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Chapter 8 - 1 ISSUES TO ADDRESS... Why are the number of dislocations present greatest in metals ? How are strength and dislocation motion related? Why.

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Presentation on theme: "Chapter 8 - 1 ISSUES TO ADDRESS... Why are the number of dislocations present greatest in metals ? How are strength and dislocation motion related? Why."— Presentation transcript:

1 Chapter ISSUES TO ADDRESS... Why are the number of dislocations present greatest in metals ? How are strength and dislocation motion related? Why does heating alter strength and other properties? Chapter 8: Deformation & Strengthening Mechanisms

2 Chapter Dislocations & Materials Classes Covalent Ceramics (Si, diamond): Motion difficult - directional (angular) bonding Ionic Ceramics (NaCl): Motion difficult - need to avoid nearest neighbors of like sign (- and +) Metals (Cu, Al): Dislocation motion easiest - non-directional bonding - close-packed directions for slip electron cloud ion cores

3 Chapter Dislocation Motion Dislocation motion & plastic deformation Metals - plastic deformation occurs by slip – an edge dislocation (extra half-plane of atoms) slides over adjacent plane half-planes of atoms. If dislocations can't move, plastic deformation doesn't occur! Adapted from Fig. 8.1, Callister & Rethwisch 3e.

4 Chapter Dislocation Motion A dislocation moves along a slip plane in a slip direction perpendicular to the dislocation line The slip direction is the same as the Burgers vector direction Edge dislocation Screw dislocation Adapted from Fig. 8.2, Callister & Rethwisch 3e.

5 Chapter Slip System –Slip plane - plane on which easiest slippage occurs Highest planar densities (and large interplanar spacings) –Slip directions - directions of movement Highest linear densities Deformation Mechanisms Adapted from Fig. 8.6, Callister & Rethwisch 3e. – FCC Slip occurs on {111} planes (close-packed) in directions (close-packed) => total of 12 slip systems in FCC – For BCC & HCP there are other slip systems.

6 Chapter Stress and Dislocation Motion Resolved shear stress,  R – results from applied tensile stresses slip plane normal, n s Resolved shear stress: RR =F s /A/A s slip direction ASAS RR RR FSFS slip direction Relation between  and RR RR = FSFS /AS/AS Fcos A/cos  F FSFS  nSnS ASAS A Applied tensile stress: = F/A  slip direction F A F

7 Chapter Condition for dislocation motion: Ease of dislocation motion depends on crystallographic orientation GPa to GPa typically Critical Resolved Shear Stress  maximum at =  = 45º RR = 0 =90°  RR =  /2 =45°   RR = 0  =90° 

8 Chapter Single Crystal Slip Adapted from Fig. 8.8, Callister & Rethwisch 3e. Adapted from Fig. 8.9, Callister & Rethwisch 3e.

9 Chapter Ex: Deformation of single crystal So the applied stress of 6500 psi will not cause the crystal to yield. = 35°  = 60°  crss = 20.7 MPa a) Will the single crystal yield? b) If not, what stress is needed?  = 6500 psi Adapted from Fig. 8.7, Callister & Rethwisch 3e.

10 Chapter Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress,  y )? So for deformation to occur the applied stress must be greater than or equal to the yield stress

11 Chapter Stronger - grain boundaries pin deformations Slip planes & directions (,  ) change from one crystal to another.  R will vary from one crystal to another. The crystal with the largest  R yields first. Other (less favorably oriented) crystals yield later. Adapted from Fig. 8.10, Callister & Rethwisch 3e. (Fig is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].) Slip Motion in Polycrystals  300  m

12 Chapter Can be induced by rolling a polycrystalline metal - before rolling 235  m - isotropic since grains are approx. spherical & randomly oriented. - after rolling - anisotropic since rolling affects grain orientation and shape. rolling direction Adapted from Fig. 8.11, Callister & Rethwisch 3e. (Fig 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  y

13 Chapter side view 1. Cylinder of tantalum machined from a rolled plate: rolling direction 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 Strategies for Strengthening Metals: 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: Adapted from Fig. 8.14, Callister & Rethwisch 3e. (Fig is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

15 Chapter Impurity atoms distort the lattice & generate stress. Stress can produce a barrier to dislocation motion. 4 Strategies for Strengthening Metals: 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 Stress Concentration at Dislocations Adapted from Fig. 8.4, Callister & Rethwisch 3e.

17 Chapter Strengthening by Alloying small impurities tend to concentrate at dislocations reduce mobility of dislocation  increase strength Adapted from Fig. 8.17, Callister & Rethwisch 3e.

18 Chapter Strengthening by Alloying large impurities concentrate at dislocations on low density side Adapted from Fig. 8.18, Callister & Rethwisch 3e.

19 Chapter Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Empirical relation: Alloying increases  y and TS. Adapted from Fig (a) and (b), Callister & Rethwisch 3e. Tensile strength (MPa) wt.% Ni, (Concentration C) Yield strength (MPa) wt.%Ni, (Concentration C)

20 Chapter Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Result: 4 Strategies for Strengthening Metals: 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 Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates formed by alloying. Adapted from chapter- opening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) 1.5  m Application: Precipitation Strengthening Adapted from chapter- opening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

22 Chapter Strategies for Strengthening Metals: 4: Cold Work (%CW) Room temperature deformation. Common forming operations change the cross sectional area: Adapted from Fig. 14.2, Callister & Rethwisch 3e. -Forging A o A d force die blank force -Drawing tensile force A o A d die -Extrusion ram billet container force die holder die A o A d extrusion -Rolling roll A o A d

23 Chapter Ti alloy after cold working: Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. Adapted from Fig. 5.11, Callister & Rethwisch 3e. (Fig is courtesy of M.R. Plichta, Michigan Technological University.) Dislocations During Cold Work 0.9  m

24 Chapter Result of Cold Work Dislocation density = –Carefully grown single crystal  ca mm -2 –Deforming sample increases density  mm -2 –Heat treatment reduces density  mm -2 Yield stress increases as  d increases: total dislocation length unit volume large hardening small hardening    y0y0  y1y1

25 Chapter Effects of Stress at Dislocations Adapted from Fig. 8.5, Callister & Rethwisch 3e.

26 Chapter Impact of Cold Work Adapted from Fig. 8.20, Callister & Rethwisch 3e. Yield strength (  y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. As cold work is increased

27 Chapter What is the tensile strength & ductility after cold working? Adapted from Fig. 8.19, Callister & Rethwisch 3e. (Fig 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.) Cold Work Analysis % Cold Work Cu yield strength (MPa)  y = 300MPa 300MPa % Cold Work tensile strength (MPa) 200 Cu ductility (%EL) % Cold Work Cu D o =15.2mm Cold Work D d =12.2mm Copper 340MPa TS = 340MPa 7% %EL= 7%

28 Chapter Results for polycrystalline iron:  y and TS decrease with increasing test temperature. %EL increases with increasing test temperature. Why? Vacancies help dislocations move past obstacles. Adapted from Fig. 7.14, Callister & Rethwisch 3e.  -  Behavior vs. Temperature 2. vacancies replace atoms on the disl. half plane 3. disl. glides past obstacle -200  C -100  C 25  C Strain Stress (MPa) disl. trapped by obstacle obstacle

29 Chapter hour treatment at T anneal... decreases TS and increases %EL. Effects of cold work are reversed! 3 Annealing stages to discuss... Adapted from Fig. 8.22, Callister & Rethwisch 3e. (Fig 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 %CW tensile strength (MPa) ductility (%EL) tensile strength ductility Recovery Recrystallization Grain Growth annealing temperature (ºC)

30 Chapter 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” RR 1. dislocation blocked; can’t move to the right Obstacle dislocation 3. “Climbed” disl. can now move on new slip plane

31 Chapter New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. Adapted from Fig (a),(b), Callister & Rethwisch 3e. (Fig (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 Recrystallization

32 Chapter All cold-worked grains are consumed. Adapted from Fig (c),(d), Callister & Rethwisch 3e. (Fig (c),(d) are courtesy of J.E. Burke, General Electric Company.) After 4 seconds After 8 seconds 0.6 mm Further Recrystallization

33 Chapter 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 Adapted from Fig (d),(e), Callister & Rethwisch 3e. (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.) Grain Growth Empirical Relation: elapsed time coefficient dependent on material and T. grain diam. at time t. exponent typ. ~ 2 Ostwald Ripening

34 Chapter TRTR Adapted from Fig. 8.22, Callister & Rethwisch 3e. º º T R = recrystallization temperature

35 Chapter Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property change 1.T m => T R  T m (K) 2.Due to diffusion  annealing time  T R = f(time) 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

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

38 Chapter Coldwork Calc Solution: Cont. For %CW = 43.8% –  y = 420 MPa –TS = 540 MPa > 380 MPa 6 –%EL = 6< 15 This doesn’t satisfy criteria…… what can we do? Adapted from Fig. 8.19, Callister & Rethwisch 3e.

39 Chapter Coldwork Calc Solution: Cont For %EL > 15 For TS > 380 MPa > 12 %CW < 27 %CW  our working range is limited to %CW = Adapted from Fig. 8.19, Callister & Rethwisch 3e.

40 Chapter 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 be calculated as follows:  Intermediate diameter =

41 Chapter Coldwork Calculations Solution Summary: 1.Cold work D 01 = 0.40 in  D f1 = m 2.Anneal above D 02 = D f1 3.Cold work D 02 = in  D f 2 =0.30 m Therefore, meets all requirements  Fig 7.19

42 Chapter Rate of Recrystallization Hot work  above T R Cold work  below T R Smaller grains –stronger at low temperature –weaker at high temperature log t start finish 50%

43 Chapter Mechanical Properties of Polymers – Stress-Strain Behavior Fracture strengths of polymers ~ 10% of those for metals Deformation strains for polymers > 1000% – for most metals, deformation strains < 10% brittle polymer plastic elastomer elastic moduli – less than for metals Adapted from Fig. 7.22, Callister & Rethwisch 3e.

44 Chapter Mechanisms of Deformation—Brittle Crosslinked and Network Polymers brittle failure plastic failure  (MPa)  x x aligned, crosslinked polymer Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e. Initial Near Failure Initial network polymer Near Failure

45 Chapter Mechanisms of Deformation — Semicrystalline (Plastic) Polymers brittle failure plastic failure  (MPa) x x crystalline block segments separate fibrillar structure near failure crystalline regions align onset of necking undeformed structure amorphous regions elongate unload/reload Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e. Inset figures along plastic response curve adapted from Figs & 8.28, Callister & Rethwisch 3e. (Figs & 8.28 are from J.M. Schultz, Polymer Materials Science, Prentice- Hall, Inc., 1974, pp ) 

46 Chapter Predeformation by Drawing Drawing…(ex: monofilament fishline) -- stretches the polymer prior to use -- aligns chains in the stretching direction Results of drawing: -- increases the elastic modulus (E) in the stretching direction -- increases the tensile strength (TS) in the stretching direction -- decreases ductility (%EL) Annealing after drawing decreases chain alignment -- reverses effects of drawing (reduces E and TS, enhances %EL) Contrast to effects of cold working in metals! Adapted from Fig. 8.28, Callister & Rethwisch 3e. (Fig is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp )

47 Chapter Compare elastic behavior of elastomers with the: -- brittle behavior (of aligned, crosslinked & network polymers), and -- plastic behavior (of semicrystalline polymers) (as shown on previous slides) Stress-strain curves adapted from Fig. 7.22, Callister & Rethwisch 3e. Inset figures along elastomer curve (green) adapted from Fig. 8.30, Callister & Rethwisch 3e. (Fig is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.) Mechanisms of Deformation— Elastomers  (MPa)  initial: amorphous chains are kinked, cross-linked. x final: chains are straighter, still cross-linked elastomer deformation is reversible (elastic)! brittle failure plastic failure x x

48 Chapter 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


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