Three kilopascals that halve a tube in bending

A long tube in bending flattens itself. The compression on one face and the tension on the other both run along a curved line once the tube has bent, both have a component pointing inward, and the circle is squeezed into an oval by the very moment it is carrying. The oval has a smaller second moment than the circle, so the moment–curvature curve bends over and reaches a peak — a limit point, with no imperfection anywhere in the story — at a flattening of exactly 2/9 of the radius, for every tube of every size in every material. Brazier found it in 1927.

That argument left one thing out, and it is the thing most tubes in service have: a pressure. A pipeline carries a fluid, a penstock carries water under a head, a flue runs at a slight suction to draw its gases, a buried culvert sits under a column of soil and groundwater, and a tank being drained can be left at a partial vacuum by a stuck vent. Each of those pressures acts on exactly the part of the tube that Brazier’s argument is about — the shape of its cross-section — and none of them appears in the bending formula a designer would write down.

The question is how much it matters, and the answer comes out in a single unit that every tube carries and almost nobody writes on a drawing.

The ring’s own stiffness, and the term the pressure adds #

Brazier’s calculation balances two energies, both per unit length of tube. One is the bending energy, $\tfrac12 E I \kappa^2$, with a second moment that falls as the section flattens: for a thin tube of radius $r$ and wall $t$ flattened by a fraction $\zeta = \delta/r$ of its radius, $I = \pi r^3 t\,(1 - \tfrac32\zeta)$. The other is the energy the ring itself stores in being bent into an oval. For the two-lobed oval $w = \zeta r \cos 2\theta$ that is $9\pi D\zeta^2/(2r)$, with $D = Et^3/12(1-\nu^2)$ the wall’s plate stiffness. Bending wants the section flat; the ring wants it round; the tube settles where the sum is least.

A uniform pressure adds a third term, and its form is already known from the ring that buckles under a pressure with no direction. For the same oval, the change in the potential of a ring under a pressure $p$ — positive inside, negative outside — is

$$V_2 = \tfrac{3\pi}{2}\,(\zeta r)^2\left(\frac{3D}{r^3} + p\right).$$

With no pressure this is the ring’s own oval energy. Under external pressure the bracket shrinks and vanishes at $p = -3D/r^3$, which is the classical buckling pressure of a long ring in its two-lobed mode. Call it $p_c$:

$$p_c = \frac{3D}{r^3} = \frac{E}{4(1-\nu^2)}\left(\frac{t}{r}\right)^3.$$

So pressure enters the problem only as a change in the ring’s stiffness against the oval, by the factor $\beta = 1 + p/p_c$. Pressure inside adds to the stiffness, because squeezing a pressurised ring into an oval reduces the area it encloses and the fluid has to be pushed out of the way; pressure outside subtracts from it, because the same reduction lets the outside pressure do work. Nothing else in Brazier’s energy moves.

The tube drawn is 1.2 m across with a 2.5 mm wall — thin, but not exotic: a spiral-welded water main, a ventilation duct or a flue liner is in that range. Each curve is found by minimising the three energies over the flattening at every curvature, not by any closed form, and the moment is read off as the derivative of the minimised energy. The curves all start along the same line, because a tube that has barely bent has barely flattened and its stiffness is just $EI$. They part as the oval grows, and each turns over at its own peak.

The peaks are the finding. A quarter of the ring’s buckling pressure inside raises the peak by 12 per cent; one whole $p_c$ inside, which for this tube is 4 kPa — the pressure at the bottom of 40 cm of water — raises it by 41 per cent. Three-quarters of $p_c$ outside, 3 kPa of suction, halves it.

Two ninths, again #

The pressure changes how high each peak is. It does not change where on the oval the peak sits.

The reason is the same short argument that produced 2/9 in the first place. Minimising the energy over the flattening gives $\zeta$ proportional to $\kappa^2$:

$$\zeta = \frac{(1-\nu^2)}{\beta}\left(\frac{\kappa r^2}{t}\right)^2,$$

which is Brazier’s expression with the stiffness factor $\beta$ underneath it. The moment is still $M = \pi E r^3 t\,\kappa\,(1 - \tfrac32\zeta)$, because the pressure does no work on the bending. The peak is where $dM/d\kappa = 0$, and with $\zeta \propto \kappa^2$ that is $1 - \tfrac32\zeta - 3\zeta = 0$, or $\zeta = 2/9$ — with $\beta$ cancelling out before it could say anything. Substituting back,

$$M_{\max}(p) = \frac{2\sqrt2}{9}\,\frac{\pi E r t^2}{\sqrt{1-\nu^2}}\,\sqrt{1 + \frac{p}{p_c}}.$$

Read across, the figure says something stronger than “the peak is at 2/9”. At every fraction of its own limit, the tube has the same oval whatever the pressure. A pressurised tube bent to half its capacity is as round as a bare tube bent to half of its; a tube under suction at nine tenths of its capacity is as flat as any other at nine tenths. The pressure has rescaled both axes of the moment–curvature curve by the same square root and left its shape exactly as Brazier drew it.

That has a practical edge. The flattening of a tube is the one thing an inspector can measure without knowing anything about the loads: two diameters at right angles with a tape. If a bent tube is found flattened by a tenth of its radius, it is at about 0.86 of its ovalisation limit — whatever pressure it is under. The pressure decides how much moment that is, not how close to the end it is.

The square root, checked #

The closed form above is a prediction; the energy minimisation does not know about it. So the peaks of the paths can be laid against it.

The dots sit on the line to the fourth figure at every pressure tried, from nine tenths of $p_c$ outside to four times it inside. The curvature at the peak follows the same square root, so a pressurised tube can be bent tighter before it gives up as well as harder.

Two ends of the line are worth reading slowly.

The left end goes to zero. At the ring’s own buckling pressure, a tube needs no bending at all to collapse — it was already on the point of ovalising under the pressure alone, and that collapse is the ring-buckling problem itself. Approaching it, the bending capacity does not fall gently. At 0.9 $p_c$ the tube keeps 0.32 of its unpressurised capacity; at 0.99, a tenth. A tube under external pressure is a tube whose bending capacity and collapse pressure are one problem, and an interaction rule that treats them as two checks to be added — so much of the capacity to the bending, so much to the pressure — is drawn as a straight line where the mechanics draws a parabola lying on its side. At half of each, the straight-line rule says the tube is exactly at its limit; the energy says it still has 0.71 of its bending left at half the pressure, and so is at 0.71 of its limit when the bending is half the bare capacity.

The right end keeps rising, but slowly. Doubling the capacity needs three times $p_c$; trebling it needs eight times. Internal pressure is an effective stiffener for a tube that is ovalising, and an increasingly expensive one.

The unit nobody writes down #

Everything so far is measured in multiples of $p_c$. The difficulty is that $p_c$ changes with the cube of the wall’s thinness, so the same pressure is a different number of units for every tube.

Over the range of tubes that get built, $p_c$ spans almost three orders of magnitude. A 1.2 m pipe with a 12 mm wall has $p_c$ of 451 kPa — four and a half atmospheres — and a few kilopascals of anything is invisible to it. The same pipe with a 2.5 mm wall has 4 kPa, and a few kilopascals are the whole of its bending capacity. Between them is the range most thin-walled steel is built in, and every one of the pressures on the right of the figure — a gust’s suction, the head of a metre of water, a vacuum left by a draining tank — crosses the curve somewhere in it.

This is the reason the pressure correction is easy to miss. In a thick pipe it is real and negligible, so the habit of leaving it out is learned on structures where leaving it out is right. In a thin one the same correction is the largest term in the calculation, and nothing in the bending formula, the section tables or the load combinations says so. A designer has to know to compute $p_c$, and nothing makes it appear by itself.

The cube is also why the thinnest tubes behave as though pressure were their structure. A fabric air beam is the limit of the argument: its wall has no plate stiffness to speak of, so $D \to 0$, $p_c \to 0$, and its resistance to ovalisation is entirely the inflation pressure. A structure held up by the air inside is a tube in which $p/p_c$ is effectively infinite — and in which, as it happens, Brazier’s limit is no longer the governing one, because a membrane that cannot carry compression wrinkles first, where the bending stress cancels the pressure’s longitudinal tension.

Where ovalisation is the failure #

Brazier’s limit is one of three ways the tube can run out of moment. It can yield; it can buckle locally on its compression face, at a stress the shell theory puts at $Et/(r\sqrt{3(1-\nu^2)})$ and a real shell reaches only a fraction of — the knockdown factor $\alpha$; or it can ovalise. With no pressure, the first essay on this effect found that the choice between the two buckling modes contains no size at all: the ratio of Brazier’s moment to the local buckling moment is $2\sqrt6/(9\alpha)$, so ovalisation wins only for a tube made well enough that $\alpha$ exceeds 0.544. Pressure undoes that cleanness.

The tube drawn, bare, ovalises first: 796 kN·m against 877 for local buckling. It is the well-made case, with a knockdown of 0.60 against the 0.544 that would tie the two. The margin is ten per cent, and it is worth almost nothing. Eight-tenths of a kilopascal inside — the pressure under eight centimetres of water — moves the tube out of ovalisation and into local buckling. Everything to the right of that is the local-buckling limit, unaffected by the pressure in this calculation, although in fact internal pressure is known to raise a cylinder’s local-buckling stress by smoothing its imperfections; the figure’s right-hand side is therefore a lower bound on what the tube carries rather than an estimate of it.

To the left, the opposite. Under suction the ovalisation limit falls away and the other two do not move. One kilopascal outside takes the limit from 796 to 691 kN·m; two kilopascals to 568; three to 409. The tube’s bending capacity at three kilopascals of suction is 41 per cent of its yield moment — a check against yield would be wrong by a factor of nearly two and a half, and a check against local buckling by more than two.

The map, and the thickness put back #

The shape of the boundary between the modes is what the pressure really changes.

The horizontal line at zero is the unpressurised tube, and along it the story is the one already told: yield up to r/t of about 187 and ovalisation beyond it, with local buckling appearing only above the line, where a pressure inside has lifted the ovalisation limit past it. Pressure inside lifts the ovalisation region off the map almost immediately — by $p_c$ inside it is gone at every thickness. Pressure outside pulls it down and to the left, so that a tube that would yield in air ovalises under suction long before it reaches its yield moment, at walls as thick as r/t = 50 when the suction approaches that tube’s $p_c$.

The axis hides one more thing, and it is the point of the whole calculation. Each row of the map is a fixed fraction of each tube’s own $p_c$, and $p_c$ goes as the cube of t/r. Read in kilopascals rather than in multiples of $p_c$, a single suction is a row very near zero for the thick tubes on the left and a row deep into the lower half for the thin ones on the right. So under a real pressure the choice between ovalisation and local buckling depends on the wall thickness after all — through the cube in $p_c$ — even though, without the pressure, the thickness cancelled out of it exactly. The knockdown still decides the bare tube; the pressure brings the geometry back.

The whole of it, by hand #

The derivation is short enough to carry about.

The tube is 1,200 mm across with a 2.5 mm wall, steel at $E$ = 205,000 N/mm² and $\nu$ = 0.3. Its plate stiffness is $D = Et^3/12(1-\nu^2) = 205{,}000 \times 15.6/10.92 = 293{,}000$ N·mm. Its ring buckling pressure is $3D/r^3 = 3 \times 293{,}000/2.16\times10^8 = 4.07\times10^{-3}$ N/mm², or 4.07 kPa. Brazier’s limit with no pressure is

$$M_0 = \frac{2\sqrt2}{9}\,\frac{\pi \times 205{,}000 \times 600 \times 2.5^2}{0.954} = 7.96\times10^8 \text{ N·mm} = 796 \text{ kN·m}.$$

Under 3 kPa of suction, $p/p_c = -0.737$ and the factor is $\sqrt{0.263} = 0.51$: the limit is 409 kN·m. Under 2 kPa inside, the factor is $\sqrt{1.491} = 1.22$ and the limit 972 kN·m — above the local-buckling moment of 877, so the tube now buckles locally first.

Three numbers, and the only one a designer would not normally compute is $p_c$. It takes one line, and for a thin tube it is the line that decides the answer.

A long tube, an elastic one, and a pressure that stays put #

The calculation rests on four assumptions, and each is a place where a real tube can differ.

The tube is long. Brazier’s oval needs a length to develop, of the order of $\sqrt{rt}$ — 39 mm here — times several, and a tube held round at its ends by flanges, rings or saddles cannot flatten near them. Ring stiffeners at close spacing suppress ovalisation altogether, which is why a penstock or a large duct is usually ringed: the stiffeners are doing for the bending what they are also doing for the external-pressure collapse, because the two are, as shown above, one problem.

The material is elastic. The steel in the tube drawn yields at 1,004 kN·m, so the elastic Brazier limit is the governing one only when it is lower than that, and the plastic interaction between ovalisation and yield — which is what governs a thicker pipe being reeled or laid — is a different calculation with the same shape.

The oval is the only shape. The two-lobed oval is the mode a bending moment drives and the one that pressure’s $p_c$ refers to. A ring under a pressure close to its buckling value can find other modes, and if the tube is buried, the ground round it makes the higher modes cheaper than the oval, which changes both the buckling pressure and the shape the tube fails into.

The pressure is uniform and stays normal to the wall. A fluid inside or outside does that. Wind on a tube does not: the pressure round a cylinder in wind is suction over most of its circumference and pressure over a narrow front, which is itself an ovalising load with a cos 2θ component, not a uniform pressure. That case adds an oval imposed from outside to the one the bending produces, and it is not the calculation drawn here.

A perfect tube, and a shell that is not one #

The figures show a perfect tube. Every limit on them is exact for the model, and the model has no imperfection — which is the virtue of Brazier’s limit and the reason it can be trusted where local buckling cannot. But a real tube arrives with some ovality from rolling, welding or handling, and under external pressure that initial oval is amplified in the way every imperfection is amplified by the load that finds it: by $1/(1 - p/p_c)$. A tube with 1 per cent ovality at 0.75 $p_c$ of suction carries 4 per cent before any bending begins, and the bending then starts from an oval rather than a circle. The perfect-tube limit drawn here is the ceiling on that, not the expected value.

They also cannot show how the internal pressure changes local buckling. That effect is real, it is in the right direction — internal pressure pushes the wall’s dimples outward and makes the shell less sensitive to them — and it has been measured on pressurised cylinders for decades. It is left out here because the figure would then be making a claim about a knockdown it cannot compute, and a lower bound stated as one is more useful than an estimate stated as a result.

Still open: a pressure that is not the same all round #

The pressure drawn here is the same at every point of the circumference, and that is what made it enter only as a stiffness. A buried tube sees more pressure from above than from the side; a tube lying in a trench half-full of water sees a pressure that grows with depth across its own diameter; a flue on a tall chimney sees wind on one side and draught suction inside. Each of those has a cos 2θ part, which is an oval imposed directly rather than a change in how stiff the ring is against one — so the bending’s oval and the pressure’s oval add, and the question becomes which way round they are. A bending plane at right angles to the ground’s heavier pressure squeezes the tube flatter in the direction the bending also flattens it; a plane in line with it may unflatten it. Whether a buried tube can be oriented, or its bending arranged, so that the soil’s oval subtracts from Brazier’s rather than adding to it is the question a uniform pressure cannot ask.