Dear Vibhor: Prof Zisman has given good information, Just to add to his comments, there is no true Mf above room temperature, i.e. in C steels there will always be some retained austenite at room temperature even in low and medium carbon steels. I address this at several places in my book STEELS PROCESSING, STRUCTURE, AND PERFORMANCE, SECOND EDITION, 2015, ASM International and recommend that you obtain a copy. The amount of martensite that forms is a function only of undercooling below Ms according to the Koistinen Marburger equation (Eq. 5.2, p 73). Therefore if MS is known or calculated (Andrews linear equation, Table 5.1) the amount of martensite, and by difference the amount of retained austenite, can be calculated. I have found this approach to be effective in medium carbon steels such as 4340. Keep in mind that Ms may be dependent not only on chemistry, but also on the effectiveness of austenitizing, since retained carbides will affect austenite chemistry and Ms. The amounts of retained austenite in medium carbon steels will be low, 2 pct in 4130, 4 pct in 4340 (see Fig 17.17, p 389. Why are you interested in retained austenite. Keep in mind it transforms to martensite by stress induced mechanisms in lath martensites (p 419), and transforms thermally at tempering temperatures above 200F to ferrite cementite mixtures ( p389). Best wishes, George Krauss     

1) Theoretically, there is no Mf since the martensite fraction in cooling approaches unity only asymptotically (exponentially), whereas the often used Mf is a kind of approximation determined on a dilatogram as a point where the L(T)  dependence  becomes visibly linear.

2) The said above is manifested, in particular, by a "paradox" of retained austenite in martensitic steel that always takes place in some quantity even if the cooling stops notably under the conventional Mf.

3) Since you could not get the PURE martensite (austenite-free), you will hardly get the thermal expansion diagram you mentioned. As to theoretical prediction of the latter, it would not be not accurate enough.

4) What is interesting indeed, how should we improve the theoretical

P mar =1 -exp{ -const*[Ms-T] } 

when taking in mind existence of ABSOLUTE ZERO=-273 C.  

 Dear Vibhor Atreya,

It seems to me that a certain volume fraction of retained austenite is necessary in view of the integrity of the specimen. So, the experimental Mf is simply the temperature of the cooling medium after quenching. The amount of retained austenite will depend on the specimen composition. The problem behind your question (residual stresses/strains and residual austenite) is dealt with in several textbooks on the martensite transformations. I am quite sure that the library of Delft University of Technology and/or of your  Faculty can deliver you the desired information. 

Thank You A.A. Zisman and P. Van Mourik. What if I want to calculate the fraction of retained austenite when the experiment stops (at room temp.)? Since the transformation isn't complete and I do not get to the stage where dilation becomes visibly linear, there seems to be no way to calculate it (otherwise I would have used extrapolation and lever rule at RT).

Dear Vibhor,

There is several empirical formulas to evaluate a M fraction and, hence, RA fraction depending on the final temperature and steel composition. See for example expressions by Koistinen and Marburger and by Magee, but do not forget each of them were fitted for certain chem. composition ranges only.

About extrapolation and stage... Let us consider APPROXIMATE Mf in use which are often HIGHER   than room temperature. With realistic accuracy of dilatometric curves such Mf is evaluated at the point of TANGENT DEFLECTION from the curve; this in no way means physically no RA but only its low quantity, say 0.1 to 1.5%, where the dilatogram non-linearity becomes hardly perceptible.  

From my experience, the mentioned formulas for Pm(T) lose accuracy at M volume fraction exceeding 95% and, unfortunately, we still have no reliable techniques to accurately measure accordingly small fractions of RA (so big fractions of M). My hope is that improved neutron diffraction techniques will solve the problem, although currently it is not the case. 

.

Dear Vibhor: Prof Zisman has given good information, Just to add to his comments, there is no true Mf above room temperature, i.e. in C steels there will always be some retained austenite at room temperature even in low and medium carbon steels. I address this at several places in my book STEELS PROCESSING, STRUCTURE, AND PERFORMANCE, SECOND EDITION, 2015, ASM International and recommend that you obtain a copy. The amount of martensite that forms is a function only of undercooling below Ms according to the Koistinen Marburger equation (Eq. 5.2, p 73). Therefore if MS is known or calculated (Andrews linear equation, Table 5.1) the amount of martensite, and by difference the amount of retained austenite, can be calculated. I have found this approach to be effective in medium carbon steels such as 4340. Keep in mind that Ms may be dependent not only on chemistry, but also on the effectiveness of austenitizing, since retained carbides will affect austenite chemistry and Ms. The amounts of retained austenite in medium carbon steels will be low, 2 pct in 4130, 4 pct in 4340 (see Fig 17.17, p 389. Why are you interested in retained austenite. Keep in mind it transforms to martensite by stress induced mechanisms in lath martensites (p 419), and transforms thermally at tempering temperatures above 200F to ferrite cementite mixtures ( p389). Best wishes, George Krauss     

Prof Zisman & Prof Krauss: Thank you very much for your valuable answers.

Prof Krauss: I am trying to compare certain properties of some martensitic microstructures. So I do not want too much retained austenite. I wanted to ascertain if I'll be needing sub zero cooling.  

Sub-zero cooling can help indeed. Recently I experimented with a low-C (0.1%) martensitic steel and have revealed the following.

1. After sufficiently long and high tempering RA is not perceptible i.e. it has a volume fraction

Please read Chapter 18 in STEELS. The answer to your question depends on carbon content and tempering and retained austenite (p419) as discussed there. Best wishes, George Krauss

Dear Mr. Atreya,

The best solution is to use a dilatometer with sub-zero quenching. However, as far as I understand, you don`t have access to such a dilatometer.

In that case, there are several work-arounds and some of those are already mentioned in the answers. The main problem of those is that they require certain assumption which may not be valid for your case. Thus, the results must be considered with a grain of salt.

I suggest the following approach , for this approach to work, you need to have :

1.  Thermal expansion coefficient of 100% martensite : The martensite should have the same composition with the tested conditions. For this, you can do an external sub-zero treatment and determine CTE in heating. 

2. CTE for Metastable Austenite : This is relatively easy , it can be acquired from quenching tests in your dilatometer. 

3. Thermal Strain for transformation from austenite to martensite : This is a little bit tricky. You also need to carefully determine the retained austenite content by a reliable method such as XRD. You can estimate the thermal strain of transformation by room temperature measurements and those. Please,refer to my latest publications and also the work of Acht et al. (which is in the references of my study.).

This method is prone to weakness of assumptions and you can accumulate errors in each measurement as each measurement has a budget of errors. The combined error will be larger. Thus, you need very precise measurements for all.

I hope that this works for your case. 

Moreover, if you have only a few samples, we can do the sub-zero tests in my institute for free. If you have many samples, we can provide service at a certain fee.

Kind Regards,