in this article we will discuss about:- 1. Geometry of the Edge Dislocations in Crystals 2. Motion of an Edge Dislocation of Crystals 3. Screw Dislocation in Crystals.
Geometry of the Edge Dislocations in Crystals:
Suppose, we have a simple cubic perfect crystal (we are taking this crystal structure for better clarity, though Polonium is the only metal having this crystal structure) in the undeformed state as illustrated in Fig, 4.18 (a). It is cut along a plane JKLM, Fig. 4.18 (b). When a shear stress is applied, atoms (like X, O, P etc.) in the upper plane move towards left, relative to atoms (like S, R, etc.) in the lower plane, Fig. 4.18 (c), till it is displaced by one interatomic distance, Fig. 4.18 (d) and by then atoms like ‘O’ and ‘S’, ‘P’ and ‘R’, etc. (atoms on either side of cut plane) form bonds between them.
The new distorted crystal lattice is shown in Fig. 4.18 (e). The crystal structure is actually almost perfect except near the line XX’. It is also clear that there is now an incomplete plane XX’ Z’ Z above the cut plane. The extra half plane ends in edge XX’.
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The elastic distortion of the crystal immediately surrounding the edge of the incomplete plane increases the energy of the crystal. The maximum distortion is centred around this edge of the incomplete plane.
This line XX’ is an edge dislocation. The displacement of atoms in the planes surrounding the dislocation line is in the direction perpendicular to the dislocation line XX’. By convention, the dislocation line XX’ is called the positive edge dislocation and is represented by the symbol, ⊥, Fig. 4.18 (f). If the extra half plane is present in the bottom of the crystal, it is called negative edge dislocation and is represented by symbol, T, Fig. 4.19 (b).
It is only a convention, because a negative edge dislocation can be converted to a positive edge dislocation by turning over the crystal. Fig. 4.20 (a) illustrates that atoms above the slip plane around the dislocation line are under compression and below the slip plane are under tension.
The nature of stress and the strain fields in a negative edge dislocation are of reverse nature. We saw in Fig. 4.18 that atoms on the right hand side of XX’ line slipped (moved) relative to atoms in the lower plane but not the atoms on the left hand side of the line. The dislocation line XX’ or in general, a dislocation line is defined as the boundary between the slipped and the unslipped parts of the slip plane.
A screw dislocation can be produced in a crystal as illustrated in Fig. 4.21. An undeformed perfect simple cubic lattice Fig. 4.21 (a) is cut along a part of the plane ABCD and a shear stress is applied, so that the crystal looks as illustrated in Fig. 4.21 (d). The displacement of the crystal on one side of ABCD relative to the other side in the direction CD or DC takes place.
X’X (or CD) is the screw dislocation. A careful examination of this model, Fig. 4.21 (d) reveals that the arrangement of atoms around the screw dislocation, CD, is like ascending a spiral stair-case as also illustrated in Fig. 4.22. The screw dislocation line (CD) itself is a pole (a line) about which this spiral-ramp circles.
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A small insect could walk from the bottom side of the crystal to the top without leaving the atomic planes. Fig. 4.23 (a) illustrates another three dimensional view of crystal having CD as the screw dislocation line about which the atomic planes are bent into a helical surface.
Fig. 4.23 (b) illustrates that the crystal may be regarded as consisting of a single atomic plane wound into a helical surface. There is incomplete shear in the crystal in a plane, but the shear vector (atoms slip in direction of screw dislocation line) and the screw dislocation are parallel, Fig. 4.23 (a) & (b), whereas in the edge dislocation, they are perpendicular.
A mixed dislocation, which has partly edge nature and partly screw nature, can be created as illustrated in Fig. 4.24. The dislocation line between points ‘A’ and ‘C’ has mixed nature. Fig. 4.25 illustrates top view of a mixed dislocation showing atom positions. Atoms above slip plane are open circles and atoms below are solid circles.
Motion of an Edge Dislocation of Crystals:
CD is a positive edge dislocation (inside the paper) in a single crystal, Fig. 4.26 (a). When a shear stress is applied as illustrated, the dislocation moves to the right, Fig. 4.26 (c). The magnitude of the shear stress should be just enough to pull [Fig. 4.26 (b)] atom number 1 further away from its neighbour, atom 2; atom 3, however simultaneously moves closer to its equilibrium distance from its neighbour, atom 4.
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These slight adjustments in just a few atoms make the dislocation to move (slip) by one step to its right [Fig. 4.26 (c)] from ‘C’ to position ‘A’. At a time, the magnitude of the shear stress required should be just sufficient to make these slight adjustments in atoms.
The dislocation moves from ‘C’ to ‘A’, and then to ‘B’ in steps, one after the other, and finally the dislocation goes out of the crystal, i.e., the extra half plane of atoms reaches the free right hand surface. It creates a slip step of one Burgers vector, or one atomic distance for the simple cubic lattice.
This unit distance, b ≈ one atomic distance is of a few Angstrom units, and thus is not visible to the naked eye (even under optical microscope). Many hundreds of dislocations move on the slip plane to produce a visible slip line. Crystal has changed shape, i.e., has yielded permanently. Theoretical calculations show that this force is of the correct order of magnitude to account for the low-yield stresses of crystals.
If the crystal had been perfect (free of dislocations) as shown in Fig. 4.27 (a), bonds between all the atoms on both sides of the slip plane, Fig. 4.27 (b) would have to be broken at the same moment, only then the atoms would have slipped to the right relative to atoms on the lower side of slip plane to cause step formation, i.e. permanent deformation.
The magnitude of the shear stress required has got to be very high, and is called theoretical yield strength. It is 103 to 105 times of the actual yield strength of real crystals, which have dislocations. A dislocation moves in steps, and the stress to cause this motion is very low.
The motion of an earthworm as illustrated in Fig. 4.11 resembles the motion of a dislocation through a crystal. As the atoms around the dislocation are symmetrically placed on opposite sides of the extra half plane, equal and opposite forces oppose (as well as assist) the motion. Thus, to a first approximation, there is no net force on the dislocation, unless applied from outside.
Screw Dislocation in Crystals:
Elastic Strain Energy of a Screw Dislocation:
Suppose in a cylindrical crystal, a screw dislocation, AB is created as illustrated in Fig. 4.40 (a) with a Burgers Vector, b. This is similar to a strip of 2 r being sheared to a displacement of b ⃗ at one end, Fig. 4.40 (c). Assuming that the Hooke’s law is being followed in this region, the shear strain, , at a distance r from the screw dislocation is,
The total elastic strain energy is obtained by integrating the strain energy between the core radiuses, r0 to R (external dimensions of the crystal when only one dislocation is present. Normally, the density of the dislocations is high in the materials and thus, R is taken as half the average spacing between the dislocations arranged at random).
Normally, r0 can be taken as a small multiple of b (= 10 A°), and R can be taken of the order of few microns, or, more. Then to an approximation,
In other words, the strain energy of a screw dislocation per unit length is directly proportional to the square of the Burgers Vector. The value of R is the average distance at which the stress field of the dislocation has decreased to the value produced by neighbouring dislocations, or this distance is equal to half the spacing of adjacent dislocations. This means, as the density of dislocations decreases, the value of R, and therefore of Eelastic increases.
Strain Energy of an Edge Dislocation:
The strain field of an edge dislocation is not as symmetric as that around a screw dislocation because the atoms on one side of the edge dislocation are under compression while the atoms on the other side of the dislocation are under tension.
If v is the Poisson’s ratio then elastic strain energy of an edge dislocation per unit length is:
This expression is similar to the expression for the energy of a screw dislocation except for the factor 1/(1 – v) Thus, the elastic strain energy per unit length of an edge dislocation is larger than the energy of the screw dislocation per unit length.
If we put, G = 0.7 x 1011 N/m2 (for iron), b = 3 x 10-10 m, R = 0.01 m, and r0 = 10-9m, the elastic strain energy of a screw dislocation per each atom plane threaded by the dislocation is about 5 eV. For the edge dislocation with v = 1/3, the elastic strain energy per atom plane threaded by the edge dislocation is about 7.5 eV.
As the energy of the core (0.5 – 1.5 eV) is much smaller than elastic strain energy of the dislocations, no great error is introduced in neglecting the core energy term while considering the total strain energy of a dislocation which is taken as:
Total strain energy per unit length = Gb2
Strain-Energy of Mixed Dislocation:
The strain-energy of a mixed dislocation can be generalised if the Burgers Vector is inclined at an angle θ to the dislocation line. The strain-energy or total energy of mixed dislocation per unit length (neglecting the core energy) is the sum of energy of screw component which has a Burgers Vector of length b cos θ, and the energy of the edge component, which has a Burgers Vector of b sin θ, i.e.,
Total energy per unit length of mixed dislocation
Problem 1:
If the density of dislocation in iron is 1012 m/m3, what is the approximate value of R in it?
Solution:
The density of dislocation = 1012 m/m3
= 1012 m2
∴ The density of dislocation can be taken as 1012 dislocations that cut through a unit cross-sectional area, i.e.,
= 106 dislocation per unit meter length
or Distance between two dislocations = 1 / 106 m
R = 10-6 m
Problem 2:
Calculate the elastic strain energy of screw dislocation per unit length for iron if r0 = 2 b and R = 2000 b.
Fig. 4.34 illustrates the motion of edge (positive and negative) and screw dislocation (positive and negative) on a plane under the shear stress ⇆. The direction of motion of the dislocation line is reversed if the shear stress is ⇄. If more dislocations move, a large plastic deformation occurs.
Let us see some more simple examples (we already know that ⊥ is a positive edge dislocation). Consider a crystal having a pair of straight edge dislocations (of opposite nature on the same slip plane). They move under the stress system (⇄) in opposite direction to move out of the crystal leaving a step on either side-surface of the crystal as illustrated in Fig 4.35.
When two screw dislocations (straight left hand (+) screw dislocation and right hand or negative screw dislocation) in a crystal in the same plane move out of the crystal, they do not leave a step on the crystal surface through which they finally leave the crystal but the step forms on the surface at which the screw dislocation terminates, because the screw dislocation moves at right angles to the direction of atomic movement as illustrated in Fig. 4.36.
The net deformation is the same whether an edge dislocation or a screw dislocation moves across a slip plane if one moves at right angles to the other. Atoms on one side of the slip plane are moved relative to those on the other by one atomic distance (B.V.). When closed circular dislocation loop expands under a shear stress, it produces steps on both sides on moving out of the crystal.
The total deformation is the same as if a pair of edge dislocations (opposite signs) or a pair of screw dislocations (opposite signs) has moved across the slip plane (compare Fig. 4.37 with Fig. 4.35 and 4.36). Steps are formed on the sides of the crystal where the edge segments of the dislocation-loop leave the crystal.
Self Check:
If the nature of shear stress in Fig. 4.37 is reversed what shall be the ultimate result by the motion of this loop of dislocation?
Self Check:
If the nature of shear stress is reversed than that given in Fig. 4.36 and the two screw dislocations of opposite sign come together, what shall be the end result?
Let us take some simple interactions between parallel edge dislocations. (a) Two parallel edge dislocations of opposite sign on the same plane as illustrated in Fig. 4.38 (a), under a stress, come together and annihilate each other leaving behind a perfect lattice.
See also Fig. 4.44 (b) If two parallel edge dislocations of opposite signs are on nearby planes as indicated in Fig. 4.38 (b), they annihilate each other but leave behind a row of vacancies and if placed as in Fig. 4.38 (c), under a stress, come closer, annihilate each other but leave behind a row of interstitialcy atoms.
In a positive edge dislocation, just above the edge of the incomplete plane, atoms are under a state of compression. A substitutional solute atom, if of smaller size than the atom of the solvent metal, tries to migrate to the compression side of the edge dislocation because by replacing a larger solvent atom, it lowers the strain energy of the crystal.
A larger substitutional solute shall likewise migrate to the tension side of the edge dislocation. Small interstitial solute atom (like of C, N etc.) tends to migrate to the tension side of the edge dislocation, because the total elastic strain energy gets reduced in this state. Such a state is called Cottrell’s atmosphere as illustrated in Fig. 4.39. Cottrell atmospheres help to understand the phenomena of sharp yield point in stress- strain curve of mild steels, strain-aging and solid solution strengthening in alloys.
Problem 3:
An optical microscope can resolve a step of minimum width 300 nm. A slip band was observed in a simple cubic crystal (a = 3 A°). How many (minimum) dislocations must have slipped out of the crystal?
Solution:
In a simple cubic crystal, Burgers Vector = 3 A° = 3 x 10-10 m
If n is the number of dislocations slipped, then,
Problem 4:
Iron showed a slip band under optical microscope which can resolve at least 300 nm. How many minimum dislocations must have passed to create this step? (Given BCC-Fe, a = 2.87 A°).
Solution:
The Burgers vector in BCC-Fe crystal is of 1/2 <111> type.
