FRF from FE Model 1

[SrinivasAluri]

I have a FE model for a deck, I obtained the displacement response at location B due to a unit sinusoidal load over a frequency range at any input location A. Can this response I obtained be called as a Frequency Response Function?

 

[GregLocock]

Yes, also known as a transfer function. It should be a complex spectrum, that is it has phase and magnitude (or real and imaginary parts).

Cheers

Greg Locock

 

[xydrex]

To be FRF or transfer function it must be output variable divided by the input force, both as functions of frequency.
Eg Mobility = velocity/force [(mm/s)/N]or Compliance =displacement/force [mm/N] or Accelerance = acceleration/force [(mm/s^2)/N] i.e. output per unit of input. Having just output, say velocity as function of freq. cannot be called a transfer function, its just a single response to a single input. Having output per unit of input allows you to predict responses to any inputs.

 

[GregLocock]

I'm sorry, you are wrong.

Neither an FRF or a transfer function has to include force. And force could go on top or underneath.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[tomirvine]

As another example, the excitation function could be in the form of base acceleration. This is also called "support motion" or "seismic loading."

An FRF can then be formed that relates the response acceleration to the base acceleration.

The FRF is a complex function of frequency, as mentioned in previous messages.

The response metric could also be relative displacement.

Tom Irvine

http://www.vibrationdata.com

 

[xydrex]

The point I was trying to make is that FRF is output spectrum divided by input spectrum; which variables they are is not as relevant, perhaps I worded it wrong.
If you obtained the displacement response at location B due to a unit sinusoidal load over a frequency range at any input location A, because its a unit force, the magnitude will be correct for FRF (divided by one), but you still need to include phase shift. Remember when you divide one complex number by another, the magnitudes divide out, but the phase angles are subtracted. Thus his output will contain the magnitude same as FRF but not the phase.

 

[djf123]

A short correction. An FRF is actually a slice through a transfer function, which is a complex plane, where the slice is along the jw=0 axis. FRF's are "commonly" referred to output/input relationships and therefore anything can go in the numerator and denominator. Experimentally there are a a number of Hx estimators that use combinations of auto and crosspower spectrums. The basic differences between these estimators is how they handle noise. Some assumptions are made on where the noise components are and the formulation will change a bit. Bendat has a number of books on this subject.