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I tested my pins in a vise and they still broke at the becket hole, which
is where they break in standard pins. If there were no becket hole,
I presume they would break at the transition, but at a higher applied force.
When some of my pins are produced with a larger transition, I'll test them
as well. I tested them with the tops of the vise jaws tightened at
roughly 5/16" below the transition shoulder and also at roughly 1/16" below
the shoulder to simulate the situation in both closed face and open face
pin blocks. I used an old extension lever just in case the lever
broke instead of the pin. Got my 24-year-old son to do the pulling.
I figure the tip on the lever is plenty tight now. I have no idea
how much force was used because I didn't want to break my torque wrench.
<p>Paul
<p>Mike and Jane Spalding wrote:
<blockquote TYPE=CITE> <font size=-1>Paul and Terry,</font> <font size=-1>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.</font> <font size=-1>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.</font> <font size=-1>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.</font> <font size=-1>Stress
concentrations can be minimized by radiusing the inside corner, by optimizing
feeds and speeds in the lathe, and polishing the radius after machining.</font> <font size=-1>Best
case, I would guess Paul's pins still have a 20% to 30% stress concentration
factor at the step.</font> <font size=-1>Hope this helps</font> <font size=-1>Mike
Spalding</font>
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<div style="FONT: 10pt arial">----- Original Message -----</div>
<div
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><b>From:</b>
<a href="mailto:larudee@pacbell.net" title="larudee@pacbell.net">larudee@pacbell.net</a></div>
<div style="FONT: 10pt arial"><b>To:</b> <a href="mailto:pianotech@ptg.org" title="pianotech@ptg.org">pianotech@ptg.org</a></div>
<div style="FONT: 10pt arial"><b>Sent:</b> Sunday, January 27, 2002 12:19
PM</div>
<div style="FONT: 10pt arial"><b>Subject:</b> Re: Tuning Pin Size</div>
Terry,
<p>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?
<p>Paul
<p>Farrell wrote:
<blockquote TYPE="CITE"><style></style>
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 <span
id=__#Ath#SignaturePos__></span>
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<div style="FONT: 10pt arial">----- Original Message -----</div>
<div
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><b>From:</b>
<a href="mailto:larudee@pacbell.net" title="larudee@pacbell.net">larudee@pacbell.net</a></div>
<div style="FONT: 10pt arial"><b>To:</b> <a href="mailto:pianotech@ptg.org" title="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>
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>
</blockquote>
</blockquote>
</blockquote>
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