Square Bar in Torsion

[mrsongs]

I have a motor that is 3 HP and 1730 RPM. It has a 102:1 gear reduction set on it. I'm having a problem with it twisting the transmition shaft, which is 1 3/8 inch solid steel. I calculate the full speed Torque to be ~930 lbft...Correct? I calculate the Maximum shear stress to be ~20600 psi...correct? Usually Shear yield is .58 of Tensile yield. So, are the bars we've been using just really weak or am I missing something.


Yes, I know it is unusual to have a transmission shaft be square

 

[electricpete]

One piece of the puzzle - a 3hp 1730 RPM electric motor can be capable of delivering much more than 3 * 5250/1730 ft-lbf of torque. The motor rating is based on thermal considerations, not max torque. For NEMA design B motor, the motor may be able to deliver approx 2.5 times that. How much torque is transmitted depends on the load torque during steady state and also upon the inertias durating transients.

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[hydtools]

Recommened design stress for shafting in pure torsion is 8000 psi (Machinery's Handbook and ASME Shafting Code).

Your calculations are right. What is the steel spec? The material yield strength should be ~42,000psi with no margin of safety.

For the 8000 psi limit your shaft size should be 1 7/8 to 2 inches square for the material you are using.

Ted

 

[mrsongs]

electricpete: I did know that there can be some transients and that during startup and coast I can exceed that torque. The thought was that the actual deformation of the shaft occured while running at speed with the operator failing to stop in time.

Hydtools: what is the ISBN for Machinery's Handbook or where can I pick that up?...Always interested in new reference material. Under what conditions does the book give for 8000 psi? The spec is a little weak, as in non-existent. I was using rule of thumb for standard steel yield strength to get an idea of what was going on. I'm relatively surprised it would have sizing for square shafts.

I've kind of shifted my focus somewhat. The problem I was brought is that they twisted the shaft. I started with the thought that I needed to redesign and to check my calculations with others, I'm one of only two ME's that work at my location and the other one is too busy on something much bigger to check this.

Now, I'm to the point of clarification: Isn't this what I want to happen? This setup has a motor rigid coupled to a gearbox, which is rigid mounted to the shaft. I certainly don't want the gearbox to fry or the motor to fry. The failure point that I want is the shaft as opposed to the other two. If I make the material stronger then I risk a much worse failure, either costly or things actually shattering and shooting out shrapnel. I think the way to go is a shear pin of some kind. The problem is I have no way of knowing for all applications how much torque I actually need.

In the end the operator failed to use the machine properly.

Thank you for your help. I'm going to have to figure out what I can do.

 

[djm883]

Could you use a controlled torque coupling? Set it at the torque you don't want to exceed and that will save your reducer and motor.

I have dealt with Falk couplings in the past.

Don

 

[Ross0684]

You need to advise what the reducer is driving as this will have some bearing on the solution i.e. machine, conveyor etc?
As electricpete said, inrush current can give 250% more torque on start up, and if load is variable this can be an issue, plus deceleration forces can come into play on conveyors etc.
Not good practice to use the shaft as the safety valve!
Suggest you do a proper evaluation of the load, both start up and running.

VSD's have electronic shear pins plus also control starting/stopping torque.

Just checking, I estimate a 2" round shaft is the minimum required to handle 3 HP at 17 rpm @8000 psi, whereas the real startup requirement could be much higher.

Do some current checks on start up and running-compare to the motor nameplate specification and this will give some indication where you are at.

 

[hydtools]

Looks like a new edition of the handbook is due out the month.

http://www.industrialpress.com/en/MachinerysHandbook/default.aspx

Ted

 

[mrsongs]

This setup is used to open bottom dump railcars. There is definitely variable torque required throughout the process. There is no way to predict how much the maximum torque required is going to be. This setup is pretty standard for the industry. The cars that are the hardest to open we definitely want the opener to be able to open because we don't want the operator having to push open a somewhat stuck slide gate. Cars like this are not maintained well.

I was trying to find a controlled torque coupling and I'll check Falk out. It still gets back to the problem that we can't predict how much load this will be under in worst conditions.

I don't want the shaft to twist but it is the cheapest member. I thought about a shear pin. I haven't calculated the needed size but I'd have to re-design the shaft because we don't have a spot where we have a round connection, they're all square. I'd also have to put some sort of a bushing in around the shear pin to protect the metal. I'm leaning this way.

I really ant to hear any suggestions anyone has, this one has me somewhat stuck since there are so many variables.

 

[djm883]

Since it is difficult to figure out the torque required, you know that the motor can provide 250%-300% of the motor torque. You can try to size the coupling to match the maximum torque the motor can put out.

Once you get the coupling and install it, you can adjust the setting (by trial and error) on the coupling to the necessary torque. If you are running your motor close to 200% for most of the operations then you may want to increase the size of the motor and/or reducer. Do you know what the reducer is rated for (HP and torque rating)?

 

[gearcutter]

Perhaps it could be done electrically, by limiting the current, using the motor's contactor. There should be an adjustable overload relay associated with contactor. By trial & error you could tweak the overload relay to trip once a certain amount of current draw has been reached effectively acting as an electronic shear pin. It would be a cheap way out and would utilize equipment already in place.

Ron Volmershausen
Brunkerville Engineering

 

[mrsongs]

I like that last suggestion. I'm going to have to try that.

 

[gearcutter]

mrsongs,
I was wondering how you solved the problem in the end. Perhaps you could share the results with us.

Ron Volmershausen
Brunkerville Engineering
Newcastle Australia

http://www.aussieweb.com.au/email.aspx?id=1194181

 

[djm883]

I'm curious too of the outcome.