Shear reinforcement, 250x500 beam, Fe500

Given: $b=250$ mm, $d=500$ mm, $V_u=110$ kN, 3-22 mm bars ($A_{st}=1140.4$ mm$^2$), M20, Fe500.

Step 1: Nominal shear stress

$$\tau_v=\frac{V_u}{bd}=\frac{110\times10^3}{250\times500}=0.880\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2$$

Step 2: Concrete shear strength

$$p_t=\frac{100\times1140.4}{250\times500}=0.912\%\ \Rightarrow\ \tau_c=0.599\ \text{N/mm}^2\ (\text{Table 19, M20})$$ $$V_c=\tau_cbd=0.599\times250\times500=74.9\ \text{kN}$$

$\tau_v>\tau_c$, so shear reinforcement is required.

Step 3: Stirrups

$$V_{us}=110-74.9=35.1\ \text{kN}$$

Using 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$), Fe500:

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times500\times100.5\times500}{35.1\times10^3}=622\ \text{mm}$$

Step 4: Spacing limits

• Minimum shear reinforcement: $s_v\le\dfrac{0.87f_yA_{sv}}{0.4b}=437$ mm
• $s_v\le0.75d=375$ mm
• $s_v\le300$ mm

The calculated spacing is larger than 300 mm, so the maximum spacing governs.

Answer: Provide 8 mm 2-legged vertical stirrups at 300 mm c/c.