Infinite Slope with Pore Pressure

Equation: FS = [c′ + (γz·cos²β − u)·tanφ′] / (γz·sinβ·cosβ). The top is the shear strength on the failure plane; the bottom is the driving shear stress. Values: c′ = 10 kPa, γ = 18 kN/m³, φ′ = 30° and u = 15 kPa are given. z = 3 m is the vertical depth to the failure plane. β = 20° is the slope angle, and the plane is parallel to the surface. Step 1: γz = 18 × 3 = 54 kPa; cos²20° = 0.883; sin20°·cos20° = 0.342 × 0.940 = 0.321. Step 2: effective normal stress, σ′ = 54 × 0.883 − 15 = 47.7 − 15 = 32.7 kPa. Step 3: shear strength, τf = 10 + 32.7 × tan30° = 10 + 32.7 × 0.577 = 28.9 kPa. Step 4: driving stress, τ = 54 × 0.321 = 17.4 kPa. Step 5: FS = 28.9 / 17.4 = 1.66. Pore pressure lowers the effective normal stress and so the friction. Ignoring u gives 2.16, the dry-slope answer.