Question

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?

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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be warmed and slipped over a brass shaft with a 2.5030 in. outside
diameter at 20.0 ?C.
A)To what temperature should the ring be warmed?
B)If the ring and the shaft together are cooled by some means
such as liquid air, at what temperature will the ring just slip off
the shaft?

A steel ring with a 2.5000-in. inside diameter at 20.0 ?C is to
be warmed and slipped over a brass shaft with a 2.5020 in. outside
diameter at 20.0 ?C.
Part A)
To what temperature should the ring be warmed?
Part B)
f the ring and the shaft together are cooled by some means such
as liquid air, at what temperature will the ring just slip off the
shaft?

A brass ring with inner diameter 2.00 cm and outer diameter 3.00
cm needs to fit over a 2.00-cm-diameter steel rod, but at 20
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To what temperature, in degrees Celsius, must the rod and ring
be heated so that the ring just barely slips over the rod?

At 20∘C, the hole in an aluminum ring is 2.100 cm in diameter.
You need to slip this ring over a steel shaft that has a
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At 20∘C, the hole in an aluminum ring is 2.200 cm in diameter.
You need to slip this ring over a steel shaft that has a
room-temperature diameter of 2.204 cm .
To what common temperature should the ring and the shaft be
heated so that the ring will just fit onto the shaft? Coefficients
of linear thermal expansion of steel and aluminum are
12×10-6K-1 and
23×10-6K-1 respectively.
Express your answer in degrees Celsius to two significant
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(a) To what temperature must the combination be cooled to
separate the two metals?
°C
Is that temperature attainable?
Yes No
(b) What if the aluminum rod were 10.04 cm in diameter?
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Is that temperature attainable?

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(a) To what temperature must the combination be cooled to
separate the two metals?
°C
(b) What if the aluminum rod were 10.02 cm in diameter?
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(b) Is that temperature attainable?
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4sig figs

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