PARADISCYL:Cylinder-Surface cross slip

Surface cross slip

ill Ryu and Wei Cai

This tutorial describes how to implement cross slip mechanism from the cylinder surface. The theoretical background is published in Computing Image Stress in an Elastic Cylinder(Proceedings of the National Academy of Sciences, 105, 14304 (2008)http://micro.stanford.edu/~caiwei/papers/Weinberger08PNAS-bccpillar.pdf (PDF)])

How to select slip plane of the surface segments

To implement surface cross slip in cylinder code, we change the slip plane of the surface nodes with respect to the magnitude of the force on the surface nodes.To do that,dislocation character of the surface nodes should be screw-like.

If there is a screw dislocation in the BCC crystal in left figure, we can see that there are three possible slip planes(red, blue green planes in the figure). From the MD calculation, we knew the image stress generate the force to move dislocation in the direction along which the length of dislocation shortens.Therefore, image force points downward for the front node, while it points upward for the back node(Figure 1). However, P-K force points in same direction both front node and back node. Taking the summation of these forces into account, the slip plane is selected as the one on which the summation of forces has maximum value.

(a)	(b)
Fig.1 (a) When the P-K force is dominant(blue plane is taken as a slip plane for the front node, while red plane is selected for the back node)(b) When the image force is dominant(green plane is taken as a slip plane) In this mechanism, straight edge dislocation has two different slip planes(See Figure 1-(a)). Algorithm 1. Searching for the surface node(node in Figure 2.(a)) 2. Find the neighbor node(nbr1 in Figure 2.(a)) 3. Check if the character of surface segment is similar to screw 1.0-$T_{ij}<=\epsilon$ } In the second test case, we consider a circular prismatic dislocation loop concentric with the cylindrical axis, discretized with 12 equally spaced nodes connected by straight segments, as shown in Fig. 3, where the dislocation loop radius is 0.8 times to the cylinder radius. (a) (b) Fig.3 (a)Schematic (b) Slip system $\mathbf {b} =a[111],\mathbf {n} _{1}=[1{\bar {1}}1],\mathbf {n} _{2}=[0{\bar {1}}1],\mathbf {n} _{3}=[{\bar {1}}01]$. For simplicity, cell is not used for the force calculatoin. To do this, modify makefile in ParaDiS/cylinder directory so that the following line is active. DEFS += -D_CYL_TEST23 Due to the symmetry of this problem, the image forces on all nodes point to the radial direction and have the same magnitude. An estimate of their relative error is plotted as a function of number of grid in the circumferential direction($\mathbf {n} _{\theta }$) in the following figure. Because this problem does not have an analytic solution, the reference value is taken to be the value obtained using Method I (Bessel) with $\mathbf {n} _{\theta }=181$. As expected, the relative error in radial image force decreases with increasing number of grids.