Expansion of a hole

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I have a very simple question. If a hole is made at the center of a metal plate and it is heated to increase its temperature, will the hole increase in size or decrease?
 

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  • #2
256bits
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The other question is:
If a hole is made at the center of a metal plate and the plate is cooled to decrease its temperature, will the hole increase or decrease in size?

If you can answer my question, you should be able to answer your original question.
 
  • #3
mathman
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I have a very simple question. If a hole is made at the center of a metal plate and it is heated to increase its temperature, will the hole increase in size or decrease?
Increase: Consider what would happen if you didn't drill the hole and heated. The area where the hole would be gets bigger.
 
  • #4
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Each and every material line within the plate grows by the same percentage when the plate is heated uniformly. This includes every differential arc length comprising the circumference of the hole.
 
  • #5
berkeman
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It kind of makes sense that if you draw a circle on the plate and heat it, the circle will grow with the plate. But if there is a hole instead, with no material in the hole to "push out" on the material that surrounds the hole, is the growth of the hole the same as the growth of the filled circle? Or could it be less growth because of the lack of the material in the middle pushing out? Just curious.
 
  • #6
AlephZero
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There are materials (e.g. many crystals, and some composite matierals) where the coefficient of thermal expansion is different in different directions. In those materials, in general the hole would not expand uniformly, and would change shape.

But apart from that:

It kind of makes sense that if you draw a circle on the plate and heat it, the circle will grow with the plate.
If the solid plate is at a constant temperature and the edges are free to expand, the stress in the plate will be zero everywhere, independent of the temperature.

So, imagine that you draw the circle, then change the temperature of the plate, then cut out the hole around the expanded or contracted shape of the line. Since the stress in the plate was zero everywhere before cutting, it will stay zero after cutting, and cutting the hole wll not change the shape of the plate.

Or more mathematically:
If there is no elastic stress in the plate, the strain field is ##\epsilon_{xx} = \epsilon_{yy} = \epsilon_{zz} = \alpha \Delta T## and the shear strains are all zero, where ##\alpha## is the corefficient of expansion and ##\Delta T## the temperature change.

Because the strain field is symmetrical in x y and z, this means that ANY two points that were originally a distance ##d## apart become a distance ##d(1 + \alpha \Delta T)## apart, indepedent of the shape of the object, and whether or not it contains holes.
 
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  • #7
berkeman
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Ah, I didn't think of the zero stress condition. Good point. Thanks AlephZero! :smile:
 
  • #8
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There are materials (e.g. many crystals, and some composite matierals) where the coefficient of thermal expansion is different in different directions. In those materials, in general the hole would not expand uniformly, and would change shape.

But apart from that:



If the solid plate is at a constant temperature and the edges are free to expand, the stress in the plate will be zero everywhere, independent of the temperature.

So, imagine that you draw the circle, then change the temperature of the plate, then cut out the hole around the expanded or contracted shape of the line. Since the stress in the plate was zero everywhere before cutting, it will stay zero after cutting, and cutting the hole wll not change the shape of the plate.

Or more mathematically:
If there is no elastic stress in the plate, the strain field is ##\epsilon_{xx} = \epsilon_{yy} = \epsilon_{zz} = \alpha \Delta T## and the shear strains are all zero, where ##\alpha## is the corefficient of expansion and ##\Delta T## the temperature change.

Because the strain field is symmetrical in x y and z, this means that ANY two points that were originally a distance ##d## apart become a distance ##d(1 + \alpha \Delta T)## apart, indepedent of the shape of the object, and whether or not it contains holes.
Very nicely explained!!!
 

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