Plastic Deformation
• Permanent, unrecovered mechanical deformation
s = F/A stress
• Deformation by dislocation
motion, “glide” or “slip”
• Dislocations
– Edge, screw, mixed
– Defined by Burger’s vector
maximum
shear
stress
– Form loops, can’t terminate
except at crystal surface
• Slip system
– Glide plane + Burger’s vector
Crystallography of Slip
• Slip system = glide plane + burger’s vector
– Correspond to close-packed planes + directions
– Why?
• Fewest number of broken bonds
• Cubic close-packed
a3
– Closest packed planes
• {1 1 1}
• 4 independent planes
– Closest packed directions
• Face diagonals
b = a/2 <1 1 0>
| b | = a/2
• <1 1 0>
a1
• 3 per plane (only positive)
– 12 independent slip systems
a2
[1 1 0]
• HCP
• “BCC”
b = a/2 <1 1 1>
| b | = 3a/2
b = a <1 0 0>
|b|=a
– Planes {0 0 1}
• 1 independent plane
– Directions <1 0 0>
• 3 per plane (only positive)
– 3 independent slip systems
– Planes {1 1 0}
• 6 independent planes
– Directions <1 1 1>
• 2 per plane (only positive)
– 12 independent slip systems
Occasionally also {1 1 2} planes in “BCC” are slip planes
Diamond structure type: {1 1 1} and <1 1 0> --- same as CCP, but slip less uncommon
Why does the number of independent slip systems matter?
s = F/A
Are any or all or some of the grains in
the proper orientation for slip to occur?
HCP
CCP
maximum
shear
stress




• Large # of independent slip systems in CCP 
at least one will be active for any particular grain
• True also for BCC
• Polycrystalline HCP materials require more stress
to induce deformation by dislocation motion
Dislocations in Ionic Crystals
viewing edge dislocations as the
termination defect of “extra half-planes”
like charges touch
2
1
like charges do not touch
long burger’s vector compared to metals
compare possible slip planes
(1) slip brings like charges in contact
(2) does not bring like charges in contact
Energy Penalty of Dislocations
bonds are compressed
E
R0
R
compression
tension
Energy / length  |b|2
bonds are under tension
attraction  annihilation
Thermodynamically unfavorable
Strong interactions
repulsion  pinning
Too many dislocations
 become immobile
Summary
• Materials often deform by dislocation glide
– Deforming may be better than breaking
• Metals
–
–
–
–
CCP and BCC have 12 indep slip systems
HCP has only 3, less ductile
|bBCC| > |bCCP|  higher energy, lower mobility
CCP metals are the most ductile
• Ionic materials/Ceramics
– Dislocations have very high electrostatic energy
– Deformation by dislocation glide atypical
• Covalent materials/Semiconductors
– Dislocations extremely rare
Elastic Deformation
• Connected to chemical bonding
– Stretch bonds and then relax back
• Recall bond-energy curve
– Difficulty of moving from R0
E
R0
– Curvature at R0
• Elastic constants
– (stress) = (elastic constant) * (strain)
– stress and strain are tensors  directional
– the elastic constant being measured depends on
which component of stress and of strain
R
Elastic Constants
Y: Young’s modulus (sometimes E)
F
A0
stress = F A  s
0
uniaxial, normal stress
material elongates: l0  l
l0
observation:
s  Y
s (stress)
l  l0
  elongation along force direction
strain =
l0
Y
(strain)
material thins/necks: A0  Ai
elongates: l0  li
true stress: use Ai; nominal (engineering) stress: use A0
true strain: use li; nominal (engineering) stress: use l0
Elastic Constants
Connecting Young’s Modulus to Chemical Bonding
R0
E
R0
Hook’s Law
F = k DR
stress*area
 k / length = Y
R
strain*length
Coulombic attraction
want k in terms of E, R0
E~
dE
F
d DR
dF
 k 
d DR DR0
d 2E
d (DR)2
 Y ~ R0
1
R
3
DR 0
observed within
some classes of
compounds
Elastic Constants
Bulk Modulus, K
• apply hydrostatic pressure
P = F/A
s = -P
hydrostatic stress
• measure change in volume
DV
V0
 DV 
• linear response s  K 

V
 0 
 2 E 
Can show: K  V0  2 
 V  V V0
Coulombic: E ~
1
R
Useful relationship:
~
1
V
1/3
analogous to Young’s modulus
K ~ V0 4/3 ~ R0 4
DV
 x y z
V0
Elastic Constants
Poisson’s ratio, n
F
• apply uniaxial stress
A
• measure ||
- elongation parallel to force
||
• measure 
- thinning normal to force
 ||
n 


l0
Rigidity (Shear) Modulus, G
A
Dl
F
f
l0
y
F
x
s = F/A
• apply shear stress t = F/A
• measure shear strain

Dlx
= tanf  f
l0, y
t  G
Elastic Constants
General Considerations
Stress, s: 3  3 symmetric tensor 
Strain, : 3  3 symmetric tensor 
6 parameters
6 parameters
In principle, each and every strain parameter
 36 elastic constants
depends on each and every stress parameter
Some are redundant
 21 independent elastic constants
in the most general case
Material symmetry
 some are zero, some are inter-related
Isotropic material
 only 2 independent elastic constants
normal stress  only normal deformation
shear stress  only shear deformation
Cubic material
 G, Y and n are independent
G
Y
2(1 n )