Mohr Circle | Mohr Circle For Active Earth Pressure

IN this article we will discuss about Mohr Circle for active 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 stresses. The points A and B in the Mohr Circle for active earth pressure diagram respectively denote these stresses at rest condition. Assume the vertical stress is held constant and the horizontal stress is decreased. By doing this the point A of the Mohr circle is shifted to position A’ and the diameter of the Mohr circle is increased.

If the process of decreasing 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 decrease 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 Active State of Plastic Equilibrium and the magnitude of horizontal stress OA” $(\sigma_h)$ at this condition is known the Active Earth Pressure and denoted by the symbol P_a.

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

$P_a = OE = OC-CE$

As $CE = CD = CB = OC sin\phi$

$P_a = OC(1-sin\phi)$ ——————–(3)

Also, $\sigma_v = OB = OC+CB = OC(1+sin\phi)$ ——————-(3.a)

From the equation (3) and (3.a) we have,

$\dfrac{P_a}{\sigma_v}=\dfrac{1-sin\phi}{1+sin\phi}$ ——————————-(3.b)

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

$P_a = K_a\sigma_v = K_a\gamma z$ —————————————-(3.c)

Where, $K_a= \dfrac{(1-sin\phi)}{1+sin\phi} = tan^2\left[45^0-\dfrac{\phi}{2}\right]$

and is known as the coefficient of active 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 $\left[45^0+ \dfrac{\phi}{2}\right]$ direction of major principal plane. The failure planes or the slip lines are shown in Fig.3 below. These lines make an angle of $\left[45^0+ \dfrac{\phi}{2}\right]$ to the major principal plane which is horizontal in this case.

One Thought to “Mohr Circle | Mohr Circle For Active Earth Pressure”

James

February 18, 2016 at 9:19 pm

Quick question regarding the real failure angle shown in figure 3 – isn’t every angle doubled in Mohr’s circle? Wouldn’t that mean that the actual angle in the soil element should be (45 +phi/2) *0.5?

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