Bent-up bars, 600 mm effective depth

Given: $b=400$ mm, $d=600$ mm, 5-25 mm bars (2 bent up at 45$^\circ$, 3 straight), $V_u=250$ kN at the support reducing linearly to 0 at mid-span (span 6 m), M20, Fe415.

Step 1: Shear force diagram. Slope $w_u=\dfrac{250}{3}=83.33$ kN/m. At $d$ from the support:

$$V_{u,d}=250-83.33\times0.6=200.0\ \text{kN},\qquad \tau_v=\frac{250\times10^3}{400\times600}=1.042\ \text{N/mm}^2\ (\text{at support})<2.8$$

Step 2: Concrete strength. Straight bars continuing to the support: 3-25, $A_{st}=1472.6$ mm$^2$.

$$p_t=\frac{100\times1472.6}{400\times600}=0.614\%,\quad\tau_c=0.516\ \text{N/mm}^2,\quad V_c=\tau_cbd=123.9\ \text{kN}$$ $$V_{us}=V_{u,d}-V_c=200.0-123.9=76.1\ \text{kN}$$

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

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

This is more than the whole requirement $V_{us}=76.1$ kN, but IS 456 cl. 40.4 limits bent-up bars to 50% of the shear reinforcement, i.e. $0.5\times76.1=38.0$ kN.

Step 4: Additional stirrups to carry $V_{us}-V_{bent}=38.0$ kN, with 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$s_v=\frac{0.87\times415\times100.5\times600}{38.0\times10^3}=573\ \text{mm}$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=227$ mm, $0.75d=450$ mm, 300 mm. Provide 8 mm 2-legged stirrups at 225 mm c/c.

Step 5: Extent of shear reinforcement. Shear equals $V_c=123.9$ kN at $x=\dfrac{V_c}{w_u}=1.49$ m from mid-span, so shear reinforcement is needed over 1.51 m from each support; beyond it, nominal stirrups (minimum) are provided.

Answer: Shear resistance of the two bent-up bars $=250.6$ kN (usable 38.0 kN); additional stirrups 8 mm 2L @ 225 mm c/c.