The flange that curls away from its stress

The bar that was bent before it was loaded found that a curved bar’s fibres were different lengths before anything was applied, so equal rotations of two plane faces strain them unequally: the stress across the section is a hyperbola, the neutral axis moves toward the centre of curvature, and a crane hook carries half as much again as the straight-beam formula reports. The wide side goes inside found that the hook’s trapezoid puts material exactly where that hyperbola is steepest. Both treated the section as rigid in its own plane: each fibre stays where it was, and only its length changes.

That is true of a hook, which is solid. It is not true of an I-beam, and an I-beam bent into a curve — a curved rafter, a ring beam, an arched canopy rolled from a standard section, the knee of a portal frame — is the case where the assumption fails first. Its flanges are thin plates, and a thin plate carrying stress round a bend is pushed sideways by that stress.

A stress round a curve is a pressure #

Carry a tension round a pulley and it presses on the pulley: a string of tension $T$ round a radius $R$ pushes radially with $T/R$ per unit length. A flange of thickness $t$ carrying a stress $\sigma$ round the curve of a beam is the same string, spread over the flange’s width: every square millimetre of it is pressed toward the centre of curvature, if the flange is in tension, or away from it, if in compression, with a pressure of $\sigma t/R$.

The web can resist that pressure where it meets the flange. It cannot resist it at the tips. The web that is crushed from inside followed the part of the radial force that the flange hands into the web, which pushes the web toward buckling. This essay follows the other part: what the pressure does to the flange’s own two outstands, which are cantilevers from the web with nothing at their tips.

Each outstand bends. Bent the way that opens the curve, the outer flange is in tension and is pulled toward the centre of curvature; the inner flange is in compression and is pushed away from it; so both are pressed toward the web, and both outstands curl toward the beam’s neutral axis. Bent the other way, both curl away from it. Either way, for the flange drawn — 300 mm wide and 15 mm thick, on a 3-metre radius — the tips move 0.8 mm.

Moving is shedding #

That movement has a consequence that makes the problem interesting rather than merely local. A fibre of the flange that moves radially by $w$ has moved to a different circle. A fibre in tension pulled inward by 0.8 mm is on a circle 0.8 mm smaller, and its circumferential strain has fallen by $w/R$ — so its stress has fallen by $Ew/R$. The outstand sheds the very stress that bends it. It bends until the loss of stress has reduced the pressure enough for its own stiffness to hold what is left.

Written for a strip of the outstand one unit long along the beam, that is one equation:

$$D\,w'''' + \frac{Et}{R^2}\,w = \frac{\sigma_0 t}{R}, \qquad D = \frac{Et^3}{12(1-\nu^2)}, \qquad \sigma = \sigma_0 - \frac{Ew}{R},$$

where $\sigma_0$ is the stress the flange would have if it did not curl. The strip is held at the web, where the flange’s symmetry keeps it from turning, and free at its tip. And the equation is one this subject has met before: a beam of stiffness $D$ on an elastic foundation of modulus $Et/R^2$ under a uniform load — the beam that sits on the ground, with the ground supplied by the curvature. The stiffer the curve, the softer the foundation; a straight beam, $R = \infty$, is a strip on no foundation carrying no load.

Hans Bleich wrote the equation down in 1933 for the flanges of curved girders. It is worth noticing whose two theories it joins. The curved-bar theory of the earlier essays and the beam on an elastic foundation both first appeared in the same book of Emil Winkler’s, in 1867; the curling flange is the place where they turn out to be one problem.

The stress across the flange #

The foundation has a characteristic length, as every beam on an elastic foundation does: the distance over which a disturbance at the web dies away across the outstand. For the curling flange it is

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

the geometric mean of the radius and the thickness, divided by a number close to 1.29. For the 15 mm flange on a 3 m radius it is 165 mm, a little longer than the 150 mm outstand, and the stress at the tip is still 0.72 of the web’s. On a 1 m radius it is 95 mm, well short of the outstand; the stress has fallen to nothing by the tip, and the flange works over 58 per cent of its width. On a 20 m radius it is 430 mm, three times the outstand, and the flange barely notices the curve.

So the whole problem has one parameter, the outstand over that length:

$$\lambda = [3(1-\nu^2)]^{1/4}\,\frac{b}{\sqrt{Rt}}.$$

How much of the flange works #

The effective share of the width, plotted against λ, is the whole of the width question for every flange and every radius. Below λ of about a half the loss is negligible. At λ = 1 the flange works over 85 per cent of its width; at 2, 46 per cent; at 4, 25. And for long outstands the curve joins $1/\lambda$ — which is to say the flange carries stress over exactly one bending length either side of the web, as a beam on an elastic foundation carries a load at its end over one characteristic length and no further. The rest of the outstand has curled until its circumferential strain is gone.

That makes the effective width a strange kind of number. It does not depend on how wide the flange is, once the flange is wide enough to curl fully. Widening the flange of a tightly curved beam adds steel that the curve will not let carry anything.

The flange bends across its width harder than along it #

The loss of width is the effect the name “effective width” invites a designer to look for, and it arrives slowly. A second effect arrives at once, and it is the one that matters first.

A cantilever bends because a moment is carried at its root, and the outstands’ root is the web. The curling bends the flange across its own width, and the bending stress that follows — tension on one face of the flange, compression on the other, acting across the beam rather than along it — is a stress no straight-beam calculation contains. At λ = 0.8, where the flange still works over 93 per cent of its width and the width loss would pass unnoticed, the transverse bending stress at the web already equals the stress along the beam. At λ = 1 it is 1.42 times that stress; at 2, 1.77.

And it has a limit that can be written down by hand. For a long outstand the moment at the root of a beam on an elastic foundation, clamped there and loaded uniformly, is the load over twice the square of $\beta$; with the load $\sigma_0 t/R$ and $\beta^2 = \sqrt{3(1-\nu^2)}/Rt$, the root stress $6M/t^2$ becomes

$$\sigma_{\text{across}} = \frac{3\sigma_0}{\sqrt{3(1-\nu^2)}} = \sqrt{\frac{3}{1-\nu^2}}\;\sigma_0 = 1.82\,\sigma_0,$$

in which the radius, the width and the thickness have all cancelled. A fully curled flange bends across its width at 1.82 times the stress it carries along the beam, whatever it is and however tightly it is bent. At the junction of flange and web, where this stress peaks, the steel carries two stresses at right angles of which the one nobody computed is the larger.

One flange, bent tighter and tighter #

For a real flange the two effects arrive at very different radii. The 300 × 15 flange keeps 95 per cent of its width down to a radius of about 5.2 m — a curve most designers would treat as gentle — but its transverse stress at the web reaches half the stress along the beam at 8.4 m and 0.82 of it at 5 m. At 2 m the width is 78 per cent and the transverse stress 1.61 times the longitudinal.

That ordering is what a check can miss. A designer who knows about flange curling as a loss of effective width will check the width at 5 m, find 96 per cent, and move on. The steel at the flange root is then carrying a transverse bending stress of more than four-fifths of the longitudinal one, at right angles to it, at the weld or the root radius of a rolled section — where a combined-stress check, or a fatigue check on the stress range, will find it.

The correction the textbook teaches is the smaller one #

A curved beam’s textbook correction is the one the earlier essays computed: fibres of different lengths, a hyperbolic stress across the depth, a peak larger than the straight-beam formula gives. For a solid section 450 mm deep bent to 3 m, that correction raises the peak bending stress by 5 per cent; bent to 1 m, by 18 per cent.

For an I-section of the same depth with the 300 × 15 flanges, the curling at 3 m takes 11 per cent off the flange’s effective width — which raises the stress in the part that works by rather more than 5 per cent, since the flanges carry most of an I-beam’s moment — and adds a transverse stress at the web of 1.26 times the longitudinal one. At 1 m it takes off 40 per cent and adds 1.82. For an I-section the curling is the larger of the two effects at every radius drawn, and it is the one a curved-beam formula does not contain, because that formula was written for sections that cannot curl.

Two reasons a flange is not all there #

“Effective width” already has a meaning in this subject, and it is a different one. A straight beam’s wide flange is not all there because the web feeds stress into the flange by shear, and shear takes a distance to spread: near a support, where the shear is large, the flange works least. That is shear lag, and its parameter is the flange’s width against the span.

Curling is a second, independent reason, and its parameter is the flange’s width against $\sqrt{Rt}$. A curved box girder or a curved plate girder with wide flanges has both: shear lag decides how much stress reaches the outer parts of the flange, and curling decides how much of what reaches them stays. The two reductions are usually applied one after the other, as factors, which assumes they do not interact; the flange in a curved girder near its support is the place where both are largest and where that assumption is least tested.

The arithmetic for one beam #

For the flange drawn, on a 3 m radius: $\sqrt{Rt} = \sqrt{3000 \times 15} = 212$ mm; divided by $[3(1-\nu^2)]^{1/4} = 1.285$, the bending length is 165 mm. The outstand is 150 mm, so $\lambda = 0.91$. The effective share of the width at that λ is 89 per cent, and the transverse stress at the web is 1.26 times the stress along the beam. With the longitudinal stress at 200 N/mm², the flange root carries about 250 N/mm² across the beam — a stress that would not exist if the same section were straight.

The remedies follow from the equation. A thicker flange raises $D$ as the cube and the foundation only in proportion, so it shortens λ as the square root of the thickness; a narrower flange shortens λ directly. Radial stiffeners — plates welded between the flanges and the web at intervals round the curve — hold the outstands and turn each panel of flange into a plate supported on three sides, whose behaviour is not the textbook plate’s either. What does not help is a wider flange, which adds outstand beyond the bending length and adds nothing to the effective width.

A bigger beam on the same curve is worse off #

The parameter hides a scale effect that runs against intuition. Write λ as $1.285\,(b/t)\sqrt{t/R}$: at a fixed slenderness of flange, $b/t$, it grows with the square root of the thickness. Scale a section up — every dimension of the flange doubled — and bend it to the same radius, and λ grows by $\sqrt{2}$.

On a 3 m radius, three flanges of the same proportions, 200 × 10, 300 × 15 and 400 × 20 mm, sit at λ = 0.74, 0.91 and 1.05. Their effective widths are 95, 89 and 83 per cent, and their transverse stresses at the web 0.92, 1.26 and 1.50 times the stress along the beam. The heaviest section, which a designer would pick for the heaviest curved member, is the one whose flanges curl most. A curve that is gentle for a small section is tight for a large one, because what the flange measures the curve against is its own bending length, and that grows only as the square root of its size.

The same effect makes the transverse stress a fatigue concern in a way the longitudinal one is not. It peaks exactly at the flange-to-web junction, where a welded girder has its fillet welds, and its range follows the moment’s range with the factor above: on a heavily curved crane girder or ring beam under repeated load, the weld toe sees a transverse stress range larger than the longitudinal one that the girder’s fatigue check was written for.

A strip, solved across the flange #

Each outstand is treated as a strip one unit long along the beam, clamped at the web and free at the tip, under the radial pressure of the stress it would carry uncurled and resting on the foundation that its own curvature supplies; the strip is solved by finite differences across the outstand with four hundred intervals. The stress along the beam at each point is then the uncurled stress less the modulus times the radial movement over the radius, and the effective width is its integral across the outstand. Two limits check the solution: a long outstand’s effective width tends to one bending length and its root stress to $\sqrt{3/(1-\nu^2)}$ times the longitudinal, both of which are closed forms for a beam on an elastic foundation, and the calculation reaches both.

A strip that ignores its neighbours, and a stress that does not vary #

The strip is one-dimensional. Each unit length of the flange is solved as if the stress along the beam were the same everywhere. Where the moment varies along the beam — near a support, or at the end of the curve — the strips pull on each other through shear in the flange’s plane, and the curling is smoother than this. The solution is the fully developed one, far from any change.

The stress is elastic. The transverse stress at the root is the first thing to yield, and once it does the outstand hinges there and curls further, shedding more of its longitudinal stress; the effective width at collapse is smaller than the elastic one.

The web does not bend. The web holds the flange’s root from turning by symmetry, which is exact for a flange symmetric about the web and a web that is itself straight. The web is also pressed by the flange’s radial force, which bends it toward buckling — the same force followed the other way.

Still open: the tapered flange of a curved knee #

A portal frame’s knee is the commonest tightly curved I-section in building, and its flange is rarely uniform: it is cut from plate to a curve, often tapering in width and sometimes in thickness from the rafter to the column. Every number here is for a flange whose outstand and thickness are the same all round the curve, so that one λ describes it. In a tapered knee λ changes along the curve, the effective width follows it, and the stress along the flange must change as the effective width does — which moves force between the flange and the web along the curve. Whether the transverse stress at the root then peaks where λ is largest, or where the effective width changes fastest, is a question about a flange whose curling is no longer the same at every section.