For the silty= clay: γ sat

(1 + w)Gs γ w (1 + 0.22)(2.72)(9.81) = = 20.36 kN/m3 (1 + wGs ) 1 + (0.22)(2.72)

= = 88.81 kN/m 2 σ ′A (2)(15.34) + (2.2)(19.45 − 9.81) + (3.5)(20.36 − 9.81)

τf = (88.81) tan 19.2° = 30.92 kN/m2

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13.5

π  2 2 a. Area of= specimen A =  (0.0635) 0.00316 m 4

τ  φ ′ = tan −1  f   σ′ 

Test No.

Normal force, N (N)

N A (kN/m2)

1

200

63.29

105

33.22

27.69

2

400

126.58

205

64.87

27.13

3

800

253.16

414

131.01

27.36

4

1600

506.33

830

263.6

27.50

σ′ =

Shear force, S (N)

S τf = A (kN/m2)

(deg)

A graph of τf vs. σ' will yield φ′ ≈ 27.5°. b. σ' = (2)(15.34) + (2.2)(19.45 − 9.81) = 51.88 kN/m2 τf = (51.88) tan 27.5° = 27 kN/m2

13.6

φ' = 28 + 0.18Dr = 28 + (0.18)(68) = 40.24°

φ′  40.24    2 σ1′ σ 3′ tan 2  45= = +  150 tan 2  45 + =  697.43 kN /m 2 2    13.7

′ σ 3′ + ∆σ d ( f = 125 + 175= 300 kN/m 2 σ 3′ = 125 kN/m 2 ; σ= 1 )  

σ1=′ σ 3′ tan 2  45 +

φ′ 

φ′  2  ;300= 125 tan  45 +  2 2 

φ ′ ≈ 24.3° 13.8

φ′   2 = σ 3′ + ∆σ 3′ σ 3′ tan 2 = ( 60.5° ) 3.12σ 3′  45 +  σ 3′ tan= 2  ∆σ 3′ ∆σ 3′ 1+ = 3.12; = 2.12 σ 3′ σ 3′

= σ 3′

18 = 8.49 lb /in 2 2.12 97

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φ′  ′ σ 3′ tan 2  45 + = σ= 1 

13.9



31   σ − ∆u ) tan  45 +  ( 2 2  3

d( f )

2

(15 − 4.8) tan 2 (60.5) = σ1′ = 31.86 lb /in 2

a. The effective principal stresses at failure are calculated as follows and the Mohr-Coulomb failure envelope is drawn from the Mohr’s circles below.

13.10

(∆σ d ) f

σ 3′= σ 3 − (∆ud ) f σ1′= σ 3′ + (∆σ d ) f (kN/m2) (kN/m2)

Test no.

σ3 (kN/m2)

(kN/m2)

(∆ud)f (kN/m2)

1

100

170

−15

115

285

2

200

260

−40

240

500

3

300

360

−80

380

740

From the graph: c' ≈ 40 kN/m2 and φ ≈ 20° b. Effective stress in the .middle of the clay layer:

σ 0′ = (2)(19 − 9.8)= 18.4 kN/m 2

τ= c′ + σ ′ tan φ ′ = 40 + 18.4 tan(20) = 46.7 kN /m 2 φ′  φ′   2 = + 33) 25 tan 2  45 +  a. σ1′ σ 3′ tan  45 +  ; (25= 13.11 2 2  

φ ′ ≈ 23.4°

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b. θ = 45 +

φ′ 2

= 45 +

23.4 = 56.7° 2

c. From Eqs. (10.8) and (10.9):

σ1′ + σ 3′ σ1′ − σ 3′ 58 + 25 58 − 25 + cos 2θ = + cos(2 × 56.7) σ ′f = 2

2

2

2

= 34.94 lb/in2 ′ tan φ ′ 34.94 tan = = 23.4 15.12 lb /in 2 τ f σ=

φ′  φ′    σ1′ σ 3′ tan 2  45 +  + 2c′ tan  45 +  13.12= 2 2   φ′  φ′   2 Specimen I: (105 + 220) = 325 = 105 tan  45 +  + 2c′ tan  45 +  2 2  

(a)

φ′  φ′   2 Specimen II: (210 + 400) = 610 = 210 tan  45 +  + 2c′ tan  45 +  (b) 2 2  

φ′   − 325 105 tan 2  45 + = Subtracting Eq. (a) from Eq. (b): 610=  ; φ ′ 27.48° 2  From Eq. (b),

c′

13.13

27.48   610 − 210 tan 2  45 +  2   = 12.48 kN /m 2 27.48   2 tan  45 +  2  

a. From Eqs. (10.8) and (10.9), θ = 40°

σ1′ + σ 3′ σ1′ − σ 3′ 105 + 325 325 − 105 + + σ′ = cos 2θ = cos(2 × 40) 2 2 2 = 234.1 kN /m

2

2

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= τ

σ1′ − σ 3′ 2

325 − 105 sin(2 = × 40) 108.32 kN /m 2 2

sin 2θ =

b. The angle of inclination of the failure plane:

θ = 45 +

φ′ 2

= 45 +

27.48 = 58.73° 2

σ1′ + σ 3′ σ1′ − σ 3′ 610 + 210 610 − 210 + + σ′ = cos 2θ = cos(2 × 58.73) 2 2 = 317.74 kN /m 2

= τ

σ1′ − σ 3′ 2

= sin 2θ

2

2

610 − 210 = 177.46 kN /m 2 sin(2 × 58.73) 2

 −1  22 + 28 0.5  φ 2 = σ σ + tan 45 ; = φ 2 13.14  tan  = 1 3    − 45 22.9° 2   22   

 −1  22 + 28 + 4 0.5  φ′  2 ′ σ1′ σ 3′ tan = + 45 ; 2 = φ  tan  =    − 45 20.48° 2   22 + 4    13.15

a. σ 3 = 150 kN/m 2 ; σ1 = 150 + 120 = 270 kN/m 2

φ  270 φ   σ1 =σ 3 tan 2  45 +  ; =tan 2  45 +  2  150 2   φ = 16.6°

φ′  2 = b. σ1′ σ 3′ tan  45 +  2 

σ1 − ∆ud ( f ) 27   = tan 2  45 + =  2.662 σ 3 − ∆ud ( f ) 2  

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270 − ∆ud ( f ) 150 − ∆ud ( f )

= 2.662

399.3 − 2.662∆ud ( f ) Or, 270 − ∆u= d( f )

∆ud ( f ) = 77.8 kN /m 2

φ′   +  150 tan 2 (45 + = 27/2) 399.4 kN/m 2 σ1′ σ 3′ tan 2  45= 13.16= 2   ∆σ d ( f ) = σ1′ − σ 3′ = 399.4 − 150 = 249.4 kN /m 2 31  φ   +  20 tan 2  45= +  62.48 lb/in 2 σ1 σ 3 tan 2  45= 13.17= 2 2  

∆σ d ( f = 62.48 − 20= 42.48 lb /in 2 )

φ′   = σ1′ σ 3′ tan 2  45 +  2  62.48 − ∆ud ( f ) 24   = tan 2  45 +  20 − ∆ud ( f ) 2   ∆ud ( f ) = −10.98 lb /in 2

A dense sand tends to expand during shear. Due to undrained condition, it creates a negative pore water pressure. A loose sand tends to contract during shear, and a positive pore water pressure is developed in undrained conditions. Therefore, for loose sand, σ′ < σ, and so, φ′ > φ.

φ′   = σ1′ σ 3′ tan 2  45 +  13.18 2  148 − ∆ud ( f ) 28   = tan 2  45 +  0 − ∆ud ( f ) 2   ∆ud ( f ) = −83.6 kN /m 2

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a.

13.19

σ1′ + σ 3′

Test no.

2

= p′ (kN/m2)

σ1′ − σ 3′

2 (kN/m2)

= q′

1

212

108

2

362

155

q=′ m + p′ tan α 108= m + 212 tan α 155= m + 362 tan α

(a) (b)

= m 41.56 = kN /m 2 , α 17.4° −1 1 b. φ ′ sin = = (tan α ) sin −= (tan17.4) 18.26°

= c′ 13.20

m 41.56 = = 43.55 kN /m 2 cos α cos(17.4)

PI = LL − PL = 39 − 20 = 19

a. Eq. (13.53): cu (VST) =+ 0.11 0.0037 PI =+ 0.11 0.0037(19) = 0.1803 σ o′

σ o′ = (3)(16) + (8)(20.5 − 9.81) = 133.52 kN/m 2 = cu (VST) (0.1803)(133.52) ≈ 24 kN /m 2

1.7 − 0.54 log( PI ) = 1.7 − 0.54 log19 = 1.009 b. Eq. (13.72): λ = = cu (design ) λ= cu (vane shear) (1.009)(24) = 24.21 kN /m 2 Eq. (13.71):

c. Eq. (13.54):  cu  c  = (OCR)0.8  u     σ o′ overconsolidated  σ o′  normally consolidated cu (overconsolidated)= (3)0.8 × 24= 57.8 kN /m 2

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13.21

LI = Liquidity index,

w − PL 28 − 20 = = 0.421 LL − PL 39 − 20

cur = a. Leroueil et al. (1983):

1 1 22.46 kN /m 2 = = 2 2 [ LI − 0.21] [0.421 − 0.21]

−2.8 22.54 kN /m 2 cur 2( LI ) −2.8 2(0.421) = = = b. Terzaghi et al. (1996):

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Chapter 14 sin φ ′ 14.1–14.4 K o= (1 − sin φ ′)(OCR)

Problem φ′ (deg) 14.1

38

0.676

14.2

32

0.679

14.3

35

0.635

14.4

30

0.5

z=

Po = 12 K oγ H 2

Ko

281.55 kN/m

H 3

2.33 m

8402.6 lb/ft

5 ft

205.7 kN/m

2m

10,500.0 lb/ft

6.67 ft

= K a tan 2 (45 − φ ′/2) 14.5–14.8

H 3

Prob.

φ′ (deg)

Ka

σ ′a ( z = H ) = K a γ H

Pa = 12 K a γ H 2

z=

14.5

30

0.333

(0.333)(105)(15) = 524.5 lb/ft2

1 2

(0.333)(105)(15)2 = 3933.6 lb/ft

5 ft

14.6

32

0.307

(0.307)(100)(18) = 552.6 lb/ft2

1 2

(0.307)(100)(18)2 = 4973.4 lb/ft

6 ft

14.7

36

0.26

(0.26)(18)(4) = 18.72 kN/m2

1 2

(0.26)(18)(4)2 = 37.44 kN/m

1.33 m

14.8

40

0.217

(0.217)(17)(5) = 18.45 kN/m2

(0.217)(17)(5)2 = 46.11 kN/m

1.67 m

1 2

Note: 1. Pressure distribution is similar to that shown in Figure 14.13a., i.e.,

σ ′a = 0 at z = 0 and σ ′a = K a γ H at z = H. 2. z = distance measured from the bottom of the wall.

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K p tan 2 (45 + φ ′/2) 14.9–14.12 =

H 3

Prob.

φ′ (deg)

Kp

σ ′p ( z = H ) = K p γ H

Pp = 12 K p γ H 2

z=

14.9

34

3.537

(3.537)(110)(8) = 3112.6 lb/ft2

(3.537)(110)(8)2 = 12,450 lb/ft

2.67 ft

14.10

36

3.852

(3.852)(105)(10) 12 (3.852)(105)(10)2 3.33 ft = 4044.6 lb/ft2 = 20,223 lb/ft

14.11

35

3.69

(3.69)(14)(5) = 258.3 kN/m2

1 2

(3.69)(14)(5)2 = 645.8 kN/m

1.67 m

14.12

30

3

(3)(15)(4) = 180 kN/m2

(3)(15)(4)2 = 360 kN/m

1.33 m

1 2

Note: 1. σ ′p ( z = 0) = 0; triangular pressure distribution.

2. z = distance measured from the bottom of the wall.

14.13

26  φ′   2 = K a tan 2  45 − =  tan  45 − =  0.39. Refer to the figure.

2 2   

σ ′a σ ′a= K a 0;= u 0 z = 0 ft: = = = 170.04 = lb/ft 2; u 0 z = 4 ft: σ ′a (109)(4)(0.39)

z = 21 ft: σ ′a= [(109)(4) + (121.5 − 62.4)(17)](0.39)= 561.87 lb/ft 2 = = 1060.8 lb/ft 2 u (62.4)(17) 106 © 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.

Area No.

Area

1

( 12 )(4)(170.04) = 340.08 lb/ft

2

(170.04)(17) = 2890.68 lb/ft

3

( 12 )(17)(561.87 − 170.04) = 3330.55 lb/ft

4

( 12 )(17)(1060.8) = 9016.8 lb/ft Pa = Σ15, 578 lb/ft

Resultant: Taking the moment about the bottom of the wall,  4   17   17   17   (340.08) 17 + 3  + (2890.68)  2  + (3330.55)  3  + (9016.8)  3           z= 15,578 = 6.47 ft

14.14

30   tan 2  45 − = K= a  0.333 2  

σ ′a qK = (16)(0.333) = 5.32 kN/m 2 ; u = 0 z = 0 m: = a [16 + (3)(13)](0.333) = 18.31 kN/m 2 ; u = 0 z = 3 m: σ ′a = [16 + (3)(13) + (18.8 − 9.81)(5)](0.333) = 33.28 kN/m 2 z = 8 m: σ ′a =

u = (9.81)(5) = 49.05 kN/m 2 107 © 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.

Area No.

Area

1

(5.32)(8) = 42.56 kN/m

2

( 12 )(3)(18.31 − 5.32) = 19.48 kN/m

3

(5)(18.31 − 5.32) = 64.95 kN/m

4

( 12 )(5)(33.28 − 18.31) = 37.42 kN/m

5

( 12 )(5)(49.05) = 122.62 kN/m Pa = Σ 287 kN/m

Location of resultant: Taking the moment about the bottom of the wall, 3  8  5 5  5  (42.56)  2  + (19.48)  5 + 3  + (64.95)  2  + (37.42)  3  + (122.62)  3             z= 287 = 2.49 m

14.15

42  36    K a= tan 2  45 − = tan 2  45 − = (1) (2)  0.198  0.259; K a = 2  2   

2

2

qK a (1) = (25)(0.259) = 6.475 kN/m ≈ 6.47 kN/m , u = 0 z = 0 m: σ ′a = [(17)(4) + 25](0.259) = 24.08 kN/m 2 z = 4 m: σ ′a =

[(17)(4) + 25](0.198) = 18.41 kN/m 2 σ ′a =

u=0

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z = 12 m: σ ′a= [(17)(4) + (23.2 − 9.81)(8)](0.198)= 34.67 kN/m 2 = u (9.81)(8) = 78.48 kN/m 2 Area No.

Area

1

(6.47)(12) = 77.64

2

( 12 )(4)(24.08 − 6.47) = 35.22

3

(8)(18.41 − 6.47) = 95.52

4

( 12 )(8)(34.67 − 18.41) = 65.04

5

( 12 )(8)(78.48) = 313.92 Pa = Σ587.34 kN/m

Location of resultant: Taking the moment about the bottom of the wall,

4  12   8 8 8 (77.64)   + (35.22)  8 +  + (95.52)   + (65.04)   + (313.92)   3  2  2 3 3 z= 587.34 = 3.72 m

14.16

26   tan 2  45 + = K= p  2.56.

2  

σ ′p 0;= u 0 z = 0 ft: = 2 σ ′p γ= (109)(4)(2.56) = 1116.16 lb/ft= ;u 0 z = 4 ft: = 1 zK p

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z = 21 ft:

σ ′p= [(109)(4) + (121.5 − 62.4)(17)](2.56)= 3688.19 lb/ft 2

= u (62.4)(17) = 1060.8 lb/ft 2 Area No.

Area

1

( 12 )(4)(1116.16) = 2232.32

2

(17)(1116.16) = 18,974.72

3

( 12 )(17)(3688.19 − 1116.16) = 21,862.25

4

( 12 )(17)(1060.8) = 9016.8 Pp = Σ52,086 lb/ft

Location of the resultant: Taking the moment about the bottom of the wall, 4    17   17   17   (2232.32) 17 + 3  + (18,974.72)  2  + (21,862.25)  3  + (9016.8)  3           z= 52,086 = 7.24 ft

a. Use Table 14.4: For a = 12° and φ′ = 28°, Ka = 0.3896

14.17

σ ′a γ= zK a (18.6)(7)(0.3896) = = 50.72 kN /m 2 = Pa b. Eq. (14.26):

1 1 2 = K aγ H 2 (0.3896)(18.6)(7) = 177.54 kN /m 2 2

Location and Direction of Resultant: At a distance of H/3 = 7/3 = 2.33 m above the bottom of the wall inclined at an angle α = 12° to the horizontal.

14.18

a. Use Eq. (14.35): For α =12° and φ′ = 28°,

Kp

cos12 + cos 2 (12) − cos 2 (28) cos12 = 2.455 cos12 − cos 2 (12) − cos 2 (28)

= σ ′a γ= zK p (18.6)(7)(2.455) = 319.64 kN /m 2

b. Eq. (14.34): 110 © 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.

1 1 2 (18.6)(7)= (2.455) 1118.74 kN /m = γ H 2K p 2 2

= Pp

Location and Direction of Resultant: At a distance of H/3 = 7/3 = 2.33 m above the bottom of the wall inclined at an angle α = 12° to the horizontal.

14.19

a. H = 8.5 m; cu = 22 kN/m2; γ = 19.6 kN/m2; φ = 0

φ  K a = tan 2  45 −  =1; σ ′a =γ z − 2cu 2  At the top (z = 0 m): −2cu = (−2)(22) σ ′a = = −44 kN /m 2

At the bottom (z = 8.5 m): = σ ′a (19.6)(8.5) − (2)(22) = 122.6 kN /m 2

The pressure diagram is shown.

zo b. Eq. (14.50): =

2cu (2)(22) = = 2.24 m γ 19.6

c. Eq. (14.52):

Pa =

1 1 γ H 2 − 2cu H = (19.6)(8.5) 2 − (2)(22)(8.5) = 334.05 kN /m 2 2 111

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d. Eq. (14.54):

Pa =

2c 2 1 γ H 2 − 2cu H + u 2 γ

=

1 (2)(22) 2 (19.6)(8.5) 2 − (2)(22)(8.5) + = 383.43 kN /m 2 19.6

z = Resultant measured from the bottom: 14.20

H − zo 8.5 − 2.24 = = 2.086 m 3 3

σ a σ o K a − 2c K a a.=

σ o =+ γ z q; K a = 1 At z = 0 m:

σ o = 20 kN/m 2 σa = −24 kN /m 2 20 − (2)(22) = At z = 8.5 m: = = + 20 186.6 kN/m 2 σ o (19.6)(8.5)

σa = 186.6 − (2)(22) = 142.6 kN /m 2 The pressure diagram is shown.

σ a 0; (γ zo + q) −= 2c 0 b. = = zo

2cu − q 44 − 20 = = 1.22 m γ 19.6

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c. Referring to the diagram in Part a,

= Pa

1 1 (7.33)(142.6) − (24)(1.22) = 509 kN /m 2 2

= d. Pa

1 = (7.28)(142.6) 519.06 kN /m 2

Location of the resultant from the bottom of the wall:

14.21

7.33 = 2.44 m 3

1 Eq. (14.26): Pa = γ H 2 K a ( R ) 2 Given: α = 10°, θ = 6°, φ′ = 30°. From Table 14.2, K a ( R ) = 0.395 . Hence,

1 2 = Pa =  (17)(6) (0.395) 120.87 kN /m 2

From Table 14.3, βa = 19.46°. Pa = acts at a distance 14.22

6 = 2 m above the base of the wall.

3

Use Eqs. (14.63) and (14.64). α = 0; θ = 10°; φ′ = 30°; γ = 18 kN/m3; H = 5 m

1 K aγ H 2 2 [Eq. (14.63)]

Pa =

Part

δ′ (deg)

Ka [Eq. (14.64)]

a

15

0.3784

85.14 kN/m

b

20

0.3769

84.80 kN/m

Pa is located at a vertical distance of 5/3 = 1.67 m above the bottom of the wall and is inclined at an angle δ′ to the normal drawn to the back face of the wall.

14.23

= Pae From Eq. (14.69),

Given :

1 γ H 2 (1 − kv ) K an 2

δ′ = 0.5; φ′ = 28°; θ = 10°; a = 5°; kh = 0.3; kv = 0 φ′

From Table 14.14, K"a = 0.825. Therefore, 113 © 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.

= = 415.33 kN /m Pae (0.5)(17.9)(7.5) 2 (1 − 0)(0.825)

Use Eq. (14.63): Pa =

1 K aγ H 2 2

From Table 14.10, Ka = 0.4327 Pa = (0.4327)(0.5)(17.9)(7.5)2 = 217.84 kN/m2 415.33 − 217.84 = 197.49 kN/m From Eq. (14.77), ∆P= ae Eq. (14.78):

 7.5  (217.84)   + (197.49)(0.6 × 7.5) 544.6 + 888.71 3   = z = = 3.45 m 415.33 415.33 Using Eqs. (14.24) and (14.27), the following table can be prepared:

14.24 θ = 10°

α = 0°

α = 10°

α = 20°

γ (kN/m3)

φ′

Ka(R)

Ka(R)γ

16.5

28°

0.402

6.633

17.7

32°

0.354

6.265

19.5

36°

0.314

6.123

16.5

28°

0.471

7.771

17.7

32°

0.414

7.327

19.5

36°

0.365

7.118

16.5

28°

0.612

10.09

17.7

32°

0.518

9.168

19.5

36°

0.447

8.716

From Eq. (14.26):

Pa = K a ( R )γ 0.5 H 2

The graph of Pa /0.5H 2 versus backfill inclination angle α is shown below.

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The above chart shows that for any backfill inclination α and any wall height H, the active force per unit length of the wall, Pa, decreases as the soil friction angle (or the compacted unit weight) increases.

For a desired level of Pa (at a given α), a compaction unit weight could be estimated from the chart for field specification.

14.25

From Eq. (14.83),

a= f

ca′ →1 c′

From Eq. (14.49), zo = 0 m From Eq. (14.84),

0.3  −1  k h  −1  β tan = =   tan    1 ± 0.15   1 ± kv 

β =14.62° for kv (↓) β =19.44° for kv (↑) From Eqs. (14.85) – (14.87), = m1 1.3140, = m2 0 = and m3 0 for kv (↓)

= m1 0.9966, = m2 0 = and m3 0 for kv (↑) From Eqs. (14.88) – (14.93), a1 = −0.3279, b1 = 1.1336, c1 = 0.3366, a2 = −0.8975, b2 = 0.2924, c2 = 0.0588 for kv ( ↓ )

a1 = −0.2536, b1 = 0.8991, c1 = 0.1764, a2 = −0.8975, b2 = 0.2924, c2 = 0.0588 for kv ( ↑ )

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From Eq. (14.94) using the values from Eqs. (14.88) – (14.93),

ηc =33.64° for kv (↓)

ηc =21.15° for kv (↑) From Eqs. (14.80) – (14.82), K aeγ 0.8597, K aec 2.1690 = = = and K 1.8784 for kv (↓) = K aeγ 1.2227, = K aec 4.8439 = and K 4.1950 for kv (↑)

From Eq. (14.79), = Pae 1959.96 kN /m for kv (↓)

= Pae 2060.35 kN /m for kv (↑)

14.26

From Eq. (14.83),

a= f

ca′ 15 = = 0.5 c′ 30

From Eq. (14.49), zo = 4.5948 m From Eq. (14.84), −1  k h  −1  0.2  = = β tan   tan    1 ± 0.1   1 ± kv 

10.30° for k v (↓) β=

and

β= 12.53° for k v (↑) From Eqs. (14.85) – (14.87), = m1 1.3164, = m2 0.1294 = and m3 0.2393 for kv (↓)

and = m1 1.0855, = m2 0.1294 = and m3 0.2393 for kv (↑)

From Eqs. (14.88) – (14.93), a1 = −0.1075, b1 = 1.2353, c1 = 0.4865, a2 = −0.8012, b2 = 0.3907, c2 = 0.1192 for kv(↓) and a1 = −0.0323, b1 = 1.0576, c1 = 0.4072, a2 = −0.8012, b2 = 0.3907, c2 = 0.1192 for kv(↑) From Eq. (14.94) using values from Eqs. (14.88) − (14.93), 116 © 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.

ηc = 40.32° for kv(↓) and ηc = 39.37° for kv(↑) From Eqs. (14.80) – (14.82), = K aeγ 1.0925, = K aec 2.1371 = and K 1.5548 for kv (↓)

and = K aeγ 1.1928, = K aec 2.2061 = and K 1.6050 for kv (↑)

From Eq. (14.79), Pae = 769.40 kN/m for kv(↓) and Pae = 618.61 kN/m for kv(↑)

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118 © 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.

Chapter 15 15.1

From Figure 15.4, for φ′ = 40° and δ′ = 20°, Kp ≈ 9.5. So,

= Pp 15.2

From Table 15.1, for φ′ = 40° and δ′ = 20°, Kp = 8.30

= Pp 15.3

1 (9.5)(18.5)(6) 2 3163.5 kN /m = 2

1 = (8.3)(18.5)(6) 2 2764 kN /m 2

Eq. (15.18): K p = K p (δ ′= 0) R Table 15.4: For φ′ = 40° and θ = 0, the value of K p (δ ′= 0) = 4.59 Table 15.5: For θ = 0, φ′ = 40°, and δ ′/φ ′ = 0.5, the value of R ≈ 2.1. So

1 (4.59 × 2.1)(18.5)(6) 2 ≈ 3210 kN /m Pp = 2 15.4

Table 15.2: For φ′ = 40° and δ′ = 20°, Kp = 8.91

= Pp 15.5

Table 15.3: For φ′ = 40° and δ′ = 20°, Kp = 9.68

Pp = 15.6

1 (9.68)(18.5)(6) 2 ≈ 3223 kN /m 2

From Table 15.1, for φ′ = 30° and δ′ = 20°, Kp = 4.68

= Pp 15.7

1 (8.91)(18.5)(6) 2 ≈ 2967 kN /m 2

1 = (4.68)(18.7)(5) 2 1094 kN /m 2

Eq. (15.18): K p = K p (δ ′= 0) R Table 15.4: For φ′ = 35° and θ = 10°, the value of K p (δ ′= 0) = 2.97 Table 15.5: For θ = 10°, φ′ = 35°, and δ ′/φ ′ = 0.6, the value of R = 1.85. So

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1 (2.97 ×1.85)(19.2)(6) 2 ≈ 1899 kN /m Pp = 2 15.8

1 Eq. (15.19): Pp = γ H12 K p 2 For θ = 0°, H1 = H = 5 m Figure 15.6: For φ′ = 30° and δ ′/φ ′ = 1, K p (δ ′=φ ′) ≈ 5.9

δ ′ 15 =′ 0.746 Table 15.6: For φ′ = 30° and = = 0.5, R φ ′ 30 K p =R′ × K p (δ ′=φ ′) =0.746 × 5.9 =4.4

= Pp

15.9

1 = (15)(5) 2 (4.4) 825 kN /m 2

1 Eq. (15.19): Pp = γ H12 K p 2 H1 For θ = 10°,=

H 5 = = 5.077 cos θ cos10

Figure 15.6: For θ = 10°, φ′ = 30° and δ ′/φ ′ = 1, K p (δ ′=φ ′) ≈ 4.8

δ ′ 15 =′ 0.746 Table 15.6: For φ′ = 30° and = = 0.5, R φ ′ 30 K p =R′ × K p (δ ′=φ ′) =0.746 × 4.8 =3.58

= Pp

15.10

1 = (15)(5.077) 2 (3.58) 692 kN /m 2

1 Eq. (15.21): Pp = γ H 2 K p 2

α 12 Step 1: = = 0.4 φ ′ 30 Step 2: From Figure 15.7: K p (δ ′=φ ′) ≈ 11.5 120 © 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.

δ ′ 15 Step 3: = = 0.5 φ ′ 30 Step 4: From Table 15.6: R′ = 0.746 0.746 ×11.5 = 8.58 Step 5: K p =R′ × K p (δ ′=φ ′) =

= Pp 15.11

1 (15)(5) 2 (8.58) ≈ 1609 kN /m 2

Given φ′ = 30°, δ′ = 15°

δ′ = 0.5 φ′ kz =1 kx

Eq. (15.23):

Pp =

1 K p ( s )γ sat H 2 2

From Fig. 15.10, K p ( s ) ≈ 4.0 1 = Pp   (4.0)(18)(5) 2 ≈ 900 kN /m 2

15.12

2.8 na = =0.4; φ ′ =30°; δ ′ =20° 7 Table 15.7:

Pa = 0.305 0.5γ H 2

Pa = (0.305)(0.5)(18)(7)2 = 134.5 kN/m

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φ′   = σ a 0.65γ H tan 2  45 −  2  15.13 38   (0.65)(18)(13.25) tan 2  45 −  2   2 ≈ 36.9 kN/m

ΣM B1 = 0

A

1  4.25   =  (36.9)(4.25)    111.08 kN/m 3  2 

= B1 (36.9)(4.25) − 111.08 = 45.75 kN/m

B= C= 2 1

(36.9)(3) = 55.35 kN/m 2

ΣM C2 = 0

1  6  = D =  (36.9)(6)    221.4 kN/m 3  2  C2 = (36.9)(6) − 221.4 = 0 kN/m 122 © 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.

Strut Loads: A = (111.08)(3) ≈ 333.24 kN B = (B1 + B2)(3) = (45.75 + 55.35)(3) = 303.3 kN C = (C1 + C2)(3) = (55.35 + 0)(3) ≈ 166.05 kN D = (221.4)(3) ≈ 664.2 kN 15.14

γ H (17.9)(12) = 2.86 < 4 a. = 75 cu Use Figure (15.17c) to determine the earth pressure envelope.

Maximum pressure intensity σ a = 0.2γ H to 0.4γ H

σ a 0.3 γ H (0.3)(17.9)(12) ≈ 64.5 kN/m 2 , the pressure envelope is shown below. = = Assuming

b. To determine the strut loads, refer to the following diagram.

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ΣM B1 = 0

3  3 A(4.5) − (0.5)(64.5)(3)  3 +  − (3)(64.5)   = 0 3  2

Therefore, A = 150.5 kN/m Also, sum of vertical forces, ΣV = 0 A + B1 = (0.5)(64.5)(3) + (64.5)(3) or 150.5 + B1 = 290.25 Therefore, B1 = 139.75 kN/m Due to symmetry, B2 = 139.75 kN/m and C = 150.5 kN/m Strut load at A = (150.5)(s) = (150.5)(3) = 451.5 kN Strut load at B = (B1 + B2)(s) = (139.75+ 139.75)(3) = 838.5 kN Strut load at C = (150.5)(3) = 451.5 kN

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Chapter 16 16.1

Eq. (16.15): c′ tan φ ′ + 2 γ H cos β tan β tan β 15 tan 20 + 2 2 (15.5)( H )(cos 30)(tan 30) tan 30 2.235 = + 0.63 2 H H = 1.63 m

Fs =

16.2

Eq. (16.17): H cr

= 16.3 γ sat

c′ 1 200 1 17.1 ft = = 2 2 γ cos β (tan β − tan φ ′) 100 (cos 22)(tan 22 − tan15)

(1900)(9.81) = 18.64 kN/m3 1000

γ ′ = 18.64 − 9.81 = 8.83 kN/m3 Eq. (16.21): Fs =

c′ 18 8.83 tan 25 γ ′ tan φ ′ + = + γ sat D sin β γ sat tan β (18.64)(8cos 20)(sin 20) 18.64 tan 20

= 0.376 + 0.607 = 0.98

= γ sat 16.4

(Gs + e)γ w (2.6 + 0.5)(62.4) = = 128.96 lb/ft 3 1+ e 1 + 0.5

γ= 128.96 − 62.4= 66.56 lb/ft 3

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c′ γ ′ tan φ ′ + γ sat D sin β γ sat tan β

= Fs

600 66.56 tan 22 + (128.96)(25cos 20)(sin 20) 128.96 tan 20 = 0.579 + 0.573 ≈ 1.15 16.5

Eq. (16.20): = Fs

 γ  tan φ ′ c′ 1 + 1 − µ w  γ D sin β  γ  tan β

γ= µγ sat + (1 − µ )γ= (0.4)(18.6) + (1 − 0.4)(16)= 17.04 kN/m3 = D H= cos β (6)(cos = 28) 5.3 m = Fs

16.6

Eq. (16.35):

= H cr 16.7

15 9.81  tan 26  + 1 − 0.4 ≈ 1.06  (17.04)(5.3)(sin 28)  17.04  tan 28

4c′  sin β cos φ ′   (4)(400)   (sin 50)(cos 25)  = = 103.1 ft γ 1 − cos(β − φ ′)   115  1 − cos(50 − 25) 

Consider a 1-m length of the wedge ABC.

Eq. (16.22):

= W

1 2 H γ (cot θ − cot = β ) (0.5)(62 )(19)(cot 32 − cot= 58) 333.6 kN 2 126

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Ta 333.6sin = = 32 176.78 kN N a 333.6 = = cos 32 282.9 kN Tr =

 538.84 1 1  6  ( ACc′ + N a tan φ ′) =   (38) + (282.9) tan 21 = Fs Fs  sin 32  Fs 

538.84 176.78 = Since Tr T= a, Fs

Therefore, Fs ≈ 3.05 16.8

Refer to the figure in the solution for Problem 16.7.

Eq. (16.22):

= W

1 2 58) 750.62 kN H γ (cot θ − cot = β ) (0.5)(92 )(19)(cot 32 − cot= 2

Ta 750.62sin = = 32 397.76 kN = N a 750.62 = cos 32 636.56 kN Tr =

1 1  9   889.73 ( ACc′ + N a tan φ ′) =   (38) + (636.56) tan 21 = Fs Fs  sin 32  Fs 

889.73 = 397.76 Since Tr T= a, Fs

Therefore, Fs ≈ 2.24 16.9

Refer to the solution for Problem 16.7.

Eq. (16.22):

= W

1 2 H γ (cot θ − cot = β ) (0.5)(62 )(22)(cot 32 − cot= 58) 386.28 kN 2

= Ta 386.28sin = 32 204.69 kN = N a 386.28cos = 32 327.58 kN Tr =

1 1  6   1053.9 ( ACc′ + N a tan φ ′) =   (75) + (327.58) tan 32  = Fs Fs  sin 32  Fs  127

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1053.9 204.69 = Since Tr T= a, Fs

Therefore, Fs ≈ 5.15 16.10

′ = F 2; c= s d

= H 16.11

4cd′  sin β cos φd′   (4)(200)   (sin 50)(cos13.12)  = = 25.94 ft   γ 1 − cos(β − φd′ )   115  1 − cos(50 − 13.12) 

m ≈ 0.192 (From Figure 16.13). Eq. (16.41): cu 30 = = 9.19 m -- Toe failure γ m (17)(0.192)

H= cr 16.12

16.13

c′ 400 tan 25  ′ tan −1  = = 200 lb/ft 2 ; φ= = d  13.12° Fs 2  2 

m ≈ 0.192 for β = 60° (Figure 16.13) c= d

cu 30 = = 15 kN/m 2 FS s 2

H =

cd 15 = = 4.6 m γ m (17)(0.192) 1 

β= tan −1  = 45°. For β = 45° and D = 1.2, m = 0.18 (Figure 16.13). 1  = cd

cu 24 = = 9.6 kN/m 2 FS s 2.5

= H

cd 9.6 = = 2.96 m γ m (18)(0.18)

16.14

H= cr

cu 24 = = 7.4 m. It is a toe circle. γ m (18)(0.18)

16.15

D a. =

12 = 1.41; γ sat = 18.5 kN/m3 . For β = 40° and D = 1.41, m = 0.175.

8.5 128

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= H cr

cu ; cu (8.5)(18.5)(0.175) = = 27.5 kN /m 2 γm

b. From Figure 16.13, midpoint circle c. From Figure 16.15, n ≈ 0.7.

Distance = nH = (0.7)(8.5) = 5.95 m 16.16

Fs =

cu M . β = 30°; H = 16 m; cu = 50 kN/m2. γH

From Figure 16.20 for kh = 0.4 and D = 1, M ≈ 5 50 = (5) 0.919 (17)(16)

= Fs

16.17

1 β tan −1 = = a.   26.57° 2

Fs 1 = = 2.75 tan φ ′ tan 20 From Figure 16.28, c′ ≈ 0.05 γ H cr tan φ ′

or, = 0.05

b.

20 = ; H cr 68.7 m (16)( H cr )(tan 20)

Fs 1  1  β tan −1  = = = 2.14.=  33.7°. tan φ ′ tan 25  1.5 

Figure 16.28: 750 c′ 0.05 ≈ 0.05; = (110)( H cr )(tan 25) γ H cr tan φ ′

H cr ≈ 292 ft

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1 2 3 β tan −1 = 16.18 =   26.57°; φ′ = 10°; c′ = 700 lb/ft ; γ = 110 lb/ft . 2

φd′ (deg)

Fφ ′ =

tan φ ′ tan φd′

m

cd′ = mγ H

Fc′ =

(lb/ft 2 )

c′ cd′

5

2.01

0.098

539

1.30

6

1.68

0.090

495

1.41

8

1.25

0.078

429

1.63

10

1.00

0.064

352

1.99

The plot of Fc′ vs. Fφ′ is shown. From the figure, Fc′ = Fφ′ = Fs = 1.5.

16.19

φd′ (deg)

n′ = 1; φ′ = 20°; c′ = 400 lb/ft2; γ = 115 lb/ft3; H = 30 ft

Fφ ′ =

tan φ ′ tan φd′

m

cd′ = mγ H

Fc′ =

(lb/ft 2 )

c′ cd′

5

4.16

0.133

458.9

0.87

10

2.06

0.105

362.3

1.1

15

1.36

0.080

276.0

1.45

20

1.00

0.058

200.1

2.0 130

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The plot of Fc′ vs. Fφ′ is shown. From the figure, Fφ′ = Fc′ = Fs = 1.36.

16.20

φd′ (deg)

 1  2 3 β tan −1  = n′ = 2.5;=  21.8°; φ′ = 12°; c′ = 24 kN/m ; γ = 16.5 kN/m ; H = 12 m  2.5 

Fφ ′ =

tan φ ′ tan φd′

m

cd′ = mγ H (kN/m2)

Fc′ =

c′ cd′

5

2.43

0.088

17.42

1.38

8

1.51

0.060

11.88

2.02

10

1.21

0.048

9.50

2.53

12

1.00

0.038

7.52

3.19

The plot of Fc′ vs. Fφ′ is shown on below.

From the figure, Fφ′ = Fc′ = Fs = 1.75.

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16.21

φ′ = 20°, β = 45°, kh = 0.4, kv = 0.2 (↑). From Table 16.5, m′ = 0.3992 Hence,

φm′ = tan −1[(tan 20)(0.3992)] = 8.26°

From Figure 16.24, m ≈ 0.11 = Fs 16.22

c′ 400 = ≈ 1.05 mγ H (0.11)(115)(30)

a. Refer to the figure.

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40 = sin 30°; x= 80 ft x 40 40 = sin 35°; R= = 69.73 ft Radius, R sin 35 With radius R = 69.73 ft, the trial surface circle has been drawn.

The following table can now be prepared.

Slice

2

Area of slices (ft )

Weight of slice Wn= A × γ (kip/ft)

αn (deg)

Wn cos an (kip/ft)

Wn sin an (kip/ft)

1

(26)(20) = 260 2

29.9

47

20.39

21.86

2

(10)(26 + 32) = 290 2

33.35

32

28.28

17.67

3

(20)(32 + 20) = 520 2

59.8

20

59.19

20.45

4

(20)(20) = 200 2

23

5

22.91

2.00

Σ127.77 Σ61.98 133 © 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.

Fs =

Rθ c′ + (ΣWn cos αn ) tan φ ′ ΣWn sin αn  π   (69.73)   (70)  (0.4) + (127.77)(tan 20)  180   = 1.3 61.98

Note: The accuracy can be increased by increasing the number of slices.

H H 5 α; x = = 10 m b. As in Part a,= sin = x sin α sin 30 x 5   = R= 7.78 m  2  = sin θ , or sin 40 R 2 With a radius R = 7.78 m, the trial surface has been drawn.

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The following table can now be prepared.

2

Slice

Area of slices (m )

Weight of slice Wn= A × γ (kN/m)

αn (deg)

33.35

54

19.60

29.98

Wn cos αn Wn sin αn (kN/m) (kN/m)

1

(2.6)(1.5) = 1.95 2

2

(2.6 + 4.2)(2) = 6.80 2

116.28

38

91.63

71.59

3

(4.2 + 2.8)(2) = 7.00 2

119.70

20

112.48

40.94

4

(3)(2.8) = 4.20 2

71.82

6

71.43

7.52

Σ295.14

Σ150.02

Fs =

Rθ c′ + (ΣWn cos αn ) tan φ ′ ΣWn sin αn  π   (7.78)   (80)  (18) + (295.14)(tan15)  180   1.83 = 150.02

Note: The accuracy will improve with smaller slices.

16.23

c′

300

= = 0.1. φ=′ 20°= ; β 26.57°; ru = 0.5; γ H (120)(25) Using Table 16.8, the following table can be prepared.

D

m″

n″

Fs = m″ − n″ru

Toe circle

1.804

2.101

0.75

1.00

1.841

1.143

1.27

1.25

1.874

1.301

1.22

1.50

2.079

1.528

1.32 135

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Fs ≈ 0.75.

16.24

c′ 6 = = 0.05. φ′ = 20°; β = 18.43°; ru = 0.5, γ H (20)(6) Using Table 16.8, the following table can be prepared.

D

m"

n"

Fs = m" − n"ru

1.00

1.840

1.387

1.15

1.25

1.834

1.493

1.09

1.50

2.011

1.705

1.16

Fs ≈ 1.09.

16.25

β = 20°; φ′ = 15°; ru = 0.5; γ = 17.5 kN/m3; c′ = 20 kN/m2; H = 15 m

c′ Fs(assumed) γ HF s

tan φ ′ φd′ (deg) Fs (calculated) = tan φ ′ d

1.2

0.0635

15.5

0.966

1.1

0.0693

14.0

1.075

1.0

0.0762

12.5

1.209

0.9

0.0847

11.5

1.317

From the plot, Fs ≈ 1.09.

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1 2 3 β tan −1 = 16.26 =   18.43°; φ′ = 25°; c′ = 12 kN/m ; γ = 19 kN/m ; ru = 0.25; H = 12.63 m 3 c′ γ HFs

φd′ (deg)

1.7

0.0294

15

1.74

1.8

0.0278

16

1.63

1.9

0.0263

17

1.53

Fs(assume d)

Fs ( calculated ) =

tan φ ′ tan φd′

From the graph, Fs ≈ 1.72.

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Chapter 17 φ=′ 25°; φ=′ 25°; N q = 12.72; N γ = 8.34 (Table 17.1)

17.1

q 1 1  q all = u =  c′N c + qN q + γ BN γ  Fs 3  2  1 1  = (600)(25.13) + (3.5)(115)(12.72) + (115)(4)(8.34)  3 2 

= 7372 lb/ft2

φ=′ 20°; N c = 17.69; N q = 7.44; N γ = 3.64 (Table 17.1)

17.2

q 1 1  qall = u =  c′N c + qN q + γ BN γ  Fs 3  2  1 1  = (14)(17.69) + (17.5)(1)(7.44) + (17.5)(1.2)(3.64)  3 2 

= 138.7 kN/m2

φ ′= 0°; N c = 5.7; N q = 1; N γ = 0 (Table 17.1)

17.3

q 1 1 qall =u = (cu N c + qN q ) = [(48)(5.7) + (17.7)(0.6)(1)] = 71 kN/m2 Fs 4 4

For a continuous foundation with vertical loading, all inclination factors and shape factors are equal to one. So,

17.4

qu 1  1  qall ==  c′N c λcd + qN q λqd + γ BN γ λγ d  Fs Fs  2 

φ=′ 25°; N c = 20.72; N q = 10.66; N γ = 10.88 (Table 17.2)  Df   3.5  1 + 0.4  1.35 = =  4   B 

λcd = 1 + 0.4 

1 + 2 tan φ ′(1 − sin φ ′) 2 λqd =

Df

 3.5  = 1 + (2)(tan 25)(1 − sin 25) 2  1.272 = B  4 

λγ d = 1

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(600)(20.72)(1.35) + (115)(3.5)(10.66)(1.272)  1  = 8248 lb/ft2 qall =  1  3 + (115)(4)(10.88)(1.0) 2  

17.5

φ=′ 20°; N c = 14.83; N q = 6.4; N γ = 5.39  1  1.333 =  1.2 

1 + 0.4  λcd =

 1  1.263 =  1.2 

1 + 2 tan 20(1 − sin 20) 2  λqd =

λγ d = 1 (14)(14.83)(1.333) + (1)(17.5)(6.4)(1.263)  1  = 158.3 kN/m2 qall = 1   3 + (17.5)(1.2)(5.39)(1.0) 2  

17.6

q= all

1  1   c′N c λcd + qN q λqd + γ BN γ λγ d  Fs  2 

φ ′= 0°; N c = 5.14; N q = 1.0; N γ = 0  Df   0.6  1 + 0.4  1.3 = = 0.8 B    

1; λcd = 1 + 0.4  λqd = λγ d =

= qall 17.7

1 [(48)(5.14)(1.3) + (0.6)(17.7) + 0] = 82.8 kN/m2 4

1 λ= λ= For continuous footing, Equation (17.31): λ= cs qs γs

λ= λ= 1 Also, for vertical load: λ= ci qi γi 1 So, qu = c′λcd N c + qλqd N q + γ Bλγ d N γ 2

φ=′ 28°. From Table 17.2, Nc = 25.8; Nq = 14.72. From Table 17.3, Nγ = 11.19.  Df   φ′  28   0.7   1 + 0.2  1.29  tan  45 +  =  tan  45 +  = 2 2   0.8    B  

λcd = 1 + 0.2 

 Df   φ′  28   0.7   1 + 0.1 1.146  tan  45 +  =  tan  45 +  = 2 2   0.8    B  

λqd = λγ d = 1 + 0.1

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q= all

17.8

(14)(25.8)(1.29) + (0.7 ×16.8)(14.72)(1.146)  qu 1   = 252.1 kN/m2 = 1  Fs 3  + (16.8)(0.8)(1.146)(11.9)  2 

qu qN q + 0.4γ ′BN γ .

Eq. (17.12): =

φ′ = 35°; Nq = 41.44; Nγ = 45.41 q = γ h + γ ′( D f − h) = (105 × 2) + (118 − 62.4)(4 − 2) = 321.2 lb/ft2

= Qall

= Qall 17.9

qu B 2 B 2 (qN q + 0.4γ ′BN γ ) = Fs Fs 52 1 = 153 kip [(321.2)(41.44) + (0.4)(118 − 62.4)(5)(45.41)] 3 1000

φ′ = 25°. From Table 17.1, Nc = 25.13; Nq = 12.72; Nγ = 8.34

γ=

(1800)(9.81) = 17.66 kN/m3 1000

q= 1.3c′N c + qN q + 0.4γ av BN γ u = q γ= D f (1.2)(17.66) = 21.19 kN/m2

γ av =

1 [γ D + γ ′( B − D)] —[Eq. (17.17a)] B

D =h − D f =2 − 1.2 =0.8 m

= γ sat = γ av

(1980)(9.81) = 19.42 kN/m3 1000 1 = − 0.8)] 13.19 kN/m3 [(17.66)(0.8) + (19.42 − 9.81)(1.8 1.8

= qu (1.3)(23.94)(25.13) + (21.19)(12.72) + (0.4)(13.19)(1.8)(8.34) = 1130.8 kN/m2 = Qall

17.10

(1130.8) B 2 (1130.8)(1.8) 2 = = 1221 kN Fs 3

qall From Eq. (17.12):=

1 (1.3c′N c + qN q + 0.4γ BN γ ) Fs

φ′ = 20°; Nc = 17.69; Nq = 7.44; Nγ = 3.64 (Table 17.1) 141 © 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.

1 [(1.3)(200)(17.69) + (3)(110)(7.44) + (0.4)(110)( B)(3.64)] 3

= qall

= 2351.5 + 53.38 B (a)

qall =

42, 260 B2

(b)

From Eqs. (a) and (b),

42, 260 = 2351.5 + 53.38 B. By trial and error, B2

B ≈ 4.1 ft 17.11

φ′ = 35°. From Table 17.2, Nq = 33.3; Nγ = 48.03

1 = qu qN q λqd λqs + γ BN γ λγ d λγ s 2 q = 321.2 lb/ft2 (see Problem 17.8). B 5  tan φ ′ = 1 +   tan 35° = 1.7 L 5

λqs = 1 + 

B 5 1 − 0.4   = 0.6 = L 5

1 − 0.4  λγ s =

1 + 2 tan φ ′(1 − sin φ ′) 2 λqd =

Df

4 = 1 + 2 tan 35(1 − sin 35) 2   = 1.204 B 5

λγ d = 1 qu

1 (321.2)(33.3)(1.204)(1.7) + (55.6)(5)(48.03)(1.0)(0.6) 2

= 25,284 lb/ft2 Qall =

17.12

(25, 284)(5) 2 = 210.7 kip (3)(1000)

1 = qu qN q λqd λqs + γ B′N γ λγ d λγ s a. For vertical load, Eq. (17.45): 2 c′ = 0, φ′ = 35°. Table 17.2: Nq = 33.3; Nγ = 37.152 (Table 17.3).

B′ =B − 2 x =5 − (2)(0.6) =3.8 ft; L′ = 5 ft Table 17.4:

φ′   B′  2   3.8  2 tan  45 +  = 1 + 0.1 1.28   tan (45 + 17.5) = 2  L′    5 

1 + 0.1 λqs = λγ s =

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 Df   φ′   3.5  tan  45 +  = 1 + 0.1 1.177   tan(45 + 17.5) = 2  3.8   B′  

λqd = λγ d = 1 + 0.1

qu

1 (110)(3.5)(1.28)(1.177)(33.3) + (1.28)(1.177)(110)(3.8)(37.152) 2

= 31,031 lb/ft2 qu B′L′ (31, 013)(3.8)(5) = Qall = = 147.3 kip (4)(1000) Fs

6 − (2)(0.5) = 5.0 ft; L′ = 6 ft. φ ′ = 25° b. . B′ = Tables 17.2 and 17.3: Nc = 20.72; Nq = 10.66; Nγ = 6.765

1 qu = c′N c λcs λcd + qN q λqs λqd + γ B′N γ λγ s λγ d 2

φ′   B′  2  5 tan  45 +  = 1 + 0.2   tan 2 (45 + 12.5) = 1.411  2  L′   6

1 + 0.2  λcs =

φ′   B′  2  5 tan  45 +  = 1 + 0.1  tan 2 (57.5) = 1.205  2  L′   6

1 + 0.1 λqs = λγ s =

 Df   φ′   4.5  tan  45 +  = 1 + 0.2  1.283   tan(57.5) = 2  5   B′  

1 + 0.2  λcd =

 Df   φ′   4.5  tan  45 +  = 1 + 0.1 1.141   tan(57.5) = 2  5   B′  

λqd = λγ d = 1 + 0.1

So, qu = (400)(20.72)(1.411)(1.283) + (120)(4.5)(10.66)(1.205)(1.141) + (0.5)(120)(5)(6.765)(1.205)(1.141) = 25,709 lb/ft2 = 25.71 kip/ft2

Qall =

(25.71)(5 × 6) = 192.8 kip 4

c. φ′ = 40°; c′ = 0. Tables 17.2 and 17.3: Nq = 64.2; Nγ = 93.69.

B′ = 3 − (2)(0.3) = 2.4 m; L′ = 3m

= γ

(1950)(9.81) = 19.13 kN/m3 1000

φ′  40   B′  2   2.4  2  tan  45 +  = 1 + 0.1 tan  45 +  = 1.368   2 2   L′    3  

λqs = λγ s = 1 + 0.1

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 Df   φ′   1.4  tan  45 +  = 1 + 0.1 1.125   tan(45 + 20) = 2  2.4   B′  

λqd = λγ d = 1 + 0.1

1 = qu qN q λqs λqd + γ BN γ λγ s λγ d 2 1 (19.13)(1.4)(64.2)(1.368)(1.125) + (19.13)(2.4)(93.69)(1.368)(1.125) 2 = 5956 kN/m2 (5956)(2.4)(3) = Qall = 3 17.13

14,294 kN

φ′ = 27°; c′ = 38 kN/m2; B = 2.25; Df = 2 m; α = 12° Using Eq. (17.31) which provides the vertical component of the inclined loading, q= all

1  1   c′N c λcs λcd λci + qN q λqs λqd λqi + γ BN γ λγ s λγ d λγ i  2 Fs  

From Table 17.2, Nc = 23.94; Nq = 13.2; Nγ = 14.47

 B   Nq   13.2  1+  1.55 = =  L   Nc   23.94 

λcs = 1+  

 Df   2  1 + 0.4  1.355 = =  2.25   B 

1 + 0.4  λcd =

2

2

α°   12   0.751 λci = λqi =− 1  =− 1  =  90°   90  B  

1 +   tan φ ′ = 1 + tan 27 = 1.509 λqs = L

1 + 2 tan φ ′(1 − sin φ ′) 2 λqd =

Df

 2  1 + 2 tan 27(1 − sin 27) 2  1.27 = = B  2.25 

B 0.6 = L

λγ s = 1 − 0.4  λγ d = 1

2

2

 α   12  λγ i = 0.308 1 −  = 1 −  =  φ   27  Since the groundwater table is above the footing level, this represents Case I in Figure 17.6.

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D = 2 −1 = 1 m q γ ( D f − D ) + γ ′D = (16.8)(2 − 1) + (19 − 9.81)(1) = 25.99 kN/m2 Eq. (17.16): =

qall =

1 (38)(23.94)(1.55)(1.355)(0.751) + (25.99)(13.2)(1.509)(1.27)(0.751)   +(0.5)(2.25)(19 − 9.81)(14.47)(0.6)(1)(0.308) 3.5  

= 558.94 kN/m2 The allowable gross vertical load is 2 ) 2829.63 kN = = Qall (= qall )( B 2 ) (558.94)(2.25

Qall(i) Therefore, the allowable inclined load,= 17.14

2829.6 ≈ 2893 kN cos12

γ = 16 kN/m3; c′ = 0; φ′ = 29°, Df = 1.3 m; B = 1.75 m; and x = 0.25 m From Table 17.2, N q = 16.44; N γ = 19.34

1 = qu′ qN q λqd λqs + γ B′N γ λγ d λγ s For vertical load, Eq. (17.44): 2

B′ =B − 2 x =1.75 − (2)(0.25) =1.25 m; L′ =1.75 m  B′   L′ 

 1.25   1.75 

1 +   tan φ ′ = 1+  1.396 λqs =  tan 29 =  B′   1.25  = 1 − 0.4  0.714  =  L′   1.75 

1 − 0.4  λγ s =

 Df   1.3  1 + 2 tan 29(1 − sin 29) 2  1.306  =  1.25   B′ 

1 + 2 tan φ ′(1 − sin φ ′) 2  λqd =

λγ d = 1 qu′

1 (16)(1.3)(1.396)(1.306)(16.44) + (16)(1.25)(19.34)(1)(0.714) 2

= 761.52 kN/m2 qu′ B′L′ (761.52)(1.25)(1.75) Qall = = = 416 kN Fs (4) 17.15

γ = 16 kN/m3; c′ = 0; φ′ = 29°, Df = 1.3 m; B = 1.75 m; and x = 0.25 m From Table 17.2, Nq = 16.44; Nγ = 19.34

1 = qu qN q λqd λqs + γ BN γ λγ d λγ s For vertical load, Eq. (17.31): 2

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e Eq. (17.53): Reduction factor, Rk = a   B

k

k  e  Eq. (17.54): Qu (eccentric) = Qu (centric) 1 − a    = Qu (centric) (1 − Rk )  B   

1   ( B L)(qu ) = ( B) 2  qN q λqd λqs + γ BN γ λγ d λγ s  Qu (centric) =× 2   B  

1 +   tan φ ′ = 1 + tan 29 = 1.554 γ qs = L B 1 − 0.4 = 0.6 = L

1 − 0.4  γγs =

 Df   1.3  1 + 2 tan 29(1 − sin 29) 2  1.218 = =  1.75   B 

λqd = 1 + 2 tan φ ′(1 − sin φ ′) 2  λγ d = 1

qu = (16)(1.3)(16.44)(1.554)(1.218) + (0.5)(16)(1.75)(19.34)(0.6)(1) = 809.7 kN/m2 2 = Qu (centric) (1.75) = (809.7) 2479.7 kN

2

B B Eq. (17.55): a =  − 1.6   + 2.13 =(1) 2 − (1.6)(1) + 2.13 =1.53 L L 2

B B Eq. (17.56):= k 0.3   − 0.56   + = 0.9 (0.3)(1) 2 − 0.56(1) + = 0.9 0.64 L L k

0.64

e  0.25  Rk a= = 0.44   (1.53)  =  B  1.75  Qu (eccentric) = Qu (centric) (1 −= Rk ) (2479.7)(1 − 0.44) ≈ 1389 kN

17.16

γ = 17 kN/m3; c′ = 0; φ′ = 34°; Df = 1.2 m; B = 1.8 m; e = 0.3 m; and a = 10 Part a: Meyerhof’s (1963) method From Table 17.2, Nq = 29.44; Nγ = 41.06 B′ = 1.8 − (2)(0.3) = 1.2 m Since it is a continuous foundation, all shape factors are equal to one.

1 = qu′ qN q λqi λqd + γ B′N γ λγ i λγ d 2 146 © 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.

λqd = 1 + 2 tan φ ′(1 − sin φ ′) 2  

λqi =− 1

α° 

2

Df  1.2  = 1 + 2 tan 34(1 − sin 34) 2  1.262 = B′  1.2  2

 10  0.79  =− 1  = 90°   90 

λγ d = 1 2

2

 α   10  λγ i = 0.498 1 −  = 1 −  =  φ   34  qu′ = (17 × 1.2)(29.44)(1.262)(0.79) + (0.5)(1.2)(17)(41.06)(1)(0.498) = 807.33 kN/m

= From Eq. (17.58): Q u ( ei )

qu′ B′ (807.33)(1.2) = = 983.74 ≈ 984 kN /m cos α cos10°

Part b: Saran and Agarwal (1991) method 1   = c′ 0,= Qu ( ei ) B  qN q ( ei ) + γ BN γ ( ei )  [Eq. (17.59)] With 2  

For φ ′ =34°,

e 0.3 = = 0.167, and α= 10°, Figures 17.20(b) and 17.21(b) yield B 1.8

N q ( ei ) ≈ 22.8 and N γ ( ei ) ≈ 14.5

1   (1.8) (17)(1.2)(22.8) + (17)(1.8)(14.5)  ≈ 1237 kN /m Qu ( ei= ) 2  

Part c: Patra et al. (2012) reduction factor method [Eq. (17.60)]   e   α  Eq. (17.54): Qu ( ei ) = Bqu 1 − 2    1 −   B   φ′  

2−( D f / B )

qu is the ultimate bearing capacity with vertical centric load. Since it is a continuous foundation, all shape factors are equal to one.

1 = qu qN q λqd + γ BN γ λγ d 2 = N q 29.44; = N γ 41.06 From Table 17.2,

λqd = 1 + 2 tan φ ′(1 − sin φ ′) 2

Df

 1.2  = 1 + 2 tan 34(1 − sin 34) 2  1.174 = B  1.8 

qu = (17 ×1.2)(29.44)(1.174) + (0.5)(17)(1.8)(41.06)(1) = 1333.3 kN/m

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  0.3    10  Qu ( ei ) = (1.8)(1333.3) 1 − (2)    1 −   1.8    34  

2 − (1.2/1.8)

= 1005.7 ≈ 1006 kN /m

Comments: Patra et al.’s method yields results fairly close to Meyerhof’s solution.

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Chapter 18 18.1

Tube 1: Di = Do – t = 50.8 – 1.24 = 49.56 mm

AR (%) = Eq. (18.6):

Do2 − Di2 (50.8) 2 − (49.56) 2 = (100) ×100 = 5.06% ≤ 10% Di2 (49.56) 2

Tube 2: Di = Do – t = 76.2 – 1.65 = 74.55 mm

AR (%) = Eq. (18.6):

Do2 − Di2 (76.2) 2 − (74.55) 2 = (100) ×100 = 4.47% ≤ 10% Di2 (74.55) 2

Tube 3: Di = Do – t = 127 – 3.05 = 123.95 mm

Do2 − Di2 (127) 2 − (123.95) 2 AR (%) = (100) = ×100 = 4.98% ≤ 10% Eq. (18.6): Di2 (123.95) 2

18.2

For all three tubes, samples can be considered undisturbed, and therefore, suitable for consolidation and unconfined compression tests. Grain size distribution and Atterberg limit tests are conducted on remolded or disturbed soils, so sample disturbance is not a factor.

Shelby tube A:

AR (%) = Eq. (18.6):

Do2 − Di2 (101.6) 2 − (98.4) 2 = (100) ×100 = 6.61% ≤ 10% Di2 (98.4) 2

Shelby tube B:

AR (%) = Eq. (18.6):

Do2 − Di2 (89) 2 − (85.7) 2 (100) = ×100 = 7.85% ≤ 10% Di2 (85.7) 2

Split spoon sampler:

Do2 − Di2 (50.8) 2 − (34.9) 2 AR (%) = = ×100 = 111% ≥ 10% (100) Eq. (18.6): Di2 (34.9) 2 Shelby tubes A and B: Samples can be considered undisturbed and, therefore, suitable for consolidation and unconfined compression tests. Grain size distribution and Atterberg limit tests are conducted on remolded or disturbed soils, so sample disturbance is not a factor.

Split spoon sampler: Samples can be considered as highly disturbed. Suitable for grain-size distribution and Atterberg limit tests, but not for consolidation and unconfined compression tests.

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18.3

σ o′

1 σ o′

Depth (m)

2

(kN/m )

C N = 9.78

1.5

23.25

2.03

8

≈16

3.0

46.5

1.43

7

≈10

4.5

69.75

1.17

12

≈14

6.0

93.0

1.01

14

≈14

7.5

116.25

9.907

13

≈12

N60 (N1)60 = CNN60

0.34

18.4

    N 60 −1   ′ φ = tan   σ o′   12.2 + 20.3     pa   

Depth (m)

σ′ο (kN/m2)

po (kN/m2)

N60

φ′ (deg) [Eq. (18.28)]

1.5

23.25

100

8

37.8

3.0

46.5

100

7

34.3

4.5

69.75

100

12

37.4

6.0

93.0

100

14

37.3

7.5

116.25

100

13

35.3

; pa ≈ 100 kN/m2

Average φ′ ≈ 36.5°

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18.5

Eq. (18.29):

= φ′

Depth (m)

15.4( N1 )60 + 20

(N1)60 φ′(deg)

1.5

16

35.7

3.0

10

32.4

4.5

14

34.7

6.0

14

34.7

7.5

12

33.6

Average φ′ ≈ 34.22° 18.6

Eq. (18.30):

= φ ′ 27.5 + 9.2 log( N1 )60 Depth (m) (N1)60 φ′(deg) 1.5

16

38.6

3.0

10

36.7

4.5

14

38.0

6.0

14

38.0

7.5

12

37.4

Average φ′ ≈ 37.74°

18.7

1.7    0.06   N 60  0.23 +   D   98   50   Eq. (18.22): Dr (%) =    9  σ o′     

0.5

(100)

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Given γ = 15.5 kN/m3. The following table can now be prepared.

Depth, z (m)

σ o′ = γ z

N60

Dr (%)

(kN/m2)

D50 (mm)

1.5

23.25

0.26

6

86.8 ≈ 87

3.0

46.5

0.26

12

86.8 ≈ 87

4.5

69.75

0.26

17

84.3 ≈ 84

6.0

93.0

0.26

21

81.2 ≈ 81

7.5

116.25

0.26

23

76

18.8

Eq. (18.23): 0.5

    N 60   Dr (%) =   (100)     ′ σ  17 + 24 o       pa    

Depth, z (m)

σ′ o (kN/m2)

N60

Dr (%)

1.5

23.25

6

51.5 ≈ 52

3.0

46.5

12

65.3 ≈ 65

4.5

69.75

17

71.0 ≈ 71

6.0

93.0

21

73.1 ≈ 73

7.5

116.25

23

71.6 ≈ 72

18.9

    N 60 −1   φ ′ = tan   σ o′   12.2 + 20.3      pa  

0.34

; pa ≈ 100 kN/m2

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Depth (m) σ′ o (kN/m2)

N60

φ′ (deg)

1

1 × 16 = 16

6

35.9

2.5

1.75 × 16 + 0.75(19.7 − 9.8) = 35.42

8

36.5

4

1.75 × 16 + 2.25(19.7 − 9.8) = 50.27

11

38.1

5.5

1.75 × 16 + 3.75(19.7 − 9.8) = 65.12

13

38.5

7

1.75 × 16 + 5.25(19.7 − 9.8) = 79.97

14

38.2

8.5

1.75 × 16 + 6.75(19.7 − 9.8) = 94.82

16

38.4

Average φ′ = 37.6° 18.10

Eq. (18.46) and Table 18.7: (qc /pa ) 0.26 = 7.64 D50 . Use pa ≈ 100 kN/m2.

N 60 D50 (mm) qc (kN/m2)

Depth (m)

N60

1

6

0.46

3746

2.5

8

0.46

4995

4

11

0.46

6868

5.5

13

0.46

8116

7

14

0.46

8741

8.5

16

0.46

9989

18.11

= cu Given: Nk ≈ 18.3. Eq. (18.42):

18.12

2B = 2 × 1.5 = 3 m

qc − σ o 1150 − (5.5)(17.8) = = 57.5 kN/m2 18.3 Nk

Cone penetration resistance and sleeve friction are averaged up to 3 m below the footing (depth of 4 m from the surface).

The following data are estimated from Figure 18.18:

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Depth (m)

qc (kN/m2) fc (kN/m2)

1

4200

90

2

3500

70

3

4500

50

4

4800

100

Avg.

4250

≈ 78

Eq. (18.47): Average friction ratio, F= r

fc 78 = = 0.018 = 1.8% qc 4250

Figure 18.13: For qc = 4.25 MN/m2 and Fr = 1.8%, the soil type is estimated as silty sand to sandy silt.

18.13

Figure 18.13: For the soil type “silty sand to sandy silt,”

(qc /pa ) =3 N 60

2 N Since average qc = 4250 kN/m2 and pa ≈ 100 kN/m= , 60

18.14

4250 ≈ 14 (100)(3)

  q  = φ ′ tan −1 0.1 + 0.38log  c   a. Eq. (18.36):  σ o′    Given γd = 17 kN/m3. The following table can now be prepared.

Depth (m) σ′ o (kN/m2) qc (kN/m2) φ′ (deg) 1

17

4200

45.26

2

34

3500

40.85

3

51

4500

40

4

68

4800

38.75

Average φ′ = 41.2° b. Eq. (18.40): Es = 3.5qc

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Depth (m) qc (kN/m2)

Es (kN/m2)

1

4200

14,700

2

3500

12,250

3

4500

15,750

4

4800

16,800

18.15

From Eqs. (18.48) and (18.49):  2.5  Recovery ratio =   (100) = 55.5%  4.5 

Rock quality designation, RQD =

2.1 = 0.467 4.5

Based on Table 18.9, the quality of the rock is poor.

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156 © 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.