Solving Force Reduction in Sheet Metal Shearing

[vin300]

1.How to find the force required in shearing sheet metal out by a cutter that does not cut at ninety degrees, but the end is inclined slightly at 90 + x degrees? The usual form is shearing stress× plate thickness× cutting length. In the new case, the tool impinges progressively from a single point over the entire area, so that the stress is concentrated at different regions at different times, thus reducing force. What would be the expression?

2. I am given an expression f×t×l÷s, where s is depth of shear, but it looks incorrect as well as dimensionally wrong.

 

[JBA]

The original formula you are given without the "thickness" is actually for the force required to shear a sheet of material with a blade that is parallel to the surface of the material.

For a blade that is perpendicular (90 degrees) to the sheet, then the Force equation is shearing stress x thickness without the length and your case falls between those two.

In determining the force required for the inclined method, draw yourself a picture of a right triangle with the cutting edge as its hypotenuse at a 90 + x degree angle from the vertical, cutting a very thick piece of material and remember that the cutting only occurs along the line where blade edge contacts the plate. After that, it is strictly a plane geometry problem.