Active thrust, 7.5 m wall, 4.5 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

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-35^\circ/2) = 0.2710 \\ K_{a2} &= \tan^2(45^\circ-30^\circ/2) = 0.3333,\quad 2c\sqrt{K_{a2}} = 2\times16\times0.5774 = 18.48\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Sand, top: $0$.
• Sand, bottom ($z=4.5$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times4.5 = 21.95$.
• Clay, top: $\sigma_v' = 18\times4.5 = 81.00$, so $p = 0.3333\times81.00 - 18.48 = 8.52$.
• Clay, bottom ($z=7.5$ m): $\sigma_v' = 81.00 + 9.69\times3 = 110.07$, so $p = 0.3333\times110.07 - 18.48 = 18.21$.
• Water pressure at the base: $u = 9.81\times3 = 29.43$ (zero at the water table).

 0 --|
     |\
sand | \
     |  \
 WT -|---21.95
     |  eff 8.52
clay |   \
     |    \ eff 18.21
 base|-----> + water 29.43

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{319.29}{133.64} = 2.39\ \text{m above the base}$$

Answer: total thrust $P = 133.6$ kN/m (water included), acting horizontally at 2.39 m above the base of the wall.