Structural Engineering & Statics
where the load goes
Stability

The buckle that will not spread out

A compression chord on a continuous restraint chooses its own number of half-waves and forgets how long it is. Give it the force it actually carries — a parabola, largest at midspan — and the number barely moves while the shape changes completely, which is the half that decides where the restraint has to be.

Assumes Held everywhere, and it forgets its length, Strong enough and still falls over and The diagram is an integral, and that is why it can be drawn by eye.

14 min read 6 figures Which failure arrives firstOne support too manyThe load must go somewhere

Held everywhere, and it forgets its length is about a member that has stopped having a buckling length. A compression chord restrained continuously — a through girder’s top chord, held by the U-frames formed of its own cross-girders and verticals — buckles into whichever number of half-waves is cheapest, and past a certain length the critical force settles at 2EIk2\sqrt{EIk} with no length in it at all.

Every number in that essay assumes the force in the chord is the same everywhere along it. It never is.

The same chord, buckling under a constant force and under a varying one. Two buckled shapes of the same 40 m compression chord on the same continuous restraint of 0.35 N/mm per mm, at the same scale. Under a constant force the buckle fills the member — 8 half-waves over 98 per cent of the length — and the critical force is 1881 kN, which is the length-free answer the closed form gives. Under the parabolic force a uniformly loaded deck delivers to it, the buckle LOCALISES: 5 half-waves over 56 per cent of it, gathered where the force is largest, and the peak force at buckling is 2113 kN. The shaded curve is the force distribution the second shape is buckling under.
Fig. 1 The same 40 m chord buckling twice, at the same scale. Under a constant force it fills the member — eight half-waves over 98 per cent of the length, at 1,881 kN, which is the length-free answer. Under the parabolic force a uniformly loaded deck actually delivers, it gathers into five half-waves over 56 per cent of the length, with a peak force of 2,113 kN. The shaded curve is the force the second shape is buckling under.

Two things have changed and they are of very different sizes. The critical force has gone up by 12 per cent. The buckled shape has gone from something that occupies the whole chord to something that occupies rather more than half of it, gathered around midspan.

The second is the useful one, and this essay is mostly about why.

Where the parabola comes from

A chord’s axial force is the girder’s bending moment divided by its depth, and nothing else. That is the whole of the relationship, and it is why the distribution is not a modelling choice.

Load, shear and moment — a simple span. The applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.
Fig. 2 A 40 m simply supported span under 30 kN/m. The moment is the second integral of the load, so it is a parabola with 6,000 kNm at midspan and nothing at the ends. On a girder 3 m deep the top chord’s compression is that divided by three: 2,000 kN at midspan, zero over the bearings, and every value between.

So the chord carries 2,000 kN where the restraint has to work hardest and nothing at all where the bearings are, and the shape of that distribution is an integral rather than an assumption. A chord under a point load carries a triangular distribution for the same reason; a continuous girder’s chord changes sign.

The constant-force model has to be handed one number to stand for all of them, and there are only two candidates: the peak and the mean. They differ by a factor of 1.5, and the answer is not the obvious one.

Which free body produced the number

The free body is the compression chord alone, cut free of the girder, with three things crossing the cut.

An axial force at each end, varying along the length in between because the cross-girders deliver load into the chord as it goes — which is what makes the distribution a parabola rather than a step.

A continuous lateral restraint along it, from the U-frames: each cross-girder and its two verticals form a portal, and its sway stiffness divided by the frame spacing is a stiffness per unit length. That is the kk in every expression here, it has units of force per length per length, and its value on this chord is 0.35 N/mm per mm.

And the chord’s own bending stiffness about the axis it would buckle across.

The equilibrium of that free body in a slightly bent position is the standard beam-column-on-a-bed equation, with one difference from the closed form: NN is a function of xx, so the equation has a variable coefficient and no sinusoidal solution. The way to answer it is to discretise — assemble the elastic stiffness and the geometric stiffness element by element, scale each element’s geometric term by the force it actually carries, and find the multiplier at which the sum stops being positive definite.

That returns a shape as well as a number, and the shape is the part the closed form cannot supply.

The number barely moves

Take the peak force as the answer’s measure, so the two models can be compared.

Three answers for the peak force a chord can carry. The peak chord force at buckling against the length of the girder, three ways. The constant-force answer flattens at 1879 kN, which is 2√(EIk) and has no length in it. The parabolic one sits above it by 12 per cent at 40 m, because the ends of the buckle sit where the force is smaller — so treating a varying force as constant at its peak is conservative, and only slightly. The third line is what a designer gets by putting the MEAN force into the constant formula: it permits a peak of 2822 kN against a true 2113, which is 34 per cent unconservative and is the error worth knowing about.
Fig. 3 The peak chord force at buckling, three ways, against the span. The constant-force line flattens at 1,879 kN — the length-free 2EIk2\sqrt{EIk}. The eigenvalue with a parabolic force sits 12 per cent above it at 40 m. The third line is what putting the mean force into the constant formula permits: a peak of 2,822 kN against a true 2,113.

The first reading is a relief and the second is not.

Treating the varying force as constant at its peak is conservative by 12 per cent, and the margin shrinks as the span grows — 8 per cent at 60 m, 38 at 12. That is a defensible way to check a chord, and it is what most design rules do.

Treating it as constant at its mean is unconservative by 34 per cent. The mean of a parabola is two thirds of its peak, and dividing a correct critical force by two thirds is not a calculation about anything: the chord does not buckle in a mode that samples the average, it buckles in a mode that has moved to where the force is largest. The average is a number the structure never sees.

That asymmetry is worth stating as a rule, because the same trap is available in every buckling problem with a varying action. An eigenvalue is not an average of the local answers. It is set by the worst region the mode can find, and the mode is free to go and find it — which is why a smaller mean with the same peak buys almost nothing, and why a smaller peak with the same mean buys nearly everything.

What the taper actually buys

The 12 per cent has a mechanical explanation and it is visible in the figure rather than in the arithmetic.

A buckle in a restrained chord costs energy in two places: bending the chord, and stretching the restraint. It releases energy through the axial force acting over the shortening the bending produces. The cheapest arrangement is a half-wave of a particular length — π(EI/k)1/4\pi(EI/k)^{1/4}, which is 5.1 m on this chord — because that is where the two costs balance.

When the force tapers, the mode faces a trade. Staying in the middle means every half-wave sits where the force is largest, which is where the release is greatest — but a mode confined to part of the length has to decay at its edges, and the decay costs bending and restraint without releasing anything. So the buckle localises as far as the decay allows and no further, which on this chord is 56 per cent of the length.

That is why the critical force moves so little. The mode gains by sitting where the force is high and pays for the confinement, and the two nearly cancel.

The same chord, buckling under a constant force and under a varying one. Two buckled shapes of the same 12 m compression chord on the same continuous restraint of 0.35 N/mm per mm, at the same scale. Under a constant force the buckle fills the member — 2 half-waves over 92 per cent of the length — and the critical force is 1968 kN, 5 per cent above the length-free 2√(EIk) because the member is too short to fit a whole number of its preferred half-waves. Under the parabolic force a uniformly loaded deck delivers to it, the buckle LOCALISES: 2 half-waves over 88 per cent of it, gathered where the force is largest, and the peak force at buckling is 2722 kN. The shaded curve is the force distribution the second shape is buckling under.
Fig. 4 A 12 m chord, where there is nowhere to localise to. The preferred half-wave is 5.1 m, so the member holds two of them and no more, and the parabolic mode is the same shape as the constant one — 88 per cent of the length against 92. The critical force rises by 38 per cent, which is the largest gap on the whole curve and comes entirely from the ends being unloaded rather than from any change of shape.

The short chord is the case where the constant-at-peak assumption is most conservative and the shape argument has nothing to say. The long chord is the reverse. Which is a division worth keeping: localisation needs room, and room means a span several times the preferred half-wave.

Why the shape matters more than the number

A restraint’s stiffness is what the eigenvalue is computed from, and its force is what the U-frame has to be designed for — and the force comes from the mode.

The restraint force at any point is kk times the lateral movement there, and the lateral movement is the buckled shape scaled by however far the chord has actually bowed. An eigenvector has no amplitude, so the honest calculation puts in an initial crookedness and computes the force it grows into, which is the same division of labour a column brace has.

But the distribution of that force is the mode shape, and the mode shape has just changed completely. Under a constant force, every U-frame along the girder is equally involved. Under the real force, the U-frames in the middle 56 per cent are carrying nearly all of it and the ones near the bearings are carrying almost nothing.

Two practical consequences follow, and neither is visible in the critical load.

The U-frames are not all the same member. Sizing them all from one restraint force distributes steel to the ends of a girder where the mode has no ordinate. That is not dangerous, and it is a quarter of the bracing on a long girder.

And a weak U-frame near midspan is much worse than a weak one near a bearing. A rule expressed as a stiffness per unit length assumes the stiffness is uniform, and a cross-girder connection that is softer than the others is a local dip in kk exactly where the mode lives.

The same chord, buckling under a constant force and under a varying one. Two buckled shapes of the same 40 m compression chord on the same continuous restraint of 1.4 N/mm per mm, at the same scale. Under a constant force the buckle fills the member — 11 half-waves over 98 per cent of the length — and the critical force is 3757 kN, which is the length-free answer the closed form gives. Under the parabolic force a uniformly loaded deck delivers to it, the buckle LOCALISES: 6 half-waves over 48 per cent of it, gathered where the force is largest, and the peak force at buckling is 4081 kN. The shaded curve is the force distribution the second shape is buckling under.
Fig. 5 The same 40 m chord with the U-frames four times stiffer. The critical peak force rises from 2,113 to 4,113 kN, and the buckle contracts further — six half-waves over 48 per cent of the length rather than five over 56. Stiffening the restraint shortens the preferred half-wave as the fourth root of kk, so a better-braced chord concentrates its mode rather than spreading it.

Stiffness concentrates the problem it solves. That is not what a designer expects from a brace, and it means the two consequences above get sharper as the bracing gets better rather than milder.

The chord under a point load

The parabola belongs to a uniformly loaded deck. A girder carrying a single heavy load — a crane runway, a bridge under one bogie — has a chord force that peaks at the load and falls off linearly either side.

The same chord, buckling under a constant force and under a varying one. Two buckled shapes of the same 40 m compression chord on the same continuous restraint of 0.35 N/mm per mm, at the same scale. Under a constant force the buckle fills the member — 8 half-waves over 98 per cent of the length — and the critical force is 1881 kN, which is the length-free answer the closed form gives. Under the triangular force a load at midspan delivers to it, the buckle LOCALISES: 4 half-waves over 46 per cent of it, gathered where the force is largest, and the peak force at buckling is 2408 kN. The shaded curve is the force distribution the second shape is buckling under.
Fig. 6 The same chord with a triangular force distribution. The peak at buckling is 2,408 kN — 28 per cent above the constant-force answer against the parabola’s 12 — and the buckle has contracted to four half-waves over 46 per cent of the length. The sharper the peak, the more the mode gains by localising and the less of the member it uses.

There is a limit implied by that trend and it is worth naming. As the distribution gets sharper, the buckle localises further and the peak force at buckling rises — but it cannot rise past the answer for a member whose whole length carries the peak, which is what a flat distribution is. So the constant-at-peak calculation is a genuine lower bound for any distribution that peaks once, and the amount it is conservative by is a measure of how sharply the force falls away.

That is a satisfying place for a design rule to sit: safe by construction, and wasteful in proportion to something a designer can see on a moment diagram.

Where the model stops

The restraint is elastic and continuous. A U-frame every 3 m on a chord whose half-wave is 5.1 m is a reasonable smearing; the same frames every 8 m are not, and the answer then is a strut with discrete springs, whose mode is decided by where the springs are rather than by the fourth root.

Nothing yields. The chord is a compression member with a real cross-section, and at 2,000 kN it is somewhere on its own column curve rather than on the elastic eigenvalue. The localisation argument survives that — the mode still goes where the force is — but the number does not.

The force distribution is the one load case. A chord force is a moment divided by a depth, so a moving load, a pattern load and a construction stage each give a different distribution and therefore a different mode. The governing case is not necessarily the one with the largest peak, since a flatter distribution with a smaller peak can be worse.

And the chord is straight. The eigenvalue says nothing about a member that was built with a bow in it, and a bow that happens to look like the localised mode is worth a great deal more than one that does not — which is what imperfection sensitivity is, and is sharper here because the mode occupies less of the member and so is easier to match by accident.

What the pictures cannot show

Both shapes are eigenvectors, so neither has an amplitude and neither carries a force. The figure showing five half-waves gathered at midspan is drawn at an arbitrary scale, and the only honest statement about its size is that it is zero until something makes it otherwise.

They also cannot show what the U-frames are doing to themselves. Each frame is a portal whose stiffness comes from a cross-girder bending and two verticals bending, and every number on this page treats that stiffness as a constant. It is not: the connection at the top of the vertical is a joint with its own rotational stiffness, and a joint is neither pinned nor rigid until it has been drawn in detail. On a real half-through bridge, the difference between the assumed and the delivered kk is larger than the 12 per cent this whole essay has been about.

The assumption the figure rests on

That the chord’s bending stiffness is constant along it.

It usually is not, for the same reason the force is not: a chord sized for its peak force is oversized at the ends, so a designer curtails it — a smaller section, or a plate that stops. That makes EIEI a function of xx as well as NN, and the two vary together, both largest at midspan.

Which way that moves the answer is not obvious and the model here can be asked. A chord whose stiffness tapers with its force has a mode that localises less, because the middle is now stiff as well as heavily loaded and the ends are cheap to bend. The two effects work against each other, and a fully stressed chord — one whose section follows the force exactly — comes close to behaving like the constant-force case again, at a critical force well below what the localisation argument would have promised.

That is a general property of optimised members and it turns up throughout stability: making every section work equally hard removes the redundancy the mode was being kept out of, which is the same finding as the best design being the most sensitive one.

The history, and the bridge that is the reason for the rule

The U-frame is a nineteenth-century device and its calculation is a twentieth-century one, which is the usual order.

A half-through girder — deck at the bottom, chords above the traffic, nothing across the top — is what a railway bridge becomes when the headroom below is fixed and the depth has to go somewhere. The arrangement is as old as wrought-iron railway bridges, and for decades the top chord’s stability was handled by proportioning rules: a chord no more slender than some number, verticals of some depth, and an outcome that was safe because the numbers came from bridges that had stood.

Engesser gave it an analysis in 1884, and the form he gave it is the one still in use: treat the discrete U-frames as a smeared elastic medium, and the chord as a strut on an elastic foundation. That substitution is the whole technique, and it is the same substitution that turns any array of discrete supports into a continuum — safe when the mode is long compared with the spacing, and quietly wrong when it is not.

What Engesser’s method does not contain is the varying force, and the reason is arithmetic rather than principle: a variable-coefficient eigenvalue problem was not solvable by hand in 1884 and the constant-force one was. So the practice that grew up around it took the chord’s maximum force and applied the constant formula, which the figures above say is conservative by between 8 and 38 per cent depending on the span — a margin nobody chose, produced by a limitation of the method rather than by a decision about safety.

That is worth carrying past this page. A conservative design rule is often a record of what could be computed at the time, and the size of its conservatism is not a measure of anybody’s caution. The way to find out is to solve the problem the rule was standing in for, and see how far apart the two answers are.

A surprising place the same shape turns up

A localising mode under a varying action is not a bridge phenomenon, and the clearest relative is one this collection has already met from the other direction.

A column that leans on its neighbours shares a storey’s stability between members that carry very different loads, and the storey buckles in a mode that is largest where the load is. The load that makes itself worse is the same statement for a single member. In each case the structure is free to choose where to fail, and choosing is what an eigenvalue does.

The general form is that an average is safe only when the mode cannot move. Where it can — a chord long enough to localise, a storey with unequal columns, a shell whose imperfection is anywhere on its surface — the answer belongs to the worst region, and averaging is not conservative in any direction that can be predicted without solving the problem.

The ladder from here

Later rungs on this anchor: discrete U-frames, where the spacing rather than the smeared stiffness decides the mode and the transition from one to the other has its own length scale. The restraint force computed properly from an assumed initial bow, which is what a U-frame is actually designed for and what no eigenvalue can supply. The chord that is also a beam, carrying local bending from the cross-girders between its restraints, where the buckling check and the bending check share a section. Inelastic buckling of a chord at a force this large, where the tangent modulus replaces EE in every fourth root on this page. And the same equation with the axial force reversed, which is a beam on an elastic foundation and a different subject with identical mathematics.

Where this goes next

The restraint chooses the buckling length, and it is not the member's. A 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. 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. A brace is a stiffness requirement, not a strength one. Critical 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. A stiffness of 60EI/L³ is marked, reaching 21.75EI/L². 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. The length at which a beam stops being a beam. Elastic critical moment against the distance between lateral restraints, with the section's plastic capacity drawn across it. The two cross at 3803 — beyond that length the beam buckles sideways before it reaches the strength its cross-section has, and the capacity is set by the restraints rather than by the steel. 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. Load, shear and moment — a simple span. The applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear. The diagram is an integral, and that is why it can be drawn by eye Load, shear and moment are one function and its two integrals. Once that is seen, the diagrams stop being things to calculate and become things to sketch.

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Bending momentBucklingCompression chordContinuous restraintCritical loadEffective lengthEigenvalueElastic foundationHalf-waveMode shapeRestraintThrough-girderU-frame