This is a multi-part message in MIME format. ---------------------- multipart/alternative attachment Well, OK Paul and Mike and others. I don't really understand it, but = I'll assume you are right with this "concentration factor". But I guess = as you pointed out Paul, that if your pins break at the becket, it kinda = makes all this moot. Your pin with the 1/0 top will have the same shear = strength as any other 1/0 pin - because the weak link is the area of the = becket. Your pins sound intriguing. Where are they available? Terry Farrell =20 ----- Original Message -----=20 From: Mike and Jane Spalding=20 To: pianotech@ptg.org=20 Sent: Sunday, January 27, 2002 2:23 PM Subject: Re: Tuning Pin Size Paul and Terry, Whenever there is an abrupt change in size/shape, there will be a = stress concentration. This is what you've got at the inside corner = where the 1/0 upper portion of your pin meets the 2/0 or larger lower = portion. Picture a pin of uniform diameter, with a straight line drawn along = it. Now twist the pin, look at the line: it's a uniform spiral, like a = barber pole, or a candy cane. Now take the stepped pin, draw the straight line along the lower = portion, in towards the center along the step, up along the upper = portion. Twist the pin, look at the line. Still generally a spiral, = but: The spiral on the upper (smaller diameter) section is faster than = the spiral on the lower portion. Where the upper spiral meets the step, = there's a distortion in the spiral which is your stress concentration. =20 Stress concentrations can be minimized by radiusing the inside corner, = by optimizing feeds and speeds in the lathe, and polishing the radius = after machining. =20 Best case, I would guess Paul's pins still have a 20% to 30% stress = concentration factor at the step. Hope this helps Mike Spalding ----- Original Message -----=20 From: larudee@pacbell.net=20 To: pianotech@ptg.org=20 Sent: Sunday, January 27, 2002 12:19 PM Subject: Re: Tuning Pin Size Terry,=20 I am not an engineer, but I consulted one while researching for my = patent, and that's how it was explaiened to me. The problem is that = when you twist or flex the top portion, the bottom portion doesn't twist = or flex as much. If you twist or flex the pin that is 2" in diameter in = the base and .276" in the top portion to the breaking point, it's going = to break at the transition point every time. If your argument were = correct, it would break randomly at any point along the top portion. = Are there any engineers who would care to elaborate? Carl?=20 Paul=20 Farrell wrote:=20 I'm trying to understand this. Let's say we have a 0.276 -in. = dia. tuning pin that is 2-in. long. Let's say it has a shear strength of = 300 inch-pounds. Meaning of course if you install the pin in a new = Baldwin, put a tuning hammer on it (or a torque wrench) and try to turn = it, when you get to a shear force of 300 inch-pounds, it will shear into = two pieces - leaving one piece in your tuning lever tip and the other in = the block. Now take a similar pin, but make it 6 inches long. Do the = same things, and it should shear at 300 inch-pounds of torque. Length = should not matter (you will of course get more twist with the longer pin = before it shears). Now take a 0.286-in. dia. tuning pin that is 2-in. = long. Let's say it has a shear strength of 350 inch-pounds. Do the same = things to it and it will shear at a torque of 350 inch-pounds. Now take = a pin with a bottom of 0.286-in. dia. and a top of 0.276-in. dia. Put it = in that same nasty Baldwin block - or a strong vice - or whatever - just = so it doesn't move - at it will shear at a torque of 300 inch-pounds. = Now take a pin with a bottom of 2-in. dia. and a top of 0.276-in. dia. = Put it in that same nasty Baldwin block - or a strong vice - or whatever = - just so it doesn't move - at it will shear at a torque of 300 = inch-pounds. The larger base would act just like the pinblock with the = constant diameter 0.276-in pin in it. They would both shear at 300 = inch-pounds. Or so it would seem to me. Concentrating shear forces? How = does it do that? Terry Farrell =20 ----- Original Message ----- From: larudee@pacbell.net To: pianotech@ptg.org Sent: Sunday, January 27, 2002 11:27 AM Subject: Re: Tuning Pin Size Terry,=20 All of what you mention affects shearing, but the bottom portion = also affects it by concentrating the shear forces at the point where the = diameter changes. In other words, the greater the difference, the more = torsion and flex will end at that point and the less those forces will = be distributed thoughout the pin.=20 Paul=20 Farrell wrote:=20 "The larger the size difference between the two portions, the = greater the risk." Why would that be? I should think the point at which = a pin would shear would depend entirely on the metal composition (let's = assume this is constant), its diameter, and the tightness of the = pin/block fit (torque). As you make any pin size fit tighter in the = block, it will get closer to its shear point. As you make any pin = smaller in diameter, you will move toward a lower shear point. Diameter = and torque - I think that is all. Why would the diameter contrast = between the top and bottom portion affect its shear strength? Is there = something about the machining process? Or do you mean (by the above = quote): 'The smaller the diameter of the top portion of the pin, the = greater the risk of shearing' (because, of course, the smaller diameter = pin will have a lower shear strength, and will shear at a lower pin = torque). How would the diameter of the bottom portion of the pin affect = the shear strength? I am assuming that the rebuilder will = drill/ream/whatever the hole to a proper diameter for the diameter of = the pin bottom portion. Terry Farrell=20 ---------------------- multipart/alternative attachment An HTML attachment was scrubbed... URL: https://www.moypiano.com/ptg/pianotech.php/attachments/e3/49/a7/e1/attachment.htm ---------------------- multipart/alternative attachment--
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