Infinite slope FoS, dry cohesive soil

Assumptions: infinite slope of dry homogeneous $c$-$\phi$ soil, inclined at angle $\beta$ to the horizontal; failure plane parallel to the slope at vertical depth $z$; no seepage; unit weight $\gamma$.

Consider a vertical slice of horizontal width $b$, so the length of its base on the failure plane is $l = b/\cos\beta$. The forces on the two sides of the slice are equal and opposite, so they cancel.

        ______ ground surface (beta)
       /|     \
      / | W    \
     /__|_____
        z   N  T  <- failure plane

Weight of slice: $W = \gamma z b$ per unit length.

$$N = W\cos\beta, \qquad T = W\sin\beta$$

Normal stress and shear stress on the plane:

$$\sigma = \frac{N}{l} = \gamma z\cos^2\beta, \qquad \tau = \frac{T}{l} = \gamma z\sin\beta\cos\beta$$

Shear strength (Mohr-Coulomb): $\tau_f = c + \sigma\tan\phi = c + \gamma z\cos^2\beta\tan\phi$

Factor of safety $F = \tau_f/\tau$:

$$\boxed{F = \frac{c + \gamma z\cos^2\beta\tan\phi}{\gamma z\sin\beta\cos\beta} = \frac{c}{\gamma z\sin\beta\cos\beta} + \frac{\tan\phi}{\tan\beta}}$$

For cohesionless soil ($c=0$), $F = \tan\phi/\tan\beta$, independent of $z$. For a given $c$, the critical (limiting) depth for $F=1$ is $z_c = \dfrac{c}{\gamma\cos^2\beta(\tan\beta - \tan\phi)}$.