Physics Friday 148

A thin bimetallic strip of thickness d is straight at a temperature T0, with length L0. The two metals have coefficients of linear thermal expansion α1 and α2, with α2>α1. If the temperature is raised to a temperature T not significantly greater than T0, what is the angle θ through which the strip bends?


We should expect the curvature to be uniform, leaving the strip to form a circular arc of angle θ. If the radius of curvature for the inner half of the strip is r, then (with θ in radians), the length of that metal is L1=rθ.
We can ignore any expansion in the direction of the thickness, so the length of the other metal is
L2=(r+d)θ.
Taking the difference, we see
L2L1=dθ,
so .

Now, the formula for thermal expansion is
(see this previous post),
so for the inner side,
,
and solving for L1,
.
Analogously,
.
Thus, the difference is
,
and so the angle θ is
.

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2 Responses to “Physics Friday 148”

  1. Dr.D Says:

    I think you have over simplified this problem. You have neglected the fact that, as a bimetalic strip, they are bonded on the interface and thus must expand/contract exactly the same amount on the interface. This is the same problem as that of bending a deck of cards that are free to shear versus being a deck of cards that have been glued into a solid block; they respond differently.

    The problem you are looking at here was treated rather we by Stephen Timoshenko in his classic work Strength of Materials, Pt. 1 Elementary Problems, Van Nostrand, 1930. The Youngs modulii and area moments of inertia for the two segments are involved because this is a beam bending problem.

    • twistedone151 Says:

      You’re right, and I should’ve known that. Checking, it seems that even assuming equal thicknesses and Young’s moduli, the above analysis is off by a factor of 3/2.

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