How to plot the activated slip plane in unit cell of FCC metal?

[LuqingCui]

Hi all，

I have the EBSD data of an FCC metal after the microhardness indentation test, and I have obtained the trace of the activated slip plane on the sample surface, as shown:

Is it possible to draw the activated slip plane in the unit cell of FCC metal using MTEX? The example is as follows:

If my description is not clear, I can provide more information.

Best Regards,
Luqing Cui

[LuqingCui]

For example, the Euler angles of a grain containing the trace of slip plane (-1 1 1) is (312.1°, 45°, 63.8°).
How could we plot the activated slip plane (-1 1 1) in the unit cell of FCC with the Euler angles of (312.1°, 45°, 63.8°)?

Looking forward to your reply!
Best regards,
Luqing Cui

[kilir]

Hi,
have a look at this issue here: #504
So, yes, you can do it, but it needs a bit of tweaking.
Cheers,
Rüdiger

[LuqingCui]

Hi Rüdiger,

Thanks very much for your reply.
I want to draw the activated slip plane {-111} in the unit cell of FCC with the Euler angle of (312.1°, 45°, 63.8°). It will be better if the sample surface can be shown in the unit cell too! I know the trace of the slip plane is the intersection of the activated slip plane with the sample surface.
I have written part of the code to achieve this goal as follows, but I don't know how to finish it:

ori = orientation.byEuler(312.1degree,45degree,63.8degree,cs)
cS = crystalShape.cube(cs)
pcS=plot(cS);
alpha(pcS,0.5);
hold on
p = cS(Miller(-1,1,1,cS.CS));
s = - p.faceCenter; % shift to the center
plot(1.3(p + s),'FaceColor','r') % potentially scale if needed
hold off

Could you help me to complete this code? Thanks very much for the help!

Looking forward to your reply!
Best regards,
Luqing Cui

[kilir]

Hi,
this shows you x-y plane as well, if I understand you correctly. You can easily adapt this for an EBSD map I guess.

cs = crystalSymmetry('cubic')
% ori = orientation.byEuler(0,0,0,cs)
ori = orientation.byEuler(312.1*degree,45*degree,63.8*degree,cs)


cS1 = crystalShape.cube(cs)
pcS=plot(ori*cS1);
alpha(pcS,0.5);
h = Miller(1,1,1,cs)
cS2 = crystalShape([h Miller(1,0,0,cs)],2.66); % magic number

% potential planes just for the example
h = symmetrise(h,'unique','antipodal')

hold on
% for some unknown reason take the second one - e.g. you found that this is your activated sS
p = ori*cS2(h(2));
s = - p.faceCenter; % shift to the center
plot(1.3*(p + s),'FaceColor','r') % arbitrary scaling
hold off

% x-y plane
[x y] = meshgrid(-0.5:0.5);
hold on
surf(x, y, zeros(size(x, 1)),'FaceColor','k','FaceAlpha',0.4)
hold off

Hope that helps.
Cheers,
Rüdiger

[LuqingCui]

Hi Rüdiger，

Thanks very much for your reply. The code runs very well.
But I have another question want to know, as follows:
The unit cell of FCC with the Euler angle of (312.1°, 45°, 63.8°) obtained by Channel 5 is as follows:

But the unit cell with the same Euler angle obtained from the MTEX using the above code is as follows:

They have different orientations.

In addition, the trace of the activated slip plane (-111) on the sample surface obtained using the above MTEX code is as follows:

The direction of slip trace is also not consistent with the SEM result, as follows:

Do you know the reason?

Best regards,
Luqing Cui

[kilir]

Hi,
this is a different topic. 1) you need to know the specimen coordinate system shown in Channel5 to compare the plots. 2) In the example I gave above, it plots (1-11).
Cheers,
Rüdiger

[LuqingCui]

Hi Rüdiger,

Thanks for your reply. I have modified the plane to (-111) during the plotting of the figure.
Yes, I also think the problem is related with the specimen coordinate system.

Best regards,
Luqing Cui