Shear at bent-up section, 8 mm stirrups

Given: $b=300$ mm, $D=500$ mm (assume overall; effective cover 50 mm, $d=450$ mm), 4-25 mm bars of which 2 are bent up at 45$^\circ$, 8 mm 2-legged stirrups @ 150 mm c/c, M20, Fe415.

Step 1: Shear capacity of concrete. At the section where the bent-up bars act, the straight bars are 2-25, $A_{st}=981.7$ mm$^2$:

$$p_t=\frac{100\times981.7}{300\times450}=0.727\%\ \Rightarrow\ \tau_c=0.553\ \text{N/mm}^2$$ $$V_c=\tau_cbd=0.553\times300\times450=74.6\ \text{kN}$$

Step 2: Stirrups ($A_{sv}=2\times50.27=100.5$ mm$^2$):

$$V_{us}=\frac{0.87f_yA_{sv}d}{s_v}=\frac{0.87\times415\times100.5\times450}{150}=108.9\ \text{kN}$$

Step 3: Bent-up bars ($A_{sb}=2\times490.87=981.7$ mm$^2$):

$$V_{us,bent}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times981.7\times0.7071=250.6\ \text{kN}$$

IS 456 cl. 40.4: bent-up bars can carry at most half of the shear reinforcement, i.e. not more than the stirrups. So the usable bent-bar contribution is $\min(250.6,\ 108.9)=108.9$ kN.

Step 4: Total shear capacity

$$V_u=V_c+V_{us}+V_{us,bent}=74.6+108.9+108.9=292.4\ \text{kN}$$

Upper limit from $\tau_{c,max}=2.8$ N/mm$^2$: $V_{max}=2.8\times300\times450=378$ kN, which is more than the above, so the section is not limited by web crushing.

Answer: Shear resisting capacity of the section at the bent-up bar $=292.4$ kN (factored). Without the 50% limit, the full bent-bar value would be $250.6$ kN.