Given Pu=1800 kN

=1800$\times 10^{3}$N

bearing strenght of concrete =0.6$\times$20

=18mpa

SBC=350 KN/m$^{2}$

Required-Design slab base

Step 1 Required area of base plate

a) Ap=$\frac{Pu}{0.6\times 20}=\frac{1800\times 10^{3}}{0.6\times 20}=150\times10^{3}mm^{2}$

AP=150$\times10^{3}mm^{2}$

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

Lc=420

Bc=350

from fig

b) 1) $\frac{2008-500}{2}$=750mm

2) $\frac{1900-450}{2}$=725mm

Lp=$(\frac{450-350}{2})+\sqrt{(\frac{420-350}{2})^{2}+150\times 10^{3}}$

L=425.87mm

provide Lp=500 mm

width of plate =$\frac{Ap}{Lp}$

=$\frac{150\times10^{3}}{500}$

Bp=300mm

provide Bp=450 mm

C0 thickness of plate (tp)=$\sqrt{2.5w\ast(a^{2}-0.3b^{2})\ast \frac{ymo}{fy}}$

w=$\frac{pu}{Ap}=\frac{1800\times 10^{3}}{(500\times 450)}\lt0.6\ast fck$

=8$\lt 0.6\ast$20

=8$\lt$12

tp=$\sqrt{2.5\times8\ast(50^{2}-0.3\times40^{2})\ast\frac{1.1}{250}}\gt fy$

tp=13.33mm$\gt$12.5mm

provide 16mm thickness $\gt$ 12.5 hence ok

Step II concrete block diagram

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

=$\frac{1.1 \times 1800}{1.5\ast 350}$

Af=3.77m$^{2}$

b) Lf=$(\frac{Lp-Bp}{2})+\sqrt{(\frac{Lp-Bp}{p})^{2}+Af}$

=$(\frac{500-450}{2})+\sqrt{(\frac{500-450}{2})^{2}+3.77\times 10^{6}}$

Lf=1.976 m=1976 mm

provide 2000 mm

c) width of footing=$\frac{Af}{Lf}=\frac{3.77\times 10^{6}}{2000}$

Bf=1885 mm

provide wf=1900 mm

d)Depth of footing=(Df)=larger projection