## Edge dislocation displacement fields — Part I

### March 4, 2007

In the previous post, I talked about a nice pedagogical paper of Eshelby in which he describes a simple derivation of the displacement fields of an edge dislocation. I will summarise the derivation in two parts. Here is part I.

Schematic from the paper:

Derivation:

The derivation proceeds in two stages. First step is to show that the elastic displacement fields of an edge dislocation can be obtained by differentiation of that of the wedge dislocation. In the second step (which will be Part II of the post), the displacement fields of the wedge dislocations are calculated.

Displacement fields of the wedge dislocations and their relationship to that of the edge dislocations:

Assume that we are dealing with a plane problem in an infinite system; or, in other words, the displacement fields perpendicular to the figure are zero.

Positive wedge dislocation creation — Figures (a) and (b) — Cut the wedge region AOB, of angle $\Omega$, remove it, and weld the lines AO and BO to close the wedge. In a similar manner, negative wedge dislocation creation is related to cutting a radial line and forcing a wedge into the cut region.

From the process described above, we can show

1. the stress fields of positive and negative wedge dislocations are equal and opposite; and,
2. the stress field of a wedge dislocation is cylindrically symmetric.

Well, how do we do that? In a material with positive wedge dislocation, any radial cut will make the material spring an opening, which can be filled by a negative wedge dislocation, leaving the material stress free (but with a positive and negative wedge dislocation, formally) — this proves claim 1. The any radial cut is the key in the above argument which tells that the stress fields are cylindrically symmetric, thus proving claim 2.

Claim 1 implies that the stresses derived from the displacement fields of the positive ($\mathbf{u}^{+}_w$) and negative ($\mathbf{u}^{-}_w$) wedge dislocation are equal; and hence, the displacements can only differ by a rigid-body displacement. Also, if the positive and wedge dislocations are made at the same radial line, $\mathbf{u}^{+}_w = - \mathbf{u}^{-}_w$.

Consider Figure (e) — here we have made a positive and negative wedge dislocation as above, except that the negative wedge dislocation is made with its apex at a distance $d$ slightly downwards. The steps involved in one such process, for example, are as follows:

• Cut a positive wedge AOB
• Slide the wedge along AO by $d$ so that the apex is at $O^{\prime}$; close the wedge leaving the material unstressed with a parallel sided radial fissure of width $b = \Omega d$
• Close the fissure. This results in an edge dislocation with Burgers vector of magnitude $b$.

Thus,

$\mathbf{u}_{e} (x,y) = \mathbf{u}^{+}_w(x,y) + \mathbf{u}^{-}_w (x, y+d) = \mathbf{u}^{+}_w(x,y) - \mathbf{u}^{+}_w (x, y+d)$

Since $b = \Omega d$,

$\mathbf{u}_{e} (x,y) = -\frac{b}{\Omega} \frac{\partial (\mathbf{u}^{+}_w)}{\partial y}$

The above expression holds a few $b$ away from the centre of the dislocations, and a relationship of the type above holds for all quantities linearly related to displacement — stress, strain, Airy stress function etc.

In the next post, I will describe how to obtain the displacement fields for the wedge dislocation.