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Modal participation factor & effective mass - semi definite system

  1. Dec 26, 2013 #1
    i have solved a torsional semi-definite system problem.
    i have uploaded the solved problem.
    solved for modal participation factor and effective mass.

    for a semi definite system we know that one of the modes is rigid mode with natural frequency 0.

    i think due to the presence of rigid mode i'm getting the modal participation factor vector as [5.3157 -3.76* 10^-3].

    is this answer right? i mean how can there be no contribution from the flexible mode?
     

    Attached Files:

  2. jcsd
  3. Dec 27, 2013 #2

    Chronos

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    It appears you are assuming an unrealistically rigid mode. There is no such thing as a perfectly rigid body.
     
  4. Dec 27, 2013 #3
    how do i compute modal participation factor then?
     
  5. Dec 27, 2013 #4

    AlephZero

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    I don't understand Chronos's comment. the zero frequency mode describes motion of the structure with no deformation, (i.e. no internal strain energy). The flexibility of the structure (or lack of it) is irrelevant. Any structure that is not fixed in some way will have zero frequency modes.

    The OP's arithmetic looks OK in terms of putting numbers into the equations, but I think the issue is that this is not the right way to use modal participation factors. A good application would be calculating the response of a building to earthquake loads. You find the normal modes with the base of the building fixed, and then use the mode participation factors to find which modes are important in the dynamics, when the base moves in a prescribed manner.

    If one of the modes corresponds to rigid body motion of the structure, it's kind of obvious that its MPF will be much bigger than the others, and that's what you got. I don't think the other MPF for the elastic mode is "zero", it's just very small number compared with MPF of the zero frequency mode.

    But if you model your rotor fixed at one end, you only have one mode anyway, so calculating its MPF won't tell you anything useful (except that if its effective mass is not equal to the real mass, you made a mistake somewhere).
     
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