Geotechnical Engineering

DHARM

LATERAL EARTH PRESSURE AND STABILITY OF RETAINING WALLS 511

this case is known to be triangular as shown by the straight line AF in Fig. 13.52 (b), the total

thrust on the wall per unit length being represented by P 0 which acts at a height of H/3 above

the base.

The wall can yield in one of two ways: either by rotation about its lower edge

[Fig. 13.52 (c)], or by sliding forward or translating away from the backfill [Fig. 13.52 (f)]. If

the wall yields sufficiently, a state of active earth pressure is reached and the thrust on the

wall in both cases is about the same (Pa). However, the pressure distribution that gives this

total thrust can be very different in each instance, as will be seen from the following discus-

sion.

H

A

B

C A

B F

Pa

H/3

A

B

C

A

A

B

Pa

H/3

Pa

R N

A

A¢

DC

D

E¢

E B

A

B JGF

A

B J

0.45 to

0.55 H

Pa

(a) At-rest condition (no wall yield) (b) At-rest pressure (c) Totally active condition

(wall yields by rotation

about heel at base)

(d) Pressure for totally

active case

(e) Force triangle (f) Arching-active

condition (wall yields

by horizontal translation)

(g) Pressure for

arching-active case

(h) Pressure diagrams

superimposed

Fig. 13.52 At rest, totally active and arching—active cases (wall friction neglected) (Taylor, 1948)
Let it be assumed that the wall yields by rotation about the heel B by an amount suffi-
cient to create the active pressure conditions. During this rotation, the wedge ABC distorts in
an essentially uniform manner throughout to the shape A′BC of Fig. 13.52 (c). The uniform
distribution leads to a φ-obliquity condition throughout and active pressures occur on the wall
over its entire height. Neglecting wall friction, the pressure distribution will appear as shown
by the straight line AG, in Fig. 13.52 (d), in a triangular shape, the pressure at any depth being
less than the at-rest value. The total thrust Pa acts at H/3 above the base. The revised positions
of the grid lines of Fig. 13.52 (a) are shown in (c). Although the explanation is somewhat
idealised in some respects, the general concept is essentially correct. This case is referred to as
the totally active case which is the same as the active Rankine state.