MT Materials Engineering (3-0-0) Instructor: Sumantra Mandal Department of Metallurgical & Materials Engg

Transcription

1 MT Materials Engineering (3-0-0) Instructor: Sumantra Mandal Department of Metallurgical & Materials Engg Chapter 9-1

2 Difference between Crystal & Lattice Chapter 9-2

3 Difference between Crystal & Lattice Chapter 9-3

4 Unit Cell Chapter 9-4

5 Crystal system and Bravais Lattice Chapter 9-5

6 Cubic Bravais Lattice Chapter 9-6

7 Lattice Defects ü Real crystals deviate from the perfect periodicity ü While the concept of the perfect lattice is adequate for explaining the structure-insensitive properties of metal, it is necessary to consider a number of types of lattice defects to explain structure-sensitive properties Chapter 9-7

8 Different Types of Lattice Defects Point defects Line defects Surface defects Volume defects Chapter 9-8

9 Point Defects Number of vacant sites (n) Vacancy Interstitial Impurity Chapter 9-9

10 Line Defects - Dislocations Edge dislocation produced by slip in cubic system Atomic arrangement near edge dislocation Chapter 9-10

11 Deformation by slip Schematic drawing of classical idea of slip Chapter 9-11

12 Slip in perfect lattice Shear displacement of plane of atoms over another Relation between shear stress and displacement The max. stress at which slip should occur Variation of shearing stress with displacement in slip direction Chapter 9-12

13 Slip by dislocation movements If dislocations don't move, deformation doesn't occur! Chapter 9-13

14 Dislocation Motion Dislocation moves along slip plane in slip direction perpendicular to dislocation line Slip direction same direction as Burgers vector Edge dislocation From Fig Callister s Materials Science and Engineering, Adapted Version. Screw dislocation Chapter 9-14

15 Critical Resolved Shear Stress for Slip Slip begins when the shearing stress on the slip plane in the slip direction reaches a threshold value called as critical resolved shear stress (CRSS) Chapter 9-15

16 Stress and Dislocation Motion Crystals slip due to a resolved shear stress, t R. Applied tension can produce such a stress. Applied tensile stress: s = F/A F A F Resolved shear stress: t R =Fs/As slip plane normal, n s t R t R Relation between s and t R t R =F S /A S A S Fcos l A/cos f F S F l F S n S f A S A t R = scos lcos f Chapter 9-16

17 Condition for dislocation motion: Crystal orientation can make it easy or hard to move dislocation t Critical Resolved Shear Stress R = scos lcos f t R > tcrss s s s typically 10-4 GPa to 10-2 GPa t R = 0 l=90 t R = s/2 l=45 f =45 t R = 0 f=90 t maximum at l = f = 45º Chapter 9-17

18 Single Crystal Slip From Fig. 10.9, Callister s Materials Science and Engineering, Adapted Version. From Fig Callister s Materials Science and Engineering, Adapted Version. Chapter 9-18

19 Ex: Deformation of single crystal f=60 l=35 a) Will the single crystal yield? b) If not, what stress is needed? t crss = 3000 psi t = s cos l cos f s = 6500 psi Adapted from Fig. 7.7, Callister 7e. s = 6500 psi t = (6500 psi) (cos35 o )(cos60 o ) = (6500 psi) (0.41) t = 2662 psi < t crss = 3000 psi So the applied stress of 6500 psi will not cause the crystal to yield. Chapter 9-19

20 Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, s y )? t crss = 3000 psi = s y cosl cosf = s y (0.41) \ s y = t coslcosf 3000 psi 0.41 crss = = 7325 psi So for deformation to occur the applied stress must be greater than or equal to the yield stress s ³ s y = 7325 psi Chapter 9-20

21 Assignment Determine the tensile stress that need to be applied along the [1-10] axis of a silver crystal to cause slip on the (1-1-1)[0-11] system. The CRSS is 6 MPa. Chapter 9-21

22 Slip Motion in Polycrystals Stronger - grain boundaries pin deformations s Slip planes & directions (l, f) change from one crystal to another. t R will vary from one crystal to another. The crystal with the largest t R yields first. Other (less favorably oriented) crystals yield later. 300 mm From Fig , Callister s Materials Science and Engineering, Adapted Version. (Fig is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].) Chapter 9-22

23 Taxonomy of Metals Metal Alloys Ferrous Steels <1.4wt%C Cast Irons 3-4.5wt%C Nonferrous Cu Al Mg Ti From Fig. 9.1 Callister s Materials Science and Engineering, Adapted Version d a ferrite 800 T( C) 600 g+l g austenite 727 C Eutectoid: C L g+fe 3 C a+fe 3 C 4.30 L+Fe 3 C Eutectic: microstructure: ferrite, graphite cementite From Fig Callister s Materials Science and Engineering, Adapted Version. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.- in-chief), ASM International, Materials Park, OH, 1990.) Fe3C cementite (Fe) C o, wt% C Chapter 9-23

24 Iron-Carbon (Fe-C) Phase Diagram 2 important points -Eutectic (A): L Þ g +Fe 3 C -Eutectoid (B): g Þ a +Fe 3 C 1600 d T( C) a 600 g g+l (austenite) R B g g g g L 1148 C A R S g+fe 3 C a+fe 3 C L+Fe 3 C S 727 C = Teutectoid Fe 3 C (ceme entite) 120 mm Result: Pearlite = alternating layers of a and Fe 3 C phases (From Fig. 7.27, Callister Adapted Version.) (Fe) 0.76 C eutectoid 4.30 Fe 3 C (cementite-hard) a (ferrite-soft) From Fig. 7.24, Callister Adapted Version. C o, wt% C Chapter 9-24

25 g g g g g g a a g g wa =S/(R+S) =(1-wa) w Fe3 C a g g a g g wa =s/(r+s) wg =(1- wa) 1600 d T( C) a Hypoeutectoid Steel g g+l (austenite) r s R S 1148 C L g + Fe 3 C a+ Fe 3 C L+Fe 3 C (Fe) C pearlite 0 w pearlite = wg 0.76 pearlite 727 C From Fig Callister s Materials Science and Engineering, Adapted Version. Fe 3 C (ce ementite) C o, wt% C (Fe-C System) 100 mm Hypoeutectoid steel From Figs and 7.29 Callister s Materials Science and Engineering, Adapted Version. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) proeutectoid ferrite Chapter 9-25

26 Fe 3 C g g g g g g g g g g g g w Fe3 C =r/(r+s) wg =(1-w Fe3 C) wa =S/(R+S) w Fe3 C =(1-w a) 1600 d T( C) a Hypereutectoid Steel g g+l (austenite) R r s 1148 C L g +Fe 3 C a +Fe 3 C L+Fe 3 C pearlite (Fe) w pearlite = wg C o pearlite S C o, wt%c Fe 3 C (ce ementite) (Fe-C System) 60 mm Hypereutectoid steel proeutectoid Fe 3 C From Figs and 7.32, Callister s Materials Science and Engineering, Adapted Version. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) From Fig. 7.33, Callister s Materials Science and Engineering, Adapted Version. Chapter 9-26

27 Alloying Steel with More Elements T eutectoid changes: C eutectoid changes: TEut tectoid ( C) Ti Mo Ni Si W Cr Mn Ceut tectoid (wt%c) Ti Si Mo Ni Cr W Mn wt. % of alloying elements From Fig Callister s Materials Science and Engineering, Adapted Version. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) wt. % of alloying elements From Fig Callister s Materials Science and Engineering, Adapted Version. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Chapter 9-27

28 Steels low carbon <0.25wt% C Low Alloy Med carbon wt% C high carbon wt% C High Alloy heat Name plain HSLA plain treatable Cr,V Cr, Ni Additions none none Ni, Mo Mo Uses auto struc. sheet bridges towers press. vessels crank shafts bolts hammers blades pistons gears wear applic. plain none wear applic. tool Cr, V, Mo, W drills saws dies austenitic stainless Cr, Ni, Mo Example Hardenability TS EL increasing strength, cost, decreasing ductility high T applic. turbines furnaces V. corros. resistant Chapter 9-28 Based on data provided in Tables 9.1(b), 9.2(b), 9.3, and 9.4, Callister s Materials Science and Engineering, Adapted Version.

29 Ferrous Alloys Iron containing Steels - cast irons Nomenclature AISI & SAE 10xx Plain Carbon Steels 11xx Plain Carbon Steels (resulfurized for machinability) 15xx Mn (10 ~ 20%) 40xx Mo (0.20 ~ 0.30%) 43xx Ni ( %), Cr ( %), Mo ( %) 44xx Mo (0.5%) where xx is wt% C x 100 example: 1060 steel plain carbon steel with 0.60 wt% C Stainless Steel -- >11% Cr Chapter 9-29

30 Cast Iron Ferrous alloys with > 2.1 wt% C more commonly wt%c low melting (also brittle) so easiest to cast Cementite decomposes to ferrite + graphite Fe 3 C à 3 Fe (a) + C (graphite) generally a slow process Chapter 9-30

31 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. From Fig , Callister s Materials Science and Engineering, Adapted Version. (Fig is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.) Hall-Petch Equation: s yield = s o + k y d -1/ 2 Chapter 9-31

32 4 Strategies for Strengthening: 2: Solid Solutions Impurity atoms distort the lattice & generate stress. Stress can produce a barrier to dislocation motion. Smaller substitutional impurity Larger substitutional impurity A C B D 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. Chapter 9-32

33 Stress Concentration at Dislocations From Fig. 10.4, Callister s Materials Science and Engineering, Adapted Version. Chapter 9-33

34 Strengthening by Alloying small impurities tend to concentrate at dislocations reduce mobility of dislocation \ increase strength From Fig , Callister s Materials Science and Engineering, Adapted Version. Chapter 9-34

35 Strengthening by alloying large impurities concentrate at dislocations on low density side From Fig , Callister s Materials Science and Engineering, Adapted Version. Chapter 9-35

36 Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Tensile st trength (MPa) wt.% Ni, (Concentration C) Yield str rength (MPa) wt.%ni, (Concentration C) From Fig (a) and (b), Callister s Materials Science and Engineering, Adapted Version. Empirical relation: Alloying increases s y and TS. s y ~ C 1/ 2 Chapter 9-36

37 4 Strategies for Strengthening: 3: Precipitation Strengthening Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Side View precipitate Large shear stress needed to move dislocation toward precipitate and shear it. Top View Unslipped part of slip plane S Slipped part of slip plane Dislocation advances but precipitates act as pinning sites with spacing S. Result: sy~ 1 S Chapter 9-37

38 Application: Precipitation Strengthening Internal wing structure on Boeing 767 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. From Fig , Callister s Materials Science and Engineering, Adapted Version. 1.5mm (Fig is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) Chapter 9-38

39 Interaction between precipitates and dislocation When the particles are small and/or soft, dislocations can cut and deform the particles as shown in the Figure Chapter 9-39

40 Interaction between precipitates and dislocation For the case of averaged non-coherent precipitates Orowan proposed the mechanism illustrated in the Fig. below. The yield stress is determined by the shear stress required to bow a dislocation line between two particles separated by a distance l, where l>r. Schematic drawing, of stages in passage of a dislocation between widely separated obstacles, based on Orowan's mechanism of dispersion hardening Chapter 9-40

41 Stage 1: A straight dislocation line approaching two particles. Stage 2: The line is beginning to bend. Stage 3: It has reached the critical curvature. The dislocation can then move forward without further decreasing its radius of curvature (R). and l = 2R, so that the shear stress required to force the dislocation between the obstacles is: Chapter 9-41

42 Stage 4: Since the segments of dislocation that meet on the other side of the particle are of opposite sign. they can annihilate each other over part of their length, leaving a dislocation loop around each particle Stage 5: The original dislocation is then free to move on Every dislocation gliding over the slip plane adds one loop around the particle. These loops exert a back stress on dislocation sources which must be overcome for additional slip to take place. This requires an increase in shear stress, with the result that dispersed non-coherent particles cause the matrix to strain-harden rapidly Chapter 9-42

43 Exercise An alurninum-4% copper alloy has a yield stress of 600 MPa. Estimate the particle spacing in this alloy. Given G ~ 27.6 GPa; b~0.25 nm. At this strength level we are dealing with a precipitation-hardening alloy that has been aged beyond the maximum strength. The strengthening mechanism is dislocation bypassing of particles. G ~ 27.6 GPa; b~0.25 nm; t 0 = 600/2 = 300 MPa Chapter 9-43

44 Typical tensile test machine Stress-Strain Testing Typical tensile specimen extensometer specimen Adapted from Fig. 6.2, Callister 7e. gauge length From Fig. 9.9, Callister s Materials Science and Engineering, Adapted Version. (Fig. 9.9 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.) Chapter 9-44

45 Linear Elastic Properties Modulus of Elasticity, E: (also known as Young's modulus) Hooke's Law: s = E e s F E Linearelastic e F simple tension test Chapter 9-45

46 Poisson's ratio, n Poisson's ratio, n: e L e n = - L e e metals: n ~ 0.33 ceramics: n ~ 0.25 polymers: n ~ n Units: E: [GPa] or [psi] n: dimensionless Chapter 9-46

47 Plastic (Permanent) Deformation (at lower temperatures, i.e. T < T melt /3) Simple tension test: engineering stress, s Elastic+Plastic at larger stress Elastic initially permanent (plastic) after load is removed e p engineering strain, e plastic strain From Fig (a), Callister s Materials Science and Engineering, Adapted Version. Chapter 9-47

48 Yield Strength, s y Stress at which noticeable plastic deformation has occurred. when e p = s y tensile stress, s s y = yield strength Note: for 2 inch sample e = = Dz/z \ Dz = in e p = engineering strain, e From Fig (a), Callister s Materials Science and Engineering Adapted Version. Chapter 9-48

49 s y Tensile Strength, TS Maximum stress on engineering stress-strain curve. TS engine eering stre ess Metals: occurs when noticeable necking starts. Polymers: occurs when polymer backbone chains are aligned and about to break. Typical response of a metal strain engineering strain From Fig Callister s Materials Science and Engineering, Adapted Version. F = fracture or ultimate strength Neck acts as stress concentrator Chapter 9-49

50 Ductility Plastic tensile strain at failure: Engineering tensile stress, s smaller %EL larger %EL L %EL = f - L o L o x 100 A L o o A f L f From Fig Callister s Materials Science and Engineering, Adapted Version.. Engineering tensile strain, e Another ductility measure: A A %RA = f x 100 A o - o Chapter 9-50

51 Toughness Energy to break a unit volume of material Approximate by the area under the stress-strain curve. Engineering tensile stress, s small toughness (ceramics) large toughness (metals) From Fig Callister s Materials Science and Engineering, Adapted Version. very small toughness (unreinforced polymers) Engineering tensile strain, e Brittle fracture: elastic energy Ductile fracture: elastic + plastic energy Chapter 9-51

52 Resilience, U r Ability of a material to store energy Energy stored best in elastic region U = ò e r 0 y sde If we assume a linear stress-strain curve this simplifies to U 1 2 s y e y From Fig Callister s Materials Science and Engineering, Adapted Version. Chapter 9-52

53 Elastic Strain Recovery From Fig Callister s Materials Science and Engineering, Adapted Version.. Chapter 9-53

54 True Stress & Strain Note: specimen area changes when sample stretched True stress True Strain st = F A i e T = ln l ( l ) i o s e T T = s = ( 1+ e) ( + e) ln 1 From Fig Callister s Materials Science and Engineering, Adapted Version. Chapter 9-54

55 Assignment Chapter 9-55

56 Hardening An increase in s y due to plastic deformation. s s y1 s y0 large hardening small hardening Curve fit to the stress-strain response: s T = K( e ) n T true stress (F/A) true strain: ln(l/l o ) e hardening exponent: n = 0.15 (some steels) to n = 0.5 (some coppers) Chapter 9-56

57 Assignment Compute the strain-hardening exponent (n) for an alloy in which a true stress of 415 MPa produces true strain of Assume a value of 1035 MPa for K. Chapter 9-57

58 Instability in Tension ü Necking generally begins at maximum load during the tensile deformation of a ductile metal. ü An ideal plastic material in which no strain hardening occurs would become unstable in tension and begin to neck just as soon as yielding took place. ühowever, a real metal undergoes strain hardening, which tends to increase the load-carrying capacity of the specimen as deformation increases. This effect is opposed by the gradual decrease in the cross-sectional area of the specimen as it elongates. Chapter 9-58

59 Instability in Tension - Necking ünecking or localized deformation begins at maximum load, where the increase in stress due to decrease in the crosssectional area of the specimen becomes greater than the increase in the load-carrying ability of the metal due to strain hardening. üthis condition of instability leading to localized deformation is defined by the condition dp = 0. P = sa dp = s da + Ads = 0 This leads to the following relationship,.(1) Chapter 9-59

60 Instability in Tension - Necking From the constancy-of-volume relationship,.(2) So that at a point of tensile instability (combining Equation 1 and Equation 2) Graphical interpretation of necking criterion Chapter 9-60

61 Exercise If the true-stress-true-strain curve is given by the relationship:, where stress is in MPa, what is the ultimate tensile strength of the material? e = Hints: hardening exponent: K( ( ) n s T = K e T s T T = s ( 1+ e) ( + e) ln 1 n = 0.15 (some steels) to n = 0.5 (some coppers) true stress (F/A) true strain: ln(l/l o ) e T = n.how you get the relationship?? Answer: 698 MPa Chapter 9-61

62 Strain rate is defined as Engineering Strain rate e& Strain Rate & e = de dt d d( L -L0)/ L0 1 dl = e = = = dt dt L dt v L 0 L 0 True Strain rate e& dt d[ln( L dt L 1 L dl dt d T T = e 0 = = = ) v L Here v is crosshead velocity of the machine Chapter 9-62

63 Relation between Stress and Strain Rate A general relationship exists between tensile stress and strain rate, at constant strain and temperature: s = C (e& ) m T T where m is known as the strain-rate sensitivity Chapter 9-63

64 Exercise The parameters obtained from tensile tests of a commercially pure aluminum are as follows at a true strain of Determine the change in flow stress for a two order of magnitude change (say 1 to 100 s -1 ) in strain rate at each of the temperatures. Chapter 9-64

65 Variability in Material Properties Elastic modulus is material property Critical properties depend largely on sample flaws (defects, etc.). Large sample to sample variability. Statistics Mean x = n S x n n Standard Deviation s = n é ê S ê ë ( x - x ) i 2 ù n -1 ú ú û 1 2 where n is the number of data points Chapter 9-65

66 Design or Safety Factors Design uncertainties mean we do not push the limit. Factor of safety, N s working = s y N Often N is between 1.2 and 4 Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5. s 220, 000N p ( d 2 / 4) working 5 = s y N d = m = 6.7 cm 1045 plain carbon steel: sy = 310 MPa TS = 565 MPa F = 220,000N d Chapter 9-66 Lo

67 4 Strategies for Strengthening: 4: Cold Work (%CW) Room temperature deformation. Common forming operations change the cross sectional area: -Forging Ao die blank -Drawing Ao die die force force Ad % CW Ad tensile force = A Adapted from Fig. 11.8, Callister 7e. o - A o A d force Ao x 100 -Rolling Ao -Extrusion ram container billet container roll roll Ad die holder extrusion die Chapter 9-67 Ad

68 Dislocations During Cold Work Ti alloy after cold working: Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. From Fig. 5.10, Callister s Materials Science and Engineering, Adapted Version. 0.9 mm (Fig is courtesy of M.R. Plichta, Michigan Technological University.) Chapter 9-68

69 Result of Cold Work Dislocation density = total dislocation length unit volume Carefully grown single crystal à ca mm -2 Deforming sample increases density à mm -2 Heat treatment reduces density à mm -2 Yield stress increases as r d increases: s s y1 s y0 large hardening small hardening e Chapter 9-69

70 Effects of Stress at Dislocations From Fig. 10.5, Callister s Materials Science and Engineering, Adapted Version. Chapter 9-70

71 Impact of Cold Work As cold work is increased Yield strength (s y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. From Fig Callister s Materials Science and Engineering, Adapted Version. Chapter 9-71

72 What is the tensile strength & ductility after cold working? 2 2 pr - p = o r CW d x 100 = 2 pro % yield strength (MPa) MPa % Cold Work s y = 300MPa Cold Work Analysis Cu 35.6% tensile strength (MPa) MPa Cu % Cold Work TS = 340MPa Do =15.2mm 60 Copper Cold Work Dd =12.2mm ductility (%EL) % Cold Work %EL = 7% From Fig , Callister s Materials Science and Engineering, Adapted Version. (Fig 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 Chapter Metals, , p. 276 and 327.) % 0 0 Cu

73 s-e Behavior vs. Temperature Results for polycrystalline iron: From Fig Callister s Materials Science and Engineering, Adapted Version. Stress (MPa) C -100 C 25 C Strain s y and TS decrease with increasing test temperature. %EL increases with increasing test temperature. Why? Vacancies help dislocations move past obstacles. 2. vacancies replace atoms on the disl. half plane 3. disl. glides past obstacle 1. disl. trapped by obstacle obstacle Chapter 9-73

74 Effect of Heating (Annealing) After Cold Working q Annealing of the cold worked structure at high temperature softens the metal and reverts to a strainfree condition. q Annealing restores the ductility to a metal that has been severely strain hardened. q Annealing can be divided into three distinct processes: ü Recovery ü Recrystallization ü Grain growth Chapter 9-74

75 Recovery Annihilation reduces dislocation density. Scenario 1 Results from diffusion Scenario 2 extra half-plane of atoms 3. Climbed disl. can now move on new slip plane 2. grey atoms leave by vacancy diffusion allowing disl. to climb 1. dislocation blocked; can t move to the right atoms diffuse to regions of tension extra half-plane of atoms Dislocations annihilate and form a perfect atomic plane. 4. opposite dislocations meet and annihilate Obstacle dislocation t R Chapter 9-75

76 Recovery Various stages in the recovery of a plastically deformed material Chapter 9-76

77 Recrystallization (a) (b) Schematic illustration of nucleation event: (a) subgrain (SG) initially grows within Grain I and reaches the critical size which allows it to overcome the capillary force; (b) subsequently it can bulge into Grain II as a new strain free grain Chapter 9-77

78 Recrystallization New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. 0.6 mm 0.6 mm 33% cold worked brass New crystals nucleate after 3 sec. at 580 C. From Fig (a),(b), Callister s Materials Science and Engineering, Adapted Version. (Fig (a),(b) are courtesy of J.E. Burke, General Electric Company.) Chapter 9-78

79 Further Recrystallization All cold-worked grains are consumed. 0.6 mm 0.6 mm After 4 seconds After 8 seconds From Fig (c),(d), Callister s Materials Science and Engineering, Adapted Version. (Fig (c),(d) are courtesy of J.E. Burke, General Electric Company.) Chapter 9-79

80 Grain Growth At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced. 0.6 mm 0.6 mm From Fig (d),(e) Callister s Materials Science and Engineering, Adapted Version. (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.) After 8 s, 580ºC Empirical Relation: exponent typ. ~ 2 grain diam. at time t. d After 15 min, 580ºC n - d n o = Kt coefficient dependent on material and T. elapsed time Chapter 9-80

81 º T R = recrystallization temperature T R From Fig , Callister s Materials Science and Engineering, Adapted Version. º Chapter 9-81

82 Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property change 1. T m => T R» T m (K) 2. Due to diffusion à annealing timeà T R = f(t) shorter annealing time => higher T R 3. Higher %CW => lower T R 4. Pure metals lower T R due to dislocation movements Dislocation can move easily in pure metals => lower T R Chapter 9-82

83 Summary Recovery : The restoration of the physical properties of the cold worked metal without any observable change in microstructure. Strength is not affected. Recrystallization : The cold worked structure is replaced by a new set of strain-free grains due to migration of high angle grain boundaries. Hardness and strength decrease but ductility increases. Grain growth : Occurs at higher temperature where some of the recrystallized fine grains start to grow rapidly. Chapter 9-83