Question

The coefficient of earth pressure for active state of plastic equilibrium in terms of sine function is _______

(a) \(K_a=1-sinφ\)

(b) \(K_a=\frac{1-sinφ}{1+sinφ} \)

(c) \(K_a=\frac{1+sinφ}{1-sinφ} \)

(d) Ka=1+sinφ

The question was asked in examination.

This key question is from Earth Pressure Introduction in chapter Failure Envelopes & Earth Pressure of Geotechnical Engineering II

Answer 1

The correct answer is (b) \(K_a=\frac{1-sinφ}{1+sinφ} \)

To explain I would say: The coefficient of earth pressure for active state of plastic equilibrium is,

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

∴ therefore in terms of sine function,

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