Held everywhere, and it forgets its length

Every buckling calculation in this collection so far has been about a length. A pin-ended column buckles at $\pi^2EI/L^2$; fixing its ends divides the length by two and multiplies the load by four; a brace at mid-height divides it again. The length is the variable, and the whole apparatus of effective lengths exists to say which length applies.

Then there are members that nothing holds at a point and everything holds a little. A through-girder bridge’s top flange is in compression and has no bracing above the deck, because the traffic is in the way; what restrains it is the stiffness of the web and the cross-girders bending as a U. A wall panel is held by the sheeting fixed to it. A truss chord is held by the deck sitting on it. A pile is held by the soil.

For all of them the answer stops being about a length, and starts being about a stiffness per unit length.

Which free body produced the number #

Assume the buckled shape is a sine with $n$ half-waves over the length, $v = a\sin(n\pi x/L)$, and write down the two energies.

Bending the member stores $\tfrac12\int EI (v'')^2\,dx$, which for that shape is $EI a^2 n^4\pi^4/4L^3$. Deflecting the foundation stores $\tfrac12\int k v^2\,dx = k a^2 L/4$. The load does work $\tfrac12 P\int (v')^2 dx = P a^2 n^2\pi^2/4L$. Set the work equal to the stored energy and the amplitude cancels:

$$P_n = \frac{n^2\pi^2 EI}{L^2} + \frac{kL^2}{n^2\pi^2}$$

The first term rises with the number of waves and the second falls. That is the whole mechanism: more waves means sharper curvature and more bending energy, but also smaller deflections and less foundation energy. The member takes whichever integer $n$ makes the sum least, and it is not usually one.

For the flange drawn, $n = 2$ and the critical load is 1,968 kN against an unrestrained 173 — a factor of 11.4 from a restraint that is doing very little at any one point.

That factor is worth reading against the curve this leaves behind. Euler’s load is the $n = 1$ term of the sum above with the foundation set to zero, so every restrained answer sits above the column curve by construction and no restrained member can ever fall below it. What the restraint does is not raise a curve. It changes which variable the answer is a function of, and the sections below are about what happens to the old variable.

The length disappears #

Treat $n$ as a real number for a moment and minimise. The derivative vanishes at $n^* = (L/\pi)(k/EI)^{1/4}$, and substituting it back gives

$$P_{cr} = 2\sqrt{kEI}, \qquad \lambda_{half} = \pi\left(\frac{EI}{k}\right)^{1/4}$$

Neither expression contains $L$. The critical load is the geometric mean of the two stiffnesses, doubled; the half-wavelength the member settles into is a property of the member and its foundation and of nothing else.

The integer constraint keeps the real member slightly above that value — 1,968 against 1,878 here, because two waves is not quite the optimum 2.33 — and the discrepancy shrinks as the member lengthens and the integers get denser. Measured: 1,925 kN at 12 m, 1,880 at 21, 1,879 at 31, 1,881 at 40. Flat to half a per cent across a range of nearly four to one in length.

A hundred metres of the same flange on the same foundation buckles at the same load as ten. The intuition that a longer compression member is a weaker one, which is correct everywhere else in this subject, is simply false here.

The curve says the load does not move. The same member drawn at the far end of it says why, which is that the mode has changed rather than the member having become stronger.

The effective length that comes back out #

Design does not want a critical load; it wants a slenderness to enter a column curve with. So the answer is turned back into a length:

$$L_e = \pi\sqrt{\frac{EI}{P_{cr}}}$$

For this flange, 3,555 mm — 0.30 of the actual member. That number is what a code check uses, and it is worth noticing what has happened to it. The effective length is no longer a property of the end conditions, as it is for a column between two supports. It is a property of the restraint’s stiffness, and it would be the same for a member of any length whatever.

The relation between $L_e$ and $k$ is a quarter power: $L_e = \pi (EI/k)^{1/4}/\sqrt{2}$. Quadrupling the restraint stiffness reduces the effective length by $\sqrt{2}$ and raises the critical load by a factor of two. That is a shallow return, and it is why restraint calculations of this kind are rarely sensitive to the exact stiffness assumed — which is fortunate, because the stiffness assumed is usually a rough estimate of a U-frame’s flexibility.

The claim is worth drawing rather than asserting, because it is the whole argument for not spending money on a stiffer U-frame.

Where the stiffness comes from #

The restraint in this calculation is a number with awkward units — force per unit deflection per unit length — and getting it is most of the work in a real check.

For a through-girder bridge, the flange is held by a U-frame: the two webs bending out of plane and the cross-girder bending in its own plane, acting as a portal without a top member. The stiffness of one such frame, divided by the spacing between them, is $k$. It is small, because it is a bending stiffness with a long lever arm, and the flexibility is dominated by whichever of the three members is weakest — usually the connection between the web and the cross-girder.

For a sheeted member, the stiffness comes from the shear stiffness of the sheeting and the flexibility of the fasteners in series, and the fasteners usually govern.

For a pile or a buried member, it is the soil’s modulus of subgrade reaction, which is the same $k$ that appears in a beam on an elastic foundation and is the least reliable number in geotechnics.

In every case the restraint’s own flexibility matters and its immobility does not, which is the part that is counter-intuitive. A U-frame moves; it moves a great deal. What it does is push back in proportion to how far the flange has gone, and that is all the derivation ever asked of it.

The mode has to be able to happen #

The derivation assumes the member is free to take whatever wave it likes. Two things can stop it, and both raise the answer.

If the member is short enough that even one half-wave is shorter than the preferred wavelength, the integer minimum is $n = 1$ and the answer is the Euler load plus a foundation term. That is the regime a short restrained member is in, and its critical load does depend on its length in the ordinary way.

If the member is restrained at points as well as continuously — cross-frames at intervals in a bridge, purlins on a sheeted rafter — the wave has to fit between them, and the discrete restraints govern whenever their spacing is below the preferred half-wavelength. Real designs sit near that boundary on purpose: putting cross-frames at about the preferred wavelength is exactly the spacing at which they stop being useful, and the check is which of the two mechanisms gives the lower load.

The method underneath all of this is guessing the shape and letting the energy decide, and it is worth naming because of where it is exact and where it is not. The sine assumed at the top of this page is the true mode for a uniform member on a uniform foundation, so nothing is being approximated there at all. Where either the section or the restraint varies along the length — a haunched girder, a U-frame stiffer near a pier — the same energy method with an assumed shape returns an upper bound on the critical load, and the accuracy is far better than the shape deserves because the error is second order in it.

Why the flange and not the girder #

There is a modelling decision buried in all of this that is worth bringing out, because it is where the idealisation is doing the most work.

A through-girder does not buckle as a strut. It buckles by lateral-torsional instability: the compression flange moves sideways, the tension flange does not, and the web twists between them. Treating the compression flange as an isolated strut with the web’s bending stiffness as its foundation is a substitution — a real one, defensible, and not exact.

What it captures is the mechanism: the flange goes sideways and something bends to let it. What it loses is the tension flange’s contribution, the web’s own resistance to the twist, and the fact that the “strut” being analysed is not a member with a definite second moment but the top part of a section whose properties depend on where the cut is imagined.

Codes handle this by defining the strut as the flange plus a stated fraction of the web — typically a third of the compression zone — which is a rule with a derivation behind it and a rounding in front of it.

One thing every number on this page assumes is worth stating separately, because the restraint’s own design depends on it. The critical load computed above is a bifurcation load for a perfectly straight member, and a real flange was never straight: it is bowed from the day it was rolled, its deflection grows from the first newton of compression with an amplification of $1/(1 - P/P_{cr})$, and there is no bifurcation to be had. The consequence for the U-frame is direct. The restraint force is $k$ times that deflection, so it is zero in the eigenvalue calculation and is not zero in the bridge — and it is the amplified bow, not the mode shape, that the cross-girder connection has to be designed to carry.

The two regimes, and where a design sits between them #

It is worth setting out the whole picture as one axis, because a member on an elastic foundation is not one problem but two with a boundary.

Short of the preferred wavelength, the member cannot fit even one of its favoured half-waves in, so it buckles in one and the foundation is a bonus term added to Euler’s load. Here the length matters in the ordinary way, halving it quadruples the answer, and everything learnt from columns applies.

Long compared with the preferred wavelength, the member has forgotten its length. It buckles at $2\sqrt{kEI}$, its effective length is a fixed 3.6 m regardless, and the only two ways to improve it are a stiffer restraint or a stiffer flange — at quarter and square-root powers respectively, so neither is a strong lever.

The boundary sits at $L \approx \pi(EI/k)^{1/4}$, which for this flange is about five metres. A through-girder bridge of any real span is far past it; a short restrained strut in a truss may not be. The check that distinguishes them costs one line — compute the preferred half-wavelength and compare it with the member — and it decides which of two quite different sensitivities the design has.

There is a practical asymmetry worth carrying. In the first regime, adding restraint stiffness helps a great deal and shortening the member helps more. In the second, shortening the member does nothing at all, and a designer who has spent money adding a cross-frame to a length that was already past the boundary has bought nothing — unless the cross-frame’s spacing is below the preferred wavelength, in which case the member has been moved back into the first regime and the money was well spent. The two statements are not in conflict and they are easy to confuse.

The formula needs the member to be long enough to use it #

The result that the length disappears is an asymptote, and it is worth knowing where the asymptote starts.

The buckle chooses a half-wavelength

$$\ell = \pi\left(\frac{EI}{k}\right)^{1/4}$$

— 5.1 m on the flange in question — and the whole argument depends on the member being able to accommodate at least one of them. A member shorter than $\ell$ cannot form the wave the restraint is being measured against. Its lowest mode is a single half-wave over its own length, the restraint contributes as a distributed spring resisting that shape rather than as a foundation the member can choose a wavelength in, and the answer is an ordinary Rayleigh calculation with the length back in it.

So the honest statement is a pair of regimes with a boundary at $L = \ell$:

Long member ($L \gg \ell$): the length has genuinely disappeared, the critical load is $2\sqrt{EIk}$, and doubling the span changes nothing at all.

Short member ($L \lesssim \ell$): the length is back, the critical load is closer to the Euler value raised by the restraint, and the member behaves like a column with springs rather than like a strut on a foundation.

And the useful check falls out of the same comparison. Whenever the computed half-wavelength approaches the member’s own length, the formula being used is the wrong one — which is a self-diagnosing failure, because $\ell$ is computed as part of the answer and comparing it with $L$ costs nothing.

It also settles the discreteness question from the same number. A restraint at spacing $a$ behaves as a continuum when $a$ is a modest fraction of $\ell$ and as a set of point braces when it is comparable with it. The flange here has $a = 4$ m against $\ell = 5.1$, which is not comfortable — a ratio nearer a third would be — and the smearing is doing real work rather than being a formality.

Where the model stops #

The foundation is linear, continuous and two-sided. Soil is none of the three, sheeting is not two-sided, and a U-frame’s stiffness falls once the connection starts to yield. Each of those makes the real restraint softer than the model at the deflections that matter.

The axial force is constant along the member. In a through-girder it is not — the flange force follows the bending-moment diagram — so the compression is largest at mid-span and vanishes at the ends, and the buckle localises where the force is high instead of repeating uniformly.

And the answer is elastic. A critical load of 1,968 kN on a flange whose squash load is smaller than that is a number about a member that yields first, and the restraint’s contribution then has to be assessed on an inelastic column curve rather than on Euler’s.

The bridge that made it a design case #

The through-girder — a deck slung between two plate girders, with traffic passing between them — exists because of a headroom constraint and not because anybody wanted it. Where a bridge has to be shallow, the girders go beside the deck rather than under it, and the top flange is then in compression with nothing above it to brace against.

Every other compression member in a bridge has bracing available. A deck girder has a slab on top of it. A truss has a top lateral system. A through-truss has portals at the ends and a lateral system between the top chords, provided there is room above the traffic. Take that room away and the U-frame is the only mechanism left.

The consequence is that this calculation is one of the few in structural engineering that was developed for a single structural form and then found to be general. Engel’s and Bleich’s work on the elastically restrained strut was written for through-girders; the same equation now covers sheeted purlins, piles in soil, rails on sleepers, buried pipes and the compression chord of any open-topped structure. The form is a niche and the mathematics is not.

It is also one of the clearest cases of a structure whose weakest element is invisible in the analysis model. A frame analysis of a through-girder returns bending moments, shears and deflections, all of them fine, and says nothing whatever about the flange going sideways — because a plane-frame model has no out-of-plane freedom to go sideways in. The check has to be recognised as necessary before it can be done, and that recognition is not prompted by anything the analysis produces.

What the pictures cannot show #

The restraint is drawn as a row of arrows because a continuum of springs cannot be drawn. In the real structure there are U-frames at four-metre centres, and whether that is a continuum depends on how it compares with the 5.1 m half-wavelength — which it does not, comfortably. The smearing that makes the mathematics clean is at its weakest exactly where the answer is most useful.

Nor can the figures show what the flange does at the ends of the member, where there is a real support and the wave has to accommodate it. The uniform repeating sine is an interior solution, and the first half-wave at each end is a different shape.

The assumption the figure rests on #

The restraint stiffness is 0.35 N/mm per mm — a U-frame of roughly 1,400 kN/m at 4 m centres. That number is the output of a small frame analysis whose largest term is the flexibility of the web-to-cross-girder connection, and connection flexibilities are estimates. The comfort is the quarter power: halving $k$ costs 29 per cent of the critical load rather than 50, and the whole calculation is far less sensitive to its least reliable input than most in this collection.

The ladder from here #

Later rungs on this anchor: the varying axial force in a through-girder, where the buckle localises and the uniform-wave answer is unconservative. The U-frame’s own stiffness, computed rather than assumed, and which of its three components governs. Discrete and continuous restraint together, and the spacing at which one takes over from the other. The restraint force — how much the foundation actually has to carry, which is an imperfection question and not an eigenvalue one. And the same equation with the sign of the axial force reversed, which is a beam on an elastic foundation and is a different subject with identical mathematics.