Field

Stability

Strong enough and still falling over: buckling, slenderness, and loads that make themselves worse.

$The column curve. Failure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.$

Strong enough and still falls over

A column can fail at a fraction of the load its material could carry, by going sideways. Buckling is a failure of stability rather than of strength, and it is decided by geometry.

$The ends decide the length that matters. Four columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.$

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.

The load that makes itself worse

A structure that has leaned carries its weight off the axis, which makes it lean further. The amplification is one over one minus the load ratio, and it runs away long before the buckling load.

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.

$A 8 mm plate, and the width it can be. The elastic critical stress of a plate in compression against its width, with the yield stress drawn across it. Below 370 mm the plate reaches yield before it buckles; above it the plate ripples first, and the fraction of the width still carrying load falls away — at 700 mm only 47 per cent of it is still working.$

The plate that ripples, and the width that is left

A wide thin plate in compression buckles at a stress that has nothing to do with the strength of the material. It then goes on carrying load — the middle drops out, and the edges work harder.

A column that was never straight. Load 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.

The column that was never straight

Euler's load is the load at which a perfectly straight column becomes indifferent to being bent. No column is perfectly straight, so no column ever reaches it — and the load it never reaches can still be measured.

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.

A load with a maximum in it, and nothing bifurcates. Load against apex movement for a two-bar frame of half-span 1000 mm and rise 150 mm. The load rises to 133.4 kN at a movement of 64 mm — well short of the 150 mm that would bring the apex level — and then falls. Past that point the frame can only be held by taking load away, so under a dead weight it goes: 260 mm of movement at constant load, arriving inverted and in tension. The minimum on the path is -133.4 kN, the exact negative of the maximum, because the geometry is symmetric about the flat position and the arithmetic knows it.

The roof that jumps

Every stability failure in this collection so far has been a bifurcation — a straight thing discovering it can be bent. A shallow frame does something else entirely. It stays perfectly symmetric, deforms steadily, and at some point the load it can carry starts to fall while it is still moving in the direction it was pushed.

Three paths out of the same critical load. Load against sideways movement past the critical load, for three systems whose critical loads are identical. The stable one climbs, so a real structure with a small crookedness reaches nearly the full load and keeps going. The unstable one falls symmetrically, so the imperfect structure has a maximum below the critical load and it matters not at all which way it leans. The asymmetric one falls one way and climbs the other, so the direction of the imperfection decides everything. All three are drawn at an imperfection of 0.02 radians.

A third of what the theory promised

A column with a small crookedness reaches almost its full Euler load. A cylinder with the same relative crookedness reaches a third of its classical one, and the theory is not wrong — what separates them is the slope of the path just past the critical load, which no calculation of the critical load itself can see.

A cruciform has three critical loads, not one. The three critical loads of a cruciform in compression, against its length, with the load it actually buckles at drawn over them. At 3000 mm the flexural loads are 921 kN about the major axis and 921 kN about the minor, while twisting about the shear centre takes 700 kN. The lowest root is 700 kN, and the column twists. The shear centre is the centroid, so the three modes are independent and the envelope is simply the lowest of them. The governing mode changes at 3442 mm: below that length the column twists, above it, it bends — because the torsional resistance keeps a term that does not grow when the member is shortened, and the flexural loads have none.

The column that twists instead of bending

Euler's column has one mode. A real column has three, and which of them governs is settled by where the shear centre sits. A cruciform strut buckles by rotating about its own length at a load that does not change no matter how short it is made.

One restraint, and several times the load. The same portal — the same columns, the same beam, the same steel — buckling with its head held against sway and with its head free to sway. The braced frame's critical load is 16.46 EI/L² and the swaying one's is 5.69 EI/L², a factor of 2.89, and the effective length factor that comes out of each eigenvalue is 0.774 against 1.317. Both are eigenvalues of the assembled frame at a beam-to-column stiffness ratio of G = 1.00; the buckled shapes are the mode vectors themselves, drawn at 18 per cent of the storey height so that the movement can be seen.

Held, and not held

One horizontal restraint at the head of a storey, carrying no vertical load whatever, moves the critical load of the columns beneath it by a factor of 2.89. Effective length is a property of the frame, not of the member.

Four guesses at one buckling mode. A 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.

Guessing the shape, and getting the load anyway

A column's buckling load can be had from a shape that is wrong everywhere, because the energy criterion is stationary at the true mode. The error in the load is the square of the error in the shape, and it is always high.

Why the curve sags, and why the two axes are not the same column. The same column curve with the sag computed rather than drawn. A hot-rolled section carries a residual compression of 30% of yield at its flange tips before anything is applied, so the tips yield first and what is left resisting a change of shape is the elastic core. About the major axis the stiffness follows the core's width; about the minor axis it follows its cube. The worst loss is 27% at λ = 74 about the minor axis against 23% about the major, and the whole effect lives between λ = 75 and λ = 89 — outside that band nothing has yielded, or everything has. No imperfection appears anywhere in this figure.

The column that had yielded before it was loaded

A real column sits below both of the two straight answers over the whole middle of the slenderness range, and the usual explanation — that it was not straight — is only half of it. The other half is that the flange tips had already yielded when it left the rolling mill.

$A buckled panel is a truss that nobody drew. A 1000 × 1000 panel of 6 mm web, at d/t = 167. It buckles in shear at 63.8 N/mm², which is 383 kN — and it then carries 696 kN, 1.82 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221.3 N per millimetre of its length. A web that never buckled at all would have reached 953 kN, so the panel ends at 73% of a stocky web's capacity on a fraction of its steel.$

The panel that carries more after it has failed

Everywhere else in this field a critical load is where the argument ends. A thin web is the exception — it buckles visibly, in waves anybody can see, and then goes on to carry nearly twice as much again by turning itself into a truss nobody drew.

The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 100 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 559 mm — 5.6 times the bearing, and 82% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to.

The load that chooses its own length

Every other load in this collection arrives over a length somebody decided. A wheel on a crane girder does not — the flange bends under it and spreads it along the web, and how far it spreads is an output of the flange's own stiffness against the web's own strength. The effective length is 5.6 times the bearing that produced it.

A built-up column has a second way to bend. A 12 m column of two chords 300 mm apart, joined by double lacing. On the left it buckles the way a solid column does, by bending; on the right the chords stay straight and the lattice racks, which a solid column cannot do at all. Neither happens alone, and the two flexibilities add rather than the two stiffnesses — so the critical load is 1287 kN against an Euler load of 1341 kN, which is 96% of it, and the column behaves as though its slenderness were 80 rather than 79.

The column made of two columns

A solid column buckles when its bending stiffness runs out. A laced one has a second way to go — the lattice shears, the chords stay straight — and the two flexibilities add rather than the two stiffnesses. A battened column reaches 23% of its own Euler load and behaves as though its slenderness were twice what it is.

How stiff a brace has to be before the frame stops swaying. The effective length factor of a swaying portal against the stiffness of a horizontal spring at its head. The curve starts at k = 1.317, the unbraced value, and falls to 0.774 — the factor for the same frame with its head held — at a brace stiffness of 23.2 EI/L³. Past that point the frame buckles in the non-sway mode, which the brace does not restrain, and further stiffness buys nothing at all. The threshold is worth stating as 1.41 N꜀ᵣ/L, which is the form the number is memorable in: for a storey carrying a thousand kilonewtons over four metres it is about 0.35 kN per millimetre of sway. Against the frame's own lateral stiffness of 12.0 EI/L³ it is a factor of 1.93.

The most dangerous day is before it is finished

A structure is analysed once, complete, with every restraint present. It spends weeks in states nobody drew — a beam landed with no deck on it holds 17% of the moment its section is worth, a frame not yet braced buckles at a third of the load it will, and a bolt not yet tightened is a pin where the analysis assumed a fixity.

The arch does not squash; it leans. The first two buckling modes of the same rib, drawn against its undeformed shape at an exaggeration of a few hundred. The first, at w_cr L³/EI = 49.7, is antisymmetric: one half rises while the other falls and the crown moves sideways. The second, at 117.1, is symmetric — the whole rib settling. The two differ by a factor of 2.36, which is why an arch is braced against sideways movement of its crown rather than against the load it is carrying, and why a tied arch with a single hanger at midspan is doing nothing for the mode that governs it.

The arch that leans instead of squashing

A masonry arch is asked whether a line of thrust fits inside it. A steel rib is asked a different question entirely: it is a column carrying an axial force along its whole length, and the mode it buckles in puts one half up and the other half down while the crown moves sideways.

A brace on the wrong flange never gets there, however stiff it is. The critical moment of an 8 m beam against the stiffness of a single midspan brace, drawn three times for the three heights the brace could sit at. On the compression flange it climbs from 143 kNm to the two-half-wave plateau of 447 — the beam braced into two 4.0 m beams — and reaches 99% of it at 447 kN/m. At the shear centre it needs 2252 kN/m, 5.0 times as much. On the tension flange it never arrives at all: at the stiffness that would have done the job on the other flange it has bought a factor of 1.068, and a stiffer brace in the same place buys the same nothing. Past the plateau the beam stops using the brace, which is where the idea of an ideal stiffness comes from.

The brace on the wrong flange

A brace on a column has one property that matters, and it is stiffness. A brace on a beam has two, and the second decides whether the first is worth anything: put the identical restraint on the tension flange and it does not reach the answer at any stiffness whatever.

A column that has nothing on it but itself. A 30 m column carrying no load except its own weight, with its buckled shape and the axial force that produced it. The force is zero at the top and largest at the base, which is why the answer is not Euler's: the eigenvalue is a load intensity and comes out as q_cr L³/EI = 7.835, a number with no π in it, computed here as the smallest eigenvalue of the same stiffness and geometric-stiffness matrices that give a tip-loaded column its Euler load. It is equivalent to a tip load of 3.18 times as much total weight — a column carries its own weight better than it carries somebody else's, because most of the weight is near the base where the buckle is not. The height limit that follows is a cube root, so a section of radius of gyration 80 mm falls over on its own at 52 m and one of twice that reaches only 82.

Too tall for nothing but itself

Every critical load on this site so far has been applied at the top of a column. A mast carries a load that is zero at the top and largest at the base, the governing equation stops being harmonic, and the answer comes out as a number with no π in it — along with a maximum height that is almost the same for steel, aluminium and wood.

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.

Not where the two loads meet. How much a column loses below the weaker of its two single-mode capacities, against the ratio of its local critical load to its global one. The received claim is that the worst place is where the two coincide; the arithmetic says otherwise. The erosion is largest at a ratio of 0.47 — 23% — sits within a per cent of that for every ratio below about a half, and at exact coincidence is only 2%. What the curve does say is the useful half of the folk claim: once the plates are stocky enough that the local critical load is twice the global one, the interaction is nothing at all, and the section is worth thickening only up to there.

Two ways of buckling at once

A thin-walled column can bow as a whole or ripple in its plates, and each has its own critical load. The received advice is that the worst arrangement is the one where the two are equal. The arithmetic says the opposite — at coincidence the interaction costs two per cent, and the expensive region is where the plates go first.

Three minima, and only two of them get a check. Elastic buckling stress against half-wavelength for a 200 × 65 × 15 × 1.5 mm lipped channel in uniform compression. The local minimum is at 200 mm and 41 N/mm²; the distortional at 689 mm and 287; the global curve falls away to the right and reaches 489 at the 1.5 m member. The distortional branch is a strut on an elastic foundation — the flange and lip rotating about the web junction, restrained by the web's own bending at 627 N·mm per radian per millimetre — so its minimum is at π(EC_w/k_φ)^¼ and its value is (2√(EC_wk_φ) + GJ)/I₀, the same closed form a continuously braced strut has. The elastic stresses are in the order local, distortional, global, and the mode that governs the strength is not the lowest of them, because they have very different amounts of post-buckling reserve.

The mode between the two that get checked

A thin-walled strut has three ways of buckling and two of them have design rules. The third has a half-wavelength several times the section depth, a shape in which the fold lines themselves move, and an elastic stress that no effective-width calculation can produce.

The deflection goes on growing, and sometimes it does not stop. Second-order deflection of a sustained-loaded concrete column against age, on a log time axis. Creep takes the effective modulus down, which takes the buckling load down with it — from 11580 kN on the day to 3309 in the long term, 29 per cent of it — so the amplifier 1/(1 − N/N_cr) grows even though nothing was added to the load. At 2200 kN the column settles: 25 mm of eccentricity on the day and 60 mm at the end, a factor of 2.4 for a load that never changed. At 5294 kN — still only 46 per cent of the day-one critical load — it does not settle, and the divergence arrives at 55 days for no new reason at all.

The column that fails years later

A concrete column under sustained load goes on straining at constant stress, so its deflection grows — and because the second-order moment is the load times that deflection, the demand grows with it. There is a load below which the two settle and one above which they never do.

The cheapest way out of being round. A ring under uniform external pressure, drawn in its first four buckling modes with the pressure each one needs underneath it, in N/mm². The pressure has no direction: it stays normal to the wall wherever the wall goes, so it does work on any change of shape that reduces the enclosed area, and the ring buckles into whichever shape is cheapest. Bare, that is the oval — n = 2 at 3EI/R³ — and the modes rise as n² − 1, so three lobes cost 2.67 times as much. Nothing in the drawing prefers any orientation, which is the point — a column has an axis to buckle about and a ring has none.

The pressure that needs no direction

Every buckling problem in this collection has had a load with a direction — a column pushed along its axis, a plate along its edge, an arch by what is on it. A buried pipe has none. The pressure is the same everywhere, it stays normal to the wall as the wall moves, and it does work on any change of shape that reduces the area inside.

A fourth power, and then a cliff. The factor of safety against rolling, against beam length, for one section hung from a roll axis 0.9 m above its centre of gravity. Nothing about the section changes along this axis. z̄ goes as the fourth power of the length — 0.236 m at 30 m becomes 0.747 m at 40 — and the factor of safety is proportional to (y_r − z̄), so it does not decline gently: it falls away and then stops existing. The working factor of 1.5 is lost at about 41 m, and past 42 m there is no hook height at all at which this beam hangs stably. Which is why long girders are lifted with the picks moved inboard, or with the beam braced, or not in one piece.

Hung from above and still unstable

A rigid body hanging from a point above its centre of gravity is a pendulum and cannot fall over. A beam is not rigid, and tilting it puts a component of its own weight sideways — which bows it, which moves its centre of gravity further out. Past a length there is no hook height at which it hangs stably at all, and the length arrives as a fourth power.

$A straight line, and the comfortable case is already two thirds down it. The capacity of a 215 mm masonry wall as a fraction of its squash load, against the eccentricity of the resultant. Nothing in this figure is a buckling calculation. A material that cannot be pulled bears on a strip of width 3(t/2 − e) under a triangular stress block, so the capacity is exactly 1.5f(t − 2e) — a straight line, zero when the resultant reaches the face, and already at 67% at the edge of the kern. The middle third is treated everywhere as the comfortable case; a wall loaded there has given away a third of its capacity before slenderness has been mentioned. The lower line is the same wall with slenderness in it, which enters as an ADDITIONAL eccentricity of 17.4 mm rather than as a reduced stress — h_ef²/2400t, for h_ef = 3000 mm. Euler's load for this wall is 9.6 times what the eccentricity rule allows, which is why no masonry calculation contains it.$

It does not buckle, it runs out of width

Every stability failure in this collection is a member that could have carried tension deciding to go sideways instead. Masonry cannot carry tension, and its failure under an eccentric load is not a bifurcation at all — the bearing area simply shrinks until it runs out. The capacity is exactly linear in the eccentricity, Euler's load is ten times anything allowed, and no material property appears until the very end.

The load that moves with the twist

A beam about to buckle sideways is beginning to rotate, and everything attached to it rotates with it. A load hung from the top flange swings out over the side and drives the rotation on; the same load hung underneath swings back and stops it. Two identical beams, two different capacities, and the only difference is a height.

A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.

The tie that spends an afternoon as a strut

A tension member is chosen by its area and nothing else. A compression member is chosen by how that area is arranged. So a member whose force reverses under some load case is not merely being asked for the same number with the other sign — it is being designed against a different variable, and the same steel can carry seventy times more or less depending on a shape nobody chose for that purpose.

$The check that everything adds up, and the error it cannot see. Four versions of the same 2-bay, 3-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 6% and 24%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 18% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure.$

The stiffness the load takes away

Buckling is usually taught as an event — a critical load, a bifurcation, a mode. Written as a matrix it stops being an event at all. A compressive load subtracts a stiffness from the structure, the subtraction grows with the load, and the critical load is simply where what is left reaches zero.

Effective length is a property of the storey. The effective length factor of the one column that resists sway, against the total gravity load on the storey as a multiple of its own. At the left-hand end it carries the storey alone and its K is 1.99 — the 2.0 every chart gives a column fixed at the base and free to sway at the top, reproduced here by a route that never mentions a chart. Then columns are added that have pinned bases and therefore no lateral stiffness whatever. They contribute load and nothing else, so they cannot buckle on their own and they lower the load at which everything buckles together. K rises as the square root of the load ratio, exactly, and at the storey drawn — three leaning columns carrying 69% of the gravity load — it is 3.57. That is off the end of every published alignment chart, and the leaning columns themselves, which a designer would take at K = 1.0 for pinned ends, are at 2.54.

The column that leans on its neighbours

A column with a pinned base and a pinned top has no lateral stiffness at all and cannot stand up alone, and yet thousands of them do. What holds them is the rest of the storey, and what it costs is paid by whichever columns do have stiffness — whose effective length rises as the square root of the load being leaned on them, straight off the end of every chart.

A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam.

The web that is crushed from inside

A plate girder's compression flange is curved by the beam's own deflection, and a curved force needs a transverse load to stay on its curve. The only thing available to supply it is the web. So a deep girder can buckle its web vertically with nothing applied to it at all, and the rule that prevents it is the only clause in the codes about a load no load case contains.

Two frequencies that meet, and a determinant that never moves. The two natural frequencies of Ziegler's two-bar column against the follower load Pℓ/k, with the determinant of its stiffness matrix drawn along the top. The determinant is k² at every load — it varies over this whole axis by 1.1e-16 of itself, which is round-off — so a static buckling analysis of this structure finds no critical load whatever and reports it as stable everywhere. The frequencies say otherwise: they approach, meet at Pℓ/k = 2.0858, the closed form (7 − 2√2)/2, and become a complex pair, which is oscillation that grows. With no damping that merge is also where the column goes unstable. Adding any internal damping at all drops the load at which that happens to 1.4643, which is 41/28 and 30% below the undamped value; the limit of the damped system is not the undamped system, which is the paradox Ziegler found in 1952 and which was taken for an arithmetic error for a decade.

The load it cannot buckle under

Every stability calculation on this site rests on an assumption nobody states: that the load has a potential, so a critical load is where a total potential energy stops being a minimum. A load that turns with the structure it is pushing has no potential, and the static analysis of such a column returns no critical load at all — a determinant that never vanishes, for a column that fails at a perfectly finite one.

The average is not the answer, and it is unsafe. Critical load of a pinned column whose middle third has been given a different stiffness, against the whole-column Euler load, with the two numbers a hand check reaches for beside it. The eigenvalue is taken from K − P·Kg over 24 elements, so nothing here is a formula for a stepped column — it is the same computation the uniform case gets. At a middle third of 0.50 times the rest the true load is 0.612 of Euler's, the arithmetic average says 0.832 and the weakest segment says 0.496. The average is high by 36% and it is high on the unsafe side, because the third of the column it is averaging over is the third where the mode has all its curvature. The weakest-segment answer is safe everywhere and wasteful by about as much.

An average stiffness is not a safe stiffness

Euler's load belongs to a column of one EI. Give the same column two, and the temptation is to average them — which is wrong, and wrong in the unsafe direction by a quarter. Buckling weights stiffness by the square of the curvature of the mode, so the middle of a pinned column decides everything and the ends decide almost nothing.

A buckling load with no compression anywhere in the member. A tube of radius 81.6 mm and wall 5 mm under pure torque — no axial load at all — buckles into a helix at T = 2πEI/L, which for the section drawn is 1877 kNm. Two things are absent from that expression and both are surprising: the length appears to the first power rather than the second, and the shear modulus does not appear at all, so the torsional stiffness of the member has nothing to do with the torque at which it buckles in torsion. The comparison is the torque that yields the same tube, 42.9 kNm — a factor of 43.8 below it. The mode arrives first only past a length of 263 m, which is πE/τ_y = 3219 radii and contains no thickness whatever.

The buckling load with no compression in it

Twist a straight bar hard enough and it snaps into a helix, with no axial load on it anywhere. The load at which that happens is 2πEI/L — first power of the length, and no shear modulus in it at all, so how stiff the bar is in torsion has nothing to do with the torque that buckles it in torsion.

A stiffener is a boundary condition, and it is bought at a threshold. The buckling stress of a 2400 × 12 mm plate with one longitudinal stiffener, against how rigid that stiffener is. Below γ the stiffener rides on the buckle and the plate takes the whole-width mode; at γ the stiffener stays straight and the plate buckles between stiffeners at 74 N/mm², 4.0 times the bare plate's 18.5. Above γ nothing further happens at all, because the sub-panel mode does not know the stiffener is there. The curve is a ramp and then a horizontal line, so a stiffener at twice γ is exactly as good as one at γ. Here γ = 31.5, which asks for an outstand of 144 mm; the 150 mm one drawn gives γ = 35.5, a margin of 1.13.

The rib that is a boundary condition

A rib on a plate is not a member carrying load. It is a line the buckle is not allowed to cross — and it becomes one at a threshold. Below the required rigidity it rides on the buckle and buys a fraction; at the threshold it stays straight and the plate buckles between stiffeners; above it, nothing further happens at all.

The tube flattens because of the bending, and then cannot carry it. Moment against curvature for a long tube of radius 300 mm and wall 4 mm. Compression on one face and tension on the other are both directed along a curved line, so each produces an inward transverse pressure and the circle is squashed into an oval by the bending it is carrying. That reduces the second moment, so the curve bends over and reaches a limit point — no bifurcation, no imperfection, nothing to be sensitive to. It arrives at an ovalisation of exactly 2/9 for every tube of every size in every material, at 1018 kNm, where the secant stiffness has fallen to 67 per cent of the undeformed value and the tangent stiffness is zero. The relaxed path reproduces the closed form to 0.004 per cent.

The tube that flattens itself

Bend a tube and the compression on one face and the tension on the other are both running along a curve, so both push inward. The circle becomes an oval, the second moment falls, and the moment–curvature curve turns over at a limit point that needs no imperfection, no bifurcation and nothing to be sensitive to.

The shape of the diagram, and not its peak

A beam's lateral-torsional capacity is quoted against uniform moment, which is the one case a beam carrying a load never has. Change the shape of the moment diagram without changing its peak and the buckling moment moves by a factor of nearly three.

The best design is where two failures arrive together. A fixed area of steel rolled into tubes of every proportion, with the three things that can end each one. Euler's load goes as r² because I = A r²/2; the local buckling stress goes as 1/r² because the wall thins as the tube grows; squashing does not care. The capacity is the lowest of the three, so it has a maximum — and the maximum is exactly where the two buckling curves cross, at r/t = 129 and 2364 kN, which the closed form r*, the fourth root of αAL² over π³β√3, reproduces to 0.52 per cent. That is the general result and it is not about tubes: the optimum of a minimum of a rising and a falling curve is always their intersection, so optimising a design against two failure modes puts both of them at the design point — which is the one configuration imperfections hurt most.

The best design is the most sensitive one

Take a fixed area of steel and roll it into a tube. Euler's load rises with the radius and local buckling falls with it, so the capacity has a maximum — and the maximum is exactly where the two failure modes arrive together, which is the one configuration imperfections hurt most.

$A 10 mm plate, and the width it can be. The elastic critical stress of a plate in compression against its width, with the yield stress drawn across it. Below 462 mm the plate reaches yield before it buckles; above it the plate ripples first, and the fraction of the width still carrying load falls away — at 900 mm only 46 per cent of it is still working.$

The coefficient that is not four

A plate's buckling stress carries a coefficient that looks like a constant and is not. It is 4 for an internal element, 0.43 for an outstand and 23.9 for a panel in shear — and the width a 10 mm plate may be runs from 152 mm to 1,130 across that range.

The load being amplified is not the load doing the amplifying. The sway amplifier 1/(1 − ΣP/P_cr) against the storey's total gravity load, as columns are added that carry load and provide no lateral stiffness. The critical load of the storey is fixed at 8203 kN by the bracing that exists, and every leaning column moves the structure along the axis without changing it. At the storey drawn the amplifier is 1.57, so the second-order sway moment is 57% on top of the first-order one — and none of that 70% of the load which is causing it appears in any stability calculation done column by column. The storey stays stable across the whole of this axis. That is the practical reason a gravity-only column is drawn on the frame model rather than designed on its own: it is not being checked, it is being counted.

Counted, not checked

A column with pinned ends and no bracing cannot buckle on its own, so nothing about it fails a stability check. It still carries load, and load with no stiffness attached lowers the buckling load of everything around it — which is why a gravity-only column is put on the frame model and never designed by itself.

A base plate, and when the bolts start working. A 550 × 450 mm plate carrying 900 kN and 140 kN·m, so the resultant sits 155.56 mm from the centre against a kern of 91.67 mm. The plate is in partial contact: bearing over 358.33 mm at a peak of 11.16 N/mm², with the holding-down bolts carrying 0 kN. The plate lifts at 82.5 kN·m and crushes at 192.95 kN·m, and the bolts are not needed until 247.5 kN·m.

The pinned base that is not pinned

A column base drawn as a pin is a plate bearing on grout, and a plate in contact over its whole length resists rotation whether anybody wanted it to or not. The stiffness it delivers depends on the axial load, so the assumption is one a frame can leave and re-enter as its loads change.

Warping stiffens a short member and nothing at all a long one. The stiffening 1/[1 − tanh(κ)/κ] against kL, both axes logarithmic, over kL from 0.05 to 200. At the low end the curve is a straight line of slope −2, because for small kL the bracket is κ²/3 and the stiffening is 3/kL²: it reaches 1201 at kL = 0.05, falls to 1.005 at the top, and every open section ever rolled sits somewhere on it. The same three plates arranged three ways are marked: the 533 by 190 mm I-section at kL 2.76 and ×1.562, the tee at kL 37 and ×1.027, the angle at kL 34 and ×1.030. A tee's warping constant is 288 times smaller than the I-section's and an angle's 236 times, because their plates meet at a point and there is no pair of flanges to bend against each other — so they have no warping resistance to offer at all, and that is the reason an angle is a poor thing to twist.

The restraint that beats the gradient

A moment-gradient factor is worth up to 2.7 on a beam's critical moment and is tabulated everywhere. Holding the ends against warping is worth more, is achieved by a detail rather than by a load case, and appears in no table at all.

A channel has three critical loads, not one. The three critical loads of a channel in compression, against its length, with the load it actually buckles at drawn over them. At 3500 mm the flexural loads are 13970 kN about the major axis and 2306 kN about the minor, while twisting about the shear centre takes 1555 kN. The lowest root is 1484 kN, and the column twists. The shear centre sits 109.7 mm from the centroid, so the modes cannot happen separately: the lowest root of the coupled problem is 4.5 per cent below the lowest of the three, and the Wagner coefficient β is 0.60. The governing mode changes at 5813 mm: below that length the column twists, above it, it bends — because the torsional resistance keeps a term that does not grow when the member is shortened, and the flexural loads have none.

The third root of the cubic

A column has three buckling loads and an Euler calculation finds two of them. The third is a twist about the shear centre, and for a section whose shear centre is not at its centroid the three cannot happen separately — so the answer is the lowest root of a cubic and can be a third below anything the two familiar modes report.

A built-up column has a second way to bend. A 24 m column of two chords 800 mm apart, joined by single lacing. On the left it buckles the way a solid column does, by bending; on the right the chords stay straight and the lattice racks, which a solid column cannot do at all. Neither happens alone, and the two flexibilities add rather than the two stiffnesses — so the critical load is 3832 kN against an Euler load of 4629 kN, which is 83% of it, and the column behaves as though its slenderness were 66 rather than 60.

The lacing decides the force it has to carry

A perfectly straight column carries no shear, so the diagonals of a built-up column are resisting a force that exists only because the column is crooked. Computing it turns out to be circular — the shear is amplified by the very flexibility the lacing supplies — and the rule of thumb that replaces the calculation is wrong by a factor that grows as the lacing gets worse.

$A buckled panel is a truss that nobody drew. A 1500 × 2000 panel of 8 mm web, at d/t = 188. It buckles in shear at 41.0 N/mm², which is 492 kN — and it then carries 1187 kN, 2.41 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 18.5° with a membrane stress of 348 N/mm² over a width of 788 mm, and it pulls on the flange at 280.2 N per millimetre of its length. A web that never buckled at all would have reached 2460 kN, so the panel ends at 48% of a stocky web's capacity on a fraction of its steel.$

The tension has to pull on something

A buckled web carries its shear on a diagonal band of membrane tension, and the band pulls sideways on the flanges and stiffeners that bound it. That pull is the design output nobody plots — it runs from 72 to 603 newtons per millimetre across ordinary panel proportions, it is largest exactly where the panel is most efficient, and at the end of the girder there is nothing beyond to take it.

What a torsional brace buys, and the stiffness it takes. The critical moment of an 8 m beam against the stiffness of a single rotational restraint at midspan — a cross-frame or a stiffened connection to a secondary beam, resisting the twist rather than the sideways movement. It climbs from an unbraced 143 kNm to the same two-half-wave plateau of 447 a lateral brace reaches, and gets to 99% of it at 253 kNm/rad. Nothing in the calculation refers to a height, which is the difference that matters: a torsional brace cannot be put on the wrong flange because it is not attached to a flange in the sense the lateral one is.

Twice the moment, four times the brace

A lateral brace has to be on the right flange and its demand is very nearly linear in the load. A torsional brace has no flange to be wrong about, and its demand is exactly quadratic — so the restraint that is indifferent to where it is attached is the one that gets expensive fastest.

The stub column a bearing stiffener makes, in plan. A plan through the girder at the bearing. The 8 mm web runs across; the pair of 100 × 12 stiffeners stands off it; and the shaded strip of web either side — 15ε t_w, or 98 mm each way — is the width that buckles with the stiffeners rather than independently of them. Together they are an area of 3962 mm² with a second moment of 9.01·10⁶ mm⁴ about the web's centreline, a radius of gyration of 48 mm over a buckling length of 900 mm — 0.75 of the depth, because the flanges hold the ends. That is a slenderness of 0.25, at which the column curve returns 0.98: the stub column reaches 98 per cent of its squash load, and the section's own strength is very nearly the whole answer.

A column nine hundred millimetres long

The patch-load check asks how much of a web a flange can spread a wheel over, and answers in a plate-buckling reduction that throws seven tenths of it away. A pair of stiffeners does not improve that answer. It replaces the question with a different one, from a different family, with a different failure in it.

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.

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.

The coefficient is an envelope, and its scallops are whole half-waves. The plate buckling coefficient against aspect ratio α = a/b. Each faint branch is one half-wave count m: k = (m/α + α/m)², a curve whose own minimum is exactly 4 at α = m. The plate takes whichever branch is lowest, so the answer is the bold envelope — four touching 4 at α = 1, 2, 3 and 4, with cusps between them at α = √(m(m+1)) = 1.41, 2.45, 3.46, 4.47, where the plate is indifferent between m and m+1 half-waves. The first cusp reaches k = 4.50 and every later one is lower — 4.17, 4.08, 4.05. Past α = 1 the envelope never exceeds 4.49, which is why the length of a plate drops out of a formula that is otherwise entirely geometry.

Four was never a fact about plates

The coefficient every plate calculation starts from is quoted as 4, derived nowhere and remembered by everyone. It is the minimum of a quantity that has nothing to do with plates in it — and what the plate supplies is not the four but the restriction that produces the scallops around it.

The coefficient a web gets depends on where its neutral axis is. The plate buckling coefficient for an internal element against ψ = σ₂/σ₁, the ratio of the stresses at the two edges of the panel. Uniform compression is ψ = 1 and k = 4; a gradient running from compression to zero is ψ = 0 and k = 7.81; pure bending is ψ = −1 and k = 23.92, six times the value a column's flange gets. The 1800 × 12 mm web drawn here starts at ψ = −1.000 and k = 23.92 and ends at ψ = −0.910 and k = 21.63, because losing width from the compressed half drops the neutral axis and deepens the compression zone. The curve is steepest exactly where a bending web sits, so a small movement of the neutral axis costs 2.29 of coefficient.

Classified by a gradient it does not have

A web in bending is the one plate whose buckling coefficient cannot be looked up. It depends on the stress gradient, the gradient depends on where the neutral axis is, and the neutral axis depends on how much of the web the coefficient has just taken away — so the answer is a fixed point, and the calculation everyone does is its first term.

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