Shrink fit of a stainless ring onto a shaft 7

[Tiogar]

Hello all,

I'm working on a new procedure at work (machinist/welding), and I've gotten conflicting answers on the formulas I need. I just started a ME program, but we haven't covered this yet.

I'm trying to shrink fit a stainless steel ring onto a stainless shaft, rather than us welding it up and machining it down.

The OD of the ring is 120mm, with an ID of 114mm and a width of 7mm. The shaft is 114 mm, with no ID. I'm trying to do the largest shrink fit I can, without causing materiel failure in the ring.

I can do the calcs for how strong the fit is, but I'm having trouble finding the calc for the max interference fit I can have without causing the ring to fail. It's a non-load bearing part. It's there to act as a corrosion barrier.

I've seen online calculators, but I'd rather know the math and theory behind it, if anyone can point me in the right direction.

Thanks!
Jeremy

 

[TugboatEng]

Yield typically begins at 0.2% elongation. Calculate your circumference, add 0.2%, then calculate the diameter. The difference between this diameter and your original is the going to give you the amount of shrink required to give maximum clamping. Any additional shrink will provide no additional force as the ring will yield.

 

[3DDave]

Since it is so thin walled, a simplifying assumption is the strain in the ring is uniform. Consider it as a straight bar that is being stretched. The amount of stretch is the difference between the inner circumference of the ring and the outer circumference of the shaft.

So you have pi*(OD-ID)/ID = strain. Since it is steel the typical range is between 25 and 30 psi of stress per PPM (parts/million) of strain. Multiply strain by 1,000,000 and then by 25 to 30 (check on the particular alloy for a more exact number) and see if that is either below the elastic limit or that the strain is below the ultimate strain (again, check on the particular alloy) These numbers are often supplied in millions of psi, but I prefer numbers that relate to loads I can lift. I can place a 25 pound load on a 1 inch square section and right away know that is 1 millionth of an inch of compression. I don't have a 25 million pound load. Anyway ....

Work your way backwards to see what the ID of the ring can be.

Example: 100 ksi elastic limit at 25 psi/ppm of strain = (100,000/25) ppm of strain = 4000 ppm. The original length is pi * 114 mm (but we're going to divide by pi to get back to diameter again so, skipping that step) (4000/1,000,000)*114 = 0.46mm diameter difference. (1,000,000 = 1e6, which is what I usually enter into a spreadsheet or calculator rather than all the zeros.)

Check my math.

If you follow this you should realize that the more stretchy the material is (lower stress/strain value) the larger the diameter difference. Like a 50 mm rubber band might easily fit onto the shaft. Likewise a lower strength means that less diameter difference is allowed.

This gets far more complex with much thicker rings - the limit is pressing a shaft into a hole in an infinite plate, but this should be enough for a first approximation.

CTE (Coefficient of Thermal Expansion) of steel is around 6.5 ppm/ degree F (check the value for the specific alloy)

To capture a steel bar 1 inch long with a 1 degreeF increase requires 12ppm/degree F * 1 degree F *25 psi/ppm. The ppms cancel and poof - 12/25 psi, ~0.5 lbf

Note how PPM simplifies the world? I am working on a universal PPM trend. Too many times seeing some material spec that says 50 and not seeing the microscopic "e-5" next to it when if it said 500 ppm ... sigh.

If you need 4000 ppm expansion then divide by 7 ppm/degree F = a bit over a 615 degree F temp increase. Wear really good gloves or put the shaft into some dry ice and split the difference. Also work really fast.

Similar thing with CTE higher CTE, lower temp to get the expansion required, but usually higher CTE = lower material strength so there's few ways to get a best choice.

 

[3DDave]

While I was typing the long answer .. 0.2% is 2000 ppm.

What is proof stress?
Proof stress is the practical limit beyond which a material would permanently deform. With further applied stress, a material can ultimately undergo fracture.

Also known as offset stress, proof stress is simply how much stress a material can withstand until it exhibits a marginal amount of plastic deformation or strain. That amount of strain is arbitrarily chosen as 0.2%. This rule-of-thumb approximation was originally based on the yield strength of steel and can be used for materials of which the yield point is difficult to define, such as aluminium, copper, and magnesium.

https://matmatch.com/learn/property/difference-between-yield-strength-tensile-strength

 

[mfgenggear]

Tioga

On YouTube this professor uses Mohr circle

Stress in press fit & rotating rings
Allowable interference or speed
TheBorn_PE
This will solve you issue Shigleys(spelling)
This added info

 

[rb1957]

3DD ... I'm not a fan. in one case "ppm" is strain ... microinch/inch, in the other case is 10^6 (as in mega.psi, as in E = 25 10^6 psi)

I see that just confusing the indices, whereas using the indices is much clearer and gets you the same result.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[rb1957]

Jeremy,

I don't follow ... "I can do the calcs for how strong the fit is, but I'm having trouble finding the calc for the max interference fit I can have without causing the ring to fail."

If doing the calcs for how strong the fit is means calcing the internal stress, then the maximum interference would have yield stress.

But I'd've thought that your fit would be "I want a positive fit, not much as it doesn't need to resist load, after thermal relaxation".

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[mfgenggear]

Rb1957
In shigley engineering using Mohr 3d point circle, it has the formulas for principle stress and the max shear stress for each member , and the interference fit.
I no expert. But I believe that would answer the op question.

 

[3DDave]

Elasticity is units of stress (psi for example) per change of length per unit of original length. By default that change of length is one X and the unit of original length is one X (where X is any length unit; feet, inches kilometers), yet few materials can withstand a 100% elongation.

Those units are usually removed/hidden for E on the grounds that lengths cancel - but they don't. They remain as change in length per original length or, cancelling change per original. One could say ppp, parts per part, but that's not common.

By using ppm the non-specific part of the unit relations are in the declaration and the change is sensible.

The 10^X notation becomes a bothersome problem when mixed use of plastics, ceramics, and metals comes in. Using ppm puts the values all on the same scale. The default isn't even scientific notation - steel should be ~1.3E7 vs aluminum ~6.5E6. They use e5 for plastics, so it isn't consistent with factors of 1000 selected for engineering notation.

 

[rb1957]

if "ppm" is a more direct/explicit term for "E6" for you then, as they say, "fill y'r boots".

if plastics analysis opts for "E5" units, that's on them !

in the bad olde days we used to have "engineering notation" as a variant of "scientific notation", where we'd step exponents in 3s (kilo, mega, giga, ...).

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[btrueblood]

Shrink fit ring will result in a crevice between the two parts, no matter how tight the shrink. As a corrosion barrier, you might worry about crevice corrosion along that seam. Weld overlay/buildup might be a better idea, if heat affected zone is not an issue.

edit: have to add, having done the excersize - a thin, floppy ring is hard to get aligned and placed before it shrinks tightly - as soon as you remove it from the heat source, it is shrinking, and there is so little mass it does so a lot faster than you would think. As a one-off for fixing a problem when you can't weld, it's ok, but be mindful of the process - have a jig to set the axial alignment of the ring, and work quickly. Good luck.

 

[3DDave]

Has brazing been considered?

 

[edison123]

tugboat

We have a case of 735 mm dia shaft with 733 mm bore steel ring (with 1 mm radial interference).

With your 0.2% limit, the maximum dia is 734.47 mm.

Is excess fit of 0.53 mm too high and prone to yield failure?



Muthu

http://www.edison.co.in

 

[TugboatEng]

That depends on the ductility of your material. Yielding the ring does not mean it has failed, yield is simply the upper limit of the friction force you can generate.

Why do you need so much friction force on such a thin ring? It can't be carrying much load of it's so thin.

 

[Stress_Eng]

I looked into thick wall cylinder theory a long time ago, and I put this template together. Although it was done out of personal interest only and has not been checked, I thought it may give you some ideas. I’ve plugged in your dimensions, with a very small ID, and created an interference fit. Steel has been assumed. I’m sure the equations used can be found in any good engineering textbook.

 

[mfgenggear]

This link

https://highered.mheducation.com/si...2833/Ch04_Section17_Press_and_Shrink_Fits.pdf

 

[edison123]

Thanks guyz for the theory and the equations.

In this case, the ring OD/ID is 825/733 mm and hence it is a 46 mm thick wall cylinder. The radial interference is 1 mm on 735 mm solid shaft.

So, no need to worry about yield failure?

Muthu

http://www.edison.co.in

 

[Stress_Eng]

Just thought I'd tweak the template I posted to allow for a solid shaft. The previous was not written to allow for that condition. The attached is the tweaked version. Hope it helps.

 

[edison123]

Thanks stress eng.



Muthu

http://www.edison.co.in

 

[Stress_Eng]

In my first response to this post, I mentioned looking into thick-wall cylinder theory. With what I found on the subject, radial and hoop stresses were well covered. In examples given in books, the only aspect in the theory that I found not covered, and just mentioned in the examples as a predefined value, was the radial distance to the mating surfaces. This radial distance was either i) the inner cylinder outer radius, ii) the outer cylinder inner radius or iii) the mean of the two.

To try and incorporate a more analytical method to determining the radial distance to the mating surfaces, the previous templates applied ‘radial force equilibrium’. The attached method is now based on solving the 2nd order differential equation for cylinder radial displacement. Based on comparing the rate of change in radial displacement for each of the cylinders, a mating radial distance can be determined and then used in the equations in deriving cylinder stresses and displacements.

As said previously, this task was conducted out of personal interest only, and the template has not been checked. But I did want to mention my findings and post this update to the subject, so that others can take onboard anything of interest or considered useful.