Principles of Foundation Engineering, 10th Edition Solutions Manual | Braja M. Das

SOLUTIONS MANUAL

PRINCIPLES OF FOUNDATION ENGINEERING
10TH EDITION

CHAPTER 2: GEOTECHNICAL PROPERTIES OF SOIL

(87.5)(9.81)
2.1 d.    17.17 kN /m 3
(1000)(0.05)

 17.17
c.     14.93 kN /m 3
1 w 1  0.15

Gs w
a. Eq. (2.12):  d 
1 e

(2.68)(9.81)
14.93  ; e = 0.76
1 e

e 0.76
b. Eq. (2.6): n    0.43
1  e 1  0.76

e. From Eq. (2.14):

Vw wGs  (0.15)(2.68) 
S    100  53%
Vv e  0.76

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

 20.1
d    16.48 kN/m3
1 w 1  0.22

Gs w Gs (9.81)
b.  d  16.48 kN/m3  
1 e 1 e

Eq. (2.15): e  wGs  (0.22)(Gs ) . So

9.81Gs
16.48  ; Gs  2.67
1  0.22Gs

,2.3 a. Eq. (2.6):

e 0.81
n   0.45
1  e 1  0.81

b. Eqs. (2.7), (2.14):

wGs  (0.21)(2.68) 
S   100  69.5%
e  0.81


c. Eq. (2.11):

Gs w (1  w) (2.68)(9.81)(1  0.21)
   17.58 kN /m3
1 e 1  .0.81

d. Eq. (2.12):

Gs w (2.68)(9.81)
   14.53 kN /m3
1 e 1  .0.81

Gs w
2.4 a. Eq. (2.12):  d 
1 e
e
Eq. (2.15): Gs 
w
e
  w
d   
w
So,
1 e

(e)(9.81)
13.5  ; e  0.98
(0.36)(1  e)

e 0.98
b. Eq. (2.6): n    0.495  0.5
1  e 1  .0.98

e 0.98
c. Eq. (2.14): Gs    2.72
w 0.36

d. Eq. (2.13):  sat   d (1  w)  (13.5)(1 0.36)  18.36 kN /m3

, emax  e
2.5 Eq. (2.23): Dr 
emax  emin

0.91  e
0.65  ; e  0.63
0.91  0.48

Gs w (2.67)(9.81)
Eq. (2.12):  d    16.07 kN /m3
1 e 1  .0.63

Eq. (2.13):    d (1  w)  (16.07)(1  0.1)  17.68 kN/m3

Gs w (1  w)
2.6 Eq. (2.11):  
1 e

(2.66)(9.81)(1  0.08)
17  ; e  0.658
1 e

emax  e
Eq. (2.23): Dr 
emax  emin

emax  0.658
0.6  ; emax  1.045
emax  0.4

Gs w (2.66)(9.81)
Eq. (2.12):  d    12.76 kN /m 3
1  emax 1  1.045

2.7 Soil A: A-1-a(0)

Soil B: A-2-6(1)

GI  0.01( F200  15)( PI  10)  0.01(33  15)(13  10)  0.54  1

Soil C: A-7-5(19)

GI  ( F200  35)[0.2  0.005( LL  40)]  0.01( F200  15)( PI  10)
 (72  35)[0.2  0.005(56  40)]  0.01(72  15)(25  10)  18.91  19

, Soil D: A-4(5)

GI  ( F200  35)[0.2  0.005( LL  40)]  0.01( F200  15)( PI  10)
 (64  35)[0.2  0.005(35  40)]  0.01(64  15)(9  10)  4.585  5

Soil E: A-2-7(1)

GI  0.01( F200  15)( PI  10)  0.01(30  15)(14  10)  0.6  1

Soil F: A-6(5)

GI  ( F200  35)[0.2  0.005( LL  40)]  0.01( F200  15)( PI  10)
 (55  35)[0.2  0.005(35  40)]  0.01(55  15)(14  10)  5.1  5

2.8 Soil A: From Table 2.9 and Figure 2.9, symbol is SM
Gravel portion = 100  92 = 8% < 15%
From Figure 2.10 group name is silty sand

Soil B: From Table 2.9 and Figure 2.9, symbol is SM
Gravel portion = 100 100 = 0% < 15%
From Figure 2.10 group name is silty sand

Soil C: From Table 2.9 and Figure 2.9, symbol is MH
Percent passing No. 200 sieve is 72 (> 30%)
Percent sand = 100  72 = 28
Percent gravel = 100  100 = 0
From Figure 2.11, group name is sandy elastic silt

Soil D: From Table 2.9 and Figure 2.9, symbol is ML
Percent passing No. 200 sieve is 64 (> 30%)
Percent gravel = 100 95 = 5
Percent sand = 9564 = 31
Group name is sandy silt (Figure 2.11)

Soil E: From Table 2.9 and Figure 2.9, symbol is SM
Gravel portion = 100  100 = 0 (< 15%)
From Figure 2.10, group name is silty sand

Soil F: From Table 2.9 and Figure 2.9, symbol is CL
Percent passing No. 200 sieve is 55 (> 30%)