Determine the maximum well of the beam can carry safely (including self weight.) use $F_{e415}$ steel and $M_{20}$ concrete. (Assume d' = 50 mm)

Data :- $b=230 \ mm \\ D=600 mm \\ 4-20 \ mm \ {\phi} \\ Ast=4{\times}314=1256 \ mm^2 \\ l=5 \ m$

Assume d'=50 mm

To find = w= ?

d = D – d’= 600 – 50 = 550 mm

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

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

For $F_{e415}=M_{umax}0.48d \\ X_{umax}=0.48{\times}550=264 \ mm \\ \therefore \ X_u \gt X_{umax} $

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

Restrict $→X_u=X_{umax}$

$M_{umax}=C_u{\times}L_a \\ M_{umax}=0.36{\times}f_ck{\times}b{\times}X_{umax}(d-0.42X_{umax}) \\ M_{umax}=0.36{\times}20{\times}230{\times}264{\times}(550-0.42{\times}264) \\ M_{umax}=192{\times}10^6Nmm \\ \therefore M_{umax}=192 KNm$

Now ,

$B.M=\frac{M_{umax}}{1.5}=\frac{192}{1.5}=128KNm \\ B.M=\frac{wl^2}{8} \\ 128=\frac{w{\times}5^2}{8} \\ \therefore w=40.96KN/m$