Shear reinforcement, 300x500 beam, 370 kN

Given: $b=300$ mm, $D=500$ mm, tension steel 5-25 mm ($A_{st}=2454.4$ mm$^2$), $V_u=370$ kN, M20, Fe415. Assume effective cover 50 mm: $d=450$ mm. (Compression steel does not affect shear design.)

Step 1: Nominal shear stress

$$\tau_v=\frac{V_u}{bd}=\frac{370\times10^3}{300\times450}=2.741\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2$$

The section size is adequate.

Step 2: Concrete shear strength

$$p_t=\frac{100\times2454.4}{300\times450}=1.818\%\ \Rightarrow\ \tau_c=0.761\ \text{N/mm}^2\ (\text{Table 19})$$ $$V_c=\tau_cbd=0.761\times300\times450=102.7\ \text{kN}$$

Step 3: Shear reinforcement

$$V_{us}=V_u-V_c=370-102.7=267.3\ \text{kN}$$

Using 10 mm 2-legged stirrups ($A_{sv}=157.1$ mm$^2$):

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times157.1\times450}{267.3\times10^3}=95\ \text{mm}$$

Step 4: Check for minimum shear reinforcement and maximum spacing

$$s_v\le\frac{0.87f_yA_{sv}}{0.4b}=\frac{0.87\times415\times157.1}{0.4\times300}=473\ \text{mm},\quad s_v\le0.75d=337.5\ \text{mm},\quad s_v\le300\ \text{mm}$$

The calculated spacing 95 mm is less than all these limits, so it governs.

Answer: Provide 10 mm 2-legged vertical stirrups at 95 mm c/c near the support. The spacing may be increased towards mid-span up to a maximum of 300 mm (minimum shear reinforcement) where the shear decreases.