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 …