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(
H
2
- z
0
2
)
- c
u
(
H - z
0
)((1
+ c
f
/c
u
)
tan
P
a
- P
w
=
½
+
cot
)
Putting
P
a
/
=
0 yields an optimum value of
. Elaboration yields
=
tan
-2
(1+
cf /cu
)
(tan
)/
=
-
(cot
)/
= -
(1/tan
)/
(tan
)/
=
(1
+ c
f
/c
u
)
-0.5
and so
tan
A
C
-
z
0
P
w
H
T
f
D
W
P
a
T
s
+
N
s
B
h
Figure 11.10
Substituting this value in the expression of
P
a
yields
(
H
2
- z
0
2
)
-
2
c
u
(
H - z
0
)(1+
c
f
/c
u
)
0.5
P
a
- P
w
=
½
Hence, the lateral active pressure at given depth
z
becomes (above
z = z
0
no
effect of wall friction)
h
=
z -
2
c
u
At the tip of the crack
z = z
0
the horizontal stress is zero, and therefore
z
0
=
2(
c
u
/
)
=
18.5 kN/m
3
, undrained strength
c
u
=
30 kPa, and adhesion
c
f
=
0.5
c
u
, one may find
For a wall height of
H =
8 m, soil weight of
(1+0.5)) = 39.2
o
=
atan(1/
z
0
= 2x30/18.5 = 3.24 m
P
w
=
½ 10 (3.24)
2
= 52.49 kN/m
P
a
=
½18.5
(8
2
-(3.24)
2
) - 2x30
(8-3.24)
(1+0.5) +
P
w
=
= 145.11 + 52.49 = 197.60 kN/m
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