help needed. vibration, calculation 1

[dho]

how to calculation the vibration amplitude of a random vibration?
one step further, if there are several superimposed sinusoidal vibrations?
i have one test requirement. the test lab told me that it exceeded their table DA.
thanks a lot.

 

[BobM3]

Random vibration is usually "shaped" in the frequency domain. You would specify (as a function of frequency) the distribution of energy (amplitude*amplitude) for each frequency.

Do you have a Power Spectral Density (PSD) plot? There is a method to take a value from the PSD and figure out the average amplitude of each frequency.

 

[dho]

thanks.
lab told me, this one exceeds 2" DA.

random
10-100 Hz, 3.01 dB/Oct
100-300 Hz, 0.01 g*g/Hz
300-500 Hz, -13.569 dB/Oct

seven sinu
4.3Hz/0.43g; 8.6Hz/0.86g; 17.2Hz/1.72g; 3.380Hz/0.338g, 6.767 Hz/0.677g, 10.150 Hz/1.015g, and 13.533 Hz/1.353g

 

[BobM3]

I'll have to think about the dB/Octave units. I haven't used them. For g*g/hz, the largest stroke would occur at the lowest frequency, 100 hertz. You need to know something about the "resolution" or "bandwidth" of the those numbers but I'll assume a couple just to come up with some numbers.

Assume a sample rate of 1024 points/sec and an FFT size of 2048 points. This gives a frame length of 2 sec or a delta f (resolution) of 1/2 or .5 hertz. For a 100 hz sine wave, a PSD value of .01 g*g/Hz would give an energy of .01 g*g/Hz * .5 hertz = .005 g*g. A square root produces a peak amplitude of .071 g's. The displacement needed would be:

x = accel/(omega**2) = .071*386/(6.28*100)**2 = .000069 inches - very small!!

Your numbers will depend on the sample rate and FFT size used to generate the numbers you gave. Doesn't appear though that this particular frequency band requires much stroke.

For your sinusoidal tests, again use x = accel/(omega**2). For the 4.3 hertz,

x = .43*386/((6.28*4.3)**2) = .227 inches - again, not much displacement.

Caution, I make a lot of math errors! Find out what test is the one they are having problems with.

 

[dho]

1) the calculation of the random will be sample rate and/or FFT size dependent. does this make sense?
2) people were talking about 3 sigma. the g*g/Hz will be three times bigger at some time.
3) the problem is the random with 7 sinusoidal SUPERIMPOSED on it. not each one individually causes excessive DA.
thanks.

 

[BobM3]

1) Yes

2) Three Sigma usually refers to Guassian distributions of random data. I think you need more info from the supplier of the accel data.

3) Add the 7 sine waves together (with the appropriate phases between them). Figure out how much stroke is needed for that acceleration sum.

I take it you are getting accel data from your customer and need to test it using your's or someone else's test equipment?

 

[dho]

yes. we got accel data from our customer. and one outside lab (calculated and) said it exceeded 2" DA and they could not run it. another lab did run it and the machine aborted minutes later (due to 3 sigma).
to educate myself, i just want to know how to calculate it.
add 7 sin is not difficult. they are two groups. one is 3.38 HZ and its multiples. the other 4.3 Hz. at one point if the test time is long enough, they will just all add up with no phase diff.
is it just a arithmatic DA sum to add sin and random together?

 

[BobM3]

I missed the two different frequencies for your discrete sinewaves so I agree that if you go long enough there may come a point where they are all phased together. I come up with plus/minus .704 inches for the 6 sine waves (worst case, all in phase).

I don't know what the 3.01 dB/Octive means for the 10-100 hz data. Seems to me they need a reference acceleration as part of part of the spec. My guess is that, because it contains the lower frequencies, that is where the stroke limitations come from. The one stroke estimate I made earlier for the 100-300 hertz band (for the stroke at just 100 hertz) was so low that I'd guess you could probably add all the strokes in that band (one for each frequency line - depends on sample rate and FFT size), assume they were all in phase (they aren't for random noise), and still have a small stroke.

See if your customer can clarify what the db/Octive means and also find out how your test lab interpreted it.

 

[IRstuff]

It means that the envelope decreases at 3.01(?) dB/octave from the .01 g^2/Hz value at 100 Hz. Similar situation on the high side.

TTFN

FAQ731-376

 

[dho]

dB=10*LOG(freq_2/freq_1)

say freq_1, 10 Hz; freq_2, 100 Hz
dB = 10

Oct=LOG(ASD_2/ASD_1)/LOG(2)
ASD_1=.001 g*g/Hz; ASD_2=.01 g*g/Hz
Oct=3.32

so the "slope" of the first leg from 10 to 100 Hz is
10/3.32=3.01 dB/Oct

(see attached.)

thanks.

 

[BobM3]

That plot makes sense. I get the stroke needed at 10 hertz to be:

x = ((.001*.5)**.5 x 386)/(6.28*10)**2 = .0022 inches single amplitude.

@20 hertz,

x = ((.002*.5)**.5 x 386)/(6.28*20)**2 = .00077 inches.

If you had 180 frequency lines between 10 and 100 hertz and they each needed .002" of stroke and they were all in phase then this portion of the random data would need 180 * .002" = .36".

These numbers assume a sampling rate of 1024 pts/sec and an FFT of 2048 points (delta f = .5 hz).

Add this to the .704" needed for the discrete sine waves and you get about plus/minus 1.1". This doesn't include the high frequency stuff (300-500 hz - probably don't need any stroke for that) but is very conservative in that it assumes all frequencies are in phase and their peaks occur at the same time.

Sure seem like small stroke values. I'm surprised your test house is having trouble. Can they send you an ASD of the drive they came up with?