I have a vague understanding of the compatibility equations for linear elasticity. They appear to be necessary in obtaining a unique displacement field. However, why is it that, in my papers I've come across, the compatibility equations aren't solved for in the linear elastic numerical formulation? Are they not a governing set of equations for linear elasticity?


1 Answer 1


A reasonably complete discussion of compatibility can be found at https://en.wikipedia.org/wiki/Compatibility_(mechanics)

There are six strain-displacement relations and only three displacements in continuum mechanics. The compatibility conditions will be needed if you wish to compute displacements given strains. However, most numerical formulations of mechanics are displacement-driven and the need for those equations typically does not arise because one uses: displacement -> strain -> stress -> force.

For force-driven formulations, one has to make sure that the displacements are compatible. That is because the direction of computations goes as: force -> stress -> strain -> displacement.

  • $\begingroup$ What about in the case of boundary conditions. If given a boundary condition in terms of traction/stress, and we are after a displacement field solution, wouldn't this necessitate the need of the compatibility equations? $\endgroup$
    – anonuser01
    Oct 2, 2018 at 20:29
  • $\begingroup$ Even in those cases we typically numerically formulate the problem as f = K u rather than u = K* f. The function spaces for the unknown fields are chosen so that they satisfy the needed integrability conditions (including jumps). It is extremely rare to see strains being integrated directly, but the assumption that they are integrable is implicit in all formulations. $\endgroup$ Oct 2, 2018 at 21:03
  • $\begingroup$ For more on the subtleties involved, see: Angoshtari, A., & Yavari, A. (2015). On the compatibility equations of nonlinear and linear elasticity in the presence of boundary conditions. Zeitschrift für angewandte Mathematik und Physik, 66(6), 3627-3644. $\endgroup$ Oct 2, 2018 at 21:09

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.