Structural Engineering & Statics
where the load goes
Stability

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.
15 min read 6 figures Which failure arrives firstGeometry beats material

A beam bent in its strong plane has one flange in compression along its whole length. That flange is a long thin strut, restrained sideways only by the web that hangs from it — and a long thin strut in compression does what long thin struts do.

It cannot simply bow sideways on its own, because the web ties it to the tension flange, which is being pulled straight. So it goes sideways and takes the section round with it, twisting as it goes. That combined motion is lateral-torsional buckling, and it happens at a moment that can be a fraction of what the cross-section could carry standing still.

The length at which a beam stops being a beamElastic 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.2000400060008000100001200002004006008001000distance between lateral restraintsthey cross at 3803the plastic capacity of the sectionelastic critical momentSt Venant torsion alone — what is left at long lengthswarping dominates here
Fig. 1 The elastic critical moment against the distance between lateral restraints, with the section’s plastic capacity drawn across it. The two cross at a length beyond which the beam buckles sideways before it reaches the strength its cross-section has.

Two mechanisms resist it, and they scale differently

What resists the twist is the section’s torsional stiffness, and an open section has two separate sources of it that behave in completely different ways.

St Venant torsion is each plate twisting about its own mid-thickness. It produces a shear flow circulating within the thickness of each plate, and its stiffness is GJGJ with J≈13∑bt3J \approx \frac{1}{3}\sum bt^3. Crucially, it does not care how long the member is: a longer beam twisted through the same angle per unit length develops the same resisting torque per unit length.

Warping torsion is the flanges bending sideways in opposite directions. Twisting an I-section makes one flange move left and the other right, so each flange is a beam bending in its own plane — and resisting that bending is a second source of torsional stiffness, with a constant EIwEI_w. This one does depend on length, and steeply: the flanges’ bending resistance falls away as the member gets longer, exactly as any beam’s does.

The critical moment combines them:

Mcr=πLEIz GJ+π2L2EIz EIw.M_{cr} = \frac{\pi}{L}\sqrt{EI_z\,GJ + \frac{\pi^2}{L^2}EI_z\,EI_w}.

Two terms under the root, one length-independent and one going as 1/L21/L^2. So the curve has two regimes. At short unrestrained lengths the warping term dominates and the capacity falls steeply. At long ones it has faded to nothing and the St Venant term is all that is left, with McrM_{cr} decaying as 1/L1/L.

The figures on this page plot both, and the faint line beneath the curve is the St Venant contribution alone — the asymptote the beam is heading for and the capacity it would have if its flanges could not bend sideways at all.

Where the crossing falls

The design question is not the shape of that curve but where it crosses the section’s own capacity, because that is the length at which the beam stops being able to reach its strength.

For an ordinary 457457 mm universal beam, the numbers work out at an unrestrained length of a little under four metres. Shorter than that and the section reaches its plastic moment; longer, and it buckles first, with the shortfall growing quickly.

Four metres is not a long way. A floor beam of eight or ten metres with no restraint between its ends would be at half its capacity or worse — which is why the question is never “is this beam long enough to worry about” but “what is holding the compression flange, and how far apart are the things that hold it”.

The usual answer is the floor. A concrete slab cast onto a beam — connected to it by studs carrying the shear flow between the two — restrains the top flange continuously, and under sagging moment the top flange is the compression one. So an ordinary composite floor beam has no lateral-torsional problem at all in its final condition — its compression flange is held along its entire length by the thing it supports.

The length at which a beam stops being a beamElastic critical moment against the distance between lateral restraints, with the section's plastic capacity drawn across it. The two do not cross anywhere in this range: the critical moment stays above the plastic capacity out to 9000, so the section reaches its full strength at any length drawn here and lateral-torsional buckling never governs.2000400060008000050100150200250300350distance between lateral restraintsthe plastic capacity of the sectionelastic critical momentSt Venant torsion alone — what is left at long lengthswarping dominates here
Fig. 2 A square hollow section of similar weight, computed the same way. Its torsion constant is two orders of magnitude larger and its warping constant is effectively nil, so the critical moment never falls to the plastic capacity at any length drawn here — the check simply does not arise.

The three cases where it governs anyway

If the floor solves it, the problem should be rare. It is not, and the exceptions are worth naming because each is a case where an assumption quietly fails.

During construction. The structure being checked is always a snapshot, and the slab that restrains the top flange does not exist until it has been poured. The beam that carries the wet concrete is a bare steel section with nothing on it at all. The construction condition is very frequently the governing check for a composite beam, and it is the one most often missed, because the beam being checked is not the beam that will exist.

Over a support. In a continuous beam the moment reverses, so near the supports the bottom flange is in compression — and the slab restrains the top flange, which is now the tension one. The stability problem sits exactly where the moment is largest, and nothing is holding the flange that needs it.

Cantilevers and their tips. A cantilever hogs throughout, so its compression flange is the bottom one along its whole length, and its tip is free to move in every direction at once. The effective length factors for cantilevers under lateral-torsional buckling exceed those for beams substantially, and a cantilever whose tip is unrestrained against twisting is a considerably worse case than one that is merely unrestrained laterally.

The pattern in all three is the same as the pattern for columns: the capacity is set by what is holding the member, not by the member. And the thing holding it is very often something whose structural role nobody wrote down.

What restraint has to do

A restraint against lateral-torsional buckling has a specific job, and it is not the obvious one.

It has to prevent the compression flange from moving sideways, or prevent the section from twisting, and either is sufficient because the buckling mode requires both to happen together. That gives two families of detail: a lateral brace to the compression flange, and a torsional restraint that stops the section rotating — a stiffener connected to something, a member framing in at both flanges, a moment connection to a secondary beam.

What does not restrain a beam is a member framing into the tension flange only. It stops the tension flange moving, which the buckling mode barely involves, and leaves the compression flange free to swing. This is the most common way a restraint exists on the drawing and not in the building.

The force required is small — of the order of one to two per cent of the flange force — and the stiffness required is not negligible, for the same reason a column brace has both requirements: a brace of insufficient stiffness allows the mode to form with the brace point moving, and the beam buckles at a length longer than the brace spacing.

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. 3 Four end conditions for a column. A beam’s compression flange is judged the same way, with the effective length being the distance between points that genuinely hold it — and the same factor of four in capacity between the best and worst arrangements.

Only part of the length is at the peak

The critical moment formula is derived for a beam under constant moment along its unrestrained length, and that is the worst arrangement there is. Almost no real beam has it.

The reason it is worst is that buckling is a property of a length rather than of a station. A beam whose moment is at its peak everywhere has its entire unrestrained length working at the peak, and there is nothing anywhere along it holding the rest back. A beam whose moment peaks in the middle and falls away has stretches at each end that are lightly loaded, and those stretches resist the buckling of the middle.

So the design capacity is the uniform-moment value multiplied by a factor that depends on the shape of the moment diagram between restraints. For a simply supported beam under a central point load the factor is around 1.351.35; for a uniformly loaded one, around 1.131.13; for a beam with equal and opposite end moments — a genuinely double-curvature case — it can exceed 2.52.5.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.1515shear15.0moment37.5 at x = 4.32the moment peaks exactly where the shear passes through zero
Fig. 4 Two symmetric point loads, with a constant-moment region between them. That middle stretch is the uniform-moment case the formula assumes, and it is the only part of an ordinary beam that genuinely has it — everywhere else the diagram falls away and the capacity is higher than the formula says.

Two practical consequences follow. Using the uniform-moment value everywhere is safe and can be wasteful by a third or more, which is why the factor is worth applying rather than ignoring. And the shape of the diagram between restraints is what matters, not along the whole beam — so adding a restraint changes two things at once: it shortens the length and it changes the segment’s moment shape, sometimes for the worse. A restraint added at the peak of the diagram leaves two segments each with a diagram falling away from one end, which is favourable. A restraint added off to one side leaves a segment with a nearly uniform moment, which is not.

Where the load is applied

There is a further effect that no cross-section property contains, and it is one of the few places in the subject where the height at which a load is applied changes the answer.

A load applied at the top flange is destabilising. As the section starts to twist, the point where the load acts swings sideways off the shear centre’s line, so the load itself begins to drive the rotation. The critical moment falls, typically by twenty to forty per cent.

A load applied at the bottom flange is stabilising, by the mirror-image argument: the load’s point of application swings the other way and generates a restoring moment. Critical values rise correspondingly.

The difference between the two is not a refinement — it can be a factor approaching two, from nothing but whether a load hangs from a beam or sits on it. A crane hook on the bottom flange is in the favourable case; a precast unit resting on the top flange during erection is in the unfavourable one; and a load applied through a bracket well above the beam is worse than either.

The same beam, cut at x = 5A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs
Fig. 5 A cut with its internal actions drawn on the exposed face. Everything about that free body is unchanged by the height at which the load was applied — which is exactly why the destabilising effect appears nowhere in an equilibrium calculation and has to be brought in from a stability one.

The general shape of this recurs across stability problems. Equilibrium is indifferent to the deformed geometry and stability is entirely about it, so any effect that depends on where something moves to is invisible in the first calculation and decisive in the second. A load’s height above the shear centre changes nothing in the moment diagram and changes the buckling load by a factor.

What the section shape does

Since McrM_{cr} contains IzI_z, JJ and IwI_w, the shape of the section decides how vulnerable it is, and the ranking is roughly the inverse of the bending ranking.

A deep narrow I-section is excellent in bending and terrible here: IzI_z is small because the flanges are narrow, and JJ is tiny because everything is thin. The more efficiently the section has been shaped for bending, the worse its lateral-torsional capacity.

A hollow section is immune in practice, for the reason a closed loop of material resists twist and an open one barely does. A closed tube’s torsion constant is a hundred and fifty times an open section’s of the same area, so GJGJ is enormous and McrM_{cr} exceeds the plastic moment at any sensible length. A rectangular hollow section used as a beam simply never has this check.

A square or nearly square section is also immune, because Iz≈IyI_z \approx I_y and a section cannot buckle sideways about an axis it is equally stiff about. The vulnerable region is precisely the useful one — deep, narrow, efficient — which is the same trade the whole section catalogue is built on.

The same material, four waysFour cross-sections of identical area, so identical weight and cost, with the second moment of area computed from each profile's own geometry. Only the arrangement differs, and the stiffest is many times the flattest.the same, laid flatI = 0.06 × 10⁶1.0× the firstsquareI = 0.75 × 10⁶13.3× the firsttall rectangleI = 10.00 × 10⁶177.8× the firstI-sectionI = 24.29 × 10⁶431.8× the firstevery section here has an area of 3000 — only the shape differsthe bar is the second moment of area, to scale
Fig. 6 Four sections of equal area, ranked by their strong-axis bending stiffness. Ranked by lateral-torsional capacity the order is very nearly reversed, since what makes a section good at bending is exactly what makes it weak about the other axis.

Where the model stops

Uniform moment. The formula above is for constant moment along the unrestrained length, which is the worst case. A beam whose moment varies — most of them — has a higher critical value, handled by a factor that depends on the shape of the moment diagram. Using the uniform-moment value everywhere is safe and can be wasteful by thirty per cent or more.

Elastic behaviour. McrM_{cr} is an elastic critical value, and a real beam of intermediate length yields partially before reaching it. Design curves interpolate between the plastic capacity and the elastic critical moment in the same way column curves interpolate, and for the same reason: residual stresses and imperfections make the transition region worse than either bound.

Doubly symmetric sections. The expression assumes the shear centre is at the centroid. For a monosymmetric section — a tee, or an I-section with unequal flanges — the offset enters the formula and the capacity depends on which flange is in compression.

Load applied at the shear centre. A load applied to the top flange is destabilising, because as the section twists the load moves outboard and drives the rotation. A load hanging from the bottom flange is stabilising for the mirror-image reason. The difference between the two is not small.

Ideal restraints. As above, and it is the assumption most likely to fail on site.

No axial force. The critical moment is derived for a member in pure bending. A beam-column carries both, and the two interact: axial compression reduces the moment at which the member buckles sideways, since it is already using up part of the minor-axis stiffness that resists the lateral movement. The combined check is a separate calculation, and treating the two actions independently is unconservative.

The figures share a limitation this whole subject has. The curve is drawn as though the capacity is a smooth function of length, and it is — but a real beam does not have a continuously variable unrestrained length. It has restraints at the places the secondary members happen to fall, so the design value jumps as the framing layout changes, and the useful reading of the curve is not “what is the capacity at this length” but “how much would be gained by adding one more restraint”.

The ladder from here

Later rungs on this anchor: the critical moment derived from the governing differential equations. Moment gradient factors. Effective lengths for beams, and the destabilising-load correction. Monosymmetric sections and the shear-centre term. Restraint requirements in strength and stiffness. Cantilever stability. Beams during construction, and the temporary condition. Plate girders, where web slenderness enters as well. And the general theory of elastic stability, in which lateral-torsional buckling is one eigenvalue problem among a family.

Prandtl and Michell independently solved the lateral buckling of a narrow rectangular beam in 1899, within months of each other and without knowing of each other’s work. The warping term — the part that matters for the I-sections everything is now built from — waited for Timoshenko in 1905.

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. 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 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 same, laid flatI = 0.06 × 10⁶1.0× the firstsquareI = 0.75 × 10⁶13.3× the firsttall rectangleI = 10.00 × 10⁶177.8× the firstI-sectionI = 24.29 × 10⁶431.8× the firstevery section here has an area of 3000 — only the shape differsthe bar is the second moment of area, to scale The same steel in a different shape, and a factor of forty Four sections of identical area, identical weight and identical cost. The stiffest is dozens of times the stiffest of the flattest, and the only thing that changed was the arrangement.