Calculating Schmid Factor for a given orientation?

[MTEXNewbie]

Say, I have a given orientation <123> for an FCC metal (the orientation is indicating direction, parallel to pulling direction). Is there a way to calculate Schmid Factor in MTEX?

[ruediger Kilian]

Hi,
maybe try to answer this question:
Given that you'd know that <123> (which is a crystal direction, not an orientation) is parallel to your x-axis, can you determine the angle between the slip plane and the x-axis and the angle between the slip direction and the x-axis?

Cheers
Rüdiger

[MTEXNewbie]

Hi Rüdiger,

There is no EBSD data, so no way to determine what you have asked.

The below code gives Maximum Schmid factor for an EBSD data and direction <123>.

Can we define the csSF without the EBSD data?

% crystal symmetry

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

% Schmid Factor

csSF = ebsd('Iron fcc').CS;

% consider fcc slip

sS = slipSystem.fcc(csSF);

% and all symmetrically equivalent variants

sS = sS.symmetrise;

% compute Schmid factor for all slip systems

SF = sS.SchmidFactor(Miller(1,2,3,csSF));

% find the maximum Schmidt factor (along the rows)

[SFmax,active] = max(SF,[],2);

SFmax

[ruediger Kilian]

Hi,
no, this has nothing to do with having data coming from ebsd or any other method, e.g. your code has not a lot to do with ebsd data, csSF is the crystal symmetry (supposedly the same as CS).
To calculate the Schmid factor, you need to know the slip plane/ slip direction with respect to your stress directions, most easily derived from a crystal orientation. If you do not have the orientation of the crystal, I do not see how you could get a Schmid factor.
Cheers,
Rüdiger

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[MTEXNewbie]

Hi Rüdiger,

Maybe I am reading the term <123> wrong, It is given for a single crystal micro-tensile test. Does it indicate the direction Or orientation with respect to pulling/loading direction? If it is the later then its possible to calculate Schmid factor?

[ruediger Kilian]

I read <123> as <uvw> which is a crystal direction.

[MTEXNewbie]

Hmm, I also read it as crystal direction. Let's assume its crystal orientation, and from the Schmid Factor, I will be able to recalculate the result in the paper and could tell what they mean by it.

Would it be possible for you to give an example code to calculate Schmid factor directly for a given orientation (123)?

[ruediger Kilian]

Again, I do not know what you mean with orientation (123), but if you'd have an orientation o, one way to get the schmid factor is: slipSystem.SchmidFactor(inv(o)*stressTensor)

Please have a look here on how to determine a Schmid factor: http://mtex-toolbox.github.io/files/doc/PlasticDeformation.html

and here on the difference between a direction and an orientation: http://mtex-toolbox.github.io/files/doc/CrystalOrientations.html and http://mtex-toolbox.github.io/files/doc/CrystalDirections.html

Hope this helps,
Rüdiger

[ruediger Kilian]

Hi,
actually after a good hint by Karsten. It is of course possible. What you were asking is exactly the example from the documentation (http://mtex-toolbox.github.io/files/doc/PlasticDeformation.html#3).
So it is of course possible.
e.g.
cs = crystalSymmetry('m-3m');
sS = symmetrise(slipSystem.fcc(cs));
sS.SchmidFactor(Miller(1,2,3,cs))
% or
SF = sS(1).SchmidFactor
eval(SF,symmetrise(Miller(1,2,3,cs)))

Cheers,
Rüdiger

[MTEXNewbie]

Hi

The first code gives 24 output, the 2nd one gives 48 outputs. Should I take the Maximum Schmid factor for both cases?

cs = crystalSymmetry('m-3m'); 

sS = symmetrise(slipSystem.fcc(cs)); 

sS.SchmidFactor(Miller(1,2,3,cs)) 

% or 

cs = crystalSymmetry('m-3m'); 

SF = sS(1).SchmidFactor 

eval(SF,symmetrise(Miller(1,2,3,cs))) 

[MTEXNewbie]

Hi Rüdiger,

Is the given code above calculates assuming the "crystal orientation is with respect to loading/pulling direction"?