Bent-up bars and stirrups, 5-25 mm bars

Given: $b=400$ mm, $d=550$ mm, 5-25 mm bars (two bent up at 45$^\circ$), $w_u=100$ kN/m on 6 m clear span, M20, Fe415.

Step 1: Design shear force

$$V_u=\frac{w_uL}{2}=\frac{100\times6}{2}=300.0\ \text{kN (at support face)}$$

Critical section at $d$ from the support face: $V_{u,d}=300.0-100\times0.55=245.0$ kN.

$\tau_v=\dfrac{245.0\times10^3}{400\times550}=1.114$ N/mm$^2<\tau_{c,max}=2.8$ N/mm$^2$, OK.

Step 2: Shear resistance of concrete. Bars continuing to the support: 3-25, $A_{st}=1472.6$ mm$^2$.

$$p_t=\frac{100\times1472.6}{400\times550}=0.669\%\ \Rightarrow\ \tau_c=0.534\ \text{N/mm}^2\ (\text{Table 19})$$ $$V_c=\tau_cbd=0.534\times400\times550=117.5\ \text{kN}$$

Shear to be carried by shear reinforcement: $V_{us}=245.0-117.5=127.5$ kN.

Step 3: Shear resistance of the bent-up bars

$$V_{us,bent}=0.87f_yA_{sb}\sin\alpha=0.87\times415\times981.7\times\sin45^\circ=250.6\ \text{kN}$$

IS 456 (cl. 40.4) allows bent-up bars to carry at most half of the shear reinforcement requirement, i.e. $0.5\times127.5=63.7$ kN. So the useful contribution of the bent bars is 63.7 kN.

Step 4: Additional stirrups. Balance shear $=127.5-63.7=63.7$ kN. For 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{63.7\times10^3}=313\ \text{mm}$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=227$ mm, $0.75d=412$ mm and 300 mm. Provide 8 mm 2-legged stirrups at 225 mm c/c near the support.

Answer: Shear resistance of bent-up bars $=250.6$ kN (limited to 63.7 kN by the 50% rule); additional stirrups 8 mm 2L @ 225 mm c/c.