This script describes how to analyze Schmid factors.

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Geometric Definition |

Stress Tensor |

Active Slip System |

The Schmid factor for EBSD data |

Let us assume a Nickel crystal

CS = crystalSymmetry('cubic',[3.523,3.523,3.523],'mineral','Nickel')

CS = crystalSymmetry mineral : Nickel symmetry: m-3m a, b, c : 3.5, 3.5, 3.5

Since Nickel is fcc a dominat slip system is given by the slip plane normal

`n = Miller(1,1,1,CS,'hkl')`

n = Miller size: 1 x 1 mineral: Nickel (m-3m) h 1 k 1 l 1

and the slip direction (which needs to be orthogonal)

`d = Miller(0,-1,1,CS,'uvw')`

d = Miller size: 1 x 1 mineral: Nickel (m-3m) u 0 v -1 w 1

For a simple shear in the z - direction

r = normalize(vector3d(1,2,3))

r = vector3d size: 1 x 1 x y z 0.267261 0.534522 0.801784

the Schmid factor for the slip system [0-11](111) is defined by

tau = dot(d,r,'noSymmetry') * dot(n,r,'noSymmetry')

tau = 0.4286

The same computation can be performed by defining the slip system as an MTEX variable

sS = slipSystem(d,n)

sS = slipSystem mineral: Nickel (m-3m) CRSS: 1 size: 1 x 1 u v w | h k l 0 -1 1 1 1 1

and using the command SchmidFactor

sS.SchmidFactor(r)

ans = 0.1750

Ommiting the tension direction r the command SchmidFactor returns the Schmid factor as a spherical function

%will be part of MTEX 4.5 % SF = sS.SchmidFactor % plot the Schmid factor in dependency of the tension direction %will be part of MTEX 4.5 %plot(SF) %will be part of MTEX 4.5 %[SFMax,pos] = max(SF)

Instead by the tension direction the stress might be specified by a stress tensor

sigma = tensor([0 0 0; 0 0 0; 0 0 1],'name','stress')

sigma = stress tensor rank: 2 (3 x 3) 0 0 0 0 0 0 0 0 1

Then the Schmid factor for the slip system `sS` and the stress tensor `sigma` is computed by

sS.SchmidFactor(sigma)

ans = 0.4082

In general a crystal contains not only one slip system but at least all symmetrically equivalent ones. Those can be computed with

`sSAll = sS.symmetrise('antipodal')`

sSAll = slipSystem mineral: Nickel (m-3m) CRSS: 1 size: 12 x 1 u v w | h k l 0 -1 1 1 1 1 1 0 -1 1 1 1 -1 1 0 1 1 1 -1 1 0 1 1 -1 -1 0 -1 1 1 -1 0 -1 -1 1 1 -1 0 -1 1 -1 1 1 -1 0 -1 -1 1 1 -1 -1 0 -1 1 1 1 0 -1 1 -1 1 -1 -1 0 1 -1 1 0 -1 -1 1 -1 1

The option `antipodal` indicates that Burgers vectors in oposite direction should not be distinguished. Now

tau = sSAll.SchmidFactor(r)

tau = Columns 1 through 7 0.1750 -0.3499 0.1750 0.0000 -0.0000 -0.0000 0.1166 Columns 8 through 12 -0.4666 -0.3499 -0.1166 -0.1750 -0.2916

returns a list of Schmid factors and we can find the slip system with the largest Schmid factor using

[tauMax,id] = max(abs(tau)) sSAll(id)

tauMax = 0.4666 id = 8 ans = slipSystem mineral: Nickel (m-3m) CRSS: 1 size: 1 x 1 u v w | h k l -1 0 -1 -1 1 1

The above computation can be easily extended to a list of tension directions

% define a grid of tension directions r = plotS2Grid('resolution',0.5*degree,'upper') % compute the Schmid factors for all slip systems and all tension % directions tau = sSAll.SchmidFactor(r); % tau is a matrix with columns representing the Schmid factors for the % different slip systems. Lets take the maximum rhowise [tauMax,id] = max(abs(tau),[],2); % vizualize the maximum Schmid factor contourf(r,tauMax) mtexColorbar

r = vector3d size: 181 x 721 resolution: 0.5° plot: true region: upper hemisphere theta: 181 x 721 double rho: 181 x 721 double

We may also plot the index of the active slip system

```
pcolor(r,id)
mtexColorMap black2white
```

and observe that within the fundamental sectors the active slip system remains the same. We can even visualize the the plane normal and the slip direction

% if we ommit the option antipodal we can distinguish % between the oposite burger vectors sSAll = sS.symmetrise % take as directions the centers of the fundamental regions r = symmetrise(CS.fundamentalSector.center,CS); % compute the Schmid factor tau = sSAll.SchmidFactor(r); % here we do not need to take the absolut value since we consider both % burger vectors +/- b [~,id] = max(tau,[],2); % plot active slip plane in red hold on quiver(r,sSAll(id).n,'ArrowSize',0.2,'LineWidth',2,'Color','r'); % plot active slip direction in green hold on quiver(r,sSAll(id).b.normalize,'ArrowSize',0.1,'LineWidth',2,'Color','g'); hold off

sSAll = slipSystem mineral: Nickel (m-3m) CRSS: 1 size: 24 x 1 u v w | h k l 0 -1 1 1 1 1 1 0 -1 1 1 1 -1 1 0 1 1 1 0 1 -1 1 1 1 -1 0 1 1 1 1 1 -1 0 1 1 1 -1 1 0 1 1 -1 -1 0 -1 1 1 -1 0 -1 -1 1 1 -1 1 0 1 1 1 -1 0 1 1 1 1 -1 1 -1 0 1 1 -1 0 -1 1 -1 1 1 -1 0 -1 -1 1 1 -1 -1 0 -1 1 1 1 0 1 -1 1 1 1 1 0 -1 1 1 0 1 -1 -1 1 1 1 0 -1 1 -1 1 -1 -1 0 1 -1 1 0 -1 -1 1 -1 1 1 1 0 1 -1 1 0 1 1 1 -1 1 -1 0 1 1 -1 1

So far we have always assumed that the stress tensor is already given relatively to the crystal coordinate system. Next, we want to examine the case where the stress is given in specimen coordinates and we know the orientation of the crystal. Lets import some EBSD data and computet the grains

mtexdata csl grains = calcGrains(ebsd); plot(ebsd,ebsd.orientations) hold on plot(grains.boundary) hold off

We want to consider the following slip systems

sS = slipSystem.fcc(ebsd.CS) sS = sS.symmetrise;

sS = slipSystem mineral: iron (m-3m) CRSS: 1 size: 1 x 1 u v w | h k l 0 1 -1 1 1 1

Since, those slip systems are in crystal coordinates but the stress tensor is in specimen coordinates we either have to rotate the slip systems into specimen coordinates or the stress tensor into crystal coordinates. In the following sections we will demonstrate both ways. Lets start with the first one

```
% rotate slip systems into specimen coordinates
sSLocal = grains.meanOrientation * sS
```

sSLocal = slipSystem CRSS: 1 size: 885 x 24

These slip systems are now arranged in matrix form where the rows corrspond to the crystal reference frames of the different grains and the rows are the symmetrically equivalent slip systems. Computing the Schmid faktor we end up with a matrix of the same size

% compute Schmid factor SF = sSLocal.SchmidFactor(sigma); % take the maxium allong the rows [SFMax,active] = max(SF,[],2); % plot the maximum Schmid factor plot(grains,SFMax) mtexColorbar

Next we want to visualize the active slip systems.

% take the active slip system and rotate it in specimen coordinates sSactive = grains.meanOrientation .* sS(active); hold on % visualize the trace of the slip plane quiver(grains,sSactive.trace,'color','b') % and the slip direction quiver(grains,sSactive.b,'color','r') hold off

We observe that the Burgers vector is in most case aligned with the trace. In those cases where trace and Burgers vector are not aligned the slip plane is not perpendicular to the surface and the Burgers vector sticks out of the surface.

Next we want to demonstrate the alternative route

```
% rotate the stress tensor into crystal coordinates
sigmaLocal = rotate(sigma,inv(grains.meanOrientation))
```

sigmaLocal = stress tensor size : 885 x 1 rank : 2 (3 x 3) mineral: iron (m-3m)

This becomes a list of stress tensors with respect to crystal coordinates - one for each grain. Now we have both the slip systems as well as the stress tensor in crystal coordiantes and can compute the Schmid factor

% the resulting matrix is the same as above SF = sS.SchmidFactor(sigmaLocal); % and hence we may proceed analogously % take the maxium allong the rows [SFMax,active] = max(SF,[],2); % plot the maximum Schmid factor plot(grains,SFMax) mtexColorbar % take the active slip system and rotate it in specimen coordinates sSactive = grains.meanOrientation .* sS(active); hold on % visualize the trace of the slip plane quiver(grains,sSactive.trace,'color','b') % and the slip direction quiver(grains,sSactive.b,'color','r') hold off

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