1
$\begingroup$

Is this a coincidence or there is a more deep connection between the notion of the harmonic oscillator and the notion of buckling?

$\endgroup$
3
  • 4
    $\begingroup$ Smoked fish are harmonic oscillators? Please consider giving more context here. $\endgroup$
    – ACuriousMind
    Jan 24, 2021 at 21:41
  • $\begingroup$ The question is about "the notion of buckling" which is more general than "bucking of axially loaded columns" (though that may be the only special case that is discussed in some engineering and physics courses!) $\endgroup$
    – alephzero
    Jan 25, 2021 at 2:01
  • $\begingroup$ @ACuriousMind The wiki page you referenced is incomplete. The panel at the bottom mentions Freshwater herrings, Longfin herrings, Round herrings and Thread herrings, but it ignores Red herrings. $\endgroup$
    – alephzero
    Jan 25, 2021 at 2:04

3 Answers 3

3
$\begingroup$

It's because the harmonic oscillator is a very general tool. If you have a displacement-dependent force which you don't know, the first approximation is to approximate it linearly as $$f=-k\delta$$ where $\delta$ is the displacement. Now, materials are often non-linear ($f$ has dependencies to $\delta^n$) but the linear (elastic) theory gives a pretty good description already!

You can model this way (as a restoring force depending on $\delta$) any phenomenon which has an equilibrium state from which it gets displaced "a little bit".

Now, when you described buckling (assume 1D) you start with a straight rod, which you assume is at equilibrium (energy $E=0$) then you deform it of an amount $\delta$ and assume the energy is increased by an amount $1/2k\delta^2$ as if it was a "spring" (remember: if $f=-k\delta$ then $E=1/2k\delta^2$). You pay an energy cost scaling with $\delta^2$ for the deformation, the same you pay to compress a spring.

The reason the linear theory is so nice is that Newton's equation for such a force reads

$$m{d^2\delta \over dt^2}=-k\delta$$

which has the very simple solution for the displacement $$\delta(t)=Acos(\Omega t+\phi)$$ where $A$ and $\phi$ depend on your problem and $\Omega$ depends on the material properties (the spring constant and the mass). Because the force is linear, if you have multiple forces acting on your rod as in buckling, where the stress distributes along the rod, you can just sum them and you get a sum of cosines which are very easy to treat with Fourier series' analysis.

Let's call $\delta(x)$ the displacement at a point $x$ along the rod. If the rod is straight at $x$ then $\delta(x)=0$, otherwise $\delta(x)$ it's the distance with respect to the equilibrium position.

The form of the force we choose for buckling has as a consequence that the main way a rod can buckle is following a cosine $\delta(x)\approx cos(\pi x/L)$ (where $L$ is the length of your rod) profile (fixed at the estremities and buckled at the center), followed by $\delta(x)\approx cos(2\pi x/L)$ with 2 "buckled" points, then three. The general solution to your buckling problem, is a sum of all these configurations.

It is a direct consequence of choosing something like $f=-k\delta$ to model your problem.

$\endgroup$
3
$\begingroup$

For small-amplitude vibrations of a structure with internal stresses, the equation of motion can be written as $$M\ddot x + \omega^2(K_e + K_\sigma)x = 0$$ where $K_e$ is the elastic stiffness of the material, and $K_\sigma$ is the additional stiffness terms caused by the internal stress. ($K_\sigma$ can be interpreted as the work done by an infinitesimal rigid body rotation of the stressed structure.)

Euler's theory of buckling is equivalent to saying that the structure buckles when the lowest frequency $\omega$ reduces to $0$.

The connection with a harmonic oscillator is rather tenuous, because this criteria doesn't really have anything to do with the mass of the structure. The real criterion is just that the internal stress distribution when $K_e + K_\sigma$ becomes singular. Since $M$ is non-singular for any real structure, $K_e + K_\sigma$ is singular if and only if $M^{-1}(K_e + K_\sigma)$ is singular, so bringing $M$ into the math doesn't change anything.

$\endgroup$
1
$\begingroup$

The only obvious connection would be as follows.

For small deflections of a slender column in compression, the resulting column curvature resembles a sine wave with zero-deflection constraints on its ends. That sine wave could in principle be associated with the natural frequency of vibration of that column, at least in the sense that if the column were loaded near its stability limit and it were then excited at that frequency, it would immediately "pop out" into its deflected shape in response to that perturbation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.