Slope stability using Taylor’s charts: A 40° slope is excavated to a depth of 10 m in saturated clay (γ = 20 kN/m^3, cu = 70 kN/m^2, φu = 0). Given Taylor’s stability coefficient S = 0.18 for φu = 0 at θ = 40°, estimate the factor of safety.

Introduction / Context:Taylor’s stability charts provide a rapid estimate of slope safety in purely cohesive (φu = 0) soils by linking soil strength, slope geometry, and unit weight through the stability number. When the stability coefficient S for a given slope angle is known, the factor of safety (F) can be obtained without a full slip-circle search.

Given Data / Assumptions:

• Slope angle θ = 40°, depth H = 10 m.
• Saturated undrained clay: cu = 70 kN/m^2, φu = 0, γ = 20 kN/m^3.
• Taylor coefficient S = 0.18 for φu = 0, θ = 40°.

Concept / Approach:

For φu = 0, the stability number is defined as S = cu / (γ H) at F = 1. For a general factor of safety F, the relationship extends to cu / (γ H) = S * F. Therefore, F = (cu / (γ H)) / S.

Step-by-Step Solution:

Verification / Alternative check:

Comparing with detailed charts or a slip surface analysis yields a factor of safety close to 2 for the chosen parameters, validating the estimate.

Why Other Options Are Wrong:

• 2.1–2.3: Slightly higher than the computed 1.94; 2.0 is the nearest correct option.
• 1.8: Too low relative to the computed ratio.

Common Pitfalls:

• Confusing S definitions; ensure S corresponds to φu = 0 charts.
• Using total stress friction angle instead of undrained parameters.

Final Answer:

2.0