Sir, I am always inspired by your way of giving lectures and I am also overwhelmed at your command in crystal dislocation concept. May Allah bless you sir.
As Franck and Read showed the gentleman behaviour to publish the theory jointly, I see the same gentle and delicate reflection in your explanation and presentation Sir 🙏 🙏
Was struggling to understand before. Very clear now. Thanks Sir. Sir's lectures are Best to understand. Sir has to use digital technics and speed up the lectures.
Sir, in your video introducing dislocation nodes, you said all the t would be either point into or away from node and sum of b is zero. In this frank-read example, we have a 3-D dislocation network and the dislocation segment is defined by nodes A and B right? But the two sense vector t on node A and B are point into and point away from node and they have same burger's vector in magnitude and direction, the sum of b seems not to be zero. Could you explain this? Thanks!
Thank you very much sir. May I ask a question? Is there any requirement on how long the length of dislocation line on the horizontal plane to start the process in the video?
Opposite sides of the loop are dislocations of the opposite character. They thus move in opposite directions under the influence of the same stress. Thus the entire loop expands.
Sir, If you please add the NPTEL official link in the comment box after uploading the videos, it will be really helpful for us to find out the transcript where the slides are shown nicely in a chronological manner.
at the point of annihilation two dislocation will be of opposite sign. But they can be either edge, screw or mixed. In my example since the initial dislocation segment (source) was edge, the two segments at the point of annihilation were screw. But if I had taken the source to be screw then the annihilating segments would be edge. And if the source is mixed then the annihilating segments are also mixed. This is because at the point of annihilation the line vector is perpendicular but the Burger vector is parallel to the corresponding vectors of the source.
Sir, At 16:37 minutes you mentioned that the force will act always normal to the line segment. Why force is acting always normal to the dislocation line?
Force is expected to play a role in the movement of the dislocation line segment. The dislocation movement is meaningful only if it moves perpendicular to itself.
Sir , 1)as you said that if shear force increase the dislocation line will keep bulging. 2) the plastic deformation is due to slip and in which if we push the plane then one half plane is attached to nearer half plane and nearerb half plane is behaved like dislocation that means dislocation is moving during the plastic deformation. These two statements are quite contradic to each other. Sir could you please explain.
Your point 2) is for slip caused by movement of a STRAIGHT EDGE dislocation. However, slip is caused by movement of all types of dislocations--edge, screw or mixed--and whether straight or curved. Whenever a dislocation line sweeps an area, the crystal on the two sides of the area have a relative slip equal to the Burgers vector. In your point 1) we have a Frank-Read source. There is a curved dislocation loop whose character is varying from point to point and is bulging. In the process of bulging it is also sweeping area on the slip plane. Thus it will also cause a relative slip of the two crystals lying on either side of the slip plane. Thus there is no contradiction. Only, the situation is more complicated in case of Frank-Read source than in the case of a straight edge dislocation.
Sir , As you explained the movement of the dislocation loop through the slip plane, What will happen to the expanding loop of it encounters a grain boundary that is lying on the slip plane. Will the dislocation loop pass through the grain boundary as well or will it stop as has been discussed previously that dislocation ends on a grain boundary. Request throw some light. Thanks and regards
You are right. A grain boundary in the path of the loop will act as an obstacle to the expanding loop. This will lead to hardening due to grain boundary.
Why does this kind of behavior happen under applied shear stress, rather than dislocation PQ breaking up into 3 separate dislocations, with dislocation AB then gliding out of the material?
We have a constraint that a dislocation line cannot end abruptly inside a crystal. th-cam.com/video/WgSx1mZjki4/w-d-xo.html Thus AB cannot break away from PA and BQ as then the three dislocations will all have abrupt endings inside the crystal: PA will end abruptly at A, AB at both A and B and BQ at B.
Sir earlier in your videos i learnt that dislocations cannot abruptly start and end inside crystal..then how come the example we have taken( AB ) can start and end inside a crystal abruptly..please correct me if iam wrong
Yes. This is called stress analysis. Given a state of stress in a body, you can calculate shear or tensile stress on any plane. The calculation of resolved shear stress on a slip plane in a slip direction for a given applied tensile stress is a special case of this general formalism. In particular, in the present case, you can show that if we apply shear stress on a given plane it will not generate any shear stress on a perpendicular plane.
A force acts on the dislocation whenever there is a shear stress on its slip plane in the direction of its Burgers vector. This force is always normal to the dislocation segment. A loop has the same Burgers vector at all its points. Thus a shear stress in the plane of the loop and in the direction of the Burgers vector will produce forces normal to each of its segments. If the force is sufficiently large, i.e., greater than the critical resolved shear stress required to move the dislocation, the loop will expand.
sir is it mean that on vertical plane dislocation will disappear or remain as it is and on the other hand on horizontal plane on which shear force is acting , there will be continuous formation and disappearance of dislocation ,whose net effect is to increase the dislocation density . sir please make a few lines comment on conclusion made from video number 116 and 117 not considering the mathematical relationship as you mentioned in lecture number 120 . i got some idea but i am not very much clear about the conclusion.
I think your statement is correct. The vertical segments will not disappear but remain stationary. That's why the end of these segments on the horizontal plane acts as pinning points for the horizontal segment. It is this horizontal segment which acts as a source of dislocations by throwing out loops.
@@diya3279__ You can take a free body with the surface of your interest as an external surface. You will find that there is no need to apply any force on this surface to keep the body in equilibrium. Thus the shear stress will be zero. If you cut the sample shown in the video along the CDEF plane (extend the cut below DE to the bottom of the sample) and consider only the back half as your free body. Then this sample will be in equilibrium due to forces on its top and bottom planes. To balance the moments there will be forces, and hence shear stress, on left and right faces as well. (Thus my claim in the above reply to Faiyaz is not entirely correct) But under these forces now the body is in equilibrium so there is no need for any force or stress on the front and back faces. The result can be obtained more directly and mathematically by the transformation of stresses. But I have tried to give a more physical explanation above. Thanks for asking.
This video blew my mind, will understand it slowly over time.
Understood well after a month 😊
Sir,
I am always inspired by your way of giving lectures and I am also overwhelmed at your command in crystal dislocation concept.
May Allah bless you sir.
As Franck and Read showed the gentleman behaviour to publish the theory jointly, I see the same gentle and delicate reflection in your explanation and presentation Sir 🙏 🙏
Perfect explanation. thanks a lot
bEST Material Science teacher. now my Concepts are clearer than ever. thanks
Wonderful lectures - always helpful! Thanks Prof. Prasad.
Was struggling to understand before. Very clear now. Thanks Sir.
Sir's lectures are Best to understand.
Sir has to use digital technics and speed up the lectures.
Thank u sir for ur clean and neat explanation.
Your lectures are awesome sir.. thank you for these efforts sir🙏🏼
great explanation.
thank you
Sir ur explanation so crystal clear.
Sir, in your video introducing dislocation nodes, you said all the t would be either point into or away from node and sum of b is zero. In this frank-read example, we have a 3-D dislocation network and the dislocation segment is defined by nodes A and B right? But the two sense vector t on node A and B are point into and point away from node and they have same burger's vector in magnitude and direction, the sum of b seems not to be zero. Could you explain this? Thanks!
u make so much sense thank you
Thank you very much sir. May I ask a question? Is there any requirement on how long the length of dislocation line on the horizontal plane to start the process in the video?
Thank you so much sir for this favour well defined sir
I am from NIT SRINAGAR ❤
Tell yours also
Let's See from where we see this legend.
IIT BHU
Sir,
Why the loop is expanding also in direction( leftwards) opposite to that of shear stress(rightwards) ?
Opposite sides of the loop are dislocations of the opposite character. They thus move in opposite directions under the influence of the same stress. Thus the entire loop expands.
@@introductiontomaterialsscience Thank you sir
Thank You Sir!
Sir, If you please add the NPTEL official link in the comment box after uploading the videos, it will be really helpful for us to find out the transcript where the slides are shown nicely in a chronological manner.
So useful for me ❤
Sir so it is a screw annihilation ? As my teacher at school stated that it is an edge annihilation by stating that symbol
at the point of annihilation two dislocation will be of opposite sign. But they can be either edge, screw or mixed. In my example since the initial dislocation segment (source) was edge, the two segments at the point of annihilation were screw. But if I had taken the source to be screw then the annihilating segments would be edge. And if the source is mixed then the annihilating segments are also mixed. This is because at the point of annihilation the line vector is perpendicular but the Burger vector is parallel to the corresponding vectors of the source.
👍
Excellent sir
Sir is this dislocation partial like that of shockleys
Sir,
At 16:37 minutes you mentioned that the force will act always normal to the line segment. Why force is acting always normal to the dislocation line?
Force is expected to play a role in the movement of the dislocation line segment. The dislocation movement is meaningful only if it moves perpendicular to itself.
Sir ,
1)as you said that if shear force increase the dislocation line will keep bulging.
2) the plastic deformation is due to slip and in which if we push the plane then one half plane is attached to nearer half plane and nearerb half plane is behaved like dislocation that means dislocation is moving during the plastic deformation.
These two statements are quite contradic to each other.
Sir could you please explain.
Your point 2) is for slip caused by movement of a STRAIGHT EDGE dislocation. However, slip is caused by movement of all types of dislocations--edge, screw or mixed--and whether straight or curved. Whenever a dislocation line sweeps an area, the crystal on the two sides of the area have a relative slip equal to the Burgers vector.
In your point 1) we have a Frank-Read source. There is a curved dislocation loop whose character is varying from point to point and is bulging. In the process of bulging it is also sweeping area on the slip plane. Thus it will also cause a relative slip of the two crystals lying on either side of the slip plane.
Thus there is no contradiction. Only, the situation is more complicated in case of Frank-Read source than in the case of a straight edge dislocation.
Sir , As you explained the movement of the dislocation loop through the slip plane, What will happen to the expanding loop of it encounters a grain boundary that is lying on the slip plane. Will the dislocation loop pass through the grain boundary as well or will it stop as has been discussed previously that dislocation ends on a grain boundary. Request throw some light. Thanks and regards
You are right. A grain boundary in the path of the loop will act as an obstacle to the expanding loop. This will lead to hardening due to grain boundary.
Why does this kind of behavior happen under applied shear stress, rather than dislocation PQ breaking up into 3 separate dislocations, with dislocation AB then gliding out of the material?
We have a constraint that a dislocation line cannot end abruptly inside a crystal.
th-cam.com/video/WgSx1mZjki4/w-d-xo.html
Thus AB cannot break away from PA and BQ as then the three dislocations will all have abrupt endings inside the crystal: PA will end abruptly at A, AB at both A and B and BQ at B.
Sir earlier in your videos i learnt that dislocations cannot abruptly start and end inside crystal..then how come the example we have taken( AB ) can start and end inside a crystal abruptly..please correct me if iam wrong
AB is only a segment of a dislocation line. Full dislocation line is PABQ with end points at P and Q. 4:27.
Thank u so much sir for ur explanation
How in real world all the thing are being done?
easy to understand
why there will be no shear stress on CDEF AND GHIJ . PHYSICALLY OR MATHEMATICALLY CAN WE PROVE IT
Yes. This is called stress analysis. Given a state of stress in a body, you can calculate shear or tensile stress on any plane. The calculation of resolved shear stress on a slip plane in a slip direction for a given applied tensile stress is a special case of this general formalism. In particular, in the present case, you can show that if we apply shear stress on a given plane it will not generate any shear stress on a perpendicular plane.
Sir what is source of that normal force that is expanding the loop you have said that it expands but there is no perfect cause of force for it
A force acts on the dislocation whenever there is a shear stress on its slip plane in the direction of its Burgers vector. This force is always normal to the dislocation segment. A loop has the same Burgers vector at all its points. Thus a shear stress in the plane of the loop and in the direction of the Burgers vector will produce forces normal to each of its segments. If the force is sufficiently large, i.e., greater than the critical resolved shear stress required to move the dislocation, the loop will expand.
@@rajeshprasad101 is there a separate name for force required to move the dislocation in the slip plane and also thanks for the previous ans sir
You can call it 'force on a dislocation'.
sir is it mean that on vertical plane dislocation will disappear or remain as it is and on the other hand on horizontal plane on which shear force is acting , there will be continuous formation and disappearance of dislocation ,whose net effect is to increase the dislocation density . sir please make a few lines comment on conclusion made from video number 116 and 117 not considering the mathematical relationship as you mentioned in lecture number 120 . i got some idea but i am not very much clear about the conclusion.
I think your statement is correct. The vertical segments will not disappear but remain stationary. That's why the end of these segments on the horizontal plane acts as pinning points for the horizontal segment. It is this horizontal segment which acts as a source of dislocations by throwing out loops.
thank you sir
Yes the vertical dislocations will not be affected by the applied shear stress for the situation shown in the diagram.
Hello sir ur explanation is so elaborate . I could understand the concept . deeply
@@introductiontomaterialsscience if loop dislocation is merges into grain boundary then how dislocation density is increased sir .please explain sir.
Sir, Why shear stress will not apply on the plane CDEF & GHIJ
A shear stress applied on any plane cannot have a component on another plane perpendicular to the first plane.
@@introductiontomaterialsscience but why so ?
@@introductiontomaterialsscience This claim is not entirely correct. See a better explanation below in repsonse to Divya's question.
@@diya3279__ You can take a free body with the surface of your interest as an external surface. You will find that there is no need to apply any force on this surface to keep the body in equilibrium. Thus the shear stress will be zero.
If you cut the sample shown in the video along the CDEF plane (extend the cut below DE to the bottom of the sample) and consider only the back half as your free body. Then this sample will be in equilibrium due to forces on its top and bottom planes. To balance the moments there will be forces, and hence shear stress, on left and right faces as well. (Thus my claim in the above reply to Faiyaz is not entirely correct) But under these forces now the body is in equilibrium so there is no need for any force or stress on the front and back faces.
The result can be obtained more directly and mathematically by the transformation of stresses. But I have tried to give a more physical explanation above.
Thanks for asking.