CHAPTER 7: DISLOCATIONS AND STRENGTHENING

1. CHAPTER 7: DISLOCATIONS AND STRENGTHENING • In Ch.6 • Plastic (permanent) & Elastic (reversible) • Yield Strength and hardness are a measure of materials resistance to deformation • In microscopic scale what is going on ? • Net movement of large number of atoms in response to an applied stress • Involves movement of dislocations, in Ch. 4 • How does strengthening happen ? • Why do some materials are stronger than others ? • How can one manipulate dislocations and their motion ?

2. Dislocations & Materials Classes • Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip. + + + + + + + + + + + + + + + + + + + + + + + + electron cloud ion cores • Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding • Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors. - - - + + + + - - - - + + + - - - + + + +

3. Dislocation Motion Dislocations & plastic deformation • Cubic & hexagonal metals - plastic deformation by plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations). • If dislocations don't move, deformation doesn't occur! Adapted from Fig. 7.1, Callister 7e.

4. DISLOCATION MOTION Plastically stretched zinc single crystal. • Produces plastic deformation ! • Bonds are incrementally broken and re-formed. Much less force is needed , why ? Adapted from Fig. 7.9, Callister 6e. (Fig. 7.9 is from C.F. Elam, The Distortion of Metal Crystals, Oxford University Press, London, 1935.) SLIP PLANE Adapted from Fig. 7.1, Callister 6e. (Fig. 7.1 is adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1976. p. 153.) Adapted from Fig. 7.8, Callister 6e. SEE THE MOVIE AGAIN ! • If dislocations don't move, plastic deformation doesn't happen! How do we generate the dislocation motion ? 3

5. DISLOCATION MOTION deformed Initial state apply force O1 O0 t3 t2 t1 t0 The motion of a single dislocation across the plane causes the top half of the crystal to move (to slip) with respect to the bottom half but we can not have to break all the bonds across the middle plane simultaneously (which would require a very large force). The slip plane (O0-O1)– the crystallographic plane of dislocation motion.

6. STRESS and Dislocation Motion Dislocation line NOT be parallel to the dislocation line !

7. Dislocation Motion • Dislocation moves along slip plane in slip direction perpendicular to dislocation line • Slip direction same direction as Burgers vector Edge dislocation Adapted from Fig. 7.2, Callister 7e. Screw dislocation

8. STRESS AND DISLOCATION MOTION

9. STRESS field around dislocations Bonds are Compressed Bonds are Stretched WHY is there a stress field ? Atoms try to relax by trying to achieve their positions for the case if there was not a dislocation in the vicinity !

10. Stress fields of dislocations interacting ! The strain fields around dislocations interact with each other. Hence, they exert force on each other. Edge dislocations, when they arein the same plane, they repel each other if they have the same sign (direction of the Burgers vector). WHY ? They can attract and annihilate if they have opposite signs. PROVE !

11. Dislocation Density • The number of dislocations in a material is expressed using the term dislocation density - the total dislocation length per unitvolume or the # of dislocationsintersecting a unitarea. Units are mm / mm3, or just / mm • Dislocation densities can vary from 103mm-2 incarefully solidified metal crystals to 1010mm-2 in heavilydeformed metals. Where do Dislocations come from, what are their sources ? • Most crystalline materials, especially metals, havedislocations in their as-formed state, mainly as a result ofstresses (mechanical, thermal...) associated with the manufacturing processes used. • The number of dislocationsincreases dramaticallyduring plastic deformation. • Dislocations spawn fromexisting dislocations, grainboundaries and surfaces and other “defects” . NICE SIMULATIONS => http://www.ims.uconn.edu/centers/simul/movie/

12. Deformation Mechanisms Slip System • Slip plane - plane allowing easiest slippage • Wide interplanar spacings - highest planar densities • Slip direction- direction of movement - Highest linear densities • FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed) => total of 12 slip systems in FCC • in BCC & HCP other slip systems occur Adapted from Fig. 7.6, Callister 7e.

13. SLIP SYSTEMS !!! • Dislocations move with ease on certain crystallographic planes and along certain directions on these planes ! • The plane is called a slip plane • The direction is called a slip direction • Combination of the plane of slip and direction is a slip system • The slip planes and directions are those of highestpacking density. • The distance between atoms is shorterthan the average…High number of coordination along the planes also important ! PLS SEE TABLE 7.1 in ur books , page 180 !!!

14. Concept Check Page 181, in 7th edition !

15. Applied tensile Relation between Resolved shear s s tR stress: = F/A and tR stress: = F /A s s F tR slip plane normal, ns FS /AS = A l f F cos A / cos tR f nS F AS slip direction l A FS FS AS slip direction F slip direction tR Stress and Dislocation Motion • Crystals slip due to a resolved shear stress, tR. • Applied tension can produce such a stress. l+f  90

16. Critical Resolved Shear Stress typically 10-4GPa to 10-2 GPa s s s tR s = 0 tR /2 = 0 tR = f l =90° l =90° =45° f =45° • Condition for dislocation motion: • Crystal orientation can make it easy or hard to move dislocation  maximum at  =  = 45º

17. Resolving the Applied Stress on a SLIP PLANE ! • Shear Stress has to be resolved on to the slip planes as; • Shear Stress is needed for dislocations to move / slip • Dislocations can only move on slip planes, and these planes are rarely on axis with the applied force ! We should resolve the force applied in a tensile test, F, on to the cross-sectional area A where the slip is going to take place;

18. Critical Resolved Shear Stress => Slip in Single X’tals Macroscopically; Q: when do materials plastically deform / yield ? Stress > YS Q: How does plastic deformation take place ? Dislocation Motion ( Elastic deformation ??) Q: On what planes does dislocations move ? Slip Planes Q: At stress = YS, what would be the minimum resolved shear stress needed to act on dislocations to initiate dislocation motion, onset of yield/plastic deformation ? HIMMM ! Lets think !!!

19. The Critical Resolved Shear Stress The minimum shear stress required to initiate slip is termed: the critical resolved shear stress

20. Plastically stretched zinc single crystal. Single Crystal Slip Adapted from Fig. 7.9, Callister 7e. A large number of dislocations are generated on slip planes, as they leave the system they form these “shear bands” Adapted from Fig. 7.8, Callister 7e.

21. Ex: Deformation of single crystal a) Will the single crystal yield? b) If not, what stress is needed? So the applied stress of 6500 psi will not cause the crystal to yield. =60° crss = 3000 psi =35° Adapted from Fig. 7.7, Callister 7e.  = 6500 psi

22. So for deformation to occur the applied stress must be greater than or equal to the yield stress Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, sy)?

23. Slip Motion in Polycrystals s 300 mm • Stronger - grain boundaries pin deformations • Slip planes & directions (l, f) change from one crystal to another. • tRwill vary from one crystal to another. • The crystal with the largest tR yields first. • Other (less favorably oriented) crystals yield later. Adapted from Fig. 7.10, Callister 7e. (Fig. 7.10 is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)

24. Plastic Deformation in Polycrystalline Materials

25. Anisotropy in sy - after rolling rolling direction - anisotropic since rolling affects grain orientation and shape. • Can be induced by rolling a polycrystalline metal - before rolling Adapted from Fig. 7.11, Callister 7e. (Fig. 7.11 is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140, John Wiley and Sons, New York, 1964.) 235 mm - isotropic since grains are approx. spherical & randomly oriented.

26. Anisotropy in Deformation 1. Cylinder of Tantalum machined from a rolled plate: 2. Fire cylinder at a target. 3. Deformed cylinder Photos courtesy of G.T. Gray III, Los Alamos National Labs. Used with permission. rolling direction plate thickness direction end view • The noncircular end view shows anisotropic deformation of rolled material. side view

27. ANISOTROPY IN DEFORMATION 1. Cylinder of Tantalum machined from a rolled plate: 2. Fire cylinder at a target. 3. Deformed cylinder Photos courtesy of G.T. Gray III, Los Alamos National Labs. Used with permission. side view plate thickness direction end view • The noncircular end view shows: anisotropic deformation of rolled material. 10

28. STRENGTHENING MECHANISMS The ability of a metal to deform depends on theability of dislocations to move with relative ease under loading conditions Restricting dislocation motion will inevitably make the materialstronger, need more force to induce same amount of deformation ! • Mechanisms of strengthening in single-phase metals: • grain-size reduction • solid-solution alloying • strain hardening • Ordinarily, strengthening mechanisms reduces ductility, why ???

29. 4 STRATEGIES FOR STRENGTHENING: 1: REDUCE GRAIN SIZE • Grain boundaries are barriers to slip. • Barrier "strength" increases with mis-orientation. High-angle boundaries are better in blocking slip ! • Smaller grain size: more barriers to slip. Grain size can be changed by processing ! Adapted from Fig. 7.12, Callister 6e. (Fig. 7.12 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.) Hall-Petch Equation : Average grain size 7

30. GRAIN SIZE STRENGTHENING: AN EXAMPLE • 70wt%Cu-30wt%Zn brass alloy • Data: Adapted from Fig. 7.13, Callister 6e. (Fig. 7.13 is adapted from H. Suzuki, "The Relation Between the Structure and Mechanical Properties of Metals", Vol. II, National Physical Laboratory Symposium No. 15, 1963, p. 524.) 8

31. 4 Strategies for Strengthening: 2: Solid Solutions • Smaller substitutional impurity • Larger substitutional impurity A C D B Impurity generates local stress at A and B that opposes dislocation motion to the right. Impurity generates local stress at C and D that opposes dislocation motion to the right. • Impurity atoms distort the lattice & generate stress. • Stress can produce a barrier to dislocation motion.

32. Strategy #2: Solid Solutions • Alloyed metals are usually stronger than their pure base metals counter parts. • Why ? Interstitial or substitutional impurities in a solutioncause lattice strain, aka distortions in the lattice • Then ? • Strain field around the impuritiesinteract with dislocation strain fields and impede dislocation motion. • Impurities tend to diffuse and segregate around thedislocation coreto find atomic sites more suited totheir radii. This reduces the overall strain energy and“anchor” the dislocation. • Motion of the dislocation core away from theimpurities moves it to a region of lattice where theatomic strains are greater, wherelattice strains due to dislocation is no longer compensated by the impurity atoms.

33. Interactions of the Stress Fields COMPRESSIVE TENSILE TENSILE COMPRESSIVE

34. Interactions of the Stress Fields

35. Stress Concentration at Dislocations Adapted from Fig. 7.4, Callister 7e.

36. Impurity Segregation Impurities tend to segregate at energetically favorable areas around the dislocation core and partially decrease the overall stress field generated around the dislocation core. However, when stress is applied more load is needed to move dislocations with impurity atoms segregated to its core !

37. Strengthening by Alloying • small impurities tend to concentrate at dislocations • reduce mobility of dislocation  increase strength Adapted from Fig. 7.17, Callister 7e.

38. Strengthening by alloying • large impurities concentrate at dislocations on low density side Adapted from Fig. 7.18, Callister 7e.

39. 180 400 120 Yield strength (MPa) Tensile strength (MPa) 300 60 200 0 10 20 30 40 50 0 10 20 30 40 50 wt.% Ni, (Concentration C) wt.%Ni, (Concentration C) Ex: Solid SolutionStrengthening in Copper • Tensile strength & yield strength increase with wt% Ni. Adapted from Fig. 7.16 (a) and (b), Callister 7e. • Empirical relation: • Alloying increases sy and TS.

40. 4 Strategies for Strengthening: 3: Precipitation Strengthening precipitate Side View Unslipped part of slip plane Top View S spacing Slipped part of slip plane • Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation “advances” but precipitates act as “pinning” sites with spacing S . • Result:

41. Application:Precipitation Strengthening 1.5mm • Internal wing structure on Boeing 767 Adapted from chapter-opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) • Aluminum is strengthened with precipitates formed by alloying. Adapted from Fig. 11.26, Callister 7e. (Fig. 11.26 is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

42. force -Forging -Rolling roll die A d A A A o blank o d Adapted from Fig. 11.8, Callister 7e. roll force -Drawing -Extrusion A o die container A d die holder tensile force A o ram A billet extrusion d force die die container 4 Strategies for Strengthening: 4: Cold Work (%CW) • Room temperature deformation. • Common forming operations change the cross sectional area:

43. Dislocations During Cold Work 0.9 mm • Ti alloy after cold working: • Dislocations entangle with one another during cold work. • Dislocation motion becomes more difficult. Adapted from Fig. 4.6, Callister 7e. (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)

44. total dislocation length unit volume s large hardening s y1 s small hardening y0 e Result of Cold Work Dislocation density = • Carefully grown single crystal  ca. 103 mm-2 • Deforming sample increases density 109-1010 mm-2 • Heat treatment reduces density  105-106 mm-2 • Yield stress increases as rd increases:

45. RESULT OF COLD WORK • Dislocation density (rd) goes up: Carefully prepared sample: rd ~ 103 mm/mm3 Heavily deformed sample: rd ~ 1010 mm/mm3 • Ways of measuring dislocation density: 40mm Micrograph adapted from Fig. 7.0, Callister 6e. (Fig. 7.0 is courtesy of W.G. Johnson, General Electric Co.) OR • Yield stress increases as rd increases: 18

46. STRENGTHENING STRATEGY 4: COLD WORK (%CW) • An increase in sy due to plastic deformation. BUT actually # of dislocations are increasing !!! • Curve fit to the stress-strain response: 22

47. Stress fields of dislocations interacting ! The strain fields around dislocations interact with each other. Hence, they exert force on each other. Edge dislocations, when they arein the same plane, they repel each other if they have the same sign (direction of the Burgers vector). WHY ? They can attract and annihilate if they have opposite signs. PROVE !

48. Effects of Stress at Dislocations Adapted from Fig. 7.5, Callister 7e.

49. IMPACT OF COLD WORK • Yield strength (s ) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases. y Adapted from Fig. 7.18, Callister 6e. (Fig. 7.18 is from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 221.) Stress % cold work Strain 21

50. Impact of Cold Work As cold work is increased • Yield strength (sy) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases. Adapted from Fig. 7.20, Callister 7e.