Data : - $b=200 \ mm \\ d=450 \ mm \\ M_{20},F_{e415} \\ M_{umax}=70 KNm$

Find Ast = ?

$M_{umax}=0.138f_ckbd^2$

$70{\times}10^6$ $=0.138{\times}20{\times}200{\times}d^2 \\ d=356.1 \\ d=360 \ mm$

But d=450 mm given (whichever big)

$\therefore$ Take $d=450 \\ \therefore{D}=d+dc \\ =450+50 \\ =500$

$Ast \ x\frac{0.5f_ckbd}{f_y}{\times}\Bigg(1-\sqrt{1-\frac{4.6M_u}{f_ckbd^2}}\Bigg) \\ Ast \ x=\frac{0.5{\times}20{\times}200{\times}450}{415}\Bigg(1-\sqrt{1-\frac{4.6{\times}70{\times}10^6}{20{\times}200{\times}450^2}}\Bigg) \\ Ast \ x =485.37 \ mm^2 \\ \\ Provide \ 3 \ - \ 16 \ mm \\ Ast \ P=3{\times}\frac{∏}{4}{\times}16^2=603.18$

Reinforcement Detail