Question

An excavation is to be made in a saturated soft clay (ϕ_u = 0) with vertical sides. What will be the maximum unsupported depth of the vertical cut?

Given that cohesion intercept = 30 kN/m², unit weight of clay = 15 kN/m³

1. 5 m
2. 8 m
3. 6 m
4. 4 m

Answer 1

Correct Answer - Option 2 : 8 m

Concept:

Critical depth/Unsupported vertical cut (Hc): It is the depth up to which vertical excavation of cohesive soil can stand without any lateral support.

\({{\rm{H}}_{\rm{c}}} = \frac{{4 \times c}}{{\gamma \times \sqrt {{K_a}} }}\)

\({{\rm{k}}_{\rm{a}}} = \frac{{1 - {\rm{sin}}\phi }}{{1 + {\rm{sin}}\phi }}\)

Where

ka = Rankine’s coefficient of active earth pressure

ϕ = Angle of internal friction

γ = Density of soil

c = cohesion

Calculation:

Given, ϕu = 0, c = 30 kN/m2, γ = 15 kN/m3

\({{\rm{k}}_{\rm{a}}} = \frac{{1 - \sin 0^\circ }}{{1 + \sin 0^\circ }} = 1\)

\({{\rm{H}}_{\rm{c}}} = \frac{{4 \times 30}}{{15 \times \sqrt {1} }} = 8{\rm{\;m}}\)