Dislocations & Strengthening (1) Engineering 45 Dislocations & Strengthening (1) Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Learning Goals Understand Why DISLOCATIONS are observed primarily in METALS and ALLOYS Determine How Strength and Dislocation-Motion are Related Techniques to Increase Strength Understand How HEATING and/or Cooling can change Strength and other Properties

Theoretical Strength of Crystals The ideal or theoretical strength of a “perfect” crystal is  E/10 For Steel, E = 200 GPa Thus the theoretical strength 20 GPa 2,000 MPa is the practical limit for steel and this is an ORDER of MAGNITUDE Less than 20,000 MPa Most commercial steels have a strength  500 MPa - Why is there such differences? Answer = Crystal Imperfections

Role of Crystal Imperfections Crystal imperfections explain why metals are weak (relative to the Theoretical) and why they are so ductile In most applications we need ductility as well as strength - so there is a plus side to the presence of imperfections The main task in deciding what strengthening process to use in metal alloys is to chose a method which minimizes the loss of ductility

Edge Dislocations Recall from Chp.4 The Crystal Imperfection of an Extra ½-Plane of Atoms Called an EDGE DISLOCATION These imperfections are the Source of PLASTIC Deformation in Xtals Extra ½-Plane of Atoms

Dislocations vs. Metals Dislocation Motion is RELATIVELY Easier in Metals Due to NON-Directional Atomic Bonding Close-Packed Crystal Planes allow “sliding” of the Planes relative to each other Called SLIP Ion Cores Electron Sea Dislocations & Slip (Deformation)

Disloc vs. Covalent Ceramics For CoValent Ceramics Dislocation Motion is RELATIVELY more Difficult Due to Directional (angular) and Powerful Atomic Bonding Examples Diamond Carbon Silicon Strong, Directional Bonds Dislocations & Slip (Deformation)

Disloc vs. Ionic Ceramics For Ionic Ceramics Dislocation Motion is RELATIVELY more Difficult Due to Coulombic Attraction and/or Repulsion Slip Will Encounter ++ & -- Charged nearest neighbors + Ion Cores − Ion Cores Dislocations & Slip (Deformation)

Dislocations vs Matl Type Metals Allow Xtal Planes to Slip Relative to Each other Relatively Low Onset of Plastic Deformation (Yield Strength, σy) Relatively High Ductility: The amount of Plastic deformation Prior to Breaking Ceramics Tend to Prevent Disloc. Slip Allow for little Plastic Deformation Failure by Brittle-Fracture (cracking)

Dislocation Motion Produces Plastic Deformation In Crystals Proceeds by Incremental, Step-by-Step Breaking & Remaking of Xtal Bonds Unit Step appears as Dislocation on the Surface WithOut Dislocation motion Plastic (Ductile) Deformation Does NOT Occur

Screw Dislocations In the EDGE configuration The axis of  is Parallel (||) to the Applied Shear Stress EDGE Dislocation SHEARING Motion A SCREW dislocation is Perpendicular to the Applied Force SCREW Dislocation TEARING Motion

Role of Imperfections in Plastic Deformation

Dislocation Motion Analogies Caterpillar LoCoMotion Disloc Carpet-Layer LoCoMotion Unit Step appears as Dislocation on the Surface

Stress and Dislocation Motion Crystals slip due to a resolved shear stress, R Applied TENSION can Produce This -Stress direction slip l F s Relation between and tR = /A cos A / f direction slip Applied tensile stress: s = F/A F A direction slip slip plane normal, ns Resolved shear stress: tR = F s /A A

Resolved Shear Stress, R (in detail) Consider a single crystal of cross-sectional area A under compression force F   angle between the slip plane normal and the compression (or Tension) axis   angle between the slip direction and the tensile axis.

Resolved Shear Stress, R cont.1 F projected on Slip Direction: The Slip Direction Slant Area, As, Relative to the Compression Area, A As Fcosλ A = Ascos

Resolved Shear Stress, R cont.2 Thus the Resolved Shear Stress As But F/A = σ; the Compression (or Tension) Stress - So Fcosλ A = Ascos

Critical Resolved Shear Stress Condition for Dislocation Motion: R>CRSS CRSS  CRITICAL Resolved Shear Stress Xtal Orientation Can Facilitate Dicloc. Motion tR = 0 l = 90° s tR = /2 l = 45° f = 45° s tR = 0 f = 90° s HARD to Slip HARD to Slip EASY to Slip

Yield Stress, y An Xtal Plastically Deforms When Thus y = 2CRSS stretched zinc single crystal. To Get Yield Strength, Need minimum → (cos cos)max

PolyXtal Disloc Motion 300 mm PolyXtal Disloc Motion Slip planes & directions (l, f) change from one crystal to another tR varies from one crystal, or Grain, to another The Xtal/Grain with the LARGEST tR Yields FIRST Other (less favorably oriented) crystals Yield LATER

Summary  Edge Dislocations Plastic flow can occur in a crystal by the breaking and reattachment of atomic bonds one at a time This dramatically reduces the required shear stress Consider how a caterpillar gets from A to B A similar mechanism applies to screw dislocations Screw & Edge dislocations often occur together

1-Phase Metal Strengthening Basic Concept Plastic Deformation in Metals is CAUSED by DISLOCATION MOVEMENT Strengthening Strategy RESTRICT or HINDER Dislocation Movement Strengthening Tactics Grain Size Reduction Solid Solution Alloying Strain Hardening Precipitation (2nd-ph)

Strengthen-1  G.S. Reduction grain boundary slip plane grain A grain B Strengthen-1  G.S. Reduction Grain boundaries are barriers to slip due to Discontinuity of the Slip Plane Barrier "Strength“ Increases with Grain MisOrientation Smaller grain size → more Barriers to slip Hall-Petch Reln → Where 0  “BaseLine” Yield Strength (MPa) ky  Matl Dependent Const (MPa•m) d  Grain Size (m)

Example  GS Reduction Calc The Hall-Petch Slope, ky, for 70Cu-30Zn (C2600, or Cartridge) Brass Find the ’s Then the Slope

Strengthen-2  Solid Solution Impurity Atoms distort the Lattice & Generate Stress Stress Can produce a Barrier to Dislocation Motion Smaller substitutional impurity Larger substitutional impurity A B C D Impurity generates local shear at A and B that opposes dislocation motion to the right. Impurity generates local shear at C & D that opposes dislocation motion to the right.

Example  Ni-Cu Solid-Soln Tensile (Ultimate) Strength, σu, and & Yield Strength, σy, increase with wt% Ni in Cu Empirical Relation: σy ~ C½ Basic Result: Alloying increases σy & σu

Strengthen-3  Strain Harden COLD WORK  Room Temp Deformation Common forming operations Change The Cross-Sectional Area: -Forging -Rolling -Drawing -Extrusion

Dislocations During Cold Work ColdWorked Ti Alloy 0.9 m Dislocations entangle with one another during COLD WORK Dislocation motion becomes more difficult

ColdWorking Consequences Dislocation linear density, ρd, increases: Carefully prepared sample: ρd ~ 103 mm/mm3 Heavily deformed sample: ρd ~ 1010 mm/mm3 Measuring Dislocation Density 40mm d = N A Area , A N dislocation pits (revealed by etching) dislocation pit r length, l 1 2 3 Volume, V = l + V r d OR σy Increases as ρd increases:

CW Strengthening Mechanism Strain Hardening Explained by Dislocation-Dislocation InterAction Cold Work INCREASES ρd Thus the Average - Separation-Distance DECREASES with Cold Work Recall - interactions are, in general, REPULSIVE Thus Increased ρd IMPEDES -Motion

Simulation – DisLo Generator Tensile loading (horizontal dir.) of a FCC metal with notches in the top and bottom surface Over 1 billion atoms modeled in 3D block. Note the large increase in Dislocation Density

-Motion Impedance Dislocations Generate Stress This Generates -Traps Red dislocation generates shear at pts A and B that opposes motion of green disl. from left to right.

ColdWork Results-Trends As Cold Work Increases Yield Strength, y, INcreases Ultimate Strength, u, INcreases Ductility (%EL or %RA) DEcreases

Post-Work Ductility is HAMMERED Cold Work Example What is the Tensile Strength & Ductility After Cold Working? s y =300MPa % Cold Work 100 3 00 5 7 Cu 2 4 6 yield strength (MPa) 300MPa Post-Work Ductility is HAMMERED

WhiteBoard Work None Today