- #1
Saladsamurai
- 3,020
- 7
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).
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 respective duplicate 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 = εtangentialE 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
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 respective duplicate 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 = εtangentialE 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