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b d b/2 y x A plane B plane

We start with two separate crystals. The lattices are identical with parameters b and d.

Initial atomic positions in the A plane are given by xA = nb

Initial atomic positions in the B plane are given by xB = nb ± b/2

The atoms in the plane shown in red in effect become an edge dislocation when the crystals are joined.

b d x y A plane B plane Δx φ

On joining the crystals to form a dislocation, the atoms relax to minimise the overall misfit energy

Misfit energy arises from:

a) Strains caused by displacements Δx within the A, B planes
  • In initial configuration, Δx = 0
b) Misalignment φ caused by displacement across the slip plane
  • In initial configuration, φ = b/2

To determine the final atomic positions we can consider only the A and B plane atoms.

Here the atoms either side of the dislocation line have been numbered.

dislocation area

The final atomic positions for the A/B planes xA' and xB' are given by xA + uA and xB + uB respectively, where u represents the displacement from original position

The displacements (shown in red) are given by uA and uB

The displacements are symmetrical either side of the displacement line

dislocation area

Far from the dislocation, the atoms are aligned such that:

  • uA = -b/4 (x = +∞)
  • uA = +b/4 (x = -∞)
  • uB = +b/4 (x = +∞)
  • uB = -b/4 (x = -∞)

The displacement u(x) varies as arctan(x) with distance from the dislocation line

tan graph

k and w are constants to be determined

dislocation area

The displacement u(x) varies as arctan(x) with distance from the dislocation line

tan graph

k and w are constants to be determined

dislocation area

tan graph dislocation area

To determine k we consider the values of u at x=0 and x=∞

At x=∞, tan-1(x/w) = π/2 and uA = -b/4

→ k = b/2π

w is the dislocation width. To find it we need to minimise the overall misfit energy. This is explained in the next animation.