mse07 - Chapter 7: Dislocations & Strengthening Mechanisms...

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1 Chapter 7 - 1 ISSUES TO ADDRESS. .. • Why are dislocations observed primarily in metals and alloys? • How are strength and dislocation motion related? • How do we increase strength? • How can heating change strength and other properties? Chapter 7: Dislocations & Strengthening Mechanisms Chapter 7 - 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 + + + + + + + + + + + + +++++ + + + + + + +
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2 Chapter 7 - 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, Callister 7e. Chapter 7 - 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, Callister 7e.
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3 Chapter 7 - 5 Slip System – Slip plane - plane allowing easiest slippage • Wide interplanar spacings - highest planar densities – Slip direction - 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, Callister 7e. Chapter 7 - 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 l i p d r e c t o 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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4 Chapter 7 - 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 Chapter 7 - 8 Single Crystal Slip Adapted from Fig. 7.8, Callister 7e. Adapted from Fig. 7.9, Callister 7e.
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5 Chapter 7 - 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, Callister 7e. Chapter 7 - 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 )?
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This note was uploaded on 03/18/2010 for the course ME 2580 taught by Professor Vandenbrink during the Spring '08 term at Western Michigan.

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mse07 - Chapter 7: Dislocations & Strengthening Mechanisms...

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