Geoscience Reference
In-Depth Information
For
B =
1 and
D =
0, loading may cause an instantaneous volume constrained
shear deformation, and if conditions allow shear to develop, yielding an
instantaneous settlement
S
0
. If pore water pressure then dissipates by drainage, this
settlement will increase at attenuating rate, according to a consolidation process, as
shown in Fig 7.8, ending in a final settlement
S
&
. For example, at the edge of a
landfill, there will be an instantaneous settlement under the edge and heave beyond
the edge, to maintain volume. In the middle, far from the edge, one-dimensional
conditions apply, since (rotational) shear deformation is not possible there and
hence no immediate settlement occurs.
application 7.1
A slide of a long slope can be predicted by simple tests and observation. In the
field, an impression of a probable slip surface angle is obtained by observation:
f
35
o
. The vertical stress is approximated on the basis of measured unit soil weight
d
=
14.5 kN/m
3
. Porosity is 0.35. The groundwater table is 2 metre
under the ground level, more or less parallel to the slope. Thus,
u =
10(
z-
2
)
kPa. A
shear test in the laboratory shows:
=
18
and
=
32
o
and cohesion
c =
5
kPa.
A
A
20
20
20
DC
1
DC
1
10
10
10
v
v
B
B
c
c
v
'
v
'
f
f
10
10
10
20
20
20
30
30
30
40
40
40
50
50
50
u
u
T
s
T
s
z
z
z
f
f
f
Figure 7.9 Practical evaluation of a potential slope slide
Consider a possible slip plane at a depth of 3 m, assume that the slip plane could
occur at
35
o
and disregard cohesion at this stage. In fact, this is a direct way to
estimate the current stress state. The construction of the corresponding circle of
Mohr allows to evaluate the slope stability. The vertical total stress at 3 m depth is
f
v
=
47 kPa and the effective stress is
v
' =
37 kPa. The Direction Centre
DC
1
must lie on the vertical through
'
v
and point
T
s
must lie on the circle and the
assumed failure line under angle 35
o
. The line through
DC
1
and
T
s
is perpendicular
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