Civil Engineering Reference
In-Depth Information
Table 6.1
Limiting suctions in saturated soil
Soil
Characteristic
grain size (mm)
Height of
saturated
zone (m)
Suction (kPa)
Clay
0.001
60
600
Medium silt
0.01
6
60
Fine sand
0.1
0.6
6
From Eq. (6.11) suctions in saturated soil vary with the inverse of the grain size. Taking
a value for
T
for water and quartz (i.e. glass) of about 7
10
−
5
kNm
−
1
and for soil
with a specific volume of 1.5 the variations of the height of the saturated zone and
the pore water suction with grain size are given in Table 6.1. This shows that even
saturated soils can develop considerable suctions.
×
6.5 Effective stress
It is obvious that ground movements and instabilities can be caused by changes of total
stress due to loading of foundations or excavation of slopes. What is perhaps not so
obvious is that ground movements and instabilities can be caused by changes of pore
pressure. For example, stable slopes can fail after rainstorms because the pore pressures
rise due to infiltration of rainwater into the slope while lowering of groundwater due to
water extraction causes ground settlements. (Some people will tell you that landslides
occur after rainfall because water lubricates soil; if they do, ask them to explain why
damp sand in a sandcastle is stronger than dry sand.)
If soil compression and strength can be changed by changes of total stress or by
changes of pore pressure there is a possibility that soil behaviour is governed by some
combination of
and
u
. This combination should be called the
effective
stress because
it is effective in determining soil behaviour.
The relationship between total stress, effective stress and pore pressure was first
discovered by Terzaghi (1936). He defined the effective stress in this way:
σ
All measurable effects of a change of stress, such as compression, distortion
and a change of shearing resistance, are due exclusively to changes of effective
stress. The effective stress
σ
is related to the total stress and pore pressure by
σ
=
σ
−
u
.
Figure 6.5 shows Mohr circles of total stress and effective stress plotted on the same
axes. Since
σ
1
σ
3
u
the diameters of the circles are the same.
The points T and E represent the total and effective stresses on the same plane and
clearly total and effective shear stresses are equal. Therefore, effective stresses are
=
σ
−
u
and
=
σ
−
1
3
σ
=
σ
−
u
(6.12)
τ
=
τ
(6.13)
