5956651.ppt

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)

Plastic
deformation
Mechanisms / Methods by which a can Material can FAIL
Fracture
Fatigue
Creep Chemical /
Electro-chemical
degradation
Physical
degradation
Wear
Erosion
Microstructural
changes
Phase transformations
Twinning
Grain growth
Elastic deformation
Particle coarsening
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.
* Beyond a certain limit
Corrosion
Oxidation
Slip
Twinning

Slip
(Dislocation
motion)
Plastic Deformation in Crystalline Materials
Twinning Phase Transformation Creep Mechanisms
Grain boundary sliding
Vacancy diffusion
Dislocation climb
+ Other Mechanisms
Note: Plastic deformation in amorphous materials occur by other mechanisms including flow (~viscous fluid) and shear
banding
 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.
Grain rotation

 Tension/Compression
 Bending
 Shear
 Torsion
Common types of deformation
Tension Compression
Shear Torsion
Deformed configuration
Bending
Note: modes of deformation in other contexts will be defined in the topic on plasticity
Tension / Compression
Torsion
Modes
of
Deformation Shear
Bending
Review

Mode I
Mode III
Modes
of
Deformation
Mode II
 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)
Peak ahead

 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.
The Uniaxial Tension Test (UTT)
0
A
P
s 
0
L
L
e


0 → initial
i → instantaneous
Subscript
Note: quantities obtained by performing an
Uniaxial Tension Test are valid only under
uniaxial state of stress

The Tensile Stress-Strain Curve
Stroke →
Load
→
e →
s
→
 →

→
Gauge Length → L0
Possible axes
Tensile specimen
Initial cross sectional area → A0
Important Note
We shall assume cylindrical specimens (unless otherwise stated)

Problem with engineering Stress (s) and Strain (e)!!
Consider the following sequence of deformations:
L0
2L0
L0
e1→2 = 1
e2→3 =  ½
e1→3 = 0
1
2
3
[e1→2 + e2→3] = ½
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.

True Stress () and Strain ()
i
A
P


0
ln
0
L
L
L
dL
L
L

 

 Ai → instantaneous area
0
A
P
s 
0
L
L
e


 The definitions of true stress and true strain are based on instantaneous values of area (Ai)
and length (Li).

Same sequence of deformations considered before:
L0
2L0
L0
 1→2 = Ln(2)
 2→3 =  Ln(2)
1→3 = 0
1
2
3
[ 1→2 +  2→3] = 0
With true strain things turn out the way they should!

Schematic s-e and - curves
Information gained from the test:
(i) Young’s modulus
(ii) Yield stress (or proof stress)
(iii) Ultimate Tensile Stress (UTS)
(iv) Fracture stress
UTS- Ultimate Tensile Strength
Subscripts:
y- yield, F,f- fracture,
u- uniform (for strain)/ultimate (for stress)
Points and regions of the curves are explained in the next slide
 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 ~10–3)
[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).
Schematics: not to scale
Note the increasing stress required
for continued plastic deformation
Neck

 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)
Sequence of events during the tension test
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

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
e)
(
ε 
 1
ln
e)
s( 
 1

Comparison between “Engineering” and “True” quantities:
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.
 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.
i
A
P


0
ln
0
L
L
L
dL
L
L

 

0
ln 1 1 ln(1+e)
L
L

 
   
 
 
0
0 0 0
1 1 (1 )
i i
i
A L L
P
s s s e
A A L L

   
      
   
   
0
0 0 i i
0
From volume constancy A L =A L i
i
A L
A L
  
Valid till
necking starts

 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 (~10–6 )
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)
 ~10–6 [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.
Where does Yielding start?
 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).
Microscopic
elastic

Macroscopic
elastic


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.

 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 < 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 (AL=A0L0).
What is meant by ductility?
0
0
(%) 100
f
f
L L
e
L

  0
0
(%) 100
f
A A
q
A

 
Note: this is ductility in
Uniaxial Tension Test
0
0
(%) 100
u
u
L L
e
L

 
0
0
1
1
A
L
L A q
 

0 0
0
0
1
1 1
1 1
L L A q
e
L A q q

     
 
0
1
q
e
q



 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’.
Comparison between reduction in area versus strain at fracture
0
0
f
f
L L
e
L


0
0
f
A A
q
A


0
1
q
e
q



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.
e)
(
ε 
 1
ln
e)
s( 
 1

Fractured surfaces
Fractured surfaces
Neck

Beyond necking
‘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.
e)
(
ε 
 1
ln
0
1
q
e
q


0
0
i
A A
q
A


0 0
ln 1 )
ε ( e
  0
1
ln 1 ln
1 1
q
ε
q q
   
  
   
 
   
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.

‘True’ values beyond necking
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
(exp=Load/Alocal), using the formula as below.
* P.W. Bridgman, Trans. Am. Soc. Met. 32 (1944) 553.
exp
2
1 ln 1
2
uniaxial
R a
A R

 
 
   
 
   
 
   
 
The Correction
Cotd..

0 1
ln ln
1
f
f f
A
A q

   
 
   
   

   
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.
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.
f
f
f
P
A
 

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.
X
elastic

Y
elastic

* 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
don’t 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’

Variables in plastic deformation T
,
,
, 

 
 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 < 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
Test
Range of strain rate
(/s)
Creep tests 10–8 to 10–5
Quasi-static tension tests (in an UTM) 10–5 to 10–1
Dynamic tension tests 10–1 to 102
High strain rate tests using impact bars 102 to 104
Explosive loading using projectiles (shock tests) 104 to 108
0
/
v L
 

 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.
,
n
plastic plastic
T
K

 
 
  
Usually expressed as (for plastic)
 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.
 Deviations from this behaviour often observed (e.g. in Austenitic stainless steel) at low
strains (~10–3) 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.
n
y plastic
K
  
 

 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.
Variables/parameters in mechanical testing

,
n
T
K

 
 
  
Variables in plastic deformation T
,
,
, 

 
When true strain is less than 1, the smaller value of ‘n’ dominates over a larger value of ‘n’
K → strength coefficient
n → 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
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
‘n’ and ‘K’ for selected materials
,
ln
ln T
n

  
  
 
 
 

,
m
T
A

 
 
  
A → a constant
m → 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)
The effect of strain rate is compared by performing tests to a
constant strain
 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.
,
ln
ln T
m

  
  
 
 
 
 Thermal softening coefficient () is also defined as below:
ln
lnT






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.

 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.
Plastic deformation by slip Click here to see overview of mechanisms/modes of plastic deformation
* Leaving the crystal part is important
** To be defined soon

Crystal Slip plane(s) Slip direction Number of slip systems
FCC {111} ½<110> 12
HCP (0001) <1120> 3
BCC
Not close packed
{110}, {112}, {123}
½[111] 12
NaCl
Ionic
{110}
{111} not a slip plane
½<110> 6
C
Diamond cubic
{111} ½<110> 12
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 Slip systems
TiO2
Rutile
{101} <101>
CaF2, UO2, ThO2
Fluorite
{001} <110>
CsCl
B2
{110}
<001>
NaCl, LiF, MgO
Rock Salt
{110} <110> 6
C, Ge, Si
Diamond cubic
{111} <110> 12
MgAl2O4
Spinel
{111} <110>
Al2O3
Hexagonal
(0001) <1120>
More examples of slip systems

Crystal Slip plane(s) Slip direction (b) Slip systems
Cu (FCC)
Fm 3m
{111} (a/2)< 1 10> 4 x 3 = 12
W (BCC)
Im 3m
{011}
{112}
{123}
(a/2)<11 1>
6 x 2 = 12
12 x 1 = 12
24 x 1 = 24
Mg (HCP)
P63/mmc
{0001}
{10 10}
{10 11}
(a/3)<1120>
1 x 3 = 3
3 x 1 = 3
6 x 1 = 6
Al2O3
R 3c
{0001}
{10 10}
(a/3)<1120>
1 x 3 = 3
3 x 1 = 3
NaCl
Fm 3m
{110}
{001}
(a/2)< 1 10>
6 x 1 = 6
6 x 1 = 6
CsCl
Pm 3m
{110} a<001> 6 x 1 = 6
Polyethylene
Pnam
(100)
{110}
(010)
c<001>
1 x 1 = 1
2 x 1 = 2
1 x 1 = 1
More examples of slip systems

 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 Schmid’s law)
Slip plane
normal
Slip direction




A’
D
Area
Force
Stress
1

















Cos
A
Cos
F
/
RSS Cos Cos
   

A








Cos
A
A'







A
F

Schmid factor
Critical Resolved Shear Stress (CRSS)
Note externally only tensile
forces are being applied
What is the connection between Peierls stress and CRSS?

Schmid’s law
CRSS
y
Cos Cos


 

RSS CRSS
 

Slip is initiated when
Yield strength of a single crystal
CRSS is a material parameter, which is determined from experiments

How does the motion of dislocations lead to a macroscopic shape change?
(From microscopic slip to macroscopic deformation  a first feel!)
Net shape change
 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).

b


Dislocation
formed by
pushing in
a plane
Step formed
when dislocation
leaves the crystal
Now visualize dislocations being punched in on successive planes  moving and finally
leaving the crystal

Net shape change
This sequence of events finally leads to deformed shape which can be approximated to a
rhombus

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.
Click here to know more about Peierls Stress







 b
w
PN e
G


2  G → shear modulus of the crystal
 w → width of the dislocation !!!
 b → |b|

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.
Forest dislocation
Strengthening mechanisms
Solid solution
Precipitate
&
Dispersoid
Grain boundary

Applied shear stress vs internal opposition
Applied shear stress () Internal resisting stress (i)
 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.

Internal resisting stress (i)
Long range obstacles (G)
Short range obstacles (T)
 Athermal    f (T, strain rate)
 These arise from long range internal stress fields
 Sources:
►Stress fields of other dislocations
►Incoherent precipitates
Note: G has some temperature
dependence as G decreases with 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
Classification of the obstacles to motion of a dislocation
Based of if the obstacle can be
overcome by thermal activation

Effect of Temperature
Equilibrium positions of a dislocation
Q
 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.
Q
kT
Ae

 
 
 

rate
Strain
dt
d






Fe
W
18-8 SS
Cu
Ni
Al2O3
Si
150
300
450
0.2 0.4 0.6
0.0
Yield
Stress
(MPa)
→
T/Tm →
Very high temperatures
needed for thermal activation
to have any effect
RT is like HT and P-N
stress is easily overcome
Fe-BCC
W-BCC
Cu-FCC
Ni-FCC
Metallic
Ionic
Covalent

 →
X

→
What causes Strain hardening? → multiplication of dislocations
 Why increase in dislocation density ?
 Why strain hardening ?
If dislocations were to leave the surface
of the crystal by slip / glide then the
dislocation density should decrease
on plastic deformation →
but observation is contrary to this
)
10
10
(
~
)
10
10
(
~ 14
12
n
dislocatio
9
6
n
dislocatio 


 

 

material
Stronger
material
Annealed work
Cold
Strain hardening
This implies some sources of dislocation
multiplication / creation should exist

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
 Frank-Read source
 Orowan bowing mechanism
 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

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

)
11
1
(
]
1
0
1
[
2
1






)
111
(
]
0
1
1
[
2
1






)
100
(
]
1
1
0
[
2
1






Formation of the Lomer-Cottrell barrier

)
100
(
]
1
1
0
[
2
1






)
111
(
]
0
1
1
[
2
1






)
11
1
(
]
1
0
1
[
2
1






+
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
]
1
1
0
[

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

Stress to move a dislocation & dislocation density


 A

 0
 0 → base stress to move a dislocation in the crystal
in the absence of other dislocations
  → Dislocation density
 A → A constant
↑ as ↑ (cold work)  ↑ (i.e. strain hardening)
Empirical relation
 (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

Grain size and strength
d
k
i
y 
 

 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]
Hall-Petch Relation

Grain size
10
1
10
)
2
(
645

 n
d
 d → Grain diameter in meters
 n → ASTM grain size number

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.

Solute Concentration (Atom %) →

y
(MPa)
→
50
100
150
10 20 30 40
200
0
Solute strengthening of Cu crystal
by solutes of different sizes
Matrix: Cu (r = 1.28 Å)
Be (1.12)
Sn (1.51)
Zn (1.31)
(Values in parenthesis are atomic radius values in Å)
Size effect
Size difference
Size effect depends on:
Concentration of the solute (c) c
y 
~

For the same size difference the
smaller atom gives a greater
strengthening effect

↑ y
Often produce Yield Point Phenomenon
Solute atoms ↑ level of  -  curve
 →
X

→
 →

→
By ↑ i (lattice friction)

Interstitial
Solute atoms
Substitutional
3Gsolute
Relative strengthening effect / unit concentration
Gsolute / 10






 
 field
distortion
Spherical







 
  field
distortion
spherical
Non
 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 of Interstitial and Substitutional atoms

Long range
(T insensitive)
Solute-dislocation interaction
Short range
(T sensitive)
Modulus
Long range order
Elastic
Substitutional → edge
Interstitial → Edge and screw dl.
Electrical
Short range order
Stacking fault
Mechanisms of interaction of
dislocations with solute atoms

The hardening effect of precipitates
Glide through the precipitate
Dislocation
Get pinned by the precipitate
If the precipitate is coherent with the matrix
 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.
A complete list of factors giving rise to hardening due to precipitates/inclusions will be considered later

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.
Incoherent
coherent
Partially
Coherent
Growth
Growth Large
Small

 


 


b
Precipitate particle
b
Schematic views
 edge dislocation glide through a coherent precipitate

Hardening effect
Part of the dislocation line segment (inside the
precipitate) could face a higher PN stress
Increase in surface area due to particle shearing
 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.
If the particle is sheared, then how does the hardening effect come about?

Pinning effect of inclusions
 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’.
Orowan bowing mechanism

Precipitate Hardening effect
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
(Complete List)

 We had noted that strain rate can vary by orders of magnitude depending on deformation
process (Creep: 10–8 to Explosions: 10–5).
 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.
Strain rate effects
d
d v
b

 
  vd → velocity of the dislocations
 d → density of mobile/glissile dislocations
 b → |b|

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

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