Modal participation factor & effective mass - semi definite system

Click For Summary

Discussion Overview

The discussion revolves around the calculation of the modal participation factor and effective mass in the context of a torsional semi-definite system. Participants explore the implications of having a rigid mode with a natural frequency of zero and its effect on the computed modal participation factors.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • The original poster (OP) presents a computed modal participation factor vector and questions the absence of contribution from the flexible mode, citing the presence of a rigid mode.
  • One participant challenges the assumption of a perfectly rigid mode, suggesting that such a scenario is unrealistic.
  • Another participant expresses confusion regarding the critique of the OP's approach, asserting that the zero frequency mode indicates motion without deformation, and that all structures not fixed will exhibit such modes.
  • This participant also suggests that the OP's arithmetic appears correct but questions the appropriateness of using modal participation factors in the given context, particularly in relation to earthquake response analysis.
  • It is noted that while the modal participation factor for the rigid mode is significantly larger, the effective mass for the elastic mode is not zero but rather a small value compared to the rigid mode.
  • Concerns are raised about the utility of calculating modal participation factors when only one mode exists, as it may not yield meaningful insights.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the rigid mode and the validity of the OP's calculations. There is no consensus on the correct interpretation of the modal participation factors or the assumptions regarding the rigid mode.

Contextual Notes

Participants highlight potential limitations in the assumptions made regarding rigid body motion and the applicability of modal participation factors in specific scenarios, such as when modeling a rotor fixed at one end.

jason.bourne
Messages
82
Reaction score
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?
 

Attachments

Engineering news on Phys.org
It appears you are assuming an unrealistically rigid mode. There is no such thing as a perfectly rigid body.
 
how do i compute modal participation factor then?
 
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).
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 2 ·
Replies
2
Views
22K
Replies
1
Views
3K
Replies
4
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
1K