Slope stability (cohesive slope) — factor of safety using Taylor stability number A deep cut of 7 m is to be excavated in a purely cohesive clay with unit weight γ = 16 kN/m^3 and cohesion c = 25 kN/m^2. The side slope is 30°. For a depth factor of 3, Taylor’s stability chart gives stability number Sn = 0.178. What is the factor of safety against slope failure?

Introduction / Context:Taylor’s stability charts provide a quick way to estimate the safety of homogeneous slopes by relating a non-dimensional stability number to slope geometry and soil strength. For purely cohesive soils (φ ≈ 0), the factor of safety (F) can be computed directly from the stability number when the slope height and unit weight are known.

Given Data / Assumptions:

• Slope height H = 7 m; slope angle = 30°.
• Soil properties: c = 25 kN/m^2, γ = 16 kN/m^3, φ = 0 (cohesive).
• Taylor’s chart gives Sn = 0.178 for the specified depth factor and slope angle.
• Undrained (total stress) analysis appropriate for short-term stability.

Concept / Approach:The stability number is defined for cohesive slopes as Sn = c / (γ * H * F). Rearranging gives F = c / (Sn * γ * H). Using the chart value of Sn ensures the geometric effect of the slope angle and depth factor is included.

Step-by-Step Solution:Compute denominator: Sn * γ * H = 0.178 * 16 * 7 = 0.178 * 112 = 19.936.Factor of safety: F = c / (Sn * γ * H) = 25 / 19.936 ≈ 1.254.Rounded to two decimal places, F ≈ 1.25.

Verification / Alternative check:F is modestly above unity, which is reasonable for a 30° slope in a low-to-moderate strength clay of the stated height.

Why Other Options Are Wrong:0.80 and 1.0: Too low; would imply near or actual failure with the given parameters.
1.1: Underestimates the computed F.
1.40: Overestimates safety for the stated Sn.

Common Pitfalls:

• Using γ_submerged instead of γ when not warranted by groundwater conditions.
• Mistyping Sn from the chart or mixing up units for c and γ.

Final Answer:1.25