Radius of curvature of a bimetallic strip

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SUMMARY

The discussion focuses on calculating the radius of curvature of a bimetallic strip subjected to different temperatures, specifically 180°C for one metal and 160°C for another. The key formula derived is r = LT/dL, where L represents the strip length, T is the thickness of one metal layer, and dL is the difference in arc lengths due to temperature-induced expansion. Participants emphasize the importance of using the coefficients of thermal expansion for the two materials to determine dL. The conversation also touches on a related scenario involving a bimaterial strip composed of glass and iron.

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Homework Statement



I need to calculate the radius of curvature of a bimetallic strip when the two strips are subjected to different temperatures. in the problem, the two metals themselves are in different temperatures. One at 180°C, other at 160°C. Anyone with good solid mechanics knowledge will do. Any suggestion will be highly appreciated.



Homework Equations



Define L = strip length, dL = L1-L2 at that temp. difference, and T = thickness of one metal layer. When bent to the required radius r, the inner-outer layer arc length difference due to a radius difference of T equals dL.

Thus theta*((r+T/2)-(r-T/2)) = dL, where theta = L/r.
Then LT/r = dL ==> r = LT/dL.

The Attempt at a Solution



Need help here
 
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welcome to pf!

hi drakierd! welcome to pf! :wink:
drakierd said:
r = LT/dL.

that's it! :smile:

you know T, and you can find dL/L from the given coefficients of expansion of the two metals (i assume they are given?) :wink:
 
Hi,

In my case, it is not a Bimetal strip, it is a bimaterial strip. And one material is glass and the other one is few microns Iron deposited on it. And finally the glass is 160C and the metal is 180C. And the question is that I need to know the radius of curvature of this thing. Any suggestions for that?
 
welcome to pf!

hi shouhardo! welcome to pf! :smile:

can't you use coefficients of expansion in exactly the same way?
 
yeah I can. You are right, that's how I have to do it.
 

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