Tuning Pin Size

[Mike and Jane Spalding]

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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

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