Mohr Circle | Mohr Circle For Passive Earth Pressure

In this article we will discuss about mohr circle for passive earth pressure, figure 1 below shows the state of stresses on the base and the side of the prism CA at depth Z in a Mohr circle. The vertical stress $\sigma_v$ is the principal stress. OP and OQ are the two Mohr envelopes satisfying the Coulomb’s equation of shear strength. Initially the stresses $\sigma_v$ and $\sigma_h$ are the major and minor principal stress.

The points A and B in mohr circle for passive earth pressure diagram respectively denote these stresses at rest condition. The Circle 1 indicates the in-situ condition, point A indicates the horizontal stress while the point B indicates the vertical stress. Assume the vertical stress is held constant and the horizontal stress is now increased. By doing this, the point A of the Mohr circle is shifted to position A’ toward right as shown in Fig.1.

The diameter of the Mohr circle is decreased in the initial phase will be zero when it equals to the vertical stress and then is increased continuously with the increase in the horizontal stress. If the process of increasing the horizontal stress is continued, a limiting condition is reached where the Mohr circle would be tangential to the Mohr envelope at a certain diameter corresponding to the point A” as shown in the Fig.1. At this condition the soil mass is at the verge of failure. After achieving this condition, it would not be possible to increase the magnitude of the horizontal stress because the soil would have been failed already before attempting to do so. This limiting condition of failure is called the Rankine’s Passive State of Plastic Equilibrium and the magnitude of horizontal stress $\sigma_h$ at this condition is known the Passive Earth Pressure and denoted by the symbol P_p

Figure 2 below shows the stress condition in a soil mass in passive state. From the figure,

$P_P = OC + CE$

$= OC + CD$

$= OC + OC sin\phi$

$= OC(1+sin\phi)$ ————————-(1)

Also,

$\sigma_v = OB = OC - BC$

$= OC-CD$

$= OC-OC sin\phi$

Therefore, $\sigma_v = OC(1-sin\phi)$ ————————–(2)

From the above equation (1) and (2) we can have,

$\dfrac{P_p}{\sigma_v}=\dfrac{1+sin\phi}{1-sin\phi}$

Or, $P_P=\left(\dfrac{1+sin\phi}{1-sin\phi}\right)\sigma_v$

$= K_p \times \sigma_v = K_p \times \gamma z$ ——————————(3)

Where, $K_p= \dfrac{(1+sin\phi)}{1-sin\phi} = tan^2\left[45^0+\dfrac{\phi}{2}\right]$ and is known as the coefficient of passive earth pressure. As in the case of water the distribution of pressure in case of soils is also triangular. In the Mohr circle shown above the line ED corresponds to the failure plane and is inclined at an angle of $\left[45^0+ \dfrac{\phi}{2}\right]$ with the direction of major principal plane which is a vertical plane in this case. The failure planes or the slip lines are shown in Fig.3 below. These lines make angle of $\left[45^0- \dfrac{\phi}{2}\right]$ to the minor principal plane, which is horizontal in the passive condition.

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