Chapter 7 - 1
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?
Chapter 7: Dislocations & Strengthening
Mechanisms
Chapter 7 - 2
Recall dislocations
Edge dislocation
Screw dislocation
Plastic deformation corresponds to motion of large number of dislocations!
Chapter 7 - 3
Dislocations & Materials Classes
• Covalent Ceramics(Si, diamond): Motion hard.
-directional (angular) bonding
• Ionic Ceramics (NaCl):Motion hard.
-need to avoid ++ and - -
neighbors.
+ + + +
+++
+ + + +
- - -
----
- - -
• Metals: Disl. motion easier.-non-directional bonding
-close-packed directions
for slip. electron cloud ion cores
+
+
+
+
+++++++
+ + + + + +
+++++++
Chapter 7 - 4
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.
Chapter 7 - 5
http://www.learner.org/jnorth/tm/monarch/LarvaLocomotion.html
From Callister 6e resource CD.
Slip: plastic deformation by dislocation motion.
Slip plane: crystallographic plane on which dislocation line traverses.
Chapter 7 - 6
Dislocation Motion
• Dislocation moves along slip plane in slip directionperpendicular to dislocation line
• Slip direction same direction as Burgers vector
Edge dislocation
Screw dislocation
Adapted from Fig. 7.2, Callister 7e.
Chapter 7 - 7
Lattice strain around dislocations
Edge dislocation:
Screw dislocation:
Pure shear strain.
Chapter 7 - 8
Stress fields from dislocations with no applied stresses.
Like ones Repel
Opposites attract
Two halves make a whole.
No strain from missing
half row.
Obstacles,
e.g. GB, twins, particles (precipitates).Larger back stress w/ many.
Dislocation pile-ups: traffic jam
Interacting Edge DislocationsInteracting Edge DislocationsInteracting Edge DislocationsInteracting Edge Dislocations
D. Johnson
Chapter 7 - 9
Different slip planes and less than 450,
like ones still repel.
Different slip planes and greater than 450,
like ones attract.
Final configuration that is stable (w/o applied stress).Low-angle grain boundary (Collection of edges).
Interacting EdgesInteracting EdgesInteracting EdgesInteracting Edges
D. Johnson
Chapter 7 - 10
Less than 450, opposites attract.
Greater than 450, opposites repel.
Final stable configuration
that minimizes local stress.
Array of opposing edges.
Interacting Oppositely Oriented EdgesInteracting Oppositely Oriented EdgesInteracting Oppositely Oriented EdgesInteracting Oppositely Oriented Edges
D. Johnson
Chapter 7 - 11
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 <110> directions (close-packed)
=> total of 12 slip systems in FCC
– in BCC & HCP other slip systems occur
Deformation Mechanisms
Adapted from Fig. 7.6, Callister 7e.
Chapter 7 - 12
Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, τR. • Applied tension can produce such a stress.
slip plane
normal, ns
Resolved shear stress: τR =Fs/As
slip
direct
ion
AS
τR
τR
FS
slip
direct
ion
Relation between σ and τR
τR=FS/AS
Fcos λ A/cos φ
λF
FS
φnS
AS
A
Applied tensile stress: = F/Aσ
slip
direct
ion
FA
F
φλσ=τ coscosR
Chapter 7 - 13
Slip System in FCC
However, there are equivalent planes and directions (i.e. families).
Therefore the actual slip system = {111}<110>
{111} = (111), (11 ), (1 1), ( 11), ( 1), (1 ), ( ), ( 1 )8 planes in this family but not all are unique.
1 1 1 11 11 111
e.g. (111) and ( ) are parallel to each other.111
In this case, we have:
(111) = ( )
( 11) = (1 )
(1 1) = ( 1 )(11 ) = ( 1)
11111
1 111
1
11
Only 4 unique
planes!
3 possible slip directions per
plane.
Total 12 slip systems in FCC
1 1
Chapter 7 - 14
FCC and BCC materials have large numbers of slip systems (at
least 12) and are considered ductile. HCP systems have few slip systems and are quite brittle.
Chapter 7 - 15
Single Crystal Slip
Adapted from Fig. 7.8, Callister 7e.
Adapted from Fig. 7.9, Callister 7e.
Chapter 7 - 16
Slip in single crystals
Even with pure tensile stress, shear component may also exist
σ
Slip direction
τR
σ’φ
λ
Note: τR will lie in the slip direction.
In general, φ + λ does not have to equal 90o.
Resolved shear stress:
λφστ coscos=R
Applied stress
Slip plane
Schmid factor
τR
σσ’
λ
φ
Normal to
slip plane
How do you obtained it?
Chapter 7 - 17
Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, tR. • Applied tension can produce such a stress.
slip plane
normal, ns
Resolved shear stress: τR =Fs/As
slip
direct
ion
AS
τR
τR
FS
slip
direct
ion
Relation between σ and τR
τR=FS/AS
Fcos λ A/cos φ
λF
FS
φnS
AS
A
Applied tensile stress: = F/Aσ
slip
direct
ion
FA
F
φλσ=τ coscosR
In metals, although there are different slip systems, once a stress axis is
specified, one slip system is most favorably oriented (for a fixed stress direction, each slip system has different f and l).
Chapter 7 - 18
• Condition for dislocation motion: CRSS τ>τR
• Crystal orientation can makeit easy or hard to move dislocation
10-4 GPa to 10-2 GPa
typically
φλσ=τ coscosR
Critical Resolved Shear Stress
τ maximum at λ = φ = 45º
τR = 0
λ=90°
σ
τR = σ/2λ =45°φ =45°
σ
τR = 0
φ=90°
σ
Chapter 7 - 19
Ex: Deformation of single crystal
So the applied stress of 6500 psi will not cause the crystal to yield.
τ = σ cos λ cos φ
σ = 6500 psi
λ=35°
φ=60°
τ = (6500 psi) (cos35o)(cos60o)
= (6500 psi) (0.41)
τ = 2662 psi < τcrss = 3000 psi
τcrss = 3000 psi
a) Will the single crystal yield?
b) If not, what stress is needed?
σ = 6500 psi
Adapted from Fig. 7.7, Callister 7e.
Chapter 7 - 20
Ex: Deformation of single crystal
psi 732541.0
psi 3000
coscoscrss ==
φλ
τ=σ∴ y
What stress is necessary (i.e., what is the
yield stress, σy)?
)41.0(cos cos psi 3000crss yy σ=φλσ==τ
psi 7325=σ≥σ y
So for deformation to occur the applied stress must be greater than or equal to the yield stress
Chapter 7 - 21
• Stronger - grain boundariespin deformations
• Slip planes & directions
(λ, φ) change from onecrystal to another.
• τR will vary from onecrystal to another.
• The crystal with the
largest τR yields first.
• Other (less favorablyoriented) crystalsyield 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].)
Slip Motion in Polycrystalsσ
300 µm
Chapter 7 - 22
Single vs. Polycrystal
τR = σ/2λ=45°φ=45°
σ
Single crystal
e.g.
σσσσy = 2ττττcrss
τR = σ/2λ=45°φ=45°
σ
polycrystalline
Center grain
For each grain,max)cos(cos λφ
τσ crss
y =
But φ and λ are different for each grain.
Which will require more stress to slip single crystal or the center grain in polycrystal?
Chapter 7 - 23
• Can be induced by rolling a polycrystalline metal
- before rolling
235 µm
- isotropic
since grains are
approx. spherical& randomlyoriented.
- after rolling
- anisotropic
since rolling affects grain
orientation and shape.
rolling direction
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.)
Anisotropy in σσσσy
Chapter 7 - 24
side view
1. Cylinder of
Tantalummachined
from a
rolled plate:
rolli
ng d
irection
2. Fire cylinder
at a target.
• The noncircular end view shows
anisotropic deformation of rolled material.
endview
3. Deformed
cylinder
plate
thickness
direction
Photos courtesyof G.T. Gray III,Los AlamosNational Labs. Used withpermission.
Anisotropy in Deformation
Chapter 7 - 25
4 Strategies for Strengthening: 1: Reduce Grain Size
• Grain boundaries arebarriers to slip.
• Barrier "strength"increases withIncreasing angle of misorientation.
• Smaller grain size:more barriers to slip.
• Hall-Petch Equation: 21 /yoyield dk −+σ=σ
Adapted from Fig. 7.14, Callister 7e.
(Fig. 7.14 is from A Textbook of Materials
Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)
Chapter 7 - 26
• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.
4 Strategies for Strengthening: 2: Solid Solutions
• Smaller substitutional
impurity
Impurity generates local stress at A
and B that opposes dislocation motion to the right.
A
B
• Larger substitutional
impurity
Impurity generates local stress at C
and D that opposes dislocation motion to the right.
C
D
Chapter 7 - 27
Stress Concentration at Dislocations
Adapted from Fig. 7.4, Callister 7e.
Chapter 7 - 28
Strengthening by Alloying
• small impurities tend to concentrate at dislocations
• reduce mobility of dislocation ∴ increase strength
Adapted from Fig. 7.17, Callister 7e.
Chapter 7 - 29
Strengthening by alloying
• large impurities concentrate at dislocations on low density side
Adapted from Fig. 7.18, Callister 7e.
Chapter 7 - 30
Ex: Solid SolutionStrengthening in Copper
• Tensile strength & yield strength increase with wt% Ni.
• Empirical relation:
• Alloying increases σy and TS.
21 /
y C~σ
Adapted from Fig. 7.16 (a) and (b), Callister 7e.
Tensile
str
ength
(M
Pa)
wt.% Ni, (Concentration C)
200
300
400
0 10 20 30 40 50 Yie
ld s
tren
gth
(M
Pa)
wt.%Ni, (Concentration C)
60
120
180
0 10 20 30 40 50
Chapter 7 - 31
• Hard precipitates are difficult to shear.Ex: Ceramics in metals (SiC in Iron or Aluminum).
• Result:S
~y
1 σ
4 Strategies for Strengthening: 3: Precipitation Strengthening
Large shear stress needed to move dislocation toward precipitate and shear it.
Dislocation “advances” but precipitates act as “pinning” sites with
S.spacing
Side View
precipitate
Top View
Slipped part of slip plane
Unslipped part of slip plane
Sspacing
Chapter 7 - 32
• Internal wing structure on Boeing 767
• Aluminum is strengthened with precipitates formedby 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.)
1.5µm
Application:
Precipitation Strengthening
Adapted from chapter-opening photograph, Chapter 11, Callister 5e.
(courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
Chapter 7 - 33
4 Strategies for Strengthening: 4: Cold Work (%CW)
• Room temperature deformation.• Common forming operations change the cross
sectional area:
Adapted from Fig. 11.8, Callister 7e.
-Forging
Ao Ad
force
die
blank
force-Drawing
tensile force
Ao
Addie
die
-Extrusion
ram billet
container
containerforce
die holder
die
Ao
Adextrusion
100 x %o
do
A
AACW
−=
-Rolling
roll
Ao
Adroll
Chapter 7 - 34
• Ti alloy after cold working:
• Dislocations entanglewith one anotherduring cold work.
• Dislocation motionbecomes more difficult.
Adapted from Fig. 4.6, Callister 7e.
(Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)
Dislocations During Cold Work
0.9 µm
Chapter 7 - 35
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 ρd increases:
total dislocation length
unit volume
large hardening
small hardening
σ
ε
σy0
σy1
Chapter 7 - 36
Impact of Cold Work
Adapted from Fig. 7.20, Callister 7e.
• Yield strength (σy) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.
As cold work is increased
Chapter 7 - 37
• 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.)
%6.35100 x %2
22
=π
π−π=
o
do
r
rrCW
Cold Work Analysis
% Cold Work
100
300
500
700
Cu
200 40 60
yield strength (MPa)
σy = 300MPa
300MPa
% Cold Work
tensile strength (MPa)
200
Cu
0
400
600
800
20 40 60
ductility (%EL)
% Cold Work
20
40
60
20 40 6000
Cu
Do =15.2mm
Cold Work
Dd =12.2mm
Copper
340MPa
TS = 340MPa
7%
%EL = 7%
Chapter 7 - 38
• Results for
polycrystalline iron:
• σy and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies
help dislocations
move past obstacles.
Adapted from Fig. 6.14, Callister 7e.
σσσσ-εεεε Behavior vs. Temperature
2. vacancies replace atoms on the disl. half plane
3. disl. glides past obstacle
-200°C
-100°C
25°C
800
600
400
200
0
Strain
Str
ess (
MP
a)
0 0.1 0.2 0.3 0.4 0.5
1. disl. trapped by obstacle
obstacle
Chapter 7 - 39
• 1 hour treatment at Tanneal...decreases TS and increases %EL.
• Effects of cold work are reversed!
• 3 Annealingstages todiscuss...
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.)
Effect of Heating After %CWte
nsile
str
eng
th (
MP
a)
du
ctilit
y (
%E
L)tensile strength
ductility
Recovery
Recrystallization
Grain Growth
600
300
400
500
60
50
40
30
20
annealing temperature (ºC)200100 300 400 500 600 700
Chapter 7 - 40
Annihilation reduces dislocation density.
Recovery
• Scenario 1Results from diffusion
• Scenario 2
4. opposite dislocations meet and annihilate
Dislocations annihilate and form a perfect atomic plane.
extra half-plane of atoms
extra half-plane of atoms
atoms diffuse to regions of tension
2. grey atoms leave by vacancy diffusion allowing disl. to “climb”
τR
1. dislocation blocked; can’t move to the right
Obstacle dislocation
3. “Climbed” disl. can now move on new slip plane
Chapter 7 - 41
• New grains are formed that:-- have a small dislocation density
-- are small
-- 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.)
33% cold
workedbrass
New crystals
nucleate after
3 sec. at 580°C.
0.6 mm 0.6 mm
Recrystallization
Chapter 7 - 42
• 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.)
After 4
seconds
After 8
seconds
0.6 mm0.6 mm
Further Recrystallization
Chapter 7 - 43
• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy)
is reduced.
After 8 s,580ºC
After 15 min,580ºC
0.6 mm 0.6 mm
Adapted from Fig. 7.21 (d),(e), Callister 7e.(Fig. 7.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.)
Grain Growth
• Empirical Relation:
Ktdd no
n =−elapsed time
coefficient dependenton material and T.
grain diam.at time t.
exponent typ. ~ 2
Ostwald Ripening
Chapter 7 - 44
TR
Adapted from Fig. 7.22, Callister 7e.
º
º
TR = recrystallization temperature
Chapter 7 - 45
Recrystallization Temperature, TR
TR = recrystallization temperature = point of
highest rate of property change
1. Tm => TR ≈ 0.3-0.6 Tm (K)
2. Due to diffusion � annealing time� TR = f(t) shorter annealing time => higher TR
3. Higher %CW => lower TR – strain hardening
4. Pure metals lower TR due to dislocation movements
• Easier to move in pure metals => lower TR
Chapter 7 - 46
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.
Chapter 7 - 47
Coldwork Calculations Solution
If we directly draw to the final diameter what happens?
%843100 x 400
3001100 x
4
41
100 1100 x %
2
2
2
..
.
D
D
xA
A
A
AACW
o
f
o
f
o
fo
=
−=
π
π−=
−=
−=
Do = 0.40 in
Brass
Cold Work
Df = 0.30 in
Chapter 7 - 48
Coldwork Calc Solution: Cont.
• For %CW = 43.8%Adapted from Fig. 7.19, Callister 7e.
540420
– σy = 420 MPa
– TS = 540 MPa > 380 MPa
6
– %EL = 6 < 15
• This doesn’t satisfy criteria…… what can we do?
Chapter 7 - 49
Coldwork Calc Solution: Cont.
Adapted from Fig. 7.19, Callister 7e.
380
12
15
27
For %EL < 15
For TS > 380 MPa > 12 %CW
< 27 %CW
∴ our working range is limited to %CW = 12-27
Chapter 7 - 50
Coldwork Calc Soln: Recrystallization
Cold draw-anneal-cold draw again
• For objective we need a cold work of %CW ≅ 12-27
– We’ll use %CW = 20
• Diameter after first cold draw (before 2nd cold draw)?
– must be calculated as follows:
100
%1 100 1%
2
02
2
2
2
02
2
2 CW
D
Dx
D
DCW ff =−⇒
−=
50
02
2
100
%1
.
f CW
D
D
−= 50
202
100
%1
.
f
CW
DD
−
=⇒⇒⇒⇒
m 3350100
201300
50
021 ..DD
.
f =
−==Intermediate diameter =
Chapter 7 - 51
Coldwork Calculations Solution
Summary:
1. Cold work D01= 0.40 in � Df1 = 0.335 m
2. Anneal above D02 = Df1
3. Cold work D02= 0.335 in � Df2 =0.30 m
Therefore, meets all requirements
20100 3350
301%
2
2 =
−= x
.
.CW
24%
MPa 400
MPa 340
=
=
=σ
EL
TS
y⇒⇒⇒⇒
%CW1 = 1−0.335
0.4
2
x 100 = 30
Fig 7.19
Chapter 7 - 52
Rate of Recrystallization
• Hot work � above TR
• Cold work � below TR
• Smaller grains
– stronger at low temperature
– weaker at high temperature
t/R
T
BCt
kT
ERtR
1:note
log
logloglog 0
=
+=
−=−=
RT1
log t
start
finish
50%
Chapter 7 - 53
• Dislocations are observed primarily in metalsand 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.
Summary
