Distance between centre to centre between columns is $2.3$. Assume width of footing as $1.5m$ and S.B.C of soil as $150 kN/m^2$. Also draw a neat sketch showing reinforcement details.

Area of footing $= \dfrac {(P_A+P_B )+10% \text { extra as self wt.}}{\text {SBC}}\\ =\dfrac{(850+1150)+\frac {10}{100}×(850+1150)}{150 } \\ A_f=14.67 m^2 \\ Now A_f=L_f×B_f \\ 14.67= L_f×1.5 \\ L_f=9.78 \\ \text { Find C.G. of load Take moment of loads @ ‘A’ }\\ x=\dfrac {[(1150×2.3)+(850×0)]}{(1150+850)}=1.325 m \\ \text { Factored upward …