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frequency response FEA

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gio1

Automotive
Jun 28, 2003
83
Hi

I have carried out a FE frequency sweep analysis of an assembly using modal superimposition. The assembly is shaken using the same acceleration amplitude across a frequency range along one direction only.

When plotting the response (acceleration) Vs frequency for several nodes there are frequency intervals where this becomes NEGATIVE.

Does a negative acceleration make sense in this case?? If so, what is the underlying physical meaning?

Thanks

Gio1
 
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Depends what you have plotted, but if you have plotted the real part or the imaginary part of the transfer function, yes it makes sense.

If you have plotted the magnitude in dB then it also makes sense, but means something completely different.

Perhaps you could save us some guesswork and post one of these puzzling plots.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight
 

All I have done is choosing the nodal response result "acceleration" and picked a node. What I get is a graph with frequency (Hz) on the X axis and acceleration (mm/s2 - consistently with my units) on the Y axis. I can clearly locate the resonances, but the line goes into negative Y a few times. Will post it as soon as I can.

Cheers

Gio1
 
If you are plotting Magnitude on the y axis and frequency on the x- axis you should only get positive values. The phase information will tell you if the value is negative or positive.

 
True, IF he is plotting linear magnitude. That's why I'd rather look at a plot than play guessing games.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight
 
I tried, tried again and then again, but that plot never came back as it was yesterday.
I think it was some bug that was affecting the session of the postprocessor I had open.
Today (after a reboot) all plots had POSITIVE acceleration magnitude - as I would expect.

I was slightly puzzled by Greg's last post. What do you mean by "linear magnitude"?

Thanks everybody!

Gio1
 
Plotting abs(v), is linear magnitude, where v is a complex vector.

It would be an unusual, but possible, choice.

abs(v)^2 is more common.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight
 
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