What size for interference fit

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I have a .375 through bore, about .150 thick. I need to bush the hole and drill new hole of center to the original bore. Through the new hole will be a shaft that will spin. This will see very light load, being turned with fingers.

What size shaft should I turn?

Everything is simple steel, either CRS or HRS equivalent. I can freeze the bushing in home freezer, heat the piece its going into in a standard kitchen oven, or use a 20T hydraulic press. Also any combination.
 
I would likely try to hit .376 and shallow taper the to allow ease of assembly. For something that thin .150” I would make a support with a slightly larger hole for the bushing. Make the bushing longer than the part and cut off excess when done.
 
An interference fit means "it ain't moving". Think you are looking for a slip fit.
 
No I need interference fit. The bushing will fix the oversize hole in the part. It needs to be permanent no movement. The I can drill the new smaller hole off center that the shaft will pass through
 
In that case, something like 0.0005" interference is probably enough. I used about 0.0005" for steel pins into aluminum for some cabinet brackets I made. The pins are locked in place. No evidence of movement after 8 months.
 
Can you bore an oversize hole at the new location and just use a flanged bushing for the shaft?
 
Firstram said:
Can you bore an oversize hole at the new location and just use a flanged bushing for the shaft?
Click to expand...
No, I'm up against an edge. I can't move it enough.

I'm going for +.0005. My son, who's about to graduate as a Mechanical Engineer said +.0003-.0006.
Now I'll have to see between my measuring tools and capabilities if I can actually hit it.
 
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