Tuning Pin Size

[Farrell]

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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 -----=20
  From: larudee@pacbell.net=20
  To: pianotech@ptg.org=20
  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

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