PARADISCYL:Cylinder-Surface cross slip: Difference between revisions
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[[Image:Schematic_view.jpg|frameless|300px|right|Fig.2. ]] |
[[Image:Schematic_view.jpg|frameless|300px|right|Fig.2. ]] |
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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 |
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 dislocations in the direction along which the length of dislocation would be shortened.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 projected total force has maximum value. |
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In this mechanism, straight edge dislocation has two different slip planes(See Figure 1-(a)). |
In this mechanism, straight edge dislocation has two different slip planes(See Figure 1-(a)). |
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==Algorithm== |
==Algorithm== |
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<math>\mathbf{F}_{i} = \mathbf{F}_{total}-(\mathbf{F}_{total} \cdot \mathbf{n}_i)\mathbf{n}_i</math> |
<math>\mathbf{F}_{i} = \mathbf{F}_{total}-(\mathbf{F}_{total} \cdot \mathbf{n}_i)\mathbf{n}_i</math> |
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6. Based on the magnitudes of <math>\mathbf{F}_{i}</math>, choose slip plane of the surface dislocation segment. |
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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. |
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|align="center"|(b) |
|align="center"|(b) |
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|colspan="2" | Fig. |
|colspan="2" | Fig.2 (a)Schematic (b) Slip system <math>\mathbf{b} = a[111],\mathbf{n}_1 = [1\bar{1}1],\mathbf{n}_2 = [0\bar{1}1],\mathbf{n}_3 = [\bar{1}01]</math>. |
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Latest revision as of 01:01, 8 December 2011
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 dislocations in the direction along which the length of dislocation would be shortened.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 projected total force has maximum value.