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).
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.
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.
Thank you so much. I've been struggling for a whole semester but thanks to you i finally understood everythig. Best regards
Had a few TTT questions on my exam and got them all correct because of this video, thank you. Well worth the watch.
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!!
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 clear explanation!!
Glad you think so!
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.
You are a legend, thank you!
Thank you🔥🔥🔥🙇♀️🙇♀️
Thaanks prof for the video. Ttt questions is nice!
very clear and great video with various of examples!
Thank you!
Wow, this was awesome! Made so much sense! Thank you so much, great video!
gracias Cesar, saludos de Argentina!!!
Much better than my material prof
Profe muy buen video. Saludos desde California
¡Muchas gracias! Un gusto saludarte.
wooooow, you've just made it simpler .AWESOME
This is VERY helful, Thanks Mr. Cesar!
thanks man great video
Thank you! Saludos de Texas.
Thank you so much for your this helpful video
Thank you Very Much!
thanks sir
Most welcome
very helpful video! thanks for sharing
U r the best, thank u very much...
awesome lecture
thank you so much. really helpful !
Thanks you from France
Very helpful! Thank u soooooo much
Golden. Thank you muchas gracias
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).
you are the man
Kansas State Mech Materials represent!
very heplful... thank you so muchhh !
very helpful， thank you！
Thanks alot sir ☺️
thank you sir
kraaal kral
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!
i got it teşekkürler
Thank You
Thank You!
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.
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
Cihangir hocama selamlar 😂😂
allah yarhamlek l walidin :D
roll tide
I still don't get it. My professor is total garbage.
Thank you!