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Cyclic Symmetry Analysis

  1. Jan 13, 2014 #1
    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
     
  2. jcsd
  3. Jan 13, 2014 #2

    AlephZero

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    That is wrong.

    The simplest way it to make the meshes match up on the cyclic faces, so that if there is a node ##n_1## at ##r, \theta, z## (in cylindrical coordinates) on one face, there is a corresponding node ##n_2## at ##r, \theta + 2\pi/ N, z## om the other face, where ##N## is the number of sectors.

    Note, it is often convenient to make the cyclic faces planes, with constant ##\theta## on each one, but that is not necessary.

    For harmonic index 0, you can then equate the corresponding displacements at each pair of nodes ##n_1## and ##n_2##, in cylindrical polar coordinates. But making them equal does not make them both zero, of course.

    For the other harmonic indexes, you can consider all the displacements in the model as complex variables instead of real, and them impose constrains on the boundary displacements something like ##u_2 =u_1 e^{i 2 \pi k /N}## for index k.

    Warning, there may be some signs and/or factors of 2 wrong here, but that's the general idea of how to do it.

    There are some variations on this theme. For example you can rewrite the transformations to work in Cartesian coordinates not cylindrical; and if your software doesn't like complex variables, you can pretend there are two separate copies of the sector model, and treat the real and imaginary parts of each complex displacement as two real displacements.
     
  4. Jan 13, 2014 #3
    That was what I was thinking. If the displacements were zero along those faces, so would the stresses which makes no sense.

    As an aside, Aleph:

    When it comes to harmonic indices, I am not too knowledgeable. For example, when we analyze the entire 360° body, without using a symmetry condition, it seems as though harmonic indices (HI) never enter the picture. But once we start using the cyclic symmetry, we start talking about HI's.

    Does that sound right? That HI's arise only as a result of using symmetry?
     
  5. Jan 13, 2014 #4

    AlephZero

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    If the structure is cyclic, you can describe the whole structure as a fourier transform of one segment, therefore you can represent any linear response as a fourier series.

    For vibration analysis there are no external loads, so the response for each fourier component (or harmonic index) is indepedent of the others. For a stress analysis, you might need to do a fourier decomposition of the loads, and then sum the response from each harmonic index - or it might be easier just to ignore the symmetry and model the complete structure.

    Of course if your model doesn't somehow how "tell" the analysis software the structure is cyclic, you will calculate the response assuming it is not cyclic. So you will get the same set of results, but not split up into different harmonic indexes.

    For nonlinear response, none of the above applies, unless the nonlinearities can also be modeled with cyclic symmetry.
     
  6. Jan 17, 2014 #5
    OK. That makes sense to some degree. I do not have a lot of Fourier Transform experience ...well none actually ... but now I have a starting point. Thanks again Aleph!
     
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