Need simple equations for vibration motion

This EE needs some help in mechanicals [not a school project, either]

I need to estimate the amount of displacement vibration will cause. Are there any simple equations to find this?

Picture a diaphragm of metal that has a piezo transducer mounted on it to make it vibrate.

Given: metal characteristics: mass density: 170 pounds per cu ft Young's modulus: ?? speed of sound is 6420 m/s thickness: 40 mils diameter of disk, mounted around outside edge approx 3 inch diam frequency: at resonance? 43KHz power: around 1 Watt

Find the displacement of the maximum displacement of the diaphragm. Assume at center

This should be easy for youse guys, eh?

- Robert -

My guess was 5 mils. How far off?

Reply to
Robert A. Macy
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Not enough data. Do this. Apply a force F to the center of the diaphragm and measure the displacement D. If you measure the force in newtons and the displacement in meters, then the force times the displacement is the work supplied in joules. Supposing the surface would take as much work to move in the other direction too, a cycle would need at least 2 X F X D joules.

At 43 kHz, 1 watt would provide 1/ 43E3 joules per cycle so F x D = 1 / 86E3 Joules And for a displacement D of 5 mil = 0.000127 meters, you would expect the displacement force F to be about

1/86E3 X 1 / 0.000127 = 0.09 newton ( 0.3 oz)

I have blithely ignored resonance which would multiply the displacement depending on the Q and I have blithely ignored other loading like air or fluid damping which would reduce the displacement and I also ignored innumerable other refinements like mounting compliance. But you wanted simple.....

Brian Whatcott Altus OK

Reply to
Brian Whatcott

Brian,

Thank you for your prompt reply.

Instead of using force to displace in the middle, how about assuming the Q is infinite, calculate the kinetic energy of the diaphragm, and then assume the Q is low and all that energy has to be replaced each cycle? How much displacement?

- Robert -

Reply to
Robert A. Macy

Reply to
Narasimham

Not responding to your suggestion, but rather, working the deflection of a circular disk of 40 mil thickness steel 3 in diameter faced with a uniform pressure of 2.8 psi gives a central deflection of 5 mils using a software code. This is much much stiffer than my prior assumption, So I imagine your guess at central deflection is way high, by a factor of 10 or more

Brian Whatcott Altus OK

Reply to
Brian Whatcott

Brian,

Thank you again, but the material is only 170 pounds per cubic foot, not steel.

But with only 2.8 psi, 5 mils is perfect. right?

How do you go from 2.8psi causing 5 mil deflection to 1 Watt of sound will only deflect 0.5 mils, not my estimate of 5 mils?

Perhaps, I should state that the sound power is not transferring from air to the diaphragm, but the diaphragm is being driven by a piezo transducer.

2.8 psi across the diaphragm is a relatively small number, yes? I mean I've worked with poisonous gas cannisters that were 2,000 psi. Don't airless paint sprayers put out a whopping 400-800 psi, at least enough to inject into flesh? Water pressure at my home was 63 psi. So, 2.8 psi is small.

These answers are EXTREMELY helpful. So far I'm within a magnitude of accuracy, would like to get closer to more like 10% though. Greatly appreciate the help.

- Robert -

Reply to
Robert A. Macy

Narasimham,

Thank you for your reply.

I don't understand your response.

Whether the Q is 10, or 1, or completely damped, doesn't make a great deal of difference to me [I think].

But you raise a great question:

When I turn the disk over, how much deflection occurs caused by gravity?

1-2 microns? Or more like 5-10 microns?

- Robert -

Reply to
Robert A. Macy

This will probably be my last offering on the topic, but it should get you close enough. But first, a word about the engineers who did NOT respond. There are several people who monitor this group who would avoid answering the question you posed, although they are well equipped to do so This is the land of the ambulance-chaser. It is the question of liability. In contrast I have been willing to answer because this is not my field of expertise, so you are at your own risk. The question I asked myself was, if I pushed a diaphragm of the kind you describe, how far would it deflect for a given force? The reason for this question is because there is a particularly simple relation between the product of force and displacement in the direction of the force - this defines work done - with no fooling around with scale factors if expressed in SI units.

Now it happens there is a rather cheap but effective code for getting deflections etc for perforated and solid disks. It is called WinPlate. It is sold by Archon Engineering, and they offer a 30 day free trial

- no strings - on a usable download. You could do that. It needs the diameter, thickness, and edge fixing method and above all the youngs modulus for the material - which essentially describes the material's stiffness. You did not give this, so I deduced it from the specific gravity that you provided, namely 2.724 (which you expressed as pounds per cubic foot.) This SG fits aluminum, glass, quartz etc. These materials happen to have comparable stiffness of 71 MPa or 1E7 psi - a third that of steel.

I could now centrally load this disk and see how the force and deflection compared. I used a central circle of 0.2 inch diam to apply a 19.1 lb force initially. I finally realised that if I gradually increase the force from 0 to 19.1 lb the deflection increases from 0 to 0.005 inches. This takes only half the effort to displace a surface with a constant force.

I mentioned previously that if you expend 1 watt at 43kHz, each cycle takes 1/43000 of that power per second. So one cycle can expend up to 1/43000 joule. This can be expended by displacing the surface by a certain amount twice, with a certain force. The displacement and force on the diaphragm scale pretty linearly down from the large initial value I tried (Remember? 0.005 in using 19.1 lb) When I scale these values to newtons and meters, I can deduce the value for deflection and force provided by the energy I mentioned above. Or to put it another way 0.005 X 19.1 X scale factors to give newtons X meters times a factor to fit this energy to the actual energy available which is 1/43000 joule.

Without troubling you with the details, it turns out that a central pressure of 3.8 psi on a circle of diameter 0.2 inches gives a central force of 0.12 lb or 1.91 oz and this provides a deflection of 0.00023 inches.

Now the real world issues - If the energy is not dissipated by this deflection, it is stored, and the deflection increases resonantly until the surface DOES dissipate the cycle's worth of energy, each cycle. So though the deflection figure is pretty respectable, it could be much bigger if the coupling to the external fluid is low.

That's as far as I go.

I enjoyed addressing the question Thanks

Brian Whatcott Altus OK

Reply to
Brian Whatcott

In order to get 5 mils displacement, you'll probably need a horn to amplify the movement of the piezo. I think if you just stuck the piezo crystal to the back of the diaphragm you would get very little motion. I have been reading an old book called "Ultrasonic Engineering" by J.R. Frederick (1965) that covers this subject well, including horn design. You may want to see if you can find a copy (or something similar).

Reply to
ms

"Robert A. Macy" wrote in news: snipped-for-privacy@g14g2000cwa.googlegroups.com:

Depends which planet you are on.

Sorry, had to say it.

If you know the resonant frequency of your disk, you know how much it will deflect under gravity.

Cheers

Greg Locock

Reply to
Greg Locock

Greg,

yes, which planet? I was thinking earth. But then again, I don't know the force on the disk either.

But from memory 2 g's is easy. 10 g's is getting up there, and 100 g's is getting closer to dropping it six feet onto concrete.

Given velocity of sound in the material, 6420 m/s and the density 170 lb/cuft I calculated the half wave resonance of a beam with its two ends fixed, as 3 inches at 43KHz

I then jumped to the conclusion [possibly erroneously] that a diaphragm that is 3 inches in diameter should act like that beam and be resonant at 43KHz.

Ok now how do I calculate the deflection?

- Robert -

Reply to
Robert A. Macy

I think that Brian Whatcott mentioned a demo program that could be used for deflection.

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seems to have what you need but I've not read it myself. You could look at
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and see if you could adapt the technique using your values but I'm not sure what the effect would be of having your transducer glued to the middle.

I found both of the above googling with "circular diaphragm" + vibration + modes so some variation on this and center/centre loaded might give you some useful hits.

Best of luck - Mike

Reply to
Mike Yarwood

I know that's a simplified calc, but I think that's a big underestimate. 2 x F x D assumes that all energy applied to the plate disipates after each half cycle. Realistically, you could displace the plate once and release it, and it will vibrate more than a half-cycle. Stored energy is not lost.

Dave

Reply to
dave.harper

Quite so

Brian Whatcott Altus OK

Reply to
Brian Whatcott

Mike,

Thank you for the URL for the paper and the detailed technique of searching.

I never seem to use the "right" terms.

The paper did have a formula that looks like what I needed.

- Robert -

Reply to
Robert A. Macy

"Robert A. Macy" wrote in news: snipped-for-privacy@g44g2000cwa.googlegroups.com:

In a single degree of freedom system the deflection due to gravity is x=mg/k

The natural frequency F=1/2/pi*sqrt(k/m)

so x=g*(2*pi/F)^2

easy!

The astonsihing thing to me is that although that derivation is for a simple spring mass system, in practice it works reasonably well for complex systems as well.

Reply to
Greg Locock

Now I understand! The disk diameter was supposed to have been set by the resonance at 43KHz.

Thanks. At least it has to be somewhere near a magnitude correct.

So if a 3 inch disk resonates at approx 43,000 Hz and g is 9.7 nt, then the deflection is 0.008 mils a REALLY small number but makes sense since the disk is rigid.

about 2000 A

that means the disk would create 4 to 5 fringes in green llight if viewed optically?

Do I have the math "about" right?

- Robert -

Reply to
Robert A. Macy

Brian,

Thank you for keeping up with this thread. This last answer verified [to me] where I was heading with this.

Took me four readings of this response and several days to soak in, but well worth it.

Enjoyed your comment explaining why "the highly qualified" are lurking. I instantly pictured the scenario after receiving "bad information" We all are on some court TV show explaining to the judge why I believed everything told to me on the 'net. ...and I win the case because my birth certificate is indeed dated the previous day.

If you're curious about the application, I can only respond directly to protect some proprietary aspects here. The Usenet is TOO public.

I can be reached at

macy ..AT.. california ..DOT.. com

Thank you again for taking the time to help me.

- Robert -

Reply to
Robert A. Macy

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