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Unread 30-01-2010, 08:20
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Re: Restrainting pin question

Quote:
Originally Posted by Tom Line View Post
Absolutely it will damage the shaft. Though most people don't realize it, metal isn't "hard". It's actually quite soft. Any time you see metal on metal interaction, for instance in the form of a pin, or a long shaft, etc, it's probably been hardened.

Machine a sacrificial pin that threads onto your shaft, and make sure that pin is supported against side loads by traveling through a hole in a plate. That way when you destroy that pin (and you will!), you can replace it and keep on rockin'.
Hmmmmmmmmmmm

Say your load is 100 pounds, you are using a .75 cylinder (.25 dia steel shaft) for the pin, and you use a double shear connection to support the shaft (a plate on both sides to ground, a plate in the middle connected to the kicker with close fit).
Cross sectional area of the shaft is just shy of 0.05 in^2. The shear force (draw a free body diagram) is 100/2 on each side or 50 pounds. That gives you a shear stress of about 1 ksi. Shear yield strength of mild steel / annealed stainless steel is a little over 20 ksi. That means you would get, oh, about a gazillion cycles before you have to worry about damaging the shaft by my calculations.

Actually there's a more complex beam in a socket calculation because there is actually minimal bending in the shaft (ideal shear doesn't occur in the real world), so you might only get half a gazillion cycles out of it.
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Last edited by Gary Dillard : 30-01-2010 at 08:23.