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Microstructure-‐Properties: I Materials Properties: Strength
27-‐301 Lecture 3
Profs. A. D. Rolle3, M. De Graef
Microstructure Properties
Processing Performance
Last modi9ied: 13th Sep. ‘15
2 Objective • The objec>ve of this lecture is to explain to you how the material property
strength is controlled by disloca>ons in duc>le materials. • Strength is defined and used to illustrate the rela>onship between materials
proper>es and microstructure. • More specifically, this lecture explains the Taylor Equa>on [Saada, Acta
metall. (1960)] that relates yield strength to disloca>on content of a material (and other obstacles to disloca>on flow):
σy = M α G b √ρ
• Movies of disloca>ons moving under stress (discrete disloca>on dynamics simula>ons): look at
• h3p://zig.onera.fr/DisGallery/ • h3p://virtualexplorer.com.au/special/meansvolume/contribs/jessell/labs/
lab1a.html
3 Notation L, l specimen length ε strain G (or µ) shear modulus b Burgers vector magnitude ρ disloca>on density ( m per m3 , or number per m2) r Par>cle size (radius) f ≡VV(α) volume frac>on (of precipitates) σ stress (macroscopic) τ shear stress (cri>cal value, in some cases) u displacement A area (cross sec>on of specimen) α geometrical constant (of order 0.5 ) φ angle between disloca>on and line perpendicular to the obstacle line <L3> mean intercept length (of precipitates) λ mean spacing (of obstacles, such as disloca>ons, precipitates) F force A area (cross sec>on of specimen) m Schmid factor (func>on of crystal orienta>on) M Taylor factor (func>on of orienta>on, texture) ∆2 nearest neighbor distance λ, φ angles between tensile axis and slip direc>on, slip plane normal, respec>vely
4 Key Concepts
• Stress, yield strength, typical values, extreme values
• Disloca>on loops between obstacles, obstacle spacings
• Cri>cal resolved shear stress, rela>onship to shear modulus, yield/modulus is o]en a small frac>on (σy ~ Y/1000) especially in pure metals
• Schmid factors, average Taylor factor for polycrystalline materials
Q&A 1 • Give examples of different ways to measure strength Tensile test, torsion test, Hopkinson bar test (dynamic), creep test • Why do disloca>ons control yield strength? Disloca>ons control yield strength because plas>c deforma>on depends on disloca>on slip, the resistance to which depends on obstacle spacing. • Explain the Orowan bowing-‐out model for strength. Disloca>ons have line tension (think rubber bands) and, under shear stress, stretch out between obstacles un>l they contact on the other side. The strength model is obtained as a force balance between the line tension and the reac>on force from the obstacles. • Explain the difference between “cri>cal resolved shear stress” (CRSS) and yield strength. The CRSS is the shear stress required to move disloca>ons 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 par>cles and disloca>ons 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 elas>c modulus? Which modulus? How does modulus relate to
mel>ng point? The line tension of disloca>ons depends on both shear modulus and the magnitude of the Burgers vector. The shear modulus, as with all moduli, scales with the mel>ng point. • What is the difference between “single slip” (Schmid factor) and “mul>ple slip” (Taylor factor)? The standard analysis of disloca>on analyzes the resolved shear stress on a single slip system, which is quan>fied with the Schmid factor. Deforming a polycrystal, however, means that each grain must deform like the overall polycrystal, which requires mul4ple slip systems to operate. This is quan>fied with the Taylor factor and is what leads to the von Mises criterion for duc>lity.
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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 ra>o of the resolved shear stress on the slip system to the applied tensile stress (stress boundary condi>ons). The Taylor factor, M, applies to deforma>on of a grain embedded in a polycrystal (strain boundary condi>ons) and is the ra>o of the magnitude of the applied stress to the cri>cal resolved shear stress. Generally M>(1/m). • Why is the square root of the disloca>on density important to strength? Disloca>ons represent obstacles to flow of other disloca>ons. Consider disloca>ons that thread through any given slip plane: the distance between neighbor pairs of disloca>ons is propor>onal to the square root of the number per unit area, which is the disloca>on density. This is quan>fied with the Taylor equa4on. • What gives rise to strain hardening? As disloca>ons move through crystals, they become entangled with each other (see the movies online) and deposit disloca>on line length which increases the disloca>on 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 effec>ve. • What is the Peierls stress and why does it explain that most ceramics are very strong but
bri3le? The Peierls stress is the intrinsic resistance to disloca>on mo>on (not discussed in this lecture), which is typically high in ceramics. Therefore the stress intensity at the >p of a crack cannot be relieved by disloca>on mo>on, which allows the crack to propagate under load.
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7 Strength • Strength is very basic to the value of a structural material. We
measure it in terms of force per unit area: σ = F/A • Strength means resistance to irreversible deforma>on or, if you
prefer, the upper limit of elas>c stress that is safe to apply to a material.
• Strength is highly dependent on microstructure because it is propor>onal to the difficulty of moving disloca>ons 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 o]en taught strength as being related to (chemical) composi>on. 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 duc<lity.
8 Comparisons
• SOFT: Lead piping (Roman!)
• HARD: Comparison of high strength (pearli>c) steels, used for bridges, tyre cord “Processing and mechanical behavior of hypereutectoid steel wires, D. Lesuer et al., Metallurgy, Processing and Applica>ons of Metal Wires, TMS, 1996.
http://www.time-‐travellers.org/Historian/Rome2001/romephotos.html
www.brantacan.co.uk/ suspension.htm
9 Types of Strength • Elsewhere in the course, we study stress and strength as tensor
quan>>es. Here, we will treat them as scalar quan>>es, i.e. a single number.
• There are different modes of loading materials: – Yield Strength: ambient condi>ons, low strain rate – Dynamic Strength: ambient condi>ons, high strain rate – Creep Strength: high temperature strength, low strain rate – Torsion Strength: strength in twis>ng – Fa>gue Strength: alterna>ng stresses
• The strength value is highly dependent on the loading mode. This indicates that strength has both complex and analy>cal a3ributes.
• Each type of strength is controlled by a variety of strengthening mechanisms.
10 Yield strength • A yield strength is a boundary between elas>c and
plas>c 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 disloca>on across its slip plane.
σ=0 σ elastic plastic Example: tensile stress
σ= σyield
11 Dislocation Motion • Disloca>ons control most aspects of strength and duc>lity in structural (crystalline) materials.
• Our objec>ve in reviewing the characteris>cs of disloca>ons is so that we can understand and control strengthening mechanisms.
• The strength of a material is controlled by the density of defects (disloca4ons, second phase par4cles, boundaries).
• For a polycrystal, for which we have to average the Taylor factors, we have the Taylor Equa>on:
σyield = <M> τcrss = <M> α G b √ρ
12 Dislocation glide • Recall the effect of disloca>on mo>on in a crystal: passage
causes one half of the crystal to be displaced rela>ve to the other. This is a shear displacement, giving rise to a shear strain.
[Dieter]
13 Dislocations & Yield
[Dieter]
• Straight lines are not a good approxima>on for the shape of disloca>ons, however: disloca>ons really move as expanding loops.
• The essen>al feature of yield strength is the density of obstacles that disloca>ons encounter as they move across the slip plane. Higher obstacle density ⇒ higher strength.
14 Why is there a yield stress?
[Dieter]
• One might think that disloca>on flow is something like elas>city: larger stresses imply longer distances for disloca>on mo>on. This is not the case: disloca>ons only move large distances once the stress rises above a threshold or cri<cal value (hence the term cri<cal resolved shear stress).
• Consider the expansion of a disloca>on loop under a shear stress between two pinning points (Frank-‐Read source).
To understand these shapes, try pulling a rubber band between two pencils
15 Orowan bowing stress
1
2
3
2r
λ
• If you consider the three consecu>ve posi>ons of the disloca>on loop, it is not hard to see that the shear stress required to support the line tension of the disloca>on is roughly equal for posi>ons 1 and 3, but higher for posi>on 2. Moreover, the largest shear stress required is at posi>on 2, because this has the smallest radius of curvature. A simple force balance (ignoring edge-‐screw differences) between the force on the disloca>on versus the line tension force on each obstacle then gives τmaxbλ = (Gb2/2), ⇔ τmax = Gb/λ where λ is the separa>on between the obstacles (strictly speaking one subtracts their diameter), b is the Burgers vector and G is the shear modulus (Gb2/2 is the approximate disloca>on line tension).
16 Orowan Bowing Stress, contd. • To see how the force balance applies,
consider the rela>onship between the shape of the disloca>on loop and the force on the disloca>on.
• Line tension = Gb2/2 Force resolved in the ver>cal direc>on = 2cosφ Gb2/2 Force exerted on the disloca>on per unit length (Peach-‐Koehler Eq.) = τb Force on disloca>on per obstacle (only the length perpendicular to the shear stress ma3ers) = λτb
• At each posi>on of the disloca>on, the forces balance, so τ = cosφ Gb2/λb
• The maximum force occurs when the angle φ = 0°, which is when the disloca>on is bowed out into a complete semicircle between the obstacle pair: τ = Gb/λ
τ
λ
Gb2/2 Gb2/2
φ
τ
λ
Gb2/2 Gb2/2
φ=0°
17 Critical resolved shear stress • It should now be apparent that disloca>ons 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 cri<cal resolved shear stress, then any disloca>on can move past all obstacles and will travel across the crystal or grain. This explains why plas>c deforma>on is highly non-‐linear.
• This analysis is correct for all types of obstacles, including second phase par>cles (precipitates) and disloca>ons (that intersect the slip plane). For weak obstacles, the shape of the cri>cal configura>on is not the semi-‐circle shown above (to be discussed later) -‐ the disloca>on does not bow out so far before it breaks through the obstacle. Since it is commonly the case that the cri>cal resolved shear stress is less than the Orowan stress, we have to include a factor, α<1, in the Taylor equa>on.
18 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 rela>vely minor difference to the cri<cal resolved shear stress, τcrss, which can be es>mated from a knowledge of the average obstacle spacing, λ, the Burgers vector magnitude, b, and the shear modulus, G, of the material, and a geometrical factor, α, that takes account of the flexibility of the disloca>ons and the finite strength of obstacles (i.e. that they do not have to bow out to the maximum stress semi-‐circular posi>on:
τcrss = αµb/λ.
-‐ What is this distance, λ? For disloca>ons that are flexible (or, the obstacles are strong), we need the nearest neighbor distance, ∆2. -‐ The geometrical factor, α, is generally taken to be 0.5. This is because disloca>ons break through obstacles, on average, at an angle, φ, less than 90°.
19 Stereology: Nearest Neighbor Distance • The 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, mul>plied by the point density , PA, mul>plied by the remaining frac>on of the cumula>ve probability.
• For strictly 1D objects such as disloca>ons, ∆2 may be used as the mean free distance between intersec>on points on a plane.
P(r)dr = 1 − P(r)dr0
r
∫[ ]PA2πrdrP(r)dr = 2πrPAe−πr
2PA
Δ 2 = rP(r)dr0
∞
∫Δ 2 =
12 PA
r
dr
Ref: Underwood, pp 84,85,185. Not examinable
20 Dislocations as obstacles
θ”
• Disloca>ons can be considered either as a set of randomly oriented lines within a crystal, or as a set of parallel, straight lines. The la3er is easier to work with whereas the former is more realis>c.
• Disloca>on density, ρ, is defined as either line length per unit volume, LV. It can also be defined by the areal density of intersec>ons of disloca>ons with a plane, PA.
• For randomly oriented disloca>ons, use standard stereology, along with the rela>onship between points and average free distance between points: λ ={LV/2})-‐1 ≡ (2√{ρ/2})-‐1. λ is the obstacle spacing in any plane.
• Straight, parallel disloca>ons: use ρ = LV = PA where PA applies to the plane perpendicular to the disloca<on lines only; ∆2=(PA)-‐1/2; thus λ = 1/√LV ≡ 1/√ρ where λ is the obstacle spacing in the plane orthogonal to the disloca<on lines only.
• Thus, we can write τcrss = α µ b √ρ [Courtney]
21 Single Crystal Deformation • To make the connec>on between disloca>on behavior and yield strength as measured in tension, consider the deforma>on of a single crystal.
• Given an orienta>on for single slip, i.e. the resolved shear stress reaches the cri>cal value on one system ahead of all others, then one obtains a “pack-‐of-‐cards” straining.
[Dieter]
22 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 tensile stress, take the component of the stress along the slip direc>on which is given by Fcosλ, and divide by the area over which the (shear) force is applied, A/cosφ. Note that the two angles are not complementary unless the slip direc>on, slip plane normal and tensile direc>on happen to be co-‐planar.
τ = F/A cosλ cosφ = σ cosλ cosφ = σ m ⇔ σ = τ / m
Schmid factor = m
[Dieter]
23 Critical Resolved Shear Stress • The experimental evidence of Schmid’s Law is that there is a
cri<cal resolved shear stress. This is verified by measuring the yield stress of single crystals as a func>on of orienta>on. The example below is for Mg which is hexagonal and slips most readily on the basal plane (all other τcrss are much larger).
“Soft orientation”, with slip plane at 45°to tensile axis
“Hard orientation”, with slip plane at ~90°to tensile axis
σ = τ/cosλcosφ
24 Polycrystal Deformation
[Dieter] Note varying orientations of slip planes
• Mul>ple 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 orienta>on. (a) undeformed (b) single slip, leading to gaps and overlaps (hypothe>cal) (c) crea>on of geometrically necessary disloca>ons (d) compa>ble deformed grains
25 Polycrystals: Taylor factor • Again, in polycrystals, each grain must deform in mul<ple slip, meaning that
several slip systems have to be ac>ve 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/cosλcosφ = 1/m. Just as for the Schmid factor, the Taylor factor depends on the orienta>on. The Taylor factors can be averaged over all the grains, which means that the average Taylor factor, <M>, is a func>on of texture.
• For a polycrystal, σyield = <M> τcrss = M α G b √ρ • Typical value of <Μ> = 3.1, i.e. the apparent hardness of the polycrystal is
approximately three >mes the cri>cal resolved shear stress. This ra>o is significantly higher than the average of the reciprocal Schmid factors for the same set of orienta>ons.
• [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 ac<vate slip on mul<ple systems, and the magnitude of these special stress states is larger than the applied (far field) stress.
26 Work Hardening Equations • Where does the stress-‐strain curve come from? Why does the flow stress
(cri>cal resolved shear stress) increase with strain? • As slip takes place in a crystal, even in cases where only one slip system appears
to be ac>ve (macroscopically), more than one system (or set of disloca>ons) is in fact ac>ve. Whenever two slip systems cross each other (intersect), the disloca>ons react with each other, leading to tangling. This tangling up of disloca>ons means that disloca>on line length is le] behind in the crystal, thus genera>ng more obstacles to disloca>on mo>on (and raising the cri>cal resolved shear stress).
• Work hardening is s>ll a very difficult theore>cal problem, so we rely on empirical descrip<ons such as the “power law” men>oned earlier:
σT = σyield + Kεn !
• Alterna>vely, the extended Voce hardening law (wri3en in terms of a single slip system) is as follows, where τ0, θ0, τ0+τ1 and Γ are the ini>al CRSS, the ini>al hardening rate, the satura>on CRSS, and the accumulated shear. We call these sorts of equa>ons “cons>tu>ve rela>ons”: !
27 Summary of Plastic Deformation • The following points are useful as a summary of important
features of plas>city from a material perspec>ve. • Single crystal behavior reflects the anisotropy of the crystal
for both elas>c (see lecture on elas>city) and plas>c behavior.
• Single crystal plas>c behavior is controlled by disloca>on movement; deforma>on twinning can supplement disloca>on glide, however, and is more common in lower symmetry crystals.
• The presence of disloca>ons that can glide at low (cri>cal resolved) shear stresses means that metals yield plas>cally at stresses far below the theore>cal strength.
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Plastic Deformation Summary, contd. • There is a cri>cal shear stress for disloca>on flow on any given slip system;
this phenomenon is known as Schmid's Law. The response is elas>c if all resolved shear stresses are less than the cri>cal value: τelastic< cosφ cosλ σapplied (or, τelastic< bσappliedn).
• Mechanical tests on single crystals generally ac>vate only one slip system and work hardening is low. The slip system that ac>vates is the one on which the resolved shear stress first reaches the cri<cal resolved shear stress. This means that the system with the largest Schmid factor is the ac>ve system.
• Larger strains in single crystal tests, or coincidence of the principal stress with a high symmetry axis leads to mul>ple 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 deforma>on assump>on discussed previously for polymers). By averaging the stresses (or strains) required for mul<ple 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.
29 Overall Summary
• The concept of strength as a material property has been explored.
• An illustra>on of the dependence of structural proper>es on microstructure has been given.
• The “Taylor Equa>on” that relates yield strength to disloca>on content of a material has been explained.
• Study the ques>ons and answers provided in addi>on to the notes themselves.
30 References • Materials Principles & Prac>ce, Bu3erworth Heinemann, Edited
by C. Newey & G. Weaver. • G.E. Dieter, Mechanical Metallurgy, McGrawHill, 3rd Ed. • D. Hull and D. J. Bacon (1984). Introduc>on to Disloca>ons,
Oxford, UK, Pergamon. • T. H. Courtney (2000). Mechanical Behavior of Materials,
Boston, McGraw-‐Hill. • P. Follansbee (2013). Fundamentals of Strength, TMS. • Scholarly ar>cle 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).
Supplement • Following slides discuss more general hardening mechanisms,
especially par>cle hardening. The development is based on Courtney’s textbook (2000).
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Hardening Mechanisms • More disloca>ons: work hardening accomplishes this (in duc>le materials). • Internal Boundaries: grain boundaries can have a strong strengthening effect. This is
classically known as the Hall-‐Petch effect. • Disloca>on Boundaries: at large strains and higher temperatures, low angle
boundaries appear as a subgrain network forms. We dis>nguish this microstructural feature from the first two categories because the [la�ce] misorienta>ons are much smaller (2-‐5°) than grain boundaries (15+°) and they are dis>nct from sta>s>cally stored disloca>ons.
• Second Phase Par>cles: whether introduced as insoluble par>cles in powder compac>on, or as precipitates in a solid state reac>on, second phase par>cles are generally the most potent strengthening agent in prac>cal high strength engineering materials. Iron-‐base, aluminum, nickel, >tanium alloys all employ second phases to achieve high strength.
• Solutes: solutes in a crystal act as obstacles to disloca>on mo>on through their elas>c and/or chemical interac>ons with disloca>ons. Most solutes are weak hardeners except for the (technologically) important class of inters>>al solutes that induce anisotropic distor>ons of the la�ce, e.g. tetragonal distor>ons of C in Fe.
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Particle Hardening • If the obstacles to disloca>on flow are primarily par>cles then we need to
adjust the rela>onship as follows. Two different types of par>cle strengthening are either par>cle cu�ng, typical of an underaged condi>on, or bowing around (impenetrable) par>cles, typical of an over-‐aged condi>on. Trea>ng the la3er first, we note that we need to determine the obstacle spacing on the slip plane, Λ. To get this we note (from Hull & Bacon) that for precipitates of volume Nb3 (N is the number of atoms per par>cle), then the average area on a sec>on plane (i.e. slip plane) is (π/6)1/3 (Nb3)2/3, then the volume frac>on of precipitates, c, must be equal to this (stereological standard result), giving us:
c = (π/6)1/3 (Nb3)2/3 / Λ2, i.e. L = b N1/3/√c
• The Orowan stress then follows from inser>ng this value into the basic looping stress equa>on:
τ = Gb/λ = Gb/(b N1/3/√c) = 0.84 G N-‐1/3 √c.
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Strengthening from Under-‐Aged Particles
• Returning to the subject of under-‐aged par>cles, which are penetrable, we need to be able to es>mate their strength. Again, there are a number of mechanisms which may control the stress required for penetra>on. A brief list is as follows.
1) Coherency Hardening: differences in density between the par>cle and the matrix give rise to elas>c stresses in the vicinity of the par>cle. 2) Chemical Hardening: crea>on of new surface when a par>cle is sheared increases the area of the interphase boundary, which increases the energy associated with the interface and hence an addi>onal force must be exerted on the disloca>on to force it through the par>cle. 3) Order Hardening: passage of a disloca>on through an ordered par>cle, e.g. Ni3Al in superalloys, results in a disordered la�ce and the crea>on of an>phase boundaries. 4) Stacking-‐fault Hardening: a difference in stacking fault energy between par>cle and matrix, e.g. Ag in Al, increases flow stress because of the different separa>on of par>al disloca>ons in the two phases. 5) Modulus Hardening: a large difference in elas>c modulus results in image forces when a disloca>on in the matrix approaches a par>cle. Consider, e.g., the difference between silver par>cles (nearly the same shear modulus) and iron par>cles (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 dis>nguish between them because there are only subtle differences between the mechanisms in terms of dependence on par>cle size etc.
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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 drama>c increase in flow stress. You must use second phase par>cles for that. The excep>on to this statement concerns inters>>al elements in bcc metals, for which the la�ce distor>on is non-‐spherical. There is not space to delve into the details, but the energy that is required to separate a disloca>on from an inters>>al solute atom is appreciable. These remarks can be quan>fied by going back to the first equa>on, i.e. the force balance between the forward mo>on and the resis>ng force. For subs>tu>onal solutes, the RHS, i.e. the reac>on force from the solute atoms is of order Gb2/120, which is a small number. This means that they are weak obstacles are disloca>ons remain nearly straight when interac>ng with them (fig. 5.2b). Inters>>als in bcc, however, can exert forces on the order of Gb2/5 to Gb2/10, which are large values. In this case, the disloca>ons bow out significantly between the atoms, and the breaking angle deviates significantly from 180°. In this case, the concentra>on dependence is easy to obtain. The spacing between inters>>als is inversely propor>onal to the (square root of the) concentra>on, and so we can insert a spacing into the standard formula to get the following, where g is a constant of order unity:
τ = γGB(√c/b) = γG√c. • The effec>ve spacing for subs>tu>onal solutes is much larger because of the weak interac>on: L ~ b/√(c{π-‐f}).
The angle π-‐φ ~ 1°, so L = 8b/√c. Combine that with the small force, and you get Eq. 5.22: τ = Gε3/2 √c / 700
• In this equa>on, ε is a parameter that is derived from both the frac>onal change in la�ce parameter with concentra>on (Eq. 5.14) and the frac>onal change in modulus (Eqs. 5.18, 5.17). However, a significant degree of empiricism is s>ll needed via Eq. 5.19 in order to make a correla>on with experimental data.
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