1. Chapter 8 - 1

2. 2 ISSUES TO ADDRESS... Why are the number of dislocations present greatest in metals ? How are strength and dislocation motion related? Why does heating alter strength and other properties?

3. Chapter 8 - 3 Deformation mechanisms proposed to explain the deformation behaviors of metals Techniques that may be used to strengthen metals in terms of these deformation mechanisms ISSUES TO ADDRESS...

4. Chapter 8 - 4 Dislocations & Materials Classes Covalent Ceramics (Si, diamond): Motion difficult - directional (angular) bonding Ionic Ceramics (NaCl): Motion difficult - need to avoid nearest neighbors of like sign (- and +) ++++ +++ ++++ --- ---- --- Metals (Cu, Al): Dislocation motion easiest - non-directional bonding - close-packed directions for slip electron cloud ion cores + + + + +++++++ + +++++ +++++ + +

5. Chapter 8 - 5 Plastic deformation: A net movement of large numbers of atoms in response to an applied stress Interatomic bonds must be ruptured and then reformed Early studies: theoretical strengths of perfect crystals are many times greater than their measured strengths!!! Plastic deformation

6. Chapter 8 - 6 During the 1930s it was theorized that this discrepancy could be explained by a type of linear crystalline defect that has since come to be known as a dislocation. It was not until the 1950s that the existence of dislocation defects was established by direct observation with TEM. Since then, a theory of dislocations has evolved that explains many of the physical and mechanical phenomena in metals [as well as crystalline ceramics

7. Chapter 8 - 7 Dislocation Motion Dislocations: edge, screw and mixed types Most often Plastic deformation corresponds to the motion of large numbers of dislocations. Figure 8.1. Atomic rearrangements that accompany the motion of an edge dislocation as it moves in response to an applied shear stress.

8. Chapter 8 - 8 Dislocation Motion An edge dislocation (extra half-plane of atoms) slides over adjacent plane half-planes of atoms. Slip: The process by which plastic deformation is produced by dislocation motion Slip plane: The crystallographic plane along which the dislocation line moves

9. Chapter 8 - 9 Dislocation Motion In response to an applied shear stress: An edge dislocation, moves parallel to the stress direction. A screw dislocation moves perpendicular to the stress direction. If dislocations can't move, plastic deformation doesn't occur! However, the net plastic deformation for the motion of both dislocation types is the same.

10. Chapter 8 - 10 Dislocation Motion The direction of motion of the mixed dislocation line is neither perpendicular nor parallel to the applied stress, but lies somewhere in between. A dislocation moves along a slip plane in a slip direction perpendicular to the dislocation line The slip direction is the same as the Burgers vector direction

11. Chapter 8 - CHARACTERISTICS OF DISLOCATIONS Dislocations are introduced during solidification, during plastic deformation, and as a consequence of thermal stresses that result from rapid cooling. Dislocation density: the total dislocation length per unit volume, or the number of dislocations that intersect a unit area of a random section. Carefully solidified metal crystals: 10 3 mm −2 Heavily deformed metals: 10 9 to 10 10 mm −2 A heat treated deformed metal: 10 5 to 10 6 mm −2 Ceramic materials: 10 2 and 10 4 mm −2 Silicon single crystals used in ICs: 0.1-1 mm −2 11

12. Chapter 8 - 12 A lithium fluoride (LiF) single crystal. The small pyramidal pits represent positions at which dislocations intersect the surface. CHARACTERISTICS OF DISLOCATIONS

13. Chapter 8 - Strain fields exist around dislocations Some atomic lattice distortion around the dislocation line  There are regions in which compressive, tensile, and shear lattice strains are imposed on the neighboring atoms. The strains magnitude decreases with radial distance from the dislocation. These influence: the mobility of the dislocations, and their ability to multiply 13 CHARACTERISTICS OF DISLOCATIONS Figure 8.4. Regions of compression (green) and tension (yellow) around an edge dislocation.

14. Chapter 8 - 14 CHARACTERISTICS OF DISLOCATIONS The strain fields surrounding dislocations may interact Two dislocations of the same sign Identical slip plane  a mutual repulsive force Two dislocations of opposite sign Identical slip plane  a mutual attraction  dislocation annihilation.

15. Chapter 8 - 15 Slip System –Slip plane - plane on which easiest slippage occurs Highest planar densities (and large interplanar spacings). –Slip directions - directions of movement Highest linear densities Deformation Mechanisms – FCC Slip occurs on {111} planes (close-packed) in directions (close-packed) => total of 12 slip systems in FCC – For BCC & HCP there are other slip systems.

16. Chapter 8 - 16 FCC or BCC crystals: a large number of slip systems (at least 12)  Extensive plastic deformation is possible along the various systems  Quite ductile HCP metals: few active slip systems  normally quite brittle

17. Chapter 8 - 17 Stress and Dislocation Motion Resolved shear stress,  R: results from applied tensile stresses slip plane normal, n s Resolved shear stress: RR =F s /A/A s slip direction ASAS RR RR FSFS SLIP IN SINGLE CRYSTALS φ: the angle between the normal to the slip plane and the applied stress direction λ: the angle between the slip and stress directions

18. Chapter 8 - 18 Condition for dislocation motion: Ease of dislocation motion depends on crystallographic orientation 10 -4 GPa to 10 -2 GPa typically Critical Resolved Shear Stress  maximum at =  = 45º RR = 0 =90°  RR =  /2 =45°   RR = 0  =90° 

19. Chapter 8 - 19 Single Crystal Slip Slip along a number of equivalent and most favorably oriented planes and directions at various positions along the specimen length Each step results from the movement of a large number of dislocations along the same slip plane. A zinc single crystal

20. Chapter 8 - 20 Ex: Deformation of single crystal So the applied stress of 6500 psi will not cause the crystal to yield. = 35°  = 60°  crss = 20.7 MPa a) Will the single crystal yield? b) If not, what stress is needed?  = 6500 psi Adapted from Fig. 8.7, Callister & Rethwisch 3e.

21. Chapter 8 - 21 Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress,  y )? So for deformation to occur the applied stress must be greater than or equal to the yield stress

22. Chapter 8 - 22 Slip planes & directions (,  ) change from one crystal to another.   R will vary from one crystal to another. The crystal with the largest  R yields first. Other (less favorably oriented) crystals yield later. Two sets of parallel intersecting lines  Two slip systems operated for most of the grains Note that during deformation, the grain boundaries usually do not come apart or open up. Stronger - grain boundaries pin deformations. Slip Motion in Polycrystals  300  m Polycrystalline copper

23. Chapter 8 - 23 Polycrystals are stronger than single-crystals! Each individual grain is constrained by its neighboring grains. A single grain may be favorably oriented, but it cannot deform until the adjacent and less favorably oriented grains are capable of slip also. Before deformation, equiaxed grains After deformation, elongated grains Slip Motion in Polycrystals

24. Chapter 8 - 24 Can be induced by rolling a polycrystalline metal - before rolling 235  m - isotropic since grains are approx. spherical & randomly oriented. - after rolling - anisotropic since rolling affects grain orientation and shape. rolling direction Anisotropy in  y

25. Chapter 8 - 25 side view 1. Cylinder of tantalum machined from a rolled plate: rolling direction 2. Fire cylinder at a target. The noncircular end view shows anisotropic deformation of rolled material. end view 3. Deformed cylinder plate thickness direction Photos courtesy of G.T. Gray III, Los Alamos National Labs. Used with permission. Anisotropy in Deformation

26. Chapter 8 - 26 8.8 DEFORMATION BY TWINNING

27. Chapter 8 - 27 Mechanisms of Strengthening in Metals Hardness and strength (both yield and tensile) are related to the ease with which plastic deformation can be made Plastic deformation depends on the ability of dislocations to move By reducing the mobility of dislocations, the mechanical strength may be enhanced Restrict dislocation motion  harder and stronger material

28. Chapter 8 - 28 During plastic deformation, slip or dislocation motion must take place across the grain boundaries Grain boundaries are barriers to slip: Barrier "strength“ increases with increasing angle of misorientation. The atomic disorder within a grain boundary region will result in a discontinuity of slip planes. Smaller grain size: more barriers to slip. 1. Grain Size Reduction Mechanisms of Strengthening in Metals

29. Chapter 8 - 29 Grain size may be regulated by the solidification rate and by plastic deformation followed by an appropriate heat treatment. d: the average grain diameter, σ 0 and k y are constants for a particular material. Hall-Petch Equation:

30. Chapter 8 - 30 2. Solid Solutions Strengthening Mechanisms of Strengthening in Metals High-purity metals are almost always softer and weaker than alloys composed of the same base metal, 24 caret gold Vs 18 caret Gold!

31. Chapter 8 - 31 Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Empirical relation: Alloying increases  y and TS.

32. Chapter 8 - 32 Substitutional impurity atoms: If smaller than the host atom  exert tensile strains on the surrounding crystal lattice If larger than the host atom  exert compressive strains on the surrounding crystal lattice, A B C D 2. Solid Solutions Strengthening

33. Chapter 8 - 33 Solute atoms tend to segregate around dislocations to reduce some of the strain in the lattice surrounding a dislocation. Smaller impurity atoms segregate above the slip plane. Larger impurity atoms segregate below the slip plane. Greater resistance to slip when impurity atoms are present because the overall lattice strain must increase if a dislocation is torn away from them. 2. Solid Solutions Strengthening

34. Chapter 8 - 34  Solute atoms, reduce mobility of dislocation! Greater applied stress is necessary to first initiate and then continue plastic deformation for solid-solution alloys  Increased strength and hardness The same lattice strain interactions will exist between impurity atoms and dislocations that are in motion during plastic deformation. Impurity atoms distort the lattice and generate stress. Stress can produce a barrier to dislocation motion. 2. Solid Solutions Strengthening

35. Chapter 8 - 35 Hard precipitates/phases are difficult to shear. Ex: Carbide/ oxide particles in metals (TiC in Iron or Al 2 O 3 Aluminum, ….) 3. Precipitation Strengthening Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation “advances” but precipitates act as “pinning” sites with spacing S Side View precipitate Top View Slipped part of slip plane Unslipped part of slip plane S Mechanisms of Strengthening in Metals

36. Chapter 8 - 36 Interaction of a dislocation with solid phase precipitates. The dislocation moves from left to right in each frame. 3. Precipitation Strengthening If dislocation can shear the precipitate

37. Chapter 8 - 37 Climbing of dislocations over an oxide particle; (b) backside pinning Pinning of a dislocations by an oxide particle

38. Chapter 8 - 38 3. Precipitation Strengthening If dislocation can’t shear the precipitate

39. Chapter 8 - 39 Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates (e.g. Al 2 Cu) formed by alloying and heat treatment. 1.5  m 3. Precipitation Strengthening

40. Chapter 8 - 40 4. Strain hardening/ Work hardening/ Cold working Strengthening and hardening of a ductile metal by plastic deformation at close to room temperature Cross sectional area changes by forming operations. Degree of plastic deformation (%CW) : -Rolling roll A o A d Mechanisms of Strengthening in Metals -Extrusion ram billet container force die holder die A o A d extrusion

41. Chapter 8 - 41 4. Strain/ Work hardening

42. Chapter 8 - 42 4. Strain/ Work hardening

43. Chapter 8 - 43 Ti alloy after cold working Dislocation density in a metal increases with deformation or cold working Dislocations entangle with one another during cold working. Dislocation motion becomes more difficult. Thus, the imposed stress necessary to deform a metal increases with increasing cold working. 0.9  m 4. Strain/ Work hardening

44. Chapter 8 - 44 In passing, for the mathematical expression relating true stress and strain, Equation 7.19, the parameter n is called the strain-hardening exponent, which is a measure of the ability of a metal to strain harden; the larger its magnitude, the greater the strain hardening for a given amount of plastic strain.

45. Chapter 8 - 45 –Carefully grown single crystal  ca. 10 3 mm -2 –Deforming sample increases density  10 9 -10 10 mm -2 –Heat treatment reduces density  10 5 -10 6 mm -2 When making hardness measurements, what will be the effect of making an indentation very close to a preexisting indentation? Why? 4. Strain/ Work hardening

46. Chapter 8 - 46 Result of Cold Work –Carefully grown single crystal  10 3 mm -2 –Deforming sample increases density  10 9 -10 10 mm -2 –Heat treatment reduces density  10 5 -10 6 mm -2 Yield stress increases as  d increases: large hardening small hardening    y0y0  y1y1 The strengthening effects due to grain size reduction and strain hardening can be eliminated or reduced by an elevated temperature heat treatment (Sections 8.12 and 8.13). Conversely, solid-solution strengthening is unaffected by heat treatment

47. Chapter 8 - 47 Tensile strength= 340 Mpa Ductility=7%

48. Chapter 8 - 48 Results for polycrystalline iron: YS and TS decrease and %El increases with increasing test temperature. Why? Vacancies help Dislocations move past obstacles.  -  Behavior vs. Temp. 2. vacancies replace atoms on the disl. half plane 3. disl. glides past obstacle -200  C -100  C 25  C 800 600 400 200 0 Strain Stress (MPa) 00.10.20.30.40.5 1. disl. trapped by obstacle obstacle 4. Strain/ Work hardening

49. Chapter 8 - 49 1 hour treatment at T anneal … decreases TS and increases %El. Effects of cold work are reversed! 3 Annealing stages! Holding time has a similar effect. Effect of Heating After %CW T R = Recrystallization Temp. = point of highest rate of property change

50. Chapter 8 - 50 Annihilation reduces dislocation density: Recovery Scenario 1 4. opposite dislocations meet and annihilate 2. grey atoms leave by vacancy diffusion allowing dislocation to “climb” RR 1. dislocation blocked; can’t move to the right Obstacle (dislocation or particle) 3. “Climbed” disl. can now move on new slip plane Scenario 2 Results from diffusion Dislocations annihilate and form a perfect atomic plane. extra half-plane of atoms extra half-plane of atoms atoms diffuse to regions of tension

51. Chapter 8 - 51 New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. 33% cold worked brass New crystals nucleate after 3 sec. at 580  C 0.6 mm Recrystallization

52. Chapter 8 - 52 All cold-worked grains are consumed. After 4 seconds at 580  C After 8 seconds at 580  C 0.6 mm Recrystallization

53. Chapter 8 - 53 T R  0.3 - 0.6 T m (K) Recrystallization

54. Chapter 8 - 54 Recrystallization

55. Chapter 8 - 55 At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced. After 10 min, 700ºCAfter 15 min, 580ºC 0.6 mm Grain Growth

56. Chapter 8 - 56 Empirical Relation (Ostwald Ripening): elapsed time Coefficient dependent on material and Temp. grain diam. at time t exponent = ~ 2 Grain Growth

57. Chapter 8 - 57

58. Chapter 8 - 58

59. Chapter 8 - 59

60. Chapter 8 - 60 TRTR º º T R = recrystallization temperature

61. Chapter 8 - 61 Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property change 1.T m => T R  0.3-0.6 T m (K) 2.Due to diffusion  annealing time  T R = f(time) shorter annealing time => higher T R 3.Higher %CW => lower T R – strain hardening 4.Pure metals lower T R due to dislocation movements Easier to move in pure metals => lower T R

62. Chapter 8 - 62 Coldwork Calculations A cylindrical rod of brass originally 0.40 in (10.2 mm) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 55,000 psi (380 MPa) and a ductility of at least 15 %EL are desired. Further more, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.

63. Chapter 8 - 63 Coldwork Calculations Solution If we directly draw to the final diameter what happens? D o = 0.40 in Brass Cold Work D f = 0.30 in

64. Chapter 8 - 64 Coldwork Calc Solution: Cont. For %CW = 43.8% 540 420 –  y = 420 MPa –TS = 540 MPa > 380 MPa 6 –%EL = 6< 15 This doesn’t satisfy criteria…… what can we do? Adapted from Fig. 8.19, Callister & Rethwisch 3e.

65. Chapter 8 - 65 Coldwork Calc Solution: Cont. 380 12 15 27 For %EL > 15 For TS > 380 MPa > 12 %CW < 27 %CW  our working range is limited to %CW = 12-27 Adapted from Fig. 8.19, Callister & Rethwisch 3e.

66. Chapter 8 - 66 Coldwork Calc Soln: Recrystallization Cold draw-anneal-cold draw again For objective we need a cold work of %CW  12-27 –We’ll use %CW = 20 Diameter after first cold draw (before 2 nd cold draw)? –must be calculated as follows:  Intermediate diameter =

67. Chapter 8 - 67 Coldwork Calculations Solution Summary: 1.Cold work D 01 = 0.40 in  D f1 = 0.335 m 2.Anneal above D 02 = D f1 3.Cold work D 02 = 0.335 in  D f 2 =0.30 m Therefore, meets all requirements  Fig 7.19

68. Chapter 8 - 68 Rate of Recrystallization Hot work  above T R Cold work  below T R Smaller grains –stronger at low temperature –weaker at high temperature log t start finish 50%

69. Chapter 8 - 69 Mechanical Properties of Polymers – Stress-Strain Behavior Fracture strengths of polymers ~ 10% of those for metals Deformation strains for polymers > 1000% – for most metals, deformation strains < 10% brittle polymer plastic elastomer elastic moduli – less than for metals Adapted from Fig. 7.22, Callister & Rethwisch 3e.

70. Chapter 8 - 70 Mechanisms of Deformation—Brittle Crosslinked and Network Polymers brittle failure plastic failure  (MPa)  x x aligned, crosslinked polymer Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e. Initial Near Failure Initial network polymer Near Failure

71. Chapter 8 - 71 Mechanisms of Deformation — Semicrystalline (Plastic) Polymers brittle failure plastic failure  (MPa) x x crystalline block segments separate fibrillar structure near failure crystalline regions align onset of necking undeformed structure amorphous regions elongate unload/reload Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e. Inset figures along plastic response curve adapted from Figs. 8.27 & 8.28, Callister & Rethwisch 3e. (Figs. 8.27 & 8.28 are from J.M. Schultz, Polymer Materials Science, Prentice- Hall, Inc., 1974, pp. 500-501.) 

72. Chapter 8 - 72 Predeformation by Drawing Drawing…(ex: monofilament fishline) -- stretches the polymer prior to use -- aligns chains in the stretching direction Results of drawing: -- increases the elastic modulus (E) in the stretching direction -- increases the tensile strength (TS) in the stretching direction -- decreases ductility (%EL) Annealing after drawing... -- decreases chain alignment -- reverses effects of drawing (reduces E and TS, enhances %EL) Contrast to effects of cold working in metals! Adapted from Fig. 8.28, Callister & Rethwisch 3e. (Fig. 8.28 is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.)

73. Chapter 8 - 73 Compare elastic behavior of elastomers with the: -- brittle behavior (of aligned, crosslinked & network polymers), and -- plastic behavior (of semicrystalline polymers) (as shown on previous slides) Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e. Inset figures along elastomer curve (green) adapted from Fig. 8.30, Callister & Rethwisch 3e. (Fig. 8.30 is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.) Mechanisms of Deformation— Elastomers  (MPa)  initial: amorphous chains are kinked, cross-linked. x final: chains are straighter, still cross-linked elastomer deformation is reversible (elastic)! brittle failure plastic failure x x

74. Chapter 8 - 74 Dislocations are observed primarily in metals and alloys. Strength is increased by making dislocation motion difficult. Particular ways to increase strength are to: -- decrease grain size -- solid solution strengthening -- precipitate strengthening -- cold work Heating (annealing) can reduce dislocation density and increase grain size. This decreases the strength. Summary

75. Chapter 8 - 75 Core Problems: Self-help Problems: ANNOUNCEMENTS Reading: