Chapter 8: Deformation & Strengthening Mechanisms. School of Mechanical Engineering Choi, Hae-Jin ISSUES TO ADDRESS

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1 Chapter 8: Deformation & Strengthening Mechanisms School of Mechanical Engineering Choi, Hae-Jin Materials Science - Prof. Choi, Hae-Jin Chapter 8-1 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 8-2 1

2 Dislocations & Materials Classes Metals (Cu, Al): + + Dislocation motion easiest - non-directional bonding close-packed directions electron cloud for slip Covalent Ceramics (Si, diamond): Motion difficult - directional (angular) bonding ion cores Ionic Ceramics (NaCl): Motion difficult - need to avoid nearest neighbors of like sign (- and +) Chapter 8-3 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, Chapter 8-4 2

3 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 Adapted from Fig. 8.2, Screw dislocation Chapter 8-5 Deformation Mechanisms 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 Adapted from Fig. 8.6, Callister & Rethwisch 3e. FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed) => total of 12 slip systems in FCC For BCC & HCP there are other slip systems. Chapter 8-6 3

4 Stress and Dislocation Motion Resolved shear stress, R results from applied tensile stresses Applied tensile stress: = F/A A F F Resolved shear stress: R =Fs/AsF slip plane normal, n s R A S Relation between and R R =F S /A S Fcos A/cos F S F n S A R F S A S R cos cos Chapter 8-7 Critical Resolved Shear Stress Condition for dislocation motion: R CRSS Ease of dislocation motion depends on crystallographic orientation R cos cos typically 10-4 GPa to 10-2 GPa R =0 =90 R = /2 =45 =45 R = 0 =90 maximum at = = 45º Chapter 8-8 4

5 Single Crystal Slip Adapted from Fig , Callister & Rethwisch 3e. Adapted from Fig. 8.8, Chapter 8-9 Ex: Deformation of single crystal = 60 a) Will the single crystal yield? b) If not, what stress is needed? =35 crss = 20.7 MPa cos cos 20.7 MPa Adapted from (45 MPa) (cos35 ) Fig. 8.7, (cos60 ) Callister & Rethwisch 3e. (45 MPa) (0.41) = 6500 psi 18.5 MPa crss 20.7 MPa So the applied stress of 6500 psi will not cause the crystal to yield. Chapter

6 Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, y )? crss 3000 psi y cos cos y (0.41) y crss coscos 3000 psi psi So for deformation to occur the applied stress must be greater than or equal to the yield stress y 7325 psi Chapter 8-11 Slip Motion in Polycrystals Stronger - grain boundaries pin deformations Slip planes & directions (, ) change from one crystal to another. R will vary from one crystal to another. Thecrystalwiththe the largest R yields first. Other (less favorably oriented) crystals yield later. 300 m 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].) Chapter

7 Anisotropy in y Can be induced by rolling a polycrystalline metal - before rolling - after rolling Adapted from Fig. 8.11, (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.) 235 m - isotropic since grains are approx. spherical & randomly oriented. rolling direction - anisotropic since rolling affects grain orientation and shape. Chapter 8-13 Strategies for Strengthening Metals Grain Size Reduction Solid Solutions Precipitation Strengthening Cold Work Chapter

8 4 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. yield 1/ 2 o kyd 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.) Hall-Petch Equation: Chapter Strategies for Strengthening Metals: 2: Solid Solutions Impurity atoms distort the lattice & generate stress. Stress can produce a barrier to dislocation motion. Smaller substitutional impurity Larger substitutional impurity A C B D 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. Chapter

9 Stress Concentration at Dislocations Adapted from Fig. 8.4, Chapter 8-17 Strengthening by Alloying small impurities tend to concentrate at dislocations reduce mobility of dislocation increase strength Adapted from Fig. 8.17, Chapter

10 Strengthening by Alloying large impurities concentrate at dislocations on low density side Adapted from Fig. 8.18, Chapter 8-19 Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Tensile strength (MP Pa) wt.% Ni, (Concentration C) Yield strength (MPa a) wt.%ni, (Concentration C) Adapted from Fig (a) and (b), Callister & Rethwisch 3e. Empirical relation: ~ C Alloying increases y and TS. y 1/ 2 Chapter

11 4 Strategies for Strengthening Metals: 3: Precipitation Strengthening Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Side View precipitate Large shear stress needed to move dislocation toward precipitate and shear it. Top View Unslipped part of slip plane S Slipped part of slip plane Dislocation advances but precipitates act as pinning sites with spacing S. Result: y~ 1 S Chapter 8-21 Application: Precipitation Strengthening Internal wing structure on Boeing 767 Adapted from chapter- opening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) Adapted from chapteropening gphotograph, p Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) Aluminum is strengthened with precipitates formed by alloying. 1.5m Chapter

12 4 Strategies for Strengthening Metals: 4: Cold Work (%CW) Room temperature deformation. Common forming operations change the cross sectional area: -Forging force -Rolling Ao die blank -Drawing Ao die die force Ad Ad tensile force % CW Adapted from Fig. 14.2, Callister & Rethwisch 3e. Ao A A o d roll Ad Ao roll -Extrusion Ao container die holder force ram billet extrusion container die x 100 Ad Chapter 8-23 Dislocations During Cold Work Ti alloy after cold working: Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. 0.9 m Adapted from Fig. 5.11, Callister & Rethwisch 3e. (Fig is courtesy of M.R. Plichta, Michigan Technological University.) Chapter

13 Dislocation density = Result of Cold Work total dislocation length unit volume Carefully ygrown single crystal ca mm -2 Deforming sample increases density mm -2 Heat treatment reduces density mm -2 Yield stress increases as d increases: y1 y0 large hardening small hardening Chapter 8-25 Effects of Stress at Dislocations Adapted d from Fig. 8.5, Callister & Rethwisch 3e. Chapter

14 Impact of Cold Work As cold work is increased Yield strength ( y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. Adapted from Fig. 8.20, Chapter 8-27 Cold Work Analysis What is the tensile strength & ductility after cold working? 2 2 r o r CW d x r o % yield strength (MPa) % tensile strength (MPa) 800 Do =15.2mm Copper Cold Work Dd =12.2mm ductility (%EL) MPa 300 Cu % Cold Work y = 300MPa MPa Cu % Cold Work TS = 340MPa % 0 0 Cu % Cold Work %EL = 7% Adapted from Fig. 8.19, (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.), Chapter

15 Effect of Heating After %CW 1 hour treatment at T anneal... decreases TS and increases %EL. Effects of cold work are reversed! annealing temperature (ºC) strength (MPa) tensile tensile strength ductility 20 du uctility (%EL) 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.) Chapter 8-29 Recrystallization New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. 0.6 mm 0.6 mm Adapted from Fig (a),(b), Callister & Rethwisch 3e. (Fig (a),(b) are courtesy of JE J.E. Burke, General Electric Company.) 33% cold worked brass New crystals nucleate after 3 sec. at 580C. Chapter

16 Further Recrystallization All cold-worked grains are consumed. 0.6 mm 0.6 mm 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 Chapter 8-31 Grain Growth At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced mm mm After 8 s, After 15 min, 580ºC 580ºC Empirical Relation: exponent typ. ~ 2 grain diam. n at time t. d d n o Kt Adapted from Fig (d),(e), Callister & Rethwisch 3e. (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.) coefficient dependent on material and T. elapsed time Ostwald Ripening Chapter

17 º T R = recrystallization temperature T R Adapted from Fig. 8.22, º Chapter 8-33 Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property p 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 Chapter

18 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. Chapter 8-35 Coldwork Calculations Solution If we directly draw to the final diameter what happens? Brass Cold Work Do = 0.40 in Df = 0.30 in A A A o f %CW A o x f x 100 A o 1 D 2 f 4 x x % D 2 o Chapter

19 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? Adapted from Fig. 8.19, Chapter 8-37 Coldwork Calc Solution: Cont For TS > 380 MPa For %EL > 15 > 12 %CW < 27 %CW Adapted from Fig. 8.19, our working range is limited to %CW = Chapter

20 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: 2 2 D 2 D 2 % CW % CW f 1 f x D D02 Df D 0. 5 D 2 % CW f 2 1 D Intermediate diameter = 100 % CW Df 1 D m 100 Chapter 8-39 Coldwork Calculations Solution Summary: 1. Cold work D 01 = 0.40 in D f1 = m %CW x Anneal above D 02 = D f1 3. Cold work D 02 = in D f 2 =0.30 m 2 Fig MPa 0 3 % 2. y CW x TS 400 MPa % EL 24 Therefore, meets all requirements Chapter

21 Mechanical Properties of Polymers Stress-Strain Behavior brittle polymer plastic elastomer elastic moduli less than for metals Adapted from Fig. 7.22, Fracture strengths of polymers ~ 10% of those for metals Deformation strains for polymers > 1000% for most metals, deformation strains < 10% 41 Chapter 8-41 Mechanisms of Deformation Brittle Crosslinked and Network Polymers (MPa) ( ) x Near Near Initial Failure brittle failure Initial Failure x plastic failure aligned, crosslinked polymer Stress-strain curves adapted from Fig. 7.22, network polymer 42 Chapter

22 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 ) Mechanisms of Deformation Semicrystalline (Plastic) Polymers (MPa) x brittle failure onset of necking unload/reload plastic failure x fibrillar structure near failure undeformed structure amorphous regions elongate crystalline regions align crystalline block segments separate 43 Chapter 8-43 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 ) Chapter

23 Mechanisms of Deformation Elastomers (MPa) initial: amorphous chains are kinked, cross-linked. x brittle failure plastic failure x elastomer x deformation is reversible (elastic)! final: chains are straighter, still cross-linked 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, Inset figures along elastomer curve (green) adapted from Fig. 8.30, (Fig is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.) Chapter 8-45 Summary Dislocations are observed primarily in metals and alloys. Strength th 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. Chapter