The length a structure was never given

Structures have lengths in them that nobody put there. A beam has a span because somebody drew one, and a depth because somebody chose it; it also has a decay length, which is the distance over which it forgets a disturbance applied at one point, and that is not a design decision at all. It falls out of the differential equation the structure obeys, and the only way to find it is to solve the equation and read it off.

This essay is about three of them, because they are three different arithmetics and the third is the strange one.

Which free body produced the number #

A vertical strip of the shell wall, one unit wide, running up from the base.

That strip bends like a beam — it has a flexural rigidity $D = Et^3/12(1-\nu^2)$ per unit width, exactly as a plate does. What resists its outward movement is not a foundation underneath it but the hoop rings it crosses: push the strip out by $w$ and every ring it passes through is stretched by $w/R$, which puts a hoop force $Etw/R$ into each of them, which pushes back on the strip with $Etw/R^2$ per unit length. That is a Winkler foundation with a spring constant $k = Et/R^2$, and it was supplied by the shell’s own geometry rather than by anything under it.

So a shell wall is a beam on an elastic foundation made of itself, and everything about it follows from the one parameter that problem has:

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

and its reciprocal is the length everything else is measured in:

$$\frac{1}{\beta} = \frac{\sqrt{Rt}}{[3(1-\nu^2)]^{1/4}} = 0.78\sqrt{Rt}$$

at $\nu = 0.3$. The beam that sits on the ground is the same equation with the springs supplied by soil; here they are supplied by the shell, and the fourth root of a stiffness ratio has collapsed into the square root of a product of two lengths.

Two boundary conditions, and the whole of the answer #

The membrane solution has the wall growing outward by $\delta = pR^2/Et$ at every height. A base slab says it does not grow here. That is all the information there is, and it is enough.

The decaying solution of the fourth-order equation is $w = e^{-\beta x}(C_1\cos\beta x + C_2\sin\beta x)$. A clamped base needs $w(0) = -\delta$ and $w'(0) = 0$, which gives $C_1 = C_2 = -\delta$ immediately, and everything after that is differentiation. Two of the results are worth having in front of the reader.

The base moment is $M_0 = p/2\beta^2$. Written out, that is $pRt/2\sqrt{3(1-\nu^2)}$ — a moment per unit width that grows with the radius and with the thickness, which is not what most people guess.

The bending stress it produces is 1.82 times the membrane hoop stress. Divide $6M_0/t^2$ by $pR/t$ and everything cancels:

$$\frac{\sigma_{bending}}{\sigma_{hoop}} = \frac{\sqrt{3}}{\sqrt{1-\nu^2}} = 1.816$$

There is no pressure in it, no radius, no thickness. A membrane structure carries eighty per cent more stress in bending at the one line where it is held than the membrane stress the entire design is about, and it does so at every size, every pressure and every wall thickness. That is the number to remember, and it is why a shell’s edges are the whole of its detailing.

Three decay lengths, three arithmetics #

The reason to put these side by side is that the rule for finding a decay length is the same in all three and the answers have nothing in common.

Saint-Venant’s is the member’s own depth. How far a wrong load reaches works it out: a self-equilibrating load applied at an end, decomposed into components of wavelength $\lambda$, dies out as $e^{-2\pi x/\lambda}$, and the longest wavelength available is the depth. So a 400 mm beam has forgotten how its end load was applied within about 400 mm, and there is one length in the problem.

A beam on the ground has a fourth root. $\ell = (4EI/k)^{1/4}$ is a ratio of two stiffnesses that belong to different objects, raised to a small power — which is why a raft’s characteristic length is stubbornly a few metres for almost any combination of raft and soil, and why a stiffer raft is so poor a way of spreading a load.

A shell has a geometric mean. $\sqrt{Rt}$ takes a radius of, say, 4,000 mm and a thickness of 12 and returns 219 — a number that is not near either of them, and that sits at the logarithmic midpoint of two quantities differing by a factor of 333. That is the property that makes shell edge effects so easy to misjudge: the intuition wants the answer to scale with the big number or with the small one, and it does neither.

The consequence is a table with a very small spread in it. A water tank wall 8 m in radius and 250 mm thick has a characteristic length of 1.09 m; a steel silo at 4 m and 8 mm has 139 mm; a pipeline at 600 mm and 12 has 66; a concrete chimney at 3 m and 200 mm has 595. Four orders of magnitude of structure, and every one of them forgets its edge in a length within a factor of twenty of the others — and every one of those lengths is 0.77 to 0.78 times the square root of the product.

What the boundary layer does on the way out #

The layer is not a simple decay. Two features of it are worth knowing because both have consequences on a drawing.

The moment changes sign at $\beta x = \pi/4$. A quarter of a characteristic length in, the wall stops hogging and starts sagging, so the reinforcement that the base moment demands on the outside face is unnecessary a few hundred millimetres up, and a different face wants a smaller amount of it a few hundred millimetres further on. A tank wall detailed with one arrangement of bars top to bottom is over-reinforced on one face and under-reinforced on the other for most of its height.

The hoop force overshoots. The wall springs back past where it was going: at about $\pi$ characteristic lengths in, the radial displacement exceeds the free membrane value by $e^{-\pi} = 4.3\%$, and the hoop force with it. It is a small number and it is exactly $e^{-\pi}$ — a constant with no shell in it at all, like the 1.82.

Beyond about three characteristic lengths the moment is under a twentieth of its edge value, and the membrane solution is recovered. That is the honest statement of what “local” means here: three lengths, where the length is the geometric mean, which for the silo above is 42 cm and for the chimney is 1.8 m.

Where the edges are, which is more places than the drawing shows #

An edge, for this purpose, is any line at which the membrane solution is contradicted. That includes several things that are not drawn as edges.

A base slab, which says the radius does not change here. A roof or a head, which says the same thing at the top and which is why a flat-ended pressure vessel is a much worse vessel than a domed-ended one. A change of thickness, which is an edge in the middle of the wall: the two halves want to grow by different amounts, and the discontinuity is the same problem with a smaller mismatch. A ring stiffener, which is an edge somebody put in on purpose and which produces a local moment for the same reason a base slab does. And a junction of two shells of different radius, where each side’s membrane solution demands a different displacement and neither can have it.

The dome’s springing is the case where the ring is doing it, and it is the one place in this collection where an edge disturbance is unavoidable rather than a detailing decision. The ring at the base of a dome is in tension, so it stretches; the shell’s own hoop force at that latitude is also tension, but of a different magnitude, so the shell wants to stretch by a different amount. The two are connected, so there is a mismatch, and the mismatch produces bending over $0.78\sqrt{Rt}$ of the meridian.

The junction, where two membrane solutions meet #

The severest case is not an edge against a foundation. It is an edge against another shell, because then both sides have an opinion about how much the radius should change and neither of them is zero.

A cylindrical tank with a conical bottom is the standard example. The cylinder under liquid pressure wants to grow by $pR^2/Et$; the cone under the same pressure wants to grow by a different amount, because its own radius is different at the junction and its meridian runs at an angle. The two are welded together, so the growths must match, and the difference is absorbed by a pair of boundary layers — one running up the cylinder over $0.78\sqrt{R t}$ and one running down the cone over its own version of the same length.

Two things follow, and both are practical.

The moment is shared in proportion to stiffness. Each side of the junction resists rotation with a stiffness of order 2eta D, so the thicker or tighter-radius side takes more of the discontinuity — which is the stiffest path takes the load at a weld line. Thickening one side of a junction to “strengthen” it drags moment into it, and the moment grows faster than the section does.

The horizontal thrust has to go somewhere. The meridional force in the cone has a radial component at the junction that the cylinder does not, and the difference is a ring load. Somebody has to carry it, and the answer is a ring — which is the same conclusion the force that is only a radius reaches for a dome and for exactly the same reason. Vessels are full of small stiffening rings at changes of shape, and every one of them is there because two membrane solutions disagreed.

The failure this prevents is a specific one and it is not collapse. A boundary layer is a local bending, so it does not reduce the structure’s capacity to carry the pressure at all — the membrane action further along is untouched. What it does is crack the concrete, leak the tank, or start a fatigue crack at the weld toe, which is the detail decides and the steel does not applied to a shell. A shell that fails at an edge nearly always fails by serviceability or by fatigue, and hardly ever by strength.

The design that removes the layer instead of reinforcing it #

There is a third response, beside computing the moment and reinforcing for it, and on a large tank it is the usual one: let the edge move.

Cast the wall on a sliding joint — a bearing strip, a membrane, a rubber pad — and the base no longer says the radius may not change. The membrane solution is then correct all the way down, the hoop force is a triangle with its peak at the bottom exactly as the textbook figure shows, and there is no boundary layer to reinforce. The price is that something else has to carry the base shear, and that the joint has to be watertight while sliding, which is a detailing problem rather than a structural one.

That is a general move worth naming. A restraint produces a force proportional to the stiffness of whatever is doing the restraining, so the cheapest way to reduce a restraint force is to remove the restraint, not to strengthen the member carrying it — the movement nobody applied is the same trade for a thermal one, and where the structure is allowed to move is it made into a scheme. A shell’s edge is a restraint like any other, and the design question is always which of the two answers is being bought.

Where the model stops #

The shell was long. The solution used only the decaying part of the general solution, which is legitimate when the far edge is several characteristic lengths away. A short cylinder — a ring, a stiffener spacing, a segment between two changes of thickness — has both edges inside each other’s boundary layer, and the two disturbances interact. That is not a small correction: at a spacing of one characteristic length the moments roughly double.

The material was elastic and the geometry linear. A concrete tank wall cracks in the boundary layer before it cracks anywhere else, and cracking softens it, which raises $\beta$ and shortens the reach. The layer moves as the structure damages, which is one of the few places in this collection where the characteristic length is not a constant.

The edge was perfectly clamped. Nothing is. A base slab has its own flexibility, and a partially restrained edge produces a moment somewhere between the clamped value and zero — which is why the honest calculation is a compatibility one between two flexibilities, a joint made of springs in series applied to a shell.

And the pressure was uniform. A tank holding liquid has a pressure that varies up the wall, so the membrane displacement varies with it and the “mismatch” at the base is against a triangle rather than a rectangle. The arithmetic changes; $\beta$ does not, because $\beta$ was never told what the load is.

The generalisation #

The idea to carry away is that a structure’s own equations hand it lengths, and those lengths decide what “local” means.

It is a habit worth applying wherever a designer is about to say that something is a local effect and can be ignored away from where it happens. The right question is not whether the effect is local but what its length is, and the length is available: it is whatever appears in the exponent when the governing equation is solved. A beam’s is its depth, a beam on springs has a fourth root of a stiffness ratio, a shell has a geometric mean of radius and thickness, a plate on a grid of supports has the support spacing, and a bar being developed into concrete has a length set by the bond strength and the bar’s own area.

The second idea is narrower and sharper: when a number comes out with none of the problem’s variables in it, it is worth stopping at. The 1.82 has no pressure, radius or thickness in it. The $e^{-\pi}$ overshoot has nothing in it at all. The $2+\pi$ in the ground is a mechanism, the 51.83° in a dome, the 1.5 shear factor of a rectangle — each is a place where the geometry has answered a question the loading was not consulted about, and each is therefore a number that will still be right when everything around it has changed.