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Rotating the coordinates to coincide the principal axes

  1. May 20, 2012 #1
    Dear all,

    We can rotate the local coordinates of the element so that the stress tensor becomes diagonal. The new coordinate system would be the principal stress axes of which are in fact the eignevectors of the stress tensor.

    Once we have the eigenvectors ( which are generally orthogonal), we can find the rotation matrix to rotate the coordinates. We usually find the rotation matrix using direction cosine or Euler angels and a bit of work is required.
    But I have found that ( I'm not sure yet) in Cartesian coordinates, the eignevectors can be readily used to construct the rotation matrix. Just set up a 3x3 matrix whose rows are the eigenvectors!

    I have used the method in my code and the results "seems" right but I need someone to confirm this. I am also worried for instances that the eigenvectors are not orthogonal: would this lead to wrong transformation ( i.e with the transformation the stress tensor wont be diagonal)?

  2. jcsd
  3. May 20, 2012 #2
    You're right, Hassan2. A useful result from linear algebra is that any real symmetric matrix has orthogonal eigenvectors. Since the stress tensor is symmetric (except in really weird situations that I don't fully understand and don't typically show up in any commonly used material), it has orthogonal eigenvectors! Hence the matrix you're constructing to do the rotation will always be orthogonal.

    Writing the rotation as a matrix whose rows are the eigenvectors is a perfectly valid way to do it. In general, the first row of a 3x3 matrix is the image of the vector (1,0,0) under the matrix transformation; the second row is the image of (0,1,0); and the third is the image of (0,0,1). So if you want to rotate the standard coordinates to the "eigenvector basis", that matrix with eigenvector rows is exactly what you want.
  4. May 20, 2012 #3
    Many thanks,

    How if the stress tensor has repeated eignenvalues. I think in this case the orthogonality of the eigenvectors is not guaranteed.
  5. May 23, 2012 #4
    Sorry, Hassan2, I keep forgetting to reply.

    You are absolutely right. An excellent point. But the eigenvectors will always be linearly independent, so if it has repeated eigenvalues you can orthonormalize the eigenvectors using Gram-Schmidt and use those orthonormalized vectors to produce your rotation matrix.

    Again, sorry for the late reply (I'm new to this forum!)
  6. May 24, 2012 #5
    Thanks medthehatta,

    I checked with Wikipedia about the principal stress which are in fact the eigenvalues of the stress tensor. For 2D space , the analytical solution is given. It seems when we have nonzero shear stress, the eigenvalues can't be equal. It is fine because for zero shear stress, we don't need to rotate the local coordinates. I'm not sure but I expect the same for there-dimensional space. I use numerical methods for finding the eigenvalues of the 3x3 stress tensor. I will check it.

    Thanks again for your help.

    Thanks again.
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