Chapter 7
Dislocations &
Strengthening Mechanisms
Chapter 7 - 1
Chapter 7: Dislocations & Strengthening
Mechanisms
ISSUES TO ADDRESS...
• How are dislocations involved in the plastic deformation
of metals/metal alloys?
• Does the crystal structure of a metal affect its mechanical
characteristics? If so, how and why?
• How are mechanical properties affected by
dislocation mobilities?
• What techniques are used to increase the
strength/hardness of metals/alloys?
• How are mechanical characteristics of deformed
metal specimens altered by heat treatments?
Chapter 7 - 2
Plastic Deformation by Dislocation
Motion
• Plastic deformation occurs by motion of dislocations
(edge, screw, mixed) – process called slip
• Applied shear stress can cause extra half-plane of atoms
[and edge dislocation line ( )] to move as follows: Fig. 7.1, Callister &
Rethwisch 10e.
• Atomic bonds broken and reformed along slip plane as
dislocation (extra half plane) moves.
Chapter 7 - 3
• So we saw that above the yield stress
plastic deformation occurs. But how? In
a perfect single crystal for this to occur
every bond connecting two planes would
have to break at once! Large energy
requirement
• Now rather than entire plane of bonds
needing to be broken at once, only the
bonds along dislocation line are broken at
once.
Chapter 7 - 4
Analogy Between Dislocation
Motion and Caterpillar Locomotion
• Caterpillar locomotion – hump formed and propelled by
lifting and shifting of leg pairs
• Dislocation motion – movement of extra half-plane of
atoms by breaking and reforming of interatomic bonds
only the bonds along dislocation line are broken at once.
rather than entire plane of bonds needing to be broken at once,
Chapter 7 - 5
Motion of Edge and Screw Dislocations
• Direction of edge disl. line ( ) motion—in direction of
applied shear stress τ.
Edge dislocation
• Direction of screw disl. line ( ) motion—perpendicular to
direction of applied shear stress.
Screw dislocation
Fig. 7.2, Callister & Rethwisch 10e.
Chapter 7 - 6
Slip Systems
Slip System—Combination of slip plane and slip direction
– Slip Plane
• Crystallographic plane on which slip occurs most
easily
• Plane with high planar density
– Slip Direction
• Crystallographic direction along which slip occurs
most easily
• Direction with high linear density
Chapter 7 - 7
Slip Systems (cont.)
• For FCC crystal structure – slip system is {111} 110
– Dislocation motion on {111} planes
– Dislocation motion in 110 directions
– A total of 12 independent slip systems for FCC
110
direction
Fig. 7.6, Callister &
Rethwisch 10e.
{111}
plane
• For BCC and HCP— other slip systems
Chapter 7 - 8
Chapter 7 - 9
Chapter 7 - 10
• For BCC and HCP, slip is possible on
more than one family of plane
Chapter 7 - 11
Slip in Single Crystals
Resolved Shear Stress
• Applied tensile stress—shear stress component exist
when slip plane oriented neither perpendicular nor
parallel to tensile stress direction
ϕ
-- From figure, resolved shear stress, τR
t
R
λ
F¢
=
A¢
• τR depends on orientation of normal to
slip plane and slip direction with direction
of tensile force F:
F¢ = F cos l
A
A¢ =
cos f
Chapter 7 - 12
Slip in Single Crystals
Resolved Shear Stress (cont.)
• Relationship between tensile stress, σ,
and τR:
ϕ
λ
F ¢ F cos l F
tR = =
= cos l cos f
A
A¢
A
cos f
  cos  cos 
Chapter 7 - 13
Slip in Single Crystals:
Critical Resolved Shear Stress
• In a single crystal there are
-- multiple slip systems
-- a variety of orientations
• One slip system for which τR is highest: τR(max) = (cos cosϕ )max
-- Most favorably oriented slip system.
• dislocation moves on the most favorable slip system when τR
reaches a critical value critical resolved shear stress τcrss
--represents the minimum shear stress required to initiate slip.
•deform or yielding occurs when τR (max) = τcrss
Yield strength of single crystal
t CRSS
sy =
(cos l cos f )
max
Chapter 7 - 14
Single Crystals
Slip—Macroscopic Scale
• Parallel slip steps form on
surface of single crystal
• Steps result from motion of
large numbers of
dislocations on same slip
plane
• Sometimes on single
crystals appear as "slip
lines" (see photograph)
Fig. 7.8, Callister &
Rethwisch 10e.
Chapter 7 - 15
•
Plastic Deformation Behavior and Mechanism of Bismuth
Single Crystals
•
hexagonal cobalt single crystal
shows the surface features
associated with slip
Chapter 7 - 16
Dislocation Characteristics
Metals
+
+
+
+
+
+
+
electron cloud
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
ion cores
• Metals:
- Examples: copper, aluminum, iron
- Dislocation motion—relatively easy
- Metallic bonding—non-directional
- Close-packed planes and directions for slip
Chapter 7 - 17
Dislocation Characteristics
Ceramics
• Ceramics—Covalently Bonded
- Examples: silicon, diamond
- Dislocation motion—relatively
difficult
- Covalent bonding—directional
• Ceramics—Ionically Bonded
- Examples: NaCl, MgO
- Dislocation motion—relatively
difficult
- Few slip systems
 motion of nearby ions of like
charge (+ and -) restricted by
electrostatic repulsive forces
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
Chapter 7 - 18
Deformation of Single Crystals
Example Problem
A single crystal of some metal has a τcrss of 20.7 MPa and is
exposed to a tensile stress of 45 MPa.
(a) Will yielding occur when ϕ = 60° and  = 35°?
(b) If not, what stress is necessary?
Solution:
(a) First calculate τR
t R = s cos l cos f
t R = (45 MPa) éëcos(35°)cos(60°)ùû
= 18.4 MPa
Since t R (18.4 MPa) < t crss (20.7 MPa) -- no yielding
Chapter 7 - 19
Deformation of Single Crystals
Example Problem (cont.)
(b) To calculate the required tensile stress to
cause yielding use the equation:
t CRSS
sy =
cos l cos f
With specified values
20.7 MPa
sy =
cos(35°)cos(60°)
= 50.5 MPa
Therefore, to cause yielding,
s ³ 50.5 MPa
Chapter 7 - 20
Slip in Polycrystalline Materials
• Polycrystalline materials—
many grains, often random
crystallographic orientations
σ
• Orientation of slip planes, slip
directions (ϕ, λ)—vary from
grain to grain.
Adapted from Fig.
7.10, Callister &
Rethwisch 10e.
(Photomicrograph
courtesy of C. Brady,
National Bureau of
Standards [now the
National Institute of
Standards and
Technology,
Gaithersburg, MD].)
• On application of stress—slip
in each grain on most favorable
slip system.
- with largest τR
- when τR > τcrss
• In photomicrograph—note slip
lines in grains have different
orientations.
300 μm
σ
Chapter 7 - 21
Slip in Polycrystalline Materials (cont.)
• Grains change shape (become distorted)—during plastic
deformation—due to slip
• Manner of grain distortion similar to gross plastic deformation
-
Grain structures before and after deformation (from rolling)
Before rolling—grains equiaxed & randomly oriented
 Properties isotropic
-
After rolling (deformation)—grains elongated in rolling direction
 Also preferred crystallographic orientation of grains
 Properties become somewhat anisotropic
- before rolling
- after rolling
Adapted from Fig. 7.11,
Callister & Rethwisch 10e.
(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 μm
rolling direction
Chapter 7 - 22
Strengthening Mechanisms for Metals
• For a metal to plastically deform—dislocations must move
• Strength and hardness—related to mobility of dislocations
-- Reduce disl. mobility—metal strengthens/hardens
-- Greater forces necessary to cause disl. motion
-- Increase disl. mobility—metal becomes weaker/softer
• Mechanisms for strengthening/hardening metals—
decrease disl. mobility
• 3 mechanisms discussed
-- Grain size reduction
-- Solid solution strengthening
-- Strain hardening (cold working)
Chapter 7 - 23
Strengthening Mechanisms for Metals
Mechanism I – Reduce Grain Size
• Grain boundaries act as barriers
to dislocation motion
• At boundary
— Slip planes change directions
(note in illustration)
— Discontinuity of slip planes
• Reduce grain size
— increase grain boundary area
— more barriers to dislocation motion
— increase yield strength, tensile
strength & hardness
Fig. 7.14, Callister & Rethwisch 10e.
(From L. H. Van Vlack, A Textbook of Materials
Technology, Addison-Wesley Publishing Co., 1973.
Reproduced with the permission of the Estate of
Lawrence H. Van Vlack.)
• Dependence of σy on average grain diameter, d:
s yield = s 0 + k y d -1/2
—σ0, ky = material constants
Chapter 7 - 24
Strengthening Mechanisms for Metals
Mechanism II – Solid-Solution Strengthening
• Lattice strains around dislocations
– Illustration notes locations of tensile, compressive
strains around an edge dislocation
Fig. 7.4, Callister
& Rethwisch 10e.
(Adapted 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.)
Chapter 7 - 25
Solid Solution Strengthening (cont.)
• Lattice strain interactions with strains introduced by impurity atoms
• Small substitutional impurities introduce tensile strains
• When located above slip line for edge dislocation as shown:
– partial cancellation of impurity (tensile) and disl. (compressive) strains
– higher shear stress required to cause disl. motion
Fig. 7.17, Callister &
Rethwisch 10e.
Chapter 7 - 26
Solid Solution Strengthening (cont.)
• Large substitutional impurities introduce compressive strains
• When located below slip line for edge dislocation as shown:
– partial cancellation of impurity (compressive) and disl. (tensile) strains
– higher shear stress required to cause disl. motion
Fig. 7.18, Callister &
Rethwisch 10e.
Chapter 7 - 27
VMSE Solid-Solution Strengthening Tutorial
Chapter 7 - 28
Solid Solution Strengthening (cont.)
• Alloying Cu with Ni increases σy and TS.
400
300
200
0 10 20 30 40 50
wt.% Ni, (Concentration C)
Yield strength (MPa)
Tensile strength (MPa)
• Tensile strength & yield strength increase with wt% Ni.
180
Adapted from Fig.
7.16 (a) and (b),
Callister &
Rethwisch 10e.
120
60
0 10 20 30 40 50
wt.%Ni, (Concentration C)
• Empirically, s y µC1/2
Chapter 7 - 29
Strengthening Mechanisms for Metals
Mechanism III – Strain Hardening
• Plastically deforming most metals at room temp. makes
them harder and stronger
• Phenomenon called "Strain hardening (or cold working)”
• Deformation—often reduction in cross-sectional area.
-Rolling
roll
Ao
Ad
roll
• Deformation amt. = percent coldwork (%CW)
Ao - Ad
%CW =
x 100
Ao
Chapter 7 - 30
Strain Hardening (cont.)
As %CW increases
• Yield strength (σy) increases.
• Tensile strength (TS) increases.
• Ductility (%EL or %AR) decreases.
Adapted from Fig. 7.20,
Callister & Rethwisch 10e.
low carbon steel
Chapter 7 - 31
Strain Hardening (cont.)
Lattice strain interactions between dislocations
Fig. 7.5, Callister &
Rethwisch 10e.
Chapter 7 - 32
Strain Hardening (cont.)
Dislocation Density and Cold Working
total dislocation length
Dislocation density =
unit volume
– Dislocation density in undeformed metal
 105-106 mm-2
– Dislocation density increases with increasing deformation
– Dislocation density in deformed (cold-worked) metal
 109-1010 mm-2
Chapter 7 - 33
Strain Hardening (cont.)
Mechanism of Strain Hardening
• Dislocation structure in Ti after cold working.
• Dislocation density increases
with deformation (cold work) by
formation of new dislocations
• As dislocation density
increases, distance between
dislocations decreases
• On average, disl.-disl. strain
interactions are repulsive
• Dislocation motion hindered by
presence of other dislocations
Fig. 4.7, Callister & Rethwisch 10e.
(Courtesy of M.R. Plichta, Michigan
Technological University.)
Chapter 7 - 34
Affect of Cold Work on Mechanical
Properties
Example Problem:
Compute the yield and tensile strengths, and ductility for a
cylindrical Cu specimen that has been cold worked by
reducing its diameter from 15.2 mm to 12.2 mm.
Copper
Cold
Work
Do = 15.2 mm
Dd = 12.2 mm
Chapter 7 - 35
Example Problem (cont.)
• Solution:
%CW =
=
%CW =
æ D ö2
æ D ö2
p çç o ÷÷ - p çç d ÷÷
è 2 ø
è 2 ø
æD ö
p çç o ÷÷
è 2 ø
2
Do2 - Dd2
Do2
x 100
x 100
(15.2 mm)2 - (12.2 mm)2
(15.2 mm)
2
x 100 = 35.6%
Chapter 7 - 36
Example Problem (cont.)
500
300
300 MPa
100
0
20
40
Cu
% Cold Work
60
σy = 300 MPa
60
800
600
400 340 MPa
Cu
200
0
20
40
60
% Cold Work
ductility (%EL)
700
tensile strength (MPa)
yield strength (MPa)
• Yield and tensile strength, and ductility (%EL) are
determined graphically as shown below for %CW = 35.6%
40
Cu
20
7%
00
20
40
60
% Cold Work
TS = 340 MPa
%EL = 7%
Fig. 7.19, Callister & Rethwisch 10e. [Adapted from Metals Handbook: Properties and Selection: Irons
and Steels, Vol. 1, 9th edition, B. Bardes (Editor), 1978; and Metals Handbook: Properties and Selection: Nonferrous
Alloys and Pure Metals, Vol. 2, 9th edition, H. Baker (Managing Editor), 1979. Reproduced by permission of ASM
International, Materials Park, OH.]
Chapter 7 - 37
Heat Treatment of Cold-Worked
Metal Alloys
600
60
tensile strength
50
500
40
400
30
ductility
20
300
100 200 300 400 500 600 700
annealing temperature (°C)
ductility (%EL)
tensile strength (MPa)
• Heat treating cold worked metals brings about changes in structure
and properties
• As a result, effects of cold work are nullified!
• This type of heat treatment sometimes termed “annealing”
• 1 hour treatment at Tanneal decreases tensile strength & increases %EL
Three Annealing stages:
1. Recovery (100-200°C)
2. Recrystallization (200500°C)
3. Grain Growth (> 500°C)
Fig. 7.22, Callister & Rethwisch 10e.
(Adapted from G. Sachs and K. R. Van Horn,
Practical Metallurgy, Applied Metallurgy
and the Industrial Processing of Ferrous and
Nonferrous Metals and Alloys, 1940.
Reproduced by permission of ASM
International, Materials Park, OH.)
Chapter 7 - 38
EXAMPLE PROBLEM 7.2
Solution
Chapter 7 - 39
Recovery
During recovery – reduction in disl. density – annihilation of disl.
• Scenario 1
• Scenario 2
extra half-plane
of atoms
atoms
diffuse
to regions
of tension
extra half-plane
of atoms
3. “Climbed” disl. can now
move on new slip plane
2. grey atoms leave by
vacancy diffusion
allowing disl. to “climb”
1. dislocation blocked;
can’t move to the right
Dislocation
annihilationhalf-planes
come together
4. dislocations of opposite
sign meet and annihilate
Obstacle dislocation
Chapter 7 - 40
Recrystallization
• New grains form that:
-- have low dislocation densities
-- are small in size
-- consume and replace parent cold-worked grains.
Recrystallized grains
Adapted from Fig.
7.21 (a),(c),
Callister &
Rethwisch 10e.
(Photomicrographs
courtesy of J.E.
Burke, General
Electric Company.)
33%CW brass before heat treatment
After 4 sec. at 580°C
Chapter 7 - 41
Recrystallization (cont.)
• All grains in cold-worked material have been consumed/replaced.
Adapted from Fig.
7.21 (d), Callister
& Rethwisch 10e.
(Photomicrographs
courtesy of J.E. Burke,
General Electric
Company.)
After 8 sec. at 580°C
Chapter 7 - 42
Recrystallization Temperature
TR = recrystallization temperature = temperature
at which recrystallization just reaches
completion in 1 h.
0.3Tm < TR < 0.6Tm
For a specific metal/alloy, TR depends on:
• %CW -- TR decreases with increasing %CW
• Purity of metal -- TR decreases with
increasing purity
Chapter 7 - 43
Cold Working vs. Hot Working
• Hot working  deformation above TR
• Cold working  deformation below TR
Chapter 7 - 44
Design EXAMPLE 7.1
Solution
Chapter 7 - 45
Grain Growth
• Grain growth occurs as heat treatment continues.
-- Average grain size increases
-- Small grains shrink (and ultimately disappear)
-- Large grains continue to grow
Adapted from Fig.
9.21 (d),(e), Callister
& Rethwisch 10e.
(Photomicrographs
courtesy of J.E. Burke,
General Electric
Company.)
After 8 sec. at 580°C
After 15 min. at
580°C
Chapter 7 - 46
Grain Size Influences Properties
• Metals having small grains – relatively strong
and tough at low temperatures
• Metals having large grains – good creep
resistance at relatively high temperatures
Chapter 7 - 47
Grain Growth (cont.)
• Empirical relationship—dependence of average grain
size (d) on heat treating time (t):
exponent typ. ~ 2
material constant
—depends on T
—independent of t
d n - don = Kt
Initial average grain
diam. before heat
treatment
Chapter 7 - 48
Recovery, Recrystallization, & Grain Growth
Summary
TR = recrystallization
temperature
annealing time = 1 h
TR
Fig. 7.22, Callister & Rethwisch 10e.
(Adapted from G. Sachs and K. R. Van Horn,
Practical Metallurgy, Applied Metallurgy
and the Industrial Processing of Ferrous and
Nonferrous Metals and Alloys, 1940.
Reproduced by permission of ASM
International, Materials Park, OH.)
º
Chapter 7 - 49
Design Problem
Description of Diameter Reduction
Procedure
A cylindrical rod of brass originally 10 mm in diameter
is to be cold worked by drawing. The circular cross
section will be maintained during deformation. A coldworked tensile strength in excess of 380 MPa and a
ductility of at least 15 %EL are desired. Furthermore,
the final diameter must be 7.5 mm. Explain how this
may be accomplished.
Chapter 7 - 50
Design Problem (cont.)
Solution:
First compute the %CW.
Brass
Cold
Work
D o = 10 mm
D d = 7.5 mm
æA -A ö
æ A ö
o
d
÷÷ x 100 = çç1- d ÷÷ x 100
%CW = çç
è Ao ø
è Ao ø
2ù
é æ
é p (D 2)2 ù
ö
7.5 mm ú
d
ê
ê
ú
= 1x 100 = 1- ç
÷ x 100 = 43.8%
2
ê è 10 mm ø ú
êë p (Do 2) úû
ë
û
Chapter 7 - 51
Design Problem Solution (cont.)
60
800
40
600
540
20
400
200
0
20
40
60
6
0
0
% Cold Work
20
40
% Cold Work
60
Fig. 7.19, Callister
& Rethwisch 10e.
• For %CW = 43.8%
– TS = 540 MPa > 380 MPa
– %EL = 6
< 15
• This doesn’t satisfy criteria… what other options are
possible?
Chapter 7 - 52
Design Problem Solution (cont.)
60
800
40
600
20
400
15
380
200
0
1220
40
% Cold Work
60
0
0
20 27 40
60
% Cold Work
For TS > 380 MPa
> 12 %CW
For %EL > 15
< 27 %CW
Fig. 7.19, Callister
& Rethwisch 10e.
To meet criteria
—deformation requirement 12 < %CW < 27
Chapter 7 - 53
Design Problem Solution (cont.)
Procedure: Cold work, anneal, then cold work again.
• To meet criteria, for 2nd deformation step: 12 < %CW < 27
– We will deform to 20%CW
• Diameter after first cold work stage (but before 2nd cold work
stage), Di, calculated as follows:
æ D2 ö
%CW = çç1- d2 ÷÷ x 100
è Di ø
0.5
æ
ö
Dd
%CW
= ç1÷
Di è
100 ø
Intermediate diameter =
Di =
Þ
Dd2
%CW
1- 2 =
100
Di
Þ Di =
Dd
æ %CW ö0.5
ç1÷
100
è
ø
7.5 mm
æ 20%CW ö
ç1÷
100 ø
è
0.5
= 8.39 mm
Chapter 7 - 54
Design Problem Summary
Stage 1: Cold work – reduce diameter from 10 mm to 8.39 mm
2ù
é æ
ö
8.39 mm ú
ê
%CW1 = 1- ç
÷ x 100 = 29.6
ê è 10 mm ø ú
ë
û
Stage 2: Heat treat (allow recrystallization)
Stage 3: Cold work – reduce diameter from 8.39 mm to 7.5 mm
2ù
é æ
ö
7.5 mm ú
ê
%CW2 = 1- ç
÷ x 100 = 20
ê è 8.39 mm ø ú
ë
û
Therefore, all criteria satisfied
%CW = 20
TS = 400 MPa
%EL = 24
Chapter 7 - 55
EXAMPLE PROBLEM 7.3
Solution
Chapter 7 - 56
Summary
• Plastic deformation occurs by motion of dislocations
• Crystallographic considerations:
-- Minimum atomic distortion from dislocation motion
- in slip planes
- along slip directions
• Deformation of polycrystals—change of grain shapes
• Strength is increased by decreasing dislocation
mobility.
• Strengthening techniques for metals:
-- grain size reduction
-- solid solution strengthening
-- strain hardening (cold working)
Chapter 7 - 57
Summary (cont.)
• Heat treatment of deformed metal specimens:
-- Processes
- Recovery
- Recrystallization
- Grain growth
-- Consequences—property alterations
- Softer and weaker
- More ductile
Chapter 7 - 58