Coulomb’s Wedge Theory | Derivation for Active Case

Coulomb’s made the following assumptions in the development of his theory and hence it is called Coulomb’s Wedge Theory.

The assumptions made by Coulomb while giving his theory was:

  • 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 cohesion-less.
  • Coulomb’s equation of shear strength is valid.

Coulomb made his derivation based on limit equilibrium approach.

Derivation of Active Case for Coulomb’s Wedge Theory

Figure 1 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

Calculation of Weight of wedge ABC

From Triangle ABC,

Area of Triangle ABC = \dfrac{1}{2}AD\times BC

weight of wedge abc

\dfrac{BC}{AB} = \dfrac{sin(\alpha + \beta)}{sin(\theta-\beta)}

Therefore, BC = AB\dfrac{sin(\alpha + \beta)}{sin(\theta-\beta)}

Again, AD = ABsin[180^0-(\alpha+\theta)] = ABsin(\alpha + \theta)

 

Area \ of\ \Delta ABC = \dfrac{1}{2}AB\ sin(\alpha + \theta) \times AB \dfrac{sin(\alpha + \beta)}{sin(\theta-\beta)}

 

\Delta ABC = \dfrac{H^2}{2sin^2a}[sin(\alpha + \theta) \dfrac{sin(\alpha + \beta)}{sin(\theta-\beta)}]

 

W = \gamma \times V = \dfrac{\gamma H^2}{2sin^2a}[sin(\alpha + \theta) \dfrac{sin(\alpha + \beta)}{sin(\theta-\beta)}]           …….(1)

Triangle of Forces for W, Pa and R | Coulomb’s Wedge Theory

From the sine rule we can write,

\dfrac{P_a}{sin(\theta-\phi)} = \dfrac{W}{sin[180^0-((\theta-\phi)+(\alpha-delta))]}         ……..(2)

Substituting the value of W from the equation (1) into (2) we get,

P_a = \dfrac{\gamma H^2}{2sin^2\alpha}[sin(\alpha + \theta) \dfrac{sin(\alpha + \beta)}{sin(\theta -\beta)}] \dfrac{sin(\theta -\phi)}{sin(180^0 -\theta + \phi -\alpha + \delta)}

triangle of force

In order to get the maximum value of Pa,

 \dfrac{\partial P_a}{\partial \theta} = 0

 

P_a = \dfrac{\gamma H^2}{2}[\dfrac{sin^2(\phi + \alpha)}{sin^2\alpha sin(\alpha -\delta(1 + \sqrt{\dfrac{sin(\phi + \delta) sin(\phi -\beta)}{sin(\alpha -\delta) sin(\alpha + \beta)}}))}]

 

P_a = K_a \dfrac{\gamma H^2}{2}

Where,

K_a = [\dfrac{sin^2(\phi + \alpha)}{sin^2\alpha sin(\alpha -\delta(1 + \sqrt{\dfrac{sin(\phi + \delta) sin(\phi -\beta)}{sin(\alpha -\delta) sin(\alpha + \beta)}}))}]

Note:

When, \beta = 0 (leveled backfilled), \delta = 0 (no wall friction), then K_a = \dfrac{1-sin\phi}{1+sin\phi}. The point of application of Pa is at a distance of H/3 above the base of the wall.

One Thought to “Coulomb’s Wedge Theory | Derivation for Active Case”

  1. Mr.Singh

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

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