Chapter 8: Deformation & Strengthening Mechanisms

1. Chapter 8: Deformation & Strengthening Mechanisms 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?

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

3. Dislocation Motion Dislocation motion & plastic deformation • Metals - plastic deformation occurs by slip – an edge dislocation (extra half-plane of atoms) slides over adjacent plane half-planes of atoms. Adapted from Fig. 8.1, Callister & Rethwisch 4e. • If dislocations can't move, plastic deformation doesn't occur!

4. Dislocation Motion • 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 Edge dislocation Adapted from Fig. 8.2, Callister & Rethwisch 4e. Screw dislocation

5. Adapted from Fig. 8.6, Callister & Rethwisch 4e. • FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed) • => total of 12 slip systems in FCC Deformation Mechanisms 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 • For BCC & HCP there are other slip systems.

6. 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 • • Resolved shearstress, tR • results from applied tensile stresses

7. 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: • Ease of dislocation motion depends on crystallographic orientation  maximum at  =  = 45º

8. Single Crystal Slip Adapted from Fig. 8.9, Callister & Rethwisch 4e. Adapted from Fig. 8.8, Callister & Rethwisch 4e.

9. Ex: Deformation of single crystal a) Will the single crystal yield? b) If not, what stress is needed? So the applied stress of 45 MPa will not cause the crystal to yield.  = 60° crss = 20.7 MPa  = 35° Adapted from Fig. 8.7, Callister & Rethwisch 4e.  = 45 MPa

10. 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)?

11. Slip Motion in Polycrystals s 300 mm • Polycrystals stronger than single crystals – grain boundaries are barriers to dislocation motion. • Slip planes & directions (l, f) change from one grain to another. • tRwill vary from one grain to another. • The grain with the largest tR yields first. • Other (less favorably oriented) grains yield later. Adapted from Fig. 8.10, Callister & Rethwisch 4e. (Fig. 8.10 is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)

12. 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. 8.11, Callister & Rethwisch 4e. (Fig. 8.11 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 equiaxed & randomly oriented.

13. 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

14. Four Strategies for Strengthening: 1: Reduce Grain Size • Grain boundaries are barriers to slip. • Barrier "strength" increases with Increasing angle of misorientation. • Smaller grain size: more barriers to slip. • Hall-Petch Equation: Adapted from Fig. 8.14, Callister & Rethwisch 4e. (Fig. 8.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

15. Four Strategies for Strengthening: 2: Form 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 lattice strains. • These strains can act as barriersto dislocation motion.

16. Lattice Strains Around Dislocations Adapted from Fig. 8.4, Callister & Rethwisch 4e.

17. Strengthening by Solid Solution Alloying • Small impurities tend to concentrate at dislocations (regions of compressive strains) - partial cancellation of dislocation compressive strains and impurity atom tensile strains • Reduce mobility of dislocations and increase strength Adapted from Fig. 8.17, Callister & Rethwisch 4e.

18. Strengthening by Solid Solution Alloying • Large impurities tend to concentrate at dislocations (regions of tensile strains) Adapted from Fig. 8.18, Callister & Rethwisch 4e.

19. VMSE Solid-Solution Strengthening Tutorial

20. 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. 8.16 (a) and (b), Callister & Rethwisch 4e. • Empirical relation: • Alloying increases sy and TS.

21. Four 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:

22. Application:Precipitation Strengthening 1.5mm • Internal wing structure on Boeing 767 Adapted from chapter-opening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) • Aluminum is strengthened with precipitates formed by alloying. Adapted from Fig. 11.45, Callister & Rethwisch 4e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

23. force -Forging -Rolling roll die A d A A A o blank o d Adapted from Fig. 14.2, Callister & Rethwisch 4e. 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 Four Strategies for Strengthening: 4: Cold Work (Strain Hardening) • Deformation at room temperature (for most metals). • Common forming operations reduce the cross-sectional area:

24. Dislocation Structures Change During Cold Working • Dislocation structure in Ti after cold working. • Dislocations entangle with one another during cold work. • Dislocation motion becomes more difficult. Adapted from Fig. 5.11, Callister & Rethwisch 4e. (Fig. 5.11 is courtesy of M.R. Plichta, Michigan Technological University.)

25. total dislocation length unit volume Dislocation Density Increases During Cold Working • Dislocation density = • Carefully grown single crystals •  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:

26. Lattice Strain Interactions Between Dislocations Adapted from Fig. 8.5, Callister & Rethwisch 4e.

27. 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. 8.20, Callister & Rethwisch 4e. low carbon steel

28. Mechanical Property Alterations Due to Cold Working Copper Cold Work Dd= 12.2 mm • What are the values of yield strength, tensile strength & ductility after cold working Cu? Do= 15.2 mm 28

29. 60 800 700 40 300 MPa 600 500 340 MPa Cu 300 Cu 20 7% 400 Cu 100 200 0 0 20 40 60 0 20 40 60 0 20 40 60 sy= 300 MPa % Cold Work TS= 340 MPa %EL = 7% % Cold Work % Cold Work Mechanical Property Alterations Due to Cold Working • What are the values of yield strength, tensile strength & ductility for Cu for %CW = 35.6%? yield strength (MPa) ductility (%EL) tensile strength (MPa) Adapted from Fig. 8.19, Callister & Rethwisch 4e. (Fig. 8.19 is adapted from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.), American Society for Metals, 1979, p. 276 and 327.) 29

30. Effect of Heat Treating After Cold Working annealing temperature (ºC) 100 200 300 400 500 600 700 600 60 tensile strength 50 500 ductility (%EL) 40 tensile strength (MPa) 400 30 ductility 20 300 Recovery Grain Growth Recrystallization • 1 hour treatment at Tanneal... decreases TS and increases %EL. • Effects of cold work are nullified! • • Three Annealing stages: • Recovery • Recrystallization • Grain Growth Adapted from Fig. 8.22, Callister & Rethwisch 4e. (Fig. 8.22 is adapted from G. Sachs and K.R. van Horn, Practical Metallurgy, Applied Metallurgy, and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.)

31. Three Stages During Heat Treatment:1. Recovery • Scenario 1 extra half-plane of atoms Dislocations Results from diffusion annihilate atoms and form diffuse a perfect to regions atomic of tension plane. extra half-plane of atoms tR 3 . “Climbed” disl. can now move on new slip plane 2 . grey atoms leave by 4. opposite dislocations vacancy diffusion meet and annihilate allowing disl. to “climb” Obstacle dislocation 1. dislocation blocked; can’t move to the right Reduction of dislocation density by annihilation. • Scenario 2

32. Three Stages During Heat Treatment:2. Recrystallization 0.6 mm 0.6 mm 33% cold worked brass New crystals nucleate after 3 sec. at 580C. • New grains are formed that: -- have low dislocation densities -- are small in size -- consume and replace parent cold-worked grains. Adapted from Fig. 8.21 (a),(b), Callister & Rethwisch 4e. (Fig. 8.21 (a),(b) are courtesy of J.E. Burke, General Electric Company.)

33. As Recrystallization Continues… 0.6 mm 0.6 mm After 8 seconds After 4 seconds • All cold-worked grains are eventually consumed/replaced. Adapted from Fig. 8.21 (c),(d), Callister & Rethwisch 4e. (Fig. 8.21 (c),(d) are courtesy of J.E. Burke, General Electric Company.)

34. Three Stages During Heat Treatment:3. Grain Growth 0.6 mm 0.6 mm After 8 s, 580ºC After 15 min, 580ºC coefficient dependent on material and T. • Empirical Relation: exponent typ. ~ 2 elapsed time grain diam. at time t. • At longer times, average grain size increases. -- Small grains shrink (and ultimately disappear) -- Large grains continue to grow Adapted from Fig. 8.21 (d),(e), Callister & Rethwisch 4e. (Fig. 8.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.)

35. TR TR = recrystallization temperature Adapted from Fig. 8.22, Callister & Rethwisch 4e. º

36. Recrystallization Temperature • TR= recrystallization temperature = temperature at which recrystallization just reaches completion in 1 h. • 0.3Tm < TR < 0.6Tm • For a specific metal/alloy, TR depends on: • %CW -- TR decreases with increasing %CW • Purity of metal -- TR decreases with increasing purity

37. Diameter Reduction Procedure - Problem A cylindrical rod of brass originally 10 mm (0.39 in) 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 380 MPa (55,000 psi) and a ductility of at least 15 %EL are desired. Furthermore, the final diameter must be 7.5 mm (0.30 in). Explain how this may be accomplished.

38. D = 10 mm o Brass Cold Work D = 7.5 mm f Diameter Reduction Procedure - Solution What are the consequences of directly drawing to the final diameter?

39. 420 540 6 Diameter Reduction Procedure – Solution (Cont.) Adapted from Fig. 8.19, Callister & Rethwisch 4e. • For %CW = 43.8% • y = 420 MPa • TS = 540 MPa > 380 MPa • %EL = 6 < 15 • This doesn’t satisfy criteria… what other options are possible?

40. 15 380 27 12 > 12 %CW < 27 %CW Diameter Reduction Procedure – Solution (cont.) Adapted from Fig. 8.19, Callister & Rethwisch 4e. ForTS > 380 MPa For%EL > 15  our working range is limited to 12 < %CW < 27

41.  Intermediate diameter = Diameter Reduction Procedure – Solution (cont.) • Cold work, then anneal, then cold work again • For objective we need a cold work of 12 < %CW < 27 • We’ll use 20 %CW • Diameter after first cold work stage (but before 2nd cold work stage) is calculated as follows:

42. Diameter Reduction Procedure – Summary • Stage 1: Cold work – reduce diameter from 10 mm to 8.39 mm • Stage 2: Heat treat (allow recrystallization) • Stage 3: Cold work – reduce diameter from 8.39 mm to 7.5 mm • Therefore, all criteria satisfied Fig 8.19 

43. Cold Working vs. Hot Working • Hot working deformation above TR • Cold working  deformation below TR

44. Grain Size Influences Properties • Metals having small grains – relatively strong and tough at low temperatures • Metals having large grains – good creep resistance at relatively high temperatures

45. Mechanical Properties of Polymers – Stress-Strain Behavior brittle polymer plastic elastomer elastic moduli – less than for metals Adapted from Fig. 7.22, Callister & Rethwisch 4e. • Fracture strengths of polymers ~ 10% of those for metals • Deformation strains for polymers > 1000% –for most metals, deformation strains < 10% 45

46. Mechanisms of Deformation—Brittle Crosslinked and Network Polymers NearFailure Near Failure Initial network polymer s (MPa) brittle failure Initial x plastic failure x e aligned, crosslinked polymer Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 4e. 46

47. Mechanisms of Deformation — Semicrystalline (Plastic) Polymers fibrillar structure near failure crystalline block segments amorphous separate crystalline regions undeformed regions align elongate structure s (MPa) brittle failure x Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 4e. Inset figures along plastic response curve adapted from Figs. 8.27 & 8.28, Callister & Rethwisch 4e. (Figs. 8.27 & 8.28 are from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.) onset of necking plastic failure x unload/reload e 47

48. 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 4e. (Fig. 8.28 is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.) 48

49. Mechanisms of Deformation—Elastomers final: chains are straighter, still cross-linked deformation initial: amorphous chains are is reversible (elastic)! kinked, cross-linked. s (MPa) brittle failure x Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 4e. Inset figures along elastomer curve (green) adapted from Fig. 8.30, Callister & Rethwisch 4e. (Fig. 8.30 is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.) plastic failure x x elastomer e • 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) 49

50. Summary • Dislocations are observed primarily in metals and alloys. • Strength is increased by making dislocation motion difficult. • Strength of metals may be increased by: -- decreasing grain size -- solid solution strengthening -- precipitate hardening -- cold working • A cold-worked metal that is heat treated may experience recovery, recrystallization, and grain growth – its properties will be altered.