The boundary line of the property is $500$ mm from the outer face of the column A. Column A and B is $400 mm \times 400 mm$ size SBCbof soil is $150KN/m^2$ .Use $M20/F_{e415} .$

Assume width of foundation as $1.5m$

Area of footing $=\dfrac {(P_A+P_B)+10 \% \text { extras self wt.}}{SBC}\\ =\dfrac {(1200+1400) +\frac {108}{100}\times 2500}{150}\\ A_f=19.06 m^2\\ Now, A_f=L_f \times B_f\\ 19.06= L_f =12.71 m$

Find C.G.of load, Take moment of load @ A

$$\overline x=\dfrac {1400\times 4 + 1200 \times 0}{2600}=2.15$$

Factored upword soil pressure (wd):

$$wd=\dfrac {1.5\times [1200+1400]}{12.71\times 1.5}=204.56 KN/m^2$$

Upword soil pressure @ Cl of longitudinal axis

$$w_d\times B_f=204.56\times 1.5=306.84 KN/m$$

$$B.M @_a =306.84\times \dfrac {4.2^2}2=2706.32 KNm$$ $ BM@_a=306.84\times \dfrac {4.5^2}2=3106.75 KNm \\ \dfrac x{511.27}=\dfrac {4-x}{716.08}\Rightarrow x=1.66\\ B.M_{max}=\Bigg[306.84\times \dfrac {(4.2+1.66)^2}2\Bigg]-1800\times 1.66=2280.38$

Two way shear

$$M_{u\space max}=0.138f_ckbd^2\\ 3106.75\times 10^6=0.138\times 20 \times 1500 dx^2\\ d_x=866.16 mm $$

Provide, d = 1000 mm

Net shear force = Pu-[(Area of critical section x Wd)]

$$= 2100 – [(1.4 x 1.4) x 204.56]\\= 1699.06KN $$

Area resisting = perimeter of critical section x d

$$ = [2 x (1.4+1.4)] x 1 = 5.6m $$

$Z_v=\dfrac {1699.06\times 10^3}{5.6\times 10^6}=0.3 N/mm^2 \\ Z_{cpermi}= Z_0^1 \times Ks\\ Z_0^1 =1.12 N/mm^2 \\ K_s=0.5+\dfrac {0.4}{0.4} = 1.5 \gt 1 \\ K_s=1\\ Z_{permi} = 1.12\times 1=1.12 N/mm^2 \gt 0.3N/mm^2 \therefore safe$

Design reinforcement:-

Column A/B

Bandwidth $= (b + 2d) = [400+2 x 1000] = 2400 mm$

$$Ast=(\dfrac {0.5\times 20\times 2400\times 1000}{415} )\times [1-\sqrt{\dfrac {4.6\times67.5\times10^6}{20\times2400\times1000^2}}]$$

$Ast=187.35mm^2\\ Ast_{min}=\dfrac {0.12}{100}bD=\dfrac {0.12}{100}\times 2400\times1100 =3168 mm^2\\ \text { Provide 16nas of 16mm $\theta$ bars}$