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<DIV><FONT size=2>Paul and Terry,</FONT></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>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></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>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></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>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></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>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></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>Best case, I would guess Paul's pins still have a =
20% to 30%
stress concentration factor at the step.</FONT></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>Hope this helps</FONT></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV><FONT size=2>Mike Spalding</FONT></DIV>
<DIV><FONT size=2></FONT> </DIV>
<BLOCKQUOTE dir=ltr
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
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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 title=larudee@pacbell.net
href="mailto:larudee@pacbell.net">larudee@pacbell.net</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A =
title=pianotech@ptg.org
href="mailto: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>
<DIV><BR></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>
<BLOCKQUOTE dir=ltr
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message -----</DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>
<A title=larudee@pacbell.net
href="mailto:larudee@pacbell.net">larudee@pacbell.net</A></DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A =
title=pianotech@ptg.org
href="mailto: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>=
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