Shear design with torsion, 250x500

Given: $b=250$ mm, $d=500$ mm, $V_u=110$ kN, $T_u=20$ kN·m, 3-20 mm bars ($A_{st}=942.5$ mm$^2$), M20, Fe500. Assume overall depth $D=550$ mm and effective cover 50 mm, so $b_1=250-100=150$ mm, $d_1=550-100=450$ mm (centre to centre of corner bars).

Step 1: Equivalent shear (IS 456 cl. 41.3.1)

$$V_e=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{20}{0.25}=238\ \text{kN}$$ $$\tau_{ve}=\frac{V_e}{bd}=\frac{238\times10^3}{250\times500}=1.904\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2\ \text{(OK)}$$

Step 2: Concrete strength. $p_t=\dfrac{100\times942.5}{250\times500}=0.754\%$, $\tau_c=0.561$ N/mm$^2$ (Table 19). $\tau_{ve}>\tau_c$, so transverse reinforcement is needed.

Step 3: Stirrup area (cl. 41.4.3)

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=\frac{20\times10^6}{150\times450\times435}+\frac{110\times10^3}{2.5\times450\times435}=0.681+0.225=0.906\ \text{mm}^2/\text{mm}$$

Check with $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.772$ mm$^2$/mm (smaller, so the above governs).

For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=\dfrac{100.5}{0.906}=111$ mm.

Step 4: Spacing limits: $s_v\le x_1=175$ mm, $\le\dfrac{x_1+y_1}{4}=162$ mm, $\le300$ mm.

Answer: Provide 8 mm dia closed 2-legged stirrups at 110 mm c/c.