A bimetallic strip is made by bonding together a 1.0-mm thick and 15.0-cm long steel strip to a similar strip made of aluminum. The strip is straight when the temperature is 20.0°C. What will the radius of curvature of the strip be when the temperature is 40.0°C? The coefficient of thermal expansion of steel is 10.5 × 10^-⁶K^-¹ and that of aluminum is 23.0 × 10^-⁶K^-¹.

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Quizplus · Accepted Answer

The radius of curvature is given by:$$R = \frac{L}{2\alpha \Delta T}$$where L is the length of the strip, $\alpha$ is the coefficient of thermal expansion, and $\Delta T$ is the change in temperature.For steel, we have:$$R_s = \frac{0.150 \ m}{2(10.5 	imes 10^{-6} K^{-1})(40.0 \ K - 20.0 \ K)} = 3.6 \ m$$For aluminum, we have:$$R_a = \frac{0.150 \ m}{2(23.0 	imes 10^{-6} K^{-1})(40.0 \ K - 20.0 \ K)} = 1.6 \ m$$The radius of curvature of the bimetallic strip is given by:$$R = \frac{R_sR_a}{R_s + R_a} = \frac{(3.6 \ m)(1.6 \ m)}{3.6 \ m + 1.6 \ m} = 2.4 \ m$$

A Bimetallic Strip Is Made by Bonding Together a 1