An excavation is to be made in a saturated soft clay (φᵤ = 0) with vertical sides. What will be the maximum unsupported depth of the vertical cut? Given that the cohesion intercept = 30 kN/m² and the unit weight of clay = 15 kN/m³.

Asked in SSC JE 30/10/2020 (Morning)

Detailed Explanation

In a clay with cohesion, the active pressure near the surface is negative (tension). Using Bell's equation for φ = 0 (K_a = 1):

\[ p_a = \gamma z - 2c \]

The tension zone extends to z_c = 2c/γ. Over a depth of twice that, the negative and positive parts of the pressure diagram cancel, so the net thrust is zero and the cut can stand without support (in theory).

\[ H_c = \frac{4c}{\gamma\sqrt{K_a}} = \frac{4c}{\gamma} = \frac{4 \times 30}{15} = 8\ \text{m} \]

Why “8 m” Is Correct

With c = 30 kN/m² and γ = 15 kN/m³, the theoretical maximum unsupported height is 8 m. (Tension cracks reduce this in practice; a common safe estimate uses 2.67c/γ.)

Unsupported vertical cut in clay (φ = 0). Near the top the clay is in tension down to z_c = 2c/γ, where tension cracks open. The net active force over the full depth becomes zero at H_c = 2z_c = 4c/γ, the theoretical maximum height the cut can stand without support.

Why the Other Options Are Wrong

A. 4 m: 4 m is the depth of the tension zone, 2c/γ, not the critical height.
B. 5 m: 5 m does not follow from 4c/γ.
D. 6 m: 6 m does not follow from 4c/γ with these values.

Key Point

Tension crack depth zc = 2c/γ; critical unsupported height Hc = 4c/γ (φ = 0).

Key Concepts for Students

With φ > 0: Hc = 4c/(γ√Ka) = (4c/γ)·tan(45° + φ/2).
Water filling tension cracks adds pressure and reduces stability.
Taylor's stability number gives Hc = c/(γ·Sn) for slopes.

Practice more: this is question 11 in Earth Pressure and Retaining Wall MCQs – Part 1. Try the whole set and check every answer instantly.