Question

At 12.25 degree C a brass sleeve has an inside diameter of 2.196 cm and a steel shaft has a diameter of 2.199 cm. It is desired to shrink-fit the sleeve over the steel shaft. To what temperature must the sleeve be heated in order for it to slip over the shaft? Alternatively, to what temperature must the shaft be cooled before it is able to slip through the sleeve? Note that it makes no difference whether you think about the circumference or the diameter of the shaft changing size since they only differ by the constant pi.

Answer 1

a)

the original length of the sleeve is

$L_o=D.\pi=2.196cm*\pi$

the final length

$L=D.\pi=2.199cm*\pi$

$\Delta L=2.199cm*\pi-2.196cm*\pi=0.003\pi$

$\Delta L=\alpha (L_o)\Delta T$

clearing delta T

$\Delta T=\frac{\Delta L}{\alpha (L_o)}$

$\Delta T=\frac{0.003\pi cm}{1.9\times 10^{-5}K^{-1}(2.196\pi cm)}=71.9ºK$

$T_f=T_o+71.9ºK\Rightarrow T_f=84.15^0C$

anserw a= 84.15ºc

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The shaft is an expansion of area. it could be calculated by the formula

$\Delta A=2\alpha (A_o)\Delta T$

$A_o=\frac{\pi}{4}D^2=1.208 \pi cm^2$

$\Delta A=\frac{\pi}{4}(2.196cm)^2-1.2089 \pi cm^2=-0.00329\pi cm^2$

Clearing Delta T

$\Delta T=\frac{\Delta A}{2\alpha (A_o)}$

$\Delta T=\frac{-0.00329\pi cm^2}{2*1.2\times10^{-5}K^{-1}(1.208\pi cm^2)}=-113.47^0K$

$T_f=T_o-113.47^0K\Rightarrow T_f=-101.22^0C$

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note: making the calculations by lineal dilatation the results isvery close

$\Delta T=\frac{\Delta L}{\alpha (L_o)}=\frac{2.196-2.199}{1.2\times 10^{-5} (2.199)}=-113.68$