Question

In a clayey soil having $100 \text{ kN/m}^2$ as unit cohesion and $15 \text{ kN/m}^3$ as unit weight, an excavation is made with a vertical face. Taking Taylor's stability number as 0.3, what is the maximum depth of excavation so that the vertical face remains stable?

• A. $22.22 \text{ m}$
• B. $7.41 \text{ m}$
• C. $11.11 \text{ m}$
• D. $44.44 \text{ m}$

Hint:

Taylor's stability number is dimensionless. Always ensure your units for $C$ ($\text{kN/m}^2$) and $\gamma \times H$ ($\text{kN/m}^3 \times \text{m}$) match!

Answer 1

The Correct Option is A

Concept: Taylor's stability number ($S_n$) relates the cohesion, unit weight, and critical height ($H_c$) of a slope. For a stable excavation, the mobilized cohesion must not exceed the available unit cohesion.

Step 1: Identify the formula for Stability Number.
The stability number $S_n$ is given by: \[ S_n = \frac{C}{\gamma H_c} \] Where:
• $C = 100 \text{ kN/m}^2$ (unit cohesion)
• $\gamma = 15 \text{ kN/m}^3$ (unit weight)
• $S_n = 0.3$ (stability number)
• $H_c = \text{Critical/Maximum depth}$

Step 2: Solve for $H_c$.
Rearranging the formula: \[ H_c = \frac{C}{\gamma \times S_n} \] \[ H_c = \frac{100}{15 \times 0.3} \] \[ H_c = \frac{100}{4.5} \] \[ H_c = 22.22 \text{ m} \]