Test Case: Dislocation Loop

Prismatic Loop

The Burgers vector of a prismatic dislocation loop is perpendicular to the plane of the dislocation loop. If such a dislocation loop moves across a solid through glide, the material that is displaced will look like a prism with the shape of the dislocation loop as its cross-section. In the absence of any external stress, a prismatic loop shrinks through climb. From an energetic point of view, the existence of a dislocation loop requires energy and a dislocation loop shrinks to minimize the total energy of the system. Let us investigate the rate of change of area of a prismatic loop in the absence of external stresses.

Consider a circular prismatic loop with radius $R\,$ and Burgers vector $\mathbf {b}$. The rate of change of area of the circular loop with respect to time $t$ is

${\frac {dA}{dt}}={\frac {d}{dt}}\pi R^{2}=2\pi R{\frac {dR}{dt}}=2\pi Rv_{R}$

where $v_{R}\,$ is the magnitude of the radial velocity for the points or line segments that constitute the loop. In our nodal representation of dislocation lines, the velocity of the nodes is calculated using a linear mobility law

$v_{R}=m_{climb}f_{R}\,$

where $f_{R}\,$ is the force per unit length experienced by a dislocation node on the loop and $m_{climb}\,$ the climb mobility of an edge dislocation. If we let $W\,$be self-energy of a dislocation loop, the force per unit length on each node is given by

$f_{R}=-{\frac {1}{2\pi R}}{\frac {dW}{dR}}$

In the non-singular continuum formalism of Cai W et. al. the self-energy of a dislocation loop $W_{a}\,$ is given by

$W_{a}=2\pi R{\frac {\mu b^{2}}{4\pi (1-\nu )}}\left(\mathrm {ln} {\frac {8R}{a}}-1\right)$

So the force per unit length within this formalism is given by

 $f_{R}=-{\frac {1}{2\pi R}}{\frac {dW_{a}}{dR}}=-{\frac {1}{2\pi R}}{\frac {\mu b^{2}}{4\pi (1-\nu )}}\mathrm {ln} {\frac {8R}{a}}$

Therefore the rate of change of area of a circular dislocation loop in the non-singular continuum formalism can be written as

${\frac {dA}{dt}}=-m_{climb}{\frac {\mu b^{2}}{2(1-\nu )}}\mathrm {ln} {\frac {8R}{a}}$

Here we will put some matlab plots.

First, loop area versus time. For different lmin = 10, 50, 100. Initial R = 500.

Second, radius versus time.