How to calculate required force, to achieve secondary (open type) hemming operation (in sheet metal Indus.)?

Usualy Hemming operation is done in two stages on press brake.


1) 30 Degree Bend (Primary Hemming = Flanging)


2) Hemming Itself (secondary Hemming.)How to calculate required force, to achieve secondary (open type) hemming operation (in sheet metal Indus.)?
There are some charts on:


http://www.krausco.biz/tonagechart.html





which give a tonnage based on several factors. Once you have a method for actually calculating, you could check your answer against this site:





exerpt:


';RULE OF THUMB formula for making multiple bends on a press brake. For shape shown, in mild steel with radii equal to metal thickness unless otherwise noted.





Multiply Metal Thickness by factor = Tons per Foot





Example for bottom bending


16ga (.062) Mild Steel. .062 x 150 = 9.3 tons per foot.';





Hope this helps!