I'm not sure how to do this problem...Can someone who has done or know how to do this help me?
3 Answers
a .75" diameter pin will not fit within a .75" diameter hole.
Well, it might fit depending on typical manufacturing tolerance ranges for a part, but designing a part that might work is not the best method. For every assembly that works, you may have many which don't function.
Tables for different classifications of fits exist based on trial and error tests.
The problem you've posted requires also reading and interpreting the table in Appendix A to determine the solution.
Some details can be found here, but using the table in your book is the best solution:
http://cobanengineering.com/Tolerances/ANSIRunningSlidingFits.asp
Bi-lateral is a dodgy practice for assigning fits. The tolerance scheme should convey intent in as clear a manner as possible. But moving on...
References:
The Machinery Handbook covers ANSI fits, as do many sites. You'll want to invest in a Machinery Handbook at some point, or at least the digital copy of it.
This PDF covers the ISO fits and is well put together: https://us.misumi-ec.com/pdf/press/us_12e_pr1261.pdf
RC4 = a close running fit
For a .7500 shaft: .7484-.7492
For a .7500 hole: .7500-.7512
These are the limit values; expressed as a 'range'.
FN2 = a medium drive fit
For a .7500 shaft: .7514-.7519
For a .7500 hole: .7500-.7508 or an H7 (ISO) tolerance
For a bi-lateral expression of the two, use the mean value and +/- to each side.
Getting to what your exercise is all about, which is probably more important than looking up the numbers, is the understanding of 'types' of fits based on the service conditions. The Misumi .pdf covers these very well whereas stating, "close running fit" etc., really doesn't do much to help a student.
Most classic example of these are bearing to shaft/journal fits. In those cases, the manufacturers guidelines, based on the service duty, are gospel.
For other things, it's good to know what type of fit to target and that it is driven, once again, by the type of service and overall function.
And then there are coefficients of thermal expansion...
