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- Thread starter eng_taha_a
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Asymptotic

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JBA

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CWatters

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jrmichler

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The clearance between punch and die has a significant effect on cut quality. Note that they assume both punch and die have sharp cutting edges.

That same source, Manufacturing Processes for Engineering Materials, by Kalpakjian, estimates peak punch force at 0.7 X UTS X A, where:

UTS = ultimate tensile strength

A = area being sheared

That same source, Manufacturing Processes for Engineering Materials, by Kalpakjian, estimates peak punch force at 0.7 X UTS X A, where:

UTS = ultimate tensile strength

A = area being sheared

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CWatters

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jrmichler

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eng_taha_a

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so what is the cutting force required ?

will it be 420875 N or less or more ?

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CWatters

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Make the drawing I suggested in #4 to work out the length of the blade in the metal.

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JBA

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eng_taha_a

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CAN YOU MAKE IT MORE CLEARMake the drawing I suggested in #4 to work out the length of the blade in the metal.

SORRY

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CWatters

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Draw the blade at the point where the cut just reaches the edge of the sheet furthest from the hinge.

Post a sketch and I will tell you if it is correct.

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jrmichler

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CWatters

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Yes that looks correct. Now think about the area in blue and the equation in post #5.

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eng_taha_a

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CWatters

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CWatters

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but if so the area of this triangle is too small because the sheet thickness is 1.5 mm so if the area is 8 mm2, UTS is 370 n/mm2 then the force = 0.7*370*8 = 2072 N is it right ?

I cannot check your calculation without more dimensions. For example I need the height of the sheet above the hinge and the depth of the blade below the hinge.

If the area of the triangle is 8mm2 and the height is 1.5mm then the base must be 10.7mm. That means the blade angle in that position is about...

Tan^-1(1.5/10.7) = 8 degrees.

The force you calculated would be the force required at the cutting point which is roughly 1300mm from the hinge. The force on the blade handle would be less due to leverage.

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jrmichler

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In a case like this, where we do not have an exact calculation, we have to rely on professional judgement. We also want to be conservative, but not ridiculously so. I would design for a cutting force of 2000 lbs = 9000 N.

Design all parts so that the deflection under load will not cause the moving blade to spring away from the fixed blade (see the diagrams in Post #5). I do not have a calculation for the sideways force pushing the moving blade away from the fixed blade, so I'll make a WAG and guess it to be half the cutting force, or 1000 lbs = 4500 N. The moving blade should not spring away from the fixed blade more than 10% of the sheet thickness, or 0.15 mm = 0.006" under a 2000 lb cutting force.

The above numbers seem realistic based on my experience attempting to cut 16 gauge sheet metal with a pair of tin snips. It worked for a very small piece, but on a larger piece I gave up and used a hacksaw.

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CWatters

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The equation I quoted from Kalpakjian is for the cutting edge of the moving blade parallel to the cutting edge of the fixed blade, not the inclined cut here.

The angle is pretty shallow so might not be far off.

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JBA

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CWatters

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Still missing the dimensions needed to check the area calculation.

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eng_taha_a

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what dimension you need ?

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CWatters

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JBA

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Also please add to your drawing the electric actuator you are envisioning.

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eng_taha_a

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this dimension is about 100 mm

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eng_taha_a

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this is the electric motor vision and ,but not the final decisionAlso please add to your drawing the electric actuator you are envisioning.

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Baluncore

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ISBN 978-0-7506-8531-3

See chapter 5; Slice-Push Ratio.

Oblique Cutting and Curved Blades, Scissors, Guillotining and Drilling.

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CWatters

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this dimension is about 100 mm

So at the end of a 1000mm cut I make the blade angle about 5.8 degrees, the length of the cut about 15mm and the triangular area under the blade about 11sq mm.

The eqns linked to by Asymptotic in #2 and jrmichler in #5 are similar and suggest the cutting force is some factor between 0.7 and 1.5 multiplied by UST multiplied by the area, so somewhere between..

0.7*350*11 = 2700N

and

1.5*350*11 = 5800N

Lets call it 6,000N.

will it be 420,875 N or less or more ?

The force required is considerably less.

The torque required is the force multiplied by the length of the blade so 6000 * 1m = 6000 Newton Meters.

this is the electric motor vision and ,but not the final decision

I notice that the arm that connects the electric actuator to the blade is shorter than the blade. Let's call the length of the actuator arm "L" (meters). Then the force the actuator would have to produce is: 6000/L Newtons.

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