1. 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

2. 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 3 per plane (only positive) –12 independent slip systems a1a1 a2a2 a3a3 Crystallography of Slip b = a/2 | b | = a/  2 [1 1 0]

3. HCP “BCC” –Planes {0 0 1} 1 independent plane –Directions 3 per plane (only positive) –3 independent slip systems –Planes {1 1 0} 6 independent planes –Directions 2 per plane (only positive) –12 independent slip systems b = a | b | = a b = a/2 | b | =  3  a/2 Occasionally also {1 1 2} planes in “BCC” are slip planes Diamond structure type: {1 1 1} and --- same as CCP, but slip less uncommon

4. Why does the number of independent slip systems matter?  = F/A Are 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

5. 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”

6. Energy Penalty of Dislocations bonds are compressed bonds are under tension R0R0 tension R E compression Energy / length  |b| 2 Thermodynamically unfavorable Strong interactions attraction  annihilation repulsion  pinning Too many dislocations  become immobile

7. 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 –|b BCC | > |b CCP |  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

8. Elastic Deformation Connected to chemical bonding –Stretch bonds and then relax back Recall bond-energy curve –Difficulty of moving from R 0 –Curvature at R 0 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 R0R0 R E 

9. Elastic Constants Y: Young’s modulus (sometimes E) l0l0 A0A0 F stress = uniaxial, normal stress material elongates: l 0  l strain =elongation along force direction observation:  (stress)  (strain) Y material thins/necks: A 0  A i elongates: l 0  l i true stress: use A i ; nominal (engineering) stress: use A 0 true strain: use l i ; nominal (engineering) stress: use l 0

10. Elastic Constants Connecting Young’s Modulus to Chemical Bonding R0R0 R E Coulombic attraction F = k  R stress*areastrain*length R0R0  k / length = Y want k in terms of E, R 0 observed within some classes of compounds Hook’s Law

11. Elastic Constants Bulk Modulus, K apply hydrostatic pressure  = -P measure change in volume P = F/A linear response Useful relationship: Can show: analogous to Young’s modulus Coulombic: hydrostatic stress

12. Elastic Constants Poisson’s ratio, apply uniaxial stress  = F/A measure  || - elongation parallel to force l0l0 A F Rigidity (Shear) Modulus, G y x measure   - thinning normal to force l0l0 ll F F apply shear stress  = F/A measure shear strain = tan     ||   A

13. Elastic Constants General Considerations  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 deformation shear stress  only shear deformation Cubic material  G, Y and are independent