Question

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?

Answer 1

R = 16.67 m

Step-by-step 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