Transverse Shear Stress (τ = VQ/Ib)

Bending stress varies linearly from the neutral axis and peaks at the extreme fibres. Shear stress does the exact opposite: it is zero at the top and bottom faces (there is nothing outside to shear against) and maximum on the neutral axis, the one place where the bending stress is nil. For a solid rectangle the algebra collapses to a memorable shortcut: $\tau_{max} = 1.5\,V/A$. A 50 × 150 mm section carrying 30 kN gives $1.5 \times 30{,}000/0.0075 = 6$ MPa, and running the full $VQ/Ib$ with $Q = 140{,}625$ mm³, $I = 14.06 \times 10^6$ mm⁴ and $b = 50$ mm returns the same 6 MPa. Two routes, one answer.

Dmitrii Zhuravskii derived this in the 1840s after timber bridges on the Saint Petersburg–Moscow railway kept splitting along their length rather than snapping across, and the split is the signature: a wooden beam overloaded in shear fails horizontally near mid-depth, right where this formula says the stress is worst. The same logic sizes the web of a steel beam, where the shortcut is different again — for a wide flange, nearly all the shear rides in the web, and designers approximate $\tau \approx V/(d\,t_w)$ and get within a few percent.

The trap is $b$: it is the width at the cut you are examining, not the overall width. Step from a flange into a web on a tee or an I-shape and $b$ drops by an order of magnitude while $Q$ barely changes, so the stress jumps discontinuously at the fillet. That step is real in the formula and smoothed in the actual part, which is why the theory overpredicts slightly at re-entrant corners and why the underlying assumption — shear uniform across the width — quietly fails in a wide, shallow section.