Coulomb derivation based on limit equilibrium approach for Passive Case

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

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, P_p and R

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 P_p,

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

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

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

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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