Chapter 7: Dislocations and Strengthening Mechanisms in Metal

1. Chapter 7: Dislocations and Strengthening Mechanisms in Metal 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. -directional (angular) bonding • Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors. - - - + + + + - - - - + + + - - - + + + +

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!

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

5. Deformation Mechanisms Slip System • Slip plane - plane allowing easiest slippage • Wide interplanar spacing - 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. 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.

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: • Crystal orientation can make it easy or hard to move dislocation  maximum at  =  = 45º

8. Single Crystal Slip Slip in Zinc single crystalspecimen after tensioned force Small step of slip on the surface parallel to one another

9. Ex: Deformation of single crystal a) Will the single crystal yield? b) If not, what stress is needed? So the applied stress of 7000 psi will not cause the crystal to yield. =60° crss = 3000 psi =35°  = 7000 psi

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 • Stronger - 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.

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

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

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. The fine-grained material is harder and stronger than the coarse-grained material.

15. 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. Impurity generates local stress at C and D that opposes dislocation motion. • Impurity atoms distort the lattice & generate stress. • Stress can produce a barrier to dislocation motion.

16. Stress Concentration at Dislocations

17. Strengthening by Alloying • Small impurities tend to concentrate at dislocations • Reduce mobility of dislocation  increase strength

18. Strengthening by Alloying • large impurities concentrate at dislocations on low density side • Partial cancellation of impurity-dislocation lattice strains

19. 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.

20. 4 Strategies for Strengthening: 3: Precipitation Strengthening precipitate Side View Unslipped part of slip plane Top View S 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 spacingS. . • Result:

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

22. force roll die A d A A A o blank o d roll force 4 Strategies for Strengthening: 4: Cold Work (%CW) • • Sometimes called “Work Hardening” • Room temperature deformation. • • Common forming operations change the cross sectional area: -Forging -Rolling

23. die A d A tensile o force die A o container die holder force ram A billet extrusion d die container 4 Strategies for Strengthening: 4: Cold Work (%CW) -Drawing -Extrusion

24. 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.

25. total dislocation length unit volume s large hardening s y1 s small hardening y0 e Result of Cold Work Dislocation density (ρd)= • 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:

26. Effects of Stress at Dislocations

27. Impact of Cold Work As cold work is increased • Yield strength (σy) increases. • Tensile strength (TS) increases. • Ductility (%EL or %RA) decreases.

28. 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 300 Cu 20 400 Cu 100 300MPa 200 0 0 20 40 60 0 20 40 60 0 20 40 60 340MPa % Cold Work % Cold Work % Cold Work %EL = 7% 7% s = 300MPa y TS = 340MPa • What is the tensile strength & ductility after cold working?

29. 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.

30. 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...

31. 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

32. 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.

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

34. 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. Ostwald Ripening • At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced.

35. TR º TR = recrystallization temperature º

36. Recrystallization Temperature, TR TR= recrystallization temperature = point of highest rate of property change • Tm => TR 0.3-0.6 Tm (K) Temperature that recrystallization just reach completion in 1 hour • 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

37. 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.

38. 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?

39. 420 540 6 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?

40. 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

41.  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:

42. Coldwork Calculations Solution Summary: • Cold work D01= 0.40 in  Df1 = 0.335 m • Anneal above D02 = Df1 • Cold work D02= 0.335 in  Df2 =0.30 m Therefore, meets all requirements Fig 7.19 

43. 50% start finish log t Rate of Recrystallization • Hot work above TR • Cold work  below TR • Smaller grains • stronger at low temperature • weaker at high temperature

44. 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.