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107
Chapter 11
11.1 Gs = 2.74; w = 9.81 kN/m3
From Eq. (11.5):
LL0274.01
88.26
100
LL
)74.2(1
)81.9)(74.2(
))(LL(1
s
ws
dG
G
11.2
c
= 84 kN/m2;
o
=62 kN/m2
Chapter 11
11.3 LL = 60; w = 25%; Z = 10 ft. From Figure 11.13, sw(free) 2%.
in. 1.73
ft 1445.0
)2)(75.475.2(
2
1
)2)(75.25.1(
2
1
)2)(5.16.0(
2
1
)2)(06.0(
2
1
100
1
4.11 S
11.5 Refer to the figure in Problem 11.4.
Depth (ft)
Total swell, S (in.)
12
0
10
072.0)12(
100
1
)2)(06.0(
2
1
0
8
324.0)12(
100
1
)2)(6.05.1(
2
1
072.0
6
834.0)12(
100
1
)2)(75.25.1(
2
1
324.0
4
734.1)12(
100
1
)2)(75.475.2(
2
1
834.0
Chapter 11
© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
109
The plot of S versus depth is shown below. From the figure, the depth of
undercut is: 5.40 – 4 = 1.4 ft below the bottom of the foundation
mm 63.5 m 0635.0
)5.0)(5.51.3(
2
1
)1)(1.35.1(
2
1
)1)(5.175.0(
2
1
)1)(75.04.0(
2
1
)1)(04.0(
2
1
100
1
6.11
S
Chapter 11
110
11.7
Depth (m)
Total swell, S (m)
6
0
5
002.0
100
1
)1)(04.0(
2
1
0
4
00775.0
100
1
)1)(4.075.0(
2
1
002.0
3
019.0
100
1
)1)(5.175.0(
2
1
00775.0
2
042.0
100
1
)1)(1.35.1(
2
1
019.0
1.5
0635.0
100
1
)5.0)(5.51.3(
2
1
042.0
The plot of S versus depth is shown. From the figure, the depth of undercut is:
2.5 – 1.5 = 1 m below the bottom of the foundation.
11.8 Eq. (11.14):
psswsZDU
tan
Chapter 11
© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
111
Eq. (11.17):
)(
4FS
22 sb
cu DD
Nc
U
)1(1.241)1(
43
)14.6)(150(
6.6174 222
bb DD
Or,
2
161.25 b
D
Db = 5.16 m
11.9 Eq. (11.17):
DU
DDNc sbcu
)(
4
)(
FS
22
Chapter 11
112
