thermal expansion: hole dia shrinks when part expands? 2

It can if the part is heated unevenly, namely, the center is heated while the outside is not. But if the entire part is heated uniformly, then I don't see how the hole would not grow along with the rest of the part.

Engineering is not the science behind building. It is the science behind not building.

thread391-246558

That was a very good explanation there, Mint! I knew the reasoning, but your explanation makes it very easy to understand.

"Good to know you got shoes to wear when you find the floor." - [small]Robert Hunter[/small]

Jump to 41:11 in the video. Wish he would have done the experiment in reverse, but he discusses the physics of it.

This reminds me of the day in High School Physics the teacher went over this problem. He explained the law of thermal expansion, did the "ring and ball" demonstration (where the copper ring will not fit over the ball, but once heated in the bunsen burner, it will), then posed this exact problem. Instead of telling the students the answer, though, he let us argue amongst ourselves for the remainder of the period, with about half the class convinced the hole would get smaller if the cylinder wall was thick enough. At the end of class, he explained it like Mint did (imagine a large piece with a small hole and a small plug in that hole, and heat it and compare that to a solid piece--why would it be any different?) and dismissed class without allowing any more discussion. At the begining of class the next day, he asked if anyone though the hole would get smaller. Nobody was wanted to make that argument any more.

rp

I understand the thought process that causes confusion. Here's an attempt at a mathmatical proof. Please go easy on me and my assumptions/simplifications.

Without going into calculus, consider a ring cut up into 360 x 1degree segments. Now look at an individual cell along the neutral axis which has initial radius R0. For small angles let's assume this cell is a square (see attached visual aid).
The initial ID0 = 2*(R0-A0/2) where A0 is the arc length.

The circumference of this neutral axis is C0 = 2*pi*R0 but we can also represent it as C0 = 360*A0 because A0 is 1/360th of the circumference. Thus,
C0 = 360*A0 = 2*pi*R0 ... A0 = R0*2*pi/360

ID0 = 2*(R0 - (R0*2*pi/360)/2) [sub in for A0 in earlier equation]
ID0 = 2*R0*(1 - pi/360) [factor out R0]
ID0 = 2*R0*(360-pi)/360 [common denominator] ***

Now consider this cell is heated and expands in all directions. For this example, A1 = A0 * E where E is the expansion multiplier. Using the same equations as before...

C1 = 360*A1 = 2*pi*R1 ... R1 = 360*A1/(2*pi)

ID1 = 2*(R1-A1/2)
ID1 = 2*(360*A1/(2*pi) - A1/2) [sub in for R1]
ID1 = 2*A1*(360/(2*pi) - 1/2) [factor out A1]
ID1 = 2*A1*(360-pi)/(2*pi) [common denominator]
ID1 = 2*E*A0*(360-pi)/(2*pi) [sub in for A1]
ID1 = 2*E*R0*(2*pi)/360*(360-pi)/(2*pi) [sub in for A0]
ID1 = 2*E*R0*(360-pi)/360 [cancel (2*pi)] ***

Now the big question, which is bigger - ID0 or ID1? Take a look at the two marked equations and you'll see that ID1 is larger than ID0 by a factor of E. Surprise!

Thanks for your replies. this confirmed my initial thoughts.

However EngineerTex might have a point. Point of discussion was the cupper contact piece, used for current transfer in GMAW (mig/mag) welding. although the instructor (sales manager) specifically said "from a certain ratio d/D", it is much more likely that the piece is heated unevenly, causing the wire to block and hence the need for overdimensioning.
(Inside of the piece is heated due to the current, outside is cooled by the shielding gases).

thanks again for your replies...

redpicker,
How could anyone argue it shrank when the teacher just got through demonstrating it expanded?

I look at it probably all too simplistically. And incoorectly.

If so, forgive me.

If it is a metal bar or rod in a straight line and you heat it up, it ought to get longer. I would therefore conclude that an arclength would also get longer. In a disk with a hole in it, if the inner arclength shrunk while the outer arclength grew, then there would have to exist a "linear expansion neutral arc" which would be unaffected by heat. While it might be argued that the metal surrounding it prevented its expansion, that wouldn't make sense to me because the arcs on the outside would be moving away from it outwards from the centre and the arcs on the inside would be moving away from it towards the centre, so where would the constraint be coming from?

From this I would conclude that the hole has to get bigger.

Regards,

SNORGY.

And...again...stupid keyboard (incoorectly...)

Regards,

SNORGY.

The example of it exapnding was with a thin-walled ring. His hypothetical was for a thick-walled tube, which allows some to think mabye the thick wall would compress the ID, as mentioned in the thread that Mint linked to.

rp

Here's a nice twist: What about casting shrinkage when dealing with a hole (cast-in feature) in the center a flat piece? Consider it to be a cast square 2ft x 2 ft with a fairly small (4") hole in the center, just to throw numbers out. Assume 4" thick.

Does ID of this hole shrink or grow once the casting cools?

Not really a pure thermal expansion question but I think it plays into the details of the discussion.

Hole grows because material volume shrinks.

Ted

Volume changes due to CTE also. A cast hole will shrink as it cools. Voids can develop inside castings when the material inside shrinks and freezes after the surfaces in contact with the mold have solidified. Shrinking liquids have no tensile strength.

Wow, this is fun.

SNORGY:
What would happen if your bar was rectangular, and the top was heated while the bottom was frozen. The netural axis remains netural. Do you think it would warp like a piece of badly cured wood?

Now, if this was in fact a thick walled tube with a small concentric hole, and the outside of the tube was contained--not allowed to expand, where would the expansion go? Would it make the hole smaller and increase the density of the tube?



Charlie