Shear design
Shear in a reinforced concrete beam shows up as diagonal cracks near the supports, where shear force is high. Unlike flexural failure, which is usually ductile and slow, shear failure can be sudden. Codes are therefore deliberate about it: they make sure a beam fails in bending, not in shear.
The design starts with the nominal shear stress τv = Vu / (b d). The code gives a design shear strength of concrete τc, which rises with the percentage of tension steel and the concrete grade, because more steel limits crack width and the uncracked compression zone carries more. There is also a maximum τc,max; if τv exceeds it, the section is too small and must be enlarged, since no amount of stirrups will save crushing of the web. If τv is less than τc, only minimum shear reinforcement is required. Otherwise, stirrups are designed for the excess shear Vus = Vu − τc b d.
Vertical stirrups resist that excess by the truss analogy: inclined cracks are crossed by stirrup legs that act as tension ties, while the concrete between cracks acts as compression struts. Spacing comes from Vus = 0.87 fy Asv d / sv. Maximum spacing rules keep at least one stirrup crossing every likely crack. Bent-up bars can share part of the shear. Near supports carrying direct loads, enhanced strength is allowed for short shear spans, and in seismic zones closer stirrups are required at beam ends.
τv = Vu / (b d)Vus = Vu − τc b dVertical stirrups: sv = 0.87 fy Asv d / Vus (fy not more than 415 MPa)Minimum: Asv / (b sv) ≥ 0.4 / (0.87 fy), fy not more than 415 MPa
- Applies to prismatic beams designed to IS 456 limit state method; τc depends on the tension steel actually continued past the section.
- Members with axial compression, deep beams and corbels need modified checks (see IS 456 cl. 40.2.2 and the deep-beam rules).
Problem. A beam 300 × 500 mm (d = 460 mm), M25, Fe 500 stirrups, carries Vu = 220 kN. Take τc = 0.55 MPa for its tension steel ratio (read from IS 456 Table 19 for the actual steel). Design 2-legged 8 mm stirrups.
- τv = 220 000 / (300 × 460) = 1.59 MPa, below τc,max = 3.1 MPa for M25, so the section size is adequate.
- Vus = 220 − 0.55 × 300 × 460 / 1 000 = 220 − 75.9 = 144.1 kN.
- IS 456 does not allow fy above 415 MPa in the stirrup design formula, so use 415 even though the bars are Fe 500.
- Asv = 2 × 50.3 = 100.6 mm². sv = 0.87 × 415 × 100.6 × 460 / 144 100 = 116 mm.
- Maximum spacing 0.75d = 345 mm or 300 mm, so strength governs; adopt 2-legged 8 mm at 110 mm c/c near the supports.
Answer. 8 mm 2-legged stirrups at 110 mm c/c near the supports
Quick quiz
1Why is shear failure treated more cautiously than flexure?
It can be sudden and brittle.
2What happens if τv exceeds τc,max?
The section must be enlarged.
3Which analogy explains stirrup action?
The truss analogy: stirrups are ties, concrete struts carry compression.
- IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
Clauses and documents to read
- IS 456:2000 cl. 40.1 (nominal shear stress)
- IS 456:2000 cl. 40.2 (design shear strength of concrete)
- IS 456:2000 cl. 40.4 (design of shear reinforcement)
- IS 456:2000 cl. 26.5.1.5 and 26.5.1.6 (spacing and minimum shear reinforcement)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- S. Unnikrishna Pillai, Devdas Menon, Reinforced Concrete Design, McGraw Hill Education (India), latest edition.
- J. K. Wight, J. G. MacGregor, Reinforced Concrete: Mechanics and Design, Pearson, latest edition.
- A. H. Nilson, D. Darwin, C. W. Dolan, Design of Concrete Structures, McGraw Hill, latest edition.
- R. Park, T. Paulay, Reinforced Concrete Structures, Wiley, 1975.
Free and official sources
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