You have asked a very interesting but at the same time a very difficult question. Honestly, I have no answer. But thanks for asking as it is forcing me to look for an answer.
I don't know the exact answer, but since Sir has yet to answer your question, I will try to answer. Every transformation definitely needs time. Take an example of eutectoid steel (0.8%C); the Iron-carbon diagram shows complete austenite will transform to pearlite at eutectoid temperature, but it will not give you how much time you need for transformation. The time comes from the TTT diagram. So if you do not provide enough time for austenite to transform to pearlite or you quench the austenite, it will transform to Martensite, not pearlite. In the same way, if you don't give enough time to cool to austenite at A3 or Acm, it will not form pro eutectoid ferrite or pro eutectoid cementite. It means if you cool it fast, it will not even start the transformation to pro-eutectoid components and stay as unstable austenite only. Please observe the alpha start line in the TTT diagram of pro-eutectoid steel. Please correct me if I am wrong.
@@pinnavarunkrishna5712 the curve will shift towards right as the carbon content increases (because both nucleation and growth rate decreases ac c content increases)
C curve will shift to left when decrease carbon content from eutectoid composition because tendency to formation of pre eutectoid ferrite will increase. But after eutectoid composition C curve will not significantly shift because here one factor leads to shifting of curve towards left again bcz of formation of pre eutectoid cementite and another factor counterbalance it. So after eutectoid composition C curve will not again significantly shift towards left. So the final answer is with increase in carbon content till eutectoid C curve will go from left to right but then not again significantly shift to left.
Austenite is unstable below the eutectoid temperature. So in equilibrium, at long time, it will disappear. But at intermediate times, as it is transforming to equilibrium phases, it can be present.
i think perhaps it is that annealing of hypereutectoid steel renders primary microconstituent of cementite which also forms the matrix rendering steel brittle
Bainite is not a single phase but a mixture of two phases: ferrite+cementite. Thus it should be called a microconstituent and not a phase. It is a nonequilibrium microconstituent.
Proeutectoid alpha has the same crystal structure as eutectoid alpha (BCC interstitial solid solution of Fe with C). However, they form at different stages of cooling of austenite. Proeutectoid alpha form first above the eutectoid temperature. Eutectoid alpha forms at the eutectoid temperature along with cementite. because of this history, their appearance in microstructure is different. Proeutectoid alpha appears more like equiaxed grains. Eutectoid alpha as lamellae of alpha sandwiched between lamellae of cementite. Thus phase-wise both are the same phase alpha. The adjectives 'proeutectoid' and 'eutectoid' describes the history of formation and the resulting morphology (shape).
To begin with we have asutenite (A). Part of it transforms to proeutectoid ferrite (Alpha, F). This transformation begins at the green line. Thus below the green line and above the black line you have A+ Alpha (F). The black line below the green line is the peralite start curve. Austenite begins to transform to pearlite on crossing this line. Thus below this line, you have proeutectiod ferrite (F), some pearilte (P) and some remaining austenite (A) which is yet to transform. Below the next black line all the austenite has transformed, so you have only P+A.
@@introductiontomaterialsscience Thanks for your answer. You mean that the end line of transformation A¬F, coincides with the beginning line of transformation A-P,
You have saved me... thank you sir....All the concepts are clear now...🤗
this is the best video I've ever seen, thank you so much for your work. Regards from Argentina
All the concept u teach us are very good and in impressive manner.. thank you sir
Thanks a lot sir , I got it after watching a lot of videos and really it cleared my all concepts
Thanks sir .....it really helped me clear all my doubts
Very informative
Sir, will the C curves for hypo-eutectoid and hyper-eutectoid steel shift to the left and right(vis a vis C curve of Eutectoid steel) respectively?
so much thankfull to you sir
Sir below the nose of the C curve why the Pro eutectoid ferrite and cementite are not formed in the hypo and hyper eutectoid steels ?
You have asked a very interesting but at the same time a very difficult question. Honestly, I have no answer. But thanks for asking as it is forcing me to look for an answer.
I don't know the exact answer, but since Sir has yet to answer your question, I will try to answer. Every transformation definitely needs time. Take an example of eutectoid steel (0.8%C); the Iron-carbon diagram shows complete austenite will transform to pearlite at eutectoid temperature, but it will not give you how much time you need for transformation. The time comes from the TTT diagram. So if you do not provide enough time for austenite to transform to pearlite or you quench the austenite, it will transform to Martensite, not pearlite. In the same way, if you don't give enough time to cool to austenite at A3 or Acm, it will not form pro eutectoid ferrite or pro eutectoid cementite. It means if you cool it fast, it will not even start the transformation to pro-eutectoid components and stay as unstable austenite only. Please observe the alpha start line in the TTT diagram of pro-eutectoid steel. Please correct me if I am wrong.
@@introductiontomaterialsscience hello sir have you found the answer? please enlighten us if you have. thanks
Love u sir.
Sir in hypoeutectoid and hypereutectoid the carbon content will differ when compared to eutectoid so will the C-curve shift left or right accordingly?
Nope
Only Ms and Mf vary
Why won't the curve shift?
@@pinnavarunkrishna5712 the curve will shift towards right as the carbon content increases (because both nucleation and growth rate decreases ac c content increases)
C curve will shift to left when decrease carbon content from eutectoid composition because tendency to formation of pre eutectoid ferrite will increase. But after eutectoid composition C curve will not significantly shift because here one factor leads to shifting of curve towards left again bcz of formation of pre eutectoid cementite and another factor counterbalance it. So after eutectoid composition C curve will not again significantly shift towards left.
So the final answer is with increase in carbon content till eutectoid C curve will go from left to right but then not again significantly shift to left.
Sir
Why are hypereutectoid steel not Furnace cooled from all austenite region
Sir between A1 and A3 will we get stable austenite or unstable austenite?
Sir, Why the curve of Pro eutectoid Alpha touches the nose? Why not it further goes down?
This is an interesting question. I do not have an answer to it.
I have a question you wrote A + p below eutectoid line as austenite is unstable below eutectoid line. So how it is possible ?
Austenite is unstable below the eutectoid temperature. So in equilibrium, at long time, it will disappear. But at intermediate times, as it is transforming to equilibrium phases, it can be present.
Can we get knowledge of cct and how it is applied practically in heat treatment??
In these videos, I have not discussed the CCT diagrams. You can look up some good books, e.g., Physical metallurgy by Reed-Hill.
Why Annealing of hyper eutectoid steel not recommended...?
Justify plz
i think perhaps it is that annealing of hypereutectoid steel renders primary microconstituent of cementite which also forms the matrix rendering steel brittle
is there a way to get pearlite without proeutecoid alpha in hypoeutectoid steel?
No some proeutectoid alpha will always form with pearlite. But you can get Bainite without proeutectoid alpha.
Bainite is an equilibrium phase or not???
Bainite is not a single phase but a mixture of two phases: ferrite+cementite. Thus it should be called a microconstituent and not a phase. It is a nonequilibrium microconstituent.
Literally a godsend. Indian tax dollars are literally funding my education.
Sir, how is pro eutectoid alpha different that from regular alpha
Proeutectoid alpha has the same crystal structure as eutectoid alpha (BCC interstitial solid solution of Fe with C). However, they form at different stages of cooling of austenite. Proeutectoid alpha form first above the eutectoid temperature. Eutectoid alpha forms at the eutectoid temperature along with cementite. because of this history, their appearance in microstructure is different. Proeutectoid alpha appears more like equiaxed grains. Eutectoid alpha as lamellae of alpha sandwiched between lamellae of cementite.
Thus phase-wise both are the same phase alpha. The adjectives 'proeutectoid' and 'eutectoid' describes the history of formation and the resulting morphology (shape).
A+P+F in that field? I don't think so....
To begin with we have asutenite (A). Part of it transforms to proeutectoid ferrite (Alpha, F). This transformation begins at the green line. Thus below the green line and above the black line you have A+ Alpha (F). The black line below the green line is the peralite start curve. Austenite begins to transform to pearlite on crossing this line. Thus below this line, you have proeutectiod ferrite (F), some pearilte (P) and some remaining austenite (A) which is yet to transform. Below the next black line all the austenite has transformed, so you have only P+A.
@@introductiontomaterialsscience Thanks for your answer. You mean that the end line of transformation A¬F, coincides with the beginning line of transformation A-P,
@@LaszloVondracsek That's right.
Sir, is it possible to make 100% pearlite for hypoeutectoid and hypereutectoid steels??
No
For hypoeutectoid,it is possible but for hypereutectoid not possible
It will depend upon composition.
Messed up
playback speed should be 10x
very slow