Press / Interference Fit - Calculate Change in Outer Diameter of Outer Member

[JCReynolds79]

With regards to a press/interference fit of cylinders, I have lots of information on how to calculate the change in inner member OD/Id and outer member ID, the interface pressure/stresses etc, but one thing I seem to be missing (unless I can't see the wood for trees) is how to calculate the CHANGE IN OUTER DIAMETER OF OUTER MEMBER (HUB).

Can anyone point me in the right direction please?

EDIT:
In actual fact, if you look at all the well known equations for calculating the strain of the HOLE ID and the SHAFT OD, they both involve "R" which is the interface radius...but with differing materials how do we know what R is to start with before we can work out the individual change in HOLE ID and SHAFT OD?

Many thanks.

Regards,

Jon Reynolds

 

[desertfox]

Hi

right off the top of my head if you calculate the interface pressure and then treat the outer member as a pressure vessel with internal pressure, I think you can calculate the change in OD from Lame’s equation.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[EdStainless]

So if you pull out your copy of Mark's (or your favorite Mech handbook) you will find a method for this. Once you have the amount of interference and the resulting pressure (knowing strength and modulus of members) it is simple to resolve the location of the resulting boundary, but most people don't care where it is. After all you are talking about ten-thousandths of an inch.

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P.E. Metallurgy, consulting work welcomed

 

[JCReynolds79]

OK, so I found in Roark's the equations for this; as you both say, treat it as a pressure vessel.

The thing that still bugs me, is that when you work out the initial interface pressure (p), you need to already know 'R' which is the interface radius. How do we know what R is to start with, as surely its the result of the two surfaces coming together but at different rates due to their different Elastic moduli? I have seen calcs done where R is assumed to just be the average in the middle of the interference zone. This may be fine for the level of detail most people want, but it just seems not quite right to me.

Am I missing something?

Regards,

Jon Reynolds

 

[desertfox]

Hi JCReynolds

I usually use the radius of the outer diameter of the inner member, so if the outer diameter was 200mm I would use this and set the interface radius to 100mm, I agree you don’t really know what the actual interface radius is, all you can do is approximate.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[BrianE22]

Isn't the deformation so small that the interference radius is basically the OD if the inner member or the ID of the outer member? Within 5% or so?

 

[GregLocock]

That's what I think. Typically you are talking about a few thou of interference on a 3 inch diameter, so in that case R is 1.5 for all practical uses. The exception would be if the shaft was very soft compared with the bore, in which case you probably wouldn't bother with Lame at all.

Cheers

Greg Locock


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[EdStainless]

I looked in Timoshenko and Lessells Applied Elasticity, while the equation starts with a thickness the difference is what you use, since Ro of the inner = Ri of the outer. The calculation ends up with a stress. You could go back and use the stress to calculate the final R, but it will be what it is.
If they both steel and the inner is solid then each 0.001" of interference per 1" of diameter will result in 30,000psi of stress. If they are hollow then there is adjust for the applies pressure.


= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[JCReynolds79]

Hmm still not happy about having a known size shaft and hole with known material properties we can't just simply calculate the resultant R before we calculate the pressures etc. I get that the sizes are so small its almost neglible but that#s not the point.

I found a reference to Precision Machine Design by Alexander Slocum. His calcualtions for this scenario seem to only refer to Diameters, but of course I now see that he uses Diameter-outer_member-ID which he uses the nominal size for...so he is basing the displacement on the pressure applied at the nominal dimension.

I'll keep looking into this otherwise I'll never sleep again ...heh

Thanks all for input.

Regards,

Jon Reynolds

 

[saplanti]

Interference fit as a subject is distributed in various pages of books unfortunatelly, you cannot find the full information under a section.

My reference is; Machine Design-An Integrated Approach, second edition, Robert L. Norton. Check the page 573 under 9.12 Interference Fit.
It says: "The amount of interference needed to create a tight joint varies with the diameter of the shaft. Approximately 0.001 to 0.002 units of diametral interference per unit of shaft diameter is typical (the rule of thousandths), the smaller amount being used with larger shaft diameters. For example, the interference for a diameter of 2 in would be about 0.004 in, but a diameter of 8 in would only about 0.009 to 0.010 in of interference. Another (aand simpler) machinist's rule of thumb is to use 0.001 in of interference up to 1 in, and 0.002 in for diameters from 1 to 4 in."

My understanding you need to select the interference fit by going through fit tables for interference. And From Fig 9-18 and equation 9.14a you need to calculate the presdure created. This will give you the pressure to calculate stresses by the formulas in tha same page.

If you try this with a spreadsheet it will give you the opportunity of selecting different fits from the table and see the results.

Hope it helps

 

[JS_]

Reference Marks.

 

[GregLocock]

Jon- so just apply the normal pressure to the shaft, or the hole, that'll give you the exact value of R.

Cheers

Greg Locock


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[3DDave]

Whatever one-step formula might exist to produce the change pf the outside diameter of a collar pressed onto a shaft based on the interface diameters and the elastic modulus of each part, it will still incorporate all the parts of the intermediates steps into one rather large equation.

 

[JCReynolds79]

Been thinking more. I agree the difference is very negligible.

GregLocock, my point is I can't just apply the normal pressure, as to calculate that normal pressure you need to start with a value for R (interference radius after fitting). So its a bit of chicken and egg situation.

3DDave, you come close and make me perhaps rephrase my question...

Given two cylinders of known size and material props, how does one determine the change in diameter when pressed together? REMEMBERING that we don't know the interface pressure without knowing the answer I am seeking first.

To further illustrate, two cylinders with a certain interference:

Scenario 1: the outer (hole) is infinitely stiff, therefore R = outer ID.
Scenario 2: the inner (shaft) is infinitely stiff, therefore R = inner OD.

There must be a way with all the sizes and material props to determine the deformation before the resultant pressure?

Thanks again.

Regards,

Jon Reynolds

 

[desertfox]

Hi JCReynolds79

I don't think you can that's the point, try using half the OD of the inner member.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[saplanti]

JCReynold79,

Assume R is the nominal radius for hub inner or shaft outer surface before the tolerance application, it does not have huge effect on the calculation of pressure at interference surfaces. After getting the max/min tolerances of shaft and hub, you need to calculate max and minimum interference. These are the important values of your calculation, and their effect are very important on the radial pressure calculation. This is very clear in the reference I had given above.

 

[JCReynolds79]

OK, thanks all.

Regards,

Jon Reynolds