The arch that turns aside before it snaps

The roof that jumps is a shallow frame of two pinned bars pressed at its apex. As it is pushed down the bars shorten, push back harder, and then — once the apex has come down far enough that the bars point more across than up — push back less. The load it can carry rises to a maximum and falls, and under a load that is only ever increased, the frame jumps. That maximum is a limit point: there is only ever one path, and it turns over.

The frame could only go one way because its bars could only stretch. Push its apex sideways and one bar lengthens while the other shortens by the same amount, which the bars resist with their full axial stiffness, so the frame has no cheap sideways mode to find. The essay ended by noting that an asymmetric imperfection “introduces a bifurcation into what was a pure limit-point problem.” That is true of a structure that has somewhere sideways to go. An arch that bends does.

Two shapes and one shortening #

The arch here is the simplest one that can show both behaviours: a shallow rib, pinned at both ends to supports that cannot move apart, its initial shape half a sine wave of rise $e$ over the span $L$, carrying a load spread in the same half sine. It is measured in its own radius of gyration $r$, because, as the arithmetic will show, the rise matters only in those units.

Two shapes carry everything. The arch can come down symmetrically, in the half sine it was made in, by $a_1$; and it can sway, in a full sine wave whose one half drops while the other lifts, by $a_2$. Bending costs energy in each separately, the full wave four times the curvature of the half and so sixteen times the energy for the same amplitude. Axial strain couples them: with the supports held, the rib’s length is fixed, and whatever the shape does to its arc length is taken up as compression. In units of $r$, with $u_1 = a_1/r$, $u_2 = a_2/r$ and $\hat e = e/r$, the potential energy is

$$\Pi = 8\left(u_1^2 + 16u_2^2\right) + S^2 - p\,u_1, \qquad S = (\hat e - u_1)^2 - \hat e^2 + 4u_2^2$$

where $S$ measures the shortening — negative is compression — and $p = 16qL^4/\pi^4 EIr$ is the load. That is the whole model. The rise appears only as $\hat e$, which is why a long shallow arch and a short one behave the same at the same rise in radii of gyration, the observation the arch that gets shorter made about thrust lost to the rib’s own strain.

The figure is an arch with a rise of six radii of gyration. Pressed symmetrically it follows the dashed path, which rises to a maximum at 374.7 and turns over, exactly as the bar frame did. But before it gets there, at 310.7, a second path leaves the first: the sideways one. Along it the arch sways, one half going down faster than the other, and the load it can carry falls steadily. The arch never reaches the top of its symmetric path: it turns aside 17 per cent below it.

Where the sideways path leaves #

Setting the derivative of the energy with respect to $u_2$ to zero gives

$$u_2\,(16 + S) = 0.$$

Either the arch does not sway, $u_2 = 0$, or the shortening has reached $S = -16$ — and that second condition is the bifurcation. It is a statement about the axial force alone. Translated back into dimensions, $S = -16$ is a thrust of exactly $4\pi^2EI/L^2$: the load at which a straight pinned strut of the same span buckles in two half-waves. The arch sways when its thrust reaches the second buckling load of the straight member it is made of, because the sideways shape is that strut’s second mode, and the arch’s shallowness makes it almost straight as far as that mode is concerned.

The symmetric path reaches that thrust only if it can. The shortening along it is $S = (\hat e - u_1)^2 - \hat e^2$, which is at most $-\hat e^2$, so the thrust can reach the bifurcation only when $\hat e > 4$. Its limit point, where the symmetric load stops rising, lies at $(\hat e - u_1)^2 = (\hat e^2 - 4)/3$ and exists only for $\hat e > 2$. The bifurcation lies at $(\hat e - u_1)^2 = \hat e^2 - 16$. Which comes first is a comparison of the two:

$$\hat e^2 - 16 > \frac{\hat e^2 - 4}{3} \quad\Longleftrightarrow\quad \hat e > \sqrt{22} = 4.69.$$

So there are three kinds of shallow arch, sorted by one number. Below a rise of two radii of gyration the arch has no limit point at all: it is so flat that its thrust never builds enough to turn the load over, and it simply stiffens into a sagging membrane. Between 2 and $\sqrt{22}$ it snaps through symmetrically, the bar frame’s behaviour, at a limit point that the sideways path cannot reach first. Above $\sqrt{22}$ it turns aside, at a load that falls further below the limit point the higher the arch: 17 per cent below at a rise of 6, 45 per cent below at 8. An analysis that holds the arch symmetric — and a single-mode analysis does, without saying so — overstates those arches by that much.

The higher arch fails earlier than its symmetric path because it has more rise to spend. Swaying lets one half come down a long way while the other goes up, so the rib’s length is used up less than if both halves came down together; the arch finds a cheaper way to shorten its span. A low arch has too little rise for that trade to pay before its thrust turns the load over anyway.

One goes down, one goes sideways #

The two failures look different at the moment they happen, and the difference is the one an engineer sees on site.

The lower arch flattens almost symmetrically. Its small imperfection has grown a little — to 0.24 radii of gyration of sway — but it reaches its maximum still essentially in its own shape, and past the maximum it snaps through to a sagging shape below the supports. The higher arch, built with the same imperfection, has swayed three times as far by its peak, with one half visibly higher than the other. Past the peak it does not go down symmetrically at all: one half drops, the other follows, and the arch ends upside down through a lopsided motion.

That lopsided motion is why the second kind of failure is the more dangerous one in practice. The sway grows from whatever asymmetry the arch already had, and a real arch always has some: an erection tolerance, a support that settled, a load that was not quite even.

A small asymmetry, a large loss #

Build the arch with a small sideways imperfection, $\delta$ radii of gyration of the full-wave shape, and the bifurcation becomes a smooth path that peaks below the perfect arch’s failure load. How far below is the whole question of imperfection sensitivity, and the answer depends on which side of $\sqrt{22}$ the arch is.

Above the crossover the loss follows a straight line on logarithmic scales with a slope of two-thirds. That is Koiter’s result for an unstable symmetric bifurcation — the same law a cylinder under axial load obeys, with its far more dramatic knockdowns — and it means small imperfections are expensive: a hundredth of a radius of gyration costs 2.9 per cent of the load, a tenth costs 12 per cent. A two-thirds power is steep near zero. Halving an imperfection that is already small saves only 37 per cent of what it cost.

Below the crossover the same imperfection barely registers. The sideways shape is not on the arch’s path to failure; the imperfection can only reach the symmetric limit point through the square of the sway it causes, so the loss goes as $\delta^2$: 0.05 per cent at a hundredth of a radius of gyration, less than a fiftieth of what the higher arch loses. Imperfection sensitivity is not a property of arches. It is a property of arches above $\sqrt{22}$, and an arch can be made insensitive by being made flatter or stiffer in bending.

The picture across rises makes the switch visible. To the left of $\sqrt{22}$ every curve is close to one: an arch that snaps symmetrically keeps nearly all its strength whatever its sideways imperfection. To the right every curve drops — to 0.88 for a tenth of a radius of gyration, 0.75 for three tenths — and stays down. The curves recover slightly at large rises because the perfect bifurcation load itself has fallen so far below the limit point that there is less for an imperfection to take away; the arch is already paying for its rise.

The sway it carries at the peak #

An imperfect arch does not bifurcate; it sways from the start, slightly, and the sway grows as the load approaches the perfect arch’s bifurcation, as a column that was never straight bows more and more as its load approaches the Euler load.

At the peak the sway is far larger than the imperfection that started it. An arch built a hundredth of a radius of gyration out of true has swayed 0.36 radii of gyration — thirty-six times its imperfection — by the time it can carry no more. That growth is the warning, and it is a better one than the symmetric arch gives: a sway that is growing faster than the crown is coming down is visible, measurable and specific to this mode. A symmetric arch near its limit point gives no comparable sign; its deflection is in the shape it already had.

It is also the reason the column analogy is only partly right. A column’s bow grows without limit as the load approaches the Euler load and the load never falls. The arch’s sway grows towards a bifurcation whose own path falls, so the imperfect arch turns over before it reaches it, at a sway that grows only as the cube root of its imperfection — 0.36 for a hundredth, 0.72 for a tenth — and the load lost goes as the square of that sway, which is where the two-thirds power comes from.

The half-model that cannot see it #

The crossover at $\sqrt{22}$ has a practical consequence that has nothing to do with arches in particular: it decides which analysis can find the failure.

A symmetric structure under a symmetric load invites a half-model. Cut the arch at its crown, put a roller there that lets the cut face move down but not sideways or round, and the analysis costs half as much and returns the symmetric path exactly. For an arch below $\sqrt{22}$ that is all there is, and the half-model’s maximum is the answer. For an arch above $\sqrt{22}$ the half-model has removed the failure. The sideways shape needs the crown to rotate and the two halves to move in opposite directions, and the roller at the cut forbids both. A half-model of a symmetric arch reports the limit point, and above a rise of $\sqrt{22}$ the limit point is a load the arch never reaches.

An eigenvalue analysis fails in the opposite direction. A linear buckling calculation on the undeformed arch looks for the load at which some shape costs nothing to add, and it finds the sideways mode readily, because that shape is the second mode of a strut whose length its ends decide. But it cannot find a limit point at all — as the bar frame showed, there is no eigenvalue at the top of a path — so below $\sqrt{22}$ it reports a sideways mode at a load the arch never reaches either, since the arch has already snapped. Each method is right on one side of the crossover and wrong on the other, and neither says which side it is on.

The analysis that is right on both sides follows the whole path with the sideways freedom left in and a small sideways imperfection put in to start it. That is the imperfect path in the figures here, and it is also the honest way to choose the size of imperfection: not as a fraction of the span drawn from a table, but as the asymmetry the arch will actually be built with. Guessing the shape makes the same point from the energy side: a buckling load is only as good as the family of shapes the calculation was allowed to try, and a symmetric family cannot find a sideways answer.

One rib, in metres #

The model has no material or size in it beyond $EI$ and $r$, so a dimensional example is a check on whether these rises are ones that occur.

Take a steel rib of 30 m span with a section like an IPE 400: $EI = 48{,}510\ \text{kN·m}^2$, radius of gyration 165 mm. A rise of six radii of gyration is 0.99 m — a thirtieth of the span, a shallow roof rib or a curved floor beam rather than an arch in the ordinary sense. Its sideways bifurcation at $p$ = 310.7 is a load of

$$q = \frac{p\,\pi^4 EI\,r}{16L^4} = \frac{310.7 \times 97.41 \times 48{,}510 \times 0.165}{16 \times 30^4} = 18.7\ \text{kN/m},$$

and the thrust there is $4\pi^2EI/L^2$ = 2,130 kN, a stress of 251 N/mm² on the section’s 8,450 mm² — inside the yield of S355. With an imperfection of a tenth of a radius of gyration, 16.5 mm on 30 m or one part in 1,800, the rib carries 16.4 kN/m: twelve per cent less, from an imperfection inside any erection tolerance. The same rib on 20 m would reach its bifurcation thrust only at 563 N/mm², past yield, so on the shorter span the material decides first — the crossover between two failures that the bar frame also had.

Two modes, and an elastic rib #

The calculation rests on choices that limit it.

Two shapes are all the arch is allowed. A real arch can deflect in every shape at once, and higher modes lower both critical loads slightly; the two-mode model is exact for the sinusoidal arch under a sinusoidal load because those loads excite nothing else, and an approximation for a parabolic arch under a uniform load, where the thresholds move by a few per cent.

The supports do not move. An abutment that spreads under the thrust releases some of the compression, raises both critical loads and moves the crossover; an arch on supports that give is a different problem in which the thrust never builds as far.

The rib stays elastic. The thrust at the bifurcation is a strut’s second buckling load, and for a stocky or short rib that is beyond yield; the bending stress from the sway adds to it. Where the material yields first, the elastic thresholds here are upper bounds.

The load stays vertical and in the half-sine shape. A load concentrated at the crown excites the sideways mode directly, through no imperfection at all, and a load on one half only pushes the arch sideways from the start; both make the sensitive regime begin at a lower rise.

The load is applied slowly. Past the peak the arch moves dynamically, and on the sideways path it does so lopsidedly, with one half overtaking the other; what it lands on and how hard it hits depends on mass and damping that the static path does not contain. A roof rib that snaps through is usually lost whichever way it went, but a rib that is part of a continuous structure may be caught by its neighbours, and then the motion matters more than the peak.

Still open: a rib that is braced sideways in its own plane #

The sideways mode needs the arch’s quarter points to move in opposite directions, so anything that ties them together — a hanger to the crown, a diagonal across the arch, a deck that the rib is connected to at its quarter points — removes the sideways path and returns the arch to the symmetric limit point, which for a rise of 8 is nearly twice the load. Whether a single member between the quarter points, sized only for the small force it carries until the arch begins to sway, can recover that whole difference — and how stiff it must be before the arch stops finding its way round it — is the question of bracing a mode rather than a member, and it belongs to the arch that was meant to be shallow.