plastic deformationeacharya.inflibnet.ac.in/data-server/eacharya... · creep mechanisms. grain...
TRANSCRIPT
PLASTIC DEFORMATIONPLASTIC 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.
A steel piece is easier to deform when heated (as compared to when it is cold).
Let us start with some observations…
Click here to know about all the mechanisms by which materials fail Click here to know about all the mechanisms by which materials fail
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
Modesof
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.
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)
0APs
0LLe
0 → initial
i → instantaneousSubscript 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 → L0Possible axes
Tensile specimen
Initial cross sectional area → A0
Important Note
We shall assume cylindrical 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 strainBut 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 ()
iAP
0
ln0
LL
LdLL
L
0A
Ps 0LLe
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
UTS-
Ultimate Tensile StrengthSubscripts:
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 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 strainTrue strain () 0.01 0.10 0.20 0.50 1.0 2.0 3.0 4.0Engineering 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
(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.
iAP
0
ln0
LL
LdLL
L
0
ln 1 1 ln(1+e)LL
0
0 0 0
1 1 (1 )i i
i
A L LP s s s eA A L L
00 0 i i
0
From volume constancy A L =A L i
i
A LA 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).
Microscopicelastic
Macroscopicelastic
Important Note
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).
Tresca
and von Mices
criterion are the two most popular ones.
Important Note
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.
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=A0
L0
).
What is meant by ductility?
0
0
(%) 100ff
L Le
L
0
0
(%) 100fA Aq
A
Note: this is ductility in Uniaxial
Tension Test
0
0
(%) 100uu
L LeL
0
0
11
ALL A q
0 00
0
11 11 1
L L A qeL A q q
0 1
qeq
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
ff
L Le
L
0
0
fA Aq
A
0 1qe
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
qeq
0
0
iA AqA
0 0ln 1 )ε ( e 01ln 1 ln
1 1qε
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
21 ln 12
uniaxial R aA R
The Correction
Cotd..
0 1ln ln1f
f f
AA 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.
ff
f
PA
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.
Xelastic
Yelastic
* 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.
,
nplastic 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.
ny 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,
lnln 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.
,
lnln T
m
Thermal softening coefficient () is also defined as below:
lnlnT
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/ 1/1m mPC A
From the definition of true strain
1 1dL dAL dt A dt
1/ 1/1 1m mdA P
A dt C A
1/
(1 ) /
1m
m m
dA Pdt 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 Pdt 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 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
TiO2Rutile {101} <101>
CaF2
, UO2
, ThO2Fluorite {001} <110>
CsCl
B2 {110} <001>
NaCl, LiF, MgO
Rock Salt {110} <110> 6
C, Ge, Si
Diamond cubic {111} <110> 12
MgAl2
O4Spinel {111} <110>
Al2
O3Hexagonal (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 = 1224 x 1 = 24
Mg (HCP)P63
/mmc
{0001}{10 10}{10 11}
(a/3)<1120>1 x 3 = 33 x 1 = 36 x 1 = 6
Al2
O3
R 3c{0001}{10 10}
(a/3)<1120>1 x 3 = 33 x 1 = 3
NaClFm 3m
{110}{001}
(a/2)< 1 10>6 x 1 = 66 x 1 = 6
CsClPm 3m
{110} a<001> 6 x 1 = 6
PolyethylenePnam
(100){110}(010)
c<001>1 x 1 = 12 x 1 = 21 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’
DAreaForceStress
1
CosACosF/
RSS Cos Cos
A
CosAA'
AF
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
bw
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 solutionPrecipitate
& 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.
QkTAe
rateStraindtd
FeW
18-8 SS
Cu
Ni
Al2
O3
Si
150
300
450
0.2 0.4 0.60.0
Yie
ld S
tress
(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-BCCW-BCC
Cu-FCC
Ni-FCC
MetallicIonic
Covalent
→
X
Strain hardening →
→
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
)1010(~
)1010(~
1412ndislocatio
96
ndislocatio
materialStrongermaterialAnnealed workCold
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
)111(
]1 0 1[21
)111(
]0 1 1[21
)100(
]1 1 0[21
Formation of the Lomer-Cottrell barrier
)100(
]1 1 0[21
)111(
]0 1 1[21
)111(
]1 0 1[21
+
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 10100 1014 0.5 10
COLD WORK ► ↑ strength
► ↓ ductility
Grain size and strength
dk
iy 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
constantsMaterial I
[MPa] k [MPa
m1/2]
Cu 25 0.11
Ti 80 0.40
Mild steel 70 0.74
Ni3
Al 300 1.70
Photo courtesy: Dr. Eswar Prasad (Johns Hopkins University, 2013)Mg alloy (AZ31, 3% Al , 1% Zn)
Grain size
101 10)2(645 nd
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(M
Pa)→
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)
Si (1.18)
Sn
(1.51)
Ni (1.25)
Zn (1.31)
Al (1.43)
(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.
Relative strengthening effect of Interstitial and Substitutional
atoms
Long range
(T insensitive)
Solute-dislocation interaction
Short range
(T sensitive)
Modulus
Long range order
ElasticSubstitutional → 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 CoherentGrowthGrowth LargeSmall
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
TWINNINGTWINNING