PARADISCYL:Cylinder-Surface cross slip - Revision history

Iryu: /* Algorithm */

2011-12-08T01:01:34Z

Algorithm

← Older revision		Revision as of 01:01, 8 December 2011
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	\|align="center"\|(b)		\|align="center"\|(b)
	\|-		\|-
	\|colspan="2" \| Fig.3 (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>.		\|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>		<HR>

Iryu: /* How to select slip plane of the surface segments */

2011-12-08T01:00:50Z

How to select slip plane of the surface segments

← Older revision		Revision as of 01:00, 8 December 2011
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	[[Image:Schematic_view.jpg\|frameless\|300px\|right\|Fig.2. ]]		[[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 ~~shortens~~ ~~their~~ ~~length~~.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.		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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Iryu: /* How to select slip plane of the surface segments */

2011-12-08T00:57:50Z

How to select slip plane of the surface segments

← Older revision		Revision as of 00:57, 8 December 2011
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	[[Image:Schematic_view.jpg\|frameless\|300px\|right\|Fig.2. ]]		[[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 ~~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.		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 shortens their length.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.

	{\|border="0" align="center"		{\|border="0" align="center"
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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)).
	<HR>		<HR>

	==Algorithm==		==Algorithm==

Iryu: /* Algorithm */

2011-12-08T00:56:21Z

Algorithm

← Older revision		Revision as of 00:56, 8 December 2011
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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>

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

Iryu: /* Algorithm */

2011-12-08T00:54:58Z

Algorithm

← Older revision		Revision as of 00:54, 8 December 2011
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	<math> 1.0-\mathbf{b} \cdot \mathbf{ \xi}<= \epsilon</math>		<math> 1.0-\mathbf{b} \cdot \mathbf{ \xi}<= \epsilon</math>
	, where <math>\epsilon</math> is a tolerance.		, 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>

	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.		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.

Iryu at 00:47, 8 December 2011

2011-12-08T00:47:56Z

← Older revision		Revision as of 00:47, 8 December 2011
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	3. Check if the character of surface segment is similar to screw		3. Check if the character of surface segment is similar to screw
	1.0-~~<math>T_~~{ij}<= \epsilon</math> }		<math> 1.0-\mathbf{b} \cdot \mathbf{ \xi}<= \epsilon</math>
			, where <math>\epsilon</math> is a tolerance.

	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.		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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	\|-		\|-
	\|colspan="2" \| Fig.3 (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>.		\|colspan="2" \| Fig.3 (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>.
	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

	[[Image:T2_relative_error.jpg\|frameless\|400px\|right\|Fig.4. ]]
	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(<math>\mathbf{n}_\theta</math>) 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 <math>\mathbf{n}_\theta=181</math>. As expected, the relative error in radial image force decreases with increasing number of grids.

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Iryu: /* Algorithm */

2011-12-08T00:43:11Z

Algorithm

← Older revision		Revision as of 00:43, 8 December 2011
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	1. Searching for the surface node(node in Figure 2.(a))		1. Searching for the surface node(node in Figure 2.(a))

	2. Find the neighbor node(nbr1 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-<math>T_{ij}<= \epsilon</math> }

	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.		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.
	{\|border="0" align="center"		{\|border="0" align="center"

Iryu: /* Algorithm */

2011-12-08T00:37:14Z

Algorithm

← Older revision		Revision as of 00:37, 8 December 2011
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	==Algorithm==		==Algorithm==

			1. Searching for the surface node(node in Figure 2.(a))
			2. Find the neighbor node(nbr1 in Figure 2.(a))
	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.		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.
	{\|border="0" align="center"		{\|border="0" align="center"
	\|[[Image:~~T2_flat_loop_Isoview~~.jpg‎ \|frameless\|~~400px~~\|caption]]		\|[[Image:cylinder_fig.jpg‎ \|frameless\|300px\|caption]]
	\|[[Image:~~T2_flat_loop_topview~~.jpg‎ \|frameless\|~~400px~~\|caption]]		\|[[Image:slip_system.jpg‎ \|frameless\|300px\|caption]]
	\|-		\|-
	\|align="center"\|(a)		\|align="center"\|(a)
	\|align="center"\|(b)		\|align="center"\|(b)
	\|-		\|-
	\|colspan="2" \| Fig.3 (a) ~~Isoview~~ (b) ~~Topview.The~~ ~~Bergurs~~ ~~vector is~~ <math>\mathbf{b} = a[~~001~~]</math>.		\|colspan="2" \| Fig.3 (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>.
	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.		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		DEFS += -D_CYL_TEST23

Iryu at 23:11, 7 December 2011

2011-12-07T23:11:46Z

@@ Line 13: / Line 13: @@
 [[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 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
+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.
 {|border="0" align="center"
 |[[Image:Force_sum1.jpg|frameless|350px|caption]]
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 |align="center"|(b)
 |-
-|colspan="2" | Fig.1 (a) When the P-K force is dominant(blue plane is taken as a slip plane)(b) When the image force is dominant(green plane is taken as a slip plane)
+|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)
 <HR>

Iryu at 23:03, 7 December 2011

2011-12-07T23:03:39Z

← Older revision		Revision as of 23:03, 7 December 2011
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	[[Image:Schematic_view.jpg\|frameless\|300px\|right\|Fig.2. ]]		[[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 ~~know~~ the image stress ~~Using~~ ~~Method I,~~the ~~relative~~ ~~error~~ in the image force is ~~computed~~ ~~from~~ the ~~image~~ ~~stress~~ ~~calcualtion.~~ ~~Right~~ ~~plot~~ ~~clearly~~ ~~show~~ ~~exponential~~ ~~decay~~ of ~~the~~ ~~relative~~ ~~error~~ ~~with~~ ~~increasing~~ ~~number~~ of ~~grids.~~ ~~The~~ ~~convergence~~ is ~~slower~~ ~~when~~ the ~~location~~ of the ~~edge~~ ~~dislocation~~ is ~~close~~ to the ~~surface.~~		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

	For simplicity, we In the first test case, image stress of a straight edge dislocation is considered, as shown in the ''Figure 1'', where dislocation is offset(<math>x0</math>) from the center of the cylinder. We have two cases of <math>x0</math>=0.5<math>R</math> and 0.9<math>R</math>.Here, <math>R</math> is the radius of the cylinder.		For simplicity, we In the first test case, image stress of a straight edge dislocation is considered, as shown in the ''Figure 1'', where dislocation is offset(<math>x0</math>) from the center of the cylinder. We have two cases of <math>x0</math>=0.5<math>R</math> and 0.9<math>R</math>.Here, <math>R</math> is the radius of the cylinder.