![Chapter 7 - 11
• Stronger - grain boundaries
pin deformations
• Slip planes & directions
(λ, φ) change from one
crystal to another.
• τR will vary from one
crystal to another.
• The crystal with the
largest τR 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].)
Slip Motion in Polycrystals
σ
300 µm](https://image.slidesharecdn.com/ch07pptscallister7e-150926091914-lva1-app6892/85/Ch07-ppts-material-11-320.jpg)
Ch07 ppts material
- 1. Chapter 7 - 1 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: Dislocations & Strengthening Mechanisms
- 2. Chapter 7 - 2 Dislocations & Materials Classes • Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding • Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors. + + + + +++ + + + + - - - ---- - - - • Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip. electron cloud ion cores + + + + +++++++ + + + + + + +++++++
- 3. Chapter 7 - 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. Chapter 7 - 4 Dislocation Motion • Dislocation moves along slip plane in slip direction perpendicular to dislocation line • Slip direction same direction as Burgers vector Edge dislocation Screw dislocation Adapted from Fig. 7.2, Callister 7e.
- 5. Chapter 7 - 5 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 Deformation Mechanisms Adapted from Fig. 7.6, Callister 7e.
- 6. Chapter 7 - 6 Stress and Dislocation Motion • Crystals slip due to a resolved shear stress, τR. • Applied tension can produce such a stress. slip plane normal, ns Resolved shear stress: τR =Fs/As slip direction AS τR τR FS slip direction Relation between σ and τR τR =FS /AS Fcos λ A/cos φ λ F FS φnS AS A Applied tensile stress: = F/Aσ slip direction F A F φλσ=τ coscosR
- 7. Chapter 7 - 7 • Condition for dislocation motion: CRSSτ>τR • Crystal orientation can make it easy or hard to move dislocation 10-4 GPa to 10-2 GPa typically φλσ=τ coscosR Critical Resolved Shear Stress τ maximum at λ = φ = 45º τR = 0 λ=90° σ τR = σ/2 λ=45° φ =45° σ τR = 0 φ=90° σ
- 8. Chapter 7 - 8 Single Crystal Slip Adapted from Fig. 7.8, Callister 7e. Adapted from Fig. 7.9, Callister 7e.
- 9. Chapter 7 - 9 Ex: Deformation of single crystal So the applied stress of 6500 psi will not cause the crystal to yield. τ=σcos λcos φ σ= 6500 psi λ=35° φ=60° τ=(6500 psi) (cos35o )(cos60o ) =(6500 psi) (0.41) τ=2662 psi <τcrss =3000 psi τcrss = 3000 psi a) Will the single crystal yield? b) If not, what stress is needed? σ = 6500 psi Adapted from Fig. 7.7, Callister 7e.
- 10. Chapter 7 - 10 Ex: Deformation of single crystal psi7325 41.0 psi3000 coscos crss == φλ τ =σ∴ y What stress is necessary (i.e., what is the yield stress, σy)? )41.0(coscospsi3000crss yy σ=φλσ==τ psi7325=σ≥σ y So for deformation to occur the applied stress must be greater than or equal to the yield stress
- 11. Chapter 7 - 11 • Stronger - grain boundaries pin deformations • Slip planes & directions (λ, φ) change from one crystal to another. • τR will vary from one crystal to another. • The crystal with the largest τR 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].) Slip Motion in Polycrystals σ 300 µm
- 12. Chapter 7 - 12 • Can be induced by rolling a polycrystalline metal - before rolling 235 µm - isotropic since grains are approx. spherical & randomly oriented. - after rolling - anisotropic since rolling affects grain orientation and shape. rolling direction 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.) Anisotropy in σy
- 13. Chapter 7 - 13 side view 1. Cylinder of Tantalum machined from a rolled plate: rollingdirection 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
- 14. Chapter 7 - 14 4 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: 21/ yoyield dk − +σ=σ Adapted from Fig. 7.14, Callister 7e. (Fig. 7.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)
- 15. Chapter 7 - 15 • Impurity atoms distort the lattice & generate stress. • Stress can produce a barrier to dislocation motion. 4 Strategies for Strengthening: 2: Solid Solutions • Smaller substitutional impurity Impurity generates local stress at A and B that opposes dislocation motion to the right. A B • Larger substitutional impurity Impurity generates local stress at C and D that opposes dislocation motion to the right. C D
- 16. Chapter 7 - 16 Stress Concentration at Dislocations Adapted from Fig. 7.4, Callister 7e.
- 17. Chapter 7 - 17 Strengthening by Alloying • small impurities tend to concentrate at dislocations • reduce mobility of dislocation ∴ increase strength Adapted from Fig. 7.17, Callister 7e.
- 18. Chapter 7 - 18 Strengthening by alloying • large impurities concentrate at dislocations on low density side Adapted from Fig. 7.18, Callister 7e.
- 19. Chapter 7 - 19 Ex: Solid Solution Strengthening in Copper • Tensile strength & yield strength increase with wt% Ni. • Empirical relation: • Alloying increases σy and TS. 21/ y C~σ Adapted from Fig. 7.16 (a) and (b), Callister 7e. Tensilestrength(MPa) wt.% Ni, (Concentration C) 200 300 400 0 10 20 30 40 50 Yieldstrength(MPa) wt.%Ni, (Concentration C) 60 120 180 0 10 20 30 40 50
- 20. Chapter 7 - 20 • Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). • Result: S ~y 1 σ 4 Strategies for Strengthening: 3: Precipitation Strengthening Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation “advances” but precipitates act as “pinning” sites with S.spacing Side View precipitate Top View Slipped part of slip plane Unslipped part of slip plane S spacing
- 21. Chapter 7 - 21 • Internal wing structure on Boeing 767 • 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.) 1.5µm Application: Precipitation Strengthening Adapted from chapter- opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
- 22. Chapter 7 - 22 4 Strategies for Strengthening: 4: Cold Work (%CW) • Room temperature deformation. • Common forming operations change the cross sectional area: Adapted from Fig. 11.8, Callister 7e. -Forging Ao Ad force die blank force-Drawing tensile force Ao Addie die -Extrusion ram billet container container force die holder die Ao Adextrusion 100x% o do A AA CW − = -Rolling roll Ao Ad roll
- 23. Chapter 7 - 23 • 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.) Dislocations During Cold Work 0.9 µm
- 24. Chapter 7 - 24 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 ρd increases: total dislocation length unit volume large hardening small hardening σ ε σy0 σy1
- 25. Chapter 7 - 25 Effects of Stress at Dislocations Adapted from Fig. 7.5, Callister 7e.
- 26. Chapter 7 - 26 Impact of Cold Work Adapted from Fig. 7.20, Callister 7e. • Yield strength (σy) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases. As cold work is increased
- 27. Chapter 7 - 27 • What is the tensile strength & ductility after cold working? Adapted from Fig. 7.19, Callister 7e. (Fig. 7.19 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.) %6.35100x% 2 22 = π π−π = o do r rr CW Cold Work Analysis % Cold Work 100 300 500 700 Cu 200 40 60 yield strength (MPa) σy = 300MPa 300MPa % Cold Work tensile strength (MPa) 200 Cu 0 400 600 800 20 40 60 ductility (%EL) % Cold Work 20 40 60 20 40 6000 Cu Do=15.2mm Cold Work Dd =12.2mm Copper 340MPa TS = 340MPa 7% %EL = 7%
- 28. Chapter 7 - 28 • Results for polycrystalline iron: • σy and TS decrease with increasing test temperature. • %EL increases with increasing test temperature. • Why? Vacancies help dislocations move past obstacles. Adapted from Fig. 6.14, Callister 7e. σ-ε Behavior vs. Temperature 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) 0 0.1 0.2 0.3 0.4 0.5 1. disl. trapped by obstacle obstacle
- 29. Chapter 7 - 29 • 1 hour treatment at Tanneal... decreases TS and increases %EL. • Effects of cold work are reversed! • 3 Annealing stages to discuss... Adapted from Fig. 7.22, Callister 7e. (Fig. 7.22 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.) Effect of Heating After %CWtensilestrength(MPa) ductility(%EL) tensile strength ductility Recovery Recrystallization Grain Growth 600 300 400 500 60 50 40 30 20 annealing temperature (ºC) 200100 300 400 500 600 700
- 30. Chapter 7 - 30 Annihilation reduces dislocation density. Recovery • Scenario 1 Results from diffusion • Scenario 2 4. opposite dislocations meet and annihilate Dislocations annihilate and form a perfect atomic plane. extra half-plane of atoms extra half-plane of atoms atoms diffuse to regions of tension 2. grey atoms leave by vacancy diffusion allowing disl. to “climb” τR 1. dislocation blocked; can’t move to the right Obstacle dislocation 3. “Climbed” disl. can now move on new slip plane
- 31. Chapter 7 - 31 • New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. Adapted from Fig. 7.21 (a),(b), Callister 7e. (Fig. 7.21 (a),(b) are courtesy of J.E. Burke, General Electric Company.) 33% cold worked brass New crystals nucleate after 3 sec. at 580°C. 0.6 mm 0.6 mm Recrystallization
- 32. Chapter 7 - 32 • All cold-worked grains are consumed. Adapted from Fig. 7.21 (c),(d), Callister 7e. (Fig. 7.21 (c),(d) are courtesy of J.E. Burke, General Electric Company.) After 4 seconds After 8 seconds 0.6 mm0.6 mm Further Recrystallization
- 33. Chapter 7 - 33 • At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced. After 8 s, 580ºC After 15 min, 580ºC 0.6 mm 0.6 mm Adapted from Fig. 7.21 (d),(e), Callister 7e. (Fig. 7.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.) Grain Growth • Empirical Relation: Ktdd n o n =− elapsed time coefficient dependent on material and T. grain diam. at time t. exponent typ. ~ 2 Ostwald Ripening
- 34. Chapter 7 - 34 TR Adapted from Fig. 7.22, Callister 7e. º º TR = recrystallization temperature
- 35. Chapter 7 - 35 Recrystallization Temperature, TR TR = recrystallization temperature = point of highest rate of property change 1. Tm => TR ≈ 0.3-0.6 Tm (K) 2. Due to diffusion annealing time TR = f(t) shorter annealing time => higher TR 3. Higher %CW => lower TR – strain hardening 4. Pure metals lower TR due to dislocation movements • Easier to move in pure metals => lower TR
- 36. Chapter 7 - 36 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.
- 37. Chapter 7 - 37 Coldwork Calculations Solution If we directly draw to the final diameter what happens? %843100x 400 300 1100x 4 4 1 1001100x% 2 2 2 . . . D D x A A A AA CW o f o f o fo = −= π π −= −= − = Do = 0.40 in Brass Cold Work Df = 0.30 in
- 38. Chapter 7 - 38 Coldwork Calc Solution: Cont. • For %CW = 43.8% Adapted from Fig. 7.19, Callister 7e. 540420 σy = 420 MPa – TS = 540 MPa > 380 MPa 6 – %EL = 6 < 15 • This doesn’t satisfy criteria…… what can we do?
- 39. Chapter 7 - 39 Coldwork Calc Solution: Cont. Adapted from Fig. 7.19, Callister 7e. 380 12 15 27 For %EL < 15 For TS > 380 MPa > 12 %CW < 27 %CW ∴ our working range is limited to %CW = 12-27
- 40. Chapter 7 - 40 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 2nd cold draw)? – must be calculated as follows: 100 % 11001% 2 02 2 2 2 02 2 2 CW D D x D D CW ff =−⇒ −= 50 02 2 100 % 1 . f CW D D −= 50 2 02 100 % 1 . f CW D D − = ⇒ m3350 100 20 1300 50 021 ..DD . f = −==Intermediate diameter =
- 41. Chapter 7 - 41 Coldwork Calculations Solution Summary: 1. Cold work D01= 0.40 in Df1 = 0.335 m 2. Anneal above D02 = Df1 3. Cold work D02= 0.335 in Df2 =0.30 m Therefore, meets all requirements 20100 3350 30 1% 2 2 = −= x . . CW 24% MPa400 MPa340 = = =σ EL TS y ⇒ %CW1 = 1− 0.335 0.4 2 x 100 =30 Fig 7.19
- 42. Chapter 7 - 42 Rate of Recrystallization • Hot work above TR • Cold work below TR • Smaller grains – stronger at low temperature – weaker at high temperature t/R T B Ct kT E RtR 1:note log logloglog 0 = += −=−= RT 1 log t start finish 50%
- 43. Chapter 7 - 43 • 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
- 44. Chapter 7 - 44 Core Problems: Self-help Problems: ANNOUNCEMENTS Reading:
Editor's Notes
- #4: So we saw that above the yield stress plastic deformation occurs. But how? In a perfect single crystal for this to occur every bond connecting tow planes would have to break at once! Large energy requirement Now rather than entire plane of bonds needing to be broken at once, only the bonds along dislocation line are broken at once.
- #17: Don’t move past one another – hardens material
- #25: Again it propagates through til reaches the edge
- #41: So after the cold draw & anneal D02=0.335m