Installing a Bearing
Consider an aluminum bearing with a maximum room-temperature outer diameter of 50.15 mm is to be installed inside of an aluminum mounting plate with a minimum room-temperature bore diameter of 50 mm. If the room temperature is 25°C and the required diametral clearance for assembly is 0.1 mm, what is the minimum temperature the mounting plate must be heated to install a room-temperature bearing? Note, aluminum’s coefficient of linear expansion is 23x10^(-6)/°C.
Expand Hint
$$$\delta_t=\alpha L(\Delta T)$$$
where
$$\delta_t$$
is the deformation caused by a change in temperature,
$$\alpha$$
is the temperature coefficient of expansion,
$$L$$
is the length of the member, and
$$\Delta T$$
is the change in temperature.
Hint 2
Required diameter change:
$$$\delta_{diameter}=d_{bearing}-d_{plate}+required\:assy\:clearance$$$
where
$$d$$
is the diameter.
Required diameter change:
$$$\delta_{diameter}=d_{bearing}-d_{plate}+required\:assy\:clearance$$$
where
$$d$$
is the diameter. Thus,
$$$\delta_{diameter}=50.15mm-50mm+0.1mm=0.25\:mm$$$
Required temperature change for the mounting plate’s bore diameter:
$$$\delta_t=\alpha L(\Delta T)$$$
where
$$\delta_t$$
is the deformation caused by a change in temperature,
$$\alpha$$
is the temperature coefficient of expansion,
$$L$$
is the length of the member, and
$$\Delta T$$
is the change in temperature.
$$$\Delta T=\frac{\delta_{diameter}}{\alpha(L)}=\frac{0.25mm^{\circ}C}{(23\times 10^{-6})(50mm)}=\frac{0.25^{\circ}C}{0.00115}=217.39^{\circ}C$$$
Minimum temperature to which the mounting plate must be heated:
$$$T=T_{room}+\Delta T=25^{\circ}C+217.39^{\circ}C=242.39^{\circ}C$$$
242°C
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