Column A is $400$ mm square and column B is $500mm$ square in size end they are placed $4m c/c.$ the property line is $600mm$ beyond the face of column A SBC of soil of soil $200 kN/m^2.$ Use $M20/Fe415.$

Assume $b_f=1.5m$

Area of footing $=\dfrac {(P_A+P_B )+10% \text {Extra as self wt}}{SBC}\\ =\dfrac {1800+\frac{10}{100}+1800}{200}=\dfrac {1980}{200}=9.9 m^2 \\ Now, A_f= L_f×B_f\\ 9.9= L_f×1.5 \\ L_f=6.6 \\ \text {Find C.G. of load Take moment @ ‘A’ } \\ \overline x=\dfrac {1000×4+800×0}{1000+800}=2.2 m \\ \text { Factored upward soil pressure }(w_d) \\ w_d=\dfrac {1.5(1800)}{6.6 ×1.5}=272.7 kN/m^2 \\ \text { Upward soil pressure @ C_Lof longitudinal axis }\\ w_d×B_f=272.7×1.5=409.05 kN/m $

$$ B.M @ a =409.05 ×\dfrac {1.1^2}2=247.47 kNm $$

$B.M @ b =409.05 ×\dfrac {1.5^2}2=460.18 kNm \\ \dfrac x{750.04}=\dfrac {4-x}{886.15}\\ x= 1.83 m \\ B.M_{max}=[409.05 ×(1.1+1.83)1902 kNm \\ \text { Two way shear: (Check for big column) }\\ M_{u\space max}=0.138 f_{ck} bd^2 \\460.18×10^6=0.138×20×1500×d^2 \\ dr=333.39 mm \\ Provide d=1000mm $

$$\text {Net shear force }= P_u-[(\text { Area of critical section } ×w_d )] $$

$=1500-[(1.5×1.5)×272.7]=886.42 kN \\ \text { Area resisting = Perimeter of critical } ×d \\ =[2×(1.5+1.5)]×1=6m^2 \\ Z_v=\dfrac {886.42×10^3}{6×10^6 }=0.147N/mm^2 \\ Z_{cpermi}=Z_c'×k_s \\ Z_c'=0.25\sqrt{f_{ck} }= 0.25\sqrt{20}=1.12 N/mm^2 \\ k_s=0.5+\dfrac {0.5}{0.5}=1.5 \gt 1 \\ k_s=1 \\ Z_{cpermi}=1.12×1=1.12N/mm^2 \gt 0.147 \space\space\space \text {Therefore Safe }\\ \text { Design Reinforcement } \\ \text { Column A }\\ \text {Bandwidth } = (b+2d)= 400+2×1000=2400mm $

$$M_u=(2.4×0.55×272.7)×\dfrac {0.55}2$$

$M_u=98.99 kNm \\ Ast =\dfrac {0.5×20×2400×1000}{415} [1-\sqrt{1-\dfrac {4.6×98.99×10^6}{20×2400×1000^2 }} \\ Ast=274.96 mm^2 \\ Ast_{min}=\dfrac {0.12}{100}×bD=\dfrac {0.12}{100}×2400×1100=3168 mm^2 \\ \text {Provide Ast } =3168 mm^2 \\ \text {Provide 16 no. of 16mm∅ bars }\\ Astp = 3217 mm^2 \\ Column \space B \\ \text {Bandwidth } = (b+2d)=500+2×1000=2500 mm$

$$M_u=(2.5×0.5×272.7)×\dfrac {0.5}2$$

$M_u=85.21 kNm \\ Ast =\dfrac {0.5×20×2500×1000}{415} [1-\sqrt{1-\dfrac {4.6×85.21×10^6}{20×2500×1000^2 }}]\\ Ast=236.58 mm^2 \\ Ast_{min}=\dfrac {0.12}{100}×bD=\dfrac {0.12}{100}×2500×1100=3300 mm^2\\ \text {Provide 17 no. of 16mm∅ bars }\\ Astp = 3418mm^2$