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 🙏 🙏
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?
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!
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.
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. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WgSx1mZjki4.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.
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 , 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.
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.
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 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
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 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.
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.
@@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.
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.
