First let us consider a wedge as shown in Figure 12.9 where for simplicity we will assume the top surface, BC, is free of boundary tractions and there is a rigid retaining wall supporting the EPS material with unit weight γ on the sloping face AB.

**Retaining Wall Solution for (EPS Material)**

First let us consider a wedge as shown in Figure
12.9 where for simplicity we will assume the top surface, BC, is free of
boundary tractions and there is a rigid retaining wall supporting the EPS
material with unit weight γ on the sloping face
AB. If the wall moves outward in the direction of the *v* arrows, the
active state of limiting equilibrium develops. A general failure surface, which
by our analysis is a combination of straight and circular segments, is assumed.
This is a combination of an upper wedge (slip field I) with plane surfaces of
sliding, a circular transition section, II, and a third wedge, III, adjoining
the wall.

Since we do not know the extent of slip fields II
and III or the stresses on the wall, we cannot solve directly but must work our
way along the slip surface* from the known boundary, BC, to the unknown
boundary, AB, using the solution [Equation (12.15)] to Kötter’s equation for ϕ= 0. Starting at point *C* we know the complete state of stress. Since τ* _{xz=}*σ

As we move to point *D*,
*x* and *z* remain principal so σ'* _{D}*
= γ

Therefore at *A*
in the orientation of the wall, the stresses on the wall from Mohr's
Circle are:

It is important to note
that the actual values of the wall stresses depend on the choice of ψ,
which specifies the extent of the circular zone II. The distribution of stress
on the wall however, is determined by the solution to be a constant shear and a
normal stress linear with depth. Any other distribution cannot be specified for
a unique solution to this Cauchy problem and maintain the unloaded boundary
conditions on the surface BC.*

The direction of the *resultant
stress* on the wall varies since the normal stress increases with depth
while the shear stress is constant. Therefore, to talk about a constant “angle
of wall friction” δ is a misnomer.
However, we can compute the components of the force resultants on the wall by
integrating the stresses, i.e.,

A specific example to
illustrate this complete stress solution at limit equilibrium for the case ψ=70 o is shown in Figure 12.10. For the passive case where the wall is forced
into the material, the failure surface is the same but the direction of
impending motion is revised. Thus the shear stress on the failure surface acts
in the opposite direction to resist the

This is the same result
as for the active case except that the constants by which the geostatic normal
stress γ*z* is modified have changed sign,
as has the shear. The shear resultants *N*, *T* are modified
accordingly as is the average angle of wall friction.

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Civil : Principles of Solid Mechanics : Slip Line Analysis : Retaining Wall Solution for (EPS Material) |

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