Ring roller power

[Crazy Old Coot]

TL;DR Summary

What is the principle of calculating the motor power and speed (thus gearbox output torque) fo drive a flat bar, or a tube, or any other section, through a set of rollers, to make a ring?

In principle, the ring is, at any instant, a beam being loaded in the centre, and in a failing state.
Thus, to drive the material continuously through the rollers, Force x Velocity = Work done/time = power.
60.P/2.pi.N = T
The units pan out, but is the concept correct?

The faster the work is done, the more power will be required. Seems logical.
The greater the roller spacing, the less Force is required, thus lower power. Seems logical.

Thank you for reading, and hopefully, bringing some order to this tortured soul!!

 

[Lnewqban]

Welcome, @Crazy Old Coot !

Your assumptions are correct.
Consider that the number of passes to reach the desired final radius of the ring determines the forces on the rollers and bearings, as well as the torque demanded from the motor.
Therefore, a small motor can achieve the same in small steps than a bigger one in less.

 

[Crazy Old Coot]

Thank you, Lnewqban, for your reply.
Your answer actually highlights the crux of the matter!

Consider that, once the "simply supported beam" has failed, further deformation is merely a question of displacement of the centre roller, with hardly any additional force.
So, under a deformation that, given any particular spacing of the two outer rollers, would generate a rolled radius of "r", the applied force "F" could be maintained and, if the velocity is maintained, the required Power should be unchanged.
Yes?

I think that taking baby steps is probably to avoid kinking?

 

[Lnewqban]

You are welcome.

Note that the very first bending is achieved by pushing the material down and at the middle point between the two bottom rollers with brute force, making it reach its yield point (plastic state) only in the cross section where the biggest moment is "felt", which is located exactly below the pushing top roller.

At that first stage, the rollers are not acting as such, but like the punch and die of a regular press brake.

The shape of the material remains straight between the central top roller, under which the initial bend happens, and the bottom ones (like a V-shape).

Then, the rolling motion starts, and from the edge of the material to the initial bend, the shape will remain untouched (note that the video below shows pre-bending of those sections for the heavy cylinder).

On the opposite direction, a continuous bending process happens as new cross-sections are forced to reach the bottom area of the top roller.

The difference from the initial bending is that the material is fed under the lowered central roller, progressively making it reach similar degree of deformation and plastic nature than the first cross-section.

As this continuous bending is progressive, it seems to use less energy or force, but in actuality, each similar length of arch requires exactly the same mechanical energy or work in order to achieve similar deformation than all the other equal lengths.

These links may guide you in a rough calculation of the needed bending and rolling forces:

https://www.mdpi.com/1996-1944/14/5/1204

https://www.harsle.com/how-to-calculate-the-bending-force-for-your-press-brake-machine

 

[jrmichler]

Consider that, once the "simply supported beam" has failed, further deformation is merely a question of displacement of the centre roller, with hardly any additional force.

No. In the case of ring rolling, the "simply supported beam" has not failed. It has, however, yielded. Yielding is also known as plastic deformation. If you load a simply supported beam to yield and keep going, the force will not increase as it deforms. Nor will the force decrease while the beam continues to bend. Since you keep pushing while the beam bends, you are exerting a force through a distance. Work is defined as a force exerted through a distance. It takes a certain amount of work to bend a beam to a shape.

Power is work divided by time. If you bend it slowly, such as in a hand pumped hydraulic press, the power is low. A powerful motor driven machine will do the same work in a short time. The total work will be the same, but the power is different because the time is different.

Lots of simplifying assumptions the above discussion, but the main point is valid.

I think that taking baby steps is probably to avoid kinking?

At some point, the force required is greater than the friction between the drive roller and steel, and the whole thing comes to a stop with the drive roller slipping. And it's difficult to fix if you roll it too far.

 

[Crazy Old Coot]

You are welcome.

Note that the very first bending ...
These links may guide you in a rough calculation of the needed bending and rolling forces:

https://www.mdpi.com/1996-1944/14/5/1204

https://www.harsle.com/how-to-calculate-the-bending-force-for-your-press-brake-machine

Thank you for this. It's a heck of a lot of maths for my old brain to process, but I'll try.

 

[Crazy Old Coot]

No. In the case of ring rolling, the "simply supported beam" has not failed....
Lots of simplifying assumptions the above discussion, but the main point is valid.


At some point, the force required is greater than the friction between the drive roller and steel, and the whole thing comes to a stop with the drive roller slipping. And it's difficult to fix if you roll it too far.

Thank you for your contribution.
I wasn't even considering the speed with which the first deformation takes place.
That is a separate system to the driving system.

However, as the material is fed into the rollers, it is akin to many continuous small manual advances and re-deforming. (I.e. repeating the initial bending process in a zillion micro steps.)

This involves distance and force, and time,

So, my initial question suggested that the feed speed and yielding force were the computational factors.
Does this somehow equate to the "many small manual advances and re-deformings?"

I have long since forgotten so much of the engineering theory of my youth... and cerebral paralysis sets in when faced with more than 100 words and two formulae!

 

[berkeman]

cerebral paralysis sets in when faced with more than 100 words and two formulae!

FYI, we do have medics available here...

 

[Crazy Old Coot]

Ha ha... a palaeontologist might be more appropriate!

 

[Baluncore]

Ha ha... a palaeontologist might be more appropriate!

I can help you there, and I have life experience.

In a roll forming line, the many small steps are taken by the many sequential stages. The total energy is the same.

 

[Crazy Old Coot]

Well, it now remains for me to acquire the gearbox that I have been offered and set up the geometry of rollers as per my rudimentary spreadsheet, and see how the real world compares to the numbers.
I can play with sprockets to get the real driving torque because speed is not an issue.

 

[Crazy Old Coot]

Thank you, guys, for your replies. I appreciate the time you took.