Beam design for flexure, shear, torsion, 650x300

Given: $b=300$ mm, $D=650$ mm, $M_u=85$ kN·m, $V_u=110$ kN, $T_u=25$ kN·m, M25, Fe415. Assume effective cover 50 mm: $d=600$ mm; $b_1=300-100=200$ mm, $d_1=650-100=550$ mm.

Step 1: Equivalent shear and moment (IS 456 cl. 41.3)

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{25}{0.3}=243.3\ \text{kN}\\ M_t&=T_u\frac{1+D/b}{1.7}=25\times\frac{1+650/300}{1.7}=46.57\ \text{kN·m}\\ M_{e1}&=M_u+M_t=85+46.57=131.57\ \text{kN·m} \end{aligned}$$

$\tau_{ve}=\dfrac{243.3\times10^3}{300\times600}=1.352$ N/mm$^2<\tau_{c,max}=3.1$ N/mm$^2$ (M25), so the section is adequate. $M_t<M_u$, so $M_{e2}$ is not required.

Step 2: Longitudinal reinforcement. $M_{u,lim}=0.138\times25\times300\times600^2=372.50$ kN·m $>M_{e1}$, hence singly reinforced:

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=646\ \text{mm}^2$$

Minimum steel $0.85bd/f_y=369$ mm$^2$. Provide 4-16 mm bars ($804$ mm$^2$) at the bottom and 2-12 mm at the top as hangers.

Side face bars: $D=650$ mm $>450$ mm, so for torsion (cl. 26.5.1.7(d)) provide $0.1\%\times300\times650=195$ mm$^2$ (98 mm$^2$ per face): 1-12 mm bar on each face at mid-depth ($113$ mm$^2$), giving a spacing of about 275 mm $<300$ mm.

Step 3: Transverse reinforcement. $p_t=0.45\%$, $\tau_c=0.46$ N/mm$^2$; $\tau_{ve}>\tau_c$.

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=0.629+0.222=0.851\ \text{mm}^2/\text{mm}$$

Code minimum: $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.739$ mm$^2$/mm, so 0.851 governs. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=118$ mm. Limits: $x_1=225$ mm, $(x_1+y_1)/4=200$ mm, 300 mm.

Answer: Longitudinal steel 4-16 mm at the bottom ($A_{st}=646$ mm$^2$ required for $M_{e1}=131.57$ kN·m), 2-12 mm at the top and 1-12 mm side face bar on each face; transverse reinforcement: closed 8 mm 2-legged stirrups at 115 mm c/c.