Influences of Crystallography and Interfaces on Stress-Strain Behavior

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1 Module #18 Influences of Crystallography and Interfaces on Stress-Strain Behavior SUGGESTED READING* COURTNEY: Ch. 4 Meyers & Chawla, 2 nd Edition: Ch. 6 Lecture Module #15 *This list does not mean that you need to read all of these chapters. It has been assembled to provide you with suggested reading from that you maybeusingorreferringtoinyourcourse. Most of these chapters cover similar material. Any required reading will be noted separately.

2 Recall F A o Dislocations move on specific planes and in specific directions Slip plane normal slip systems A s F Slip direction cos cos F A o This is what leads to slip traces in single crystals and polycrystals How does X-tal. orientation influence stress-strain behavior?

3 Characteristic tensile stress-strain strain curve for a single crystal Characteristic vs. for single crystals depends on crystallographic ap c orientation of the stress axis. II III I CRSS Stage I Stage II Stage III 3 stages of work hardening for single crystals Stage I Stage II Stage III Easy Linear Parabolic Glide Hardening Hardening

4 Deformation of single crystals (1) Characteristic vs. for single crystals depends on crystallographic orientation of the stress axis. II III I CRSS Stage I Stage I: After yielding, the shear stress for plastic deformation is essentially constant. t There is little or no work hardening. This is typical when there is a single slip system operative. Dislocations do not interact much with each other. Easy glide Active slip system is one with maximum Schmid factor (i.e., m = cos cos)

5 Deformation of single crystals (2) Characteristic vs. for single crystals depends on crystallographic orientation of the stress axis. II III I Stage II: CRSS Stage II The shear stress needed d to continue plastic deformation begins to increase in an almost linear fashion. There is extensive work hardening ( G/300). This stage begins when slip is initiated on multiple slip systems. Work hardening is caused by interactions between dislocations moving on intersecting slip planes.

6 Deformation of single crystals (3) Characteristic vs. for single crystals depends on crystallographic orientation of the stress axis. II III I Stage III: CRSS There is a decreasing rate of work hardening. Stage III This decrease is caused by an increase in the degree of cross slip resulting in a parabolic shape to the curve.

7 Deformation of single crystals (4) Characteristic vs. for single crystals depends on crystallographic orientation of the stress axis. T 1 T 2 CRSS T 2 > T 1 Stage I Stage II Stage III Effect of Temperature: Increasing T results in a decrease in the extent of Stage I and Stage II and in a decrease in τ CRSS. Stage I: Initiation of secondary slip systems is easier Stage II: Cross slip is easier

8 Deformation of single crystals (5) Characteristic vs. for single crystals depends on crystallographic orientation of the stress axis. II III I Stacking Fault Energy (SFE): FCC metals CRSS Stage I Stage II Stage III Decrease SFE, decrease cross slip (faults get wider). This increases the stress level needed to have a transition from Stage II to Stage III. Example: Cu-Zn: Cu-30 at.% Zn has low SFE, extends Stage II to high stress levels.

9 Outline for Discussions of Plastic Deformation in Single Crystals and Polycrystals 1. Deformation of single crystals a. Crystallography of slip b. Stress-strain curves c. Influence of # of slip systems d. Crystal rotation during deformation e. Temperature dependence of stress 2. Deformation of polycrystals a. Implications of grain boundaries b. Stress-strain curves c. Influence of # of slip systems d. Influence of # of deformation modes

10 Orientation of slip plane and slip direction relative to the loading axis F Stress normal to slip plane: A o 2 N cos Slip plane normal A s Shear stress acting on slip plane: Slip direction coscos RSS Shear stress acting on slip plane, in slip direction F 90 is not necessarily 90 cos cos CRSS

11 What happens to a single crystal during a tensile test? (a) Consider a single crystal oriented for slip on planes oriented degrees from the tensile axis. (b) Ideally, crystal planes will glide over one another without changing their relative orientation to the load axis.. Orientation of crystal slip plane during tensile deformation. (a) before deformation; (b) after deformation w/o grip constraint; (c) after deformation with grip constraint [Scanned from R.W. Hertzberg, Deformation & Fracture Mechanics of Engineering Materials, 4th ed., (John Wiley & Sons, New York, 1996) p. 95]. (c) However, during tensile testing, the ends of the tensile bar are constrained. Thus the crystal planes cannot glide freely. They are forced to rotate t towards the tensile axis ( i < o ).

12 The slip plane re-orients as the length of the specimen changes. The glide shear stress and shear strain can be determined from the initial orientation of the slip plane ( o) and slip direction ( o) and the extension of the specimen (L i /L o ). 2 1/2 1 L i 2 sin o cos o sin o L o 1/2 2 P sin o sin o 1 A Li Lo Crystal rotation can be traced with the aid of a stereographic projection.

13 Standard Triangle slip systems 2 planes; 1 direction in each slip systems 3 planes; 2 directions in each (FCC) 1 slip system 2 slip systems 1 plane; 2 directions slip systems 4 planes; 2 directions in each 2 slip systems 2 planes; 1 direction in each 4 slip systems 2 planes; 2 directions in each Crystal Orientation: Determines how many slip systems are active (through the Schmid factor).

14 Example Problem Consider a cylindrical single crystal of goldwitha5mm diameter and its axis parallel to [144]. This crystal begins to deform plastically in compression at aloadof100n of N. What are operative slip systems and what is the CRSS on each of the operative slip systems?

15 Solution to Example Problem 2 Gold has an FCC crystal structure. Thus the slip system is There are 12 distinct slip systems for FCC. For each we must compute and tabulate the corresponding angles and, as well as the Schmid factors. This is done via the cosine law. hh 1 2 k2 k1 ll 1 2 coshkl 1 1 1hk 2 2l h1 k1 l1 h2 k2 l2 Slip direction Slip plane u v w h k l cos cos Schmid factor Shear Stress operative slip systems!

16 Solution to Example Problem 2 Silver has an FCC crystal structure. Thus the slip system is There are 12 distinct slip systems for FCC. For each we must compute and tabulate the corresponding angles and, as well as the Schmid factors. Thisisdoneviathecosine law. hh 1 2 k2 k1 ll 1 2 coshkl hk l h1 k1 l1 h2 k2 l2 P 100 N CRSS coscos coscos 5.09 MPa m 2 A (5 mm ) / 4 max On [110](111) and [101](111), 2.21 MPa CRSS For the right crystal orientation, more than one slip system can be active! Compare w/ the examples in module #14

17 Why does the number of active slip systems change? Orientation of crystal slip plane during tensile deformation. (a) before deformation; (b) after deformation w/o grip constraint; (c) after deformation with grip constraint. Note rotation in gage to χ i. [Scanned from R.W. Hertzberg, Deformation & Fracture Mechanics of Engineering Materials, 4th ed., (John Wiley & Sons, New York, 1996) p. 95] During plastic deformation, the single crystal will either: Undergo a translation of the upper and lower sections relative to each other due to slip OR When (as is usual) testing constrains the upper and lower ends keeping them aligned, the crystal will rotate such that the angle between the stress axis and the slip direction decreases. THUS the Schmid factor changes! This can lead to the initiation of slip on a different system.

18 Why does the number of active slip systems change? Primary slip direction (001) Stereographic projection showing lattice rotation for an FCC crystal during tensile elongation. [Adapted from R.W. Hertzberg, Deformation & Fracture Mechanics of Engineering Materials, 4th ed., (John Wiley & Sons, New York, 1996) p. 98] Primary slip plane This figure illustrates lattice rotation for a crystal oriented for single slip (orientation P). This figure is explained on the next page.

19 Why does the number of active slip systems change? As the crystal rotates, the angle between the slip plane and the tensile axis decreases. The orientation of the tensile axis changes. When the rotation causes the tensile orientation to change to P,, the Schmid becomes equal for two slip systems. Slip will occur on two slip systems (i.e., dislocations will move on both systems) Overshoot caused by latent hardening 112 P Final stress axis orientation [112] [101] [011] Initial stress axis orientation The crystal will continue to rotate with deformation occurring on alternating slip systems. This will continue until the load axis reaches [112] where the crystal will neck down until failure without changing orientation. Figure Lattice rotation of an FCC crystal involving overshoot of primary and conjugate slip systems. [Adapted from R.W. Hertzberg, Deformation & Fracture Mechanics of Engineering Materials, 4th ed., (John Wiley & Sons, New York, 1996) p. 99]

20 What about compression? Slip- plane normal o Compression axis See p in Hosford Compression axis In a compression test, the compression axis rotates towards the slipplane normal. The viewgraph on the next page shows the rotations on a stereogram.

21 100 Standard Triangle slip systems 2 planes; 1 direction in each 113 (FCC) slip system slip systems 1 plane; 2 directions 8 slip systems 4 planes; 2 directions in each 2 slip 4 slip systems 2 planes; 1 direction in each systems 2 planes; 2 directions in each Lattice Rotation: Compression axis rotates towards [111] until it reaches the [001]-[011] boundary. Duplex slip begins there which induces a net rotation towards [011].

22 RECALL Characteristic tensile stress-strain strain curve for a single crystal Characteristic vs. for single crystals depends on crystallographic orientation of the stress axis. II III I What would this look like in compression? CRSS Stage I Stage II Stage III 3 stages of work hardening for single crystals Stage I Stage II Stage III Easy Linear Parabolic Glide Hardening Hardening

23 Influence of stress axis orientation The stress axis orientation plays a major role in the stress-strain behavior of a single crystal 6 slip systems FCC Crystal 2 slip systems 1 slip system Why is crystallographic orientation so important? NOTE: there is no stage I for crystals that contain multiple slip systems? WHY? The stress axis orientation controls the number of active slip systems. Recall: Slip occurs when the Schmid factor is maximum. More slip systems means a harder material.

24 Table. Slip systems for crystals at room temperature, unless otherwise stated. Crystal type and examples Class Lattice Direction Plane Remarks FCC metals and solid m3m F solutions Al, Cu, -CuZn BCC metals Fe, Nb m3m I Predominant system Ta, W, Na, K pencil glide frequent 111 a Diamond C, Si, Ge m3m F At temperatures above one-half the melting temperature Sodium chloride NaCl, LiF, MgO, NaF, AgCl, Ni4I, KI, UN, LiCl, LiBr, KCl, NaBr, RbCl, KBr, NaI, AgBr m3m F TiC, UC m3m F At high temperature PbS, PbTe m3m F not certain for PbS Cesium Chloride CsCl, m3m P CsBr, NH4Cl, NH4Br, AuZn, AuCd CuZn m3m P Flourite CaF 2, UO2 m3m F Al2 O3 3m R Above 1000 C Above 1200C Indices refer to triply-primitive hexagonal cell Graphite 6/mmm P HCP metals 6/mmm P Predominant Zn, Cd, Mg a Table - cont'd. Crystal type and examples Class Lattice Direction Plane Remarks Ti, Zr 6/mmm P Predominant 1011 Zr shows only slip Te 32 P Be and AgMg 6/mmm P 1120 Predominant Ga mmm A a Sphalerite -ZnS, InSb, 4 3m F Slip direction not GaAs certain for -ZnS -Sn 4/mmm I Rare Rutile 4/mmm P TiO2 100 Bi 3m R Primitive cell Hg 3m -U mmm C R a a Primitive cell Rare Al2 O3-MgO spinel m3m F In non-stoichiometric i 110 crystals, with excess Al2O 3, 110 is the predominant slip plane a Indicates systems where reversing the slip direction does not produce crystallographically equivalent motions a [Adapted from Kelly, Groves, & Kidd, p ]

25 # Independent slip systems Critical parameter for plastic deformation of metals. 5 independent slip systems required to maintain strain compatibility across grain boundaries. In general, # geometric slip systems > # independent slip systems

26 HCP structure a 3 a 2 An arbitrary displacement in one direction can be duplicated by a suitable combination of the other two. a 1 Basal plane (0001) 3 non-parallel slip directions noted. Only 2 are independent. a a a Geometrical : 3 Independent : 2

27 Is there a required # slip systems? Materials w/ < 5 independent slip systems: polycrystalline form of material behaves in a brittle manner extensive plastic deformation is impossible. Ductility may be possible through activation of secondary slip systems or other deformation mechanisms. Must have at least 5 independent d slip systems. This is known as the von Mises criterion*. * R. Von Mises, Plastic deformation of crystals, Zeitschrift fur Angewandte Mathematik und Mechanik, 8 (1929)

28 Von Mises criteria Recall: an arbitrary deformation has six independent components of the strain tensor: ε xx, ε yy, ε zz, ε xy, ε xz, ε yz. Deformation occurs at constant volume. From lecture module #7, the volume strain is: xx yy zz dxdydz dxdydz dxdydz xx yy zz If assume small strains, we can neglect the products of strains (i.e., ε ii ε jj ), yielding: 0 xx yy zz

29 Von Mises criteria cont d Since only the three components in the main diagonal of the strain tensor are related, only two of them are independent, which reduces the strain tensor to five independent components Example: Thus five independent slip systems or deformation modes must be present to produce pure strain. Groves and Kelly [1,2] outline a method for determining the number of active slip systems. A more recent work by Cotton et al. [3] outlines a simplified method. 1. A. Kelly, G.W. Groves, and P. Kidd, Crystallography yand Crystal Defects, Revised Edition, Wiley, 2000,,pp G.W. Groves and A. Kelly, Independent slip systems in crystals, Philosoophical Magazine 8 (1963) J.D. Cotton, M.J. Kaufman, and R.D. Noebe, A simplified method for determining the number of independent slip systems in crystals, Scripta Metallurgica et Materialia, 25 (1991)

30 [Courtney]

31 Do the same things occur in polycrystal? ABSOLUTELY! All grains will rotate. See the next page. The result is the development of texture which refers to the development of preferred grain orientations. ti Table 7.2. Comparison of theoretical end orientations with preferred orientations. Single crystals: End orientations Polycrystals: Preferred orientations Crystal structure & slip systems Tension Compression Tension Compression FCC {111} & BCC pencil glide & & 111 HCP (0001) [0001] 1120 [0001] HCP {1100} 1120 & {1101} [0001] [0001] Table adapted from Hosford, Mechanical Behavior, 2 nd Edition, (Cambridge University Press, 2009) p. 126

32 Grains rotate and elongate during plastic deformation. Has implications on strength. What about elastic modulus? [Figure copied from W.D. Callister, Jr, Materials Science and Engineering: An Introduction, 7 th Edition, (Wiley, New York, 2007) p.186.

33 Implications for polycrystalline materials Plastic deformation within an individual grain is constrained by the neighboring grains. Because strains along ggrain boundaries must be the same for each grain, the grains will deform in a cooperative manner. If this is not the case, catastrophic failure occurs. z x y A B Since plastic deformation of a single grain is restrained by its neighboring grain, a polycrystalline material will have an intrinsically greater resistance to plastic flow than would a single crystal. A x A z A xz B x B z B xz Required to maintain continuity of the grain boundary

34 Texture Hardening Though we will discuss strengthening mechanisms a little bit later, it is worth discussing texture hardening now. It arises because grains are not randomly oriented in a textured material meaning that some grains are oriented more favorably for slip than others. Each grain has its own characteristic Schmid factor. Just as with single crystals, the ones with the lowest Schmid factors tend to deform last. It is reasonable to relate the tensile yield stress to the CRSS: ys M * where M is called the Taylor factor. CRSS *Note the similarity to Schmid s law [τ CRSS = σ YS (cosλcosφ) = σ YS m]

35 Texture Hardening cont d The Taylor factor represents the average of the Schmid factor values for all of the grains constituting the polycrystal. NOTE: this is not a geometrical average as it reflects the greater constraint provided by the least-favorably oriented grains. Ranges for Taylor factors: FCC: to 2.228; usually quoted as ~3.06 BCC: to 2.08; usually quoted as ~2.75 This tells you that work hardening is generally going to be greater in polycrystals than single crystals.

36 ys M CRSS Figure from Hosford, Mechanical Behavior, 2 nd Edition, (Cambridge University Press, 2009) p. 128

37 What does it all mean? Since in general one grain (either A or B) will have a higher resolved shear stress ( RSS ), the plastic deformation of that grain will be restricted. THUS: Higher yield stress for polycrystal versus single crystal. Greater work hardening for polycrystal versus single crystal. Figure 4.15 This is the basis for texture hardening. Room-temperature tensile (force/stress)-strain curves of single crystal, bicrystals and polycrystals y of (a) Nb and (b) NaCl. Higher stress levels and lower ductilities of polycrystals compared to single crystals are due to the restraints that adjacent crystals in a polycrystal place on the deformation of each other. The level of the flow stress of a bicrystal depends on the relative misorientation of the two crystals comprising it. (Data from C.S. Pande and Y.T. Chou, Treat. Matls. Sc. Tech., ed., H. Herman, 8, 43, 1975; for NaCl from R.J. Stokes, Proc. Br. Cer. Soc., 6, 189, 1966.). [Adapted from Courtney]

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