Lecture11-12 - Dislocations & Materials Classes • Metals:...

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1 1 Lecture 11-12 Mechanisms of plastic deformation and strengthening mechanisms 2 Dislocations & 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 cloud ion cores + + + + + + + + + + + ++++++ + + + + + + + 3 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). • If dislocations don't move, deformation doesn't occur! Adapted from Fig. 7.1, Cal ister 7e. 4 Dislocation Motion • Dislocation moves along slip plane in slip direction perpendicular to dislocation line • Slip direction same direction as Burgers vector Edge dislocation Screw dislocation Adapted from Fig. 7.2, Cal ister 7e. 5 Slip System – Slip plane - plane allowing easiest slippage • Wide interplanar spacings - highest planar densities –S l ip d i rec t ion - 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 Deformation Mechanisms Adapted from Fig. 7.6, Cal ister 7e. 6 Stress and Dislocation Motion • Crystals slip due to a resolved shear stress, τ R . • Applied tension can produce such a stress. slip plane normal, n s Resolved shear stress: τ R = F s / A s s lip d ir e c t io n A S τ R τ R F S Relation between σ and τ R τ R = F S / A S F cos λ A /cos φ λ F F S φ n S A S A Applied tensile stress: = F / A σ F A F φ λ σ = τ cos cos R
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2 7 • Condition for dislocation motion: CRSS τ > τ R • Crystal orientation can make it easy or hard to move dislocation 10 -4 GPa to 10 -2 GPa typically φ λ σ = τ cos cos R Critical Resolved Shear Stress τ maximum at λ = φ = 45º τ R = 0 λ =90° σ τ R = σ /2 λ =45° φ =45° σ τ R = 0 φ =90° σ 8 Single Crystal Slip Adapted from Fig. 7.8, Cal ister 7e. Adapted from Fig. 7.9, Cal ister 7e. 9 Ex: Deformation of single crystal So the applied stress of 6500 psi will not cause the crystal to yield. τ=σ cos λ cos φ σ = 6500 psi λ =35° φ =60° τ= (6500 psi) (cos35 o )(cos60 o ) = (6500 psi) (0.41) 2662 psi crss = 3000 psi τ crss = 3000 psi a) Will the single crystal yield? b) If not, what stress is needed? σ = 6500 psi Adapted from Fig. 7.7, Cal ister 7e. 10 Ex: Deformation of single crystal psi 7325 41 . 0 psi 3000 cos cos crss = = φ λ τ = σ y What stress is necessary (i.e., what is the yield stress, σ y )? ) 41 . 0 ( cos cos psi 3000 crss y y σ = φ λ σ = = τ psi 7325 = σ σ y So for deformation to occur the applied stress must be greater than or equal to the yield stress 11 • Stronger - grain boundaries pin deformations • Slip planes & directions ( λ , φ ) change from one crystal to another.
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This note was uploaded on 09/15/2010 for the course MATSCIE 250 taught by Professor Yalisove during the Fall '08 term at University of Michigan.

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Lecture11-12 - Dislocations & Materials Classes • Metals:...

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