Civil Engineering Reference
In-Depth Information
Table 19.2
Work done by internal stresses for mechanism in Figure 19.19
Slip plane
Shear stress
Length
Displacement
δ
W
=
s
u
L
δ
w
√
2
w
f
1
oa
s
u
√
2
B
s
u
B
δ
w
f
√
2
w
f
1
√
2
ob
s
u
s
u
B
δ
w
f
B
Fan
s
u
-
-
π
s
u
B
δ
w
f
fa
B
δ
w
f
0
Total
(
2
+
π
)
s
u
B
δ
w
f
Equating
δ
E
and
δ
W
, the upper bound for the collapse load is
V
u
=
(2
+
π
)
Bs
u
(19.43)
(b) Lower bound with stress fans
Figure 19.20(a) shows a state of stress with two stress fans in regions II and IV. As
before, the state of stress is symmetric about the centre-line and Eqs. (19.25) and
(19.26) apply in regions I and III respectively. Figure 19.20(b) shows Mohr circles of
total stress for elements at A and C and the points a and c represent the stresses on the
outermost discontinuities in the fan in region II. From the geometry of Fig. 19.20, the
fan angle is
90
◦
=
π
θ
f
=
/2 and from Eq. (19.36) the change of stress through the fan is
=
θ
f
=
π
s
u
(19.44)
s
s
u
From the geometry of Fig. 19.20(b),
q
l
+
γ
z
=
(2
+
π
)
s
u
+
γ
z
(19.45)
and hence a lower bound for the collapse load is
V
l
=
(2
+
π
)
Bs
u
(19.46)
Strictly, we should examine the state of stress in region VI where the stress fans overlap.
It is intuitively fairly clear that the stresses in region VI will be less critical than those
near the edges of the foundation and that the conditions in the overlapping stress fans
will tend to cancel each other out.
Notice that the upper and lower bounds given by Eqs. (19.43) and (19.46) are equal
and so they must be an exact solution. We have been very fortunate to obtain an exact
solution with such simple upper and lower bound solutions; normally you would only
be able to obtain unequal bounds.
