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Natural Science Forum / Physics / Acoustics / June 2005Does This Make Sense?
Seems to me it's a whole lot easier finding technical specs for a nuclear weapon than it is finding the formula to determine the resonant frequency of unimorphs. For brief moment I was happy when I found the two formulas here at http://www.americanpiezo.com/technical_updates/unimorph.pdf on page 1. But when I took a closer look at it my poor brain short-circuited; I have to say that math has never been one of my talents and the formulas don't make sense to me. Two variables are there that just seem to pop out of nowhere. If somebody has a better grasp of math-- assuming there's nothing wrong with the formula-- would they take a moment to explain this to me? Ron -- "You see me now a veteran, of a thousand psychic wars. I've been living on the edge so long where the winds of limbo roar." > Seems to me it's a whole lot easier finding technical specs for a > nuclear [quoted text clipped - 11 lines] > wrong with the formula -- would they take a moment to explain > this to me? Hi Ron The conversion from MS Word to Adobe pdf had problems. The formulas are correct, but the text has got wrong letters: lambda = wavelength, rho = density, nu = poison's ratio or modulus and E = Young's modulus. Cheers Eberhard Sengpiel Forum for microphone recordings and sound studio techniques http://www.sengpielaudio.com >> Seems to me it's a whole lot easier finding technical specs for a >> nuclear [quoted text clipped - 21 lines] > Cheers > Eberhard Sengpiel Equation 3, for the resonant frequency of an isotropic circular plate that is clamped around its circumference, is of the form Fres=KRQ,where K represents all of the numerical constants, R = t/(a^2), and Q = sqrt [Y/(p(1-n^2))], where t is the thickness, a is the radius, Y is Youngs Modulus, p is the density and n is Poisson's ratio. What is interesting about the generalized equation Fres=KRQ is that for a circular plate of a specific material and dimensions, R and Q are determined, and that there is a unique relationship between the multiplicative constant K and the boundary condition (eg, circumference clamped, circumference free to translate but not rotate, free-free, node mounted, center mounted). Since I have some experience with piezo bimorph design, a value for K of 0.941 in equation 3 (irrespective of boundary conditions) struck me as being too large. This lead me to my personal library and to a fascinating discovery about the value of K for the clamped circumference boundary condition. First to Olson (Acoustical Engineering) who claims that K=0.467; and then to Morse & Ingard (Theoretical Acoustics), who claim that K=0.934. Since Morse & Ingard disagreed with Olson, I wondered how Kinsler & Frey (Fundamentals of Acoustics), Lord Rayleigh (Theory of Sound) and Blevins (Natural Frequencies and Mode Shapes) weigh in on the issue. Here's a summary of what I found. Olson..............K=0.467 Kinsler & Frey.....K=0.47 Blevins............K=0.470 Lord Rayleigh......K=0.941 Morse & Ingard.....K=0.934 equation (3).......K=0.934 At this point it is worth noting that the ratio of 0.94/0.47=2.00. So, at least on the surface it would appear that the two groups of authorities differ on the value of K for the clamped circumference case by almost exactly a factor of two. This is good news because mistakes often involve factors of 2, sqrt 2, 10, etc. The only question is, which group of authorities is right and which group of authorities is wrong. In order to answer this question I fabricated a brass plate (25.4mm dia and 0.038mm thick) and clamped it between two stainless steel toroids (25.4mm OD, 12.7mm ID, 7mm thick. Therefore the exposed diameter of the brass plate was 12.7mm. One of the stainless steel toroids was threaded to accommodate a 1/2" B&K 4134 microphone capsule. When the microphone capsule was threaded into the toroid, it created a small air volume between its diaphragm and the clamped brass disk. Because the pressure in that volume is directly proportional to the displacement of the brass disk, the output of the microphone provided an indirect measurement of the vibration of the brass disk. The exposed side of the brass disk was exposed to broadband noise produced by a wide-range speaker and the output of the microphone was analyzed using a Bruel & Kjaer 2032 FFT analyzer to determine the fundamental resonant frequency of the clamped brass disk. Without going into the details, suffice it to say that the calculated resonant frequency of the clamped brass disk was Fres=3548K. By comparing the calculated with the measured resonant frequency it was possible to experimentally determine the value for K. And the answer is....K=0.---. Before I disclose the answer, I'd like to take a little survey. It's Olson, Kinsler & Frey and Blevins vs Lord Rayleigh, Morse & Ingard and the author of equation (3). Time to vote. >>> Seems to me it's a whole lot easier finding technical specs for a >>> nuclear [quoted text clipped - 87 lines] > Olson, Kinsler & Frey and Blevins vs Lord Rayleigh, Morse & Ingard > and the author of equation (3). Time to vote. You have mis-read some of the books. Lord Rayleigh and Morse/Ingard define t as the half-thickness. >>>> Seems to me it's a whole lot easier finding technical specs for a >>>> nuclear [quoted text clipped - 90 lines] > You have mis-read some of the books. Lord Rayleigh and Morse/Ingard > define t as the half-thickness. Not only I, but also the authors of the paper at http://www.americanpiezo.com/technical_updates/unimorph.pdf Their equation 3 is clearly copied directly from Morse & Ingard as is evidenced by the (1.015)^2 term. Apparently, the authors copied the formula from Morse & Ingard but incorrectly define h as the thickness instead of the half thickness of the plate. Consequently, equation 3 is indeed incorrect with the variables as defined and will predict a resonant frequency twice as high as it actually is. I am particularly greatful to you for pointing out my oversight regarding Morse & Ingard's definition of h as the half thickness. I have no doubt that this is also responsible for a previous factor of two discrepancy that I encountered when comparing Roark's equation for static volume displacement of a clamped cirular plate with the low-frequency sinusoidal steady state result that is given in Morse & Ingard. > You have mis-read some of the books. Lord Rayleigh and Morse/Ingard define > t as the half-thickness. I found a formula for the plate with free edges, one nodal circle in R. Roark and W. Young, 5th edition, page 378, case 12. For our unimorph case, K=9.08 when calculated by: f = [K/2pi]sqrt[Dg/(wr^4)] where D = Et^3/[12(1-v^2)] w = weight per unit area. (w/g is the same as mass, m) v = Poisson's ratio I don't have the enthusiasm for calculating it out.. But FYI, the multiplier, K depends on mode shape and mounting conditions as: K Mode Shape Free Edge Condition (floating): 5.25 "two nodal diameters" 9.08 "one nodal circle" (possible deployment wish) 12.2 "three nodal diameters" (6 piece pizza cut) 20.5 "one nodal diameter and one nodal circle" Edge Simply Supported ("hinged"): 4.99 "fundamental" (usual deployment wish) 13.9 "one nodal diameter" 25.7 "two nodal diameters" 29.8 "one nodal circle Edge fixed (cemented solid): 10.2 "fundamental" 21.3 "one nodal diameter" 34.9 "two nodal diameters" 39.8 "one nodal circle That's all the cases that Roark & Young treated. Angelo Campanella snip.....snip > I found a formula for the plate with free edges, one nodal circle in > R. Roark and W. Young, 5th edition, page 378, case 12. > > For our unimorph case, K=9.08 when calculated by: snip....snip That brings up another important issue which is the applicability of standard equations for the resonant frequency of circular "isotropic" plates to circular unimorphs. It has been my experience that the standard equations are reasonably accurate (within 20%) in predicting the resonant frequencies of bimorphs, but that they are grossly in error in predicting the resonant frequencies of unimorphs. With bimorphs, the structure is primarily piezoelectric material with two very thin layers of epoxy and a thin metal vane. As a result, a reasonable prediction is obtained using the Youngs modulus, density and Poisson's ratio as if the entire structure. were made of piezoelectric ceramic material. However, with unimorphs, the thickness of the metal base plate is generally equal to or greater than the thickness of the piezo element. So, which Youngs modulus and density do you use....or should you take an average? In addition, the diameter of the piezo element is generally considerably less than the diameter of the base plate. So what diameter do you use....the diameter of the baseplate....the diameter of the piezo element....or something midway between the two? In summary, the equation for resonant frequency of circular isotropic plates which is given in the paper at http://www.americanpiezo.com/technical_updates/unimorph.pdf is not only in error, it is not applicable to piezoelectric unimorphs. > > You have mis-read some of the books. Lord Rayleigh and Morse/Ingard define > > t as the half-thickness. [quoted text clipped - 42 lines] > > Angelo Campanella Uh, okay... But back to the original equations; does either equations 3 or 4 work. And for the mathematically weak-minded such as myself, are the values "p" and "v" something I should be aware of? Thanks. Ron > Uh, okay... But back to the original equations; does either equations 3 or 4 > work. And for the mathematically weak-minded such as myself, are the values "p" > and "v" something I should be aware of? Thanks. "p" is an effort to write "rho", the density of the material; piezoelectric ceramic in this case. The corresponding factor in Roark's formulation is w, the weight per unit area. To cut to the quick, one needs to weigh the wafer (probably a few grams), then convert according to units learned in college physics. "v", I think is a best effort to represent "nu", the Poisson's ratio. This is the ratio of lateral strain that occurs as a result of a given amount of longitudinal strain. That is, if you compress a body, it will also spread itself outwards-sideways to the extent of about o.1 to o.33 of that compression distance. v does not change much among materials, being o.1 to o.2 for inorganic rocks & concrete (probably ceramics as well), etc, to o.25 for iron and to about o.33 for aluminum. Angelo Campanella > > Uh, okay... But back to the original equations; does either equations 3 or 4 > > work. And for the mathematically weak-minded such as myself, are the values "p" [quoted text clipped - 12 lines] > concrete (probably ceramics as well), etc, to o.25 for iron and to about > o.33 for aluminum. Great, Angelo. Thanks! I can't tell you how long and hard I have been looking for that formula. Ron > > Uh, okay... But back to the original equations; does either equations 3 or 4 > > work. And for the mathematically weak-minded such as myself, are the values "p" [quoted text clipped - 12 lines] > concrete (probably ceramics as well), etc, to o.25 for iron and to about > o.33 for aluminum. Thanks, Angelo; much appreciated. You don't know how long and hard I looked for a working unimorph formula. I had the complete specs for four commercial unimorphs (3.0 kHz, 4.0 kHz, 6.3 kHz, and 7.5 kHz) and I used those specs as references to see if the formula produced the predicted results from the data at hand. And I am happy to say that it works perfectly. There were slight variations-- formula predicted 3.82 kHz for a 4 kHz transducer-- but I suspect the unimorph's manufacturer may have rounded up in that one case-- but otherwise, it's pretty accurate for rough estimations. One unimorph was stated to be 7.5 kHz and the formula predicted 7.65 kHz; a 0.1 Hz difference I can happily live with. By the way, that h value is as it should be; if you halved it as somebody suggested the results would be wrong. Ron > > Uh, okay... But back to the original equations; does either equations 3 or 4 > > work. And for the mathematically weak-minded such as myself, are the values "p" [quoted text clipped - 12 lines] > concrete (probably ceramics as well), etc, to o.25 for iron and to about > o.33 for aluminum. Once again, thanks, Angelo; much appreciated. I had the complete specs for four commercial unimorphs (3.0 kHz, 4.0 kHz, 6.3 kHz, and 7.5 kHz) and I used those specs as references to see if the formula produced the predicted results from the data at hand. And I am happy to say that it works perfectly. There were slight variations-- formula predicted 3.82 kHz for a 4 kHz transducer-- but I suspect the unimorph's manufacturer may have rounded up in that one case-- but otherwise, it's pretty accurate for rough estimations. One unimorph was stated to be 7.5 kHz and the formula predicted 7.65 kHz; a 0.1 Hz difference I can happily live with. By the way, that "h" value is as it should be; if you halved it as somebody suggested the results would be wrong. Ron > By the way, that "h" value is as it should be; if you halved it as > somebody suggested the results would be wrong. Whether you care to accept it or not, the fact of the matter is that Gordo is indeed correct in pointing out that "h" in equation (3) is the half thickness, and not the thickness (as stated), of a rigidly mounted isotropic circular plate. Consequently, any agreement between the predictions of equation (3) and manufacturer's specs is purely coincidental and simply demonstrates what can happen when one applies an incorrect equation to a situation to which it does not apply. A piezoelectric unimorph consists of piezoelectric ceramic that is (generally) bonded to a brass baseplate and is no way an isotropic structure. Furthermore, manufacturer's of unimorphs specify resonant frequencies for the free-free case, not for a rigidly mounted case. Fortunately Mother Nature has a way of confronting blissful ignorance with reality, as you will soon learn. > And I am happy to say that it works perfectly. There were slight variations-- > formula predicted 3.82 kHz for a 4 kHz transducer-- but I suspect the unimorph's > manufacturer may have rounded up in that one case-- but otherwise, it's pretty > accurate for rough estimations. One unimorph was stated to be 7.5 kHz and the > formula predicted 7.65 kHz; a 0.1 Hz difference I can happily live with. Great! Which edge condition and mode shape did you use; I gave many conditions and shapes... > By the way, that "h" value is as it should be; if you halved it as somebody > suggested the results would be wrong. I usually see that dimension, thickness in this case, as "t". Roark's book "Formulas for Stress and Strain" is very useful in calculating the breaking strength for any body, and for vibrations analysis (vibration is treated only in the last chapter). I have the 5th edition, but I have used earlier editions. I used that book for seismic analysis of equipment that should want to stay in place during earthquakes, etc. The chore is to prove that the "g" forces from the quake will not push any part near the yield limit. Also for vibration, the chore is to prove that the equipment has no resonance below the "plateau frequency" of earthquakes, typically around 20 Hz. The beauty of Roark's book is that he has analyzed every imaginable shape for strain maxima. so it's easy to analyze most any equipment with a little imagination. (Roark died before the 5th edition was completed. Young, I think, was a math prof that had worked with Roark. Young edited the manuscript to insert a lot of variations from the basic shapes on which Roark had catalogued the strain formulae. As such, Young's text and tables are a little hard to read and follow. The earlier editions by Roark alone were easier to use. Ang. C. > > And I am happy to say that it works perfectly. There were slight variations-- > > formula predicted 3.82 kHz for a 4 kHz transducer-- but I suspect the unimorph's [quoted text clipped - 6 lines] > Which edge condition and mode shape did you use; I gave many conditions > and shapes... While they really really aren't clamped, the formula that seems most applicable to unimorphs is the formula for fundamental resonant frequency (flexural mode) of isotropic circular membranes that are edge clamped; formula number 3 in that .pdf file. http://www.americanpiezo.com/technical_updates/unimorph.pdf That formula takes into consideration all aspects of the vibration plate which determines the resonance frequency more than anything else in unimorph design while the characteristics of the ceramic itself are totally unimportant in that regard, but the thickness of the ceramic determines how much voltage and power you can pump into a unimorph before the ceramic cracks. I hadn't tried to verify the accuracy of formula number 4 since it doesn't seem applicable to anything useful. > > By the way, that "h" value is as it should be; if you halved it as somebody > > suggested the results would be wrong. > > I usually see that dimension, thickness in this case, as "t". Yes, you're right there. I'll re-write that part to use the more standard "t" to avoid future confusion when I write a quick DOS program to do the math for me. But I see why the authors used "h" since further on in the article (Figure 6), "t" was used to denote the thickness of the complete unimorph rather than just the vibration plate by itself. > Roark's book "Formulas for Stress and Strain" is very > useful in calculating the breaking strength for any body, and > for vibrations analysis (vibration is treated only in the last > chapter). Maybe, but it seems like the unimorph formula is as guarded as the formula for Coke; I had spent endless hours of research going through old patents and as many abstracts that I could see without having to pay money for them, trying to find that one simple formula. I wonder how the authors in the article came about it? Oh, well... :-) Ron > While they really really aren't clamped, the formula that seems most applicable to > unimorphs is the formula for fundamental resonant frequency (flexural mode) of [quoted text clipped - 9 lines] > can pump into a unimorph before the ceramic cracks. I hadn't tried to verify the > accuracy of formula number 4 since it doesn't seem applicable to anything useful. The "plane extension" mode represented by equation (4) there must be radial stretch along the plane of the ceramic wafer. I can't see where this is of direct interest. It would take two such wafers, one on each side of a metal membrane, acting in opposition (push-pull) to make that system act like a unimorph. More explanation is needed here, not that it matters for now. > I'll re-write that part to use the more standard "t" to > avoid future confusion when I write a quick DOS program to do the math for me. But [quoted text clipped - 6 lines] > Maybe, but it seems like the unimorph formula is as guarded > as the formula for Coke; The main problem is to determine the K-factor, as it really is matter of Bessel Functions, and likely their integral... I see now that the problem is complicated by the fact that the ceramic wafer is attached to a metal plate that acts as the sound emitting diaphragm. One expects that the resonance frequency is controlled by the diaphragm following the formulas I sent, plus a central weight of the ceramic's material. Roark & Young did not treat that general case, since their book still dealt with the vast majority of static stress cases ("strength of materials"). > I had spent endless hours of research going through old > patents and as many abstracts that I could see without having to pay money for > them, trying to find that one simple formula. I wonder how the authors in the > article came about it? Oh, well... :-) Whoever wrote that Unimorph PDF used a reference book like Roark & Young's. Many "books" have been written by Russian and other former Soviet Republic authors that catalog many special cases of many engineering areas., so it is likely that that was what was done. FWIW: I suppose if someone had a mind to, they could use Roark's formulations and approaches, and compute the special case of a diaphragm of w mass per unit area loaded by a point weight, W, on its center, or stretched out like Figure 6 shows in detail.... Angelo Campanella > I see now that the problem is complicated by the fact that the ceramic > wafer is attached to a metal plate that acts as the sound emitting > diaphragm. One expects that the resonance frequency is controlled by > the diaphragm following the formulas I sent, plus a central weight of > the ceramic's material. That's an interesting suggestion, especially since the applicable formula for the resonant frequency of a circular plate with a central concentrated mass is available in at least two reference texts of which I am aware. Initially I was inclined to dismiss your suggestion out of hand, but after looking at the specifics I think that it may well have have some merit. For most available unimorphs, the thickness of the piezo element is almost exactly the same as the thickness of the brass baseplate, and the area of the piezo element is approximately half the area of the brass baseplate. Since the densities of brass and the piezo element are approximatly the same (within about 10%), the mass of the piezo element amounts to about 30% of the total mass of the unimorph structure. At 30%, it strikes me as somewhat of a wobbler, especially in view of the area ratio. Nonetheless, it would be interesting to know what level of accuracy in predicting resonant frequency is achieved by lumping the distributed mass of the piezo element at the center of the brass disk. The formula exists and the experiment is trivial, and I have some unimorph elements on order from Mouser. > Whoever wrote that Unimorph PDF used a reference book like Roark & > Young's. The equation in the Unimorph PDF is verbatim from Morse (Vibration & Sound), but with the incorrect stipulation of h as the thickness. > > Seems to me it's a whole lot easier finding technical specs for a > > nuclear [quoted text clipped - 18 lines] > lambda = wavelength, rho = density, nu = poison's ratio > or modulus and E = Young's modulus. Thanks, Eberhard, that helps some, but I was doing so well on formulas 3 and 4 when all of a sudden those "v" and "p" variables suddenly came from out of nowhere as none were previously defined. Was that "p" supposed to mean Poisson's or something else instead? Ron |
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