Structural Engineering & Statics
where the load goes
Stability

Held everywhere, and it forgets its length

A brace at a point divides a member's buckling length. A restraint spread along the whole member does something else — the member chooses its own number of half-waves, and past a few of them the critical load stops depending on the length at all.

Assumes The ends decide the length that matters, Strong enough and still falls over and The beam that sits on the ground.

20 min read 8 figures Which failure arrives firstThe load must go somewhereGeometry beats material

Every buckling calculation in this collection so far has been about a length. A pin-ended column buckles at π2EI/L2\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.

The restraint chooses the buckling length, and it is not the member'sA compression flange 12 m long held sideways not at points but everywhere, by a restraint of 0.35 N/mm per mm of length. Unrestrained it would buckle at 173 kN in a single half-wave, drawn faintly. Restrained it buckles at 1968 kN — 11.4 times as much — in two half-waves, because the sum n²π²EI/L² + kL²/n²π² has its minimum there and every other n is worse. The effective length that answer implies is 3555 mm, which is 0.30 of the member and is a property of the restraint rather than of the span.1968 kNthe restraint: 0.35 N/mm per mmtwo half-waves, each 6000 mmunrestrained 173 kN in one half-wave, drawn faintly · effective length 3555 mm
Fig. 1 A compression flange twelve metres long, held sideways not at points but everywhere, by a third of a newton per millimetre of deflection per millimetre of length. Unrestrained it would buckle at 173 kN in one half-wave, drawn faintly. Restrained it buckles at 1,968 kN in two — because the sum of the two energy terms is least there, and every other number of waves is worse.

Which free body produced the number

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

Bending the member stores 12∫EI(v′′)2 dx\tfrac12\int EI (v'')^2\,dx, which for that shape is EIa2n4π4/4L3EI a^2 n^4\pi^4/4L^3. Deflecting the foundation stores 12∫kv2 dx=ka2L/4\tfrac12\int k v^2\,dx = k a^2 L/4. The load does work 12P∫(v′)2dx=Pa2n2π2/4L\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:

Pn=n2π2EIL2+kL2n2π2P_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 nn makes the sum least, and it is not usually one.

For the flange drawn, n=2n = 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.

The column curveFailure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither
Fig. 2 The curve this leaves behind. Euler’s load is the n=1n = 1 term with no foundation at all, and every restrained answer sits above it. What the restraint does is not raise a curve but change which variable the answer is a function of.

The length disappears

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

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

Neither expression contains LL. 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.

Past a certain length it stops mattering how long it isThe critical load of a continuously restrained strut against its own length, with the restraint held constant. The faint curve is the unrestrained Euler load, falling as the square of the length in the way every column on this site does. The solid one is the restrained member, which falls at first and then **stops**: at 40 m it is 1881 kN against 1878 kN for a member of infinite length. The member has chosen 8 half-waves instead of one, and the half-wavelength it settles on — 5146 mm — is π(EI/k)^¼, a length with the member's own nowhere in it.5101520253035400500100015002000restrained length (m)critical load (kN)2√(kEI) = 1878unrestrainedhalf-wavelength π(EI/k)^¼ = 5146 mm, whatever the length is
Fig. 3 The critical load against length, with the restraint held. The unrestrained curve falls as 1/L21/L^2 for ever; the restrained one falls, flattens, and lands on 2kEI2\sqrt{kEI} — the horizontal line the member is heading for as soon as it has room for its preferred wave.

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:

Le=πEIPcrL_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 LeL_e and kk is a quarter power: Le=π(EI/k)1/4/2L_e = \pi (EI/k)^{1/4}/\sqrt{2}. Quadrupling the restraint stiffness reduces the effective length by 2\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.

A brace is a stiffness requirement, not a strength oneCritical load against brace stiffness for a pinned column braced at mid-height. The curve climbs from the unbraced Euler load of 9.87EI/L² and flattens at 39.48EI/L², which is the Euler load of the braced segment — past that the column buckles in a shape the brace does not obstruct, and further stiffness buys nothing. The knee is at about 159EI/L³.05010015020025001020304050brace stiffness (units of EI/L³)critical load (units of EI/L²)ideal stiffness ≈ 159 EI/L³39.5 — braced9.87 — unbraced
Fig. 4 The discrete version of the same question. A single brace at mid-height has a threshold stiffness past which more buys nothing, because the member has switched modes and the brace is now at a node. A continuous restraint has no threshold and no knee: the return is smooth and it is a quarter power for ever.

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 kk. 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 kk 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 ground pushes back hardest where the beam has gone down furthestA strip 16.4 m long and 1 m wide on ground of subgrade modulus 50 × 10³ kN/m³, carrying 1000 kN at its centre. The beam settles 3.90 mm under the load and the ground pushes back in proportion — the arrows are k times the settlement above them, peaking at 195 kN per metre — so the pressure diagram is the settlement bowl and not an assumed distribution. The characteristic length 1/β is 2.56 m: the bowl crosses zero at 6.04 m, which is 3π/4 of it, and beyond that the arrows reverse because the beam has lifted off. By 8.05 m — one π/β — the disturbance is 4.3% of what it was, which is why the moment 641 kNm and the peak pressure 195 kN/m contain no length at all. The settlement is drawn 217 times full size — the real bowl is 3.90 mm deep over 16.4 m, about 1 in 4207 — and at true scale the beam would be a straight line.1000 kN1/β = 2.56 mthe beam lifts off at 6.04 msettlement 3.90 mm · contact pressure 195 kN/m · moment 641 kNm, none of which contains the length of the beam
Fig. 5 The same kk in the same equation, without the axial force. A beam on an elastic foundation under a point load decays in a length π/β\pi/\beta with β=(k/4EI)1/4\beta = (k/4EI)^{1/4} — the same quarter power on the same ratio. Add compression and the decaying exponential becomes a repeating sine, but the characteristic length is the same one.

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=1n = 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.

Four guesses at one buckling modeA pin-ended column, with four assumed shapes and the load each of them gives. The reference is a ten-term Ritz expansion solved as an eigenvalue problem, at 9.8696 EI/L² — which is π², as it must be. a half sine gives 9.870, its own sag shape gives 9.882, a mid-span sag gives 10.000, a parabola gives 12.000. Every one of them is high and none of them is low, because an assumed shape is a constraint on the column and a constraint can only stiffen it.a half sine9.870 EI/L²exactits own sag shape9.882 EI/L²0.13% higha mid-span sag10.000 EI/L²1.32% higha parabola12.000 EI/L²21.59% highreference9.8696 EI/L²ten Ritz terms,as an eigenvalue problemevery guess is anupper boundP
Fig. 6 Guessing the shape and letting the energy decide, which is what the derivation above does. The sine assumed here is exact for a uniform member on a uniform foundation; where either varies, the same energy method with an approximate shape gives an upper bound, and the accuracy is much better than the shape deserves.

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.

A column that was never straightLoad against lateral deflection at mid-height, for a column starting with an initial bow of 0.002. There is no critical value to reach: the deflection grows from the first increment, slowly at first and then without bound as the ratio approaches 1.00 — which is the Euler load, and which the column therefore never attains. The perfect column, drawn for comparison, sits on the vertical axis until it arrives there and then has no answer at all.00.0050.010.0150.020.02500.20.40.60.81lateral deflection at mid-heightload ÷ P꜀ᵣP ÷ P꜀ᵣ = 1.00, approached and never reachedinitial bow: δ₀ = 0.002
Fig. 7 What an initial crookedness does to any of this. The critical load computed above is a bifurcation load for a perfectly straight member; a real flange is bowed, and the deflection grows from the first newton with an amplification of 1/(1−P/Pcr)1/(1 - P/P_{cr}). The restraint’s force follows that deflection, which is what a U-frame actually has to be designed for.

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 2kEI2\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≈π(EI/k)1/4L \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.

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 kk 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 ends decide the length that mattersFour columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.K = 0.5both ends fixedK = 0.7one fixed, one pinnedK = 1both ends pinnedK = 2fixed at the base, free at the topsame column, same section, four ways of holding the endsthe load at which each buckles goes as 1 ÷ K² — a factor of sixteen across this row
Fig. 8 The idea this displaces. Four columns of the same section buckling at loads sixteen times apart, because of what holds their ends. A continuously restrained member has no ends worth naming — the answer would be the same if it were twice as long — and the effective length it reports is a translation for the benefit of a code check rather than a description of anything.

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.

Where this goes next

K = 0.5both ends fixedK = 0.7one fixed, one pinnedK = 1both ends pinnedK = 2fixed at the base, free at the topsame column, same section, four ways of holding the endsthe load at which each buckles goes as 1 ÷ K² — a factor of sixteen across this row The ends decide the length that matters Four columns of identical height and section, buckling at loads sixteen times apart. Nothing differs but what is holding the two ends. 05010015020025001020304050brace stiffness (units of EI/L³)critical load (units of EI/L²)ideal stiffness ≈ 159 EI/L³21.839.5 — braced9.87 — unbraced The brace that need not be strong A brace holding a column at mid-height carries almost no force. What it has to be is stiff — and the stiffness required is exact, large, and reached at a knee past which more buys nothing at all. 1000 kN1/β = 2.56 mthe beam lifts off at 6.04 msettlement 3.90 mm · contact pressure 195 kN/m · moment 641 kNm, none of which contains the length of the beam The beam that sits on the ground Every other beam in this collection is held at points. A footing is held everywhere, by something that pushes back in proportion to how far it is pushed — and that single change hands the structure a length it did not choose. Two or three of those lengths from the column, nothing knows the load happened. 2000400060008000100001200002004006008001000distance between lateral restraintsthey cross at 3803the plastic capacity of the sectionelastic critical momentSt Venant torsion alone — what is left at long lengthswarping dominates here The beam that fails sideways A deep narrow beam bending in its strong plane can, at a moment well below its capacity, swing out of that plane and twist. The failure has nothing to do with how much it can carry and everything to do with what is holding it.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

BracingBucklingCompression flangeContinuous restraintCritical loadEffective lengthEigenvalueElastic foundationHalf wavelengthImperfectionLateral torsional bucklingMode shapeSlendernessStiffnessU frame