Modal participation factor & effective mass - semi definite system

AI Thread Summary
A torsional semi-definite system problem has been solved, focusing on the modal participation factor and effective mass. The presence of a rigid mode with a natural frequency of zero raises questions about the contribution from flexible modes, leading to a modal participation factor vector of [5.3157, -3.76*10^-3]. The discussion highlights that while the arithmetic appears correct, the application of modal participation factors may be misapplied, particularly in scenarios like earthquake response analysis. It is noted that the rigid mode's participation factor will dominate, suggesting that the flexible mode's contribution is minimal but not zero. Proper modeling is essential, as fixing the rotor at one end limits the modes available for analysis.
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).
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Back
Top