It is used as a simply supported beam over an effective span of 5.5m Use $M_{20}/F_{e415}$

Data:- $b=230 \ mm\\ D=550 \ mm$

Assume $d_c=50 \ mm$

$d=D-d_c=550-50=500 \ mm \\ Ast=4-20 \ mm \ {\phi} \\ Ast=4{\times}314=1256 \ mm^2 \\ M_{20}F_{e415}$

To find = w = ?

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

$C_u=T_u \\ 0.36{\times}f_ckbX_u=0.87{\times}f_ykAst \\ 0.36{\times}20{\times}230{\times}X_u=0.87{\times}415{\times}1256 \\ X_u=273.8 mm$

For $F_{e415}=X_{umax} \ 0.48 d \\ X_{umax}=0.48{\times}500=240 \ mm \\ \therefore X_u \gt X_{umax}$

Hence, it is over reinforced section not permitted as per IS : 456

Restrict →$X_u = X_{umax}$

$M_{umax}=C_u{\times}L_a \\ M_{umax}=0.36{\times}20{\times}230{\times}(500-0.42{\times}240) \\ \therefore M_{umax}158.65 KNM$

Now,

$B.M=\frac{M_{umax}}{1.5}=\frac{158065}{1.5}=105.76 KNm \\ B.M=\frac{wl^2}{8} \\ 128=\frac{w{\times}5.5^2}{8} \\ \therefore w=27.96 KN/m$