CHAPTER 7: DISLOCATIONS AND STRENGTHENING

1
CHAPTER 7 DISLOCATIONS AND STRENGTHENING

• In Ch.6
• Plastic (permanent) Elastic (reversible)
• Yield Strength and hardness are a measure of
  materials resistance to deformation
• In microscopic scale what is going on ?
• Net movement of large number of atoms in response
  to an applied stress
• Involves movement of dislocations, in Ch. 4
• How does strengthening happen ?
• Why do some materials are stronger than others ?
• How can one manipulate dislocations and their
  motion ?

2
Dislocations Materials Classes
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.
4
DISLOCATION MOTION
Plastically stretched zinc single crystal.
Produces plastic deformation ! Bonds are
incrementally broken and re-formed. Much less
force is needed , why ?
Adapted from Fig. 7.9, Callister 6e. (Fig. 7.9 is
from C.F. Elam, The Distortion of Metal Crystals,
Oxford University Press, London, 1935.)
SLIP PLANE
Adapted from Fig. 7.1, Callister 6e. (Fig. 7.1 is
adapted from A.G. Guy, Essentials of Materials
Science, McGraw-Hill Book Company, New York,
1976. p. 153.)
Adapted from Fig. 7.8, Callister 6e.
SEE THE MOVIE AGAIN !
If dislocations don't move, plastic
deformation doesn't happen!
How do we generate the dislocation motion ?
3
5
DISLOCATION MOTION
deformed
Initial state
apply force
O1
O0
t0
t1
t2
t3
The motion of a single dislocation across the
plane causes the top half of the crystal to move
(to slip) with respect to the bottom half but we
can not have to break all the bonds across the
middle plane simultaneously (which would require
a very large force). The slip plane (O0-O1) the
crystallographic plane of dislocation motion.
6
STRESS and Dislocation Motion
Dislocation line
NOT be parallel to the dislocation line !
7
Dislocation Motion

• Dislocation moves along slip plane in slip
  direction perpendicular to dislocation line
• Slip direction same direction as Burgers vector

Edge dislocation
Adapted from Fig. 7.2, Callister 7e.
Screw dislocation
8
STRESS AND DISLOCATION MOTION
9
STRESS field around dislocations
Bonds are Compressed
Bonds are Stretched
WHY is there a stress field ? Atoms try to relax
by trying to achieve their positions for the case
if there was not a dislocation in the vicinity !
10
Stress fields of dislocations interacting !
The strain fields around dislocations interact
with each other. Hence, they exert force on each
other. Edge dislocations, when they are in the
same plane, they repel each other if they have
the same sign (direction of the Burgers vector).
WHY ? They can attract and annihilate if they
have opposite signs. PROVE !
11
Dislocation Density

• The number of dislocations in a material is
  expressed using the term dislocation density -
  the total dislocation length per unit volume or
  the of dislocations intersecting a unit area.
  Units are mm / mm3, or just / mm
• Dislocation densities can vary from 103 mm-2 in
  carefully solidified metal crystals to 1010 mm-2
  in heavily deformed metals.
• Where do Dislocations come from, what are their
  sources ?
• Most crystalline materials, especially metals,
  have dislocations in their as-formed state,
  mainly as a result of stresses (mechanical,
  thermal...) associated with the manufacturing
  processes used.
• The number of dislocations increases dramatically
  during plastic deformation.
• Dislocations spawn from existing dislocations,
  grain boundaries and surfaces and other defects
  .

NICE SIMULATIONS gt http//www.ims.uconn.edu/cente
rs/simul/movie/
12
Deformation Mechanisms

• 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
  lt110gt directions (close-packed)
• gt total of 12 slip systems in FCC
• in BCC HCP other slip systems occur

Adapted from Fig. 7.6, Callister 7e.
13
SLIP SYSTEMS !!!

• Dislocations move with ease on certain
  crystallographic planes and along certain
  directions on these planes !
• The plane is called a slip plane
• The direction is called a slip direction
• Combination of the plane of slip and direction is
  a slip system
• The slip planes and directions are those of
  highest packing density.
• The distance between atoms is shorter than the
  averageHigh number of coordination along the
  planes also important !

PLS SEE TABLE 7.1 in ur books , page 180 !!!
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Concept Check
Page 181, in 7th edition !
15
Stress and Dislocation Motion
Crystals slip due to a resolved shear stress,
tR.
Applied tension can produce such a stress.
lf ? 90
16
Critical Resolved Shear Stress
Condition for dislocation motion
Crystal orientation can make it easy or
hard to move dislocation
? maximum at ? ? 45º
17
Resolving the Applied Stress on a SLIP PLANE !

• Shear Stress has to be resolved on to the slip
  planes as
• Shear Stress is needed for dislocations to move /
  slip
• Dislocations can only move on slip planes, and
  these planes are rarely on axis with the applied
  force !

We should resolve the force applied in a tensile
test, F, on to the cross-sectional area A where
the slip is going to take place
18
Critical Resolved Shear Stress gt Slip in Single
Xtals
Macroscopically Q when do materials plastically
deform / yield ? Stress gt YS Q How does
plastic deformation take place ? Dislocation
Motion ( Elastic deformation ??) Q On what
planes does dislocations move ? Slip Planes Q
At stress YS, what would be the minimum
resolved shear stress needed to act on
dislocations to initiate dislocation motion,
onset of yield/plastic deformation ? HIMMM !
Lets think !!!
19
The Critical Resolved Shear Stress
The minimum shear stress required to initiate
slip is termed the critical resolved shear stress
20
Single Crystal Slip
Plastically stretched zinc single crystal.
Adapted from Fig. 7.9, Callister 7e.
A large number of dislocations are generated on
slip planes, as they leave the system they form
these shear bands
Adapted from Fig. 7.8, Callister 7e.
21
Ex Deformation of single crystal
a) Will the single crystal yield? b) If not,
what stress is needed?
?60
?crss 3000 psi
?35
Adapted from Fig. 7.7, Callister 7e.
? 6500 psi

• So the applied stress of 6500 psi will not cause
  the crystal to yield.

22
Ex Deformation of single crystal
What stress is necessary (i.e., what is the
yield stress, sy)?
23
Slip Motion in Polycrystals
Stronger - grain boundaries pin
deformations Slip planes directions (l,
f) change from one crystal to another. tR
will vary from one crystal to another.
The crystal with the largest tR yields
first. Other (less favorably oriented)
crystals yield later.
Adapted from Fig. 7.10, Callister 7e. (Fig. 7.10
is courtesy of C. Brady, National Bureau of
Standards now the National Institute of
Standards and Technology, Gaithersburg, MD.)
24
Plastic Deformation in Polycrystalline Materials
25
Anisotropy in sy
Can be induced by rolling a polycrystalline
metal
- before rolling
Adapted from Fig. 7.11, Callister 7e. (Fig.
7.11 is from W.G. Moffatt, G.W. Pearsall, and J.
Wulff, The Structure and Properties of Materials,
Vol. I, Structure, p. 140, John Wiley and Sons,
New York, 1964.)
235 mm
- isotropic since grains are approx.
spherical randomly oriented.
26
Anisotropy in Deformation
side view
27
ANISOTROPY IN DEFORMATION
1. Cylinder of Tantalum machined
from a rolled plate
2. Fire cylinder at a target.
3. Deformed cylinder
Photos courtesy of G.T. Gray III, Los
Alamos National Labs. Used with permission.
side view
plate thickness direction
end view
The noncircular end view shows anisotropic
deformation of rolled material.
10
28
STRENGTHENING MECHANISMS

• The ability of a metal to deform depends on the
  ability of dislocations to move with relative
  ease under loading conditions
• Restricting dislocation motion will inevitably
  make the material stronger, need more force to
  induce same amount of deformation !
• Mechanisms of strengthening in single-phase
  metals
• grain-size reduction
• solid-solution alloying
• strain hardening
• Ordinarily, strengthening mechanisms reduces
  ductility, why ???

29
4 STRATEGIES FOR STRENGTHENING 1 REDUCE
GRAIN SIZE
Grain boundaries are barriers to slip.
Barrier "strength" increases with
mis-orientation. High-angle boundaries are
better in blocking slip ! Smaller grain
size more barriers to slip. Grain size
can be changed by processing !
Adapted from Fig. 7.12, Callister 6e. (Fig. 7.12
is from A Textbook of Materials Technology, by
Van Vlack, Pearson Education, Inc., Upper Saddle
River, NJ.)
Hall-Petch Equation
Average grain size
7
30
GRAIN SIZE STRENGTHENING AN EXAMPLE
70wtCu-30wtZn brass alloy
Data
Adapted from Fig. 7.13, Callister 6e. (Fig. 7.13
is adapted from H. Suzuki, "The Relation Between
the Structure and Mechanical Properties of
Metals", Vol. II, National Physical Laboratory
Symposium No. 15, 1963, p. 524.)
8
31
4 Strategies for Strengthening 2 Solid
Solutions
Impurity atoms distort the lattice generate
stress. Stress can produce a barrier to
dislocation motion.
32
Strategy 2 Solid Solutions

• Alloyed metals are usually stronger than their
  pure base metals counter parts.
• Why ? Interstitial or substitutional impurities
  in a solution cause lattice strain, aka
  distortions in the lattice
• Then ?
• Strain field around the impurities interact with
  dislocation strain fields and impede dislocation
  motion.
• Impurities tend to diffuse and segregate around
  the dislocation core to find atomic sites more
  suited to their radii. This reduces the overall
  strain energy and anchor the dislocation.
• Motion of the dislocation core away from the
  impurities moves it to a region of lattice where
  the atomic strains are greater, where lattice
  strains due to dislocation is no longer
  compensated by the impurity atoms.

33
Interactions of the Stress Fields
COMPRESSIVE
TENSILE
TENSILE
COMPRESSIVE
34
Interactions of the Stress Fields
35
Stress Concentration at Dislocations
Adapted from Fig. 7.4, Callister 7e.
36
Impurity Segregation
Impurities tend to segregate at energetically
favorable areas around the dislocation core and
partially decrease the overall stress field
generated around the dislocation core. However,
when stress is applied more load is needed to
move dislocations with impurity atoms segregated
to its core !
37
Strengthening by Alloying

• small impurities tend to concentrate at
  dislocations
• reduce mobility of dislocation ? increase
  strength

Adapted from Fig. 7.17, Callister 7e.
38
Strengthening by alloying

• large impurities concentrate at dislocations on
  low density side

Adapted from Fig. 7.18, Callister 7e.
39
Ex Solid SolutionStrengthening in Copper
Tensile strength yield strength increase
with wt Ni.
Adapted from Fig. 7.16 (a) and (b), Callister 7e.
Empirical relation
Alloying increases sy and TS.
40
4 Strategies for Strengthening 3
Precipitation Strengthening
Hard precipitates are difficult to shear.
Ex Ceramics in metals (SiC in Iron or Aluminum).
Large shear stress needed
to move dislocation toward
precipitate and shear it.
Dislocation
advances but
precipitates act as
pinning sites with
S
.
spacing
Result
41
Application Precipitation Strengthening
Internal wing structure on Boeing 767
Adapted from chapter-opening photograph, Chapter
11, Callister 5e. (courtesy of G.H. Narayanan and
A.G. Miller, Boeing Commercial Airplane Company.)
Aluminum is strengthened with precipitates
formed by alloying.
Adapted from Fig. 11.26, Callister 7e. (Fig.
11.26 is courtesy of G.H. Narayanan and A.G.
Miller, Boeing Commercial Airplane Company.)
42
4 Strategies for Strengthening 4 Cold Work
(CW)
Room temperature deformation. Common
forming operations change the cross
sectional area
43
Dislocations During Cold Work
Ti alloy after cold working
Dislocations entangle with one another
during cold work. Dislocation motion
becomes more difficult.
Adapted from Fig. 4.6, Callister 7e. (Fig. 4.6
is courtesy of M.R. Plichta, Michigan
Technological University.)
44
Result of Cold Work

• Dislocation density
• Carefully grown single crystal
• ? ca. 103 mm-2
• Deforming sample increases density
• ? 109-1010 mm-2
• Heat treatment reduces density
• ? 105-106 mm-2

Yield stress increases as rd increases
45
RESULT OF COLD WORK
Dislocation density (rd) goes up Carefully
prepared sample rd 103 mm/mm3 Heavily
deformed sample rd 1010 mm/mm3
Ways of measuring dislocation density
40mm
Micrograph adapted from Fig. 7.0, Callister 6e.
(Fig. 7.0 is courtesy of W.G. Johnson, General
Electric Co.)
OR
Yield stress increases as rd increases
18
46
STRENGTHENING STRATEGY 4 COLD WORK (CW)
An increase in sy due to plastic
deformation. BUT actually of dislocations are
increasing !!!
Curve fit to the stress-strain response
22
47
Stress fields of dislocations interacting !
The strain fields around dislocations interact
with each other. Hence, they exert force on each
other. Edge dislocations, when they are in the
same plane, they repel each other if they have
the same sign (direction of the Burgers vector).
WHY ? They can attract and annihilate if they
have opposite signs. PROVE !
48
Effects of Stress at Dislocations
Adapted from Fig. 7.5, Callister 7e.
49
IMPACT OF COLD WORK
Yield strength (s ) increases. Tensile
strength (TS) increases. Ductility (EL or
AR) decreases.
y
Adapted from Fig. 7.18, Callister 6e. (Fig. 7.18
is from Metals Handbook Properties and
Selection Iron and Steels, Vol. 1, 9th ed., B.
Bardes (Ed.), American Society for Metals, 1978,
p. 221.)
Stress
cold work
Strain
21
50
Impact of Cold Work
As cold work is increased
Yield strength (sy) increases.
Tensile strength (TS) increases.
Ductility (EL or AR) decreases.
Adapted from Fig. 7.20, Callister 7e.
51
Cold Work Analysis
What is the tensile strength ductility
after cold working?
Adapted from Fig. 7.19, Callister 7e. (Fig.
7.19 is adapted from Metals Handbook Properties
and Selection Iron and Steels, Vol. 1, 9th ed.,
B. Bardes (Ed.), American Society for Metals,
1978, p. 226 and Metals Handbook Properties
and Selection Nonferrous Alloys and Pure
Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.),
American Society for Metals, 1979, p. 276 and
327.)
52
s- e Behavior vs. Temperature
Results for polycrystalline iron
Adapted from Fig. 6.14, Callister 7e.
sy and TS decrease with increasing test
temperature. EL increases with increasing
test temperature. Why? Vacancies help
dislocations move past obstacles.
53
Effect of Heating After CW
1 hour treatment at Tanneal...
decreases TS and increases EL. Effects of
cold work are reversed!
3 Annealing stages to discuss...
Adapted from Fig. 7.22, Callister 7e. (Fig. 7.22
is adapted from G. Sachs and K.R. van Horn,
Practical Metallurgy, Applied Metallurgy, and the
Industrial Processing of Ferrous and Nonferrous
Metals and Alloys, American Society for Metals,
1940, p. 139.)
54
Recovery
Annihilation reduces dislocation density.
Scenario 2

55
Recrystallization
New grains are formed that -- have a
small dislocation density -- are smalller
then initial cold worked ones -- consume
cold-worked grains.
Adapted from Fig. 7.21 (a),(b), Callister 7e.
(Fig. 7.21 (a),(b) are courtesy of J.E. Burke,
General Electric Company.)
56
Further Recrystallization
All cold-worked grains are consumed.
Adapted from Fig. 7.21 (c),(d), Callister 7e.
(Fig. 7.21 (c),(d) are courtesy of J.E. Burke,
General Electric Company.)
57
Grain Growth
At longer times, larger grains consume smaller
ones. Why? Grain boundary area (and
therefore energy) is reduced.
Adapted from Fig. 7.21 (d),(e), Callister 7e.
(Fig. 7.21 (d),(e) are courtesy of J.E. Burke,
General Electric Company.)
58
º
TR recrystallization temperature
Adapted from Fig. 7.22, Callister 7e.
º
59
Coldwork Calculations

• A cylindrical rod of brass originally 0.40 in
  (10.2 mm) in diameter is to be cold worked by
  drawing. The circular cross section will be
  maintained during deformation. A cold-worked
  tensile strength in excess of 55,000 psi (380
  MPa) and a ductility of at least 15 EL are
  desired. Further more, the final diameter must
  be 0.30 in (7.6 mm). Explain how this may be
  accomplished.

60
Coldwork Calculations Solution

• If we directly draw to the final diameter what
  happens?

61
Coldwork Calc Solution Cont.
Adapted from Fig. 7.19, Callister 7e.

• For CW 43.8

• ?y 420 MPa

• TS 540 MPa gt 380 MPa

• EL 6 lt 15

• This doesnt satisfy criteria what can we do?

62
Coldwork Calc Solution Cont.
Adapted from Fig. 7.19, Callister 7e.
For TS gt 380 MPa
For EL lt 15
? our working range is limited to CW 12-27
63
Coldwork Calc Soln Recrystallization

• Cold draw-anneal-cold draw again
• For objective we need a cold work of CW ? 12-27
• Well use CW 20
• Diameter after first cold draw (before 2nd cold
  draw)?
• must be calculated as follows

64
Coldwork Calculations Solution

• Summary
• Cold work D01 0.40 in ? Df1 0.335
  m
• Anneal above D02 Df1
• Cold work D02 0.335 in ? Df 2 0.30 m
• Therefore, meets all requirements

Fig 7.19
?
65
Rate of Recrystallization

• Hot work ? above TR
• Cold work ? below TR
• Smaller grains
• stronger at low temperature
• weaker at high temperature

No Strain hardening occurs

• TR 0.3 Tm-0.7 Tm
• TR depends on CW
• Decrease with increasing CW

66
Summary
Dislocations are observed primarily in metals
and alloys.
Strength is increased by making dislocation
motion difficult.
Particular ways to increase strength are to
--decrease grain size --solid solution
strengthening --precipitate strengthening
--cold work
Heating (annealing) can reduce dislocation
density and increase grain size. This
decreases the strength.
67
ANNOUNCEMENTS
Reading Chapter 7
Core Problems 7.6, 7.7, 7.13 (see page 183-4),
7.14, 7.23, 7.31, 7.41 (figure 7.25)
Bonus Problems 7.40, 7.36, 7.39 Due date is
April 3th
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