Stability

The arch that leans instead of squashing

A masonry arch is asked whether a line of thrust fits inside it. A steel rib is asked a different question entirely: it is a column carrying an axial force along its whole length, and the mode it buckles in puts one half up and the other half down while the crown moves sideways.

Assumes The hinge put in on purpose, Strong enough and still falls over and The load that makes itself worse.

The masonry arch’s question is whether a line of compression exists inside the ring. It is a good question and it is the right one for a thick stubby structure built out of blocks: if such a line can be found, the arch stands, and nothing about the material’s stiffness enters the answer.

A steel or concrete rib is the opposite shape. It is slender, it carries a large axial compression along its entire length, and it will run out of stiffness long before it runs out of strength or out of room inside its own depth. That is a column’s question, and the answer is a column’s answer — an eigenvalue.

Two things make it more than a column. The thrust is generated by the load rather than applied to the arch, so the critical quantity is a load intensity and it scales as EI/L3EI/L^3 rather than EI/L2EI/L^2. And the rib has two competing modes that are not the same shape.

The arch does not squash; it leansThe first two buckling modes of the same rib, drawn against its undeformed shape at an exaggeration of a few hundred. The first, at w_cr L³/EI = 46.2, is antisymmetric: one half rises while the other falls and the crown moves sideways. The second, at 106.4, is symmetric — the whole rib settling. The two differ by a factor of 2.31, which is why an arch is braced against sideways movement of its crown rather than against the load it is carrying, and why a tied arch with a single hanger at midspan is doing nothing for the mode that governs it.first · antisymmetric · C 46.2second · symmetric · C 106.4dashed is the rib before it moved · drawn hugely exaggerated
Fig. 1 The two modes of a 60 m parabolic rib at a rise of 12 m. The first is antisymmetric — the crown moves sideways and the two halves go opposite ways — at wcrL3/EI=46w_{cr}L^3/EI = 46. The second, the symmetric squash that looks like the load, is 2.3 times higher. The mode that governs is not the one the loading suggests.

Which free body produced the number

There is no closed form here and none is quoted. The rib is a plane frame of straight segments, and the answer comes out of two solves.

The first is linear: apply the load, get the axial force in every segment. For a parabolic arch under a uniform load that force is very nearly constant along the rib and equal to H/cosθH/\cos\theta — the horizontal thrust divided by the cosine of the local slope — and H=wL2/8fH = wL^2/8f from the crown free body.

The second is an eigenproblem. Each segment’s axial force generates a geometric stiffness KgK_g, which subtracts from the elastic stiffness KK; the rib becomes unstable when KλKgK - \lambda K_g loses positive definiteness; and λ\lambda is the factor the reference load has to be multiplied by.

(KλKg(N))x=0\left(K - \lambda K_g(N)\right)\mathbf{x} = 0

That is the same pair of matrices every other stability figure on this site is built from, assembled for a curved member instead of a straight one. The machinery reproduces π2EI/L2\pi^2EI/L^2, π2EI/4L2\pi^2EI/4L^2 and 20.19EI/L220.19EI/L^2 for a straight column to seven figures, which is what makes the arch numbers worth quoting.

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. 2 The column the rib is a bent version of. Everything about the mechanism is the same — axial force, geometric stiffness, an eigenvalue — and the difference is that a column is handed its load while an arch manufactures its own.

Why the antisymmetric mode wins

The symmetric mode requires the rib to shorten along its own axis, because a symmetric downward movement of an arch is very nearly a uniform shortening of the arc. Axial stiffness is enormous compared with bending stiffness, so that mode is expensive.

The antisymmetric mode requires nothing of the kind. One half goes up, the other goes down, the crown slides sideways, and the arc length is very nearly preserved throughout — the movement is inextensional, and it costs only bending.

So the arch has one cheap mode and one expensive one, and the cheap one has nothing to do with the direction the load is pointing.

This is the single most useful fact about arch stability and it dictates how arches are braced. A tied arch is braced by holding the crown laterally in the plane of the rib, by making the deck stiff enough to resist a sway of the hangers, or by using two ribs and cross-bracing them. Holding the crown down achieves nothing at all.

A three-pinned arch, rise 12 on span 60A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 225.00, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch.crown hinge — no moment here, by constructionH = 225.0H = 225.0180.0180.0thrust line and axis coincide — the definition of funicular
Fig. 3 The thrust the load generates, which is the input to the eigenvalue rather than the answer to it. H=wL2/8fH = wL^2/8f from a crown free body, and it is the quantity that the geometric stiffness is proportional to — so anything that reduces the thrust raises the buckling load.

There is a best rise, and it is about a third of the span

Two things fight as the rise grows. A taller arch has less thrust — HH goes as 1/f1/f — which reduces the destabilising term. And a taller arch is longer, which increases the length the buckle has to happen over.

The result is a maximum. Swept over rise-to-span ratios from 0.05 to 0.5 on the same rib:

f/Lf/L wcrL3/EIw_{cr}L^3/EI
0.05 15.6
0.11 30.6
0.16 41.5
0.22 47.7
0.275 49.7
0.33 48.7
0.39 45.9
0.50 38.3

The optimum is at 0.275 and the curve around it is very flat: anything from about 0.18 to 0.40 is within a tenth of the best. That flatness is worth as much as the maximum, because it means the rise can be chosen for the road level, the headroom or the look, and stability will not object.

There is a best rise, and it is not the tallest archThe dimensionless buckling load w_cr L³/EI of a two pinned parabolic rib, against its rise divided by its span. A flat arch buckles at almost nothing because the same load generates an enormous thrust in it; a very tall one buckles at less than its best because the rib has become long. The maximum is at f/L = 0.28, where the coefficient reaches 49.7, and the curve is flat enough on either side that anything from about 0.15 to 0.4 is within a tenth of it. At the rise drawn the coefficient is 46.2 and the mode is antisymmetric. Nothing here is read off a table: each point is the smallest eigenvalue of the rib's own stiffness against the geometric stiffness its own thrust produces.0.100.200.300.400.5001020304050rise ÷ spanw_cr L³ ÷ EI49.7 at f/L = 0.28every pointan eigenvaluethe rib drawn
Fig. 4 The whole sweep. Every point on it is the smallest eigenvalue of a rib’s own stiffness against the geometric stiffness its own thrust generates, computed rather than read off a table — and the shape of the curve is the competition between a falling thrust and a rising length.

What the supports are worth

Restraining the ends against rotation roughly doubles the answer: the same rib fixed at both springings gives 103 instead of 46. That is the arch’s version of the factor of four between a pinned column and a fixed one, and it arrives for the same reason — the buckled shape is forced into a shorter half-wave.

A three-pinned arch, on the other hand, gives 46.2. Adding a hinge at the crown changes the in-plane buckling load by four parts in a thousand, which is a genuinely surprising result and has a clean explanation: the antisymmetric mode has its point of contraflexure at the crown anyway, so the hinge is being inserted at a section that was carrying no moment in that mode.

Which means the three-pinned arch, so much easier to analyse and so much kinder about settlement, gives away nothing in stability. It gives away a great deal in the symmetric mode, and the symmetric mode was not governing.

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. 5 The end conditions, for a straight column. The arch’s version has the same character and different numbers, and the crown hinge is the case with no counterpart there: a release at a section that the governing mode was not using.

When it actually governs

Being a stability problem does not make it the governing problem. Put real numbers in.

A 60 m rib at a 12 m rise, in a steel box giving EI=6.3×106EI = 6.3 \times 10^6 kNm² and an area of 60,000 mm², buckles at wcr=1,346w_{cr} = 1{,}346 kN/m and squashes at 451. Strength governs by a factor of three, and the eigenvalue is of no interest at all.

Take the same span and rise down to a much lighter rib — EI=2.5×105EI = 2.5\times10^5 kNm², area 12,000 mm², radius of gyration 316 mm — and buckling arrives at 54 kN/m against a squash load of 90. Now stability governs, by a factor of 1.7.

The switch happens at a radius of gyration somewhere around 400 mm on this span, which is a slenderness of about 165 measured along the arc. That is a useful way to hold the whole subject: an arch rib is a column of length equal to about half the arc, and the ordinary column question applies to it.

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. 6 What slenderness costs, on the axis this whole comparison sits on. An arch rib enters that graph at a slenderness computed from its arc rather than its span, which is why a shallow arch is a much worse column than its span suggests.

The tie, the deck and the hangers

Most arches built now are tied: the horizontal thrust is taken by a tension member along the springing line rather than by the ground, and a deck hangs from the rib rather than sitting on it. Every part of that arrangement has an opinion about the antisymmetric mode.

The tie does almost nothing for it. An antisymmetric mode moves the two springings hardly at all — the crown slides, the springings stay — so the tie is barely strained and its stiffness barely enters.

The deck does a great deal, if it is connected to the rib in a way that lets it. An antisymmetric sway of the rib requires the hangers to lean, and a deck stiff in its own plane resists that lean by acting as a beam. That is the usual bracing mechanism for a tied arch and it is a stiffness rather than a strength requirement.

The hangers decide whether the deck’s stiffness is available at all. Vertical hangers pin-connected at both ends transmit only their own axial force, so a rib swaying sideways drags them along and the deck resists only through the small angle change. Inclined hangers form a truss with the rib and the deck, and the sway becomes a shear deformation of that truss instead — which is very much stiffer.

How much of a point load the cable ends up takingThe fraction of a mid-span point load that reaches the cable, against μ = L√(H/EI) — how many characteristic lengths of girder fit in the span. A stiff girder gives a small μ and takes most of the load itself; a limp one gives a large μ and hands nearly all of it over. At μ = 9.9 the cable has 99% of it. What the curve does not show, and the shapes view does, is that the girder's real job is not on this axis at all: even where it carries almost nothing it is still the thing that turns a kink into a curve.0204060800%20%40%60%80%100%μ = L√(H/EI)share the cable takesthe whole loadwhat the cable takesleft is a stiff girder,right a limp one
Fig. 7 The deck as the stiffening element, which is the same argument for a suspension bridge. There the deck stiffens a cable against a load distribution it cannot resist by shape; here it stiffens a rib against a buckling mode it cannot resist by axial stiffness. Two different failures, one member fixing both.
The further it deflects, the harder it pulls backTotal load against midspan sag for a 30 m cable of 1000 mm² prestressed to 500 kN, carrying 5 kN/m. The cubic H³ − T₀H² − w²L²EA/24 = 0 was bisected at every point of the curve, so the sag at the full 150 kN is 0.740 m rather than the 1.125 m the flat-cable formula WL/8T₀ gives — the straight dashed line, which is the tangent to this curve at the origin and nothing more. Its slope is the initial stiffness 8T₀/L = 133.3 kN/m; at the marked point the tangent has reached 341.2 kN/m, 2.56 times as stiff, and the horizontal component of the tension has risen from 500 kN to 760 kN. Nothing about the steel changed. The geometry got better at the job.00.20.40.60.811.2020406080100120140160midspan sag (m)total load on the cable (kN)the design load, 150 kNsolved 0.740 m1.125 mtangent here 341.2 kN/mk₀ = 8T₀/L = 133.3 kN/mthe flat-cable law
Fig. 8 And the stiffness that comes from the shape, which is the arch’s own mechanism read the other way up. A cable is stable because its geometry changes to suit the load; an arch is unstable for the same reason, because its geometry can change and the compression pays for the change.

The load that turns with the structure

There is a closed form for one case, and this machinery does not reproduce it — which turned out to be more interesting than agreeing would have been.

For a two-pinned circular arch of half-angle α\alpha under uniform radial pressure, Timoshenko gives

qcrR3EI=(πα)21\frac{q_{cr}R^3}{EI} = \left(\frac{\pi}{\alpha}\right)^2 - 1

The finite-element answer comes out 7.6% higher, at every mesh from 24 segments to 160, so it is not a discretisation error.

The difference is an assumption in the closed form. That expression is derived for hydrostatic pressure — a load that stays perpendicular to the arch as the arch moves, the way water pressure would. What is applied here is a dead load that keeps its direction.

A load that turns with the structure it is destabilising is worth more to the buckle than one that does not, and 7.6% is the price. It is a reminder worth carrying past this page: a critical load is a property of the load’s behaviour as well as of the structure’s, and two loads with the same magnitude and the same initial direction can give different eigenvalues.

Three paths out of the same critical loadLoad 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.stable symmetric — a columnan imperfection is a nuisancecritical89%unstable symmetric — a shellan imperfection is a demolitioncritical77%asymmetric — a frameand it matters which waycritical
Fig. 9 What happens after the eigenvalue. An arch is an imperfection-sensitive structure of the worst kind — its post-buckling path falls away — so the computed critical load is an upper bound that a real rib with a real out-of-straightness never reaches.

The second-order form, which is what a designer actually does

Nobody computes an arch’s eigenvalue on a Tuesday afternoon. What is done instead is a second-order analysis: apply the load to a rib that has been given an initial out-of-straightness in the shape of the governing mode, solve including the geometric stiffness, and check the stresses that come out.

That route needs the eigenvalue anyway — the imperfection has to be shaped like the mode, and the amplification factor 1/(1w/wcr)1/(1 - w/w_{cr}) needs wcrw_{cr} — but it produces a stress rather than a load factor, which is what a member check wants.

The relationship between the two is the one every second-order argument on this site uses. A rib at half its critical load has its imperfection doubled; at three quarters, quadrupled; and the moment that follows is the axial force times the amplified offset, which is a two-force member’s moment written for a curved member.

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×first-order analysis says the answer is always 1×one over one minus the ratio
Fig. 10 The amplification, which is where the eigenvalue actually gets used. The critical load is rarely the answer to anything on its own; it is the denominator in the factor that turns an imperfection into a moment.

Where the model stops

Everything here is in the plane of the arch. A real rib also buckles sideways out of it, which is a lateral-torsional problem with the rib’s own torsional stiffness in it, and for a rib with no lateral bracing that mode is very often the lower one. Nothing on this page can see it.

The load is uniform and stays uniform. A load over half the span produces a thrust line that leaves the arch axis, and the arch then has a real bending moment before it buckles at all — which turns a bifurcation problem into a second-order stress problem where the two effects multiply.

And the rib is elastic. At the slendernesses where buckling governs it very nearly is, but the transition region — where the critical stress is near the yield stress — has the same character as the inelastic column and the same reduction.

What the pictures cannot show

The modes are drawn at a huge exaggeration, and their amplitude is meaningless: an eigenvector has no scale. What the drawing shows is a shape, and what the number beside it gives is a load — and the two together say nothing whatever about how far the arch actually moves.

Nor can they show the imperfection. A real rib is not on the axis it was drawn on, so its response to load is a growing deflection rather than a sudden bifurcation, and the eigenvalue is an asymptote it approaches without reaching.

The assumption the figure rests on

The thrust is taken from a first-order analysis and then held constant while the eigenvalue is found. That is the standard linearised-buckling assumption and it is the one every number here rests on. A real arch’s thrust changes as it deflects — the crown drops, the rise falls, the thrust rises — so the destabilising term grows with the load faster than linearly, and the true critical load is below the eigenvalue for that reason as well as for the imperfection one.

The history, and why the tables have so many numbers in them

Arch buckling coefficients occupy a great deal of space in older handbooks: tables of a dimensionless factor against rise-to-span ratio, end condition, load pattern and sometimes the ratio of rib stiffness to deck stiffness, running to several pages.

That is what the eigenvalue on this page replaces, and it is worth being clear about why the tables were so large. Every one of those parameters changes KK or KgK_g, and before the matrices could be assembled and solved in a second, each combination had to be solved once — by series solutions of the governing differential equation, by energy methods with assumed shapes, or by measurement — and then written down.

The tables are still right. What has changed is that the two numbers a designer needs, the coefficient and the mode shape, now come from the same arithmetic that produced the thrust, on the actual rib, with the actual loading, rather than from the nearest row of somebody else’s problem.

Four guesses at one buckling modeA pin-ended column, with four assumed shapes and the load each of them gives. The reference is a ten-term Ritz expansion solved as an eigenvalue problem, at 9.8696 EI/L² — which is π², as it must be. a half sine gives 9.870, its own sag shape gives 9.882, a mid-span sag gives 10.000, a parabola gives 12.000. Every one of them is high and none of them is low, because an assumed shape is a constraint on the column and a constraint can only stiffen it.a half sine9.870 EI/L²exactits own sag shape9.882 EI/L²0.13% higha mid-span sag10.000 EI/L²1.32% higha parabola12.000 EI/L²21.59% highreference9.8696 EI/L²ten Ritz terms,as an eigenvalue problemevery guess is anupper boundP
Fig. 11 The energy method the tables were built with: assume a shape, compute the load at which its strain energy is exhausted, and get an upper bound. It is how nearly every coefficient in this subject was first obtained, and it is right to within a per cent or two whenever the assumed shape is nearly the real one.

The ladder from here

Later rungs on this anchor: out-of-plane buckling of a rib, where torsional stiffness enters and a tied arch’s hangers are the only lateral restraint there is. Asymmetric loading, and the second-order arch that never bifurcates because it was bending from the start. Snap-through of a shallow arch, which is the other stability mode an arch has and the one that governs below about a tenth of the span. The tied arch as a system, where the tie’s stiffness and the hangers’ arrangement enter the eigenvalue. And the effective-length approach used in practice, which packages all of the above into a length factor and a column curve, and what it is quietly assuming when it does.

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.

Antisymmetric modeArchBracingBucklingCritical loadEffective lengthEigenvalueGeometric stiffnessLine of thrustRise to spanSlendernessSnap throughThrust