Chapter 2 Manual for... · 2019. 10. 27. · sat(clay) 18.55 kN/m 1 (0 3478)(2. 74) (1 )...

1

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Chapter 2

2.1 d. 3kN/m 17.17

)05.0)(1000(

)81.9)(5.87(

c. 3

kN/m 14.93

15.01

17.17

1 wd

a. Eq. (2.12): 0.76

e

ee

G ws ;1

)81.9)(68.2(14.93 .

1

b. Eq. (2.6): 0.43

76.01

76.0

1 e

en

e. Eq. (2.14): 53%

)100(

76.0

)68.2)(15.0(

e

Gw

V

VS s

v

w

2.2 a. From Eqs. (2.11) and (2.12), it can be seen that,

3

kN/m 16.48

22.01

1.20

1 wd

b. e

G

e

G sws

1

)81.9(

1kN/m 16.48 3

Eq. (2.14): ).)(22.0( ss GwGe So,

2.67

s

s

s GG

G;

22.01

81.948.16

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Chapter 2

2


2

2.3 a. 0.55

e

ee

wG ws ;1

)12.01)(4.62)(65.2(119.5 .

1

)1(

b. 0.355

55.01

55.0n

c. 57.8% 10055.0

)65.2)(12.0(

e

GwS s

d. 3

lb/ft 106.7

12.01

5.119

1 wd

2.4 a. w

eGs . 0.97

ee

e

e

w

ew

d ;1

)4.62(36.0

85.43 .1

)(

b. 0.49

97.01

97.0

1 e

en

c. 2.6936.0

97.0

w

eGs

d. 3

lb/ft 115.9

97.01

)4.62)(97.069.2(

1

)(sat

e

eG ws



Chapter 2

3


2.5 From Eqs. (2.11) and (2.12): 3lb/ft 108

08.01

64.116

d

Eq. (2.12): 53.0 ;1

)4.62)(65.2(108 ;

1

e

ee

G wsd

Eq. (2.23): 0.94

max

max

max

minmax

max ;44.0

53.082.0 e

e

e

ee

eeDr

3lb/ft 85.2

94.01

)4.62)(65.2(

1 max

(min) e

G wsd

2.6 Refer to Table 2.7 for classification.

Soil A: A-7-6(9) (Note: PI is greater than LL 30.)

GI = (F200 – 35)[0.2 + 0.005(LL – 40)] + 0.01(F200 – 15)(PI – 10)

=

=

(65 – 35)[0.2 + 0.005(42 – 40)] + 0.01(65 – 15)(16 – 10)

9.3 9

Soil B: A-6(5)

GI =

=

(55 – 35)[0.2 + 0.005(38 – 40)] + 0.01(55 – 15)(13 – 10)

5.4 5

Soil C: A-3-(0)

Soil D: A-4(5)

GI =

=

(64– 35)[0.2 + 0.005(35 – 40)] + 0.01(64 – 15)(9 – 10)

4.585 ≈ 5

Soil E: A-2-6(1)

GI = 0.01(F200 – 15)(PI – 10) = 0.01(33 – 15)(13 – 10) = 0.54 ≈ 1

Soil F: A-7-6(19) (PI is greater than LL 30.)

GI =

=

(76– 35)[0.2 + 0.005(52 – 40)] + 0.01(76 – 15)(24 – 10)

19.2 ≈ 19



Chapter 2

4


4

2.7 Soil A: Table 2.8: 65% passing No. 200 sieve.

Fine grained soil; LL = 42; PI = 16

Figure 2.5: ML

Figure 2.7: Plus No. 200 > 30%; Plus No. 4 = 0

% sand > % gravel – sandy silt

Soil B: Table 2.8: 55% passing No. 200 sieve.

Fine grained soil; LL = 38; PI = 13

Figure 2.5: Plots below A-line – ML

Figure 2.7: Plus No. 200 > 30%

% sand > % gravel – sandy silt

Soil C: Table 2.8: 8% passing No. 200 sieve.

% sand > % gravel – sandy soil – SP

Figure 2.6: % gravel = 100 – 95 = 5% < 15% – poorly graded sand

Soil D: Table 2.8: 64% passing No. 200 sieve

Fine grained soil; LL = 35, PI = 9

Figure 2.5 – ML

Figure 2.7: % sand (31%) > % gravel (5%) – sandy silt

Soil E: Table 2.8: 33% passing No. 200 sieve; 100% passing No. 4 sieve.

Sandy soil; LL = 38; PI = 13

Figure 2.5: Plots below A-line – SM

Figure 2.6: % gravel (0%) < 15% – silty sand

Soil F: Table 2.8: 76% passing No. 200 sieve; LL = 52; PI = 24

Figure 2.5: CH

Figure 2.7: Plus No. 200 is 100 – 76 = 24%

% gravel > % gravel – fat clay with sand



Chapter 2

5


2.8 43.01117

)4.62)(68.2(1e ;

1

d

wswsd

G

e

G

Eq. (2.37):

43.01

43.0

63.01

63.0

22.0;

1

1

3

3

2

2

32

1

31

2

1

k

e

e

e

e

k

k; k2 = 0.08 cm/s

2.9 From Eq. (2.41):

n

nn

e

e

e

e

k

k)6316.0(1667.0 ;

9.1

2.1

2.2

9.2

1091.0

102.0or ;

1

16

6

2

1

1

2

2

1

898.31996.0

778.0

)6316.0log(

)1667.0log(

n

cm/s 10 0.075 6-

)10216.0(9.1

9.0

1

10216.02.1

)2.2)(102.0()1(

6998.3

3

6

998.3

6

1

11

e

eCk

e

ekC

n

n



Chapter 2

6


6

2.10 The flow net is shown.

/m/sm 1017.06

36-

8

)4)(25.5(

10

105.6

So, m. 25.575.17 cm/s; 105.6

2

4

21max4

q

HHhk

2.11 a.

7825.03

2

7825.03

210

6.01

6.0)2.0(4622.2

)1(4622.2

e

eDk cm/s 0.041

b. 32.2

6.0332.2

106.0

3

)2.0(2.0

4.0

6.01

6.0)35()(

135

DC

e

ek u cm/s 0.171



Chapter 2

7


2.12 3

dry(sand) kN/m 16.8455.01

)81.9)(66.2(

1

e

G ws

3

sat(sand) kN/m 81.0248.01

)48.066.2(81.9

1

e

eG wws

3

sat(clay ) kN/m 18.55)74.2)(3478.0(1

)3478.01)(81.9)(74.2(

1

)1(

s

ws

wG

wG

At A: σ = 0; u = 0; σ = 0

At B: σ = (16.84)(3) = 50.52 kN/m2

u = 0

σ = 50.52 kN/m2

At C: σ = σB + (20.81)(1.5) = 50.52 + 31.22 = 81.74 kN/m2

u = (9.81)(1.5) = 14.72 kN/m2

σ = 81.74 – 14.72 = 67.02 kN/m2

At D: σ = σC + (18.55)(5) = 81.74 + 92.75 = 174.49 kN/m2

u = (9.81)(6.5) = 63.77 kN/m2

σ = 174.49 – 63.77 = 110.72 kN/m2

2.13 Eq. (2.54): Cc = 0.009(LL – 10) = 0.009(42 – 10) = 0.288

Eq. (2.65):

mm 87.2

110

155log

82.01

mm) 10007.3)(288.0(log

1 o

o

o

ccc

e

HCS

2.14 Eq. (2.69):

mm 56.69

128

155log

82.01

)3700)(288.0(

110

128log

75.01

mm) 3700(5

288.0

log1

log1 c

o

o

cc

o

c

o

csc

e

HC

e

HCS



Chapter 2

8


8

2.15 a. Eq. (2.53): 0.392

150

300log

792.091.0

log1

2

21

eeCc

From Eq. (2.65): o

o

o

ccc

e

HCS

log

1

Using the results of Problem 2.12,

2kN/m 87.88)81.955.18(

2

5)81.981.20(5.1)84.16)(3( o

953.0)74.2)(3478.0( so wGe

mm 194.54

87.88

5087.88log

953.01

mm) 5000)(392.0(cS

b. Eq. (2.73): 2H

tCT v

v . For U = 50%, Tv = 0.197 (Table 2.11). So,

days 609 sec 105262;cm) 500(

1036.9197.0 4

2

4

tt

2.16 a. Eq. (2.53): 0.377

120

360log

64.082.0

log1

2

21 eeCc

b. 0.736

2

2

1

2

21 ;

120

200log

82.00.377 ;

log

eeee

Cc



Chapter 2

9


2.17 Eq. (2.73): 2H

tCT v

v . For 60% consolidation, Tv = 0.286 (Table 2.11).

Lab time: min 6

49 min 8

61 t

/minin. 0788.0;)5.1(

6

49

286.0 2

2

v

v

C

C

Field: U = 50%; Tv = 0.197

days 6.25 min 9000;

2

1210

)0788.0(197.0

2

t

t

2.18 5.060

30U

tt

H

tCT

v

v6

221

)1(

)1( 102

2

10002

))(2(

tt

H

tCT

v

v6

222

)2(

)2( 108

2

10001

))(2(

So, Tv(1) = 0.25Tv(2). The following table can be prepared for trial and error

procedure.

Tv(1) Tv(2) U1 U2

UHH

HUHU

21

2211 (Figure 2.22)

0.05

0.10

0.125

0.1125

0.2

0.4

0.5

0.45

0.26

0.36

0.40

0.385

0.51

0.70

0.76

0.73

0.34

0.473

0.52

0.50

So, Tv(1) = 0.1125 = 2 10-6

t; t = 56,250 min = 39.06 days



Chapter 2

10


10

2.19 Eq. (2.84): .2H

tCT cv

c tc = 60 days = 60 24 60 60 sec;2

2H m = 1000 mm.

0415.0)1000(

)60602460)(108(2

3

cT

After 30 days: 0207.0)1000(

)60602430)(108(2

3

2

H

tCT v

v

From Figure 2.24 for Tv = 0.0207 and Tc = 0.0415, U = 5%. So

Sc = (0.05)(120) = 6 mm

After 100 days: 069.0)1000(

)606024100)(108(2

3

2

H

tCT v

v

From Figure 2.24 for Tv = 0.069 and Tc = 0.0415, U 23%. So

Sc = (0.23)(120) = 27.6 mm

2.20

N

S1tan

Normal force, N (lb) Shear force, S (lb)

NS1tan (deg)

50

110

150

43.5

95.5

132.0

41.02

40.96

41.35

From the graph, 41

2.21 Normally consolidated clay; c = 0.

245tan2

31 ; 38

;

245tan309630 2



Chapter 2

11


2.22

245tan2

31 ; 30

;

245tan204020 2

2.23 c = 0. Eq. (2.91): 2kN/m 387.8

2

2845tan140

245tan 22

31

2.24 Eq. (2.91):

245tan2

24531 c

245tan2

245tan140368 2 c (a)

245tan2

245tan280701 2 c (b)

Solving Eqs. (a) and (b), = 24; c = 12 kN/m2

2.25 25

1332

1332sinsin 1

31

311

31

311sin

23

23 lb/in. 5.75.513 ;lb/in. 5.265.532

34

5.75.26

5.75.26sin 1

Normally consolidated clay; c = 0 and c = 0

2.26

245tan2

31 . 22

1 kN/m 305.92

2045tan150



Chapter 2

12


12

2kN/m 61.9

u

u

u ;

2

2845tan

150

9.305 ;

245tan 22

3

1

2.27 a. )log(6.14.01026 50DCD ur

30.7 )]13.0)[log(6.1()1.2)(4.0()53.0)(10(26

b. bae

1

327.209.0

21.0097.0101.2097.0101.2

15

85

D

Da

081.0)327.2)(398.0(845.0398.0845.0 ab

33.67

081.0)68.0)(327.2(

1



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Chapter 2 Manual for... · 2019. 10. 27. · sat(clay) 18.55 kN/m 1 (0 3478)(2. 74) (1 )...

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