Dislocations & Materials Classes. Dislocation Motion. Dislocation Motion. Lectures 9 and 10

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Lectures 9 and 10 Chapter 7: Dislocations & Strengthening Mechanisms Dislocations & Materials Classes Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip.

How can heating change strength and other properties? Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding Ionic Ceramics (NaCl): Motion hard.

1 Lectures 9 and 10 Chapter 7: Dislocations & Strengthening Mechanisms Dislocations & Materials Classes Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip. electron cloud ion cores 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? Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors 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). Dislocation Motion Dislocation moves along slip plane in slip direction perpendicular to dislocation line Slip direction same direction as Burgers vector Edge dislocation 7.2, Screw dislocation If dislocations don't move, deformation doesn't occur! 7.1, 3 4 1

Deformation Mechanisms Slip System Slip plane - plane allowing easiest slippage Wide interplanar spacings - highest planar densities Slip direction - direction of

other slip systems occur 7.6, 5 Stress and Dislocation Motion Crystals slip due to a resolved shear stress, τ R. Applied tension can produce such a stress.

and τ R τ R =F S /A S Fcos λ F λ = cos λcos φ F S n S A/cos φ φ A S A 6 Critical Resolved Shear Stress Condition for dislocation motion: τ R > τcrss Single Crystal Slip

2 Deformation Mechanisms Slip System Slip plane - plane allowing easiest slippage Wide interplanar spacings - 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 7.6, 5 Stress and Dislocation Motion Crystals slip due to a resolved shear stress, τ R. Applied tension can produce such a stress. Applied stress: = F/A A slip direction F F Resolved shear stress: τ R =Fs/As slip plane normal, n s τ R slip direction τ R F S τ R A S slip direction Relation between and τ R τ R =F S /A S Fcos λ F λ = cos λcos φ F S n S A/cos φ φ A S A 6 Critical Resolved Shear Stress Condition for dislocation motion: τ R > τcrss Single Crystal Slip Crystal orientation can make it easy or hard to move dislocation τr= cos λcos φ typically 10-4 GPa to 10-2 GPa 7.9, τ R = 0 λ=90 τ R = /2 λ=45 φ =45 τ R = 0 φ=90 τ maximum at λ = φ = 45º 7 7.8, 8 2

Ex: Deformation of single crystal φ=60 Adapted from Fig. 7.7, = 65 MPa λ=35 a) Will the single crystal yield? b) If not, what stress is needed?

30 MPa 9 Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, y )? τ = 30 MPa = cosλ cosφ = (0.41) crss y y τ 30 MPa = = cosλ cosφ 0.41 crss = y = 73.

2 MPa So for deformation to occur the applied stress must be greater than or equal to the yield stress 10 Slip Motion in Polycrystals Stronger - grain boundaries pin deformations Slip

Other (less favorably oriented) crystals yield later. 300 µm 7.10, (Fig. 7.10 is courtesy of C.

3 Ex: Deformation of single crystal φ=60 Adapted from Fig. 7.7, = 65 MPa λ=35 a) Will the single crystal yield? b) If not, what stress is needed? τ crss = 30 MPa τ = cosλ cosφ = 30 MPa τ = (65 MPa) (cos 35 )(cos60 ) = (65 MPa) (0.41) τ = 26.6 MPa < τ crss = So the applied stress of 65 MPa will not cause the crystal to yield. 30 MPa 9 Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, y )? τ = 30 MPa = cosλ cosφ = (0.41) crss y y τ 30 MPa = = cosλ cosφ 0.41 crss = y = 73.2 MPa y 73.2 MPa So for deformation to occur the applied stress must be greater than or equal to the yield stress 10 Slip Motion in Polycrystals Stronger - grain boundaries pin deformations Slip planes & directions (λ, φ) change from one crystal to another. τ R will vary from one crystal to another. The crystal with the largest τ R yields first. Other (less favorably oriented) crystals yield later. 300 µm 7.10, (Fig is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].) 11 Can be induced by rolling a polycrystalline metal - before rolling 235 µm - isotropic since grains are approx. spherical & randomly oriented. Anisotropy in y - after rolling rolling direction - anisotropic since rolling affects grain orientation and shape. 7.11, (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.) 12 3

1. Cylinder of Tantalum machined from a rolled plate: rolling direction Anisotropy in Deformation 2. Fire cylinder at a target. 3. Deformed cylinder side view Photos courtesy of G.T. Gray III, Los Alamos National Labs.

Smaller grain size: more barriers to slip. 7.14, (Fig. 7.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.

end view plate thickness direction 13 14 4 Strategies for Strengthening: 2: Solid Solutions Stress Concentration at Dislocations Impurity atoms distort the lattice & generate stress.

4 1. Cylinder of Tantalum machined from a rolled plate: rolling direction Anisotropy in Deformation 2. Fire cylinder at a target. 3. Deformed cylinder side view Photos courtesy of G.T. Gray III, Los Alamos National Labs. Used with permission. 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. 7.14, (Fig is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.) The noncircular end view shows anisotropic deformation of rolled material. end view plate thickness direction Strategies for Strengthening: 2: Solid Solutions Stress Concentration at Dislocations 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. 7.4,

5 Strengthening by Alloying small impurities tend to concentrate at dislocations reduce mobility of dislocation increase strength Strengthening by alloying large impurities concentrate at dislocations on low density side 7.17, 7.18, Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Tensile strength (MPa) wt.% Ni, (Concentration C) Yield strength (MPa) Alloying increases y and TS wt.%ni, (Concentration C) 7.16 (a) and (b), 19 4 Strategies for Strengthening: 3: Precipitation Strengthening Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Side View Top View precipitate Unslipped part of slip plane S Slipped part of slip plane Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation advances but precipitates act as pinning sites with spacing S. The closer the spacing between precipitates the higher y and TS 20 5

Application: Precipitation Strengthening 4 Strategies for Strengthening: 4: Cold Work (%CW) Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates formed by alloying. 1.

Common forming operations change the cross sectional area: -Forging force -Rolling die blank -Drawing die die force Ad Ad tensile force 11.

9 µm Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. 4.6, (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.

6 Application: Precipitation Strengthening 4 Strategies for Strengthening: 4: Cold Work (%CW) Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates formed by alloying. 1.5µm Adapted from chapteropening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) 11.26, (Fig is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) 21 Room temperature deformation. Common forming operations change the cross sectional area: -Forging force -Rolling die blank -Drawing die die force Ad Ad tensile force 11.8, A % o A CW = d force -Extrusion x 100 ram container billet container roll roll Ad die holder extrusion die 22 Ad Dislocations During Cold Work Ti alloy after cold working: 0.9 µm Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. 4.6, (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.) 23 Dislocation density = Result of Cold Work Carefully grown single crystal ca mm -2 Deforming sample increases density mm -2 Heat treatment reduces density mm -2 Yield stress increases as ρ d increases: total dislocation length unit volume y1 y0 large hardening small hardening ε 24 6

7 7.5, Effects of Stress at Dislocations Impact of Cold Work As cold work is increased Yield strength ( y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. 7.20, Cold Work Analysis What is the tensile strength & ductility after cold working? 2 2 πr π % = o r CW d x 100 = 35.6% 2 πro yield strength (MPa) MPa 300 Cu % Cold Work y = 300MPa tensile strength (MPa) MPa Cu % Cold Work TS = 340MPa Do =15.2mm ductility (%EL) % Cold Work %EL = 7% 7.19, (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.), American Society for Metals, 1979, p. 276 and 327.) % 0 0 Copper Cold Work Cu Dd =12.2mm 27 -ε Behavior vs. Temperature Results for polycrystalline iron: 6.14, Stress (MPa) Strain y and TS decrease with increasing test temperature. %EL increases with increasing test temperature. Why? Vacancies help dislocations move past obstacles vacancies replace atoms on the disl. half plane -200 C -100 C 3. disl. glides past obstacle 25 C obstacle 1. disl. trapped by obstacle 28 7

Effect of Heating After %CW 1 hour treatment at T anneal... decreases TS and increases %EL. Effects of cold work are reversed!

Annealing stages to discuss... 7.22, (Fig. 7.22 is adapted from G. Sachs and K.R.

) Recovery Annihilation reduces dislocation density. Scenario 1 Results from diffusion Scenario 2 extra half-plane of atoms 3. Climbed disl. can now move on new slip plane 2.

dislocation blocked; can t move to the right atoms diffuse to regions of tension extra half-plane of atoms Dislocations annihilate and form a perfect atomic plane. 4.

8 Effect of Heating After %CW 1 hour treatment at T anneal... decreases TS and increases %EL. Effects of cold work are reversed! tensile strength (MPa) tensile strength annealing temperature (ºC) Recovery ductility Recrystallization ductility (%EL) Grain Growth 3 Annealing stages to discuss , (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.) Recovery Annihilation reduces dislocation density. Scenario 1 Results from diffusion Scenario 2 extra half-plane of atoms 3. Climbed disl. can now move on new slip plane 2. grey atoms leave by vacancy diffusion allowing disl. to climb 1. dislocation blocked; can t move to the right atoms diffuse to regions of tension extra half-plane of atoms Dislocations annihilate and form a perfect atomic plane. 4. opposite dislocations meet and annihilate Obstacle dislocation τ R Recrystallization New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. 0.6 mm 0.6 mm Further Recrystallization All cold-worked grains are consumed. 0.6 mm 0.6 mm Adapted from Fig (a),(b), (Fig (a),(b) are courtesy of J.E. Burke, General Electric Company.) Adapted from Fig (c),(d), (Fig (c),(d) are courtesy of J.E. Burke, General Electric Company.) 33% cold worked brass New crystals nucleate after 3 sec. at 580 C. After 4 seconds After 8 seconds

9 Grain Growth º At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced. 0.6 mm 0.6 mm T R = recrystallization temperature After 8 s, 580ºC After 15 min, 580ºC Adapted from Fig (d),(e), (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.) T R 7.22, Empirical Relation: exponent typ. ~ 2 grain diam. n at time t. d d n o = Kt coefficient dependent on material and T. elapsed time Ostwald Ripening 33 º 34 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(t) 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 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

STRENGTHENING MECHANISM IN METALS

Background Knowledge Yield Strength STRENGTHENING MECHANISM IN METALS Metals yield when dislocations start to move (slip). Yield means permanently change shape. Slip Systems Slip plane: the plane on which