Toughness?

[ghost squire]

Besides, you wouldn't really want permanent deformation so I fail to see, why a large area from the yield point to fracture would be of much interest, or we would be using brass or copper for knives.

You just have to use common sense, and when you look at a stress-strain graph, see if that is the balance you're looking for. You cannot just look at an arbitrary number and say "oh, thats how tough it is", no you must calculate what balance you desire between strength and ductility.

I'm sorry its not any simpler then that, but look at the line going up, then look at the line going sideways, and decide how long you want either one to be for that knife.

That area is, by the way, far from non-existing in glass. Glass is able to hold only small amounts of strain to fracture, but it does so at huge stesses. It is an extremely steep rise in stess-strain curve and it is practically ideally linear in the first part. Especially if thin glass fibers are tested.

Actually it is almost non-existent, which is what I said. Again, applying common sense, compared to something like 1084 tempered for a good balance of ductility and strength, glass isn't even going to be close toughness wise.

why a large area from the yield point to fracture would be of much interest

I don't know, why don't you tell me since I never said that. You have to take into consideration all parts of the chart. Otherwise lead would be the toughest material.

But that is exactly my point. Batoning (impact) has nothing to do with a semi-static stress-strain experiment.

Actually thats exactly my point, it DOES. A stress-strain chart will tell you how well something will take batoning. You look at a chart for glass and it will tell you that it will likely shatter with little give.

I see there is no longer really a point in me trying to convince you of the merit of this method for determining toughness. Glass and lead both have relatively tiny surface areas beneath their stress strain charts. Regardless of that you have to use common sense. You obviously want a good balance in a knife, and depending on what its application is that balance is going to shift.

 

[HoB]

Actually, you don't have to convince me. I agree that the stress-strain curve is a very useful tool. But you are now arguing something different than you did before. The definition of toughness IS just a number: the integrated area under the stress-strain curve. If someone gives you the toughness value such calculated, you have absolutely no idea how the graph looks like and the number is pretty useless. That is all I said. Lucky for you if you have stess-strain curves for all common bladesteels at all possible hardnesses. I find them pretty difficult to find. You are usually lucky if you find a single number relating to toughness.

And therefore that single number can be badly misused. If somebody wants to sell a really "tough" blade with a large toughness calculated by the area under the stress strain curve, he could make one where that area is largely dominated by the plastic range beyond the yield strength. The blade would be very tough but may be pretty useless nontheless.

So as I said before, toughness as defined by the area under the stress-strain curve does not seem to me as the most useful definition.

Finally, I still maintain that shock and strain are physically different animals. The stress-strain curve has no dynamic component to it. It is semi-static and pretty much as far from batoning with a framing hammer as you can get.

 

[searcher]

The behavior of a material stressed past it's elastic yield point (plastic range) is often usefull. In bending a part can be partially stressed into the plastic range and still not considered to have failed (plastic stress design). In this way materials can be used to their full potential, as opposed to earlier engineering practices when any yielding meant failure. A large strain between yield point to fracture is often particularly usefull for giving a larger safety factor before fracture occurs, like for elevator cables. Of course it is very interesting to people who use manufacturing processes that permanently deform metal (forming, bending, rolling, forging)

In knives it can mean that if you are ever in a do or die situation when you have to push your blade to its limits and beyond you may still be able to continue using or at least salvage the blade.

 

[Cliff Stamp]

The behavior of a material stressed past it's elastic yield point (plastic range) is often usefull.

The advantages of extreme ductility is one of the largest conflicts in heavy use knives, there are three camps, Fowler who leaves the spines soft and the blades can bend severely before breaking, Swamp Rat who spring hardens the spine so they bend a lot and are stiffer and then Strider who uses air hardening steels which won't bend very much at all, but at very difficult to flex significantly. You usually don't gain ductility without losing strength so it is a question of what is more valuable, a higher yield point or a greater ductility. Kevin Cashen was one of the first ones I saw note that many times by increasing ductility the actual ultimate tensile point is lowerer and this has problems in extremes because the blade could be more likely to break in an overstress, this has been discussed in detail on swordforums complete with stress/strain graphs for different temper draws.

-Cliff

 

[searcher]

The advantages of extreme ductility is one of the largest conflicts in heavy use knives, there are three camps, Fowler who leaves the spines soft and the blades can bend severely before breaking, Swamp Rat who spring hardens the spine so they bend a lot and are stiffer and then Strider who uses air hardening steels which won't bend very much at all, but at very difficult to flex significantly. You usually don't gain ductility without losing strength so it is a question of what is more valuable, a higher yield point or a greater ductility. Kevin Cashen was one of the first ones I saw note that many times by increasing ductility the actual ultimate tensile point is lowerer and this has problems in extremes because the blade could be more likely to break in an overstress, this has been discussed in detail on swordforums complete with stress/strain graphs for different temper draws.

-Cliff

Up to the lowest yield stress of the steels being compared, all steel bends the same amount for a given load ,with very little variation (same stiffness).

 

[Joe Talmadge]

searcher, you've said this before, and I want to make sure I understand what you're saying. Given two pieces of steel of the exact same dimensions, but one hardened to 60 Rc and one at 40 Rc, you're saying it will be exactly as difficult to bend the 60 Rc one as the 40 Rc one?

 

[Cliff Stamp]

Yes, if you check the stress strain curves, the linear part, which is where the blades will return to true if released, is near idential. However the softer steels yield almost immediately and thus if you press on a knife with a annealed spine vs a spring one, the annealed spine bends much easier, because it is only linear for a very short time, they have no flexibility, high ductility, but no flexibility. This is compounded by the fact that as you bend a knife, in most cases due to issues of leverage it gets easier to put more force into it for a given effort, so what you feel exertion wise is even more lopsided. Stick a cheap 420 blade into a piece of wood and a ATS-34 one and see which one you would call easier to bend.

-Cliff

 

[HoB]

Up to the lowest yield stress of the steels being compared, all steel bends the same amount for a given load ,with very little variation (same stiffness).

That is very interesting. I didn't know that the slope in the elastic region is nearly the same for all steels regardless of the hardness. Does that even hold comparing highly alloyed steels and basic carbon steels?

However, isn't that a pointless comparison? That would imply that in practice the blade never gets bend beyond the maximum yield of the weaker steel. In that case toughness is completely irrelevant since the plastic deformation region is never entered.

 

[searcher]

That is very interesting. I didn't know that the slope in the elastic region is nearly the same for all steels regardless of the hardness. Does that even hold comparing highly alloyed steels and basic carbon steels?

However, isn't that a pointless comparison? That would imply that in practice the blade never gets bend beyond the maximum yield of the weaker steel. In that case toughness is completely irrelevant since the plastic deformation region is never entered.

For any steel, the ratio of tensile (or compressive) stress / strain is a constant up to its yield point. This is known as Hooke's law. This constant of proportionality is known as Young's Modulus of Elasticity (symbol "E") and for carbon steels it is around 30,000,000 PSI (strain is "dimensionless" as it is change in length per unit length for (Eg. inches of stretch per inch of length). For stainless it is around 27,500,000 PSI. In torsion the constant is called the "shear modulus of elasticity" or the "modulus of rigidity" with symbol "G." For steels G = about 12,000,000 PSI

Since the yield points of different steels vary the stress past which it will take a permanent set (yield stress) varies. Weaker steel is easier to bend to the point of permanent set. Hope I didn't just muddy the water further.

 

[Joe Talmadge]

Yes, if you check the stress strain curves, the linear part, which is where the blades will return to true if released, is near idential. However the softer steels yield almost immediately and thus if you press on a knife with a annealed spine vs a spring one, the annealed spine bends much easier, because it is only linear for a very short time, they have no flexibility, high ductility, but no flexibility. This is compounded by the fact that as you bend a knife, in most cases due to issues of leverage it gets easier to put more force into it for a given effort, so what you feel exertion wise is even more lopsided. Stick a cheap 420 blade into a piece of wood and a ATS-34 one and see which one you would call easier to bend.

-Cliff

That makes sense, gotcha

 

[Cliff Stamp]

Joe, to quantify it look at the following :

http://www.panix.com/~alvinj/graphStrength.jpg

Note how much earlier the blades take a set (move off linear) as the temper is draw, note the last one corrosponds to a 50 HRC blades, the annealed spines are 20+ hrc and they are much weaker still, they have almost no linear region at all, as soon as they are flexed they deform plastically and thus the force plateaus. Thus the force it takes to bend them to even a moderate angle like 30 degrees can be a small fraction of what it takes with a flexible steel which will return to true from this angle.

-Cliff

 

[me2]

Sometimes I use Dieter's Mechanical Metallurgy. For instance, I never understood why roller levelling works to flatten steel strip until I read about the Bauschinger (sp?) effect and how a stress-strain hysteresis loop is formed by bending a strip back and forth thru rolls that actually serves to dissipate internal stresses in the strip.

Looks like I'm gonna have to dig in again. We mainly used it for dislocation theory and failure modes (fatigue, stress rupture, etc). Sounds like there is a lot more in there that could be useful. This could take a while, thats a big ass book.

 

[HoB]

Cliff, I can not access that page or is that just me?

Speaking of failure modes: I wonder if the key in a knife isn't to adjust the yield strength to the transition from a plain-strain to a plain-stress fracture (or at least towards the plain-stress fracture mode). From the few broken blades that I have seen (which are admittedly not very many) the better ones seem to be breaking pretty much by a plain-strain fracture if you can really tell by inspection of the fracture, while the cheap stainless steels are clearly plain-stress. I tried to calculate the theoretical condtion for plain-strain but I couldn't find the nessessary numbers. Anybody know K_I and yieldstrength for the same steel at a resonable hardness (for a blade that is)?

Makes perfect sense, searcher. I had seen graphs for steels at different hardness and noticed that they are running essentially colinear left of the yield point, but I had always assumed that there would be a much larger variation in the slope (Young's modulus, sorry but I tend to think graphically) for different steels.

That has a pretty important consequence, if I am not mistaken: If you can ensure (by adding thickness to the blade and increasing the yield strength by hardening) that you never enter the plastic region, any steel, no matter whether it is SS or high or low carbon steel should essentially perform the same.

What strikes me though, is that as far as I have been able to access the data there seems to be a large difference between the impact toughness of high and low carbon steel even if the failure is a plain-strain failure, which according to the going theory should be depending pretty much only on the elastic range of the material (essentially pure cleaving). The problem is that the data that I found often omits hardness, so I can not be sure, if they simply tested high carbon steel at high hardness and low carbon steel at low hardness.

 

[Cliff Stamp]

If you can ensure (by adding thickness to the blade and increasing the yield strength by hardening) that you never enter the plastic region, any steel, no matter whether it is SS or high or low carbon steel should essentially perform the same.

When you increase thickness to get strength you decrease the flexibility because for a given stress the strain is greater as the thickness increases, essentially consider the difference between the length of the outside of a blade and the inside when bent through a specific curvature. This is the main reason why thick tacticals can't bend much at all before they break, but thin blades out of the same steel can be flexed massively.

-Cliff

 

[HoB]

Yes, I see, silly me. And you are more likely to end up in the plain-strain failure mode as well. Mmmmh, got to think that through one more time .

 

[Cliff Stamp]

You also of course lose cutting ability (in general, there are complications), so the benefits of a higher yield point is multi-fold. What is interesting in regards to the relation of thickness on strain is that since you can take the high HRC edges lower because they are stronger they can often be more ductile than weaker steels due to cross section at the edge inducing essentially minimal strain under a given stress. The very thin edges on the full hard 1095 blade I have for example will deform heavily and not crack because they are so thin.

-Cliff

 

[searcher]

When you increase thickness to get strength you decrease the flexibility because for a given stress the strain is greater as the thickness increases, essentially consider the difference between the length of the outside of a blade and the inside when bent through a specific curvature. This is the main reason why thick tacticals can't bend much at all before they break, but thin blades out of the same steel can be flexed massively.

-Cliff

Hooke's Law says, "As the stress, so the strain." If the stress is constant (given) then so is the strain. This is the case for Tension/Compression eg.) pounding on the handle. If you are talking about bending the blade (as in prying) then you would consider the blade to be a cantilever beam (simplyfying the fulcrum action (couple) at the tip of blade to a single applied force. As an example, if you bent a blade to a certain angle with tip clamped in a vise, then bent the same shaped blade but twice as thick to the same angle, the maximum bending stress would double. But it would take 8 times the force to do it.

Ref. Roark's Formula's for Stress and Strain sixth ed. beam deflection formulae, case 1a) pp. 100 (fixed end cantilever beam)

 

[Cliff Stamp]

Hooke's Law says, "As the stress, so the strain."

Prying with a knife doesn't have all parts under equal deformation, as would be the case if you were pulling it apart.

As an example, if you bent a blade to a certain angle with tip clamped in a vise, then bent the same shaped blade but twice as thick to the same angle ...

They don't bend to the same angle, that was the point. As you increase the thickness to increase strength you will directly decrease the ability to bend because the stress is greater at a given angle, this is why you don't get the cubic strength increase because it breaks earlier and thus it is quadratic.

This is one of the central problems in large utility knives, some of the ones with very thick cross sections do have a very high break point in regards to maximal force, but the stress is much higher at a given angle and thus they will crack at no warning at very low angles so are problematic for prying.

It is similar to the issue of hardening to get strength vs drawing to get ductility as noted previously. There are makers (and users) on both sides of the design. It is basically thick ground stainless tacticals vs slim and spring spine drawn distal tapered forged blades.

The issue is compounded by lack of impact toughness at the high hardness levels, and generally of the type of steels used in modern tacticals which tend to make them poor prying tools in general because they are fairly easy to crack unlike an actual prybar which is actually much softer and weaker.

-Cliff

 

[searcher]

Prying with a knife doesn't have all parts under equal deformation, as would be the case if you were pulling it apart.



They don't bend to the same angle, that was the point. As you increase the thickness to increase strength you will directly decrease the ability to bend because the stress is greater at a given angle, this is why you don't get the cubic strength increase because it breaks earlier and thus it is quadratic.

This is one of the central problems in large utility knives, some of the ones with very thick cross sections do have a very high break point in regards to maximal force, but the stress is much higher at a given angle and thus they will crack at no warning at very low angles so are problematic for prying.

It is similar to the issue of hardening to get strength vs drawing to get ductility as noted previously. There are makers (and users) on both sides of the design. It is basically thick ground stainless tacticals vs slim and spring spine drawn distal tapered forged blades.

The issue is compounded by lack of impact toughness at the high hardness levels, and generally of the type of steels used in modern tacticals which tend to make them poor prying tools in general because they are fairly easy to crack unlike an actual prybar which is actually much softer and weaker.

-Cliff

pry with a knife till you reach a certain stress. Now double the thickness and you can apply four times the force before you get to the same stress level. None of this has anything to do with toughness. If you want to design for shock absorbtion you want to "spread the stress out" so that more of the material takes the stress, in effect averaging out the stress over more material. this can even mean using less material overall to get more shock resistance

 

[Cliff Stamp]

None of this has anything to do with toughness.

That depends on how you define it and there are many definations. For many applications the fact that the thicker knife will snap earlier is a problem, especially if by thickening it you have reduced the angle so much that it won't bend at all now and/or everything is focused at one spot. The angles to set and then crack are increased when the cross section is lower, the force needed and thus strength is much higher and this is usually a common tradeoff in alloy as well as geometry.

Speaking of how people use knives, a blade which can bend a lot more would be considered tougher by many simply because it was harder to break when prying and is far less likely to be accidently overstressed and broken, though far more easily bent. You can see the same thing in edges, edge angles which are very thin and acute can pass "toughness" tests that thicker edges fail, the brass rod test for example.

In knives the application is further compounded beyond the basic materials data because as you thicken a knife to get strength it loses cutting ability and often general control and thus it sees much greater and uncontrolled forces in use. Often times it is readily possible for an edge to become stronger and tougher *in use* by reducing the edge angle, but any materials data would show differently. But you have to consider the use of the tool, not simply the raw ability of the steel.

-Cliff