Passive thrust on retaining wall back

Assumptions: smooth vertical wall back, horizontal ground, no water. Rankine passive pressure $p_p = K_p\sigma_v + 2c\sqrt{K_p}$, with total height 10 m of soil (5 m + 5 m) acting on the back.

Coefficients

$$\begin{aligned} K_{p1} &= \tan^2(45^\circ+15^\circ) = 3.000 \\ K_{p2} &= \tan^2(45^\circ+7.5^\circ) = 1.6984,\quad \sqrt{K_{p2}} = 1.3032 \\ 2c\sqrt{K_{p2}} &= 2\times30\times1.3032 = 78.19\ \text{kN/m}^2 \end{aligned}$$

Pressures (kN/m$^2$)

• Top layer: $z=0$: $p=0$; $z=5$ m: $\sigma_v=18\times5=90$, $p = 3.000\times90 = 270.0$.
• Bottom layer top: $p = 1.6984\times90 + 78.19 = 231.05$.
• Bottom layer base ($z=10$ m): $\sigma_v = 90+20\times5 = 190$, $p = 1.6984\times190 + 78.19 = 400.89$.

 0 |
   |\
   | \
 5 |--270.0
   |--231.05
   |     \
10 |------400.89

Forces (arm above the base of the 10 m wall)

$$\bar y = 8095.78/2254.85 = 3.59\ \text{m}$$

Answer: total passive thrust $P_p = 2255$ kN/m, acting horizontally at 3.59 m above the base (bottom of the wall).