Torque Transmission Of An Interference Fit Joint 1

[desertfox]

I am used to calculating interference fits between a hub and a shaft and their torque capability when both materials are stressed within their elastic limit. I was looking at something the other day which may lead to the shaft being stressed beyond its yield value, while the hub material would still be in the elastic range, my gut feeling is that if such a joint where made it would have a reduced torque capability because one material has yielded and therefore reduced the pressure at the joint interface.
Anyone else any thoughts on this?

Regards

desertfox

 

[GregLocock]

Yes, the maximum normal pressure is going to be limited by the yield of the shaft. So, the rough answer is to work out the torque capacity at the elastic limit of the shaft, if you are assuming constant post yield stress.

Cheers

Greg Locock


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[israelkk]

If the hub was yielded the pressure is still larger than at yield point because tensile strength is larger than yield strength. The only thing is that if you increase the interference with the same parts and assuming a flat tensile strength graph the pressure will still be the same.

 

[tbuelna]

Don't know if I would actually describe the torque capacity as being "reduced". As GregLocock notes, peak interface pressure is at the yield point. So a more apt description would seem to be "maximized".

A similar situation that comes to mind are "torque to yield" cylinder head bolts.

 

[electricpete]

I'm not an ME, so bear with me. Isn't the shaft in something like hydrostatic compression. if yes, does yield really apply?

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(2B)+(2B)' ?

 

[electricpete]

I guess a long hub would create a state of hydrostatic compression more than a short hub

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(2B)+(2B)' ?

 

[desertfox]

Hi All

Thanks for all your comments.

I think what Greg is saying if I go to the yield stress value but not above it, then the maximum pressure at the interface will be the yield stress figure and that cannot be increased, however if I go beyond yield stress of the shaft but not the hub then the interface pressure will be reduced and thereby reduce the torque capability of the joint. Greg please correct if I've read to much into your post.

Israelkk
Its definitely the shaft that would yield and not the hub, so do you agree with my reasoning?

tbuelue
I with what you're saying upto the yield point but if the shaft goes beyond the yield point then what happens?
Your reference to the torque to yield bolts in a cylinder head is a good example and thats originally what got me thinking, imagine your tightening a bolt but you exceed the yield strength of the clamped material, so now the preload that was just achieved at the point of yield as now reduced because the clamped material as deformed permanently.

Hi electricpete
Yes the shaft would be assumed to be in a state of uniform compression but I'm missing your point about the yield stress value.

regards

desertfox

 

[dhengr]

Desertfox:
I’m not quite sure what you are doing or trying to prove or disprove... but some food for thought. I made essentially the same post over in the M.E. forum, and then saw there was more, new action here.

How are you achieving your interference fit? That interference fit contact force is in compression in the shaft, it’s akin to a bearing stress on the shaft. The yield stress in bearing is actually higher than the yield stress in tension, due in part to confinement by surrounding material. (Epete’s hydrostatic compression?) Furthermore, we typically pick (or the material spec., ASTM min. is) a tensile yield stress at a .2% strain offset, but the stress strain curve continues to climb to ultimate for many steels. So, I think the yield stress would be a lower limit for the normal stress in your torque calc. if you are already at Fy. Some steels have a fairly horizontal plastic range for some amount of strain increase, with little change in stress. Then Fy would be a max. normal stress/force as GregL suggests. But, I think most alloy steels have a sloping/curving upward strain hardening range with no horiz. plateau, where stress increases with increased strain. Obviously, your upper limit must be short of Fult. Also, at those kinds of contact stresses you could probably consider a fairly high coef. of friction, since you would probably actually be getting some mechanical or almost atomic/molecular level bonding or forge welding bonding btwn. the two surfaces, I would think, but gotta think about this some more.

Check out a slight taper press fit btwn. the shaft/axle and the hub/wheel, just as the railroads mount wheels to axles. They use a little white lead lubricant, press to some very high pressing forces and can develop some significant torques, as on locomotive drive wheels. I would think that this method also develops significant mechanical or almost atomic/molecular level bonding or forge welding bonding btwn. the two surfaces.

 

[desertfox]

Hi dhengr

The interference fit will be achieved by heating and shrinking the hub onto the shaft after which the joint will be used to transmit a heavy torque in the order from memory of about 2.5MNm.
All that concerned me was the formula used to do these calculations always assume's the hub and shaft remain within the elastic limit and not go over the yield value unlike autofrettage in a pressure vessel.
Looking at the information I had it appeared possible that the shaft might go over the yield stress value after the hub is shrunk on, which didn't seem a very good idea to me but before I raise the query I thought I would check and see if anyone else had come across this situation.
The way I see it is if the shaft permanently deforms then some of the interference is lost and therefore the interface pressure between hub and shaft is reduced, which in turn would affect the torque capability of the joint or am I missing something?

regards

desertfox

 

[GregLocock]

I don't think torque capacity is lost as such, in the ideal case where the plastic stress = the elastic limit. It just won't increase.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[electricpete]

Hi electricpete
Yes the shaft would be assumed to be in a state of uniform compression but I'm missing your point about the yield stress value.

My understanding is that yield does not occur in pure hydrostatic compression. Yield is associated with shear stress (in certain failure theories) which does not exist in pure hydrostatic compression.

Shigley's Mechanical Design 8th Ed equation 5.8:
Distortion energy ud = u − uv =
ud = (1 + ν)/(3E)*[(σ1 − σ2)^2 + (σ2 − σ3)^2 + (σ3 − σ1)^2] (5–8)

For hydrostatic case σ1=σ2= σ3 and ud=0



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(2B)+(2B)' ?

 

[electricpete]

Also the 3-d Mohr's circle collapses to a point when sigma1=sigma2=sigma3, which explains how we can have no shear stress.

If it were close to pure hydrostatic compression, I'd think there would be no need to consider yield.

If we were considering yield of the hub instead of the shaft, based on the other comments I'd think the conservative approach would be consider only the interference that brings the hub to yield. I see your logic that it would be tough to take credit beyond that.

For shaft in realistic situation of not completely hydrostatic compression, it seems like the true answer lies somewhere between but again you might lean towards conservatism of only taking credit for amount of interference that brings you to yield.... if that's something that can be calculated.... sounds tough to calculate.



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(2B)+(2B)' ?

 

[desertfox]

Hi All

Ah I see what you mean now EP thanks.

I had a word with some stress engineers about the joint and they said neither shaft or hub should go above yield and if it does it may cause a problem.
I did the calculations and although the shaft had a much lower yield value than the hub they were both within the elastic area, so its okay now.

Thanks for all your contributions.

Regards

desertfox