Find out ultimate moment of resistance of doubly reinforced rectangular section of size $250 \ mm \ {\times} \ 550 mm$ (effective) having tensile reinforcement $(3054 \ mm^2 )$ and compressive reinforcement $(982 \ mm^2 )$

The material are $M_{20}$ concrete and TOR steal.

Refer Table 1 for the values of $F_{sc}$ for various $\frac{d_c}{d}$ ratio.

Data :- $b=250 \ mm \\ D=550 \ mm$

$Ast=3054 \ mm^2 \\ Asc=982 \ mm^2 \\ f_ck=20 N/mm^2 \\ TOR \ steal =F_{e415} \\ i.e f_y=415N/mm^2 \\ Assume \ d_c=40 mm$

Now,

$\frac{d_c}{d}=\frac{40}{550}=0.072$

$\therefore f_{sc}=353.69N/mm^2 \\ f_{cc}=0.446{\times}20=8.92N/mm^2$

To find Depth of actual N.A

$C_u=T_u \\ (C_{u1}+C_{u2})=(T_{u1}+T_{u2}) \\ (0.36f_ckbX_u)+(f_{sc}+f_{cc})Asc=0.87f_yAst \\ (0.36{\times}20{\times}250{\times}X_u)+(353.69-8.92){\times}982=(0.87{\times}415{\times}3054) \\ \therefore X_u=424.49 mm \\ X_{u \max}=0.48d=0.48{\times}550=264 mm^2 \\ \therefore X_u \gt X_{u \max}$

Hence it is an over reinforced section.

Not permitted as per IS:456

Hence restrict $X_u= X_{u \max}$

$X_{u \max}=(C_{u1}{\times}L_{a1})+(C_{u2}{\times}L_{a2}) \\ X_{u \max}=[(0.36f_ckb X_{u \max}){\times}(d-0.42 X_{u \max})]+\bigg[[(f_{sc}-f_{cc}){\times}Asc]{\times}(d-d_c)\bigg] \\ X_{u \max}=[(0.36{\times}20{\times}250{\times}264){\times}(550-0.42{\times}264)]+\bigg[[(353.69-8.92){\times}892]{\times}(550-40)\bigg] \\ \therefore X_{u \max}=381.33KNm$