The surface that carries by being curved

Take a sheet of paper by one edge and hold it out horizontally. It flops. Curve it slightly across its width, between finger and thumb, and it stands out straight and carries its own weight and a good deal more. Nothing has been added — same sheet, same paper, same thickness — and the change was made in a second by a hand.

That trick is not a curiosity at the edge of the subject. It is the largest single efficiency available anywhere in structures — larger than the gain from putting material far from the middle, larger than the gain from depth — and it is available to any surface that can be persuaded to curve. A flat plate carries transverse load by bending: the two faces work in opposite directions across a lever arm of a fraction of the thickness, while the material near the middle does very little. A curved surface carries the same load as direct force in its own surface, uniformly through the thickness, every particle at the same stress. There is no lever arm to be short, because there is no lever.

The claim, stated once #

A shell in membrane action needs no bending stiffness at all, and therefore needs almost no thickness. The forces it carries are per unit width — kilonewtons per metre, not kilonewton-metres per metre — and the thickness required is nothing more than that force divided by an allowable stress. Halve the load and the thickness halves with it.

A plate in bending obeys a different exponent. The section modulus of a strip of unit width is $t^2/6$, so

$$t = \sqrt{\frac{6M}{\sigma}}.$$

The membrane thickness is linear in the load; the bending thickness is a square root of it. That difference in exponent, rather than any constant, is what makes the comparison lopsided — and it means the advantage grows as the load falls, which is the reverse of the intuition that curvature is a trick for heavy structures.

The price of bending, in millimetres #

The comparison is worth doing with everything but the geometry held fixed.

Five millimetres against a hundred and fifty. That factor is not a claim about steel being better in tension than in bending, since the same steel and the same allowable stress appear on both sides; it is a claim about a lever arm that does not have to exist. And the closed form $\sqrt{3\sigma/p}$ says where the trick pays best: at a tenth of the pressure the factor is not 30 but 95.

The flat alternative does not have to be a naive strip, either. A plate can be made to span two ways at once, and that helps — but only up to a point, and the point arrives quickly.

Spanning both ways is a redistribution, and a good one. Curving the same slab is a change of mechanism.

Which free body produced the number #

Everything above rests on two equations, and both come out of a cut that can be described in a sentence.

The first free body is the cap. Slice the dome on a cone of half-angle $\varphi$ with its apex at the centre of the sphere, and lift off everything above the cut. That cap has a surface area of $2\pi R^2(1 - \cos\varphi)$ and therefore a weight of $w$ times it. What holds it up is the meridional force $N_\varphi$ acting all the way round a circle of radius $R\sin\varphi$, directed along the tangent to the meridian — so its vertical component per unit length is $N_\varphi \sin\varphi$. Vertical equilibrium of the cap, and nothing else:

$$N_\varphi \sin\varphi \cdot 2\pi R \sin\varphi = -w \cdot 2\pi R^2 (1 - \cos\varphi)$$

$$N_\varphi = -\frac{wR(1-\cos\varphi)}{\sin^2\varphi} = -\frac{wR}{1 + \cos\varphi}.$$

The cancellation in that last step is why the answer is so tidy. At the crown $N_\varphi = -wR/2$; at the equator $-wR$. Compression everywhere, on any dome, under any uniform surface load.

The second free body is an infinitesimal patch of the surface, and the equation is equilibrium normal to it. A force per unit width running along a curved path pushes inward at a rate of force divided by radius of curvature, so the two membrane forces supply $N_\varphi/R + N_\theta/R$ against the normal component of the load, $w\cos\varphi$. That gives $N_\varphi + N_\theta = -wR\cos\varphi$, and with $N_\varphi$ already known,

$$N_\theta = wR\left(\frac{1}{1+\cos\varphi} - \cos\varphi\right).$$

Two equations, two unknowns, no stiffness, no material property, no compatibility. The membrane state of a dome is statically determinate, which is why it can be written down at all and why these figures can be trusted before any elastic assumption has been made. The free body was a choice, and choosing the cone rather than the wedge is what made the arithmetic collapse.

The angle at which the hoops change their mind #

A dome invites being read as a set of arches leaning against each other round a circle, and that reading gets a great deal right. Each meridian is in compression from crown to base, exactly as an arch is, and its thrust arrives at the base leaning outward and has to be caught.

The reading breaks at one point, and the break is the whole of what a shell is. A dome has hoops and an arch does not. Cut an arch into slices and each slice stands alone; cut a dome along its meridians and the slices fall in or fall out, because the ring forces holding the meridians to their spacing have been severed. Those hoop forces are the second equation above, and they do something an arch has no vocabulary for: they change sign.

Above the parallel where $N_\theta = 0$ the hoops are in compression, squeezing each ring smaller. Below it they are in tension and the rings are being pulled apart. Setting $N_\theta$ to zero and clearing the fraction gives

$$\cos^2\varphi + \cos\varphi - 1 = 0,$$

whose root is $\cos\varphi = (\sqrt5 - 1)/2$ — the reciprocal of the golden ratio, arriving in a statics problem with no proportion, no aesthetics and no rectangle in sight. The angle is $51.827292°$. The solver behind these figures finds it by bisecting the hoop force between the crown and the equator and then checks it against the closed form to nine decimals, on the principle that a result this pretty should be made to survive a test rather than admired.

That parallel is not an abstraction: it is a line that can be walked to on the inside of any large old dome, because it is where the cracks are. Masonry has no tensile strength worth the name, so below 51.8° the hoops cannot exist as drawn; the shell splits along its meridians and stops being a shell at all, becoming a set of independent arch slices, each of which must find a line of thrust inside its own thickness.

This is what the eighteenth century worked out at St Peter’s. The dome had cracked; Le Seur, Jacquier and Boscovich reported in 1743 on a mechanism analysis of the split slices, and Poleni, five years later, hung a chain loaded to represent the weights of a slice and found its inverted shape lay inside the masonry. The remedy built was iron: chains round the base and the haunches, tightened to supply the hoop tension the stone could not. Nobody added stone to the crown, and the reason is the next section.

Thickness does not help #

Under self-weight the surface load is $w = \gamma t$. Put that into the meridional force and divide by the thickness to get a stress:

$$\sigma = \frac{N_\varphi}{t} = \frac{\gamma t R}{t\,(1 + \cos\varphi_0)} = \frac{\gamma R}{1 + \cos\varphi_0}.$$

The thickness cancels. Not approximately, not for slender shells, not to first order — exactly, and for the same reason a column carrying only its own weight has a stress that depends on its height and not on its area.

Three things follow, and the third is the useful one. A masonry dome cannot be made safe by being made heavier. Its stress is set by its radius and its material, so scale is what threatens it and thickness is not. And since the stress is trivially small in any case — half a megapascal in a stone that will take twenty — the governing question about a masonry dome was never its stress and never its thickness. It is whether the hoop tension can be carried and whether a thrust line fits, both of which are questions about geometry rather than material.

The whole price is at the edge #

Nothing so far has been free; the bill has simply not been presented. It is presented at the boundary, in two instalments.

The first is the ring. The meridional compression reaches the base leaning outward, and its horizontal component is a radial line load on whatever is there. A radial outward load $H$ on a ring of radius $r$ puts $Hr$ of tension in it — the free body being half the ring, cut on a diameter, with the radial load integrated round it.

Two searches, on two different functions, arrive at the same angle — and they must, because differentiating the ring force produces $2\cos\varphi\,(1 - \cos\varphi - \cos^2\varphi)$ over a positive denominator, and the bracket is the hoop-force equation with its signs flipped. The angle at which the hoops stop helping is the angle at which the ring is asked for most: a dome cut anywhere near $52°$ demands the maximum from the one component that must be made of something other than masonry.

The second instalment is the reason membrane theory is a first chapter rather than a whole book. A membrane solution generally violates the boundary condition. The membrane state has a definite radial growth at the edge — the shell wants to move out by $(r/Et)(N_\theta - \nu N_\varphi)$ — and a ring beam, a stiff foundation or a continuous adjacent shell does not permit it. Compatibility then requires bending, which membrane theory had assumed away, and the bending comes back in from the edge.

Half a megapascal of edge bending against six-tenths of membrane stress: at the boundary, the bending the theory denied is almost as large as the direct force the theory was built to find. And it is local. Over four and a quarter metres of a thirty-metre shell the whole disturbance dies, which is why the standard treatment is to design the shell for membrane forces, design the edge for bending, and let the two overlap in a strip a few times $\sqrt{Rt}$ wide.

The same equation as a beam sitting on the ground #

Near a restrained edge, a strip of the shell running along a meridian behaves as a beam. Its flexural rigidity is $D = Et^3/12(1-\nu^2)$. What resists its radial displacement is the hoops: pushing a ring of radius $R$ inward by $y$ strains it by $y/R$, which puts $Ety/R$ of force in it, which pushes back on the meridian at $Ety/R^2$ per unit length. That is a spring of stiffness $k = Et/R^2$, distributed continuously along the strip. So the shell edge solves

$$D\,y'''' + \frac{Et}{R^2}\,y = q,$$

which is $EI\,y'''' + k\,y = q$ — the equation of a beam on an elastic foundation, where $k$ is the springiness of the soil. The two structures could hardly be less alike: one is a curved surface whose stiffness comes from its own geometry, the other is a footing lying on dirt. The equation does not care. Its solution decays as $e^{-\beta x}$ times a sinusoid, with

$$\beta = \left(\frac{k}{4D}\right)^{1/4} = \frac{[3(1-\nu^2)]^{1/4}}{\sqrt{Rt}},$$

and the distance at which $\beta x = \pi$ — where the disturbance is down to $e^{-\pi}$, about four per cent — is $2.4440\sqrt{Rt}$. The textbook figure of “about $2.45\sqrt{Rt}$” is a rounding of exactly that constant, and the constant is $\pi/[3(1-\nu^2)]^{1/4}$.

The generalisation is not that shells resemble footings. It is that a fourth-order equation with a restoring term proportional to displacement is a shape of problem, and its answer is always the same: a disturbance with a characteristic length, dying in a few of them, with the same fixed stations along the way. It turns up wherever something stiff in bending is held by something that pushes back in proportion to how far it has moved — a rail on sleepers, a pipe on a bed, and, with the restoring force supplied by tension rather than by material, a cable whose stiffness comes entirely from its geometry.

Where the model stops #

The membrane state is a possibility, not a fact. Membrane theory is a lower-bound argument of the same family as the safe theorem for arches: it exhibits internal forces in equilibrium with the load. Whether the shell takes up that state depends on its edges permitting it, and a great many real edges do not.

Buckling is the failure mode, and none of these figures contains it. A shell compressed to half a megapascal is nowhere near crushing and may be very near buckling, because the critical stress of a spherical shell goes as $Et/R$ and falls with the same thinness that made the shell efficient. Worse, it is the extreme case of imperfection sensitivity: a real shell reaches a fraction of its theoretical buckling load, and the fraction is decided by dents nobody measured. That is the price of the whole trick — every particle at the same stress means every particle available to go unstable at once.

Loads must be smooth, and the surface complete. A membrane has no mechanism for carrying a concentrated load; the theory returns infinite forces under a point load, which is a correct statement that bending must appear. A crown lantern is a ring load at the top and changes the meridional force everywhere below it. Openings, ribs, valleys and edge beams are all local violations, each launching its own disturbance.

The load here is per unit area of surface. Snow and wind are per unit area of plan, which changes both functions and moves the sign-change angle. The figures on this page carry self-weight and say so.

The edge case shown is fully restrained, which is an upper bound. A real ring beam is softer than infinitely stiff, so it takes some of the growth and the shell takes correspondingly less bending. The 0.52 MPa is a ceiling on the edge stress, not a prediction of it.

What these pictures cannot show #

Every figure here is a curve on a graph or a section through a meridian, drawn for a shell already built, sitting still, at one instant.

None shows the shell in three dimensions, which is the only place buckling lives — the failure mode is a dimple, and a dimple has no representation in a plot of $N_\varphi$ against $\varphi$. None shows the construction sequence, and a structure is not complete until it is complete: a dome part-built has no closed hoop, so the ring forces these figures rely on do not yet exist. None shows time, and a concrete shell’s creep both relaxes the edge bending and magnifies the deflection that seeds the buckle. And none shows a crack: the tension below 51.8° is drawn as a smooth positive curve, which is what an elastic isotropic shell does, while masonry answers by splitting.

The assumption underneath every one of the drawings is worth naming last: the shell is thin enough that stress may be taken as uniform through its thickness. That is the definition of membrane action rather than a consequence of it, and it is exactly the assumption the edge disturbance destroys — which is why the honest shape of the subject is a membrane solution plus a correction rather than a theory.

The ladder from here #

Later rungs on this anchor: the general shell of revolution, and what happens to the two forces when the meridian is not a circle. Hyperbolic paraboloids, and the ruled surfaces that let a doubly curved shell be built on straight formwork. The funicular shell, found by hanging a cloth and inverting it — the funicular arch taken into two dimensions. Shell buckling and the imperfection knockdown, the anchor’s hardest rung. The cylindrical barrel vault, where one of the two curvatures is zero. Openings, edge beams, and the disturbance each one launches. Prestressed ring beams and the load put on backwards, which is how a modern dome catches its thrust. And the gridshell, where the surface is a mesh of bars and the question is which of these results survive discretisation.

The Zeiss works in Jena built the first modern thin-shell dome in 1922: a hemisphere of 25 m diameter in 30 mm of shotcrete on a light steel mesh, a thickness-to-radius ratio of about one in four hundred. Dischinger and Bauersfeld had the membrane theory already. What was new was the nerve to build at a thickness that looked like an error.