How do you become University lecturer? You don't know the subject and you are teaching and spoiling the students life in your University. 😤😤🤬🤬 Please 🙏 quit your job. It's my kind regards.
@@haran2023 Did it occur to you that he's interested in someone else's teaching method? Or maybe he lectures a different subject? This is such an ignorant comment.
@@vijayr1485 you fool he says he says he is lecturer and yet he is learning from UA-cam videos. Do you know why you are here 😂 to watch this video?? Because lecturer's like him is in your college and teaching you.
@@haran2023 You are what they call 'scum' of the internet. There was no need to ridicule a complete stranger on the internet who did no harm to you whatsoever. But You had to let the gutters flow, right? I'm sure you must have patted yourself on the back for being an obnoxious troll when you went to bed with your lonely and worthless existence.
This video makes a great explanation of this concept, I am going to explain this in class tomorrow based on the examples you presented. Greetings, from México!
I'm an Australian student and I've seen many professors on Coursera and UA-cam for big American universities and Australian universities, and none of them explain as clearly as you do - thank you so much!
Great introductory video about creep. Well explained, concise, supported by drawings. Also, excellent handwriting and good level of academic English. Subscribed.
Sir it requires lot of patience to tie camera on forehead and shoot such marvellous videos for 1-2 hrs straight...Hats off your content and dedication.
Great Lecture! Why Spring 1 did not elongate and Spring 2 did? Seems the set up was the same for both, apparently same material and same load. Thanks for sharing this amazing content!
sir, this is a query i wanted to raise in grain boundary lecture as well. sir, why do we prefer single crystal material for gas turbine blades(working at very high temperature.). Sir, i came across an explanation that at high temperature creep/fracture propagates through grain boundaries (as they are weaker region) hence we try to minimise the no of grain boundaries to avoid fracture.i wanted to ask if this is the correct explanation and if yes then why grain boundaries are weaker at higher temperature(which ultimately result in intergranular fracture). because the explanation of the same was not mentioned in the book. i have completed all the lectures before raising this question. I will be very thankful to you if you could respond to my query . Thank you , sir
I was asked this question by my professor. He asked me to explain it in today's lecture, and I am proudly going to explain it today with examples. 😀Thanku sir
Stress rupture data are sometimes correlated with the Dorn parameter, θ = t exp[-Q/(RT )], where t is the rupture time, T is absolute temperature, and θ is assumed to depend only on stress. If this parameter correctly described a set of data, then a plot of log(t) versus 1/T for data at a single level of stress would be a straight line. If the Larson-Miller parameter correctly correlates the data, a plot of data at constant stress (therefore constant P) of log (t) versus 1/T also would be a straight line. A. If both parameters predict straight lines on log (t) versus 1/T plots, are they really the same thing? B. If not, how do they differ? How could you tell from a plot of log (t) versus (1/T ) which parameter better correlates a set of stress rupture data?
The equation θ = t exp[-Q/(RT )] is essentially an Arrhenus rate equation with Q as the activation energy. Both Q and θ are constant at a given stress. Thus if Q and θ are determined for a given stress the rupture time t (or time for a fixed amount of strain) can be calculated at any given temperature. The important point is that Q and θ are constant at a given stress. What hapens if the stress is changed. Well, in reality both Q and θ may change. The two parameters are based on two simplifying assumptions: 1. Larsen-Miller para,eter assumes that Q varies with stress whereas θ is constant. The L-M parameter is essentially the value of Q. 2. Dorn parameter or Sherby-Dorn parameter or Orr-Sherby-Dorn parameter assumes that Q is constant and θ varies with stress. Thus although, both are based on the same Arrhenius equation at a given stress, their assumptions regarding the variation of constants in Arrhenius equation with stress is different. Thus their predictions will also be different. A good discussion of this with worked-out problems can be found n N.E. Dowling, Mechanical Behaviour of materials.
its pb/sn alloy and its temp ration beyond 0.4Tm. so its behave on high temperature with its own weight constant stress. inside the metal different mechanisms of creep happens. dislocation creep always occur and diffusion creep occur because of high temp and grain boundaries are sliding. this cause to elongate the entire spring.
This professor failed me twice in one of his cources and didn't even allow me to do the course during summer break. I wouldn't argue the fact that I was weak in studies but what I would like to point out is that professors at IITs have no intentions of helping weak students who actually need it. They instead just love the students who excel and are very cordial and helpful to them. The good students become their friends and the weak ones are left to become an outlet of their collective frustration. Good luck to his "friend Srikanth" but my heart goes out to those weak students whos voices are often unheard and even misinterpreted and they're labeled lazy, incapable and what not.
@@himanshushekhar3365 Everyone can be kind to the strong and the smart ones. I very vividly remember telling him that I couldn't afford the extra fee for the extended degree but he just didn't allow me to do that solitary course I needed to complete my degree in the summer.
@@himanshushekhar3365 one of my close family member had met with a serious accident and I missed one too many classes because of that and he didn't even to listen to me. So much for a well behaved and interactive professor.
High or low temperature for a material is relative to its melting point. For snow even a temperature of -20 C is rather close to its meting point and so it feels hot!
Dear sirr...... I have to tell u a little fundamental confusion that creep is the accumulation of plastic strain wrt time under its self weight ( it will be time depending function) but when we are calculating deformation due to self weight i.e ∆l=(√*L^2)/(2E) there will be no any time function ...... Why ?? ..Is It Also one kind creep deformation.. please tell me more on this sir?????
Thank you very much sir.......it's a very pleasure for .me💜 But sir ...elastic deformation Doesn't depending upon time ...... So.... Every material have some kind Small elongated at ( ∆t=O ) so including its actual length...... So can I say sir every material as see in day to day life , length of That's material doesn't it actual length it must be include small amount of inertial deformation.????
Nice explanation with simple example. I wanted to understand is Temp. ratio >0.5 for metals? Does it also applicable to plastics. if not for plastic, what is the ratio for different types of plastics. i will be grateful to you, if you can share table with ratios for different type of materials.
According to Principles of Polymer Engineering by McCrum, Buckley and Bucknall (2nd. edn.), p125, polymers creep at all temperatures above minus 200 degrees C. Thus even at room temperature polymers creep.
If you do a fixed load test then you will not get constant stress. The stress will increase as the cross-sectional area decreases, as you have correctly pointed out. But in more careful tests, it is possible to adjust the load to keep the stress constant.
Respected sir I have some doubts after listening this video. Why we will choose the 0.5Tm as creep temperature? How we will decide one material creep temperature? Below this temperature will creep occur or not?
This is just a rough rule based on experience. In reality, creep happens at all temperatures. But the rate depends on temperature. It increases exponentially with temperature. So at low temperatures it is hardly observable. Creep rates become significant at about 0.5 Tm.
Sir please help me solve this doubt🙄 Let's consider tungsten filament with Tm=3410°C. 0.5Tm=1705°C. So creep would be significant only at a temperature greater than this I guess. Even this is a Very High temperature and all other electrical components will melt at this temperature. So I believe tungsten undergoes creep certainly at a temperature lower than this I'm totally confused with this. Please help to solve this confusion/misunderstanding
Respected sir, Please go through the definition of Atomic Energy mentioned in section 2(a) of the Atomic Energy Act 1962. Unluckily, this definition covers only Nuclear Energy related Matters and the matters related to atomic energy are out of scope of the Act 1962. Our Act 1962 is available on internet to entire world so the scientifically developed country know the facts regarding funds been spend on nano technology.
This an interesting topic. Perhaps I should make a new video on this. One can do several things to make a component creep resistant: 1. Choose an alloy of high melting point 2. Solid solution strengthening 3. Precipitation strengthening 4. align grain boundaries along the stress axis (directional solidification) 5. remove grain boundaries altogether (single crystal), 6. use internal cooling to keep the temperature of the component low 7. use thermal barrier coating to insulate the component from heat. There is a video on this channel that discusses some of the techniques with respect to making creep-resistant turbine blades: ua-cam.com/video/SHf_KiIwmyo/v-deo.html
I am university's lecturer myself and i find your UA-cam video contents are very good and i learn a lot from it. keep up the good work!
🤣🤣🤣
How do you become University lecturer? You don't know the subject and you are teaching and spoiling the students life in your University.
😤😤🤬🤬
Please 🙏 quit your job.
It's my kind regards.
@@haran2023 Did it occur to you that he's interested in someone else's teaching method? Or maybe he lectures a different subject? This is such an ignorant comment.
@@vijayr1485 you fool he says he says he is lecturer and yet he is learning from UA-cam videos.
Do you know why you are here 😂 to watch this video??
Because lecturer's like him is in your college and teaching you.
@@haran2023 You are what they call 'scum' of the internet. There was no need to ridicule a complete stranger on the internet who did no harm to you whatsoever. But You had to let the gutters flow, right? I'm sure you must have patted yourself on the back for being an obnoxious troll when you went to bed with your lonely and worthless existence.
This video makes a great explanation of this concept, I am going to explain this in class tomorrow based on the examples you presented. Greetings, from México!
I'm an Australian student and I've seen many professors on Coursera and UA-cam for big American universities and Australian universities, and none of them explain as clearly as you do - thank you so much!
Great introductory video about creep. Well explained, concise, supported by drawings. Also, excellent handwriting and good level of academic English. Subscribed.
IIT
Amazing! Big respect to you, sir, to your students and to your work!
Are you preparing for any exam?
@Dark Knight No.
Sir it requires lot of patience to tie camera on forehead and shoot such marvellous videos for 1-2 hrs straight...Hats off your content and dedication.
Fantastic, pedagogical explanation & lecture. Thank you very much indeed for uploading it.
Great Lecture! Why Spring 1 did not elongate and Spring 2 did? Seems the set up was the same for both, apparently same material and same load. Thanks for sharing this amazing content!
What an explanation sir.
Lacture made me understand the creep .
Thankyou so much sir.
Love you😘😘
sir, this is a query i wanted to raise in grain boundary lecture as well. sir, why do we prefer single crystal material for gas turbine blades(working at very high temperature.). Sir, i came across an explanation that at high temperature creep/fracture propagates through grain boundaries (as they are weaker region) hence we try to minimise the no of grain boundaries to avoid fracture.i wanted to ask if this is the correct explanation and if yes then why grain boundaries are weaker at higher temperature(which ultimately result in intergranular fracture). because the explanation of the same was not mentioned in the book.
i have completed all the lectures before raising this question. I will be very thankful to you if you could respond to my query .
Thank you , sir
Sir it would be so great if you make videos on metallurgical thermodynamics
The experiment and the lecture was really helpful thankyou!😀
I was asked this question by my professor. He asked me to explain it in today's lecture, and I am proudly going to explain it today with examples. 😀Thanku sir
Thank you for a great explanation with a demo... But pls teach faster...
Thanks. Others have also commented on slow speed. May be nest time would be faster.
The teaching speed is absolutely fine Sir. No need to go fast
You gave us some good example of creep.
Nicely explained! . Much appreciated!
Wow great demonstration
👍👍👍 *Thanks Sir, your explanation is easy to understand*
I love these videos ! Helps me so much
good illustration
Stress rupture data are sometimes correlated with the Dorn parameter, θ = t exp[-Q/(RT )], where t is the rupture time, T is absolute temperature, and θ is assumed to depend only on stress. If this parameter correctly described a set of data, then a plot of log(t) versus 1/T for data at a single level of stress would be a straight line. If the Larson-Miller parameter correctly correlates the data, a plot of data at constant stress (therefore constant P) of log (t) versus 1/T also would be a straight line.
A. If both parameters predict straight lines on log (t) versus 1/T plots, are they really the same thing?
B. If not, how do they differ? How could you tell from a plot of log (t) versus (1/T ) which parameter better correlates a set of stress rupture data?
The equation θ = t exp[-Q/(RT )] is essentially an Arrhenus rate equation with Q as the activation energy. Both Q and θ are constant at a given stress. Thus if Q and θ are determined for a given stress the rupture time t (or time for a fixed amount of strain) can be calculated at any given temperature.
The important point is that Q and θ are constant at a given stress. What hapens if the stress is changed. Well, in reality both Q and θ may change. The two parameters are based on two simplifying assumptions:
1. Larsen-Miller para,eter assumes that Q varies with stress whereas θ is constant. The L-M parameter is essentially the value of Q.
2. Dorn parameter or Sherby-Dorn parameter or Orr-Sherby-Dorn parameter assumes that Q is constant and θ varies with stress.
Thus although, both are based on the same Arrhenius equation at a given stress, their assumptions regarding the variation of constants in Arrhenius equation with stress is different. Thus their predictions will also be different.
A good discussion of this with worked-out problems can be found n N.E. Dowling, Mechanical Behaviour of materials.
Thank you so much sir ❤️
Awesome lecture
Sir please tell the reason why curve is like this? What phenomena dominates in all three regions?
Thank you sir for such nice lecture on Creep.
Thank you so much sir 🙏🏼
Love from anavkar sir ❤️
Thankyou shrikant
bIG Fan of you sir,
KEEP UP THE GOOD WORK HAPPY INTERNATIONAL TEACHERS DAY
Sir may I know y the second spring undergone deformation. Any load was applied on it
its pb/sn alloy and its temp ration beyond 0.4Tm. so its behave on high temperature with its own weight constant stress. inside the metal different mechanisms of creep happens. dislocation creep always occur and diffusion creep occur because of high temp and grain boundaries are sliding. this cause to elongate the entire spring.
Watch 14.00
Thank you so much!
never met sir like you in ms
I like your detailed explanation of Heat treatment process sir. Could you explain about post heat treated of steel stress and strain curve behavior?
This professor failed me twice in one of his cources and didn't even allow me to do the course during summer break. I wouldn't argue the fact that I was weak in studies but what I would like to point out is that professors at IITs have no intentions of helping weak students who actually need it. They instead just love the students who excel and are very cordial and helpful to them. The good students become their friends and the weak ones are left to become an outlet of their collective frustration. Good luck to his "friend Srikanth" but my heart goes out to those weak students whos voices are often unheard and even misinterpreted and they're labeled lazy, incapable and what not.
Bro he’s one of the best professors of IITD in terms of behaviour and interactions.
@@himanshushekhar3365 Everyone can be kind to the strong and the smart ones. I very vividly remember telling him that I couldn't afford the extra fee for the extended degree but he just didn't allow me to do that solitary course I needed to complete my degree in the summer.
@@himanshushekhar3365 one of my close family member had met with a serious accident and I missed one too many classes because of that and he didn't even to listen to me. So much for a well behaved and interactive professor.
If creep is considered a high temperature phenomenon, how is it a factor of glacier movement?
High or low temperature for a material is relative to its melting point. For snow even a temperature of -20 C is rather close to its meting point and so it feels hot!
Dear sirr......
I have to tell u a little fundamental confusion that creep is the accumulation of plastic strain wrt time under its self weight ( it will be time depending function) but when we are calculating deformation due to self weight i.e ∆l=(√*L^2)/(2E) there will be no any time function ...... Why ?? ..Is It Also one kind creep deformation.. please tell me more on this sir?????
The formula you are suggesting will give elastic elongation. It cannot give plastic deformation.
Thank you very much sir.......it's a very pleasure for .me💜
But sir ...elastic deformation Doesn't depending upon time ...... So.... Every material have some kind Small elongated at ( ∆t=O ) so including its actual length...... So can I say sir every material as see in day to day life , length of That's material doesn't it actual length it must be include small amount of inertial deformation.????
@@sibammajumdar2749 I will agree with you on that.
Thanks u sirr..........
Nice explanation with simple example. I wanted to understand is Temp. ratio >0.5 for metals? Does it also applicable to plastics. if not for plastic, what is the ratio for different types of plastics. i will be grateful to you, if you can share table with ratios for different type of materials.
According to Principles of Polymer Engineering by McCrum, Buckley and Bucknall (2nd. edn.), p125, polymers creep at all temperatures above minus 200 degrees C. Thus even at room temperature polymers creep.
DEAE Please help me to give answer , question is , it nut tighten a metal is creep after long years
1 Question:
Why didn't you consider Gravitational force along with self weight on the spring that underwent some plastic deformation???????
The spring does deform under its own weight. It has to be considered.
Sir, How constant stress acts during creep when area of cross section is reducing continuously ? Thank you in advance
If you do a fixed load test then you will not get constant stress. The stress will increase as the cross-sectional area decreases, as you have correctly pointed out. But in more careful tests, it is possible to adjust the load to keep the stress constant.
Thank u so much sir
Respected sir I have some doubts after listening this video.
Why we will choose the 0.5Tm as creep temperature?
How we will decide one material creep temperature?
Below this temperature will creep occur or not?
This is just a rough rule based on experience. In reality, creep happens at all temperatures. But the rate depends on temperature. It increases exponentially with temperature. So at low temperatures it is hardly observable. Creep rates become significant at about 0.5 Tm.
Thank you sir.
Super lecture sir..
What's the answer for those three questions???
11:33 really, r those pens erasable ?
Sir please help me solve this doubt🙄
Let's consider tungsten filament with Tm=3410°C. 0.5Tm=1705°C. So creep would be significant only at a temperature greater than this I guess. Even this is a Very High temperature and all other electrical components will melt at this temperature. So I believe tungsten undergoes creep certainly at a temperature lower than this I'm totally confused with this. Please help to solve this confusion/misunderstanding
The high temperature is in filament only and not in other components. The temperature in the filament has to be high for it to glow and give light.
love u from bangladesh
Respected sir,
Please go through the definition of Atomic Energy mentioned in section 2(a) of the Atomic Energy Act 1962. Unluckily, this definition covers only Nuclear Energy related Matters and the matters related to atomic energy are out of scope of the Act 1962. Our Act 1962 is available on internet to entire world so the scientifically developed country know the facts regarding funds been spend on nano technology.
Tq sir
hi. introduction to creep pdf i need. can you send me pdf of this book?
What are the ways to prevent this creep failure
This an interesting topic. Perhaps I should make a new video on this. One can do several things to make a component creep resistant: 1. Choose an alloy of high melting point 2. Solid solution strengthening 3. Precipitation strengthening 4. align grain boundaries along the stress axis (directional solidification) 5. remove grain boundaries altogether (single crystal), 6. use internal cooling to keep the temperature of the component low 7. use thermal barrier coating to insulate the component from heat.
There is a video on this channel that discusses some of the techniques with respect to making creep-resistant turbine blades: ua-cam.com/video/SHf_KiIwmyo/v-deo.html
@@rajeshprasadlectures great work by you sir your videos are very helpful for all of us.Thanks
what a creep