So far, we have come across a couple of equations that include the magnitude of Burgers vector in this chapter. The concept of Burgers vector was first introduced in Section 2.2. Now we need to understand how the magnitude of the Burgers vec­tors can be determined. Burgers vector is basically the shortest lattice vector joining one lattice point to another. This type of Burgers vector is associated with disloca­tions termed as perfect or unit dislocation. For example, in a BCC crystal, the shortest lattice vector is the distance between a corner atom and the body center. We can resolve the vector by a0/2 length from the origin along X-, Y-, and Z-axes, where a0 is the lattice constant. The standard notation for Burgers vector is then [(a0/2), (a0/2), (a0/2)] or (a0/2)[111]. The magnitude (strength) of the Burgers vector ofa perfect dislocation in a BCC crystal is

b — ^4 + 0° + a0 (i’6’’ half °f the body diagonal).

Similarly, in FCC metals, the Burgers vector is (a0/2) [110], that is, half of the face diagonal. The magnitude of the vector is a0//2. In a cubic crystal, we can follow the following procedure to find out the Burgers vector of a dislocation (does not need to be a perfect dislocation). If a dislocation has Burgers vector of q[uvw] with x being a fraction or a whole number, the magnitude of the Burgers vector is q-(u2 + v2 + w2)1/2. For instance, Burgers vector of a0/3 [112] dislocation has a mag­nitude of (a0/3)(12 + 12 + 22)1/2 — (a0/3)p6.

Determination of the Burgers vector of a perfect dislocation in a HCP crystal is much simpler than the cubic metals. The strength of the Burgers vector is given by the edge of the base, that is, a0.

■ Example 4.3

An FCC crystal (lattice constant a0 — 0.286 nm) contains a total dislocation density of 109 m~2, of which only half are mobile (i. e., glissile). If we assume that the Burgers vector of all the mobile dislocations is a0/2 [110] and the shear strain rate is 10-1 s_1, what is the average dislocation velocity?

Solution

To solve the above problem, we need to use Eq. 4.8.

Given are the shear strain rate (у) = 10-1 s-1, mobile dislocation density (pm) = 0.5 x 109 cm-2, and the magnitude of the Burgers vector = a0//2 = 0.202 nm. Therefore, the average dislocation velocity is given by

10-1 s—1

Vd =

4.1.5