Chapter 7 - ANNOUNCEMENTS - 3/1 Reading: Chapter 7 Homework...

Info iconThis preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 7 - 1 ANNOUNCEMENTS - 3/1 Reading: Chapter 7 Homework #5 due today (3/1)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 - 2 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
Background image of page 2
Chapter 7 - 3 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 + + + + + + + + + + + + + + + + + + + + + + +
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 - 4 In a perfect crystal . . . 2a x τ . . . b τ m ax ) 2 sin( max a x π τ τ= ) 2 ( max a x ) ( ) 2 ( max b x G a x = = When x is small, Hooke’s Law for elasticity, τ = G ε = G (x/b) this is the theoretical strength for a perfect crystal For metals, τ y ~ 10 -5 – 10 -3 G !! G G G G b a 1 . 0 10 2 ) 2 )( ( max = =
Background image of page 4
Chapter 7 - 5 Yield strength in crystals τ y ~ 10 -5 – 10 -3 G !! Metals τ th (= G/30), GPa τ exp , MPa τ exp / τ th Al 0.9 0.78 0.00087 Cu 1.4 0.49 0.00035 Ni 2.6 3.2 0.007 -Fe α 2.6 27.5 0.011 Ag 1.0 0.37 0.00037
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 - 6 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.
Background image of page 6
Chapter 7 - 7 Dislocation Motion Dislocations & plastic deformation Dislocation movement requires only the switch of local bonds much less energy Edge dislocation Screw dislocation
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 - 8 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.
Background image of page 8
Chapter 7 - 9 Slip System Slip plane - plane allowing easiest slippage Wide interplanar spacings - highest planar densities (smoothest) Slip direction - direction of movement - Highest linear densities (shortest interatomic distance) 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.
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 - 10 Stress and Dislocation Motion • Crystals slip due to a resolved shear stress, τ R . • Applied tension can produce such a stress. slip plane Resolved shear stress: τ = F s / A s slip direction A τ τ F Relation between σ and τ R = F F cos λ A /cos φ λ F F φ A Applied tensile stress: = F / A σ F A F φ λ σ = τ cos cos R φ λ cos cos : factor Schmid
Background image of page 10
Chapter 7 - 11 • 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 = 0 λ =90° σ = σ /2 λ =45° φ =45° σ = 0 φ =90° σ
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 - 12 Single Crystal Slip Adapted from Fig. 7.8, Callister 7e.
Background image of page 12
Image of page 13
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 54

Chapter 7 - ANNOUNCEMENTS - 3/1 Reading: Chapter 7 Homework...

This preview shows document pages 1 - 13. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online