# Linear Thermal Expansion

• posted
How do I calculate an amount of thermal expansion for a given shape if you
have Coefficient of Linear Thermal Expansion (7.2x105 in my case)
Thanks,
Alex
• posted
Sorry, wrong coefficient. It is 1.1E-04
• posted
Isn't that pretty high? At least for metals...
Depends on the units... is the coeff. in cm/°C or inches/°F or..?
In any case it's a differential thing, if it's 5 inches at 20° and 5.1" at 50°...
Tim
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• posted
Right. Key is to look at the units of the thermal contraction.
It's most often quoted in dimensionless units, per degree.
Jim
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• posted
length x coefficent x temp change
Alex wrote:
• posted
First you must know the units for the coefficient. ie... in/ft/deg. C in/in/deg F. depending upon the material. The material expands exactly as stated in the coefficient. Take the distance from point one to point two and apply the coefficient to it.
John
• posted
One possible point of confusion that has not been mentioned was implied in the original question--it has to do with the shape of the object. The shape does not matter. For example, if you were calculating the change in height of a letter "S," you would do it exactly the same as a letter "I." The spaces in a piece expand with the same coefficient as the surrounding material.
• posted
I would of thought it would be some function of the relation between OD and ID. But maybe not pi*D is linear so the relation would be the same wouldn't it? The change in ID may depend still on the ratio of OD to ID, ie a large circle verses a small circle.
I can tell I'm going to have to get the calculator out and do some figuring.
John
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• posted
Actually not. I had remembered an old parlor trick where one had two iron discs with holes cut in them, and an iron ball on a stick.
The ball would go thru one hole, but not thru the other.
Heating both plates would cause the ball to pass by the one that it formerly stuck in, and to *not* pass thru the one that it used to go thru!
The key here is the a) the expansion did indeed depend on the OD/ID ratio of hole to stock in the plates, and b) the most important one - the plates are heated non-uniformly. The inner rim of the hole gets hotter than the exterior.
For *uniform* temperature of the material, the OD/ID ratio does not matter.
Jim
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• posted
Long ago when I was in grade school there was a blacksmith shop across the highway. I can still remember the old gentleman beating the hell out of various pieces of glowing red iron he pulled out of his forge. I also remember seeing him make horse shoes starting with a straight piece of iron and finishing by putting a stinking hot horseshoe on a live horse and watching the smoke emerge from the hoof. Talk about getting a good fit... that horseshoe burned its own place on the hoof.
One procedure which was particularly fascinating was to watch him put an iron rim on a wooden wagon wheel. He heated the sucker up so that the parts were glowing red and then put the sucker on a wooden wheel. He quickly dowsed it with water which created a bunch of steam, smoke and probably prevented the wooden parts from going up in smoke. At any rate the rim was there to stay presumably because of a decrease in the internal diameter of the iron rim.
Now my observation would be that heating the rim increased the interior diameter (it now was large enough to fit over the wheel) and quenching it tightened the sucker tight enough so the rim didn't come off the wheel as soon as the customer left the shop. If this weren't factual he wouldn't get a whole lot of business putting iron rims on wooden wheels.
Now those are facts. To convince me that if the rim was thicker the hole would get smaller on heating would take a whole lot of theoretical physics.
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I know I'm getting in on this late, but a piece of metal with a hole in it expands when heated just as though there was no hole in it. The wagon tires that we have installed expanded about 1/8" per foot in diameter between about room temperature and about 750 degrees F. The tires are a tight fit when properly measured and welded, but in my limited experience, the wheelwright also bolts them to the felloes in several places.
Pete Stanaitis
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