Differential Tempering

If you think of the knife as a spring, it makes sense. Springs are rated in pounds per inch or some similar measure. Each inch of bend takes an increase of force at the rate of the spring. Once you exceed the yield point, the force no longer increases, as it breaks or bends. All same shaped blades flex at the same amount of force until exceeding the yield point.

A common example is motor sports. People mistakenly believe a chromoly frame is stiffer than a mild steel frame. There may be a slight difference based on alloying, but the two frames will feel basically identical, until reaching failure. At that point the chromoly frame will be able to spring back, while the mild steel frame will take a set. It is stronger in the total load it can take, but both have the same feel at loads up to the max load the mild steel frame can handle.
 
Bo T said:
So, in the stress/strain diagrams is the strain correlated with the amount of bend?
Click to expand...

Stress, strain and deformation are all directly related. The other factors involved are the material stiffness (youngs modulus or modulus of elasticity), and the shape of the cross section.

For any given configuration of applied load, shape of material and point at which deflection is measured, there will be an equation that defines the behavior in terms of stress, deflection, youngs modulus and material shape. Provide any 3 and you can calculate the 4th. Likewise there will be an equation that relates stress and strain and would allow you to express the first equation in terms of strain instead of stress.



Willie71 said:
If you think of the knife as a spring, it makes sense. Springs are rated in pounds per inch or some similar measure. Each inch of bend takes an increase of force at the rate of the spring.
Click to expand...

And really any material that can behave somewhat linearly below its elastic limit can function as a spring. There is really nothing special about "spring steel", it is just steel that is hardened to a yield strength beyond the stress that it will work at as a spring. Simple mild steel can function as a spring but it doesn't have much strength and in use it would be very limited in what you could do with it before you would permanently deform it. But up until that point it would be indistinguishable from a spring made from much higher strength (hardness) of steel.
 
In looking at stress/strain diagrams I stumbled on to a true stress strain/strain diagram. This curve gave results that showed a different slope for different alloys or points of carbon. Does this relationship start at the elastic limit?
 
There is a good video on YT that explains these concepts very well. While it is dealing with Aluminum, it should still apply. It covers the crystalline structure, work hardening, tempering, etc and shows what happens to the crystals. It explains the elastic limit and plastic deformation.
It's an old video, from 1945 in B&W but it's pretty good.
http://www.youtube.com/watch?v=BBFBJ1zJlmA

There's also a part 2 but I haven't watched it.
 
WOOOOOW!!!

I wish I had "computer time" every day, but at this time of year there's no way.

I can't believe the amount of scientific knowledge here. You guys blow me away with it.

I thought I was asking a simple question. I should know there is no such thing here.

I'm going to go through this with a fine tooth comb, hopefully tonight. I definitely don't have this figured out yet.

It seems unthinkable that two pieces of the same steel, let's say 1/8" thick by 1" wide, would take the same amount of force to bend if one is dead soft, and the other is full hard. I've got to be misunderstanding something.

Someone asked me if I was differentially hardening blades, and full quenching them. I don't know how you'd know that, but yes, that is part of it. I like the naturally flowing hardening lines after etching. I just think that's a way to make a beautiful, original looking blade. Doing that is kind of fun for me.

It doesn't make a strong enough blade IMHO, so after I get the pretty hardening lines, I use my kiln to bring the whole blade to an exact temperature, and quench it exactly.

That doesn't make a tough enough blade, IMHO. It will bend to whatever degree you temper it, but you're taking away cutting performance to gain toughness.

I don't want to settle for that either. That's why I want to differentially temper a martensitic blade.

Everyone who thinks this is a lot of fooling around is right. I don't really care though.

I simply want the best looking highest performance blade I can make.

I don't want to make a bunch of knives. I've done that. Nothing very special about them.

I want outstanding.

Y'all help a LOT, and I'm eternally grateful!

I'm completely enthralled with all the things I'm learning about this. Thank you all so very much for all your selfless contributions! I'll be studying this thoroughly tonight.
 
grizzled gizzard said:
It seems unthinkable that two pieces of the same steel, let's say 1/8" thick by 1" wide, would take the same amount of force to bend if one is dead soft, and the other is full hard. I've got to be misunderstanding something.
Click to expand...


Define what you mean by "bend". The term can be used in different ways. Depending on how you are using the term, "same amount of force" might or might not be true.
 
Believe it or not, for a particular steel type, Young's modulus (AKA - modulus of elasticity) is the same hard or soft. What is different is whether the steel flexes and returns ...breaks....or stays bent.

If you clamped two identical 12" bars of the same size stock in a vise, one hardened and the other not - The same amount of force is needed to deflect the other end the same distance ( up to the point where failure is reached). Flex/stiffness is merely a factor of geometry....thinner flexes more....thicker is stiffer.
 
bladsmth said:
Believe it or not, for a particular steel type, 1) Young's modulus (AKA - modulus of elasticity) is the same hard or soft. What is different is whether the steel flexes and returns ...breaks....or stays bent.

If you clamped two identical 12" bars of the same size stock in a vise, one hardened and the other not - 2) The same amount of force is needed to deflect the other end the same distance ( up to the point where failure is reached)..
Click to expand...

(1) is easy to prove & agree with - YM is the same for hard & soft for identical steel composition.

However for (2) - my subjective experiences seem different. In structural, I think, for different elastic deflection limits, Elastic Strain Curves are non intersecting except at the origin. Mathematically the attach figure is easy for my mind to grasp but I can't figure out how can 3 curves are the same (super-imposed) for 3 different limits of elastic deflection?

YoungModulus.png
 
I wonder if where getting off topic?
This is starting to hurt my head........but honestly this is great!
 
bluntcut said:
Mathematically the attach figure is easy for my mind to grasp but I can't figure out how can 3 curves are the same (super-imposed) for 3 different limits of elastic deflection?
Click to expand...

I don't know where that graph came from but it doesn't have any meaning with respect to steel.


chknifemaker said:
I wonder if where getting off topic?
This is starting to hurt my head........but honestly this is great!
Click to expand...

If I start trying to explain the reason that differential tempering (or laminated steel) works, it would probably make your head explode. :o
 
I drew that graph to visualize/rationalize in my head how steel phrases and structure (micro & grain) could be varying from Young Modulus static equation. Take grain boundary strengthening as an example - http://en.wikipedia.org/wiki/Grain_boundary_strengthening - of how yield gain (stronger) should translate in less deflection angle and in addition to extending the elastic limit.

I just am curious on how could YM singularity factor apply for all steel protons & electrons in vast combination of configurations? even General Relativity has to dealt with Quantum Physics...

bdmicarta said:
I don't know where that graph came from but it doesn't have any meaning with respect to steel.
Click to expand...

OP - sorry about my OT posts in your thread.
 
Perhaps there is a difference but we do not look at the stress strain relationship with enough precision to elucidate the differences. i.e. the differences for a given steel are so small as to have no practical effect.
 
Greetings to all people on this forum. I need some help, can please someone give me direct link about clay coating knives? I want to make dagger from a saw blade. Thank you very much.
 
nocke,
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For the same steel the lines overlap until the yield point...where starts the plateau.
passed the yield point is where the steel can't return true
the martensite has a yield point WAY HIGHER then the pearlite, so you can add way more weight and still return true
the pearlite has lower yield point so with way less weight the blade will take a set permanently

The higher the yield point the shorter the plateau... the ht decides the yield point location
 
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Willie71 said:
If you think of the knife as a spring, it makes sense. Springs are rated in pounds per inch or some similar measure. Each inch of bend takes an increase of force at the rate of the spring. Once you exceed the yield point, the force no longer increases, as it breaks or bends. All same shaped blades flex at the same amount of force until exceeding the yield point.
Click to expand...

Exactly, and with martensite the yield point is higher, so you can continue adding pound per inch, flex and still return true, whereas with pearlite the blade is already permanently bent beacause the yield point has been reached before and you are in the "plateau" zone.
In the end you'll reach the yield point with the martensite and it will snap... let's say with 180 kg, while the other pearlite blade of the same thickness has become a "u-shape" knife with only 80 kg.
 
I do believe that the non linear elastic region reflects the variation in section driven by the flex itself...still remaining in the "return to true" zone ;)
You can better visualize it if you apply to tensile experiments vs. flex experiments.
Be aware I may be wrong...

Stefano
 
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