Question

The coefficient of earth pressure for active state of plastic equilibrium is given by ___________

(a) \(K_a=\frac{1}{tan^2 (45°+\frac{φ}{2})} \)

(b) \(K_a=tan^2 (45°+\frac{φ}{2}) \)

(c) \(K_a=sin^2 (45°+\frac{φ}{2}) \)

(d) \(K_a=\frac{1}{cot^2 (45°+\frac{φ}{2})} \)

The question was asked in examination.

My question comes from Earth Pressure Introduction in chapter Failure Envelopes & Earth Pressure of Geotechnical Engineering II

Answer 1

The correct option is (a) \(K_a=\frac{1}{tan^2 (45°+\frac{φ}{2})} \)

To explain: For the case of active state, the major principal stress denoted by σ1  is in vertical direction and the minor principal stress σ3  is in horizontal direction.

∴ \(\frac{σ_1}{σ_3} =\frac{σ_v}{σ_h} =K_a=\frac{1}{tan^2 (45°+\frac{φ}{2})}. \)