Figure 1 shows the Mohr’s circle in which point B indicates the vertical stress and point E represents the passive pressure. The wall carries a uniform surcharge q. The circle is tangent to the failure envelope. For this case the relation between Pa and Ïƒv is given by:
Pp = KpÏƒv + qKp + 2c√Kp ———-(1)
Where, Ka = (1 + sinÉ¸)/(1 – sinÉ¸) = tan2[450 + (É¸/2)]
Pressure Distribution of passive case in cohesive soil – Figure 2 given below shows the pressure distribution behind a wall retaining a cohesive backfill. The total resultant passive earth pressure Pp is obtained by integrating Equation 1. Thus:
For, É¸ = 0, the equation (2) reduces to:
For soils below the Water table, the submerged unit weight is to be used.