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)
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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
Modesof
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 (σ) and True Strain (ε) (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)
0A
Ps =
0L
Le
∆=
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 →
Loa
d →
e →
s →
ε →
σ →
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
Σ[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 (σ) and Strain (ε)
iA
P=σ
0
ln0
L
L
L
dLL
L
== ∫ε
� Ai → instantaneous area
0A
Ps =
0L
Le
∆=
� 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
ε 1→2 = Ln(2)
ε 2→3 = − Ln(2)
ε1→3 = 0
1
2
3
Σ[ε 1→2 + ε 2→3] = 0
With true strain things turn out the way they should!
Schematic s-e and σ-ε 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 σy 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 σ-ε 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 σ) 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 (ε) 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( += 1σ
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.
iA
P=σ
0
ln0
L
L
L
dLL
L
== ∫ε0
ln 1 1 ln(1+e)L
Lε
= + − =
0
0 0 0
1 1 (1 )i i
i
A L LPs s s e
A A L Lσ
= = = + − = +
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).
Microscopicelasticε
Macroscopicelasticε
Important Note
� σy 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. σy < σf).
� 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
(%) 100f
f
L Le
L
−= × 0
0
(%) 100fA Aq
A
−= ×
Note: this is ductility in Uniaxial Tension Test
0
0
(%) 100uu
L Le
L
−= ×
0
0
1
1
AL
L A q= =
−0 0
00
11 1
1 1
L L A qe
L A q q
−= = − = − =
− −0 1
qe
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 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 σ-ε 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 ‘ε’ and ‘s’ to ‘σ’ 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, σF) has to be
corrected for: (i) triaxial state of stress, (ii) correct cross sectional area.
e)(ε += 1lne)s( += 1σ
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 += 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.
‘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 (σuniaxial) is calculated from the stress from the experiment (σexp=Load/Alocal), using the formula as below.
* P.W. Bridgman, Trans. Am. Soc. Met. 32 (1944) 553.
exp
21 ln 1
2
uniaxial R a
A R
σσ =
+ +
The Correction
Cotd..
0 1ln ln
1f
f f
A
A qε
= = −
Calculation of fracture stress/strain: � To calculate true fracture stress (σF) 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
P
Aσ =
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 ~σX. Because of strain hardening* the yield strength of the specimen is higher.
Xelasticε
Yelasticε
* 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 ~σX (i.e. as we don’t know the ‘prior history’ of the specimen loading, we would call ~σX as yield stress of the specimen).
The blue part of the curve is also called the ‘flow stress’
Variables in plastic deformation T , , , εεσ &
� 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 σ-ε 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.
ε&
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 Lε =&
� In the σ-ε 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
TK
εσ ε = &
Usually expressed as (for σplastic)
� 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 (τE) � Onset of microscopic plastic deformation above a stress of τA. � For Mo a comparison of these quantities is as follows: τE = 0.5 MPa, τA = 5 MPa and τ0 (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 plas ticKσ σ ε= +
� 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
εσ ε = &
Variables in plastic deformation T , , , εεσ &
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
nε
σ ε σε σ ε
∂ ∂= =
∂ ∂ &
,
m
TA
εσ ε = &
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
mε
σ ε σε σ ε
∂ ∂= =
∂ ∂
&
& &
� Thermal softening coefficient (ν) is also defined as below:
ln
lnT
σν ∂=
∂
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
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 normal
Slip direction ϕϕϕϕ
θθθθ
σσσσ
ττττ
A’
DArea
ForceStress
1
=
=θα
τCosA
CosF
/
RSS Cos Cosτ σ θ ϕ=
A
=θCos
AA'
=A
F σ
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
τσ
θ ϕ=
RSS CRSSτ τ≥Slip is initiated when
Yield strength of a single crystal
τCRSS 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).
br
�
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 eGπ
τ2
� 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 (τ) Internal resisting stress (τi)
� 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 (τi)
Long range obstacles (τG)
Short range obstacles (τT)
� Athermal ⇒ τ ≠ f (T, strain rate) � These arise from long range internal stress fields � Sources: ►Stress fields of other dislocations ►Incoherent precipitates
Note: τG has some temperature dependence as G decreases with T
� Thermal ⇒ τ = 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
kTAeε − =&
rateStraindt
d →=
εε&
Fe
W
18-8 SS
Cu
Ni
Al2O3
Si
150
300
450
0.2 0.4 0.6 0.0
Yie
ld S
tres
s (M
Pa)
→
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
ε →
X
σ →
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(~
1412
ndislocatio
96
ndislocatio − →
− ρρmaterialStrongermaterialAnnealed
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[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
)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
]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
ρττ A+= 0
� τ0 → base stress to move a dislocation in the crystal in the absence of other dislocations
� ρ → Dislocation density
� A → A constant
ρ↑ as ε↑ (cold work) ⇒ τ↑ (i.e. strain hardening)
Empirical relation
τ (MN / m2) ρ ( m/ m3) τ0 (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
kiy += σσ
� σy → Yield stress [N/m2]
� σi → 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
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 %) →
σy (
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 effect Size difference
Size effect depends on:
Concentration of the solute (c) cy ∝~ σ
For the same size difference the smaller atom gives a greater strengthening effect
↑ σy
Often produce Yield Point Phenomenon
Solute atoms ↑ level of σ - ε curve
ε →
X
σ →
ε →
σ →
By ↑ σi (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
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 CoherentGrowthGrowth LargeSmall
→ →
b
Precipitate particleb
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
dd vb ρε =& � vd → velocity of the dislocations
� ρd → 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
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