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Elasticity of diving board 
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#1
Sep109, 08:14 AM

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Trying to find please equation(s) to calculate the "elasticity", "stiffness" of a homotropic cantilever diving board. Am seeking some analogue of the spring constant of Hooke's law, F= kx, for this 3D system. I'm unclear about whether Young's modulus applies here given the force ( or pressure ) of the diver being normal to the long axis of the board. Am seeking a treatment insofar as possible at a level of advanced high school physics, ie for the "ideal" board with assumptions of homotropic material, isothermal, elastic limit not exceeded etc / forego differential equations please.
Although it be a topic for another day, I am looking at the diving board as a model for my ultimate goal to measure the elasticity of a woodwind reed as a function of age ( use ). Thanks in advance. 


#2
Sep109, 03:09 PM

P: 12

Beaubello
Maybe this previous thread can get you started http://www.physicsforums.com/showthread.php?t=278252 


#3
Sep709, 11:27 PM

P: 6

Having read said thread it sounds like a diving board, at least as it's described here with freedom to pivot, is not a good model for my reed which is not free to pivot at its clamped end on the mouthpiece in situ ( although I'd probably study it clamped onto a table ). I'm interested in stressing the reed at its free end perpendicular to the "plane" of the reed in order to determine its "elasticity". I'm familiar with Young's modulus applied where stress is in the same direction as the body's long axis ( e.g. wire ) but am unclear how it might apply here. To simpify the discussion for now let's presume that the reed has uniform thickness from clamp to tip ( although it in fact thins from clamp to tip ). Do I understand correctly that without freedom to rotate the moments of inertia from the above diving board thread would not apply here? Any help with formula(s) for "elasticity" of said reed before I can find a method to make the measurement?



#4
Sep809, 08:24 AM

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Elasticity of diving board
There are four independent ways to deform a beam (which is a pretty good model for a reed, though a plate might also be appropriate): axial (pulling on the end), torsion (twisting the end around the long axis), transverse (a load perpendicular to the long axis), and bending moment (the load that turns a straight beam into a smile or a frown). It sounds like your load will predominantly excite the transverse and bending moment modes. These are well described in any mechanics of materials book (e.g., Beer and Johnston) as well as numerous places online. The resulting deflection of the beam is a linear function of the compliance (i.e., the inverse stiffness, or Young's elastic modulus) of the material. Does this help?



#5
Sep909, 06:28 PM

P: 6

Given my exposure to only intro physics texts your reference to a book of mechanics of materials is surely helpful. Although I've not yet checked out the Beer & Johnston perhaps you would be good enough to clarify a few points. The transverse and bending deformations sound very similar 1) is the distinction in the point of load application, central vs end? 2) Do I understand correctly that I should be able to calculate Young's modulus of the reed if I can somehow find a method to measure stress/ strain = Y ? If so, I know how to express the stress as F/A but 3) how does one express strain in terms of delta x, the vertical displacement of the tip ( which I was thinking to measure )? I've seen Young's modulus only applied to axial deformation in which strain = delta L / L . However I can't recognize a counterpart of the axial length, L in the case of axial deformation, for my case of transversebending deformation of a reed. Help?



#6
Sep909, 07:06 PM

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In your case, however, it seems like you may be interested in finding only the relative change in stiffness due to use. Simply mount the reed before and after use (in any configuration!), apply a load, and measure the deflection. As long as the material is loaded in the linear regime* and the dimensions and mounting conditions are the same for both measurements, the ratio of the two deflections is the inverse of the ratio of the two Young's modulus values. That's the great benefit of linear elasticity. (Note that it may be difficult to replicate exactly the mounting conditions, though.) *You'll know this is the case if an x% larger load gives an x% larger deflection (for a measurable value of x). 


#7
Sep1009, 06:41 PM

P: 6

Ok, from the Calculator I find the formula for "cantilevered beam with uniform load", y= WL^3/ 3YI. Substituting W=ky and solving for k gives the same constant, 3YI/ L^3, supplied by Phantom Jay in the thread to which YellowPeril referred me. Having come full circle let me confirm the understanding I've gained along the way. (1) Said constant applies to my reed loaded within the linear regime? Although I'm familiar with the term spring ( stiffness ) constant for k=F/x applied to a spring (2) how do engineers customarily refer to such a k as above for a beam?
Letting y_1, i.e. y subone, = reed deflection preuse and y_2 = deflection postuse, then the Calculator equation simplifies to y_2/ y_1 = Y_1/ Y_2 [ assuming no change in W, L, I ]. (3) Is this whence one derives your indication that the ratio of the 2 deflections is the inverse of the ratio of the 2 Young's moduli? If so, I would add that in my experiment reed thickness t likely changes because of hydration effects of playing. Hopefully though I wouldn't even need to deal with the difficulty of measuring reed thickness because I trust your assessment is right on that I'm interested in finding only the relative stiffness due to use. (4)Shouldn't I be able to make only the easier measurements of y and W to calculate k=W/y and forego the more difficult measures of its factors Y and t? Is there Eureka upon your next response, Mapes? 


#8
Sep1009, 08:28 PM

P: 6

P.S. to my last post. Although you probably surmised it, I forgot to add my point to mentioning change in reed thickness was that it would change I= bt^3/12, also supplied by Phantom Jay in the aforementioned thread.



#9
Sep1109, 06:45 AM

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