Heat Treating for Flex

Flexibility isn't determined by HT. It's determined by geometry.

ETA: grind your knife, then flex it side to side, like you would when using it. If it has the proper geometry it will snap back straight. If it stays bent, you need to change your geometry.
The point that it bent, would be the point it at which it broke, if you had done the HT.
 
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I'm sorry, but Mr. McPherson is wrong in his explanation.

Flex is a property of geometry. It is the thinness and shape of a blade the determines how much it will flex. Unhardened steel of the identical dimensions will flex as much as fully hardened steel. The difference where HT comes in is whether it will return to straight, take a set ( permanently bend), or break. This is where steel selection and HT regime are important. Tempering on the higher side will lower the hardness, but increase the amount of flex allowed without a break. This is why higher tempers are often called a "spring temper".

For a filet knife in CPM154, try this HT regime:
pre-heat/equalize at 1400F for 10 minutes
Ramp to 1950F and soak for 30 minutes
plate quench and allow to cool to room temp in the plates
temper at 400F for 1 hour
cryo in LN for 6 hours
temper at 500F for 1 hour
 
I think I was trying to say the same thing... sorry if I was incorrect or unclear.

Stacy,
Just for the sake of clarity and to make sure I don't give people bad advise anymore, here's how I understand things. Please correct me if I am wrong!

So you have three pieces of the same steel, ground exactly the same way. One is fully soft, one is fully hardened, and one is tempered to a spring temper.

When you clamp them in a vice and start bending they all bend the same up to a certain point, let's say 45* just for a round number.
At 45 degrees, they all take a permanent deformation.

This is where my understanding is a bit more fuzzy...

The soft one bends and stays bent at that angle, the hard one snaps and the tempered one returns to almost straight, but still has deformed?

Thanks and sorry for the bad info!
 
A.McPherson said:
Flexibility isn't determined by HT. It's determined by geometry.

ETA: grind your knife, then flex it side to side, like you would when using it. If it has the proper geometry it will snap back straight. If it stays bent, you need to change your geometry.
The point that it bent, would be the point it at which it broke, if you had done the HT.
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I have heard something like this many times yet still don't get it.

If you have a thin knife made of 52100 have it hardened without tempering with as quenched hardness of 66HRC and another one that has been triple tempered to 59HRC. Do you really think these who knife going to flex the same? I doubt the first knife gonna break as soon as you try to flex it.

So saying flexibility isn't determined by HT doesn't sound make sense for me.
 
shqxk said:
I have heard something like this many times yet still don't get it.

If you have a thin knife made of 52100 have it hardened without tempering with as quenched hardness of 66HRC and another one that has been triple tempered to 59HRC. Do you really think these who knife going to flex the same? I doubt the first knife gonna break as soon as you try to flex it.

So saying flexibility isn't determined by HT doesn't sound make sense for me.
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I think what is being confused is flexibility is actually rigidity. A simple way to understand it is like such. Take your 52100 as an exzample. One at 66RC and another at 59rc and another that's annealed and soft. Thy are identical except the heat treat. clamp them to the edge of a bench so exzactly the same amount of each blade is sticking out. Now hang a weight on the end of the blade. All 3 blades will flex the exzact same amount. Hardness does not change how rigid the steel is, it just changes its elasticity. But as soon as we exceed the steels elasticity we move into plastic deformation. The state where the steel is permently deformed is call its yeald strength. If you keep going past that point it will fracture and this is called tensile strength or ultimate tensile strength. All we do by heat treating is move these "points" around. We as knife makers are going for the perfect Balance between yeald strength and tensile strength. But like I said above, HT will not change how much the blade flexes with X weight as long as we stay within the elastic zone.
 
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BCC/ferrite has different bending property/mechanical than BCT/martensite. At simplistic view - hardened steel has combination of majority BCT% and minority BCC%. Keep it simple - Higher HRC = increased BCT%, decreased BCC%.

* RA/FCC has unidirectional bending behavior similar to BCC... let's ignore this complexity ;)

cH5o466.jpg


Let's use common sense to interpret 3 HRC points at 3000 MPa load

54rc: Yieded, deflected total ~7 mm. This sample is permanently bent (some mm).

58rc: Deflected total ~4 mm. This sample returned straight since load is below bend yield strength 3.2GPa.

62rc: Deflected total ~1.5 mm. This sample returned straight since load is below bend yield strength 3.8GPa.

At 5000 MPa load

54rc: Fractured, deflected beyond 10 mm. This sample is broken.

58rc: Deflected total ~6.5 mm. This sample stays bent (~1-2mm).

62rc: Deflected total ~4 mm. This sample stays bent (~1-2mm)
 
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just so your not worried i have a show off knife that is cpm154 filet knife its rc62 and i can bend it no problem 90degrees
 
I have the greatest respect for you bluntcut and your expertise in cutlery steels, but I just can’t read that explanation in that graph. I can’t help but think that you may be misinterpreting what they are trying to show. I’ll also say that it isn’t real obvious to me either, but I will have a go:

@54Rc – fractured after it had bent by a total of ~9mm plus a smidge, at some applied load which is unspecified but which calculates back in a simplistic way to an ultimate stress of 5000MPa in the inner and outermost imaginary fibres. Of that 9mm, 6.5mm took place after yield, in the plastic zone. So, if you had stopped bending just before it broke, it would have sprung back about 2.5mm leaving you with a 6.5mm permanent bend. It first yielded, or started to take a set at a force or load equating to 2500MPa, so about half the load that it broke at.

@58Rc – ditto but final bend before fracture of about 7mm (equating to 5600MPa in the outermost fibres), of which slightly more than 4mm was permanent. It first yielded at a load equating to 3200MPa or almost 30% more than that which it took to leave the 54Rc sample with a (tiny) bend but then broke at only 12% higher force.

@62Rc – same again but a final deflection of only 5mm (maybe 5700MPa) and only 2mm of which was permanent. This one didn’t start to take a set until a load which equated to 4000MPa or 60% higher than the 54Rc sample and 25% higher than the 58Rc one, but it broke at only a little if any higher force than the 58Rc one.

Everyone saying that the harder material in the same geometry takes the same load to bend the same amount is totally correct up to the point at which the permanent set starts. For these particular samples around 2.5 to 3mm. Up to that time, they would be indistinguishable. After that point, the softer one will bend further for a given load, thus seeming less stiff and most people will realise they have pushed too hard, like when you strip a thread, and it will then stay somewhat bent whereas the harder one will continue to push back just as stiffly as it did and end up putting up with a higher load even though it will, in this case, actually finally break with less total deflection. But by that time you will have been able to flex the 62Rc sample about 60% further than the 54Rc one went before it was permanently bent. I guess it may depend on your criteria for failure as to which you prefer.
 
Thanks. I like your analysis.

Vanadis4E plasticity range is around 3.8mm, thus maximum bend (when almost fracture-load is removed) is 3.8mm. 'Bend Yield Strength' line is corresponding to 'Plastic Deformation' line, saying at that load, it will bend that much, if higher load then steel will yield[=enter (beyond elastic range) into plasticity zone. e.g. 62rc take 3.8GPa to bend 2.2mm, same 3.9GPa load will bend 54rc 7.5mm and it won't return straight (maybe 1 mm permanent bend).

'Bend Fracture Strength' corresponds to 'Total Deflection' = it takes this much load to bend this far.

I've done quite a bit of bend tests. Here is one of my few public videos bend testing

sandgrouper said:
I have the greatest respect for you bluntcut and your expertise in cutlery steels, but I just can’t read that explanation in that graph. I can’t help but think that you may be misinterpreting what they are trying to show. I’ll also say that it isn’t real obvious to me either, but I will have a go:

@54Rc – fractured after it had bent by a total of ~9mm plus a smidge, at some applied load which is unspecified but which calculates back in a simplistic way to an ultimate stress of 5000MPa in the inner and outermost imaginary fibres. Of that 9mm, 6.5mm took place after yield, in the plastic zone. So, if you had stopped bending just before it broke, it would have sprung back about 2.5mm leaving you with a 6.5mm permanent bend. It first yielded, or started to take a set at a force or load equating to 2500MPa, so about half the load that it broke at.

@58Rc – ditto but final bend before fracture of about 7mm (equating to 5600MPa in the outermost fibres), of which slightly more than 4mm was permanent. It first yielded at a load equating to 3200MPa or almost 30% more than that which it took to leave the 54Rc sample with a (tiny) bend but then broke at only 12% higher force.

@62Rc – same again but a final deflection of only 5mm (maybe 5700MPa) and only 2mm of which was permanent. This one didn’t start to take a set until a load which equated to 4000MPa or 60% higher than the 54Rc sample and 25% higher than the 58Rc one, but it broke at only a little if any higher force than the 58Rc one.

Everyone saying that the harder material in the same geometry takes the same load to bend the same amount is totally correct up to the point at which the permanent set starts. For these particular samples around 2.5 to 3mm. Up to that time, they would be indistinguishable. After that point, the softer one will bend further for a given load, thus seeming less stiff and most people will realise they have pushed too hard, like when you strip a thread, and it will then stay somewhat bent whereas the harder one will continue to push back just as stiffly as it did and end up putting up with a higher load even though it will, in this case, actually finally break with less total deflection. But by that time you will have been able to flex the 62Rc sample about 60% further than the 54Rc one went before it was permanently bent. I guess it may depend on your criteria for failure as to which you prefer.
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bluntcut said:
A tactical metallurgy punch would be more effective... ;)
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At some point once discussion has shown to not be effective, charlatan trolls should simply be pointed out and ignored. I don't feed trolls.
 
I don't think Blunt Cuts info is really any help.

JT has the exact explanation I was going to describe.

While there can be a slightly different Modulus of elasticity for different alloys and structures, the simple answer for most knife steel is that the HT does not affect the flexibility to any detectable degree in use.
 
HT to harden steel for more strength. Please explain why we want this 'strength' when strength is not useful(or applicable) in conjunction with elasticity of blade & edge? Nifty sure, car springs are made out of hardened, not annealed steels or iron.
 
bluntcut said:
HT to harden steel for more strength. Please explain why we want this 'strength' when strength is not useful(or applicable) in conjunction with elasticity of blade & edge? Nifty sure, car springs are made out of hardened, not annealed steels or iron.
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As noted above, annealed steel takes a set much quicker than hardened steel. As you go harder, the steel breaks, rather than reaching its elastic limits. In the lower hardness ranges, steel will take a set when bent, before breaking. The amount of force needed to flex the steel doesn’t change, that’s down to geometry. The point it takes a set, or breaks completely changes as different hardness levels. This is one of the most confused principles when working with steel.
 
*bold blue text* - fracture can't occurs before yield. Elasticity range is between zero and yield load.

*bold orange text* - only when steel has similar microstructure. While, Ferrite Matrix (annealed) is very different than Martensite Matrix(hardened), therefore this assertion is false.

Since samples are identical in dimension/geometry, so leave it out.

Maybe I get inspire to conduct a simple experiment with 2 points bending test. A cantilevers of annealed and hardened bars of identical dimension and setup. Hang same weight/load on each cantilever. Start with weight1 = near annealed's yield strength. Then weight2 = near hardened yield strength. So if *bold orange* is correct, both bars would bend the same amount/deflection each time.

Similar setup as this video: youtube.com/watch?v=GXEJMjZ0JHM

But wait, annealed steel has yield strength around 240 MPa and UTS around 900-1000 MPa. Hardened steel YS roughly around 1400 MPa and UTS 1700MPA. So when weight2 is 1200 MPa, annealed bar break. All we need is this/one case to disprove the *orange bold text*. Nevertheless, next time doing bend test video, I will try to remember to include 1-2 samples of annealed steel.

Willie71 said:
As noted above, annealed steel takes a set much quicker than hardened steel. As you go harder, the steel breaks, rather than reaching its elastic limits. In the lower hardness ranges, steel will take a set when bent, before breaking. The amount of force needed to flex the steel doesn’t change, that’s down to geometry. The point it takes a set, or breaks completely changes as different hardness levels. This is one of the most confused principles when working with steel.
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