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Determining strength of shrink fit

crossthread

Titanium
Joined
Aug 5, 2004
Location
Richmond,VA,USA
Is there a non destructive method for determining how much torque can be transmitted through a shrink fit. I am assuming if you know the coefficient of expansion for the two materials, the amount of interference, the amount of contact area etc., a pretty close approximation could be found, but what if you have to be more accurate? I asked my wife, who retired as a materials engineer, and being the smart ass that she is she said "they are all tight if you put enough in them". Very helpful dear. I am thinking the only way is by destructive testing but not sure.
 
...."they are all tight if you put enough in them".....

Hmm, I get that all the time from my wife.

There's a lot of variability here, depending on even what the assembly lube was. One way to do this is to press the assembly
together, note the force value, and then press it apart, again note that value. Then repeat a few times. Ideally the forces
are pretty repeatable. I recall seeing RR wheel/axle assemblies done that way, the best lube was tallow. Worse was lube oil.

I'd probably start with the surface area of the joint, and try to figure out the hoop stress induced by the press, and then
use the frictional coefficients for the materials in use. You'd be lucky to get it within an order of magnitude.

Your destructive testing approach really sounds best.
 
Surface finish of the mating parts and the material will also be factors. Seems you could have more predictable and repeatable results - if those are the goals - by adding an appropriate Loctite adhesive.
 
Is there a non destructive method for determining how much torque can be transmitted through a shrink fit.
There is, but I don't know how to do it, so not much help :) I was surprised to learn that the lobes on large diesel engines are just shrunk on, and also the main gear drives on some. The bore of the lobe is a negative fit on the shaft, with a groove centered in it. Lobe is slid onto a slightly smaller area at the beginning of the shaft and high-pressure oil pumped into the groove to expand the bore. Then slide the lobe to its resting place, remove hydraulic pressure, and viola, lobe installed where it belongs. Much easier than trying to grind a giant camshaft.

One would need a pretty good mathematical solution to this problem before feeling confidence in this, if the engine quits running in a ship 3,000 miles at sea someone is going to be pissed, so pretty sure the answer to your question exists.
 
As Rozen suggests, coefficient of friction is one of the largest unknowns and it's difficult to narrow this down without lots of testing and a well controlled process.

You can look up common friction coefficients to get a rough estimate, which combined with other factors will get you a number. But it's unwise to depend on a precise prediction of the strength of the fit. Better to conservatively assume a coefficient of friction on the low end of the range, then carry a large factor of safety commensurate with the application.

Another (obvious) factor is tolerances in the fit. Good practice as a designer is to size the strength of the joint based on the LMC of the mating parts. For a 1-off you get to cheat with as-measured values of course.
 
Who is "viola"?

There is, but I don't know how to do it, so not much help :) I was surprised to learn that the lobes on large diesel engines are just shrunk on, and also the main gear drives on some. The bore of the lobe is a negative fit on the shaft, with a groove centered in it. Lobe is slid onto a slightly smaller area at the beginning of the shaft and high-pressure oil pumped into the groove to expand the bore. Then slide the lobe to its resting place, remove hydraulic pressure, and viola, lobe installed where it belongs. Much easier than trying to grind a giant camshaft.

One would need a pretty good mathematical solution to this problem before feeling confidence in this, if the engine quits running in a ship 3,000 miles at sea someone is going to be pissed, so pretty sure the answer to your question exists.
 
Not an engineer so I maybe talking out my ass here. But I've always thought the strength of a shrink fit would depend on the surface area of the mating parts, surface finish (finer == more area) and yield strength of the members (depends on cross-sectional area).

The idea is to achieve a large fraction of the yield strength, without exceeding it. After all, once it yields, it is no longer shrunk and could slide right off.

Keeping in mind that large locomotive and engine parts have been built this way for well over a century. Also the breeches of artillery.
 
I believe a simple, but probably not quite accurate, answer is the same as the classic engineering fixed-fixed beam thermal expansion problem.

Stress = Young's modulus of the holder's material (E) * coefficient of thermal expansion of the holder's material (alpha) * change in temperature of the tool holder of how hot it is when the tool can come out of the holder, not how hot it gets completely, and room temperature (delta T)

And stress = force / area
 
While you could calculate a resisting torque based on forces normal to the surface and their coefficient of friction, the number will be quite low. Compared to an actual test to failure.

The reason a shrink fit is so strong is that much of the torque resistance is gained by shear of the high spots on each surface. A hard smooth surface does not offer any shear in the joint and you rely on solely on friction. The holding strength of 2 rougher surfaces offer plenty of hill and valley contact and nesting. These high spots on each surface must be sheared off to break the joint loose.

So you have the friction component plus the shear component combining to give a stronger joint. Stainless is even better as it would be happy to weld itself
 
In the old D series Ford truck 'suitcase gearboxes",the countershaft gears are all a shrink fit with no other driving device,such as a keyway......very ocassionally a gear would break loose from the shaft ,but only in extrme circumstances.......The downside was that the countershaft was not considered repairable,and the whole assy had to be replaced ....The terminally cunning could fix them sometimes,by cutting off the faulty gear ,and salvaging a gear from another damaged shaft by blowing out shaft,thereby shrinking it to free the gear.Only worked with the front drive gear tho,which in any case took most wear.
 
I think besides destructive testing, you could use FEA to simulate the contact, but it really depends on how much error you can allow in your measurements. It would probably be cheaper and faster to go the destructo route TBH.
 
You calculate the hoop stress. I forget the formula it is the cube root of the properties, geometry and dimensions of the members. You can search for the formula if you like. If you are just curious, you can search for "shrink fits class FN 5," and get a range of values. If you are needing an absolute, perform with out fail, solution, let me know and I will give you assistance.
I have used the formula once to eliminate failures of a hydraulic cylinder shaft clevis hub fixture joint. The shaft was 8 inches in diameter, don't remember the hub dimensions or temperatures required. I had to travel to the rebuilder in Mobile to certify the specifications and procedure. The shop specialized in rebuilding Hydraulic cylinders for the offshore and inland machines.
 
quoting myself from earlier post:

"look at this or maybe find a better calculator:

https://www.tribology-abc.com/sub23.htm


Pa= f*3.142*d*L*Pc

where f=friction coefficient
Pc=contact pressure between the two members
d= nominal shaft dia
L=length of external member.
Pa= axial force required to interference fit

to calculate Pc for a given interference use the formula:-

Pc=x/[Dc*[((Dc^2+Di^2)/(Ei(Dc^2-Di^2))+((Do^2+Dc^2)/(Eo*(Do^2-Dc^2))-((Ui/Ei)+Ui/Eo))]

where x = total interference
Dc=dia of shaft
Di = inner dia of shaft(this is zero for solid shaft)
Do= outside dia of collar
Uo=poissons ratio for outer member
Ui=poissons ratio for inner member
Eo=modulus of elasticity for outer member
Ei=modulus of elasticity for inner member

This formula for Pc will simplify if the materials are the same."

as mentioned the problem will be the friction coeff. for steel its somewhere between 0.05 and 0.5. would be interesting to calculate this with loctite (e.g. 648).
 
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Viola, Bob's your uncle, and don't call me Shirley. As a design issue wouldn't one use any number of easily calculated options to insure torque transmission such as keys, tapers, splines.
 
Viola, Bob's your uncle, and don't call me Shirley. As a design issue wouldn't one use any number of easily calculated options to insure torque transmission such as keys, tapers, splines.

1. press/shrink fit is cheap
2. most others you listed create stress risers in the shaft
 
Thanks to everyone for the help. I am working on the computations and will more or less verify the results with destructive testing. I can tell right now that this is not going to be easy and may be kind of a crap shoot. There are just too many variables. It seems like even a slight change in the surface finish can drastically alter the end results etc. Not to hijack my own thread, but if you were not doing a shrink fit and just an interference fit, could you determine how much force it would take to separate two parts by how much force it takes to press them together? For instance, my press has a strain gauge with a digital readout. If it takes 4.2 tons to push a shaft into something, will it withstand a 4.2 ton pull. All this back and forth has just got me thinking about it.
 
Thanks to everyone for the help. I am working on the computations and will more or less verify the results with destructive testing. I can tell right now that this is not going to be easy and may be kind of a crap shoot. There are just too many variables. It seems like even a slight change in the surface finish can drastically alter the end results etc. Not to hijack my own thread, but if you were not doing a shrink fit and just an interference fit, could you determine how much force it would take to separate two parts by how much force it takes to press them together? For instance, my press has a strain gauge with a digital readout. If it takes 4.2 tons to push a shaft into something, will it withstand a 4.2 ton pull. All this back and forth has just got me thinking about it.

Maybe ask over at eng-tips.com
 
Thanks to everyone for the help. I am working on the computations and will more or less verify the results with destructive testing. I can tell right now that this is not going to be easy and may be kind of a crap shoot. There are just too many variables. It seems like even a slight change in the surface finish can drastically alter the end results etc. Not to hijack my own thread, but if you were not doing a shrink fit and just an interference fit, could you determine how much force it would take to separate two parts by how much force it takes to press them together? For instance, my press has a strain gauge with a digital readout. If it takes 4.2 tons to push a shaft into something, will it withstand a 4.2 ton pull. All this back and forth has just got me thinking about it.

You could try pushing apart a shrink fit for comparison. I read somewhere that the holding power for shrink was about 3x that of pressing together with the same interference. When you press something, you are shearing off the hills and therefore losing a lot of the grip possible with the shrink fit. This would probably only apply to non-hardened parts, where the mating surfaces are soft enough to deform as compression occurs.

Shrink fits make some fits possible, as with comparatively large interferences which are easy on larger diameter components. It's the little bastards that will stick on you.
 








 
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