Infinite Slope Safety with Cohesion

Equation: FS = [c′ + (γz·cos²β − u)·tanφ′] / (γz·sinβ·cosβ). The top is the soil's shear strength on the failure plane and the bottom is the shear stress driving the slide. Values: c′ = 10 kPa, γ = 18 kN/m³, φ′ = 30° and u = 0 are given. z = 3 m is the vertical depth from the surface to the failure plane. β = 20° is the slope angle, and the failure plane has the same angle because it is parallel to the surface. Step 1: vertical stress at the plane, γz = 18 × 3 = 54 kPa. cos20° = 0.940, so cos²20° = 0.883, and sin20° = 0.342. Step 2: effective normal stress on the plane, σ′ = γz·cos²β − u = 54 × 0.883 − 0 = 47.7 kPa. Step 3: shear strength, τf = c′ + σ′·tanφ′ = 10 + 47.7 × 0.577 = 37.5 kPa. Step 4: driving shear stress, τ = γz·sinβ·cosβ = 54 × 0.342 × 0.940 = 17.4 kPa. Step 5: FS = 37.5 / 17.4 = 2.16. Leaving out the cohesion c′ gives 27.5 / 17.4 = 1.58, and using cosβ instead of cos²β overstates the normal stress.