Beam design for bending, torsion, shear, 425x550

Given: $b=425$ mm, $D=550$ mm, $M_u=100$ kN·m, $T_u=28$ kN·m, $V_u=100$ kN, M20, Fe415, mild exposure. Assume effective cover 50 mm: $d=500$ mm; $b_1=425-100=325$ mm, $d_1=550-100=450$ mm.

Step 1: Equivalent forces (IS 456 cl. 41.3)

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=100+1.6\times\frac{28}{0.425}=205.4\ \text{kN}\\ M_t&=T_u\frac{1+D/b}{1.7}=28\times\frac{1+550/425}{1.7}=37.8\ \text{kN·m}\\ M_{e1}&=M_u+M_t=100+37.8=137.8\ \text{kN·m} \end{aligned}$$

$M_t<M_u$, so $M_{e2}=M_t-M_u$ is negative and no compression-face steel is required.

Step 2: Section check. $\tau_{ve}=\dfrac{205.4\times10^3}{425\times500}=0.967$ N/mm$^2$ $<\tau_{c,max}=2.8$ N/mm$^2$, OK. $M_{u,lim}=0.138f_{ck}bd^2=293.17$ kN·m $>M_{e1}$, so the section is singly reinforced.

Step 3: Longitudinal steel

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=831\ \text{mm}^2$$

Provide 3-20 mm bars ($942$ mm$^2$, $p_t=0.44\%$). Minimum steel $0.85bd/f_y=435$ mm$^2$, OK. Provide 2-12 mm bars at the top corners as hangers.

Side face bars: $D=550$ mm $>450$ mm, so cl. 26.5.1.7(d) requires extra longitudinal bars of $0.1\%\times425\times550=234$ mm$^2$ (117 mm$^2$ per face). Provide 1-16 mm bar on each face at mid-depth ($201$ mm$^2$), giving a vertical spacing of about 225 mm $<300$ mm.

Step 4: Stirrups. $\tau_c=0.45$ N/mm$^2$; since $\tau_{ve}>\tau_c$:

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=0.530+0.246=0.776\ \text{mm}^2/\text{mm}$$

This is more than the code minimum $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.609$ mm$^2$/mm. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=129$ mm. Limits: $x_1=350$ mm, $(x_1+y_1)/4=206$ mm, 300 mm.

Answer: Longitudinal steel $A_{st}=831$ mm$^2$ (3-20 mm bars at the bottom), 2-12 mm at the top, 1-16 mm side face bar on each face; closed 8 mm 2-legged stirrups at 125 mm c/c.