![]() The problem is that the energy of suchĪ dislocation loop increases rapidly as the diameter d loop grows. You gain free energy in removing them,Īnd the (energy) costs of having small dislocation loops instead of lots of individual pointĭefects my well be worth it. Now we also have the nucleation and growth of a dislocation loop enclosingĪ stacking fault as a way to get rid of unwanted point defects.Īre around, starting such a loop is great for nirvana. I only gave you grain boundaries as "sinks" for point defect, as places where theyĬould disappear. Problems), it could easily gobble up all the excess point defects, growing in diameter Small dislocation loop of this kind has formed (with the usual nucleation Must be a Frank partial dislocation (with a Burgers vector of b Fr They typically form a dislocation loop enclosing the stacking fault. Or some internal interface, by needs must be bound by a partial dislocation.ĭefects agglomerate on a plane they start small and thus cannot be bound by interfaces allĪround. Here, it becomes clear that a stacking fault that doesn't end at the surface ![]() Mentioned the word " partial dislocation" for Vacancies or self-interstitial atoms on one plane. But how is the crystal doing it?Īs you know from the backbone, by condensing It is easy to insert or to remove atomic planes That necessitates the movement of all the atoms in the stacking On its formal glide plane (the plane containing Burgers and line vector) would require that Note that a Frank partial dislocation cannot move without the help of point defects. One extra plane has been inserted or removed, generating a Frank dislocation with a stacking Red line denotes the direction along which you count the stacking sequence. On the left is the perfect lattice / crystal. What one gets is shown below.ĭislocations and stacking faults in a fcc lattice / crystal. Of course, we can do the same thing by moving the other way and The particular kind we made in fcc crystals is called That we get something that looks very much like an edge dislocation but with a stacking fault on one side! Now between an A-layer and B-layer you can only insert a C-layer,īetween a B-layer and C-layer only an A-layer fits, and so on. Vector, we have precisely the space to insert exactly oneĪtomic layer. If we make the cut and then move up just 1/3 of the length of a proper lattice translation it is indicated in the bottom part of the figure above. Of a fcc lattice / one-atom-base crystal. We only consider fcc lattices and crystals now. That implies that we don't move the cut part up by a full translation vector after the cutīut only by a partial one. So how about filling in just one atomic layer? We fill in as many atomic layers as needed and produce a simple perfect edge dislocation once more. Well, taking into account this little problem, To make that a bit clearer, let's look at an example for a diamond kind of crystal (which is based on the fcc lattice). In our example we need to fill in three planes of atoms, the whole stacking sequence ABC. Lattice plane might mean to fill in several atomic planes. Depending on the geometry, filling in one That's a point where the distinction between lattice and crystal might cause considerableĬomplications. To fill in lattice points and therefore also atoms now. Push the two halves apart by the "right" Burgers vector as shown. Vector, perpendicular to the cutting plane. We just do the simple thing of cutting and moving the right distance, i.e. ![]() Let's cut and move upwards just 1/3 of the shortest possibleīut now you have a problem! You generated a stacking fault. ![]() Then you must insert three of those atomic layers to make Side denote the hexagonally densely packed atomic layers on the planes of the crystal. aluminum (Al)) and that the faint green lines at its Let's do that now - and be prepared for some mind-boggling results.Ĭut and move upwards a translation vector Part we have not looked at the dislocations produced when the shift has a component perpendicular ![]()
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