1 Microstructure-Properties: I - PowerPoint Presentation

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Microstructure-Properties: I Materials Properties: Strength

27-301Lecture 3Profs. A. D. Rollett, M. De Graef

Microstructure

Properties

Processing

Performance

Last modified: 13

th

Sep. ‘15Slide2

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ObjectiveThe objective of this lecture is to explain to

you how the material property strength is controlled by dislocations in ductile materials. Strength is defined and used to illustrate the relationship between materials properties and microstructure.

More specifically, this lecture explains the Taylor Equation

[Saada

, Acta metall. (1960)] that relates yield strength to dislocation content of a material (and other obstacles to dislocation flow): sy = M

a G b √rMovies of dislocations moving under stress (discrete dislocation dynamics simulations): look athttp://zig.onera.fr/DisGallery/

http://virtualexplorer.com.au/special/meansvolume/contribs/jessell/labs/lab1a.htmlSlide3

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Notation

L, l specimen lengthe strainG (or µ) shear modulus

b

Burgers vector magnitude

r dislocation density ( m per m3 , or number per m2)

r Particle size (radius)f V

V(a

) volume fraction (of precipitates)s

stress (macroscopic)t shear stress (

critical value, in some cases)u displacement

A area (cross section of specimen)a

geometrical constant (of order 0.5 )f

angle between dislocation and line perpendicular to the obstacle line<L3

> mean intercept length (of precipitates)

l mean spacing (of obstacles, such as dislocations, precipitates)F

forceA area (cross section of specimen)

m

Schmid

factor (

function of

crystal orientation)

M

Taylor

factor (function of orientation, texture)

∆

2

nearest neighbor distance

l, f

angles between tensile axis and slip direction, slip plane normal, respectivelySlide4

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Key ConceptsStress, yield strength, typical values, extreme values

Dislocation loops between obstacles, obstacle spacingsCritical resolved shear stress, relationship to shear modulus, yield/modulus is often a small fraction (sy ~ Y/1000) especially in pure metalsSchmid factors, average Taylor factor for polycrystalline materialsSlide5

Q&A 1

Give examples of different ways to measure strengthTensile test, torsion test, Hopkinson bar test (dynamic), creep test

Why do dislocations control yield strength?Dislocations control yield strength because plastic deformation depends on dislocation slip, the resistance to which depends on obstacle spacing.Explain the Orowan bowing-out model for strength.Dislocations have line tension (think rubber bands) and, under shear stress, stretch out between obstacles until they contact on the other side. The strength model is obtained as a force balance between the line tension and the reaction force from the obstacles.

Explain the difference between “critical resolved shear stress” (CRSS) and yield strength.

The CRSS is the shear stress required to move dislocations on their slip plane and there is a geometrical factor between this value and the macroscopic (or far field) stress that must be applied to the material.

Explain how particles and dislocations control the CRSS (via Orowan bowing-out).The answer to this is the same as for the Orowan bowing strength.

Why does strength scale with elastic modulus? Which modulus? How does modulus relate to melting point?The line tension of dislocations depends on both shear modulus and the magnitude of the Burgers vector. The shear modulus, as with all moduli, scales with the melting point.What is the difference between “single slip” (Schmid factor) and “multiple slip” (Taylor factor)?The standard analysis of dislocation analyzes the resolved shear stress on a

single slip system, which is quantified with the Schmid factor

. Deforming a polycrystal, however, means that each grain must deform like the overall polycrystal, which requires multiple slip systems to operate. This is quantified with the Taylor factor

and is what leads to the von Mises criterion for ductility.

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Q&A 2

Explain what the Schmid factor is; how does the Schmid factor differ from the Taylor factor?For a tensile test on a single crystal, the Schmid factor,

m, is the ratio of the resolved shear stress on the slip system to the applied tensile stress (stress boundary conditions). The Taylor factor, M, applies to deformation of a grain embedded in a polycrystal (strain boundary conditions) and is the ratio of the magnitude of the applied stress to the critical resolved shear stress. Generally

M

>(1/m

).Why is the square root of the dislocation density important to strength?Dislocations represent obstacles to flow of other dislocations. Consider dislocations that thread through any given slip plane: the distance between neighbor pairs of dislocations is proportional to the square root of the number per unit area, which is the dislocation density. This is quantified with the Taylor equation.What gives rise to strain hardening

? As dislocations move through crystals, they become entangled with each other (see the movies online) and deposit dislocation line length which increases the dislocation density, which in turn increases the flow stress.What kinds of microstructure give high strength in metallic alloys?Any high density of obstacles, but fine precipitates are most effective.What is the Peierls stress and why does it explain that most ceramics are very strong but brittle

?The Peierls stress is the intrinsic resistance to dislocation motion (not discussed in this lecture), which is typically high in ceramics. Therefore the stress intensity at the tip of a crack cannot be relieved by dislocation motion, which allows the crack to propagate under load.

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StrengthStrength is very basic to the value of a

structural material. We measure it in terms of force per unit area: s = F/AStrength means resistance to irreversible deformation or, if you prefer, the upper limit of elastic stress that is safe to apply to a material.Strength is highly dependent on microstructure because it is proportional to the difficulty of moving dislocations through (and between) the grains.

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

Engineers are often taught strength as being related to (chemical) composition. Materials engineers study

strengthening mechanisms and therefore understand how to control strength.Strength is typically measured in a tension test, but we will also examine this test when we discuss ductility.Slide8

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Comparisons

SOFT: Lead piping (Roman!)

HARD:

Comparison of high strength (pearlitic) steels, used for bridges,

tyre cord“Processing and mechanical behavior of hypereutectoid steel wires, D. Lesuer et al., Metallurgy, Processing and Applications of Metal Wires, TMS, 1996.

http://

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

www.brantacan.co.uk/ suspension.htmSlide9

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Types of StrengthElsewhere in the course, we study stress and strength as tensor quantities. Here, 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 rateDynamic Strength: ambient conditions, high strain rateCreep Strength: high temperature strength, low strain rateTorsion Strength: strength in twistingFatigue Strength: alternating stressesThe strength value is highly dependent on the loading mode. This indicates that strength has both complex and analytical attributes.

Each type of strength is controlled by a variety of

strengthening mechanisms.Slide10

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Yield strengthA yield strength is

a boundary between elastic and plastic flow. This means that it corresponds to a threshold of some kind and therefore it is a complex property. We now establish that it is directly related to how difficult it is to move a dislocation across its slip plane.

s

=0

s

elastic

plastic

Example: tensile stress

s

=

s

yieldSlide11

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Dislocation MotionDislocations control most aspects of strength and ductility in structural (crystalline) materials.

Our objective in reviewing the characteristics of dislocations is so that we can understand and control strengthening mechanisms.The strength of a material is controlled by the density of defects (dislocations, second phase particles, boundaries).For a polycrystal, for which we have to average the Taylor factors, we have the Taylor Equation:

s

yield = <M> t

crss = <M> a G b

√rSlide12

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Dislocation glideRecall the effect of dislocation motion in a crystal: passage causes one half of the crystal to be displaced relative to the other. This is a shear displacement, giving rise to a

shear strain.

[Dieter]Slide13

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Dislocations & Yield

[Dieter]Straight lines are not a good approximation for the shape of dislocations, however: dislocations really move as expanding loops

.

The essential feature of yield strength is the density of obstacles that dislocations encounter as they move across the slip plane. Higher obstacle density

 higher strength.Slide14

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Why is there a yield stress?

[Dieter]One might think that dislocation flow is something like elasticity: larger stresses imply longer distances for dislocation motion. This is

not the case: dislocations only move large distances once the stress rises above a threshold or critical

value (hence the term critical resolved shear stress

).Consider the expansion of a dislocation loop under a shear stress between two pinning points (Frank-Read source).To understand these shapes, try pulling a rubber band between two pencilsSlide15

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Orowan bowing stress

If you consider the three consecutive

positions of the dislocation loop, it is

not hard to see that the shear stress

required to support the line tension of

the dislocation is roughly equal for positions 1 and 3, but higher for position 2. Moreover, the largest shear stress required is at position 2, because this has the smallest radius of curvature. A simple force balance (ignoring edge-screw differences) between the force on the dislocation versus the line tension force on each obstacle then gives 

tmax

bl = (

Gb2/2),

 tmax

= Gb/

lwhere

l is the separation between the obstacles (strictly speaking one subtracts their diameter), b

is the Burgers vector and G is the shear modulus (Gb

2/2 is the approximate dislocation line tension).Slide16

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Orowan Bowing Stress, contd.To see how the force balance applies, consider the relationship between the shape of the dislocation loop and the force on the dislocation.

Line tension = Gb2/2Force resolved in the vertical direction = 2cosf Gb2

/2

Force exerted on the dislocation per unit length (Peach-Koehler Eq.) = t

bForce on dislocation per obstacle (only the length perpendicular to the shear stress matters) = ltbAt each position of the dislocation, the forces balance, so t = cos

f Gb2/lbThe maximum force occurs when the angle f = 0°, which is when the dislocation is bowed out into a complete semicircle between the obstacle pair:

t = Gb/

l

t

l

Gb

2

/2

Gb

2

/2

f

t

l

Gb

2

/2

Gb

2

/2

f=0°Slide17

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Critical resolved shear stressIt should now be apparent that dislocations will only move short distances if the stress on the crystal is less than the

Orowan bowing stress. Once the stress rises above this value, which we call the critical resolved shear stress, then any dislocation can move past all obstacles and will travel across the crystal or grain. This explains why plastic deformation is highly non-linear.This analysis is correct for all types of obstacles, including second phase particles (precipitates) and dislocations (that intersect the slip plane). For weak obstacles, the shape of the critical configuration is not the semi-circle shown above (to be discussed later) - the dislocation does not bow out so far before it breaks

through the obstacle. Since it is commonly the case that the critical resolved shear stress is less than the

Orowan stress, we have to include a factor,

a<1, in the Taylor equation.Slide18

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Arrays of Obstacles

[Hull & Bacon]

In reality, obstacles are not uniformly distributed and so there is a spectrum of obstacle strengths. Again, it turns out that this makes a relatively minor difference to the

critical resolved shear stress

, tcrss, which can be estimated from a knowledge of the average obstacle spacing, l, the Burgers vector magnitude, b, and the shear modulus, G, of the material, and a geometrical factor, a,

that takes account of the flexibility of the dislocations and the finite strength of obstacles (i.e. that they do not have to bow out to the maximum stress semi-circular position: tcrss = a

µb/l.

- What is this distance,

l?For dislocations that are flexible (or, the obstacles are strong), we need the nearest neighbor distance, ∆

2.- The geometrical factor, , is generally taken to be 0.5. This is because dislocations break through obstacles, on average, at an angle,

, less than 90°.Slide19

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Stereology: Nearest Neighbor DistanceThe nearest neighbor distance

(in a plane), ∆2, can be obtained from the point density in a plane, PA.The probability density, P(r), is given by considering successive shells of radius, r: the density is the shell area, multiplied by the point density ,

P

A, multiplied by the remaining fraction of the cumulative probability.For strictly 1D objects such as dislocations, ∆

2 may be used as the mean free distance between intersection points on a plane.

r

dr

Ref: Underwood,

pp

84,85,185.

Not examinableSlide20

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Dislocations as obstacles

q

”

Dislocations can be considered either as a set of randomly oriented lines within a crystal, or as a set of parallel, straight lines. The latter is easier to work with whereas the former is more realistic.

Dislocation density, r, is defined as either line length per unit volume, LV. It can also be defined by the areal density of intersections of dislocations with a plane, P

A.For randomly oriented dislocations, use standard stereology, along with the relationship between points and average free distance between points: l ={

LV/2})-1

 (

2√{r

/2})-1.

l is the obstacle spacing in any plane.

Straight, parallel dislocations: use r

= L

V = PA

where PA

applies to the plane perpendicular to the dislocation lines only; ∆

2

=(P

A

)

-1/2

;

thus

l

= 1/

√

L

V



1/

√

r

where

l

is the obstacle spacing in the plane orthogonal to the dislocation lines only

.

Thus, we can write

t

crss

=

a

µ

b

√

r

[Courtney]Slide21

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Single Crystal DeformationTo 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 resolved shear stress reaches the critical value on one system ahead of all others, then one obtains a “pack-of-cards” straining.

[Dieter]Slide22

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Resolved Shear StressGeometry of slip: how big an applied stress is required for slip?

To obtain the resolved shearstress based on an applied tensilestress, take the component ofthe stress along the slip directionwhich is given by Fcosl

, and divide by the area over which the (shear)

force is applied, A/

cosf. Note that the two angles are not complementary unless the slip direction, slip plane normal and tensile direction happen to be co-planar. t = F/A

coslcosf = s coslcosf =

sm⇔

s = t/

m

Schmid factor = m

[Dieter]Slide23

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Critical Resolved Shear StressThe experimental evidence of Schmid’s Law is that there is a

critical resolved shear stress. This is verified by measuring the yield stress of single crystals as a function of orientation. The example below is for Mg which is hexagonal and slips most readily on the basal plane (all other tcrss are much larger).

“Soft orientation”,

with slip plane at

45°to tensile axis

“Hard orientation”,

with slip plane at

~90°to tensile axis

s =

t/

coslcosfSlide24

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Polycrystal Deformation

[Dieter]Note varying orientations of slip planes

Multiple slip is needed in each grain in order to have the grains in the polycrystal hang together.

Consider

how a polycrystal deforms with slips in individual grains, each of which has a different orientation.(a) undeformed(b) single slip, leading to

gaps and overlaps (hypothetical)(c) creation of geometricallynecessary dislocations(d) compatible deformedgrainsSlide25

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Polycrystals: Taylor factorAgain, in polycrystals, each grain must deform in

multiple slip, meaning that several slip systems have to be active at once in order for an individual grain to change shape in the same way as the bulk material.Each grain has a Taylor factor, M, which is analogous to (but generally larger than) the reciprocal of the Schmid factor, 1/cos

lcos

f =

1/m. Just as for the Schmid factor, the Taylor factor depends on the orientation. The Taylor factors can be averaged over all the grains, which means that the average Taylor factor, <M>, is a function of texture.For a polycrystal, syield =

<M> tcrss = M a G b √

rTypical value of , i.e. the apparent hardness of the polycrystal is approximately three times the critical resolved shear stress. This ratio is significantly higher than the average of the reciprocal Schmid factors for the same set of orientations.

[not examinable] The reason that the Taylor factor (~ 3) is larger than the corresponding reciprocal Schmid factor (which has a maximum of 2) is because the stress in each grain has to take very specific tensor values in order to activate slip on multiple systems, and the magnitude of these special stress states is larger than the applied (far field) stress.Slide26

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Work Hardening Equations

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 than one system (or set of dislocations) is in fact active. Whenever two slip systems cross each other (intersect), the dislocations react with each other, leading to tangling. This tangling up of dislocations means that dislocation line length is left behind in the crystal, thus generating more obstacles to dislocation motion (and raising the critical resolved shear stress).Work hardening is still a very difficult theoretical problem, so we rely on empirical descriptions

such as the “power law” mentioned earlier:

sT

= syield + Ken

Alternatively, the extended Voce hardening law (written in terms of a single slip system) is as follows, where t0, q0, t0

+t1 and  are the initial CRSS, the initial hardening rate, the saturation CRSS, and the accumulated shear. We call these sorts of equations “constitutive relations”: Slide27

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Summary of Plastic DeformationThe following points are useful as a summary of important features of plasticity from a material perspective.

Single crystal behavior reflects the anisotropy of the crystal for both elastic (see lecture on elasticity) and plastic behavior.Single crystal plastic behavior is controlled by dislocation movement; deformation twinning can supplement dislocation glide, however, and is more common in lower symmetry crystals.The presence of dislocations that can glide at low (critical resolved) shear stresses means that metals yield plastically at stresses far below the theoretical strength. Slide28

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Plastic Deformation 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 is elastic if all resolved shear stresses are less than the critical value:

telastic

<

cosf coslsapplied (or,

telastic< bsappliedn).

Mechanical tests on single crystals generally activate only one slip system and work hardening is low. The slip system that activates is the one on which the resolved shear stress first reaches the

critical resolved shear stress. This means that the system with the largest Schmid factor

is the active system.Larger strains in single crystal tests, or coincidence of the principal stress with a high symmetry axis leads to multiple slip (slip on more than 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 previously for polymers). By averaging the stresses (or strains) required for

multiple slip in each crystal, an average for the Taylor factor,

can be obtained whose value is 3.07 for cubic materials deformed in tension or compression with {111}<110> (or {110}<111>) slip systems.Slide29

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Overall SummaryThe concept of

strength as a material property has been explored.An illustration of the dependence of structural properties on microstructure has been given.The “Taylor Equation” that relates yield strength to dislocation content of a material has been explained.Study the questions and answers provided in addition to the notes themselves.Slide30

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ReferencesMaterials 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

.P. Follansbee (2013). Fundamentals of Strength

, TMS.Scholarly article about

work hardening: Kocks, U. F. & Mecking, H., 'Physics and phenomenology of strain hardening: the FCC case', Progress in Materials Science, 48, 171-273 (2003).Slide31

Supplement

Following slides discuss more general hardening mechanisms, especially particle hardening. The development is based on Courtney’s textbook (2000).31Slide32

Hardening Mechanisms

More dislocations: work hardening accomplishes this (in ductile materials).Internal Boundaries: grain boundaries can have a strong strengthening effect. This is classically known as the Hall-Petch effect.Dislocation Boundaries: at large strains and higher temperatures, low angle boundaries appear as a subgrain

network forms. We distinguish this microstructural feature from the first two categories because the [lattice] misorientations are much smaller (2-5°) than grain boundaries (15+°) and they are distinct from statistically stored dislocations.Second Phase Particles: whether introduced as insoluble particles in powder compaction, or as precipitates in a solid state reaction, second phase particles are generally the most potent strengthening agent in practical high strength engineering materials. Iron-base, aluminum, nickel, titanium alloys all employ second phases to achieve high strength.Solutes: solutes in a crystal act as obstacles to dislocation motion through their elastic and/or chemical interactions with dislocations. Most solutes are weak hardeners except for the (technologically) important class of interstitial solutes that induce anisotropic distortions of the lattice, e.g. tetragonal distortions of C in Fe

.

32Slide33

Particle Hardening

If the obstacles to dislocation flow are primarily particles then we need to adjust the relationship as follows. Two different types of particle strengthening are either particle cutting, typical of an underaged condition, or bowing around (impenetrable) particles, typical of an over-aged condition. Treating the latter first, we note that we need to determine the obstacle spacing on the slip plane, . To get this we note (from Hull & Bacon) that for

precipitates of volume Nb3 (N is the number of atoms per particle), then the average area on a section plane (i.e. slip plane) is (π/6)1/3 (Nb3)2/3, then the volume fraction of precipitates, c, must be equal to this (stereological standard result), giving us

: c = (π/6)

1/3 (Nb3)

2/3 / 2, i.e.L = b N1/3/√cThe Orowan stress then follows from inserting this value into the basic looping stress equation:t = Gb/

l = Gb/(b N1/3/√c) = 0.84 G N-1/3 √c.33Slide34

Strengthening from Under-Aged Particles

Returning to the subject of under-aged particles, which are penetrable, we need to be able to estimate their strength. Again, there are a number of mechanisms which may control the stress required for penetration. A brief list is as follows.1) Coherency Hardening: differences in density between the particle and the matrix give rise to elastic stresses in the vicinity of the particle.

2) Chemical Hardening: creation of new surface when a particle is sheared increases the area of the interphase boundary, which increases the energy associated with the interface and hence an additional force must be exerted on the dislocation to force it through the particle.3) Order Hardening: passage of a dislocation through an ordered particle, e.g. Ni3Al in superalloys, results in a disordered lattice and the creation of antiphase boundaries. 4) Stacking-fault Hardening: a difference in stacking fault energy between particle and matrix, e.g. Ag in Al, increases flow stress because of the different separation of partial dislocations in the two phases

.

5) Modulus Hardening: a large difference in elastic modulus results in image forces when a dislocation in the matrix approaches a particle. Consider, e.g., the difference between silver particles (nearly the same shear modulus) and iron particles (much higher shear modulus) in aluminum

.Unfortunately, in real materials, there may be more than one mechanism that could be operable and it can be hard to distinguish between them because there are only subtle differences between the mechanisms in terms of dependence on particle size etc.34Slide35

Solute Hardening

Most solutes have only a rather weak effect on strength. In other words, even if you put several per cent of a soluble atom into another element, you will not see a dramatic increase in flow stress. You must use second phase particles for that. The exception to this statement concerns interstitial elements in bcc metals, for which the lattice distortion is non-spherical. There is not space to delve into the details, but the energy that is required to separate a dislocation from an interstitial solute atom is appreciable. These remarks can be quantified by going back to the first equation, i.e. the force balance between the forward motion and the resisting force. For

substitutional solutes, the RHS, i.e. the reaction force from the solute atoms is of order Gb2/120, which is a small number. This means that they are weak obstacles are dislocations remain nearly straight when interacting with them (fig. 5.2b). Interstitials in bcc, however, can exert forces on the order of Gb2/5 to Gb2/10, which are large values. In this case, the dislocations bow out significantly between the atoms, and the breaking angle deviates significantly from 180°. In this case, the concentration dependence is easy to obtain. The spacing between interstitials is inversely proportional to the (square root of the) concentration, and so we can insert a spacing into the standard formula to get the following, where g is a constant of order unity

:

t

= gGB(√c/b) = gG√c. The effective spacing for substitutional solutes is much larger because of the weak interaction: L ~ b/√(c{π-f}). The angle π-f ~ 1°, so L = 8b/√c. Combine that with the small force, and you get Eq. 5.22

:= Ge3/2 √c / 700In this equation, e is a parameter that is derived from both the fractional change in lattice parameter with concentration (Eq. 5.14) and the fractional change in modulus (Eqs. 5.18, 5.17). However, a significant degree of empiricism is still needed via Eq. 5.19 in order to make a correlation with experimental data.

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