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Chapter 7:. Dislocations & Strengthening Mechanisms. WHY STUDY Dislocations and Strengthening Mechanisms?. Understand underlying mechanisms of the techniques used to strengthen and harden metals and their alloys. Design and tailor mechanical properties of materials.

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    1. Chapter 7: Dislocations & Strengthening Mechanisms Dr. Mohammad Abuhaiba, PE

    2. WHY STUDY Dislocations and Strengthening Mechanisms? • Understand underlying mechanisms of the techniques used to strengthen and harden metals and their alloys. • Design and tailor mechanical properties of materials. Dr. Mohammad Abuhaiba, PE

    3. WHY STUDY Dislocations and Strengthening Mechanisms? • Strengthening mechanisms will be applied to the development of mechanical properties for steel alloys. • Why heat treating a deformed metal alloy makes it softer and more ductile and produces changes in its microstructure. Dr. Mohammad Abuhaiba, PE

    4. Learning Objectives Describe edge and screw dislocation motion. Describe how plastic deformation occurs by motion of edge & screw dislocations in response to applied shear stresses. Define slip system. Describe how grain structure of a polycrystalline metal is altered when it is plastically deformed. Explain how grain boundaries impede dislocation motion and why a metal having small grains is stronger than one having large grains. Dr. Mohammad Abuhaiba, PE

    5. Learning Objectives Describe & explain solid-solution strengthening for substitutional impurity atoms in terms of lattice strain interactions with dislocations. Describe and explain phenomenon of strain hardening in terms of dislocations and strain field interactions. Describe recrystallization in terms of both alteration of microstructure and mechanical characteristics of the material. Describe phenomenon of grain growth from both macroscopic and atomic perspectives. Dr. Mohammad Abuhaiba, PE

    6. 7.2 BASIC CONCEPTSDislocation Motion Cubic & hexagonal metals - plastic deformation is by plastic shear or slip where one plane of atoms slides over adjacent plane by dislocations motion. Dr. Mohammad Abuhaiba, PE

    7. 7.3 Characteristics of Dislocations • When metals are plastically deformed: • Around 5% of deformation energy is retained internally • remainder is dissipated as heat • The major portion of stored energy is strain energy associated with dislocations. Dr. Mohammad Abuhaiba, PE

    8. 7.3 Characteristics of Dislocations Figure 7.4: some atomic lattice distortion exists around the dislocation line because of presence of extra half-plane of atoms. Dr. Mohammad Abuhaiba, PE

    9. 7.3 Characteristics of Dislocations • Atoms immediately above and adjacent to dislocation line are squeezed together. As a result, these atoms may be thought of as experiencing a compressive strain relative to atoms positioned in the perfect crystal and far removed from the dislocation. • Directly below the half-plane, the lattice atoms sustain an imposed tensile strain. Dr. Mohammad Abuhaiba, PE

    10. 7.3 Characteristics of Dislocations • Shear strains also exist in the vicinity of the edge dislocation. • For a screw dislocation, lattice strains are pure shear only. • These lattice distortions may be considered to be strain fields that radiate from the dislocation line. • The strains extend into the surrounding atoms, and their magnitude decreases with radial distance from the dislocation. Dr. Mohammad Abuhaiba, PE

    11. 7.3 Characteristics of Dislocationsstrain fields surrounding • Consider two edge dislocations that have the same sign and the identical slip plane (Figure 7.5a). • Compressive and tensile strain fields for both lie on the same side of the slip plane; there exists between these two isolated dislocations a mutual repulsive force that tends to move them apart. Dr. Mohammad Abuhaiba, PE

    12. 7.3 Characteristics of Dislocations • Two dislocations of opposite sign and having same slip plane will be attracted to one another (Figure 7.5b), the two extra half-planes of atoms will align and become a complete plane. • These strain fields and associated forces are important in the strengthening mechanisms for metals. Dr. Mohammad Abuhaiba, PE

    13. 7.4 Slip Systems • Slip plane - plane allowing easiest slippage • Wide inter-planar spacing - highest planar densities • Slip direction - direction of movement - Highest linear densities • FCC Slip occurs on {111} planes in <110> directions • => total of 12 slip systems in FCC • in BCC & HCP other slip systems occur Dr. Mohammad Abuhaiba, PE

    14. Table 7.1: Slip Systems for FCC, BCC, and HCP Metals Dr. Mohammad Abuhaiba, PE

    15. 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 7.5 Slip in Single Crystals • • Crystals slip due to a resolved shear stress, tR. • • Applied tension can produce such a stress. • l = angle between stress direction & Slip direction •  = angle between normal to slip plane & stress direction Dr. Mohammad Abuhaiba, PE

    16. 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 Dr. Mohammad Abuhaiba, PE

    17. 7.5 Slip in Single Crystals Resolved  (shear stress) is maximum at  =  = 45º And CRSS = y/2 for dislocations to move (in single crystals) Dr. Mohammad Abuhaiba, PE

    18. Determining  and  angles for Slip in Crystals • For cubic xtals, angles between directions are given by: • For metals we compare Slip System (normal to slip plane is a direction with exact indices as plane) to applied tensile direction using this equation to determine the values of  and  to plug into the R equation to determine if slip is expected Dr. Mohammad Abuhaiba, PE

    19. EXAMPLE PROBLEM 7.1 Resolved Shear Stress and Stress-to-Initiate-Yielding Computations Consider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction. • Compute the resolved shear stress along a (110) plane and in a direction when a tensile stress of 52 MPais applied. • If slip occurs on a (110) plane and in a [-111] direction, and the critical resolved shear stress is 30MPa, calculate the magnitude of the applied tensile stress necessary to initiate yielding. Dr. Mohammad Abuhaiba, PE

    20. 7.6 Plastic Deformation ofPolycrystalline Materials s 300 mm • Stronger since 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 (slip) later. Dr. Mohammad Abuhaiba, PE

    21. 7.6 Plastic Deformation of Polycrystalline Materials • Ordinarily ductility is sacrificed when an alloy is strengthened. • The relationship between dislocation motion and mechanical behavior of metals is significance to the understanding of strengthening mechanisms. • The ability of a metal to plastically deform depends on the ability of dislocations to move. • Virtually all strengthening techniques rely on this simple principle: Restricting or Hindering dislocation motion renders a material harder and stronger. • We will consider strengthening single phase metals by: • grain size reduction • solid-solution alloying • strain hardening Dr. Mohammad Abuhaiba, PE

    22. 7.8 Strengthening by Grain Size Reduction • Grain boundaries are barriers to slip. • Barrier "strength" increases with Increasing angle of mis-orientation. • Smaller grain size: more barriers to slip. Dr. Mohammad Abuhaiba, PE

    23. 7.8 Strengthening by Grain Size Reduction Hall-Petch equation: Dr. Mohammad Abuhaiba, PE

    24. 7.8 Strengthening by Grain Size Reduction Grain Size Reduction Techniques: • Increase Rate of solidification from the liquid phase. • Perform Plastic deformation followed by an appropriate heat treatment. • Notes: • Grain size reduction also improves toughness of many alloys. • Small-angle grain boundaries are not effective in interfering with the slip process because of the small crystallographic misalignment across the boundary. • Boundaries between two different phases are also impediments to movements of dislocations. Dr. Mohammad Abuhaiba, PE

    25. 7.9 Solid-solution Strengthening • 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. Dr. Mohammad Abuhaiba, PE

    26. 7.9 Solid-solution Strengthening • small impurities tend to concentrate at dislocations on the “Compressive stress side” • reduce mobility of dislocation  increase strength Dr. Mohammad Abuhaiba, PE

    27. 7.9 Solid-solution Strengthening • Large impurities concentrate at dislocations on “Tensile Stress” side – pinning dislocation Dr. Mohammad Abuhaiba, PE

    28. 180 400 120 Tensile strength (MPa) Yield 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) 7.9 Solid-solution Strengthening • Tensile strength & yield strength increase with wt% Ni. • Empirical relation: • Alloying increases sy and TS. Dr. Mohammad Abuhaiba, PE

    29. 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 7.10 Strain Hardening • Room temperature deformation. • Common forming operations change the cross sectional area: Dr. Mohammad Abuhaiba, PE

    30. 0.9 mm 7.10 Strain Hardening • Ti alloy after cold working: • Dislocations entangle and multiply • Thus, Dislocation motion becomes more difficult. Dr. Mohammad Abuhaiba, PE

    31. total dislocation length unit volume s large hardening s y1 s small hardening y0 e 7.10 Strain Hardening 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 rdincreases: Dr. Mohammad Abuhaiba, PE

    32. 7.10 Strain Hardening • As cold work is increased • Yield strength (sy) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases. Dr. Mohammad Abuhaiba, PE

    33. Cold Work Analysis Copper Cold Work D D =12.2mm o d =15.2mm • What is the tensile strength & ductility after cold working? Dr. Mohammad Abuhaiba, PE

    34. tensile strength (MPa) yield strength (MPa) ductility (%EL) 60 800 700 40 600 500 Cu 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 TS = 340MPa %EL = 7% Cold Work Analysis What is the tensile strength & ductility after cold working to 35.6%? YS = 300 MPa Dr. Mohammad Abuhaiba, PE

    35. 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. Dr. Mohammad Abuhaiba, PE

    36. Recovery, Recrystallization,and Grain Growth 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... Dr. Mohammad Abuhaiba, PE

    37. 7.11 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 Dr. Mohammad Abuhaiba, PE

    38. 7.12 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 low dislocation density • are small • consume cold-worked grains. Dr. Mohammad Abuhaiba, PE

    39. 0.6 mm 0.6 mm After 8 seconds After 4 seconds 7.12 Recrystallization • All cold-worked grains are consumed. Dr. Mohammad Abuhaiba, PE

    40. 7.12 Recrystallization Dr. Mohammad Abuhaiba, PE

    41. TR º TR = recrystallization temperature Dr. Mohammad Abuhaiba, PE º

    42. DESIGN EXAMPLE 7.1 Description of Diameter Reduction Procedure A cylindrical rod of non-cold-worked brass having an initial diameter of 6.4 mm is to be cold worked by drawing such that the cross-sectional area is reduced. It is required to have a cold-worked yield strength of at least 345 MPa and a ductility in excess of 20%EL; in addition, a final diameter of 5.1 mm is necessary. Describe the manner in which this procedure may be carried out. Dr. Mohammad Abuhaiba, PE

    43. 7.13 Grain Growth 0.6 mm 0.6 mm After 8 s, 580ºC After 15 min, 580ºC exponent typ. ~ 2 elapsed time • At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced. • Empirical Relation: coefficient dependent on material & Temp. grain dia. At time t. Dr. Mohammad Abuhaiba, PE This is: Ostwald Ripening

    44. Recrystallization Temperature, TR TR= recrystallization temperature = point of highest rate of property change • 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 Dr. Mohammad Abuhaiba, PE

    45. 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. Dr. Mohammad Abuhaiba, PE

    46. 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? Dr. Mohammad Abuhaiba, PE

    47. 420 540 6 ColdworkCalc Solution: Cont. • For %CW = 43.8% • y = 420 MPa • TS = 540 MPa > 380 MPa • %EL = 6 < 15 Dr. Mohammad Abuhaiba, PE • This doesn’t satisfy criteria…… what can we do?

    48. 15 380 27 12 > 12 %CW < 27 %CW ColdworkCalc Solution: Cont. ForTS > 380 MPa For%EL > 15  our working range is limited to %CW = 12 – 27% Dr. Mohammad Abuhaiba, PE

    49. Intermediate diameter = This process Needs an Intermediate 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: Dr. Mohammad Abuhaiba, PE

    50. Coldwork Calculations Solution Summary: • Cold work D01= 0.40 in  Df1 = 0.335 in • Anneal above Ds2 = Df1 • Cold work Ds2= 0.335 in  Df2 =0.30 in Therefore, meets all requirements  Fig 7.19 Dr. Mohammad Abuhaiba, PE