The buckling load with no compression in it

Every buckling load in this collection so far has had a compression in it somewhere. Euler’s column is a compression along the member. A beam that fails sideways has a compression flange. A plate that ripples is in compression across its width, and a ring that goes out of round is under a hoop compression. The mechanism is always the same: a force that was doing something useful, once the member has moved, starts doing something else.

There is a case where nothing is in compression at all and the same thing happens.

Why a torque can buckle something #

Take the bar straight, and apply a torque $T$ about its own axis at each end. The torque vector points along the bar. Nothing about that configuration is unstable, and nothing in the bar is in compression.

Now let the bar bow, by a very small amount, into a shape $y(z)$. The ends have rotated, because the ends of a bowed bar are no longer perpendicular to the original axis, and the torque vector — which is applied about a fixed direction, or about the bar’s own end tangent, depending on the mechanism applying it — is no longer parallel to the deflected bar. It now has a component transverse to the bar.

A transverse component of a torque is a bending moment about an axis at right angles to the plane of the bow, and it therefore bends the bar in the perpendicular plane. That bow acquires its own transverse torque component, which bends the bar back in the first plane. The two bending planes drive each other, and the deflected shape they sustain between them is a helix.

Writing the two coupled equations,

$$EI\,y'''' = -T\,z''', \qquad EI\,z'''' = T\,y''',$$

and combining them in the complex variable $u = y + iz$ gives $EI\,u'''' = iT\,u'''$, whose non-trivial solution for a bar pinned at both ends against bending but free to rotate about its axis is

$$T_{cr} = \frac{2\pi EI}{L}.$$

Greenhill wrote it in 1883, the year after he did the same for a column under its own weight — the problem this site has already drawn — and the second result is much less known than the first.

What is not in the expression #

There is no $G$ and no $J$. The critical torque of a member in torsion contains only its bending stiffness. That is worth stopping on, because it is entirely counter-intuitive and it is exactly right: what buckles is a bending mode, and the torque is merely the agent driving it. The bar’s resistance to being twisted is irrelevant, because it is not being asked to twist any more than it already is.

The length appears to the first power. Euler’s load goes as $1/L^2$ and this goes as $1/L$, which means the two do not scale together: the ratio $T_{cr}/(P_E L)$ is a constant of the section, and a longer member loses axial capacity faster than it loses torsional capacity. That is the opposite of what the shared word “buckling” suggests.

And the constant is $2\pi$. No square roots, no numerical fitting, no Poisson’s ratio. The end conditions change it in a very simple way — 2 for a bar pinned at both ends, 4 for one clamped at both, and 1.4303 for a cantilever with a torque at its free end.

Which free body produced the number #

The free body is a length of the bar cut at one end, with the applied torque on the cut face and the bar deflected. Two things are on it: the torque, drawn as a vector along the original axis, and the internal bending moments on the cut. The whole instability is contained in the observation that those two are not parallel once the bar has moved, and the component of the first about the second’s axis is what the equations balance.

That is the same structure of argument as second-order analysis in general: a force whose direction relative to the member changes as the member moves. What is different here is that the force is a couple rather than a push, and that the equilibrium being disturbed is one in which no member was under any threat.

The mode it is not #

There is a second thing called torsional buckling in this subject and it is a different phenomenon with an unfortunately similar name.

Telling them apart is easy once the question is what is applied. In flexural-torsional buckling an axial force finds a twisting mode. Here a torque finds a bending mode. The first is a property of the section’s shape — it is why channels and angles are awkward and hollow sections are not — and the second is a property of nothing but $EI$ and $L$.

The number that ends it #

A critical load is only a design case if it arrives before something else. Compare the critical torque with the torque that simply yields the same section, $T_y = \tau_y J/r$:

$$\frac{T_{cr}}{T_y} = \frac{2\pi E \pi r^3 t / L}{\tau_y \cdot 2\pi r^3 t / r} = \frac{\pi E r}{L\,\tau_y}.$$

Every dimension has cancelled except the radius, and the thickness has gone entirely. Setting the ratio to one gives the length at which the mode becomes reachable:

$$L^* = \frac{\pi E}{\tau_y}\,r \approx 3{,}142\,r$$

for any structural steel, since $E/\tau_y$ is about a thousand for all of them. The 168 mm tube drawn would have to be 256 m long before it buckled under torque rather than yielding — and at 6 m the factor is 42.7.

So this is a real instability, exactly computed, that governs nothing anybody would build. It matters in exactly one place, and there the length is not a problem: a drill string is a tube several kilometres long, driven by torque from the surface, and helical buckling is its central design difficulty.

The half that does matter #

The result that a structural engineer can use is not the critical torque but its effect on something else.

$$\left(\frac{T}{T_{cr}}\right)^2 + \frac{P}{P_E} = 1$$

is the linear-theory result for pinned ends, and it is worth reading for what it says at small $T$: because the torque term is squared, a torque of a third of the critical value costs about 11 per cent of the axial capacity and one of a tenth costs 1 per cent. The interaction is flat where it matters, which is the reason nobody checks it.

That flatness is the real conclusion of this essay. The mode exists, the expression is exact and elegant, the coupling is genuine — and the practical instruction it produces is ignore it, arrived at by computing it rather than by never having asked.

There is one place the flatness stops helping, and it is where a member is already close to its axial capacity. At $P/P_E = 0.9$ the remaining torque capacity is $\sqrt{0.1} = 0.32$ of $T_{cr}$ — so a nearly-buckled column has lost two thirds of a capacity it never knew it had. Whether that matters depends on whether anything is applying a torque, which in a building frame it very rarely is.

What it shares with the rest of the subject #

Strip away the torque and the helix, and this result belongs to the same family as everything else in this field.

The general statement is that a load applied to a member does two things at once: it produces stresses, and it removes stiffness. The removal is the geometric stiffness, a term proportional to the force already present, and a buckling load is the load at which it has removed all of the structural stiffness there was. Which mode it removes the stiffness from depends on how the load moves when the member does, and a torque moves in a way that attacks bending.

That framing also explains the absent $G$ in one line. Geometric stiffness is subtracted from whatever stiffness resists the mode being driven; the mode here is bending; so the stiffness that appears is the bending one. The torsional stiffness resists a mode nobody is driving.

The drill string, which is where it is not academic #

Every structure in this collection is short in the sense that matters here. A drill string is not: it is a continuous tube two to eight kilometres long, driven from the surface by a torque large enough to turn a bit through rock, and carrying almost no axial compression because most of its weight is held up in tension from above.

Put 3,000 m into $L^* = 3{,}142\,r$ and the radius at which the mode arrives is 955 mm — vastly larger than a drill pipe, which is about 60 mm. So a drill string is not merely past the threshold; it is past it by more than an order of magnitude, and helical buckling is not a check but the normal operating state of the lower part of the string.

Two things change once the mode is not merely reachable but unavoidable. The helix has a borehole wall to lie against, so the deflection is bounded and the problem becomes one of contact rather than of stability — what is wanted is the pitch of the helix and the friction it generates, not a critical load. And the friction the helix generates against the wall absorbs torque that was meant for the bit, which is why the torque at the bit is a fraction of the torque at the surface and why the fraction is not known.

Neither of those questions is a structural-statics question, and both are downstream of an expression written in 1883 about a bar with nothing pushing on it.

Where the model stops #

The bar is prismatic, straight and elastic. Every one of those is doing work. A bar with an initial bow has no bifurcation at all; the helix grows from the first newton-metre, and the amplification is one over one minus the load ratio, exactly as an imperfect column’s deflection does.

The torque is applied in a stated way. “Semi-tangential”, “quasi-tangential” and “axial” torques are three different loadings with three different critical values, differing by up to a factor of two — because a torque is a couple, and a couple applied through a universal joint is not the same load as one applied through a gear. That sensitivity is the same one a follower force has, and it means the answer depends on the machine at the end of the member.

The section is doubly symmetric. For anything else the bending and twisting are coupled before the torque arrives, and the problem becomes a general flexural-torsional one with the applied torque as an extra term.

The comparison with yield uses a thin tube. For an open section the elastic torsional stress at a given torque is far higher, so $T_y$ falls and the reachable length falls with it. An open section still buckles at $2\pi EI/L$ and yields very much sooner, so the mode is further out of reach rather than nearer.

And the drawing is of the mode, not of the path. The helix has an undetermined amplitude, as every eigenvector does, and no stress can be computed from it. What happens past the critical torque — whether the helix grows stably or the bar wraps itself into a lock-up — is a large-displacement question this linear result cannot be asked.

The ladder from here #

Later rungs on this anchor: the drill string proper, where the tube is confined inside a borehole so the helix has a wall to lie against, and the buckling is a contact problem whose answer is a pitch rather than a load. Coiled tubing, which is the same problem with residual curvature in the member before it starts. The combined torque, axial load and internal pressure case, which is the real drilling condition. Torsional buckling of a bar with an initial helical imperfection, and the amplification factor it obeys. Lateral-torsional buckling read as a member of this family, where the destabilising agent is a bending moment rather than a torque and the mode is the same coupled bending-and-twisting. And the historical thread: Greenhill’s two 1883 papers between them opened the whole subject of non-conservative and geometrically-coupled instability, and everything on this site about second-order effects is downstream of them.