1.

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. \(K_a=\frac{1}{cot^2 (45°+\frac{φ}{2})} \)
Answer» B. \(K_a=tan^2 (45°+\frac{φ}{2}) \)


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