Structural Engineering & Statics
where the load goes
Stability

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.

Assumes The beam that fails sideways, The section that cannot stay flat and The point that is not in the section.

20 min read 11 figures Which failure arrives firstGeometry beats materialThe statics of things that do not move

A beam bending in its strong plane, at a moment well below its capacity, can swing sideways and twist. That is lateral-torsional buckling, and the usual account of it involves a length between restraints, a torsional constant and a warping constant.

There is a term in the same expression that the usual account leaves out, and it can be worth a factor of two. It is the height at which the load is attached.

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 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.20004000600080001000012000140000200400600800distance between lateral restraintsthey cross at 4086the plastic capacity of the sectionelastic critical momentSt Venant torsion alone — what is left at long lengthswarping dominates here
Fig. 1 The length at which a beam stops being a beam. Beyond the crossing the section’s own strength is unreachable, and the capacity is decided by the restraints instead of by the steel.

Which free body produced the number

Take the beam at the instant it begins to buckle. It has rotated by a small angle θ\theta about its longitudinal axis, and the axis it rotates about is the shear centre — the point that is not in the section — because that is the point about which a transverse force produces no twist.

Now ask where the load is. If it is applied at a height aa above the shear centre, then the rotation has carried its point of application sideways by aθa\theta. The load is vertical and it has not changed; but it is now acting at a horizontal offset from the axis of rotation, so it delivers a torque WaθWa\theta about that axis.

That torque is in the same direction as the rotation that produced it. It is a destabilising moment proportional to the rotation, which is the signature of a stability problem, and it enters the eigenvalue exactly as a negative stiffness.

Hang the same load a distance aa below the shear centre and every sign reverses. The load swings back under the axis, the torque opposes the rotation, and the beam is stiffer against twisting than it would be with the load applied at the shear centre.

Two beams, one difference

The consequence is that “the load on this beam” is not a complete description of the load.

A joist carrying a floor slab bearing on its top flange has its load applied at +d/2+d/2. The same joist carrying a load hung from its bottom flange — a monorail, a services tray, a ceiling — has it at −d/2-d/2. Everything else about the two is identical, and the elastic critical moments differ by roughly a factor of two on an ordinary rolled section at an ordinary unrestrained length.

The factor is not a constant, and this is the part that makes the effect awkward. It depends on the ratio of the load height to a length built out of the section’s own torsional and warping stiffnesses, so it is largest on the sections that are already worst — deep, narrow, and torsionally weak.

The twist a warping restraint takes awayTwist along the same I-section, drawn twice: once as GJ alone predicts and once with the flanges helping. GJ alone gives 24.39° at the worst section; the real answer is 19.51°, a stiffening of 1.250. The whole of that difference is the bracket 1 − tanh(κ)/κ with κ = 5.00, and it is a property of the member's length rather than of its material — the same section at ten times the length would be stiffened by almost nothing, because the restraint reaches only about a decay length into it.024680510152025distance along the member (m)twist (degrees)×1.250GJ alone: 24.39°with warping: 19.51°
Fig. 2 Where the twisting stiffness comes from. The flanges bending in their own plane supply a large part of a beam’s resistance to twist, and how much depends on the length rather than on the material.

The same argument for a restraint

If the height of a load matters, so does the height of a restraint, and for exactly the same reason: a brace prevents motion at the point it is attached to, and the point it is attached to is moving by the sideways displacement of the shear centre plus aθa\theta.

A brace on the wrong flange never gets there, however stiff it isThe critical moment of a 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 366 kN/m. At the shear centre it needs 2252 kN/m, 6.2 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 1.022, 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.02004006008000100200300400brace stiffness (kN/m)critical moment (kNm)compression flangeshear centretension flange366 kN/m is enougha rigid brace buys 3.12, not two
Fig. 3 One midspan brace at three heights, with the critical moment against the brace’s stiffness. The three curves are the same beam and the same brace, moved up and down the web.

The three curves in that figure are worth reading one at a time.

On the compression flange the brace works. The critical moment climbs from the unbraced 143 kNm to 447 — the beam has become two shorter beams — and it reaches 99% of that plateau at 366 kN/m of stiffness. Past the plateau it stops helping, because the beam has found a mode with a node where the brace is and the brace no longer moves.

At the shear centre the same plateau is reached, and it costs 2,252 kN/m: 6.2 times the stiffness, for the identical result. The brace is now restraining a point that barely moves in the buckled shape, so it has to be very stiff to do anything at all.

On the tension flange it never arrives. At the stiffness that finished the job on the compression flange it has bought a factor of 1.022 — two per cent — and stiffening it further buys the same nothing, because the buckled shape simply rotates about the braced point and carries on. That is the brace on the wrong flange, and it is the most consequential detailing error available in steelwork: a member exists, it is connected, it is strong, and it does not restrain anything.

Why the plateau exists at all

Both the load-height effect and the brace-height effect are the same term in one energy expression, and the plateau is the evidence for it.

A brace raises the critical load by forcing the buckled shape to have a node where the brace is. Once the beam is buckling in the shape with that node — two half-waves instead of one — additional brace stiffness restrains nothing, because the node does not move. The critical load is then the critical load of the shorter segment and no arrangement of braces can raise it further.

A brace is a stiffness requirement, not a strength oneCritical load against brace stiffness for a pinned column braced at 40% of its height. The curve climbs from the unbraced Euler load of 9.87EI/L² and flattens at 27.42EI/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 107EI/L³.050100150200250300350400010203040brace stiffness (units of EI/L³)critical load (units of EI/L²)ideal stiffness ≈ 107 EI/L³27.4 — braced9.87 — unbraced
Fig. 4 The same knee in the simpler problem. A brace on a column climbs to the Euler load of the braced segment and then flattens exactly, because past that point the column buckles in a shape the brace does not obstruct.

That is the general shape of every restraint problem on this site, and it is why a brace need not be strong. What the brace has to be is stiff enough to reach the knee, and where it has to be is somewhere that moves.

What it costs when the section is torsionally weak

The height effect scales with how easily the section twists, so it is largest where the trouble already is.

A channel has three critical loads, not oneThe three critical loads of a channel in compression, against its length, with the load it actually buckles at drawn over them. At 6000 mm the flexural loads are 4754 kN about the major axis and 785 kN about the minor, while twisting about the shear centre takes 811 kN. The lowest root is 754 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 3.9 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.4000600080001000012000140000500100015002000length of the column (mm)critical load (kN)flexural about ytorsionalflexural about zthe mode changes at 5813 mmthe lowest root — what the column actually does
Fig. 5 A channel has three critical loads rather than one, and the shear centre sitting away from the centroid couples them, so the governing root is below any of the three.

A channel is the extreme case in ordinary construction, because its shear centre is outside the section entirely — here 109.7 mm from the centroid — so every load is applied at an eccentricity from the axis of rotation, whether or not anybody chose to apply it there. The three uncoupled critical loads at 6 m are 4,754 kN, 785 kN and 811 kN; the coupled answer is 754 kN, 3.9% below the lowest of them, and the mode that governs changes with length.

An I-section is much better behaved, because its shear centre is at the centroid by symmetry — the property a section that will not stay flat trades on, so the load height is measured from the middle of the depth and a load on either flange is the same distance from it in opposite directions. A section that is symmetric about neither axis has no such convenience.

The lifted beam, where the same term has an asymptote

The clearest available demonstration is not a floor beam. It is a beam hanging from a crane.

Raising the hook helps, and only past a height that the beam decidesThe factor of safety against rolling over, against the height of the roll axis above the centre of gravity, for a 30 m beam of 330 kN. There is a vertical asymptote at y_r = 0.236 m, which is z̄ — the sideways deflection of the beam's own centre of gravity under its own weight applied laterally — and below it there is no equilibrium at any tilt whatever. A rigid body would have no asymptote: hang it from anywhere above its centre of gravity and it is a pendulum. The beam is not rigid, and the difference between the two answers here is a factor of 1.90.00.511.522.53010203040roll-axis height above the centre of gravity (m)factor of safetyz̄ = 0.236 ma working factoras lifted: 5.3if the beam were rigidthe rigid answer is optimistic by a factor of 1.90
Fig. 6 Factor of safety against rolling over, against the height of the roll axis above the centre of gravity. There is a vertical asymptote, and below it there is no equilibrium at any tilt whatever.

Everyone knows the rigid-body rule: hang a body from a point above its centre of gravity and it is a pendulum, stable at any hook height above zero. A beam is not a rigid body. Tilt it, and the component of its own weight acting across its weak axis deflects it sideways by an amount zˉ\bar{z}, which moves the centre of gravity further out.

The stability then depends on the difference yr−zˉy_r - \bar{z} — a hook height minus a lateral sag — and that difference goes through zero at a finite hook height. For this 30 m beam it happens at yr=0.236y_r = 0.236 m, and the answer a rigid-body calculation gives is out by a factor of 1.90.

This is hung from above and still unstable in one line: the attachment height matters because the thing being attached to is flexible, and the flexibility supplies a length that competes with the height.

At thirty degrees each leg carries the whole loadA 100 kN lift on two legs at 30 degrees to the horizontal. Each leg is a two-force member, so its force is the vertical share it carries divided by the sine of its angle: T = W/(2 sin β) = 100.0 kN, which is 1.00 times the whole load in each. The curve on the right is that division, and it is the reason the rule about sling angles exists rather than a convention: at sixty degrees a leg carries 0.58 W, at forty-five 0.71, at thirty exactly 1.00, and at fifteen 1.93. And the horizontal components do not disappear — they run through the thing being lifted, which here carries 173.2 kN of compression between the two pick points. That is what a spreader beam is for: it takes the compression as a designed strut so the legs above it can stand up.hook100 kN100.0 kN100.0 kN30°173.2 kN of compression, in the load2040608000.511.52sling angle (degrees)leg force ÷ load30°
Fig. 7 The other half of a lift. Sling angle decides the force in each leg, and the horizontal components run through the thing being lifted — which is why the pick points and their height are a stability question as well as a strength one.

What is really being computed

Every result above is an eigenvalue, and the load height enters it as a term that is linear in the rotation. That is worth stating plainly because it explains why the effect refuses to be captured by a single factor.

A destabilising term linear in θ\theta sits alongside a restoring term that is also linear in θ\theta, and their ratio depends on the section, the length, the restraint condition and the shape of the moment diagram. The published expressions handle this with a coefficient — one factor for the moment diagram’s shape, another for the load height — multiplied together as though they were independent. They are not independent, and the product is an approximation to a coupled eigenvalue, accurate to a few per cent over the range it was fitted on and unreliable outside it.

One coefficient, and nothing else in itThe deflected shapes of one beam under four load cases, each scaled so that its mid-span deflection is the same, with the tangent at the left-hand support drawn on each. The end rotation is that deflection times a coefficient that depends only on the shape of the load: 3.20 for a uniform load, 3.00 for a load at mid-span, 2.99 for a triangular load, 3.60 for a load on half the span. Every material property, every second moment and the span itself cancel out of the ratio θL/δ, so a beam at any deflection limit has an end rotation that is known before anything about it is: at L/360 it is 8.89 milliradians, or 0.51 of a degree.0246810-0.15-0.1-0.0500.050.10.15along the span (m)deflection, each scaled to the same mid-span drop3.20 · a uniform load3.00 · a load at mid-span2.99 · a triangular load3.60 · a load on half the spanθ = C · δ/L, and the same δ gives 1.21× the rotation across these
Fig. 8 A reminder of how little of this is about materials. The ratio of end rotation to midspan deflection depends only on the shape of the load; the modulus, the second moment and the span all cancel.

Where it decides a design

Four cases, all ordinary, in which the height of an attachment is the design.

A crane runway beam. The wheel load sits on the top flange, at the top of the rail, which is above the top flange again. It is also free to move along the beam, so there is no restraint at the point of application at all, and the beam has to be checked with the load at its worst position and its worst height simultaneously. Runway beams are the reason the load-height term appears in the standards.

A portal rafter under wind uplift. Under gravity the purlins restrain the top flange and the top flange is in compression, which is the happy arrangement. Under uplift the moment reverses, the bottom flange goes into compression, and the purlins are restraining the tension flange — the third curve in the figure above, the one that buys two per cent. That is why portal frames carry fly braces from the purlins down to the inside flange, and why a frame checked only for gravity is not checked.

A monorail or a services load hung below. Here the height effect is favourable, and worth about a factor of two on a section that would otherwise be governed by lateral buckling. Almost nobody claims it, because it takes an explicit calculation to claim and the conservative answer is free.

A beam during erection. Before the deck is on, a beam has no restraint anywhere along it and carries its own weight applied at its own centroid. That is the most dangerous day in the life of the member, and the unrestrained length is the whole span rather than the purlin spacing the finished design assumed.

The common feature of all four is that the load height is decided by a detail — a rail, a purlin cleat, a hanger, a lifting lug — drawn by somebody who is not doing the stability check.

Where the model stops

Everything above is elastic and perfect. A real beam is bowed, twisted and has residual stresses, so it never reaches an elastic critical moment; it follows a curve that approaches it asymptotically, exactly as a bowed column does. Load height changes the asymptote and therefore changes the whole curve, but the design capacity is somewhere below it and the ratio is not preserved.

A column that was never straightLoad against lateral deflection at mid-height, for a column starting with an initial bow of 0.006. 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.00.020.040.060.0800.20.40.60.81lateral deflection at mid-heightload ÷ P꜀ᵣP ÷ P꜀ᵣ = 1.00, approached and never reachedinitial bow: δ₀ = 0.006
Fig. 9 Why an elastic critical value is never reached. The imperfect member deflects from the first increment and approaches its critical load without arriving.

The load may not stay where it was put. A load applied by a slab bearing on the top flange stays on the top flange while the beam twists a little and does not once it twists a lot, because the slab is stiff and the bearing rotates. Everything here assumes the load’s point of application is carried around with the section, which is right for a hanger and questionable for a bearing.

Restraint is rarely a spring at a point. Decking, a slab, or a purlin line restrains a beam continuously and partially, and at a height that varies with the detail. The single-brace model is a stand-in for that, and the useful part of it is the ordering rather than the numbers.

How high it starts decides how badly it endsLoad against apex movement for four frames of the same half-span and the same bars, differing only in rise. The maximum load goes as the cube of the rise — 6% giving 8.7 kN, 10% giving 40.0 kN, 15% giving 133.4 kN, 22% giving 410.4 kN — and every one of them has a load it cannot pass and a region of movement in which it can carry nothing at all. The paths are exact: no shallow-arch approximation is used to draw any of them.0100200300400500600-400-2000200400movement of the apex (mm)load at the apex (kN)rise 6%rise 10%rise 15%rise 22%
Fig. 10 The reason none of this can be settled by a load factor. Each of these frames has a load it cannot pass, and the maximum goes as the cube of a geometric quantity — small changes in a shape moving an answer by more than any material property would.

What the picture cannot show

None of the figures shows the beam twisting, because a critical moment is a statement about the instant before anything happens. The buckled shape drawn on an eigenvalue plot has an arbitrary amplitude — it is a mode vector — and it says which way the beam goes and nothing about how far.

Nor do they show the thing that decides most real cases: whether the restraint that is drawn is actually connected to anything that can take a force away. A brace on the compression flange is worth a factor of three, and only if the other end of it reaches a stiff point. A line of braces all connected to each other and to nothing else restrains the beams to one another and lets the whole set buckle together, which is a mode no single-beam calculation contains.

What is left of the flange to resist a change of shapeA flange carrying a residual compression of 45% of yield at its tips, shaded where it has yielded, at four levels of applied stress. The yielded part still carries load and contributes no stiffness at all, so what resists buckling is the elastic core: at 0% of yield the core is 100% of the width, at 50% of yield the core is 100% of the width, at 80% of yield the core is 44% of the width, at 95% of yield the core is 11% of the width. About the major axis the flanges are lever arms and the stiffness follows the core's width; about the minor axis each flange bends about its own centre, so it follows the CUBE of it — 100.0%, 100.0%, 8.8%, 0.1% against 100%, 100%, 44%, 11%.0%core 100% · major 100% · minor 100.0%50%core 100% · major 100% · minor 100.0%80%core 44% · major 44% · minor 8.8%95%core 11% · major 11% · minor 0.1%applied stress, and the flange it leavesthe shaded ends have yielded: they carry load and no stiffness
Fig. 11 What is left of a flange to resist a change of shape. About the weak axis the stiffness follows the cube of the elastic core’s width, which is why the same residual stress costs so much more in the mode that matters here.

The generalisation

The habit worth carrying is a question to ask of any restraint or any load: what point of the section is it attached to, and does that point move in the mode being prevented?

It is a question about the mode rather than about the member, and it has a general answer. A restraint is worth something in proportion to how far its attachment point moves in the shape it is trying to prevent; a load is destabilising in proportion to how far its attachment point moves in the same shape. They are the same quantity with opposite signs, which is why one figure in this essay answers both.

The rule extends past beams. A tuned mass damper works because it is attached where the mode has its largest amplitude and would do nothing at a node. A guy is worth more on a mast at the height where the mast wants to move. And an outrigger on a tall building makes the columns work precisely because it connects a core to columns at a level where the core’s rotation is large — the same argument, three orders of magnitude up.

The failure mode is always the same too. Something has been provided, it is adequate in every check made of it, and it is attached to a point that does not move.

Where this goes next

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 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. 020040060080010000100200300400brace stiffness (kN/m)critical moment (kNm)compression flangeshear centretension flange447 kN/m is enougha rigid brace buys 3.12, not two 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. 1015202530354045500510152025beam length (m)factor of safety against rolling1.5, a working factorpast 42 m30 m: 13.3 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. web centrelineshear centree = 31.7no twisttwiststhe flange flows are equal, opposite, and separated — which is a coupleand nothing about the section's 20.19 × 10⁶ second moment predicts it The point that is not in the section A channel loaded down its web twists. To stop it, the load must be applied through a point outside the steel entirely — in the air beside the section, where nothing can be attached.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

BracingCompression flangeCritical loadEigenvalueImperfectionLateral torsional bucklingLoad heightRestraintRiggingSecond orderShear centreStabilityTorsional constantTwistWarping