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Deformation in sheared material for coupon development 1

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Citizen of the Stars

Citizen of the Stars

Materials
Apr 11, 2023
3
I am tasked with purchasing a shearing machine that will help my lab cut coupons for tensile and quality testing but I am scared to death that I'll buy a machine that will introduce significant bending, chamfering, or twisting.

I need the coupons that I cut to be flat or as close to flat without introducing work hardening or deflection. I will be working with mostly aluminum Series 3000, 5000, and 6000 with different heat treatments. I believe that the most aggressive aluminum I work with consistently would be 6061-T6 @ 3.5mm thickness, 38.1mm x 101.6mm.

I've just been cutting these up on a band saw and grinding away (slowly!) the burrs, which takes up a lot of my day.

The machines I'm looking at have an adjustable rake angle which I can move to better suit the materials ultimate shear strength but the higher the angle, the higher probability of bending.

So my questions today are:

1) Would a machine with a higher tonnage effectively shear a flat sheet at a lower rake angle? Is that line of thinking correct?

2) When calculating the force needed to shear, do I factor in the cross sectional area of the WHOLE sheet or just at the length of blade contact?

3) Are there any tricks of the trade that I can incorporate to reduce bending and work hardening?

Any help from fellow coupon makers would be amazing. Thanks in advance.

This is also my first post on this forum. Let me know anything I'm doing wrong and I will correct it.

 
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I am far more experienced with forming than cutting so I can't help you much with questions 1 & 3 but I may for Q2. So lets say you have a 6ft long sheet, but you are only actually cutting 3ft of it, and you are wondering which number to use. I may be misunderstanding the question but I would think 3ft would be the obvious answer if you are only making a 3ft cut, the rest of the sheet is irrelevant. Feel free to correct me if I am misunderstanding the scenario.
 
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2) When calculating the force needed to shear, do I factor in the cross sectional area of the WHOLE sheet or just at the length of blade contact?

To clarify a little better, lets use that 3ft sheet as an example:

On a sheet of Aluminum that is 3ft in length, 3.5mm thick, and in the throat of the shear press, from how I'm reading the formulas is that the entire area of the cross sectional length of the sheet (3.5mm x 914.4mm[3ft]) needs to be in the equation. To me, a layman, that doesn't make sense since the blade is only going to be in contact with a fraction of the aluminum plate. So shouldn't it be the length of the blade that is in contact with the sheet as the cutting blade comes down?

Hope that helps
 
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The shears that I have used have very little contact angle (across the width).
So the blade was still within the material at the starting edge while it was contacting the trailing edge.
So you use the entire width of cut.
You have to cold work the edges to shear, that is how it works.
Either grind the edges after or use large enough coupons so that the shallow worked material does not change the measured properties.
Most shops that do a lot of testing either punch the coupons in a die, cut them on an NC mill, or waterjet them.
We have an old mill that just cuts coupons in our lab.


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P.E. Metallurgy, consulting work welcomed
 
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After a few quick google searches, it looks like there are a couple styles of shears. One is basically a giant pair of scissors where the blade makes contact on one side and then cuts across as it goes down. The other style is a straight blade the makes contact with the full 3ft width all at once. The forces in these two scenarios are going to be drastically different so it will depend on your machine.

The 3.5 x 914.4 you gave I think would be correct for the full width cut, while the blade is only cutting through part of that cross section at any given moment, the deformation, friction and other forces are still going to be at play, so you will likely need to stick to that full cross section equation, at least as a rule of thumb.

If you have the other style of shear, the cross section and thus cutting force will be significantly smaller, but may be harder to measure as the amount of blade that is engaging the material at a time will depend on a number of factors that I am not qualified to speak on.
 
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@EdStainless - Oof.. punching coupons. is there a lot of waste? I have also been using a Mill to get what I need as well. usually when the tolerances are tight and I need to be exact.

@lucky-guesser - Ah, yeah. I believe the first shears you described are called Guillotine shears(I think). They are normally mechanical/manual shears that have a huge curve to them which assists in the hand operated shearing action. I have a smaller style of these shears in my machine shop and they bend the coupons quite a lot. Its nice for trimming small sections on thinner materials but horrible for the coupon sizes that I need.
_________________________

It's looking like I might need to find someone with these machines so I can demo them. I doubt that anyone's going to let me waltz into their shop and throw a 1ft x 1ft sheet of 3.5mm 6000 AL into their 14 gauge shear press. Although that would be kinda fun if I can run fast enough not to get caught.
 
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Didn't understand how guillotine shears could optimize bending operations.
Only one is actual - to cut without burrs because burrs reduce the life of tooling for bending.
 
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citizen

discuss with sales engineer that sells these shears. he will set you up with the correct equipment. and get three different quotes from three different suppliers.
also an old technique is to use thin carbon 1018 sheets as a backing to prevent or minimize the bend and burr.
an other tip is to investigate the exact clearance required between the shear blade, an old timer taught me this tip when I was young, but to many years have passed.
I don't remember the specifics. again talk to the sales engineers.
I use to make and test tensile coupons when I was young. but I would get something like a router, or mill that is cnc controlled to mill the coupons exact.
 
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Today I would waterjet these.
I have been to SS mills where all of the coupons are stamped (fineblanked) using an electrical punch press.
The counterpressure of the fineblanking keeps things very flat.
They had 2 or 3 set up for various thickness/strength ranges.
Of course they are doing 200-400/day.
The full width panel comes into the lab (24"-60" wide).
A piece with the label is lopped off in a shear to a std size.
They are loaded into a feed, marked, stamped.
A robot picks them up, reads the information marked on them, hold it for a hardness test, and loads them into the tensile machine.

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P.E. Metallurgy, consulting work welcomed
 
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ed
Kindly please advise how close of tolerance water jet can hold with new equipment
 
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y'd better to have a blade clearance adjustable machine. The basic idea is to balance between shear and fracture. Too tight clearances result in deeper shear zones and smaller fracture zones, if the clearance is too wide, the sheet will be torn apart rather than sheared. This can be evaluated the cross section, shear zone is burnished and shiny, fractured zone is dull because compressive forces exceed the ultimate strength of the sheet.

Rule of thumb is to set the tolerance 4-8% of sheet thickness.
 
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Waterjet sizes will vary slightly as you change material hardness and thickness.
Not a big deal as you will hand smooth the edges and then take real measurements before testing I hope.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed
 
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