Shear stress in beams
A beam carries vertical shear force as well as bending moment. The shear is not spread evenly over the depth. It is zero at the top and bottom faces, which are free surfaces, and largest near the neutral axis. The distribution comes from horizontal equilibrium of a slice of beam: as bending moment changes along the span, the bending stresses on two faces of a slice differ, and horizontal shear on the cut must balance the difference.
The result is the shear formula τ = V Q / (I b). V is the shear force, I the second moment of area of the whole section, b the width at the level considered, and Q the first moment of the area beyond that level about the neutral axis. For a rectangle this gives a parabola with a peak of 1.5 times the average V / (b d). For a circle the peak is 4/3 of the average. In an I-section the web carries nearly all the shear at an almost uniform stress, while the flanges carry little vertical shear, which is why steel designers check web shear as V / (d t_w).
The same horizontal shear flow q = V Q / I tells us how much force the bolts, welds or shear studs must carry to make built-up and composite beams act together. Reinforced concrete departs from this elastic picture once it cracks, so concrete codes use empirical shear strengths and stirrups instead.
τ = V Q / (I b)Rectangle: τ_max = 1.5 V / (b d)Circle: τ_max = (4/3) V / AShear flow: q = V Q / I; connector spacing s = F_connector / q
- τ = VQ/(Ib) assumes linear elastic, uncracked, prismatic beams and gives the average shear stress across the width b at that level.
- It is accurate for narrow rectangles; for wide, shallow sections the stress varies across the width and the formula underestimates the peak.
- In thin-walled open sections, shear flow also runs along the flanges; the formula is then applied along the wall thickness.
Problem. A timber beam 100 × 250 mm carries a maximum shear of 18 kN. Find the maximum shear stress and the stress 50 mm above the neutral axis.
- Average = 18 000 / (100 × 250) = 0.72 MPa; maximum = 1.5 × 0.72 = 1.08 MPa.
- I = 100 × 250³ / 12 = 130.2 × 10⁶ mm⁴.
- Area above y = 50 mm: 100 × 75 = 7 500 mm², centroid at 87.5 mm, so Q = 656 250 mm³.
- τ = 18 000 × 656 250 / (130.2 × 10⁶ × 100) = 0.91 MPa.
Answer. 1.08 MPa at the neutral axis; about 0.91 MPa at 50 mm above it
Quick quiz
1Where is shear stress zero in a rectangular beam?
At the top and bottom faces.
2Ratio of maximum to average shear in a rectangle?
1.5.
3Which part of an I-beam carries most vertical shear?
The web.
- IS 800 (India): General construction in steel: limit state design of steel members and connections.
- IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
Clauses and documents to read
- IS 800:2007 cl. 8.4 (shear)
- IS 456:2000 cl. 40 (shear in RC)
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.
- J. M. Gere, B. J. Goodno, Mechanics of Materials, Cengage, latest edition.
- F. P. Beer, E. R. Johnston, J. T. DeWolf, Mechanics of Materials, McGraw Hill, latest edition.
- E. P. Popov, Engineering Mechanics of Solids, Prentice Hall, 2nd edition.
Free and official sources
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