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Craig's Soil Mechanics 7th Edition

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the overall pore water pressure (u) is
equal to the sum of the static and excess components, i.e.

u ¼ us þ ue ð3:4Þ

The reduction of excess pore water pressure as drainage takes place is described as
dissipation and when this has been completed (i.e. when ue ¼ 0) the soil is said to be in
the drained condition. Prior to dissipation, with the excess pore water pressure at its
initial value, the soil is said to be in the undrained condition. It should be noted that the
term ‘drained’ does not mean that all water has flowed out of the soil pores: it means
that there is no stress-induced pressure in the pore water. The soil remains fully
saturated throughout the process of dissipation.
As drainage of pore water takes place the solid particles become free to take up new

positions with a resulting increase in the interparticle forces. In other words, as the
excess pore water pressure dissipates, the effective vertical stress increases, accompan-
ied by a corresponding reduction in volume. When dissipation of excess pore water

74 Effective stress

pressure is complete the increment of total vertical stress will be carried entirely by the
soil skeleton. The time taken for drainage to be completed depends on the permeability
of the soil. In soils of low permeability, drainage will be slow, whereas in soils of high
permeability, drainage will be rapid. The whole process is referred to as consolidation.
With deformation taking place in one direction only, consolidation is described as
one-dimensional.
When a soil is subject to a reduction in total normal stress the scope for volume

increase is limited because particle rearrangement due to total stress increase is largely
irreversible. As a result of increase in the interparticle forces there will be small elastic
strains (normally ignored) in the solid particles, especially around the contact areas,
and if clay mineral particles are present in the soil they may experience bending. In
recoverable compression due to increases in interparticle forces, especially if there is
face-to-face orientation of the particles. When a decrease in total normal stress takes
place in a soil there will thus be a tendency for the soil skeleton to expand to a limited
extent, especially in soils containing an appreciable proportion of clay mineral particles.
As a result the pore water pressure will be reduced and the excess pore water pressure
will be negative. The pore water pressure will gradually increase to the static value,
flow taking place into the soil, accompanied by a corresponding reduction in effective
normal stress and increase in volume. This process, the reverse of consolidation, is
known as swelling.
Under seepage (as opposed to static) conditions, the excess pore water pressure is

the value above or below the steady seepage pore water pressure (uss), which is
determined, at the point in question, from the appropriate flow net.

Consolidation analogy

The mechanics of the one-dimensional consolidation process can be represented by
means of a simple analogy. Figure 3.2(a) shows a spring inside a cylinder filled with
water and a piston, fitted with a valve, on top of the spring. It is assumed that there can
be no leakage between the piston and the cylinder and no friction. The spring

Figure 3.2 Consolidation analogy.

Response of effective stress to a change in total stress 75

represents the compressible soil skeleton, the water in the cylinder the pore water and
the bore diameter of the valve the permeability of the soil. The cylinder itself simulates
the condition of no lateral strain in the soil.
Suppose a load is now placed on the piston with the valve closed, as in Figure 3.2(b).

Assuming water to be incompressible, the piston will not move as long as the valve is
closed, with the result that no load can be transmitted to the spring; the load will be
carried by the water, the increase in pressure in the water being equal to the load
divided by the piston area. This situation with the valve closed corresponds to the
undrained condition in the soil.
If the valve is now opened, water will be forced out through the valve at a rate

governed by the bore diameter. This will allow the piston to move and the spring to be
compressed as load is gradually transferred to it. This situation is shown in Figure
3.2(c). At any time the increase in load on the spring will correspond to the reduction
in pressure in the water. Eventually, as shown in Figure 3.2(d), all the load will be
carried by the spring and the piston will come to rest, this corresponding to the drained
condition in the soil. At any time, the load carried by the spring represents the effective
normal stress in the soil, the pressure of the water in the cylinder the pore water
pressure and the load on the piston the total normal stress. The movement of the
piston represents the change in volume of the soil and is governed by the compress-
ibility of the spring (the equivalent of the compressibility of the soil skeleton). The
piston and spring analogy represents only an element of soil since the stress conditions
vary from point to point throughout a soil mass.

Example 3.1

A layer of saturated clay 4m thick is overlain by sand 5m deep, the water table being
3m below the surface. The saturated unit weights of the clay and sand are 19 and
20 kN/m3, respectively; above the water table the unit weight of the sand is 17 kN/m3.
Plot the values of total vertical stress and effective vertical stress against depth. If sand
to a height of 1m above the water table is saturated with capillary water, how are the
above stresses affected?
The total vertical stress is the weight of all material (solids þ water) per unit area

above the depth in question. Pore water pressure is the hydrostatic pressure corres-
ponding to the depth below the water table. The effective vertical stress is the differ-
ence between the total vertical stress and the pore water pressure at the same depth.
Alternatively, effective vertical stress may be calculated directly using the buoyant unit
weight of the soil below the water table. The stresses need to be calculated only at depths
where there is a change in unit weight (Table 3.1).

Table 3.1

Depth (m) �v (kN/m
2) u (kN/m2) �0v ¼ �v � u (kN/m2)

3 3� 17 ¼ 51:0 0 51.0
5 (3� 17)þ (2� 20) ¼ 91:0 2� 9:8 ¼ 19:6 71.4
9 (3� 17)þ (2� 20)þ (4�19) ¼ 167:0 6� 9:8 ¼ 58:8 108.2

76 Effective stress

The alternative calculation of � 0v at depths of 5 and 9m is as follows:

Buoyant unit weight of sand ¼ 20� 9:8 ¼ 10:2 kN=m3
Buoyant unit weight of clay ¼ 19� 9:8 ¼ 9:2 kN=m3
At 5m depth: �0v ¼ ð3� 17Þ þ ð2� 10:2Þ ¼ 71:4 kN=m2
At 9m depth: �0v ¼ ð3� 17Þ þ ð2� 10:2Þ þ ð4� 9:2Þ ¼ 108:2 kN=m2

The alternative method is recommended when only the effective stress is required. In
all cases the stresses would normally be rounded off to the nearest whole number. The
stresses are plotted against depth in Figure 3.3.

Effect of capillary rise

The water table is the level at which pore water pressure is atmospheric (i.e. u ¼ 0).
Above the water table, water is held under negative pressure and, even if the soil is
saturated above the water table, does not contribute to hydrostatic pressure below the
water table. The only effect of the 1m capillary rise, therefore, is to increase the total unit
weight of the sand between 2 and 3m depth from 17 to 20kN/m3, an increase of 3 kN/m3.
Both total and effective vertical stresses below 3m depth are therefore increased by
the constant amount 3� 1 ¼ 3:0 kN/m2, pore water pressures being unchanged.

Example 3.2

A 5m depth of sand overlies a 6m layer of clay, the water table being at the
surface; the permeability of the clay is very low. The saturated unit weight of the

Figure 3.3 Example 3.1.

Response of effective stress to a change in total stress