Procedure:

Data:- Dimension, $M_d , T_u , V_u .$

Find: Ast, Asc, shear-reinforcement

$M_u=M_d+M_t\\ If \space M_d \gt M_t \to \text { singly reinforced section }\\ Ast=(\dfrac {0.5f_ckbd}{f_y})\times [1-\sqrt{1-\dfrac {4.6M_u}{f_ckb^2}}] \\ If \space M_d \lt M_t \to \text { Doubly reinforced section } \\ Asc= \dfrac {(M_t-M_d)}{(f_{sc}-f_{cc})\times (d-d_c)} \\ Ast=(\dfrac {0.5f_ckbd}{f_y})\times [1-\sqrt{1-\dfrac {4.6M_u}{f_ckb^2}}]\\ Shear :- (V_e) =V_u+\dfrac {1.6T_u}b\\ Z_{uc}=\dfrac {V_e}{bd} \lt Z_{c\space max}(2.8 N/mm^2) \\ pt=\dfrac {100\times Ast_p}{bd0}$

Zuc by interpolation.

$V_{u\space min}=0.4bd\\ If \space V_e \lt (V_{uc} + V_{u\space min})$

Provide minimum shear reinforcement

Spacing

$S_2=0.75d\\ S_3=300 mm\\ If \space V_e \gt (V_{uc}+V_{u\space min})$

Design & provide shear reinforcement

Spacing :

$S_2=0.75d\\ S_3=300 mm\\ If \space V_e \gt (V_{uc}+V_{u\space min})\\ asv=(\dfrac {T_u\times S_u}{b_1d_10.87f_y})+(\dfrac {V_uS_u}{2.5\times d_1\times (0.87\times f_y)})\\ asv=\dfrac {Zue-zuc}{0.87f_y}\times b\times S_s$