Shear reinforcement of T-beam, 250 kN

Given (from figure): $b_f=1200$ mm, $D_f=125$ mm, $b_w=250$ mm, $D=500$ mm, 4-20 mm bars ($A_{st}=1256.6$ mm$^2$), $V_u=250$ kN, M25, Fe500. Assume effective cover 50 mm: $d=450$ mm. In a T-beam, shear is resisted by the web, so $b_w$ is used.

Step 1: Nominal shear stress

$$\tau_v=\frac{V_u}{b_wd}=\frac{250\times10^3}{250\times450}=2.222\ \text{N/mm}^2<\tau_{c,max}=3.1\ \text{N/mm}^2\ (\text{M25})$$

Step 2: Concrete shear strength

$$p_t=\frac{100A_{st}}{b_wd}=\frac{100\times1256.6}{250\times450}=1.117\%\ \Rightarrow\ \tau_c=0.668\ \text{N/mm}^2$$ $$V_c=\tau_cb_wd=75.2\ \text{kN}$$

Step 3: Shear taken by stirrups

$$V_{us}=V_u-V_c=250-75.2=174.8\ \text{kN}$$

Using 10 mm 2-legged stirrups ($A_{sv}=157.1$ mm$^2$), Fe500:

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times500\times157.1\times450}{174.8\times10^3}=176\ \text{mm}$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b_w}=683$ mm, $0.75d=337.5$ mm and 300 mm; the calculated value governs.

Answer: Provide 10 mm 2-legged vertical stirrups at 175 mm c/c near the support.