Deformation and strengthening

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MSE 280: Introduction to Engineering Materials
Reading: Chapter 8 Callister
Deformation and Strengthening
g p
• Plastic deformation at the atomic level.
• Why is dislocation motion primarily in metals
(why not in ceramics and covalent solids)?
• Relation between strength and dislocation
MSE280© 2007, 2008 Moonsub Shim, University of Illinois
1
motion.
• Strategies for strengthening.
© 2007 University of Illinois Board of Trustees. All rights reserved.
Recall dislocations
Edge dislocation
2
Edge dislocation
Screw dislocation
Plastic deformation corresponds to motion of dislocations!

2
Why are dislocations important in
plastic deformation?
Shear force
All bonds on this
plane need to be
Perfect crystal
plane need to be
broken and new
ones made at the
same time….
With a
dislocation
3
dislocation,
break and make
bond along one
line at a time….
Slip: plastic deformation by dislocation motion.
Slip plane: crystallographic plane on which dislocation line traverses.
http://www.learner.org/jnorth/tm/monarch/LarvaLocomotion.html
From Callister 6e resource CD.
4

3
Why mainly in metals and not other materials?
+
++
+++++++
+ + + + + +
• Metals: Disl. motion easier.
-non-directional bonding
-close-packed directions for slip.
electron cloud
++++++++
close packed directions for slip.
-dislocation density ~ 103 to 1010 mm-2).
ion cores
• Covalent Ceramics
(Si, diamond): Motion hard.
-directional (angular) bonding.
-dislocation density ~ 0.1 to 1 mm-2 for
single crystal Si.
5
• Ionic Ceramics: Motion hard.
-need to avoid ++ and --
neighbors.
-dislocation density ~ 102 to 104 mm-2.
+ + + +
+++
+ + + +
- - -
----
- - -
From Callister resource CD
Lattice strain around dislocations
Edge dislocation:
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Screw dislocation:
Pure shear strain.

4
Stress fields from dislocations with no applied stresses.
Like ones Repel
Interacting Edge Dislocations
Opposites attract
Two halves make a whole. No
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strain from missing half row.
Obstacles,
e.g. GB, twins, particles
(precipitates).
Larger back stress w/ many.
Dislocation pile-ups: traffic jam
Different slip planes and less than 450,
like ones still repel.
Interacting Edges
Different slip planes and greater than 450,
like ones attract.
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Final configuration that is stable (w/o applied stress).
Low-angle grain boundary (Collection of edges).
D. Johnson

5
Less than 450, opposites attract.
Final stable configuration
Interacting Oppositely Oriented Edges
Greater than 450, opposites
repel.
g
that minimizes local stress.
9
Array of opposing edges.
Slip system
Slip system = slip plane + slip direction
z
Slip plane
Slip direction
e.g. slip in
FCC
y
10
x
Slip on (111) along direction indicated
Slip system = (111) [1 0 ]1
For metals: slip plane is the
most close-packed plane.
(Why?) – Burger’s vector

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Slip 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 1 1
y
e.g. (111) and ( ) are parallel to each other.111
In this case, we have:
(111) = ( )
( 11) = (1 )
(1 1) = ( 1 )
(11 ) = ( 1)
111
11
1 1
11
1
1
1
Only 4
unique slip
planes!
11
( ) ( ) p
3 possible slip directions
per plane.
Total 12 slip systems in
FCC
12
Many more slip systems then HCP.
FCC and BCC are typically more
ductile then HCP.

7
Slip in single crystals
13
Even with pure tensile stress, shear component may also exist
σ
Note: τR will lie in the slip direction.Applied stress
Slip direction
τR
σ’
φ
λ
In general, φ + λ does not have to
equal 90o.
Slip plane
τR
σσ’
λ
φ
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Resolved shear stress:
λφστ coscos=R
Schmid factor

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σ = F/Ao
τR = Fs/As
Fs=Fcosλ
τR = F cosλ/ (Ao/cosφ)
= F/Ao (cosλ cosφ)
Ao
σ’
Slip direction
φ
φ
As
F
Fs
λ
As = Ao/cosφ
τR = σ cosλ cosφ
As τR
λ
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Note: 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 φ and λ).
φ
Ao
Critical resolved shear stress (τcrss): minimum shear stress required
to initiate slip: This is when yielding begins (i.e. yield strength).
C diti f di l ti ti τCondition for dislocation motion:
τR > τcrss max)cos(cos λφ
τ
σ crss
y =
• Crystal orientation can make it easy or difficult to move dislocations.
0
σ
/2
σσ
a) b) c)
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τR = 0
φ=90°
τR = σ/2
λ=45°
φ=45°
τR = 0
λ=90°
What happens in cases a and c (w.r.t. plastic deformation)?

9
Example1
Example 2
Determine the Schmid factor for an FCC single crystal oriented
with its [100] direction parallel to the loading axis.
Tensile stress is applied along [010] direction of a single crystal of BCC
iron.
a) Calculate the resolved shear stress along slip system (110)[ 11] when
tensile stress = 52 MPa.
b) Calculate the magnitude of applied stress to initiate yielding if critical
resolved shear stress is 30 MPa.
Example 2
1
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Example 3
For an FCC Ag single crystal, slip initiates at 1.1 MPa of
applied tensile stress along [001] direction.
Calculate the critical resolved shear stress.
Plastic deformation in
polycrystalline materials
• Similar to single crystals,
slip systems that have theslip systems that have the
most favorable orientation
w.r.t. applied stress are where
dislocation motions occur.
• More complex then single
crystals: direction of slip
varies from grain to grain due
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to random crystallographic
orientation.
Which do you expect to be
stronger: single crystalline or
polycrystalline?

10
Single vs. polycrystal
σ
Single crystal
e.g. σ
polycrystalline
τR = σ/2
λ=45°
φ=45°
e.g.
τR = σ/2
λ=45°
φ=45°
Center grain
For each grain,
τcrss
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σy = 2τcrss
max)cos(cos λφ
σ 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?
Before & After Plastic Deformation
20
Grains elongate in the direction of stress but grain boundaries remain intact.
In order for the center grain in the previous slide to slip,
surrounding grains, which may have higher yield strengths,
also have to slip.

11
Strengthening mechanisms
• Grain size reduction
• Solid solutions
• Precipitation strengthening
• Strain hardening (cold working)
21
1. Strengthening via grain size reduction
• Grain boundaries are barriers to slip.
• Barrier "strength“ increases with
misorientation.
• Smaller grain size: more barriers to slip
slip plane
rain
B
• Smaller grain size: more barriers to slip.
• Grain size may be controlled by solidification
process and by plastic deformation followed
by appropriate heat treatment.
grainboundary
grain A
gra
2/1−
+= dkyoy σσ
Hall-Petch Equation:
22
What happens at very small or
very large grain sizes?
Material-dependent constants
Average grain diameter
Fig. 7.13 Callister 6e

12
2. Strengthening via solid solutions
Smaller impurity atoms
• Impurity atoms distort the lattice & generate stress.
• Stress can produce a barrier to dislocation motion.
Smaller impurity atoms
Larger impurity atoms
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g p y
2. Strengthening via solid solutions
24

13
3. Strengthening via precipitation
Large shear stress needed
precipitate
g
to move dislocation toward
precipitate and shear it.
Side View
Top View
Slipped part of slip plane
Uns lipped part of slip plane
S
Dislocation
“advances” but
precipitates act as
“pinning” sites with
spacing S.
25
• Result: σy ~
1
S
Simulation
• Precipitate volume fraction: 10%
A i it t i 64 b (b 1 t i li di t )
Simulation courtesy of Volker
Mohles, Institut für Materialphysik
der Universitåt, Münster, Germany
(http://www.uni-munster.de/physik
/MP/mohles/). Used with
permission.
• Average precipitate size: 64 b (b = 1 atomic slip distance)
26

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Example: aluminum alloys used for airplanes
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4. Strengthening via strain hardening
• Strengthening by plastic deformation.
• Called cold working because deformation is carried out at a
“low” temperature (i e below melting temperature)low temperature (i.e. below melting temperature).
• Usually at RT for metals.
Percent cold work = %CW =
Ao − Ad
Ao
x100
28

15
Forging Rolling
Extrusion Drawing
29
30

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Example problem
• Of the 3 metals from previous slide (Cu,
b d 1040 t l) hi h b ldbrass and 1040 steel), which can be cold
worked to give yield strength > 345 MPa
and %EL > 20%?
31
Recovery
Annihilation reduces dislocation density.
• Scenario 1 extra half-plane
of atoms Disl.
• Scenario 2
atoms
diffuse
to regions
of tension
extra half-plane
of atoms
annhilate
and form
a perfect
atomic
plane.
3. “Climbed” disl. can now τR
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1. dislocation blocked;
can’t move to the right
obstacle dislocation
2. grey atoms leave by
vacancy diffusion
allowing disl. to “climb”
4. opposite dislocations
meet and annihilate
move on new slip plane
From Callister resources CD

17
Dislocations can avoid obstacles.
• Scenario 2
-- Dislocation is block by obstacle (precipitate or another dislocation).
-- Edge moves to another slip plane by vacancy-assisted climb, which is
Recovery by Edge “Climb” Process
temperature-dependent process.
-- Climbed dislocation can now move on parallel slip plane and annihilate, as in
scenario 1.
33
Inclusion,
Precipitate.
t=0 t= Δt1 t= Δt2
D. Johnson
Can screw dislocations annihilate one another?
• Recall two edges dislocations (positive and negative)
• Screws also have positive and negative sense.
Disl. Motion
34
Burger’s Vector
What’s left when the two screw dislocations meet?
Line direction of screw
D. Johnson

18
Recrystallization
• New crystals are formed that:
--have a small disl. density
--are smallare small
--consume cold-worked crystals.
Adapted from
Fig. 7.19 (a),(b),
Callister 6e.
(Fig. 7.19 (a),(b)
are courtesy of
0.6 mm 0.6 mm
35
33% cold
worked
brass
New crystals
nucleate after
3 sec. at 580C.
are courtesy of
J.E. Burke,
General Electric
Company.)
Further recrystallization
• All cold-worked crystals are consumed.
0 6 mm0 6 mm
Adapted from
Fig. 7.19 (c),(d),
Callister 6e.
(Fig. 7.19 (c),(d)
are courtesy of
J E Burke
0.6 mm0.6 mm
36
After 4
seconds
After 8
seconds
J.E. Burke,
General Electric
Company.)

19
Grain Growth
• At longer times, larger grains consume smaller ones.
• Why? Grain boundary area (and therefore energy)
i d d
0.6 mm 0.6 mm
Adapted from
Fig. 7.19 (d),(e),
Callister 6e.
(Fig. 7.19 (d),(e)
are courtesy of
J.E. Burke,
General Electric
Company.)
is reduced.
37From Callister resources CD
After 8 s,
580C
After 15 min,
580C
dn
− do
n
= Kt
elapsed time
coefficient dependent
on material and T.
grain diam.
at time t.
exponent typ. ~ 2
Effects on mechanical properties
38

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Example problem
10.2 mm 7.6 mm
Brass Brass
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How do we achieve the above change while maintaining
σy>380MPa & ductility > 15%EL?
Deformation in Ceramics
• Crystalline: plastic deformation via dislocation
motion However more difficult for dislocationsmotion. However, more difficult for dislocations
to move and smaller number of them to start
with.
– Ionic ceramics: electrostatic repulsion makes it
difficult to generate and to move dislocations.
– Covalent ceramics: relatively strong bonds, limited
number of slip systems, and dislocation structures are
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number of slip systems, and dislocation structures are
often complex.
• Noncrystalline: plastic deformation by viscous
flow (like liquids).

21
Deformation in polymers
• Semicrystalline polymers
– Elastic deformation:
• Elongation of polymer chains along load axis.
• Bending and stretching of bonds (some displacement of
neighboring molecules).
• Elastic Modulus: combination of crystalline and
amorphous regions.
– Plastic deformation:
1. Separation of crystalline regions via elongation of
amorphous regions.
41
p g
2. Distortion of crystalline regions (alignment along load axis).
3. Segmentation of crystalline blocks.
4. Alignment of crystalline blocks along load axis.
Note: may be reversible by heating to recover original
configuration.
Semicrystalline polymer deformation
42

22
Factors that influence mechanical properties
• Molecular weight
– Tensile modulus: usually independent of chain length.
– Tensile strength: longer chains tend to have more entangled
h i hi h i TSchains which increases TS.
• Degree of crystallinity
– Crystalline regions are more densely packed: more secondary
bonds.
– Tensile modulus increases with increasing crystallinity.
– Tensile strength also increases with increasing crystallinity
nM
A
TSTS −= ∞ TS in the “infinite”
MW limit.
43
Tensile strength also increases with increasing crystallinity.
– Brittleness increases with crystallinity.
– e.g. Polyethylene.
Factors that influence mechanical properties
• Predeformation by Drawing
– Neck extension
(continued)
– Neck extension.
– Analogous to strain hardening.
– Anisotropic properties:
• Tensile modulus parallel can icrease up to ~3x
• TS parallel can increase ~ 2 to 5x
• TS perpendicular can decrease by about 1/3 ~ ½
• Heat treating (annealing)
44
Heat treating (annealing)
– Increase modulus & yield strength.
– Decrease ductility.
– Opposite effects to annealing metals.

23
Deformation of elastomers
Elastomers
Driving force for elastic deformation: ENTROPY
Disorder Ordered
45
Criteria to be an elastomer:
1. Amorphous
2. Easy chain bond rotation (e.g. single vs. double bond)
3. Delayed onset of plastic deformation (e.g. via crosslinking preventing
“slipping” of chains)
4. Glass transition temperature (above it: otherwise becomes brittle).
Cis vs. trans isoprene
C CH
CH3
1
42
CH
C
C
CH3
H
C C
H2
CH3
CH2
C
C
H
CH2
trans-isoprene
3
CH2
C
CH2
cis-isoprene
C
CH3
H
H
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trans-1,4-polyisoprene
C
H2
C
C
H
C
n n
C
C
H
H
H
H
cis-1,4-polyisoprene
Which one will be a “better” elastomer?

24
Vulcanization
Crosslinking process in elastomers.
Enhances – elastic modulus
– tensile strengthtensile strength
– resistance to degradation
Typically ~1 – 5 wt% sulfur in useful rubbers.
47
Concepts to remember
• Dislocation motion and plastic deformation.
• Why is dislocation motion primarily in metals?y p y
• Lattice distortions around dislocations.
• Interaction between dislocations.
• Slip system.
• Single crystal slip.
• Single crystalline vs. polycrystalline deformation.
• Strengthening mechanisms.
48
• Recovery, recrystallization, and grain growth.
• Deformation in ceramics.
• Deformation in polymers (different factors that contribute
to mechanical properties).

Deformation and strengthening

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