I have tried solving this issue by using the relation τ =Gb/l.However, in this context I do not understand how to get the value for b because it is a well given relation where I can substitute τ as a function of G.however is the value of b something that is explicitly defined?Please help
The b is Burgers vector. It defines the magnitude and direction of the slip. For example, in fcc materials the preferred slip direction is [110], and in bcc materials it is [111]. When you consider the slip in the [110] direction, the shortest displacement of atom will be 1/2[110]. Physically it means that the atom moves from the corner of the (100) plane to the center of the (100) plane (face center). The value of b is defined as a0[u2 + v2 + w2]0.5. Here a0 = lattice parameter, u, v, w are the coordinates of the displacement. For fcc, typical uvw will be 1/2, 1/2, 0. For bcc, typical uvw will be 1/2, 1/2, 1/2 (from corner to body center). Therefore, for fcc material the Burgers vector will be a0/(20.5). For bcc, b = a0/(30.5)
If you know the lattice parameter, and the direction of slip, then you can easily calculate the value of b using geometric relation. For cubic lattice the above relations are valid. Standard relations are available for other structures such as hcp, tetragonal etc.
Krishnan Raja Sir, Why do we have 1/2 as a magnitude in multiplied form that will be associated with the shortest lattice transition.Why not something else like 1/6 or 1/12?Secondly this magnitude seems extremely relatable to that of perfect dislocation and partial dislocation.Can you shed some light in this regard as to why partial dislocations and perfect dislocations need to be classified so distinctly when both are dislocations?
The quotient 1/2 is there because that specifies the next equilibrium position of the atom in a fcc or bcc cell. To understand it better, consider the coordinates of atoms in a face of a fcc lattice. You will write those as (000), (100), (010), (110) for corner atoms, and (1/2 1/2 0) for the face centered atom. When the (000) atom slips, the next nearest equilibrium position is (1/2 1/2 0). If it moves to (100) or (010) this is not the shortest distance and it is called as dislocation with unit strength. This will require a larger energy than that required for moving to the next shortest equilibrium position at (1/2 1/2 0). As you know, the energy per unit length of dislocation is approximated by the relation Gb2. This is the status of perfect dislocation.
Your question is why not 1/6 or 1/12 quotients. Now you are talking about partial dislocations. The movement of atom from corner to face center can occur in two steps with appropriate 1/6[112] steps. Now you have to imagine a 3-D lattice structure and spaces available in between the stacked atoms. Again, the two step movements occur because of energy considerations. It takes less energy to move the atoms in two steps rather than a single step.
Krishnan Raja Sir,I am extremely thankful for your reply but can you tell me why dislocations doesn't move after it assumes 1/2 quotient.Generaally perfect dislocations in many cases become sessile. Why is it so?
It is not correct that dislocation does not move after assuming the next equilibrium position. It will continue to move with incremental steps of 1/2[110] or 1/2[111] or 116[112] etc until it is pinned by some microstructural features including grain boundaries. For example, during tensile testing, the plastic deformation occurs by incremental movement of dislocations from one equilibrium position to another continuously. When the dislocations are pinned and/or when they become seessile by interaction of two dislocations, strain hardening of the material occurs.
In order to move a dislocation, it should be present on a high density crystallographic plane such as (111) for fcc. If the Burgers vector of an edge dislocation does not lie on the slip plane, you cannot move the atom from one position to another. This type of situation arises when dislocations interact with each other and the resultant dislocation lies in a plane whose Burgers vector is normal to the slip plane. Now the interacted dislocations become sessile. This is one example (Lomer-Cottrell barrier). There are different types of mechanisms to form sessile dislocation such as Frank partial dislocation, Shockley partial etc.
If you consider a set of dislocations and their elastic interactions, the yield strength is typically proportional to the square root of the dislocation density. For instance, for FCC metals, you can find detailed calculation of the mean free path l in Prof. Kubin's paper:
B. Devincre, T. Hoc, L. Kubin
Dislocation Mean Free Paths and Strain Hardening of Crystals
Science 27 Jun 2008: Vol. 320, Issue 5884, pp. 1745-1748
I suggest you to read about basic dislocation theory in the book:
Hull, Bacon,
Introduction to Dislocations
Best Regards,
Nicolò
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