Chapter 7: Dislocations & Strengthening Mechanisms

1. Chapter 7: 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?

2. 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. - very hard, Tm>3550C -directional bonding • Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors. - non-directional bonding - - - + + + + - - - - + + + - - - + + + + Dislocation types: edge and screw PD corresponds to the motion of large numbers of dislocations

3. Dislocation Motion Plastic deformation mechanisms are 1. slipping 2. twinning 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!

4. Dislocation Motion Slip is the process by which plastic deformation is produced by dislocation motion • Dislocation moves along slip plane in slip direction perpendicular to dislocation line( slip system) • Slip direction have the same direction as Burgers vector Edge dislocation Screw dislocation

5. Slipping Mechanism Slip System • Slip plane – plane along which the dis. line traverses • plane allowing easiest slippage • 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

6. The slip planes are usually the most densely packed planes. Slip is favored on closed packed planes since a lower shear stress for atomic displacement is required than for less densely packed planes. Slip in the closed packed direction is also favored since less energy is required to move the atoms from one position to another if the atoms are closer together.

7. Slip: The process by which plastic deformation is produced by dislocation motion The slip system depends on the crystal structure of the metal and is such that the atomic distortion that accompanies the motion of a dislocation is a minimum dislocation density: (The number of dislocations).is expressed as the total dislocation length per unit volume, or equivalently, the number of dislocations that intersect a unit area of a random section. The units of dislocation density are millimeters of dislocation per cubic millimeter or just per square millimeter. Dislocation densities as low as 10 3 mm -2 All metals and alloys contain some dislocations that were introduced during 1.solidification, 2.during plastic deformation, and as a consequence of 3.thermal stresses that result from rapid cooling.

8. 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 • Crystals slip due to a resolved shear stress, tR. • Applied tension can produce such a stress.

9. 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: Critical resolved shear stress: is the minimum shear stress required to initiate slip and is a property of the material that determine when yielding occurs • Crystal orientation can make it easy or hard to move dislocation  maximum at  =  = 45º

10. For The crystal ordinarily fractures rather than deforming plastically

11. Single Crystal Slip

12. Ex: Deformation of single crystal a) Will the single crystal yield? b) If not, what stress is needed? So the applied stress of 6500 psi will not cause the crystal to yield. =60° crss = 3000 psi =35°  = 6500 psi

13. 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)?

14. Slip Motion in Polycrystals s 300 mm • Stronger metals- grain boundaries pin deformations • Slip planes & directions (l, f) change from one crystal to another. • tRwill vary from one crystal to another. • The crystal with the largest tR yields first. • Other (less favorably oriented) crystals yield later.

15. Anisotropy in sy - after rolling rolling direction - anisotropic since rolling affects grain orientation and shape. Isotropic. Having identical values of a property in all crystallographic directions. • Can be induced by rolling a polycrystalline metal - before rolling 235 mm - isotropic since grains are approx. spherical & randomly oriented.

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

17. Twinning mechanism • A part of the atomic lattice is deformed so that is forms a mirror image of the un-deformed lattice next to it. • Twinning plane: is the plane between the un-deformed and deformed parts of the metal lattice

18. Comparison between slip and twinning mechanisms

19. Mechanisms of strengthening in metals • The ability of a metal to plastically deform depends on the ability of dislocations to move. • ductility is sacrificed when an alloy is strengthened • restricting or hindering dislocation motion renders a material harder and stronger. • strengthening mechanisms for single phase metals, by grain size reduction, solid-solution alloying, and strain hardening(cold work).

20. 4 Strategies for Strengthening: 1: Reduce Grain Size • Grain boundaries are barriers to slip. Change direction Discontinuity of slip planes • Barrier "strength" increases with Increasing angle of mis-orientation. • Smaller grain size: more barriers to slip. • Hall-Petch Equation: d = average grain diameter σy - yield strength σ0, ky are constant for particular material

21. Materials are a fine grained materials (one that has small grains) and a coarse grained. Fine grained material is harder and stronger than coarse grained. Grain size may be regulated by the rate of solidification from the liquid phase, and also by plastic deformation followed by an appropriate heat treatment ( depends on time and temperature). The greater tem. And time the greater grain size.

22. 4 Strategies for Strengthening: 2: 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 stress. • Stress can produce a barrier to dislocation motion.

23. Stress Concentration at Dislocations

24. Strengthening by Alloying • small impurities tend to concentrate at dislocations • reduce mobility of dislocation  increase strength

25. Strengthening by alloying • large impurities concentrate at dislocations on low density side

26. 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. • Empirical relation: • Alloying increases sy and TS.

27. force -Forging -Rolling roll die A d A A A o blank o d 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 4 Strategies for Strengthening: 3: Cold Work (%CW) Strain hardening: a ductile metal becomes harder and stronger as it is plastically deformed( the effect of CW may be removed by annealing HT) • Room temperature deformation. • Common forming operations change the cross sectional area:

28. Dislocations During Cold Work 0.9 mm • Ti alloy after cold working: • Dislocations entangle with one another during cold work. • Dislocation motion becomes more difficult.

29. total dislocation length unit volume s large hardening s y1 s small hardening y0 e 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 rd increases:

30. Effects of Stress at Dislocations

31. Impact of Cold Work As cold work is increased • Yield strength (sy) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases.

32. Cold Work Analysis Copper Cold Work D =15.2mm D =12.2mm o d tensile strength (MPa) yield strength (MPa) ductility (%EL) 60 800 700 40 600 500 Cu 300MPa 300 Cu 20 400 340MPa Cu 7% 100 200 0 0 20 40 60 0 20 40 60 0 20 40 60 % Cold Work % Cold Work % Cold Work s = 300MPa TS = 340MPa %EL = 7% y • What is the tensile strength & ductility after cold working?

33. s-e Behavior vs. Temperature 800 -200C 600 -100C Stress (MPa) 400 25C 200 0 0 0.1 0.2 0.3 0.4 0.5 Strain 3 . disl. glides past obstacle 2. vacancies replace atoms on the obstacle disl. half 1. disl. trapped plane by obstacle • Results for polycrystalline iron: • sy and TS decrease with increasing test temperature. • %ELincreases with increasing test temperature. • Why? Vacancies help dislocations move past obstacles.

34. 4 Strategies for Strengthening: 4: 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:

35. Application:Precipitation Strengthening 1.5mm • Internal wing structure on Boeing 767 • Aluminum is strengthened with precipitates formed by alloying.

36. Note: The strengthening effects due to grain size reduction and strain hardening can be eliminated or at least reduced by an elevated temperature heat treatment. Conversely, solid solution strengthening is unaffected by heat treatment

37. Recovery, Recrystallization, and Grain Growth • These properties and structures may revert back to the precold-worked states by appropriate heat treatment (sometimes termed an annealing treatment). Such restoration results from two different processes that occur at elevated temperatures: recovery and recrystallization, which may be followed by graingrowth.

38. Effect of Heating After %CW 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 reversed! • 3 Annealing stages to discuss...

39. 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 Annihilation reduces dislocation density. • Scenario 2

40. 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 a small dislocation density -- are small -- consume cold-worked grains.

41. Further Recrystallization 0.6 mm 0.6 mm After 8 seconds After 4 seconds • All cold-worked grains are consumed.

42. TR º TR = recrystallization temperature º

43. Recrystallization Temperature, TR TR= recrystallization temperature = point of highest rate of property change • Tm => TR 0.3-0.6 Tm (K) • Due to diffusion  annealing time TR = f(t) shorter annealing time => higher TR • Higher %CW => lower TR– strain hardening • Pure metals lower TR due to dislocation movements • Easier to move in pure metals => lower TR

44. Grain Growth • grain growth: After recrystallization is complete, the strain-free grains will continue to grow if the metal specimen is left at the elevated temperature thisphenomenon is called grain growth. • Grain growth does not need to be preceded by recovery and recrystallization; it may • occur in all polycrystalline materials, metals and ceramics alike • As grains increase in size, the total boundary area decreases, yielding an attendant • reduction in the total energy; this is the driving force for grain growth. • Grain growth occurs by the migration of grain boundaries. Obviously, not all • grains can enlarge, but large ones grow at the expense of small ones that shrink. • The directions of boundary movement and atomic motion are opposite to each • other growth via atomic diffusion

45. 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, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced.

46. 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.

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

48. 420 540 6 Coldwork Calc Solution: Cont. • For %CW = 43.8% Adapted from Fig. 7.19, Callister 7e. • y = 420 MPa • TS = 540 MPa > 380 MPa • %EL = 6 < 15 • This doesn’t satisfy criteria…… what can we do?

49. 15 380 27 12 > 12 %CW < 27 %CW Coldwork Calc Solution: Cont. ForTS > 380 MPa For%EL < 15  our working range is limited to %CW = 12-27

50.  Intermediate diameter = 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: