Structural Engineering & Statics
where the load goes
Concept

Imperfection sensitivity — where it appears

How much of its theoretical capacity a structure loses to a small initial deviation from its perfect shape. It is decided by the slope of the post-buckling path: a column keeps nearly all of its load and a cylinder keeps a third.

Named by 7 essays across one field — each of them below, with the objects they name alongside it.

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.

stability · Imperfection sensitivity 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.

stability · Mode interaction 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.

stability · Distortional buckling 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.

stability · Ring buckling 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.

stability · Follower force 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.

stability · Stepped column 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.

stability · Wall optimum

Named alongside it

The objects these essays reach for when they reach for this one.

Critical loadSlendernessBucklingEigenvalueLocal bucklingOptimisationPost-bucklingSecond momentEffective widthElastic foundationEquilibrium pathMode interaction

All concepts