Civil Engineering Reference
In-Depth Information
q
=
50 kPa
P
q
6 m
P
a
h
2
h
3
58.8
24.5
(a) The problem
(b) Pressure distribution (kPa)
Fig. 7.22
Example
7.9
.
Since soil surface behind wall is horizontal,
p
=
K q
=
0 49 50
.
× =
24 5
.
kPa
q
a
The pressure diagram is now plotted (Fig.
7.22b
).
1
2
+
Total thrust Area of pressure diagram P
=
= + = ×
P
58 8 6
.
×
(
24 5 6
.
×
)
a
q
. . kN
The point of application of this thrust is obtained by taking moments of forces about
the base of the wall, i.e.
=
176 4 147
+
=
323 4
6
3
323 4
.
× =
h
147 3 176 4
× +
.
×
793 8
323 4
.
.
⇒ =
h
=
2 45
.
m
Resultant thrust acts at 2.45 m above base of wall.
Example 7.10:
Uniform surcharge (ii)
A vertical retaining wall is 5 m high and supports a soil, the surface of which is horizontal
and level with the top of the wall and carrying a uniform surcharge of 75 kPa.
The properties of the soil are:
φ
=
20°; c
′
=
10 kPa;
γ
=
20 kN/m
3
;
δ
=
φ
′
/2.
Determine the value of the maximum horizontal thrust on the back of the wall:
(a) by the Culmann line construction;
Solution:
c
′ <
50
kPa
⇒ = ′ =
c
c
10
kPa
w
′
2
c
h
c
=
tan(
45
°+ ′ =
φ
/
2
)
1 43
.
m
γ
