Griffith's Criterion
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- เผยแพร่เมื่อ 1 ต.ค. 2024
- Correction: At 13:11 I have incorrectly written surface energy as 2 a B gamma. It should be 4 a B gamma. I have detected this mistake later at 19:15 and have corrected it. So the final expression is correct.
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At 13:11 I have incorrectly written surface energy as 2 a B gamma. It should be 4 a B gamma. I have detected this mistake later at 19:15 and have corrected it. So the final expression is correct.
Someone, please give him the Nobel prize. Such splendid Explanation!. Hats off Professor
Sir, You are explaining in very well manner
Please upload more lecture on material science
Thank you sir
Sir you explained in the most crisp and lucid way. Thanks!
AAAAAAAAAAAAAAH GRIFFIIIIIIIIIIIIIIIIIIIIIIITH!!!!
Making difficult things easy is a quite valuable competence, thanks so much sir.
Sir u cleared my all doubts. Infact i didnot get this much knowledge by reading rolfe and barsom. Hats off to ur explanation
At 17:23, change in energy (not the energy) is dropping. As long as the graph is above x-axis, change in energy is positive. After cutting the x-axis, change in energy becomes negative, i.e. energy is now decreasing. So, shouldn't the critical crack length be the point where the graph crosses the x-axis instead of the point of maxima ?
This is a nice point and a subtle one. The reference of energy here is the uncracked component, i.e. energy is per when a=0. You have rightly pointed out that from a=0 to the point where the curve cuts the x-axis (let call this value of a as ax. the energy is positive, and only beyond this point it becomes negative. This means that all components with a
Sir i did the simulation of hydrogen embrittlement in nickel. For pure nickel the crack tip was sharp and growing smoothly in x-direction. When i simulated nickel with 2% and 4% hydrogen , the crack tip was blunt and it shows parabolic nature. But when i increased the hydrogen concentration upto 10%, the crack tip started moving towards y direction instead of moving in x direction? I don't understand why is this happening? If i have explained the problem well , could you please tell me the possible reason for this?
Too difficult a question for me!
This is why IIT students succeed later in life...All due to such great professors!
Reason is he is also IIT-BHU alumni
Thanks for your interesting lecture. I had problem in understanding the Griffith criterion for crack propagation and your lecture totally solved my issues. You are a great Teacher!
Sir is this crack propagation is the way in which fracture occurs in ductile and brittle?or does fracture in ductile occurs some other way?
Griffith's analysis is primarily based on brittle materials like glasses. For ductile fracture, Irwin has further extended this approach.
@@TheSourav77 can u provide me a link borther?🙏
@@ankushsaha007 You may follow the online course by Prof. Ramesh, IITM on engineering fracture mechanics. nptel.ac.in/courses/112/106/112106065/
@@TheSourav77 thanx a lot bro🙏
So great Professor... Nice lecture :)
There are around 27k views and only 262 likes....shame:(
at least appreciate this good initiative taken...
really nice lecture. I am really thankful to you sir, because teacher of this subject in our college are not that good.
I hate Griffith for what he did to guts
Hello sir , i want to ask you about m.tech programme in iit delhi. I scored decent mark in gate and i want to join mtech in material science . So will this course provide good career opportunities in future ?
I cleared my all doubts from your videos . Dhanyawad guruji 🙏
sir in plotting graph you wrote surface energy as 2abY instead of 4abY as derived. why?
Thanks. Indeed it is a mistake. It should be 4ab gamma.
Excellent explanation, thanks you so much, Great Professor
Very helpful explanation
Wah sir mja aa gya....
Sir, here we are assuming that the object already has a crack in it. I would like to know how a crack is formed initially, from material science point of view. Is there any object which has no cracks in it (ideal case)?
Generally when the material is subjected to stress, Initially the material absorbs energy according to its toughness and when the value of the stress crosses the ultimate shear stress, the crack is formed. And when this crack meets another surface,. Shearing of material takes place. And in this way punching or blanking is performed. Propagation of shearing is called tearing.
@@navdiprathavi3692 sir, what is ultimate shear stress? I have heard maximum shear stress for Tresca yield surface, but I guess that term is ultimate tensile strength and for brittle material doesn’t even go beyond yield stress as it gets fractured at yield limit. Correct me if I am wrong.
Sir how to find the quantitative value of surface energy of a crack?
Surface energies are always difficult to determine and find. You have to estimate it by consulting databases.
@@introductiontomaterialsscience thank you sir
Sir, In 12:50 you have mentioned the change in surface energy as 4(a)(B)(gamma) however in the next slide for calculating total energy you have mentioned the change in surface energy as 2(a)(B)(gamma).
I guess he plotted the general definition considering the crack length equal to "a" and not to "2*a" like in the previous slide
This is a mistake. Fortunately, the mistake is at this point only. I have used the correct value 4 a B gamma before and after this point. The final result is therefore correct. I have added a correction comment and have included it in the description also. Thanks for pointing this out.
Dear Sir, Thank you for the video. I want to ask you that I have two samples contain the same quantities of H ions, then I annealed the samples one at 150 deg for 2h and the second at 350 deg for 2h. I found blisters in both samples but the sample annealed at high temperature has higher blister size. My question could I use this concept to analyse and explain this finding?
Thank you and with my best regards
I will guess that at higher temperature you have more H comes out and forms larger blisters.
@@introductiontomaterialsscience thank you for your answer, it's right
You are great sir
Thanku....
Sir, in my book they mentioned that,the elastic energy available per unit width is 2 x ( sigma square x pi x C square )/ 2 E. Why did they multiply elastic strain energy by 2...????
Actually tsrain energy is not being multiplied by 2. You can see that in your expression 2 is coming both in the numerator and the denominator, and so they cancel giving the expression as the one used in my slide. In fact I have not derived the expression. And I think your book also must not have derived it. Derivation requires more advanced concepts from elasticity theory. So we simply take the final expression and use it for further derivation of the Griffith's criterion.
According to previous standards (12th or from strength of materials)
Strain energy = 1/2 *(stress) *(strain)
Now for that 2
According to the book I refer
It is Bcz the crack is throughout, so u have two surfaces which will try to resist the propagation of crack
If you wanted to find the critical stress where crack would propogate (rather than critical crack length), you might be able to differentiate the total Energy with respect to stress and not crack length. However you will find the critical stress is 0. Please explain?
is it the potential energy of the system that is decreasing? Because according to the law of conservation of mechanical energy if an isolated system is subject only to conservative forces, then the mechanical energy is constant.
This is indeed a difficult question. Let me think about it.
Such a nyz lucture...
Fall in ❤ with Materials Science
Sir at 6:40 time-stamp, shouldn’t we write uncracked-cracked, because when crack propagates the cracked part of the energy increases and make the total change in mechanical energy negative?
good vedio
sir pls upload vedio for fatigue creep interaction.
Respected sir, hope you are in good health.Sir, I have a question: is fracture stress sensitive to temperature? I mean if we look for Yield strength, it decreases with temperature but I don't know how it is for fracture stress. Kindly help me sir, this doubt is making me restless😔
Thank you for your help sir 🙏🏻
Wow sir ,you explain it in a detailed way ...Thanks
Than you sir...
Griffith!!!!!
Thanks. This video cleared my doubt in nucleation topic too.
👍🤗
Woow
Sir English and Hindi mix study karye to jyga samjh ata
why do we take B as the thickness for the new surfaces created (B is the overall thickness of the material, and the crack is not present throughout the thickness of the material)?
The geometry we are considering is a plate with a through-thickness crack. So the width of the crack is the same as the thickness of the plate.
Introduction to Materials Science and Engineering ,Thank you for the reply sir . Just one more thing , Is the direction of propogation of crack along it's length (i.e. 2a) ?
Yes. So we are analysing a very idealised situation where we have a flat-faced through-thickness crack which propagates in its own plane. This is to keep the analysis simple. A real crack need not have flat faces. And it need not propagate in a single direction.
In this problem we are considering through-thickness crack. Thus breadth of the crack is the same as the thickness of the plate.
Hi sir. The surface energy of the new formed surface after crack is the barrier for crack to occur. If the crack is propagating along the length of crack 2a, how the new surface area formed due to crack will be 2×2a×b. it should be 2×total length of crack(after crack)× b.
Sir what is the reason of crack initiation
Clear ideas, very helpful. Thank you
Sir at 17:26 u said if crack size(say a1) is less than critical size it dont propagate,my doubt is how this crack of size a1 gets formed in the first place?it had to have been created from no crack state,and if we increase crack size from 0 then process is unfavourable as energy increases so system resist this formation of crack,so how we are getting this pre existing crack of size a1?did it occured due to a higer applied stress tht we applied before?
This is indeed a difficult question. The Griffith theory presented here takes care of crack propagation. It cannot handle crack initiation. As far as I know, there is no well-established theory of crack initiation.
@@rajeshprasad101 thanx a lot sir🙏
👍👍👍👍👍👍👍
Best 💯
Excellent
nice lecture, Thanks
Sir, as we know that near the crack the stress gets concentrated. Hence, in the formula of "critical Crack size for fracture" which stress are we considering - the applied stress or the concentrated stress?
As what we can apply and what we can control is the applied stress, Griffith's formulation is in terms of applied stress.
Sir it is the derivation of griffth theory or not?
Could you please explain what exactly is your confusion?
thank you sir,super
Nothing is understanding from your time consuming lecture
Watch the before lectures from beginning to understand this,ur basic arnt clear obviously,just like to understand calculus u must first know to add and subtract etc
What happens when the crack is inclined at an angle to the tensile axis. How would I find critical stress? please
The tensile component, perpendicular to the crack face, of the applied stress should be taken as the relevant stress in the Griffith's formula. Thus if the crack face is inclined at an angle 'theta' to the applied tensile stress 'sigma' axis then the relevant tensile stress perpendicular to the crack face will be sigma*cos^2(theta). Thus, other factors being the same, the critical stress will be enhanced by a factor (1/cos^2 theta) with respect to case where the crack face is perpendicular to applied tensile stress.