Coulomb derivation based on limit equilibrium approach for Passive Case

Figure 1 below shows the cross section of a retaining wall.

coulomb's passive earth pressure

Weight of Wedge ABC

From Triangle ABC,

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

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

Therefore: – BC = AB\dfrac{sin(\alpha + \beta)}{sin(\alpha-\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^2 a}[sin(\alpha + \theta) \dfrac{sin(\alpha + \beta)}{sin(\theta-\beta)}]

 

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

Triangle of Forces for W, Pp and R

triangle of force in passive case

From the sine rule,

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

Substituting the value of W from the equation (3.34) into (3.35) we get,

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

In order to get the maximum value of Pp,

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

 

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

 

 P_P = K_P \dfrac{\gamma H^2}{2}

Where,

K_p = [\dfrac{sin^2(\alpha -\phi)}{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) and \alpha = 90^0 (vertical interface between wall and backfill), then, K_p = \dfrac{1-sin\phi}{1+sin\phi}. The point of application of Pp is at a distance of H/3 above the base of the wall.

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