Test Case: Dislocation Loop - Revision history

Caiwei at 19:49, 19 December 2007

2007-12-19T19:49:02Z

← Older revision		Revision as of 19:49, 19 December 2007
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	The Burgers vector of a prismatic dislocation loop is perpendicular to the plane of the dislocation loop. If such a dislocation loop moves across a solid through glide, the material that is displaced will look like a prism with the shape of the dislocation loop as its cross-section. In the absence of any external stress, a prismatic loop shrinks through climb. From an energetic point of view, the existence of a dislocation loop requires energy and a dislocation loop shrinks to minimize the total energy of the system. Let us investigate the rate of change of area of a prismatic loop in the absence of external stresses.		The Burgers vector of a prismatic dislocation loop is perpendicular to the plane of the dislocation loop. If such a dislocation loop moves across a solid through glide, the material that is displaced will look like a prism with the shape of the dislocation loop as its cross-section. In the absence of any external stress, a prismatic loop shrinks through climb. From an energetic point of view, the existence of a dislocation loop requires energy and a dislocation loop shrinks to minimize the total energy of the system. Let us investigate the rate of change of area of a prismatic loop in the absence of external stresses.

	===Energy and Kinetics of a Circular Prismatic Loop==		===Energy and Kinetics of a Circular Prismatic Loop===
	Consider a circular prismatic loop with radius <math>R \,</math> and Burgers vector <math>\mathbf{b}</math>.		Consider a circular prismatic loop with radius <math>R \,</math> and Burgers vector <math>\mathbf{b}</math>.
	The rate of change of area of the circular loop with respect to time <math>t</math> is		The rate of change of area of the circular loop with respect to time <math>t</math> is
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	===Comparision with DDLab simulations==		===Comparision with DDLab simulations===
	Here we will put some matlab plots.		Here we will put some matlab plots.

Caiwei: /* Prismatic Loop */

2007-12-19T19:48:31Z

Prismatic Loop

← Older revision		Revision as of 19:48, 19 December 2007
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	The Burgers vector of a prismatic dislocation loop is perpendicular to the plane of the dislocation loop. If such a dislocation loop moves across a solid through glide, the material that is displaced will look like a prism with the shape of the dislocation loop as its cross-section. In the absence of any external stress, a prismatic loop shrinks through climb. From an energetic point of view, the existence of a dislocation loop requires energy and a dislocation loop shrinks to minimize the total energy of the system. Let us investigate the rate of change of area of a prismatic loop in the absence of external stresses.		The Burgers vector of a prismatic dislocation loop is perpendicular to the plane of the dislocation loop. If such a dislocation loop moves across a solid through glide, the material that is displaced will look like a prism with the shape of the dislocation loop as its cross-section. In the absence of any external stress, a prismatic loop shrinks through climb. From an energetic point of view, the existence of a dislocation loop requires energy and a dislocation loop shrinks to minimize the total energy of the system. Let us investigate the rate of change of area of a prismatic loop in the absence of external stresses.

			===Energy and Kinetics of a Circular Prismatic Loop==
	Consider a circular prismatic loop with radius <math>R \,</math> and Burgers vector <math>\mathbf{b}</math>.		Consider a circular prismatic loop with radius <math>R \,</math> and Burgers vector <math>\mathbf{b}</math>.
	The rate of change of area of the circular loop with respect to time <math>t</math> is		The rate of change of area of the circular loop with respect to time <math>t</math> is
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	<math>W_a = 2\pi R\frac{\mu b^2}{4\pi (1-\nu)}\left(\mathrm{ln}\frac{8R}{a}-1\right)</math>		<math>W_a = 2\pi R\frac{\mu b^2}{4\pi (1-\nu)}\left(\mathrm{ln}\frac{8R}{a}-1\right)</math>
	So the force per unit length within this formalism is given by		So the force per unit length within this formalism is given by
	<math>f_R=-\frac{1}{2\pi R}\frac{dW_a}{dR}=-\frac{1}{2\pi R} \frac{\mu b^2}{~~4\pi~~ (1-\nu)}\mathrm{ln}\frac{8R}{a}</math>		<math>f_R=-\frac{1}{2\pi R}\frac{dW_a}{dR}=-\frac{1}{2\pi R} \frac{\mu b^2}{2(1-\nu)}\mathrm{ln}\frac{8R}{a}</math>
	Therefore the rate of change of area of a circular dislocation loop in the non-singular continuum formalism can be written as		Therefore the rate of change of area of a circular dislocation loop in the non-singular continuum formalism can be written as
	<math>\frac{dA}{dt}=-m_{climb}\frac{\mu b^2}{2(1-\nu)}\mathrm{ln}\frac{8R}{a}</math>		<math>\frac{dA}{dt}=-m_{climb}\frac{\mu b^2}{2(1-\nu)}\mathrm{ln}\frac{8R}{a}</math>



			===Comparision with DDLab simulations==
	Here we will put some matlab plots.		Here we will put some matlab plots.

Caiwei at 18:39, 19 December 2007

2007-12-19T18:39:38Z

← Older revision		Revision as of 18:39, 19 December 2007
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	Second, radius versus time.		Second, radius versus time.

			We would like to get analytic solution of R(t) (e.g. from Maple) and compare with DD simulation.

			What is the purpose of this comparison?

			1. The nodal speed and the total life time of the loop

			2. What is the accuracy of approximating the area of a circle by a polygon.

			How can we calculate the radius of the polygon?

Caiwei at 18:20, 19 December 2007

2007-12-19T18:20:54Z

New page

==Prismatic Loop==
The Burgers vector of a prismatic dislocation loop is perpendicular to the plane of the dislocation loop. If such a dislocation loop moves across a solid through glide, the material that is displaced will look like a prism with the shape of the dislocation loop as its cross-section. In the absence of any external stress, a prismatic loop shrinks through climb. From an energetic point of view, the existence of a dislocation loop requires energy and a dislocation loop shrinks to minimize the total energy of the system. Let us investigate the rate of change of area of a prismatic loop in the absence of external stresses.

Consider a circular prismatic loop with radius <math>R \,</math> and Burgers vector <math>\mathbf{b}</math>.
The rate of change of area of the circular loop with respect to time <math>t</math> is
<math>\frac{dA}{dt}=\frac{d}{dt}\pi R^2=2\pi R\frac{dR}{dt}=2\pi Rv_R</math>
where <math>v_R \,</math> is the magnitude of the radial velocity for the points or line segments that constitute the loop. In our nodal representation of dislocation lines, the velocity of the nodes is calculated using a linear mobility law
<math>v_R = m_{climb}f_R\,</math>
where <math>f_R \,</math> is the force per unit length experienced by a dislocation node on the loop and <math>m_{climb}\,</math> the climb mobility of an edge dislocation. If we let <math>W\,</math>be self-energy of a dislocation loop, the force per unit length on each node is given by
<math>f_R=-\frac{1}{2\pi R}\frac{dW}{dR}</math>
In the non-singular continuum formalism of Cai W ''et. al.'' the self-energy of a dislocation loop <math>W_a \,</math> is given by
<math>W_a = 2\pi R\frac{\mu b^2}{4\pi (1-\nu)}\left(\mathrm{ln}\frac{8R}{a}-1\right)</math>
So the force per unit length within this formalism is given by
<math>f_R=-\frac{1}{2\pi R}\frac{dW_a}{dR}=-\frac{1}{2\pi R} \frac{\mu b^2}{4\pi (1-\nu)}\mathrm{ln}\frac{8R}{a}</math>
Therefore the rate of change of area of a circular dislocation loop in the non-singular continuum formalism can be written as
<math>\frac{dA}{dt}=-m_{climb}\frac{\mu b^2}{2(1-\nu)}\mathrm{ln}\frac{8R}{a}</math>

Here we will put some matlab plots.

First, loop area versus time. For different lmin = 10, 50, 100. Initial R = 500.

Second, radius versus time.