Torque to load in conveyor application
Torque to load in conveyor application
(OP)
Hello guys, i have a question reguarding my resulting force on a shaft that a motor is driving.
See attachment for clarification.
My issue is calculating the total load. I would then use that determined load and use it in a beam calc between the bearings. Static - Lets say my arm for my motor is 4" and my distance from the arm to the center of gravity to my motor is 8" and my motor weighs 30 lbs. I determined static load by 30lb * 8in = 240in/lb divided by 4in to get 60lbs. Dynamic lets say my motor outputs 800 in/lb of torque. I take 800in/lb divided by 4in to get 200lb. This would be the load at the end of the torque arm. What would be the load on the shaft itself linearly? When I calculated this before i just took my max 200lb (dynamic) + 60lb (static) = 260 lb max load. This cant be 100% acurate but if i were to figure worse case this would be it right?
Thanks for your guys help
See attachment for clarification.
My issue is calculating the total load. I would then use that determined load and use it in a beam calc between the bearings. Static - Lets say my arm for my motor is 4" and my distance from the arm to the center of gravity to my motor is 8" and my motor weighs 30 lbs. I determined static load by 30lb * 8in = 240in/lb divided by 4in to get 60lbs. Dynamic lets say my motor outputs 800 in/lb of torque. I take 800in/lb divided by 4in to get 200lb. This would be the load at the end of the torque arm. What would be the load on the shaft itself linearly? When I calculated this before i just took my max 200lb (dynamic) + 60lb (static) = 260 lb max load. This cant be 100% acurate but if i were to figure worse case this would be it right?
Thanks for your guys help





RE: Torque to load in conveyor application
I can see an arm off the motor connected to the frame via a single bolt, but how is the rest of the motor supported?
If the motor is only supported by the arm with the bolt and the rest of the motor weight is reacted by the drive shaft then your moment calculation regarding the dead weight appears to be correct, but I don't think its a very good idea to support the motor in that way.
When you apply torque on the motor, the bolt via the arm is the only thing stopping the motor casing from rotating.
Can you give us a bit more detail please.
desertfox
RE: Torque to load in conveyor application
This is not a school problem, is it?
RE: Torque to load in conveyor application
I picture if i were to take a pen and push it against the table, if the table is solid and rigid the force i apply will also come back to my hand.. I was wondering if the 200lb force against he block could be transmited back to the motor.. would it be more? less?
RE: Torque to load in conveyor application
(Try to get the reducer closer to the bearing and step the shaft to get more diameter away from the reducer. You may want to investigate the prospect of shaft whip. Also, make sure that the maintenance guys are going to be happy with you when they have to disassemble the conveyor to repair the reducer.)
RE: Torque to load in conveyor application
Consider this.
The torque at the motor shaft,T1, drives the inlet to the reducer and the output torque, T2, of the reducer connects to the power shaft you are concerned with.
Your static analysis is OK but when the motor spins, the reducer torques, T1 and T2 must be absorbed by the power shaft, so you have two orthogonal torques on the shaft in addition to the force you have correctly identified coming from the restraint torque, T2.
In addition, the pin supporting the motor/reducer has to bear the twisting torque of the motor, T1, which ids the reaction torque of the stator.
In my opinion, this design will result in binding and significant bearing problems. You should redesign it and NOT depend on the power shaft to support the extraneous torques and forces.
RE: Torque to load in conveyor application
We used to use Leeson motors and reducers. We are moving to a more positive and better solution with SEW Gearmotors. This box in a hollow shaft mount does not have mounting holes on it, anywhere. On the initial design there was a channel that ran across underneth and we bolted the reducer directly to the channel from the bottom. Now with SEW we dont have that option.
So lets say i had to change the design to support motor from bottom. What would i need to due to ensure that next time if i wanted to use a motor in this application this way? What would be the way to calculate the loads the output shaft takes?
thanks for the input guys
RE: Torque to load in conveyor application
If you support the body of the motor on the underside and bolt the motor to that support, that will take care of the torque reaction of the motor, if there are no holes in the reducer box then the torque arm should react the box torque, although it would also help to have a support plate under the reducer box.
So your drive shaft then only as the torque on it required to drive whatever it is your driving.
desertfox
RE: Torque to load in conveyor application
RE: Torque to load in conveyor application
By the way, this conveyor is a chain driven conveyor that is driving an empty pallet (100lb load). Im not sure if that helps you guys in terms of invisioning how much load this thing is actually taking.
RE: Torque to load in conveyor application
A lot better than what you have; answers your motor/gearbox replacement problem and if positioned properly with respect to the takeoff chain, will mitigate the stresses in the power shaft.
RE: Torque to load in conveyor application
Can you give us the SEW references of the parts your using?
desertfox
RE: Torque to load in conveyor application
1/2 = 16lb * 51.30 *.9 (eff) = 738.7 true in/lb torque
That is the spec on the motor being used.
Carrying 100 lb pallet and 14 lb of chain on a 17T 60p sprocket @ 60fpm. Roller chain on steel keystock.
That is the spec on the motor being used.
Also to make it more clear, this chain conveyor is on a pivoting carrage and pops up and down thru another chain conveyor to make a 90° transfer. See attachment
RE: Torque to load in conveyor application
1) if the motor assembly weighs 30 lb the static force on the shaft camnnot be greater than 30 lb.
It should be
Fs=Wm*L1/L
Wm weight of motor/gearbox assembly
L1= dist from bolt to center of mass of assembly
L = dist between bolt and shaft
2) If the dynamic torque is 800 in lb , how can you get 200 lb on the shaft. That would inmply that the distance between the pivot bolt and the shaft is 4". I don't think so. It should be
T/L
L= dist from bolt to shaft
So, your forces appear to be smaller than my calculation would suggest.Also, the small orthogonal torque from the stator probably can be neglected and I must correct my first
description to read that that torque is distributed between the bolt end and the shaft,mostlt to the shaft.
RE: Torque to load in conveyor application
description to read that that torque is distributed between the bolt end and the shaft,mostlt to the shaft"
My bad,
The socalled orthogonal torque is an internal torque and does NOT manifest itself in any torque on either the bolt or the
the main shaft.
The force sought is simply
T/L
RE: Torque to load in conveyor application
I was trying to find a manual on your SEW parts but I can't seem to find them, do you now which manuals from the SEW site to look at? I found geared motors type DR but your reference is DRS.
I thought it might show mounting positions and recommendations which would help in understanding better your problem rather than people making assumptions.
desertfox
RE: Torque to load in conveyor application
Not sure about the manuals but i could discuss it with my SEW supplier and have his take. If not he would beable to guide me to a manual like u discribe.
Zekeman.. the reason i was saying the load on the shaft would be MORE then its weight is only if the center of gravity of the motor is beyond the shaft.. the static load on the shaft is from a torque at the end of the torque arm.. so i translated it from that.. would this be incorrect?
if i were to take a plate thats 8 inches long and put 100 lbs at the end, pin the plate at the opposite end and tried to hold it up from the center of the plate (4 inches away from pin) wouldnt it be much harder to hold then if i moved it under the 100 lb load? (in this case twice as hard do to the rotating moment on the pinned end) that is what i visualize.
I rounded my numbers but the torque arm is somewhere around 4 inches.. So if i took the torque of 800/4 that gives me 200lb at the torque arm.. the fixture would have to hold that weight.. I am trying to determine the reaction against the shaft when holding the 200lb load steady.. in this case i accounted if the shaft took the same load 200lb + static.. and used in a deflection calc on my shaft.. the load was fine, however if i produce more load on my shaft then i can see it possibly being a problem.. could it be producing more load? How can i find this information acturately is what brings the question.
thanks for your repsonses i appreciate them
RE: Torque to load in conveyor application
I think talking to SEW is the best thing you could do and get there advice on mounting the motor.
With your static calculation on the load of the shaft due to mass of motor, I believe you are correct, however I have never seen a shaft being driven that has to suppport the mass of motor driving it.
desertfox
RE: Torque to load in conveyor application
You keep saying the dynamic force on the shaft is the torque delivered divided by the socalled "torque arm" which you say is 4 inches. But the force on the shaft is the torque delivered by the reducer divided by the distance between the bolt and the shaft, and that distance looks much greater than 4 inches.
I guess I still don't understand what you mean by " torque arm".
RE: Torque to load in conveyor application
Yes the shaft will take up the torque on the shaft too.. (800in/lbs) but that is thru rotation. I was trying to determine linear deflection on the shaft since it is supporting the motor as fox says.
I will investigate with my SEW rep and see what he thinks.
Still would like to know how it might be determined.. There has to be some way to determine what the linear force on the shaft generated from the motor torque. It may be very low, but do all of you agree there should be some extra force when running the motor?
RE: Torque to load in conveyor application
From your 1st drawing it does not look like 4 inches-- looks like motor length plus 1/2 the reducer + a link to the bolt-- to me more than 4 inches.
As I have told you, the dynamic force is equal to the torque , 800 in lb divided by that distance, L, I'm asking you to get. It's simple physics.
You might also revisit your static result, since your explanation of having the motor CM on the other side of the shaft doesn't make sense to me. From your drawing, it is clear to me that the static force must be less than the weight of the motor/gearbox.
As a final note, you should add a dynamic safety factor of at least 2 for the dynamic force. Also,another reason that supporting the motor with the driven shaft is not a good idea is that,vibrations coming from the motor dive could resonate through the shaft.
RE: Torque to load in conveyor application
Yes its a very small box.. its only about 5 inches in height and the arm is 4.33
the true center of gravity isnt 8" out. but i know in this case if it was it would be less thus my static load would be less.. im taking this into worse case senario's.. reason center of gravity is not acurate is because the motor is a solid mass.. so thats how solidworks generates it.. im sure its more like 5" or 6" in realtion to the end fo torque arm but again, if i consider 8" then it results in a higher load when calculated.
Yes i agree with you dynamic load would be 800/4.33 = 185lb. this is on the reducer torque mount. right?
I was trying to determine a reaction fore of that load if any onto the shaft in a linear motion.
RE: Torque to load in conveyor application
I'd listen to the SEW rep like your alluding to and I agree with him, if you re-read my first post its not a good idea to support the motor on the shaft its supposed to be driving.
Once you have established your mounting position for the motor and you need further help feel free to come back.
PS let us know how you get on.
regards
desertfox
RE: Torque to load in conveyor application
I have come into this discussion a bit late.
However, we build many conveyors in a given year using SEW and other brand hollow shaft gearmotors with torque arm mounting.
I surmise that this application is for a lift up transfer in a pallet handling system and would guess that the conveyor only runs in the raised position??? Service factor would be well above minimum recommendations.
I have never had any ongoing issues with mounting a drive in the manner you have shown-perhaps you could move the torque pick up point to the underside which will take some of the static weight as well as place the gearmotor as close to the bearing as possible.
Isn't any torque generated by the motor reacted directly by the torque on the sprocket driving the chain(which in your case is very small)
I have had issues with gearmotor suppliers trying to prove to me that the conveyor design is too light to handle the capability of a selected gearmotor but they lose sight of the fact that the gearmotor will only generate enough torque to drive the load ( unless of course the conveyor can drive the load into a solid structure and this is easily controlled by the use of torque limiters or thermal overload protection).
A good test to satisfy your mind is to briefly start the conveyor with a spring balance in place of the torque arm mount and you will be surprised to see how little reaction there is being generated.
With regard to the SEW SA 37 range, one of the downfalls of this box is the small shaft size available.
If they are available have a look at the WA series if you can get the necessary ratio as they are available with a 25 mm (or 1" I guess in the US) bore.
Regards
Ross
RE: Torque to load in conveyor application
Well, the reason i have started this thread is because i was questioned on the design as it was built. I did the calcuations above and found it to be okay. It is built and the motor is mounted as we speak. However it has yet to be run. I will be sure to let all of you know how it turns out and what i need to do. Meanwhile i will talk with my SEW rep because i have mounted reducers between bearings like that ALL the time. Im not sure what affect that would cause which is why i will contact my rep and post some info that we discuss as well.
As a side note, yes the motor is over rated. When we switched to SEW my boss wanted us to use them for their engineering and have them spec our motors. I have come back with possibly using a 1/4 hp and it would be just fine. 36 * 1/4 = 9 lbs out * 51 reductions = 459 * .9 (eff) = 413 in/lb. I am using a 17 T sprocket = 4.082 PD 2.041" radius. 413/2.041 = 202 lb out. Not to mention my load is running on roller chain. Well above even with a 1/4 HP
I will keep you guys posted.
Thanks for the replies!
RE: Torque to load in conveyor application
Maybe you can tell me from what you know, what this so-called "torque arm" is.
Is it the connection from the bolt pivot to the centerline of the hollow shaft ( i.e. the output drive shaft), or what? Or better still, tell me what you think is the distance between the bolt and the hollow shaft.
I suspect that I'm missing something here, since my take on this is that independent of what you call a torque arm, the only thing of interest here is the distance between the bolt pivot and the hollow shaft.Are they the same?
And where is this so-called "torque pickup point" located?
Isn't it at the hollow shaft?
RE: Torque to load in conveyor application
Its bolted too the side of the reducer with 4 bolts. Its hard mounted to the gearbox and the arm extends out from center to center 4.33". It has rubber inside the end of it so it has some flex at the end of the arm where it is hard mounted with a bolt as shown.
Talked to SEW rep, actually had him in my manufacturing plant to examine the part. He said they dont recommend putting reducers between bearings because of the shafts not being perfectly straight. It doesnt hurt but it can put wear on the shaft. As long as the motor is allowed to float (as it does in this design) its okay. He looked at the application and claimed it would be just fine but its not typically recommended. He suggested using a flange mount motor on that side as a substitude for the bearing if possible.
All fine and dandy but still doesnt solve my original thought of how much load gets produced by driving the motor. I wish i was able to do some testing like ross recommended but i dont have the time at work to set that up (or the tools.. lol)
Thanks for posts guys!
RE: Torque to load in conveyor application
Well I was waiting for the mounting details before posting how it loaded the shaft or otherwise, however going back to your original post I imagined from the information you gave that the shaft would see bending and torsion, bending from supporting the mass of the motor and torsion from the operational load.
So you could calculate the shaft stresses using a Mohr circle for example.
A starting point might be this site:-
ht
I also loaded a file for your reference and I think you got the 60lb static load on the motor by envisaging it the way I have in my sketch, anyway its just for reference.
Perhaps when we get the true mounting details we can give a better picture.
desertfox
RE: Torque to load in conveyor application
I will spend sometime examining the site you posted.
See my attachment on mounting.
RE: Torque to load in conveyor application
I can only see a bending load from motor mass, the torque will be governed by whatever your driving, so other than that I can only see the loads as per my sketch.
Have a look at this link:-
http://www.freestudy.co.uk/d209/t9.pdf
go to page 19
desertfox
RE: Torque to load in conveyor application
I think your SEW rep should have another think about what he said:
""dont recommend putting reducers between bearings because of the shafts not being perfectly straight.""
If it sitting on the shaft it must be straight-the flexible mount within the torque arm will take up any discrepancies between the mounting shaft and the torque arm pick up point.
Also:
""He suggested using a flange mount motor on that side as a substitude for the bearing if possible.""
This then requires the plate to which the gearbox flange is attached to be perfectly square to the centreline of the monting shaft-not easily achieved without introducing machining of the conveyor frame into the equation(with substantial costs).
This is the very reason torque arm mounts are utilised.
After over 45 years of being in conveyor manufacture I become very cynical about some of the advice equipment reps hand out. Granted SEW give their people good training but reps should also listen to their customers and the practicality of actually manufacturing something.
I will get down off my soap box now.
Regards
Ross
RE: Torque to load in conveyor application
Like the document that fox posted says how if the bearings arnt aligned the shaft can flex. I suppose thats why they dont recommend using it between bearings because it can flex.. not sure why as you said the torque arm takes misalignment up.
Im going to have to review the documents you posted fox and get a better understanding of them.. I will still post the results of the design for everyone who might like to know how it turns out.
thanks guys!
RE: Torque to load in conveyor application
Below is the details of what I was trying to say
"The torque arm concept is the most preferred, since it enables a hollow shaft reducer to hang from
a solid shaft and to be totally supported by that shaft. A torque arm is simple to install and guards
against potential misalignment caused by foot mounting or flange mounting. When correctly
installed, the reducer experiences zero binding and zero overhung load, even with a load shaft that
is out-of-round. In fact, it is quite common for the reducer to slightly wobble during operation."
ie if there is any flexure in the shaft the flexible mount within the torque arm absorbs it.
Also if you are using self aligning bearings( which I would be very surprised if you wern't), the bearings automatically become aligned.
Ross
RE: Torque to load in conveyor application
Now, as for the dynamic force load on the shaft, it is precisely what you got in your first post, since the torque delivered by the hollow shaft must be taken by a couple whose
"arm" (distance from bolt pivot to hollow shaft)is 4.33 inches and so
F=800/4.33=185 lb
In addition,there is a small ignorable twisting torque on the main shaft due to the angularity of the line from the bolt centerline to the gripping point of the hollow shaft.
Both ends of the torque arm also have 185 lb of force but the end connected to the reducer carries the 800 in-lb of torque to be delivered to the reducer housing. It is not a requirement that the torque arm be located as shown; it could be anywhere on the reducer housing and will produce the same dynamic 800in-lb of torque and 185 of force on the main shaft.
Another way doing this to consider the whole assembly rigid with equal external forces on the "link" at the bolt and the hollow shaft and the 800in-lb of torque delivered by this "link" which is a solid connection starting at the bolt and ending at the hollow shaft.
The total load is the static force plus the dynamic torque and force on the main shaft.
RE: Torque to load in conveyor application
Fox, havnt reviewed those documents yet but will at some point. I appreciate you posting them for me.
Ross, yes your assumption would be correct, i would be nuts not to use self aligning bearings lol. Yea when i said misalignment thats what i was refering too. I suppose i should have put it if the shaft wobbles the torque arm allows the reducer to move and prevent it from taking that load.
Thanks for the posts. Been helpful!