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Thermal Expansion in 4140 - Solid Disk vs Ring

Joined
Dec 21, 2016
Hi PM,

Calculating thermal expansion in a 4140 HT&Q, 28-32 Rc ring: 6" OD x 4.75" ID by 1.00" Thick. Using a spreadsheet based calculator I found on line it says a 20 degree temperature differential I should compensate the ID of 4.775 by .0006" (.00057"). Is this correct?

Also I was wondering if a thin walled ring would change more, less or be the same as a solid disk of the otherwise same dimensions. Clearly the disk grows 'linearly' across the diameter. A ring is the same as a bar , right? So unroll it and you have a bar the length of the circumference. Run the calc for the bar length & rewrap it to find the delta diameter. But would one use the ring OD, ID, centerline to calc the bar length? TIA

William
 
This feels like one of those facebook math questions lol. Looking at it objectively 4140 bar or solid is still 4140. So I would assume thickness wouldn't matter, beisdes expansion time, whether ring or solid. At the end of the day you still have 4140 with the same thermal components for expansion(im totally guessing) I've seen in real time a ring type part expand more throughout a hot day but quicker than a solid part. Really good question and im more or less interested in a correct answer, I am no engineer and probably no help here lol.
 
I've experienced ring parts expand more than solid parts as well. We run a few jobs out of UHMW plastic which has a very high coefficient of thermal expansion. A solid part will grow a small amount with temperature, but a ring of a similar diameter will grow a TON. It's so much that we avoid holding the part when measuring it because the heat from our hands will make it grow. So, I definitely think there is something to it.

I can't explain why this happens where the ring grows more than the solid part, but I'd always imagined that it was because the ring didn't have that material in the middle to hold it in tension and prevent it from growing even more.

I like your idea to calculate the growth as if it were a bar. I'd use the centerline because it makes for a good average.

But this makes me wonder at what point do you treat a round part with a hole in it as a ring? And when do you treat it as a solid? What if your 6" OD part only has a 1/2" hole in it? Still a ring?
 
Think about the thermal mass.

It takes less energy to heat a smaller volume of material.

Your cut is probably similar thick or thin part. Therefore heat going in is the same.

Thermal conductivity of material will have an impact. Specific heat capacity will also.
 
I think that the rings grow more as well, although it has now been several years now since it's come up first hand, that I second guess myself on this.


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Think Snow Eh!
Ox
 
A sold disk and a circular ring (or a tube) of the same outside diameter and same material--That is have the same coefficient of expansion will expand or contract as the temperature changes. The diameters will be the same when he tempuratre has stabilized.

A similar question is: what will the ID of a circular ring do when itm is heated. Get larger or smaller.

Lost
 
A similar question is: what will the ID of a circular ring do when itm is heated. Get larger or smaller?

Larger -- in the first order, holes act like solids for the purpose of thermal expansion.

Otherwise, the "ball going through the ring" experiment in Physics class would have behaved the opposite to the way it does.

Regards.

Mike
 
Hi PM,

Calculating thermal expansion in a 4140 HT&Q, 28-32 Rc ring: 6" OD x 4.75" ID by 1.00" Thick. Using a spreadsheet based calculator I found on line it says a 20 degree temperature differential I should compensate the ID of 4.775 by .0006" (.00057"). Is this correct?

Also I was wondering if a thin walled ring would change more, less or be the same as a solid disk of the otherwise same dimensions. Clearly the disk grows 'linearly' across the diameter. A ring is the same as a bar , right? So unroll it and you have a bar the length of the circumference. Run the calc for the bar length & rewrap it to find the delta diameter. But would one use the ring OD, ID, centerline to calc the bar length? TIA

William


I think you're confusing the linear coefficient of expansion with the NON-linear coefficient of expansion. Now, where did I leave my tables of non-linear coefficients laying around...damn it, wmpy, did you not send those back to me ;)
 
I think you're confusing the linear coefficient of expansion with the NON-linear coefficient of expansion. Now, where did I leave my tables of non-linear coefficients laying around...damn it, wmpy, did you not send those back to me ;)

Must be lost in the mail...
 
Using a spreadsheet based calculator I found on line it says a 20 degree temperature differential I should compensate the ID of 4.775 by .0006" (.00057"). Is this correct?

Depends on what you're measuring with. If your measuring device has a similar coefficient of expansion, was calibrated at 68°F, and is the same temperature as the workpiece when you measure, don't fudge at all.
 
Hi PM,

Calculating thermal expansion in a 4140 HT&Q, 28-32 Rc ring: 6" OD x 4.75" ID by 1.00" Thick. Using a spreadsheet based calculator I found on line it says a 20 degree temperature differential I should compensate the ID of 4.775 by .0006" (.00057"). Is this correct?

Also I was wondering if a thin walled ring would change more, less or be the same as a solid disk of the otherwise same dimensions. Clearly the disk grows 'linearly' across the diameter. A ring is the same as a bar , right? So unroll it and you have a bar the length of the circumference. Run the calc for the bar length & rewrap it to find the delta diameter. But would one use the ring OD, ID, centerline to calc the bar length? TIA

William

1. Let's assume infinitely thin ring (just chain of single atoms), diameter D, changing its circumference length due to temperature raise ΔT proportional to expansion coefficient Ω of the material. The length of the ring at initial temperature L(i)= Π*D. The length of the ring at final temperature L(f)=Π*D+Π*D*ΔT*Ω=Π*D*[1+ΔT*Ω]. The diameter of the expanded ring D(f)=D*[1+ΔT*Ω].
2. Now let's assume the disk of same initial diameter as above mentioned ring. Lets cut infinitely thin rectangular (one atom wide) just from the center of the disk. The length L of this rectangular is of course equal to D, the disk diameter. Now, it is heated exactly as it was above. Due to thermal expansion, it grows. Its final length L(f)=D(f)=D*[1+ΔT*Ω].
Conclusion: The diameter of both disk and ring made from same material grows equally with the temperature.

In order to make it more intuitive, let's think about the external, very thin (1 atom) "skin" around both ring and disk of same diameter. As temperature grows and stabilizes at target value, the length of this skin grows equally on both the ring and the disk, as they are made from same atoms.
 








 
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