Question

A vinyl siding panel for a house is installed on a day when the temperature is 15.4 C

If the coefficient of thermal expansion for vinyl siding is 55.8×10−6K−155.8×10−6K−1, how much room (in mmmm) should the installer leave for expansion of a 3.76-mm length if the sunlit temperature of the siding could reach 48.6 C?

Answer 1

Change in temperature = $\Delta$ T = 48.6^o C - 15.4^o C = 33.2^o C = 33.2 K.

Actual length of the sliding, at 15.4^o C = L = 3.76 mm.

The coefficient of thermal expansion for the sliding = $\alpha$ = 55.8 x 10^-6 K^-1.

Hence, expansion of the sliding at 48.6^o C is $\Delta$ L = $\alpha$ L $\Delta$ T = 55.8 x 10^-6 K^-1 x 3.76 mm x 33.2 K ~ 7 x 10^-3 mm.

So, the installer should leave about 7 x 10^-3 mm room for the expansion of the sliding.