Force required to bend 5" plate x 10' wide

[pq102]

I want to cold roll a cylinder (on a pyramid plate bending rolls):

Plate thckness: 5"
Plate width: 10'0"
Material: Mild Steel
Diameter of cylinder to be rolled: 13'0"

The machine I intend to use is capable (according to the manual) of 3" thickness over 12'0" However, the build of the machine suggest that it should do considerably more.

Can anyone help ?

The machine is mechanical driven bottom rolls (rolling) 190HP. Pinch of the bottom rolls is mechanical 75 HP each.

All rolls are 800mm dia.

Working time is not a consideration as long as we can do the job.

Appreciate any advise.

Peter Quigley

 

[metengr]

There is an excellent reference document that pertains to cold forming of steel plates. The source of this information is from Bethlehem Lukens. The equation for determining press load for roll forming is as follows;

P = (0.833 * U * t^2 * L)/W

where U = ultimate tensile strength Ksi
t = plate thickness (in)
L = length of plate to bend (inches)
W = width of die opening (inches)

This document can be obtained from the web site below;

http://www.intlsteel.com/content/products/prod_plate_lit.aspx

 

[JStephen]

I'm not certain where the 0.833*U comes from in that equation. But note particularly, that if you decrease the length to 5/6 of the original value, and increase the thickness to 5/3 of the original value, the required force goes up by a factor of 2.3.