# Rankine’s Earth Pressure in Cohesive Soil for Active Case

Figure 1 shows the Mohr’s circle in which point B indicates the vertical stress and point E represents the active pressure. The circle is tangent to the failure envelope. For this case the relation between Pa and Ïƒv is given by:

$P_a=K_a\sigma_v-2c\sqrt{K_a}$———-(1)

$P_a=K_a\gamma z-2c\sqrt{K_a}$  ———-(2)

Where, $K_a=\dfrac{1-sin\phi}{1+sin\phi}=tan^245^0-\dfrac{\phi}{2}$

In the above equation (2) where z = 0

$P_a=-2c\sqrt{K_a}$

The negative sign indicates that the pressure is negative and tensile, As a result there would be gap between backfill and wall. The tensile stress decreases with the increasing value of z. When z = zc say Pa=0. Then,

$0=K_a\gamma z_c-2c\sqrt{K_a}$

Or, $0=\gamma z_c\sqrt{K_a}-2c$

$2c=\gamma z_c\sqrt{K_a}$

$Z_c=\dfrac{2c}{\gamma\sqrt{K_a}}$———-(3)

The depth zc is the depth of gap between backfill and wall, and is known as tension crack. The negative pressure eventually results in the formation of tension crack along the length of wall to a depth as defined by the equation. (3)

Pressure Distribution in cohesive soil for active case – The figure 2 given below shows the pressure distribution behind a wall retaining a cohesive backfill. The total resultant active earth pressure Pa is given by:

Pa = Êƒ (Ka Î³z – 2c√Ka)dz

———- (4.a)

———-(4.b)

For, É¸ = 0, the equation reduces (4.b) to:

———-(4.c)

For soils below the water table, the submerged unit weight is to be used.

Height of Unsupported Cut– Figure 2 shows that the pressure is negative in the top region. It becomes zero at depth zc. If the height of the Wall is 2Zc, the total earth pressure is zero and it is given by the relation:

———(5)

For, É¸ = 0, the equation reduces (5) to:

Hc = 4c/Î³  ———-(6)