2 bars of different materials separated by 0.50mm when the temperature is 20 deg

Question

2 bars of different materials separated by 0.50mm when the temperature is 20 degrees C. At what temperture would they touch?

materials: Brass C36000   L=1.9m    a= 20.5 x 10^-6                stainless steel   L= 2.8m   a=10.4 x 10^-6                   initial gap = .05mm 2 bars of different materials separated by 0.50mm when the temperature is 20 degrees C. At what temperture would they touch?

materials: Brass C36000   L=1.9m    a= 20.5 x 10^-6                stainless steel   L= 2.8m   a=10.4 x 10^-6                   initial gap = .05mm

Explanation / Answer

the 2 bars were seperated by 0.5 mm considering this as initial gap, we have


for Brass C36000 L=1.9m
a= 20.5 x 10^-6
for stainless steel L= 2.8m
a=10.4 x 10^-6
Let length expanded by brass = x1 = La(T2-T1) = 1.9 x 20.5 x 10^-6 (T2-20)
Let length expanded by steel = x2 = La(T2-T1) = 2.8 x 10.4 x 10^-6 (T2-20)

gap = x1 + x2 = 0.5 mm = 5 x 10^-4 m

so
(3.895x10^-5 + 2.912x10^-5)(T2-20) = 5 x 10^-4
hence
T2 = 20 + 7.345 = 27.345 C

at the end of the question you have mentioned some other value as initial gap by it is not clear.

Hope my procedure helped you!

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