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Homework #1 Solution
1.
Given:
For Sand,
γ
m1
116.4pcf
:=
γ
sat1
121.4pcf
:=
K
1
0.35
:=
For Clay,
γ
m2
112.4pcf
:=
γ
sat2
117.4pcf
:=
K
2
0.50
:=
γ
w
62.4pcf
:=
σ
v
γ
w
8ft
()
⋅
γ
sat1
10ft
⋅
+
γ
sat2
8ft
⋅
+
2.652 ksf
⋅
=
:=
u
γ
w
26ft
⋅
1.622 ksf
⋅
=
:=
σ
'
v
σ
v
u
−
1.03 ksf
⋅
=
:=
2.
σ
'
v
γ
sat1
γ
w
−
10ft
⋅
γ
sat2
γ
w
−
8ft
⋅
+
1.03 ksf
⋅
=
:=
3
σ
v
γ
m1
10ft
⋅
γ
m2
2ft
⋅
+
γ
sat2
6ft
⋅
+
2.093 ksf
⋅
=
:=
u
γ
w
6ft
⋅
0.374 ksf
⋅
=
:=
σ
'
v
σ
v
u
−
1.719 ksf
⋅
=
:=
σ
'
h
K
2
σ
'
v
⋅
0.859 ksf
⋅
=
:=
σ
h
σ
'
h
u
+
1.234 ksf
⋅
=
:=
4. Given:
γ
sat
23
kN
m
3
:=
γ
w
9.81
kN
m
3
:=
K
0.5
:=
(a)
σ
v
γ
sat
7m
⋅
161 kPa
⋅
=
:=
u
γ
w
⋅
68.67 kPa
⋅
=
:=
σ
'
v
σ
v
u
−
92.33 kPa
⋅
=
:=
σ
'
h
K
σ
'
v
⋅
46.165 kPa
⋅
=
:=
σ
1
σ
'
v
92.33 kPa
⋅
=
:=
σ
3
σ
'
h
46.165 kPa
⋅
=
:=
(b)
τ
max
σ
1
σ
3
−
2
23.082 kPa
⋅
=
:=
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View Full Document5. Given:
R
6ft
:=
q
5000
lbf
ft
2
:=
Using P & B Chart:
aR6
f
t
⋅
=
:=
r0
:=
at
z
1.5ft
:=
z
a
0.25
=
r
a
0
=
I
0.99
:=
Δσ
v
qI
⋅
4.95 ksf
⋅
=
:=
at
z
3ft
:=
z
a
0.5
=
r
a
0
=
I
0.9
:=
Δσ
v
⋅
4.5 ksf
⋅
=
:=
at
z
6ft
:=
z
a
1
=
r
a
0
=
I
0.65
:=
Δσ
v
⋅
3.25 ksf
⋅
=
:=
at
z
9ft
:=
z
a
1.5
=
r
a
0
=
I
0.42
:=
Δσ
v
⋅
2.1 ksf
⋅
=
:=
at
z
12ft
:=
z
a
2
=
r
a
0
=
I
0.28
:=
Δσ
v
⋅
1.4 ksf
⋅
=
:=
Using Boussinesq formula:
B2
R
⋅
12 ft
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 Fall '08
 Staff

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