show me your bend tested knife..

[butcher_block]

Can you make a shaving razor thats springy, with a rc hardness of 60 ?


Palmer...

matter of fact yes you can and even more so tho not in a bend test like you have showed

the much called for brass rod test was used in razor factory's to test the edge flex of a blade and it was not the HT that was being tested it was how thin they were ground by the grinding station

a well ground full hollow razor can flex to a great degree and they are often called singing or extra hollow razors as when they are stroped they hum and "sing" at a high pitch. when you shave with them they ping when they cut the hair its liek driving a lihgt fast car around your face (lots of fun but can get you hurt fast too )
i have 52100 blades at 62 hardness and a cpm154 test blade at 63.5rc
the cpm blade was not a good shaver till i bumped the temper a bit and brought the hardness down to a little more sane 62 but both of these blades had no problem passing the "flex test "


now with all that out of the way a razors impact resistance is bad as one drop on a hard sink the wrong way can take a huge chip out of the blade (read that as up to half of the hight and 1/2 inch wide at the edge )

 

[Kevin R. Cashen]

Springs stay in the elastic region of the stress vs strain diagram. If you have permanent deformation, that's in the inelastic/plastic range. In other words, that blade you have pictured is not springy. A blade that springs back to it's original shape is springy.

Thank you Destraal! Heat treat, heat treat, heat treat, the entire blademaking community loves to talk about heat treat when discussing the "flexing" of blades, when so much more is about the cross sectional thickness. I guess we will never have this discussion enough for people to get what you just touched on Destraal.

You often see a blade with a dead soft spine and a very hard edge go to 90 degrees, well if heat treatment is so critical and the softness of the spine so important to the bend... then how is it that you can have very hard edges that do not crack at 90??? Both parts went to 90 degrees yet both had very different heat treatments. The difference is that the edge was thin enough to accomplish it at a much higher hardness.

Destraal seems to have a good grasp of the proportional range of the stress strain curve and I would wager that he also has a pretty good grasp of area moment of inertia. A thicker blade, when soft, has a greater chance of bending, when hard it has a greater chance of breaking, due to the forces generated in pulling it over. If you want a knife that will flex to 90 with no problem and still be fully hardened, make it 1/16" at the spine.

Do any heat treat you like, but in "flexing" within the elastic range of the deformation, it means nothing; here cross section rules absolute. Move outside of the elastic range to where the steel yields any bit and now the heat treat takes over, and that yield will occur very quickly in a softer steel that bends than it will in a hard steel that breaks, but breaking will require almost 10 times the amount of load as the bent one. It is basic undeniable physics that not even the ABS can get around.

If you could longitudinally cut right at the hardening line one of the bent ABS blades that had an intact edge, you would probably see the edge spring back almost to true, but if things were reversed and somehow the spine could be ground thinner than the edge the dead soft spine would surprise you on how much it would return to true due to the lesser amount it thin cross section would have pushed beyond the yield point.

 

[Kevin R. Cashen]

matter of fact yes you can and even more so tho not in a bend test like you have showed

the much called for brass rod test was used in razor factory's to test the edge flex of a blade and it was not the HT that was being tested it was how thin they were ground by the grinding station

a well ground full hollow razor can flex to a great degree and they are often called singing or extra hollow razors as when they are stroped they hum and "sing" at a high pitch. when you shave with them they ping when they cut the hair its liek driving a lihgt fast car around your face (lots of fun but can get you hurt fast too )
i have 52100 blades at 62 hardness and a cpm154 test blade at 63.5rc
the cpm blade was not a good shaver till i bumped the temper a bit and brought the hardness down to a little more sane 62 but both of these blades had no problem passing the "flex test "


now with all that out of the way a razors impact resistance is bad as one drop on a hard sink the wrong way can take a huge chip out of the blade (read that as up to half of the hight and 1/2 inch wide at the edge )

You posted while I was typing, but thank you as well Butch, you certainly have the understanding of this with your experience in thin edges and how they flex, but your point about the difference between gradual loading (bending, flexing) and sudden loading (whack!) is very important as well. The famous brass rod test is claimed to be inspired by an old German cutlery industry practice, I would suspect that Butch is probably also aware of what guy the fellow who did the flex testing stood behind to see to it the job was being done correctly. It wasn't the heat treater, it was the grinder! The blade was flexed over a little tubular ring on his thumb to insure the edge cross section was ground correctly. Those Germans were pretty sharp, so I would imagine they had just an efficient test to accurately check the heat treatment.

 

[Dmitry.M]

"well if heat treatment is so critical and the softness of the spine so important to the bend... then how is it that you can have very hard edges that do not crack at 90???"

Karl!

If you will correctly answer your own question you will understand a LOT;

 

[Virgilio]

matter of fact yes you can and even more so tho not in a bend test like you have showed

the much called for brass rod test was used in razor factory's to test the edge flex of a blade and it was not the HT that was being tested it was how thin they were ground by the grinding station

a well ground full hollow razor can flex to a great degree and they are often called singing or extra hollow razors as when they are stroped they hum and "sing" at a high pitch. when you shave with them they ping when they cut the hair its liek driving a lihgt fast car around your face (lots of fun but can get you hurt fast too )
i have 52100 blades at 62 hardness and a cpm154 test blade at 63.5rc
the cpm blade was not a good shaver till i bumped the temper a bit and brought the hardness down to a little more sane 62 but both of these blades had no problem passing the "flex test "


now with all that out of the way a razors impact resistance is bad as one drop on a hard sink the wrong way can take a huge chip out of the blade (read that as up to half of the hight and 1/2 inch wide at the edge )

When trying to make a point, it helps to keep within the standards of the subject we are currently talking about. if you are going to go off subject and inject a bunch of other standards. then enough said, this thread could go on for ever.

my point was, I believe that one steel is not the answer, one type of heat treatment is not the answer.

and to your "YES" in the above statement, make one "straight razor" that will pass the bend test, I will purchase it from you, but, it has to have an edge hardness of 60 like I stated above..

good luck.

palmer

 

[Karl B. Andersen]

"well if heat treatment is so critical and the softness of the spine so important to the bend... then how is it that you can have very hard edges that do not crack at 90???"

Karl!

If you will correctly answer your own question you will understand a LOT;

You only copied part of Kevin's description of the process, that's cheating!
Following in the very same paragraph,
"Both parts went to 90 degrees yet both had very different heat treatments. The difference is that the edge was thin enough to accomplish it at a much higher hardness."

If you can understand how to accomplish that, you can pass "the Test"!

This could go on ad nauseum.
Back to the original topic, the "test" is to show that you can do about whatever you want.
There are just far too many variable to discuss in one thread.
And who would want to?

 

[Dmitry.M]

Hint.
Steel elongation on the outside surface of bending bar in coarse approximation is DeltaL=R*Sin Alfa. Where R is the bar thickness and Alfa is the angle of bend. In plain words the thicker the blade the more stretch is applied to the outside surface layers.
Elongation at break has negative correlation with hardness.

 

[Karl B. Andersen]

"well if heat treatment is so critical and the softness of the spine so important to the bend... then how is it that you can have very hard edges that do not crack at 90???"

Karl!

If you will correctly answer your own question you will understand a LOT;

I see - that wasn't my question. That was Kevin's post. Which he answered in the very next sentence.
Dmitry!

 

[Virgilio]

Springs stay in the elastic region of the stress vs strain diagram. If you have permanent deformation, that's in the inelastic/plastic range. In other words, that blade you have pictured is not springy. A blade that springs back to it's original shape is springy.

so what you are saying is that a blade is really only springy if it returns to it's original shape. so only after it is bent 90 degrees and then returns to original shape it is then considered a "true" spring..

if it doesn't return to it's original shape it is then considered "inelastic/plastic".

I did not know that...! Thanks..

palmer..

 

[Kevin R. Cashen]

I see - that wasn't my question. That was Kevin's post. Which he answered in the very next sentence.
Dmitry!

And it was a rhetorical question at that I probably didn't do a very good job at clearly answering it myself.

 

[Karl B. Andersen]

And it was a rhetorical question at that I probably didn't do a very good job at clearly answering it myself.

You should be more careful, Kevin.
Go finish your roof before winter.

 

[Destraal]

Mr. Cashen, you have guessed correctly. I'm presently a senior at GA Tech studying to be a civil engineer, with a strong emphasis in structural engineering. Although to be honest, I wasn't even thinking of the role of blade geometry.

Unfortunately, I don't have the time this instant to go into the long-winded explanation of stress in flexure and how this pertains to knifemaking. I'll give a very long and detailed post on that later today though. Strangely I'm really excited about it.

 

[Virgilio]

"well if heat treatment is so critical and the softness of the spine so important to the bend... then how is it that you can have very hard edges that do not crack at 90???"

Karl!

If you will correctly answer your own question you will understand a LOT;

THAT is funny....!

Palmer..

 

[deloid]

Mr. Cashen, you have guessed correctly. I'm presently a senior at GA Tech studying to be a civil engineer, with a strong emphasis in structural engineering. Although to be honest, I wasn't even thinking of the role of blade geometry.

Unfortunately, I don't have the time this instant to go into the long-winded explanation of stress in flexure and how this pertains to knifemaking. I'll give a very long and detailed post on that later today though. Strangely I'm really excited about it.

I look forward to reading this Destraal.

 

[JasonSCarter]

so what you are saying is that a blade is really only springy if it returns to it's original shape. so only after it is bent 90 degrees and then returns to original shape it is then considered a "true" spring..

if it doesn't return to it's original shape it is then considered "inelastic/plastic".

I did not know that...! Thanks..

palmer..

Tell me your not serious...Please!?!?!?!?!

 

[Destraal]

Before I can really describe any of this, I have to explain a few background topics.

Stress:
Stress is defined as a force per area. For example, if you have a 1 inch x 1 inch bar (cross section, length of the bar isn’t important for this example) of steel and you put 4 lbs of force on it, the value of stress would be 4 lbs/square inch. If it was instead a 2 inch x 2 inch bar with a 4 lbs force (or load) on it, the value of stress would be 1 lbs/square inch.

In the previous examples the stress was axial stress, which is usually represented by the lower case greek letter sigma. There’s also shear stress, represented by the a lower case tau, but we won’t worry about that.

The concept of stress is important because it’s under a certain stress that a material will fail, not under a specific load. So something like A36 steel, which was once used in most structural applications, fails, or yields, at 36 ksi (36000 lbs/ square inch). This means that regardless as to whether it’s a huge I-beam, or something like a knife, it’ll still fail at 36 ksi, though the actual load will be very different.

There are a few different values of stress that you should know about, which are dependent on the material being used. A side note: the terms stress and strength are frequently used interchangeably.

Yield stress – This is the point at which the material begins to deform in such a manner that it won’t return to it’s original shape when unloaded. This is what’s used for calculating the loads that a building can withstand. Also called yield strength.

Ultimate stress – This is the largest stress that a material can withstand. After yield stress, the material can frequently take more stress, but this is not without permanent deformation. Also called ultimate strength.

Proportional limit – This is the point at which the stress stops being directly related to strain, which I’ll define later. It’s important to note that this point marks the limit of the linear elastic range, but there’s still a region that is nonlinear elastic. In both of these regions, after the load has been removed the material will return to its original shape.

Rupture stress – This is the stress that steel experiences at failure. It’s important to know that rupture stress is not synonymous with ultimate stress. This is probably the least important of the stress points from a design standpoint. Also called rupture strength.

Strain:
Strain is defined as the change in length due to a load divided by the original length. For example, if a bar is originally 10 inches long and is pulled to 12 inches long, the strain would be 2 inches / 10 inches, or 0.2. Sometimes this is referred to as 20% strain instead. Note that strain is unitless.

Relation between Stress and Strain:
There are 3 ranges of the Stress vs Strain graph that I’ll talk about here. Those aren’t the only ranges, but they’re the most relevant ones to knives.

Linear elastic – This is the range for which stress is proportional to strain. In other words, stress is equal to strain multiplied by a constant. This constant is called Young’s Modulus and is denoted as E. For most structural steels E = 29,000 ksi and most steels hover around this value. Within this range, strain will go straight back to 0 once the load is removed, which is the same as saying that the material will go back to it’s original shape. This range extends to the proportional limit.

Nonlinear elastic – For this range the stress is no longer proportional to strain. However within this range, the strain will still go back to 0 once the load is removed. This range starts at the proportional limit and ends at the yield stress.

Inelastic/Plastic – In this range you will have permanent deformation, meaning even when the load is removed, strain will not go back to 0. This range is past the yield stress.

There’s a picture of a stress vs strain curve in the following article:
http://en.wikipedia.org/wiki/Yield_(engineering)

In the next installment I’ll describe how the stress due to a bending moment is calculated. This is the type of stress that a knife is subjected to during a bending test.

 

[Destraal]

The equation to find the axial stress at the outer fibers (think outer strips, not literal fibers) of a beam is as follows:

Sigma = M * y / I

Simple looking equation, but what the heck do those letters mean?

M – is the bending moment applied to the beam. A bending moment is a force applied in such a way that it causes a free object to rotate. For example, when you put your blade in a vice, you apply a force perpendicular to the blade near the butt to get it to bend. The bending moment can be found by multiplying the distance from the vice to the butt by the force you apply at the butt.

y – is the distance from the neutral axis (more about this below) to the fiber who’s stress you want to know. So the outer most fiber is usually the most important since the largest value of y will result in the largest value of sigma, with M and I remaining constant that is.

I – is the moment of inertia of the cross section. It’s by far the hardest thing for me to explain here. For the sake of simplicity, I will say that it is a measure of how much area in the cross section is how far from the neutral axis. It’s worth noting that this changes based on the direction that the bending moment is applied.

The neutral axis is defined as the imaginary line running through a beam at which the stress is 0. In a symmetric beam, it would be the line of symmetry.

As you can see from the equation, a greater M and a greater y both result in larger values of stress (sigma). However, a larger value of I results in a lower value for stress.

I’m sure some of you already see how this relates to the bending test. If you don’t see it yet, don’t feel bad, I’ll explain.

How you can use this equation:
The only variable you have much control of is y in this case, you can’t really control M and changing I is going to be really awkward. y is essentially half the thickness of your blade at any given point along the blade. From that you can see that a thicker blade will result in a larger stress value. Since failure is a function of stress (basically), a higher stress means your blade is more likely to fail.

Changing I:
An example of changing I is the famous I-beam in buildings. What the I shape does is concentrate the cross-sectional area at the extremities of the beam, resulting in a larger moment of inertia (I). I personally can’t think of a good way to apply this to blades, other than a fuller. Unfortunately, a fuller would not help in the direction that the typical bending test is done, but it could help with the type of load that chopping with the knife would impart.

In the next and final installation I’ll explain the effects of loading rates and hardness on the material properties.

 

[Destraal]

As I should've mentioned to begin with, these posts do have quite a number of simplifications, so keep that in mind.

Edit: I should also mention that I will likely finish tomorrow, not today.

 

[Dmitry.M]

AFAIR the strain changes the sign over the neutral line. The inner parts are compressed, the outer stretched. Correct me if I am wrong.

 

[Destraal]

AFAIR the strain changes the sign over the neutral line. The inner parts are compressed, the outer stretched. Correct me if I am wrong.

Yes that's also true. I should've mentioned something about how one side of a beam in flexure is in compression and the other in tension. Although unless you're seriously worried about buckling, that's not too big of a deal. For stuff like reinforced concrete and fiber reinforced polymers, that's a huge deal.