<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=us-ascii"> <META content="MSHTML 5.50.4522.1800" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=#ffffff> <DIV>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.</DIV> <DIV> </DIV> <DIV>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).</DIV> <DIV> </DIV> <DIV>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.</DIV> <DIV> </DIV> <DIV>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. <DIV> </DIV> <DIV>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.</DIV> <DIV> </DIV> <DIV>Or so it would seem to me.</DIV></DIV> <DIV> </DIV> <DIV>Concentrating shear forces? How does it do that?</DIV> <DIV> </DIV> <DIV>Terry Farrell <SPAN = id=__#Ath#SignaturePos__></SPAN> </DIV> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style="FONT: 10pt arial">----- Original Message ----- </DIV> <DIV style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: = black"><B>From:</B> <A title=larudee@pacbell.net href="mailto:larudee@pacbell.net">larudee@pacbell.net</A> </DIV> <DIV style="FONT: 10pt arial"><B>To:</B> <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">pianotech@ptg.org</A> </DIV> <DIV style="FONT: 10pt arial"><B>Sent:</B> Sunday, January 27, 2002 = 11:27 AM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Tuning Pin = Size</DIV> <DIV><BR></DIV>Terry, <P>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. <P>Paul <P>Farrell wrote: <BLOCKQUOTE TYPE="CITE"> "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 <SPAN id=__#Ath#SignaturePos__></SPAN></BLOCKQUOTE></BLOCKQUOTE></BODY></HTML= >