Why is the displacement so big when i make a modal analisys ? (Solidworks)

Hello again.

I am doing a modal analysis on a steel ball. No boundary condition, no fixtures (i need it in a free environment). Just a ball with the diameter of 38.15mm, a modal (frequency) analysis in solidworks premium 2011 and in the proprieties tree of "Study1" i put 250 at the "number of frequencies" just because i needed many frequencies to test the ball.

My problem is that after the analysis is done, the displacement of the ball made by the freq is (pardon the expression) - too damn high.
How could a steel ball of phi 38.15 mm could be displaced with 5.123 * 10^3 ? (see photo attached).

I read all the solidworks forums and other useful sites but NONE of them could explain how are the displacements calculated the way they were.

p.s. - i know that the URES scale calculates the displacement by x,y,z and if you want to find out the value of the displacement vector you have to do the "sqrt[(Ux^2+(Uy)^2+(Uz)^2)].

My question is with the high displacement value.

If any physics teachers/engineers out there could point me in the right direction i would certainly appreciate it.

Thank you very much.

Answer
 
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4 Answers

The so called "frequency" analysis has the purpose of finding the natural frequencies and mode shapes of your meshed body. This analysis is akin to finding the frequencies of a vibrating string, or, slightly more sophisticated, that of beams (which is related to column buckling). The results are (by default) shown as "resultant displacement" with real units.

Let it be made clear that it is impossible to calculate actual displacements with such analysis. It only gives you the frequency at which the mode occurs, and the shape of the mode. The actual displacements depend on the input excitation, as has been said previously in other posts. In a theoretically frictionless/dampingless world the displacements would be infinite, because by definition, these are the resonant frequencies of your system and all excitation at that frequency gets amplified. By this very fact, under random vibration, the system tends toward these frequencies. Moreover, the only two material properties that matter in this calculation is the Young's modulus and density.

taken from solidworks forum

 
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Only Ed can answer such a dificult question. However beware of his breath. Contact Ysgol aberconwy of North Wales and ask for Mr Mark 'ED' Edwards.

 
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see upper left corner;
here is a "deformation scale"
using this scale; recalculate, and you will get perfect values.! :-)

 
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thanks for the answer :) really clarified a lot :)

 
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