Tuning Pin Size

[Ron Nossaman]

>
> 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.


OK. Let's also assume for illustration that if the pin had held together, it
would have turned at 325"lb.


>
> 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).


It shouldn't turn any farther than the short pin before shearing if the PSI
friction load of the block against the pin is the same. If the pin sheared, the
bottom of the short pin never moved either. Since there is so much more surface
area in contact with the block with this pin, it would take maybe 900"lb to
turn it in the block. Just like the above situation, the yield point of the pin
was surpassed before the block moved. 



>
> 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.


OK, and might have moved at 375 if the pin hadn't failed.


>
> 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.


Correct, but again, the same fit sheared off the 0.276 pin with only a 0.276
diameter in the block. If it takes 375"lb to turn the 0.286 pin, but the 0.276
fails at 300"lb, the pin isn't going to turn.  


>
> 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?


Simple. The weakest link fails. The fit in the block, whatever the pin size,
determines what it takes to turn the pin. The diameter of the part (and the
becket hole) above the block determines the shear strength. If the pin shears
before it moves in the block, there's no way of knowing, and it doesn't really
matter if the block fit is 0.04 PSI too high, or 14,921 PSI too high. The
touchy part of fitting oversized pins is the narrower margin for error in
drilling and pin uniformity. A 0.001" difference in hole size will make more
difference in "lb torque readings with a 2/0 pin than with a 1/0 because the
surface area in contact with the block is larger with a 2/0 pin than with a
1/0. The bigger the pin, the greater the affect of a 0.001" change in hole size
relative to pin diameter.


Ron N