Infinite clay slope, submerged and seepage

Data: $\gamma_t = 18$, $\gamma' = 9$ kN/m³, $c = 25$ kN/m², $\phi' = 28^\circ$, $\beta = 35^\circ$. The depth of the failure plane is taken as $z = 12$ m (the slope height is measured as vertical depth). For seepage, the saturated unit weight is taken as $\gamma_{sat} = \gamma_t = 18$ kN/m³.

Useful values: $\sin\beta\cos\beta = 0.4698$, $\tan\phi'/\tan\beta = 0.5317/0.7002 = 0.7594$.

(i) Slope fully submerged (static water, no seepage)

Normal stress on the plane: $\sigma' = \gamma' z\cos^2\beta$. Shear stress: $\tau = \gamma' z\sin\beta\cos\beta$.

$$\begin{aligned} F &= \frac{c'}{\gamma' z\sin\beta\cos\beta} + \frac{\tan\phi'}{\tan\beta} \\ &= \frac{25}{9(12)(0.4698)} + 0.7594 \\ &= 0.493 + 0.759 = 1.25 \end{aligned}$$

$F = 1.25 > 1$, so the slope is stable (but its margin is small; $F < 1.5$).

(ii) Seepage parallel to the slope (water table at the surface)

Normal effective stress: $\sigma' = \gamma' z\cos^2\beta$. Shear stress: $\tau = \gamma_{sat} z\sin\beta\cos\beta$.

$$\begin{aligned} F &= \frac{c'}{\gamma_{sat} z\sin\beta\cos\beta} + \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi'}{\tan\beta} \\ &= \frac{25}{18(12)(0.4698)} + \frac{9}{18}(0.7594) \\ &= 0.246 + 0.380 = 0.63 \end{aligned}$$

$F = 0.63 < 1$, so the slope is unstable with seepage parallel to it.

Answer: (i) $F = 1.25$, stable; (ii) $F = 0.63$, unstable.