The web that is crushed from inside

A plate girder two metres deep is loaded until it forms a plastic hinge at midspan. Photograph it afterwards and the web at the hinge has buckled — not in shear, and not under a bearing, but vertically, in a wave down the depth of the panel, with the compression flange pressed into it.

Nothing was applied there. There is no load on the top flange at that section, no stiffener, no bearing. The web buckled under a force that the beam generated by bending.

Which free body produced the number #

A short length of the compression flange.

It carries a force $F = A_{fc}\sigma_f$, and the beam it belongs to has bent, so the flange follows a curve of curvature $\kappa$. The load that comes from changing direction gives the transverse force any curved force path demands:

$$q = F\kappa$$

per unit length, directed toward the centre of curvature — which, for a sagging beam whose top flange is in compression, is downward, into the web.

The web has to supply it. There is nothing else there.

Two lines of arithmetic put numbers on it. The flange’s curvature is the beam’s, and at a section where the flange is at a strain $\varepsilon$ the curvature is about $2\varepsilon/h_w$. So

$$q = A_{fc}\sigma_f\cdot\frac{2\varepsilon}{h_w}$$

and for a flange at yield with $\varepsilon = f_y/E$ that is a few newtons per millimetre — small compared with a bearing reaction, and applied continuously along a panel whose resistance to a vertical load is very small indeed.

The rule, derived #

The web resists that force as a strip in compression down its depth. Its plate buckling stress is

$$\sigma_{cr} = \frac{k_\sigma\pi^2E}{12(1-\nu^2)}\left(\frac{t_w}{h_w}\right)^2 = C\,E\left(\frac{t_w}{h_w}\right)^2$$

and the force per unit length it can take is $\sigma_{cr}t_w$. Set the two equal:

$$\frac{2A_{fc}f_y^2}{Eh_w} = C E \frac{t_w^3}{h_w^2}$$

Substitute $A_w = h_wt_w$, let $x = h_w/t_w$, and it rearranges to

$$x^2 = \frac{C}{2}\left(\frac{E}{f_y}\right)^2\frac{A_w}{A_{fc}} \quad\Longrightarrow\quad \frac{h_w}{t_w} \le \sqrt{\frac{C}{2}}\cdot\frac{E}{f_{yf}}\sqrt{\frac{A_w}{A_{fc}}}$$

which is exactly the form the codes use, arrived at without them. Four lines, one free body, and no fitting anywhere.

It is worth pausing on what has and has not been assumed, because the derivation is short enough to look like a trick.

Nothing about the material appears except $E$ and $f_y$, and they appear as a ratio — which is why the rule is written in $E/f_y$ and why a higher-grade steel makes it worse: the same flange at a higher yield stress is strained further before it yields, so it curves more, so it pushes harder. That runs against every intuition about strength and it follows in one line from the derivation.

Nothing about the span appears at all. The demand is a curvature and the resistance is a plate stress, and neither knows how long the girder is. A 20 m girder and a 60 m girder of the same section have exactly the same check.

And nothing about the applied load appears. The check is a statement about a state — a flange at a stated strain — rather than about a load case, which is why it is written as a limit on a proportion rather than as a comparison of forces.

The constant, which is not the same one #

The derivation gives $k = \sqrt{C/2}$, and with a simply supported plate coefficient $k_\sigma = 4$ that is 1.34.

The codes use 0.3 for a plastic hinge, 0.4 for an elastic-plastic member and 0.55 for an elastic one. A factor of between two and four and a half.

That gap is not a safety margin, and reading it as one loses the interesting part. Look at where the strain entered: $q$ is proportional to $\varepsilon$, so $C$ in the derivation is inversely proportional to it, and $k$ goes as $1/\sqrt{\varepsilon}$.

$$\frac{k_{elastic}}{k_{code}} = \sqrt{\frac{\varepsilon_{code}}{\varepsilon_{yield}}}$$

Put 1.34 and 0.3 into that and the implied strain is about twenty times yield — around 3%, which is the strain-hardening range of mild steel and exactly what a fully rotated plastic hinge asks of its flange.

The code’s constant is not conservative. It is a different beam. The 0.55 case is a girder whose flange never passes yield; the 0.3 case is one at a plastic hinge with the flange strained into hardening. The three constants are three curvatures, and each is right for its own member.

That is worth having because it makes the rule usable outside the cases it was written for. A girder designed to remain elastic under all loads has a genuine limit near 0.55; a girder in a seismic frame that will be asked for large rotations has one nearer 0.3 or below.

Why a thin web is worse twice over #

The web thickness appears on both sides of the inequality, and that is the reason this failure has such a sharp threshold.

The demand, $h_w/t_w$, goes as $1/t_w$. The limit, $k(E/f)\sqrt{A_w/A_{fc}}$, contains $A_w = h_wt_w$ and therefore goes as $\sqrt{t_w}$. So the utilisation goes as $t_w^{-3/2}$: reduce the web thickness by 20% and the utilisation rises by 40%.

Which means the check is not one a girder is marginally inside or outside. A girder with a 14 mm web is comfortably fine and the same girder with a 10 mm web can be well past the limit, with nothing else changed.

It also explains why the rule bites on exactly one class of member. A rolled beam has a stocky web and never approaches it. A shallow welded girder has a small $h_w$ and does not either. The failure belongs to deep, thin-webbed, heavily flanged girders — plate girders and box girders in bridges — and to those same members at plastic hinges.

Why the rule is written as a proportion #

There is a stylistic point in the rule’s shape that is worth drawing out, because it is what makes it survive.

$h_w/t_w \le k(E/f_y)\sqrt{A_w/A_{fc}}$ contains no load, no span, no bending moment and no section modulus. It is a comparison of two dimensionless groups, and either side of it can be evaluated from a section drawing alone. That is why it can be checked in ten seconds by somebody who has not seen the analysis, and it is why it appears in fabrication guides as a proportioning rule rather than in the design chapter as a check.

Rules of that shape are rare and worth collecting. A span-to-depth ratio, a flange outstand-to-thickness ratio, a bar spacing in diameters, an aspect ratio for a shear panel: each is a whole calculation collapsed into a proportion, and each is a check that a drawing can be tested against before anything has been analysed. Depth is the cheapest strength is the same idea used positively — a proportion that tells a designer what to reach for rather than what to avoid.

The price of the form is that the physics is invisible. Nobody reading $0.3E/f_y\sqrt{A_w/A_{fc}}$ would guess that it is a flange pressing into a web, and a rule nobody can reconstruct is a rule that gets applied outside its range without anybody noticing that it has been.

Where else the same force appears #

The radial force from a curved flange is the same quantity in three other places in this collection, and putting them together is worth doing because they look unrelated.

A curved girder in elevation. A haunched or arched girder has curvature built in rather than produced by loading, so the radial force exists at first application of load and is very much larger — which is why a curved-in-elevation girder needs web stiffeners at a spacing set by the radius rather than by the shear.

A tension flange curving the other way pushes outward, away from the web, which puts the web into tension and does nothing harmful at all. The failure is one-sided, and only the compression flange has it. That asymmetry is a small thing with a practical consequence: a girder that has been erected upside down by mistake, or one in a continuous span where the flange in compression is the bottom one, has its flange-induced problem at the other face — and the stiffener that was provided for it is on the wrong side.

And a flange that is not straight in plan. A girder curved in plan has its flange force turning horizontally as well, and the radial force is then transverse to the web rather than along it — which is a torsion problem and belongs with bending that arrives as twist.

What the numbers look like on a real girder #

Take the girder drawn: a web 2,000 mm deep and 12 mm thick, flanges 500 by 35, in S355.

The flange force at yield is $500\times35\times355 = 6{,}210$ kN. The curvature at first yield of the flange is $2\varepsilon_y/h_w$, and $\varepsilon_y = 355/210{,}000 = 0.00169$, so $\kappa = 1.69\times10^{-6}$ per millimetre — a radius of 590 metres. The radial force is $F\kappa = 10.5$ N per millimetre of girder.

Ten newtons per millimetre is ten kilonewtons per metre, which sounds like nothing at all: it is less than the girder’s own weight. But it is applied to the top edge of a plate 2,000 mm deep and 12 mm thick, and that plate’s own buckling resistance as a vertical strut is $\sigma_{cr}t_w$ with $\sigma_{cr} = C E(t/h)^2 = 3.6\times210{,}000\times(0.006)^2 = 27$ N/mm² — so $qR = 327$ N/mm, thirty times the demand.

Comfortable, at first yield. Now take the flange to a plastic hinge and strain it to 3%, eighteen times yield. The curvature goes up by eighteen and the demand with it, to 190 N/mm, and the margin has fallen from thirty to under two. Thin the web to 10 mm and the resistance falls to 189 — and the girder is at its limit, in a member nobody would look at twice.

That is the whole of the phenomenon in four numbers: a force that is negligible in service, a resistance that is small because the plate is thin, and a demand that goes up with the rotation the member is asked for.

The remedy, and why it is a stiffener #

Once the mechanism is understood the fix is obvious and it is the one the codes give: a transverse stiffener interrupts the web strip, shortens its buckling length, and multiplies its resistance by a large factor.

That is a different job from the one stiffeners are usually there for. A bearing stiffener carries a reaction; an intermediate stiffener anchors a tension field; a longitudinal stiffener raises the shear or bending buckling stress of a panel. A stiffener provided against flange-induced buckling is holding a flange onto its curve, and it can be needed at a section where the shear is zero.

The asymmetry between the two failures is worth noticing. The panel that carries more after it fails is about a web buckling in shear and finding a second mechanism. A web buckling under the flange has no second mechanism: once it has moved out of plane it cannot hold the flange, the flange loses its restraint and its own local buckling length grows, and the two failures feed each other.

Where the model stops #

The web was treated as a plate in uniform compression. It is not: the radial force is applied at one edge and the web is simultaneously carrying shear and bending stresses of its own, so the real buckling coefficient depends on a combination of stress fields the derivation ignores.

The flange was assumed to be at a stated strain everywhere. It is at that strain only at the hinge, and the curvature falls away either side — so the demand is local and the resistance is a panel property, and the two are being compared at different scales.

Nothing here is about the flange’s own stability. A compression flange that loses its web restraint is a plate supported along one edge, and its local buckling and its lateral-torsional buckling both get worse. The beam that fails sideways is the member-level consequence, and it arrives at the same section.

The web was assumed to be flat. A web with an initial out-of-plane bow, which every welded web has, is already partly deflected and the flange’s push amplifies it from the first newton — so the real behaviour is a growing deflection rather than a bifurcation, in exactly the way a third of what the theory promised describes for shells. The rule is a bifurcation criterion applied to a member that does not bifurcate.

And the failure is very rare. Almost no girder in service has ever failed this way, which is a compliment to the rule rather than an argument against it — the limit is easy to satisfy, and the members that could violate it are designed by people who know the clause exists.

The generalisation #

The lesson is about where loads come from.

A load schedule lists things that are applied: weights, pressures, reactions, temperature. A structure’s internal forces are computed from that list, and every check compares an internal force against a capacity. The scheme is complete and it has one gap: a force that arises from the deformed shape appears on no list, because it did not exist until the structure moved, and the structure moved because of the loads that were listed.

Flange-induced buckling is the clearest small example. The load that makes itself worse is the large one, where a sway generates a moment that generates more sway. Both are the same category: a second-order force, generated by displacement, and found only by asking what the deformed structure is doing rather than what was put on the undeformed one.

There is a second reason these forces are hard to find, and it is about how analysis is organised rather than about mechanics. A frame model computes displacements from forces and then computes member forces from the displacements, and stops. Nothing in that chain asks what the displacements demand, because the demand is a load and the loads were the input. The free body is a choice is the remedy: choose a free body that is a piece of the deformed structure, and the second-order forces appear on it as ordinary equilibrium terms.

The habit that finds them is to draw the structure deflected — really deflected, with the curvature exaggerated — and then to look along every large force path for a place where it is no longer straight. Wherever it is not, $F\kappa$ per unit length is being asked of something, and the question is what.