
RMS displacements in Linear Dynamic Analysis
Kirby Meyer Apr 2, 2009 9:38 PM (in response to Kirby Meyer)Hello All
I just found that the RMS displacement provides values in units length over the frequency span. When flagging 'show PSD value' under Edit Definition, the probe response data is in^2/Hz. The integral of this curve meets the RMS displacement value.
Kirby

Re: RMS displacements in Linear Dynamic Analysis
1L5M1MV Dec 10, 2009 9:08 AM (in response to Kirby Meyer)Kirby,
You seem to have a lot of experience in this field, so hopefully you can make this a little clearer for me. I have been asked to reduce the "vibration" of a camera mount (not vague at all, yeah). Anyway, I specified a sensor and have been analyzing the displacement, since I'm more concerned with image stability than failure of the electronics.
I understand the RMS value of the displacement, but I am having some trouble interpreting the displacement response graphs (PSD displacement?). Here are some of my results for peak displacement:
Config 1: 0.509 mm^2/Hz at 27.1 Hz
Config 2: 0.150 mm^2/Hz at 36.0 Hz
Config 3: 0.060 mm^2/Hz at 35.8 Hz
I am trying to figure out how to compare these configurations. I hesitate to compare the peak response displacement, because each occurs at a different frequency. I tried "normalizing" these results by multiplying each PSD displacement result by the frequency and then taking the square root, but I don't quite understand what this value will represent. Will it be the 1s value when the bracket is excited to that frequency?
Thanks for your help!
Bryson Cook

Re: RMS displacements in Linear Dynamic Analysis
Anthony Botting Dec 10, 2009 11:15 AM (in response to 1L5M1MV)HI Bryson: Just saw this and believe I can help. Been studying this for a bit now. Apparently, the PSD response as a function of frequency is to be interpreted as the "energy content" or "energy of vibration" at particluar frequencies. The higher the PSD value, the more vibrational energy is present in the structure at that frequency. That's it.
The evaluator is supposed to look at both the PSD response vs. frequency to identify the frequencies at which a large amount of vibration energy is present (and a contour plot to observe where on the structure this is happening), AND use the RMS contour plot of displ/vel/acceleration to get the magnitude. The RMS is the best you can do to represent the statistical response of the structure over the input PSD. I have seen aerospace applications spec. the assembly to 4sigma, for example, and design to withstand that RMS. There is no way to backout the actual time history (i.e., displacement response as a function of time so you could actually observe the real values) as the time history information is absent from the start of the analysis (since a PSD curve is used for the excitation). I hope this helps. Write if you find out any more interesting tidbits!. Tony

Re: RMS displacements in Linear Dynamic Analysis
1L5M1MV Dec 10, 2009 2:17 PM (in response to Anthony Botting)Tony,
I was under the impression that the "vibration energy" as the area under the PSD curve was only valid for PSD acceleration. I am not trying to create an artificial timehistory of the acceleration input, nor am I trying to determine a peak value of the resultant displacement. Part of my problem is that I'm not designing a part to any particular specification.
I think that the difficulty I am having is with the units. I understand that RMS displacement is a statistical representation of actual displacement over the whole operating range, with units of length. I can quantify improvements by reduction in RMS displacement, but I also want to be able to interpret the meanings of the peaks on the graph of PSD displacement (mm^2/Hz) vs. frequency. Is there a physical meaning to these values? If I multiply the peak PSD displacement by the frequency at which it occurs, then take the square root, what will my results (which will have units of mm) mean?
I assume that the result will also be a statistical representation of the actual displacement, just at that given frequency. Still, this doesn't seem 100% concrete and I can't quite put my finger on why. Thanks for the help, and hopefully you (or anyone) have some more information for me.
Bryson Cook


