Calculating Area of Thermal Expansion for Non-Isotropic Materials

AI Thread Summary
To calculate the area of thermal expansion for a non-isotropic material, the fractional increase in length varies in different directions due to differing coefficients of thermal expansion. Given initial lengths of 6.1 cm and 6.0 cm, and coefficients of expansion of 10×10^-5 K^-1 and 40×10^-5 K^-1, the material expands until it forms a perfect square. The area of the resulting square is determined to be 38 cm². The challenge lies in finding the temperature change necessary for the rectangular dimensions to equalize. Without a specific temperature, the equations for linear expansion cannot be directly applied to find the final area.
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Homework Statement


A non-isotropic material is heated, fractional increase in length is different along different directions. With length l1=6.1 cm and l2=6.0 cm. In direction of l1, the coefficient expansion is \alpha1= 10×10-5 K-1 and direction l2 is \alpha2= 40×10-5K-1. The plate is heated until it becomes a perfect square. What is the area of the square? (The answer is 38 cm2) I just don't know how use the equations without a given temperature.


Homework Equations


ΔL=\alphaL0T
Lx=Lx0+Lx0\alphaΔT
Ly=Ly0+Ly0\alphaΔT
A0=Lx0Ly0
A=LxLy

The Attempt at a Solution


Not sure what to do with the equations without the temperature.
 
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You need to find the temperature from the condition that the rectangular becomes square.
 

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