Structural Engineering & Statics
where the load goes
Stability

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.
10 min read 7 figures Which failure arrives firstOne support too many

The Euler formula contains a length, and the length in it is not the length of the column. It is the length of the piece of the buckled shape that looks like a pin-ended column — and that depends entirely on what is holding the ends.

The ends decide the length that mattersFour columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.K = 0.5both ends fixedK = 0.7one fixed, one pinnedK = 1both ends pinnedK = 2fixed at the base, free at the topsame column, same section, four ways of holding the endsthe load at which each buckles goes as 1 ÷ K² — a factor of sixteen across this row
Fig. 1 Four columns of identical height and section, buckling under four sets of end conditions. The factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.

The shape says the length

A pin-ended column buckles into a half sine wave: zero deflection at each end, maximum in the middle, and no curvature reversal anywhere. The length of that half wave is the length in the formula.

Change the ends and the shape changes, and the half wave that appears in it is a different length.

Both ends pinned. The whole column is one half wave. K=1K = 1, by definition — this is the reference case.

Both ends fixed. The ends cannot rotate, so the shape has points of contraflexure a quarter of the way in from each. Between those two points the column is doing exactly what a pin-ended column does, and that stretch is half the height. K=0.5K = 0.5, and the capacity is four times the pinned case.

One end fixed, one pinned. An intermediate shape with one contraflexure point. K=0.7K = 0.7, capacity twice the pinned case.

Fixed at the base, free at the top. The worst case. The buckled shape is a quarter wave, so the half wave it belongs to is twice the height. K=2K = 2, and the capacity is a quarter of the pinned case — and a sixteenth of the fixed-fixed case.

Pcr=π2EI(KL)2.P_{\mathrm{cr}} = \frac{\pi^2 EI}{(KL)^2}.

A factor of sixteen across four arrangements of the same column. Nothing else available to a designer moves capacity that far.

The ends decide the length that mattersFour columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.K = 0.5both ends fixedK = 2fixed at the base, free at the topsame column, same section, four ways of holding the endsthe load at which each buckles goes as 1 ÷ K²
Fig. 2 The two extremes side by side. Same column, same section, same material — and one carries sixteen times what the other does, because of what is at the ends.

Free to sway is the expensive word

The fourth case is worth separating out, because it is the difference between two kinds of building frame and it is not primarily about rotation.

A column whose top is held against sideways movement — by a brace, a shear wall, a core, all of which are redundant restraints — can only buckle by bowing between its ends. KK is at most 1 and usually less.

A column whose top can move sideways buckles by leaning. The whole storey goes over together, and KK exceeds 1 — often substantially. That is a sway frame, and its columns are worth a fraction of what the same columns are worth braced.

The distinction is the reason bracing exists and the reason a building with a stiff core can use much lighter columns than one relying on frame action. It is also why removing a wall during refurbishment is a structural act even if the wall carries no vertical load: the wall may have been what stopped the frame swaying, and its removal doubles every column’s effective length.

Length costs more than it looksThe same column section at four lengths, with the buckling capacity of each drawn as a bar. Capacity falls as the inverse square of the length, so a column three times as long carries a ninth as much.1× the length100% of the capacity1.5× the length44% of the capacity2× the length25% of the capacity3× the length11% of the capacityidentical section, identical material, identical end conditions
Fig. 3 Capacity against length for one section. Effective length enters this curve directly, so a KK of 2 rather than 1 moves a column two steps to the right — the same as making it twice as long.

The standard values are idealisations

The four values assume perfect restraint or none, and reality is always in between.

A “fixed” end is fixed by whatever it frames into, and that thing has finite stiffness. A column welded to a beam that is itself flexible has a partly restrained end, with an effective length somewhere between the theoretical values. Design codes handle this with alignment charts or with a stiffness ratio at each end, and both are ways of interpolating.

The consequences of getting it wrong are asymmetric. Assuming more fixity than exists overestimates the capacity, which is unsafe. Assuming less is conservative and expensive. Codes therefore recommend values slightly worse than theory — the fixed-fixed case is usually designed at K=0.65K = 0.65 rather than 0.5, and the sway cantilever at 2.1 rather than 2.0 — precisely because perfect fixity does not exist.

There is a second trap. The restraint can differ between the two axes, and a section’s two radii of gyration differ too. A column braced about one axis at mid-height by a beam framing in, and unbraced about the other, has two different effective lengths and two different slendernesses, and the governing one is whichever gives the larger λ\lambda. That is often not the axis with the smaller radius of gyration, and checking only the obvious axis is a well-known way to miss the real answer.

Restraint does not have to be at the ends

The effective length is set by the distance between points of restraint, and those points can be anywhere.

Adding a single lateral restraint at mid-height of a pinned column halves the buckling length and quadruples the capacity — the same reciprocal-square return working in the useful direction for once. That is an extraordinary return for one small member, and it is why bracing is the cheapest structural steel on any project.

The restraint has to be able to do its job, which means being both stiff enough and strong enough. The force it must resist is small — codes typically require something around 1 to 2.5 percent of the load in the member being restrained — but it is not zero, and a brace connected to something that itself moves is not a restraint.

This is where a great many failures live. A compression member restrained by a purlin that is connected to a roof sheet that is fixed to nothing in particular has a restraint on the drawing and not in the building.

The column curveFailure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither
Fig. 4 The column curve. Reducing the effective length moves a member left along this axis, and the gain is steepest exactly in the slender region where the curve is falling fastest.

What buckles is the whole frame

For a single column the end conditions are an input. For a frame, they are an output — every member restrains its neighbours, and the frame buckles as a whole.

The load that makes itself worseThe 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.00.20.40.60.80246810applied load ÷ buckling load1.3×1.7×2.5×5.0×10.0×first-order analysis says the answer is always 1×one over one minus the ratio
Fig. 5 Amplification against the ratio of applied load to buckling load. For a frame, the buckling load in that ratio is the frame’s, not any individual column’s, and the amplification applies to the whole storey’s sway.

That is why modern analysis computes an elastic critical load factor for the entire structure rather than an effective length for each column. If the factor is large the frame is stiff and effective lengths near 1 are fine; if it is small the frame is sway-sensitive and second-order effects have to be included explicitly.

The effective-length method is a way of packaging that whole-frame behaviour into a single number per member, and it works well for regular frames and poorly for irregular ones. Where it fails, the honest answer is to analyse the frame with its imperfections included and let the amplification appear on its own.

Restraint costs something to provide

A restraint is a member, and it has to be designed like one.

A beam, its loads and its reactionsA free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other.1289.510.5ΣM about one support gives the other reaction; ΣF then gives the first
Fig. 6 A free body with its reactions. A lateral restraint appears in exactly this way — as a force that has to be resisted by something, which itself has to reach the ground.

The load must go somewhere: a brace resisting one or two percent of a column’s load has to deliver that force into a stiff element, and a brace connected to something flexible is not a restraint at all. Most restraint failures are failures of the load path beyond the brace rather than of the brace itself.

A Warren truss of 6 panelsA Warren truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 11 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 7 A braced bay. Bracing is a truss, and its members are sized by exactly the same joint equilibrium as any other truss — which is why the forces in it can be computed rather than assumed small.

A braced bay is a vertical truss, and treating it as one rather than as an afterthought is what turns a nominal restraint into a real one.

Where the model stops

Perfect restraint. As above.

Uniform axial force. The formula assumes the load is constant along the column, which a free body cut at two heights shows it is not. A column in a multi-storey frame carries more at the bottom than the top, and its effective length is not a standard case.

Prismatic members. Tapered or stepped columns have a buckling length that is not a simple multiple of anything, and the counting arguments that work for a frame do not help.

Elastic behaviour. Effective length modifies the Euler curve, and the Euler curve is only the whole answer for slender members.

One member at a time. A frame buckles as a system; treating each column separately is a convenience that has to be checked.

The figures share an honest limitation: each buckled shape is drawn at a visible amplitude, which no column at its critical load has. The linear theory that produces KK says the shape and says nothing at all about the amplitude — the deflection is indeterminate at the critical load, which is exactly what makes it critical. Drawing a definite curve implies a definite state that the mathematics does not contain.

The ladder from here

Later rungs: the four end conditions derived from the differential equation. Alignment charts and stiffness ratios. Sway and non-sway frames. The elastic critical load factor. Bracing forces and how much a restraint must resist. Restraint stiffness requirements. Buckling of frames as systems. Effective lengths in codes, and why they differ from theory. And the second-order analysis that removes the need for the concept altogether.

The effective-length factor is a nineteenth-century device for making a twentieth-century calculation possible with a slide rule. It has survived the arrival of the computer, largely because it packages an intuition that engineers still want.

Where this goes next

5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither 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. 00.20.40.60.80246810applied load ÷ buckling load1.3×1.7×2.5×5.0×10.0×first-order analysis says the answer is always 1×one over one minus the ratio 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. simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment One support too many, and what it costs to know Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.