Torque to load in conveyor application

[CavemanJones]

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

 

[desertfox]

Hi Cavemanjones

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

 

[zekeman]

I see an unsupported gearhead motor so you only have a small part of the problem solved. What about the torque to the gearbox and how about the takeoff torque at the shaft to the load?
This is not a school problem, is it?

 

[CavemanJones]

No, this is at my work.. Yes, its a single bolt so that is why i was refering to it as a rotational point along it.. above i said, lets say i start that motor at 800 in/lbs if my arm is 4in then my load on the mounting block is 200lbs. so i figured the max load against the SHAFT/BLOCK would be 200lbs plus my static load of 60lbs totaling 260 lbs.. if i use that in a beam formula my max deflection on the shaft is next to nothing meaning i shouldnt have any trouble using the shaft to support my motor. Right?

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?

 

[dvd]

Try replacing the torque with a couple to be able to resolve the forces.

(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.)

 

[zekeman]

What happened to the torque the motor shaft delivered to the reducer? Did it magically disappear?
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.

 

[CavemanJones]

Here is how the problem presented itself.

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

 

[desertfox]

Hi CavemanJones

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

 

[dvd]

Shaft mounted reducers are quite common. One can design a shaft that will support the gearmotor and transmit the torque. If an auxiliary support on the reducer is added, there may be an induced load on the shaft unless care is made to avoid misalignment. Have you read the SEW literature on designing shaft mounts?

 

[CavemanJones]

We only refrenced it to for the style of their torque arms. Using a fixed bolt on the end. They have a rubber bushing inside for misalignment/shock load. The motor is as close to the bearing as we could get it based off the frame that this carrage pops down into. Its about 5 inches away from the bearing. With my above calculation if my loads resulted in 260lb max against the shaft i wouldnt see any problems in deflection.

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.

 

[zekeman]

Why nor use a chain drive for the last stage?
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.

 

[desertfox]

Hi Cavemanjones

Can you give us the SEW references of the parts your using?


desertfox

 

[CavemanJones]

SEW #SA37TDRS71S4; 1/2 HP; 3/4" HOLLOW SHAFT; TORQUE ARM; 51.30:1 RATIO; 33 RPM; M1 MOUNT; T BOX 0 R; CB@0

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

 

[zekeman]

OK, so you have 800 inlb at the dirve shaft and my cross torque is only 800/51=16 in-lb which is negligible, I'm having trouble with two things:

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.

 

[zekeman]

"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"

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

 

[desertfox]

Hi CavemanJones

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

 

[CavemanJones]

SEW is crazy with their numbers.. DR is the new energy efficent motors they just started supplying. Any motor less then 1 HP is called a DRS and anything 1 and above is DRE.. its wierd

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

 

[desertfox]

Hi Cavemanjones

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

 

[zekeman]

Caveman,
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".

 

[CavemanJones]

It is actually 4.33" (110mm)

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?