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Chapter 2
3
)81.9)(5.87(
76.0
e
V
v
2.2 a. From Eqs. (2.11) and (2.12), it can be seen that,
Chapter 2
wG ws ;
)12.01)(4.62)(65.2(
)1(
Chapter 2
2.5 From Eqs. (2.11) and (2.12):
3
lb/ft 108
08.01
64.116
d
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
4
4
2.7 Soil A: Table 2.8: 65% passing No. 200 sieve.
Fine grained soil; LL = 42; PI = 16
Chapter 2
2.8
43.01
117
)4.62)(68.2(
1e ;
1
d
wsws
d
G
e
G
Chapter 2
2.10 The flow net is shown.
8
8
2.15 a. Eq. (2.53):
0.392
150
300
log
792.091.0
log
1
2
21
ee
Cc
2.16 a. Eq. (2.53):
0.377
120
360
log
64.082.0
log
1
2
21 ee
Cc
9
2.17 Eq. (2.73):
2
H
tC
Tv
v
. For 60% consolidation, Tv = 0.286 (Table 2.11).
2.18
5.0
60
30 U
Chapter 2
10
10
2.19 Eq. (2.84):
.
2
H
tC
Tcv
c
tc = 60 days = 60 24 60 60 sec;
2
2
H
m = 1000 mm.
Sc = (0.23)(120) = 27.6 mm
2.20
N
S
1
tan
2.21 Normally consolidated clay; c = 0.
Chapter 2
11
2.22
2
45tan2
31
;
30
;
2
45tan204020 2
2.26
2
45tan2
31
.
22
1kN/m 305.9
2
20
45tan150
Chapter 2
12
12
2
u
u ;
28
9.305
1
2.27 a.
)log(6.14.01026 50
DCD ur
