PLASTIC DEFORMATIONMechanisms of Plastic DeformationThe Uniaxial Tension TestMechanisms of Plastic DeformationMechanical MetallurgyGeorge E Dieter McGraw-Hill Book Company, London (1988)

Plastic deformationMechanisms / Methods by which a can Material can FAILFractureFatigueCreepChemical / Electro-chemical degradationPhysical degradationWearErosionMicrostructural changesPhase transformationsTwinningGrain growth Elastic deformationParticle coarseningIf 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 limitCorrosionOxidationSlipTwinning

Slip (Dislocation motion)Plastic Deformation in Crystalline MaterialsTwinningPhase TransformationCreep MechanismsGrain boundary slidingVacancy diffusionDislocation climb+ Other MechanismsNote: Plastic deformation in amorphous materials occur by other mechanisms including flow (~viscous fluid) and shear bandingPlastic 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/CompressionBendingShearTorsionCommon types of deformationTensionCompressionShearTorsionDeformed configurationBendingNote: modes of deformation in other contexts will be defined in the topic on plasticityReview

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

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

The Tensile Stress-Strain CurveGauge Length L0Possible axesTensile specimenInitial cross sectional area A0Important NoteWe shall assume cylindrical specimens (unless otherwise stated)

Problem with engineering Stress (s) and Strain (e)!!Consider the following sequence of deformations:L02L0L0e12 = 1e23 = e13 = 0123[e12 + e23] = 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 Clearly there is a problem with the use (definition) of Engineering strain Hence, a quantity known as True Strain is preferred (along with True Stress) as defined in the next slide.

True Stress () and Strain () Ai instantaneous areaThe definitions of true stress and true strain are based on instantaneous values of area (Ai) and length (Li).

Same sequence of deformations considered before:L02L0L0 12 = Ln(2) 23 = Ln(2)13 = 0123[ 12 + 23] = 0With true strain things turn out the way they should!

Schematic s-e and - curvesInformation gained from the test:Youngs modulusYield stress (or proof stress)Ultimate Tensile Stress (UTS)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 ~103) [E is measured in GPa and y in MPa thus giving this small strains] the linear portion is practically vertical and stuck to the Y-axis (when efracture and eelastic is drawn to the same scale).Schematics: not to scaleNote the increasing stress required for continued plastic deformationNeck

O unloaded specimenOY 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 lengthN 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 stressF Fracture of specimen (many polycrystalline materials like Al show cup and cone fracture)Sequence of events during the tension testNotes: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 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.Valid till necking starts

Comparison between true strain and engineering strainTrue strain ()0.010.100.200.501.02.03.04.0Engineering strain (e)0.010.1050.220.651.726.3919.0953.6

Yielding can be defined in many contexts. Truly speaking (microscopically) it is point at which dislocations leave the crystal (grain) and cause microscopic plastic deformation (of unit b) this is best determined from microstrain (~106 ) experiments on single crystals. However, in practical terms it is determined from the stress-strain plot (by say an offset as described below).True elastic limit (microscopically and macroscopically elastic where in there is not even microscopic yielding) ~106 [OA portion of the curve]Microscopically plastic but macroscopically elastic [AY portion of the curve]Proportional limit the point at which there is a deviation from the straight line elastic regimeOffset 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).

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

Slip is competing with other processes which can lead to failure.In simple terms a ductile material is one which yields before failure (i.e. y < f).Ductility depends on the state of stress used during deformation.We can obtain an measure of the ductility of a material from the uniaxial tension test as follows (by putting together the fractured parts to make the measurement): Strain at fracture (ef) (usually expressed as %), (often called elongation, although it is a dimensionless quantity) Reduction in area at fracture (q) (usually expressed as %) q is a better measure of ductility as it does not depend on the gauge length (L0); while, ef depends on L0. Elongation/strain to necking (uniform elongation) can also be used to avoid the complication arising from necking. Also, q is a more structure sensitive ductility parameter.Sometimes it is easier to visualize elongation as a measure of ductility rather than a reduction in area. For this the calculation has be based on a very short gauge length in the necked region called Zero-gauge-length elongation (e0). e0 can be calculated from q using constancy of volume in plastic deformation (AL=A0L0).What is meant by ductility?Note: this is ductility in Uniaxial Tension Test

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

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.Fractured surfacesFractured surfacesNeck

Beyond neckingTrue values beyond neckingCalculation 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.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 neckingCalculation 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.The CorrectionCotd..

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 strainThe 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.

Unloading the specimen during the tension testIf the specimen is unloaded beyond the yield point (say point X in figure below), elastic strain is recovered (while plastic strain is not). The unloading path is XM.The strain recovered ( ) is more than that recovered if the specimen was to be unloaded from Y ( ) i.e. in this sense the material is more elastic in the plastic region (in the presence of work hardening), than in the elastic region!If the specimen is reloaded it will follow the reverse path and yielding will start at ~X. Because of strain hardening* the yield strength of the specimen is higher.* Strain Hardening is also called work hardening The material becomes harder with plastic deformation (on tensile loading the present case) We will see later that strain hardening is usually caused by multiplication of dislocations.If one is given a material which is at point M, then the Yield stress of such a material would be ~X (i.e. as we dont know the prior history of the specimen loading, we would call ~X as yield stress of the specimen).The blue part of the curve is also called the flow stress

Variables in plastic deformationIn 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 experimentsTestRange of strain rate (/s)Creep tests108 to 105Quasi-static tension tests (in an UTM)105 to 101Dynamic tension tests101 to 102High strain rate tests using impact bars102 to 104Explosive loading using projectiles (shock tests)104 to 108

In the - plot the plastic stress and strain are usually expressed by the expression given below. Where, n is the strain hardening exponent and K is the strength coefficient.Usually expressed as (for plastic)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=0.5MPa, A=5MPa and 0(macroscopic yield stress in shear)=50MPa.Deviations from this behaviour often observed (e.g. in Austenitic stainless steel) at low strains (~103) and/or at high strains (~1.0). Other forms of the power law equation are also considered in literature (e.g. ). An ideal plastic material (without strain hardening) would begin to neck right at the onset of yielding. At low temperatures (below recrystallization temperature- less than about 0.5Tm) strain hardening is very important to obtain good ductility. This can be understood from the analysis of the results of the uniaxial tension test. During tensile deformation instability in the form of necking localizes deformation to a small region (which now experiences a triaxial state of stress). In the presence of strain hardening the neck portion (which has been strained more) hardens and the deformation is spread to other regions, thus increasing the ductility obtained.

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

Variables in plastic deformationWhen true strain is less than 1, the smaller value of n dominates over a larger value of nK 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 ~ 0.15n and K for selected materials

MaterialnK (MPa)Annealed Cu0.54320Annealed Brass (70/30)0.49900Annealed 0.5% C steel0.265300.6% carbon steel Quenched and Tempered (540C)0.101570

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 strainAt 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. Thermal softening coefficient () is also defined as below:

Funda CheckWhat is the important of m and nWe 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 slipClick here to see overview of mechanisms/modes of plastic deformation* Leaving the crystal part is important** To be defined soon

Slip systemsIn 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).

CrystalSlip plane(s)Slip directionNumber of slip systemsFCC{111}12HCP(0001)3BCC Not close packed{110}, {112}, {123}[111]12

NaCl Ionic{110} {111} not a slip plane6C Diamond cubic{111}12

More examples of slip systems

CrystalSlip plane(s)Slip directionSlip systemsTiO2Rutile{101}CaF2, UO2, ThO2Fluorite{001}CsCl B2{110}

NaCl, LiF, MgO Rock Salt{110}6C, Ge, Si Diamond cubic{111}12MgAl2O4Spinel{111}Al2O3Hexagonal(0001)

More examples of slip systems

CrystalSlip plane(s)Slip direction (b)Slip systemsCu (FCC)Fm 3m{111}(a/2)< 1 10>4 x 3 = 12W (BCC)Im 3m{011} {112}{123}(a/2)6 x 2 = 1212 x 1 = 1224 x 1 = 24Mg (HCP)P63/mmc{0001}{10 10}{10 11}(a/3)1 x 3 = 33 x 1 = 36 x 1 = 6Al2O3R 3c{0001}{10 10}(a/3)1 x 3 = 33 x 1 = 3NaClFm 3m{110}{001}(a/2)< 1 10>6 x 1 = 66 x 1 = 6CsClPm 3m{110}a6 x 1 = 6PolyethylenePnam(100){110}(010)c1 x 1 = 12 x 1 = 21 x 1 = 1

As we saw plastic deformation by slip is due to shear stresses.Even if we apply an tensile force on the specimen the shear stress resolved onto the slip plane is responsible for slip.When the Resolved Shear Stress (RSS) reaches a critical value Critical Resolved Shear Stress (CRSS) plastic deformation starts (The actual Schmids law)Slip plane normalSlip directionAASchmid factorCritical Resolved Shear Stress (CRSS)Note externally only tensile forces are being appliedWhat is the connection between Peierls stress and CRSS?

Schmids lawSlip is initiated whenYield strength of a single crystalCRSS 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!)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).

Dislocation formed by pushing in a planeStep formed when dislocation leaves the crystalNow visualize dislocations being punched in on successive planes moving and finally leaving the crystal

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

Stress to move a dislocation: Peierls Nabarro (PN) stressWe 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 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 dislocationStrengthening mechanismsSolid solutionPrecipitate & DispersoidGrain boundary

Applied shear stress vs internal oppositionApplied 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 precipitatesNote: 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 atomsClassification of the obstacles to motion of a dislocationBased of if the obstacle can be overcome by thermal activation

Effect of TemperatureEquilibrium positions of a dislocationMotion 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.

FeW18-8 SSCuNiAl2O3Si1503004500.20.40.60.0Yield Stress (MPa) T/Tm Very high temperatures needed for thermal activation to have any effectRT is like HT and P-N stress is easily overcomeFe-BCCW-BCCCu-FCCNi-FCCMetallicIonicCovalent

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 thisThis 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 transformationFrank-Read sourceOrowan bowing mechanismIt 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) IrradiationSome specific sources/methods of formation/multiplication of dislocations include

Strain hardeningWe 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 hardeningOne 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

Formation of the Lomer-Cottrell barrier

+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 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 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 relationCOLD WORK strength ductility

(MN / m2) ( m/ m3)0 (MN / m2)A (N/m)1.510100.51010010140.510

Grain size and strength y Yield stress [N/m2] i Stress to move a dislocation in single crystal [N/m2] k Locking parameter [N/m3/2] (measure of the relative hardening contribution of grain boundaries) d Grain diameter [m]Hall-Petch Relation

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

Solute Concentration (Atom %) y (MPa) 50100150102030402000Solute strengthening of Cu crystal by solutes of different sizesMatrix: Cu (r = 1.28 )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 )Size effectSize differenceSize effect depends on:Concentration of the solute (c)For the same size difference the smaller atom gives a greater strengthening effect

yOften produce Yield Point PhenomenonSolute atoms level of - curve X By i (lattice friction)

InterstitialSolute atomsSubstitutional3GsoluteRelative strengthening effect / unit concentrationGsolute / 10Interstitial 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 interactionShort range (T sensitive)ModulusLong range orderElasticSubstitutional edgeInterstitial Edge and screw dl.ElectricalShort range orderStacking faultMechanisms of interaction of dislocations with solute atoms

The hardening effect of precipitatesGlide through the precipitateDislocationGet pinned by the precipitateIf the precipitate is coherent with the matrixPrecipitates 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 precipitateOnly 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.

bSchematic views edge dislocation glide through a coherent precipitate

Hardening effectPart of the dislocation line segment (inside the precipitate) could face a higher PN stressIncrease in surface area due to particle shearingWe 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 inclusionsDislocations 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 effectThe 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 shearingStacking-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 particleCoherency Strengthening: due to elastic coherency strains surrounding the particleOrder 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: 108 to Explosions: 105).Strain rate effects become significant (on properties like flow stress) only when strain rate is varied by orders of magnitude (i.e. small changes in flow stress do not affect the value of the properties much).Strain rate can be related to dislocation velocity by the equation below.Strain rate effects vd velocity of the dislocations d density of mobile/glissile dislocations b |b|

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

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