Calculating young's modulus and inclination angle histogram

[Chamara Sandun Sarath Kumara]

Hi,

I am very new to MTEX.

I need some help to calculate following things using MTEX for an EBSD data set (attached).

1. I want to plot inclination angle (of each crystal orientation) histogram, relative to the normal direction of the specimen acquisition surface.

2. I want to calculate young's modulus from the EBSD data. But it has to be parallel to the specimen acquisition surface and normal to the surface. 

Can anyone help me with this?

Many thanks in advance :)

[grandr...@gmail.com]

For part 1.
First define a vector 3d perpendicular to the surface.  This will depend on how you loaded your data, but will probably be something like (001).  See manual 'Specimen Directions'
dir= vector3d(0,0,1)
Next decide which crystal axis you're interested in.  Let's say magnesium (0001) axis.  For each EBSD point, calculate the angle between the (0001) axis and the sample axes.  (see manual 'to plot histograms of misoreintation to a mode')

cs=ebsd ('Magnesium').CS

delta= ebsd('Magnesium').orientations*Miller(0,0,0,1,cs);
Next,

%calculate the angles between the two

ang=angle(dir, delta,'antipodal')/degree;

%and plot a histogram

figure;histogram(ang);

The manual can be found online at https://www.researchgate.net/profile/Jessica_Hiscocks  scroll down to 'Featured research'.


As for your second question 

2. I want to calculate young's modulus from the EBSD data. But it has to be parallel to the specimen acquisition surface and normal to the surface. 

First, I'm not certain this is even  theoretically possible (maybe someone can correct me on that)  Was the sample under load? What sort of experiment was this?

But it has to be parallel to the specimen acquisition surface and normal to the surface.  

I'm not certain what this means.  What is the 'specimen acquisition surface'?

Jessica

[Rüdiger Kilian]

Hi Chamara,
if you know the elastic stiffness tensor for your phase, you can probably do something like the following (example for quartz), just use instead of yvector your desired reference direction.


    % Elastic constants
    % Ogi, H., Ohmori, T. Nakamura, N. and Hirao M., 2006. Elastic, anelastic and piezoelectric
    % coefficients of alpha-quartz determined by resonance ultrasound spectroscopy.
    % Journal of Applied Physics, 100, 053511.
  

    % Define Cartesian Tensor crystal symmetry & frame
    cst = crystalSymmetry('-3m',[  4.9134  4.9134  5.4052],...
        [  90.0000  90.0000 120.0000]*degree,'x||a','z||c',...
        'mineral','Alpha-Quartz 2006');
    % elastic Cij stiffness tensor (GPa) as matrix M
    M =....
        [[   86.76    6.868   11.85  -18.02    0.00    0.00];...
        [     6.868   86.76   11.85  +18.02    0.00    0.00];...
        [    11.85    11.85  105.46    0.00    0.00    0.00];...
        [   -18.02   +18.02    0.00   58.14    0.00    0.00];...
        [     0.00     0.00    0.00    0.00   58.14  -18.02];...
        [     0.00     0.00    0.00    0.00  -18.02   39.946]];
    %
    % M as stiffness tensor C
    C = tensor(M,cst);

    % orientations of phase q from ebsd
    go=ebsd('q').orientations;
    % assuming both, your ebsd and cst above have not the same reference
    % frame use transformReferenceFrame
    go = transformReferenceFrame(go,cst); % in case
    % other wise

    % rotate the compression direction which is yvector in this case, into crystal coordiantes
    yc=inv(go).*yvector;
    % youngs modulous paralle yvector
    Yy =YoungsModulus(C,yc);

    % now you can for example plot it
    plot(ebsd('q'),Yy)
   
Hope that helps.
Cheers,
Rüdiger

[MTEXNewbie]

Hi Rüdiger,

I tried to use the above code for FCC crystal but it did not work.

% Define Cartesian Tensor crystal symmetry & frame

CS = crystalSymmetry('m-3m',[3.6599 3.6599 3.6599],[90.0000 90.0000 90.0000]*degree,'x||a','z||c','mineral','Iron fcc');

I suspect the frame part is wrong, but could not figure out with my limited knowledge.

[Ralf Hielscher]

Your line of code defining the crystal symmetry seems to be ok. Could you please explain in detail what "did not work" means?

Ralf.

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[MTEXNewbie]

I have only one grain (grainId 1) -

% M as stiffness tensor C

C = tensor(M,CS);

% orientations of phase q from ebsd

go=grains(1).meanOrientation;

% rotate the compression direction which is xvector in this case, into crystal coordiantes

yc=inv(go).*xvector;

% young's modulous parallal xvector

Yy =YoungsModulus(C,yc);

% now you can for example plot it

plot(grains(1),Yy)

I am getting a plot of the grain, no values.

[MTEXNewbie]

Adding mtexColorbar at the bottom colors the plot with Yy value.

Is there some other way I can get the Yy value for a certain grain (this is a general question involving mtexColorbar usage because it is difficult to figure out the exact value of variables from colorbar)?

[David Mainprice]

Here is an example script for plotting Young’s Modulus for FCC Alpha-Iron with a function for plotting elastic wave speeds

all the best David

[MTEXNewbie]

Hi David,

Thank you for the script. But it seems it is for BCC Fe, so the Elastic constants should be different for FCC Fe.

% Define Cartesian Tensor crystal crystalSymmetry & frame

cs_Tensor = crystalSymmetry('m-3m',[  2.874  2.874  2.874],...

 [  90.0000  90.0000  90.0000]*degree,'x||a','z||c',...

 'mineral','Alpha-Iron');

Any way to update the script for FCC Fe?

[David Mainprice]

Yes this is BCC alpha iron, I do not have the elastic constants of FCC in my data base

Here is an example script for plotting Young’s Modulus for BCC Alpha-Iron with a function for plotting elastic wave speeds

all the best David

[David Mainprice]

[David Mainprice]

Dear All,

There was an error in my elastic constants for FCC Fe-15Ni-15Cr Alloy, I have now corrected. I put all the scripts here.

[MTEXNewbie]

Hi Ralf,

If I define the Cartesian Tensor crystal symmetry & frame by using the default MTEX import definition, the elastic modulus is still the same. Can I use any one of these two or is there any reason to use the Rüdiger's template one explicitly for Elasticity calculations?

Default MTEX import definition -

CS = crystalSymmetry('m-3m', [3.6599 3.6599 3.6599], 'mineral', 'Iron fcc', 'color', 'light blue')

From Rüdiger's template above -

[Ralf Hielscher]

There is no difference between those lines.

Ralf.

[MTEXNewbie]

Hi Ralf,

In MTEX 5.0.3, using the two (2) crystal symmetries from the above discussion is causing this warning -

Warning: Symmetry missmatch! The following crystal frames

seem to be different

  Iron fcc (m-3m) and Iron fcc (m-3m)

 

> In orientation/transformReferenceFrame (line 34)

  In Test_01 (line 407) 

Now, I have changed the 2nd crystal symmetry definition to the 1st one but still got the same warning message. Additionally, the Young's Modulus pole figure plot is different than MTEX 4.5.2.

Maybe there is some issue regarding having 2 crystal symmetry definitions in the same script??

[Ralf Hielscher]

Hi MTEXNewbie,

I have checked the demo ElasticityTensor.m with MTEX 4.5.2 and MTEX 5.0.3 and found no difference in the scaling of the YoungsModulus. Could you therefore provide a reduced code that show up the difference.

Ralf.

[MTEXNewbie]

The actual value of Young's modulus is correct (~197 GPa). Perhaps this is causing the problem?

% set some colormap well suited for tensor visualisation

setMTEXpref('defaultColorMap',blue2redColorMap);

figure

plot(C,'PlotType','YoungsModulus','complete','upper');

mtexColorbar

[MTEXNewbie]

Hi Ralf,

Here is the code I am using to get Young's modulus, but the pole figure shows a high value of E in MTEX 5.0.3. Would it be possible for you to comment on what is going wrong?

Edit:

I have found out what is causing the problem.

If I use plot(C,'PlotType','YoungsModulus','complete','upper') than E is higher. But if I use plot(C.YoungsModulus,'complete','upper') from this example then E is correct - https://mtex-toolbox.github.io/files/doc/ElasticityTensor.html#5

Is that a bug or the second code is a new addition?