Design steps of doubly reinforced section

A doubly reinforced beam is used when the factored moment $M_u$ exceeds the limiting moment $M_{u,lim}=0.138f_{ck}bd^2$ (Fe415) of the section whose size is restricted.

  b                strain     stress        forces
 +---------+ d'     0.0035   0.446fck
 | o  Asc  |         |\     fcc|==|     Cc = 0.36fck.b.xu
 |         |         | \xu    |  |     Cs = Asc(fsc-fcc)
 |         |        -+--NA    |  |
 |         |         |  \
 | o o  Ast|         |   \es   T = 0.87fy.Ast
 +---------+

Steps

1. Assume $b$, $D$, cover $d'$; find $d$ and factored moment $M_u$.
2. Compute $M_{u,lim}=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})$ using $x_{u,max}=0.53d$, $0.48d$ or $0.46d$ for Fe250, Fe415, Fe500.
3. If $M_u>M_{u,lim}$, find the extra moment $M_{u2}=M_u-M_{u,lim}$, to be carried by a steel couple.
4. Find strain in compression steel at the limiting NA: $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_{u,max}}\right)$. Get the stress $f_{sc}$ from the stress-strain curve (IS 456 Fig. 23 / SP16 Table F). Concrete stress displaced by steel $f_{cc}=0.446f_{ck}$.
5. Compression steel: $$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}$$
6. Tension steel: $$A_{st}=A_{st1}+A_{st2},\quad A_{st1}=\frac{0.36f_{ck}b\,x_{u,max}}{0.87f_y},\quad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}$$
7. Check limits: $A_{st}\ge0.85bd/f_y$; $A_{st}$ and $A_{sc}$ not more than 4% each of $bD$; $d'/d\approx0.1$-$0.2$.
8. Detailing: provide stirrups spacing $\le$ the least of $16\phi$ of compression bar, 300 mm and $b$, so that compression bars do not buckle. Check shear and deflection (with compression steel factor $k_c$).