Slope Stability – Critical Height from Stability Number for Purely Cohesive Clay For a clay slope (cohesive soil) with cohesion C = 2.5 t/m^2, unit weight γ = 2.0 t/m^3, and stability number Sn = 0.05, compute the critical height Hc of the slope at limiting equilibrium.

Introduction / Context:
For purely cohesive slopes (phi ≈ 0), charts or formulas express stability in terms of a stability number Sn which relates soil strength and slope height. Designers often back-calculate the maximum (critical) height that can stand without failure for given C and γ.

Given Data / Assumptions:

• Cohesion C = 2.5 t/m^2.
• Unit weight γ = 2.0 t/m^3.
• Stability number Sn = C / (γ * Hc) = 0.05.
• Phi ≈ 0 assumption implicit in Sn usage.

Concept / Approach:

For cohesive slopes at limiting equilibrium, Sn = C / (γ * Hc). Rearranging gives Hc = C / (Sn * γ). This provides a quick estimate from stability charts without iteration when Sn is specified.

Step-by-Step Solution:

Verification / Alternative check:

The order of magnitude is reasonable: with γ = 2 t/m^3 and C = 2.5 t/m^2, a few tens of metres is expected for stability numbers around 0.05.

Why Other Options Are Wrong:

4.0 m, 12.5 m, and 15.0 m result from arithmetic mistakes (e.g., using Sn as 0.5 or misplacing decimals). They do not satisfy Sn = C / (γ * H).

Common Pitfalls:

Confusing units (t/m^2 vs t/m^3); using actual slope height instead of critical height; mixing up φ = 0 assumptions.

Final Answer:

25.0 m