Force required for cutting sheet metal

[eng_taha_a]

hi every one
can in you help me in finding the required force to cut sheet metal electrically as shown blew

 

[Asymptotic]

Sheet metal shearing force calculation

 

[JBA]

One problem with the design shown is that the blade angle to the cutting surface is not constant (as it is in conventional metal shears); and, as a result the required cutting force will increase as the cutting proceeds through the shearing process.

 

[CWatters]

Think the worse case will be just as the blade starts to cut the last bit. Make a drawing of the blade in that position and you can work out how much of the blade is still in the metal (and how much has already cut through it). Put that length into the equation for the shear force.

 

[jrmichler]

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

 

[CWatters]

How do they define the area being sheared? Is it Length * Clearance? That would imply the force tends to zero as the clearance tends to zero.

 

[jrmichler]

Length of material in the cut * material thickness. The discussion implies that cutting force is affected by the clearance, although they do not provide any numbers.

 

[eng_taha_a]

thanks for reply ,but if the sheet to be cut is 1.25 mm thickness and 1300 mm and have UTS of 350 n/mm2 and the angle blade is 15 degree
so what is the cutting force required ?
will it be 420875 N or less or more ?

 

[CWatters]

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

 

[JBA]

eng_taha_a: You have been given the necessary information to perform the calculation so it is time for you to do so, rather than to ask for someone else to perform this for you.

 

[eng_taha_a]

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

CAN YOU MAKE IT MORE CLEAR
SORRY

 

[CWatters]

When you start cutting a sheet only a small part of the blade (near the hinge) makes contact. As the cut continues more of the blade will be cutting. The part doing the cutting moves away from the hinge. Eventually it reaches the edge of the sheet that is furthest from the hinge. In that position some of the blade will have gone right through the metal so it will no longer be cutting. The force required on the handle will increase as the cutting point moves away from the hinge.

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.

 

[jrmichler]

When you are not sure what to do (especially on this forum), make a sketch with your best guess, and write down your thoughts about it. Discuss where and why you are not sure. If you have a second best guess, sketch that, label it Second Guess, and discuss it also. That way, you are showing us that you are trying, what you do and do not understand, and where we are having difficulty communicating with you.

 

[eng_taha_a]

this is my best guess
wish you help me more
thanks

 

[CWatters]

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

 

[eng_taha_a]

the area in blue will change from start from 1300 mm at the start of cutting process and ends at the end of the cutting process and the change will be in mm , and according to this equation the value of shear force will change as the area changes ?

 

[CWatters]

I've had a closer look at your drawing and there is a small mistake. We want the area shown below in blue. See how the moving blade has reached the end of the sheet. In this position it is cutting furthest from the hinge.

 

[eng_taha_a]

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 ?

 

[CWatters]

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.

 

[jrmichler]

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. I did not find any calculation for this particular case, so we have to do a SWAG (that's SCIENTIFIC Wild Assed Guess). I suggest using 1.5 mm X 10.7 mm as the area, and adding an additional factor to account for the additional work to bend the cutoff piece down. Make the additional factor about 1.4, and the calculation becomes 1.5 X 10.7 X 370 = 5900 N = 1300 lbs.

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.

 

[CWatters]

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.

 

[JBA]

With respect to the above comment on the lever length vs. the distance from the pivot to the point of of cutting. Please explain how you intend to "electrically" power this cutter so the mechanical advantage issues can be properly addressed.

 

[eng_taha_a]

this is the design of the shear machine

 

[CWatters]

Still missing the dimensions needed to check the area calculation.

 

[eng_taha_a]

what dimension you need ?

 

[CWatters]

See #19.

The cutting edge of the top blade is not in line with the hinge. When the cutter is down the edge of the blade is below the hinge. How far below?

 

[JBA]

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

 

[eng_taha_a]

this dimension is about 100 mm

 

[eng_taha_a]

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

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

 

[Baluncore]

Get a copy of; “The Science and Engineering of Cutting”. 2011. By Tony Atkins.
ISBN 978-0-7506-8531-3
See chapter 5; Slice-Push Ratio.
Oblique Cutting and Curved Blades, Scissors, Guillotining and Drilling.

 

[CWatters]

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.