Reply With QuoteNo, the forces applied and transferred to the shrink disk is some vector not normal to the applied forces.Originally Posted by mhayes01
You will also need to consider the friction. I'll need to think about the math - it's been awhile.
I am designing a shaft coupler and I am trying to figure out the best way (theoretically) to determine the clamping force. The torque on the bolts are 120 in-lbs per bolt. There are 12 bolts. What I am doing is bolting 2 angled plates over a shrink disc and the shrink disc is squeezing the shaft as the plates are moving closer. Am I correct to assume that the clamping force created during the tightening of the bolts is the same as the clamping force of the shrink disc? I attached a cut-away of my drawing as a reference. Thanks.
No, the forces applied and transferred to the shrink disk is some vector not normal to the applied forces.
You will also need to consider the friction. I'll need to think about the math - it's been awhile.
Tell me and I forget. Teach me and I remember. Involve me and I learn.
The clamping force will be the length of the taper divided by its height i.e. the cotangent of the taper angle. After determining this force you should divide this force by the contact circumference area between the tube and the sleeve to determine the clamping pressure and resulting compressive pressure stress on the tube. This is important because the wall thickness of the tube can be a limiting factor on the amount of clamping force you can apply before exceeding the yield strength of the tube and plastically neck down the tube diameter without applying any actual increase in clamping force.
As far as the longitudinal gripping force, this will be the clamping force multiplied by the friction factor for the mating ring and tube materials; and, is also the friction between the clamping ring taper and the clamping plates, which will also be expressed as an increase in the tensile loading and stress on the clamping bolts.
Thanks. I am working on the answer now.
It looks like my focus has been changed. Now I need to figure out how much force needs to be applied to the shaft (by the shrink disc) so that the shrink disc will rotate with the shaft. Basically, how hard does the shrink disc need to squeeze the shaft in order for the shrink disc to rotate along with the shaft. Also, I am using a solid shaft now instead of a hollow shaft.
Is there some reason you aren't using tapered ring hubs available commercially? Several sources and styles are available and they've done all the math. And the real world testing!
I thought about that initially, but we want to keep the machining in-house. The diameter is very large and purchasing from an outside source would add to the cost of the product I am helping to design.
I understand, but to me the facts you use as justification not to "go outside" are the very ones I would use to justify doing exactly that. One thing that obtaining my gray hair has taught me is the - the value of experience learned in the real world can not be underestimated. Think about the questions for which you had to resort to asking an on-line forum. If you're willing to take the advice of some guys on line who, though thoroughly competent and perfectly well meaning, still do not have that all important virtue of having suffered through AND LEARNED FROM real world failures in designing and building exactly the kind of device you are building, why not just go ahead get the advantage gained by someone else's personal experience? Pay for his failures instead of your own. Get the advice of experts. To my way of thinking, the larger the device is, the more costly failure would be, and the more important it would be to guarantee success!
When the machine fails, do you really want to say, "Well, these guys on line said it would work."? Think about it.
At least pay an expert to design it, even if you machine it in house.
Just sayin...
That makes perfect since to me. I'm going to pitch it to my manager one more time from the perspective that you just gave me. Thanks.
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