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Eigenvalue formulation to find the principal stresses, directions

  1. Jan 7, 2009 #1
    1. The problem statement, all variables and given/known data

    3.) Stress analysis at a critical point in a machine member gives the three-dimensional state of stress in MPa as the following:

    y =

    [ 105 0 0
    0 -140 210
    0 210 350 ]
    Using the eigenvalue formulation to find the principal stresses (eigenvalues) and principal directions (eigenvectors). Show these stresses on a properly oriented element. Use the three-dimensional Mohr’s circle to obtain the maximum shear stress and show this on a properly oriented element. Repeat this problem using MATLAB.


    2. Relevant equations

    I can do all except finding the directions. I'm not sure how to find that. Do i just plug back my eigenvalues into mhy homogeneous eqn?



    3. The attempt at a solution


    skipping some steps:

    det( 105 - s 0 0
    0 -140 - s 210 = [0]
    0 210 350 - s )

    Solving for S (using characteristic eqn): S = 105, 427.68, -217.68
    These are my eigenvalues.

    In a previous step, my homogeneous eqn was:

    [ 105 - s 0 0
    0 -140 - s 210 * [l;m;n] = [0]
    0 210 350 - s ]

    If i plug my S values back in, i dont know how to solve.
     
  2. jcsd
  3. Jan 8, 2009 #2

    nvn

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    Science Advisor
    Homework Helper

    As you mentioned, the equation for, e.g., your third eigenvalue, S3 = -217.684056, is (A - S3*I)*X = 0, where I = identity matrix, A is your given matrix under item 1, and X = eigenvector column vector (x1,x2,x3). Therefore, solve for x1, x2, and x3 in the following. The equation (A - S3*I)*X = 0, for your third eigenvalue, is as follows.

    [322.684056 0.0 0.0]
    [0.0 77.684056 210.0]*(x1,x2,x3) = (0,0,0)
    [0.0 210.0 567.684056]

    First, you need to get the above system of equations in row-reduced form. If you are not familiar with row-reduced form, look it up in your linear algebra text book. In row-reduced form, the above system of equations becomes as follows.

    [1.0 0.0 0.0]
    [0.0 1.0 2.7032574]*(x1,x2,x3) = (0,0,0)
    [0.0 0.0 0.00000]

    Therefore, e.g., let x2 = t. Thus, the second equation (row) shown above, 0*x1 + x2 + 2.7032574*x3 = 0, becomes 0 + t + 2.7032574*x3 = 0; and solving for x3 gives x3 = -0.369924*t. The first equation (row) shown above, x1 + 0*x2 + 0*x3 = 0, becomes x1 + 0*t + 0*t = 0; and solving for x1 gives x1 = 0*t. Normalizing the current results, we have (0.0,1.0,-0.369924)*t/sqrt(0.0^2 + 1.0^2 + 0.369924^2), for t not equal to zero. Therefore, the eigenvector corresponding to your third eigenvalue (S3 = -217.684056) is X = (0.0,0.937885,-0.346946).
     
  4. Jan 8, 2009 #3
    ah that rings a bell. So i also have to do thv same thing for my other two eigenvalues too?
     
  5. Jan 8, 2009 #4

    nvn

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    Science Advisor
    Homework Helper

    Yes.
     
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