With regard to this particular phenomenon that occurs there are a couple of things that need to be addressed.Firstly,Why do I have a region of stacking fault in that particular junction. I mean why is it shared between the two planes?Secondly,There is an unique plane given by index (100) which is the plane of new dislocation. Now,why is that going to have the said index?Thirdly,Is it that for the intersecting planes I can have any combination from the {111} family?(Like (1,-1,1) instead of (-1,1,1)along with the other intersecting plane which will remain (1,1,1) as usual?)
(1) In the LC lock, the perfect dislocations first dissociate into partials, before they intersect (this is the feature that distinguishes LC lock from a Lomer lock). So the stacking faults shown on your diagram were formed before the intersection, one on each plane. So it isn't a single fault shared between planes, those are two different faults (stacking faults are planar defects, so they can't exist on 2 planes at once).
(2) Now, at the intersection of the two planes, the two leading partial dislocations (one from each plane) interact with each other to form a new dislocation. Combination of dislocations follows the same rule as dissocation; the resultant should have a Burgers vector equal to the sum of the combining partials. This way, the Burgers vector can be found*. Cross-product of the line direction (shown in diagram as [0-11] and Burgers vector will give the indices of the plane which the resultant dislocation lies on. This is how (100) is found.
(3) You can take any combination of {111} planes, so long as you write the correct dissociation and combination reactions. The resultant dislocations will also be different, but of the same type. All reactions will not be identical in terms of energy; some will form stronger locks than others.
* There can be some ambiguity about deciding which partial is leading and which is trailing - the lock is formed by leading partials only. Some conventions are followed to resolve this.
Aashranth B. In point (1) is it that the Stacking fault is generated because the line of intersection causes one line of atoms to miss the plane and thereby generating a stacking fault?I mean what is the cause of the fault?
J.B. Yang Please give an example of a partial dislocation in the diagram that I have provided.Secondly Why is it that Stacking Faults should be bounded by partial dislocations and vice-versa?please elaborate on this.
Rajorshi Koyal - A stacking fault forms as soon as a partial dislocation is created, regardless of intersection. Perfect dislocations are called 'perfect' because they don't affect the periodic arrangement of atoms as they move. Partial dislocations, break this periodic arrangement of atoms when they move. These breaks in periodicity are the stacking faults.
The fault forms simply because it is energetically favourable for a perfect dislocation to dissociate into partials (and partials require stacking faults).
Intersection of dislocations on the two {111} planes has nothing to do with formation of the fault. In this example of LC lock, the intersection happens well after partial dislocations (and hence stacking faults) are created.
Aashranth B. Another thing is the Cross Product formation for the common plane for perfect dislocation doesn't seem to work for me.
Rajorshi Koyal - Are you unsure about how/why it works, or are you trying to do the calculation and not getting the right answer?
If its the former, well an edge dislocation must lie on the plane containing its Burgers vector and line direction. If we know the vectors for these two, performing the cross-product gives us the mutually perpendicular direction, which is also the direction of the plane normal containing these two vectors.
If its the second, you're either using the wrong Burgers' vector or making a calculation mistake. Finding the Burgers vector is not trivial due to the ambiguity mentioned, and requires a little more familiarity with the geometry of slip planes/directions than you may have at the moment.
Aashranth B. Could you name the Burger's Vector so that I can try and work it out for myself.
[011], ignoring the scalar multiplier. Similarly, remember that in the product [nx ny nz] = n[x y z], which is the same direction as [x y z]. The scalar multiplier 'n' only indicates the length of vector (which is not relevant to us when we want to find the plane), and does not change the direction.
Aashranth B. How did you find out the vector [011].Please tell me the steps.Should I only take into account the dislocation reaction. Is there any energy constraint while calculating the energy of the resultant dislocation and to check whether one dominates over the other?
I already mentioned the concept in point (2) of my first answer. Like I said, the actual calculation requires knowledge of some conventions and a little more familiarity with slip geometry than you may have at the moment. I'd suggest reading up a bit more (standard texts, nothing fancy) before you try solving this.
Aashranth B. Thanks a lot for replying.Can I get in touch with you over fb?
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