The pressure that needs no direction

A column has an axis to buckle about. A plate has a direction its edges are being pushed in. An arch has a load on it, and the load’s distribution decides what shape the arch prefers to go into.

A ring under uniform external pressure has none of that. The pressure is the same at every point of the wall, it stays perpendicular to the wall wherever the wall moves to, and it does positive work on any deformation that reduces the enclosed area. So the ring does not need a direction to buckle in. It needs only to stop being round, and it will take whichever shape is cheapest. That is a different relationship with imperfection from a column’s, where an out-of-straightness in one plane decides which plane the column goes in.

Which free body produced the number #

Take the ring’s deformation as $w = w_n \cos n\theta$ — a shape with $n$ lobes — and compare the energy it costs against the work the pressure does.

The bending energy of a ring whose radial displacement varies that way is proportional to $(n^2 - 1)^2$, because the change of curvature involves $w'' + w$ and for a cosine that brings a factor $(1 - n^2)$. The work done by the pressure is proportional to the area lost, which for the same shape is proportional to $(n^2 - 1)$. Setting them equal,

$$p_n = (n^2 - 1)\frac{EI}{R^3}$$

The lowest $n$ that is not a rigid-body motion is two — $n = 0$ is a uniform contraction and $n = 1$ is a translation, neither of which is buckling — so a bare ring collapses into an oval at $3EI/R^3$.

Taking the wall as a plate in plane strain, $I = t^3/12$ per unit length of pipe and $E \to E/(1-\nu^2)$, and the result becomes

$$p_{cr} = \frac{2E}{1 - \nu^2}\left(\frac{t}{D}\right)^3$$

which is Levy’s 1884 result and is what every pipe standard in the world is built on.

The cube, and why there is nothing to do about it #

The exponent is the whole design.

Nothing else in this collection punishes thinness that hard, and more to the point, nothing else leaves so little to do about it. A beam’s bending strength goes as the square of its depth — but the depth can be increased. A column’s Euler load goes as the square of its radius of gyration — but the material can be moved outwards. A ring’s critical pressure goes as the cube of the wall thickness, and there is nowhere further out to move the material, because the geometry is already fixed by whatever the pipe is carrying.

The consequence is that a pipe against external pressure is specified by exactly one number — its diameter-to-thickness ratio, which the industry calls SDR — and that number does more work than any other single parameter in this collection. It is also why manufacturing tolerance matters so much here: a wall ten per cent thin has lost a quarter of its resistance before anything else has gone wrong. That sensitivity is the scale argument in an unusually raw form — a small proportional change in one dimension moving a capacity by a large factor.

What the soil does, which is not merely to help #

Bury the pipe and every outward lobe has to push the surrounding soil away. That is a Winkler foundation — exactly the model a beam on the ground uses — and adding it changes the arithmetic in a way that is worth working through.

The foundation adds an energy term proportional to $\int w^2$, which for the cosine shape is independent of $n$. Balancing it against the same pressure work gives

$$p_n = (n^2 - 1)\frac{EI}{R^3} + \frac{k_s R}{n^2 - 1}$$

The second term falls with $n$. That is the surprise. A long-wavelength lobe displaces a lot of soil and a short one displaces little, so the restraint penalises exactly the shape the bare ring prefers.

Minimising over $n$ — treating $n^2 - 1$ as a continuous variable — gives

$$p_{cr} = \frac{2\sqrt{EI\,k_s}}{R}, \qquad n^* = \sqrt{1 + R^2\sqrt{\frac{k_s}{EI}}}$$

For a 600 mm plastic pipe in ordinary granular backfill, that is eight times the bare pressure and $n^* \approx 3.7$: the buried pipe buckles into four lobes, not two.

Two things about that are worth separating. The first is the factor of eight, which is why buried pipes are so much thinner than submerged ones and why a restraint’s stiffness rather than its strength is what is being bought. The second is more interesting: the restraint has changed the shape of the failure, not merely its magnitude. A designer expecting an oval and inspecting for one will not recognise what a buried pipe actually does.

Reading the soil term, which is not a stiffness anybody measures #

The expression $2\sqrt{EI k_s}/R$ has one number in it that is not a property of the pipe, and it is the weakest link in the whole calculation.

$k_s$ is a modulus of soil reaction: the pressure the soil exerts back per unit radial movement of the pipe wall, in force per unit area per unit length. It is not a soil property. It depends on the soil, on how well it was compacted, on the width of the trench, on the stiffness of the trench walls, and on how far the pipe has already moved — and it is conventionally taken from a table with entries like “coarse-grained soil, 85 per cent compaction: 7 MPa”, which is one significant figure over a range of three.

Two consequences follow from that, and both are practical.

The first is that the answer’s precision is illusory. $p_{cr}$ goes as $\sqrt{k_s}$, so a factor of three uncertainty in the soil is a factor of 1.7 in the collapse pressure — which is larger than any of the refinements this essay has discussed. Quoting a buried pipe’s critical pressure to three figures is a statement about the arithmetic, not about the pipe.

The second is that the square root is a mercy. Because the dependence is weak, a soil that turns out to be half as stiff as assumed costs 30 per cent rather than 50 — and the same square root appears on the pipe’s own $EI$, so a thicker wall is also worth only its square root once the soil is doing most of the work. Past a certain point the two contributions are interchangeable, and adding wall to compensate for poor compaction is an expensive way to buy something the compactor buys cheaply.

The interchangeability is easiest to see by drawing the same sweep twice, once for the pipe above and once for a pipe of half the wall, and reading the two ends of each curve against each other.

The same substitution, in a problem that looks nothing like it #

That result is not new to this site. A strut on an elastic foundation does exactly the same thing: the bending term rises with the number of half-waves, the foundation term falls with it, and the minimum over the two gives a critical load $2\sqrt{EI k}$ with no length in it at all — the strut has forgotten how long it is and buckles into a wavelength the foundation chooses.

The ring is that problem wrapped round on itself. The only differences are that the wave number has to be an integer, because the shape has to close, and that the length is a circumference rather than a span. Everything else — the falling foundation term, the square-root critical load, the mode number set by the ratio of restraint to stiffness — is identical.

That is a genuinely useful piece of transfer. A great many problems in this subject reduce to the same competition between a term that rises with wave number and one that falls, and once it is recognised the answer can be written down: the critical load is twice the geometric mean of the two coefficients, and the mode is where they cross.

The three pressures a buried pipe actually sees #

It is worth naming the loads, because a buried pipe’s design case is unusual in this collection: none of the three is a structural load in the ordinary sense.

Groundwater. A pipe below the water table is subject to the full hydrostatic head at its crown, acting uniformly round it — which is exactly the loading in every figure above. Three metres of head is 30 kPa, and that is the commonest cause of a plastic pipe collapsing after installation.

Vacuum. A pipe carrying liquid can be put into internal vacuum by a pump tripping, a valve closing, or a siphon breaking. The external pressure is then atmospheric — 100 kPa, three times the groundwater case — applied over the whole length in a fraction of a second, and it is the case that governs most large-diameter pipelines.

Soil and traffic. Overburden is not uniform: it presses down at the crown and is resisted upward at the invert, with lateral support from the sides. That is an ovalising load rather than a compressive one, and its main effect is to supply the initial ovality that the collapse calculation is so sensitive to. So the soil appears twice and with opposite signs — as a restraint that multiplies the critical pressure and as a load that erodes it.

Only the first two are uniform, which is worth noticing given that every formula in this essay assumes uniformity. A pipe that has been ovalised five per cent by a badly compacted trench is a different structure from the one Levy’s formula describes, and the standard practice of limiting installed deflection to five per cent is the rule that keeps the formula approximately honest.

Where the model stops, which is where the pipe actually fails #

A real pipe never reaches any of the pressures above, and the reason is the one imperfection sensitivity always gives.

A pipe is not round. It leaves the factory with an ovality of a per cent or two and acquires more from handling and from the backfill going in unevenly. That initial ovality is amplified by the pressure — the classic $1/(1 - p/p_{cr})$ — and the amplified out-of-roundness produces a bending moment in the wall, which produces a stress, which reaches yield well before the elastic critical pressure is approached.

Solving that as a quadratic gives the collapse pressure, and for two per cent ovality it is around 40 per cent of the elastic answer.

So the design pressure of a buried pipe is set by three things in sequence: the elastic critical pressure from the cube law, a soil enhancement that can be a factor of several, and an imperfection reduction that gives back a large part of it. The number that comes out is not far from the bare ring’s, which is a coincidence worth being suspicious of and is why pipe design has stayed empirical much longer than most of this subject.

The one that is not a ring at all #

There is a limiting case that the ring model quietly assumes away: it assumes the pipe is long enough that every cross-section behaves alike.

A short cylinder — a pressure vessel between stiffening rings, a tank between its floor and its roof, a submarine hull between frames — cannot ovalise freely, because its ends hold it round. It buckles instead into a pattern with lobes round the circumference and half-waves along the length, and the critical pressure is higher, given by von Mises’s much longer expression. The design of a submarine’s pressure hull is very largely the spacing of the ring frames that make each bay short enough.

The competition is the same one, counted twice over. A plate under edge compression has a wave number along its length and another across its width, and its critical stress is set by whichever combination of the two is cheapest; a short cylinder has a lobe count round the circumference and a half-wave count along the axis, and the minimum runs over both. In each case a restraint that shortens one of the two dimensions forces the buckle to find its wavelength in the other, at a higher price.

That is the same relationship an effective length has with a column, arriving in a problem with two wave numbers instead of one. The rings do not carry the pressure; they shorten the length over which the wall is free to choose its own shape.

Where it decides something above ground #

The ring problem is not only about pipes, and two of its other homes are worth naming because the arithmetic transfers exactly.

A circular silo or tank under partial vacuum is the same problem with a much larger radius and no soil at all. Emptying a sealed tank faster than air can enter it is the standard way of collapsing one, and the fact that the failure is invariably a set of lobes round the circumference rather than a general squashing is the ring result made visible at building scale. Anti-vacuum valves on tank roofs exist entirely because of the cube law: at $D/t$ of five hundred, a few kilopascals will do it.

A tunnel lining is a ring in soil at very large $k_s$, and the arithmetic above says what happens: the mode number climbs, the critical pressure becomes almost entirely a soil term, and buckling stops being the governing check at all. A segmental lining is designed for thrust and bending from ground loading, not for stability, and the reason is visible in the formula — with $k_s$ large the ring cannot find a cheap shape anywhere.

What it is designed for instead runs the opposite way to every stiffness argument in this essay, and the same generator will draw it. The ground round a tunnel is already stressed before the hole exists, and the vertical and horizontal stresses are not equal; their difference tries to squash the opening into an ellipse whether or not anything is in the way. That is an imposed deformation rather than a load, and a lining resisting an imposed deformation collects moment in proportion to how hard it refuses.

So a ring in the ground is asked two different questions and answers them with opposite signs of the same property. Stiffness raises the pressure at which the ring collapses and raises the moment the ring has to carry, and the buried pipe and the tunnel lining sit at the two ends of that trade.

Between the two sits the case that catches people: a large-diameter thin steel pipe during construction, before backfilling, with rainwater in the trench. It has a tank’s slenderness and a pipe’s exposure and none of the soil support the design assumed, and the collapse pressure at that moment is the bare ring’s rather than the buried one’s — a factor of eight below what the finished condition allows. It is the same erection-stage problem every other structure on this site has, in a member that has no obvious erection stage.

A vessel is not rated for the pressure it is rated for #

Put the two directions of pressure side by side and the asymmetry is larger than anything else in this essay.

Internal pressure produces hoop tension, resisted by the wall’s area. The capacity is

$$p_i = \frac{2f_y t}{D} = \frac{f_y t}{R}$$

— linear in the thickness.

External pressure produces buckling, resisted by the wall’s bending stiffness, and the capacity goes as the cube:

$$p_e \approx \frac{2E}{1-\nu^2}\left(\frac{t}{R}\right)^3$$

Take their ratio and the geometry survives as a square:

$$\frac{p_e}{p_i} = \frac{2}{1-\nu^2}\cdot\frac{E}{f_y}\left(\frac{t}{R}\right)^2$$

For steel with $E/f_y \approx 590$ and $t/R = 1/250$, that is $2.2 \times 590 \times 1.6\times10^{-5} = 0.021$.

Two per cent. A vessel rated for 10 bar internal will collapse at about 0.2 bar external — a fifth of an atmosphere, which is a partial vacuum that a cooling vessel, a blocked vent or a pump on the wrong line will produce without anybody intending it.

That is why a vacuum rating is a separate design and not a fraction of a pressure rating, and why the vessels that implode are so often ones that were adequate for their process and were emptied, cooled or steam-cleaned. The two capacities are governed by different powers of the same thickness, so no factor applied to one describes the other, and a nameplate quoting a pressure says nothing at all about a vacuum.

The same asymmetry decides how the two are made safe. Internal pressure is protected against by a relief valve, which is a device that lets the excess out and whose failure mode is a leak. External pressure is protected against by a vacuum breaker, which is a device that lets air in and whose failure mode is a collapse. One of the two is a nuisance when it sticks and the other is not, and they are the same size of fitting on the same tank.

The generalisation #

The habit worth carrying out of this is about what a buckling load is a property of.

For a column, the answer is comfortable: it is a property of the member, its material and its end conditions. For a ring, and for every problem on an elastic foundation, it is a property of the member and of what surrounds it, and the surroundings decide not only the magnitude but the shape. That is a different kind of dependence, and it makes the calculation depend on a soil modulus — which is the least well known number in any of these designs, quoted to one significant figure and varying by a factor of three between the same soil compacted well and compacted badly.

Set the two side by side and the difference is in what has to be known before the calculation can start. A column curve is drawn from a modulus, a yield stress, a slenderness and a pair of end conditions, and every one of those is a number somebody has measured to three figures. A buried ring’s curve is drawn from those and from a modulus of soil reaction, which is a number read off a table with a range of three in it — so the ring’s answer inherits a factor of 1.7 of uncertainty that the column’s does not have, before any of the structural approximations are argued about.

Which produces the field’s characteristic instruction, and it is not a structural one: compact the backfill. A pipe’s resistance to collapse is mostly bought by the person operating the plate compactor beside the trench, and no amount of wall thickness substitutes for it — because the wall thickness enters as a square root once the soil is there, and the soil enters as a square root too.