Bending stresses act along the beam axis; transverse shear stresses act across the section and balance the shear force. Understanding their distribution explains why beams split horizontally near supports, why I-section webs are checked for shear, and how built-up beams are bolted together. Here is the derivation and the worked examples.
The Shear Stress Formula
τ = V · Q / (I · b)| Symbol | Meaning |
|---|---|
| τ | Shear stress at the level considered |
| V | Transverse shear force at the section |
| Q | First moment of area (A·ȳ) of the portion above (or below) the level, about the neutral axis |
| I | Moment of inertia of the entire cross-section about the NA |
| b | Width of the section at the level considered |
Derivation (Equilibrium of a Beam Element)
Consider two sections dx apart where the bending moment changes from M to M + dM. The bending stresses on the two faces differ, so the top "block" of the section above a level y₁ has an unbalanced horizontal force. That imbalance is resisted by horizontal shear on the cut at y₁.
The unbalanced force = ∫(dM·y/I) dA over the area above y₁ = (dM/I)·Q, where Q = ∫y dA = A·ȳ.
This must equal the shear stress times the area it acts on, τ·b·dx:
τ · b · dx = (dM / I) · Q → τ = (dM/dx) · Q / (I b) = V Q / (I b)since dM/dx = V (the shear force). By the principle of complementary shear, the horizontal shear stress equals the vertical shear stress at the same point.
Rectangular Section — The Parabolic Distribution
For a rectangle b × d, at distance y from the NA:
τ = (6V / b d³) · ( d²/4 − y² )- At the top/bottom (y = ±d/2): τ = 0.
- At the neutral axis (y = 0): τmax = 1.5 · V/A.
So the maximum shear stress is 50% higher than the average V/A — a result every designer memorises.
Worked Example 1 — Rectangular Beam
A 150 mm wide × 250 mm deep beam carries a shear force of 75 kN. Find the average and maximum shear stress.
- A = 150 × 250 = 37 500 mm²
- τavg = V/A = 75 000 / 37 500 = 2.0 MPa
- τmax = 1.5 × 2.0 = 3.0 MPa (at the neutral axis)
Worked Example 2 — Shear at a Glued Joint
A rectangular timber beam 100 mm wide × 200 mm deep is made of two 100 mm planks glued at mid-depth. The shear force is 20 kN. Find the shear stress at the glue line.
- I = bd³/12 = 100 × 200³ / 12 = 6.667 × 10⁷ mm⁴
- Q for the top half about NA: A = 100 × 100 = 10 000 mm²; ȳ = 50 mm; Q = 500 000 mm³
- τ = VQ/(Ib) = (20 000 × 500 000)/(6.667×10⁷ × 100) = 1.5 MPa
The glue must safely resist 1.5 MPa of horizontal shear — exactly the "shear flow" that holds built-up beams together.
I-Sections — Why the Web Governs
Because τ ∝ Q/b, the shear stress jumps sharply where the width changes from the wide flange to the thin web. The flanges carry very little shear; the web carries the overwhelming majority. This is why IS 800 (and AISC) shear design uses the approximation:
τ_avg(web) ≈ V / (d · t_w)and why a beam can fail by web buckling or web shear yielding near heavily loaded supports.
Practical Consequences
- Timber beams can split horizontally along the neutral axis near supports (high shear, low bending there).
- Bolt/weld spacing in built-up girders is set by the shear flow q = VQ/I.
- Short, heavily loaded beams may be governed by shear, not bending.
Common Mistakes
- Using the width b of the whole section instead of the local width at the level of interest.
- Computing Q for the wrong portion — it is the area beyond (outside) the level, not the whole section.
- Assuming τ is uniform; it is parabolic in rectangles and far from uniform in flanged sections.