shrink fitting?

Question

I have to shrink fit some brass rings onto a shaft. how much smaller than the shaft diameter do i make them. the shaft is 4.25 inches diameter

Answer 1

Degree of interference is based on operational requirements, including torque transmission. It's also based on whether a key is used; judging by how you've worded your query, I'll assume you aren't employing one.

For heavy-duty applications, the typical interference is 0.0015 inches per inch of nominal shaft diameter, although in some specific cases, this could go as high as 0.0025 inches per inch of nominal shaft diameter.

For light duty operations, it could be as loose as 0.0005 inches per inch of nominal shaft diameter.

This means your ring's nominal inner diameter could be anywhere from a "loose" fit (4.2479 inches diameter) to a "tight" fit (4.2436 inches) to an "extra-tight" fit (4.2394 inches).

Also remember to account for machining tolerances ... there will always be some allowable variation. Typcial practice for something with a two place decimal (such as your 4.25 diameter) will be +/- 0.002 inches. This means your actual shaft OD is 4.248 - 4.252 inches. Make sure that the chosen "fit" has enough grip at both ends of the range ... especially since your ring ID will ALSO have a similar tolerance on it.

Answer 2

.003 to .005" should do it

Are you planning on dipping the shaft in liquid nitrogen, and heating the ring as well?

remember if you do that, you had better have a fixture or some other "positive" stop method do locate the ring exactly where you want it, because when it shrinks, its stuck wherever it winds up!

Answer 3

Have no clue. Good luck with that.

Answer 4

4.20