The coefficient that is not four

A plate ripples rather than buckling as a column does, and the stress at which it does so is

$$\sigma_{cr} = k\,\frac{\pi^2 E}{12(1-\nu^2)}\left(\frac{t}{b}\right)^2$$

Everything in that expression except $k$ is either a material constant or a dimension. $k$ is the whole of the structural content, it is written as a single letter, and it is quoted so often as 4 that it reads like part of the formula.

What the coefficient is #

$k$ is the eigenvalue of a plate’s buckling problem in dimensionless form, and it depends on three things and nothing else.

How the longitudinal edges are held. A plate supported along both edges buckles into a shape with a node in the middle of each half-wave; one supported along one edge and free along the other has nothing to hold its free edge, and its buckle is a cantilever curl.

How the stress is distributed across the width. Uniform compression is one case; a linear gradient — compression on one edge, tension on the other, as in the web of a beam in bending — is another, and the compressed strip is narrower so the plate buckles later.

What kind of stress it is. Shear buckles a plate into diagonal waves rather than longitudinal ones, and the coefficient for that mode is much larger.

A factor of 7.4 in the permissible width, from one number. That is the practical importance of the coefficient and it is why a flange outstand and a web of the same steel and the same thickness are governed by limits that look unrelated.

The three figures also show something the coefficient alone does not: what the plate is worth past its limit. At 900 mm the internal element still delivers 46 per cent of its width, the outstand 16 per cent and the shear panel 91. The coefficient decides where the fall begins and the same coefficient decides how steep it is, because both are functions of the ratio of the actual width to the limiting one — so a plate that buckles early also loses effectiveness fast, and the outstand is punished twice.

That is the reason a wide flange outstand is such a poor arrangement and why every efficient compression element in structural steel is an internal one: a box, a tube, a stiffened panel or a flange with a lip on it. Adding an edge is the cheapest structural act available on a plate, and it is what a lip, a return, a stiffener and a closed section all are.

Which free body produced the number #

The free body is the whole plate, and the question asked of it is not equilibrium.

Give it a small out-of-plane displacement $w(x,y)$ consistent with its supports. Two energies follow: the strain energy of bending it into that shape, which is a property of its flexural rigidity $D = Et^3/12(1-\nu^2)$, and the work the in-plane stress does as the plate shortens along its own surface, which is a property of $\sigma$ and of the shape’s slope.

At the critical stress those are equal for some shape, and the smallest stress for which that is possible is the answer. $k$ is the ratio of the two integrals for the best available shape, and it depends on which shapes the boundary conditions permit and on how the in-plane stress is distributed over them.

That is why the coefficient is dimensionless and why it is an eigenvalue rather than a formula: it is the outcome of a minimisation over shapes, and the shape it selects is the buckle everybody photographs.

Two consequences follow from the free body being the whole plate rather than a strip of it.

A plate is not a wide column. A column’s buckling load falls as the inverse square of its length; a plate’s critical stress falls as the inverse square of its width and does not depend on its length at all, because a long plate simply buckles into more half-waves of the same size.

And the buckle does not have to be the failure. A buckled plate can go on carrying load in the strips near its supported edges, which is the effective width the section is left with and has no counterpart in a column.

The class limits are the same expression rearranged #

Section classification looks like a table of numbers and is one equation with $k$ substituted.

Set $\sigma_{cr} = f_y$ and solve for $b/t$: the limiting ratio is $\sqrt{k\pi^2 E/12(1-\nu^2)f_y}$, which for any $k$ is a constant times $\sqrt{E/f_y}$ — and $\sqrt{235/f_y}$ is exactly the $\varepsilon$ that every steel code multiplies its limits by.

That constant ratio is the finding. The derivation and the quoted limit both go as $1/\sqrt{f_y}$, so their ratio is fixed — which is what a knockdown factor looks like rather than a different theory. The derivation is for a perfect plate; the quoted limit is for a rolled one, carrying residual stress and not quite flat, and the difference between them is 33 per cent of width.

A section that cannot reach its own strength is a section one of whose plates exceeds that limit, and which plate it is decides the class of the whole.

The values worth knowing, and what moves them #

Four coefficients cover most of structural steelwork, and the reasons they differ are worth carrying rather than the numbers.

0.43 — an outstand in uniform compression. One edge supported, one free. The buckle is a curl of the free edge, there is no node anywhere across the width, and the plate has only the restraint at its root to work against.

4.0 — an internal element in uniform compression. Both edges supported. The buckle has to have zero displacement at each edge, which forces a shorter half-wave across the width and a much stiffer shape.

23.9 — a panel in pure shear. Both edges supported. The buckle runs diagonally, so its half-wave measured across the panel is shorter again, and the mode is stiffer still.

Up to 23.9 for a web in bending — both edges supported with a stress gradient from compression to tension. Only the compressed part of the width can drive the buckle, so the effective width doing the buckling is a fraction of the plate’s and the coefficient rises accordingly.

The pattern is one sentence: the coefficient measures how short a half-wave the boundary conditions and the stress pattern force the buckle into, and every one of the four is an instance of it. That is more useful than the table, because it says which way an unlisted case will go.

A stiffener is a boundary condition bought at a threshold #

If $k$ is set by the edges, adding an edge is the strongest move available.

The shape of that curve is the whole design rule. A stiffener is not a contribution; it is a switch. Below the threshold it is carried along by the buckle and does nothing; at the threshold it becomes a node in the buckled shape and the plate’s width is effectively halved; above it the plate has stopped noticing, because the sub-panel mode has no displacement where the stiffener is.

A rib is a boundary condition, and the design question is therefore not how stiff but stiff enough or not — which is unusual, and it means a stiffener at twice the required rigidity is precisely as good as one at exactly the required rigidity, and one at 90 per cent of it is worth very little.

What more of them cost #

Since one stiffener multiplies the buckling stress by four, the temptation is to add several.

A gain that goes as the square against a price that goes as the fourth power has a clear best answer, and it is a small number. One or two stiffeners are excellent value; four have put more steel into the ribs than there is in the plate, and at that point the honest comparison is against a thicker plate with no stiffeners at all.

That comparison is why orthotropic decks look the way they do — closely spaced small ribs on a thin plate — and why the ribs are closed troughs rather than flats: a closed trough has a far larger $I_s$ for its own area, which is the quantity $\gamma^*$ is a demand on.

Where the threshold comes from, and why it is so large #

The rigidity a stiffener needs is the least intuitive number on this page, and it is worth seeing where it comes from because the size of it decides whether stiffening is worth doing at all.

The stiffener has to be a node in the buckled shape — a line along which the plate does not move out of plane. To be that, it must be stiff enough that the energy of bending it into the whole-width mode exceeds the energy the plate saves by buckling in the narrower sub-panel mode instead. That is a comparison of two energies, and the plate’s flexural rigidity $D$ is in both of them.

The threshold that comes out, $\gamma^* = EI_s/bD$, is dimensionless and is typically in the hundreds. Converting it back to a physical stiffener: on the 3,000 × 14 mm plate above, $\gamma^* = 226.8$ asks for an outstand of 332 mm, on a plate 14 mm thick. The rib is twenty-four times the plate’s thickness deep, which looks wrong on a drawing and is right.

The reason the number is so large is that $D$ goes as $t^3$ while $I_s$ goes as $h_s^3$ for a flat, so the two are competing on the same power and the ratio of dimensions has to do the whole job. A closed trough beats a flat here by a wide margin, because its $I_s$ comes from an enclosed area rather than from a depth — which is why every orthotropic deck in the world uses one.

What happens past the critical stress #

The coefficient decides when a plate buckles and says nothing about what happens afterwards, and the two are only loosely related.

A flat plate is the stable symmetric case, which is why it has a post-buckling reserve at all: the buckled shape stretches the plate in its own plane, which resists, so the load can keep rising. That is what makes the effective-width idea legitimate and it is a property of the plate’s geometry rather than of its coefficient.

A web in shear does the same thing on a larger scale, reorganising into a diagonal tension field after it has buckled and carrying several times the critical load. So the $k = 23.9$ that made the shear panel comfortable is not even the interesting number for that case — the panel is allowed past it.

A cylindrical shell is the unstable symmetric case, its imperfect strength is a fraction of its critical stress, and that is the reason shell buckling is a different subject rather than a harder version of this one.

Why the aspect ratio drops out #

One feature of the formula deserves attention because it is the difference between a plate and everything else in stability.

The length of a plate does not appear. A long plate buckles into a chain of half-waves each about as long as the plate is wide, so lengthening it adds waves rather than lowering the stress. The coefficient’s dependence on aspect ratio is a shallow scalloped curve with minima at integer ratios, and past an aspect ratio of about one it never departs from 4 by more than a few per cent.

That has two consequences a designer uses without noticing.

A transverse stiffener does almost nothing for a long panel in compression. Dividing a plate of aspect ratio 4 into four squares changes the coefficient hardly at all, because the plate was already buckling in squares. Transverse stiffeners on a compression flange are there to carry a transverse load or to stabilise a longitudinal stiffener, not to raise $k$.

And a transverse stiffener does a great deal for a panel in shear. The shear coefficient does depend on aspect ratio — strongly, because the diagonal buckle’s wavelength is set by the panel’s proportions — so dividing a long web into shorter panels raises $k_\tau$ substantially. That asymmetry is why the stiffeners on a plate girder are transverse and the stiffeners on a wide compression flange are longitudinal, and it comes entirely from which of the two coefficients notices the panel’s length.

What to carry away #

$k$ is the structural content of the expression and it moves by an order of magnitude. 0.43 for an outstand, 4 for an internal element, 23.9 for shear, and values in between for stress gradients.

The class limits are this equation with $\sigma_{cr}$ set to $f_y$, times a knockdown of about 1.34 for the fact that real plates are not flat. There is nothing else in them.

And a stiffener is a threshold rather than a contribution. Reach $\gamma^*$ and the plate’s width halves; fall short and almost nothing has been bought; exceed it and nothing more is available.

Where the model stops #

The plate is perfectly flat and free of residual stress. It is neither, and the 1.34 between the derived and quoted limits is what that is worth.

The edges are idealised as simply supported or free. A real flange is elastically restrained by the web it is attached to, so the true coefficient for an outstand is above 0.43 and for the web below 4, and the two errors are in opposite directions.

The stress is assumed uniform along the length. In a member under a moment gradient it is not, and a plate under a varying longitudinal stress has a coefficient that no table contains.

Nothing here treats interaction between modes. A stiffened panel can buckle locally between stiffeners, globally as a strut, or in both at once, and the coincidence of two modes is dangerous rather than efficient.

The stiffener’s own stability is not checked here. A stiffener at $\gamma^*$ is a member in compression along with the plate, and it can buckle as a strut, twist, or buckle locally in its own outstand — three modes that the rigidity threshold has nothing to say about and that decide whether the threshold is achievable.

And the material is assumed elastic. For a stocky plate the critical stress computed here exceeds yield, at which point the plate does not buckle elastically at all and the tangent modulus rather than $E$ decides.

There is one more reading of the class table worth having. Because both the derivation and the quoted limit scale as $1/\sqrt{f_y}$, a higher-grade steel does not let a plate be wider — it makes the same plate slenderer. Going from 235 to 460 raises the strength by 96 per cent and drops the limiting width-to-thickness ratio by 29, so the section that was class 1 in mild steel may be class 3 in high-strength steel with no dimension changed. That is the single most common surprise in substituting a grade, and it is one line of algebra away from being obvious.

The coefficient is a boundary condition wearing a number, which is why two other essays about plates are really about their edges. A rib is a boundary condition rather than a member, and what is left after a plate ripples is what the post-buckling reserve does with the width the coefficient decided.

The ladder from here #

Later rungs on this anchor: the plate buckling equation derived from the energy statement, with the half-wave count falling out of the minimisation. Coefficients for a stress gradient, which is what a web in bending actually has. Winter’s formula for effective width, and where its exponent came from. Post-buckling stability of a stiffened panel, where local and overall modes arrive together. Shear buckling and tension-field action worked properly. Stiffened plates and orthotropic decks, where the ribs are smeared into the plate’s properties. Cold-formed sections, which live almost entirely past the limit. And shell buckling, where the mild imperfection sensitivity of a plate becomes savage.

The coefficient is Bryan’s, from 1891, and the plate problem is one of the very few in structural stability that was solved before it was needed. What arrived later was everything on either side of it: the effective width in the 1930s, the class limits in the 1960s, and the stiffener rigidity threshold in between — three design rules built on one eigenvalue that nobody has improved on.