Shrink fit (INTERFERENCE)mild steel parts

[lrouxcm]

I need an interference fit between a 50 mm shaft and a piece of mild steel 250 x 100 x 40 mm. This is for a part of an old 6 m elliot shapers that broke down. The part of the shaper is the stroke adjusting slider that drives the ram of the shaper., The shaft is BMS BRIGHT SHAFTING STEEL and the slider is mild steel ( I presume it is ansi 1050 mild steel.) The force that will be induced on this part is unknown, I can only presume it is fairly large. I can heat up the piece with the hole to about 300 deg c The question is to what size must I machine the hole in order to have a shrink fit that will withstand the force of the shaper The Coefficient of linear expansion α(1/ºC) of steel is indicated as 12x10-6. is this correct?????
ANY DO,s AND DONT,s will be appreciated

 

[kb58]

Well, this is a classic situation where you're asked, "well what does your Machinist's Handbook say?"

The general rule of thumb I'm told was 0.001" interference fit for every inch of diameter. That's for "normal" press fit. For shrink fit, meaning heating and-or cooling the parts... yeah, I'd look at what the Handbook recommends. I don't see why the rule of thumb wouldn't work though...

 

[benmychree]

Correct! the allowance for shrink fits and press fits is the same, .001 per inch of diameter. The shrink fit is 3-5 times more effective than a press fit, because, the small irregularities on the surfaces are smeared smooth on the press fit, not so on the shrink fit. I was told that in some cases where maximum holding was an issue, that the shaft would be lightly oiled and dusted with grit before assembly.

 

[SLK001]

You could always drill and ream for a taper pin to pound into place after the two are pressed together.

 

[pacifica]

I need an interference fit between a 50 mm shaft and a piece of mild steel 250 x 100 x 40 mm. This is for a part of an old 6 m elliot shapers that broke down. The part of the shaper is the stroke adjusting slider that drives the ram of the shaper., The shaft is BMS BRIGHT SHAFTING STEEL and the slider is mild steel ( I presume it is ansi 1050 mild steel.) The force that will be induced on this part is unknown, I can only presume it is fairly large. I can heat up the piece with the hole to about 300 deg c The question is to what size must I machine the hole in order to have a shrink fit that will withstand the force of the shaper The Coefficient of linear expansion α(1/ºC) of steel is indicated as 12x10-6. is this correct?????
ANY DO,s AND DONT,s will be appreciated

I've had good luck with a 300 degree F. difference between the 2 pieces. You do need to be ready to quickly and smoothly press one piece into the other. Definitely can't stop halfway and then it might get stuck(a big headache).Accurate measurement of diameter differentials is crucial. I also use
.001" for every inch diameter for a 300 degree difference.Makes sense to use oil, but I have never tried using fine grit(should grit be harder or softer than the steel?). I usually ream or bore the hole and have shaft as round as possible.

If I did the math right using C. it is .183 mm difference between the 50mm hole and shaft. If the shaft is 50mm than the hole should be 49.82 to 49.85 diameter. That is with a difference in temp between the 2 pieces of 300 C. You might want to monitor speed of cool down for best accuracy.

 

[Winegrower]

The math is wrong. It’s about a 2” hole, that would be, by the previous suggestions, 0.002 difference.
That is about 0.05 mm.

 

[lrouxcm]

Thanks for all the replies and comments
I found this formula from another forum .
is this a valet quotation
δ = R×α×(T - To)
Where,
δ = change in radius
R = radius at room temperature
α = coefficient of thermal expansion (11x10-6)
T = elevated temperature
To= room temperature
If I use this, then the delta diameter is 0.165 mm

 

[WCraig]

Note that there has been some mixing of units of measure in this thread.

A 300 C temperature change equals 518 F.
A 300 F temp change is 148.9 C.
The coefficient of thermal expansion for mild steel is 0.0000122 per degree C v. 0.0000068 per degree F [1].
50 mm is 1.969 inches.

A 300 F temp change on 2" steel is 0.004 inches (or 0.102 mm).
A 520 F temp change on 2" steel is 0.007 inches (or 0.180 mm).
A 300 C temp change on 50 mm steel is 0.183 mm (or 0.007 inches).
A 150 C temp change on 50 mm steel is 0.092 mm (or 0.004 inches).

[1] https://amesweb.info/Materials/Linear_Thermal_Expansion_Coefficient_of_Steel.aspx

HTH

Craig

 

[rwm]

Correct! the allowance for shrink fits and press fits is the same, .001 per inch of diameter. The shrink fit is 3-5 times more effective than a press fit, because, the small irregularities on the surfaces are smeared smooth on the press fit, not so on the shrink fit. I was told that in some cases where maximum holding was an issue, that the shaft would be lightly oiled and dusted with grit before assembly.

Please clarify for me. Does the shrink fit diameter take into account the diameter at the hot temperature? Or do you NOT adjust for temperature?
It looks in this case like the change in diameter with heating (.007") is well in excess of the prescribed interference (.002")?
Robert

 

[benmychree]

Please clarify for me. Does the shrink fit diameter take into account the diameter at the hot temperature? Or do you NOT adjust for temperature?
It looks in this case like the change in diameter with heating (.007") is well in excess of the prescribed interference (.002")?
Robert

The larger allowance is for ease of assembly, to insure that the two parts do not seize together before final seating. If they do seize before final seating, and need to be pressed apart and tried again, massive galling will likely result, ruining the parts.