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)

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)
 size: 1 x 1
  U    V    T    W  | H    K    I    L CRSS
  2   -1   -1    0    0    1   -1    0    1

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)
 size: 1 x 1
  U    V    T    W  | H    K    I    L CRSS
  1    1   -2    0    0    0    0    1    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)
 size: 3 x 1
   U    V    T    W  | H    K    I    L CRSS
   1    1   -2    0    0    0    0    1    1
   1   -2    1    0    0    0    0    1    1
  -2    1    1    0    0    0    0    1    1

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

sSBasalSym.b.norm

ans =
    3.0000
    3.0000
    3.0000

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Rotating slip systems

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