I've been reading the Comsol post on geometric nonlinearity. The article introduces situations where the small angle approximation and linear analysis of structures gives inaccurate results. For example, the case of a beam attached from both ends is discussed, specifically when the deformation is so large that the horizontal supporting force must be taken into account and the stiffness of the beam is no longer constant. Buckling also comes up. Here is a quote:

Buckling, or the loss of stability when the load reaches a certain critical value, is caused by geometrically nonlinear effects.

Why is buckling caused by geometrically nonlinear effects? What is meant by this statement?

  • $\begingroup$ Your title and your question aren't exactly the same. nonlinear effects aren't the same as nonlinear geometry. $\endgroup$
    – Eric S
    Commented Dec 31, 2019 at 0:49
  • $\begingroup$ Linear implies that the deformation continues in a predictable change so when load is doubled so is also the deformation. Buckling is a quick change in a otherwise linear otherwords not linear. Loss of stability means much the same sudeen change sudden change means nonlinear. Now this follows simply from the usage of the words in context they just describe things that can not be expected to work linearily (we wouldbt call it loss of stability if it would be linear, thats what stable means) $\endgroup$
    – joojaa
    Commented Dec 31, 2019 at 8:12

1 Answer 1


The real world doesn't know anything about approximations you make in mathematical models of it. If always behaves in a fully 3-dimensional way.

For Euler buckling of a column, you are really studying perturbations about the steady stress condition, which is just an axial direct stress in the column. You want to find the end load at which a small perturbation becomes unstable.

If you try to make a mathematical model that only includes first-order approximations to the behaviour of the perturbed structure, the results say that the column is stable for any compressive load. Since that doesn't correspond to what happens in real life, the over-simplified model is just wrong, and useless.

As an example of why first-order approximations don't always work, consider a bar of length $L$ along the $x$ axis, from $x = 0$ to $x = L$, with no loads applied to it. The strain and stress in the bar is obviously zero.

Now, rotate the bar through 180 degrees, so the end at $x = L$ is now at $x = -L$. Obviously the strain and stress in the bar is still zero.

But if you analyse this as a simple column, you have changed the original length of $+L$ to a length of $-L$, creating an axial strain of $-2$, and therefore there is a huge compressive stress in the bar. Clearly something went wrong with the math!

To make a model with the correct behaviour (or at least, closer to the correct behaviour than "completely wrong") you have to include second-order effects. For example, if the column is perturbed into a curved shape, the point of application of the end load position moves corresponding to the deformed shape of the column. If the end of the column moves "sideways" from the axis, that means the end load is now applying a bending moment to the base of the column as well as an axial load.

Also, since the perturbed shape of the column is a curve, and the length of the center line of the column does not change measured along the curve, that implies that the end of the column moves axially as well as sideways. That means that the applied load does work (= the appled force $\times$ the axial component of the displacement).

The problem with understanding most "simple" derivations of Euler buckling is that you meet them before you have learned the general concept of how to model arbitrary (large) displacements of a body. Therefore, the derivation often seems to be "pulling rabbits out of a hat" in order to get to the accepted answer, but doesn't provide a convincing explanation of why it was done that way.

An article like your blog link doesn't really help. It doesn't say anything that is wrong, but it can't replace a university-level course on continuum mechanics that is a follow up to the "small displacements and small strains" approximations that are made in most first courses.

  • $\begingroup$ Ok, this is a good answer. However it could be more accessible if you would recount the simplification that is made to make these computations linear. $\endgroup$
    – joojaa
    Commented Jan 2, 2020 at 7:36

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