Slip Systems

How to analyze slip transmission at grain boundaries

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Burgers vector and normal directions
Rotating slip systems

Burgers vector and normal directions

Consider hexagonal symmetry of alpha-Titanium

cs = crystalSymmetry('622',[3,3,4.7],'x||a','mineral','Titanium (Alpha)')
 
cs = crystalSymmetry  
 
  mineral        : Titanium (Alpha) 
  symmetry       : 622              
  a, b, c        : 3, 3, 4.7        
  reference frame: X||a, Y||b*, Z||c
 

Then basal slip is defined by the Burgers vector (or slip direction)

b = Miller(2,-1,-1,0,cs,'UVTW')
 
b = Miller  
 size: 1 x 1
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
  U  2
  V -1
  T -1
  W  0

and the slip plane normal

n = Miller(0,1,-1,0,cs,'HKIL')
 
n = Miller  
 size: 1 x 1
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
  h  0
  k  1
  i -1
  l  0

Accordingly we can define a slip system in MTEX by

sSBasal = slipSystem(b,n)
 
sSBasal = slipSystem  
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
 CRSS: 1
 size: 1 x 1
  U   V   T   W | H   K   I   L
  2  -1  -1   0   0   1  -1   0

The most important slip systems for cubic, hexagonal and trigonal crystal lattices are already implemented into MTEX. Those can be accessed by

sSBasal = slipSystem.basal(cs)
 
sSBasal = slipSystem  
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
 CRSS: 1
 size: 1 x 1
  U   V   T   W | H   K   I   L
  1   1  -2   0   0   0   0   1

Obviously, this is not the only basal slip system in hexagonal lattices. There are also symmetrically equivalent ones, which can be computed by

sSBasalSym = sSBasal.symmetrise('antipodal')
 
sSBasalSym = slipSystem  
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
 CRSS: 1
 size: 3 x 1
   U   V   T   W | H   K   I   L
   1   1  -2   0   0   0   0   1
   1  -2   1   0   0   0   0   1
  -2   1   1   0   0   0   0   1

The length of the burgers vector, i.e., the amount of displacment is

sSBasalSym.b.norm
ans =
    3.0000
    3.0000
    3.0000

Rotating slip systems