"Bell" modes? for a circular tube [Son of Chime] 2

[Dinosaur]

Guys,

I'm back to discuss the design of a chime, a circular tube to make a sort of bell like sound during a church service. For you old timers, we talked about this about four years ago. Back then, my time ran out and I decided to order a single chime tube from an orchestral chime producer. Makes me feel like an engineering wimp that I couldn't do the calculations, but a wise man once said (e.g. Clint Eastwood) "A man's got to know his limitations." But seriously folks ...

It is my belief that to make a proper chime, you cut a tube to the length corresponding to the acoustic resonance length for an open-open tube resonating at the requisite frequency. For a concert A of 220 Hz for example,

344 m/s / (2 x 220 Hz) = 0.782 m -or- 30.78 inches

The exact frequency is not important, but let us assume we will be shooting for the concert "A" for the purpose of this discussion. (220 Hz = 1382.3 rad / s)

So the tube will be 30.78 inches long. However, my experiments led me to conclude that the mode of vibration I am seeking in the design is not the vibration of a free-free rod in its second mode, which I originally believed, but rather a "bell" mode as suggested in this forum years ago. I interpret that to mean a shell mode of vibration of the tube wall through the thickness.

If this is the case, then I guess that if the tube were considered narrow relative to its length, the only variables we are working with are the material properties of the tube (the mass density and Young’s modulus) as well as the diameter and the wall thickness.

I still have some aluminum tube from the original experiment so I would like to stick with the aluminum for now. (E=10,000,000 psi & p=165 pcf) The diameter of the tube is around one inch. Using these values, what would the wall thickness have to be to get a good "Bell" mode? Is it possible to describe the arithmetic in a post? Thanks for everyone's help. - Dinosaur

 

[electricpete]

Den Hartog's "Mechanical Vibrations" has a formula for frequenies of vibration a cylinder which is sometimes applied to motor cores. I'll post that formula tonight.

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[electricpete]

I shouldn't call it cylinder - it was called a ring mode as you say not dependent upon length

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[electricpete]

Also Tom Irvine (Vibrationdata.com) is a frequent contributor at this site and has some great material available for a small fee. His file on ring vibration modes shows:

Ring frequency = sqrt(E * mu) / (Pi * d)
Where mu is mass density
This is the speed where wavelenght = circumference.
For aluminum this frequency in hertz is 63,790 / d when d is in units of inches.


He also details in-plane bending modes and out-of-plane bending modes. I think out-of-plane bending modes are not relevant unless the cylinder length is much less than it’s diameter.

For in-plane bending modes:

f = [n*(n^2 – 1)*sqrt( E*t^2 /{3mu}) ] / [pi * d^2 * ( sqrt(n^2+1)]

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[electricpete]

One correction:
"Ring frequency = sqrt(E * mu) / (Pi * d)"
should have been
"Ring frequency = sqrt(E / mu) / (Pi * d)"

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[electricpete]

On a similar subject, I have a question for folks here:

For the mode described by
Ring frequency = sqrt(E / mu) / (Pi * d)

What is the mode shape? I picture this as a traveling wave rotating around the ring. From my perspective we could add a mode number n and say
Ring frequency Fn = n* sqrt(E / mu) / (Pi * d)
This corresponds to any integral number n of wavelengths in one circumference.

Then the deformation would be given by y(theta,t) = cos ( w*pi*Fn * t – theta / n)
For example n = 2 would be an oval deformation rotating at twice the fundamental ring frequency.

Does this sound right to you guys? The rotating mode shape may or may not be interesting for chimes (what do you guys think?) but is very interesting for motors.

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[electricpete]

Last correction (hopefully):
"Then the deformation would be given by y(theta,t) = cos ( w*pi*Fn * t – theta / n)"
should have been
"Then the deformation would be given by y(theta,t) = cos ( w*pi*Fn * t – theta * n)"

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[Dinosaur]

I am confused (but, of course not for the first time). Your formula does not have the wall thickness, t, as a variable. This is counterintuitive. Am I missing anything?

 

[GregLocock]

pete I think you may be posting the tangential compression/rarefaction mode, as Dinosaur says, there ought to be a t^3/t in there for a bending mode.

According to Blevins the flexural modes of a slender ring are

i=2,3....

f=i(i^2-1)/(2piR^2(i^2+1))*sqrt(EI/m)

i is the number of lobes of the shape.

HOWEVER

a tube is not a slender ring. I very much doubt that this would be a low frequency mode for a tube.




Cheers

Greg Locock

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

 

[electricpete]

Just to clear up the clutter of my previous comments, let me start again:

Tom Irvine's paper mentions 3 types:
1 - ring frequency
2 - in-plane bending modes (take a circle and squash it into an oval or a clover).
3 - out-of-plane bending modes (take a circle and bend it like a saddle).

#3 I rejected for this scenario on the basis that a long tube does not seem likely to undergo this deformation.

#1 - (ring frequency) - he gives frequency as f = sqrt(E / mu) / (Pi * d)
where mu is mass density in his terms. His derivation is that this is based on the time it takes to travel one circumference at the speed of sound in the material sqrt(E/mu). He also seems to describes this as a compression/expansion mode. I added the stuff about higher order modes on the basis that it sounded more like a traveling wave to me...let's forget that for now.

I don't see this equation in any other references. You guys object on the basis of no dependence on thickness? Could be... I don't know.

# 2 - (in-plane bending modes) he gives
f = [n*(n^2 – 1)*sqrt( E*t^2 /{3mu}) ] / [pi * d^2 * ( sqrt(n^2+1)]

When I look at the derivation I see there are two errors:
I = (1/12) * h*t^3
and
mu = mu * h *t

Now one more to consider: Den Hartog gives the following for in-plane bending mode:
f = (1/[2Pi])*i*(i^2-1)*sqrt(E*I1/(mu*R^4)) / (sqrt(i^2+1))

If I compare with Greg I see two discrepancies: mu=m/L in Den Hartog's formula takes the place of m in Greg's formula. That seems ok as long as Greg's m is intended to mean mass per length.

The other discrepancy is that Greg's denominator has (i^2+1) whereas Den Hartog's denominator has sqrt(i^2+1).

I have two other sources which show the sqrt(i^2+1) in the denominator: one is Tom Irvine's paper (before the rectangular substitutions). The other is an article by Alger. Based on the above I am inclined to believe Den Hartog's formula is correct for #2 (in-plane bending). Greg can you check whether perhaps you have omitted a square root from around i^2+1 in your denominator?

I have one more question (if Dinosaur doesn't object to me messing up his thread). Is this in-plane bending mode associated with the above frequency equations restricted to a standing wave where the nodes never change? Or can it be a traveling wave where the modes rotate around the circumference?

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[GregLocock]

Ok, I wouldn't have called (1) a ring frequency (but there's nothing wrong with it as a name, and it can be a very important mode), but that is essentially a one dimensional plane wave, hence the absence of t.

for (2) yes I forgot the square root.

In my experience out of plane bending modes (type 3) in brake rotors sometimes rotate with the structure, and sometimes lock to the brake pads, and let the structure pass underneath them. I suspect type 2s would be just as annoying.

For this tubular bell I strongly suspect that the beam flexural modes are far more important than the in plane bending modes (type 2), and I also doubt that a conventionally proportioned tubular bell would exhibit pure type (2) modes. A quick FEA would confirm that the high order modes are dispersed in length and theta, forming a patchwork, typical of high frequency modes in thin walled structures. These are relatively inefficient radiators as adjacent patches are in antiphase and so cancel each others contributions.













Cheers

Greg Locock

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[Dinosaur]

E-pete,

No problem with the discussion going where it wants to go. We had a lot of fun with this thread the first time around including discussions of boundary conditions, location of nodes, etc. It is a fun theoretical discussion with a real world problem most of us can relate to. If we get too far off track, I'll just restate the question.

Regarding the ring mode ...

It sounds to me like a compression wave in-plane traveling around the circumference. That certainly is a mode, but I doubt it would have enough energy to excite the air within the tube to create much sound.

As I said earlier, the original thread from four or more years back included much speculation that the important modes would be "Bell Modes" while I was attempting to compute beam modes. I am not sure what a bell mode is but I have an idea. I assume it is a shell mode of the tube vibrating thru the thickness, but the pattern of crests and troughs forms a kind of checkerboard over the surface if you were to unroll it. Since the length of the tube is so much greater than the diameter, it seems to me that it would approach an infinitely long medium.

I could be all wet on this. Heck, it might not even be a bell mode. All I know how to do is calculate beam modes, and I don't think that is the answer.

Anything you have to offer in the discussion of vibration of a long circular tube is relevant and welcome. Thanks again. - Dinosaur

Greg-L

It will take me a moment to digest what you are saying. You just posted while I was in the middle of this.

 

[electricpete]

A lot to read. I wanted to make one quick clarification. I may have jumped to conclusions when I said Tom Irvine's formula was in error. Those substitutions of "I" and mu I recognize as rectangular. I initially thought they were wrong because we are dealing with a tubular geomtery. But thinking again for in-plane bending maybe it is correct?? After all side-view of a tube looks like a rectangle.

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[electricpete]

To add to previous comment, the derivation uses I (called Iy) as Iy = (1/12)*h*t^3 with comment that this applies to a thin ring. Then later m = mu*h*t with no comment.

I guess if you slit the tube axially, unroll it (sincd it's thin and therefore radius of curvature large)and look into the cross section of the slit, then it looks like you're looking into the end of a rectangular beam.

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[electricpete]

Dinosaur - it might be cheating, but how about working the problem backwards? If look at the pitch of two different chimes then we can calculate the frequency ratio (octave = 2:1, fifth = 3:2, fourth = 4/3, major third = 5/4, other intervals can be determined as well). If you look at the the dimensions then we can predict what the frequency ratio of those two sets of dimensions would be under various types of vibration: #1, #2, #3, or the open/closed air column vibration modes (perhaps there is combination of metal vibratio mode and air column mode at approx the same frequency?). Anyway, that might give a clue which type of mode is responsible for the sound.

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[electricpete]

I forgot to mention beam bending modes as another candidate (Greg's suggestion).

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[tomirvine]

Thank you for mentioning my work. Here are some resources:

http://www.vibrationdata.com/WindChimes.htm

http://www.vibrationdata.com/tutorials2/beam.pdf

http://www.vibrationdata.com/beam.exe

Please let me know if you have further questions.

Tom Irvine

 

[electricpete]

Talk about a good response. Great info Tom. It sounds like Greg was on-target with the beam-bending modes.

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[Dinosaur]

Tom,

Your website resource is exactly what I needed. I had tried to compute the beam modes before, and I guess I had messed up the conversion of the mass from pcf to slugs/cu. ft. or something because the answers I had then did not correlate with my test results. Back then I thought the beam modes were the important sturctural modes and the challenge was to match a beam mode with an acoustic mode for the tube length.

I have a piece of one in dia. galvanized steel conduit 63" long. I have computed several modes for various lengths shorter than the 63". Now I am going to locate a hang point at 22% L from one end and cut the tube to length to match a mode. If it resonates, then I will cut it down to the next length, and see if it resonates again. If I am successful, it will prove the math you provided me is correct. Thankfully, I have Mathcad and can use this to "design" other chimes.

The next step will be to experiment with the materials available in my area to find a material that gives good tone. After that I will make a chime (one note) for my church. When I get to that point, that I have verified your math and can "handle it", I will become a supporter of your work at your website. You've earned it. Thanks - Dinosaur

 

[MikeyP]

Has it really been 4 years? Man, I'm getting OLD!

Perhaps the most interesting point from Tom's analysis is that there appears to be no equality between the "beam" longitudinal and bending modes and the acoustical modes inside. (Although, you could argue that the 2nd bending modes are possibly exciting the 4th acoustic modes)

It also suggests that ring frequency and cylinder modes (c.f. beam modes) will be too high frequency to be of any significance.

A very interesting experiment would be to find out what happens when the 2nd beam mode does match the fundamental acoustic mode. Perhaps, Dinosaur, you could make one that matches and one that doesn't and let us know the difference between the sounds.

Tom, does the longest chime sound different to the others because the initial response is dominated by the "untuned" 3rd beam mode instead of the "tuned" 2nd beam mode? It could just be that the position of the striker is such that it is closer to a 3rd bending antinode than a 2nd bending antinode.

It is interesting to see that the (mostly) dominant 2nd beam mode has a much shorter RT than the 1st beam mode. Is this because it is a better radiator and has higher radiation damping? That seems like a big difference in damping to be explained by radiation alone. Perhaps the suspension has more damping imfluence than one might initially think.

The good news for you, Dinosaur, is that a simple Euler-Bernoulli beam model will be plenty accurate enough for your design and will save an awful lot of mucking about with tedious elasic cylinder theories and FE models!

M

--
Dr Michael F Platten