I was wondering that if a proper process of heat treatment (heating to a temperature above upper critical temperature, then soaking and then quenching) is applied to a metal like steel, then what happens within the metal itself on microscopic level which makes it more strong and hard but decreases its toughness?

I mean I know this that when I increase temperature (or in other words, heat) of a metal like steel, then Elastic Modulus and strength decreases (because the metallic bonds become weaker and it is now easier to deform the metal and cause a slip between the layers of metal atoms) and the material becomes softer, but we are also applying heat to the metal in heat treatement but it becomes more hard and strong. How does these two things work in opposite ways?

  • 4
    $\begingroup$ Steve Mould using steel balls to represent crystal structure and how dislocations in that make it harder: youtube.com/watch?v=xuL2yT-B2TM Heating just lets the crystal lattice change, but how it cools basically determines how the crystal structure forms. So it's not application of heat that is really deciding things. It's the progression crystal changing as it cools and you trying to freeze it at the right time to keep it frozen at some part in that progression. $\endgroup$
    – DKNguyen
    Commented Sep 23, 2021 at 14:17
  • 2
    $\begingroup$ @DKNguyen With a little elaboration on the effects of cooling and tempering, this will be a valuable answer to the OP's concerns. $\endgroup$
    – r13
    Commented Sep 23, 2021 at 17:25
  • 1
    $\begingroup$ answers here are right on, but this is several weeks of learning in a Strength of Materials undergrad engineering class. $\endgroup$
    – Tiger Guy
    Commented Sep 23, 2021 at 19:16
  • $\begingroup$ @r13 Unfortunately, that is almost everything I know and anything I would just basically be the difference between annealing, tempering, and hardening all of which involve heat...unless that is what the OP is actually asking. $\endgroup$
    – DKNguyen
    Commented Sep 23, 2021 at 23:39
  • 1
    $\begingroup$ @RameezUlHaq All I have to say about that is that it has nothing to do with metallurgy but the fact you think being hot right now is the same as being hot in the past even if it is cold now. It should be should be evident why these are not the same now that I have pointed it out. If it still isn't clear, I will further point out that heat treated metal still gets weaker at elevated operating temperatures and that these operating temperatures are all lower than what you heat the metal up to during heat treatment. All operating temperatures are cold relative to the heat treatment temperature. $\endgroup$
    – DKNguyen
    Commented Sep 24, 2021 at 15:44

4 Answers 4


In short the heat treatments in steel change the phase of iron between the following phases:

  • Austenite
  • Cementite
  • Martensite
  • Bainite
  • Ferrite
  • Perlite.

(Actually quenching does not allow low temperature phase changes to occur, so effectively the phases are sort of "frozen" in their high temperature equivalents).

enter image description here

Figure 1 : example of continuoous cooling transformation diagram of low alloy steel (sourceindustrialheating)

Of course this is also a function of the Fe-C phase diagram

enter image description here

Figure 2: FE-C phase diagram (source calphad.com)

Each phase has its own characteristic properties. There are several online resources that describe the differences (e.g. Interpretation of the microstructure of STeels).

  • $\begingroup$ OP asked about the physical mechanism, and gave an example of bond weakening during heating. I don't see physical intuition in this answer. It has some graphs, but no explanation. $\endgroup$
    – Craeft
    Commented Sep 27, 2021 at 23:55

The effects of heating-quenching a metal is explained below

Transformation hardening is the heat-quench-tempering heat treatment cycle addressed earlier in this article. It's used to adjust strength and ductility to meet specific application requirements. There are three steps to transformation hardening:

  1. Cause the steel to become completely austenitic by heating it 50 to 100 degrees F above its A3-Acm transformation temperature (from that steel's iron-carbon diagram). This is called austenitizing.

  2. Quench the steel; that is, cool it so fast that the equilibrium materials of pearlite and ferrite (or pearlite and cementite) can't form, and the only thing left is the transitional structure martensite. The idea here is to form 100 percent martensite.

  3. Reduce brittleness by tempering the martensitic steel, which requires heating it, but keep temperatures below A1. Typically, this means temperatures are between 400 and 1,300 degrees F, which allows some of the martensite to turn into pearlite and cementite. Then allow the piece to air-cool slowly.

By using the proper heat treatment and choosing a steel with just the right amount of carbon, you can get just about any combination of hardness and ductility to meet a specific requirement. Remember, the more pearlite and cementite that forms, the more ductile and less brittle the steel will be. Conversely, more martensite means less ductility but more hardness.

One topic has been ignored up to this point is grain structure changes during precipitation hardening. A steel's grain size depends on the austenitizing temperature. When a steel that will transform is heated to slightly above its A3temperature and then cooled to room temperature, grain refinement takes place. Fine grain size offers better toughness and ductility.

Metallurgy Matters: Making steels stronger

  • 1
    $\begingroup$ "earlier in this article" — if you post a quote, please format it as such. Otherwise this looks as if you wrote all this yourself, and just added an "additional info" link at the end, which smells like plagiarism even if it isn't. $\endgroup$
    – Ruslan
    Commented Sep 25, 2021 at 8:35
  • $\begingroup$ I don't have enough clarity in my mind about this yet to write an answer, but I can imagine that, even in the absence of phase transitions, the period at elevated temperature could create a large population of dislocations on different slip systems by thermionic excitation, and the rapidity of the cooling would then stop those dislocations from going away as they would during an anneal; the presence of that large and varied population of dislocations would then lead to an earlier onset of work-hardening during any subsequent plastic deformation of the sample. $\endgroup$ Commented Sep 25, 2021 at 12:05

You are mixing apples and oranges. Many steels harden by rapid cooling, but very few other metals do that; specifically, only aluminum bronze and certain titanium alloys. Many metals will strengthen by age hardening; Rapid cool softens, and then time at a lower temperature strengthens them.

There are a myriad of combinations, like HSS (high-speed steels). They harden by rapid cooling and then age harden during tempering. And many other secrets you can only learn by joining the secret order of metallurgists.

  • $\begingroup$ How many years does it take to progress from initiate to the next level? $\endgroup$
    – NMech
    Commented Sep 23, 2021 at 16:15
  • 1
    $\begingroup$ Four years ( BS) to get into the club , but then there is always something new being developed. Such as micro alloys with Nb ( Cb for old people) ; high strength, very high toughness, very low C, seems to combine martensite with age hardening with ( warm) cold work. $\endgroup$ Commented Sep 23, 2021 at 16:42
  • 1
    $\begingroup$ Micro-alloy are also called TMCP ( thermo mechanical control process). But don't believe all that you find on the net because I am sure some are written by persons not in the secret order of metallurgists. $\endgroup$ Commented Sep 23, 2021 at 16:52
  • 2
    $\begingroup$ IMO, you have misunderstood the OP's concerns. First of all, he mentioned "steel" at the very beginning of the text body. Second, he's curious about why heating (a process that weakening the steel) then fast cooling can add hardness to the steel. Third, what are the metallurgical changes in each phase of the entire process of quenching? Please note that the OP and I both are not a member of the secret society, so please be patient with our ignorances. Nevertheless, the question is a good one. $\endgroup$
    – r13
    Commented Sep 23, 2021 at 17:05
  • $\begingroup$ The first paragraph of the book you ask for is : A conventional steel transforms to austenite ( FCC) at high temperature. If cooled, the austenite is unstable and transforms to ferrite (BCC) and iron carbide in layers ( pearlite). These are relatively slow nucleation and growth reactions. If cooled very rapidly there is not time for nucleation and growth so the austenite instantly transforms ( in each grain) to martensite ( BC TETRAGONAL) , with very small iron carbides throughout. The tetragonal crystal with dispersed carbides has very limited slip planes available so is strong/hard. $\endgroup$ Commented Sep 24, 2021 at 1:11

The other answers describe the "materials science" mechanisms of iron vs. temperature. I'm going to add this:

Matter "tries" to reach a minimum energy state whenever possible. In general, then, if you cool something as slowly as possible, you'll come closest to a solid which is a perfect crystalline structure. See "annealing." If you cool it a bit faster, dislocations and perhaps plane slips start to happen. And if you quench or otherwise cool it quickly, none of the molecules can settle into the low-energy state, and you get a glass. For fun, take a look at videos of Prince Rupert's Drops.

For iron, then, the overall strength in all 3 dimensions, flexibity vs. shatter point, etc., varies across the defined structures. You pick the cooling sequence you want based on the performance requirements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.