A.) Derive an analytical expression for the radius of curvature of a bimetallic

Question

A.) Derive an analytical expression for the radius of curvature of a bimetallic strip made by fusing together two metals when its temperature is raised by T. The initial length of both the metals is Lo, the initial thickness of each metal is do, and the coefficients of thermal expansion of the metals are 1 and 2 with 1 > 2 and the width of the strip is w.

B.) Now suppose that one of the metals is aluminum and the other one is iron. Assume Lo = 20 cm and do = 0.5 mm and and the width of the strip is w = 1 cm. Calculate the radius of curvature at different temperatures T = 300 K and T = 100 K.

Explanation / Answer

Part 1

You may assume that each metal has thickness x/2 (w = x/2), and you may assume that x<<R

We assume that the initial length is Lo. After heating, the lengths of the mid-lines of the two metallic strips are respectively
L1= L0(1 + 1 T)
L2= L0(1 + 2 T)

So, from both the equations,

L2 - L1 = L0 * T * (1 - 2) --1

Assuming that the radius of curvature is R, the subtending angle of the strip is Theta, and the change of thickness is negligible, we have

L2 = (R + x/4) *Theta
L1 = (R - x/4) *Theta

L2 - L1 = (x/2) * Theta
= (x/2) * (L1 + L2 / 2R)

L2 - L1 = x * L0 / 4 R [ 2 + (1 + 2) T ] --2

From equations 1 & 2, we get,

x * L0 / 4 R [ 2 + (1 + 2) T ] = L0 * T * (1 - 2)

R = ( x /4) * [ 2 + (1 + 2) T ] / (1 - 2)T

Part 2
R = ( x /4) * [ 2 + (1 + 2) T ] / (1 - 2)T

Aluminium 1 = 25
Iron 2 = 11.7

So, R at temp 300 = (2*1 / 4 ) * [ 2 + (25 + 11.7) * 300] / (25 - 11.7)*300

R = 0.5 * (11012 / 3990 )

R = 1.379 cm

So, R at temp 100 = (2*1 / 4 ) * [ 2 + (25 + 11.7) * 100] / (25 - 11.7)*100

R = 0.5 * (3672 / 1330 )

R = 1.380 cm

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