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If I have a metal ring and I heat it up, I will expect thermal expansion.

From my point of view of expansion, the metal will expand to free space. In that case, the inner radius of the ring should be smaller and the outer diameter of the ring should be bigger.

I understand that is not the case, but could anyone explain me why?

nicael
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5 Answers5

18

Let's suppose that the inner radius is actually becoming smaller.

This would mean that the molecules on the inner radius would have to be "packed" closer.

(also applies to any molecules that reside on the radius smaller than $\dfrac{R_1 + R_2}{2}$)

But then it would contradict with the fact the metal is expanding. Hence the inner radius will have to remain at least of its initial size.

Below is the pic of the expansion you suppose to happen (it won't happen this way, for the reasons described above).

nicael
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16

Another way to answer this question is based on the assumption that we are dealing with a linear expansion.

This means that the distance between any two points in a heated body will increase by the same percentage.

So if our two points are the end points of the internal diameter of a heated ring, we'll have to conclude that the internal diameter will expand... by the same percentage as the external diameter or the distance between any other two points of the ring.

V.F.
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8

Yet another way to think about this situation, in addition to the good answers so far, is: imagine you are uniformly heating a solid disk that has many concentric circles inscribed upon it. As you heat the disk, all the circles get bigger. Moreover, the disk does not buckle or fold anywhere from an internal stress. It smoothly and uniformly gets bigger everywhere.

Now suppose you cooled the disk and removed the material inside one of the circles. If you heated up the now-ring again, the same thing would happen as before; all the remaining circles would get bigger, including the one that is now the interior edge of the ring. The ring doesn't "know" that there's no center in the disk anymore; it gets bigger just like it did before.

Eric Lippert
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4

what you describe is a standard job interview question traditionally inflicted upon materials scientists. here is the "standard" answer:

Imagine you slice the ring into two half arcs spanning 180 degrees each. Each has radius "R".

place the two arcs back into contact, and heat them up. for each half-hoop, the radius increases to R + delta r where delta r comes from the thermal expansion.

the diameter of the pair of arcs has been increased thereby by 2*(delta r).

From this you can see that the diameter of the hole in the ring increases when the ring is heated, thereby proving what every machinist knows: to get a firm fit of a collar on a shaft, heat the collar, slide it on over the shaft, and allow to cool. the result is called a "shrink fit".

niels nielsen
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4

I don't believe any of the other answers so far have mentioned something important: it is the rigid structure of the metal ring that makes this work. A gas, liquid, or a flexible solid will expand inward the way your intuition suggests.

A YouTube video of a guy blowing up a donut-shaped balloon provides an example:

Uninflated donut balloon Inflated donut balloon

Robert
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