Structural Engineering & Statics
where the load goes
Stability

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.

Assumes The ends decide the length that matters, Strong enough and still falls over and The frame that leans, and what stops it.

24 min read 11 figures Which failure arrives firstOne support too manyGeometry beats material

Two portal frames, drawn side by side. Same columns, same beam, same steel, same height. The only difference is a hatched block against the head of the left-hand one, holding it against sideways movement — a restraint that carries no vertical load at all, and which on a drawing looks like an afterthought.

One restraint, and several times the loadThe 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.the storey that cannot driftk = 0.774N꜀ᵣ = 16.46 EI/L²the storey that cank = 1.317N꜀ᵣ = 5.69 EI/L²the same column, 2.9 times the load — the restraint is the whole of the difference
Fig. 1 The same portal, buckling twice. Held against drift its critical load is 16.46 EI/L²; free to drift it is 5.69 EI/L², a factor of 2.89. The effective length factor coming out of each eigenvalue is 0.774 against 1.317. Both are the lowest root of the assembled frame at a beam-to-column stiffness ratio of G = 1.00; the shapes are the mode vectors themselves, drawn at 18 per cent of the storey height so the movement can be seen at all.

The column on the right carries a third of what the column on the left carries. Nothing about the column changed.

The four that are on every chart, and describe nothing

The textbook opens with four cases and closes the subject with them. Pinned at both ends, fixed at both ends, fixed at one and pinned at the other, and fixed at the base with the top free — factors of 1, 0.5, 0.7 and 2, and a span of sixteen in capacity across the row.

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. 2 The four end conditions the subject is usually taught from. Each factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square, so the row spans a factor of sixteen.

Every one of those four is an eigenvalue of an isolated member with an idealised end, and no column in any building has an idealised end. A real column runs into beams, and a beam is a spring — it resists the column’s end rotation with a stiffness of its own, somewhere between the nothing of a pin and the everything of a built-in support. Assuming more restraint than exists overestimates capacity, which is the unsafe direction.

That much is well known and codes handle it by interpolating. What the four cases hide is larger and is not an interpolation at all: two of them describe a column whose head cannot move sideways and two describe one whose head can, and that distinction is worth more than the rotational restraint at either end. The factor of 2.89 in the figure above comes entirely from it, at a beam stiffness held constant.

The claim

Effective length is a property of the frame.

Not of the member, not of its ends, and not of anything visible on a drawing of the column alone. It is set by two things: the rotational stiffness the beams supply at the joints, and whether the storey is free to translate. The second is a yes-or-no question about the building, and a column has no way of knowing the answer.

This is why removing a wall is a structural act even when the wall carries nothing, why a stiff core lets a tower use much lighter columns than frame action would allow, and why the sway classification is the single judgement in stability design that changes the most numbers downstream.

The alignment chart, recovered rather than read

The profession’s device for all this is the alignment chart: two scales of relative stiffness, a straight edge laid across them, and a factor read off the middle. It is a solved problem presented as a lookup, and everything in it is recoverable by assembling the frame and asking for the eigenvalue.

The alignment chart, computed rather than looked upThe effective length factor against G = (EI/L) of the column ÷ (EI/L) of the beams, for a storey held against sway and for one free to sway. Every point on both curves is the lowest eigenvalue of the assembled frame, swept over 25 beam stiffnesses — not a nomogram, and nothing here is read off a chart. The non-sway curve runs from k = 0.505 at G = 0.01, where the beams are stiff enough to be built-in, to k = 0.990 at G = 40, where they are soft enough to be pins: the whole of it lies between a half and one. The sway curve starts at k = 1.003 and has no upper bound at all, reaching 5.81 at the same G — so the braced frame carries 34.4 times the load of the unbraced one at its worst point on this sweep.0.010.11100123456G = (EI ÷ L) of the column ÷ (EI ÷ L) of the beamseffective length factor kfree to swayheld against swayk = 1.0 — a pin-ended columnk = 0.5 — beams rigid, sway preventedk = 0.505k = 0.990
Fig. 3 The effective length factor against G, the ratio of column stiffness to beam stiffness, for a storey held against sway and for one free to sway. Every point on both curves is the lowest eigenvalue of the assembled frame, swept over 25 beam stiffnesses. The non-sway curve runs from k = 0.505 to k = 0.990 across the whole sweep; the sway curve starts at k = 1.003 and reaches 5.81, giving a braced frame 34.4 times the load of the unbraced one at the worst point drawn.

Two features of that picture carry the whole argument.

The held curve is trapped between a half and one. However soft the beams get, a storey that cannot drift has k≤1k \le 1. The worst it can do is behave like a pin-ended column, and the best it can do is 0.5. All the rotational restraint in the world buys a factor of four, and that is the entire range available.

The sway curve has no ceiling. It starts at 1.0 where the beams are rigid — a swaying storey with infinitely stiff beams is still no better than a pin-ended column — and climbs without limit as the beams soften. At the right-hand end of the sweep it is at 5.81, meaning a critical load of π2EI/(5.81L)2\pi^2EI/(5.81L)^2, which is a thirty-fourth of what the same steel carries braced.

Softening the beams cannot make a braced frame worse than a pin-ended column. It can make an unbraced one arbitrarily bad. That asymmetry is the reason bracing exists, and it is invisible on a row of four end conditions because the row contains one sway case and prices it at 2.0.

A portal frame swaying under 20A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 22.9. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.20H 10.0 M 22.9H 10.0 M 22.9the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 4 The frame the eigenvalue is taken on, at the same G = 1 — a six-metre beam of I = 1.5 on four-metre columns of I = 1 — solved under a horizontal load instead. The base shears come out at 10.0 each, adding to the applied 20, and the peak moment is 22.9. The split between the columns came from stiffness rather than statics — the same reason a shared displacement is settled by relative stiffness — and the sway is drawn hugely exaggerated: a real frame at this load moves a fraction of a millimetre.

Which free body produced the number

Cut a horizontal plane through the storey, just below the beam, and take everything above it as the free body.

That body carries the vertical loads ∑N\sum N, and in the buckled state it has moved sideways by Δ\Delta. Moments about a point on the cut give a destabilising couple ∑N⋅Δ\sum N \cdot \Delta — the vertical loads riding their own lateral displacement, which is the load making itself worse written as one term. Restoring it are the column shears at the cut, which for a storey of lateral stiffness SS and height hh amount to a couple SΔ⋅hS\Delta \cdot h.

Equilibrium of that free body is possible in a displaced position only when

∑N⋅Δ=SΔh,\sum N \cdot \Delta = S \Delta h,

and Δ\Delta cancels. It always does; that cancellation is what a stability problem is. The storey goes critical at

∑Ncr=Sh.\sum N_{\mathrm{cr}} = S h.

The computed frame has a lateral stiffness of 12.0 in units of EI/L3EI/L^3, and two columns share the load, so this free body predicts 6.0 EI/L26.0\,EI/L^2 per column. The assembled eigenvalue says 5.69. The one-line free body is within 5 per cent, and it is 5 per cent high, because it treats the columns as rigid bars and ignores their own bowing, which spends flexibility the model has not accounted for.

That is worth pausing on. The whole of sway stability sits in a single moment equation on a single cut, and the eigenvalue solver exists to correct it by a twentieth. The equation says what matters: the storey’s lateral stiffness, the height, and the total vertical load — not the load in any one column. A lightly loaded column in a swaying storey is not safe; it goes when the storey goes, because the load must go somewhere and the somewhere is the storey as a whole.

The endpoints are the check

An eigenvalue solver that returns plausible numbers is not evidence of anything. What makes these trustworthy is that the frame model, given degenerate inputs, reproduces the four textbook cases it was written to argue past — none supplied, all recovered.

One restraint, and several times the loadThe 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 20.19 EI/L² and the swaying one's is 2.47 EI/L², a factor of 8.18, and the effective length factor that comes out of each eigenvalue is 0.699 against 2.000. Both are eigenvalues of the assembled frame at a beam-to-column stiffness ratio of G = infinity, there being no beam at all; the buckled shapes are the mode vectors themselves, drawn at 18 per cent of the storey height so that the movement can be seen.the storey that cannot driftk = 0.699N꜀ᵣ = 20.19 EI/L²the storey that cank = 2.000N꜀ᵣ = 2.47 EI/L²the same column, 8.2 times the load — the restraint is the whole of the difference
Fig. 5 The same machinery with the beam deleted and the feet built in, which is no longer a frame but two isolated columns. Held against drift it returns k = 0.699 and a critical load of 20.19 EI/L² — the fixed-pinned case, whose exact factor is the transcendental root 0.6992. Free to drift it returns exactly k = 2.000 and 2.47 EI/L², which is the cantilever, π²/4. A factor of 8.18 between two conditions that differ only in whether the head can move.

The fixed-pinned value is the interesting recovery. It has no closed form — it comes from tan⁡kL=kL\tan kL = kL, whose first root is 4.4934, giving π/4.4934=0.6992\pi/4.4934 = 0.6992 — and the assembled frame lands on 0.699 without ever having been told the equation exists.

The alignment chart, computed rather than looked upThe effective length factor against G = (EI/L) of the column ÷ (EI/L) of the beams, for a storey held against sway and for one free to sway. Every point on both curves is the lowest eigenvalue of the assembled frame, swept over 21 beam stiffnesses — not a nomogram, and nothing here is read off a chart. The non-sway curve runs from k = 0.502 at G = 0.01, where the beams are stiff enough to be built-in, to k = 0.698 at G = 100, where they are soft enough to be pins: the whole of it lies between a half and one. The sway curve starts at k = 1.002 and has no upper bound at all, reaching 1.95 at the same G — so the braced frame carries 7.8 times the load of the unbraced one at its worst point on this sweep.0.010.111010000.511.522.5G = (EI ÷ L) of the column ÷ (EI ÷ L) of the beamseffective length factor kfree to swayheld against swayk = 1.0 — a pin-ended columnk = 0.5 — beams rigid, sway preventedk = 0.502k = 0.698
Fig. 6 The same sweep on a portal with built-in feet rather than a beam at the base, taken out to G = 100 over 21 points. The held curve runs 0.502 to 0.698 — from fixed-fixed to fixed-pinned — and the sway curve runs 1.002 to 1.95, which is the cantilever appearing as a limit rather than as an assumption. The whole range of the swaying case is now bounded, at 7.8 times, because a fixed base is itself a restraint against the mode.

Five known answers fall out of one solver: 0.5, 1.0, 0.6992 and 2.0 from the degenerate frames, and 1.0 again as the swaying storey’s best case. The site’s gate for this family also runs the same frames through Livesley’s stability functions, written independently from the slope-deflection equations rather than from an assembled matrix, and the two agree to 2.4 parts in ten thousand.

The brace is tiny, and that is the finding

If the difference between 0.774 and 1.317 is one horizontal restraint, the obvious question is how stiff that restraint has to be. The answer is small enough to be worth stating carefully, because it is easy to disbelieve.

How stiff a brace has to be before the frame stops swayingThe 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 1.93.02040608000.20.40.60.811.21.41.6brace stiffness (units of EI/L³)effective length factor k23.2 EI/L³ = 1.41 N꜀ᵣ/L0.77 — held1.32 — freepast the threshold the frame buckles in a mode the brace does not hold
Fig. 7 The effective length factor of the swaying portal against the stiffness of a horizontal spring at its head. The curve starts at k = 1.317, falls to 0.774 — the held value for the same frame — at a brace stiffness of 23.2 EI/L³, and past that point is exactly flat. Stated against the load it holds, the threshold is 1.41 N꜀ᵣ/L; against the storey’s own lateral stiffness of 12.0 EI/L³ it is a factor of 1.93.

Take a storey four metres high carrying a thousand kilonewtons. The threshold works out at about 0.35 kilonewtons per millimetre of sway. A diagonal rod across a four-by-four-metre bay delivers EALdcos⁡2θ\frac{EA}{L_d}\cos^2\theta of lateral stiffness, which for steel at that geometry is 18.6 newtons per millimetre for every square millimetre of rod. The threshold therefore asks for 19 mm² of steel — a rod about 5 mm across.

That is not a design recommendation and nothing here sizes anything. It is a statement about proportion: the member that trebles a storey’s capacity is smaller than the bolts holding it on. Real braces are far larger, for three reasons all worth separating from the stiffness question — the singularity in the brace force near the threshold, the flexibility of the connections in series with the rod, and the wind the brace is also there to resist. Those push the practical section up by a factor of three or so. None changes the fact that the stability requirement itself was met by 19 mm².

The free-body model predicts the threshold too. Raising the sway critical load from 5.69 to 16.46 per column means raising SS from 12.0 to 2×16.46=32.92 \times 16.46 = 32.9, so the brace supplies 20.9 in units of EI/L3EI/L^3 against the computed 23.2. The same optimism as before, from the same neglected bowing, and now worth 10 per cent rather than 5 because it lands on the difference between two numbers instead of on one.

Past the threshold, exactly nothing

The flat portion of that curve is genuinely flat, and the reason is a mode change. Below the threshold the frame buckles by swaying, and the brace obstructs that mode, so stiffening it raises the load. At the threshold the sway mode has become as expensive as the non-sway mode. Above it the frame buckles in the non-sway mode instead — the columns bow in opposite directions and the head does not move, so the brace is not stretched and cannot contribute anything at all.

How stiff a brace has to be before the frame stops swayingThe effective length factor of a swaying portal against the stiffness of a horizontal spring at its head. The curve starts at k = 2.036, the unbraced value, and falls to 0.916 — the factor for the same frame with its head held — at a brace stiffness of 19.0 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 1.61 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 0.40 kN per millimetre of sway. Against the frame's own lateral stiffness of 4.8 EI/L³ it is a factor of 3.96.020406000.511.522.5brace stiffness (units of EI/L³)effective length factor k19.0 EI/L³ = 1.61 N꜀ᵣ/L0.92 — held2.04 — freepast the threshold the frame buckles in a mode the brace does not hold
Fig. 8 The same sweep with the beams four times softer, at G = 4. The unbraced frame is now at k = 2.036 and the held one at 0.916, so the brace is worth more; the threshold rises to 19.0 EI/L³, which against this frame’s much smaller lateral stiffness of 4.8 EI/L³ is a factor of 3.96. Softer beams make the brace both more valuable and relatively larger, and the plateau past the threshold is flat here too.

This is the same shape of argument as the brace that need not be strong, and the difference between the two is worth being precise about, because the family resemblance hides genuinely different bodies.

A brace is a stiffness requirement, not a strength oneCritical 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 158EI/L³. A stiffness of 23EI/L³ is marked, reaching 14.49EI/L².02040608010012014016018020001020304050brace stiffness (units of EI/L³)critical load (units of EI/L²)ideal stiffness ≈ 158 EI/L³14.539.5 — braced9.87 — unbraced
Fig. 9 The member-level version of the same knee, with the storey’s threshold marked on it. Critical load against brace stiffness for a pinned column braced at mid-height: it climbs from 9.87 EI/L² to 39.48 EI/L² — the Euler load of the half length — with the ideal stiffness near 158 EI/L³. The 23 EI/L³ that took the whole storey to its held value reaches only 14.5 EI/L² here. Same mechanism, different free body: this brace restrains one member against its own bowing.

The member brace acts at mid-height, against the column’s own half sine wave, and its ceiling is the Euler load of the half length. The mode it stops using is the full sine wave, which has a node exactly where the brace sits. Its ideal stiffness comes out near 16π2EI/L316\pi^2 EI/L^3, a number in the hundreds.

The storey brace acts at the head, against a rigid-body lean of everything above the cut, and its ceiling is the non-sway eigenvalue of the frame. The mode it stops using is the frame’s own double-curvature mode. Its threshold comes out at 23.2 in the same units — an order of magnitude smaller, because leaning is a much softer thing to stop than bowing.

One braces a member against itself. The other braces a storey against a rigid-body motion of the whole floor. They share a knee, a plateau and an eigenvalue crossing; they do not share a free body, a mode, or an order of magnitude.

What these pictures cannot show

Each buckled shape on this page is a mode vector, and a mode vector has no amplitude. The linear eigenvalue problem returns a direction in displacement space and says nothing whatever about how far along it the structure goes — which is precisely what makes the load critical. The captions say the shapes are drawn at 18 per cent of the storey height because that is a drafting decision, not a result, and any figure that draws a definite deflected curve is implying a definite state the mathematics does not contain.

Nor does any of it show what happens before the critical load, which is where every real structure lives.

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 load1.3×2.0×4.0×10.0×first-order analysis says the answer is always 1×one over one minus the ratio
Fig. 10 Amplification of a sway deflection against the ratio of applied load to critical load, marked at a quarter, a half, three quarters and nine tenths. A storey at half its buckling load has already drifted 2.0 times as far as a first-order analysis says; at nine tenths, 10.0 times. This is what makes the sway classification a serviceability question as well as a strength one.

And the chart’s horizontal axis is a single number GG for a whole storey. A real building has a different GG at every joint, a different vertical load in every column, and columns that lean on each other through the floor plate — so the storey’s weakest column is held up by its neighbours, and its strongest may be doing the holding.

Where the model stops

Elastic behaviour. Every number here is an elastic eigenvalue. A stocky column reaches the material’s limit long before it reaches these loads, and the effective length then only positions the member on the column curve.

A perfect frame. The eigenvalue assumes no initial lean. Real frames are erected out of plumb, and an imperfection turns a bifurcation into an amplification that starts at zero load. For sway frames this matters more than for braced ones, because the knock-down from imperfections is largest exactly where the mode is a rigid-body one.

Rigid joints. The beams are assumed continuous into the columns, so the whole moment goes round the corner. A semi-rigid connection supplies a fraction of that restraint, which moves GG to the right along the chart — and on the sway curve, the right is where the curve is steepest.

One storey, one mode. A multi-storey frame buckles as a system. The critical mode may involve several storeys at once, and the storey-by-storey calculation each column’s GG implies is a decomposition of convenience.

One buckling mode per member. The frame model bends its columns in plane. A column can also fail by twisting about its shear centre, which no plane-frame eigenvalue will ever report.

An I-section has three critical loads, not oneThe three critical loads of an I-section in compression, against its length, with the load it actually buckles at drawn over them. At 3000 mm the flexural loads are 19014 kN about the major axis and 1454 kN about the minor, while twisting about the shear centre takes 2589 kN. The lowest root is 1454 kN, and the column bends about its minor axis. The shear centre is the centroid, so the three modes are independent and the envelope is simply the lowest of them. The torsional load stays above both flexural loads at every length drawn, so this section never twists in preference to bending and the mode is one a design check may leave out.2000300040005000600070000200040006000length of the column (mm)critical load (kN)flexural about ytorsionalflexural about zthe lowest root — what the column actually does
Fig. 11 The mode the plane frame cannot see. For an I-section at 3000 mm the flexural critical loads are 19014 kN about the major axis and 1454 kN about the minor, while twisting about the shear centre takes 2589 kN. Minor-axis bending governs here — but the ordering depends on the section, and a frame analysis reports only one of the three.

A brace that reaches the ground. The threshold is a stiffness seen by the storey. Everything between the brace and the foundation is in series with it, and a structure part-way through erection may have the diagonal in place and nothing beneath it.

The generalisation

The pattern here — a quantity that looks like a member property but is a system property — turns up wherever members share a displacement instead of a force. The stiffest path takes the load is the same statement about force distribution; moment distribution is the same statement about joint rotation, iterated; one support too many is the same statement about reactions. In each case a designer’s instinct to reason about one member in isolation gives an answer that is not merely imprecise but categorically incomplete, because the missing information is not in the member.

The most consequential instance is on the drawings rather than in the analysis. A column removed at a lower level, so that everything above it lands on a transfer member, changes the storey’s lateral stiffness as well as its load path — and the columns that remain have had their effective lengths altered by a decision made about architecture. Nothing on the column schedule will say so.

The ladder from here

Later rungs on this anchor: the sway index αcr\alpha_{cr} and where the threshold between a braced and an unbraced classification actually sits. Leaning columns, and how a storey’s weak members are carried by its strong ones. Multi-storey critical modes, and why a storey-by-storey check can miss them. The notional horizontal force, and its derivation from erection tolerance rather than from wind. Amplified sway methods against direct second-order analysis. The stiffness of a real bracing system, assembled from diagonals, gussets, floor diaphragms and foundations in series. Base fixity, and how much of it a nominally pinned base actually delivers. And the interaction of sway stability with plastic hinge formation, where the frame’s stiffness falls during the very event that is loading it.

The four end conditions survive because they are memorable and because they were once the only calculation available with a slide rule. What the assembled eigenvalue shows is that they are not four cases of a general rule — they are four degenerate corners of a two-parameter surface, and every real column sits somewhere in the middle of it, on whichever sheet the frame around it has chosen.

Where this goes next

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. 05010015020025001020304050brace stiffness (units of EI/L³)critical load (units of EI/L²)ideal stiffness ≈ 159 EI/L³21.839.5 — braced9.87 — unbraced 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. 20H 10.0 M 22.2H 10.0 M 22.2the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre The frame that leans, and what stops it A rectangle of pinned bars folds flat. Make the corners rigid instead of adding a diagonal and it does not — which buys an unobstructed opening and costs bending in every member of it. 00.20.40.60.80246810applied load ÷ buckling load1.1×1.3×1.4×2.0×3.3×10.0×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.

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.

BracingBuckled mode shapeCritical loadEffective lengthEigenvalueSecond order effectsStiffnessSway frame