The infinite saturated sand slope shown is at impending sliding failure. Its angle is 20°, vertical layer thickness is 5 m and saturated unit weight is 21 kN/m³. The water table coincides with the ground surface and $\gamma_w=9.81$ kN/m³. Find the effective friction angle in degrees, to one decimal place.
For a cohesionless infinite slope with seepage parallel to the surface, $F=[(\gamma_{sat}-\gamma_w)/\gamma_{sat}](\tan\phi\prime/\tan\beta)$. Set $F=1$ for impending failure. Then $\tan\phi\prime=[21/(21-9.81)]\tan20^\circ=0.6830$. Hence $\phi\prime=34.33^\circ$, rounding to 34.3°. Layer thickness cancels from the force ratio.
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