Lateral Earth Pressure - Elements of Soil Mechanics - page 215

Maximum P a due to submerged soil = 16 kN
Horizontal component of P a = 16.0 × cos 20° = 15 kN
Horizontal thrust from water pressure

2

4
2

= 9 81 . × = 78 5 . kN

Total horizontal thrust = 93.5 kN/m run of wall

(ii) With vertical drain on back of wall
The flow net for steady seepage from the flooded surface of the soil into the drain
is shown in Fig. 7.29c . From this diagram it is possible to determine the distribution
of the excess hydrostatic head, h w , along the length of the failure surface of each of
the four trial wedges. These distributions are shown in Fig. 7.29d and the area of
each diagram times the unit weight of water gives the upward force, P w , acting at
right-angles to each failure plane.
The tabulated calculations are:

Wedge Saturated weight (kN) P w (kN)

1 40 8

2 80 18

3 120 30

4 160 45

The force diagrams and the Culmann line construction are shown in Fig. 7.29e .
From the force diagram, maximum P a = 45 kN.
⇒ Maximum horizontal thrust on wall = × 45 cos 20 ° = 42 kN/m

(iii) With inclined drain
As has been shown earlier, for all points in the soil above the drain there can be no
excess hydrostatic heads. The force diagram is therefore identical with Fig. 7.29e
except that, as P w is zero for all wedges, it is removed from each polygon of forces.
When this is done it is found that the maximum value of P a is 30 kN.

⇒ Maximum horizontal thrust on back of wall = × 30 cos 20 ° = 28 kN/ m

7.11 Influence of wall yield on design

A wall can yield in one of two ways: either by rotation about its lower edge (Fig. 7.30b ) or by sliding
forward (Fig. 7.30c ). Provided that the wall yields sufficiently, a state of active earth pressure is reached
and the thrust on the back of the wall is in both cases about the same (P a ).
The pressure distribution that gives this total thrust value can be very different in each instance, however.
For example, consider a wall that is unable to yield (Fig. 7.30a ). The pressure distribution is triangular and
is represented by the line AC.
Consider that the wall now yields by rotation about its lower edge until the total thrust = P a (Fig. 7.30b ).
This results in conditions that approximate to the Rankine theory and is known as the totally active case.
Suppose, however, that the wall yields by sliding forward until active thrust conditions are achieved (Fig.
7.30c ). This hardly disturbs the upper layers of soil so that the top of the pressure diagram is similar to
the earth pressure at rest diagram. As the total thrust on the wall is the same as in rotational yield, it
means that the pressure distribution must be roughly similar to the line AE in Fig. 7.30c .