If a member is very slender then its length will be large compared to its cross-sectional area. I've heard and read that slender members can potentially cause the stiffness matrix of a beam/frame element to become ill-conditioned in FEM.

Question: How/why can a slender member cause the stiffness matrix to become ill-conditioned?

Here are my own thoughts regarding this question, which may or may not be correct - I'd like some confirmation. A long slender member will have larger axial stiffness than flexural stiffness. In the stiffness matrix, the axial stiffness term has the length of the member ($L$) in the denominator. However, the flexural stiffness terms typically have $L^2$ and $L^3$ in their denominators. Therefore, given constant cross-sectional properties ($A$ and $I$), as $L$ grows the flexural stiffness terms approach zero much quicker than axial stiffness. Is this the reason for the potential for ill-conditioning? Or is there something I'm missing?

  • $\begingroup$ To clarify, you're talking about modelling a beam (or column) with finite elements instead of using a unidimensional bar element, correct? I'm no FEM expert so I can't give an actual answer, but I took a class once where I was taught that the FEM equations somewhat "break down" when an element is badly proportioned, such as if one of its dimensions is more than 4x the size of the others. The FEM software STAAD throws a warning if this "4x" threshold is passed. This would imply that smaller elements would stop the matrix from becoming ill-conditioned. As to the why, I honestly don't remember. $\endgroup$
    – Wasabi
    Commented Sep 5, 2015 at 12:41
  • $\begingroup$ Yes that's right. Sorry I should have stated that. Finite elements with beam elements. Okay that's interesting. I'll lookup the user manual for the software and maybe it will provide some reasoning as to why a factor of 4 is chosen. $\endgroup$
    – pauloz1890
    Commented Sep 5, 2015 at 12:45
  • $\begingroup$ Wait, are the slender beams modeled with unidimensional bar elements? The 4x threshold I mentioned is for 2D or 3D elements, not for 1D. $\endgroup$
    – Wasabi
    Commented Sep 5, 2015 at 12:53
  • $\begingroup$ The beam element is 2D but like any element it can be anywhere in 3D space. See here $\endgroup$
    – pauloz1890
    Commented Sep 5, 2015 at 12:56
  • $\begingroup$ @Wasabi you are right that having elements with very different sizes can cause numerical problems, but the ill-conditioning of the OP's problem is caused by the behaviour of the whole physical structure, not by how it is discretised. Modelling a single beam-like component of the structure with several equal-sized elements joined end-to-end can make the numerical conditioning worse, not better, because with $n$ elements you have a factor of $n^3$ between the bending stiffness of one beam element, and the stiffness of the whole structure. $\endgroup$
    – alephzero
    Commented Sep 6, 2015 at 1:47

2 Answers 2


You have the right idea, but it doesn't really make sense to say "the ratio of bending to axial stiffness depends on $L^2$" because $L$ isn't dimensionless. If you measured the length of same beam in mm and km you would reach two very different conclusions about the ratio.

A better version of the argument is to say that the axial stiffness is of order to $A/L$ and the bending stiffness $I/L^3$. For a typical beam cross section, $I$ is of order $At^2$ where $t$ is the "thickness" or "diameter" of the beam. So the ratio of axial to bending stiffness is of the order of $t^2/L^2$ (or $A/L^2$ if you prefer that) which is a dimensionless quantity.

In some special situations this doesn't matter at all. For example, if the axis of the beam is aligned with one of the coordinate axes of the model, the axial and bending stiffness terms are not cross-coupled in any way, and there is nothing to degrade the answers.

In practice, even with FE software that uses a simple-minded equation solver working in 64-bit arithmetic (double precision, 16 significant figures), there shouldn't be any serious problems with a ratio of the order of $10^6$, but if you exceed $10^8$ you are starting to live dangerously and at $10^{10}$, all bets are off unless you have some solid evidence that the results are meaningful.

If you have structural analysis FE code that works entirely in 32-bit arithmetic, my advice would be "don't use it for anything except toy problems" - but that advice may be over-pessimistic for some non-structural applications of FE.

  • $\begingroup$ Yes, thank you for this. So it's a problem with round-off and truncation errors in numerical computation as I thought. $\endgroup$
    – pauloz1890
    Commented Sep 6, 2015 at 1:57

Yes, you're right. As you imply, in a long, slender beam the bending stiffnesses will be orders of magnitude less than the axial stiffness and mixing numbers that are orders of magnitude different in any numerical calculation can cause problems.


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