Embankment 36°, method of slices FoS

Data (per metre run): $\Sigma T = 450$ kN, $\Sigma N = 900$ kN, $\Sigma U = 216$ kN (neutral force, pore pressure force), arc length $L = 27$ m, $c' = 20$ kN/m², $\phi' = 18^\circ$.

Effective normal force: $\Sigma N' = \Sigma N - \Sigma U = 900 - 216 = 684$ kN.

(a) Factor of safety with respect to shear strength

The factor of safety is the ratio of the available shear strength to the shear force:

$$\begin{aligned} F_s &= \frac{c'L + (\Sigma N-\Sigma U)\tan\phi'}{\Sigma T} \\ &= \frac{20(27) + 684\tan18^\circ}{450} \\ &= \frac{540 + 222.2}{450} = 1.69 \end{aligned}$$

(b) Factor of safety with respect to cohesion

Here the full friction is mobilised ($F_\phi = 1$) and only the cohesion is reduced to the value needed for equilibrium:

$$\Sigma T = c_m L + (\Sigma N - \Sigma U)\tan\phi' \;\Rightarrow\; c_m = \frac{450 - 222.2}{27} = 8.44\ \text{kN/m}^2$$ $$F_c = \frac{c'}{c_m} = \frac{20}{8.44} = 2.37$$

Answer: (a) $F_s \approx 1.69$; (b) $F_c \approx 2.37$.