Burgers displacement

Figure 5.1 Schematic elevation view of the center of a kink on a screw disocation in the diamond crystal structure. D0 is the bond length, b is the Burgers displacement. The black circles are in the central plane of the figure. The white circles lie in a plane slightly in front of the central plane, while the gray circles lie in a plane slightly behind the central plane.

Figure 5.8 Projection of the diamond structure so the (111) glide planes (AB) are perpendicular to the plane of the figure. Then the covalent bonds connecting the atoms in planes (AB) and (A ) are perpendicular to the (111) planes.The glide plane spacing, a of the figure corresponds to the bond length AA. The Burgers displacement, b corresponds to the atomic spacing along A or A. ...

Path of layer 1 cell s center (four partial Burgers displacements)... 

The crystal structure of NiAl is the CsCl, or (B2) structure. This is bcc cubic with Ni, or A1 in the center of the unit cell and Al, or Ni at the eight comers. The lattice parameter is 2.88 A, and this is also the Burgers displacement. The unit cell volume is 23.9 A3 and the heat of formation is AHf = -71.6kJ/mole. When a kink on a dislocation line moves forward one-half burgers displacement, = b/2 = 1.44 A, the compound must dissociate locally, so AHf might be the barrier to motion. To overcome this barrier, the applied stress must do an amount of work equal to the barrier energy. If x is the applied stress, the work it does is approximately xb3 so x = 8.2 GPa. Then, if the conventional ratio of hardness to yield stress is used (i.e., 2x3 = 6) the hardness should be about 50 GPa. But according to Weaver, Stevenson and Bradt (2003) it is 2.2 GPa. Therefore, it is concluded that the hardness of NiAl is not intrinsic. Rather it is determined by an extrinsic factor namely, deformation hardening. 

The distance that the small segment of a dislocation line moves when a kink moves is called the Burgers displacement, b. Figure 11.2 illustrates it for the case of quartz. It determines the amount of work that is done by the advance of a kink (per unit width of the kink) which is acted upon by the virtual force generated by the applied shear stress, x. This force is xb per unit length of the dislocation line. Letting the kink width be b since the displacement is b, the work done is xb3. This is resisted by the strength, U (eV) of a Si-O bond which... 

The intrinsic energy band-gap of YAG is about 6.6 eV., and the Burgers displacement is about half the unit cell size, or 6 A. Then, if a kink volume is taken to be 6 x 3 x 3 = 54 A3, the bond modulus is 0.11 eV/A3, or 1800 kg/mm2. Given how little is known about dislocation motion in garnet, this agreement with the room temperature hardness value is largely fortuitous. 

At low temperatures, A1203 is hard and brittle, but it can be plastically deformed at high temperatures. The primary glide plane is the basal (0001) plane, and the Burgers displacement at low temperatures is 5.84 A. When the Al atoms become mobile at high temperatures this shortens to about 2.76 A. 

Figure 1.17. (a) An HRTEM image of the intergrowths and dislocations in a complex block structure using GaNbn029 as an example (b) a schematic diagram of the structure in the area shown in (a) (c) a structural model of the misfit fault in D2 and (d) the idealized structure of the dislocation D6. The Burgers (displacement) vector and mismatch of cation levels due to the dislocation are shown (after Gai P L and Anderson J S 1976 Acta Cryst. A 32 157). 

Burger s vector A measure of the crystal lattice displacement resulting from the passage of a dislocation. 

Dislocation motion produces plastic strain. Figure 9.4 shows how the atoms rearrange as the dislocation moves through the crystal, and that, when one dislocation moves entirely through a crystal, the lower part is displaced under the upper by the distance b (called the Burgers vector). The same process is drawn, without the atoms, and using the symbol 1 for the position of the dislocation line, in Fig. 9.5. The way in... 

Figure 4.2 Quasi-hexagonal dislocation loop lying on the (111) glide plane of the diamond crystal structure. The <110> Burgers vector is indicated. A segment, displaced by one atomic plane, with a pair of kinks, is shown a the right-hand screw orientation of the loop. As the kinks move apart along the screw dislocation, more of it moves to the right.

Figure 14.1 Schematic comparison of dislocation lines in a crystalline and a glassy structure. Dashed line indicates the center of a dislocation line. The vectors indicate the displacement of the atoms in the next level above the plane of the figure. At (a) the displacement (Burgers) vectors In the periodic crystal have a constant value. At (b) the displacements in the glass fluctuate in both magnitude and direction.

The Burgers vector of most dislocations is neither perpendicular nor parallel to the dislocation line. Such a dislocation has an intermediate character and is called a mixed dislocation. In this case the atom displacements in the region of the dislocation are a complicated combination of edge and screw components. The mixed edge and screw nature of a dislocation can be illustrated by the structure of a dislocation loop,... 

To illustrate this, take the situation in a very common and relatively simple metal structure, that of copper. A crystal of copper adopts the face-centered cubic (fee) structure (Fig. 2.8). In all crystals with this structure slip takes place on one of the equivalent 111 planes, in one of the compatible <110> directions. The shortest vector describing this runs from an atom at the comer of the unit cell to one at a face center (Fig. 3.10). A dislocation having Burgers vector equal to this displacement, i <110>, is thus a unit dislocation in the structure. 

Figure 3.12 Partial dislocations in copper (a) a unit dislocation, Burgers vector bl (b) initially slip is easier in the direction represented by the Burgers vector of the partial dislocation b2 than bl (c) the result of the movement in (h) is to generate a stacking fault and (d) the combined effect of displacements by the two partial dislocations b2 and b3 is identical to that of the unit dislocation, but the partials are separated by a stacking fault.

Stack of lamellar crystals generated by spiral growth at one or more screw dislocations. Note The axial displacement over a full turn of the screw (Burgers vector) is usually equal to one lamellar thickness. 

In extended defects, the displacement vector b (or R) associated with them can be defined from the Burgers Circuit shown in figure 2.4(a), for a simple cubic system (Frank 1951, Cottrell 1971, Amelinckx et al 1978). In the defective crystal (A), a sequence of lattice vectors forms a clockwise ring around the dislocation precisely the same set of lattice vectors is then used to make a second... 

Figure 2.4. Definition of a displacement (Burgers or shear) vector b (a) a Burgers vector around a dislocation (defect) A in a perfect crystal there is a closure failure unless completed by b (b) a schematic diagram of a screw dislocation—segments of crystals displace or shear relative to each other (c) a three-dimensional view of edge dislocation DC formed by inserting an extra half-plane of atoms in ABCD (d) a schematic diagram of a stacking fault. (Cottrell 1971 reproduced by the courtesy of Arnold Publishers.)...

Dislocations are line defects. They bound slipped areas in a crystal and their motion produces plastic deformation. They are characterized by two geometrical parameters 1) the elementary slip displacement vector b (Burgers vector) and 2) the unit vector that defines the direction of the dislocation line at some point in the crystal, s. Figures 3-1 and 3-2 show the two limiting cases of a dislocation. If b is perpendicular to s, the dislocation is named an edge dislocation. The screw dislocation has b parallel to v. Often one Finds mixed dislocations. Dislocation lines close upon themselves or they end at inner or outer surfaces of a solid. 

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