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

2. 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

3. 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

4. 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

5. 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

6. 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)

7. 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)

8. 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)

9. 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)

10. 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

11. 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

12. Role of Imperfections in Plastic Deformation

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

14. 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

15. 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.

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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)

23. 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)

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

25. 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.

26. 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

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

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

29. ColdWorking Consequences
Dislocation linear density, ρd, increases:
Carefully prepared sample: ρd ~ 103 mm/mm3
Heavily deformed sample: ρd ~ 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:

31. 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

32. 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

33. -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.

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

35. 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

36. WhiteBoard Work
None
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