Calculate the ultimate moment of resistance using ultimate load method(IS Method). Take $σ_{cu}=\frac{20N}{mm^2} \ and \ σ_{sy}=425N/mm^2$

b=250mm

d = 450mm

$Ast=4-18mm∅ = 1017.87 mm^2$

$$σ_{cu}=f_ck=20N/mm^2 \\ σ_{sy}=f_y=425N/mm^2$$

To find = $M_u$= ?

To find depth of actual N.A.

$$C_u=T_u \\ \frac{2}{3} f_{ck} ba= f_y Ast \frac{2}{3}×20×250×a= 425×1017.87$$ a=129.77mm $$a_{max}=\frac{d}{2}=\frac{450}{2}=225mm$$

Here $a \lt a_{max}→$ Under reinforced section

$$M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ M_u=415×1017.87×\Big(450-\frac{129.77}{2}\Big) \\ M_u=166.59 kNm$$