Shear design for 230x450 beam, 5 m span

Given: $L=5$ m, $b=230$ mm, $D=450$ mm, central point load 30 kN (taken as service load), 4-16 mm bars with one bent up (3 bars, $A_{st}=603.2$ mm$^2$, continue to the support), M20, Fe415, 8 mm 2-legged stirrups. Assume effective cover 40 mm, $d=410$ mm.

Step 1: Design shear force

$$\begin{aligned} \text{Self weight}&=0.23\times0.45\times25=2.587\ \text{kN/m},\quad w_u=1.5\times2.587=3.881\ \text{kN/m}\\ P_u&=1.5\times30=45\ \text{kN}\\ V_u&=\frac{w_uL}{2}+\frac{P_u}{2}=32.20\ \text{kN (at support)} \end{aligned}$$

At the critical section $d$ from the support: $V_{u,d}=V_u-w_ud=32.20-3.881\times0.41=30.61$ kN.

Step 2: Nominal shear stress

$$\tau_v=\frac{V_{u,d}}{bd}=\frac{30.61\times10^3}{230\times410}=0.325\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2$$

Step 3: Shear strength of concrete. $p_t=\dfrac{100A_{st}}{bd}=\dfrac{100\times603.2}{230\times410}=0.640\%$, so $\tau_c=0.525$ N/mm$^2$ (Table 19, M20). Since $\tau_v<\tau_c$, no stirrups are required by calculation, but minimum shear reinforcement is provided (cl. 26.5.1.6):

$$\frac{A_{sv}}{b\,s_v}\ge\frac{0.4}{0.87f_y}\ \Rightarrow\ s_v\le\frac{0.87\times415\times100.5}{0.4\times230}=395\ \text{mm}$$

Other limits: $s_v\le0.75d=308$ mm and $\le300$ mm.

Step 4: Contribution of the bent-up bar (available as extra capacity): $V_{us}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times201.1\times0.707=51.3$ kN. Total capacity $=\tau_cbd+V_{us,bent}=49.5+51.3$ kN $>V_u$.

Answer: Provide 8 mm 2-legged vertical stirrups at 300 mm c/c throughout (minimum shear reinforcement governs); the bent-up bar adds an extra 51.3 kN.