PLASTIC DEFORMATION

1
PLASTIC DEFORMATION

• Mechanisms of Plastic Deformation
• The Uniaxial Tension Test
• Mechanisms of Plastic Deformation

Mechanical Metallurgy George E Dieter
McGraw-Hill Book Company, London (1988)
2
If failure is considered as change in desired
performance- which could involve changes in
properties and/or shape then failure can occur
by many mechanisms as below.
Mechanisms / Methods by which a can Material can
FAIL
Elastic deformation
Chemical /Electro-chemicaldegradation
Creep
Physicaldegradation
Fatigue
Plastic deformation
Fracture
Microstructuralchanges
Twinning
Wear
Slip
Twinning
Erosion
Corrosion
Phase transformations
Oxidation
Grain growth
Particle coarsening
Beyond a certain limit
3

• Plastic deformation in the broadest sense means
  permanent deformation in the absence of external
  constraints (forces, displacements) (i.e.
  external constraints are removed).
• Plastic deformation of crystalline materials
  takes place by mechanisms which are very
  different from that for amorphous materials
  (glasses). The current chapter will focus on
  plastic deformation of crystalline materials.
  Glasses deform by shear banding etc. below the
  glass transition temperature (Tg) and by flow
  above Tg.
• Though plasticity by slip is the most important
  mechanism of plastic deformation, there are other
  mechanisms as well. Many of these mechanisms may
  act in conjunction/parallel to give rise to the
  observed plastic deformation.

Plastic Deformation in Crystalline Materials
Slip(Dislocation motion)
Twinning
Phase Transformation
Creep Mechanisms
Grain boundary sliding
Other Mechanisms
Vacancy diffusion
Grain rotation
Dislocation climb
Note Plastic deformation in amorphous materials
occur by other mechanisms including flow
(viscous fluid) and shear banding
4
Common types of deformation
Review

• Tension/Compression
• Bending
• Shear
• Torsion

Bending
Deformed configuration
Torsion
Shear
Tension
Compression
Note modes of deformation in other contexts will
be defined in the topic on plasticity
5
Peak ahead

• In addition to the modes of deformation
  considered before the following modes can be
  defined w.r.t fracture.
• Fracture can be cause by the propagation of a
  pre-existing crack (e.g. the notches shown in the
  figures below) or by the nucleation of a crack
  during deformation followed by its propagation.
• In fracture the elastic energy stored in the
  material is used for the creation of new surfaces
  (when the crack nucleates/propagates)

Mode I
ModesofDeformation
Mode II
Mode III
6
The Uniaxial Tension Test (UTT)

• One of the simplest test which can performed to
  evaluate the mechanical properties of a material
  is the Uniaxial Tension Test.
• This is typically performed on a cylindrical
  specimen with a standard gauge length. (At
  constant temperature and strain rate).
• The test involves pulling a material with
  increasing load (force) and noting the elongation
  (displacement) of the specimen.
• Data acquired from such a test can be plotted as
  (i) load-stroke (raw data), (ii) engineering
  stress- engineering strain, (iii) true stress-
  true strain. (next slide).
• It is convenient to use Engineering Stress (s)
  and Engineering Strain (e) as defined below ? as
  we can divide the load and change in length by
  constant quantities (A0 and L0). Subscripts 0
  refer to initial values and i to instantaneous
  values.
• But there are problems with the use of s and
  e (as outlined in the coming slides) and hence
  we define True Stress (?) and True Strain (?)
  (wherein we use instantaneous values of length
  and area).
• Though this is simple test to conduct and a
  wealth of information about the mechanical
  behaviour of a material can be obtained (Modulus
  of elasticity, ductility etc.) ? However, it must
  be cautioned that this data should be used with
  caution under other states of stress.

0 ? initial
Note quantities obtained by performing an
Uniaxial Tension Test are valid only under
uniaxial state of stress
Subscript
i ? instantaneous
7
The Tensile Stress-Strain Curve
Tensile specimen
Gauge Length ? L0
Possible axes
Initial cross sectional area ? A0
Important Note
We shall assume cylindrical specimens (unless
otherwise stated)
8
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 (definition) of
Engineering strain ? Hence, a quantity known as
True Strain is preferred (along with True
Stress) ? as defined in the next slide.
9
True Stress (?) and Strain (?)

• The definitions of true stress and true strain
  are based on instantaneous values of area (Ai)
  and length (Li).

• Ai ? instantaneous area

10
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!
11
Schematic s-e and ?-? curves

• These are simplified schematics which are close
  to the curves obtained for some metallic
  materials like Al, Cu etc. (polycrystalline
  materials at room temperature).
• Many materials (e.g. steel) may have curves which
  are qualitatively very different from these
  schematics.
• Most ceramics are brittle with very little
  plastic deformation.
• Even these diagrams are not to scale as the
  strain at yield is 0.001 (eelastic 103)E is
  measured in GPa and ?y in MPa ? thus giving this
  small strains ? the linear portion is
  practically vertical and stuck to the Y-axis
  (when efracture and eelastic is drawn to the same
  scale).

Note the increasing stress required for continued
plastic deformation
Schematics not to scale

• Information gained from the test
• Youngs modulus
• Yield stress (or proof stress)
• Ultimate Tensile Stress (UTS)
• Fracture stress
• UTS- Ultimate Tensile Strength
• Subscripts y- yield, F,f- fracture,
• u- uniform (for strain)/ultimate (for stress)

Neck
Points and regions of the curves are explained in
the next slide
12
Sequence of events during the tension test

• O ? unloaded specimen
• OY ? Elastic ? Linear Region in the plot
  (macroscopic linear elastic region)
• Y ? macroscopic yield point (there are many
  measures of yielding as discussed later)Occurs
  due to collective motion of many dislocations.
• YF ? Elastic Plastic regime ? If specimen is
  unloaded from any point in this region, it will
  unload parallel to OY and the elastic strain
  would be recovered. Actually, more strain will be
  recovered than unloading from Y (and hence in
  some sense in the region YF the sample is more
  elastic than in the elastic region OY).? In
  this region the material strain hardens ? flow
  stress increases with strain.? This region can
  further be split into YN and NF as below.
• YN ? Stable region with uniform deformation along
  the gauge length
• N ? Instability in tension ? Onset of necking
  True condition of uniaxiality broken ? onset of
  triaxial state of stress (loading remains
  uniaxial but the state of stress in the
  cylindrical specimen is not).
• NF ? most of the deformation is localized at the
  neck? Specimen in a triaxial state of stress
• F ? Fracture of specimen (many polycrystalline
  materials like Al show cup and cone fracture)

• Notes
• In the ?-? plot there is no distinct point N and
  there is no drop in load (as instantaneous area
  has been taken into account in the definition of
  ?) in the elastic plastic regime (YF)
• The stress is monotonically increasing in the
  region YF ? true indicator of strain hardening

13
Comparison between Engineering and True
quantities

• In engineering stress since we are dividing by a
  constant number A0 (and there is a local
  reduction in area around the neck)
• Engineering and true values are related by
  the equations as below.
• At low strains (in the uniaxial tension test)
  either of the values work fine.
• As we shall see that during the tension test
  localized plastic deformation occurs after some
  strain (called necking). This leads to
  inhomogeneity in the stress across the length of
  the sample and under such circumstances true
  stress should be used.

Valid till necking starts
Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain Comparison between true strain and engineering strain
True strain (?) 0.01 0.10 0.20 0.50 1.0 2.0 3.0 4.0
Engineering strain (e) 0.01 0.105 0.22 0.65 1.72 6.39 19.09 53.6
Note that for strains of about 0.4, true and
engineering strains can be assumed to be equal.
At large strains the deviations between the
values are large.
14
Where does Yielding start?

• Yielding can be defined in many contexts. ?
  Truly speaking (microscopically) it is point at
  which dislocations leave the crystal (grain) and
  cause microscopic plastic deformation (of unit
  b) ? this is best determined from microstrain
  (106 ) experiments on single crystals. However,
  in practical terms it is determined from the
  stress-strain plot (by say an offset as described
  below).
• True elastic limit (microscopically and
  macroscopically elastic ? where in there is not
  even microscopic yielding) ? 106 OA portion
  of the curve
• Microscopically plastic but macroscopically
  elastic ? AY portion of the curve
• Proportional limit ? the point at which there is
  a deviation from the straight line elastic
  regime
• Offset Yield Strength (proof stress) ? A curve is
  drawn parallel to the elastic line at a given
  strain like 0.2 ( 0.002) to determine the yield
  strength.

• In some materials (e.g. pure annealed Cu, gray
  cast iron etc.) the linear portion of the curve
  may be limited and yield strength may arbitrarily
  determined as the stress at some given strain
  (say 0.005).

15
Important Note

• ?y is yield stress in an uniaxial tension test
  and should not be used in other states of stress
  (other criteria of yield should be used for a
  generalized state of stress).
• Tresca and von Mices criterion are the two most
  popular ones.

16
What is meant by ductility?

• Slip is competing with other processes which can
  lead to failure.
• In simple terms a ductile material is one which
  yields before failure (i.e. ?y lt ?f).
• Ductility depends on the state of stress used
  during deformation.
• We can obtain an measure of the ductility of a
  material from the uniaxial tension test as
  follows (by putting together the fractured parts
  to make the measurement)? Strain at fracture
  (ef) (usually expressed as ), (often called
  elongation, although it is a dimensionless
  quantity)? Reduction in area at fracture (q)
  (usually expressed as )
• q is a better measure of ductility as it does
  not depend on the gauge length (L0) while, ef
  depends on L0. Elongation/strain to necking
  (uniform elongation) can also be used to avoid
  the complication arising from necking.Also, q
  is a more structure sensitive ductility
  parameter.
• Sometimes it is easier to visualize elongation as
  a measure of ductility rather than a reduction in
  area. For this the calculation has be based on a
  very short gauge length in the necked region ?
  called Zero-gauge-length elongation (e0). e0
  can be calculated from q using constancy of
  volume in plastic deformation (ALA0L0).

Note this is ductility in Uniaxial Tension Test
17
Comparison between reduction in area versus
strain at fracture

• We had seen two measures of ductility? Strain
  at fracture (ef) ? Reduction in area at fracture
  (q)
• We had also seen that these can be related
  mathematically as
• However, it should be noted that they represent
  different aspects of material behaviour.
• For reasonable gauge lengths, e is dominated by
  uniform elongation prior to necking and thus is
  dependent on the strain hardening capacity of the
  material (more the strain hardening, more will be
  the e). Main contribution to q (area based
  calculation) comes from the necking process
  (which is more geometry dependent).
• Hence, reduction in area is not truly a
  material property and has geometry dependence.

18
What happens after necking?

• Following factors come in to picture due to
  necking
• Till necking the deformation is uniform along
  the whole gauge length.
• Till necking points on the ?-? plot lie to the
  left and higher than the s-e plot (as below).
• After the onset of necking deformation is
  localized around the neck region.
• Formulae used for conversion of e to ? and
  s to ? cannot be used after the onset of
  necking.
• Triaxial state of stress develops and uniaxiality
  condition assumed during the test breaks down.
• Necking can be considered as an instability in
  tension.
• Hence, quantities calculated after the onset of
  necking (like fracture stress, ?F) has to be
  corrected for (i) triaxial state of stress, (ii)
  correct cross sectional area.

Neck
Fractured surfaces
Fractured surfaces
19
True values beyond necking

• Calculation of true strain beyond necking
• True strain values beyond necking can be
  obtained by using the concept of
  zero-gauge-length elongation (e0). This involves
  measurement of instantaneous area.

Beyond necking
Note Further complications arise at strains
close to fracture as microvoid nucleation
growth take place and hence all formulations
based on continuum approach (e.g. volume
constancy) etc. are not valid anymore.
20
True values beyond necking
Cotd..

• Calculation of true Stress beyond necking
• Neck acts like a diffuse notch. This produces a
  triaxial state of stress (radial and transverse
  components of stress exist in addition to the
  longitudinal component) ? this raises the value
  of the longitudinal stress required to case
  plastic flow.
• Using certain assumptions (as below) some
  correction to the state of stress can be made
  (given that the state of stress is triaxial, such
  a correction should be viewed appropriately).
• Assumptions used in the correction ? neck is
  circular (radius R), ? von Mises criterion can
  be used for yielding, ? strains are constant
  across the neck.
• The corrected uniaxial stress (?uniaxial) is
  calculated from the stress from the experiment
  (?expLoad/Alocal), using the formula as below.

The Correction
P.W. Bridgman, Trans. Am. Soc. Met. 32 (1944)
553.
21
Fracture stress and fracture strain

• The tensile specimen fails by cup cone
  fracture ? wherein outer regions fail by shear
  and inner regions in tension.
• Fracture strain (ef) is often used as a measure
  of ductility.

• Calculation of fracture stress/strain
• To calculate true fracture stress (?F) we need
  the area at fracture (which is often not readily
  available and often the data reported for
  fracture stress could be in error).
• Further, this fracture stress has to be corrected
  for triaxiality.
• True strain at fracture can be calculated from
  the areas as below.

22
Unloading the specimen during the tension test

• If the specimen is unloaded beyond the yield
  point (say point X in figure below), elastic
  strain is recovered (while plastic strain is
  not). The unloading path is XM.
• The strain recovered ( ) is more than
  that recovered if the specimen was to be unloaded
  from Y ( ) ? i.e. in this sense the
  material is more elastic in the plastic region
  (in the presence of work hardening), than in the
  elastic region!
• If the specimen is reloaded it will follow the
  reverse path and yielding will start at ?X.
  Because of strain hardening the yield strength
  of the specimen is higher.

Strain Hardening is also called work hardening
?The material becomes harder with plastic
deformation (on tensile loading the present
case) ? We will see later that strain hardening
is usually caused by multiplication of
dislocations.
If one is given a material which is at point
M, then the Yield stress of such a material
would be ?X (i.e. as we dont know the prior
history of the specimen loading, we would call
?X as yield stress of the specimen).
The blue part of the curve is also called the
flow stress
23
Variables in plastic deformation

• In the tension experiment just described the
  temperature (T) is usually kept constant and the
  sample is pulled (between two crossheads of a
  UTM) at a constant velocity. The crosshead
  velocity can be converted to strain rate ( )
  using the gauge length (L0) of the specimen.
• At low temperatures (below the recrystallization
  temperature of the material, T lt 0.4Tm) the
  material hardens on plastic deformation (YF in
  the ?-? plot ? known as work hardening or strain
  hardening). The net strain is an important
  parameter under such circumstances and the
  material becomes a partial store of the
  deformation energy provided. The energy is
  essentially stored in the form of dislocations
  and point defects.
• If deformation is carried out at high
  temperatures (above the recrystallization
  temperature wherein, new strain free grains are
  continuously forming as the deformation
  proceeds), strain rate becomes the important
  parameter instead of net strain.

Range of strain rates obtained in various experiments Range of strain rates obtained in various experiments
Test Range of strain rate (/s)
Creep tests 108 to 105
Quasi-static tension tests (in an UTM) 105 to 101
Dynamic tension tests 101 to 102
High strain rate tests using impact bars 102 to 104
Explosive loading using projectiles (shock tests) 104 to 108
24

• In the ?-? plot the plastic stress and strain are
  usually expressed by the expression given below.
  Where, n is the strain hardening exponent and
  K is the strength coefficient.

Usually expressed as (for ?plastic)

• Deviations from this behaviour often observed
  (e.g. in Austenitic stainless steel) at low
  strains (103) and/or at high strains (1.0).
  Other forms of the power law equation are also
  considered in literature (e.g.
  ).
• An ideal plastic material (without strain
  hardening) would begin to neck right at the onset
  of yielding. At low temperatures (below
  recrystallization temperature- less than about
  0.5Tm) strain hardening is very important to
  obtain good ductility. This can be understood
  from the analysis of the results of the uniaxial
  tension test. During tensile deformation
  instability in the form of necking localizes
  deformation to a small region (which now
  experiences a triaxial state of stress). In the
  presence of strain hardening the neck portion
  (which has been strained more) hardens and the
  deformation is spread to other regions, thus
  increasing the ductility obtained.

• For an experiment done in shear on single
  crystals the equivalent region to OY can further
  be subdivided into ? True elastic strain
  (microscopic) till the true elastic limit (?E)?
  Onset of microscopic plastic deformation above a
  stress of ?A.? For Mo a comparison of these
  quantities is as follows ?E  0.5 MPa,
  ?A  5 MPa and ?0 (macroscopic yield stress in
  shear)  50 MPa.

25
Variables/parameters in mechanical testing

• Process parameters (characterized by parameters
  inside the sample)? Mode of deformation, Sample
  dimensions, Stress, Strain, Strain Rate,
  Temperature etc.
• Material parameters? Crystal structure,
  Composition, Grain size, dislocation density, etc.

26
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
When true strain is less than 1, the smaller
value of n dominates over a larger value of n
n and K for selected materials
Material n K (MPa)
Annealed Cu 0.54 320
Annealed Brass (70/30) 0.49 900
Annealed 0.5 C steel 0.26 530
0.6 carbon steel Quenched and Tempered (540?C) 0.10 1570
27

• At high temperatures (above recrystallization
  temperature) where strain rate is the important
  parameter instead of strain, a power law equation
  can be written as below between stress and strain
  rate.

The effect of strain rate is compared by
performing tests to a constant strain
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)

• Thermal softening coefficient (?) is also defined
  as below

28
Funda Check

• What is the important of m and n

• We have seen that below recrystallization
  temperature n is the important parameter.
• Above recrystallization temperature it is m
  which is important.
• We have also noted that it is necking which
  limits the ductility in uniaxial tension.
• Necking implies that there is locally more
  deformation (strain) and the strain rate is also
  higher locally.
• Hence, if the locally deformed material becomes
  harder (stronger) then the deformation will
  spread to other regions along the gauge length
  and we will obtain more ductility.
• Hence having a higher value of n or m is
  beneficial for obtaining good ductility.

29
Plastic deformation by slip
Click here to see overview of mechanisms/modes of
plastic deformation

• As we noted in the beginning of the chapter
  plastic deformation can occur by many mechanisms
  ? SLIP is the most important of them. At low
  temperatures (especially in BCC metals) twinning
  may also be come important.
• At the fundamental level plastic deformation (in
  crystalline materials) by slip involves the
  motion of dislocations on the slip plane ?
  finally leaving the crystal/grain (creating a
  step of Burgers vector).
• Slip is caused by shear stresses (at the level of
  the slip plane). Hence, a purely hydrostatic
  state of stress cannot cause slip (or twinning
  for that matter).
• A slip system consists of a slip direction lying
  on a slip plane.
• Under any given external loading conditions, slip
  will be initiated on a particular slip system if
  the Resolved Shear Stress (RSS) exceeds a
  critical value the Critical Resolved Shear
  Stress (CRSS).
• For slip to occur in polycrystalline materials, 5
  independent slip systems are required. Hence,
  materials which are ductile in single crystalline
  form, may not be ductile in polycrystalline form.
  CCP crystals (Cu, Al, Au) have excellent
  ductility.
• At higher temperatures more slip systems may
  become active and hence polycrystalline materials
  which are brittle at low temperature, may become
  ductile at high temperature.

Leaving the crystal part is important To be
defined soon
30
Slip systems

• In CCP, HCP materials the slip system consists of
  a close packed direction on a close packed plane.
• Just the existence of a slip system does not
  guarantee slip ? slip is competing against other
  processes like twinning and fracture. If the
  stress to cause slip is very high (i.e. CRSS is
  very high), then fracture may occur before slip
  (like in brittle ceramics).

Crystal Slip plane(s) Slip direction Number of slip systems
FCC 111 ½lt110gt 12
HCP (0001) lt11?20gt 3
BCCNot close packed 110, 112, 123 ½111 12
NaClIonic 110111 not a slip plane ½lt110gt 6
CDiamond cubic 111 ½lt110gt 12
31
More examples of slip systems
Crystal Slip plane(s) Slip direction Slip systems
TiO2 Rutile 101 lt10?1gt
CaF2, UO2, ThO2 Fluorite 001 lt1?10gt
CsClB2 110 lt001gt
NaCl, LiF, MgORock Salt 110 lt110gt 6
C, Ge, SiDiamond cubic 111 lt110gt 12
MgAl2O4 Spinel 111 lt1?10gt
Al2O3 Hexagonal (0001) lt11?20gt
32
More examples of slip systems
Crystal Slip plane(s) Slip direction (b) Slip systems
Cu (FCC) Fm ?3m 111 (a/2)lt 1 ?10gt 4 x 3 12
W (BCC) Im ?3m 011 112 123 (a/2)lt11 ?1gt 6 x 2 12 12 x 1 12 24 x 1 24
Mg (HCP) P63/mmc 0001 10 ?10 10 ?11 (a/3)lt11?20gt 1 x 3 3 3 x 1 3 6 x 1 6
Al2O3 R ?3c 0001 10 10 (a/3)lt11?20gt 1 x 3 3 3 x 1 3
NaCl Fm ?3m 110 001 (a/2)lt 1 ?10gt 6 x 1 6 6 x 1 6
CsCl Pm ?3m 110 alt001gt 6 x 1 6
Polyethylene Pnam (100) 110 (010) clt001gt 1 x 1 1 2 x 1 2 1 x 1 1
33
Critical Resolved Shear Stress (CRSS)
What is the connection between Peierls stress and
CRSS?

• As we saw plastic deformation by slip is due to
  shear stresses.
• Even if we apply an tensile force on the specimen
  ? 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
Note externally only tensile forces are being
applied
34
Schmids law
Slip is initiated when
?CRSS is a material parameter, which is
determined from experiments
Yield strength of a single crystal
35
How does the motion of dislocations lead to a
macroscopic shape change? (From microscopic slip
to macroscopic deformation ? a first feel!)

• When one bents a rod of aluminium to a new shape,
  it involves processes occurring at various
  lengthscales and understanding these is an
  arduous task.
• However, at the fundamental level slip is at the
  heart of the whole process.
• To understand how slip can lead to shape
  change? we consider a square crystal deformed
  to a rhombus (as Below).

36
Step formed when dislocationleaves the crystal
Dislocation formed bypushing ina plane
?
Now visualize dislocations being punched in on
successive planes ? moving and finally leaving
the crystal
37
(No Transcript)
38
This sequence of events finally leads to deformed
shape which can be approximated to a rhombus
Net shape change
39
Stress to move a dislocation Peierls Nabarro
(PN) stress

• We have seen that there is a critical stress
  required to move a dislocation.
• At the fundamental level the motion of a
  dislocation involves the rearrangement of bonds ?
  requires application of shear stress on the slip
  plane.
• When sufficient stress is applied the
  dislocation moves from one metastable energy
  minimum to another.
• The original model is due to Peierls Nabarro
  (formula as below) and the sufficient stress
  which needs to be applied is called
  Peierls-Nabarro stress (PN stress) or simply
  Peierls stress.
• Width of the dislocation is considered as a basis
  for the ease of motion of a dislocation in the
  model which is a function of the bonding in the
  material.

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

Click here to know more about Peierls Stress
40
How to increase the strength?

• We have seen that slip of dislocations weakens
  the crystal. Hence we have two strategies to
  strengthen the crystal/material ? completely
  remove dislocations ? difficult, but dislocation
  free whiskers have been produced (however, this
  is not a good strategy as dislocations can
  nucleate during loading)? increase resistance to
  the motion of dislocations or put impediments to
  the motion of dislocations ? this can be done in
  many ways as listed below.

• Solid solution strengthening (by adding
  interstitial and substitutional alloying
  elements).
• Cold Work ? increase point defect and dislocation
  density (Cold work increases Yield stress but
  decreases the elongation, i.e. ductility).
• Decrease in grain size ? grain boundaries provide
  an impediment ot the motion of dislocations
  (Hall-Petch hardening).
• Precipitation/dispersion hardening ? introduce
  precipitates or inclusions in the path of
  dislocations which impede the motion of
  dislocations.

Strengthening mechanisms
Precipitate Dispersoid
Forest dislocation
Solid solution
Grain boundary
41
Applied shear stress vs internal opposition

• PN stress is the bare minimum stress required
  for plastic deformation. Usually there will be
  other sources of opposition/impediment to the
  motion of dislocations in the material. Some of
  these include ? Stress fields of other
  dislocations? Stress fields from
  coherent/semicoherent precipitates? Stress
  fields from low angle grain boundaries? Grain
  boundaries? Effect of solute atoms and
  vacancies? Stacking Faults? Twin boundaries?
  Phonon drag ? etc.
• Some of these barriers (the short range
  obstacles) can be overcome by thermal activation
  (while other cannot be- the long range
  obstacles).
• These factors lead to the strengthening of the
  material.

Applied shear stress (?)
Internal resisting stress (?i)
42
Based of if the obstacle can be overcome by
thermal activation
Classification of the obstacles to motion of a
dislocation

• Athermal ? ? ? f (T, strain rate)
• These arise from 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

43
Effect of Temperature

• Motion of a dislocation can be assisted by
  thermal energy.
• However, motion of dislocations by pure thermal
  activation is random.
• A dislocation can be thermally activated to cross
  the potential barrier Q to the neighbouring
  metastable position.
• Strain rate can be related to the temperature (T)
  and Q as in the equation below.
• This thermal activation reduces the Yield stress
  (or flow stress).
• Materials which are brittle at room temperature
  may also become ductile at high temperatures.

Equilibrium positions of a dislocation
44
Metallic
Very high temperaturesneeded for thermal
activationto have any effect
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
45
What causes 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
46
Dislocation sources

• It is difficult to obtain crystals without
  dislocations (under special conditions whiskers
  have been grown without dislocations).
• Dislocation can arise by/form? Solidification
  (errors in the formation of a perfect crystal
  lattice)? Plastic deformation (nucleation and
  multiplication)? Irradiation

Some specific sources/methods of
formation/multiplication of dislocations include

• 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
• Frank-Read source
• Orowan bowing mechanism

47
Strain hardening

• We had noted that stress to cause further plastic
  deformation (flow stress) increases with strain ?
  strain hardening. This happens at
• 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

48
Formation of the Lomer-Cottrell barrier
49

(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

50
Impediments (barriers) to dislocation motion

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

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

51
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)
? (MN / m2) ? ( m/ m3) ?0 (MN / m2) A (N/m)
1.5 1010 0.5 10
100 1014 0.5 10
COLD WORK ? ? strength ? ? ductility
52
Grain size and strength
Hall-Petch Relation

• ?y ? Yield stress N/m2
• ?i ? Stress to move a dislocation in single
  crystal N/m2
• k ? Locking parameter N/m3/2
  (measure of the relative hardening contribution
  of grain boundaries)
• d ? Grain diameter m

53
Grain size

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

54
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? more the size difference,
  more the hardening effect? 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
• As shown in the plots in the next slide,
  increased solute concentration leads to an
  increased hardening. However, this fact has to be
  weighed in the backdrop of solubility of the
  solute. Carbon in BCC Fe has very little
  solubility (0.08 wt.) and hence the approach of
  pure solid solution strengthening to harden a
  material can have a limited scope.

55
For the same size difference thesmaller atom
gives a greaterstrengthening effect
Size effect
Size difference
Size effect depends on
Concentration of the solute (c)
200
Matrix Cu (r 1.28 Å)
Sn (1.51)
Be (1.12)
150
Solute strengthening of Cu crystal by solutes of
different sizes
Si (1.18)
Ni (1.25)
Al (1.43)
100
?y (MPa) ?
Zn (1.31)
50
(Values in parenthesis are atomic radius values
in Å)
Solute Concentration (Atom ) ?
40
10
20
30
0
56
By ? ?i (lattice friction)
X
? ?y
? ?
Solute atoms
? level of ? - ? curve
? ?
Often produce Yield Point Phenomenon
57
Relative strengthening effect of Interstitial and
Substitutional atoms

• Interstitial solute atoms have a non-spherical
  distortion field and can elastically interact
  with both edge and screw dislocations. Hence they
  give a higher hardening effect (per unit
  concentration) as compared to substitutional
  atoms which have (approximately) a spherical
  distortion field.

Relative strengthening effect / unit concentration
Interstitial
3Gsolute
Solute atoms
Gsolute / 10
Substitutional
58
(No Transcript)
59
Interstitial ? Edge and screw dl.
Elastic
Substitutional ? edge
Long range(T insensitive)
Modulus
Long range order
Mechanisms of interaction of dislocations with
solute atoms
Solute-dislocation interaction
Stacking fault
Short range (T sensitive)
Electrical
Short range order
60
The hardening effect of precipitates

• Precipitates may be coherent, semi-coherent or
  incoherent. Coherent ( semi-coherent)
  precipitates are associated with coherency
  stresses.
• Dislocations cannot glide through incoherent
  precipitates.
• Inclusions behave similar to incoherent
  precipitates in this regard (precipitates are
  part of the system, whilst inclusions are
  external to the alloy system).
• A pinned dislocation (at a precipitate) has to
  either climb over it (which becomes favourable at
  high temperatures) or has to bow around it.

Glide through the precipitate
If the precipitate is coherent with the matrix
Dislocation
Get pinned by the precipitate
A complete list of factors giving rise to
hardening due to precipitates/inclusions will be
considered later
61
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
  (though it is usually higher as precipitates can
  be intermetallic compounds).
• 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.

62
Schematic views ? edge dislocation glide through
a coherent precipitate
b
63
If the particle is sheared, then how does the
hardening effect come about?

• We have seen that as the dislocation glides
  through the precipitate it is sheared.
• If the precipitate is sheared, then how does it
  offer any resistance to the motion of the
  dislocation? I.e. how can this lead to a
  hardening effect?
• The hardening effect due to a precipitate comes
  about due to many factors (many of which are
  system specific). The important ones are listed
  in the tree below.

Increase in surface area due to particle shearing
Hardening effect
Part of the dislocation line segment (inside the
precipitate) could face a higher PN stress
64
Pinning effect of inclusions
Orowan bowing mechanism

• Dislocations can bow around widely separated
  inclusions. In this process they leave
  dislocation loops around the inclusions, thus
  leading to an increase in dislocation density.
  This is known as the orowan bowing mechanism as
  shown in the figure below. (This is in some
  sense similar to the Frank-Read mechanism).
• The next dislocation arriving (similar to the
  first one), feels a repulsion from the
  dislocation loop and hence the stress required to
  drive further dislocations increases.
  Additionally, the effective separation distance
  (through which the dislocation has to bow)
  reduces from d to d1.

65
Precipitate Hardening effect
(Complete List)

• The hardening effect of precipitates can arise in
  many ways as below
• Lattice Resistance the dislocation may face an
  increased lattice friction stress in the
  precipitate.
• 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 (when
  dislocations are split into partials)
• 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

66
Strain rate effects

• We had noted that strain rate can vary by orders
  of magnitude depending on deformation process
  (Creep 108 to Explosions 105).
• Strain rate effects become significant (on
  properties like flow stress) only when strain
  rate is varied by orders of magnitude (i.e. small
  changes in flow stress do not affect the value of
  the properties much).
• Strain rate can be related to dislocation
  velocity by the equation below.

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

67
In a UTT why does the plot not continue along OY
(straight line)?
Funda Check

• When stress is increased beyond the yield stress
  the mechanism of deformation changes.
• Till Y in the s-e plot, bond elongation
  (elastic deformation) gives rise to the strain.
• After Y, the shear stress resulting from the
  applied tensile force, tends to move dislocations
  (and cause slip) ? rather than stretch bonds ? as
  this will happen at lower stresses as compared to
  bond stretching (beyond Y).
• If there are not dislocations (e.g. in a whisker)
  (and for now we ignore other mechanism of
  deformation), the material will continue to load
  along the straight line OY ? till dislocations
  nucleate in the crystal.

68
TWINNING