The section has two areas

Every first calculation of a shear stress divides the shear force by the area of the section, and every such calculation is wrong everywhere in the section.

$$\tau_{avg} = \frac{V}{A}$$

is an average over a distribution that is zero at the top face, zero at the bottom face, and largest somewhere in between. For a rectangle the distribution is a parabola and its peak is exactly

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

— one and a half times the average, with nothing fitted about the number. It is one of the very few results in this subject that can be written down in closed form and is almost universally ignored.

The ratio is called the form factor, and different sections have wildly different ones. That fact is the essay.

Which free body produced the number #

The horizontal one. Take a slice of beam of length $dx$ and cut it again with a horizontal plane at height $y$; the free body is the piece above the cut, between the two vertical cuts.

The bending stresses on the two vertical faces are not equal, because the moment changes along the beam. The difference is a net horizontal force, and the only thing available to balance it is a shear on the horizontal cut. Equating gives

$$q = \frac{VQ}{I}, \qquad \tau = \frac{VQ}{Ib}$$

with $Q$ the first moment of the area above the cut. The shear nobody draws is the essay about that horizontal shear flow and about the fact that it is the same shear as the vertical one, by complementarity.

What matters here is what the expression contains. There is a $Q$, which grows as the cut moves toward the neutral axis and vanishes at the free faces, and there is a $b$ in the denominator, which is the width at the cut. So the shear stress is large where there is a lot of area above the cut and where the section is narrow — and for a section that has both at once, it is very large indeed.

Why an I-section is two members #

A rolled I-section has flanges that are wide and thin and a web that is narrow and deep. Run the shear flow formula down it and something striking happens at the junction: $Q$ is continuous, $I$ is a constant, and $b$ drops from the flange width to the web thickness — a factor of twenty or more. The shear stress jumps by that factor.

The result is that the web carries around ninety per cent of the shear on about a fifth of the area, and the flanges carry nearly all of the moment. The two are not merely different; they are almost disjoint.

This is why the two design checks on the same beam look at two different objects. A moment check uses the whole section, and the flanges dominate it. A shear check uses $A_v$, the shear area, which for a rolled section is conventionally the web at the overall depth — and the flanges are simply absent from it.

Both at once and neither matters is the essay about what happens when the two arrive together, and the answer there — that the interaction is negligible over most of the range and bites only in the corner — is a consequence of exactly this separation. Two checks on two different pieces of steel do not interact until one of them has used up nearly all of its own piece.

There is a practical corollary that follows from the separation and is worth stating on its own, because it explains a detailing rule that otherwise looks arbitrary. If the flanges carry the moment and the web carries the shear, then a hole in the web costs shear capacity and very little moment capacity, and a hole in the flange costs the opposite. That is why service openings in beams are cut through the web at mid-depth — where the moment contribution is smallest — and why they are kept away from the supports, where the shear is largest. The hole that costs nothing and everything is that trade drawn properly, and the reason it is a trade at all is the separation above.

It also explains why a beam’s web is so thin. A designer sizing a web is sizing it for shear, which is largest at the supports and zero at midspan on a uniformly loaded beam, while the flanges are sized for moment, which does the opposite. Two quantities with opposite distributions along the member, carried by two pieces of the section that barely overlap — the member is very nearly two structures sharing an outline, and each is thickest where the other is thinnest.

The two shear areas, which are not the same number #

Here is the part that surprises people who have used both.

A shear area appears in two completely different calculations, and the value that belongs in each is different for the same section.

For strength, the shear area is the area that will be at yield when the section fails in shear. The plastic shear capacity is $A_v f_y/\sqrt{3}$, and $A_v$ is the area that can actually reach the shear yield stress — the web, taken at the full depth.

For deflection, the shear area is an energy quantity. The shear strain energy in a length $dx$ is

$$dU = \frac{V^2}{2GA_s}dx$$

and $A_s$ is defined so that this expression gives the same energy as integrating $\tau^2/2G$ over the real distribution. That definition produces a different number: for a rectangle it is $\tfrac{5}{6}A$, not $\tfrac{2}{3}A$, and the two differ by a quarter.

For a rectangle the two are $0.667A$ and $0.833A$. For an I-section they are much closer, because the web really does carry nearly all the shear and the energy integral and the plastic area very nearly agree. So the discrepancy is worst exactly where the section is simplest, which is the opposite of the usual pattern and is why it is so easy to miss: anybody checking the idea on an I-beam will find the two areas agree to a few per cent and conclude that the distinction is pedantry.

It is not. The deflection that is not bending is the essay about when the shear term matters at all — short members, deep members, sandwich sections, wall panels — and every one of those cases is a case where the section is stubby and the two areas differ.

The reason the two definitions disagree is worth stating because it is a general fact about how a “equivalent area” is arrived at. The plastic area answers a question about a limit: which material can reach the yield stress. The energy area answers a question about an integral: how much strain energy the real distribution stores. A limit is decided by the largest value in a distribution and an integral by the mean of its square, and those two summaries of the same curve are never the same number unless the curve is flat. The only section for which the two shear areas coincide exactly is one whose shear stress is uniform — which is a thin closed tube, and nothing else.

That is the same argument one deflection without solving everything makes about virtual work: a deflection is an energy statement, and energy statements weight a distribution differently from strength statements. Two calculations on the same member using the same symbol for two different quantities is a trap the notation sets, and the tables do not help — a section table gives $A_v$ and calls it the shear area, without saying which of the two it is.

The number 1.5, and where it comes from #

The rectangle’s form factor is worth deriving because it is one of the few numbers in structural mechanics that is exactly a small fraction.

For a rectangle of breadth $b$ and depth $h$, at a height $y$ from the neutral axis, the area above the cut has first moment

$$Q = \frac{b}{2}\left(\frac{h^2}{4} - y^2\right)$$

and $I = bh^3/12$, so

$$\tau = \frac{VQ}{Ib} = \frac{6V}{bh^3}\left(\frac{h^2}{4} - y^2\right)$$

which is a parabola, zero at $y = \pm h/2$ and maximum at $y=0$ where it is $3V/2bh$. Divide by the average $V/bh$ and the answer is $3/2$, with no dimension left in it.

A circle gives $4/3$. A thin tube gives $2$. A diamond gives $9/8$. Every one of them is a pure number and every one is a property of the shape.

Where the separation breaks down #

The tidy picture — flanges for moment, web for shear — holds for a rolled section and fails in four places, each of which is a different member.

A plate girder has a web so slender that it buckles in shear before it yields, at which point the shear is carried by a diagonal tension field across the panel rather than by a uniform shear stress. The shear area stops being an area at all and becomes a band. The panel that carries more after it fails is what happens next.

A corrugated web has almost no axial stiffness along the beam, so it carries essentially all of the shear and none of the moment — the separation taken to its logical end, deliberately.

A box section has two webs, so the shear area is doubled and the shear flow goes round a closed loop — which changes the torsion problem completely and leaves the shear problem very nearly the same. The closed loop is worth a moment’s attention because it is the one case where a shear flow does something a shear stress cannot: it can circulate. An open section’s shear flow starts and ends at free edges and has to integrate to the applied shear; a closed one can carry a constant flow all the way round in addition, which is exactly what a torque is. So a box has a shear area and a torsion constant, and the same web thickness appears in both, doing two jobs that an open section does with one.

And a channel or an angle has a shear flow whose resultant does not pass through the centroid, which is the point that is not in the section. The shear area is still meaningful; the line of action of the shear is not where the section is.

Where the model stops #

The distribution assumed the shear stress is constant across the width. It is not: in a wide flange the shear stress varies across the breadth and the formula’s $\tau = VQ/Ib$ is a width-average. For a rectangle of aspect ratio near one the error is a few per cent; for a wide flat bar it is larger.

Plane sections were assumed to stay plane, and shear is exactly the assumption’s failure. A member carrying shear warps out of plane, which is why a shear deflection exists at all, and the section that warps is the section the bending calculation assumed did not. The two calculations rest on contradictory assumptions and are added together, which works because both errors are small.

Nothing here is about a section that has already yielded. The plastic shear capacity assumes the whole shear area is at the shear yield stress at once, which is a rectangular distribution rather than a parabolic one — so the plastic form factor is 1.0 where the elastic one was 1.5, and a rectangle has a shear shape factor of exactly 1.5.

Nothing here is about how the shear got into the section. A shear force at a section is a resultant, and how it is delivered — through a bearing, a bolt group, a welded end plate — decides the stress in the first member depth, which is a discontinuity region and not a section at all. A shear area is a property of a member away from its ends.

And the conventional web area is generous. Taking the web at the overall depth credits the section with the small pieces of web inside the flange thickness, which is a two or three per cent optimism that every shear check in the world contains.

The generalisation #

The habit worth taking away is to be suspicious of any calculation that divides a force by “the area”.

A section is not one number. It has a gross area, a net area, an effective area, a shear area for strength, a shear area for energy, a bearing area, an area at a bolt line and an area after local buckling has removed part of it — and every one of those is the right answer to a different question. Reaching for $A$ is a decision about which question is being asked, and it is very often made without being noticed.

There is a second habit that goes with it, and it is about reading a formula rather than choosing a number. Whenever a section property appears in a check, ask what integral it came from. A gross area came from $\int dA$; a second moment from $\int y^2 dA$; a shear area from an energy integral or a yield condition; a torsion constant from a warping problem. Each of those integrals weights the material differently, so each ranks the same set of shapes differently — which is why the same steel in a different shape has as many answers as there are questions, and why a single figure of merit for a section does not exist.

The deeper point is the one geometry beats material keeps making. Nothing in this essay is about steel. A form factor of 1.5, a web carrying ninety per cent of the shear, two shear areas differing by a quarter — all of them are consequences of where the material is, computed from the same free body, and all of them hold for a section of any material whatever. The section shape decides how the force is distributed inside it, and the distribution is what a stress is.