Craig's Soil Mechanics 7th Edition
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Craig's Soil Mechanics 7th Edition

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6.27(a)); they can
thus be eliminated from the calculations.
If the water table levels are different and if steady seepage conditions have devel-

oped and are maintained, the distributions on the two sides of the wall will be
unbalanced. The pressure distributions on each side of the wall may be combined
because no pressure coefficient is involved. The net distribution on the back of the wall
could be determined from the flow net, as illustrated in Example 2.1. However, in most
situations an approximate distribution, ABC in Figure 6.27(b), can be obtained by
assuming that the total head is dissipated uniformly along the back and front wall
surfaces between the two water table levels. The maximum net pressure occurs
opposite the lower water table level and, referring to Figure 6.27(b), is given by

uC ¼ 2ba
2bþ a �w

In general, the approximate method will underestimate net water pressure, especially if
the bottom of the wall is relatively close to the lower boundary of the flow region (i.e. if
there are large differences in the sizes of curvilinear squares in the flow net). The
approximation should not be used in the case of a narrow excavation between two
lines of sheet piling where curvilinear squares are relatively small (and seepage pressure
relatively high) approaching the base of the excavation.
In Figure 6.27(c), a depth of water is shown in front of the wall, the water level being

below that of the water table behind the wall. In this case the approximate distribution
DEFG should be used in appropriate cases, the net pressure at G being given by

uG ¼ ð2bþ cÞa
2bþ cþ a �w

A wall constructed mainly in a soil of relatively high permeability but penetrating
a layer of clay of low permeability is shown in Figure 6.27(d). If undrained conditions
apply within the clay the pore water pressure in the overlying soil would be hydrostatic
and the net pressure distribution would be HJKL as shown.
A wall constructed in a clay which contains thin layers or partings of fine sand or silt

is shown in Figure 6.27(e). In this case it should be assumed that the sand or silt allows
water at hydrostatic pressure to reach the back surface of the wall. This implies
pressure in excess of hydrostatic, and consequent upward seepage, in front of the wall.
For short-term situations for walls in clay (e.g. during and immediately after

excavation), there exists the possibility of tension cracks developing or fissures
opening. If such cracks or fissures fill with water, hydrostatic pressure should be
assumed over the depth in question: the water in the cracks or fissures would also
result in softening of the clay. Softening would also occur near the soil surface in front
of the wall as a result of stress relief on excavation. An effective stress analysis would
ensure a safe design in the event of rapid softening of the clay taking place or if work
were delayed during the temporary stage of construction; however, a relatively low
factor of safety could be used in such cases. Padfield and Mair [15] give details of a
mixed total and effective stress design method as an alternative for short-term situations

202 Lateral earth pressure

in clay, i.e. effective stress conditions within the zones liable to soften and total stress
conditions below.

Seepage pressure

Under conditions of steady seepage, use of the approximation that total head is
dissipated uniformly along the wall has the consequent advantage that the seepage
pressure is constant. For the conditions shown in Figure 6.27(b), for example, the
seepage pressure at any depth is

j ¼ a
2bþ a �w

The effective unit weight of the soil below the water table, therefore, would be
increased to �0 þ j behind the wall, where seepage is downwards, and reduced to
�0 � j in front of the wall, where seepage is upwards. These values should be used in
the calculation of active and passive pressures, respectively, if groundwater conditions
are such that steady seepage is maintained. Thus, active pressures are increased and
passive pressures are decreased relative to the corresponding static values.

Anchorages

Tie rods are normally anchored in beams, plates or concrete blocks (known as dead-
man anchors) some distance behind the wall (Figure 6.28). Ultimate limit states are
pullout of the anchor and fracture of the tie. The serviceability limit state is that
anchor yield should be minimal. The tie rod force is resisted by the passive resistance
developed in front of the anchor, reduced by the active pressure on the back, both
calculated either (a) by applying a lumped factor of safety to gross or net passive
resistance or (b) by applying partial factors to the shear strength parameters, to ensure
that the serviceability limit state is satisfied. The passive resistance should be calcu-
lated assuming no surface surcharge and the active pressure calculated assuming that
at least a minimum surcharge pressure of 10 kN/m2 is imposed. To avoid the possibility

Figure 6.28 (a) Plate anchor and (b) Ground anchor.

Embedded walls 203

of progressive failure of a line of ties, it should be assumed that any single tie could fail
either by fracture or by becoming detached and that its load could be redistributed
safely to the two adjacent ties. Accordingly, it is recommended that a load factor of at
least 2.0 should be applied to the tie rod force. If the height (b) of the anchor is not less
than half the depth (da) from the surface to the bottom of the anchor, it can be
assumed that passive resistance is developed over the depth da. The anchor must be
located beyond the plane YZ (Figure 6.28(a)) to ensure that the passive wedge of the
anchor does not encroach on the active wedge behind the wall.
The equation of equilibrium is

1

2
ðKp � KaÞ�d2a l � Kaqdal � Ts ð6:29Þ

where

T ¼ tie force per unit length of wall;
s ¼ spacing of ties;
l ¼ length of anchor per tie;
q ¼ surface surcharge pressure:

Ties can also be anchored to the tops of inclined piles. Tensioned cables, attached to
the wall and anchored in a mass of cement grout or grouted soil (Figure 6.28(b)), are
another means of support. These are known as ground anchors and are described in
Section 8.8.

Example 6.7

The sides of an excavation 2.25m deep in sand are to be supported by a cantilever
sheet pile wall, the water table being 1.25m below the bottom of the excavation. The
unit weight of the sand above the water table is 17 kN/m3 and below the water table
the saturated unit weight is 20 kN/m3. Characteristic parameters are c0 ¼ 0, �0 ¼ 35�
and � ¼ 0. Allowing for a surcharge pressure of 10 kN/m2 on the surface, (a) determine
the required depth of embedment of the piling to ensure a factor of safety of 2.0 with
respect to gross passive resistance and (b) check that the rotational and translational
limit states would be satisfied for the above embedment depth, using appropriate
partial factors.

(a) For �0 ¼ 35� and � ¼ 0,Ka ¼ 0:27 and Kp ¼ 3:7.
Below the water table the effective unit weight of the soil is (20� 9:8) ¼ 10:2 kN/m3.
To allow for possible over-excavation the soil level should be reduced by 10% of the

retained height of 2.25m, i.e. by 0.225m. The depth of the excavation, therefore,
becomes 2.475m, say 2.50m, and the water table will be 1.00m below this level.
The design dimensions and the earth pressure diagrams are shown in Figure 6.29.

The distributions of hydrostatic pressure on the two sides of the wall balance and can
be eliminated from the calculations. The procedure is to equate moments about C, the

204 Lateral earth pressure

point of application of the force representing the net passive resistance below the point
of rotation. The forces, lever arms and moments are set out in Table 6.5(a), forces (5),
(6) and (7) being divided by the specified factor of safety.
Equating the algebraic sum of the moments about C to zero produces the following

equation