Thermal expansion amounts steel vs aluminum

Hoping some one here understands these equations, I sure don't.

Wanting to find out the ID change of an aluminum tube 1"OD X 3/4"ID X 2" length say at 20ºf/-6.6ºc compared to 165ºf./74ºc

Wanting to find the OD change of a solid steel shaft 3/4" OD X 12" length say at 20ºf/-6.6ºc compared to 165ºf./74ºc

Both dimensions are originally dimensioned at 70ºf/21ºc



Coefficient of Linear
Thermal Expansion (a)
Steel:
7.0 x 10^-6 in/in-deg F
3.0 x 10^-6 m/m-deg C

Aluminum:
13 x 10^-6 in/in-deg F
23 x 10^-6 m/m-deg C

Thanks for any input

Ken

Using the inch equations (your metric number for steel is wrong; it should be 13e-6), it means the steel will expand .000007" for each inch of original length for each degree F increase, or 0.001015 inches for each inch of diameter with a 145°F increase, so the diameter increase will be .00076" for the 0.75" diameter (.000007*145*.75). Over the same temperature range, the aluminum sleeve will change .000013*145*.875 (.875 is the average diameter of the sleeve) or .0016".

If you're cooling the shaft and heating the tube to make a shrink fit, adjust the numbers by the actual temperature change for each part.

Thanks Dana,
Am trying to sort it out but still confused about the numbers.
We are talking then about only a .001 difference between them over the 145ºf temp range?

Was a copy paste on the equations, found it on the internet, has to be right :D'

It is not a heat/shrink fit but a free sliding member with as close a fit as reasonable with out the possibility of seizing from temperature expansion.
.

What is the actual interfacing diameter of both parts? If they are both exactly 0.75", then it will not be free-sliding anyway (at least until some wear occurs :) ) In reality, neither part is going to be exactly 0.75", and whether or not the thermal expansion matters at all is going to depend on the difference between the OD of the rod and the ID of the tube. You'll probably want a few thousandths clearance anyway so you'll have a true free-sliding fit, and if you have that, then the thermal expansion difference is not going to matter.

Of course the parts are not exactly .750 but I wasn't wanting to confuse my question with 10ths :)

What I was hoping for is some properly calculated figures that show exactly the thermal expansion through out the temp range.
Dana's I believe does that but the wording confuses me.
May be simpler just to make a few test samples of the aluminum tube, heat those installed on the shaft in the oven to the upper limit and see what we have..

Thanks

If you have an aluminum tube and a steel shaft, then heating them up isn't really going to tell you anything, as the aluminum tube will expand more than the steel shaft. The loosest fit will be at the highest temperature. The tightest fit will be at the lowest temperature, so put both pieces in your freezer if you want to see if they will bind up at low temperatures.

I think you're making a mountain out of a molehill here...I think Dana has his numbers mixed up; the aluminum part will change in diameter by about 0.0016" through the entire temperature range you desire, and the steel part will change about 0.0009" over the entire temperature range, but still, this means that the fit changes by less than one thou over the temperature range. Given that you'll need a clearance of at least 2 or 3 thousandths for a slip fit, the difference in thermal expansion between the two parts is far less than the clearance needed for the slip fit, and all of this is irrelevant.

I use 160 differential degF to give .001 in/in expansion difference between steel and aluminum. Same in mm/mm or ft/ft for 160 degF.

Remember the parts also have to be straight and round..........!

"I think you're making a mountain out of a molehill here"
LOL, yes, I do tend to over think and consequently over complicate things.

Thanks for all the input, it has been very helpful and my question has been completely answered.

One last one on the equation,

Can one of you detail that equation in understandable terms using basic math, meaning add, subtract multiply and divide?


Ken

I corrected the numbers in my post above and [hopefully] made the math clearer.

Quote:

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Originally Posted by Blue Chips

"I think you're making a mountain out of a molehill here"
LOL, yes, I do tend to over think and consequently over complicate things.

Thanks for all the input, it has been very helpful and my question has been completely answered.

One last one on the equation,

Can one of you detail that equation in understandable terms using basic math, meaning add, subtract multiply and divide?


Ken

---

Ken,
Its pretty simple.
Change in length = Original lengh x COE x Change intemperature

That helped a bunch Dana, ran the figures a couple time to be sure and it came out the same.

Aaron, yes, quite simple once you know how :)

The "7.0 x 10^-6" is what threw me a curve, did not know what that meant.
Also did not know to average the tube diameter.

This is saved so next time it can be done with out asking for help.


Thanks again!
Ken