I wrote a UEL for one element and compared to a reference UEL (FEM), where the results agreed satisfactorily, except the reaction force. The stress, strain, strain energy all were pretty accurate.

I am unable to understand this, how does Abaqus calculate reaction force?

I am simply restraining the element in the bottom(two nodes) and applying displacement(incrementally) at the top(two nodes).

As far as I know, the STIFFNESS matrix is used to calculate reaction force in these cases, the stiffness values are quite similar in my UEL and the FEM UEL. Also if this was the case, the other parameters wouldn't have had the same values.

This is probably leading to convergence issues when I am using my UEL for more than one element, where the solutions cannot even converge for the first increment.

Would be grateful if anybody could help me figure this out.

  • $\begingroup$ Abacus probably assumes that for every force there is an equal and opposite force , so are you sure you have got everything correct and accounted for? Have you made an incorrect assumption? $\endgroup$ – Solar Mike Apr 21 '18 at 7:55
  • $\begingroup$ @SolarMike: I take it that Abaqus takes the equal and opposite of the RHS vector? My results for the single element match perfectly with the standard FEM solution for Stress-Strain and other parameters. But the reaction force is only different and the difference is quite big. All assumptions and approximations made are same for both the FEM and my UEL. $\endgroup$ – Schneider Apr 21 '18 at 15:50
  • $\begingroup$ I would also like to mention that I am applying an all displacement BC, in the single element. $\endgroup$ – Schneider Apr 21 '18 at 16:09
  • $\begingroup$ cross posted : scicomp.stackexchange.com/q/29344/5090 $\endgroup$ – agentp Apr 22 '18 at 16:14
  • $\begingroup$ +1. As one of this site's top users in the Abaqus tag, I wonder if you've seen this question and/or if you might know the answer to it? mattermodeling.stackexchange.com/q/2055/5 $\endgroup$ – Nike Dattani Nov 14 '20 at 3:37

Your Answer

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

Browse other questions tagged or ask your own question.