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113
Chapter 12
12.1 a. c′ = 0. From Eq. (12.3),
φ
τ
′
′
=
tanσ
f
12.2
( ) ( )
48.170
1000071.0
4
675
2
=
=
′
π
σ
kN/m
2
≈ 170.5 kN/m
2
114
© 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.
determines the center, C, of the Mohr’s circle. With the center at C and the radius
as OC, the Mohr’s circle is drawn by trial and error such that the circle is tangent
to the failure envelope at O. From the graph:
a.
2
kN/m 122≈
′
3
σ
;
2
kN/m 284
≈
′
1
σ
b. The horizontal line OP drawn from O determines the pole, P. Therefore, the
orientation of the major principal plane with the horizontal is given by the
angle, α ≈ 58°
°°
°.
12.3 For
σ
′ = 22 lb/in.
2
,
τ
f
= 22
tan
41° = 19.12 lb/in.
2
2
12.4 a. Area of specimen
2
2
m 00316.0
1000
5.63
4=
=
π
A
Test
No.
Normal
force, N
(N)
A
N
σ=
′
(kN/m
2
)
Shear
force, S
(N)
A
S
f
=
τ
(kN/m
2
)
′
=
′
−
σ
f
τ
φ
1
tan
(deg)
b. For the silty sand:
34.15
72
.
0
1
)81.9)(69.2(
1
=
+
=
+
=
e
γG
γ
ws
d
kN/m
3
f
115
12.5 a. Area of specimen
22
m 00316.0)0635.0(
4=
=
π
A
Test
No.
Normal
force, N
(N)
A
N
=
′
σ
(kN/m
2
)
Shear
force, S
(N)
A
S
f
=
τ
(kN/m
2
)
′
=
′
−
σ
τ
φ
f
1
tan
(deg)
b.
88.51)81.945.19)(2.2()34.15)(2(
=
−
+
=
′
σ
kN/m
2
kN/m 27=°= 5.27tan)88.51(
f
τ
12.6 c′ = 0. From Eq. (12.8):
°=
′
′
+
′
=
′2.19 ;
2
45tan
2
31
φ
φ
σσ
(failure) 1 3
d
12.7
°=
′
′
+
′
=
′2.19 ;
2
45tan
2
31
φ
φ
σσ
fd
12.8 a. From Eq. (12.4): °=+=
′
+= 54.6
2
2.19
45
2
45
φ
θ
116
© 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.
5.51
2
105208
Radius =
−
== OC
τ
α
= (CD) sin 60º = 51.5 sin 60º =
44.6 kN/m
2
2
kN/m 182.25=°++=
′60cos5.515.51105
α
σ
12.9
φ
′ = 28 + 0.18D
r
= 28 + (0.18)(88) = 43.84°
2
+=
′
84.43
2
22
31
φ
12.10
2 2
3 1 3 ( )
145 kN/m ; 145 198 343 kN/m
d f
σ σ σ σ
′ ′ ′
= = + ∆ = + =
12.11
3
2
3
2
3)failure(3
282.2
2
23
45tan
2
45tan σσσσσ
d
′
=
+
′
=
′
+
′
=∆+
′
φ
27
2
117
12.12
( )
2 2
1 3 3 ( )
23
tan 45 tan 45
2 2
d f
u
ϕ
σ σ σ
′
′ ′
= + = − ∆ +
12.13
44.7)2.410(24.13)(
31
=
−
−
=
′
−
′
=
∆
σ
σ
σ
fd
lb/in.2
∆
)(
fd
f
σ
12.14 a. The effective principal stresses at failure are calculated as follows and the
Mohr-Coulomb failure envelope is drawn from the Mohr’s circles.
Test
No.
σ
3
(kN/m2)
(∆σ
d
)
f
(kN/m2)
(∆u
d
)
f
(kN/m2)
fd
uσσ )(
33
∆
−
=
′
(kN/m2)
fd
σσσ )(
31
∆
+
′
=
′
(kN/m2)
b. Effective stress in the middle of the clay layer:
118
12.15 a.
′
+
′
=
′2
45tan
2
31
φ
σσ ;
′
+=+ 2
45tan35)4235(
2
φ
2
2
f
12.16
′
+
′
+
′
+
′
=
′2
45tan2
2
45tan
2
31
φφ
σσ
c
′
′
2
φφ
+
2
09.26
45tan2
12.17 a. From Eqs. (10.8) and (10.9):
θ
= 33°
2
kN/m
179
=
119
© 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.
1 3
225 70
sin 2 sin(2 33)
2 2
f
σ σ
τ θ
′ ′
−
−
= = × =
2
70.8 kN / m
b. The angle of inclination of the failure plane
26.09
45 45 58.04
2 2
ϕ
θ
′
= + = + =
°
°°
°
12.18 °
°°
°25.03=
−
+
=
+= 45
15
2215
tan2 ;
2
45tan
5.0
1-2
31
φ
φ
σσ
+
915
2
31
12.19 a.
σ
3 = 225 kN/m2;
σ
1 = 225 + 112 = 337 kN/m2
b.
′
+
′
=
′2
45tan
2
31
φ
σσ
120
12.20
′
+
′
=
′2
45tan
2
31
φ
σσ
= 225 tan2
+2
21
45 = 476.33 kN/m2
31
fd
12.21
2 2 2
1 3
33
tan 45 14 tan 45 47 48 lb / in.
2 2 .
ϕ
σ σ
= + = + =
2
12.22
′
+
′
=
′
2
45tan
2
31
φ
σσ
12.23
192039
=
−
=
−
=
PLLLPI
o
121
c
u
(VST) = (0.1803)(133.52) ≈ 24 kN/m
2
b.
Eq. (12.62):
1.7 0.54log( ) 1.7 0.54log19 1.009
λ PI
= − = − =
2
12.24 Liquidity index, 421.0
20
39
2028
=
−
−
=
−
−
=
PL
LL
PLw
LI
2
1
1
CRITICAL THINKING PROBLEM
12.C.1 Task 1
1
1
At the end of consolidation, 0;
3
=
′
=
′
oo
qp
σ
σ3
(kN/m2)
o
p
′
(kN/m2)
ffd
qσ
=
∆
)(
(kN/m2)
f
p
′
(kN/m2)
122
The drained stress paths and the failure line are shown below.
Task 2
At point O:
p
′
= 0; q = 0
