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Large Diameter Interference FIts: Calculating Removal Load

Large Diameter Interference FIts: Calculating Removal Load

Hello. I have 2 large diameter cylindrical shells that are "rabbetted" together. I am trying to determine the approximate load required to pull them apart without much success. The parameters are as follows (all length dimensions in inches) and result in a .005" (diametric) interference:

Outer Shell (female):
OD2 = 40.525
ID2 = 40.175
t2 = .350

Inner Shell (male):
OD1 = 40.180
ID1 = 39.98
t1 = .200

The axial engagement length of the fit is:
L = .125
The materials are the same for inner & outer shell:
v = 0.3
E = 221 GPa

I have tried a couple of approaches that yield very different numbers. I first tried just considering the force required to deflect the inner shell by the interference value (radially .0025") and then using this "normal force" to determine the frictional load to overcome. I used a formula in Roarks to determine this force. This requires a significant amount of force compared to using a "shaft/hub" press fit calculator such as the one at The difference in answers is from about 1200 lbf to 20 lbf, respectively.

I think a couple of things are happening here, but could use some guidance:

1) I believe that the Roarks approach might be overly conservative since it does not consider the outer shell and that it will expand a little to accommodate the inner shell.

2) For a shaft/hub equation, I am guessing that they might not be too accurate for "shafts/hubs" as large as the ones I am looking at. But maybe someone could comment on this?

NX7.5.5.4 - Teamcenter 8
ANSYS Workbench 14.5

RE: Large Diameter Interference FIts: Calculating Removal Load

Hating to look things up ... I'd analyze this based on the fact that the pressure load is the same on both parts and then iterate/solve for the mutual pressure at the interface that is required to change each part to have a matching mating diameter. For thin walls and small amounts of interference over the diameters given, it won't be much.

The biggest contributors to uncertainty is the friction and whether any corrosion has increased the amount of interference between the parts and with it the pre-load.

RE: Large Diameter Interference FIts: Calculating Removal Load

The word 'rabbet' implies that at least one of the parts has a stepped cut in one 'corner', but your description does not detail such a thing.

Actually assembling a joint as described requires a chamfer or special tooling. A 45 deg chamfer won't do. An appropriate chamfer would take up ~half the specified length of engagement.

Even with no chamfer and some kind of magical special tooling to assemble the joint, I'd tend to believe it would take only 20 pounds of force to knock it apart. Or it would separate under gentle handling.

I sure hope this is an academic exercise; I hate to see such, er, used food, in actual product designs. No offense intended.

Mike Halloran
Pembroke Pines, FL, USA

RE: Large Diameter Interference FIts: Calculating Removal Load

how much force is required to mate the two parts ? 0.005" interference is a lot ... are they freeze fitted ??

i'd look up the irish stress theoretican ... timoshenko, probably plates and shells.

Quando Omni Flunkus Moritati

RE: Large Diameter Interference FIts: Calculating Removal Load

@MikeHalloran: You are correct. I oversimplified the problem for the sake of uploading a picture or being too wordy. The male part is in fact stepped. The OD of the inner part indeed the rabbet (step) diameter and the engagement depth is the axial length of the 'step.'

@GregLocock: No CAD here...that's all natural talent ;) I listed the t's for convenience ... guess I need to work on my definition of convenient.

@ Dave & RB: Thanks for that. There's no relative heating for the fit. I am a contractor for a major aviation company and it appears they do these kinds of fits all the time. Unfortunately we are discouraged from asking them questions, so I am left to fend for myself. I thought that maybe someone here had done this before and had a nice formula obviating the need to re-derive something. I am all for derivations, but this seemed like it may have been a common issue. Maybe not though.

Thanks for the replies :)

NX7.5.5.4 - Teamcenter 8
ANSYS Workbench 14.5

RE: Large Diameter Interference FIts: Calculating Removal Load

there's probably a lot of resistence from the rabbets, at a minimum a lot more surface area.

i'm assuming the SRM/MM doesn't go into any detail ... possibly just "sent the ass'y back to us for maintenance". I suspect that they have a proprietary process for disassembly (without wrecking the parts) and that's why they're not talking to you ...

go carefully ...

40" diameter ... axle or engine shaft ?

Quando Omni Flunkus Moritati

RE: Large Diameter Interference FIts: Calculating Removal Load

Hi rb: I am not sure I understand the question, but it is neither. It is a shell structure. You can imagine that a jet engine is just made up of a bunch of cylindrical shell bodies of varying complexity. For example, the outer-most diameter on the compressor case in the schematic at Wikipedia here. Imagine another "case" concentric with the compressor case 'fitted' inside of it. It would of course be bolted in place via a flanged bolted joint, but the rabbet fit is there to align things.

Thanks again.

NX7.5.5.4 - Teamcenter 8
ANSYS Workbench 14.5

RE: Large Diameter Interference FIts: Calculating Removal Load

i was just guessing what might be 40" dia and heavy interference. sounds like it the inner wall of the (HP?) compressor stage ... ie not the reaaly big diameter LP compressor (that by-passes the engine core). this sounds like it's part of the engine ... i'd be hesitant to take it apart ... something i think the engine manufacturer should do ...

Quando Omni Flunkus Moritati

RE: Large Diameter Interference FIts: Calculating Removal Load

In aluminum it's only a 10F delta T to eliminate .005 interference. Just warm the outer part a little and it drops in place. A chamfer would still be nice on both parts to prevent forming a burr if the parts bump each other.
.005 = XdegF*13ppm/degF*40;

X = .005/40/(13/1000000)degF

RE: Large Diameter Interference FIts: Calculating Removal Load

So for the sake of argument you've got a quarter inch thick ring 40" diameter fitted inside another one by 1/8", with a little interfernece

One tap with a hammer should release it.


Greg Locock

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