The arch that leans instead of squashing

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 $EI/L^3$ rather than $EI/L^2$. And the rib has two competing modes that are not the same shape.

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 $H/\cos\theta$ — the horizontal thrust divided by the cosine of the local slope — and $H = wL^2/8f$ from the crown free body.

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

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

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 $\pi^2EI/L^2$, $\pi^2EI/4L^2$ and $20.19EI/L^2$ for a straight column to seven figures, which is what makes the arch numbers worth quoting.

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.

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 — $H$ goes as $1/f$ — 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:

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.

The far end of that curve is the part intuition gets wrong, so it is worth drawing the arch that sits there rather than reading its coefficient off the sweep. A semicircular rib on the same span has a rise of 30 m, a rise-to-span of a half, and a thrust barely a third of the shallow arch’s:

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.

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 $EI = 6.3 \times 10^6$ kNm² and an area of 60,000 mm², buckles at $w_{cr} = 1{,}346$ 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 — $EI = 2.5\times10^5$ 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.

Which is also the answer to where the rib belongs on the ordinary column curve. What slenderness costs is measured against a length, and the length an arch rib brings to that graph is computed from its arc rather than from its span — which is why a shallow arch is a much worse column than its span suggests, and why the switch above lands at a radius of gyration a designer would have thought generous.

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.

That the deck earns its keep this way is the same argument as the deck is not there to carry the load on 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 — and both are versions of the stiffness that comes from the shape read in opposite directions, because a cable is stable for exactly the reason an arch is not: its geometry changes to suit the load, and the arch’s compression pays for the same change.

What the tie itself does is a separate calculation, and it is a statics one rather than a stability one:

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 $\alpha$ under uniform radial pressure, Timoshenko gives

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

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.

What happens after the eigenvalue is the other half of that warning, and an arch is a third of what the theory promised in the same sense a cylindrical shell is. Its post-buckling path falls away rather than rising, so the computed critical load is an upper bound that a real rib with a real out-of-straightness never reaches, and the shortfall grows with the imperfection rather than saturating.

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 $1/(1 - w/w_{cr})$ needs $w_{cr}$ — 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.

So 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, and a rib at 60 per cent of it has a crown offset two and a half times the one it was drawn with.

The rib’s bending comes from things that are not loads #

An arch shaped to its load carries that load as pure thrust: the funicular shape has no bending moment in it at all. Which raises the question of what the rib’s bending stiffness is for — and the answer is that it is there for the actions that are not the design load, and they are the ones nobody draws.

Temperature. An arch is held between abutments, so a temperature rise cannot lengthen it and produces a thrust instead. For a two-hinged parabolic rib the unit-load method gives

$$H_t = \frac{15\,EI\,\alpha\,\Delta T}{8f^2}$$

which is worth reading twice. There is no span in it. And it goes as the inverse square of the rise, so a flat arch develops far more thermal thrust than a deep one — for a 60 m concrete rib at 3 m rise and a 30 degree change, of the order of 2,000 kN, against a load thrust of perhaps 30,000.

That thrust acts on the arch’s own eccentricity and produces a moment $H_t\,y$, which at the crown is $H_t f$. On the same rib that is around 6,000 kNm — and since the load’s own moment is nominally zero, the thermal moment is essentially the whole of the rib’s bending.

Shrinkage does the same thing with the same expression, replacing $\alpha\Delta T$ by a shrinkage strain of 300 to 500 microstrain — which is larger than a 30 degree temperature change and permanent.

And a settling abutment does it in reverse. An abutment that spreads by a few millimetres relieves the thrust, which sounds helpful and is: it is the arch shedding a self-stress, and it is why a masonry arch on soft ground cracks into three hinges and then stands quite happily.

The reading to take from that is the same one the thermal thrust gives: the arch’s bending is a residue of quantities that nearly cancel, so it is the difference between two large numbers rather than a fraction of a small one.

There is a consequence for the stability argument that is exact and slightly deflating. The thermal thrust is proportional to $EI$ — and so is the buckling load. So the fraction of the critical load consumed by a temperature change,

$$\frac{H_t}{P_{cr}} \;\propto\; \alpha\,\Delta T\left(\frac{L}{f}\right)^2$$

contains no rib stiffness whatever. Stiffening the rib raises the critical load and raises the thermal thrust by exactly the same factor, and the margin between them is unchanged. The only levers on that ratio are the rise-to-span proportion and the temperature range — one of which is architecture and the other weather.

Which is the general shape of the thing and it recurs throughout this collection. An imposed deformation generates a force proportional to the stiffness resisting it, so stiffening a structure against a self-strain buys nothing at all. The arch is the case where that is most visible, because the load it was shaped for contributes no bending and everything left over came from somewhere else.

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 $K$ or $K_g$, 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.

The energy method those tables were built with is worth naming, because it is the one that can still be done on paper: guess the shape, compute the load at which its strain energy is exhausted, and take the result as 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 — which, for an arch, means guessing the antisymmetric mode and not the symmetric 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.