Transverse Shear in Beams (τ = VQ/Ib, Shear Flow, Fastener Spacing)

stress

Bending gets all the attention, but a loaded beam is also being sheared — the transverse force $V$ tries to slide each horizontal layer past the one below it. That longitudinal sliding is resisted by shear stress inside the beam, and for a beam built up from separate pieces it is exactly what the nails, screws, or bolts have to carry. This page is the shear formula $\tau = VQ/Ib$, the parabolic stress it predicts, and the fastener spacing that falls out of it — the same section whose bending the simply-supported beam and cantilever beam pages handle.

$\tau = \frac{V Q}{I\,b}, \qquad \tau_{max} = \frac{3V}{2A}\ \text{(at the neutral axis)}, \qquad q = \frac{V Q}{I} = \tau\,b$

Here $Q$ is the first moment, about the neutral axis, of the part of the cross-section beyond the cut you are checking; $I$ is the second moment of the whole section, and $b$ the width at the cut.

The parabola, and why the middle is hottest

Bending stress is largest at the top and bottom fibers and zero at the neutral axis. Transverse shear is the exact opposite: it is zero at the top and bottom surfaces — there is nothing beyond them to shear off, so $Q = 0$ — and largest at the neutral axis, where $Q$ is biggest. For a rectangle the distribution is a clean parabola, and the peak works out to

$\tau_{max} = \frac{3V}{2A} = \tfrac{3}{2}\,\tau_{avg},$

exactly 1.5 times the naive “force over area” value, independent of size or load. Drive $V$ and the section knobs in the widget and watch the parabola breathe — always peaked in the middle, always dead at the edges. (Note how large the safety factor usually is: transverse shear rarely governs a slender beam. Where it bites is short deep beams — the widget warns you there — and, always, the connections.)

Shear flow sizes the fasteners

Multiply the shear stress by the width it crosses and you get the shear flow $q = \tau b = VQ/I$: the longitudinal force per unit length of beam that any horizontal cut must transmit. Build a beam by nailing two planks together along the neutral axis and the nails have to carry that flow. Each fastener, spaced $s$ apart, resists

$F_{fast} = q\,s,$

which is the equation that actually sets the nail spacing in a plywood box beam or the bolt pitch in a plate girder — tighten $s$ and each connector carries less. Convention on this page: $F_{fast}$ is the force at one interface per spacing $s$; if you put a row of two paired fasteners at each station, each carries half. (The widget evaluates $q$ at the neutral axis, its maximum — the critical joint; a seam located up at a flange carries the smaller $q = VQ_{flange}/I$ of that flange’s own first moment.)

Three quantities, one dimension — why kinds exist

This page is the catalog’s clearest illustration of invariant 2. The shear flow $q$ is measured in newtons per metre. So is a beam’s distributed load $w$ (the simply-supported beam carries one). So is a spring rate $k$. All three share the identical SI dimension vector $[\mathrm{M\,T^{-2}}]$ — and yet feeding one into another would be physical nonsense: a shear flow is not a load you can hang on a structure, and a stiffness is neither. The engine refuses those connections not by dimension — which would allow them — but by a separate quantity_kind tag (shear_flow, line_load, stiffness). Same units, three meanings, kept apart on purpose.

There is even a second, unrelated “shear flow” in the catalog: Bredt’s torsional shear flow in a closed thin-walled tube, $q = T/2A_m$, which runs around a closed cell rather than along a beam. Same words, different physics — the open beam here versus the closed tube there is one of the sharpest contrasts on the site.

A strength-only material axis

Only one material property enters, and it enters weakly: $\sigma_y$ sets the shear-yield warning ($\tau_{max}$ vs $\sigma_y/2$, the Tresca criterion) and the safety factor. Stiffness $E$ and density $\rho$ do not appear at all — the elastic shear-stress distribution in a homogeneous prismatic beam is pure statics and geometry. That is the honest picture, not a padded one: swapping steel for aluminium changes nothing here except the margin. The build’s physics test proves the $VQ/Ib$ derivation from slice equilibrium, the $3/2$ peak, and the theorem that the parabola integrates back to precisely $V$; the /verification/ page records what is machine-proven and what rests on citation.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget answers a narrow elastic question — given this section and this shear force, what is the transverse shear stress and the shear flow? — and turns the shear flow into a fastener force. The ways a real beam gets into trouble in shear mostly live in what the page deliberately leaves out.

• The formula assumes the shear is uniform across the width — for a wide section it is not. $\tau = VQ/Ib$ quietly assumes the shear stress is the same all the way across the width $b$ at a given height. That is excellent for a tall narrow rectangle but increasingly wrong as the section gets wide and flat: the true stress bulges toward the edges, and the flanges of an I-beam or wide-flange carry shear in a way this one-dimensional formula only approximates. Wide sections need the full two-dimensional elasticity solution the page does not attempt.

• Unsymmetric and open thin-walled sections have a shear center — miss it and you add torsion. For a channel, an angle, or any section without two axes of symmetry, the transverse load must pass through a particular point — the shear center, generally not the centroid — or the beam twists as well as bends, superposing a torsional shear on the transverse one. This page assumes a doubly-symmetric section loaded through its centroid, so the shear center never comes up; it is its own topic (and its own future THING).

• Shear deflection is computed nowhere here — only warned about. Euler–Bernoulli beam theory (the simply-supported and cantilever pages) ignores the deflection this very shear causes, which is fine for slender beams and not fine for short deep ones. The page flags $L < 10h$ with a warning but does not compute the extra sag — Timoshenko beam theory does, and that is out of scope.

• The fastener readout is force, not a complete connection design. $F_{fast} = q\,s$ tells you the shear a connector carries, but a real joint check also involves the fastener’s own shear and bearing capacity, edge distances, the direction the flow reverses near supports and load points (where $V$ flips sign and the spacing may need to tighten), and — for glued or welded seams — the strength per unit length rather than per fastener. And mind the paired-fastener bookkeeping the overview flags: counting a double row as a single connector halves the force you think each one sees.

• Static and elastic only. No fatigue at the connectors (where cyclic shear actually cracks built-up members), no plastic shear redistribution once the neutral-axis fibers yield, no stress concentration at bolt holes, no combined bending-plus-shear interaction — near a support a beam sees its largest $V$ and a large $M$ at once. Read the page as where the shear goes and what the connection must carry, not as a finished design.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §5.8–5.11 (shear stresses in beams of rectangular cross section, the shear formula τ = VQ/Ib, shear flow, and built-up beams / fastener spacing).
• Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §5-4 (maximum-shear-stress theory, S_sy = 0.5 S_y).