These dimensions exist in an ambient temperature of 68 degrees F and RH of 45%
What happens to the hole diameter upon raising the ambient temperature? Will the hole expand or contract, and will this deformation ever reverse upon further temperature rise?
This message has been edited. Last edited by: DorianGreyed,
The hole diameter will expand if the temperature is raised.
Matter, unless it undergoes a state change (and usually even then) will expand if the temperature changes (assuming the pressure, etc. doesn't change).
I've tested the theory to convince myself, but since I can't show you that, you'll have to settle for a theoretical explanation.
The simplest way to think this through is to think of a ring of atoms. If the material is heated, the atoms must move apart, making the ring bigger. The steel disk can be thought of as a series of these rings inside each other, each one expanding with the heat.
That's what I thought methos,............at first. But now think of the ring as a ring of raw dough, like a doughnut. Drop it in hot oil and every characteristic swells, in effect reducing the diameter of the hole.
In a single ring of atoms, or even molecules, the expansion may only be circular or tangential, but in a complex such as my steel ring, won't the molecules expand in every direction, making the hole smaller at some point? (It's my own mental conflict on the matter that caused me to ask if the resulting deformation would at some point reverse).
Yeah, as I said, I had to test the theory to convince myself.
The difference between dough and metal is that the dough can easily deform. The hole on a bagel doesn't simply symmetrically shrink, it forms all sorts of creases to do so. The dough is also in a more fluid state than metal at the beginning of the process. To compicate things, there are chemical changes taking place when cooking.
For a hole to shrink in an object without significantly distorting it, the atoms near the hole would have to move closer together. As long as the metal is homogeneous and isn't brought to its melting point, this sort of distortion shouldn't occur.
OK. So I’m sure we will agree that the OD, or outside diameter of my ring will increase with a rise in temperature, and does so without significant distortion. (We’re only talking hundredths of mm, or 0.01) I’m guessing this is because the material has room to expand that isn't counter to itself, (as in the forces at work at the keystone in an arch). This leads me to believe that the shape of the hole is significant. Now imagine the hole is square, or perhaps even the sides of the square hole are slightly convex to opposing faces. If we now calculate the area of this profile, and again after heating, will the area be reduced?
As an aside and practical example, mechanics rebuilding close tolerance machinery will often heat up a doughnut-shaped part in hot oil to facilitate, say, slipping it back over a shaft.
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Here is the proof: Originally Posted by Rohin.T.Narayan When a metal plate with a circular hole at its centre, is heated, definitely along with the areal expansion of the plate the diametre of the circular hole also increases .But can you give a mathematical proof for this using the differential equations of coefficients of expansions You don't need to worry about differential equations. It is a linear expansion.
The sides of the plate are A and B. Divide it into quarters. The length of the sides with the quarter hole taken out of them are A-r and B-r. heat it up. The A side and holed A side will be: r = original radius of hole r' = new radius of hole c = coefficient of thermal expansion T = temperature change
(1) A’ = A + AcT (2) A’ – r’ = (A – r) + (A – r)cT
Subtracting (2) from (1): (3) r’ = A + AcT – ((A - r) + (A – r)cT) (4) r’ = r + rcT
Do the same thing for the B and B-r sides.
So the hole radius increases at the same linear rate as the metal.
