I am designing an engine driven system of belts and pulleys. The engine manufacturer has specified that the maximum torque that can be imposed on the engine by the system is 100 in-lb. After consulting with my coworkers, the overtorque protection method that I am leaning toward is to machine a groove in the splined shaft that will be driving this system to a diameter that will fracture if the torque exceeds 100 in-lb. This seems too simple to me. Can it be as easy as calculating the maximum tortional shear stress with Tc/J for a specific shaft material? What am I missing? Is there a better way to do this? Any input would be helpful.
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Overtorque protection 3
- Thread starter RUFUS2K
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- #2
You could look at the possibility of incorporating commercial shear pins into your design. The challenge with your idea is having the detailed knowledge of the actual mechanical properties of the splined shaft. Commercially available shear pins have closely defined mechanical properties and more predictable performance.
RichLeimbach
Mechanical
Rufus,
Seems to me that fracturing your shaft is an expensive way to protect the engine. I'd lean more towards a shear pin (as carburize has suggested) or a commercially available slip clutch. You may be able to find a number of good products on the Stock Drive website that would help you:
Seems to me that fracturing your shaft is an expensive way to protect the engine. I'd lean more towards a shear pin (as carburize has suggested) or a commercially available slip clutch. You may be able to find a number of good products on the Stock Drive website that would help you:
- Thread starter
- #4
Thanks for the input. I have been looking at some sort of shear pin design today. As for some type of slip clutch design, the problem that I see is that they all seem to have some sort of integral bearing in them and what is to protect the engine in the case of that bearing failing? Of course, that is unlikely (especially if you talk to the factory reps) but I don't think that it is any more unlikely than the other bearings in the system failing. And that is what I have to design against in the first place.
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- #5
Hi RUFUS2K
I would give the same advice as the two earlier writers, ie:
try to use something thats comercially availble if possible.
However if you go down the road of failing your shaft, then
your right it is not as simple as the formula Tc/J.
Firstly when you hit the yield stress for the shaft it is only the outerskin that as yielded, you now have to apply more torque to get the yield stress to occur not only at
maximum "c" but at all values of "c".
So your now into fully plastic torque theory on your shaft this can be 33% greater than the maximum elastic torque first put on the shaft to yield the outerskin.
Furthermore the shaft material may under go work hardening which will also increase your failure torque.
The formula for the torque to produce a fully plastic shaft in torsion is as follows:-
Tfp = (2*3.142/3)* R^3* yielding shear stress.
If you use this formula it should get you into the right ball park for the failure torque.I would then suggest that you try practical tests on various shafts and compare theory with practice and adjust the groove according to your practical results. Also bear in mind that you will need
to have good mechanical properties which are consistent for all future shafts using this overload safety principle.
Hope this helps
regards
Desertfox
I would give the same advice as the two earlier writers, ie:
try to use something thats comercially availble if possible.
However if you go down the road of failing your shaft, then
your right it is not as simple as the formula Tc/J.
Firstly when you hit the yield stress for the shaft it is only the outerskin that as yielded, you now have to apply more torque to get the yield stress to occur not only at
maximum "c" but at all values of "c".
So your now into fully plastic torque theory on your shaft this can be 33% greater than the maximum elastic torque first put on the shaft to yield the outerskin.
Furthermore the shaft material may under go work hardening which will also increase your failure torque.
The formula for the torque to produce a fully plastic shaft in torsion is as follows:-
Tfp = (2*3.142/3)* R^3* yielding shear stress.
If you use this formula it should get you into the right ball park for the failure torque.I would then suggest that you try practical tests on various shafts and compare theory with practice and adjust the groove according to your practical results. Also bear in mind that you will need
to have good mechanical properties which are consistent for all future shafts using this overload safety principle.
Hope this helps
regards
Desertfox
- Thread starter
- #6
DESERTFOX,
Thanks for the great info. I really think that this is the direction that I am going to have to take. It turns out that we have built systems in the past that employed this design but we never did the development in house. One question though, in your formula, is "R" the radius of the shaft or some other variable?
Thanks again
Thanks for the great info. I really think that this is the direction that I am going to have to take. It turns out that we have built systems in the past that employed this design but we never did the development in house. One question though, in your formula, is "R" the radius of the shaft or some other variable?
Thanks again
Rufus2k; You probably have your equipment built by now, but a few comments. If you require much accuracy and repeatability you might consider other methods, mainly because of material properties variations. A shear pin would be better, but for best accuracy and repeatability it should be notched to concentrate stress and minimize variation due to material variation. A friction clutch would be better, but not highly repeatable because coefficient of friction changes with use. A spring-loaded detent clutch would be best if you can find one with end connections and trip torque range that you need. Good luck. Dale100.
strokersix
Mechanical
If you do decide to use a necked-down shaft for protection, consider drilling out the center. This will reduce the strain concentration on the surface and give you a more predictable/reliable design.
Mike
Mike
Try for technical information about a commercial product that will disengage the load from the driver at a predictable torque level. This component does not destroy shaft or shear pins.
- Thread starter
- #11
Wow!, all this activity after such a long time.
Thank you all for your input. Let me give you an update on this project since I am almost ready to begin testing on the first batch of sample pieces. I had hoped to incorporate some sort of commercial product to protect the engine driving this system. What became evident over the course of my research was that the majority of these products incorporate some type of internal bearing which would always lead to the question "well, what if THAT bearing failed?" , "What will protect the engine in that case?" The final design had to take into consideration that if something did happen and the device was engaged, it might still be a matter of hours before the engine could be shut down. This, in my mind, eliminated the majority of the friction-type devices. When it all boiled down, it became clear that to protect the engine in the event of a failure at ANY of the points of bearing contact, the protection must be placed between the point of connection and the first point of contact. This left some some sort of shear coupling as my option. Finally, to reduce the overall size of the installed assembly, the necked down "shear section" in the drive shaft was chosen as the final design. Granted, this means that the shaft will be destroyed in the event of some over torque event, but the only events that I am protecting against are catastrophic failures of the driven components In which case, replacing the drive shaft is of minor consequence.
Thank you all again for you input. I have learned alot in this process!
Rufus2k
Thank you all for your input. Let me give you an update on this project since I am almost ready to begin testing on the first batch of sample pieces. I had hoped to incorporate some sort of commercial product to protect the engine driving this system. What became evident over the course of my research was that the majority of these products incorporate some type of internal bearing which would always lead to the question "well, what if THAT bearing failed?" , "What will protect the engine in that case?" The final design had to take into consideration that if something did happen and the device was engaged, it might still be a matter of hours before the engine could be shut down. This, in my mind, eliminated the majority of the friction-type devices. When it all boiled down, it became clear that to protect the engine in the event of a failure at ANY of the points of bearing contact, the protection must be placed between the point of connection and the first point of contact. This left some some sort of shear coupling as my option. Finally, to reduce the overall size of the installed assembly, the necked down "shear section" in the drive shaft was chosen as the final design. Granted, this means that the shaft will be destroyed in the event of some over torque event, but the only events that I am protecting against are catastrophic failures of the driven components In which case, replacing the drive shaft is of minor consequence.
Thank you all again for you input. I have learned alot in this process!
Rufus2k
Hi RUFUS2K
Thanks for the feedback its always nice when people let us know how their problems turnout.
Did you make use of my formula? I would be interested to know how close the theory was to the practical tests perhaps you will let me know.
regards
desertfox
Thanks for the feedback its always nice when people let us know how their problems turnout.
Did you make use of my formula? I would be interested to know how close the theory was to the practical tests perhaps you will let me know.
regards
desertfox
- Thread starter
- #13
Hi,
I have 2 solutions for your problem.
1> Provide a torque sensor and range the torque for cutoff limit.
2> Use a fluid coupling between engine and the driven members(ie provide a fluid coupling at the output shaft of the engine)
Both the solutions are very reliable but need to work out the cost.
I dont believe in shear pin since this will lead production downtime.Also the slip clutch may work for 2-3 ocassions and after sometime it never transmit torque in the favorable position too. These designs are not recommended for such applications. Since you need to protect engine , I hope you have to go for aforesaid options only. Reliability and operability is one of the most important thing what a designer should look for. If these things are poor then your product will not be marketed very well
I can give you further details if you like the options mentioned.
Thanks and best of luck
Sudev Nair
I have 2 solutions for your problem.
1> Provide a torque sensor and range the torque for cutoff limit.
2> Use a fluid coupling between engine and the driven members(ie provide a fluid coupling at the output shaft of the engine)
Both the solutions are very reliable but need to work out the cost.
I dont believe in shear pin since this will lead production downtime.Also the slip clutch may work for 2-3 ocassions and after sometime it never transmit torque in the favorable position too. These designs are not recommended for such applications. Since you need to protect engine , I hope you have to go for aforesaid options only. Reliability and operability is one of the most important thing what a designer should look for. If these things are poor then your product will not be marketed very well
I can give you further details if you like the options mentioned.
Thanks and best of luck
Sudev Nair
- Thread starter
- #15
Desertfox,
Well, here are the results. I had calculated that if I wanted the shaft to fracture at 400 in-lb of torque, the shear section of the shaft would have to be .290" in diameter. I tested 4 sample pieces and the actual diameter is closer to .260" dia. I was very happy with those results. Especially since diameter was larger than I needed so I was able to remove the excess material a little at a time to achieve the desired value instead of having to make whole new pieces!
Thanks again for your help!
Rufus2k
Well, here are the results. I had calculated that if I wanted the shaft to fracture at 400 in-lb of torque, the shear section of the shaft would have to be .290" in diameter. I tested 4 sample pieces and the actual diameter is closer to .260" dia. I was very happy with those results. Especially since diameter was larger than I needed so I was able to remove the excess material a little at a time to achieve the desired value instead of having to make whole new pieces!
Thanks again for your help!
Rufus2k
Hi RUFUS2K
Thanks for the feedback,I calculate the difference you found
in the shear area's to be just less than 10.5%, which I think is good correlation between theory and practice.
Its nice to know how peoples problems turn out.
Thanks again
Regards Desertfox
Thanks for the feedback,I calculate the difference you found
in the shear area's to be just less than 10.5%, which I think is good correlation between theory and practice.
Its nice to know how peoples problems turn out.
Thanks again
Regards Desertfox
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