Question

At 18.10 ∘C a brass sleeve has an inside diameter of 2.22063 cm and a steel...

At 18.10 ∘C a brass sleeve has an inside diameter of 2.22063 cm and a steel shaft has a diameter of 2.22333 cm. It is desired to shrink-fit the sleeve over the steel shaft.

A.) To what temperature must the sleeve be heated in order for it to slip over the shaft?

B.) Alternatively, to what temperature must the shaft be cooled before it is able to slip through the sleeve?

Homework Answers

Answer #1

As we know that the expression for the thermal expansion
∆L/L = α∆T, α is linear thermal expansion coef
L = L₀(1 + α∆T)

and,
linear thermal expansion coefficient of steel = 11.8 x 10^-6 /K
linear thermal expansion coefficient of brass = 19 x 10^-6 /K

(A) Here we want to heat the brass so that it's diameter expands from 2.22063 to 2.22333 cm

Difference = 2.22333 - 2.22063 = 0.0027

now -
∆L/L = α∆T, α is linear thermal expansion coef
0.00270 / 2.22063 = (19 x 10^-6) /K ∆T

=> ∆T = 63.99ºC

So, T = 18.10 + 63.99 = 82.09ºC

(B) ∆L/L = α∆T,
∆L/L = α∆T,
0.00270 / 2.22333 = (11.8 x 10^-6) /K ∆T
∆T = 102.91

therefore -

T = 18.10 – 102.91 = -84.81ºC

therefore, the shaft must be cooled to 84.81ºC.

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