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 …