How to derive the coefficient of thermal expansion?

  • Thread starter tony_engin
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  • #1
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Hi!
How to relate the Young's Modulus with the coefficient of linear thermal expansion? It is known that they are inversely proportional to each other, but how to prove this?
 

Answers and Replies

  • #2
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Hello,

It can be derived from basic principles. You can start by finding the strain caused by
the thermal expansion. The change in length for this part can be found by subbing in the thermal expansion formuale. Hence we have found the strain.

Y=stress/strain
=Fl/A(dl) where dl= change in length
I cannot proceed further than this due to lack of information on which kind of shape it have. ( to calculate A) However, the coefficient of thermal expansion could be seen to be inversely proportional to Y after subbing in the expression for dl.
 
  • #3
538
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Ok...For this you need to have a good grip on basic formulae

Since you are dealing with linear Young's modulus.First of all Young's Modulus for a wire of a specific material is given by:

[itex]Y= \frac{FL}{A (dL)}[/itex]

where dL=Change in Length

Now change is length due to linear expansion is given by:

[itex]dL=L (a) (dT)[/itex]

a=coefficient of linear expansion
dT=Change in Temperature

Now putting the values in Young's modulus equation:

[itex]Y= \frac{F}{A (a) (dT)}[/itex]

Therefore Y is inversely proportional to a...easy isnt it?
 

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