torque for horizontal chain and sprocket drive 1

[d1eng]

Hi,
I have 4 vertical power screws which are all linked together by a horizontal chain and sprocket drive. So i have 4 sprockets of the same pitch diameter (one on each screw) and 1 chain (running on the periphery of all the the sprockets). There is a handwheel ontop one of the power screws, therefore turning the handwheel turns all the screws at the same time.
I am building a lifting device that uses these 4 power screws. I already calculated the torque required to lift the load to be 15 Nm. But my question is, how do i determine the maximum torque required to initiate the chain drive? Im sure the answer is really simple but i just dont see it.
There is no slack side on the chain, so my 1st instinct would be to take the intial tension on the chain and multiply it by the pitch diameter. then multiply this value by 4 to obtain my final value. is this correct? Or should i consider the final speed of the chain drive and the friction.
I would appreciate any suggestions

 

[dvd]

The tension in your chain will increase in each span of chain to match the torque required to turn each power screw. The tension is additive around your chain loop. Draw a sketch of the four sprockets with a force vector applied at each pitch diameter and then sum them up around the system stopping at the crank.

You will have a slack side to the chain, it will be the span after your crank handle. Depending on the center distance between the power screws and how much your chain stretches you may have problems with the chain staying on the sprockets, particularly the span after the crank. You can make drawings which show the chain taut on all spans, but the chain is going to stretch and you will get a catenary sag that can cause the link plates on the chain to rub the sprockets and wear. You should consider chain tensioners on the spans adjacent to the crank handle, since this will be a reversing drive.

Also, depending on how you attach the power screws to the load you may end up with a lot higher torque requirement than you anticipated. (Assuming that you are locating a bearing very close to the sprockets so that the power screws don't deflect when you turn the crank.) If your sprocket bearing centers and the power screw centers are not matched very closely you will have high side loads applied through the nuts on the power screws, as I have experienced firsthand. Provide some type of floating attachment for the nuts and hopefully provide some type of linear guide besides the power screws.

 

[d1eng]

Thank you for replying. I did consider chain tensioners but was unsure which span to place it on. Now that i know where to put it, wouldn't this eliminate the slack on the span after the crank? I am still unsure as to what approach should i take to determine the starting torque of the chain drive.
The power screws are all vertical. and will remain vertical during operation of the lift. Each power screw is braced at the top and near the bottom by radial ball bearings. The threaded length is between the two radial ball bearings, therefore any moments or deflection generated by the load is canceled out at the radial bearing supports. At the bottom of the power screws there are thrust bearings that supports the direct trust of the load. Now the chain and sprocket drive is between the lower radial and trust bearings. (5cm apart). Im hoping that the tension in the chain drive isnt so much as to rip trough the trust bearings or signficantly increase the final required torque. This is why i really have to know how to determine the maximum torque in the chain drive. Any ideas?

 

[dvd]

Yes, the point in using a tensioner is to eliminate the slack. I'm sorry but I don't understand your question about the starting torque. Since this is a hand operated machine you won't be accelerating the components greatly so my opinion is that your final torque required will be very close to your starting torque. How much is the load you are trying to lift (or compress), what is your screw diameter and pitch, and how fast are you trying to move this?

 

[d1eng]

Yeah, I see your point about the final torque. The lifting device would have a capacity of 450lbs. The outer screw diameter is 1 1/4". and the inner screw dia. is 1.05" and the pitch is 5 threads per inch (acme threads). The final velocity can be anywhere between 40 to 50 rpm.

 

[dvd]

If the work to lift your 450 lb load 1 inch is 450 lb-in, then the torque required with no frictional loss is 450 lb-in/(5*2*pi radians) = 14.3 lb-in. So it looks like you are using an efficiency for the screws of around 0.11. It sounds like a nice conservative estimate for the torque.

As for the thrust bearing, are you going to lock the radial bearing to the shaft with a press fit or shaft collars or some other means? If so the radial bearing is going to see the thrust first, you could consider using a tapered roller bearing along with a shoulder on your power screws and eliminating the thrust bearing.

 

[raviteja]

I HAVE DEVELOPED THE SIMILAR MACHINE FOR SEPARATING THE CAKES.IHAVE GIVEN FOUR SCREW WITH FOUR GUIDE RODS WITH BRONZE BUSHES.U SHOUL ENSURE THAT NO RADIAL LOADS COMES ON TO THE SCREW THIS IS ACHIVED BY PROVIDING THE LESSER CLEARENCE BETWEEN BRONZE BUSH AND GUIDEROD THAN THE PLAY BETWEEN NUT NAD SCREW.ALWAYS KEEP THE GUIDE ROD BIGGER SIZE THAN SCREW ROD DIA. MY M/C THE SEPARATING LOAD(LIFTING LOAD IS ATOUND 500 KG. WITH A TRAVEL OF 500 MM.THE M/C IS RUNNING VERY WELL.WEE HAVE CONNECTED THE FOUR SCREWRODS BY MEANS OF A BEVEL GERBOX WHICH IN TURN CONEECTED TO MOTOR BY PROPELLER SHAFT.I HOPE THIS WILL HELP U A LITTLE IN UR DESIGN

RAVITEJA
DESIGN ENGINEER
HYDERABAD

 

[d1eng]

Thanks alot for your replies.
DVD, that torque is the torque required by the power screws and not the chain drive. The radial brgs have a sliding fit on the shaft and is press fitted to its housing. no need for tapered brgs.

I guess, it would be best to estimate the starting torque of the chain drive as the running torque

 

[d1eng]

thanks alot for ur help dvd

 

[SERVOCAM]

I have sized applications like this for servo or stepper motors. When calculating torque, you can't just use static / holding torque. You need to take in consideration the torque it will take to accelerate the load to you desired velocity. This will be your Acceleration Torque, Ta. You will also need to calculate your torque from friction and gravity. Then you can calculate you deceleration torque. Then you can calculate you slew torque and RMS torque.

Regarding chain tension, it should have no effect on the torque required for the application, as long as bearings can handle the tension. If they can't, then you have other worries to think about. A large amount of slack will cause a bang and eventually break the chain.

Now for coming up with torque, we first need to calculate inertia. Why?

Acceleration Torque = inertia x rotational acceleration. (Most just think of force x radius)

Inertia Pulley: Jpulley

Wr^2
Jsprocket = ------
2g

Where:
Jsprocket = Inertia of Sprocket (lb-in-sec^2)
W = weight of sprocket
r = pitch radius of sprocket
g = gravity = 386 in/sec^2

---------------------------------------------

Inertia Chain: Jchain

Wr^2
Jchain = ------
g

Where:
Jchain = Inertia of Chain (lb-in-sec^2)
W = weight of chain
r = pitch radius of sprocket
g = gravity = 386 in/sec^2

--------------------------------------------


I need to run, I will finish my post later Cameron Anderson - Sales & Applications Engineer
Aerotech, Inc. -

http://www.aerotech.com

"Dedicated to the Science of Motion"