15
Chapter 3
e
+
γe
γG
γeG
wwsws
)(
eSγ
w
)/(
γwSe
γG
wws
3.2 a.
3
kN/m 15.6=
=
=)14.01(
8.17
γ
γ
69
.
0
e
1
ss
W
V
W
w
w
γ
3
2
)58.4)(4)(4/( ==
π
17
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
d.
3
sat
lb/ft 8.130
568
.
0
1
)4.62)(568.072.2(
1
)( =
+
+
=
+
+
=
e
γeG
γ
ws
Water to be added =
lb 0.316=
−=− )033.0(
033.0
4
8.130)(
sat
Vγγ
3.5 a.
3
lb/ft 100.9
11
=
=wV
W
γ
3.6 a.
09.1)73.2)(4.0(
=
=
=
s
wGe
3.7 a.
12.5
0.1
W
γ
V
= = =
3
125 lb / ft
18
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
c.
0.54=−=−= 1
64.109
)4.62)(71.2(
1
d
ws
γ
γG
e
d.
0.35=
+
=
+
=
54
.
0
1
54.0
1
e
e
n
e.
70.2%==== 702.0
54
.
0
)71.2)(14.0())((
e
Gw
S
s
f. Volume of water =
3
ft 0.024≈
−
=
−
4.62
)1.0)(64.109125(
γ
)(
w
d
Vγγ
3.8 a.
0.8=−=−= 1
8.14
)81.9)(72.2(
1
d
ws
γ
γG
e
+
+
8
.
0
1
1
sat
e
3.9 a.
3
kN/m 75.18
8
.
0
1
)81.9)](8.0)(9.0()81.9)(72.2[(
1
)( =
+
+
=
+
+
=
e
γSeG
γ
ws
3.10 a.
3
kN/m 18.82=
+
+
=
+
+
=)81.9)(68.2(
55.01
11.01
1
1
ws
γG
e
w
γ
+
+
55
.
0
1
1
e
d
19
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
c.
53.6%==== 536.0
55
.
0
)68.2)(11.0())((
e
Gw
S
s
d.
20.5%==== 205.0
68.2
55.0
sat
s
G
e
w
3.11 a.
3
3
2
ft 015.0
12
1
)18()375.1(
4=
=
π
V
3
85.1
W
71
.
0
e
3.12 a.
3
lb/ft 125.89=
+
+
=
+
+
=
71
.
0
1
)4.62)(71.074.2(
1
)(
sat
e
γeG
γ
ws
20
3.13 a.
S
γGw
γGw
γ
wsws
+
=
+
=
)1(
)1( ;
72.0
)4.62)()(26.01(
108
s
G
+
=;
2.72
=
G
3.14 a.
3
kN/m 17.67=
+
=
+
=
166
.
0
1
6.20
1
w
γ
γ
d
52
.
0
e
3.15 a.
3
kN/m 7.20
52
.
0
1
)81.9)(52.0)(9.0()81.9)(74.2(
1
)( =
+
+
=
+
+
=
e
γSeG
γ
ws
3.16 a.
3
kg/m 1639.8=
+
=
+
=
18
.
0
1
1935
1
w
d
ρ
ρ
8.1639
d
ρ
+
+
0.65
1
1
e
21
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
c.
74.7%==== 747.0
65
.
0
)7.2)(18.0(
e
wG
S
s
d.
3
sat kg/m 2030
65
.
0
1
)1000)(65.07.2(
1
)( ≈
+
+
=
+
+
=
e
eG
ws
ρ
ρ
Water to be added = 2030 – 1935 =
95 kg/m
3
3.17 a.
3
lb/ft 112
=
+
=
+
=)098.01(25.0
75.30
1w
γ
γ
d
c. Volume of water =
ft 0.044
≈
=
4.62
w
d
γ
3.18 a.
0.56=
−
=
−
=
36
.
0
1
36.0
1
n
n
e
3.19
6701
65.15
)81.9)(67.2(
1.
γ
γG
e
d
ws
≈−=−=
67
.
0
e
3.20 a.
3
lb/ft 124.7=
+
+
=
+
+
=
+
+
=
72.0
)71.2)(148.0(
1
)4.62)(71.2)(148.01(
1
)1(
1
)1(
S
wG
γGw
e
γGw
γ
s
wsws
22
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Volume of water =
soil of /ftft 0.258
3
=
−
=
−
4.62
)1)(6.1087.124()(
w
d
γ
Vγγ
3.21 a.
e
γSeG
γ
ws
+
+
=
1
)(
;
e
eG
s
+
+
=
1
)4.62)(55.0(
106
3.22 For the borrow pit,
ws
wsws
γG
γG
e
γG
γ476.0
1.111 borrow
borrowd =
+
=
+
=
−
wsws
γG
γG
3.23 a.
3
kN/m 14=
+
+
=
+
+
=1.11
)81.9)(7.2)(11.01(
1
)1(
borrow
borrow e
γGw
γ
ws
3.24 a.
3
d borrow
111
90.24 lb/ft
1 1 0.23
γ
γ
w
−
= = =
+ +
23
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
3
d embankment
118 lb/ft
γ
−
=
Volume of the first lift:
[
]
/ftft 93)3)(6)(5.0(2)3)(25(
3
1
=+=V
Volume of borrow soils =
( )
3
embankment
1
borrow
118
(93) 121.6 ft /ft
90.24
d
d
γV
γ
−
−
= =
Weight of borrow soils = (121.6)(111) = 13,498 lb/ft
b.
0.8461
24.90
)4.62)(67.2(
1
borrow
≈−=−=
d
ws
γ
γG
e
3
sat lb/ft 85.118
846
.
0
1
)4.62)(846.067.2(
1
)( =
+
+
=
+
+
=
e
γeG
γ
ws
Weight of borrow soils = (121.6)(118.85) = lb 14,452
3.25
minmax
max
ee
ee
D
r
−
−
=
;
52
.
0
75
.
0
75.0
65.0 −
−
=e
;
0.6
=
e
+
+
6
.
0
1
1
e
d
3.26
minmax
max
ee
ee
D
r
−
−
=
;
41
.
0
77
.
0
77.0
90.0 −
−
=e
; e ≈ 0.45
3.27
3
kN/m 76.15
11
.
0
1
5.17
1
=
+
=
+
=
w
γ
γ
d
24
3.28
e
γGw
γ
ws
+
+
=
1
)1(
;
e
+
+
=
1
)4.62)(68.2)(16.01(
122
;
59.0
=
e
−
−
−
−
42.085.0
59.085.0
minmax
max
ee
ee
r
3.29
minmax
max
ee
ee
D
r
−
−
=
;
42
.
0
85
.
0
85.0
88.0 −
−
=e
;
47.0
=
e
3
+
+
47
.
0
1
)4.62)(68.2(
1
e
γG
ws
d
CRITICAL THINKING PROBLEMS
3.C.1 a.
s
v
V
V
e=
;
s
V
V
e
1
1
1=+
92.1
1
V
V
b.
92.192.011
1
)1(
wswsws
d
γGγG
e
γG
γ=
+
=
+
=
;
65
.
1
)2(
ws
d
γG
γ=
(2) (1)
(1) (1)
1 1
∆1.65 1.92 0.163
1
1.92
d d
d
d d
γ γ
γ
γ γ
−
−
c.
92.0
1
1
ss
wG
e
wG
S==
;
65
.
0
2
s
wG
S=
2 1
1 1
1 1
0.65 0.92 0.415
1
0.92
S S
S
S S
−
−
∆= = = =
25
3.C.2 a.
minmax
max
ee
ee
D
r
−
−
=
+
702
.
0
1
d
b.
047.0
786.01
702.0786.0
1
∆∆
1
=
+
−
=
+
=e
e
H
H
3.C.3 a. The graph is shown.
26
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
The maximum dry unit weight tends to increase with increasing finer
fractions until an optimum level (maximum value) is reached. For the
binary soil mixes considered in this study, the maximum dry unit weight
occurs when the fine soil content is about 40% by volume.
Median grain size ratios may be used as a criterion for mixing and
compacting granular materials in field applications.