Coulomb’s Wedge Theory | Derivation for Active Case

Coulomb’s made the following assumptions in the development of his theory.

The assumptions are:

  • The soil is isotropic and homogeneous.
  • The surface of rupture is a plane.
  • The failure wedge is a rigid body.
  • There is friction between and the wall.
  • Back of wall need not be vertical.
  • Failure is two dimensional.
  • The soil is cohesionless.
  • Coulomb’s equation of shear strength is valid.

Coulomb made his derivation based on limit equilibrium approach.

Active Case

Figure 3.25 below shows the cross section of a retaining Wall. Equilibrium analysis of failure wedge ABC involves:

  • Weight of wedge ABC (magnitude and direction known)
  • Pa (direction known, magnitude unknown)
  • R (direction known, magnitude unknown)

Hence, from the triangle of forces can be drawn and Pa can be determined.

coulomb's active earth pressure

Weight of wedge ABC

From Δ ABC,

Area of Δ ABC = ½ AD × BC

weight of wedge abc

BC/AB = {sin(α + β)/ sin(Ï´  – β)}

BC = AB {sin(α + β)/ sin(Ï´  – β)}

Again, AD = AB sin[1800 – (α + Ï´)] = AB sin(α + Ï´)

 columb active earth pressure formula 1

columb active earth pressure formula 2

 

 

columb active earth pressure formula 3

 

 

Triangle of Forces for W, Pa and R

From the sine rule,

columb active earth pressure formula 4

 

Substituting the value of W from the equation         into         we get,

columb active earth pressure formula 5

 

 

triangle of force

In order to get the maximum value of Pa,

formula of active pressure 1

 

 

columb active earth pressure formula 6

 

 

 

columb active earth pressure formula 8

 

 

Where,

columb active earth pressure formula 7

 

 

 

Note:

When, β = 0 (leveled backfilled), δ = 0 (no wall friction), then ka = Ka = (1 – sinɸ)/(1 + sinɸ). The point of application of Pa is at a distance of H/3 above the base of the wall.

  • Mr.Singh

    My text book had nothing about calculating the weight of the wedge… This web page is a life saver!!!!