How much force to hold down a steel sheet?

[carter213]

I have a process where an uncoiled steel strip travels down a conveyor line. The strip width varies (30~80") as well as the thickness (.25~.625"). The strip has some natural waviness to it, so it does not always rest perfectly flat against the conveyor line.

There is one point on the line where I need to hold the strip flat against the conveyor using a cylindrical roller on top of the strip across its width. The roller does not need to permanently deform the strip to lay flat. A temporary elastic deformation will suffice.

Is there any way to calculate how much force the roller needs to apply to the strip to hold it flat momentarily? I'm thinking that it will involve the modulus of rigidity, but I want to get the opinion of more learned men than I before proceeding.

 

[corus]

Treat it as a beam, simply supported at its ends, and calculate the load required to give you that deflection to make it flat. ie. from

delta = (PL^3)/(48EI)

Where:
delta = Deflection
P = Load
E = Modulus of Elasticity (Young's Modulus)
I = Moment of Inertia (width.thickness^3/12

corus

 

[MiketheEngineer]

Pretty much agree with corus - but you might check with manufacturers of "level" lines that acutally flatten steel coils. Check out Braner.

The force will not be all that much if you just want to temporarily level it. The 5/8'' might be a bit tough - especially if stainless or other type material

 

[zekeman]

You are better off with 2 tandem rollers separated at a distance less than 1/2 the pitch of the waviness; then you have at least the distance from the valley to the peak of the deformation.
Then you can apply the deformation formulas above using a deflection equal to the maximum delta of the strip between the rollers and the beam distance equal to the separation distance of the rollers.

 

[corus]

Levellers plastically deform a coil but still use beam deflection as the basis of the theory. Remember, of course, that you also have self weight which will remove some of the deflection so the force you'll calculate from the above will be higher than actually needed. Very thin plate may not need any force at all as it'll just fall flat.

corus