The oval the pressure brings

Three kilopascals that halve a tube in bending put a pressure on Brazier’s tube — a long thin tube that, bent, squashes its own section into an oval until the oval can carry no more. A uniform pressure turned out to enter only as a change in the ring’s stiffness against the oval: inside it stiffens the ring, outside it softens it, by the factor $1 + p/p_c$, where $p_c$ is the pressure at which the ring buckles on its own. The limit moment then scales as the square root of that factor, and the flattening at the limit stays at exactly two ninths of the radius, as the tube that flattens itself found for a tube with no pressure at all.

That essay ended on its own simplification. Its pressure was the same at every point round the tube, which is why it could only change a stiffness. A buried tube carries more soil on its crown than on its sides; a free-standing steel flue in wind is pushed on its windward face and sucked on its flanks; a tube lying half full of water carries a pressure that grows with depth across its own diameter. Each of those pressures has a part that varies as $\cos 2\theta$ round the tube — an oval imposed directly, rather than a change in how stiff the ring is against one. The bending’s oval and the pressure’s oval then add, and the question is which way round they are.

The pressure’s oval does work on the bending’s #

Brazier’s calculation balances energies per unit length of tube. Bending stores $\tfrac12 EI_0(1 - \tfrac32\zeta)\kappa^2$, with a second moment that falls as the section flattens by a share $\zeta$ of its radius. The ring stores $\tfrac12 K\zeta^2$ in being bent into the oval, with $K = 3\pi r^2(3D/r^3 + p)$: the ring’s own stiffness, changed by any uniform pressure $p$. The tube settles at the flattening where their sum is least.

Write the varying part of the pressure as $q_2\cos 2(\theta - \varphi)$: pushing in on the diameter at an angle φ to the bending plane, and out on the one at right angles. The bending flattens the tube by moving its extreme fibres in, a radial movement $-\zeta r\cos 2\theta$ with θ measured from the bending plane. The pressure does work on that movement, and the work is

$$W = \pi\,q_2\,r^2\,\zeta\cos 2\varphi,$$

which enters the energy with a minus sign. Minimising over the flattening gives

$$\zeta = \zeta_0 + \frac{3EI_0}{4K}\,\kappa^2, \qquad \zeta_0 = \frac{\pi q_2 r^2\cos 2\varphi}{K} = \frac{q_2\cos 2\varphi}{3(p_c + p)}.$$

The pressure gives the tube a flattening of its own, $\zeta_0$, before it has bent at all, and the bending adds to it exactly as it did before. The second moment and the moment carried, $M = EI_0\kappa(1 - \tfrac32\zeta)$, see only the total. A part of the pressure that varies as $\sin 2\theta$ makes an oval at 45°, which changes the section’s product of inertia but not its second moment about the bending axis, so to first order it does not enter.

Already partly flattened #

The tube is the earlier essay’s: steel, 1.2 m across with a 2.5 mm wall, $r/t = 240$, its ring buckling under 4.07 kPa. Let a pressure push in on one diameter and out on the other with a second harmonic of a quarter of that, 1.02 kPa.

With the pressure’s oval in the bending plane the tube starts already flattened by 0.083 of its radius, and its moment–curvature curve starts lower and turns over sooner: its limit is 651 kN·m against 796. With the oval across the bending plane the pressure has bulged the tube in the bending plane before the bending begins, the bending has to undo that first, and the limit rises to 949 kN·m. All three curves start along the same line — at small curvature the section is nearly whatever the pressure made it, and its stiffness is nearly $EI_0$ — and part as the bending’s oval grows.

Which way it lies #

The limit, minimising the energy along each path and finding its peak, lands on a closed form:

$$M_{lim} = M_0\sqrt{1 + p/p_c}\,\left(1 - \tfrac32\zeta_0\right)^{3/2}.$$

Its angular dependence is all in $\zeta_0 \propto \cos 2\varphi$, so the limit is lowest with the pressed diameter in the bending plane, highest with it across, and exactly the round tube’s at 45°, whatever the size of the oval. The same pressure on the same tube is a weakness or a strength depending only on which way the tube is bent. The curves are not symmetric about one: an oval half the ring’s buckling pressure takes 35 per cent off in the bending plane and adds 40 across it, because a section already flattened loses second moment faster than a bulged one gains it.

This is the property that makes the non-uniform case worth having in hand. A buried pipe whose ground has settled under it bends in the vertical plane, sagging into the settlement, and the soil’s extra weight on its crown pushes in on the vertical diameter: the two ovals are aligned, and they add. The same pipe bent sideways by ground moving horizontally — a slope creeping, a trench backfilled unevenly — has the soil’s oval across its bending plane, and is strengthened by it. A designer who checks the vertical case has checked the worse one; one who checks the horizontal case, because it was the larger movement, has checked the case in which the pressure is helping.

Two ninths was the flattening of a round tube #

The earlier essays found Brazier’s limit at a flattening of exactly 2/9 of the radius, whatever the tube, whatever its material and whatever uniform pressure was on it — a number with no dimension in it. With a pressure oval it becomes

$$\zeta_{lim} = \frac29 + \frac23\,\zeta_0.$$

The bending’s own share of the flattening at the limit is $(1 - \tfrac32\zeta_0)/4.5$, less than 2/9 when the pressure has flattened the tube in the bending plane and more when it has bulged it; the total moves by two thirds of the pressure’s oval. So 2/9 was never a property of the bending. It was the flattening at which a round tube’s loss of second moment balances its gain in curvature, and a tube that starts out of round reaches that balance somewhere else.

The figure carries a second line, which the earlier essays did not need: the flattening at which the tube’s wall yields. Ovalling bends the ring itself, and the two-lobed oval changes the wall’s curvature by $3\zeta/r$. Its bending stress is then

$$\sigma_\theta = \frac{18D\zeta}{r\,t^2} = \frac{3E\,t\,\zeta}{2r(1-\nu^2)},$$

which at the round tube’s 2/9 is already 313 N/mm² for this tube, 0.88 of its 355 N/mm² yield. The wall yields at a flattening of 0.252. A pressure oval in the bending plane of only 0.045 of the radius pushes the limit’s flattening past it.

A fraction of the ring’s own buckling pressure #

The scale is set by the ring’s buckling pressure, and for a thin tube that is small: 4.07 kPa here, the pressure at the bottom of 40 cm of water. A pressure difference between two diameters of a few kilopascals is not exotic. A tube’s pressure oval of a quarter of $p_c$ is a second harmonic of 1 kPa: in the pressure that needs no direction, a uniform pressure four times as large was what buckled the ring outright. The oval that takes a fifth off a tube in bending is a pressure difference a ring would hardly notice on its own.

A free-standing thin steel flue in wind is the clearest case. In flow round a cylinder the windward face is pushed in and the flanks are sucked out, a second harmonic of the pressure with its pressed diameter in the wind’s direction; in the ideal flow that has no separation the harmonic is twice the wind’s dynamic pressure, and in real flow it is less but of the same sign. The flue bends in the wind’s direction too. Its two ovals are aligned by the physics of the wind, not by bad luck, and a wind of 30 m/s, whose dynamic pressure is about 0.55 kPa, can supply a harmonic of the order of a quarter of this tube’s $p_c$ on its own.

Where else a pressure has an oval #

A tube lying half full of water carries an internal pressure that is nothing above the surface and grows with depth below it, $\gamma_w r\cos\theta'$ at an angle θ′ from the bottom. Its second harmonic, worked out over the wetted half, is $(2/3\pi)\gamma_w r$ pushing outward on the vertical diameter — the same as pushing inward on the horizontal one. For the 600 mm tube that is 1.25 kPa, 0.31 of its ring’s buckling pressure, with the pressed diameter across the vertical. A half-full pipe spanning between supports sags in the vertical plane under the water’s weight, so the water’s oval lies across its bending plane: it raises the tube’s limit by a factor of 1.24. The water that bends the pipe also props it against the bending.

A buried pipe has the opposite arrangement. The ground’s vertical pressure on it is the weight of the cover, and its lateral pressure is that times a coefficient of earth pressure at rest of about a half, so the difference — pushing in on the vertical diameter — is about a quarter of the cover’s weight per square metre of crown. The ground pressure is a load the pipe has to carry before any oval is counted, as the pipe that decides what the soil weighs found, and its oval lies in the plane in which a settling trench bends the pipe. And wind on a flue is mostly a matter of its flanks, which most of it is suction found of every building face the wind runs along: the suction pulls the flanks out and the windward push presses the face in, an oval whose pressed diameter points into the wind.

What the three have in common is that the oval and the bending are produced by the same thing — the water’s weight, the ground’s, the wind’s — and so are not independent. A designer cannot choose the angle between them. The physics chooses it, aligned in the buried pipe and the flue, across in the half-full pipe.

Suction multiplies it #

The pressure’s own oval is $q_2/3(p_c + p)$: a uniform pressure outside softens the ring, and a softer ring is ovalled further by the same harmonic. Under half the ring’s buckling pressure of suction a uniform-pressure-only tube loses 29 per cent of its limit, and an aligned harmonic of a quarter of $p_c$ then takes 35 per cent of what remains, where with no suction it took 18. The two effects multiply rather than add, and the tube under both keeps 0.46 of its round, unpressured limit.

So the order in which pressures are considered matters to anyone checking a tube one effect at a time. It is the same arithmetic as the load that makes itself worse: a softening that is harmless alone amplifies everything that leans on it. A uniform external pressure checked alone and an oval checked alone each leave a margin; together they leave much less, because each makes the other worse.

From one failure to another #

For this tube, round and unpressured, ovalisation governs: its limit is 0.91 of the moment that would buckle it locally at a knockdown of 0.6 and 0.79 of the moment that would first yield it. An aligned oval lowers the ovalisation limit, and from a harmonic of 0.15 of $p_c$ something else gets there first: the wall yields in ring bending before the section stops carrying moment. At an oval of half $p_c$ the tube carries 0.58 of its round limit, and what ends it is the wall’s own bending, not the section’s.

That changes what kind of failure it is. An ovalisation limit is elastic: unload the tube before it passes the peak and it springs back round. A wall that has yielded in ring bending keeps some of its oval when the load comes off, so the next load finds a tube already flattened — a pressure oval of its own making, aligned with the bending that made it.

An imperfection put in by the load #

A tube with a pressure oval is a tube with an initial imperfection — out of round by $\zeta_0$ — that a load has put there, and its behaviour is that of every imperfect structure: the closer it starts to the shape it fails in, the sooner it fails. The column that was never straight carried its initial bow into the same kind of formula, a capacity falling with the bow in the direction of the buckle and rising against it. The difference is that a column’s bow is a tolerance nobody chose, and a tube’s pressure oval is a calculation anyone can make: it is $q_2/3(p_c + p)$, and its direction is known.

That makes the out-of-roundness tolerance a fabricator works to — a few per cent of the radius for a rolled tube — the same quantity as a pressure oval of a few per cent of $p_c$ times three. A tube rolled to 1 per cent out of round, with the flat side in the bending plane, has lost $1 - (1 - 0.015)^{3/2} = 2.2$ per cent of its limit; one rolled with the flat side across the bending plane has gained as much. Fabrication and pressure are one effect, and they add.

The oval and the limit by hand #

The ring’s buckling pressure is $p_c = E(t/r)^3/4(1-\nu^2) = 205{,}000 \times (2.5/600)^3/(4 \times 0.91) = 0.00407$ N/mm². A second harmonic of a quarter of it, 1.02 kPa, in the bending plane, flattens the ring by $\zeta_0 = 0.25/3 = 0.083$ of its radius, 50 mm, before the tube bends.

The round tube’s limit is $M_0 = (2\sqrt2/9)\,\pi E r t^2/\sqrt{1-\nu^2} = 796$ kN·m. With the oval, $(1 - 1.5 \times 0.083)^{3/2} = 0.875^{3/2} = 0.818$, so 651 kN·m, reached at a flattening of $0.222 + 0.667 \times 0.083 = 0.278$. Across the bending plane, $1.125^{3/2} = 1.193$: 949 kN·m at 0.167. The wall’s ring-bending stress at 0.278 would be $313 \times 0.278/0.222 = 391$ N/mm², past its yield, so the wall yields first, at a flattening of 0.252 and 647 kN·m.

What the energy leaves out #

One oval, the two-lobed one. The pressure’s harmonic and the bending both excite the $\cos 2\theta$ shape, and the tube is allowed nothing else. A real pressure has harmonics of every order; the fourth and higher are resisted by the ring far more stiffly and hardly couple to the bending’s oval, so the second harmonic is the part that matters, and the others become local shell imperfections, which reduce the local buckling moment rather than the ovalisation one.

A long tube. The oval takes a length of the order of $\sqrt{rt}$ to develop away from anything that holds the section round, as the tube that flattens itself found. A pressure oval that varies along the tube — a load of soil heavier in one place — makes the limit local to where it is largest.

The pressure does not follow the oval. Soil pressure on a ring that moves into the soil grows, and on a ring that moves away from it falls: the ground is a spring as well as a load. A buried tube’s soil both makes the oval and resists it, and which wins depends on the soil’s stiffness against the ring’s.

Elastic to the limit. The wall yields in ring bending near the round tube’s own limit at this $r/t$, so beyond a small aligned oval the elastic count gives the moment at first yield of the wall rather than a collapse, and the tube’s real capacity is set by how far its yielded wall can go on ovalling.

Still open: the ground that pushes back #

The buried tube was the first example here and the one this count is least fair to. Its ground loads the ring with an oval and then resists the oval it has made: soil pushed outward by a flattening ring presses back harder, and soil the ring moves away from presses less, so the soil acts on the oval as a spring of a stiffness set by its own modulus. A spring of that kind adds directly to $K$, which reduces the pressure’s own oval and raises the bending limit at the same time. Whether a buried thin tube’s bending limit is then set by the soil’s weight or by its stiffness — and whether a stiff, well-compacted backfill that carries more weight on the crown is better or worse for a pipe that has to bend into a settling trench than a loose one that carries less — is the question the ground puts back to the pressure.