which function is right to get GND density and how to seperate contributions from screw and edge dislocations?

[huihu...@hotmail.com]

Please be as detailed as possible, explain what you want to do, what kind of data you are using. Attaching simplified code and data is always useful. Click the the button { } to inline code and finally remove these lines before posting :)

I am using MTEX to calculate GND density as mentioned in  http://mtex-toolbox.github.io/files/doc/GND_demo.html. there are two problems which i don't know how to solve? 

Question1: 

 I saw there are two ways to sovle GND, as follows:

1. gnd = factor*sum(abs(rho .* dSRot.u),2)  ?

2. gnd = calcGND(ebsd)   ?

which choice is right? please give an explanation as possisble as clearly. thanks a lot!

Question2: 

how to seperately get screw and edge dislocation density in gnd calculation?   

GND_screw + GND_edge =GND

Thank you very much!

Huihui

[ruediger Kilian]

Hi Huihui,

with respect to your first question: 1. is what is actually happening ins calcGND. You can inspect the function using "edit calcGND".

With respect to your second question: Are you looking for a solution based on only edge/screw dislocations or do you want the proportion of each while fitting both at the same time?

Cheers,
Rüdger

[huihu...@hotmail.com]

Dear Rüdger,

Thank you so much for your answers. 

The gnd density per pixel is computed by command:   gnd = factor*sum(abs(rho .* dSRot.u),2) . Yes, many thanks.

question2:

With respect to your second question: Are you looking for a solution based on only edge/screw dislocations or do you want the proportion of each while fitting both at the same time? 

I would like to get the distributions(proportions) of two kinds of dislocations (including screw and edge) based on total (gnd) dislocation density.  I really hope there are three images, including total gnd cloud map, screw cloud map, and edge dislocation cloud map. Could you give me an example for this question.

Thanks a lot!

 Huihui

[ruediger Kilian]

Hi Huihui,

my first try would be the lines below. However, please be aware that the result in all cases depends on .u (dislocation energy) which in the example is simply set to 2 and 1 which is certainly off the guess using e_edge/e_screw = 1/(1-poissonsratio) or any realistic value which - to me - seems not trivial to derive.

% prepare some data
mtexdata single
ebsd = ebsd(inpolygon(ebsd,[6 8 3 3]));
[grains,ebsd.grainId] = ebsd.calcGrains;
f=halfQuadraticFilter; f.alpha = [0.05 0.05];
ebsd=smooth(ebsd,f);
ebsd = ebsd.gridify;
%%
cs = ebsd('a').CS;
dS = dislocationSystem.fcc(cs)

% compute curvature tensors
kappa = ebsd.curvature;
% rotate dislocationSystem into specimen coordinates
dSRot = ebsd.orientations * dS;
% fit dislocationSystem to curvature tensor
[rho,factor] = fitDislocationSystems(kappa,dSRot);

% total
gnd_tot = factor*sum(abs(rho .* dSRot.u),2);
% gnd screw (13-18)
gnd_screw = factor*sum(abs(rho(:,13:18) .* dSRot(:,13:18).u),2);
% gnd edge (1-12)
gnd_edge = factor*sum(abs(rho(:,1:12) .* dSRot(:,1:12).u),2);
plot(ebsd,gnd_tot)
nextAxis
plot(ebsd,gnd_screw)
nextAxis
plot(ebsd,gnd_edge)
mtexColorbar



Hope that helps.
Cheers,
Rüdiger

[huihu...@hotmail.com]

Dear Rüdger,

Very sorry not to reply promptly to your e-mail beacuse of the delay of celebrating Chinese New Year.

Thank you for your kind help!

Thank you for including the code to calculate GND. You help me a lot and I am starting to understand the gnd_demo.  

[Mustafa Rifat]

Dear Rüdger,

I am working on the same GND demo right now with my own data (fcc-Ni superalloy). I have 2 questions regarding this.

1. The default dislocation energy is 1 and 2 for screw and edge dislocations respectively. It is said that depending on the material it can vary. How can I know what is the actual dislocation energy for my material? 

2. By plotting:

 plot(ebsd,factor*sum(abs(rho .* dSRot.u),2),'micronbar','off')

we get the dislocation energy density, according to the demo. But how can we get actual dislocation density like fig. 7 or 8 from this paper 

Influence of plastic deformation heterogeneity on development of geometrically necessary dislocation density in dual phase steel?

Thanks a lot!

[Rüdiger]

Hi,
How can you know the actual dislocation energy? I think this is not
possible.
You can simply make an assumption and hope that your assumption isn't
too bad.

In the demo, it is simply assumed 1 and 0.7 for a screw. This is simply
an example. If you read the comment, a first approach would be to assume
a crystal to be elastically isotropic and use the form provided e.g. in
Hull& Bacon. If one doesn't like elastically isotropic crystals, Hirth &
Lothe p.443ff (chapter 13-3 Straight dislocations in anisotropic media /
Stresses and Energies) provide a procedure.

The result you get is in m^-2.

Hope this helps.

Cheers,
Rüdiger

[Mustafa Rifat]

Thank you for your reply. I think I could not express my question #2 clearly. I wanted to ask that the last plot in the demo, what we are getting can we call it dislocation density plot? Or this is dislocation energy density? I am referring to this image.

 If it is not dislocation density, is it really possible to get a single dislocation density plot? Because it is a tensor and it has 9 components, it is not a single value for every pixel.

[MOHAMED OU LAHCEN EDDAHBI .]

Hi Mustafa,

The above plot, yes, it gives an estimation of distribution of dislocation density, but it is valid only for the dislocation system and dislocation energy supposed in the example, as Rüdiger mentioned.

Of course, this is not a general rule. You must know what happens in your material and then impose the appropriate dislocation system, energy etc...(deformation and/or annealed state of your material; i.e. deformation, disorientation, stacking fault energy, etc.. )

Salam

Mohamed

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Mohamed OuLahcen, Eddahbi

Researcher in Department of Physics

 Engineering School - University Carlos III of Madrid

Av. de La Universidad 11, Leganés-Madrid, 28911 Spain

[Mustafa Rifat]

Mohamed, thanks a lot for your clarification. My material is fcc, so there will be change in dislocation systems. There will be change in Poissions Ratio too. Other than that, I can not think of any more changes. 

I mainly got confused because the image I attached is titled as total dislocation energy but it's unit is (1/m^2), there is not energy unit attached. I guess it is because we assumed the screw and edge dislocation energy 0.7 and 1 respectively, without any unit. 

But I understand your point. Though it is dislocation energy density, but it is a good estimation of dislocation density.   

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[Mustafa Rifat]

Hello All,

In the demo page,it is written, U_screw=(Gb^2/4pi)ln R/r0

and, U_edge=(1-nu)U_screw

Should not it be: U_edge=U_screw/(1-nu)?