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- 1. Chapter 5 Dislocations and Plastic Deformation
- 2. The Frank-Read Source Dislocation can just be generated at the surface of the crystal The most important mechanism that dislocation is generated in the bulk of a crystal – Frank-Read mechanism
- 3. We have a segment of dislocation firmly anchored at two points (red circles). The force f=b⋅τres The dislocation segment responds to the force by bowing out. If the force is large enough, the critical configuration of a semicircle may be reached. τmax=Gb/l l If the shear stress is higher than Gb/l, the radius of curvature is too small to stop further bowing out. The dislocation is unstable and the following process now proceeds automatically and quickly.
- 4. 欲使長度l，端點被釘住之直 線差排打彎，長度增加，必 需有力作用於差排，否則差 排將因本身的線張力(Γ)而恢 復原長，即作用於長度l之差 θ 排總力(τbl)必須大於等於差 l 排張力在施力反方向之分力 (2sinθ) Γ Γsinθ Γsinθ f = τbl Gb 2 Γ= 2 τbl = Gb 2 sin θ Gb τ= sin θ l Gb θ → 90o τ max = l
- 5. The two segments shortly before they touch. Since the two line vectors at the point of contact have opposite signs, the segments on contact will annihilate each other. It will immediately form a straight segment after contact, and a “nice” dislocation loop which will expand under the influence of the resolved shear stress. The regained old segment will immediately start to go through the whole process again, and again,… as long as the force exist. Stable configuration after the process. The loop is free to move.
- 6. One of the main mechanisms for dislocation multiplication under stress is the Frank-Read mill or Frank-Read source. The operation of a Frank-Read source can be observed on a dislocation segment pinned at its ends.
- 7. Two interacting Frank-Read sources When a Frank-Read source interacts with other dislocations, its critical stress for dislocation multiplication is modified. Interactions between two sources illustrate this property. The critical stress for dislocation multiplication is decreased or increased when two repulsive or attractive dislocations are respectively considered. Two repulsive sources Two attractive sources
- 8. Nucleation of Dislocation If dislocations are not formed by dislocation generators (Frank-Read source), then they must be created by a nucleation process. Nucleation process created in two ways: Homogeneous nucleation formed in a perfect lattice by the action of a simple stress, no agency other than stress being required. requires extremely high stresses. Heterogeneous nucleation the dislocations are formed with the help of defects present in the crystal, perhaps impurity particles. Defects make the formation of dislocations easier by lowering the applied stress required to form dislocations If dislocations are not formed by Frank-Read sources, then they must be nucleated heterogeneously.
- 9. The dislocations have moved The pits connected with the dislocation always have pointed extremities. Observing dislocation in crystal by etching reagent, which forms an etch pit on the surface of a crystal at each point where a dislocation intersects the surface. The velocity of a dislocation moving under a fixed applied stress = distance that a dislocation moved by the time
- 10. (c) 2003 Brooks/Cole Publishing / Thomson Learning 4-4 Observing Dislocations A sketch illustrating dislocations, slip planes, and etch pit locations.
- 11. Optical image of etch pits in silicon carbide (SiC). The etch pits correspond to intersection points of pure edge dislocations a with Burgers vector < 1120 >and the dislocation line direction 3 along [0001] (perpendicular to the etched surface). Lines of etch pits represent low angle grain boundaries
- 12. Bend Gliding Zero at the neutral axis The bending of crystals can be explained in terms of Frank-Read or other sources. Let equal couples (of magnitude M) be applied to the ends of the crystal. The effect of these couples is to produce a uniform bending moment (M) throughout the length of the crystal. The stress distribution across any cross-section is My σx = I M : the bending moment y : the vertical distance measured from the neutral axis I : the moment of inertia of the cross - section
- 13. The stress distribution on slip planes corresponding to the elastic deformation. The shear stress component parallel to the slip plane. The sense of the shear stress changes its sign as it crosses the neutral axis The shear stress is zero at the neutral axis and a maximum at the extreme ends of the slip plane. upper: compressive stress; lower: tensile stress
- 14. Effect of the stress distribution on the movement of dislocations. Distribution of the excess edge dislocations in a plastically bent crystal. positive edge : move toward the surface (high stress region) and disappear negative edge : move toward the specimen’s neutral axis (decreasing shear stress) neutral axis : free of dislocations, not be stressed above the elastic limit, deformation will be elastic and not plastic
- 15. Rotational Slip Type of deformation due to dislocations: simple shear, bending, rotational slip. Rotational slip Can be explained in terms of screw dislocations lying on the slip plane. Required more than one set of dislocations (slip plane must contain more than one slip direction) Ex. FCC , and HCP
- 16. An array of parallel screw dislocations
- 17. A double array of screw dislocations. This array does not have a long-range strain field.
- 18. Slip Planes and Slip Directions Slip 滑移 The process by which plastic deformation is produced by dislocation motion Slip plane 滑移平面 The crystallographic plane along which the dislocation line traverses Slip direction 滑移方向 Direction along which dislocation motion Slip system 滑移系統 The combination of slip direction and slip plane Dislocation density 差排密度 It is expressed as the total dislocation length per unit volume (cm/cm3) or, equivalently, the number of dislocations that intersect a unit area of a random section
- 19. q a c c c c c a m n a r Two ways in which a simple cubic lattice can be sheared while still maintaining the lattice symmetry Shear in a close-packed direction (mn) Shear in a nonclose-packed direction (qr) However, strain energy of a dislocation is proportional to b2. Thus, a dislocation with a b equal to the spacing of atoms in a close-packed direction would be unique.
- 20. Critical Resolved Shear Stress - Schmid’s Law The relationship between shear stress, the applied stress, and the orientation of the slip system F τr = = resolved r shearstressin theslipdirection τ r = σ cosφ cosλ A F σ = = unidirecti stressapplied thecylinde onal to A0 A = A0 / cosφ (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
- 21. Critical resolved shear stress the yield stress for crystals of a shear stress resolved on the slip plane and in the slip direction. • When a crystal possesses several crystallographically equivalent slip systems, slip will start first on the system having the highest resolved shear stress. • It is not possible to produce slip on a given plane when the plane is either parallel or perpendicular to the axis of tensile. • The maximum shear stress occurs when both λand φ equal 45o. τ r = 0.5σ
- 22. Critical resolved shear stress For a given crystallographic plane, it is independent of the orientation of the crystal for some metal. It is sensitive to changes in composition and handling. It is a function of temperature. For FCC, the temperature dependence may be small. BCC, HCP and rhombohedral show a larger temperature effect.
- 23. Slip Systems in Different Crystal Forms FCC slip system Close-packed plane: {111} octahedral plane Close-packed directions: <110> Slip systems: 4x3=12 Large number of equivalent slip system
- 24. a curve several slip systems have nearly equal resolved shear stresses. b curve 1 : only one slip plane Easy glide 2 : multiple glide on intersecting slip planes Crystal hardens rapidly with increasing strain 3 : decreasing the rate of increase of the dislocation density
- 25. HCP slip system Close-packed plane : (0001) Close-packed direction : < 1120 > (c) 2003 Brooks/Cole Publishing / Thomson Learning™
- 26. Titanium (0001) 110 Titanium (1010) 49 Beryllium (0001) 39 Zirconium (1010) 6.2 Zinc, cadmium and magnesium possess both a low critical resolved shear stress and a single primary slip plane (basal plane)
- 27. The idea ratio c/a is 1.632.
- 28. BCC slip system Close-packed direction : <111> Slip plane (contains a close-packed <111> direction) : {110},{112},{123} The lack of close-packed plane is the high critical resolved shear stress for slip. Ex. Fe: 28 MPa

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