Active thrust, 7.5 m wall, 3 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 ($z=3$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times3 = 14.63$.
• Clay, top: $\sigma_v' = 18\times3 = 54.00$, so $p = 0.2710\times54.00 - 16.66 = -2.02$.
• Clay, bottom ($z=7.5$ m): $\sigma_v' = 54.00 + 9.69\times4.5 = 97.60$, so $p = 0.2710\times97.60 - 16.66 = 9.79$.
• Water pressure at the base: $u = 9.81\times4.5 = 44.15$ (zero at the water table).

The effective pressure at the top of the clay is negative (-2.02), so a tension crack is assumed; the negative part is ignored. The pressure becomes zero at $z_0 = 4.5\times2.02/(9.79+2.02) = 0.77$ m below the water table (interface) and the clay diagram is a triangle from there.

 0 --|
     |\
sand | \
     |  \
 WT -|---14.63
     |  eff -2.02
clay |   \
     |    \ eff 9.79
 base|-----> + water 44.15

Forces per metre run and lever arms about the base

$$\bar y = \frac{\sum M}{\sum F} = \frac{292.41}{139.53} = 2.10\ \text{m above the base}$$

Answer: total thrust $P = 139.5$ kN/m (water included), acting horizontally at 2.10 m above the base of the wall.