The column that twists instead of bending

Take four flat plates, each a hundred millimetres wide and six thick, and weld them into a cross. Stand the cross on end and press. It does not bow sideways. It rotates — every arm sweeping round about the line where the four meet, the two ends staying exactly where they were put, the member’s axis staying dead straight throughout.

Nothing in Euler’s formula describes that. The formula is about a column that leaves its axis; this one keeps its axis and abandons its orientation instead. And the load at which it happens has a property no flexural buckling load has: it is the same at every length. A cruciform two metres long and a cruciform ten metres long twist at the identical load, and shortening one buys nothing at all.

The formula that has one mode, and the column that has three #

The column curve every reader meets first is drawn against slenderness, and it contains two failure modes: squashing, and buckling about the weaker principal axis. The second of those is written $\pi^2 EI/(KL)^2$, and the $I$ in it is the smaller of $I_y$ and $I_z$.

Taking the smaller of two numbers is already an act of counting modes: a column has two ways to bend and takes the cheaper. The claim here is only that the count is wrong. A cross-section has three degrees of freedom it can leave equilibrium in — displacement in each of the two principal directions, and rotation about the longitudinal axis — and each is a buckling mode with a critical load of its own:

$$N_{cr,y} = \frac{\pi^2 E I_y}{(K_y L)^2}, \qquad N_{cr,z} = \frac{\pi^2 E I_z}{(K_z L)^2}, \qquad N_{cr,T} = \frac{1}{i_0^2}\left(GJ + \frac{\pi^2 E I_w}{(K_w L)^2}\right).$$

The third is the one nothing in an introductory treatment mentions, and its shape is different from the other two in a way that matters more than its unfamiliarity. It has two terms, and only one of them contains the length. $GJ$ is St Venant torsion — each plate twisting about its own mid-thickness — and it is a stiffness per unit length of member, so a shorter member gets no more of it. $EI_w$ is warping torsion, the flanges bending against each other, and that term does behave like the flexural ones, falling as $1/L^2$.

So a section with a healthy warping constant has a torsional load that climbs as the member is shortened, in company with the flexural ones. A section with no warping constant has a torsional load that is a constant, and the flexural loads climb past it.

Which body was cut, and about which point #

The number 700 kN in the hero figure is not quoted from anywhere. It comes out of an equilibrium statement, and the whole argument of this essay is contained in the choice of point that statement is written about.

The free body is a slice of the column of length $dx$, cut on two planes normal to the axis, with the axial stresses on both faces and the internal torque on both faces. The equilibrium equation written is moment about the shear centre, not about the centroid.

That choice is forced. The shear centre is the point about which the section’s shear resultants have no moment, so it is the point about which a twisting section’s internal resistance is naturally expressed. Write moments about the centroid instead and the internal shear flows contribute a torque that has to be carried as a correction; write them about the shear centre and they contribute nothing.

Having chosen that pole, the axial load’s own line of action acquires a lever arm, since $N$ is applied through the centroid by definition. If the two points coincide, the lever arm is zero, the twisting equation contains no displacement term, and the column takes the lowest of three independent numbers. If they do not, a sideways displacement moves the centroid off the shear centre’s line, the axial load acquires a moment about the pole, and the section twists; and twisting moves the centroid again.

Written out, the three equations are a matrix eigenvalue problem $(\mathbf{K} - N\,\mathbf{K}_g)\,\mathbf{v} = \mathbf{0}$ in which $\mathbf{K}$ is diagonal — the three uncoupled loads — and the geometric matrix $\mathbf{K}_g$ carries the shear-centre offsets $y_0$ and $z_0$ off its diagonal. The critical loads are its eigenvalues, found here by the same negative-pivot count that locates an ordinary column’s, and they are the roots of a cubic rather than the smallest of three separate answers. The solver’s check on itself is exactly the limit this paragraph describes: drive the offset to zero and the three coupled roots must come back to the three uncoupled loads.

A cross has nothing to warp with #

The cruciform is the case that makes the point cleanly, because its warping constant is not small — it is zero, exactly.

The sectorial coordinate $\omega$ is twice the area swept by a radius running from the shear centre to a point walking along the midline. A wall aimed straight at the pole sweeps no area, so $\omega$ stays at zero along it. All four arms of a cross point at the crossing, the crossing is the shear centre, and $I_w = \int \omega^2 t\,ds$ is therefore zero rather than merely small. A tee and an angle follow for the same reason.

What is left is $GJ = 2.33 \times 10^9$ N·mm², divided by the polar radius $i_0^2 = 3329$ mm², giving 700 kN — a load with no length in it anywhere. The flexural load equals it at 3442 mm, which corresponds to a slenderness of 84, thoroughly ordinary. Below that length the cross twists; above it, it bends. At two metres the Euler calculation returns 2073 kN and the truth is 700 kN, so the familiar formula is 2.96 times the answer.

A cruciform is not a common column, but the family it heads is: any section whose plates all meet at a point has $I_w = 0$, and single angles are used as struts by the thousand.

Where the two points differ, the modes are not separate #

Move one flange of an I-section to the other side of the web and the section becomes a channel. The plates are unchanged, so the area is unchanged at 5315 mm², and the torsion constant is unchanged at $1.59 \times 10^5$ mm⁴ — $J$ is $\frac{1}{3}\sum bt^3$ and does not care how the plates are arranged. What changes is where the shear centre goes.

That distance is the whole of the coupling. It enters twice, and the two entries are worth separating because they are often run together.

First, it changes $i_0^2$, the polar radius. The $i_0$ in the torsional load is measured about the shear centre, not about the centroid, so an offset pole inflates it and depresses $N_{cr,T}$ before any coupling happens at all. Second, it puts the off-diagonal terms into the geometric matrix, and those are what make the problem a cubic. The single number that reports the second effect is the Wagner coefficient

$$\beta = 1 - \frac{y_0^2 + z_0^2}{i_0^2},$$

which is 1.00 when the two points coincide and falls toward zero as they separate. The channel’s is 0.60.

That offset is a shear-flow result, obtained by integrating the flow along the walls and taking moments — a calculation with no columns and no buckling anywhere in it. The number a section-properties exercise produces turns out to be the number that settles a stability question three fields away.

What the coupling costs, and which pair pays it #

The coupling does not touch all three modes. A channel is symmetric about one axis, its shear centre lies on that axis, and so one of the two offsets is zero. The mode that pairs with twist is bending about the axis of symmetry; the other flexural mode is left exactly alone.

Four point two per cent is not a dramatic loss, and the modest size of it is part of the point. The mode that couples with twist here is the major-axis one at 19014 kN, far too stiff to be in danger, so the coupling drags a little off the torsional root and gives it to a mode nothing was going to use. It always redistributes that way — from the lowest root to the highest — which is why it can never help, and why the loss is largest when two of the three uncoupled loads sit close together.

The offset cannot be varied on its own, which is why that figure sweeps a flange width rather than a distance. The shear centre is a property of the section, so moving it means building a different section, with different second moments and a different $J$ as well. Every point on those curves is a real channel.

The tee is where the same arithmetic bites hard. Its shear centre sits 80.6 mm from its centroid, its warping constant is zero, and at 3000 mm its coupled critical load is 349 kN against 478 kN for the lowest uncoupled mode — a 27 per cent loss to coupling alone.

The section where the mode never governs, and the inequality that says so #

A reader has to learn where a mode does not matter as firmly as where it does, and the I-section is the case. Its torsional load stays above both flexural loads at every length there is, so the check simply does not arise — and one inequality settles it.

Set the torsional load equal to the lower flexural load and solve for the length. The crossover exists only when

$$\frac{I_w}{i_0^2} < \min(I_y, I_z),$$

because the warping term and the flexural term both fall as $1/L^2$, and if warping wins that race at one length it wins at all of them. For the I-section here, $I_w/i_0^2 = 7.90 \times 10^6$ mm⁴ against $I_z = 6.31 \times 10^6$ mm⁴. The left side is larger, the inequality fails, and there is no crossover: the minor axis governs from zero length to infinity.

That is why the mode is left out of most teaching and most quick checks. Hot-rolled I-sections and universal columns are the shapes a designer meets first, they have generous warping constants and coincident shear centres, and for them $\pi^2EI_z/L^2$ is not an approximation but the exact answer. The formula’s reputation is deserved within its hypotheses; what has been lost is the hypotheses. Change to a cold-formed channel, a lipped C, a single angle or a cruciform and the inequality flips — which is why cold-formed design rules run so much longer than hot-rolled ones for what looks like the same member.

Four sections at three metres #

Four sections of unremarkable proportions, one length, one steel, and the ratio between the textbook answer and the answer runs from 1.00 to 2.08 — a spread caused by none of the things a designer normally varies, only by where the plates were put relative to each other.

Every curve coming home to one is the reassuring half of the finding. Torsional and flexural–torsional buckling is a short-column phenomenon. The stubby braces, the short posts between floors, the strut whose slenderness looked comfortable — those are where it lives, which is precisely where nobody looks, because a short column reads as a safe one.

Where the model stops #

Pinned ends, free to warp. The three loads above use effective length factors of one throughout, including $K_w$ for the warping term. Ends that are prevented from warping — a heavy end plate, a member built into a thick base — halve that factor and raise the torsional load by a factor of four in its warping part. The detail that restrains warping is rarely drawn as a structural component and is almost never in the calculation.

Elastic behaviour. These are elastic critical loads, and a real short column of this kind yields before it reaches one. The design curve interpolates between the squash load and the critical load in the same way the column curve does, and the interpolation region is where most real members sit.

A straight member. The critical load is a bifurcation of a perfect column. A real one is never straight and never untwisted, so it twists from the first increment of load rather than at a threshold, and the imperfection sensitivity of the coupled mode is worse than that of a pure flexural one because two imperfections feed each other.

Thin walls that stay flat. Every plate is treated as a line with a thickness. A 100 mm arm 6 mm thick has a width-to-thickness ratio of 16.7, below the point where the plate ripples on its own — but only just, and a thinner cruciform fails by the section not reaching its own strength before any of this applies. Local and torsional buckling interact, and the interaction is not the lower of the two.

Axial load alone. A member carrying moment as well is a different problem, in which the same offset reappears as the monosymmetry term in lateral-torsional buckling, and the two checks combine rather than being taken separately.

Nothing attached along the length. Sheeting, a slab or a bracing member changes the torsional restraint completely, and a member held against twist at intervals has an effective length for that mode which is a property of what holds it, exactly as the flexural one is.

What these pictures cannot show #

Every figure on this page plots a load, and none of them draws the column. That is a real limitation and it hides the most interesting object in the problem.

The eigenvector is not on the page. Each critical load has a mode shape — the sideways displacement of the shear centre in two directions and the twist — and the coupled modes are mixtures whose proportions change with length. The channel at 3000 mm buckles in a shape that is mostly twist with a little major-axis bending; at 8000 mm the same section buckles in a shape that is mostly bending. The curves show the loads crossing; they cannot show that the mode has been changing continuously the whole way, so the crossing looks like a switch when it is a blend.

The section figures draw the shear centre as a dot in empty space beside the steel, which invites the reading that something is there. Nothing is there. It marks a line of action, in the same way the resultant of a distributed load acts where no force is applied.

One assumption behind all of them deserves saying rather than implying: the section is assumed to keep its shape. Every plate can displace and the whole section can rotate, but the angles between the plates never change. Drop that and the flanges start to move relative to the web, which is distortional buckling — a fourth family, with its own critical loads, sitting between the local and the global ones.

The ladder from here #

Later rungs on this anchor: the governing differential equations and where the cubic comes from. Effective lengths for the torsional mode, and the warping-restraint factor $K_w$. Monosymmetric sections in general, and the sign convention that decides which flange helps. Fully unsymmetric sections, where all three modes couple and no root is exact. Distortional buckling, and the three-family interaction. Cold-formed lipped sections, where lips exist to move the shear centre. Design curves for the flexural–torsional mode, and why they borrow the flexural imperfection factors. Built-up and battened members, where the shear flexibility of the lacing enters the same eigenvalue problem. And the general theory of elastic stability, in which all of these are one pencil with different matrices in it.

Wagner published the torsional buckling of open sections in 1929, working on aircraft structures, where thin unsymmetric open shapes were unavoidable and a strut failing at a third of its calculated load was a live problem rather than a curiosity. Vlasov’s thin-walled beam theory followed in the 1940s and put the warping function, the shear centre and the sectorial coordinate into one formulation. The sections the check applies to — cold-formed, lipped, thin — only became ordinary in buildings decades later. The mode was understood well before the shapes that need it were common, which is the reverse of the usual order.