The section that cannot stay flat

Take a length of I-section, weld one end to something immovable, and twist the other. A circular shaft treated this way does something simple: every cross-section rotates about the axis and stays where it was, flat and undisturbed. An I-section does nothing of the kind. As it twists, the top flange slides one way along the member and the bottom flange the other, and the cross-section — a plane before the load arrived — is no longer plane at all. It has dished.

That dishing is called warping, and on its own it is free: let the member warp uniformly along its length and no fibre changes length, no longitudinal stress appears, and the torque is carried entirely by shear. The trouble starts the moment something stops it — a welded end plate, a symmetry condition at mid-span, or simply a torque larger at one section than another. Then adjacent sections want to warp by different amounts, the fibres between them have to stretch and shorten to allow it, and the member has found a second way to resist being twisted.

Plane sections stay plane, except in the one field they do not #

The plane-sections assumption is a statement about bending, and it survives an astonishing amount of abuse there. In torsion it is false for every section that is not a circle or a circular tube, and Saint-Venant’s 1855 achievement was working out what shape the section takes instead.

The out-of-plane displacement at a point is $u = -\omega\,\theta'$, where $\theta'$ is the rate of twist along the member and $\omega$ is a geometric property of the point called the sectorial coordinate: the area swept by the line from the shear centre to that point as it traces the section’s mid-line. It has units of area — 12,169 mm² at a flange tip of the section above.

Two consequences follow, and between them they are the whole subject. If $\theta'$ is constant, every section warps by the same amount, no fibre gets longer or shorter, no longitudinal stress arises, and the member is in uniform torsion — the case the torsion constant $J$ describes. If $\theta'$ varies along the member, sections a small distance apart want to warp by different amounts and the fibres between them must strain to allow it. Longitudinal strain means longitudinal stress, and a longitudinal stress varying along the member has to be balanced by shear — a second torque-carrying mechanism, running alongside the first.

The second mechanism is only visible because the first is so weak #

None of this would matter if open sections were stiff in torsion.

An open section throws away almost all of its torsional stiffness, for the reason that essay sets out: the shear can no longer run round a closed loop, so it closes on itself within the thickness of each plate. What is left is small enough that a mechanism which would be a rounding error beside a box turns out to be the larger half of the answer.

For the I-section above the two rigidities are $GJ = 10.57$ kN·m² and $EI_w = 34.21$ kN·m⁴, and their units differ by a length squared. The ratio of the two is an area, and its square root is a length.

$$k = \sqrt{\frac{GJ}{EI_w}} = 0.556\ \text{m}^{-1}, \qquad \frac{1}{k} = 1.80\ \text{m}$$

That 1.80 m is the distance over which warping restraint reaches into the member — 30% of a 6 m length, and the number that decides whether any of this is worth computing.

Which free body, and what crosses the cut #

The split between the two mechanisms is not a modelling convention. It comes from cutting the member at a station and asking, as any internal force is found, what must cross the cut for the free body beyond it to be in equilibrium. Two quite different things do.

A circulating shear. Within the thickness of each plate the shear runs one way near one surface and the other way near the other, closing on itself. Its resultant force is zero and its moment about the member axis is $T_{sv}$. This is Saint-Venant’s mechanism.

A pair of flange shears. Each flange, bending in its own plane, carries a transverse shear $V_f$ — the top flange’s pointing one way and the bottom flange’s the other. Two equal and opposite forces a distance $h$ apart are a couple, and this couple turns about the member axis — a torque assembled out of ordinary transverse shears.

At the built-in end the arithmetic is direct. The two flange shears are 1.69 kN each and the flange centroids are $h = 295$ mm apart, so the couple is $1.69 \times 0.295 = 0.500$ kN·m — the entire applied torque. The circulating shear there carries nothing at all, because the section is not allowed to warp and $\theta'$ is zero.

The solver behind that figure refuses to return an answer if the two shares do not add to the applied torque at every station — a closure broken at once by a wrong wavenumber, a wrong boundary condition or a missed factor of a thousand in the units.

A stress resultant with nowhere to be plotted #

The flange shears are half of the pair. Each flange also carries a bending moment in its own plane — 3.04 kN·m at the held end — and again the two are equal and opposite. Their sum is zero, so they are not a moment. Their difference, times the distance between them, is

$$B = M_f \times h = 3.04 \times 0.295 = 0.897\ \text{kN·m}^2$$

and that quantity is the bimoment. It is to a self-cancelling pair of moments what a moment is to a self-cancelling pair of forces: a resultant one level further up.

The units are the tell. Everything else revealed by a cut is a force (kN) or a moment (kN·m); a bimoment is kN·m², and no diagram in this subject has an axis it could be plotted against. Reduce the four flange-tip forces it represents to their resultants and every one vanishes: no net axial force, no net shear either way, no net moment about either bending axis, no net torque. A free body can satisfy all six equations of equilibrium exactly and still have a bimoment in it — so the six equations a drawing hides are not the complete inventory of what crosses a cut in a thin-walled member.

That puts the bimoment in company it is not usually kept in: a self-equilibrating stress field invisible to every resultant is exactly what a residual stress is, and what a restrained temperature change leaves behind. Those two arrive without a load and this one is produced by the applied torque, but the reason none of the three shows in a member force diagram is the same, and it is about resultants rather than about stress.

Stiffness that comes from the end condition #

The first practical effect, before any stress is computed, is that the member is stiffer than its torsion constant says.

A 43% stiffening is a lot to find in a member whose section properties have not changed, and an end condition produced it:

$$\theta = \frac{TL}{GJ}\left[1 - \frac{\tanh \kappa}{\kappa}\right], \qquad \kappa = kL$$

For large $\kappa$ the bracket goes to one and the member behaves as the torsion constant says. For small $\kappa$ it goes to $\kappa^2/3$ and the twist collapses. The deciding parameter is $kL$, and $L$ is in it — so a torsional stiffness quoted as a section property is only ever a partial answer, the same species of result as effective length being a property of the ends rather than of the member.

The length $1/k$ is not local to this subject either. It is a characteristic length produced by dividing one stiffness by another, and the family is large: a beam on an elastic foundation has its own $(4EI/k)^{1/4}$, past which nothing knows the load happened, and a shell’s edge disturbance dies out over a distance proportional to $\sqrt{Rt}$. Each is a boundary condition smeared over a length nobody chose — Saint-Venant’s principle with a number attached.

The stress that nothing draws #

The bimoment would be a curiosity if the stress it produced were small.

The stress nothing draws is the larger of the two, by nearly a factor of two, and it is of the wrong type for any shear check to notice. Worse, it lands at the flange tip: the point furthest from the section’s own bending axis, where the stress from the couple that carries the moment is already largest, so the two add directly at the same fibre. A torsion check that computes $\tau = Tt/J$, compares it with a shear capacity and stops is not being conservative — it has looked at the smaller stress in the wrong place, and what would have told it so lives in a quantity whose units the check has no slot for.

The agreement of the two routes matters. $B\omega/I_w$ runs through the sectorial coordinate and the warping constant, unfamiliar enough that an error in either would be hard to spot; $M_f/Z_f$ divides the flange’s own bending moment by its own minor-axis modulus, $t_f b^2/6 = 45{,}375$ mm³, and involves nothing anyone would need to look up. Both give 67.0 N/mm².

Held at both ends, with nowhere for the decay to go #

The decay length is fixed by the section and the material. The member’s length is not, and shortening it puts the whole member inside its own boundary layer.

The twist follows the same bracket with $\kappa = kL/4$, here 0.417. Saint-Venant alone would give 4.065°; the real answer is 0.220°, a stiffening of ×18.5. A member eighteen times stiffer than its torsion constant says is not a correction to a theory — it is a different theory, with the original as a limiting case this member is nowhere near.

The sign change at mid-span is the other thing worth taking. Both halves are held flat at their outer ends, and mid-span is held flat by symmetry rather than by a support. That is why a member with both ends entirely free to warp still has a bimoment in it: symmetry is a restraint, and nothing need be attached for it to act.

The sections that have none of this to offer #

Warping resistance is not a property of open sections generally. It belongs to those with a pair of flanges that can bend against each other, and many rolled shapes have no such pair.

The reason is geometric, and worth stating exactly, because “an angle has no warping stiffness” is usually offered as a fact to be memorised. The sectorial coordinate is a swept area measured from the shear centre, and for an angle both legs’ mid-lines pass through the intersection of the legs — which is where the shear centre of an angle sits, because that is where the two shear flows meet. A radius drawn from a point to a line through that same point sweeps no area at all. So $\omega = 0$ everywhere, the warping displacement is zero everywhere, and there is nothing for a restraint to prevent.

What is left above is secondary warping — each plate warping across its own thickness — three orders of magnitude smaller, and the reason the constant is not exactly zero. An angle twisted at its ends has the stiffness its torsion constant says and no more, one more entry on the list of things an angle does not do well.

That same constant is the hinge of a quite different failure. A column that twists instead of bending resists that mode with $GJ + \pi^2EI_w/L^2$, so a section with no warping constant has only $GJ$ to offer — and $GJ$ for an open section is what the slit box above was left with after losing 432 times over. The cruciform, the extreme case, has zero warping constant by the same geometric argument as the angle and buckles torsionally at a load Euler’s formula never mentions. The same $I_w$ that decides how much of a torque the flanges carry decides whether a column has a third buckling mode at all, and the sections worst at one are worst at the other.

Restraint attracts the very thing it resists #

There is a sting in this for anyone reaching for warping restraint as a fix.

A torque that exists only because two members must rotate together is shared by stiffness, and warping restraint is stiffness. Welding an end plate onto a spandrel to stop its ends warping makes it 1.427 times stiffer in torsion, and a stiffer member attracts more of the torque it competes for — the stiffest path takes the load, arriving where the stiffness was added on purpose. For equilibrium torsion that is a straightforward gain; for compatibility torsion it can be a loss, since the extra torque comes with the flange stresses the restraint itself produces.

Where the model stops #

The cross-section keeps its shape. The theory assumes the section is rigid in its own plane and only warps out of it. A member with slender unstiffened plates also distorts, the flanges rotating relative to the web — a third mechanism with its own decay length, and the reason box girders carry diaphragms.

The restraint is perfect or absent. Every number here comes from a section either free to warp or wholly prevented. Real details sit between: a bolted end plate offers partial restraint known to perhaps a factor of two, which is the difficulty of a connection that is neither pinned nor rigid in a degree of freedom nobody classifies.

Nothing else is happening. A real spandrel bends as well, and the warping stress adds to the bending stress at a flange tip while the two shear fields add on one face of the flange and subtract on the other. The governing point is found by combining, not by checking each alone. Nor does anything here yield: a flange tip past yield sheds torque back to the shearing mechanism.

The sectorial coordinate is measured from the shear centre and nowhere else. Using the centroid as the pole gives an $\omega$ wrong everywhere and an $I_w$ wrong by an amount depending on how far apart the two points are — which for a channel is a long way, and outside the steel entirely.

What the pictures cannot show #

The figures share a limitation worth saying plainly: not one of them draws the warping. The displacement is out of the plane of the cross-section, along the member axis, and every view here is a section or a plot against distance. The flange-plan view comes closest, and what it draws is the consequence of warping rather than the dishing itself, at 20 times its true size against a real ratio of about 1 in 205.

The bimoment fares worse. It has no diagram anywhere in the subject, and the figure carrying its name plots the stress instead, because a stress in N/mm² can share an axis with another stress and a quantity in kN·m² can share an axis with nothing. The reason a bimoment goes unnoticed is the reason it is hard to plot — the ordinary inventory of internal actions has no slot of the right shape. The split into two mechanisms is a decomposition rather than an observation, too: a strain gauge at the flange tip measures 67.0 N/mm² and cannot report which mechanism produced it. That the two shares sum to the applied torque at every station is the whole of the evidence for it.

History, and why it arrived so late #

Saint-Venant solved uniform torsion in 1855, and the warping of a non-circular section was the centrepiece of what he found. The theory of what happens when that warping is prevented took another eighty years, and it came from aircraft, where thin-walled open sections carrying torsion were unavoidable rather than a nuisance. Wagner’s 1929 work on the torsional buckling of open struts introduced the machinery, and Vlasov’s Thin-Walled Elastic Beams — circulated in Russian through the 1930s and 1940s, and not in English until 1961 — set out the sectorial coordinate, the bimoment and

$$EI_w\,\theta''' - GJ\,\theta' = -T$$

in the form still used. The order of events is the one the torsion field records elsewhere: the easy case is the circular shaft, which structures never use, and what structures need arrived as its correction, from an industry with a different problem.

The ladder from here #

Later rungs on this anchor: the sectorial coordinate derived properly, with the shear-centre pole and the normalisation that makes $\int\omega\,dA = 0$. The warping constant of a channel, where the shear centre is outside the section and the arithmetic stops being symmetric. The governing equation solved for distributed torque and every standard end condition. Combined bending and torsion at a flange tip, where the design check actually lives. Warping in cold-formed sections, where $kL$ is small enough that the flanges carry nearly everything. Partial warping restraint, and what an end plate is really worth. The distortional mode, the next mechanism down. And warping torsion in the plastic range, where the flanges hinge in their own planes.