How to calculate the force necessary to achieve the secondary (open type) Hemming operation (in the sheet metal of the Indus.)?
Normally Hemming operation is performed in two stages the press brake. 1) 30 Degree Bend (Elementary Hemming = Flanging) 2) Hemming Itself (secondary Hemming.)
There are some Graphs on: http://www.krausco.biz/tonagechart.html giving a tonnage based on several factors. Once you have a method to calculate effectively, you can check response against this site: excerpt: "golden rule for making the formula multiple curves in a press brake. It is shown, in mild steel with radius equal to the thickness of metal unless otherwise noted. Multiplying by the thickness of the metal factor Pie tons = eg for bending the bottom 16GA (.062) of mild steel. 062 X 150 = 9.3 tons per foot. "I hope this helps!
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