Lecture 09 Dislocations & Strengthening Mechanisms

Transcription

1 Lecture 09 Dislocations & Strengthening Mechanisms Chapter 7-1

2 Dislocations & Strengthening Mechanisms ISSUES TO ADDRESS... Why are dislocations observed primarily in metals and alloys? How are strength and dislocation motion related? How do we increase strength? How can heating change strength and other properties? Chapter 7-2

3 Mechanical Behavior & Structure Controlled by Chemical bond Chemistry Controlled by Processing via defects, crystal structure and microstricture Chapter 7-3

4 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 Chapter 7-4

5 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! Chapter 7-5

6 Dislocation Motion (Cont d) Caterpillar b Chapter 7-6

7 Dislocation Motion Dislocation moves along slip plane in slip direction perpendicular p to dislocation line Slip direction same direction as Burgers vector Edge dislocation Screw dislocation Chapter 7-7

8 Comparison of Dislocation Motion Edge vs Screw Dislocations Chapter 7-8

9 Slip System Deformation Mechanisms 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 Chapter 7-9

10 Deformation Mechanisms Common Slip Systems Chapter 7-10

11 Slip Systems in Common Metals After Dieter, Mechanical Metallurgy (1990) Chapter 7-11

12 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 F S t R A S Relation between s and t R t R =F S /A S Fcos l F l F S n S A/cos f f A S A S Schmid s Law R cos cos Chapter 7-12

13 Critical Resolved Shear Stress Condition for dislocation motion: Crystal orientation can make it easy or hard to move dislocation cos cos R CRSS R s s s typically 10-4 GPa to 10-2 GPa t R = 0 l =90 l =45 f =45 t R = s /2 t R = 0 f =90 maximum at = = 45º Chapter 7-13

14 Single Crystal Slip Single crystal wire Chapter 7-14

15 =60 Ex: Deformation of single crystal =35 a) Will the single crystal yield? b) If not, what stress is needed? crss = 3000 psi cos cos 6500 psi (6500 psi) (cos35 )(cos60 ) (6500 psi) (0.41) 2662 psi crss 3000 psi = 6500 psi So the applied stress of 6500 psi will not cause the crystal to yield. Chapter 7-15

16 Ex 1: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, y )? crss 3000 psi y cos cos y (0.41) 3000 psi crss y cos cos psi So for deformation to occur the applied stress must be greater than or equal to the yield stress y 7325 psi Chapter 7-16

17 Ex 2: Deformation of Single crystal After Dieter, Mechanical Metallurgy (1990) Chapter 7-17

18 Slip Motion in Polycrystals Stronger - grain boundaries pin deformations Slip planes & directions change from one crystal tlto another. R will vary from one crystal to another. The crystal with the largest R yields first. Other (less favorably oriented) crystals yield later. 300 mm Chapter 7-18

19 Anisotropy in y Can be induced by rolling a polycrystalline metal - before rolling - after rolling 235 mm rolling direction - isotropic - anisotropic since grains are approx. spherical & randomly oriented. since rolling affects grain orientation and shape. Chapter 7-19

20 Anisotropy in Deformation 1. Cylinder of Tantalum machined from a rolled plate: 2. Fire cylinder at a target. 3. Deformed cylinder side view direction rolling d The noncircular end view shows anisotropic deformation of rolled material. end view plate thickness direction Chapter 7-20

21 Deformation by Mech l Twinning Twins also re-orient slip planes and contribute to dislocation slip indirectly Twinning is an alternate (plastic) deformation mechanism observed when the strain rate is very high or at low T (dislocation slip is suppressed) It is a strain-relief mechanism. Very common in BCC and HCP metals when slip is restricted! Chapter 7 -

22 Mech l Twinning vs. Dislocation Slip Homogeneous (Twin Band) After Dieter. Mech l Metallurgy Chapter 7 -

23 Strategies for Strengthening: 1. Rd Reduce Grain Size Grain boundaries are barriers to slip. Barrier "strength" increases with increasing angle of misorientation. Smaller grain size (d): more barriers to slip. Hall-Petch Equation: yield o k y d 1/ 2 Chapter 7 -

24 Strategies for Strengthening: 1. Reduce Grain Size (Cont d) Very difficult to synthesize polycrystalline materials with d<100 nm yield o k y d 1/ 2 Chapter 7 -

25 Strategies for Strengthening: 1. Reduce Grain Size (Cont d) Primary mechanism for strengthening by grain boundaries Chapter 7-25

26 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 7-26

27 Stress Concentration at Dislocations What makes a dislocation move & interact? The stress field Around the dislocation interacts with the applied stress (via R ) and the Strain field of structural imperfections such as grain boundaries and solute atoms Chapter 7 -

28 Strengthening by Alloying Small impurities tend to concentrate at dislocations Reduce mobility of dislocation; increase strength Chapter 7-28

29 Strengthening by alloying (Cont d) Large impurities concentrate at dislocations on low density side Chapter 7 -

30 Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. 180 Pa) Tensile st trength (M wt.% Ni, (Concentration C) wt.%ni, (Concentration C) ength (MP Pa) Yield str Empirical relation: y ~ C 1/ 2 Alloying increases s y and TS. Chapter 7 -

31 Dislocation Sources: Frank-Read Sources Chapter 7-31

32 Strategies for Strengthening: 3P 3. Precipitation iitti Strengthening th 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 it t 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: y~ y 1 S The smaller the S, the larger the dislocation line needs to bow out, the more surface energy needs to be spend! Chapter 7 -

33 Application: Precipitation Strengthening Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates formed by alloying. Black spots are the ppts. 1.5mm Chapter 7-33

34 Strategies for Strengthening: 4CldW 4. Cold Work k(%cw CW) Room temperature deformation. Common forming operations change the cross sectional area: -Forging force -Rolling Ao die blank -Drawing Ao die die force Ad Ad tensile force % CW A A o A d A o force Ao Ao -Extrusion ram x 100 container billet container roll roll Ad die holder extrusion die Chapter 7 - Ad

35 Dislocations During Cold Work Ti alloy after cold working: Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult mm Strain hardening due to dislocation-dislocation interactions cutting through a forest of dislocations i Chapter 7 -

36 Dislocations During Cold Work Edge and screw dislocation interaction leads to strain hardening After Dieter. Mech l Metallurgy Chapter 7-36

37 Dislocation density = Result of Cold Work total dislocation length unit volume Carefully grown single crystal ca mm -2 Deforming sample increases density mm -2 Heat treatment reduces density mm -2 Yield stress increases as r d increases: y1 y0 large hardening small hardening Chapter 7 -

38 Effects of Stress at Dislocations Chapter 7 -

39 Impact of Cold Work As cold work is increased: Yield strength ( y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. Chapter 7 -

40 What is the tensile strength & ductility after cold working? 2 2 r o r CW d x ro % Cold Work Analysis 35.6% Do =15.2mm Copper Cold Work Dd =12.2mm 700 yield strength (MPa) 800 tensile strength (MPa) 60 ductility (%EL) MPa Cu MPa Cu Cu 100 7% % Cold Work % Cold Work % Cold Work y = 300MPa TS = 340MPa %EL= 7% Chapter 7 -

41 - Behavior vs. Temperature Results for polycrystalline iron: Stress (M MPa) C -100C 25C Strain y and TS decrease with increasing test temperature. %EL increases with increasing test temperature. Why? Vacancies help dislocations 2. vacancies move past obstacles. replace atoms on the disl. half plane 1. disl. trapped 3. disl. glides past obstacle by obstacle obstacle Chapter 7 -

42 Effect of Heating After %CW 1 hour treatment at T anneal decreases TS & increases %EL. Effects of cold work are reversed! tens sile streng gth (MPa) Annealing temperature (ºC) tensile strength ductility ductility (%EL) 3 Annealing stages to discuss... Chapter 7 -

43 Recovery Annihilation reduces dislocation density. Scenario 1 Results from diffusion Scenario 2 extra half-plane of atoms atoms diffuse to regions of tension extra half-plane of atoms Dislocations annihilate and form a perfect atomic plane. 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 1. dislocation blocked; Obstacle dislocation can t move to the right t R Chapter 7 -

44 Recrystallization New grains are formed that: - Have a small dislocation density - Are small - Consume cold-worked grains. 06mm mm % cold New crystals worked nucleate after brass 3 sec. at 580C. Chapter 7 -

45 Further Recrystallization All cold-worked grains are consumed. 0.6 mm 0.6 mm After 4 seconds After 8 seconds Chapter 7 -

46 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 After 8 s, 580ºC After 15 min, 580ºC Empirical Relation: Eexponent typ. ~ 2 Grain diam. at time t. d n d n o Kt Coefficient dependent on material and T. Elapsed time Ostwald Ripening Chapter 7 -

47 Grain Growth º T R = Recrystallization temperature T R Cold work state, is A metastable state. The energy stored Due to cold work is the driving for Recrystallization. Recrystallized grain are strain free. º Chapter 7 -

48 Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property change 1. T m => T R 0306T m (K) 2. Due to diffusion annealing time T R = f(t) 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 Chapter 7 -

49 Recrystallization Temperature, T R Chapter 7-49

50 Stages of Recrystallization 33% CW Brass 3 s 580 o C Initial Recryst. Chapter 7 -

51 Stages of Recrystallization 4 s 580 o C Recryst. Cont d 8 s 580 o C Recryst. Complete Chapter 7-51

52 Stages of Recrystallization 15 min 580 o C Grain Growth 10 min 700 o C Grain Growth Chapter 7-52

53 Cold Work 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 maybe accomplished. Chapter 7 -

54 Cold Work Calculations Solution If we directly draw to the final diameter what happens? Brass Cold Work Do = 0.40 in Df = 0.30 in Ao A A f f % CW x x A o A o D 2 f 2 D o x x % Chapter 7 -

55 Coldwork Calc Solution: Cont For %CW = 43.8% y = 420 MPa TS = 540 MPa > 380 MPa %EL =6 <15 This doesn t satisfy criteria what can we do? Chapter 7 -

56 Coldwork Calc Solution: Cont For TS > 380 MPa >12%CW For %EL < 15 < 27 %CW Our working range is limited to %CW = Chapter 7-56

57 Coldwork Calc Soln: Recrystallization Cold draw-anneal-cold draw again For objective we need a cold work of %CW We ll use %CW = 20 Diameter after first cold draw (before 2 nd cold draw)? must tbe calculated l as follows: 2 2 D 2 D 2 % CW % CW f 1 f x D D % CW D 1 f 2 D Df 2 D Intermediate t diameter = 1 % CW D 1 D f m 100 Chapter 7-57

58 Cold Work Calculations: Solution Summary: 1. Cold work D 01 = 0.40 in D f1 = m %CW x Anneal above D 02 = D f1 3. Cold work D 02 = in D f2 =0.30 m %. CW x Therefore, meets all requirements y TS % EL 340 MPa 400 MPa 24 Chapter 7 -

59 Rate of Recrystallization E logr logt logr 0 start 50% kt B logt C 1 T finish T R note : R 1/ t Hot work above T R log t Cold work below T R Smaller grains stronger at low temperature weaker at high temperature Chapter 7-59

60 Summary 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. Chapter 7-60

61 Homework Problems: 7.9, 7.12, 7.13, 7.15, 7.17, , 7.29, 7.31, 7.32, 7.36, 7.41, 7.D3, 7.D4. Due date:. Chapter 7-61

62 Advanced Topics in Dislocation Phenomena Chapter 7-62

63 Slip in a Perfect Lattice First hint as to why there should be crystalline defects involved!!! Chapter 7-63

64 Slip in a Perfect Lattice Chapter 7-64

65 Slip in a Perfect Lattice Chapter 7-65

66 Atomic movements near dislocation in slip (plastic deformation) Chapter 7-66

67 Why ceramics do not exhibit plastic deformation at ambient Temperatures? Chapter 7-67

68 Why ceramics do not exhibit plastic deformation at ambient Temperatures? Chapter 7-68

69 Why ceramics do not exhibit plastic deformation at ambient Temperatures? Chapter 7-69

70 Stages of Dislocation slip in FCC Metals Chapter 7-70

71 Stages of Dislocation slip in FCC Metals Chapter 7-71

72 Stages of Dislocation slip in FCC Metals Chapter 7-72

73 Stresses around an Edge Dislocation Dislocation line passes through O and extends out of the plane of the slide Chapter 7-73

74 Stresses around an Edge Dislocation Chapter 7-74

75 Forces on Dislocations Chapter 7-75

76 Forces on Dislocations Chapter 7-76

77 Dislocation Sources Chapter 7-77