PARADISCYL:Cylinder-Surface cross slip: Difference between revisions

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This tutorial describes how to implement cross slip mechanism from the cylinder surface. test cases to check if cylinder code works well, especially for the image stress calculation. We provide Matlab files which generates ParaDiS inputs and plot the results. The theoretical background is published in Computing Image Stress in an Elastic Cylinder(''Journal of the Mechanics and Physics of Solids'', '''55''', 2027 (2007)[http://micro.stanford.edu/~caiwei/papers/Weinberger07jmps-image.pdf (PDF)])
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.
[[Image:Schematic_view.jpg|frameless|300px|right|Fig.2. ]]

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.

{|border="0" align="center"
|[[Image:Force_sum1.jpg|frameless|350px|caption]]
|[[Image:Force_sum2.jpg|frameless|350px|caption]]
|-
|align="center"|(a)
|align="center"|(b)
|-
|colspan="2" | 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)).
<HR>

==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
<math> 1.0-\mathbf{b} \cdot \mathbf{ \xi}<= \epsilon</math>
, where <math>\epsilon</math> is a tolerance.

4. Given burgers vector, there are three possible slip planes. For example, if <math>\mathbf{b} = a[111]</math>, then possible slip planes are <math>\mathbf{n}_1 = [1\bar{1}1],\mathbf{n}_2 = [0\bar{1}1],\mathbf{n}_3 = [\bar{1}01]</math>(See figure 2(b))

5. Compute projected forces on each plane.
<math>\mathbf{F}_{i} = \mathbf{F}_{total}-(\mathbf{F}_{total} \cdot \mathbf{n}_i)\mathbf{n}_i</math>

6. Based on the magnitudes of <math>\mathbf{F}_{i}</math>, choose slip plane of the surface dislocation segment.
{|border="0" align="center"
|[[Image:cylinder_fig.jpg‎ |frameless|300px|caption]]
|[[Image:slip_system.jpg‎ |frameless|300px|caption]]
|-
|align="center"|(a)
|align="center"|(b)
|-
|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>.
<HR>

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.

Fig.2.

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.

caption caption
(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

   

, where is a tolerance.

4. Given burgers vector, there are three possible slip planes. For example, if , then possible slip planes are (See figure 2(b))

5. Compute projected forces on each plane.

6. Based on the magnitudes of , choose slip plane of the surface dislocation segment.

caption caption
(a) (b)
Fig.2 (a)Schematic (b) Slip system .