Tuning Pin Size

[larudee@pacbell.net]

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Terry,

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?

Paul

Farrell wrote:

> 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
>
>      ----- Original Message -----
>      From: larudee@pacbell.net
>      To: pianotech@ptg.org
>      Sent: Sunday, January 27, 2002 11:27 AM
>      Subject: Re: Tuning Pin Size
>       Terry,
>
>      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.
>
>      Paul
>
>      Farrell wrote:
>
>     >  "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
>

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