If the angle of shearing resistance is 60° then the product of active earth pressure

If the angle of shearing resistance is 60°, then the product of active earth pressure coefficient and passive earth pressure coefficient will be

A) 0

B) 1/3

C) 1

D) 3

Coefficient of Active Earth Pressure Coefficient :-

Ka=1−sin(ϕ)1+sin(ϕ)=tan2(45−ϕ2)$K_a=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}={\tan }^2\left(45-\frac{\varphi }{2}\right)$

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Coefficient of Passive Earth Pressure Coefficient :-

Kp=1+sin(ϕ)1−sin(ϕ)=tan2(45+ϕ2)$K_p=\frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}={\tan }^2\left(45+\frac{\varphi }{2}\right)$

Calculations:

Point to be remembered:

For any value of shearing angle, the product of active earth pressure coefficient and passive earth pressure coefficient will always equal to 1.

Ka×Kp=1−sin(ϕ)1+sin(ϕ)×1+sin(ϕ)1−sin(ϕ)=1$K_a\times K_p=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}\times \frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}=1$

Alternate Solution:

For ϕ = 60°

Ka=tan2(45−602)=0.0718$K_a={\tan }^2\left(45-\frac{60}{2}\right)=0.0718$

Ka=tan2(45+602)=13.928$K_a={\tan }^2\left(45+\frac{60}{2}\right)=13.928$

Therefore,

Ka × Kp = 0.0718 × 13.928 = 1