Equilibrium of an embedded retaining wall

• Before excavation begins, the pressure is equal on both sides of the wall. K = K₀

Here K₀ is taken to be (1-sinf) ÖR_o
Where Ro = the over consolidation ration (i.e. the current vertical stress / maximum previous vertical stress)
If f =30° and R_o = 1 then  K₀ = (1-sin30) Ö1 = 0.5
Unit weight of soil = 18 kN/m³
s_h' @ 10m depth  =  K * unit weight * depth = 0.5 * 18 * 10  =  90 kPa
Force on wall =  ½ max pressure * depth = ½ 90 * 10  =  450kN per m length of wall

• Excavating on one side reduces the pressure on that side causing the wall to move.
•  If there is a ground anchor or prop preventing rotation at one position then the movement will be a rotation.
•  Balance is reestablished because K decreases on one side (wall moves away & soil expands), and increases on the other (soil is compresses).


Failure occurs when K reaches the limiting values
Ka = (1-sinf) / (1+sinf) = (1- 0.5) / (1+ 0.5) = 0.333 if f=30°
Kp = (1+sinf) / (1-sinf) = (1+ 0.5) / (1- 0.5) = 3  if f=30°

• Wall tends to rotate about the base
• This results in two forces which may be equal but not in line
• Rotational equilibrium is not satisfied
• Wall actually rotates about a point above the base
• "kick back" below point of rotation creates 3^rd force
• Horizontal & rotational equilibrium are now satisfied.