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Coil hold down forces and energy

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Christian62

Materials
May 21, 2002
17
Hello Fellow mechanical eng.
I need an easy way to calculate the hold down pressure or strapping strength required to keep a coil wrapped.

The variables are: Coil diameter, Thickness of material, Width of material, Yield strength, Ultimate strength, Modulus of elasticity, Poisson's ratio, Friction coefficient.

I know It's a lot but maybe someone has created an Excel macro for that or I could if I'm given some good math on the problem.

Extra: It would be nice to get the residual stress also.. but that's because I'm curious.


Christian Filion
Atlas Stainless Steels
Supervisor Laboratories
 
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You have not identify the problem well. What type of a wrapped coil do you want strapped and why must it be strapped? Is it in a container to be pressurized? Is it to be transported or lifted? What loads are imposing on the coil in order to strap it? Need more explicit details
 
Seems like this problem is somewhat similar to a power spring (sometimes called a clock spring). Maybe you could try to modify some of the equations for used that?
 
Well,
We take 0.250" thick by usually 48" wide strip of stainless steels and run it throw a cold mill to bring it to a lower thickness and improve its mechanical properties (yield, UTS,...). At the end of the mill the strip is wrapped around a mandrel and builds up until the strip is all processed (we have coil that are about 20 tones and can build up to 69" in diameter). At the end of the wrapping there is a set of rollers that pushes on the processed strip to keep the end of the strip down. This hold down forces keeps the coil tight while the operator straps the coil. The coil is then lifted to the next operation. Did I answer your question ?
 
I could simplify the problem further (maybe to much?)

We process flat material so it would be more like a spring clock (has suggested by Rich) with the last wrap functional. The other wrappes would not be functional because the internal friction of the steel keeps it from pulling. So I am faced with one loop of flat "spring" steel pressed against a mandrel. How do I keep this flat spring against the mandrel ?
 
Hi Christian!
If you try to solve your problem theoreticaly you can spend lots of time, apply different theoretical approach, but the results will never be correct.
The problem is that for the same type of coil, material etc, the spring effect is different and the way the line is set up - close loop or open, line speed, type of operation -stop and go or continious, makes segnificant difference. That's why the coils are holded using air or hydraulic cylinders with pressure regulating valves. This way the operator can adjust the pressure accordingly.
For example 15,000Lb - 40,000Lb are buid with holdon hydrauilc cylinder 3 1/2-4" bore and hydraulic motor for the Holdon Roll drive
Dako
 
This is what I would do: observe carefully the amount of projecting stock at completion of process that is pushed down by the rollers to keep the coil tight; then determine the ammount of tip deflection and use the beam formula to determine the applied end force; make several such observations and you may be able to come up with an average value for each stock thickness.
There probably will a tangential force to keep the stock from unravelling and that value can be guesstimated from the force needed to keep the coild tight during the process.
I dont think that you can bank on the frictional force created by the normal force calculated by the beam formula.
The coil is probably oily diminishing any frictional gain therefore you would need the tangential force.


 
Thank you all for your suggestions,

I've done some theoretical calculations that are hmm questionable. I might finally go to an instrumented piston or ... I'm still looking.
As soon as I get some good results I'll let you know. (I'll look at the deflection)

Christian

P.S. if you ear anything, feel free to inform us.
 
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