The resistance which a crystal lattice offers to a gliding dislocation in the absence of other imperfections and impurities depends on the width of the dislocation. In a positive edge dislocation, the atoms above and below the slip plane, in the vicinity of the dislocation lines are displaced from their equilibrium positions of the perfect lattice.
The extent of the displacement from the equilibrium position decreases with increasing distance from the core of the dislocation. Fig. 4.83 (a) illustrates a stiff dislocation (extreme case), where no relaxing displacements have taken place in the adjacent planes of atoms surrounding the dislocation region.
Here, the bond lengths have normal value around the dislocation except in the region below the incomplete plane, where the bonds are virtually broken and strain is concentrated. Fig. 4.83 (b) illustrates a relaxed dislocation.
The strains are distributed more evenly among the bonds around the dislocation line, resulting in some compression above and some tension below. A dislocation is said to be wide if the displacements are distributed over an appreciable distance.
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The width, w, of the dislocation, denotes the distance along the glide plane in the direction parallel to the Burgers vector over which the displacement of atoms is greater than half the maximum displacement. As the maximum displacement can be b/2, hence, w is the length of portion on both sides of the dislocation, where the displacement exceeds b/4.
If the atoms of the crystal have little compressibility, or if the shear resistance of the bonds across the slip plane is small, the width of the dislocation is large, (because the same strain due to absence of plane of atoms gets distributed over larger number of atoms). Alternatively, if the atoms of the crystal have high compressibility, or the shear strength of the bond is high, the misfitting portion is confined to a small width.
The nature of chemical bonding in the crystal determines the extent of relaxation, and thus, the width of the dislocation. Estimates of ‘w’ depends on the form of force-distance relationship between atoms. The values of ‘w’ are generally considered to lie in the range of b to 10b.
When a wide dislocation moves, (the adjustment that takes place in bond lengths in the dislocation region is distributed over a large number of bonds in the region, and the change in any one bond length is very small), it requires a low magnitude of stress.
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In a narrow dislocation, such as in Fig. 4.83 (a), the row of atoms below the slip plane has to move through one full interatomic distance. Thus, narrow dislocations are more difficult to move (i.e., requires much higher stress) than wide dislocations.
Peierls and Nabarro, basing the calculations on first principles, derived an equation to find out the shear stress required to move a dislocation through a crystal lattice in the absence of other defects and solute atoms.
Although the equation is unable to predict the CRSS of different metallic crystals as compared to the actual observed values, i.e., it does not appear to yield accurate results quantitatively, but helps to appreciate the differences in plastic deformation behaviour of crystals in a qualitative way.
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Peierls and Nabarro stress to move a dislocation:
where,
pn is the Peierls-Nabarro shear stress,
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G is the shear modulus of the crystal,
w is the width of the dislocation,
b is the Burgers vector of the dislocation.
The effect of the width of the dislocation (w), by putting different values of w can be seen from the following table:
For a stiff dislocation as illustrated in Fig. 4.83 (a), the width of a dislocation, w ≈ 0, and thus, a high value of stress equal to ‘G’ is required to move a dislocation. This value is almost equal to the stress required to deform a perfect crystal. The above table illustrates that as the width of dislocation increases, the stress required to move a dislocation decreases drastically, such that for w = 10b, the stress is negligibly small.
This is because, the stress is a very sensitive function of width of the dislocation (as well as the Burgers vector), and thus, a very precise determination of the width of the dislocation is a necessary requirement. As there are inherent problems to measure practically the precise width of dislocations in crystals, it has not been done.
A qualitative explanation of the dependence of shear stress on the nature of bonding in crystals is briefly given here:
(a) Covalently-Bonded Crystals:
Diamond has covalent bonding and is brittle. This is because, covalent bonding is strong but directional, which results in little relaxation of bonds. The dislocation is stiff and narrow, i.e., V is very small, and thus, Peierls-Nabarro shear stress is very high. This stress is so high that before this could be applied to move a dislocation, the brittle fracture of the covalently bonded crystal occurs by crack propagation, and as no plastic deformation occurs, the material fails by brittle fracture.
(b) Metallic Crystals:
Gold is a metal and is highly ductile. Gold leaf or foil is obtained by reducing its thickness hundred times of its original thickness by cold rolling without fracture. Actually metallic bond such as in Gold is not so a strong bond as well as is non-directional. Thus, the dislocation is really relaxed and wide, i.e., the width of the dislocation is large and thus, the Peierl-Nabarro stress is low, resulting in large ductility.
Iron, a typical transition metal has some covalent bonding in it due to d orbital bonding. As a covalent bond is directional and strong, its presence makes iron much harder and stronger, but with reduced ductility, i.e., cannot be cold rolled as gold.
(c) Ionic-Bonded Crystals:
NaCl is ionic-bonded substance. Ionic bonding is of moderate strength as well as non-directional, but still NaCl does not undergo plastic deformation. Invariably, the crack on the crystal surface leads to brittle fracture.
In ionic crystals, the Burgers vector of the dislocations is invariably large because only those dislocations can cause slip which will not bring two cations as nearest neighbours (as two neighbouring cations is not a stable state but repel each other), i.e., the Burgers vector of the dislocation is equal to its lattice constant.
It has got to be a unit dislocation. For example, the Burgers vector of a full dislocation in NaCl is 3.95 A°, but in Gold is 2.98 A°. As the Burgers vector increases, the Peierls-Nabarro stress increases as per above equation, and thus, the surface crack propagates to cause brittle fracture.