Pre-Quench Soak at Critical or No?

[Dave Stephens]

Hi All:

 

I'm a bit confused about grain growth and quenching.

 

Oh, I should state at the beginning that all my questions/comments in this post refer to my usual mix of 1095/15n20 pattern weld. And, a dummy disclaimer: I know just enough about the underlying science of heat treating to be dangerous, and my head still spins a bit when people start rattling off austentite, pearlite, etc. etc.

 

When I was doing shorter blades, warps on quench were hardly ever a problem. Triple normalize, quench in oil, temper, and voila! Now that I'm doing longer swords (and after realizing how friggin THIN these blades are supposed to be), I'm trying to do everything I can to mimize warping during the quench.

 

So, here's my question: Is it beneficial to soak a blade at critical (1500F) for any period of time before the quench, or does one simply quench the minute the steel is evenly up to temp? I am using an electric kiln so I do have the ability to hold the steel at temp. precisely and not over-heat. Also, I would have performed a full normalization sequence on the blade prior to this.

 

I read a thread over on bladeforums.com in which some seemed to be advocating for a 10 minute soak at critical or slightly below critical prior to quenching. Elsewhere I have read that soaking at critical causes grain growth.

 

On all my other blades I have not soaked, so I know one can achieve proper hardness without a soak. Again, I'm interested in it's effect on warpage, if any.

 

Thanks for your advice. And remember, use small words . . . I get confused easily!

 

Dave

[Brian S. Pierce]

Dave,

I'm not an expert, but I believe Kevin Cashen recommends a 5-10 minute soak(for simple carbon/low alloy steels) provided you can maintain a consistent temperature. My understanding is that grain growth is largely a function of temperature, not time. There was a post over at Bladeforums where they analyzed O-1 soaked at critical for 5 hours (IIRC) which showed no increase in grain size vs. O-1 soaked 10 minutes at critical. Of course, decarburization was an issue in an atmospheric forge....

 

Link to Original Thread

 

Link to 5-Hour soak results (pictures missing unfortunately)

 

Another thread on grain size, time vs. temp.

Edited July 29, 2012 by Brian S. Pierce

[Doug Lester]

Dave, there is no simple answer to that question. It is dependent on the alloy that we are talking about. I will assume that we are not talking about any of the high temperature tool steels or stainless steels with high amounts of tungsten and vanadium in them that resist going into solution. With some of those we would be talking about long soaks at 1800-1900°. From the information that I have, and let me say now that I make no claim on being a metallurgist, simpler steels with about 75-80 points of carbon in them, or less, can be taken to just above critical temperature for that alloy and needs to be held only long enough to heat evenly throughout. Above 75-80 points of carbon and steels are more subject to grain growth and needs to have a soak at between lower and upper critical temperatures.

 

It will also depend on whether or not silicon or aluminum is used to deoxidize the steel. Steel deoxidized with aluminum (fully killed), has less of a problem with grain growth. Other alloying elements will also effect grain growth. A little vanadium or tungsten will put a physical drag on grain growth but, then again, there are limits to this. It's the high amounts of vanadium and/or tungsten in high temperature tool steel and some stainless steels that prevent grain growth when soaking at 1800° for 30-60 minutes. You just have to know the heat treating requirements of the steel that you are using.

 

Doug

Edited July 29, 2012 by Doug Lester

[Doug Lester]

Dave, go to KnifeDogs.com >Knife Makers Area >Heat Treating Forum. Kevin Cashen just posted an article on hardening that deals with what you are asking about. It's worth printing out and reading.

 

Doug

[Rory Carter]

I do heat treatment of mostly tool steels for complex injection molds. Disclaimer... I have never heat treated a sword blade. However we noticed that we were warping our mold components a lot ( this was with s-7 steel) so I looked up the heat treat info sheet from the manufacturer and found that when hardening the components a pre-heating cycle should be used. I had previously thought that this ment that you put the parts in the oven at room temperature and then let them heat as the oven warmed... this was erroneous.. what it ment is that I needed to put the parts in at room temp and let it heat to 1400 deg F then let them soak for 1 hour per inch of thickness and then boost the temp to the critical temp for hardening which is 1750 deg F then let the temp heat the part evenly and then soak for another hour per inch of thickness. this solved all of our warping issues with this steel.. So to sum up this is not a totally similar case but I think the principles may apply. Look up the manufacturer's data and see if it recommends a preheat cycle. This cycle will be significantly lower temp than critical temp for hardening. What it does is it causes less stress in the metal when it goes through its phase changes. This should not cause any decarburization either. Obviously the times of heating the blade will not be the same as mold parts but again I think this would be worth pursuing.

[Dave Stephens]

Thanks everyone.

 

Doug -- Kevin's article is exactly the info I needed. Thanks!

 

Here's a link to it: http://knifedogs.com/showthread.php?25737-Hardening-Part-1

 

--Dave

[Jerrod Miller]

I have long had an issue with soak times. Mainly due to the lack of concrete data on them. As Kevin points out in his article that Dave links (still works as of today), the time and temp are dependent on several things. I finally came across a chart that shows some data on the subject and figured this was as good of a place as any to put it. Now I just have to find one that includes chrome carbides. On a related note, the ASM Heat Treater's Guide (2nd Ed.) is awesome.

Sorry for the slightly crooked scan.

[stuart davenport]

Jerrod, I need HELP reading the chart ! Is the time in seconds or minutes? What is it we are after, the shaded area, or the area that says "homogenous austenite"? Thanks for posting the chart, once I get straightened out on what it is telling us, I'll be all set!

 

Of course, this chart says "eutectoid steel"....something along the lines of 1080 or 1084.

 

I think the time is in seconds, and what we want is the homogenous austenite. I "think" the chart is saying that at 1475, austenite is completely formed after 6 seconds (once it reaches 1475), then at just under 100 seconds, austenite with carbides dissolved, then from about 100 seconds to just over 1000 seconds (16.6 minutes) we may have clumps of carbide here, clumps of carbide there, but after 16.6 minutes (make it 20 minutes because the curve is a little past the 16.6 minute mark), all the carbide concentrations should be dispersed evenly across the blade.

 

I hope that is right.

Edited June 20, 2014 by stuart davenport

[Jerrod Miller]

You've got it! I never noticed that the units weren't there on the time axis. It is generally agreed that phase transformation happens at the speed of sound in the material. So as your carbides dissolve (which takes some time to go from carbide to loose Fe and C) the atoms go from free Fe and C to austenite at about 20,000 ft/s. It is the pesky carbide bond that takes some time/energy to break, not the phase change itself.

 

Keep in mind that is from a sample that was previously normalized from 1605F. Different things happening to the steel prior to your soak will make things behave a little differently. For example, soaking lets diffusion happen so you get homogeneity of your carbon in solution. This in turn spreads out your carbides and keeps them smaller (depending on how you cool). If you keep your carbides small they will dissolve faster and you will become homogeneous quicker than if left big.

 

Now, do you NEED to get everything dissolved and homogeneous? This gets into the territory of "diminishing returns". Short answer: I'd say no. By the time any given smith gets his/her steel it is already pretty good in terms of homogeneity (the awesome folks over in Bloomers and Buttons are the exception to this rule, and they are well aware of that). If we go with Kevin's assertion (again, in the link Dave posted) that anything over 0.60% C has little effect on hardness, and I'm inclined to completely agree with him on this, then you can leave 0.20% of the carbon in your eutectoid steel in their fine distribution of carbides. So that makes it about a 20 second soak is all you probably really need for 1080. This is just my opinion and assumes that your steel is good to begin with. I've tried to present my reasoning, so make your own choices based on data presented and, more importantly, your own trials.

[stuart davenport]

Very imformative. I've followed Kevin's advice all along. Even if the steel I receive is from reputable sources, I like to KNOW how the structure is set up. In hypereutectoids, I normalize them all at 1650F and only air cool. That is always followed by grain refinement cycling at lower temps, usually only three more, from 1550F down to 1450F down to 1400F or so air cool only. Then I usually will spherodize if grinding is needed to be done in the soft state. If the knives are thin, I go straight to hardending and then grind them hard. In the future, with 52100, I would like to try quenching after the normalizing heat instead of air cool only, and quench during the thermal cycling heats as well, just to see if I can tell a difference in quenching vs air cool only. I know that quenching from 1650F is precarious....micro fracturing and what have you becoming an issue. Great thread. Learned quite a bit here from you chaps.

 

One thing I have never quite understood is the difference between carbon and carbides. I understand carbon is an element all itself, and carbides are carbon bonded with another element, like iron or vanadium or chromium, etc. What I don't get is this....take a piece of steel from your favorite vendor in annealed condition, let's say 1.0% carbon W2 with .2% Vanadium and .2% chromium. When you receive the steel as annelaed pearlite, is ALL the carbon alone, and all the vanadium alone and all the chromium alone (not bonded to anything)? When we then harden this steel, I don't understand exactly how much of chromium gets used in carbides, and how much vanadium gets used in carbides. I would think ALL the vanadium and chromium gets tied up with the carbon, right? There isn't really a situation where you have LEFT OVER vanadium, or LEFT OVER chromium, right? I can see if your chromium levels are on par with D2 or something, then maybe some chromium gets tied up in chromium carbides, will some chromium is left over (and the left over chromium is what actually aids in the corrosion resistance...not the chromium carbide, right?). When a steel is hardened, only about .8% carbon bonds to iron, right? If we have 1.0% carbon, that still leaves us with .2% carbon. I assume that 2% bonds with the chromium and vanadium....but if there is only .2% carbon to bond with .2% chromium and .2% vanadium......does the chromium get the carbon before the vanadium....or is it automatically evenly ditributed between the chromium and vanadium?

 

Forgive the rambling questions. I suppose this carbide thing and how they make their bonds has me still confused.

[Jerrod Miller]

There isn't really a quick answer, and I am hesitant to bog this particular thread down with all of it. Short answer is that carbon in solution (not tied up in carbides) sits in-between the bigger atoms (like iron). This positioning is called the "interstitial" location. Bigger "extra" alloying elements that aren't tied up as carbides replace iron atoms in the lattice. Carbides form on/cause grain boundaries and are not in the lattice. I like to think about it as stacking fruit in the grocery store. Think of stacking oranges. The natural way to stack them would be in layers with the first one in a grid, the second layer sitting in the valleys created by the first layer, and the third layer sits on top of that - exactly in line with the first layer. Now imagine all the gaps that are between the oranges. Could you fit a grape in there? (Hint: the answer is "yes".) That is what is going on with carbon in solution. Now imagine switching one of your oranges out for a grapefruit. It is a little bigger, but things still generally work out. That is like putting Chrome (or vanadium, or manganese, or whatever). This works with slightly smaller things too. Now with all that in mind read up on the different crystal structures; face center cubic (FCC), body center cubic (BCC), and body center tetragonal (BCT). The orange example configuration noted here is BCC. And remember that carbides do not fit in the stack. They sit at the edges between stacks.

 

Also of note: Carbides are not all the same, even when it is all the same metal atom base. There are several common structures: MC, M₃C, M₃C₂, M₂₃C₆, and M₇C₃ being some of the most common (where M represents the metallic atom such as Fe, Cr, V, W, etc.). This greatly affects the amount of carbon and/or alloying content that gets tied up in various forms of carbide.

 

And yes, that WAS the short answer.

[stuart davenport]

The short answer, huh? I think I am starting to derail this thread a bit, you're right. Thanks for the help, Jerrod! I need to do more reading of Verhoven and gang to understand all this, especially the different lattice structures. I'm still in the dark as to all of that.

[Jan Ysselstein]

This is a wonderful thread my gratitude to Jerrod and Kevin for the clarification of the phenomena we all need to be familiar with.

 

Jan