PARADISCYL:Cylinder-Remove surface debris: Difference between revisions
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There is remaining issue realted to the time step due to the small dislocation segments.This tutorial describes how to remove tiny dislocation segment in cylinder code. |
There is remaining issue realted to the time step due to the small dislocation segments.This tutorial describes how to remove tiny dislocation segment in cylinder code. |
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When two dislocations meet near the surface and form a junction, it would generate very tiny segments, which occur rapid drop in time step. To overcome this issue,we remove these small dislocation segment according to the following algorithm. |
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==Algorithm== |
==Algorithm== |
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1. Searching for the surface node(node in Figure |
1. Searching for the surface node(node in Figure 1.(a)) |
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2. Find the neighbor node(nbr1 in Figure |
2. Find the neighbor node(nbr1 in Figure 1.(a)) |
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3. Check if |
3. Check if ''seglength'' is less than ''minSeg'' |
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<math> 1.0-\mathbf{b} \cdot \mathbf{ \xi}<= \epsilon</math> |
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, where <math>\epsilon</math> is a tolerance. |
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4. Count the number of independent slip planes of inside arms. |
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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)) |
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''Case I'':if nbr1 has a independent slip plane and nbr1 has 3, 2 of which are linked to surface |
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-> Project nbr1 to the surface along the line(See figure 1.(b)) |
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''Case II'':if nbr1 has 2 independent slip planes |
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Project nbr1 to surface along the intersection of two slip planes(See figure 1.(c)) |
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5. Compute projected forces on each plane. |
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<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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{|border="0" align="center" |
{|border="0" align="center" |
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|[[Image: |
|[[Image:Remove_surface_debris1.jpg |frameless|200px|caption]] |
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|[[Image: |
|[[Image:Remove_surface_debris2.jpg |frameless|200px|caption]] |
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|[[Image:Remove_surface_debris3.jpg |frameless|200px|caption]] |
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|align="center"|(a) |
|align="center"|(a) |
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|align="center"|(b) |
|align="center"|(b) |
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|align="center"|(c) |
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|colspan="2" | Fig. |
|colspan="2" | Fig.1 (a)Schematic diagram (b) Case ''I'' (c) Case ''II''. |
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Latest revision as of 08:14, 8 December 2011
Remove surface debris
ill Ryu and Wei Cai
There is remaining issue realted to the time step due to the small dislocation segments.This tutorial describes how to remove tiny dislocation segment in cylinder code.
When two dislocations meet near the surface and form a junction, it would generate very tiny segments, which occur rapid drop in time step. To overcome this issue,we remove these small dislocation segment according to the following algorithm.
Algorithm
1. Searching for the surface node(node in Figure 1.(a))
2. Find the neighbor node(nbr1 in Figure 1.(a))
3. Check if seglength is less than minSeg
4. Count the number of independent slip planes of inside arms.
Case I:if nbr1 has a independent slip plane and nbr1 has 3, 2 of which are linked to surface -> Project nbr1 to the surface along the line(See figure 1.(b))
Case II:if nbr1 has 2 independent slip planes Project nbr1 to surface along the intersection of two slip planes(See figure 1.(c))
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| (a) | (b) | (c) |
| Fig.1 (a)Schematic diagram (b) Case I (c) Case II.
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