The Schmid Factor edit page

Let us assume a Nickel crystal

CS = crystalSymmetry('cubic',[3.523,3.523,3.523],'mineral','Nickel')
CS = crystalSymmetry
 
  mineral : Nickel       
  symmetry: m-3m         
  elements: 48           
  a, b, c : 3.5, 3.5, 3.5

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

n = Miller(1,1,1,CS,'hkl')
n = Miller (Nickel)
  h k l
  1 1 1

and the slip direction (which needs to be orthogonal)

d = Miller(0,-1,1,CS,'uvw')
d = Miller (Nickel)
  u  v  w
  0 -1  1

For tension in direction 123

r = normalize(vector3d(1,2,3))
r = vector3d
         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 (Nickel)
 
  u    v    w  | h    k    l CRSS
  0   -1    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

SF = sS.SchmidFactor

% plot the Schmid factor in dependency of the tension direction
plot(SF)

% find the tension directions with the maximum Schmid factor
[SFMax,pos] = max(SF)

% and annotate them
annotate(pos)
SF = S2FunHarmonic
 bandwidth: 64
 antipodal: true
SFMax =
    0.5000
 
pos = vector3d
 antipodal: true
         x          y          z
  0.408421 -0.0919374   0.908152

Stress Tensor

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

sigma = stressTensor.uniaxial(vector3d.Z)
sigma = stressTensor (xyz)
  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

Active Slip System

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 (Nickel)
 size: 12 x 1
 
   u    v    w  | h    k    l CRSS
   0   -1   -1   -1    1   -1    1
  -1   -1    0   -1    1   -1    1
   1    0   -1   -1    1   -1    1
   0   -1    1   -1   -1   -1    1
   1   -1    0   -1   -1   -1    1
   1    0   -1   -1   -1   -1    1
   0   -1    1    1   -1   -1    1
  -1    0   -1    1   -1   -1    1
  -1   -1    0    1   -1   -1    1
   0   -1   -1    1    1   -1    1
   1   -1    0    1    1   -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
    0.2916    0.1750    0.1166   -0.1750    0.1750    0.3499   -0.1166
  Columns 8 through 12
    0.4666    0.3499   -0.0000   -0.0000   -0.0000

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 (Nickel)
 
   u    v    w  | h    k    l CRSS
  -1    0   -1    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

We may also plot the index of the active slip system

pcolor(r,id)

mtexColorMap(vega20(12))

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 absolute 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,'LineWidth',2,'Color','r');

% plot active slip direction in green
hold on
quiver(r,sSAll(id).b.normalize,'LineWidth',2,'Color','g');
hold off
sSAll = slipSystem (Nickel)
 size: 24 x 1
 
   u    v    w  | h    k    l CRSS
  -1    0    1   -1    1   -1    1
  -1   -1    0   -1    1   -1    1
   0    1    1   -1    1   -1    1
   0   -1   -1   -1    1   -1    1
   1    1    0   -1    1   -1    1
   1    0   -1   -1    1   -1    1
  -1    0    1   -1   -1   -1    1
  -1    1    0   -1   -1   -1    1
   0   -1    1   -1   -1   -1    1
   0    1   -1   -1   -1   -1    1
   1   -1    0   -1   -1   -1    1
   1    0   -1   -1   -1   -1    1
  -1   -1    0    1   -1   -1    1
  -1    0   -1    1   -1   -1    1
   0   -1    1    1   -1   -1    1
   0    1   -1    1   -1   -1    1
   1    0    1    1   -1   -1    1
   1    1    0    1   -1   -1    1
  -1    0   -1    1    1   -1    1
  -1    1    0    1    1   -1    1
   0    1    1    1    1   -1    1
   0   -1   -1    1    1   -1    1
   1   -1    0    1    1   -1    1
   1    0    1    1    1   -1    1

If we perform this computation in terms of spherical functions we obtain

% ommiting |r| gives us a list of 12 spherical functions
tau = sSAll.SchmidFactor

% now we take the max of the absolute value over all these functions
contourf(max(abs(tau),[],1),'upper')
mtexColorbar
tau = S2FunHarmonic
 size: 24 x 1
 bandwidth: 64
 antipodal: true

The Schmid factor for EBSD data

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. Let's import some EBSD data and compute the grains

mtexdata csl

% take some subset
ebsd = ebsd(ebsd.inpolygon([0,0,200,50]))

grains = calcGrains(ebsd);
grains = smooth(grains,5);

plot(ebsd,ebsd.orientations,'micronbar','off')
hold on
plot(grains.boundary,'linewidth',2)
hold off
ebsd = EBSD
 
 Phase   Orientations  Mineral         Color  Symmetry  Crystal reference frame
    -1  154107 (100%)     iron  LightSkyBlue      m-3m                         
 
 Properties: ci, error, iq, x, y
 Scan unit : um
 
 
ebsd = EBSD
 
 Phase  Orientations  Mineral         Color  Symmetry  Crystal reference frame
    -1  10251 (100%)     iron  LightSkyBlue      m-3m                         
 
 Properties: ci, error, iq, x, y
 Scan unit : um

We want to consider the following slip systems

sS = slipSystem.fcc(ebsd.CS)
sS = sS.symmetrise;
sS = slipSystem (iron)
 
  u    v    w  | h    k    l CRSS
  0    1   -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 (xyz)
 CRSS: 1
 size: 71 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
sigma = stressTensor.uniaxial(vector3d.Z)
SF = sSLocal.SchmidFactor(sigma);

% take the maxium allong the rows
[SFMax,active] = max(SF,[],2);

% plot the maximum Schmid factor
plot(grains,SFMax,'micronbar','off','linewidth',2)
mtexColorbar location southoutside
sigma = stressTensor (xyz)
  rank: 2 (3 x 3)
 
 0 0 0
 0 0 0
 0 0 1

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 = inv(grains.meanOrientation) * sigma
sigmaLocal = stressTensor (iron)
  size: 71 x 1   
  rank: 2 (3 x 3)

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

Strain-based analysis on the same data set

eps = strainTensor(diag([1,0,-1]))

epsCrystal = inv(grains.meanOrientation) * eps

[M, b] = calcTaylor(epsCrystal, sS);

plot(grains,M,'micronbar','off')
mtexColorbar southoutside
eps = strainTensor (xyz)
  type: Lagrange 
  rank: 2 (3 x 3)
 
  1  0  0
  0  0  0
  0  0 -1
 
epsCrystal = strainTensor (iron)
  size: 71 x 1   
  type: Lagrange 
  rank: 2 (3 x 3)
[ bMax , bMaxId ] = max( b , [ ] , 2 ) ;
sSGrains = grains.meanOrientation .* sS(bMaxId) ;
hold on
quiver ( grains , sSGrains.b)
quiver ( grains , sSGrains.trace)
hold off