Infinite slope seepage, max angle for FoS 2

Data: $c' = 0$, $e = 0.7$, $G = 2.7$, $\phi = 35^\circ$, required $F = 2.0$, seepage parallel to the slope with water at the surface.

Step 1: Unit weights

$$\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{3.4(9.81)}{1.7} = 19.62\ \text{kN/m}^3, \qquad \gamma' = 19.62 - 9.81 = 9.81\ \text{kN/m}^3$$ $$\frac{\gamma'}{\gamma_{sat}} = \frac{G-1}{G+e} = \frac{1.7}{3.4} = 0.5$$

Step 2: Maximum slope angle for $F=2$

$$F = \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi}{\tan\beta} \;\Rightarrow\; 2.0 = 0.5\,\frac{\tan35^\circ}{\tan\beta}$$ $$\tan\beta = \frac{0.5\times0.7002}{2.0} = 0.1750 \;\Rightarrow\; \beta = 9.93^\circ \approx 9.9^\circ$$

Step 3: Factor of safety if the water level falls well below the surface (dry or moist soil, no seepage force):

$$F = \frac{\tan\phi}{\tan\beta} = \frac{0.7002}{0.1750} = 4.0$$

Answer: maximum slope angle $\approx 9.9^\circ$; factor of safety rises to 4.0 when the water level drops well below the surface.