An air arch pays for its thrust out of its pressure

A straight inflated tube is a beam for as long as its pressure’s pull along the axis outweighs the push a bending moment puts on the inside of the bend. An inflated beam wrinkles at a moment with no cloth in it found that moment exactly — πpr³/2, seventy-nine newton metres for a tube a fifth of a metre across at half a bar — and found no property of the cloth in it.

It ended on the case that most air beams actually are. A shelter’s frame, an inflatable boat’s bow, a play tunnel: the tube is curved before anything loads it, and the essay expected the curve to matter from the start. “A curved tube under pressure is not uniformly pre-tensioned,” it said: the outside of the bend has more wall than the inside, so an arch should begin with its two faces pulled differently, and its first wrinkle should form where that difference and the load’s moment combine worst.

The first half of that expectation is wrong, and the reason it is wrong moves the question somewhere more interesting.

A curved tube is a piece of a torus #

Bend a tube round a curve of radius $R$ and its wall is a piece of a doughnut. The membrane forces in a pressurised torus are a textbook result, and they are simple. Measure the angle $\varphi$ round the tube’s section from the outside of the bend, so a point of the wall sits at a distance $\rho = R + r\cos\varphi$ from the centre of the curve. Then the pull per unit length is

$$n_{\text{along}} = \frac{pr}{2}, \qquad n_{\text{around}} = \frac{pr}{2}\cdot\frac{2R + r\cos\varphi}{R + r\cos\varphi}.$$

The pull along the tube is the straight tube’s pr/2, at every point of the section, on every bend. It does not vary from the inside of the curve to the outside at all. Everything the curve changes is in the other direction: the pull around the tube is larger on the inside of the bend and smaller on the outside.

Both forces together satisfy Laplace’s equation for a curved membrane, $n_{\text{around}}/r + n_{\text{along}}\cos\varphi/\rho = p$, at every point, and the uniform along-pull is what makes that possible. The inside of the bend has less wall, but it also has the smaller radius $\rho$, and the geometry makes up the difference in the hoop direction rather than along the tube.

So the arch starts even #

The lead the earlier essay followed was the right instinct applied to the wrong direction. It reasoned from a tube of one size pressing the calf harder than the ankle, where a circumference that varies along a limb does change the force the cloth carries. But that is the hoop direction, where the torus does vary. The direction a wrinkle is about — the pull along the tube, which a bending moment has to cancel — is the one direction in which a curved tube is exactly as even as a straight one.

An air arch therefore begins, before any load, with the straight tube’s pre-tension everywhere. Its first wrinkle is not placed by any unevenness in that pre-tension, because there is none. If the first wrinkle forms somewhere other than where the moment peaks, it has to be for another reason.

Where the curve does enter #

The curve enters through bending, and it enters small.

A curved bar does not carry a moment with the straight-line stress distribution a straight one does. Plane sections still stay plane, but the fibres on the inside of the curve are shorter to begin with, so the same rotation strains them more. Winkler’s curved-bar theory gives the wall force from a moment as proportional to $(\rho - \rho_n)/\rho$ rather than to the distance from the axis, where for a thin ring section $\rho_n = \sqrt{R^2 - r^2}$ is the radius at which the moment’s force passes through zero.

The result is a factor on the wrinkling moment that depends on which face the moment compresses. A moment that compresses the inside of the bend meets the pre-tension there sooner than the straight tube’s πpr³/2; one that compresses the outside meets it later. Integrated round the section, the factors are 0.975 and 1.025 on a bend of twenty tube radii, 0.899 and 1.101 on a bend of five, and 0.828 and 1.172 on a bend of three.

The curve costs about r over 2R, twice #

The two effects of the curve have the same size and it is worth seeing why. Both come from the inside of the bend being shorter than the outside by a share $r/R$ of its length, and each takes half of that share.

On the hoop, the inside’s pull rises to $(2R - r)/(2(R - r))$ times pr: 2.6 per cent more on a shelter arch of twenty tube radii, 12.5 per cent on a bend of five. It is the same geometry that makes a band press where the limb turns: a tension carried round a tighter curve has to be larger to hold the same pressure, and on a bent tube the tighter curve is the inside of the bend. That matters for how hard the tube can be pumped. The essay on the straight beam found the cloth entering the strength only through the pressure it will hold, its hoop strength over its radius; on a bend the inside reaches that strength first, so a tight bend holds a little less pressure and has a little less budget to spend.

On the wrinkling moment, the same share appears through Winkler’s factor, and its sign depends on the load. For a gentle arch both effects are a couple of per cent. The curve is a correction, not a cause.

The arch spends its pressure twice #

What an arch has that a beam does not is thrust. A beam carries its load in bending alone; an arch pushes along its own length, into its feet, and the push is most of how it carries anything. That is why masonry arches stand and why a stone beam of the same span does not.

An air arch cannot push, because a cloth cannot carry a push. What it can do is let the thrust take some of the pressure’s pull away, exactly as the bending moment does. At a section carrying a moment $M$ and a thrust $N$, the most compressed fibre’s pull along the tube is what the pressure gave it less what each of them takes:

$$\frac{pr}{2} - \frac{N}{2\pi r} - k\,\frac{|M|}{\pi r^2},$$

with $k$ the curved-bar factor for the face the moment compresses. It wrinkles when that reaches nothing, which is

$$k\,|M| + \frac{N r}{2} = \frac{\pi p r^3}{2}.$$

The straight beam’s whole budget, πpr³/2, is now shared between the moment and the thrust. The thrust spends it at a rate of half a tube radius per newton, and the first wrinkle forms where the two together peak — which need not be where the moment does.

Where a shelter arch wrinkles under snow #

Take the arch in the opening figure: a semicircle four metres across, pinned at both feet, of a 200-millimetre tube at 50 kilopascals, so that the straight tube’s budget is 78.5 newton metres and the pressure pulls along the wall with a total of 1,571 newtons. Below wrinkling its bending stiffness is the same everywhere, so the one unknown the pins leave — the horizontal thrust at the feet — follows from virtual work, and for snow lying evenly over the span it is the classical $4wa/3\pi$.

The first wrinkle forms 25° above a foot, at 167 newtons per metre of span — 669 newtons of snow in all. There the moment is 60 newton metres and compresses the inside of the curve, and the thrust is 336 newtons, which has spent 21 per cent of the budget. Judged by its moment alone, the arch would have carried 218 newtons per metre; the thrust costs it nearly a quarter of that.

The curve’s own share is small, as the last section promised. With the curved-bar factor switched off the arch wrinkles at 171 newtons per metre rather than 167: the moment at the haunch compresses the inside of the bend, which is the face the curve weakens, by 2.5 per cent.

Half the snow wrinkles the other side #

Snow drifts, and the load that governs a real arch is often snow on one side only.

The first wrinkle forms on the unloaded side, 40° above its foot, at 223 newtons per metre of the loaded half. It is the classical behaviour of a two-hinged arch under an unbalanced load, and it is worth saying plainly because it is where nobody looks: the loaded side sags and the unloaded side is pushed outward and up, and the larger moment is on the side with nothing on it. The thrust there is half the full-snow thrust and takes 9 per cent of the budget.

Wind wrinkles the windward side, and pulls #

Wind on a semicircle is a sideways load, and a crude one is enough to see what it does: a horizontal push spread over the windward quarter.

The first wrinkle forms 34° above the windward foot, at 105 newtons per metre of height, where the moment compresses the outside of the curve. And here the axial force has the other sign: the windward side is being pulled rather than pushed, so the tension adds to the pressure’s pull and lends the section 4 per cent of budget. The arch carries 105 where its moment alone would have allowed 98.5.

A tight arch spends most of its pressure on thrust #

The thrust’s share depends on how tight the arch is, and not a little.

The reason is scale. The moment in an arch grows as the load times the span squared; the thrust grows as the load times the span; and the thrust’s cost is the thrust times half a tube radius. So the thrust’s share of the budget goes roughly as the tube radius over the arch radius. A shelter arch of twenty tube radii spends a fifth of its pressure on thrust under snow, an arch of five spends half, and one of three spends three fifths. An inflatable boat’s bow, a pool toy, a small play arch are all tight in these terms, and for them the straight beam’s rule — all of the budget on the moment — overstates what they will hold by a factor that approaches two.

On a tight arch the first wrinkle moves #

And on a tight enough arch, the first wrinkle leaves the place the moment chooses. Where the thrust’s share is large, the section with the greatest thrust — near a foot — overtakes the section with the greatest moment.

Under snow on one half, the first wrinkle jumps from the unloaded side to 14° above the loaded foot once the arch is tighter than 3.8 tube radii. Under a load at the crown it leaves the crown for 32° above a foot below 3.1 tube radii. Under wind it jumps from the windward side to 49° above the leeward foot below 7.4 tube radii — and there the curved-bar factor is part of the cause, because the leeward section’s moment compresses the inside of the curve, the face the curve weakens, while the windward one’s compresses the outside.

That is the place the earlier essay’s question has a yes for an answer. The first wrinkle does form somewhere other than where the load’s moment is largest. But the cause is the arch’s thrust, not an uneven pre-tension, and for an arch of shelter proportions it does not happen at all.

What a funicular air arch would do #

The shape that carries a load by thrust alone is the load’s funicular: for snow spread evenly over the span it is a parabola, and a parabolic arch under that snow carries no moment anywhere. For masonry it is the ideal. For an air arch it is a different kind of failure.

With no moment, the whole budget goes on thrust, and the thrust is largest at the feet. A parabola four metres across and two tall under snow thrusts horizontally at $wL^2/8f$, equal to $w$ in newtons per metre, and pushes along its tube at the feet with $\sqrt{5}\,w$. The feet go slack when $\sqrt{5}\,w \cdot r/2$ reaches 78.5 newton metres: at 702 newtons per metre, four times the semicircle’s 167. But a tube whose pull has gone to nothing all round its section at once has not wrinkled on one side; it has lost its bending stiffness entirely at that point, and an arch with a hinge-like foot and a thrust on it buckles. The funicular trades the semicircle’s early, local, survivable wrinkle for a late and sudden one.

The model named #

The wall is the torus membrane solution, with the pull along the tube $pr/2$ and the pull around it varying as stated, checked against Laplace’s equation at 24 points round the section. The moment is carried by Winkler’s curved bar for a thin ring section, with the neutral radius $\sqrt{R^2 - r^2}$ and the factors integrated round the section at 3,600 points; plane sections stay plane and the section stays round. The arch is a two-hinged semicircle with uniform bending stiffness, its thrust found by virtual work from bending alone, and checked against the two classical values: $4wa/3\pi$ for a load per metre of span and $P/\pi$ for a load at the crown. The wrinkle is the first section where $k|M| + Nr/2$ reaches $\pi p r^3/2$.

The tube and the pressure are the straight-beam essay’s: 200 millimetres across at 50 kilopascals. The loads are idealisations — snow spread evenly, snow on one half, one point at the crown, and a horizontal push over the windward quarter for the wind, which is the crudest of the four, since real wind on a curved roof is a suction over most of it.

What was counted #

Four loads on the four-metre arch, each at 720 sections, and each at 600 sections on 49 arches from three to sixty tube radii; the section with the largest spent budget was taken as the first wrinkle, with twin peaks of a symmetric load resolved to the first found. The jumps quoted are where that section moves by more than 8° between neighbouring arches in a finer sweep at steps of a twentieth of a tube radius.

What the picture cannot show #

The section stays round. A bent tube’s section flattens — for a thin curved tube the flattening is large and changes its stiffness several-fold — and the pressure is what resists it. The straight-beam essay’s folding moment already assumed a round section; here the assumption also carries the curved-bar factor, which for a flattening section is not the thin-ring one. The factor’s size, about r/2R, is the part to trust; its third figure is not.

Nothing buckles. A two-hinged semicircle under a spread load buckles in its own plane at about $3EI/R^3$, which for this tube and an illustrative cloth of one meganewton per metre is some seven times the snow load that wrinkles it; a stiffer cloth moves it further off and a much softer one could bring it close. And the arch is assumed not to move out of its own plane at all, which is the first thing a real shelter’s guy-lines are for.

The feet are pins. A clamped foot moves the thrust and the moments, and a real anchor is neither. The cloth is a free membrane. Almost every air beam is coated, and coated is a state: a film bonds the crossings, so the wall resists shear and some compression, and the slack side of a coated tube does not go slack as cleanly as the argument assumes. Its budget is a little larger than πpr³/2 and its wrinkle a little less sudden. The pressure does not change as the arch deflects, and the cloth enters only through the hoop pull the inside of the bend must hold, which is where a membrane is cut smaller than it is begins to matter: an arch patterned from flat cloth is not exactly the torus its pressure wants, and a woven cloth’s bias lets it shear towards one.

Who found it, and when #

The torus’s membrane forces are nineteenth-century shell theory, in every text on pressure vessels; Winkler’s curved bar is from 1858; the two-hinged arch’s thrust by virtual work is the standard nineteenth-century method. The thrust entering an inflated beam’s wrinkling condition as an axial force is how the engineering literature on air beams has treated compression since inflatable structures were first analysed.

What is done here is to put them together for an arch, and to correct a statement this account made: a curved tube is not unevenly pre-tensioned along its length. The consequence — that an air arch’s first wrinkle is placed by its thrust and not by its curve, and that its thrust takes a share of the budget set by the tube radius over the arch radius — follows once that is straight.

Still open: what a stiffened section does #

Every tube here is round and made of one cloth, so it has one budget, pr/2 along the wall, and every load shares it. An inflated cylinder wants an unbalanced cloth found that the hoop direction needs twice what the axial one does, and a woven tube is usually built balanced anyway; a braided one instead settles at the angle a hose wants, where the two directions are served in that same two-to-one proportion by fibre at a single angle.

An arch has a better use for the spare axial fibre than a straight tube does. The inside of the bend is where the hoop pull rises and where, under snow, the moment compresses; a stripe of stiffer or pre-tensioned cloth along the inside of the curve, or a laced strap along it as some shelters carry, adds pull exactly where the budget is spent first. Whether a strap of a given stiffness buys more load on the haunch than the same weight of cloth spread evenly, and whether it moves the first wrinkle somewhere worse, is the question that would turn this arithmetic into a specification.