Calculate Flywheel KE and press Tonnage?

[AaronH]

All,

I am working on a project to roughly estimate the tonnage developed in a press that uses a flywheel and screw. The flywheel is attached to a fast lead screw which then drives the ram down. Given enough information I thought this would be a straight forward calc, but for some reason I can't come up with results that make sense. Here's what I have.

Moment of Inertia of flywheel and screw is estimated at 12,905 lbs-ft^2, weight of flywheel and screw is about 6070 lbs. I calcualted the Radius of Gyration at 8.2739 ft. The press strokes down about 18" in 1.55 seconds (therefore the ram is traveling at about .968 ft/sec at the moment of impact). I do not know the lead of the screw, but visually the flywheel makes about 1 and a half revolutions in one down stroke. So, we're talking about 1 rev/s.

The formula I am using is KE=W*V^2/2*g, where weight is in pounds, velocity is the tangential velocity at radius of gyration in ft/sec and g is 32.2 ft/s^2. I get 238,565 lb-ft of energy. The distance traveled after initial contact is approx .1 inches (.008333 ft). I then calculated force applied by dividing the KE by the distance traveled and I get 14,314 tons. The press is only rated at 1,000 tons. I am off by a factor of 10!! Anyone have any idea where my calcs went wrong?

Thanks in advance!
Aaron

p.s. I have attempted to attach my spreadsheet with calcs. We'll see how successful I am.

 

[AaronH]

Ok, as I suspected.... I failed the attachment test.

Here's attempt number B. I think I've got it this time.

 

[Compositepro]

Energy is force times distance. In terms of a press there is no direct link between stored energy and tonnage. Tonnage is limited by the design of the press structure but is controlled by the material that is being processed. The force is the resistance to being "worked". With a lever I can develop 1000 tons of force with my pinky.

 

[dvd]

It may be that you are assuming that the total kinetic energy of the flywheel is being absorbed. In practice, some small percentage of the kinetic energy is used to develop the tonnage of the press. I looked at your spreadsheet quickly and didn't notice that you had initial and final angular velocity of the flywheel, so this may be your problem, or not.

 

[AaronH]

C-Pro,

I'm not sure I follow your analogy. Tell me the length of the lever you are using and I can calculate the force you are applying with your pinky. If a known amount of energy is stored in a flywheel and that flywheel is brought to a complete stop over a known amount of distance and velocities and masses are known (or calcualted) then a correlation should be able to be made to applied force, no?

DVD,

In the design of our presses the flywheel is driven by friction part way through the stroke (both down and up). At bottom of stroke the flywheel comes to a complete stop, then rotation is reversed to lift the ram back up. I suppose it is possible that part of the up stroke is actually a bounce from the down stroke so not all of the flywheel energy is actually transfered into force applied on the down stroke. The whole cycle is 3 seconds +/-. There is no way to perceive a bounce in that time visually. I'm thinking we would have to disable the up stroke power from the machine and measure the height the ram achieves from the bounce. Then we could calculate that energy and subtract from the total flywheel energy to get a net energy that was actually used to generate a force. That is assuming that bounce is actually occuring.

The only other way I could think of comming at this one would be to convert the flywheel energy into a torque and factor that with the lead of the screw to calculate a linear force applied. This, of course, would be dependent on whether or not all energy is used on the down stroke.

Aaron

 

[Compositepro]

I've never seen a flywheel on a press reverse to move back to open (except if the press gets jammed). That would be contrary to the pupose of the flywheel - to store energy.

Would you expect the press to generate the same tonnage shearing a sheet of paper as opposed to 1/2" steel plate? How do you calculations account for that?

 

[AaronH]

These are screw forging presses. They don't look or function like your typical crank style press. Are you familiar with the Great Seal Press? Screw forging presses are very similar, but much larger and powered instead of manually operated. See link for a sketch of a screw press.

As far as paper versus 1/2" steel plate... That's an excellent question. I think the answer is that the linear force generated is calculated based on flywheel energy dissipated over a known time/distance. That is to say, when the ram is coming down there is zero tonnage. Nothing is opposing the the ram so no force is generated. All flywheel energy is conserved. Once the ram and top die comes in contact with the part a resistance is met. Now you begin to develop tonnage. As the part is worked, more flywheel energy is converted to linear force. The amount of energy used is based on the starting part geometry versus final part geometry and the force required to make said change. Once the two die halves come in contact with each other (the dies do have positive shut height stops) all remaining energy of the flywheel is converted to linear force and is absorbed completely by the machine base. Right now, I am not trying to figure out how much energy is put into the part versus how much is absorbed by the machine base. calcualting part forming energy would require advanced simulation software, I believe. I am basically looking for total energy (I think). I dunno, maybe I'm looking at it the wrong way. I do agree that the piece of paper will absorb very little of energy versus the 1/2" steel plate. I shall ponder this further.

Aaron

 

[desertfox]

Hi AaronH

Won't the tonnage be limited by the size of screw?
Anyway what the other posters are trying to say and actually you have said it yourself in your last post:-
The ram is moving down there is no force and by your calculation its moving at constant velocity ie no acceleration and no force, now introduce resistance to prevent ram moving down, now a force is generated due to deceleration,the amount of force is dependant on the resistance ie the ram will only impart the amount of force to overcome the resistance depending on whether its a thin sheet of paper or a piece of steel your calculation is assuming that all the energy is absorbed from the flywheel but as others have said it isn't.

regards

desertfox

 

[AaronH]

I believe the machine capabilites are determined not only by the screw, but also by the flywheel specifications, mass of the ram, and what the machine base will handle without failing.

I agree that the force imparted into the part being formed is dependent on the work being done to the part. However, there is more energy stored in the system than what is required by the forming operation. The flywheel does come to a stop, then reverses. The dies have a positive stop at shut height. Therefore, unless the ram is bouncing (rebounding), when the flywheel stop that means the flywheel has given up all of it's energy. Thus, whatever energy not used to form the part is put into the die stops. That energy would ultimately be absorbed by the machine base. Like I said, I'm not trying to figure out how much is absorbed by the forming process versus the machine base. I would need specialized FEA to calculate the energy required to form the part. These machines have a tonnage rating. There must be a mathmatical correlation between operating parameters versus tonnage.

During setup of the machine they will cycle the press without forming the part. The die stops will come in contact with each other and the flyhweel will stop and reverse as I have previously described. In that case, with no part all energy was absorbed by the machine. The machine generated a force that was trasmitted though the die stops. There must be a matmatical model to represent that force.

Aaron

 

[TVP]

Aaron,

There is an excellent section on Machine Tools for Metal Forming (Chapter 8) of the book Handbook of Metal Forming which describes the force and energy characteristics of screw presses. If you are serious about working in this area, I strongly encourage you or employer to pay $75 for this book-- it is the most comprehensive reference on the subject of metal forming, including coverage on all processes (sheet metal, hot forging, cold extrusion, etc.), materials, presses, etc. Another option would be to visit the engineering library of a university that owns this book. The following links will help you with either of these options:

http://www.sme.org/cgi-bin/get-item.pl?PI2288&2&SME&

http://www.amazon.com/Handbook-Metal-Forming-Kurt-Lange/dp/0872634574

http://www.worldcat.org/oclc/9971590&referer=brief_results

 

[Compositepro]

Interesting presses. Sort of like Ben Franklin's printing press. In the web site you provided is this statement:

The force of stopping the flywheel would produce a theoreticaly infinite force, however it is limited by the stretch of the heavy frame.

 

[AaronH]

C Pro,

Well, there you have it. I hadn't actually read that site, I just googled for a fancy pic. I had perviously considered as the impact time moves to zero your deceleration moves to infinity and thus force devloped would move to infinity. My calcs actually demonstrate that. So, in the end, the machine base will flex to absorb the left over energy up to it's elastic limit, then it will fail. I had thought about machine base failure, but I didn't think about flex as a means to dissipate energy. Makes perfect sense now. You said it earlier that it would be limited by the press structure. Seeing it worded differently with "flex" flipped the switch for me.

I guess in the end my calcs are working more or less ok. It's just a matter of what it is you are trying to describe. Knowing the overall tonnage that the flywheel can generate probably isn't as useful as I first thought. To a certain degree I can compare KE's of known situations. I think I could actually calculate tonnage to work a part if I have very precise velocity data of the flywheel. I would need to know velocity just before hitting the part and the velocity after the part working is complete. The second velocity (something greater than zero) must be calculated just before the positive stop of the die. Then, the delta V could be used to arrive to the part working tonnage.

TVP,

Thanks for the book reference. I will definately check that out.

Thanks for all of the input guys. I don't get to work on these types of things very often. It's nice to take the brain out for a walk now and then, even if to find out you've gone the long way around to end up right back where you started.

Aaron

 

[dvd]

Something about these doesn't make sense to me. If the flywheel stops every 3 seconds, then it needs to be accelerated back up to full speed every 1.5 seconds, assuming that the downward travel takes 1.5 seconds. I am not going to do the calc, but I wonder how much power it takes to accelerate the flywheel from rest back to full speed. It seems like there is no point in having a flywheel. If I was building this machine, I would look at sizing the drive motor to provide full torque.

I also would not let stop blocks on a die absorb the unused kinetic energy. The mathematical model that would describe that is most likely for a plastic deformation.

I could see some system that transferred clockwise rotation of one flywheel into counterclockwise rotation of the opposite side flywheel. But the system shows two smallish flywheels that appear to be in contact with the screw plate. Both flywheels on a common shaft turn in the same direction, so what gives? How does the screw plate contact both flywheels and have a friction drive? The flywheels would have to turn in opposite directions for the friction drive to work.

If there is anyone listening who can explain this press I would appreciate being enlightened.

 

[desertfox]

Hi dvd

Look at this site, the second press that is discussed looks like the one the op talks about.

http://www.shirepost.com/ShopTour.html

desertfox

 

[AaronH]

DVD,

The answer is this. The two large discs are both flywheels in themselves. They do share a common shaft and thus turn in the same direction. What you can't tell from the picture is that they can slide on the shaft. The shaft has a spline. Those drive discs are pushed by large pneumatic cylinders. The screw is driven by one disc at a time, not both. As you have figured out, two friction drive discs on either side of the flywheel in contact at the same time would not work. Another important fact is these types of presses are not generally designed to operate at the same cycle rates that a typical flywheel driven stamping press runs at. Generally, we're talking 30 seconds to a minute between cycles, not a few seconds.

The screw press cycle is as follows:

At the start of cycle the ram is at top of stroke, neither disc is contact with the screw flywheel, the screw flywheel is at rest, and the drive discs are spinning at their proper speed. The operator cycles the press. A large pneumatic cylinder pushes one of the drive discs in contact with the screw flywheel. That friction drive accelerates the screw flywheel to speed. At some point as determined by machine settings the air cylinder is retracted allowing the drive disc to disengage. The amount of time the drive disc is engaged determines the max speed at which the flywheel will spin. Being that the ram is coupled to the flywheel via a threaded shaft and nut the ram is moved downward at a rate determined by the lead of the screw. The screws are large (about 12" diameter) and have multiple thread starts, so they are very fast lead (about 12" lead). At the bottom of the stroke the die stops come in contact with each other and all energy remaining that wasn't used to form the part is transfered to the machine base. As the ram decelerates, so does the flywheel until both are at rest. This happens very fast! You get a loud bang and large vibration as evidence. Now, a different pneumatic cylinder is energized and the other drive disc is pushed over until it contacts the flywheel. It, by means of friction drive, accelerates the flywheel in the reverse direction, there by lifting the ram to top of stroke. This all happens in 3 seconds!!

There are some differences in presses. Our presses the screw and flywheel travel with the ram. So, the flywheel position relative to the centerline of the drive discs changes as the ram moves. This means the tangeantial velocity that the drive discs are accelerating the flywheel changes with ram position. When the ram is up the flywheel is very close to the centerline of the drive disc. This inherently gives you could torque to get the flywheel started spinning, but not a lot of speed. At the ram and flywheel drop the contact point on the drive disc moves closer and close to the rim of the drive disc. So, you are continuing to speed up the flywheel as long as you are in contact with the drive disc. Think Pi*D. There are presses that are designed with cone drives. In this design the flywheel and screw is stationary. The flywheel speed is then controlled by varying the speed of the drive cones via. a VFD.

The reason why the large drive discs are required is because it just isn't possible, we'll maybe practical is a better word, for a motor to generate those kind of energies in such a short amount of time. When a machine is started up it takes a good couple of minutes before the drive discs are up to full speed. You just couldn't afford to wait that long in between strokes. Not only that, it's inefficient. It would put a pretty big hurt on the light bill.

Aaron

 

[marcopress]

All,
I'm working on a project were we are producing a part on a Platarg mechanical trasfer press (it's like a Waterbury farrel). I can produce the component one at a time but I can't run it in automatic because the press energy is not enough. I have no idea how much energy the press really has and looking on the net tells me that the builder (who is no longer in buisness) should have told us the energy of the model.
Can anyone help me figure out how much energy I have available (formulas and spreadsheets are really appreciated)?

 

[AaronH]

Marco,

While I am not familiar with your particular press, here is what I can tell you. Mechanical presses, like crank presses, will have a tonnage curve graph. That graph would likely have been included in the manual for the machine. If it's a hydraulic press, the max tonnage is calculated by multiplying the maximum hydraulic pressure generated by the power unit times the area of the rams' hydraulic cylinder times the number of cylinders. You also might look around on the net and see if you can locate another machine that's the same as yours and see if the owner would be willing to pass along any info they have. I have some old equipment myself that did not come with any doccumentation when I aquired the machine. I've been fairly successful finding shops running the same machine who have been very helpful in providing any information they had.

Aaron