This document provides an introduction to materials science chapter on dislocations and strengthening mechanisms in metals. It addresses why dislocations are primarily observed in metals, how strength and dislocation motion are related, and how strength can be increased. It discusses dislocation motion, slip systems, factors affecting slip, and strategies to strengthen metals including reducing grain size, solid solution strengthening, and cold working. Cold working increases yield strength and hardness but decreases ductility due to increased dislocation density and entanglement. Heating allows recovery which reduces dislocation density and reverses effects of cold working.

1. 26/03/2012
INTRODUCTION TO MATERIALS Chapter 7:
SCIENCE Dislocations and Strengthening
Mechanisms in Metal
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?
AP Dr. Puteri Sri Melor bt. Megat Yusoff
Semester January 2012 1 PSMMY Jan 2012
PSMMY Jan 2012 Chapter 7 - Chapter 7 - 2
Learning Outcomes Learning Outcomes
To describe edge and screw dislocation motion from To explain how grain boundaries impede dislocation
an atomic perspective. motion and why a metal having small grains is
To describe how plastic deformation occurs by the stronger than one having large grains.
motion of edge and screw dislocations in response to
f To describe and explain solid-solution strengthening
applied shear stress. for substitutional impurity atoms in terms of lattice
To define slip system and cite one example; relate the strain interactions with dislocations.
number of slip systems to the material properties. To describe and explain the phenomenon of strain
To describe how the grain structure of a hardening or cold working in terms of dislocations
polycrystalline metal is altered when it is plastically and strain field interactions.
deformed.
PSMMY Jan 2012 PSMMY Jan 2012
Chapter 7 - 3 Chapter 7 - 4
Dislocations & Materials Classes
• Metals: Disl. motion easier.
+ + + + + + + +
Learning Outcomes -non-directional bonding + + + + + + + +
-close-packed directions + + + + + + + +
To describe and explain the phenomenon of grain size for slip. electron cloud ion cores
reduction in terms of dislocations and strain field
interactions. • Covalent Ceramics
To describe recrystallisation in terms of both the (Si, diamond): Motion hard.
alteration of microstructure and mechanical
characteristics of the material. -directional (angular) bonding
To describe the phenomenon of grain growth from
both macroscopic and atomic perspective. • Ionic Ceramics (NaCl):
+ - + - + - +
To apply the Schmid Law to solve for critical resolved Motion hard.
shear stress in a single crystal slip process. - + - + - + -
-need to avoid ++ and - -
neighbors. + - + - + - +
PSMMY Jan 2012 PSMMY Jan 2012
Chapter 7 - 5 Chapter 7 - 6
1

2. 26/03/2012
Dislocation Motion Dislocation Motion
Dislocations & plastic deformation • Dislocation moves along slip plane in slip direction
• Cubic & hexagonal metals - plastic deformation by perpendicular to dislocation line
plastic shear or slip where one plane of atoms slides • Slip direction same direction as Burgers vector
over adjacent plane by defect motion (dislocations).
Edge dislocation
Adapted from Fig. 7.2,
Callister 7e.
Screw dislocation
• If dislocations don't move, Adapted from Fig. 7.1,
Callister 7e.
deformation doesn't occur! Chapter 7 - 7 Chapter 7 - 8
PSMMY Jan 2012
Deformation Mechanisms Stress and Dislocation Motion
Slip System • Crystals slip due to a resolved shear stress, τR.
– Slip plane - plane allowing easiest slippage • Applied tension can produce such a stress.
• Wide interplanar spacings - highest planar densities
Applied tensile Resolved shear Relation between
– Slip direction - direction of movement - Highest linear stress: σ = F/A stress: τR =Fs /A s σ and τR
densities
Adapted from Fig. A
F slip plane
τR τR =FS /AS
7.6, Callister 7e.
normal, ns
AS Fcos λ
F A/cos φ
A/
FS
F nS φ
λ A
τR FS AS
F
– FCC Slip occurs on {111} planes (close-packed) in <110>
directions (close-packed)=> total of 12 slip systems in FCC
– in BCC: {110}<111>, {211}<111>, {321}<111>: 48 systems
τR = σ cos λ cos φ
PSMMY Jan 2012
Chapter 7 - 9 PSMMY Jan 2012 Chapter 7 - 10
Critical Resolved Shear Stress Single Crystal Slip
• Condition for dislocation motion: τR > τCRSS
• Crystal orientation can make typically Adapted from Fig.
it easy or hard to move dislocation 7.9, Callister 7e.
10-4 GPa to 10-2 GPa
τR = σ cos λ cos φ
σ σ σ
τR = 0 τR = σ/2 τR = 0
λ=90° λ =45° φ =90°
φ =45°
τ
Adapted from Fig. 7.8, Callister 7e.
maximum at λ = φ = 45º Chapter 7 - 11 PSMMY Jan 2012 Chapter 7 - 12
2

3. 26/03/2012
Factors Affecting Slip Motion Ex: Deformation of single crystal
a) Will the single crystal yield?
b) If not, what stress is needed?
Factor FCC BCC HCP φ=60°
τcrss = 3000 psi
τ crss (psi) 50–100 5,000-10,000 50-100 λ=35°
τ = σ cos λ cos φ
No.
No of slip systems 12 48 3 σ = 6500 psi
p
Cross slip can occur can occur no Adapted from
Fig. 7.7,
τ = (6500 psi) (cos 35o )(cos 60o )
= (6500 psi) (0.41)
Callister 7e.
Properties ductile strong relatively ductile
τ = 2662 psi < τcrss = 3000 psi
σ = 6500 psi
So the applied stress of 6500 psi will not cause the
PSMMY Jan 2012 13 crystal to yield. PSMMY Jan 2012
Chapter 7 - Chapter 7 - 14
Ex: Deformation of single crystal Slip Motion in Polycrystals
• Stronger - grain boundaries
σ
What stress is necessary (i.e., what is the
pin deformations
yield stress, σy)?
• Slip planes & directions
τcrss = 3000 psi = σ y cos λ cos φ = σ y (0.41) (λ, φ) change from one
Adapted from Fig.
7.10, Callister 7e.
(Fig. 7.10 is
crystal to another. courtesy of C.
Brady, National
τcrss 3000 psi • τR will vary from one
Bureau of
∴ σy = = = 7325 psi Standards [now the
National Institute of
cos λ cos φ 0.41 crystal to another. Standards and
Technology,
Gaithersburg, MD].)
• The crystal with the
So for deformation to occur the applied stress must
largest τR yields first.
be greater than or equal to the yield stress
• Other (less favorably
σ ≥ σ y = 7325 psi oriented) crystals
300 μm
yield later.
PSMMY Jan 2012 Chapter 7 - 15 Chapter 7 - 16
PSMMY Jan 2012
3 Strategies for Strengthening: 3 Strategies for Strengthening:
1: Reduce Grain Size 2: Solid Solutions
• Grain boundaries are • Impurity atoms distort the lattice & generate stress.
barriers to slip. • Stress can produce a barrier to dislocation motion.
• Barrier "strength" • Smaller substitutional • Larger substitutional
increases with impurity impurity
Increasing angle of
misorientation. Adapted from Fig. 7.14, Callister 7e. A C
(Fig. 7.14 is from A Textbook of Materials
• Smaller grain size: Technology, by Van Vlack, Pearson Education,
Inc., Upper Saddle River, NJ.)
more barriers to slip. B D
Impurity generates local stress at A Impurity generates local stress at C
and B that opposes dislocation and D that opposes dislocation
motion to the right. motion to the right.
PSMMY Jan 2012
Chapter 7 - 17 PSMMY Jan 2012 Chapter 7 - 18
3

4. 26/03/2012
Stress Concentration at Dislocations Strengthening by Alloying
Lattice distortion
• small impurities tend to concentrate at dislocations
Lattice strain
• reduce mobility of dislocation ∴ increase strength
Tensile lattice strain Partial cancellation of impurity-
dislocation lattice strain
Adapted from Fig. 7.4,
Callister 7e.
PSMMY Jan 2012 Adapted from Fig. PSMMY Jan 2012
Chapter 7 - 19 7.17, Callister 7e. Chapter 7 - 20
Ex: Solid Solution
Strengthening by alloying Strengthening in Copper
• large impurities concentrate at dislocations on low • Tensile strength & yield strength increase with wt% Ni.
density side
180
trength (MPa)
rength (MPa)
Adapted from Fig.
400 7.16 (a) and (b),
Callister 7e.
120
300
Tensile st
Yield str
200 60
0 10 20 30 40 50 0 10 20 30 40 50
wt.% Ni, (Concentration C) wt.%Ni, (Concentration C)
Compressive lattice strain Partial cancellation of impurity-
dislocation lattice strain • Empirical relation: σy ~ C 1/ 2
• Alloying increases σy and TS.
Adapted from Fig. PSMMY Jan 2012 PSMMY Jan 2012
7.18, Callister 7e. Chapter 7 - 21 Chapter 7 - 22
3 Strategies for Strengthening: Characteristics of Dislocations
3: Cold Working/strain hardening/work hardening Dislocations During Cold Work
• Ti alloy after cold working:
• a concept where metals become harder as they are plastically
deformed
• room temperature deformation.
• Dislocations entangle
• common forming operations change the cross sectional area: with one another
during cold work.
-Forging force -Rolling
roll • Dislocation motion
die Ad
Ao becomes more difficult.
difficult
Ao blank Ad
Adapted from Fig.
11.8, Callister 7e.
roll
force
-Drawing -Extrusion
Ao Adapted from Fig.
die Ad container die holder
4.6, Callister 7e.
Ao tensile force (Fig. 4.6 is courtesy
force ram billet extrusion Ad of M.R. Plichta,
Michigan
die container die Technological
0.9 μm University.)
Ao − Ad
%CW = x 100
Ao PSMMY Jan 2012
Chapter 7 - 23 Chapter 7 - 24
4

5. 26/03/2012
Characteristics of Dislocations Characteristics of Dislocations
Dislocation Density Increases During Cold Working Lattice Strain Interactions Between Dislocations
Recall:
Dislocation density = total dislocation length
unit volume
Existing dislocations,
– Carefully grown single crystals grain boundaries,
ca. 103 mm-2 scratches (other
– Deforming sample increases density internal stress
contributors) are
109-1010 mm-2 possible sites for
– Heat treatment reduces density dislocation formation
105-106 mm-2
• Yield stress increases as ρd
increases: σ
σy1 large hardening
σy0 small hardening
Adapted from Fig.
PSMMY Jan 2012
ε Chapter 7 - 25
7.5, Callister &
Rethwisch 8e. PSMMY Jan 2012 Chapter 7 - 26
Impact of Cold Work Cold Work Analysis
As cold work is increased • What is the tensile strength & Copper
• Yield strength (σy) increases. ductility after cold working? Cold
Work
• Tensile strength (TS) increases. πr 2 − πrd2
• Ductility (%EL or %AR) decreases. %CW = o x 100 = 35.6%
2
πro Do =15.2mm Dd =12.2mm
yield strength (MPa) tensile strength (MPa) ductility (%EL)
60
700 800
Adapted from Fig. 7.20,
Callister 7e.
500 600 40
300
300MPa Cu
Cu 400 340MPa 20
Cu 7%
100
0 20 40 60 200 00
0 20 40 60 20 40 60
% Cold Work % Cold Work % Cold Work
σy = 300MPa TS = 340MPa %EL = 7%
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
Chapter 7 - 27 (Managing Ed.), American Society for Metals, 1979, p. 276 and 327.) Chapter 7 - 28
Effect of Heating After %CW Recovery
• 1 hour treatment at Tanneal... Annihilation reduces dislocation density.
decreases TS and increases %EL.
• Effects of cold work are reversed!
extra half-plane
annealing temperature (ºC) of atoms Dislocations
100 200 300 400 500 600 700 annihilate
tensile strength (MPa)
600 60 atoms
tensile strength and form
diffuse
ductility (%EL)
50 to
t regions
i a perfect
500 of tension atomic
• 3 Annealing plane.
40 extra half-plane
stages to of atoms
400 30 discuss...
Adapted from Fig. 7.22, Callister 7e. (Fig.
ductility 20 7.22 is adapted from G. Sachs and K.R. van
300 Horn, Practical Metallurgy, Applied Metallurgy,
and the Industrial Processing of Ferrous and
Nonferrous Metals and Alloys, American
Society for Metals, 1940, p. 139.)
PSMMY Jan 2012
PSMMY Jan 2012 Chapter 7 - 29 Chapter 7 - 30
5

6. 26/03/2012
Recrystallization Further Recrystallization
• New grains are formed that: • All cold-worked grains are consumed.
-- have a small dislocation density
-- are small 0.6 mm 0.6 mm
-- consume cold-worked grains.
0.6 mm 0.6 mm
Adapted from
Adapted from Fig. 7.21 (c),(d),
Fig 7 21 (c) (d)
Fig. 7.21 (a),(b), Callister 7e.
Callister 7e. (Fig. 7.21 (c),(d)
(Fig. 7.21 (a),(b) are courtesy of
are courtesy of J.E. Burke,
J.E. Burke, General Electric
General Electric Company.)
Company.)
After 4 After 8
33% cold New crystals seconds seconds
worked nucleate after
brass 3 sec. at 580°C.
PSMMY Jan 2012
PSMMY Jan 2012 Chapter 7 - 31 Chapter 7 - 32
º
Grain Growth
• At longer times, larger grains consume smaller ones.
• Why? Grain boundary area (and therefore energy) TR = recrystallization
is reduced. temperature
0.6 mm 0.6 mm
Adapted from TR
Fig. 7.21 (d),(e),
Callister 7e.
(Fig. 7.21 (d),(e)
are courtesy of
t f
J.E. Burke,
General Electric
Company.)
After 8 s, After 15 min, Adapted from Fig.
580ºC 580ºC 7.22, Callister 7e.
PSMMY Jan 2012
º
Chapter 7 - 33 Chapter 7 - 34
Recrystallization Temperature, TR Coldwork Calculations
TR = recrystallization temperature = point of A cylindrical rod of brass originally 0.40 in (10.2 mm)
highest rate of property change 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
1. TR ≈ 0.3-0.6 Tm (K) of 55,000 psi (380 MPa) and a ductility of at least 15
2. Due to d us o
ue o diffusion a ea g
annealing time TR = f(t)
e (t) %EL are desired. Further more, the final diameter
shorter annealing time => higher TR must be 0.30 in (7.6 mm). Explain how this may be
3. Higher %CW => lower TR accomplished.
4. Pure metals lower TR due to dislocation
movements
• Easier to move in pure metals => lower TR
PSMMY Jan 2012 PSMMY Jan 2012
Chapter 7 - 35 Chapter 7 - 36
6

7. 26/03/2012
Coldwork Calculations Solution Coldwork Calc Solution: Cont.
If we directly draw to the final diameter
what happens?
Brass 420 540
Cold
Work
6
Do = 0.40 in Df = 0.30 in
⎛ A − Af ⎞ ⎛ A ⎞ • For %CW = 43.8%
Adapted from Fig.
%CW = ⎜ o
⎜ ⎟ x 100 = ⎜1 − f ⎟ x 100
⎟ ⎜
7.19, Callister 7e.
⎝ Ao ⎠ ⎝ Ao ⎟
⎠ – σy = 420 MPa
⎛ πD 2 4 ⎞ ⎛ D2 ⎞ ⎛ 2
⎞ – TS = 540 MPa > 380 MPa
= ⎜1 − ⎟ x 100 = ⎜1 − 2 ⎟ x 100 = ⎜1 − ⎛ 0.30 ⎞ ⎟ x 100 = 43.8%
⎜ ⎜ 0.40 ⎟ ⎟
f f
⎜ πD 2 4 ⎟ ⎜ D ⎟ ⎝ ⎠ ⎠ – %EL = 6 < 15
⎝ o ⎠ ⎝ o ⎠ ⎝
PSMMY Jan 2012 • This doesn’t satisfy criteria…… what can we do?
Chapter 7 - 37 Chapter 7 - 38
PSMMY Jan 2012
Coldwork Calc Solution: Cont. Coldwork Calc Soln: Recrystallization
Cold draw-anneal-cold draw again
• For the objective, we need a cold work of %CW ≅ 12-27
– We’ll use %CW = 20
• Diameter after first cold draw, Df1 is also the diameter before 2nd
cold draw, D02. Hence: Df1 =D02
– calculated as follows:
380 15
⎛ D 2⎞ D
2
%CW
%CW = ⎜1 − f 2 2 ⎟ x 100 ⇒ 1 − f 2 2 =
⎜ D
⎟
27 ⎝ 02 ⎠ D02 100
12
0. 5 Df 2
Df 2 ⎛ %CW ⎞ D02 =
Adapted from Fig. = ⎜1 − ⎟ ⇒ ⎛ %CW ⎞
0 .5
For TS > 380 MPa > 12 %CW 7.19, Callister 7e. D02 ⎝ 100 ⎠ ⎜1 − ⎟
⎝ 100 ⎠
For %EL >,= 15 < 27 %CW 0 .5
⎛ 20 ⎞
Intermediate diameter = D f 1 = D 02 = 0 . 30 ⎜ 1 − ⎟ = 0 . 335 in
∴ our working range is limited to %CW = 12-27 ⎝ 100 ⎠
PSMMY Jan 2012 PSMMY Jan 2012
Chapter 7 - 39 Chapter 7 - 40
Coldwork Calculations Solution Summary
Summary:
1. Cold work D01= 0.40 in Df1 = 0.335 in • Dislocations are observed primarily in metals
and alloys.
⎛ ⎞
⎛ 0.335 ⎞2 ⎟ • Strength is increased by making dislocation
%CW1 = ⎜1− ⎜ x 100 = 30
⎜ ⎝ 0.4 ⎟ ⎟ ⎠ motion difficult.
⎝ ⎠
2. Anneal at TR. The effect of cold work from (1) is • Particular ways to increase strength are to:
removed. Now: D02 = Df1 --decrease grain size
d i i
--solid solution strengthening
3. Cold work D02= 0.335 in Df 2 =0.30 in --cold work
⎛ ⎛ 0 .3 ⎞ 2 ⎞ σ y = 340 MPa
%CW2 = ⎜1 − ⎜ ⎟ x 100 = 20 Fig 7.19
⎜ ⎝ 0.335 ⎟ ⎟
⎝ ⎠ ⎠ ⇒ TS = 400 MPa • Heating (annealing) can reduce dislocation density
and increase grain size. This decreases the strength.
%EL = 24
Therefore, meets all requirements.
PSMMY Jan 2012 PSMMY Jan 2012
Chapter 7 - 41 Chapter 7 - 42
7