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# Craig's Soil Mechanics 7th Edition

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```¼ �wb2 sin ���h�wb ¼ vector de
Boundary water force on BC (static case)

¼ �wb2 sin � ¼ vector dc
Resultant boundary water force (seepage case)

¼ vector be
Resultant boundary water force (static case)

¼ �wb2 ¼ vector bc
Seepage force ¼ �h�wb ¼ vector ce
Resultant body force (seepage case)

¼ vector ae
Resultant body force (static case)

¼ vector ac ¼ �0b2

The resultant body force can be obtained by one or other of the following force
combinations:

1 Total (saturated) weightþ resultant boundary water force, i.e. vector abþ vector be.
2 Effective (buoyant) weight þ seepage force, i.e. vector ac þ vector ce.

Only the resultant body force contributes to effective stress. A component of seepage
force acting vertically upwards will therefore reduce a vertical effective stress compon-
ent from the static value. A component of seepage force acting vertically downwards
will increase a vertical effective stress component from the static value.
A problem may be solved using either force combination 1 or force combination 2,

but it may be that one combination is more suitable than the other for a particular
problem. Combination 1 involves consideration of the equilibrium of the whole soil
mass (solids þ water), while combination 2 involves consideration of the equilibrium
of the soil skeleton only.

The quick condition

Consider the special case of seepage vertically upwards. The vector ce in Figure 3.6(c)
would then be vertically upwards and if the hydraulic gradient were high enough the
resultant body force would be zero. The value of hydraulic gradient corresponding to

82 Effective stress

zero resultant body force is called the critical hydraulic gradient (ic). For an element of
soil of volume V subject to upward seepage under the critical hydraulic gradient, the
seepage force is therefore equal to the effective weight of the element, i.e.

ic�wV ¼ �0V

Therefore

ic ¼ �
0

�w
¼ Gs � 1

1þ e ð3:9Þ

The ratio �0/�w, and hence the critical hydraulic gradient, is approximately 1.0 for
most soils.
When the hydraulic gradient is ic, the effective normal stress on any plane will be

zero, gravitational forces having been cancelled out by upward seepage forces. In the
case of sands the contact forces between particles will be zero and the soil will have no
strength. The soil is then said to be in a quick condition (quick meaning ‘alive’) and if
the critical gradient is exceeded the surface will appear to be ‘boiling’ as the particles
are moved around in the upward flow of water. It should be realized that ‘quicksand’ is
not a special type of soil but simply sand through which there is an upward flow of
water under a hydraulic gradient equal to or exceeding ic. In the case of clays, the quick
condition may not necessarily result when the hydraulic gradient reaches the critical
value given by Equation 3.9.

face of a sheet pile wall. Figure 3.7 shows part of the flow net for seepage under a sheet pile
wall, the embedded length on the downstream side being d. A mass of soil adjacent to the
pilingmay become unstable and be unable to support thewall.Model tests have shown that
failure is likely to occur within a soil mass of approximate dimensions d � d/2 in section
(ABCD in Figure 3.7). Failure first shows in the form of a rise or heave at the surface,
associated with an expansion of the soil which results in an increase in permeability. This in
turn leads to increased flow, surface ‘boiling’ in the case of sands and complete failure.
The variation of total head on the lower boundary CD of the soil mass can be obtained

from the flow net equipotentials, but for purposes of analysis it is sufficient to determine
the average total head hm by inspection. The total head on the upper boundary AB is zero.
The average hydraulic gradient is given by

im ¼ hm
d

Since failure due to heaving may be expected when the hydraulic gradient becomes ic,
the factor of safety (F ) against heaving may be expressed as

F ¼ ic
im

ð3:10Þ

Influence of seepage on effective stress 83

In the case of sands, a factor of safety can also be obtained with respect to ‘boiling’ at
the surface. The exit hydraulic gradient (ie) can be determined by measuring the
dimension �s of the flow net field AEFG adjacent to the piling:

ie ¼ �h
�s

where �h is the drop in total head between equipotentials GF and AE. Then, the
factor of safety is

F ¼ ic
ie

ð3:11Þ

There is unlikely to be any appreciable difference between the values of F given by
Equations 3.10 and 3.11.
The sheet pile wall problem shown in Figure 3.7 can also be used to illustrate the two

methods of combining gravitational and water forces.

Figure 3.7 Upward seepage adjacent to sheet piling.

84 Effective stress

1 Total weight of mass ABCD ¼ 1
2
�satd

2

Average total head on CD ¼ hm
Elevation head on CD ¼ �d
Average pore water pressure on CD ¼ (hm þ d )�w
Boundary water force on CD ¼ d

2
(hm þ d )�w

Resultant body force of ABCD ¼ 1
2
�satd

2 � d
2
ðhm þ d Þ�w

¼ 1
2
ð�0 þ �wÞd2 � 1

2
ðhmd þ d2Þ�w

¼ 1
2
�0d2 � 1

2
hm�wd

2 Effective weight of mass ABCD ¼ 1
2
�0d2

Average hydraulic gradient through ABCD ¼ hm
d

Seepage force on ABCD ¼ hm
d
�w

d2

2

¼ 1
2
hm�wd

Resultant body force of ABCD¼1⁄2� 0d2� 1⁄2hm�wd as in method 1 above.
The resultant body force will be zero, leading to heaving, when

1

2
hm�wd ¼ 1

2
�0d2

The factor of safety can then be expressed as

F ¼
1
2
�0d2

1
2
hm�wd

¼ �
0d

hm�w
¼ ic
im

If the factor of safety against heaving is considered inadequate, the embedded
length d may be increased or a surcharge load in the form of a filter may be placed
on the surface AB, the filter being designed to prevent entry of soil particles. If the
effective weight of the filter per unit area is w0 then the factor of safety becomes

F ¼ �
0d þ w0
hm�w

Example 3.3

The flow net for seepage under a sheet pile wall is shown in Figure 3.8(a), the saturated
unit weight of the soil being 20 kN/m3. Determine the values of effective vertical stress
at A and B.

Influence of seepage on effective stress 85

1 First consider the combination of total weight and resultant boundary water
force. Consider the column of saturated soil of unit area between A and the soil
surface at D. The total weight of the column is 11�sat (220 kN). Due to the change in
level of the equipotentials across the column, the boundary water forces on the sides
of the column will not be equal although in this case the difference will be small. There
is thus a net horizontal boundary water force on the column. However, as the
effective vertical stress is to be calculated, only the vertical component of the
resultant body force is required and the net horizontal boundary water force need
not be considered. The boundary water force on the top surface of the column is only
due to the depth of water above D and is 4�w (39 kN). The boundary water force on
the bottom surface of the column must be determined from the flow net, as follows:

Number of equipotential drops between the downstream soil surface and A ¼ 8:2.
There are 12 equipotential drops between the upstream and downstream soil
surfaces, representing a loss in total head of 8m.

Total head at A, hA ¼ 8:2
12

� 8 ¼ 5:5m.
Elevation head at A, zA ¼ �7:0m.
Pore water pressure at A, uA ¼ �w(hA � zA)

¼ 9:8(5:5þ 7:0) ¼ 122 kN=m2
i.e. boundary water force on bottom surface ¼ 122 kN.
Net vertical boundary water force ¼ 122� 39 ¼ 83 kN.
Total weight of the column ¼ 220 kN.
Vertical component of resultant body force ¼ 220� 83 ¼ 137 kN
i.e. effective vertical stress at A ¼ 137 kN/m2.

(a) (b)

Figure 3.8 Examples 3.3 and 3.4.

86 Effective stress

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