Active thrust, 7.5 m wall, 5 m sand

Assumptions: smooth vertical wall, horizontal backfill, Rankine theory; sand above the water table is taken at the given unit weight; the clay is saturated, so the effective stress uses $\gamma' = \gamma_{sat} - \gamma_w = 19.50 - 9.81 = 9.69\ \text{kN/m}^3$.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-35^\circ/2) = 0.2710 \\ K_{a2} &= \tan^2(45^\circ-35^\circ/2) = 0.2710,\quad 2c\sqrt{K_{a2}} = 2\times16\times0.5206 = 16.66\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Sand, top: $0$.
• Sand, bottom ($z=5$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times5 = 24.39$.
• Clay, top: $\sigma_v' = 18\times5 = 90.00$, so $p = 0.2710\times90.00 - 16.66 = 7.73$.
• Clay, bottom ($z=7.5$ m): $\sigma_v' = 90.00 + 9.69\times2.5 = 114.22$, so $p = 0.2710\times114.22 - 16.66 = 14.30$.
• Water pressure at the base: $u = 9.81\times2.5 = 24.53$ (zero at the water table).

 0 --|
     |\
sand | \
     |  \
 WT -|---24.39
     |  eff 7.73
clay |   \
     |    \ eff 14.30
 base|-----> + water 24.53

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{310.60}{119.16} = 2.61\ \text{m above the base}$$

Answer: total thrust $P = 119.2$ kN/m (water included), acting horizontally at 2.61 m above the base of the wall.