Structural Engineering & Statics
where the load goes
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. The amplification of a deflection against the ratio of applied load to buckling load. A structure at half its buckling load deflects twice as far as first-order analysis predicts, and the curve runs away well before the load is reached.

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 wcr 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 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 4086 — 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 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.

 It is the square of the diagram that destabilises. Four moment diagrams normalised to the same peak, and the buckling factor each one earns. Eliminating the lateral displacement from the coupled buckling equations leaves one functional in the twist, and its destabilising side is ∫M(z)²φ²/EI_z — the SQUARE of the moment, weighted by where the beam wants to twist. A diagram with a peak over a short length has a much smaller weighted square than a flat one of the same maximum, so it buckles at a higher peak: uniform 1.00, uniformly distributed load 1.13, central point load 1.36, cantilever 1.71. The root-mean-square of each diagram, printed beside it, very nearly predicts the order — which is as close to an intuition for C₁ as the subject has.

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.

 The same restraint, spread and gathered. The buckled shape of a 24.0 m compression chord with the same smeared restraint, 0.35 N/mm per mm, delivered by U-frames at two spacings. With frames every 2.0 m the chord buckles in half-waves of about 5.1 m that ignore the frames — the smeared shape — at 1892 kN, matching the smeared answer. With frames every 4.0 m it buckles between them, with a node at every frame, at 1548 kN: 18 per cent below the smeared 1897 and at the Euler load of one bay. The dots are the frames.

A row of frames is not a foundation

The top chord of a half-through girder is held sideways by U-frames, and the standard calculation smears them into a continuous elastic foundation. That is right while the frames are close and quietly wrong once they are not. The crossover sits at seven-tenths of the buckle's own half-wavelength, whatever the frames' stiffness — and beyond it the chord buckles between frames at a load no stiffening of the frames can raise.

 Four numbers read from a moment diagram. The moment diagram of the fixed-ended beam under a central point load, scaled to a largest value of one, with the four values the quarter-point formula reads: the peak, and the magnitudes at a quarter, a half and three quarters of the span — 1.00, 0.00, 1.00 and 0.00. The formula turns them into a gradient factor of 1.923. Solving the buckling problem for the whole diagram, on a beam 8 m between lateral restraints, fork-supported at its ends, gives 1.723: the formula is 12 per cent too high, on the unsafe side.

The formula that reads four numbers

The factor that credits a beam for the shape of its moment diagram is usually taken from a formula that reads the diagram at four places — its peak and its quarter points — and nowhere else. On the straight-line diagrams it was built around it is safe. On a fixed-ended beam under a central load, whose moment is zero exactly where the formula looks, it is twelve per cent unsafe, and it would give the same answer for a diagram that deserves twenty-two per cent less.

 The bow the frames have to hold. A chord 24 m long, EI 2.52 × 10¹² N·mm², held by U-frames every 4 m, carrying 930 kN critical at 1548 kN, drawn with its sideways movement exaggerated; its critical load is 1.66 times the chord force. Dashed, the worst of the bows tried: four half-waves, each 12.0 mm high, a five-hundredth of its own length. Solid, the chord once the compression has grown the bow; each arrow is the force the frame there must supply, the largest 13.6 kN — 1.47 per cent of the chord force.

What the frames hold is the bow

A row of U-frames is sized for a force, and the rules give that force as a percentage of the compression chord's force that nobody derives. Derived, it is not a property of the frames at all. It is the chord's initial bow, grown by how near the chord works to buckling — and it is least, not greatest, near the stiffness at which the chord stops using the frames. Stiff frames tend to a limit that is π²/500 of the chord force: the familiar two per cent, arriving from a bow of one five-hundredth.

 At the tip it is the tension flange that swings. The buckled shape of a 3.00 m cantilever of 457 mm universal beam carrying a load at its tip at the shear centre, root built in: slope and warping held, seen from above: the sideways movement of the top flange and of the bottom flange along the length, scaled so the larger is one. The cantilever hogs, so its bottom flange is in compression, and that flange bows out furthest 0.58 of the way from the root, where it moves 0.15 — and has come back to 0.00 by the tip. The top flange, in tension, swings out to the full 1.00 at the tip. The section twists as it goes, so the flange that is doing the buckling is not the flange that moves most at the end. Critical moment at the root: 1266 kN·m.

The flange that swings is in tension

A cantilever hogs, so its bottom flange is the one in compression and the one that buckles. At its tip the bottom flange barely moves: the section turns about it, and the flange that swings out is the top one, in tension. So the restraint that matters at a cantilever's tip is on the flange every rule for spans calls the wrong one — and the root, which a drawing shows as a single line, moves the critical moment by a factor of thirty-four.

 Two quotients from one guessed shape. The critical load of a pin-ended column whose outer quarters keep 10% of the middle's flexural rigidity from four guessed shapes, each worked two ways: Rayleigh's quotient, strain energy of the guess's own curvature over the work of the load, and Timoshenko's, which uses the curvature the guess's bending moment would cause instead. The exact load is 3.225 EI₀/L². a half sine: 8.256 by Rayleigh and 3.745 by Timoshenko; its own sag shape: 8.041 by Rayleigh and 3.712 by Timoshenko; a mid-span sag: 8.875 by Rayleigh and 3.847 by Timoshenko; a parabola: 6.600 by Rayleigh and 3.492 by Timoshenko. Both are upper bounds, and from the same shape the second is never the worse of the two.

The bound from underneath

Rayleigh's quotient turns a guessed shape into a buckling load that is always too high. Divide the same guess differently — use the curvature its bending moment would cause instead of its own — and a parabola that was 21.6 per cent high is 1.3 per cent high. Add one more number that needs no guess at all and the load is caught from below as well, which is the only side of it an amplifier can safely use.

 A column curve with no plateau. The tangent-modulus buckling stress of perfect pin-ended columns of an aluminium alloy of 0.2 per cent proof stress 250 N/mm² and modulus 70,000, as a share of the proof stress, against the slenderness λ̄ — the square root of the proof stress over the Euler stress — for Ramberg–Osgood exponents of 5, 10, 20 and 40; dashed, the sharp envelope a material with a plateau at the proof stress would give, the lesser of the proof stress and Euler's. n = 5: 1.54 at λ̄ = 0.2, 0.66 at 1, 0.41 at 1.5; n = 10: 1.16 at λ̄ = 0.2, 0.74 at 1, 0.45 at 1.5; n = 20: 1.04 at λ̄ = 0.2, 0.82 at 1, 0.45 at 1.5; n = 40: 1.00 at λ̄ = 0.2, 0.88 at 1, 0.45 at 1.5. Stocky columns buckle above the proof stress, because the curve goes on rising past it; slender ones follow Euler; in between every curve sags below the envelope.

The column with no plateau

A steel column's curve is the lesser of two lines, its yield stress and Euler's, because mild steel keeps its whole stiffness up to yield and then has a plateau to stop on. Aluminium and stainless steel have no plateau: their stiffness starts falling well below the proof stress and never stops. Their column curves are not the steel curve moved; they are a different shape — four tenths below the corner for a stainless steel, above the proof stress for a stocky column — and the shape is set by two numbers, only one of which describes the knee.

 Wrong in both directions until the building racks. Across buildings from a pure bending tower to a pure racking frame (αH on a logarithmic scale), each 20 storeys with gravity putting its own critical factor at 5.0: solid, the least storey sway factor over the building's own, which is what taking the least storey as the building's costs; dashed, the ground storey's second-order increase estimated from its own sway factor, over the true increase. At αH = 0.1 the least storey factor is 0.77 of the true one and the ground storey's estimate 0.09 of the true increase; at αH = 10, 0.86 and 0.37; at αH = 100, 0.94 and 0.98. The first error is on the safe side and the second is not, and both vanish only for a building that racks.

The storey that cannot see the building lean

A sway check made storey by storey asks each storey how much it drifts under a push and how much gravity sits on it, and reads a critical load factor for the storey from the two. For a frame whose storeys rack like a stack of shelves that is exact. For a building that bends — a braced core, a wall — it is wrong twice. Taking the least storey as the building's reads a critical factor of 3.9 for a building whose own is 5. And amplifying each storey by its own factor finds a 2 per cent second-order increase at the ground storey, where the truth is 20, because the ground storey hardly drifts and carries the lean of everything above it.

 The rib yields long before it would buckle. The largest combined stress, axial plus bending, anywhere in a 60 m two-pinned parabolic rib at a 12 m rise (EI 250,000 kN·m²) with a 12,000 mm² section 1,000 mm deep, against a factor on 15 kN/m of dead load over the span and 6 kN/m of live load over the left half together. Dashed: first order; solid: second order; dotted: the 355 N/mm² yield stress. First order, the rib reaches yield at a factor of 1.73; second order, at 1.18. The same load pattern, spread uniformly, would bifurcate the rib at a factor of 2.97, the right-hand edge — which is the number the eigenvalue reports and a number the rib never reaches.

The weight that bends a rib it cannot bend

A parabolic rib carries its own uniform dead load as pure thrust, with no bending in it at all. Put a live load on half the span and the rib bends in the shape of its own buckling mode from the first kilonewton, and the whole thrust, most of it from the dead load, multiplies that bending. The rib never reaches its buckling load. It yields well short of it, at a load its first-order check says it can carry.

 The buckling mode moves up to where the spine is not. The building's first buckling mode, each floor's sideways movement as a fraction of the roof's, up the height of a 20-storey building of a bending spine and a racking frame (αH 1.0), its gravity scaled to an elastic critical factor of 5.0, under an equal lateral load at every floor: the spine to the roof, and stopping at storeys 15 and 10. With the spine to the roof the mode is the bending tower's, growing all the way up. With the spine stopping, at storey 15 its top moves less than a thousandth as far as the roof, and at storey 10 its top moves less than a thousandth as far as the roof: the building buckles in the first storey of frame above the spine, every floor above that storey moving with it as one, and the spine takes no part.

The storey above the core decides

A building whose bending core runs to the roof shares one sway mode between its core and its frame, and the storey-by-storey stability check reads it wrongly twice. Stop the core five storeys short and the check becomes exact: the first storey of bare frame above the core reads the building's own critical factor to three figures, because the whole building buckles there and nowhere else. The check stays wrong below the core, and it goes wrong again when the frame above is more than about twice as stiff.

 Two models of one buckled web. The shear resistance of a web 1500 mm deep and 8.0 mm thick in steel of 355 N/mm², as a share of its plastic shear 2,460 kN, against the spacing of its intermediate stiffeners as a multiple of the depth, by Basler's tension field, by the rotated stress field with a rigid end post and with a non-rigid one, and by buckling alone; the dots at the right-hand edge are the same web with no intermediate stiffeners. With stiffeners at the depth Basler gives 0.71 and the rotated field 0.50; at three depths 0.38 and 0.42; with none, 0.14 — buckling alone — and 0.41. The two cross at a spacing of about 2.6 depths. Basler's reserve is a band the stiffeners anchor and it narrows to nothing as they move apart; the rotated field's is spread over the whole web and anchored at its ends, and the spacing reaches it only through the buckling stress.

Two reserves for one buckled web

A slender web keeps carrying shear after it buckles, and design uses two models of how: Basler's band of tension anchored on the stiffeners, and the rotated stress field spread over the whole web and anchored at its ends. Given the same 1,500 × 8 mm panel they agree about the buckling and disagree about everything after it — and in particular about what an intermediate stiffener is for. One says stiffeners save 37 per cent of the steel; the other, 1 per cent.

 Two formulas that agree until the shear is soft. The buckling load over the Euler load, by Engesser (dashed) and Haringx (solid), against the shear stiffness over the Euler load on a logarithmic scale. Marked: double lacing, 23.48: Engesser 0.959, Haringx 0.961; battens, 0.30: Engesser 0.232, Haringx 0.419; rubber bearing, 6.6e-4: Engesser 0.001, Haringx 0.025. They agree within 5 per cent while the shear stiffness is more than about 4 times the Euler load, and for a shear stiffness a thousandth of it Haringx's is 31 times Engesser's.

The bearing Engesser says has buckled

A member that is flexible in shear buckles below its Euler load, and there are two classical formulas for how far below. For a laced column they agree to a fraction of a per cent. For a battened column one gives nearly twice the other. For a laminated rubber bearing, Engesser's says it buckles at a fourteenth of the load it carries every day and Haringx's says it is safe by a factor of nearly three. They are not two theories. Put the axial load into Engesser's shear stiffness and it becomes Haringx's, exactly — so the whole disagreement is about which shear stiffness was measured.

 Two theories, one plate, and a column beside it. The buckling stress of a long plate of an austenitic stainless steel of proof stress 230 N/mm² and modulus 200,000 (Ramberg–Osgood exponent 6), simply supported on its edges and compressed along them, as a share of the proof stress, against its slenderness λ̄p, the square root of the proof stress over the elastic plate's buckling stress. Flow theory: 2.30 at λ̄p = 0.6 and 0.86 at 1. Deformation theory: 0.99 and 0.71. A column of the same material at the same slenderness, by its tangent modulus: 0.76 and 0.58. Dashed, the sharp envelope, the lesser of the proof stress and the elastic stress. Flow theory runs far above the proof stress on stocky plates — 5.29 at 0.4 — which no plate test has reached.

The plate the wrong theory gets right

A column of stainless steel buckles by its tangent modulus, the one stiffness its rounded curve has left. A plate needs three — along the load, across it, and in twist — and the two classical theories of plasticity give it different ones. Flow theory, the one whose physics is right, keeps most of the plate's elastic stiffness and puts a stocky stainless plate's buckling stress at twice its proof stress. Deformation theory, whose physics is wrong, softens every direction and agrees with the tests. Two thirds of the gap is the twist alone, and either way a rounded curve costs a plate much less than it costs a column.

 The limit moment goes as the square root of the ring's stiffness, and pressure is part of it. Brazier's limit moment for a steel tube of radius 600 mm and wall 2.5 mm (r/t = 240), as a multiple of its value with no pressure, against the pressure as a multiple of the ring's own buckling pressure, pc = 3D/r³ = 4.07 kPa; positive inside. The dots are the peaks of the moment–curvature path found by minimising the energy at each curvature; the line is √(1 + p/pc). Half pc outside leaves 0.71 of the bending capacity, and the capacity is gone at pc; pc inside gives 1.41, three times pc 2.00. The pressure enters only by adding to the ring's stiffness against the oval, (3D/r³ + p) where it had 3D/r³.

Three kilopascals that halve a tube in bending

A long thin tube in bending flattens until its section can carry no more, at an oval of exactly two ninths of its radius. Put a pressure on it and nothing about that shape changes. What changes is how much bending it takes to get there — as the square root of one plus the pressure over the ring's own buckling pressure — and for a steel tube 1.2 m across with a 2.5 mm wall that unit is 4 kPa. Three kilopascals of suction halve its bending capacity; a fifth of an atmosphere inside doubles it and hands the tube to another failure entirely.

 Below its ceiling, a row of braces is only its total. The critical moment of a beam 24 m long whose unbraced critical moment in uniform bending is 837 kN·m, held by one, two, three or five torsional braces equally spaced along it, against the braces' total rotational stiffness. At 4,336 kN·m/rad in all, the rows of two, three and five give 3,815, 3,726, 3,578 kN·m: how the stiffness is divided hardly matters, while one has already stopped. Each row then stops at its own ceiling, where the braces are stiff enough to be nodes and the beam buckles between them in one more half-wave than it has braces: 2,637 kN·m for one, 5,585 kN·m for two, 9,704 kN·m for three, 21,466 kN·m for five. The dashed line is 5,627 kN·m, the moment the beam is asked to carry; one and two cannot reach it however stiff they are.

The stiffness a row of braces shares out

One torsional brace at midspan has a ceiling, the moment at which the beam gives up twisting it and buckles in two half-waves instead. A row of braces raises the ceiling with every brace added, and the stiffness each must have to reach its ceiling rises with the count — seventy-four times as much for eight braces as for one. That is the right answer to the wrong question. Asked for the moment the beam has to carry rather than for its ceiling, a row of braces needs very nearly the same total stiffness however many it is divided between, so each brace gets softer as the count goes up. The count has one job, which is to put the ceiling above the moment; the stiffness has the other.

 The arch turns aside before it reaches the top of its path. The load parameter p = 16qL⁴/π⁴EIr against the crown's deflection, in radii of gyration, for a shallow pinned arch whose rise is 6.0 times its radius of gyration. Kept symmetric (dashed) it would carry the load to a limit point at p = 374.7. It never gets there: at p = 310.7 a sideways path branches off (solid), along which the load falls as the arch sways — the failure is a bifurcation, 17 per cent below the limit point. With an antisymmetric imperfection of 0.10 radii of gyration (thin line) the arch follows neither path and peaks at p = 272.6, 12 per cent below the perfect arch.

The arch that turns aside before it snaps

A shallow frame of pinned bars can only go one way when it is pressed: down, symmetrically, over a maximum. An arch that bends has a second way out — sideways, one half dropping while the other lifts — and above a rise of about four and a half radii of gyration it takes that way before it ever reaches the maximum. The switch decides more than the load. Below it an arch is almost indifferent to a sideways imperfection; above it the load an arch carries falls as the two-thirds power of how crooked it was built, and a tenth of a radius of gyration costs it an eighth of its strength.

 Hung from its deck, the rib's bending is not amplified at all. The bending moment along a 60 m parabolic rib at a 12 m rise (EI 250,000 kN·m²) under 25 kN/m of dead load and 10 kN/m of live load on the left half. Standing free on pinned springings (solid) it peaks at 1,303 kN·m, against 576 at first order (dashed): the thrust amplifies the half-span bending 2.26 times. The same rib hung over a deck 0.10 times as stiff as the rib on vertical hangers (thick) peaks at 524 kN·m, 0.99 times its first-order 528, and the deck (thin) carries 59. The deck closes the thrust in tension, and the tension it carries straightens the deck as much as the thrust bends the rib.

The rib that cannot lean on its own tie

A free arch rib under a half-span live load bends in the shape of its own buckling mode, and its thrust amplifies the bending more than twofold. Hang a deck from the same rib on vertical hangers and close the thrust in the deck, and the amplification disappears: the thrust and the tie's pull move together, so the couple they make keeps its lever arm, and the only second-order term left is the small difference between the rib's deflection and the deck's. The critical load rises sixteenfold, and the deck's share of the bending is simply its share of the stiffness.

 The deck moves the frames' feet, and the chord goes with them. A chord 24 m long, EI 2.52 × 10¹² N·mm², held by U-frames of 1,400 N/mm every 4 m, carrying 930 kN, in plan, sideways movements drawn 6 times the scale of the length: each frame's foot pushed 7.5 mm inward by its cross-girder's rotation under the deck (arrows), the chord's movement under that push alone (solid), and under the worst bow alone, four half-waves at 12.0 mm (dashed). The chord follows the push, and under its compression goes a little further — 9.1 mm at its middle frames against the 7.5 mm the feet moved — because a chord 24 m long is far softer sideways than the frames holding it; the frames that carry most are the end ones, beside the chord's fixed ends.

The frames the deck pushes first

A half-through girder's U-frames are the ends of the cross-girders that carry its deck. Load the deck and every cross-girder sags, its ends rotate, and every frame's foot swings inward by millimetres before the compression chord has asked the frames for anything. It looks like a second imperfection on top of the chord's bow. It is a much weaker one: the frames all move together and the chord simply goes with them, so a 7.5 mm push adds 1.4 kN to frames that the bow loads with 13.6, and stiffer frames make it smaller, not larger. What the push leaves is mostly in the frames' own members, not in what they ask of the chord.

 Three braces do more where the moment is. The 24 m plate girder, its minor-axis second moment 2.67 × 10⁸ mm⁴ and its warping constant 1.00 × 10¹⁴ mm⁶, under a uniform load, held by three torsional braces asked to bring its peak moment to 5,627 kN·m: above, the moment diagram; below it, the twist of the buckled girder with the braces equally spaced, 6.0 m apart, and clustered 4.0 m either side of midspan, each at the stiffness that just reaches the moment. Equally spaced the row needs 8,537 kN·m/rad in all; clustered, 6,325, 26 per cent less. The clustered braces sit where the girder twists most, and the bays they leave at the ends carry little moment.

Three braces where the moment is

A row of torsional braces is spaced equally because that is how it is drawn, and under a uniform moment equal spacing is very nearly the best there is. Under a moment that peaks at midspan it is not. Moving the outer two of three braces toward the middle cuts the total stiffness a 24 m girder needs by a quarter under a uniform load and a third under a point load, and the bays it leaves at the ends cost nothing, because they carry so little of the moment.

 Braces that cannot move can still be made stiffer where the moment is. The 24 m plate girder, its minor-axis second moment 2.67 × 10⁸ mm⁴ and its warping constant 1.00 × 10¹⁴ mm⁶, under a uniform load, held by braces at midspan and 6.0 m either side and asked to bring its peak moment to 5,627 kN·m: above, the moment diagram; below it, the twist of the buckled girder with the three braces equally stiff and with the middle one given 61 per cent of the row's stiffness, each row at the total that just reaches the moment; the ticks are drawn thicker for a stiffer brace. Equal, the row needs 8,537 kN·m/rad, 2,846 in each brace; graded, 7,067, 4,280 at midspan and 1,394 at each side, 17 per cent less.

The braces that stay where the framing puts them

Three cross-frames on a girder usually go where the floor beams or the bridge's bays put them, and cannot be moved toward the moment. They can still be made unequal. Giving the middle one three-fifths of the row's stiffness saves a sixth of the total under a uniform load and more than a quarter under a point load — most of what moving them would save. Moved and graded together, they save no more than moved alone, because both are ways of making every brace twist by the same amount.

 A brace on the sideways mode gives back the snap. The load parameter against the crown's deflection, both in radii of gyration, for a shallow pinned arch whose rise is 8.0 radii of gyration, built 0.1 of a radius of gyration out of shape in its sideways mode, unbraced and with a brace on the sideways mode of 0.88, 1.75, 3.50 times the rib's own antisymmetric bending stiffness. Unbraced it peaks at 410; with 0.88 it peaks at 658; with 1.75 it peaks at 794; with 3.50 it peaks at 840. The perfect arch's limit point, which a brace can only return it to, is 844; unbraced, it turns sideways at 461.

The brace that holds a mode

A shallow arch rising eight radii of gyration turns sideways at a load of 461 when its symmetric snap would not come until 844. A member that only stops the quarter points moving in opposite senses — a brace on the sideways mode, not on any one point — gives the whole difference back once it is 1.75 times as stiff as the rib's own resistance to that mode, and nothing more after that. The stiffness needed grows as the square of the rise. And the brace is not the light member the argument suggests: an arch built a tenth of a radius of gyration out of shape puts an eighth of its whole load through it at the peak.

 A hat makes the top storeys look stiff. Each storey's own sway factor α up the height of a 20-storey building of a core to the roof and a racking frame (αH 1.0), its gravity scaled to an elastic critical factor of 5.0, under an equal lateral load at every floor: with no hat, with a hat truss at the roof holding the core's rotation with 10 times the core's EI/H, and with the same restraint at storey 10; the building's critical factor 5.0 as a vertical line. With no hat the top storeys read the least, 4.06 at the roof. With the hat the roof storey reads 15.47 and the least, 4.14, has moved down to storey 9. With the outrigger at mid-height the least is back at the top, 3.85.

The hat that hides the lean

A hat truss on a core-and-frame building holds the top of the core, nearly doubles the building's critical load, and makes its top storeys look three or four times as stiff as they did. The storey check believes them. Under the hat the roof storey reads a sway factor of 15 and estimates 7 per cent of second-order drift; the building still buckles leaning, and the roof storey's true increase is 24 per cent. The hat turns the storey check from pessimistic to unsafe at exactly the storeys it was built to stiffen.

 Caught from both sides without a mode shape. The elastic critical load factor of a one-bay frame of six storeys, 3.5 m high and 6 m wide, columns of 40,000 kN·m² and beams of 60,000, 400 kN on every floor, 14.96 (level line), and the bounds the traces of its flexibility give with no assumed shape, against how many traces are taken: from below, 5.12, 10.82, 13.11, 14.07 for the first four; from above, 22.90, 19.23, 17.39, 16.39. Dotted, the storey estimate of EN 1993-1-1, 12.86, and Rayleigh's quotient on the first-order sway, 15.88.

The trace that adds up every storey

A column's buckling load can be bounded from below by the traces of its flexibility, with no shape guessed. A sway frame has a flexibility too — one number per pair of floors — and its traces bound the frame's critical load factor from below and from above, closing in as more are taken: on a six-storey frame whose factor is 14.96, the first four give 14.07 to 16.39. But the first trace is Dunkerley's sum over the storeys, counting each as though its sway mode were the frame's, and it bounds a well-balanced frame worst of all: one over the number of storeys when every storey is alike, which is exactly when Eurocode 3's storey-by-storey estimate is exact.

 The knee moves out as the braces multiply. The critical load of a pin-ended column, over the Euler load of the segments between its braces, against the braces' stiffness in units of that load over the segment's length, for one, two and four braces. Each curve rises to the segment's Euler load and stays there; the knee where it arrives is at 2.00, 3.00, 3.62. With no braces at all the column buckles at 1/(n + 1)² of the segment's load: 0.25, 0.11, 0.04.

The braces that have to hold each other

One brace at mid-height must have a stiffness of twice the column's load over the length it halves. Spread the same column over four braces and each one must be stiffer, not softer: 3.62 times the load over its segment, rising towards four with every brace added, because the shape the braces have to stop is a zigzag in which every brace point is pushed by both its neighbours. Braces that tie the column to itself instead of to the ground need one, however many there are. A row with surplus stiffness can carry one brace at two thirds of the ideal, but not one that is missing, and the force each brace has to resist grows with the count.

 Braces that pull one way never brace the girder to its moment. The critical moment of the 24 m plate girder under a uniform load, braced at midspan and 6.0 m either side, against each brace's stiffness (logarithmic), when every brace resists twist both ways, when the three alternate in the direction they resist, when two face one way and one the other, and when all three face the same way; dotted, the 5,627 kN·m the girder must reach. Two-way braces of 2,846 kN·m/rad reach it. At that stiffness the alternating row reaches 3,087, the two-and-one row 1,734 and the row facing one way 947, the girder's unbraced moment; with rigid braces, 4,063, 1,891 and 947.

The braces that pull only one way

A brace made of a single tie, or a cross-frame whose diagonals can only pull, resists a girder twisting one way and not the other. A buckling girder has no preference between the two directions, so it takes the one its braces leave free. Three such braces all facing the same way leave a 24 m girder exactly as weak as if they were not there. Facing alternately, they reach 3,087 kN·m at the stiffness that takes two-way braces to 5,627, and no stiffness takes them past 4,063. Opposing the outer pair does not help; one one-way brace anywhere in the row is enough to cap it. Every brace has to resist both ways, and made of ties that costs twice the steel.

 A stiff enough crown brace stops the arch snapping. The load parameter against the crown's deflection for a shallow pinned arch of rise 6.0 radii of gyration under a sinusoidal load, built with an antisymmetric imperfection of 0.10 radii of gyration, with a crown brace of κc = 0, 1, 2, 4; the dots mark the first peak of the load, where the arch snaps. κc = 0: peaks at 273; κc = 1: peaks at 299; κc = 2: peaks at 327; κc = 4: no peak, the load rising throughout.

The prop at the crown of the arch

A shallow arch that turns sideways before it snaps can be braced at its quarter points, which holds the sideways mode and carries almost nothing. A single prop at the crown is cheaper and does something else. The crown does not move in the sideways mode, so the prop cannot stop it; it raises the load at which the arch turns aside only by taking part of the load itself. But behind a prop about three times as stiff as the rib's own first mode, the sideways branch stops falling, and the arch no longer snaps at all — at every rise, by the same prop.

 A concrete rib creeps into snapping long before its prop matters. For the shallow pinned arch of rise 6.0 radii of gyration under a sinusoidal load, propped at its crown by a concrete column of stiffness κc = 3.5, the creep coefficient at which a load that stays on loses equilibrium, against that load as a share of the day's critical load: a steel rib on the creeping column (solid), which loses only the column's share, so that by a creep coefficient of 3 no load below 85 per cent snaps; and a concrete rib whose column creeps with it (dashed), whose lasting load acts as if (1 + φ) times larger, so that it snaps at φ = pcr/p − 1; dotted, a creep coefficient of 2. At half the critical load the concrete arch snaps at a creep coefficient of 1.00, at 70 per cent at 0.43; the steel arch on its column does not snap at either.

The column that creeps under the arch

A shallow arch propped at its crown by a stiff enough column stops snapping — the sideways branch it would have fallen down turns upwards. If the column is concrete, it creeps under the force it holds, and after a decade it holds that force as a spring a third as stiff, well on the side where the arch snaps. But a load that arrives in seconds does not meet the crept column; it meets the column's instantaneous stiffness, from a rest length that creep has merely moved down, and the arch's branch still rises. Creep reaches only the load that stays on, and there it can take back only the column's own share — 13 per cent of it. A concrete rib is another matter: it creeps itself, and a lasting load at half its critical load snaps it before its creep coefficient reaches one.

 A milliradian of slack costs the brace more than a milliradian of crookedness. For the 24 m plate girder under a uniform load, braced at midspan and 6.0 m either side by braces of 5,691 kN·m/rad — 2.0 times the 2,846 that just reaches 5,627 kN·m — erected with 5.0 mrad of initial twist: the moment in the middle brace when the girder reaches 5,627 kN·m, against each brace's slack (solid), and against the same amount added instead to the girder's initial twist, the braces tight (dashed). With tight braces and no extra twist, 34.1 kN·m; each milliradian of slack adds 8.0, each milliradian of initial twist 6.8. At 10.0 mrad, 114.0 against 102.4.

The braces with slack in their bolts

A cross-frame bolted into clearance holes resists nothing until the girder it braces has twisted far enough to take up the clearance, and then resists fully. It is a two-way brace that starts late. The girder, erected a little crooked, twists freely into its braces while the load is still small, and from then on the braces hold a girder that is more crooked than the one they were fitted to. The slack acts as an imperfection, and a costly one: each milliradian of it adds more to the middle brace's force than a milliradian of extra crookedness would, and no stiffness takes it back out. On a 24 m plate girder braced at twice the stiffness it needs, three milliradians of slack — a pair of ordinary clearances over its depth — raise the middle brace's force by 70 per cent.

 Which way the oval lies decides whether the pressure helps or harms. The limit moment of a steel tube of radius 600 mm and wall 2.5 mm (r/t = 240), whose ring buckles under 4.07 kPa, over its limit with no pressure (796 kN·m), against the angle between the bending plane and the diameter a non-uniform pressure pushes in on, for second harmonics of 0.10, 0.25 and 0.50 of that pressure; lines the closed form, dots the minimised energy. At 0.10, 0.93 aligned and 1.08 across; at 0.25, 0.82 aligned and 1.19 across; at 0.50, 0.65 aligned and 1.40 across; at 45° the oval does not touch the limit at all.

The oval the pressure brings

A uniform pressure changes how stiff a tube's ring is against the oval that bending squashes it into, and nothing else: the tube still gives way at a flattening of two ninths of its radius. A pressure that is larger on one diameter than the other has an oval of its own, and that oval does work on the bending's directly. Lying in the bending plane it has done some of the flattening before the bending starts, and a pressure difference a quarter of the ring's own buckling pressure takes a fifth off the tube's bending capacity; lying across it, it adds a fifth. At forty-five degrees it does nothing at all. And the two ninths turn out to have been a property of a round tube.

 Where real roots sit on the warping curve. The elastic critical moment of a 3.0 m cantilever of 457 mm universal beam with its tip load at the shear centre, with its root holding the lateral slope and resisting warping through a spring, against the spring as a multiple of EIw/L on a logarithmic scale — from 521 kN·m free to warp to 1,266 built in — with four real roots placed on it. Top storey, column ends: 2.5 times EIw/L, 791 kN·m, 36% of the way; 15 mm plate, column runs on: 5.9 times EIw/L, 945 kN·m, 57% of the way; 40 mm plate, column runs on: 6.8 times EIw/L, 970 kN·m, 60% of the way; 15 mm plate, column boxed: 19.4 times EIw/L, 1,124 kN·m, 81% of the way.

The root that belongs to the column

A cantilever's root must hold its flanges against warping at about fifty times the cantilever's own warping stiffness to count as built in, and doubles its buckling moment if it does. A flush end plate is the usual root, so it ought to be the plate's thickness that decides. It is not. The flanges warp by turning in plan in opposite directions, the plate passes that to the column as two opposite torques, and an open column twists between them more easily than the plate bends: three quarters of the root's flexibility is the column. An ordinary root is worth about six times the cantilever's stiffness, halfway up the curve, and a thicker plate barely moves it. Boxing the column does.

 A floor that shears lowers the storey's buckling load before the core does. The buckling factor of a storey of eleven lines of leaning columns, ten bays, each column carrying 3,000 kN over 3.6 m and built a three-hundredth out of plumb, held by a floor to a core sized for a buckling factor of 4.0 were the floor rigid, against the floor's shear stiffness per bay as a multiple of a column's sway softening P/h, logarithmic, with the core at one end, in the middle, or split between both ends; dotted, the rigid floor's 4.0. At 100 times the factor is 1.52, 2.80 and 3.02; at 1,000 times, 3.49, 3.84 and 3.88; a concrete slab is several thousand times.

The floor the columns lean on

A storey of pin-ended columns, every one a little out of plumb, leans on its floor, and the floor leans on a core. Count the floor as rigid and the core needs the sum of the columns' demands. A floor that shears in its own plane is a brace in series with the core, and the far columns lean on it before their push arrives. With a floor a hundred times stiffer per bay than a column's own sway softening, the storey's buckling factor falls from 4.0 to 1.5 and the core is asked for twice as much.

All essays