Active pressure, 8 m wall, table of layers

Assumptions: smooth vertical wall, horizontal backfill, no groundwater (none is given), Rankine theory.

Coefficients

$$\begin{aligned} K_{a1} &= \tan^2(45^\circ-15^\circ) = 0.3333,\quad \sqrt{K_{a1}} = 0.5774 \\ K_{a2} &= \tan^2(45^\circ-17^\circ) = 0.2827 \end{aligned}$$

Layer 1 (0-4 m, $c=10$, $\phi=30^\circ$, $\gamma=18$): $p = K_{a1}\gamma z - 2c\sqrt{K_{a1}}$

• $z=0$: $p = -2\times10\times0.5774 = -11.55$ kN/m$^2$ (tension).
• Zero at $z_0 = 2c/(\gamma\sqrt{K_{a1}}) = 1.925$ m; the tension zone is ignored (tension crack).
• $z=4$ m (top layer): $p = 0.3333\times72 - 11.55 = 12.45$ kN/m$^2$.

Layer 2 (4-8 m, $c=0$, $\phi=34^\circ$, $\gamma=20$): $\sigma_v$ at 4 m $= 18\times4 = 72$ kN/m$^2$.

• Top: $p = 0.2827\times72 = 20.36$ kN/m$^2$.
• Base ($z=8$ m): $\sigma_v = 72 + 20\times4 = 152$, $p = 0.2827\times152 = 42.97$ kN/m$^2$.

 0  |  (tension zone, 0 to 1.92 m, ignored)
    |
1.92|\
    | \
    |  \
 4m |---12.45
    |-- 20.36
    |   \
    |    \
 8m |-----42.97

Forces per metre run (arm above base)

$$\bar y = 283.79/139.58 = 2.03\ \text{m above the base}$$

Answer: resultant active thrust $P_a = 139.6$ kN/m, horizontal, acting 2.03 m above the base (distribution as shown).