Civil Engineering Reference
In-Depth Information
For
P
a
7 .
:
P
=
0 49 24 4 5
.
× ×
.
+
23 5
.
=
76 4
.
kPa
a
7 5
.
The active pressure diagram is shown in Fig.
7.7
b and the value of the total active thrust
is simply the area of this diagram:
3
2
4 5
2
.
16
× +
23 5 4 5 52 9
.
×
.
+
.
×
=
248 8
. kN
Example 7.3:
Rankine active pressure; presence of groundwater
A vertical retaining wall 6 m high is supporting soil which is saturated and has a unit
weight of 22.5 kN/m
3
. The angle of shearing resistance of the soil,
φ
′
, is 35° and the
surface of the soil is horizontal and level with the top of the wall. A groundwater level
has been established within the soil and occurs at a level of 2 m from the top of the
wall.
Using the Rankine theory calculate the significant pressure values and draw the
diagram of pressure distribution that will occur on the back of the wall.
Solution:
Figure
7.8a
illustrates the problem and Figs
7.8b and 7.8c
show the pressure distribution
due to the soil and the water.
1
−
sin
sin
35
°
K
a
=
°
=
0 27
.
1
+
35
Although there is the same soil throughout, there is a change in unit weight at a depth
of 2 m as the unit weight of the soil below the GWL is equal to the submerged unit
weight. The problem can therefore be regarded as two layers of different soil, the upper
having a unit weight of 22.5 kN/m
3
and the lower (22.5 - 9.81)
=
12.7 kN/m
3
.
Consider the upper soil:
At depth
=
2
m p
:
=
K h
γ
=
0 27 22 5 2
.
×
.
× =
12 2
.
kPa
a
a
2
2 m
12.2
4 m
25.9
39.2
(a) The problem
(b) Earth pressure (kPa)
(c) Water pressure (kPa)
Fig. 7.8
Example
7.3.
