Thank you very much for your comment. Re-heating martensite to temperatures below 350° or above 600°C (always below the eutectoid temperature and avoiding the 375C-575°C Tempered Embrittlement range) will create tempered martensite, increasing toughness and lowering hardness. For spherodization, a perlitic microstructure should be re-heated below the eutectoid temperature for spherodites to form from the lamellar perlite. I haven't come accross a TTT Diagram showing these but typical aging curves (Property vs time for different temperatures).
Sorry i didn’t understand why u proceed the following steps all from 0 second , since ur calculation of percentage takes accounts of previous step loss of austenite ,they transformed into others, then to calculate whats left and what they are going to change further should start from the point of last step ending , The shape of transformation should be a staircase isnt it?
Excellent point! This is something which is not as straight forward as it might seems to be at first. Whatever we can discuss about this, keep in mind that TTT diagrams are made by quenching to ONE single temperature and waiting until full transformation; when using the TTT diagram in the way we do (quenching to several different temperatures) we are pushing the concept a little bit too far; none the less, it’s still a good first approach. Now, also keep in mind that, when quenching a material, heat transfer takes place at the surface of the material, so the center of the material is the last part to cool down; finally, we are working with a log-log scale, so that “numbers” at the far left are orders of magnitude smaller than numbers to the right; finally, control of temperature is not perfect in practice. Taking all this into account, although your idea is “quite right” it needs a small correction (and it would require to assume that the full material is at the same temperature everywhere at any time, which requires a very small sample with a very high relationship Area to Volume and a very high heat transfer coefficient); if that is the case, it wouldn’t be a “stair” but rather a “straight line” with a slope equivalent to the “cooling speed” of the quenching process departing from the starting cooling temperature. That is way, assuming that your quenching is fast enough (“very high” cooling speed), then the assumption of a straight vertical line (infinite slope) is the other option you have, and this is the one I’m using. In practice, what we need to do is a simulation of the heat transfer and analyzing the thermal path of each part of the material. That is way we also have the CCT Diagrams which it is a more practical option.
Is it possible to calculate the resulting hardness of say example 2? would it be correct to say 50% Bainite at 41HRC and 50% Martensite with 57HRC results in a hardness of 20.5+28.5=49HRC? Thanks
On average, yes. What you propose is known as "The proportions rule" and it applies to this example in the way you present it. Excelent! It is important to recognize that, in reality, you will have regions with different microstructures and properties; hence, this average value might be of significance or not, depending on the purpose of your answer. For example, it might be that you have a harder surface than the core, which will be benefitial in applications where you need certain ductility for the whole element with a hard surface that might give it stiffness. In such case, this average might no be as imprtant as the distribution of microstructures.
Had a few TTT questions on my exam and got them all correct because of this video, thank you. Well worth the watch.
Great video. I needed just this to ace my materials final exam. Thank you for sharing this.
final? this is on my midterm
Your voice makes me feel like ronaldo is teaching
thank you !!! Why cant the profs in my school be as straight forward as you about this concepts.
Very helpful!! watched this 2hrs before my final paper and managed to grasp the concept
Thank you so much Dr. Núñez!! This makes so much sense now!
thanks a lot...now my math concept is cleared.
Wow, this was awesome! Made so much sense! Thank you so much, great video!
Thaanks prof for the video. Ttt questions is nice!
You are a legend, thank you!
Very clear explanation!!
Glad you think so!
Profe muy buen video. Saludos desde California
¡Muchas gracias! Un gusto saludarte.
Thank you! Saludos de Texas.
very helpful! I just wish one example had one with tempered martensite or spheroidite, but a great video nonetheless
Thank you very much for your comment. Re-heating martensite to temperatures below 350° or above 600°C (always below the eutectoid temperature and avoiding the 375C-575°C Tempered Embrittlement range) will create tempered martensite, increasing toughness and lowering hardness. For spherodization, a perlitic microstructure should be re-heated below the eutectoid temperature for spherodites to form from the lamellar perlite. I haven't come accross a TTT Diagram showing these but typical aging curves (Property vs time for different temperatures).
This is VERY helful, Thanks Mr. Cesar!
very helpful video! thanks for sharing
thank you sir
Sorry i didn’t understand why u proceed the following steps all from 0 second , since ur calculation of percentage takes accounts of previous step loss of austenite ,they transformed into others, then to calculate whats left and what they are going to change further should start from the point of last step ending ,
The shape of transformation should be a staircase isnt it?
Excellent point! This is something which is not as straight forward as it might seems to be at first. Whatever we can discuss about this, keep in mind that TTT diagrams are made by quenching to ONE single temperature and waiting until full transformation; when using the TTT diagram in the way we do (quenching to several different temperatures) we are pushing the concept a little bit too far; none the less, it’s still a good first approach. Now, also keep in mind that, when quenching a material, heat transfer takes place at the surface of the material, so the center of the material is the last part to cool down; finally, we are working with a log-log scale, so that “numbers” at the far left are orders of magnitude smaller than numbers to the right; finally, control of temperature is not perfect in practice. Taking all this into account, although your idea is “quite right” it needs a small correction (and it would require to assume that the full material is at the same temperature everywhere at any time, which requires a very small sample with a very high relationship Area to Volume and a very high heat transfer coefficient); if that is the case, it wouldn’t be a “stair” but rather a “straight line” with a slope equivalent to the “cooling speed” of the quenching process departing from the starting cooling temperature. That is way, assuming that your quenching is fast enough (“very high” cooling speed), then the assumption of a straight vertical line (infinite slope) is the other option you have, and this is the one I’m using. In practice, what we need to do is a simulation of the heat transfer and analyzing the thermal path of each part of the material. That is way we also have the CCT Diagrams which it is a more practical option.
kraaal kral
you are the man
Very helpful! Thank u soooooo much
thanks sir
Most welcome
Thanks alot sir ☺️
i got it teşekkürler
Thank you so much. I've been struggling for a whole semester but thanks to you i finally understood everythig. Best regards
YOU ARE A SAVIOUR!!!!!!! I was sick at that lecture and i wasn't able to understand this from slides. Thank you so much, really really appreciated!!
Much better than my material prof
Thank you🔥🔥🔥🙇♀️🙇♀️
thanks man great video
I still don't get it. My professor is total garbage.
Thank you Very Much!
thank you so much. really helpful !
Cihangir hocama selamlar 😂😂
Thank you so much for your this helpful video
wooooow, you've just made it simpler .AWESOME
U r the best, thank u very much...
Thanks you from France
very heplful... thank you so muchhh !
Golden. Thank you muchas gracias
very helpful， thank you！
Thank You
awesome lecture
Thank you!
Thank You!
Kansas State Mech Materials represent!
gracias Cesar, saludos de Argentina!!!
Is it possible to calculate the resulting hardness of say example 2?
would it be correct to say 50% Bainite at 41HRC and 50% Martensite with 57HRC results in a hardness of 20.5+28.5=49HRC?
Thanks
On average, yes. What you propose is known as "The proportions rule" and it applies to this example in the way you present it. Excelent! It is important to recognize that, in reality, you will have regions with different microstructures and properties; hence, this average value might be of significance or not, depending on the purpose of your answer. For example, it might be that you have a harder surface than the core, which will be benefitial in applications where you need certain ductility for the whole element with a hard surface that might give it stiffness. In such case, this average might no be as imprtant as the distribution of microstructures.
Thank you for your detailed answer and quick reply! very helpful!
very clear and great video with various of examples!
Thank you!
allah yarhamlek l walidin :D
water boils at 100C which is equivalent to 212 F (Not 100F). does that not the first answer?
Kiarash Amini not really. If you see, the graph ends at about 200F
roll tide