Civil Engineering Reference
In-Depth Information
Figure 22.8
Combined loading on a shallow foundation.
1
2
π
H
=
s
u
B
and it will continue to do so as
V
is increased. When
V
/
s
u
B
=
(1
+
)
the horizontal force to cause failure starts to decrease and when
V
/
s
u
B
) the
foundation cannot support any horizontal load. If the envelope in Fig. 22.8(b) is a
plastic potential the directions of movement
=
(2
+
π
δ
h
and
δ
v
are given by the directions of
the arrows which are normal to the envelope.
Figure 22.8(c) shows a failure envelope for a simple foundation for drained loading.
Most of the features in Fig. 22.8(b) for undrained loading are also in Fig. 22.8(c) for
drained loading. If
V
0 because the shearing resistance is frictional. For
small values of
V
failure by increasing
H
will cause dilation and the foundation will
heave and slide sideways. Because Figs. 22.8(b) and (c) have been normalized with
respect to the failure load
V
c
the upper parts correspond to states on the dry side of
critical (dense sand and overconsolidated clay) and the lower parts correspond to states
on the wet side (loose sand and normally consolidated clay).
Figures 22.8(b) and (c) illustrate what you should do if you are in a car, off road,
and you come to a hill. There will have to be shear stresses between the tyres and the
ground to get you up the hill so some horizontal load
H
must be mobilized. If you
are at point A on loose sand or normally consolidated clay the wheels are spinning
and the tyres are sinking into the ground. You should unload the car to reduce
V
and
then drive up the hill slowly to keep
H
small. If you are at point B in dense sand in
=
0 then
H
=
