Solid Ring Expansion (within Elastic Limit) / Interference fit

[laxhuskie9]

I have two solid rings of the SAME MATERIAL, the lengths of which, are not important. The inner ring has a "bump" on the OD, and the outer ring has a "bump" on its ID, thus creating a slight interference (0.016 difference in Diameter). I need to calculate both the force required to push the outer ring past the inner ring, and the pressure/stress that the inner ring induces on the outer ring. Here are my dims:

Outer Ring OD: ODo=18.59in
Outer Ring ID: IDo=17.91in (this includes the bump)
Inner Ring OD: ODi=17.926in (this includes the bump)
Inner Ring ID: IDi=16.72in

Width of each bump = 0.18in
Each side of bump has 30deg chamfer leading to dia.

Yield Strengths = 75000psi
Modulus of Elasticity = 30x10^6psi
Poisson's Ratio = 0.295

I am treating the inner ring as a solid ring, therefore requiring the outer ring's bump to deflect outward to the inner ring's "bump" OD, allowing the outer ring to slip past the inner ring bump. Temperature is not an object here so do not take into consideration thermal expansion.

I've found information regarding interference fits on this forum, but it doesn't seem to apply to my problem (to my knowledge).

Sorry this is so long, but I really need to figure this out. Let me know if I have left any information out.

Thanks!

 

[djm883]

As far as getting the outer ring passed the inner ring, could you heat the outer ring and cool the inner ring so there is no interference during mating?

From your drawing, it looks like there is only 1 point of contact between the 2 rings. What is the interference at this area?

Equation to determine stress due to interference fit:

Online calculator:

http://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-calculator.htm

Equations:

http://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-equations.htm

 

[laxhuskie9]

The point is actually to purposely have this interference, therefore requiring a specific load to get passed it. Sort of like a detent. It also cannot shear, as I need it to be able to do this more than once, so the stresses need to stay within the elastic limit.

Yes, there is only one contact point, which is the little detent "bumps". Interference between those is a diameter interference of 0.016".

I've tried using an interference fit calculation but what I have are two thin-walled rings, while the equations were for thick-walled. So, I have the hoop stress equation, but can't figure out how to get my pressure.

 

[GregLocock]

Can't see the drawing, what is docx?

Work out the elastic stiffness of the inner ring for a concentrated load (I assume the rings are long compared with the bump?). Same for the outer ring. Then set it so that the pressure on each is equal and enough to stretch the outer, and compress the inner to the same diameter. That force, times mu, is the longitudinal force required.

OK, that is easy enough, but frankly I think it answers the wrong question.

I think it will wear very quickly. How many times does this thing need to work? Is the exact force important?

Also the force required to push the two detents past each other will be INCREDIBLY sensitive to dirt, filings, oil, water, etc etc.





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

i think you're trying to do something that's really difficult ... stretching something over a point without wear and with low enough stresses to do this lots of times.

how feasible is it to have a spring loaded ball bearing (instead of a fixed point) ?

 

[laxhuskie9]

Greglocock - docx is for Word 2007. I've attached a version compatible with Word 2003-2007. Thank you for the explanation, and yes, the rings are long compared to the bumps.

The reason I mentioned that the rings need to be able to move past each other more than once, is for retrieval. These rings are a part of a seal assembly and the outer ring will be forced past the inner ring to set the seal. The outer ring needs to be able to be pulled back past the inner ring in case the assembly needs to be retrieved.

This probably doesn't make much sense as it is not a typical situation for most designers (in most industries), but is necessary for my application.

If the assembly is designed correctly, it shouldn't need to be retrieved, so the rings shouldn't have to move back and forth many times at all. Just need to have the ability to.


rb1957 - a spring loaded ball bearing is a little too complicated for this application and would prove too costly. I also don't have a whole lot of space to work with so wall thickness may be an issue.

THANKS FOR THE REPLIES!

 

[mikedee42]

Hi there,

To be honest, pressing those two together with 0.016" interference seems excessive to me. You will end with severe galling on the interfering surfaces. Also, judging by the location of the locking shoulders, they will not move very much when pressure is applied as you have the inside one located over what looks to be a solid mass and the outside one is located directly below a shoulder.

Even if you heat one and cool the other, there will still be significant interference considering that steel(assuming you are using steel) moves about 0.000086"/in/deg F.

Could you not place one of the locking shoulders on a removable flange? Install with sockethead caps? Take it in and out as many times as you please after that.

Good luck

 

[prex]

Well it's not that unfeasible, though it's an unusual application and I agree with all the above comments about wear and friction.
If you want to crunch some numbers, here is the simple route.
First consider that the length of the rings is important, unless they are longer than say 10 times their thickness: in this case only a portion of their length will participate in the deformation.
Assuming the rings are not that long (can't see your drawings, the link doesn't work at this moment), let's take a length w=0.6" (about the same as the thickness of the inner ring). With such a length both rings will deform with a constant circumferential stress over the respective sections.
Now calling p the linear radial force at the interference, D,t₁ and t₂ the diameter at the interference and the thicknesses, the relative change in diameter of each ring is
[Δ]D_i/D=pD/Ewt_i
and
[Δ]D₁+[Δ]D₂=.016"=pD²(1/t₁+1/t₂)/Ew
From that p=200 lb/in
This is a reasonable value. Assuming no friction (not true of course) and your 30 deg slope, it would give some 6500 lb of axial force to snap over the interference.
The stress in the outer ring is 17 ksi, well below the yield stress.
However if the length of the rings is higher than what I assumed, the forces will be correspondingly higher. If you could machine locally the rings to reduce the thickness near the bump (on the opposite face of each ring of course), you should be able to control the force to an even lower value with respect to the result above, but how much will be the force is definitely a matter of testing.
On the other side, if the rings are relatively short, you could put the bump near the end of each instead of at center, in this way they would deform also in bending (so called false torsion) and would be much more flexible. Also this kind of behavior can be calculated with simple formulae.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[rb1957]

to me, 8 thou (radial) interference is Huge ...
are the raised portions complete around the radius, or only bumps (like i interpreted from the OP) ? could you offset the interference around the circumference, so you could slide the two pieces together, then rotate them and then they'd interfere ?

am i right in thinking that the interference is a one-time thing, during assembly ?

IMHO, i don't think this'll work.

could you use a big, heavy cer-clip on the open end ?

 

[racookpe1978]

Second problem: If the there are only two "bumps" then the inner ring is going to be significantly !!! displaced sideways (radially) from the center of the outer ring.

Can't tell if this matters for your application, but the whole assembly better NOT be rotating. Turbine repairs regularly use a "close fit" (interference fit) and heat outside/cool inside to simply silde the inner and outer rings over each other. The circumferences are kept radially aligned that way. Using 4 or 6 "bumps" still would not assure you of alignment - since one or two of them are going to preferencially rub down while the other two remain as high spots.

and, as noted above, there will be no galling and scarring as the two silde into each other if you use heat/cold.

 

[laxhuskie9]

prex - The lengths of the outer and inner rings are 5.5" and 11.5", respectively. Where can I get the derivations for the equations you supplied previously? That equation will work, as long as I can include the length of each ring (even though they are > 10x their thickness. I'm aiming for around 25-30kips of axial force to snap over the interference so once I can figure out the axial force with the current interference, I can go back and tweak the interference to create the axial force I'm looking for.

How did you arrive at 6500 lbs? I'm unclear about what area to use.

Everyone else - your ideas are excellent, however, will not work for this application. The whole assembly will NOT be rotating, but that's besides the point. For clarification, the "bumps" do in fact go all the way around the circumference. I'm really not looking for solutions to get one ring past the other, but rather calculating the force necessary...and it WILL WORK

Thanks again to all

 

[rb1957]

i guess i'm still confused (what is is new ?)

"... tweak the interference to create the axial force I'm looking for." ... you want the pieces to break out over the bumps ??

if you want a large force (and presumably better controlled) why not hace a blunt mating face ?

do you really want to assemble by pushing over ths bump ? why, if you can slide the two pieces together then rotate and align the bump ??

 

[laxhuskie9]

I am still in design stages, so there are no physical parts we're dealing with.

I'm confused with what you're talking about with "pieces to break out over the bumps". I can adjust the interference if I calculate that the current force to snap over the interference is too low.

It is not during assembly that the rings are pushed past one another. That is just a function of the seal assembly AFTER installation. Like I mentioned before, your recommendation of rotating and aligning the bump, will not work for my application.

Thanks

 

[prex]

The formula comes from
[Δ]D/D=[ε]=[σ]/E and from equilibrium for a ring with a uniformly circumferentially stressed section A
[σ]=pD/2A (er...it appears that I forgot the factor 2 above!)
The axial force comes from [π]pD tan 30° (and its value doubles due to that error).
I've now seen your drawing and it appears that the outer ring is more closely represented as a long tube with D_mean=18.25" and t=0.34" (plus a small ring due to the change in diameter that I'll discard for the moment).
This can be calculated with the formulae for long (or even short) thin walled cylindrical shells that you can find in the Roark (also available in the first site below).
Under a radial linear load the displacement at the load for a long tube is given by:
[Δ]R=p/8D[λ]³ with
D=Et³/12(1-[ν]²) and
[λ]=(3(1-[ν]²)/R²t²)^1/4
By rearranging the formula you can find that the deflection is equivalent to that of a solid ring uniformly stresses with length w=8t and A=0.925 sq in
So now the linear load required to have [Δ]R=0.008" is
p=EA[Δ]R/R²=2700 lb/in and
[σ]=pR/A=27 ksi , plus a local bending value at the load that's almost double, so we are much closer to yield. Also the axial load, not counting for friction, is now 4 times what you was expecting.
I'll join the others in saying that you can't do that in that way.
Worries additional to those already raised are:
- buckling stability of the outer ring, that's really thin
- need for a perfectly circumferentially distributed pushing and pulling load, a minimum deviation and you jam the assembly
As you discarded all proposed variants, I guess you can't cut longitudinally one of the rings...

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[prex]

Sorry, your outer ring has a thickness of 0.16", not 0.34" . The equivalent area now becomes A=0.3 sq in and the linear radial load p=900 lb/in, while the stress remains the same. If you account for the actual length of the ring, the result doesn't change much (you can see the calculation at Xcalcs).
Now the axial force without friction would be of the order of 30 kips.
If you can solve all the problems raised above, good luck!

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads