Craig's Soil Mechanics 7th Edition
458 pág.

Craig's Soil Mechanics 7th Edition

Disciplina:Mecânica dos Solos 2652 materiais5.233 seguidores
Pré-visualização50 páginas
77

sand is 19 kN/m3 and that of the clay is 20 kN/m3. A 4m depth of fill material of
unit weight 20 kN/m3 is placed on the surface over an extensive area. Determine the
effective vertical stress at the centre of the clay layer (a) immediately after the fill
has been placed, assuming this to take place rapidly and (b) many years after the
fill has been placed.
The soil profile is shown in Figure 3.4. Since the fill covers an extensive area it can be

assumed that the condition of zero lateral strain applies. As the permeability of the
clay is very low, dissipation of excess pore water pressure will be very slow; immedi-
ately after the rapid placing of the fill, no appreciable dissipation will have taken place.
Therefore, the effective vertical stress at the centre of the clay layer immediately after
placing will be virtually unchanged from the original value, i.e.

�0v ¼ ð5� 9:2Þ þ ð3� 10:2Þ ¼ 76:6 kN=m2

(the buoyant unit weights of the sand and the clay, respectively, being 9.2 and
10.2 kN/m3).
Many years after the placing of the fill, dissipation of excess pore water pressure

should be essentially complete and the effective vertical stress at the centre of the clay
layer will be

�0v ¼ ð4� 20Þ þ ð5� 9:2Þ þ ð3� 10:2Þ ¼ 156:6 kN=m2

Immediately after the fill has been placed, the total vertical stress at the centre of the
clay increases by 80 kN/m2 due to the weight of the fill. Since the clay is saturated and
there is no lateral strain there will be a corresponding increase in pore water pressure
of 80 kN/m2 (the initial excess pore water pressure). The static pore water pressure is
(8� 9:8) ¼ 78:4 kN/m2. Immediately after placing, the pore water pressure increases
from 78.4 to 158.4 kN/m2 and then during subsequent consolidation gradually
decreases again to 78.4 kN/m2, accompanied by the gradual increase of effective
vertical stress from 76.6 to 156.6 kN/m2.

Figure 3.4 Example 3.2.

78 Effective stress

3.4 PARTIALLY SATURATED SOILS

In the case of partially saturated soils part of the void space is occupied by water and
part by air. The pore water pressure (uw) must always be less than the pore air pressure
(ua) due to surface tension. Unless the degree of saturation is close to unity the pore air
will form continuous channels through the soil and the pore water will be concentrated
in the regions around the interparticle contacts. The boundaries between pore water
and pore air will be in the form of menisci whose radii will depend on the size of the
pore spaces within the soil. Part of any wavy plane through the soil will therefore pass
through water and part through air.
In 1955 Bishop proposed the following effective stress equation for partially satur-

ated soils:

� ¼ �0 þ ua � �ðua � uwÞ ð3:5Þ

where � is a parameter, to be determined experimentally, related primarily to the
degree of saturation of the soil. The term (ua � uw) is a measure of the suction in the
soil. For a fully saturated soil (Sr ¼ 1), � ¼ 1; and for a completely dry soil
(Sr ¼ 0), � ¼ 0. Equation 3.5 thus degenerates to Equation 3.1 when Sr ¼ 1. The value
of � is also influenced, to a lesser extent, by the soil structure and the way the
particular degree of saturation was brought about. Equation 3.5 is not convenient
for use in practice because of the presence of the parameter �.
A physical model may be considered in which the parameter � is interpreted as the

average proportion of any cross-section which passes through water. Then, across
a given section of gross area A (Figure 3.5) total force is given by the equation

�A ¼ �0Aþ uw�Aþ uað1� �ÞA ð3:6Þ

which leads to Equation 3.5.
If the degree of saturation of the soil is close to unity it is likely that the pore air will

exist in the form of bubbles within the pore water and it is possible to draw a wavy
plane through pore water only. The soil can then be considered as a fully saturated soil
but with the pore water having some degree of compressibility due to the presence of

Figure 3.5 Partially saturated soil.

Partially saturated soils 79

the air bubbles; Equation 3.1 may then represent effective stress with sufficient
accuracy for most practical purposes.

3.5 INFLUENCE OF SEEPAGE ON EFFECTIVE STRESS

When water is seeping through the pores of a soil, total head is dissipated as viscous
friction producing a frictional drag, acting in the direction of flow, on the solid
particles. A transfer of energy thus takes place from the water to the solid particles
and the force corresponding to this energy transfer is called seepage force. Seepage
force acts on the particles of a soil in addition to gravitational force and the combin-
ation of the forces on a soil mass due to gravity and seeping water is called the resultant
body force. It is the resultant body force that governs the effective normal stress on a
plane within a soil mass through which seepage is taking place.
Consider a point in a soil mass where the direction of seepage is at angle � below the

horizontal. A square element ABCD of dimension b (unit dimension normal to the
paper) is centred at the above point with sides parallel and normal to the direction of
seepage, as shown in Figure 3.6(a), i.e. the square element can be considered as a flow
net field. Let the drop in total head between the sides AD and BC be �h. Consider the
pore water pressures on the boundaries of the element, taking the value of pore water
pressure at point A as uA. The difference in pore water pressure between A and D is
due only to the difference in elevation head between A and D, the total head being the
same at A and D. However, the difference in pore water pressure between A and either
B or C is due to the difference in elevation head and the difference in total head
between A and either B or C. The pore water pressures at B, C and D are as follows:

uB ¼ uA þ �wðb sin ���hÞ
uC ¼ uA þ �wðb sin �þ b cos ���hÞ
uD ¼ uA þ �wb cos �

The following pressure differences can now be established:

uB � uA ¼ uC � uD ¼ �wðb sin ���hÞ
uD � uA ¼ uC � uB ¼ �wb cos �

These values are plotted in Figure 3.6(b), giving the distribution diagrams of net
pressure across the element in directions parallel and normal to the direction of flow.
Therefore, the force on BC due to pore water pressure acting on the boundaries of

the element, called the boundary water force, is given by

�wðb sin ���hÞb

or

�wb
2 sin ���h�wb

80 Effective stress

and the boundary water force on CD by

�wb
2 cos �

If there were no seepage, i.e. if the pore water were static, the value of �h would
be zero, the forces on BC and CD would be �wb

2 sin � and �wb
2 cos �, respectively,

and their resultant would be �wb
2 acting in the vertical direction. The force �h�wb

represents the only difference between the static and seepage cases and is there-
fore called the seepage force (J ), acting in the direction of flow (in this case normal
to BC).
Now, the average hydraulic gradient across the element is given by

i ¼ �h
b

hence,

J ¼ �h�wb ¼ �h
b
�wb

2 ¼ i�wb2

or

J ¼ i�wV ð3:7Þ

where V is the volume of the soil element.

Figure 3.6 Forces under seepage conditions. (Reproduced from D.W. Taylor (1948) Funda-
mentals of Soil Mechanics, ª John Wiley & Sons Inc. [2], by permission.)

Influence of seepage on effective stress 81

The seepage pressure ( j) is defined as the seepage force per unit volume, i.e.

j ¼ i�w ð3:8Þ

It should be noted that j (and hence J ) depends only on the value of hydraulic gradient.
All the forces, both gravitational and forces due to seeping water, acting on the

element ABCD, may be represented in the vector diagram (Figure 3.6(c)). The forces
are summarized below.

Total weight of the element ¼ �satb2 ¼ vector ab
Boundary water force on CD (seepage and static cases)

¼ �wb2 cos � ¼ vector bd
Boundary water force on BC (seepage case)