Calculate the MR of a doubly reinforced R.C beam of rectangular section of size $300 mm {\times} 450 mm$ deep reinforced with 6 No. 20 mm dia. bars on tension side and case:-

1. 4 no. 20 mm ${\phi}$ on comp side.

2. 5 no. 20 mm ${\phi}$ on comp side.

Assume effective cover of 40 mm on both sides. Use $M_{20}/F_{e415}$

Data:- $b=300 \\ D=450 \ mm[ \because deep] \\ d_c=40 \ mm \\ d=D-d_c=450-40 \\ \hspace{2cm}=410 \ mm$

$f_ck=20N/mm^2,f_y=415N/mm^2 $

Case 1:-

$Ast=6{\times}\frac{\pi}{4}{\times}20^2=6{\times}314=1884mm^2 \\ Asc=4{\times}\frac{\pi}{4}{\times}20^2=4{\times}314=1256mm^2 \\ \frac{d_c}{d}=\frac{40}{410}=0.097$

from table, use interpolation

$0.05\hspace{1cm}355 \\ 0.097\hspace{0.8cm}f_{sc}=? \\ 0.1\hspace{1.2cm}353 \\ \therefore f_{sc}=353.12N/mm^2 \\ f_{sc}=0.446{\times}20=8.92N/mm^2$

To find Depth of actual N.A

$C_u=T_u \\ (C_{u1}+C_{u2})=(T_{u1}+T_{u2}) \\ (0.36f_ckbX_u)+(f_{sc}+f_{cc})Asc=0.87f_yAst \\ (0.36{\times}20{\times}300{\times}X_u)+(353.12-8.92){\times}1256=(0.87{\times}415{\times}1884) \\ \therefore X_u=114.76 \ mm \\ X_{u \max}=0.48 \ d=0.48{\times}410=196.8 \ mm \\ \therefore X_u \lt X_{u \max}$

Hence it is an under reinforced section.

$M_u=(C_{u1}{\times}L_{a1})+(C_{u2}{\times}L_{a2}) \\ M_u=[(0.36f_ckbX_u){\times}(d-0.42X_u)]+[(f_{sc}-f_{cc}){\times}Asc]{\times}(d-d_c) \\ M_u=[(0.36{\times}20{\times}300{\times}114.76){\times}(410-0.42{\times}114.76)]+\bigg[[(353.12-8.92){\times}1256]{\times}(410-40)\bigg] \\ \therefore M_u=249.64Knm$

Case 2:-

$Ast=6{\times}\frac{\pi}{4}{\times}20^2=6{\times}314=1884mm^2 \\ Asc=5{\times}\frac{\pi}{4}{\times}20^2=5{\times}314=1256mm^2 \\ \therefore f_{sc}=353.12K/mm^2 \\ f_{cc}=8.92N/mm^2 $

To find Depth of actual N.A

$C_u=T_u \\ (C_{u1}+C_{u2})=(T_{u1}+T_{u2}) \\ (0.36f_ckbX_u)+(f_{sc}+f_{cc})Asc=0.87f_yAst \\ (0.36{\times}20{\times}300{\times}X_u)+(353.12-8.92){\times}1570=(0.87{\times}415{\times}1884) \\ \therefore X_u=64.73 \ mm \\ X_{u \max}=0.48 \ d=0.48{\times}410=196.8 \ mm \\ \therefore X_u \lt X_{u \max}$

Hence it is an under reinforced section.

$M_u=(C_{u1}{\times}L_{a1})+(C_{u2}{\times}L_{a2}) \\ M_u=[(0.36f_ckbX_u){\times}(d-0.42X_u)]+[(f_{sc}-f_{cc}){\times}Asc]{\times}(d-d_c) \\ M_u=[(0.36{\times}20{\times}300{\times}64.73){\times}(410-0.42{\times}64.73)]+\bigg[[(353.12-8.92){\times}1570]{\times}(410-40)\bigg] \\ \therefore M_u=253.46Knm$