Question

The coefficient of earth pressure for passive state of plastic equilibrium is given by ________

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

(b) \(K_p=tan^2 (45°+\frac{φ}{2})\)

(c) \(K_p=sin^2 (45°+\frac{φ}{2})\)

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

I got this question in an international level competition.

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

Answer 1

Correct option is (b) \(K_p=tan^2 (45°+\frac{φ}{2})\)

The best I can explain: For the case of passive state, the major principal stress denoted by σ1 is in horizontal direction and the minor principal stress σ3  is in vertical direction.

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