Lateral Earth Pressure-Rankine’s Case From C.R. Scott Book
Applications of Rankine’s Method: The applications of Rankine’s method can best be seen in a
series of examples. First, consider the simplest possible case – a wall supporting a uniform
cohesionless soil with horizontal surface (Fig. 9.3 (a)). For the moment, the pore pressure (u)
will be assumed to be zero everywhere.
Example 9.1
This is equal to the area of the pressure diagram ABC. The line of action of the force Pa is at the
centroid of the pressure diagram, which is 3.33 m above the base.
Now consider a uniform surcharge load, q per unit area, applied over the whole of the soil
surface. The vertical effective stress (𝝈′𝒛 ) is everywhere increased by q, and the effective active
pressure ((𝒑′𝒂 )) is therefore increased by kaq.
Example 9.2
9.5 Application of Rankine's method where the pore pressure (u) is not zero: Where the pore
pressure is not zero, the effective stress and pore pressure resultants must be computed
separately. The derivation of Rankine's method requires that the stresses are constant at any
given depth below the ground surface, which implies that the water table is parallel to the ground
surface. Where the soil surface is horizontal the method can therefore be used only for
hydrostatic conditions. The pore pressure is then defined by the position of the water table, at
which u = 0. Above the water table, water may be held in the pores by capillary forces, and the
pore pressure in this capillary zone is negative. This negative pore pressure increases the
effective stress, and consequently increases the shear strength also. However, this increase in
strength is unreliable, for two reasons. Firstly, the maximum height of the capillary zone depends
on the capillary suction.
This is a function of the pore size, and is generally unknown. Secondly, downward seepage of
surface water towards the water table may increase the pore pressure in the capillary zone above
the value computed for the strictly hydrostatic condition. It is therefore usual to take a
conservative view, and to assume that the pore pressure is zero at all points above the water
table. Below the water table, the body force must be taken to be (𝛾sat - 𝛾w) (see Section 8.6) when
calculating the effective stress.
For a soil having a density of y, supporting a uniform surcharge load q per unit area of the surface, p' a is
zero at a depth zo where
Since the soil cannot sustain tension indefinitely, it is usual to assume that cracks develop to a depth z0 ,
and to ignore the negative part of the pressure diagram.
Cohesive soils commonly have low permeabilities. If surface water can drain into a tension crack, it may
be some time before it drains away, and the crack may be completely filled. This water therefore exerts
an additional thrust on the wall, although the pore pressures within the soil are not immediately affected.