Civil Engineering Reference
In-Depth Information
slope fails. A mechanism could be a straight slip surface at a depth
z
as shown,
and the forces on the block with length
L
down the surface are marked on the
diagram. If the slope is very long,
F
1
and
F
2
are equal and opposite. The normal
and shear forces on the slip surface are
T
=
τ
L
and
N
=
σ
L
and the weight is
W
Lz
cos
i
c
. Figure 20.1(b) is a polygon of these forces which closes (i.e. the forces
are in equilibrium) when
=
γ
T
N
=
τ
σ
=
tan
i
c
(20.3)
Hence, from Eq. (20.2), the limit equilibrium solution is
=
φ
i
c
(20.4)
Strictly we should consider other possible mechanisms with combinations of curved
and straight slip surfaces, but it is fairly obvious that the mechanism illustrated in
Fig. 20.1 is one of the most critical. The solution
i
c
=
φ
can also be obtained as an
upper bound and as a lower bound so it is an exact solution.
Figure 20.2(a) shows a section of a foundation with width
B
and unit length out of
the page so that the width
B
is equal to the foundation area
A
. The foundation is loaded
undrained and the undrained strength of the soil is
s
u
. The problem is to determine the
collapse load
V
c
or the ultimate bearing capacity
q
c
=
V
c
/
A
. A mechanism could be a
circular slip surface with centre O at the edge of the foundation. The rotating block of
soil is in equilibrium when the moments about O balance and
s
u
B
ST
1
2
B
V
c
×
=
(20.5)
where ST
B
is the length of the arc ST. Notice that the lines of action of the weight
W
of the soil block and the normal stresses on the circular slip surfaces act through O
and so their moments about O are zero. From Eq. (20.5) we have
=
π
V
c
=
2
π
Bs
u
(20.6)
As before we should now consider other possible mechanisms with combinations of
straight and curved slip surfaces to seek the minimum value of
F
c
, which will be the
limit equilibrium solution. Figure 20.3 shows a circular slip surface with its centre at a
height
h
above the ground surface. Readers should show that the minimum value for
this mechanism is
V
c
0.58; one way to do this is to take trial
values of
h
and plot
V
c
against
h
to determine the minimum value of
V
c
.
Remember that in Chapter 19 we obtained equal upper and lower bound solutions
(i.e. an exact solution) for a foundation on undrained soil as
V
c
=
5.5
Bs
u
when
h
/
B
=
)
Bs
u
(see
Eq. 19.46) and so, in this case, the best limit equilibrium solution with a circular arc
slip surface overestimates the true solution by less than 10%.
=
(2
+
π
