Steels

Reader Question – O1 vs 80CrV2

Larrin,

Someone said austempered o1 would be as tough or tougher then martempered 80crv2. Could you help me wrap my head around that?

I have an affinity for fine grained simple steels, and o1 being precision ground and available in all different sizes is great, but if 80crv2 has a finer grains structure and is tougher I’d be sold. It’s for a run of belt hawks that will be used primarily for woods and hunting duties not destruction tools. But I’d also like to focus on one steel for a while and curious if for small to medium belt knives which you’d recommend? I’ve played with most of the high wear and too steels, but haven’t messed with too many of the high carbon steels as I hated 1095.

Anyways ramble over appreciate your time Larrin.

Tim

Hi Tim,

Definitions: austempering involves quenching to an intermediate temperature (such as 400-900F) and holding for an extended amount of time to form a phase called bainite rather than quenching to room temperature to form martensite and tempering. This allows a one-step process, and lower bainite has sometimes been reported to have better toughness than tempered martensite. Martempering, also called marquenching, is also performed by quenching to an intermediate temperature, though usually at somewhat lower temperature, held long enough for the part to equalize at the quenchant temperature, and then allowed to air cool. This allows martensite formation to occur more evenly throughout the part rather than only the surface followed later by the core. Uneven martensite formation causes internal stresses that can lead to cracking.

O1 is an oil hardening steel with approximately 1.0C-1.2Mn-0.5Cr-0.5W. The high manganese in combination with chromium is what makes it an oil hardening steel. 80CrV2 has 0.8C-0.4Mn-0.6Cr-0.2V. The chromium addition provides it some more hardenability vs a steel like 1095 but is not a true oil hardening steel. With the thin cross sections used with knives quenching in fast oil is sufficient, however.

It would be very difficult to predict which would be tougher between austempered O1 and martempered 80CrV2. For one thing, good studies comparing austempering, martempering, and conventional quenching are surprisingly tough to find. A lot of the sources cited are from many decades ago with little replication. And there are many variables such as length of time of austempering, the temperature selected, whether it was tempered after austempering, etc. Austempering or martempering may lead to superior toughness (emphasis on may), but they also may lead to higher retained austenite and/or lower hardness, which may not be desirable. Testing the different alloys with different heat treatment parameters would be required for a full comparison.

O1 and 80CrV2 both should have a relatively fine carbide structure, though the grain size is greatly controlled by the heat treatment. The vanadium addition may help to maintain a fine grain size in 80CrV2 but nothing is guaranteed. Furthermore, the specific processing (forging, normalizing, annealing, etc) of the as-delievered steel will affect the grain size of either, and O1 can come from many sources.

As far as comparing the toughness of the two with the same heat treatment, toughness is greatly controlled by carbon content especially with simple steels so 80CrV2 is likely to have greater toughness in general. The higher carbon content of the O1 means that it has a greater volume of carbides for superior wear resistance. Neither steel has particularly high wear resistance, however, in line with most other low alloy steels.

It’s hard to give specific recommendations for steel choice because it depends on so many factors, such as the way you like to make knives (forged vs stock removal for example), the equipment that you are using, the intended customers, how thin the edges are intended to be, what levels of wear resistance and toughness are required, other specific properties that are important to you (ease in finishing or others), and more factors I am forgetting. If it were me, for example, I might like a belt knife made with AEB-L due to its corrosion resistance combined with ease in sharpening and ability to hold fine edges. But that is unlikely to be suitable for a bladesmith who forges knives.

The higher hardenability of O1 can make it “easier” to heat treat when using a well-controlled furnace because it allows the use of relatively slow oil for limiting distortion and cracking. 80CrV2 is easier for forgers in part because the low hardenability means that normalization is easy with air cooling and annealing occurs relatively rapidly. Normalizing of O1 in air sometimes leads to a martensitic microstructure (not the goal of normalizing) and annealing is difficult even when cooling slowly in something like vermiculite.

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The “featured” micrograph of O1 was taken from here: https://hocktools.wordpress.com/2010/06/19/o1-v-a2-one-more-time/

10 thoughts on “Reader Question – O1 vs 80CrV2”

  1. Hello.
    Hope it’s not too late, but a comparison of alloying elements in solution and carbide volume at “normal” austenitizing temperatures would also be interesting, while you are at it :). I’m talking of the values calculated with JMatPro and ThermoCalc, if you can take the time…

    PS: in Europe 80crv2 is lovingly referred to as “1.2235” and usually contains small amounts of Si and Mo too. Even some Ni sometimes.

    Br, Dorin

      1. If not much extra work, estimated retained austenite would also be awesome.
        I think this kind of simulation generated data is really nice when discussing steels on a theoretical level.
        Br, dorin

          1. Hello. The numbers definitely seem… Interesting. First of all – where is all the Vanadium?
            Then thr RA numbers – whoa! Quite a lot…
            And lastly, at 1486F (807C) everything is already in solutiom for 80crv2? The recommended austenitizing intervals are usually higher, i’ve even seen 840-880 C (1545-1615 F)….
            But first things first: where’s the vanadium? Carbides? Solution?
            Ow, and thank you for the simulations :).

          2. The calculations are at thermodynamic equilibrium (infinite hold time) so the numbers will be higher than what is achieved with austenitizing at the same temperature. That results in retained austenite numbers being higher as well. This seems to be the case particularly with low alloy steels, whereas calculations with high alloy steels seem to match relatively closely with experiment. Perhaps because hold times are often short with low alloy steels.

            I didn’t include vanadium in the calculations because the amount of carbide is small and the vanadium carbides wouldn’t be dissolving at such low temperature. The amount of VC would be roughly equal to the amount of vanadium, and almost all of it would be tied up in carbides.

  2. Think the .12 difference in carbon content would make a difference in wear resistance and cutting ability?

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