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Published byTobias Daniel Modified about 1 year ago

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PLASTICITY

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Dislocations and Materials Classes Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip. electron cloudion cores

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Dislocation Motion Dislocations and 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). Cubic & hexagonal metals - plastic deformation by plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations). If dislocations don't move, deformation doesn't occur! Adapted from Fig. 7.1, Callister 7e.

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Dislocation Motion Dislocation moves along slip plane in slip direction perpendicular to dislocation line Dislocation moves along slip plane in slip direction perpendicular to dislocation line Slip direction same direction as Burgers vector Slip direction same direction as Burgers vector Edge dislocation Screw dislocation Adapted from Fig. 7.2, Callister 7e.

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If a material is subjected to a load of sufficient magnitude it shows permanent (irrecoverable) deformation. It is result of the permanent displacement of atoms and molecules from their original position. If the deformation is continuously increasing then the phenemenon is called FLOW. Slip plane P x’ x P PnPn PsPs Slip direction

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Slip System: Slip plane and direction Slip plane - plane allowing easiest slippage Slip plane - plane allowing easiest slippage - Highest planar densities Slip direction - direction of movement - Highest linear densities Slip direction - direction of movement - Highest linear densities Deformation Mechanisms

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FCC FCC Slip occurs on {111} planes (close-packed) in directions (close-packed) => total of 12 slip systems in FCC

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(111) Parallel

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(111), [101] or [101]---1 (111), [110] or [110]---2 (111), [011] or [011] So, for (111) plane: Therefore, for an FCC structure: {111} -, there are 12 slip systems.

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To sum up: Slip phenemenon is used to explain the plastic behaviour of materials. Slip occurs along certain crystal planes and directions. Slip planes & slip directions make slip systems. For an FCC structure {111} -, there are 12 slip systems. For an BCC structure {110} - → 12 slip systems. For an HCP structure → 3 slip systems.

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The stress that initiates slip is known as the critical resolved shear stress. Ø: Angle between the normal to the slip plane and the applied stress directions. λ: Angle between the slip and stress directions. AsAs A F τ CR = σ. cos λ. cos Ø

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Stress and Dislocation Motion Crystals slip due to a resolved shear stress, R. Applied tension can produce such a stress. slip plane normal, n s Resolved shear stress: RR =F s /A/A s slip direction ASAS RR RR FSFS slip direction Relation between and RR RR = FSFS /AS/AS Fcos A/cos F FSFS nSnS ASAS A Applied tensile stress: = F/A slip direction F A F

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F s = F cos λ (shear force along the slip direction) τ CR = σ cos λ cos Ø A s = A / cos Ø (shearing area) cos λ. cos Ø F AA / cos Ø F cos λ FsFs A sA s τ CR == = σ

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Single Crystal Slip Adapted from Fig. 7.8, Callister 7e. Adapted from Fig. 7.9, Callister 7e.

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Ex: Deformation of single crystal So the applied stress of 65 MPa will not cause the crystal to yield. =35° =60° crss = 30 MPa a) Will the single crystal yield? b) If not, what stress is needed? = 65 MPa Adapted from Fig. 7.7, Callister 7e.

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Ex: Deformation of single crystal What stress is necessary (i.e., what is the yield stress, y )? (!We will learn about yield stress later on!) So for deformation to occur the applied stress must be greater than or equal to the yield stress

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Ex: Consider a single crystal of BCC iron oriented such that a tensile stress is applied along [010] direction. a) Compute the resolved shear stress along a (110) plane and in a [111] direction when a tensile stress of 52 MPa is applied. b) If slip occurs for the above plane and direction, and the critical resolved shear stress is 30 MPa, calculate the magnitude of the applied tensile stress necessary to initiate yielding.

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a) Ø=45° angle b/w normal [110] & stress [010] σ= 52MPa (110) [110] z y x σ= 52 cos Ø = 1*0 + 1*1 + 0*0 ( ) ( ) = 1 √2 Ø = 45° λ: angle b/w slip direction [111] & stress direction [010] cos λ = -1*0 + 1*1 + 1*0 ( ) (1) = 1 √3 λ = 54.7° τ CR = 52 * cos 45 * cos 54.7 = 21.3 MPa

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b) σ y = τ CR cos Ø cos λ = 30 cos 45 cos 54.7 σ y = 73.4 MPa

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Stronger - grain boundaries pin deformations Slip planes & directions (, ) change from one crystal to another. R will vary from one crystal to another. The crystal with the largest R yields first. Other (less favorably oriented) crystals yield later. Adapted from Fig. 7.10, Callister 7e. (Fig is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].) Slip Motion in Polycrystals 300 m

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Condition for dislocation motion: Crystal orientation can make it easy or hard to move dislocation GPa to GPa typically Critical Resolved Shear Stress maximum at = = 45º RR = 0 =90° RR = /2 =45° RR = 0 =90°

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DISLOCATIONS & σ-ε CURVES

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I II III

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Elastic Region: σ y is the stress required to start plastic deformation. It is called the yield stress. From σ y & on → plastic deformation starts. Stage I: dσ/dε ≈ 0 (work hardening rate) is low because only the primary slip systems are active. The slip planes are parallel to each other and only these parallel planes will slip and they do not intersect themselves, i.e. dislocations are moving along parallel planes.

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Stage II: Other dislocations will start to move along intersecting planes (more than one slip system becomes active). Therefore they form barriers to one another’s motion. It becomes harder to further deform the material. This stage is known as Work Hardening Stage. Stage III: The geometry of the planes have so changed that the planes will accelerately slip and failure will occur.

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