The tube that flattens itself

Every stress in a bent tube is running along a curve, and a force that follows a curve pushes sideways. That is the whole of this essay, and everything else is arithmetic.

Take a circular tube and bend it. The wall above the neutral axis is in compression along a line curved concave-down, so its resultant transverse pressure points downward — inward, toward the axis. The wall below is in tension along a line curved the same way, so its resultant points upward — also inward. Both faces are being squeezed toward the middle by the very bending they are carrying, and the circle flattens into an oval.

The 2/9, which is a pure number #

Write the flattening as $\zeta = \delta/r$, the fractional reduction of the radius across the section. Brazier’s energy solution of 1927 gives two statements:

$$\zeta = \left(\frac{\kappa r^2 \beta}{t}\right)^2, \qquad M = \pi E r^3 t\, \kappa\left(1 - \tfrac{3}{2}\zeta\right),$$

with $\beta = \sqrt{1-\nu^2}$. The first says the flattening grows as the square of the curvature. The second says the moment is the undeformed one reduced by the second moment the flattening has cost.

Differentiate the second and use the first — since $\zeta \propto \kappa^2$, $d\zeta/d\kappa = 2\zeta/\kappa$ — and

$$1 - \tfrac{3}{2}\zeta - 3\zeta = 0 \quad\Longrightarrow\quad \zeta = \tfrac{2}{9}.$$

Exactly two ninths, for every tube of every size in every material. No radius, no thickness, no modulus, no Poisson’s ratio. A steel pipeline, an aluminium mast, a bamboo culm and a drinking straw all reach their limit at the same fractional flattening, and the only thing the properties decide is what curvature and what moment that corresponds to.

Substituting back:

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

The curvature it arrives at is $\kappa = (\sqrt2/3)\,t/(r^2\beta)$, which for the tube drawn is $2.2\times10^{-5}$ per mm — a radius of curvature of 45.5 m, or about 150 diameters. A tube reaches its ovalisation limit while still visibly straight, and that is worth knowing because it means there is no warning: the section has flattened by 22 per cent across its diameter at a bend nobody would look at twice.

The thickness is squared. That is the second surprise: an elastic section modulus goes as $t$, so a limit that goes as $t^2$ falls off much faster with thinness — because thinning the wall both weakens the section and makes it flatten more readily, and the two multiply.

An instability with nothing to be sensitive to #

Everything else in this field’s stability essays is a bifurcation. A column carries load along a straight path until a second path appears; a shell does the same and then loses two thirds of the answer to a dent nobody can measure. The characteristic of a bifurcation is that the perfect structure is in equilibrium on both paths and needs a nudge to choose.

This is not one. The perfect tube’s ovalisation is not a deviation from the path — it is the path. There is one equilibrium configuration at every curvature, it is stable up to the peak and unstable beyond, and no imperfection appears anywhere in the derivation.

That makes Brazier’s result one of the few numbers in shell stability anyone can trust. A cylinder in axial compression is theoretically capable of $Et/(r\sqrt{3(1-\nu^2)})$ and reaches perhaps half of it, unpredictably. A tube in bending reaches $(2\sqrt2/9)\pi Ert^2/\beta$, and reaches it.

What actually happens first #

Three things can end a tube in bending, and they run at different rates.

For a 600 mm steel tube at $r/t = 75$ — an ordinary structural CHS — the numbers are 401 kNm at yield, 401 kNm at local buckling, and 1,018 kNm from ovalisation. Brazier’s limit is two and a half times away, and nothing about it will ever be reached.

The comparison that matters is between the two buckling modes, and it produces a result with nothing in it:

$$\frac{M_{brazier}}{M_{local}} = \frac{(2\sqrt2/9)\pi E r t^2/\beta}{\alpha\,\dfrac{Et}{r\beta\sqrt3}\,\pi r^2 t} = \frac{2\sqrt6}{9\alpha}.$$

No radius, no thickness, no length, no material. Only $\alpha$, the knockdown factor that says how far short of the classical stress a real cylinder falls. Ovalisation governs a tube made well enough that $\alpha > 2\sqrt6/9 = 0.5443$; local buckling governs one that is not.

It needs length #

Brazier’s tube is infinitely long, and the assumption is doing work.

Ovalisation is a change of shape of the cross-section, and the ends of a real tube are held round by whatever they are connected to — a flange, a weld to a plate, a bulkhead. The flattening therefore has to grow into the span from each end, over the shell decay length $\sqrt{rt}$ that everything on a cylinder uses.

For the tube drawn, $\sqrt{rt} = 34.6$ mm, so four decay lengths at each end is 277 mm of a 12 m tube — 2 per cent, and the assumption is safe. For a 600 mm tube one metre long it is 55 per cent, and the tube cannot ovalise at all; it will reach a considerably higher moment and fail by local buckling instead.

Where it does govern #

The list of structures whose tubes are thin enough, long enough and made of the right material is short, and it is worth naming because it is the reason the result is still taught.

A pipeline being installed. A steel pipe going over a lay barge’s stinger, or being unwound from a reel, is bent to a radius of tens of metres by an imposed geometry rather than by a load — which is precisely the curvature-controlled test the descending branch requires. Wall thicknesses of $r/t$ around 20 are typical there and the material yields first, but the ovality left behind after each bending pass is a Brazier ovality and it accumulates.

A composite or bamboo tube. $E/f_y$ decides whether ovalisation arrives before yield, and for a fibre-reinforced tube or a plant stem it is very much higher than for steel. A bamboo culm’s nodes are ring stiffeners, spaced to suppress exactly this — which is either a very good coincidence or several hundred million years of selection.

A large-diameter thin shell in bending. A silo wall, a chimney, a wind-turbine tower. There $r/t$ runs to several hundred and local buckling is the check that appears in the codes, with ovalisation folded into the imperfection allowance rather than checked separately.

Which free body produced the number #

The transverse pressure is the free body worth naming. Take a slice of the wall subtending an angle $d\theta$, carrying a longitudinal stress $\sigma$ over a thickness $t$, and bend it to a curvature $\kappa$. The longitudinal force $\sigma t$ enters one face and leaves the other at an angle $\kappa\,ds$ different, so the resultant is $\sigma t \kappa$ per unit length, directed toward the centre of curvature.

That is the same deviation force a curved tendon puts on the concrete around it, and the same one that makes a cable’s transverse load equal to $H\kappa$. What is unusual here is only that the force is generated by the member’s own bending rather than by anything applied, and that it acts inward from both faces at once because compression on a concave-down curve and tension on the same curve push the same way.

The energy method then balances the work that pressure does in flattening the section against the strain energy of flattening it, and Brazier’s $\zeta$–$\kappa$ relation falls out.

The stiffness it loses on the way #

The moment is the headline and the stiffness is the part that gets used.

At the limit point the tangent stiffness is zero by definition — that is what a maximum means — but the secant stiffness, the one a deflection calculation would use, has only fallen to 67 per cent of the undeformed value. That gap is worth reading: the tube is still returning a moment two thirds of what its geometry originally promised, and it is on the point of failing.

The practical form of that is a deflection calculation that is wrong in the unsafe direction. A tube bent close to its limit deflects more than $ML/EI$ predicts, by up to half again, and the extra is invisible to any calculation that uses the undeformed section — which is every calculation anybody does. The same thing is true of a column near its critical load, where the load has removed stiffness that the first-order analysis still assumes is there.

Where the model stops #

Everything is elastic. For steel it is not: at $r/t = 75$ the section yields at 401 kNm, long before the ovalisation limit, so the real path leaves this one early. Brazier’s result is the right answer for a material with a very high $E/f_y$ — a composite tube, a bamboo culm, a glass-reinforced pipe — and an upper bound for steel.

The ovalisation is a pure $\cos 2\theta$ shape. It is very nearly, and the assumption is what makes the closed form possible. The exact shape has higher harmonics in it and the exact limit moment is a per cent or two below.

The tube is long and unsupported. Any ring stiffener, any internal diaphragm and any change of section resets the decay and reduces the ovalisation over a length either side of it. A tube with ring stiffeners at four decay lengths never ovalises appreciably at all — which is why pipelines being reeled onto a lay barge are the place this matters and pipelines in service, buried and supported, are not.

The moment is uniform along the tube. Under a moment gradient the curvature varies along the length, so the ovalisation does too, and the flattening at the worst section is restrained by the less-flattened material either side of it. Uniform bending is the worst case, and it is the one Brazier solved.

The knockdown factor is one number. It is not: it depends on the loading, and a cylinder in bending is measurably less imperfection-sensitive than the same cylinder in uniform axial compression, because only a small part of the circumference is at peak stress and a dent has to be in that part to matter. Using an axial-compression α in the mode comparison is conservative toward local buckling and therefore toward the wrong conclusion.

And the drawing shows a path that nothing traverses. Past the peak the tube is on a descending branch and the test that finds it must be curvature-controlled — bend the tube by an imposed rotation, not by hanging a weight on it. Under a moment applied by a load, the peak is the end: the moment cannot be sustained, the curvature runs away, and the tube kinks in a single local fold within a diameter or so of wherever it started.

The number a designer would actually use #

None of this appears as a clause. What appears is a slenderness limit on the section — a maximum $d/t$ beyond which the tube is Class 4 and its capacity is computed on an effective section rather than a gross one — and the limit is set at $50\varepsilon^2$ for a hollow section in bending, which for S355 is $d/t = 33$, or $r/t = 16.5$.

That is far stockier than anything on this page’s axis, and it is set by neither of the two buckling modes: it is set by the requirement that the section reach its plastic moment and hold it through enough rotation to redistribute. A classification limit is about ductility, not about strength, and the modes discussed here are two or three classes further out.

So the honest place of Brazier’s result in structural design is not as a check but as an explanation. It says why a very thin tube’s capacity falls faster than its section modulus does, why the codes’ effective-section rules for circular hollow sections are more severe than a stress argument would justify, and why the classification limit for a tube is written in $d/t$ squared while every other section’s is written in $c/t$.

The ladder from here #

Later rungs on this anchor: ovalisation combined with internal or external pressure, which is the pipeline case — internal pressure resists the flattening and raises the limit, external pressure adds to it and can halve it. The reeling problem in full, where a pipe is bent past yield onto a drum and straightened again, and where the residual ovality after each pass accumulates. Ring stiffeners and the spacing at which they suppress the effect, which is a decay-length calculation of exactly the kind an edge disturbance uses. Plastic ovalisation, where the wall yields as it flattens and the limit moment falls further. The kink that forms after the limit, which is a localisation problem and not a stability one. And the same effect in a section that is not circular — a rectangular hollow section flattens too, and its flanges pull in toward one another, which is the reason a very thin RHS in bending fails at the middle of its compression flange rather than at its corners.