The load is assumed to be distributed in a ratio of 1H:4V on the equivalent mat at a distance of (2/3)rd of the pile length ie (2/3*9 =) 6m. Ref. fig.5.46.

Width of equivalent map = 3 + 2 * (6/4) = 6m

Length of equivalent map = 2.1 + 2 * (6/4) = 5.1m

Udl acting on the equivalent map = 150/(6 * 5.1) = 4.902

Below the equivalent map it is analyzed as shallow foundation, by dividing it into no. of layers as shown in fig. 5.47.

The Effective overburden pressure at center of each layer:

= (16.4 x 2) + (2 x 19.1) + (5.5 x (19.1-9.81)) = 122.095 KN /

= 122.095 + (3 x (19.1-9.81)) = 149.965 KN /

= 149.965 + (1.5 x (19.1-9.81)) + (1.25 x (20-9.81)) = 176.64 KN /

= 176.64 + (1.25 x (20-9.81)) = 189.375 KN /