Infinite slope, cohesionless, steady seepage

Conditions: infinite slope of cohesionless soil ($c=0$), slope angle $\beta$, steady seepage parallel to the slope with the water table at the ground surface. Saturated unit weight $\gamma_{sat}$, submerged unit weight $\gamma'$.

Consider a slice of depth $z$ and width $b$. Seepage is parallel to the slope, so the equipotential lines are perpendicular to the slope and the pore pressure on the plane is

$$u = \gamma_w z\cos^2\beta$$

Total weight of the slice: $W = \gamma_{sat} z b$.

$$\sigma = \gamma_{sat} z\cos^2\beta, \qquad \tau = \gamma_{sat} z\sin\beta\cos\beta$$

Effective normal stress:

$$\sigma' = \sigma - u = (\gamma_{sat} - \gamma_w) z\cos^2\beta = \gamma' z\cos^2\beta$$

Shear strength $= \sigma'\tan\phi'$, so the factor of safety is

$$F = \frac{\gamma' z\cos^2\beta\tan\phi'}{\gamma_{sat} z\sin\beta\cos\beta} = \boxed{\frac{\gamma'}{\gamma_{sat}}\,\frac{\tan\phi'}{\tan\beta}}$$

Remarks:

• As $\gamma'/\gamma_{sat}\approx 0.5$, the factor of safety with seepage is about half the dry-slope value $\tan\phi'/\tan\beta$.
• The limiting slope angle with seepage ($F = 1$) is therefore given by $\tan\beta \approx 0.5\tan\phi'$, which is roughly half the friction angle.
• If the water table is at height $m z$ above the plane (with $0<m<1$), and the same unit weight $\gamma_{sat}$ is taken above it, $F = \left(1 - \dfrac{m\gamma_w}{\gamma_{sat}}\right)\dfrac{\tan\phi'}{\tan\beta}$.