PLASTIC DEFORMATION

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PLASTIC DEFORMATIONCREEP

• PLASTICITY
• CREEP

Mechanical Metallurgy George E Dieter
McGraw-Hill Book Company, London (1988)
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Mechanisms / Methods by which a can Material can
FAIL
Elasticdeformation
Chemical /Electro-chemicaldegradation
Creep
Physicaldegradation
Fatigue
Fracture
Slip
Microstructuralchanges
Wear
Twinning
Erosion
Phase transformations
Grain growth
Particlecoarsening
Failure can be considered as change in desired
performance- which could involve changes in
properties and/or shape
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Tension / Compression
ModesofDeformation
Torsion
Bending
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Mode I
ModesofDeformation
Mode II
Mode III
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The Tensile Stress-Strain Curve
Tensile specimen
Gauge Length ? L0
Possible axes
Initial cross sectional area ? A0
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Engineering Stress (s) and Strain (e)
0 ? initial
Subscript
i ? instantaneous
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Problem with engineering Stress (s) and Strain (e)
Consider the following sequence of deformations
L0
1
e1?2 1
2
2L0
e1?3 0
?e1?2 e2?3 ½
e2?3 ?? ½
3
L0
It is clear that from stage 1 ? 3 there is no
strain But the decomposition of the process into
1 ? 2 2 ? 3 gives a net strain of ½ ? Clearly
there is a problem with the use of Engineering
strain
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True Stress (?) and Strain (?)
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Same sequence of deformations considered before
L0
1
? 1?2 Ln(2)
2
2L0
?1?3 0
?? 1?2 ? 2?3 0
? 2?3 ?? Ln(2)
3
L0
With true strain things turn out the way they
should
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Comparison between Engineering and True
quantities
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Due to collective motionof many dislocations
Yield point
X
Necking begins
Strain hardening ?
UTS
X
s ?
? ?
? ?
e ?
Elastic region
Usually expressed as (for ?plastic)
Plastic region
Fracture
X
UTS- Ultimate Tensile Strength
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Variables in plastic deformation
K ? strength coefficientn ? strain / work
hardening coefficient ? Cu and brass (n 0.5)
can be given large plastic strain more
easily as compared to steels with n 0.15
A ? a constantm ? index of strain rate
sensitivity ? If m 0 ? stress is independent
of strain rate (stress-strain curve would be
same for all strain rates) ? m 0.2 for common
metals ? If m ? (0.4, 0.9) the material may
exhibit superplastic behaviour ? m 1 ?
material behaves like a viscous liquid (Newtonian
flow)
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Slip systems
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• All plastic deformation by slip is due to shear
  stresses
• The shear stress resolved onto the slip plane is
  responsible for slip
• When the Resolved Shear Stress (RSS) reaches a
  critical value ? Critical Resolved Shear Stress
  (CRSS) ? plastic deformation starts (The
  actual Schmids law)

?
A
?
Slip plane normal
Slip direction
?
?
A
Schmid factor
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Schmids law
?CRSS is a material parameter
Slip is initiated when
Yield strengh of a single crystal
Read slides from dislocations chapter refer
table 11.2 from textbook for shear strength of
perfect and real crystals Also read section on
whiskers from book
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Plastic Deformation in Crystalline Materials
Creep Mechanisms
Slip(Dislocation motion)
Twinning
Phase Transformation
Grain boundary sliding
Vacancy diffusion
Dislocation climb
Other Mechanisms
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How does the motion of dislocations lead to a
macroscopic shape change? (From microscopic slip
to macroscopic deformation ? a first feel!)
Step formed when dislocationleaves the crystal
Dislocation formed bypushing ina plane
?
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Net shape change
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Stress to move a dislocation Peierls Nabarro
stress

• Width of the dislocation as a basis for the ease
  of motion of a dislocation

• Unrelaxed condition- stiff
• Smaller width of displacement fields
• ? atomic adjustments required (for any
  one atom) for dislocation motion are
  large

• Relaxed condition
• Large width of displacement fields
• ? atomic adjustments required for
  dislocation motion are small

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Peierls Nabarro stress (?PN) ? stress to move a
dislocaiton

• G ? shear modulus of the crystal
• w ? width of the dislocation !!!
• b ? b

Hence, ? narrow dislocations are more difficult
to move than wide ones ? dislocations with
larger b are more difficult to move
Peierls Nabarro stress (?PN) ? P-N stress ?
Lattice Friction
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Width of the dislocation

• Nature of chemical bonding in the crystal
  determines the ? extent of relaxation the
  width of the dislocation

Covalent crystals

• Strong and directional bonds ? small relaxation
  (low w) ? high ?PN
• Fail by brittle fracture before ?PN is reached

Metallic crystals

• Weaker and non-directional bonds ? large
  relaxation (high w) ? low ?PN
• Cu can be cold worked to large strains
• Transition metals (e.g. Fe) have some covalent
  character due to d orbital bonding ? harder
  than Cu

Ionic crystals

• Moderate and non- directional bonds
• Surface cracks usually lead to brittle fracture
• Large b (NaCl b 3.95Å) ? more difficult to
  move

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Intermetallic compounds / complex crystal
structures

• Intermetallic compounds and complex crystal
  structures (Fe3C, CuAl2) do not have good slip
  systems ? favorable planes directions ?
  usually brittle
• Ordered compounds may have very large b
• In CuZn (an ordered compound) dislocations move
  in pairs to preserve the order during slip
• Quasicrystals have 4, 5 or 6 dimensional b and
  the 3D component is not sufficient to cause slip
  in the usual sense

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Applied shear stress vs internal opposition
Applied shear stress (?)
Internal resisting stress (?i)
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• Athermal ? f (T, strain rate)
• Long range internal stress fields
• Sources ?Stress fields of other
  dislocations ?Incoherent precipitates

Long range obstacles (?G)
Note ?G has some temperature dependence as G
decreases with T
Internal resisting stress (?i)
Short range obstacles (?T)

• Thermal f (T, strain rate)
• Short range 10 atomic diameters
• T can help dislocations overcome these obstacles
  
• Sources ?Peierls-Nabarro stress
  ?Stress fields of coherent precipitates solute
  atoms

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Effect of Temperature
Motion of a dislocation can be assisted by
thermal energy
Equilibrium positions of a dislocation
Motion of dislocations by pure thermal activation
is random

• vd ? velocity of the dislocations
• ?d ? density of mobile/glissile dislocations
• b ? b

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Very high temperaturesneeded for thermal
activationto have any effect
Metallic
Ionic
Fe-BCC
Fe
W-BCC
W
450
Al2O3
Covalent
Si
300
Yield Stress (MPa) ?
18-8 SS
Ni-FCC
150
Ni
Cu-FCC
Cu
0.6
0.0
0.2
0.4
T/Tm ?
RT is like HT and P-Nstress is easily overcome
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Strain hardening ? multiplication of dislocations

• Why increase in dislocation density ?
• Why strain hardening ?

If dislocations were to leave the surfaceof the
crystal by slip / glide then the dislocation
density should decrease on plastic deformation ?
but observation is contrary to this
This implies some sources of dislocationmultiplic
ation / creation should exist
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Dislocation sources

• Solidification from the melt
• Grain boundary ledges and step emit dislocations
• Aggregation and collapse of vacancies to form a
  disc or prismatic loop (FCC Frank partials)
• High stresses ? Heterogeneous nucleation at
  second phase particles ? During phase
  transformation

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Double Ended Frank-Read Source
Initial configuration
A
B
Dislocation line segment pinnedat A and B

• Pinning could be caused by
• Dislocation in the plane of the paper intersects
  dislocations in other planes
• Anchored by impurity atoms or precipitate
  particles
• Dislocation leaves the slip plane at A and B

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Application of stress on dislocation segment
?
A
B
Force ? b
Line tension
d?
r
?
ds
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Line tension force opposes the applied stress
For equilibrium in curved configuration
?max ? rmin
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L
? b
Direction of dislocation motion is ? to the
dislocation line (except at A and B)
Increasing stress
? b
semicircle? corresponds to maximum stress
required to expand the loop
After this decreasing stress to expand the loop
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• Frank-Read dislocation source ?
• Can operate from a single source producing a
  loop each time
• This loop produces a slip of 1b each time on
  the slip plane
• The maximum value of shear stress required is
  when the bulge becomes a semi-circle (rmin
  L/2) ? ?max Gb/L ? ?? as L? i.e. The
  longest segments operate first ? When the long
  segments get immobilized shorter segments
  operate with increasing stress ? work
  hardening
• If the dislocation loops keep piling up on the
  slip plane the back stress will oppose the
  applied stress
• When the back-stress gt ?max the source will
  cease to operate
• Double ended F-R sources have been observed
  experimentally they are not frequent ? other
  mechanisms must exist

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Strain hardening

• Dislocations moving in non-parallel slip planes
  can intersect with each other ? results in an
  increase in stress required to cause further
  plastic deformation ? Strain Hardening / work
  hardening
• One such mechanism by which the dislocation is
  immobilized is the Lomer-Cottrell barrier

Dynamic recovery

• In single crystal experiments the rate of strain
  hardening decreases with further strain after
  reaching a certain stress level
• At this stress level screw dislocations are
  activated for cross-slip
• The RSS on the new slip plane should be enough
  for glide

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Formation of the Lomer-Cottrell barrier
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(111)
(-111)
(100)

• Lomer-Cottrell barrier ?
• The red and green dislocations attract each
  other and move towards their line of
  intersection
• They react as above to reduce their energy and
  produce the blue dislocation
• The blue dislocation lies on the (100) plane
  which is not a close packed plane ? hence
  immobile ? acts like a barrier to the motion of
  other dislocations

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Impediments (barriers) to dislocation motion

• Grain boundary
• Immobile (sessile) dislocations ?
  Lomer-Cottrell lock ? Frank partial dislocation
• Twin boundary
• Precipitates

• Dislocations get piled up at a barrier and
  produce a back stress

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Stress to move a dislocation dislocation density

• ?0 ? base stress to move a dislocation in the
  crystal in the absence of other dislocations
• ? ? Dislocation density
• A ? A constant

Empirical relation
?? as ?? (cold work) ? ?? (i.e. strain
hardening)
COLD WORK ? ? strength ? ? ductility
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Strengthening Mechanisms

• Remove dislocations- produce whiskers Problem
  with this ? dislocations can nucleate in service
• Cold Work (Cold work increases Yield stress but
  decreases the elongation, i.e. ductility)
• Decrease in grain size
• Solid solution strengthening
• Precipitation hardening

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Grain size and strength

• ?y ? Yield stress
• ?i ? Stress to move a dislocation in single
  crystal
• k ? Locking parameter measure of the
  relative hardening contribution of grain
  boundaries
• d ? Grain diameter

Hall-Petch Relation
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Grain size
Important Please read section on ASTM grain size
and grain size measurement

• d ? Grain diameter in meters
• n ? ASTM grain size number

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Effect of solute atoms on strengthening

• Solid solutions offer greater resistance to
  dislocation motion than pure crystals (even
  solute with a lower strength gives
  strengthening!)
• The stress fields around solute atoms interact
  with the stress fields around the dislocation to
  leading to an increase in the stress required for
  the motion of a dislocation
• The actual interaction between a dislocation and
  a solute is much more complex
• The factors playing an important role are ?
  Size of the solute ? Elastic modulus of the
  solute (higher the elastic modulus of
  the solute greater the strengthening effect)
  ? e/a ratio of the solute
• A curved dislocation line configuration is
  required for the solute atoms to provide
  hindrance to dislocation motion

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Solute strengthening of Cu crystal by solutes of
different sizes
200
Sn (1.51)
Be (1.12)
Matrix Cu (r 1.28 Å)
150
Si (1.18)
Al (1.43)
100
?y (MPa) ?
Ni (1.25)
Zn (1.31)
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(Values in parenthesis are atomic radius values
in Å)
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10
20
30
0
Solute Concentration (Atom ) ?
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By ? ?i (lattice friction)
X
? ?y
? ?
Solute atoms
? level of ? - ? curve
? ?
Often produce Yield Point Phenomenon
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Relative strengthening effect / unit concentration
3Gsolute
Interstitial
Solute atoms
Gsolute / 10
Substitutional
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Size effect
For the same size difference thesmaller atom
gives a greaterstrengthening effect
Size difference
Size effect depends on
Concentration of the solute (c)
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Interstitial ? Edge and screw dl.
Elastic
Substitutional ? edge
Long range(T insensitive)
Modulus
Long range order
Solute-dislocation interaction
Stacking fault
Short range (T sensitive)
Electrical
Short range order
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The hardening effect of precipitates
Glide through the precipitate
If the precipitate is coherent with the matrix
Dislocation
Get pinned by the precipitate
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Dislocation Glide through the precipitate

• Only if slip plane is continuous from the matrix
  through the precipitate ? precipitate is
  coherent with the matrix
• Stress to move the dislocation through the
  precipitate is that to move it in the matrix
• Usually during precipitation the precipitate is
  coherent only when it is small and becomes
  incoherent on growth
• Glide of the dislocation causes a displacement
  of the upper part of the precipitate w.r.t the
  lower part by b ? cutting of the precipitate

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Slip plane
b
Precipitate particle
b
Increase in surface area due to particle shearing
Hardening effect
Part of the dislocation line segment (inside
theprecipitate) could face a higher PN stress
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Pinning effect of the precipitate
Can act like a Frank-Reed source
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Precipitate Hardening effect
(Complete List)

• Chemical Strengthening arises from additional
  interface created on shearing
• Stacking-fault Strengthening due to difference
  between stacking-fault energy between particle
  and matrix when these are both FCC or HCP
• Modulus Hardening due to difference in elastic
  moduli of the matrix and particle
• Coherency Strengthening due to elastic
  coherency strains surrounding the particle
• Order Strengthening due to additional work
  required to create an APB in case of
  dislocations passing through precipitates which
  have an ordered lattice

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CREEP
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High-temperature behaviour of materials

• Increased vacancy concentration
• High diffusion rate ? diffusion controlled
  processes become important
• Phase transformations can occur
• Grain related ? boundary weakening ? boundary
  migration ? recrystallization / growth
• Dislocation related ? climb ? new slip
  systems can become active ? change of slip
  system
• Overaging of precipitate particles and particle
  coarsening
• Enhanced oxidation and intergranular penetration
  of oxygen

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CREEP ? Permanent deformation of a material
under load as a function of time

• Appreciable only at T gt 0.4 Tm

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Constant load creep curve
I
II
Strain (?) ?
III
?0 ? Initial instantaneous strain
?0
t ?

• The distinguishability of the three stages
  strongly depends on T and ?

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Constant Stress creep curve
II
I
Strain (?) ?
III
?
?
t ?
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Stages of creep
I

• Creep rate decreases with time
• Effect of work hardening more than recovery

II

• Stage of minimum creep rate ? constant
• Work hardening and recovery balanced

III

• Absent (/delayed very much) in constant stress
  tests
• Necking of specimen start
• specimen failure processes set in

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Effect of stress
Strain (?) ?
Increasing stress
?0 increases
?0
t ?
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Effect of temperature
Strain (?) ?
E? as T?
Increasing T
? ?
?0 increases
?0
? ?
t ?
As decrease in E with temperature is usually
small the ?0 increase is also small
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Creep Mechanisms of crystalline materials
Cross-slip
Dislocation climb
Creep
Vacancy diffusion
Grain boundary sliding
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Cross-slip

• In the low temperature of creep ? screw
  dislocations can cross-slip (by thermal
  activation) and can give rise to plastic strain
  as f(t)

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Dislocation climb

• Edge dislocations piled up against an obstacle
  can climb to another slip plane and cause
  plastic deformation as f(t), in response to
  stress
• Rate controlling step is the diffusion of
  vacancies

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Nabarro-Herring creep ? high T ? lattice diffusion
Diffusional creep
Coble creep ? low T ? Due to GB diffusion

• In response to the applied stress vacancies
  preferentially move from surfaces/interfaces
  (GB) of specimen transverse to the stress axis
  to surfaces/interfaces parallel to the stress
  axis? causing elongation
• This process like dislocation creep is
  controlled by the diffusion of vacancies ? but
  diffusional does not require dislocations to
  operate

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Grain boundary sliding

• At low temperatures the grain boundaries are
  stronger than the crystal interior and impede
  the motion of dislocations
• Being a higher energy region, the grain
  boundaries melt before the crystal interior
• Above the equicohesive temperature grain
  boundaries are weaker than grain and slide past
  one another to cause plastic deformation

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Creep Resistant Materials

• Higher operating temperatures gives better
  efficiency for a heat engine

High melting point ? E.g. Ceramics
Dispersion hardening ? ThO2 dispersed Ni (0.9 Tm)
Creep resistance
Solid solution strengthening
Single crystal / aligned (oriented) grains

• Cost, fabrication ease, density etc. are other
  factors which determine the final choice of a
  material
• Commonly used materials ? Fe, Ni, Co base alloys

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• Cost, fabrication ease, density etc. are other
  factors which determine the final choice of a
  material
• Commonly used materials ? Fe, Ni, Co base alloys
• Precipitation hardening (instead of dispersion
  hardening) is not a good method as particles
  coarsen (smaller particles dissolve and
  larger particles grow ? interparticle separation
  ?)
• Ni-base superalloys have Ni3(Ti,Al) precipitates
  which form a low energy interface with the
  matrix ? low driving force for coarsening
• Cold work cannot be used for increasing creep
  resistance as recrystallization can occur which
  will produced strain free crystals
• Fine grain size is not desirable for creep
  resistance ? grain boundary sliding can cause
  creep elongation / cavitation ? Single crystals
  (single crystal Ti turbine blades in gas
  turbine engine have been used) ? Aligned /
  oriented polycrystals

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