1. Chapter 5 Dislocations and Plastic Deformation

2. The Frank-Read SourceDislocation 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=bresThe dislocation segment responds to the force by bowing out. If the force is large enough, the critical configuration of a semicircle may be reached. max=Gb/ll 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. l()l (bl) l(2sin) sin sin f = blGb 2 = 2 bl = Gb 2 sin Gb =sin l Gb 90o 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 sourcesWhen 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 DislocationIf 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 movedThe pits connected with the dislocationalways 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 DislocationsA 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 theneutral axisThe 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 Myx = 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 specimens 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 SlipType of deformation due to dislocations: simple shear, bending, rotational slip. Rotational slipCan be explained in terms of screwdislocations lying on the slip plane.Required more than one set ofdislocations (slip plane must containmore 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 DirectionsSlip The process by which plastic deformation is produced by dislocation motion Slip plane The crystallographic plane along which the dislocation line traverses Slip direction Direction along which dislocation motion Slip system The combination of slip direction and slip plane Dislocation density 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 ar 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 - Schmids LawThe relationship between shear stress, the applied stress, and the orientation of the slip systemF r = = resolvedr shearstressin theslipdirection r = cos cosA F = = unidirecti stressapplied thecylinde onalto A0A = A0 / cos(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 and equal 45o. r = 0.5 22. Critical resolved shear stressFor a given crystallographic plane, it is independent of theorientation 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 FormsFCC slip system Close-packed plane: {111}octahedral plane Close-packed directions: Slip systems: 4x3=12 Large number of equivalent slip system 24. a curve several slip systems havenearly equal resolved shearstresses. b curve1 : only one slip planeEasy glide2 : multiple glide onintersecting slip planesCrystal hardens rapidly withincreasing strain3 : decreasing the rate ofincrease of the dislocationdensity 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)6.2Zinc, 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 : Slip plane (contains a close-packed direction) : {110},{112},{123} The lack of close-packed plane is the high critical resolved shear stress for slip. Ex. Fe: 28 MPa