Force to cold-bend steel plate

[DavidCR]

I need help to estimate how much force is needed to bend a plate. It is for a 11m diameter welded surge tank for a hydroelectric project.

Plate material: Carbon steel, tensile strenght: 70000 psi (There is also an option of using a 50000 psi plate)
Thickness: 50mm (1.97").
Wide of plate 3m (118").
Diameter (Bending diameter): 11m (433")

The guy who is asking this says that they think to perform this in cold with a roller over the plate between two rollers spaced 0.3m (12"), but he does not know what is the force needed to be sure is they can do it.

So I´d appreciate a lot if somebody can help me with a formula to calculate the force required, or with an estimation of the force also is somebody wants to help with recomendations on how to bend the plate in an easy way, that would be welcome.

Hi and blessings to everybody there.

 

[sreid]

Modeled as a beam with a point force at the center, the force to reach the yeild stress would be over one million lb.

 

[metengr]

DavidCr;
I can try to take a stab at this from a "practical" view. The force required in metal working operations is basically the flow stress of material X the deformation area because stress X area = force.

I would "estimate" that the flow stress for this plate material is going to be somewhere around 45 Ksi, as any good number that is above yield and allows for some strain hardening during cold working.

The contact area will be the width of the roller contact on the surface of the plate X the thickness of material that is being deformed.

This is a crude approach but could at least get you in the ball park in terms of load.

To reduce load, you could heat the plate and hot bend to take advantage of the reduced flow stress because the yield strength will decrease as a function of increasing metal temperature.

Good Luck

 

[deanc]

What you are trying to do is not that uncommon. The radius is the key as long as the pinch rolls will handle the thickness. Call around for quotes-you may be surprized.

http://www.bertschrolls.com

http://www.roundo.com

 

[JoeTank]

DavidCR,

At the risk of embarrassing myself, I'll take a whack it an answer. If I remenber my strength of materials class correctly, M=EI/R. Using E=30,000,000 psi, I=t^3/12 and R=217 inch radius... I obtain the moment necessary to elastically bend a 2 inch thick plate to be 92,400 inch-pounds per inch of plate width. For a 118 wide plate the total moment required is 10,903,200 inch-pounds. For the rolls to be spaced at 12 inches c-c a roll force of 3,634,400 pounds is required. I used M=PL/4. This equates to a roll force of 1820 tons.

However, the 50 ksi plate will begin to yield at a roll force of about 650 tons or about 1/3 of the force noted above. This suggests to me that a roll force much lower that 1820 tons will be required. I'm just not smart enough to figure out the rest, but I would guess that a fully plastic moment would be enough to form the plate. This would be occur at a roll force of about 1000 tons.

Now, please don't tell me that this was a homework problem.

Steve Braune
Tank Industry Consultants

http://www.tankindustry.com

 

[metengr]

Here is a web site from one of the major steel companies that specifically addresses press load requirements for forming steel plate. The equation that is referenced is as follows;

Press load = 0.833 X Ultimate Tensile Strength X thickness ^2 X Length/width of the die opening

dimensions are in inches.

The web site is below. The attachment is "Guidelines for Fabricating and Processing Plate Steel". This is excellent reference material to keep handy.

http://www.intlsteel.com/PDFs/products/duracorravail.pdf