Effect of Sloping Surcharge in Passive Case

What are Effect of Sloping Surcharge in Passive Case?

In previous article we had already posted Effect of Sloping Surcharge in Active Case, now in this article we are dealing with Effect of Sloping Surcharge in Passive Case. The Mohr Circle for the passive case of a retaining wall carrying a sloping backfill is shown in Fig.1. Consider an element subjected to a stress in the \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.

passive earth pressure mohr circle in sloping backfill

The equivalent vertical stress-acting acting parallel to the surface of backfill is given by

\sigma_{ve} = \sigma_vcosi = \gamma z cosi

Referring to the figure the coefficient of passive earth pressure can be written as:

K_p=\dfrac{\sigma_x}{\sigma_{ve}}=\dfrac{\sigma_p}{\sigma_vcosi}

 = \dfrac{OP}{OA}

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

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

We can also write:

    CD = OC sini

    OD = OC cosi

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

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

R = OC sin\phi

By substitution these values in the expression for Kp we get:

K_p=\dfrac{OCcosi+\sqrt{OC^2sin^2\phi-OC^2sin^2i}}{OCcosi-\sqrt{OC^2sin^2\phi-OC^2sin^2i}} K_p=\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}} P_P = K_p\gamma z cosi

K_P = K_P cosi ———–(1)

Where,

K_p=cosi\left[\dfrac{cosi+\sqrt{cos^2i-cos^2\phi}}{cosi-\sqrt{cos^2i-cos^2\phi}}\right]

And is known as the coefficient of passive earth pressure.

If, \alpha_1 and \alpha_2 are the inclination of failure planes with horizontal as shown in figure 2 then,

Passive Earth Pressure Slip Surface in sloping Backfill

\theta_{f2} = \alpha_1-\theta_2 \alpha_1 = \theta_{f2} + \theta_2 \psi = -i+2\theta_2 \theta_2=\dfrac{(\psi-i)}{2}

But, \theta_{f2}=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}{CF}=\dfrac{OCsini}{OCsin\phi}CD normal to OA

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

The pressure diagram is shown in figure 3 for both active as well as passive cases.

Pressure Diagram in Active as Well as Passive Case in Sloping Backfill

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