PARADISCYL:Cylinder-test

Test Cases in cylinder code

ill Ryu and Wei Cai

This tutorial describes 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)(PDF)) and Dislocation Dynamics Simulations in a Cylinder(Proceedings of the Dislocations 2008 International Conference, IOP Conference Series: Materials Science and Engineering, vol 3,012007 (2009).(PDF)).

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TEST1 : Straight edge dislocation along the cylinder axis

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

(a)	(b)
Fig.1 (a) Isoview (b) Topview.The Bergurs vector is $\mathbf {b} =a[100]$. Since the image stress field of this configuration had been solved analytically[Eshelby, J.D., 1979],we compare the image force between the simulation result and analytic solution.The number of grids in circumferential direction is same as the one along the cylinder axis.To consider periodic images along the cylinder axis under PBC, analytical stress field of edge dislocation is implemented in the function of "AllSegmentStress_no_cell_test1".To use this function, modify makefile in ParaDiS/cylinder directory so that the following line is active. DEFS += -D_CYL_TEST1 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. TEST2 : Circular prismatic dislocation loop 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. (a) (b) Fig.3 (a) Isoview (b) Topview.The Bergurs vector is $\mathbf {b} =a[001]$. 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 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($\mathbf {n} _{\theta }$) 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 $\mathbf {n} _{\theta }=181$. As expected, the relative error in radial image force decreases with increasing number of grids. TEST3 : Circular inclined glide dislocation loop In the third test case, we consider more general shaped dislocation loop, discretized with 10 equally spaced nodes connected by straight segments, as shown in Fig. 5. (a) (b) Fig.5 (a) Isoview (b) Topview.The Bergurs vector is $\mathbf {b} =\mathbf {a} /{\sqrt {2}}[{\bar {1}}10]$ and the slip plane is $\mathbf {n} =1/{\sqrt {6}}[{\bar {1}}{\bar {1}}2]$ Likewise, 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 The orientation and magnitude of the image force on every node is different. Fig. 10 plots the component of the image force in the direction away from the center of the loop for all 10 nodes with different values of $\mathbf {n} _{\theta }$ using Method I. With more than 45 number of grids, they converge to almost same value. TEST4 : Circular glide dislocation loop In the final test case, we consider time evolution of glide dislocation loop, discretized with 50 equally spaced nodes connected by straight segments, as shown in the following Figure. (a) (b) Fig.7 (a) Isoview (b) Topview.The Bergurs vector is $\mathbf {b} =\mathbf {a} [100]$ and the slip plane is $\mathbf {n} =[001]$ In general, the energy of an edge dislocation is greater than that of a screw, $\mathbf {E} _{e}>\mathbf {E} _{s}$ and the line tension of an edge dislocation is smaller than that of a screw, $\mathbf {T} _{e}<\mathbf {T} _{s}$. This apparent anomaly arises because when an edge dislocation is bowed out,less energetic screw segments are created, while when a screw dislocation is bowed, out more energetic edge segments are created. For this reason it is easier to bow an edge dislocation compared to a screw dislocation.One consequence of the differences in line tension between the edge and screw is that a dislocation segment does not take a circular shape when it is bowed out by the action of a stress.(Nix,W.D. Course note on MATSCI 206) With this in mind, we put circular glide dislocation loop inside cylinder and watch the time-evolution of dislocation loop.In this model, we use $\mathbf {n} _{\theta }=\mathbf {n} _{z}=101$. As expected,The loop initially becomes oval as the edge components are pulled towards the surface and the screw components are pulled inward. The edge components then leave the cylinder and the dislocation loop becomes two truncated segments, which ultimately leave the cylinder as well .