Plastic Deformation• Permanent, unrecovered mechanical deformation

= F/A stress• Deformation by dislocation

motion, “glide” or “slip”• Dislocations

– Edge, screw, mixed– Defined by Burger’s vector– Form loops, can’t terminate

except at crystal surface

• Slip system– Glide plane + Burger’s vector

maximum shear stress

• Slip system = glide plane + burger’s vector– Correspond to close-packed planes + directions– Why?

• Fewest number of broken bonds

• Cubic close-packed– Closest packed planes

• {1 1 1} • 4 independent planes

– Closest packed directions• Face diagonals• <1 1 0>• 3 per plane (only positive)

– 12 independent slip systems

a1

a2

a3

Crystallography of Slip

b = a/2 <1 1 0>| b | = a/2

[1 1 0]

• HCP • “BCC”

– 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

b = a <1 0 0>| b | = a

b = a/2 <1 1 1>| b | = 3a/2

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?

= F/AAre any or all or some of the grains in the proper orientation for slip to occur?

HCP

CCP

• 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

maximum shear stress

Dislocations in Ionic Crystals

like charges touch

like charges do not touch

long burger’s vector compared to metals

1

2

(1) slip brings like charges in contact

(2) does not bring like charges in contact

compare possible slip planes

viewing edge dislocations as the termination defect of “extra half-planes”

Energy Penalty of Dislocationsbonds are compressed

bonds are under tension

R0

tension

R

E

compression

Energy / length |b|2

Thermodynamically unfavorable Strong interactions

attraction annihilation 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

– 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

R0

R

E

Elastic ConstantsY: Young’s modulus (sometimes E)

l0

A0

F stress = 0

FA uniaxial, normal stress

material elongates: l0 l

strain = 0

0

l ll

elongation along force direction

observation: Y

(s

tress

)(strain)

Y

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 ConstantsConnecting Young’s Modulus to Chemical Bonding

R0

R

E

1~ER

Coulombic attraction

F = k R

stress*area strain*length

0

R

dF kd R

dEFd R

R0

k / length = Y

want k in terms of E, R0

2

20

( ) R

d Ed R

30 ~ Y R

observed within some classes of compounds

Hook’s Law

Elastic ConstantsBulk Modulus, K

•apply hydrostatic pressure

= -P

•measure change in volume0

VV

P = F/A

•linear response0

VKV

Useful relationship:0

x y zV

V

Can show:0

2

0 2V V

EK VV

analogous to Young’s modulus

Coulombic: 1/31 1~ ~ E

R V 4/3 4

0 0~ ~ K V R

hydrostatic stress

Elastic ConstantsPoisson’s ratio, •apply uniaxial stress = F/A

•measure ||

||

0,

x

y

ll

- elongation parallel to force

l0

AF

Rigidity (Shear) Modulus, G

y

x

•measure - thinning normal to force

l0

lF

F

•apply shear stress = F/A

•measure shear strain

= tan

G

||

A

Elastic ConstantsGeneral Considerations

2(1 )YG

6 parametersStress, : 3 3 symmetric tensor

In principle, each and every strain parameter depends on each and every stress parameter

Strain, : 3 3 symmetric tensor 6 parameters

36 elastic constants

21 independent elastic constants in the most general case

Some are redundant

Material symmetry some are zero, some are inter-related

Isotropic material only 2 independent elastic constants

normal stress only normal deformationshear stress only shear deformation

Cubic material G, Y and are independent