The effective cover provided to the tensile reinforcement is 40 mm and the reinforcement comprises of are M-20 concrrte and HYSD steal of grade $F_{e415}$ .

Data :- $b=150 \ mm \\ D=360 \ mm \\ d_c=50 \ mm \\ \therefore d=D-d_c=360-50 \\ \ \ \ \ \ \ =310 \ mm$

$f_ck=20N/mm^2\ \ \& f_y=415N/mm^2 \\ M_{20},F_{e415}$

Ast $=3-16mm \\ =3{\times}\frac{\Pi}{4}{\times}16^2=603.18mm^2$

Steps 1:- To find depth of actual N.A

$C_u=T_u \\ 0.36f_CkbX_u=0.87{\times}f_y{\times}Ast \\ 0.36{\times}20{\times}150{\times}X_u=0.87{\times}415{\times}603.18 \\ X_u=201.64 mm$

For $F_{e415}X_{umax}=0.48 d \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =148.8 mm \\ \therefore X_u \gt X_{umax}$

Hence, It is over reinforced section.

Not permitted as per IS : 456

Restrict $X_u=X_{umax}$

$M_{umax}=C_u{\times}L_u \\ 0.36{\times}20{\times}150{\times}148.8{\times}(310-0.42{\times}148.8) \\ M_{umax}=39.69 KNm.$