Civil Engineering Reference
In-Depth Information
A limit equilibrium solution for the active force on a wall retaining dry soil was
found in Sec. 20.3. The mechanism and the polygon of forces were shown in Fig. 20.6
and the solution is
H
2
tan
2
45
◦
−
φ
1
2
1
2
P
a
=
γ
(24.2)
φ
is the appropriate friction angle discussed in Sec. 24.10.
Assuming that the effective active pressure
where
σ
a
increases linearly with depth the earth
pressures corresponding to this limit equilibrium solution are
σ
a
=
σ
v
tan
2
45
◦
−
2
φ
1
K
a
σ
v
=
(24.3)
σ
v
is the vertical effective stress and
K
a
is called the active earth pressure
coefficient. It is quite easy to show that the solution for the passive pressure is
where
σ
b
=
σ
v
tan
2
45
◦
+
2
φ
1
K
p
σ
v
=
(24.4)
where
K
p
is called the passive earth pressure coefficient.
These solutions are for a smooth vertical wall with a level ground surface. A more
general case is shown in Fig. 24.9 where the ground surface and the back of the wall
are both inclined and the wall is rough. Shear stresses between the soil and the wall
are given by
τ
s
=
σ
n
tan
δ
(24.5)
σ
n
is the normal stress for the appropriate active or passive pressure and
δ
where
is
<δ
<φ
the critical angle of wall friction. Obviously 0
and a value commonly taken
2
3
φ
. The general case was considered in Sec. 20.3 (see Fig. 20.7).
Tables and charts are available giving values for
K
a
and
K
p
for various combinations
of
δ
for design is
=
φ
,
δ
,
α
and
β
.
Figure 24.9
Earth pressures on a rough wall with a sloping face and with sloping ground.
