Question

A temperature controller, designed to work in a steam environment, involves a bimetallic strip constructed of brass and steel, connected at their ends by rivets (Figure 1). Each of the metals is 3.0 mm thick. At 10 ∘C, the strip is 10.0 cm long and straight.Find the radius of curvature r of the assembly at 105 ∘C.

Answer 1

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.

So look up the thermal linear expansion coefficients of brass and steel, take the difference and multiply by delta temperature; the result is dL/L. Go from there.

I found coefficients for brass = 19E-6 /deg K and steel = 13E-6 /deg K.

Then dL/L = 6E-6*(105-10) = 5.7E-4, and r = 0.003/5.7E-4 = 5.26 m (answer).