Design reinforcement under torsion, 400x750

Given: $b=400$ mm, $D=750$ mm, $T_u=150$ kN·m, $M_u=400$ kN·m (hogging), $V_u=110$ kN, M25, Fe500, mild exposure. Assume effective cover 50 mm: $d=700$ mm. Centre-to-centre distances of the corner bars: $b_1=400-2\times50=300$ mm, $d_1=750-2\times50=650$ mm.

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

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{150}{0.4}=710.0\ \text{kN}\\ M_t&=T_u\,\frac{1+D/b}{1.7}=150\times\frac{1+750/400}{1.7}=253.7\ \text{kN·m}\\ M_{e1}&=M_u+M_t=400+253.7=653.7\ \text{kN·m (on the tension face, top)} \end{aligned}$$

Since $M_t=253.7$ kN·m is less than $M_u=400$ kN·m, $M_{e2}=M_t-M_u$ is negative, so no extra steel is needed on the compression face (provide nominal bars there).

Step 2: Check the section. $\tau_{ve}=\dfrac{V_e}{bd}=\dfrac{710.0\times10^3}{400\times700}=2.54$ N/mm$^2$ $<\tau_{c,max}=3.1$ N/mm$^2$ (M25). The section size is adequate.

For Fe500, $x_{u,max}/d=0.46$: $M_{u,lim}=0.36\times0.46(1-0.42\times0.46)f_{ck}bd^2=0.1336f_{ck}bd^2=654.67$ kN·m $\ge M_{e1}=653.7$ kN·m, so the section can (just) be singly reinforced for $M_{e1}$.

Step 3: Longitudinal steel (tension face, top)

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=2647\ \text{mm}^2$$

Provide 6-25 mm dia bars ($2945$ mm$^2$) at the top, $p_t=1.05\%$.

Side face bars: since $D=750$ mm $>450$ mm, cl. 26.5.1.7(d) requires extra longitudinal bars: $0.1\%\times400\times750=300$ mm$^2$, i.e. 150 mm$^2$ per face. Provide 2-10 mm bars on each face ($157$ mm$^2$), so the bar spacing up the side is less than 300 mm.

Step 4: Transverse (stirrup) reinforcement (IS 456 cl. 41.4.3). $\tau_c=0.65$ N/mm$^2$ for $p_t=1.05\%$ (Table 19). Since $\tau_{ve}>\tau_c$, closed stirrups are needed:

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.768+0.156=1.924\ \text{mm}^2/\text{mm}$$

Also $\dfrac{A_{sv}}{s_v}\ge\dfrac{(\tau_{ve}-\tau_c)\,b}{0.87f_y}=1.732$ mm$^2$/mm. The first value governs.

Using 12 mm 2-legged closed stirrups ($A_{sv}=226.2$ mm$^2$): $s_v=118$ mm. Maximum spacing: $\le x_1=325$ mm, $\le\dfrac{x_1+y_1}{4}=250$ mm and $\le300$ mm.

Provide 12 mm closed 2-legged stirrups at 115 mm c/c.

Answer: Top steel 6-25 mm dia ($A_{st}=2647$ mm$^2$ required for $M_{e1}=653.7$ kN·m); bottom 2-16 mm bars (nominal, at the corners); side face bars 2-10 mm on each face; closed stirrups 12 mm @ 115 mm c/c.