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Anonymous Poster

Shrink Fits

05/06/2010 2:03 PM

I need to create an interference fit between a Ø.7500-7505 shaft consisting of alum 6061 t-6511 and a Ø.7495-.7500 journel consisting of 17-4 stainless. In freezing the 6061 t-6511 how much shrinkage will occur and will it be enough to where i wont have to heat the 17-4 journal?
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lyn
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Join Date: Oct 2008
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#1

Re: shrink fits

05/06/2010 2:31 PM

Thermal expansion (and contraction) of Al 6061 is 23.4x10-6/°C.
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Labyguy
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#2

Re: Shrink Fits

05/06/2010 4:18 PM

For the math impaired, 23.4 X 10^-6 = 0.0000234, so if you can freeze the shaft to say -25c your math would look like this:

-25c + 20c (room temp) = -45c (Delta change in temp)

so we have -45 x 0.0000234 = - .001053 change in size

so .7505 - .001053 = .749447

and it will clear.

to answer your question... go to -25c

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Anonymous Poster
#3
In reply to #2

Re: Shrink Fits

05/07/2010 11:11 AM

THANK YOU FOR THE INFO IT WAS EXACTLY WHAT I NEEDED
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lyn
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#4
In reply to #3

Re: Shrink Fits

05/07/2010 3:35 PM

I'll give Labyguy a GA for you. That's customary when you get the answer you needed.
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Tornado
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#5
In reply to #4

Re: Shrink Fits

05/07/2010 9:32 PM

Well, it should be customary; but alas it isn't....

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