chapter 9b plasticity
DESCRIPTION
mechanical propertyTRANSCRIPT
PLASTIC DEFORMATION
Modes of Deformation
The Uniaxial Tension Test
Mechanisms underlying Plastic Deformation
Strengthening mechanisms
Mechanical MetallurgyGeorge E Dieter
McGraw-Hill Book Company, London (1988)
An Al rod when bent through a large angle does not come back to its original shape.
Steel is more difficult to deform as compared to Al.
A steel piece is easier to deform when heated (as compared to when it is cold).
‘Chinese’ Clay when deformed does not regain its original shape.
‘Silly putty’ deforms like Chinese clay when slowly deformed. However, when one
bounces a ball of silly putty it bounces like a rubber ball.
Let us start with some observations… Revise: the mechanisms by which materials fail
There is no volume change during plastic deformation (by slip/twinning).
Shear stresses lead to plastic deformation in metallic materials Pure hydrostatic stresses
cannot cause plastic deformation (metals).
Fracture strain is strongly influenced by hydrostatic stresses.
Plastic deformation by slip (motion of dislocations leaving the crystal/grain) involves
shear stresses at the level of the slip plane (i.e. even if we apply tensile forces, certain
planes may feel shear stresses, which can lead to slip).
Amorphous materials can deform by ‘flow’ (e.g. glass blowing of heated glass), etc. →
these are not the focus of the current chapter.
Important points to be kept in mind
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
A body can be deformed using many modes:
Tension/Compression
Bending
Shear
TorsionIt is important to note that these are macroscopically defined with respect to a body of given geometry (even in tensile loading inclined
planes will be subjected to shear stress)
Common types of deformation
Tension Compression
ShearTorsion
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
DeformationMode 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
The following aspects need to be understood to comprehend plasticity*:
External process parameters (Temperature, strain rate etc.)
Macroscopic and Microscopic aspects of plasticity
Continuum and Discrete views of plasticity
Plasticity in single crystals
Plasticity in polycrystals
Plasticity in multiphase materials
Plasticity in nanomaterials
Path to understanding plasticity
* Some of these aspects will be covered in the current chapter
One of the simplest test which can performed to evaluate the mechanical properties of a
material is the Uniaxial Tension Test. The force/load applied is uniaxial.
This is typically performed on a cylindrical specimen with a standard ‘gauge length’. (At
constant temperature and strain rate). Other types of specimens are also used. Usually the specimen is polycrystalline.
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, 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)
0A
Ps
0L
Le
0 → initial
i → instantaneous
SubscriptNote: quantities obtained by performing an
Uniaxial Tension Test are valid only under
uniaxial state of stress
The Tensile Stress-Strain Curve
Stroke →
Lo
ad
→
e →
s →
→
→
Gauge Length → L0
Possible axes
Tensile specimen
Initial cross sectional area → A0
Important Note
We shall assume cylindrical polycrystalline specimens (unless
otherwise stated)
Note that L0 is NOT the length of the
specimen, but the gauge length
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 (for large
strains as in the example above).
► Hence, a quantity known as ‘True Strain’ is preferred (along with True Stress)
as defined in the next slide.
True Stress () and Strain ()
iA
P
0
ln
0L
L
L
dLL
L
0A
Ps
0L
Le
The definitions of true stress and true strain are based on instantaneous values of area (Ai)
and length (Li) and not on the original measures (as for engineering stress and strain).
Ai → instantaneous area
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
(v) Toughness (area under the curve)
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
* It better to determine the Young’s modulus from sound propagation experiments,
than from UTT experiments.
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
(the stress to cause continued plastic
deformation is called flow stress)
Neck
Polycrystalline Specimen
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 finally leaving the grain boundary or crystal surface.
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 Plastic 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 1ln
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, the original area (while
there is a local reduction in area around the neck). This leads to an underestimate of the stress (the flow stress).
‘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.
iA
P
0
ln
0L
L
L
dLL
L
0
ln 1 1 ln(1+e)L
L
0
0 0 0
1 1 (1 )i i
i
A L LPs s s e
A A L L
00 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 Points
Yielding begins when a stress equal to y (yield stress) is applied; however to cause further
plastic deformation increased stress has to be applied i.e. the material hardens with
plastic deformation → known as work hardening/strain hardening.
Beyond necking the state of stress becomes triaxial (in a cylindrical specimen considered).
Technically the yield criterion in uniaxial tension cannot be applied beyond this point.
We shall keep our focus on plastic deformation by slip.
y is yield stress in an uniaxial tension test (i.e. plastic deformation will start after crossing yield stress
only under uniaxial tensile loading) and should not be used in other states of stress (other criteria of
yield should be used for a generalized state of stress).
I.e in uniaxial tension the yield criterion is very simple:
Yielding starts when: y .
Hydrostatic pressure (leading to hydrostatic stress in a material) does not lead to yielding in
a continuous solid (usually!).
This implies that the stress deviator holds the key to yielding.
For an isotropic material the yield criterion will not be independent of the choice of the axes
(i.e. a invariant function). Hence, for an isotropic material the yield criteria (we will note that there are more than
one) will be a function of the invariants of the stress deviator.
Two commonly yield criteria are:
Von Mices or Distortion-Energy Criterion
Maximum shear stress or Tresca Criterion. (We will consider these later).
Plastic deformation (= yielding) and state of stress
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, eu) 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
(%) 100f
f
L Le
L
0
0
(%) 100fA A
qA
Note: this is ductility in
Uniaxial Tension Test
0
0
(%) 100uu
L Le
L
0
0
1
1
AL
L A q
0 00
0
11 1
1 1
L L A qe
L A q q
01
qe
q
We had seen two measures of ductility:
Strain at fracture (ef)
Reduction in area at fracture (q)
We had also seen that strain based measure can be related to area based measure 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 Le
L
0
0
fA Aq
A
01
qe
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 1lne)s( 1
Fractured surfacesFractured 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 1ln0
1
qe
q
0
0
iA Aq
A
0 0ln 1 )ε ( e 0
1ln 1 ln
1 1
qε
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.
SEM top view of fractured specimen showing shear lips
SEM micrograph: typical features observed in the case of ductile
fracture (microvoid formation)
‘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
21 ln 1
2
uniaxialR a
A R
The Correction
Cotd..
0 1ln ln
1f
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
The UTT test as described was done on a ductile material, then why did it
fracture? (don’t brittle materials fracture?)Funda Check
A ductile material is one whose fracture stress is above its yield stress: y < f.
Two factors contribute to this fracture:
(i) the triaxial state of stress introduced by the necking (like at a crack tip).
(ii) the progressive work hardening, which makes y > f (i.e. the yield stress becomes more
than the fracture stress).
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’.
Unloading followed by re-loading
Usually, there is a small hysteresis during unloading and reloading (as in the figure below).
The unloading takes place along the path XM, while the reloading is along the path MR.
I.e. during reloading, the curve joins the flow curve YF (that obtained during uninterrupted
loading) at R. (XR is the missing part of the flow curve YF).
This hysteresis is often neglected.
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
TestRange 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 plasticT
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 plasticK
There are variables, which we have to ‘worry about’, when we do mechanical testing
(for the test and to interpret the results):
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
TK
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 (before
fracture) 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 (540C)0.10 1570
‘n’ and ‘K’ for selected materials ,
ln
ln T
n
,
m
TC
C → 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
In some materials (due to structural condition or high temperature) necking is prevented by
strain rate hardening.
Further aspects regarding strain rate sensitivity
,
m
TC
m PC
A
1/ 1/1
m mP
C A
From the definition of true strain
1 1dL dA
L dt A dt
1/ 1/1 1
m mdA P
A dt C A
1/
(1 ) /
1m
m m
dA P
dt C A
If m < 1→ smaller the cross-sectional area, the
more rapidly the area is reduced.
If m = 1 → material behaves like a Newtonian
viscous liquid → dA/dt is independent of A.
1/
(1 ) /
1m
m m
dA P
dt C A
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) <1120> 3
BCC
Not close packed{110}, {112}, {123}
½[111] 12
NaCl
Ionic{110}
{111} not a slip plane½<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} <101>
CaF2, UO2, ThO2
Fluorite{001} <110>
CsCl
B2{110}
<001>
NaCl, LiF, MgO
Rock Salt{110} <110> 6
C, Ge, Si
Diamond cubic{111} <110> 12
MgAl2O4
Spinel{111} <110>
Al2O3
Hexagonal(0001) <1120>
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)<1120>
1 x 3 = 3
3 x 1 = 3
6 x 1 = 6
Al2O3
R 3c
{0001}
{10 10}(a/3)<1120>
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
normalSlip direction
A’
DArea
ForceStress
1
CosA
CosF
/
RSS Cos Cos
A
Cos
AA'
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
CRSSy
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 eG
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
&
DispersoidGrain 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 dislocationBased of if the obstacle can be
overcome by thermal activation
Effect of Temperature
Equilibrium positions of a dislocation
Q
It is common observation that metals can be easily deformed when heated→ Ironsmith
gives shape to steel after heating it.
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
kTAe
rateStraindt
d
Fe
W
18-8 SS
Cu
Ni
Al2O3
Si
150
300
450
0.2 0.4 0.60.0
Yie
ld
Str
ess
(MP
a) →
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
Note in the figure below that the metals which start out stronger (have a high yield stress)
become severely weakened on heating.
For covalent and ionic materials the curves are not drawn at low temperatures → at low
temperatures these materials fracture (are brittle).
What causes Strain hardening? → multiplication of dislocations
Hence two important questions need to be answered:
How does the dislocation density increase?
What leads to strain hardening?
6 9 12 14
dislocation dislocation
~ (10 10 ) ~ (10 10 )
Cold workAnnealed material Stronger material
Strain hardening
→
X
Strain hardening →
→
→
X
Strain hardening →
→
The increase in flow stress with strain is called strain hardening (or work hardening).
Plastic deformation is caused by dislocations moving and leaving the crystal that
dislocation density should decrease with plastic deformation. However, it is observed that
dislocation density increases with strain! Further, if dislocations are the agent weakening the
crystal then with increased dislocation density the material should get weaker! But, we
observe on the contrary an increase in strength!
This implies some sources of dislocation multiplication /
creation should exist &
More dislocations is somehow causing a ‘traffic jam’ kind
of scenario and leading to strengthening.
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/work 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
Dislocations can also be rendered immobile as in the formation of the Lomer-Cottrell
barrier.
Dynamic recovery
The word ‘dynamic’ as an adjective implies (in the context of deformation) that the process
to which it is prefixed occurs during deformation.
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.
)111(
]1 0 1[2
1
)111(
]0 1 1[2
1
)100(
]1 1 0[2
1
Formation of the Lomer-Cottrell barrier
]1 1 0[
)100(
]1 1 0[2
1
)111(
]0 1 1[2
1
)111(
]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
Impediments (barriers) to dislocation motion
Grain boundaries (especially high angle grain boundaries).
Immobile (sessile) dislocations
► Lomer-Cottrell lock
► Frank partial dislocation.
Twin boundaries.
Precipitates (especially finely distributed incoherent precipitates) and inclusions.
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
Other dislocations present in the material which intersect (or be present in) the slip plane
can impede the motion of dislocations. This implies that increase in dislocation density can
lead to an increase in the strength of the material (as given by the empirical relation as below).
Dislocation can intersect with each other and ‘react’ to give jogs and kinks (apart from
sessile dislocations).
Cold work will increase the strength of the material via an increase in dislocation density,
but will lead to a decrease in the ductility (as expected!).
Dislocation ReactionsClick here to know more about dislocation interactions
to give rise to jogs and kinks
Grain size and strength
d
kiy
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
Hall-Petch constants
Material I [MPa] k [MPa m1/2]
Cu 25 0.11
Ti 80 0.40
Mild steel 70 0.74
Ni3Al 300 1.70101 10)2(
645
n
d d → Grain diameter in meters
n → ASTM grain size number
Grain size is related to the ASTM grain size number
by the following formula
Studies across many materials has shown that on decreasing the grain size the yield
strength of the material increases. The precise formula for this relation is the famous Hall-
Petch relation.
At vary small grain sizes this formula is expected to break down as we will approach the
theoretical shear strength of the material. In some nanostructured materials even inverse
Hall-Petch relation has been found (i.e. decrease in strength with decrease in grain size).
The usual mechanism suggested for this behaviour is the dislocation ‘pile-up’ mechanism.
Photo courtesy: Dr. Eswar Prasad (Johns Hopkins University, 2013). Mg alloy (AZ31, 3% Al , 1% Zn)
Dislocations get piled up at a barrier (say grain boundary) and produce a back stress. This
back stress impedes the motion of other dislocations on the same slip plane (towards the
boundary).
Dislocation pile up at grain boundary
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
(MP
a)→
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 effectSize difference
Size effect depends on:
Concentration of the solute (c) cy ~
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 fielddistortionSpherical
fielddistortionsphericalNon
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.
However, it should be noted that interstitial atoms usually have very limited solubility
(e.g. 0.01 wt.% C is soluble in BCC Fe at RT). On the other hand, substitutional atoms can
have complete solubility (e.g. isomorphous Au-Ag system→ satisfies Hume-Rothery
rules).
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.
IncoherentcoherentPartially Coherent
GrowthGrowthLargeSmall
b
Precipitate particleb
Schematic views
edge dislocation glide through a coherent precipitate
Hardening effectPart 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
dd vb 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
TWINNING