Principles Of Foundation Engineering, 8th Edition Solution Manual

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An ’s Solutions Manual to Accompany
PRINCIPLES OF FOUNDATION
ENGINEERING, 8TH EDITION, SI EDITION
BRAJA M. DAS

Contents
Chapter Page
2.................................................................................................................................1
3...............................................................................................................................13
4...............................................................................................................................25
5...............................................................................................................................37
6...............................................................................................................................49
7...............................................................................................................................55
8...............................................................................................................................69
9...............................................................................................................................75
10..............................................................................................................................95
11............................................................................................................................107
12............................................................................................................................113
13............................................................................................................................121
14............................................................................................................................133
15............................................................................................................................147
16............................................................................................................................157
Note that there are no end of chapter questions for Chapter 1.

1
Chapter 2
2.1 d.3
kN/m17.17 )05.0)(1000(
)81.9)(5.87(
c.3
kN/m14.93




 15.01
17.17
1 w
d
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
e
n
e. Eq. (2.14):53%



 )100(
76.0
)68.2)(15.0(
e
Gw
V
V
S s
v
w
2.2 a. From Eqs. (2.11) and (2.12), it can be seen that,3
kN/m16.48




 22.01
1.20
1 w
d
b.e
G
e
G sws




 1
)81.9(
1
kN/m16.48 3
Eq. (2.14):).)(22.0( ss GwGe  So,2.67

 s
s
s G
G
G ;
22.01
81.9
48.16
2.3 a.0.55





 e
ee
wG ws ;
1
)12.01)(81.9)(65.2(
18.79.
1
)1(

2
b.0.355

 55.01
55.0
n
c.57.8% 100
55.0
)65.2)(12.0(
e
Gw
S s
d.3
kN/m16.78




 12.01
79.18
1 w
d
2.4 a.w
e
Gs  .0.97















 e
e
e
e
w
e w
d ;
1
)81.9(
36.0
18.43.
1
)(
b.0.49



 97.01
97.0
1 e
e
n
c.2.69 36.0
97.0
w
e
Gs
d.3
kN/m18.23





 97.01
)81.9)(97.069.2(
1
)(
sat e
eG ws
2.5 From Eqs. (2.11) and (2.12):3
kN/m98.16
08.01
34.18 

d
Eq. (2.12):53.0;
1
)81.9)(65.2(
16.98;
1 




 e
ee
G ws
d
Eq. (2.23):0.94





 max
max
max
minmax
max ;
44.0
53.0
82.0 e
e
e
ee
ee
Dr3
kN/m13.4




 94.01
)81.9)(65.2(
1 max
(min) e
G ws
d

3
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
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

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4
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
2.843.01
39.18
)81.9)(68.2(
1e;
1 






d
wsws
d
G
e
G
Eq. (2.37):





























43.01
43.0
63.01
63.0
22.0
;
1
1
3
3
2
2
3
2
1
3
1
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
102.0
or;
1
1 6
6
2
1
1
2
2
1 




























 
898.3
1996.0
778.0
)6316.0log(
)1667.0log( 


n

Loading page 7...

5cm/s100.075 6-






















)10216.0(
9.1
9.0
1
10216.0
2.1
)2.2)(102.0()1(
6
998.3
3
6
998.3
6
1
11
e
e
Ck
e
ek
C
n
n
2.10 The flow net is shown./m/sm1017.06 36-









 




8
)4)(25.5(
10
105.6
So,m.25.575.17cm/s;105.6
2
4
21m ax
4
q
HHhk
2.11 a.7825.0
3
2
7825.03
2
10 6.01
6.0
)2.0(4622.2
)1(
4622.2 





















 e
e
Dkcm/s0.041
b.32.2
6.03
32.2
10
6.0
3
)2.0(
2.0
4.0
6.01
6.0
)35()(
1
35 



















 DC
e
e
k ucm/s0.171

Loading page 8...

6
2.123
dry(sand) kN/m16.84
55.01
)81.9)(66.2(
1 




 e
G ws  3
sat(sand) kN/m81.02
48.01
)48.066.2(81.9
1 





 e
eG wws3
sat(clay ) kN/m18.55
)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):mm87.2











 110
155
log
82.01
mm)10007.3)(288.0(
log
1 o
o
o
cc
c e
H

Loading page 9...

7
2.15 a. Eq. (2.53):0.392
150
300
log
792.091.0
log
1
2
21 

















 ee
Cc
From Eq. (2.65):o
o
o
cc
c e
HC
S 


 log
1
Using the results of Problem 2.12,2
kN/m87.88)81.955.18(
2
5
)81.981.20(5.1)84.16)(3( o953.0)74.2)(3478.0(  so wGemm194.54




 

 87.88
5087.88
log
953.01
mm)5000)(392.0(
cS
b. Eq. (2.73):2
H
tC
T v
v  . For U = 50%, Tv = 0.197 (Table 2.11). So,days609sec105262;
cm)500(
1036.9
197.0 4
2
4




t
t
2.16 a. Eq. (2.53):0.377


















120
360
log
64.082.0
log
1
2
21 ee
Cc
b.0.736

















 2
2
1
2
21 ;
120
200
log
82.0
0.377;
log
e
eee
Cc

Loading page 10...

8/mincm5057.0;
)cm8.3(
6
49
286.0 2
2 






 v
v
C
C
Field: U = 50%; Tv = 0.197days6.29min9060;
cm
2
10005.3
)5057.0(
197.0 2 





 
 t
t
2.185.0
60
30 Ut
t
H
tC
T v
v
6
22
1
)1(
)1( 102
2
10002
))(2( 






 
t
t
H
tC
T v
v
6
22
2
)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 U2U
HH
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  10-6t; t = 56,250 min = 39.06 days

Loading page 11...

90415.0
)1000(
)60602460)(108( 2
3




cT
After 30 days:0207.0
)1000(
)60602430)(108( 2
3
2 



H
tC
T 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
tC
T 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
S1
tan
Normal force, N (N) Shear force, S (N)




 
N
S1
tan (deg)
222.4
489.3
667.2
193.5
424.8
587.1
41.02
40.96
41.35
From the graph,   41
2.21 Normally consolidated clay; c = 0.




 
 2
45tan2
31
;38




 
 ;
2
45tan2074.662207 2

Loading page 12...

10
2.23 c = 0. Eq. (2.91):2
kN/m387.8




 




 
 2
28
45tan140
2
45tan 22
31
2.24 Eq. (2.91):




 





 
 2
45tan2
2
4531 c




 





 
 2
45tan2
2
45tan140368 2 c
(a)




 





 
 2
45tan2
2
45tan280701 2 c
(b)
Solving Eqs. (a) and (b),  = 24; c = 12 kN/m2
2.2525















 
7.898.220
7.898.220
sinsin 1
31
311







 
31
311
sin

Loading page 13...

Loading page 14...

Loading page 15...

13
Chapter 3
3.1 Eq. (3.3):8.96%



 100
73
732.76
100(%) 2
22
2
22
i
io
R D
DD
A
3.2
Depth
from
ground
surface
(m) N60
cu (kN/m2)
[Eq. (3.8)]o (MN/m2) OCR
[Eq. 3.9)]
3.0 5 92.403854.0
)]81.919)(5.1()5.16)(5.1[(
1000
1

 5.51
4.5 8 129.60490.0
)81.98.16)(5.1(
1000
1
03854.0

 6.46
6.0 8 92.40595.0
)81.98.16)(5.1(
1000
1
0490.0

 5.65
7.5 9 141.107.0
)81.98.16)(5.1(
1000
1
0595.0

 5.48
9.0 10 152.20805.0
)81.98.16)(5.1(
1000
1
07.0

 5.35
Average cu = 121.5 kN/m2 Average OCR = 5.69

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Civil Engineering

Principles Of Foundation Engineering, 8th Edition Solution Manual

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