MSE306Chap4Dislocations-3

MSE306Chap4Dislocations-3 - Preview Dislocations have an...

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Preview Dislocations have an elastic strain field surrounding them. We know approximately what the displacement field should be (based upon the Burgers vector), so we can calculate the strain field and resulting stress field. Given this stress field, there is strain energy associated with a dislocation. The “self-energy” of a dislocation represents an increase in the free energy of the system. We will discuss this in more detail next, next week…for now you should remember 2 things 2 Gb ξ In general…for a dislocation reaction to proceed, the energy of the system must be reduced
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General Rules Perfect dislocations have Burgers vectors stretching from one lattice position to another. Burgers vectors of perfect dislocations are nearest neighbor distances along closest packed directions. (shortest possible, WHY? ) The slip planes could, in theory, be any plane containing the above Burgers vectors. In practice, they are the closest packed planes, with few exceptions. WHY? Tests like “Elementary Dislocation Theory” were written when there was no proven way to predict slip planes. Now atomistic simulations give insight regarding “core spreading”.
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b1 + b2 = b3; b3 – b4 = b5 What does this tell us about where dislocations lines can end? *** What about dislocation reactions? Continuity very important concept! often confuses students* Analogous to Kirchhoff’s law of current flow… Burgers vector of a dislocation (or collection of dislocations) is continuous Net Burgers vector equal on opposite sides of nodes b1 b2 b3 b4 b5 ?
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Exam Considerations We have now covered Chapters 1, 2, 3, and much of 4 For now, ignore sections 4.7-14 Use homeworks (and commentary therein) for the type of question . The subject matter will not be restricted to homework questions. Mainly short, pointed questions At least one question will be aimed at making you think (a computationally simple extension of a concept we have explored.)
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Dislocation junctions Continuity of the Burgers vector b 1 + b 2 = b 3 Simple Energetic Considerations Based upon the simple relation b 1 2 + b 2 2 >?=?< b 3 2 2 2 b μ ξ
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Possible Reactions Perfect dislocations within the plane might interact according to the following schemes. b1 + b2 = a/2[01-1] + a/2[-101] = a/2[-110] = b3 b1 - b2 = a/2[01-1] - a/2[-101] = a/2[11-2] Perfect dislocations on different planes, e.g. (111) and (11-1), might interact according the following schemes. b1 + b5 = a/2[01-1] + a/2[101] = a/2[110] The resulting dislocation is contained by neither slip plane? (use ) For cubics, you may use dot products between the plane’s indices and the Burger’s vectors, in order to determine if they are perpendicular. n t b ˆ ˆ ˆ = ×
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Partial Dislocations Only three possible perfect dislocation reaction result in a decrease in strain energy. ( the two already mentioned and annihilation ) FCC crystals are composed of {111} close pack planes of atoms with an ABCABCAB… type stacking sequence (as opposed to the
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