Civil Engineering Reference
In-Depth Information
in Fig. 20.5(b). The shear force on the wall
S
w
is given by
S
w
=
s
w
H
(20.10)
where
s
w
is the shear stress between the soil and the wall; obviously
s
w
must be in
the range 0
s
u
depending on the roughness of the wall. Free water in the
excavation applies a total force
P
w
to the wall, given by
≤
s
w
≤
1
2
w
H
w
P
w
=
γ
(20.11)
where
H
w
is the depth of water in the excavation. For the undrained case the pore
pressures in the soil do not come into the calculation and will not be in equilibrium
with the water pressures in the excavation. Again the only unknowns are the mag-
nitudes of the forces
N
and
P
a
, so the problem is statically determinate. The limit
equilibrium solution is the maximum value of
P
a
and coincides with the critical slip
surface.
The case shown in Fig. 20.6 is similar to that in Fig. 20.4 except that the soil is
drained and dry so pore pressures are zero. The forces on the triangular wedge are
shown in Fig. 20.6(a). There are now three unknown forces,
T
,
N
and
P
a
, but the
forces
T
and
N
are related by Eq. (20.2) so the resultant of
T
and
N
, shown by the
broken line, is at an angle
to the direction of
N
. (The primes are added to these
forces because they are associated with the effective stresses in the dry soil.) This now
provides sufficient information to construct the force polygon shown in Fig. 20.6(b) to
calculate the magnitude of
P
a
. To obtain the limit equilibrium solution you must vary
the angle
φ
1
2
45
◦
+
φ
, and the
α
to find the critical value for
P
a
. This occurs when
α
=
limit equilibrium solution is
H
2
tan
2
45
φ
1
2
1
2
P
a
=
γ
−
(20.12)
This solution was developed by Rankine in about 1850 (but in a different way) and is
really a case of the Coulomb wedge analysis.
Figure 20.6
Coulomb wedge analysis for a smooth wall for drained loading.
