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Bent-up bars in shear strength

Short answer

Bent-up bars resist diagonal tension with a vertical component Vus = 0.87fy Asb sinα. They contribute to shear resistance but code limits them to at most 50% of total shear reinforcement; the rest is carried by stirrups.

How do the bent-up bars contribute in shear strength of beam? Explain it.

Bent-up bars are part of the main tension bars that are bent upwards at 45∘^\circ (or 30∘^\circ to 60∘^\circ) near the supports, where the bending moment is small but shear is large. They carry the diagonal tension across the inclined cracks, as well as serving as flexural steel at mid-span.

   crack
     \       bent-up bar (alpha = 45 deg)
 ====o\=====o=====
        \        \
         \_______/ stirrup

Contribution to shear

The tension T=0.87fyAsbT = 0.87f_yA_{sb} in a bar inclined at angle α\alpha to the beam axis has a vertical component:

Vus=0.87fyAsbsin⁡αV_{us}=0.87f_yA_{sb}\sin\alpha

Here AsbA_{sb} is the total area of the bent-up bars in one plane. For α=45∘\alpha=45^\circ, Vus=0.615fyAsbV_{us}=0.615f_yA_{sb}.

Code provisions (IS 456 cl. 40.4)

  • Bent-up bars should not carry more than 50% of the total shear to be resisted by shear reinforcement; the rest is carried by vertical stirrups.
  • For a single bar or a single group of parallel bars all bent at the same cross-section, VusV_{us} is as above; for a series of bars bent at different sections, spacing is also considered.
  • Bent-up bars must be properly anchored and bent at gradual radius; bends are placed where they are no longer needed for flexure.

Total shear resistance: Vu=Vc+Vus(stirrups)+Vus(bent-up)V_u=V_c+V_{us}(\text{stirrups})+V_{us}(\text{bent-up}), with Vc=τcbdV_c=\tau_cbd.

✓ Answer checked by Claude Opus 5.5 (high effort)

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