l3 strength 10sep07

1ObjectiveExampleStrengthHardnessDuctilityOrowanBowingSingleCrystalYieldPolyxtalYield

Microstructure-Properties: IMaterials Properties:

Strength, Ductility

27-301Lecture 3Fall, 2007

Profs. A. D. Rollett,M. de Graef

Microstructure Properties

Processing Performance


Objective

The objective of this lecture is to remind you of what amaterial property is.

Strength and ductility are defined and used to illustrate therelationship between materials properties and microstructure.

The measurement of a stress-strain curve is described. More specifically, this lecture explains the Taylor Equation

that relates yield strength to dislocation content of a material(and other obstacles to dislocation flow):

y = M G b

Look at www.steeluniversity.org, orhttp://www.steeluniversity.org/content/html/eng/default.asp?catid=1&pageid=1016899460,and specifically Tensile Test, Hardness Test, for self-learning guides


NotationL, l := specimen length := strainG (or ) := shear modulusb := Burgers vectorr := Particle size (radius)f VV() := volume fraction (of precipitates) := stress (macroscopic) := shear stress (critical value, in some cases)u := displacementA := area (cross section of specimen) := geometrical constant (~1) := angle between dislocation and line perpendicular to the obstacle line := mean intercept length (of precipitates) := mean spacing (of dislocations, precipitates)F := forceA := area (cross section of specimen)m := Schmid factorM := Taylor factor2 := nearest neighbor distance, := angles between tensile axis and slip direction, slip plane normal,

respectively


Key Concepts Stress, yield strength, typical values, extreme values Strain, engineering versus logarithmic strain Stress-strain curves Ductility, necking limit, relationship to hardening parameters,

Considres Criterion Dislocation loops, obstacle spacings Critical resolved shear stress, relationship to shear modulus Schmid factors, average Taylor factor for polyxtal


What is a Material Property? A Material Property is some quantifiable behavior of

a material. For a property to be a material property, it should

be a characteristic of the material, not theconfiguration in which it is used.

Example: the load carrying capacity of a beamdepends on the cross-section of the beam,therefore is not a material property.

The yield strength is a material property because itis the same no matter how the material is tested.


Properties & Microstructure Why are [some] properties dependent on microstructure? Many properties are controlled by the propagation of defects

within the material. The defect propagation is an example of a mechanism that

controls the property. Example: yield strength measures the resistance to plastic

flow, which is controlled by the mechanism of dislocationmotion. Dislocations are line defects whose motion is moresensitive to precipitates, grain boundaries etc. than to thelattice. The latter constitutes microstructure, as previouslydiscussed.


Issues, new ideas, so far The following new ideas or concepts have been

introduced.1. Strength2. Hardness3. Ductility4. Military non-diffusional transformations5. Martensite (a lower symmetry crystal structure,

formed as a result of a military transformation)6. The Fe-C phase diagram (not completely new)7. Diffusional transformations, decomposition8. Pearlite (a two-phase structure, formed as a result

of a diffusional transformation)9. Tempering

Properties

Processes


Strength Strength is very basic to the value of a structural material. We

measure it in terms of force per unit area: = F/A Strength means resistance to irreversible deformation or, if

you prefer, the upper limit of elastic stress that is safe to applyto a material.

Strength is highly dependent on microstructure because it isproportional to the difficulty of moving dislocations through(and between) the grains.

Typical values? Most useful structural metals have strengthsin the range 100-1000 MPa; ultra-high strength steel wire canbe produced up to 5,500 MPa!

Engineers are often taught strength as being related to(chemical) composition. Materials engineers studystrengthening mechanisms and therefore understand how tocontrol strength.

Strength is typically measured in a tension test, but we willalso examine this test when we discuss ductility.


Comparisons

SOFTSOFT: Lead piping(Roman!)

HARD: Comparison of highstrength (pearlitic) steels,used for bridges, tyre cord

Processing and mechanical behaviorof hypereutectoid steel wires, D.Lesuer et al., Metallurgy, Processingand Applications of Metal Wires, TMS,1996.

http://www.time-travellers.org/Historian/Rome2001/romephotos.html

www.www.brantacanbrantacan.co..co.ukuk/ suspension./ suspension.htmhtm

10

ObjectiveExampleStrengthHardnessDuctilityOrowanBowingSingleCrystalYieldPolyxtalYield

Types of Strength Later in the course, we will study stress and strength as tensor

quantities. For now, we will treat them as scalar quantities,i.e. a single number.

There are different modes of loading materials: Yield Strength: ambient conditions, low strain rate Dynamic Strength: ambient conditions, high strain rate Creep Strength: high temperature strength, low strain rate Torsion Strength: strength in twisting Fatigue Strength: alternating stresses

The strength value is highly dependent on the loading mode. Each type of strength is controlled by a variety of

strengthening mechanisms.

11


Yield strength A yield strength is boundary between elastic and

plastic flow.

=0 elastic plasticExample: tensile stress

= yield

12


Ductility Ductility measures the ability of a material to undergo plastic

deformation without fracture intervening. Ductility is the hallmark of structural materials because it

makes structures damage tolerant. If one element of astructure is overloaded, it will deform before it breaks and thusnot jeopardize the entire structure.

We cannot discuss ductility without first defining strain andthen examining stress-strain behavior.

Ranges of ductility: most oxides break (in tension) before theyyield plastically. Useful structural metals have at least 5%ductility. Superplastic materials (not just metals!) can exhibitenormous ductilities, >500%!

http://hightc.mtl.kyoto-u.ac.jp/english/laboratory/microstructure/microstructure.htm

High strain rate superplasticity of an Fe-Cr-Ni-Mo dual-phase stainless steel. Grain refinement of (+) duplexstructure up about 1m has established a largeelongation over 1000% even at high-strain rates in theorder of 0.1 s-1.

13


Strain Strain measures the change in shape of a body. If

you apply a force to a body, naturally there is achange in size. By normalizing the change in agiven dimension by the original dimension, onearrives at a quantity that again can be used tocharacterize the properties of a material. For now,we'll simply state that strain, properly described isalso a second rank tensor.

strain = [change in length]/[original length]

=L/L0=(L-L0)/L0.

Reminder: strain is a tensor because a body canchange shape in all three directions at once.

14


Strain - diagram

Courtney

15


Strain - notes A better definition of strain is that of a gradient in displacement

of points in a body. Take a tensile strain as an example: if wefix one end of the body and apply a tensile force, then thefixed point does not move. The point at the other end of thebody moves the most. The change in position, i.e. thedisplacement, is then proportional to the distance away fromthe fixed point. The strain can then be defined as the gradientin displacement, u; = du/dx, where x is the position along thebody.

In order to measure strain, one must choose points on aspecimen, measure their spacing, perform the test, and thenre-measure.

Since strain is always a ratio of lengths then it isdimensionless. Per-cent (%) is useful because manymaterials have ductilities less than 50%. Fractional strain isalso used, however.

16


Stress-Strain: measurement The measurement of the stress-strain characteristics of a

material, which we will perform in the second laboratory in301, requires us to examine some practical aspects.

At ambient conditions and easily attained strain rates (roomtemperature, one atmosphere of air, strain rates between 10-5and 100 per second), the most straightforward test is thetensile test. A bar of constant cross section [area] is stretchedat controlled displacement rate. The load required for thestretch is recorded.

Essentially all materials exhibit a maximum strain, beyondwhich failure (fracture) occurs.

Note that, although strength is a tensor quantity, one can onlymeasure in one direction at once. In many cases, it isreasonable to assume isotropy.

17


Stress-strain curves If one applies a large enough load to a ductile

material (of uniform cross-section) plasticdeformation will result in the following (typical)behavior.

Elastic

Plastic

Courtney

The elastic strain can besubtracted from the total strain inorder to produce a curve of stressversus plastic strain only, which isuseful for many problems. Alinear stress-(elastic) strainresponse is assumed; for eachdata point, the elastic straincorresponding to that stress(stressmodulus) is subtracted(translate to the left, parallel tothe strain axis).

This procedure can also be usedto correct for machinecompliance.

18


Compliance Correction Often, a tensile testing machine is

not perfectly stiff and the lack ofstiffness is evident in the test resultsas an apparent elastic modulus thatis lower than the expected value (i.e.what you find in a handbook). Thereason is that applying a load to thespecimen produces elasticdisplacements in the machine aswell as in the specimen.Displacement is measured at thecross-head and so additional,apparent strain occurs. This can becorrected for in a straightforwardmanner by measuring the differencein slope between the measured,Emeasuredl, and the known elasticmodulus, Ematerial. The permits amachine displacement to becomputed at any given load, and theresulting strain subtracted from themeasured strain value.

!

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= "measured

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= "measured

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19


Stress-strain characteristics The initial part of the curve represents the elastic regime of the

material. If the load is released, the strain of the specimen will returnto zero and no permanent deformation occurs. The slope of this partof the curve is called Young's modulus or Modulus of Elasticity.

Further imposed strain results in a drastic change in slope of thecurve which signals the onset of permanent plastic deformation. Theyield strength is a measure of the stress required for permanentplastic flow. The usual definition of this property is the offset yieldstrength determined by the stress corresponding to the intersectionof the curve and a line parallel to the elastic part but offset by aspecific strain (usually 0.2%). Beyond this point, the material workhardens until the ultimate tensile strength is attained. At this point,the incremental increases in stress due to decrease in cross-sectional area becomes greater than the increase in load carryingability due to strain hardening. Starting at this point, all further strainis concentrated in the "necked" portion of the specimen.

20


Ductility measures The reduction of area at fracture and the elongation to fracture

are used as percent reduction of the original area andpercentage increase of the original gage length. Thepercentage reduction of area at fracture is only slightlyaffected by the shape of the tensile test specimen. As long asthe ratio of the width to thickness does not exceed about 5:1,for a rectangular cross-section, the percent reduction of arearemains the same as for circular cross-sections.

Elongation to failure = f = (lfinal-l0)/l0 x 100% Reduction in Area = (Afinal-A0)/A0 x 100% The reduction of area is usually larger than the elongation to

fracture.

21


Derived Quantities The Gauge Length is the length between the shoulders of the

specimen Elongation to fracture is usually measured by fitting the broken

specimen back together and measuring the distance betweenpunch or scribe marks. Elongation may also be calculatedfrom the load-extension diagram; the two do not necessarilyagree. Elongation is so much affected by the gauge lengthover which it is measured that the gauge length must alwaysbe specified when reporting data.

Tensile Strength, or, Ultimate Tensile Strength (UTS) is themaximum stress that the material experiences during the test.

Work Hardening or Strain Hardening is the increase in stressduring the test.

22


Engineering Stress, Strain One important technical issue in tensile (and compression)

tests arises from the change in area. Load-displacement curves are all that can be measured in a

tensile test. Load must be divided by area to arrive at stress.Displacement must be divided by an initial length (such as agauge length) to arrive at a strain.

If the initial cross-sectional area, A0, is used to calculatestress, then this is known as nominal or engineering stress.

n = F/A0 Engineering stress -strain plots are useful because they show

the maximum load carrying capacity of the material by thechange in sign of the slope (peak in the curve at dn/d=0).

Similarly, use of linear strain based on the initial length isknown as nominal or engineering strain.

n = l / l0

23


True Stress, Logarithmic Strain If the stress is divided by the current area, A, then the current,

or true stress is obtained. The current area is easily obtainedfrom the length.

Constancy of volume: it is an experimental fact that thevolume change experienced in ductile flow is negligible. Thisis a result of plastic flow being accommodated by shear/slip.Therefore,

Al = A0l0 This permits us to write, = F / A = Fl / A0l0 = F(l0+l) / A0l0 = F(1+n) / A0 =n(1+n)

True, or logarithmic strain is defined as,

! =dl

ll0

l

" = lnl

l0

#

$ % %

&

' ( ( ! = ln 1+ !n( )

24


Why Ductility? Why do materials exhibit ductility? The reason for ductility in

metals is that they work harden. The key concept is the re-distribution of strain. That is to say,

if one sub-region of a material hardens as a result of theaccumulation of dislocations then its load carrying capacity ishigher than that of the neighboring regions. More specifically,the flow stress is lower in the non-hardened regions than in thehardened region. Therefore plastic flow is larger in the non-hardened region(s) and, in effect, the strain is redistributed to adifferent part of the specimen.

Many polymer systems also exhibit bulk ductility because the longchain molecules are present in folded form, either regularly arrangedas in the semi-crystalline polymers, or irregularly as in theamorphous polymers. This conformation of the long chain moleculesallows for considerable stretching during plastic deformation andoften to a few hundred percent.

25


Why is Ductility limited? Ductility is limited because the rate of work hardening

decreases with strain. Straining does not continue indefinitely. There are several ways

in which plastic deformation will cease; collectively, the variousphenomena are discussed as fracture. One limit to strainingcomes when the material exhausts its ability to redistribute strain.This exhaustion is dependent on the geometry of the test,however. For example, the tensile deformation results in asteady decrease in the cross section which sets up a competitionbetween strain hardening and geometric softening from theperspective of load carrying capacity of a given element ofmaterial. When the strain hardening no longer "keeps up with"the geometric softening then strain redistribution ceases and aneck will start to form.

26


Analysis of ductility Consider the load on the tensile specimen:

F = A

Differentiate:

dF = dA + dA

The criterion for instability is that the increase in load in anygiven element of material is less than or equal to zero, dF=0.The load increase is positive from work hardening (anddominates at first) but negative from the change in area. Notethat we must work with current values, i.e. the true stress.

27


Analysis, contd.dF = 0 dA + dA = 0 dT/T = -dA/A

= ln(A0/A) d = -dA/A

dT/T = d dT/d = T

In words, the hardening rate (of the true stress) is equal to the(true) stress at the point at which the material can no longersupport an increasing load. Beyond this point on the stress-strain curve, the deformation will tend to localize in a (diffuse)neck.

28


Considres Criterion

Dieter: fig. 8-8, p290.

Considre developed anelegant geometricalconstruction for determiningthe maximum load in atensile test. The true stressis plotted against theengineering strain. A straightline is drawn through thepoint A, (-1,0), and tangent tothe curve. The stress at thetangent point is the maximumstress/load.

If the stress-strain curve canbe described as a power-lawrelationship with exponent n, T = Kn ,then the engineering strain atthe maximum load, eu = n.

29


Ductility-Microstructure How does microstructure influence ductility? Provided that dislocations move easily through the

material and macroscopic instabilities (such asnecking) do not intervene, ductility can be very large.

Any microstructural element that leads to localcracking will tend to lower ductility by decreasing theload carrying capacity of the material.

Inclusions, second phase particles, grain boundaries,for example, are all potential fracture sites.

Qualitatively, cleaner, purer materials have higherductility.

30


Example Problem

[Courtney]

31


Summary (intermediate) Tensile strength and ductility have been explained. Standards methods of calculating these quantities

from the load-displacement curve from a tensiletesting machine have been described.

32


Dislocation Motion

Dislocations control most aspects of strength andductility in structural (crystalline) materials.

Our objective in reviewing the characteristics ofdislocations is so that we can understand and controlstrengthening mechanisms.

The strength of a material is controlled by thedensity of defects (dislocations, second phaseparticles, boundaries).

For a polycrystal:

yieldyield = = crsscrss = G b G b

33


Dislocation glide Recall the effect of dislocation motion in a crystal:

passage causes one half of the crystal to bedisplaced relative to the other. This is a sheardisplacement, giving rise to a shear strain.

[Dieter]

34


Dislocations & Yield

[Dieter]

Straight lines are not a good approximation for theshape of dislocations, however: dislocations reallymove as expanding loops.

The essential feature of yield strength is the densityof obstacles that dislocations encounter as theymove across the slip plane. Higher obstacledensity higher strength.

35


Why is there a yield stress? One might think that dislocation flow is something like

elasticity: larger stresses imply longer distances for dislocationmotion. This is not the case: dislocations only move largedistances once the stress rises above a threshold or criticalvalue (hence the term critical resolved shear stress).

Consider the expansion of a dislocation loop under a shearstress between two pinning points (Frank-Read source).

[Dieter]

36


Orowan bowing stress

1

2

3

2r

!

If you consider the three consecutive positions of the dislocation loop, it isnot hard to see that the shear stressrequired to support the line tension of the dislocation is roughly equal for positions 1 and 3, buthigher for position 2. Moreover, the largest shear stressrequired is at position 2, because this has the smallest radiusof curvature. A simple force balance (ignoring edge-screwdifferences) between the force on the dislocation versus theline tension force on each obstacle then gives maxb = (Gb2/2), max = Gb/where is the separation between the obstacles (strictlyspeaking one subtracts their diameter), b is the Burgers vectorand G is the shear modulus (Gb2/2 is the approximatedislocation line tension).

37


Orowan Bowing Stress, contd. To see how the force balance applies,

consider the relationship between theshape of the dislocation loop and theforce on the dislocation.

Line tension = Gb2/2Force resolved in the vertical direction= 2cos Gb2/2Force exerted on the dislocation perunit length (Peach-Koehler Eq.) = bForce on dislocation per obstacle (onlythe length perpendicular to the shearstress matters) = b

At each position of the dislocation, theforces balance, so = cos Gb2/b

The maximum force occurs when theangle = 0, which is when thedislocation is bowed out into acomplete semicircle between theobstacle pair: = Gb/

Gb2/2 Gb2/2

Gb2/2 Gb2/2

=0

MOVIES: http://www.gpm2.inpg.fr/axes/plast/MicroPlast/ddd/

38


Critical stress It should now be apparent that dislocations will only move

short distances if the stress on the crystal is less than theOrowan bowing stress. Once the stress rises above this valuethen any dislocation can move past all obstacles and willtravel across the crystal or grain. This explains why plasticdeformation is highly non-linear.

This analysis is correct for all types of obstacles, includingsecond phase particles (precipitates) and dislocations (thatintersect the slip plane). For weak obstacles, the shape of thecritical configuration is not the semi-circle shown above (to bediscussed later) - the dislocation does not bow out so farbefore it breaks through.

39


Arrays of Obstacles

Hull & Bacon

In reality, obstacles are not uniformly distributed and so there is aspectrum of obstacle strengths. Again, it turns out that this makes arelatively minor difference to the critical resolved shear stress, crss,which can be estimated from a knowledge of the average obstaclespacing, , the Burgers vector magnitude, b, and the shear modulus,G, of the material, and a geometrical factor, , that takes account ofthe flexibility of the dislocations (i.e. that they do not have to bow outto the maximum stress semi-circular position:

crss = Gb/.

- What is this distance, ?For dislocations that are flexible (or, theobstacles are strong), we need thenearest neighbor distance, 2.- The geometrical factor, , is generallytaken to be 0.5. This is a becausedislocations break through obstacles, onaverage, at an angle, , less than 90.

40


Stereology: Nearest Neighbor Distance The nearest neighbor distance

(in a plane), 2, can beobtained from the point densityin a plane, PA.

The probability density, P(r), isgiven by consideringsuccessive shells of radius, r:the density is the shell area,multiplied by the point density ,PA, multiplied by the remainingfraction of the cumulativeprobability.

For strictly 1D objects such asdislocations, 2 may be usedas the mean free distancebetween intersection points ona plane.

P(r)dr = 1 ! P(r)dr0

r

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Ref: Underwood, pp 84,85,185.Not examinable

41


Dislocations as obstacles Dislocations can be considered either as a set of

randomly oriented lines within a crystal, or as aset of parallel, straight lines. The latter is easierto work with whereas the former is more realistic.

Dislocation density, , is defined as either linelength per unit volume, LV. It can also be definedby the areal density of intersections ofdislocations with a plane, PA.

For randomly oriented dislocations, use standardstereology: = LV = 2PA; 2 = (2PA)-1/2;thus = (2{LV/2})-1 (2{/2})-1. is the obstacle spacing in any plane.

Straight, parallel dislocations: use = LV = PAwhere PA applies to the plane perpendicular to thedislocation lines only; 2=(PA)-1/2;thus = 1/LV 1/where is the obstacle spacing in the planeorthogonal to the dislocation lines only.

Thus, we can write crss = G b

42


Single Crystal Deformation To make the connection between dislocation

behavior and yield strength as measured in tension,consider the deformation of a single crystal.

Given an orientation for single slip, i.e. the resolvedshear stress reaches the critical value on onesystem ahead of all others, then one obtains apack-of-cards straining.

[Dieter]

43


Resolved Shear Stress Geometry of slip: how big an

applied stress is required for slip? To obtain the resolved shear

stress based on an applied tensilestress, take the component ofthe stress along the slip directionwhich is given by Fcos, and divideby the area over which the (shear)force is applied, A/cos. Note thatthe two angles are not complementary unless theslip direction, slip plane normal and tensile directionhappen to be co-planar.

= F/A cos cos = cos cos = m

Schmid factor = m

44


Critical Resolved Shear Stress The experimental evidence of Schmids Law is that there is a

critical resolved shear stress. This is verified by measuringthe yield stress of single crystals as a function of orientation.The example below is for Mg which is hexagonal and slipsmost readily on the basal plane (all other crss are muchlarger).

Soft orientation,with slip plane at45to tensile axis

Hard orientation,with slip plane at~90to tensile axis

= /coscos

45


Determining Schmid Factors This brief review of single slip in crystals raises the question of how to

determine the Schmid factor for an arbitrarily oriented crystal. Recall the general formula for how to resolve a general (tensor) stress onto a

slip plane: = b n = bi ij nj Then simplify this formula for the case where the stress is a tensile stress

parallel to a direction, A: = bA nA = cos cos It is best to use a spreadsheet (e.g. Excel), or a Math program such as Mathematica or

Maple, and make a list of all possible combinations of slip plane (111, -111, 1-11, -1-11) and slip direction (e.g. 111 is orthogonal to 110, -101 and 0-11), taking onlypositive versions of each (unit) vector. This will give you a table with 12 rows, onefor each slip system (3 X 4 = 12 combinations of plane and direction). Then calculatethe dot products of the tensile axis, A, with each combination of plane+direction inturn in order to obtain cos and cos respectively (2 more columns). Then calculatethe Schmid factor as cos*cos (1 more column). Finally, identify the row with thelargest absolute value of the Schmid factor in it (i.e. positive or negative).

You can expand the table to include the negatives of each slip direction in addition:this will give you 24 rows (e.g. 111 is orthogonal to 110, -101, 0-11, 1-10, 10-1 and01-1). If you use the 24 row version, you will find that you obtain a pair of positiveand negative Schmid factors for each pair of positive and negative slip directions.This positive/negative pairing corresponds to positive and negative directions of slip.

46


Polycrystal Deformation Consider how a polycrystal deforms with slips in

individual grains, each of which has a differentorientation.(a) undeformed(b) single slip, leading togaps and overlaps(hypothetical)(c) creation of geometricallynecessary dislocations(d) compatible deformedgrains

[Dieter]

Note varying orientations of slip planes

47


Polycrystals: Taylor factor In a later discussion, we will see in more detail how slip at the

single crystal level is related to (ductile) deformation of apolycrystal.

In polycrystals, each grain must deform in multiple slip,meaning that several slip systems have to be active at once inorder for an individual grain to change shape in the same wayas the bulk material.

Each grain has a Taylor factor, M, which is analogous to (butgenerally larger than) the reciprocal of the Schmid factor, 1/coscos = 1/m. The Taylor factors can be averaged over all thegrains.

For a polycrystal, yield = crss = M G b Typical value of = 3.1, i.e. the apparent hardness of the

polycrystal is approximately three times the critical resolvedshear stress.

48


Work Hardening Where does the stress-strain curve come from? Why does the

flow stress (critical resolved shear stress) increase with strain? As slip takes place in a crystal, even in cases where only one

slip system appears to be active (macroscopically), more thanone system (or set of dislocations) is in fact active. Whenevertwo slip systems cross each other (intersect), the dislocationsreact with each other, leading to tangling. This tangling up ofdislocations means that dislocation line length is left behind inthe crystal, thus generating more obstacles to dislocationmotion (and raising the critical resolved shear stress).

Work hardening is still a very difficult theoretical problem, sowe rely on empirical descriptions such as the power lawmentioned earlier: T = yield + Kn

49


Summary of Plastic Deformation The following points are useful as a summary of important

features of plasticity from a material perspective. Stress-strain curves provide a straightforward way to measure

yield stress, ultimate tensile stress and ductility. The maximum load and maximum uniform elongation are

predictable from the stress-strain curve (e.g. power law),which is known as Considres construction.

Single crystal behavior reflects the anisotropy of the crystal forboth elastic (see lecture on elasticity) and plastic behavior.

Single crystal plastic behavior is controlled by dislocationmovement; deformation twinning can supplement dislocationglide, however, and is more common in lower symmetrycrystals.

The presence of dislocations that can glide at low (criticalresolved) shear stresses means that metals yield plastically atstresses far below the theoretical strength.

50


Summary, contd. There is a critical shear stress for dislocation flow on any given slip

system; this phenomenon is known as Schmid's Law. The response iselastic if all resolved shear stresses are less than the critical value: elastic< coscos applied (or, elastic< bappliedn).

Mechanical tests on single crystals generally activate only one slipsystem and work hardening is low.

Larger strains in single crystal tests, or coincidence of the principalstress with a high symmetry axis leads to multiple slip (slip on morethan one system); in this case the stress-strain behavior is polycrystal-like.

A polycrystals can be thought of as a composite of single crystals.The appropriate model for this composite is the iso-strain model(equivalent to the affine deformation assumption discussed previouslyfor polymers). By averaging the stresses (or strains) required formultiple slip in each crystal, an average for the "inverse Schmidfactor", or (more usually) "Taylor factor", can be obtained whose valueis 3.07 for cubic materials deformed in tension or compression with{111} (or {110}) slip systems.

51


Summary The concept of material property has been

explored. An illustration of the dependence of structural

properties on microstructure has been given. Basic mechanical properties (strength, hardness,

ductility) have been defined and illustrated withrespect to practical methods for measurement.

The Taylor Equation that relates yield strength todislocation content of a material has beenexplained.

52


Sample Problem From Dieter, p219 (adapted): Question: Al-4%Cu (by wt.) has a yield stress of 600MPa. Estimate the

particle size and spacing. Solution: recognize that this stress relates to age hardening beyond the

peak hardness. Therefore use the Orowan bowing stress to estimatethe stress.

= crss = Gb/ G=27.6GPa; b=0.25nm; =3.1:

spacing = 3.1*27,600*0.25.10-9/ 600= 35.7 nm Now we must estimate the volume fraction of particles for which we use

the phase diagram, assuming that we are dealing with the equilibriumphase, , which is 54 w/o Cu, and the in equilibrium with it, 0.5 w/oCu.

Wt. % Al = (54-4)/(54-0.5) = 93.5; wt. % = 4-0.5/(54-0.5)=6.5 Volume of = 93.5gm/2.7 gm/cm3 =34.6 cm3 Volume of = 6.5/ 4.443 gm/cm3 = 1.5 cm3 Volume fraction of = 0.96; volume fraction of = 0.04. Use =4r(1-f)/3f (slide 22): r =3*0.04*35.7/4/(1-0.04) = 1.12 nm.

53


Mean Free Distance

If the dislocations are relatively inflexible and therefore straight (or theobstacles are weak) then we need the mean free distance, , betweenobstacles, instead of the nearest neighbor distance. This applies to anykind of obstacle (dislocations or particles).

The mean free distance, , between particle edges is given by the aboveequation. Note that it is closely related to the mean intercept distance,. Finite volume fraction, (Vv), or f, decreases the distance betweenparticle edges.

Alpha () represents the particle phase. Thus, for mono-disperse spheres we can write for the c.r.s.s.:

crss = Gb3f/2D. Compare with the nearest neighbor distance for particles:

crss = Gb2(6f/)/D. These two formulae differ only by a numerical factor, and the presence of

the volume fraction in a square root term.

Ref: Underwood, pp 80-85.

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54


References Materials Principles & Practice, Butterworth

Heinemann, Edited by C. Newey & G. Weaver. G.E. Dieter, Mechanical Metallurgy, McGrawHill,

3rd Ed. D. Hull and D. J. Bacon (1984). Introduction to

Dislocations, Oxford, UK, Pergamon. T. H. Courtney (2000). Mechanical Behavior of

Materials, Boston, McGraw-Hill.

55


Summary: 3 The variation of mechanical behavior with temperature and strain

rate depends on the kind of obstacle that dislocations have to movepast. In fcc metals, yield is dominated by other dislocations (the"forest hardening model") such that the strain rate/temperaturevariation is dominated by the (weak) variation in shear modulus(with temperature) through the "Taylor equation", =MGb.

In bcc metals, yield at low temperatures is dominated by latticefriction (i.e. the Peierls stress) and large strain rate/temperaturesensitivities are observed.

Most ceramics follow the bcc model because they too have highlattice frictions at low temperatures (but become plastic and ductileat elevated temperatures).

Single crystals are important because many high temperatureapplications require single crystal or coarse polycrystals in order tomaximize creep resistance, i.e. by minimizing grain boundary area.Microelectronic applications use single crystals of Si where theabsence of grain boundaries is not important unless MEMS devicesare being designed.

56


Summary: 4 The work hardening behavior of single crystals is summarized by four

stages: stage I is known as "easy glide"; stage II as "linear, athermalhardening"; stage III as "dynamic recovery"; and stage IV as "linearhardening".

For a polycrystal to exhibit ductility, it must be possible for every grainto deform plastically in an arbitrary manner. This is summarized as vonMises criterion which states that a minimum of five independent systemsare required for ductility. This can be understood most easily byconsidering that an arbitrary strain has five independent components:there is an equation (linear) that links the slip on an individual slipsystem (or twinning system) to the macroscopic shape change (i.e.strain); therefore five independent systems are needed in order to satisfythe five independent strain components.

57


Summary: 5 Dislocation flow in a polycrystal is quite heterogeneous. Dislocations get

entangled in one another as they expand over their slip planes. The majorconsequence of this is that any dislocation motion (over a distance larger than themean spacing) leaves behind a certain amount of dislocation; this is calleddislocation storage and hardens the crystal. By a combination of collapse oftangles and cross-slip (switching of slip planes by screw-configurationsegments), however, dislocations of opposite sign can meet and annihilate; this iscalled dynamic recovery (because it only happens during continuing straining)and decreases the hardening rate (i.e. the net storage rate of dislocationsdecreases because of dynamic recovery). Eventually dynamic recovery balancesstorage and the flow stress saturates, or nearly so.

At high temperatures, dynamic recovery occurs early on in straining and, withthe ease of non-conservative motion (climb), the work hardening becomesnegligible. With rapid dynamic (and static) recovery, the dislocation structurebecomes a sub-grain structure with well defined, low angle boundaries. If asingle crystal is bent, then the dislocations left behind after the deformation tendto re-arrange themselves into walls of edge dislocations of the same type andsign. Such a recovered or polygonized structure is a clear example ofgeometrically necessary dislocations.

l3 strength 10sep07

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