A bimetallic strip consists of two 1-mm thick pieces of brass and steel bonded together. At 20 C, the strip is completely straight and has a length of 10 cm. At 30 C, the strip bends into a circular arc. What is the radius of curvature of this arc?

R = 16.67 m

Explanation:

Given:

- Initial Temperature T_i = 20 C

- Thickness of both strips t = 0.001 m

- Final Temperature T_f = 30 C

- Length of the strip L = 0.1 m

- coefficient of linear expansion for brass a_b = 19*10^-6

- coefficient of linear expansion for steel a_s = 13*10^-6

Find:

What is the radius of curvature of this arc R?

Solution:

- The radius of curvature R in relation to dT and a_b, a_s:

R = t / dT * (a_b - a_s)

R = 0.001 / (30-20) * (19-13) * 10^-6

R = 16.67 m