Use whitney’s method. Use $M_{20}/F_{e415}$.

Given: b=230 mm

d = 600mm.

$$σ_{cu}=f_ck=20N/mm^2 \\ σ_{sy}=f_y=415 N/mm^2$$

$Ast = 3-16 mm∅ = 3×\frac{Π}{4}×16^2=603 mm^2$

To find

$1) M.R.= ? \\ 2) W_{udl}= ?$

Case 1:

To find depth of actual N.A.

$$C_u=T_u \\ \frac{2}{3} f_{ck} ba= f_y Ast \\ \frac{2}{3}×20×230×a= 415×603$$

a=84.44mm

$$a_{max}=\frac{d}{2}=\frac{550}{2}=275mm$$

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

$$M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ M_u=415×603×\Big(550-\frac{81.6}{2}\Big) \\ M_u=127.42 kNm$$

Case 2:

$$M_u=127.42 kNm \\ M_u=\frac{w_d l^2}{8} \\ 127.42=\frac{w_d ×5^2}{8} \\ w_d=40.77 \\ w_{self}=0.23×0.6×25=3.45kN.m \\ w_d=1.5 D.L+(2.2 ×L.L) \\ 40.77=1.5 3.45+(2.2 ×L.L) \\ L.L=1.61 kN/m$$