The load it cannot buckle under

There is an assumption underneath every buckling calculation in this collection, and it is so nearly always true that it is never written down: the load has a potential.

A weight hangs downward whatever the structure does. So the work it does depends only on where the structure ends up, not on how it got there, and a potential energy exists — which is what makes the whole apparatus of energy methods legitimate. A critical load is where the second variation of the total potential stops being positive; guessing the shape gets a good answer from a bad guess because the answer is a minimum; and a static analysis finds the load at which the stiffness matrix becomes singular, because that is the same condition.

Take the potential away and every one of those statements fails at once.

Which free body produced the number #

Two rigid bars, each of length $\ell$, with a rotational spring of stiffness $k$ at the base and another at the knee, and a force $P$ at the tip that stays tangential to the upper bar.

That last clause is the whole of it. A rocket’s thrust stays aligned with its nozzle; a jet of fluid leaves a pipe along the pipe; a friction drive pushes along the surface it bears on; a moving load on a deforming track pushes along the track. In each case the force’s direction is a function of the structure’s displacement, so the work it does depends on the path taken and not only on the endpoint.

Write the stiffness matrix of the two-degree-of-freedom system in the two bar rotations and it comes out

$$\mathbf{K} = k\begin{bmatrix}2 - p & -1 + p\\ -1 & 1\end{bmatrix}, \qquad p = \frac{P\ell}{k}$$

and the first thing to notice is that it is not symmetric. A conservative system’s stiffness matrix is the Hessian of a potential and is symmetric by Clairaut’s theorem; an unsymmetric one is a certificate that no potential exists.

The second thing to notice is its determinant: $(2-p)k \cdot k - (-1+p)k(-k) = 2k^2 - pk^2 - k^2 + pk^2 = k^2$. Constant. Every $p$ cancels.

A static analysis that finds nothing #

A static buckling analysis looks for the load at which a non-trivial equilibrium configuration first becomes possible — the load at which $\det \mathbf{K} = 0$. Here it never does. The analysis runs, converges, and reports that the column is stable at every load whatever, up to and past the point where a real one would have destroyed itself.

That is not a numerical difficulty and it is not a modelling error. It is the correct answer to the question a static analysis asks, and the question is the wrong one. There is no adjacent equilibrium configuration, so divergence — the buckling everything else in this field does — is not the failure mode available.

What is available is found by putting the masses back. Give the joint a mass $2m$ and the tip a mass $m$, and the free vibration problem $\det(\mathbf{K} - \omega^2\mathbf{M}) = 0$ gives

$$2\Omega^4 - (7 - 2p)\Omega^2 + 1 = 0$$

with $\Omega^2 = \omega^2 m\ell^2/k$. Two real roots at $p = 0$ — two natural frequencies, 0.386 and 1.831. As $p$ rises they approach one another, and at

$$p = \frac{7 - 2\sqrt{2}}{2} = 2.0858$$

the discriminant vanishes, the two frequencies merge, and beyond it the roots are a complex conjugate pair. A complex frequency is an oscillation with an exponentially growing amplitude. The column flutters.

Flutter, and why it is not resonance #

Flutter is worth distinguishing carefully from the other ways a structure can shake itself apart, because the words overlap and the mechanisms do not.

Resonance needs a driver at a frequency the structure has. Something outside is supplying energy at the right rate; remove it and the motion decays. The train that arrives in time with itself is the case.

Self-excitation needs a force in phase with the velocity, which subtracts from the damping; past a critical wind speed or a critical crowd the net damping is negative and the motion grows out of nothing. The motion that feeds itself is that mechanism, and galloping and pedestrian lock-in are its instances.

Flutter needs neither. It is a coupling of two modes: the follower force feeds energy from one mode into the other, and past the critical load the pair exchanges energy in a way that grows. There is no external frequency and no velocity-dependent force; the instability is in the stiffness, and it is invisible to a static analysis because the stiffness is not symmetric.

The practical difference is that flutter has no threshold speed to stay below and no damping level that fixes it — which is the part of this that was thought to be an error.

Ziegler’s paradox #

Add damping to the two springs — a dashpot in parallel with each, of any size at all — and recompute. The characteristic polynomial becomes a quartic in $s$ rather than a quadratic in $\omega^2$, and the stability condition is Routh–Hurwitz rather than “two real positive roots”.

The answer is $p = 41/28 = 1.4643$.

Not 2.0858 reduced a little. Thirty per cent below it, and independent of how small the damping is. At $c = 10^{-6}$ it is 1.4643; at $10^{-4}$ it is 1.4643; at $10^{-2}$ it is 1.4643. The limit as the damping goes to zero is not the undamped system’s answer, and the undamped system is not the limit of the damped ones.

That is Ziegler’s paradox, published in 1952, and for about ten years it was widely assumed to be a mistake in the algebra. It is not. The undamped problem is degenerate: its characteristic polynomial is even, its roots come in symmetric quadruples, and an arbitrarily small damping breaks that symmetry and moves a pair of roots across the imaginary axis. A system whose stability depends on an exact symmetry is a system whose stability is not robust, and the physical reading is that the undamped answer was never the answer; it was an artefact of a model with a coincidence in it.

Past the discontinuity, damping behaves again: at $c = 1$ the critical load is 1.96, at $c = 2$ it is 3.46. So more damping helps, and any damping hurts, and both of those are true.

Reading the unsymmetric matrix #

It is worth spending a paragraph on what the asymmetry of $\mathbf{K}$ actually means, because it is the diagnostic and it costs nothing to look at.

$K_{12}$ is the moment at the base caused by a unit rotation at the knee; $K_{21}$ is the moment at the knee caused by a unit rotation at the base. For a structure with a potential these are equal, and that equality is Maxwell’s reciprocal theorem — the theorem that swaps the question round is the general statement of it, and it is a consequence of the strain energy being a function of state.

Here they are $-1 + p$ and $-1$. They differ by $p$, and the difference is exactly the moment the follower force supplies when the upper bar rotates and the lower one does not. Reciprocity has failed, and the amount by which it has failed is the load.

That gives a cheap test for any structural model, and it is one worth running on anything unusual. Assemble the stiffness matrix and subtract its transpose. If the result is not zero to round-off, the model contains a non-conservative effect — a follower load, an aerodynamic coupling, a friction force that tracks the motion, a geometric stiffness derived inconsistently — and every energy-based conclusion about it is suspect. Most finite element programs do not report it; some will not even assemble it, because they store only the upper triangle on the assumption that there is nothing else.

The energy that goes round in a circle #

There is a physical picture of where the growing motion’s energy comes from, and it is worth having because “no potential” is a negative statement.

Take the tip of the column through a small closed loop — out, up, back, down, returning exactly to where it started. A dead weight does zero net work on that loop, by definition of a potential. The follower force does not: because its direction changed during the loop in a way that depended on the path, the work it does going one way round is not minus the work it does going the other.

So a structure executing a closed cycle of motion — which is what an oscillation is — can extract net work from the load once per cycle, and if that work exceeds what the damping dissipates, the amplitude grows. That is the same energy accounting as the motion that feeds itself and as a hysteresis loop, run the other way: here the loop’s area is positive and it is the structure that gains.

Which also explains the flutter mechanism’s need for two modes. A single-degree-of-freedom motion is a line traversed back and forth, and a line encloses no area — so a one-mode system under a follower force cannot extract net work and cannot flutter. It takes two coordinates to draw a loop, which is why Ziegler’s column has two bars and why the simplest flutter is always a coupling.

Where this occurs, and where it is claimed to occur #

The honest position is that the pure follower force is rare, and the literature’s enthusiasm for it has been criticised for exactly that reason.

Where it is real. A rocket under its own thrust. A flexible pipe conveying fluid, where the momentum flux leaving the free end acts along the pipe — the standard demonstration, and a garden hose flailing is a genuine flutter. A cantilever in an axial flow. A grinding or cutting tool pushed along the surface it is machining, where the friction force follows the tool. A drill string, where the bit’s reaction follows the hole.

Where it is claimed and is doubtful. A great many textbook applications assume a force that stays tangential to a member with no mechanism to make it do so, and if there is no such mechanism then the force is conservative and the whole apparatus is unnecessary. The test is simple and worth applying: is there a physical reason the force’s direction is tied to the structure’s displacement? If nobody can name one, the load has a potential.

Where it is real and not called this. Aeroelastic flutter of a wing or a bridge deck is the same mathematics — two modes, a non-symmetric coupling supplied by the aerodynamics, and a critical speed at which the eigenvalues collide. The motion that feeds itself treats the single-mode version; the two-mode version is flutter proper, and the Tacoma Narrows deck is its most famous instance.

Where the model stops #

The model was linear. A growing oscillation grows until something limits it, and what limits it is non-linear — yielding, large displacements, a change in the mechanism supplying the follower force. So the critical load is the load at which the amplitude stops being zero, and what happens above it needs a different analysis. A subcritical flutter can produce a finite-amplitude limit cycle at loads below the linear critical value, which is imperfection sensitivity reappearing in a dynamic problem.

The masses were lumped. A continuous follower-force column — Beck’s column — gives a critical load of $20.05\,EI/L^2$ against the conservative cantilever’s $2.467$, so the pattern is the same and the numbers are entirely different. The two-bar model is a demonstration, not a design tool.

The damping was internal. External damping — a dashpot to ground rather than across a joint — behaves differently, and can raise the critical load rather than lower it. The paradox is a property of a particular damping distribution, not of damping as such, which is why it took a decade to become uncontroversial.

And the structure was perfect. A real column has an imperfection, so the response below the critical load is not zero — it is a forced response that grows as the critical load is approached, which is the only warning available and which a static analysis cannot supply either. That is the same relationship the column that was never straight sets out for a conservative column, where the imperfection turns a bifurcation into a growth and the growth is what a test measures.

And the springs were linear and elastic. A joint that yields under a growing oscillation dissipates energy, and dissipation is exactly what decides whether an unstable mode grows or settles into a limit cycle — the only thing that stops it is the general accounting, and here the amount of it decides the amplitude rather than the threshold.

The generalisation #

The idea to carry away is not about follower forces. It is that a method carries assumptions it does not state, and the assumptions become visible only when a problem violates them.

The energy method’s unstated assumption is that a potential exists. The static buckling analysis’s unstated assumption is the same one. Both are satisfied by every load a building sees, which is why both are taught as though they were the definition of stability rather than as a technique for a class of problems. The follower force is valuable chiefly because it is a member of the complement of that class, and it shows what the class was.

There is a smaller version of the same point that is worth having on its own. The symmetrised matrix gives 2.000 and the true answer is 2.0858 — four per cent apart, which is close enough that a designer comparing the two would conclude the approximation was fine. It is not fine; it is a coincidence of this particular set of masses and springs, and the two numbers are answers to different questions about different structures. A method that is wrong in kind can be right in magnitude, and the magnitude is not evidence.

Two habits follow. Ask what a method’s failure mode is, not only its answer — a method that returns “stable at every load” for a structure that fails is failing silently, and the only defence is knowing what would make it do that. And be suspicious of a result that depends on an exact symmetry: the undamped column’s 2.0858 is a correct answer to a problem whose symmetry no physical system has, and the thirty per cent that separates it from the real answer is the price of the coincidence.