Find $M_u$ for the following limit state.

1) Analysis:-

$C_u=T_u \\ C_{u1}+C_{u2}=T_{u1}+T_{u2} \\ [(0.36{\times}f_ck{\times}b{\times}X_u)+(F_{cs}-f_{cc}){\times}Asc]=0.87{\times}f_y{\times}Ast \\ X_u=?$

Now,

$X_{u \max}=0.53 \ d →F_{e250} \\ X_{u \max}=0.48 \ d→F_{e415} \\ X_{u \max}=0.46 \ d→F_{e500}$

$\therefore X_uX_{u \max}→$ under reinforced section

$M_u=(C_{u1}{\times}L_{a1})+(C_{u2}{\times}L_{a2}) \\ M_u=[(0.36{\times}f_ckbX_u){\times}(d-0.42x_0)]+[(f_{sc}-f_{cc})A_{sc}{\times}(d-d_c)] \\ M_u=?$

If, $X_u \gt X_{u \max}→$over reinforced section

Not permitted as per IS: 456

Restrict $X_u=X_{u \max}$

$M_{u \max}=(C_{u1}{\times}L_{a1})+(C_{u2}+L_{a2}) \\ M_{u \max}=[(0.36{\times}f_ckbX_{u \max}){\times}(d-0.42X_{u \max})]+[(f_{sc}-f_{cc})A_{sc}{\times}(d-d_c)] \\ M_{u \max}=?$

$F_{cc}=0.446f_ck \\ F_{sc}=0.87f_y\rightarrow only \ for \ F_{e250}$

2) Design:-

Data:- Dimension, $M_d,f_ckl,f_y$

Find:- Ast, $A_{sc}$

Step:-

$M_{u \max}=0.149f_ckbd^2\rightarrow F_{e250} \\ M_{u \max}=0.138f_ckbd^2 \rightarrow F_{e415} \\ M_{u \max}=0.133f_ckbd^2 \rightarrow F_{e500} \\ If \ M_d=M_{u \max}$

Singly reinforced section is sufficient

$\therefore Ast=\frac{0.5f_ckbd}{f_y}{\times}\Bigg(1-\sqrt{1-\frac{4.6M_d}{f_ckbd^2}}\Bigg)$

$If \ M_d=M_{u \max}$

Design a doubly reinforced section

$\therefore M_{u1} = M_{u \max.\sin \ gly} \\ M_{u2}=M_d-M_{u1} \\ M_{u1}=T_{u1}{\times}L_{a1}.............(1) \\ M_{u1}=0.87{\times}f_yAst_1{\times}(d-0.42X_{u \max}) \\ Ast_1=\frac{M_{u1}}{0.87{\times}f_y{\times}(d-0.42X_{u \max})} \\ Ast_1=? \\ M_{u2}=T_{u2}{\times}L_{u2}..............(2) \\ M_{u2}=0.87f_yAst_2{\times}(d-d_c) \\ Ast_2=? \\ \therefore Ast=Ast_1+Ast_2 \\ M_{u2}=C_{u2}{\times}{L_{a2}}...........(3) \\ M_{u2}=(f_{sc}-f_{cc}){\times}A_{sc}{\times}(d-d_c) \\ \therefore Asc=?$