Bent-up bars shear, 550 mm effective depth

Given: $b=300$ mm, $d=550$ mm, 4-20 mm bars (2 bent up at 45$^\circ$), $V_u=425$ kN at the support, $L=6$ m, M20, Fe415.

Step 1: Shear force. Reaction $=w_uL/2=425$ kN, so $w_u=\dfrac{2\times425}{6}=141.67$ kN/m. At $d$ from the face of support:

$$V_{u,d}=425-141.67\times0.55=347.1\ \text{kN}$$

Check: $\tau_v=\dfrac{425\times10^3}{300\times550}=2.576$ N/mm$^2<\tau_{c,max}=2.8$ N/mm$^2$, so the section is adequate.

Step 2: Concrete strength. Straight bars: 2-20, $A_{st}=628.3$ mm$^2$.

$$p_t=\frac{100\times628.3}{300\times550}=0.381\%,\quad\tau_c=0.423\ \text{N/mm}^2,\quad V_c=69.8\ \text{kN}$$

$V_{us}=V_{u,d}-V_c=347.1-69.8=277.3$ kN.

Step 3: Shear resistance of the bent-up bars ($A_{sb}=2\times314.16=628.3$ mm$^2$):

$$V_{us,bent}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times628.3\times0.7071=160.4\ \text{kN}$$

By IS 456 cl. 40.4 this should not exceed half of the total shear reinforcement load: $0.5\times277.3=138.7$ kN. Hence bent-up bars are credited with 138.7 kN.

Step 4: Additional stirrups for $V_{us}=277.3-138.7=138.7$ kN, using 10 mm 2-legged stirrups ($A_{sv}=157.1$ mm$^2$):

$$s_v=\frac{0.87\times415\times157.1\times550}{138.7\times10^3}=225\ \text{mm}$$

Limits: $0.75d=412$ mm, 300 mm, $\dfrac{0.87f_yA_{sv}}{0.4b}=473$ mm.

Answer: Shear resistance of bent-up bars $=160.4$ kN (limited to 138.7 kN by the code); additional stirrups: 10 mm 2-legged @ 220 mm c/c near the supports.