I am trying to understand a little further how software such as ANSYS implements cyclic symmetry in an analysis. A colleague of mine spoke to a support engineer and I think that he may have misinterpreted what was said. He is now under the impression that when we invoke a cyclic symmetry condition, the two circumferential faces of the wedge are "fixed" (i.e. all degrees of freedom (DOF) on those faces have zero displacement).(adsbygoogle = window.adsbygoogle || []).push({});

I cannot seem to accept that this is the case. If the face DOFs were fixed, then that implies that the full 360° body has surfaces along which the stresses and strains are zero (which seems physically unreasonable to me). I am under the impression that what really happens is that the bounding faces of the wedge are "constrained" to their respectiveduplicate sector faces.

Secondly, he is under the impression that in a body that is cyclic symmetry, such as a cylinder, that the circumferential displacement is everywhere zero. I can't buy this either. That would imply that hoop stress = pR/t = ε_{tangential}E is also everywhere zero (which we know to be false).

Any input would be appreciated. I feel like my logic is sound, but could use a sanity check.

-Thanks

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# Cyclic Symmetry Analysis

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