Trapezoidal masonry wall, sliding and overturning

Geometry and assumptions: top width 1 m, base width 3 m, height 4 m. The soil face (back) is vertical and the front face is inclined, so the toe is the front bottom corner. The backfill slopes at $\beta = 20^\circ$, with $\phi = 30^\circ$ and $\gamma = 20\ \text{kN/m}^3$. The only $\delta$ given, $24^\circ$, is used both as the wall friction (for Coulomb's $K_a$) and as the base friction (for sliding). Passive resistance is neglected.

1. Coulomb active thrust

$$K_a = \frac{\sin^2(\theta+\phi)}{\sin^2\theta\,\sin(\theta-\delta)\left[1+\sqrt{\dfrac{\sin(\phi+\delta)\sin(\phi-\beta)}{\sin(\theta-\delta)\sin(\theta+\beta)}}\right]^2} = 0.4162\quad(\theta = 90^\circ)$$ $$P_a = \tfrac12 K_a\gamma H^2 = 0.5\times0.4162\times20\times4^2 = 66.59\ \text{kN/m}$$

It acts at $H/3 = 1.333$ m above the base, at $24^\circ$ below the horizontal:

$$P_h = P_a\cos24^\circ = 60.83,\qquad P_v = P_a\sin24^\circ = 27.08\ \text{kN/m}$$

2. Weights and moments about the toe

Overturning moment: $M_o = P_h\times1.333 = 60.83\times1.333 = 81.11$ kN m/m.

3. Base pressure

$$\bar x = \frac{M_r - M_o}{\sum V} = \frac{449.25-81.11}{219.08} = 1.680\ \text{m}$$ $$e = \frac{B}{2}-\bar x = 1.5-1.680 = -0.180\ \text{m},\qquad |e| = 0.180\ \text{m} < B/6 = 0.5\ \text{m}$$ $$q = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right)$$

The resultant falls toward the heel, so $q_{max}$ is at the heel and $q_{min}$ is at the toe:

$$q_{max} = 99.4\ \text{kN/m}^2,\qquad q_{min} = 46.7\ \text{kN/m}^2$$

Both are compressive, so there is no tension.

4. Factors of safety

$$FS_{ot} = \frac{M_r}{M_o} = \frac{449.25}{81.11} = 5.54\ (>2,\ \text{safe})$$ $$FS_{s} = \frac{\sum V\tan24^\circ}{P_h} = \frac{219.08\times0.4452}{60.83} = 1.60\ (>1.5,\ \text{safe})$$

Answer: $q_{max} = 99.4$ and $q_{min} = 46.7$ kN/m$^2$; $FS_{ot} = 5.54$; $FS_{sliding} = 1.60$.