Four was never a fact about plates

The coefficient is the whole of the structural content of a plate’s buckling stress, and that essay took it as given — 4 for an internal element, 0.43 for an outstand, 23.9 for a panel in shear — and drew what each is worth. It said the number is an eigenvalue and stopped there, because the consequences of the number are a design subject and its origin is not.

Its origin turns out to be the more interesting half, for a reason that is not visible from the table. The four is not a fact about plates. It is the minimum of a quantity with no plate in it, and every structural feature of the problem lives in the scallops around that minimum rather than in the minimum itself.

Two energies, and no amplitude in either of them #

The derivation is an energy argument and it needs one guess: a shape.

Take a rectangular plate, length $a$, width $b$, thickness $t$, supported along all four edges and compressed along its length. Give it a small out-of-plane displacement that respects those supports —

$$w = A \sin\frac{m\pi x}{a}\,\sin\frac{\pi y}{b}$$

— one half-sine across the width, because that is the only shape the two supported edges leave available, and $m$ half-sines along the length, with $m$ a whole number for the same reason at the ends.

Two quantities follow. Bending the plate into that shape costs strain energy, which depends on the plate’s flexural rigidity $D = Et^3/12(1-\nu^2)$ and on how sharply the surface curves. And as the plate bows out of its plane it becomes shorter along its own surface, so the in-plane stress at the ends does work, which depends on $\sigma$, on the thickness, and on how steeply the surface slopes.

Both are proportional to $A^2$, so the amplitude cancels. That single fact is what makes the answer a stress rather than a deflection, and it is also why nothing here can say how far the plate moves: the calculation has been arranged so that the one quantity a reader might want is the one thing it has divided out.

The tangency is the answer, and it is worth dwelling on because it is not the shape of argument most stability calculations have. A column’s critical load is the smallest eigenvalue of a linear problem. This is a minimisation over shapes: the plate buckles at the lowest stress for which some admissible shape can pay for itself, which is where the work first reaches the bending cost anywhere at all.

The cheapest shape to bend is not the shape that buckles #

The figure contains a result that reads as a mistake until it is checked.

The bending curve has a minimum of its own. Its lowest point is at a half-wave 1.73 times the plate’s width — a long, lazy bulge, which is genuinely the cheapest surface to push a plate into among all the shapes on the axis. That is not where the plate buckles. The tangency sits at a half-wave of exactly one width, a buckle a third shorter than the cheap one.

The reason is that the work term has a length in it too, and it points the other way. A short wave slopes more steeply for the same amplitude, so the plate shortens more along its own surface and the stress does more work. The buckle is a compromise between a shape that is cheap to bend and a shape that pays well, and it lands nearer the paying end than the cheap one.

This is the same structure of argument as guessing a buckling load from an assumed shape, where the error in the load is second-order in the error in the shape. It is why the crude assumption of a single half-sine across the width — which is exact here only because both edges are simply supported — is forgiving elsewhere, and why the coefficient for an elastically restrained edge can be estimated well from a shape that is visibly wrong.

What the algebra reduces to, and where the four comes from #

Divide the two energies and the plate disappears from the comparison.

Write $r$ for the half-wave length as a multiple of the width. Bending goes as $r + 2/r + 1/r^3$ and the work goes as $S/r$, where $S$ is the stress in units of $\pi^2 D / b^2 t$. Setting them equal and solving for $S$ gives

$$S = \left(r + \frac{1}{r}\right)^2$$

and that expression is the buckling coefficient. Everything structural has gone: no modulus, no thickness, no Poisson’s ratio, no width. One number, $r$, and one function of it.

The minimum of $(r + 1/r)^2$ is 4, for every positive $r$, and it is 4 for a reason that predates structural engineering by two millennia. The arithmetic mean of $r$ and $1/r$ is at least their geometric mean, which is 1; so $r + 1/r \geq 2$, with equality only when $r = 1$. Square it and the floor is 4.

That is the whole derivation of the famous coefficient. It is an instance of the arithmetic–geometric mean inequality, applied to a quantity and its reciprocal, and it would be 4 for any problem whose energy ratio has that shape. Nothing in it distinguishes a plate from anything else.

The same shape of answer, three fields away #

A quantity plus its own reciprocal is what appears whenever two costs pull in opposite directions on one dimension, and it has appeared before under other names.

Depth is the cheapest strength a beam has because the force in a chord falls as the inverse of the depth while the material in the web rises with it, so the total has a minimum at a definite proportion and is flat around it. Halving a truss panel shortens the strut and adds two more of them, which is the same trade at a smaller scale and produces the same flat-bottomed answer. In both cases the interesting statement is not where the minimum sits but how little is lost by missing it — which is the property $(r + 1/r)^2$ has and is the reason the plate can be a whole half-wave away from square and pay three per cent.

The flatness is worth one line of arithmetic. Near $r = 1$ the coefficient is $4 + 4(r-1)^2$ to leading order, so a ten per cent error in the half-wave length costs four hundredths of one per cent of coefficient. A quantity at a minimum is insensitive to its argument by construction, and every design rule that quotes a proportion rather than deriving one is relying on it, usually without saying so.

What the plate actually contributes #

The plate’s own contribution arrives at the next line, and it is a restriction rather than a value.

$r$ cannot be chosen. The half-wave length is the plate’s length divided by a whole number of half-waves, so $r = \alpha/m$ where $\alpha = a/b$ is the aspect ratio and $m$ is an integer. A plate of aspect ratio 2.4 may have a half-wave of 2.4 widths, or 1.2, or 0.8, or 0.6 — and nothing between them.

So the answer is not the minimum of the curve. It is the smallest value the curve takes at one of a discrete set of points, and the discreteness is the only part of the calculation that is about a plate. The rest is arithmetic.

That inversion is worth carrying, because it tells a reader which way an unfamiliar case will go. Change the material and the coefficient does not move. Change the edge conditions and the assumed shape across the width changes, so the whole expression changes and 4 stops being the floor — which is why an outstand gets 0.43 and no rearrangement of half-waves will recover it. Change the length and only the set of available points moves, which is a small effect and is the subject of the rest of this essay.

The scallops, and what is at the bottom of each #

Sweeping the aspect ratio and asking the same question at every value produces one curve per half-wave count.

Every branch has the same minimum and the same value at it. The $m$-th branch is $(m/\alpha + \alpha/m)^2$, which is the same function of $\alpha/m$ as the first branch is of $\alpha$, stretched along the axis by a factor of $m$. So the whole family is one curve at four scales, and the envelope is a chain of identical scallops getting wider.

The crossings have a closed form worth knowing. Branch $m$ and branch $m+1$ are equally good when $\alpha = \sqrt{m(m+1)}$, which puts the first cusp at $\sqrt{2} = 1.414$ and the next at $\sqrt{6} = 2.449$. At those aspect ratios the plate is genuinely indifferent between two buckled shapes — a real degeneracy rather than a rounding, and the one place where a plate’s mode shape is not determined by its geometry.

Two shapes that are equally good, and what decides between them #

A cusp is a genuine degeneracy and it is the one place in this problem where the geometry stops determining the answer.

At $\alpha = \sqrt{2}$ a plate has two buckled shapes with identical critical stresses, one with a half-wave 1.414 widths long and one with two of 0.707. Linear theory says both, offers no way to choose, and is right: for a perfectly flat plate at exactly that aspect ratio the buckling mode is a two-dimensional space rather than a shape, and any combination of the two is also a mode.

What breaks the tie is the plate’s own out-of-flatness. A plate rolled or welded with a gentle single bow will take the one-wave mode because that is the shape its imperfection already resembles; one with a ripple from a weld run will take two. The mode a real plate at a cusp takes is a property of its manufacture rather than of its design, which is not a comfortable sentence and is the honest one.

It has a practical edge. A longitudinal stiffener placed at the middle of the length is a node for the two-wave mode and an antinode for the one-wave mode, so its usefulness at $\alpha = \sqrt{2}$ depends on which shape the plate was going to take — and the plate has not decided. Away from a cusp the question does not arise, because the two competing modes differ in critical stress by enough that the imperfection cannot outvote the geometry.

Why the length drops out, stated as a number #

The reason a plate’s length is absent from every design expression is usually given as a sentence — a long plate buckles into more waves of the same size — and the sentence is true and does not say how much the effect is worth.

The arithmetic is exact and short. The envelope exceeds 4.2 in only one interval above unit aspect ratio, from 1.248 to 1.602 — a window 0.35 wide in the whole of the positive axis. It exceeds 4.1 in three windows and 4.05 in four, each narrower than the last, and by an aspect ratio of four the scallops are 1.3 per cent deep.

A design formula that writes 4 and omits the length is therefore wrong by at most 12.5 per cent in stress and 6.1 per cent in permissible width, and only at $\alpha = \sqrt{2}$. That is the honest size of the simplification, and it is small enough that no code has ever thought it worth a column of table.

There is a second reason the worst case is rarer than the envelope makes it look. A plate girder’s web panel between transverse stiffeners is usually proportioned at an aspect ratio between 1 and 2 for reasons that have nothing to do with this curve — the stiffener spacing follows from shear buckling and from fabrication — and a compression flange between longitudinal stiffeners runs to aspect ratios in the tens. The first of those straddles the only expensive window on the whole axis; the second is far out on the flat part where the scallops are worth one per cent. So the case where knowing the aspect ratio would be worth something is exactly the case in which a designer has least freedom to choose it.

The error has a sign, and that is why nobody minds #

There is a stronger statement available and it is the one that actually justifies the practice.

The scallops are all on one side. Four is not a value near the middle of the envelope — it is the infimum of it, attained exactly at integer aspect ratios and exceeded everywhere else. So a designer writing $k = 4$ is not making an approximation that might go either way. They are computing a critical stress that is always at or below the true one, by an amount that vanishes at integer aspect ratios and never exceeds an eighth.

Six per cent of width is not nothing, and it is roughly the accuracy with which the rest of the calculation is known. The gap between the derived limit and the quoted one is 33 per cent, and that gap is residual stress and out-of-flatness. An aspect-ratio refinement of six per cent sits inside it, which is the real reason it is never made.

A column’s length does not drop out, and the contrast says why #

The habit of ignoring a member’s length is peculiar to plates, and setting it beside the member it is usually confused with makes the mechanism visible.

A column’s ends decide an effective length and the load then falls as the inverse square of it, without limit. The reason is that a pin-ended column has exactly one shape available at its lowest eigenvalue and the shape spans the whole member: lengthening it lengthens the buckle. A plate’s buckle does not have to span the plate, because the supported edges supply a length scale that the ends do not. The plate divides itself into squares and the number of squares is the only thing that changes.

The general statement is that a structure forgets its length exactly when it has a second length scale of its own to buckle at. A beam restrained continuously along its length forgets its span in the same way, and for the same reason: the restraint sets a wavelength, the member takes a whole number of them, and the span becomes a rounding. A rib that is stiff enough to be a node is that mechanism applied deliberately, halving the width rather than the length.

What the assumed shape is doing, and what it is not #

The single half-sine across the width is not a simplification here and is one everywhere else, and the distinction matters.

For a plate simply supported along both longitudinal edges, $\sin(\pi y/b)$ is the exact eigenfunction: it satisfies the plate equation and both boundary conditions, so the coefficient of 4 is exact rather than an upper bound. Nothing about the derivation above is approximate except the arithmetic.

For every other edge condition it is an assumption, and the energy method returns an upper bound on the coefficient rather than the coefficient. That is the right direction for safety in a deflection calculation and the wrong direction here, because an overestimated coefficient is an overestimated critical stress. It is the reason published coefficients for restrained edges are computed rather than assumed, and the reason the 0.43 for an outstand is not a member of the family this essay has been describing at all.

Where this calculation cannot be pushed #

Nothing here is a failure load. The tangency locates a bifurcation of a perfect flat plate. A real plate is not flat, so it has no bifurcation at all — it deflects from the first newton, and what is left after it ripples is a separate calculation with a separate empirical curve in it.

The amplitude is undetermined by construction. $A$ divided out, and with it every stress, every deflection and any statement about whether the buckle is a nuisance or a collapse. A panel in shear carries several times its critical load afterwards; a cylindrical shell reaches a third of it. The eigenvalue is the same kind of object in both cases and predicts neither outcome.

The ends are assumed to hold the plate straight. That is what makes $m$ an integer, and it is the only reason the scallops exist. A plate whose ends are free to warp out of plane has no such restriction, its buckle is not periodic, and the whole envelope argument lapses.

The stress is uniform along the length. A plate in a member under a moment gradient has a stress that varies from one end to the other, so no single $\sigma$ appears in the work integral and the minimisation is over a different problem.

And the plate is one plate. A stiffened panel has a local mode and an overall one, and the coincidence of two modes is a hazard rather than an efficiency — the branches drawn here cross harmlessly because both belong to the same plate and neither is a collapse.

Still open: the coefficient a web has rather than the one it is given #

Every coefficient in this essay belongs to a stress pattern that was known before the plate was analysed. Uniform compression is one such pattern, and it is the only one for which that is unconditionally true.

A web in bending is the case where it is not. Its coefficient depends on the gradient of stress across it, the gradient depends on where the elastic neutral axis is, and the neutral axis depends on how much of the web is still working — which is what the coefficient was being computed to find out. The three quantities close a loop, and the answer is a fixed point rather than a formula. A section already knows how to lose part of itself; what it has not yet had to do is lose part of itself and then be re-measured because of it.

Beyond that, Winter’s expression for the effective width, whose exponent is an empirical fit through test data and has survived every attempt to derive it; and the coefficient for a shear panel, which is the one case where the aspect ratio does not drop out, because a diagonal buckle’s wavelength is set by the proportions of the panel and not by its width alone.