1
$\begingroup$

IANAE

I want to make an 1/8" steel rotary switch shaft into 1/4" so that it will accept the much more common 1/4" shaft knobs.

If I use McMaster-Carr 92510A422:

mcmaster-carr 92510A422

which is an aluminum unthreaded spacer that has an ID of 0.115", will heating it expand it enough to fit the 0.125" shaft?

I have a 13" drill press but otherwise, I do not have any ability to mill things so I need to start with off-the-shelf parts.

UPDATE:

I just realized, the nut that goes over the shaft has an ID of only 0.2175". So it would not fit over the new shaft.

$\endgroup$
2
$\begingroup$

Shrink fitting depend on the fact that you have a shaft and collar which are sized to an interference fit when cold.

One problem with aluminium is that you have a fairly narrow window for heating it before it first anneals and then melts.

A rough calculation suggests that that change in diameter with heating to 200C is in the order of thousandths of an inch which is somewhat less that the tolerance of your part (although this is back of the envelope stuff so I may be wrong but is suspect that its not too far out).

In any case I think that for a knob you will be much better off using an adhesive, epoxy or a bearing retaining adhesive should be plenty strong enough and a lot more convenient.

| improve this answer | |
$\endgroup$
1
$\begingroup$

The required strain is:

 (0.125 - 0.115)/0.115 = 0.087

the thermal expansion coefficient of aluminum is ~ 22 10^-6/C, so to expand that much you need to heat to

 0.087/(22 10^-6) or ~4000C

ie You will melt the aluminum long before it expands that much. Note the drawing tolerance is -0/+0.01, so maybe buy a bunch and you will get lucky :-)

Hunt around for a split tube..

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.