Effect of Sloping Surcharge in Active Case

What are the Effect of Sloping Surcharge in Active Case?

in this article we will discuss about Effect of Sloping Surcharge in Active Case. A wall carrying a sloping backfill is shown in Fig.1. Consider an element subjected to a stress \sigma_v in the vertical direction and \sigma_x in the direction parallel to the backfill. As these stresses in one plane are parallel to the direction of another plane, these stresses are conjugate stresses and the planes are the conjugate planes. The equivalent vertical stress acting parallel to the surface of backfill is given by:

\sigma_{ve} = \sigma_v cosi  ———- (1)

\sigma_{ve} = \gamma z cosi

The Mohr Circle, which represents this case of inclined backfill is shown in the Fig 1.

Active earth pressure slip surface in sloping backfill

Referring to the figure above (figure 1), the coefficient of active earth pressure can be written as:

K_a=\dfrac{\sigma_x}{\sigma_{ve}}=\dfrac{P_a}{yz cosi}

=\dfrac{OB}{OA}=\dfrac{OP}{OA}

=\dfrac{OD-PD}{OD+AD}

=\dfrac{OD-PD}{OD+PD}

 We can also write:

CD = OC sini

OD = OC cosi

active earth pressure mohr circle in sloping backfill

 

PD=\sqrt{PC^2-CD^2}

PD=\sqrt{R^2-CD^2}  Because PC = R

 = OC sin\phi

By substitution these values in the expression for Ka we get;

K_a=\dfrac{OCcosi-\sqrt{OC^2sin^2\phi-OC^2sin^2i}}{OCcosi+\sqrt{OC^2sin^2\phi-OC^2sin^2i}}

K_a=\dfrac{cosi-\sqrt{sin^2\phi-sin^2i}}{cosi+\sqrt{sin^2\phi-sin^2i}}=\dfrac{cosi-\sqrt{cos^2i-cos^2\phi}}{cosi+\sqrt{cos^2i-cos^2\phi}}

Therefore, P_a = K_a\gamma z cosi = K_A\gamma z

Where,

K_A = K_acosi = K_acosi[\dfrac{cosi-\sqrt{cos^2i-cos^2\phi}}{cosi+\sqrt{cos^2i-cos^2\phi}}] and is known as the coefficient of active earth pressure.

If, \alpha_1 and \alpha_2 are the inclination of failure planes with horizontal as shown in figure 3

Active earth pressure slip surface in sloping backfill

\theta_{fa} = \alpha_1 + \theta_1

\alpha_1 = \theta_{f1}-\theta

\psi = i+2\theta_1

\theta_1=\dfrac{(\psi-i)}{2}

But, \theta_{f1}=45^0+\dfrac{\phi}{2}

Hence, \alpha_1=45^0+\dfrac{\phi}{2}-\dfrac{\psi-i}{2}

And Similarly,  \alpha_2=45^0+\dfrac{\phi}{2}+\dfrac{\psi-i}{2}

Where, sin\psi=\dfrac{CD}{CP}=\dfrac{CD}{CR}=\dfrac{OCsini}{OCsin\phi}CD normal to OA

and \psi=sin^{-1}\dfrac{sini}{sin\phi}

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