The basic function of this is to distribute the concentrated load from the column over a longer area initially the column load is distributed over a base plate & then to supporting concrete & finally to the sail below such that the bearing capacity of soil is not exceeded .

The size of the steel plate is selected such that its projection beyond the column perifery.

Design procedure are of base plate

Step 1 Cal required are of base plate

a) Ap=$\frac{column \ load}{Bearing \ strength \ of \ concrete}$

=$\frac{Pu}{0.6\ast fck}$ bearing strength of concrete=0.6+fck

b)length of plate(Lp)=$(\frac{Lc-Bc}{2})+\sqrt{(\frac{Lc-Bc)^{2}}{2}+Ap}$

c) width of plate Bp=$\frac{Ap}{Lp}$

Lc = length of column or longer side of column

Bc=width of column or shorter side

d) thickness of plate(ts)=$\sqrt{2.5w(a^{2}-0.3b^{2})\ast \frac{ymo}{fy}}\lt fy$

w=$\frac{Pu}{Ap}$

a=longer projection

b=smaller projection

w=intensity of px between sail & plate $\lt 0.6\ast fck$

Step II concrete block design

a) Area of footing (Af)=$\frac{10% column \ load}{ultimate \ soil \ bearing \ capacity}$

b) length of footing(Lf)=$(\frac{Lp+Bp}{2})=\sqrt{(\frac{Lp+Bp}{2})^{2}+Af}$

c) width of footing=$\frac{Af}{Lf}$

d) Depth of footing(De)=larger projection of footing by providing 1.1. slope