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Eindhoven University of TechnologyDepartment of Mechanical Engineering

Supervisors:Dr. ir. R.H.J. PeerlingsDr.-Ing. C.B. Hirschberger

A discrete analysis ofdislocation pile-up

T.W.J. de Geus

August 2009MT 09.19

Contents

Introduction 1

1 Problem description and assumptions 51.1 Problem description . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Basic assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Governing equations 92.1 Dislocation motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Stress on a dislocation . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Single slip plane . . . . . . . . . . . . . . . . . . . . . . . . . 112.2.2 Infinite vertical walls of dislocations on equidistant slip planes 13

2.3 Model summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Numerical implementation and verification 193.1 Numerical implementation . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Dislocation density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.3 Basic validation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.3.1 No dislocation interaction . . . . . . . . . . . . . . . . . . . . 233.3.2 Pile-up of one mobile dislocation . . . . . . . . . . . . . . . . 24

4 Results for a single slip plane 274.1 Review of the literature . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 Parameter values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.3 Results of the discrete numerical model . . . . . . . . . . . . . . . . 28

5 Results for infinite vertical walls 335.1 Limit case: large vertical spacing . . . . . . . . . . . . . . . . . . . . 335.2 Results for different vertical spacing . . . . . . . . . . . . . . . . . . 335.3 Review of the literature . . . . . . . . . . . . . . . . . . . . . . . . . 38

Conclusion 43

I

Appendix 47

A Secant method 47A.1 Numerical iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47A.2 Numerical iteration applied . . . . . . . . . . . . . . . . . . . . . . . 49

B Validation 51

C Additional results 53C.1 Complete discrete analysis . . . . . . . . . . . . . . . . . . . . . . . . 53C.2 Only nearest neighbor interaction . . . . . . . . . . . . . . . . . . . . 57C.3 Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Bibliography 61

II

List of symbols

Symbol Description Unitν Poisson-ratio [-]σ Stress [Pa]τ Applied shear stress [Pa]b Burgers vector [m]B Viscous drag coefficient [N·s/m]G Shear modulus [Pa]Ḡ Material parameter

(= G2π(1−ν)

)[Pa]

h Vertical spacing between slip planes [m]n Number of slip planes [-]x Horizontal position [m]y Vertical position [m]dt Time step [s]Estop Critical “kinetic energy”

(= v2

)[m2/s2]

III

Introduction

When metallic materials are loaded beyond their elastic limit, it can be observedthat after unloading some deformation is remaining. This so-called plastic behaviorcan be understood by looking at the underlying atomic structure of metals.

Metals consist of a crystalline microstructure with a regular arrangement of atomswithin each crystallite. Whenever a certain threshold load on this atomic lattice isexceeded, atomic bonds may be broken and defects may start to propagate throughthe crystal. Moving line defects – that are denoted as dislocations – are directlycorrelated to what is macroscopically observed as plasticity.

For a more detailed introduction on dislocation the reader is referred to for in-stance Callister (2003). One kind of dislocations, which we will restrict our focusto, are so called edge dislocations. They are characterized by an extra half plane ofatoms that is inserted in the atomic lattice of the metal, as is illustrated in figure 1.Characteristic for dislocations is that they have a stress field surrounding them,because they cause an incompatibility in the lattice.Chapter 4 / Imperfections in Solids

Burgers ye • Ficuni: 4.3 The atom positions around anedge dislocation; extra half-plane of atoms

~ shown in perspective. (Adapted from A. G.

• Guy, Essentials of Materials Science,~ McGraw-Hill Book Company, New York,

•~ ~jIll~IiII 1976, p. 153.)Edge ~

dislocation

~L1*I~~JIi!’

there is some localized lattice distortion. The atoms above the dislocation line inFigure 4.3 are squeezed together, and those below are pulled apart; this is reflectedin the slight curvature for the vertical planes of atoms as they bend around this extra half-plane. The magnitude of this distortion decreases with distance away fromthe dislocation line; at positions far removed, the crystal lattice is virtually perfect.Sometimes the edge dislocation in Figure 4.3 is represented by the symbol , whichalso indicates the position of the dislocation line. An edge dislocation may also be

Interactive MSE .~ Modules formed by an extra half-plane of atoms that is included in the bottom portion of~ Dislocations the crystal; its designation is a

Another type of dislocation, called a screw dislocation, exists, which may be

_____ thought of as being formed by a shear stress that is applied to produce thedistortion shown in Figure 4.4a: the upper front region of the crystal is shifted~ Dislocation-Screwone atomic distance to the right relative to the bottom portion. The atomicdistortion associated with a screw dislocation is also linear and along a dislocation line, line AB in Figure 4.4b. The screw dislocation derives its name from thespiral or helical path or ramp that is traced around the dislocation line by the

~ISE atomic planes of atoms. Sometimes the symbol C is used to designate a screw~ Dislocations dislocation.

Most dislocations found in crystalline materials are probably neither pure edgenor pure screw, but exhibit components of both types; these are termed mixed dislocations. All three dislocation types are represented schematically in Figure 4.5~

~ Dislocation-Mixedthe lattice distortion that is produced away from the two faces is mixed, having varying degrees of screw and edge character.

The magnitude and direction of the lattice distortion associated with a dislocation is expressed in terms of a Burgers vector, denoted by a b. Burgers vectorsare indicated in Figures 4.3 and 4.4 for edge and screw dislocations, respectively.Furthermore, the nature of a dislocation (i.e., edge, screw, or mixed) is defined bythe relative orientations of dislocation line and Burgers vector. For an edge, theyare perpendicular (Figure 4.3), whereas for a screw, they are parallel (Figure 4.4);they are neither perpendicular nor parallel for a mixed dislocation. Also, eventhough a dislocation changes direction and nature within a crystal (e.g., from edgeto mixed to screw), the Burgers vector will be the same at all points along its line.For example, all positions of the curved dislocation in Figure 4.5 will have theBurgers vector shown. For metallic materials, the Burgers vector for a dislocationwill point in a close-packed crystallographic direction and will be of magnitudeequal to the interatomic spacing.

Figure 1: Edge dislocation (taken from Callister (2003))

1

164 Chapter 7 / Dislocations and Strengthening Mechanisms

Inleruetive NISE~ Modules~ Dislocations

~ Motion-Edge

which also defines the dislocation line (Figure 4.3). A screw dislocation may bethought of as resulting from shear distortion; its dislocation line passes through thecenter of a spiral, atomic plane ramp (Figure 4.4). Many dislocations in crystallinematerials have both edge and screw components; these are mixed dislocations(Figure 4.5).

Plastic deformation corresponds to the motion of large numbers of dislocations. An edge dislocation moves in response to a shear stress applied in adirection perpendicular to its line; the mechanics of dislocation motion are represented in Figure 7.1. Let the initial extra half-plane of atoms be plane A. Whenthe shear stress is applied as indicated (Figure 7.la), plane A is forced to theright; this in turn pushes the top halves of planes B, C, D, and so on, in the samedirection. If the applied shear stress is of sufficient magnitude, the interatomicbonds of plane B are severed along the shear plane, and the upper half of planeB becomes the extra half-plane as plane A links up with the bottom half of planeB (Figure 7.lb). This process is subsequently repeated for the other planes, suchthat the extra half-plane, by discrete steps, moves from left to right by successiveand repeated breaking of bonds and shifting by interatomic distances of upperhalf-planes. Before and after the movement of a dislocation through some particular region of the crystal, the atomic arrangement is ordered and perfect; itis only during the passage of the extra half-plane that the lattice structure isdisrupted. Ultimately this extra half-plane may emerge from the right surface ofthe crystal, forming an edge that is one atomic distance wide; this is shown inFigure 7.lc.

The process by which plastic deformation is produced by dislocation motionis termed slip; the crystallographic plane along which the dislocation line traverses is the slip plane, as indicated in Figure 7.1. Macroscopic plastic deformationsimply corresponds to permanent deformation that results from the movement

S.. tI!IltiIPI4~~S~• eeaeeu~~~ 0—\\~flSøtS1IJI

Slip plane ~. \~4lf//S~u~,IIIiiIIIEdge

dislocation [~SHShIiII!4jIIIne - -

Shearstress

A B C •D

Shearstress

Shearstress

A B C D A B C D

It flLH!~iIUnit step

of slip

(I,)

Atomic rearrangements that accompany the motion of an edgedislocation as it moves in response to an applied shear stress. (a) The extrahalf-plane of atoms is labeled A. (b) The dislocation moves one atomic distanceto the right as A links up to the lower portion of plane B; in the process, theupper portion of B becomes the extra half-plane. (c) A step forms on thesurface of the crystal as the extra half-plane exits. (Adapted from A. G. Guy,Essentials of Materials Science, McGraw-Hill Book Company, New York,1976, p. 153.)

Figure 2: Slip of an edge dislocation under influence of shear stress. This illustrationis copied from Callister (2003)

When a certain shear stress is applied, dislocations will to glide – i.e. plasticallyslip – along their slip planes, as illustrated in figure 2. Most metals posess a poly-crystalline microstructure, which is composed of grains with an atomic lattice ori-entation which differs from grain to grain. The resulting discontinuity of slip planesat the boundaries between grains acts as a barrier for the motion of dislocations.Besides the grain boundaries, different phases and impurities may also obstruct theglide of dislocations through the lattice. This is one of the processes that underlystrengthening of the metal.

The most rigorous influence of these heterogeneities would be exerted by a hardobstacle which allows no dislocation to penetrate. Under influence of an externallyapplied shear stress – causing equally oriented dislocations to move towards thisbarrier – these dislocations will pile-up against the barrier. The pile-up has a finitesize because dislocations of the same orientation – or rather sign – repel each otherwith an interaction force which increases asymptotically as the distance betweenthem approaches zero, and therefore never coalesce. Dislocation pile-up has beenthe subject of much research, because under its influence cracks can initiate, yieldingmay take place and size effects can occur.

The influences of heterogeneities at the atomic level is not taken into account inconventional continuum or crystal plasticity modeling. In fact, for large specimensthey are negligible, but for very small specimens – as they appear in for examplemicro-manufactured electronic devices – they are not. Thus, attempts have beenmade to account for dislocation-induced size effects in crystal plasticity by for ex-ample Evers et al. (2004) and Yefimov et al. (2004). These approaches take intoaccount the dislocations on individual slip systems by means of dislocation densityfields. The interaction between them is captured by means of a back stress term

2

that is derived from the stress fields around the individual dislocations. However,these back stress terms are based on certain simplifying assumptions that do notalways appear to accurately capture the underlying physics.

In this work the interaction of discrete dislocations is studied considering ideal-ized pile-up situations. Using a numerical algorithm the equilibrium state of thepile-up is determined. The resulting dislocation density fields are evaluated for var-ious geometries. These density fields can be used as reference solutions for crystalplasticity models. A comparison is made to observations reported in the literature.

As to the structure of this report, first the problem considered will be demarcatedin chapter 1; here also the assumptions made are discussed. In the next chapterthe physics involved with dislocations are recapitulated. In chapter 3 the numericalalgorithm is discussed, together with some simple cases used for validation. In thefollowing 2 chapters the results are presented for the two problems defined in chaptertwo. The report closes with a conclusion.

3

Chapter 1

Problem description andassumptions

In this work the pile-up of dislocations against an impenetrable obstacle is studied.A constant shear stress field is applied to a number of dislocations, such that theymove towards the obstacle until they reach an equilibrium state.

This chapter covers the assumptions made to demarcate a problem that is simpleenough to model. In the first section, the problem that is subject of this research isdescribed. In the second section, the basic assumptions are stated.

1.1 Problem description

The problem that is being investigated throughout this report is the problem offinding the equilibrium positions in a pile-up of dislocations. In this equilibriumstate an applied shear stress is in equilibrium with the mutual repelling forces ofdislocations. Two different configurations – discussed below – are investigated usinga numerical algorithm.

1.1.1 Configurations

The two configurations that are investigated throughout this study are as follows. Inthe first configuration, the pile-up of dislocations on a single slip plane is considered– as sketched in in figure 1.1. In the second configuration, an infinite number ofparallel slip planes are considered. This problem will be referred to as “verticalwalls of dislocations on equidistant slip planes” or simply “walls of dislocations”.This designation will become more apparent after reviewing the assumptions, whichare defined in the next section.

5

→ τ

x

y

← τ

Figure 1.1: Configuration one: single slip plane

→ τ

x

y

h

h

h

h

← τ

Figure 1.2: Configuration two: walls of dislocations

6

1.2 Basic assumptions

For both configurations – defined in the previous section – a number of assumptionsare made. These assumptions are listed below for reference purposes.

(a) Only straight infinite dislocation lines are considered which are parallel to eachother and to the z-axis of a Cartesian coordinate system. The governing equa-tions therefore depend solely on the remaining x- and y-coordinates.

(b) Only edge dislocations are considered, thus no screw dislocations (or mixeddislocations) are allowed. The Burgers vectors of these dislocations are parallelto the x-axis and the glide planes are perpendicular to the y-axis.

(c) Only dislocations of the same sign – only positive ones – are considered.

(d) An infinite medium with uniform isotropic elastic properties is considered. Thisassumption allows the dislocations to be described in the same manner through-out to domain considered.

(e) Dislocations stay on their slip plane at all times. In other words, dislocationscan only glide and no climb can take place.

(f) The hard obstacle against which the dislocations will pile-up is impenetrable tothe dislocations. Furthermore it is not affected by the dislocations piling-up toit.

(g) The slip plane(s) is (are) perpendicular to the hard obstacle against which thedislocations pile-up.

(h) The slip plane(s) is (are) straight, and infinitely large. This means that dislo-cations are only bounded at the hard obstacle; at the other end of the domainthe dislocations are not bounded in any way.

(i) The stress that is applied is a simple shear stress.

For the second configuration – that of vertical walls of dislocations containing mul-tiple slip planes – three additional assumptions are listed below.

(j) The slip planes are equidistant with a constant vertical spacing h.

(k) The walls are infinite in y-direction, which means that an infinite number of slipplanes are considered at once.

(l) Dislocations within one wall remain directly above each other; they thereforehave the same horizontal coordinate at all times.

7

Chapter 2

Governing equations

In the previous chapter, two pile-up problems were defined by introducing a num-ber of assumptions. Using these assumptions, in this chapter the physics involvedare reviewed. First, dislocation motion is discussed. Second, the stress needed todescribe dislocation motion is analyzed; here both problems defined in chapter 1 areconsidered. Concluding this chapter, an outline of the most important relations isgiven together with the implications thereof.

2.1 Dislocation motion

Based on the assumption that dislocations are mobile, a linear drag relation betweenthe glide velocity of dislocation i, v(i), and the driving force for the motion, calledPeach-Koehler force, f (i), is employed

v(i) =1Bf (i) (2.1)

where B denotes the drag coefficient.

The Peach-Koehler force can be calculated using the stress tensor σ(i) at the posi-tion of the dislocation, together with the normal vector on the slip plane, and theBurgers vector as

f (i) = ~n(i) · σ(i) ·~b(i). (2.2)

The stress σ(i) is determined in the next section. The assumptions made in chapter1 imply that the normal vector ~n and the Burgers vector ~b are equal to:

~n = ~ey and ~b = b~ex (2.3)

Substitution in (2.2) shows that only the shear component of the stress tensor σ(i),i.e. σxy, is relevant.

9

2.2 Stress on a dislocation

The stress acting on a dislocation, used in equation (2.2) to find the Peach-Koehlerforce, is examined in this section.

Superposition holds in calculating the stress σ(i) on dislocation i. This means thatthe stress on this dislocation is the stress caused by interaction with all other dislo-cations σint, combined with the externally applied stress σext

σ(i) =n∑j 6=i

σ(j)int + σext (2.4)

where n is the number of dislocations.

As to the externally applied stress, recall the assumptions made in chapter 1. Hereit was stated that the externally applied stress is a shear stress. In the previoussection it was shown that σxy is the only relevant component of the stress tensoranyway. The external stress tensor can thus be written as

σext = τ (~ex~ey + ~ey~ex) (2.5)

The internal stress tensor σint is constructed below in two subsections. There, onlythe interaction stress of one dislocation on another is reviewed. From equation (2.4)it follows that – when n dislocation are considered – the total interaction stress ona dislocation due to the interactions with all other n− 1 dislocations is the sum ofthe interaction stresses calculated individually.

In the second problem defined, each dislocation wall is considered as an entity –see section 2.2.2; n in this case denotes the number of walls of dislocations.

10

2.2.1 Single slip plane

The first problem defined in chapter 1 is that of a single slip plane. Recall thatsingle slip, under the assumptions made, means that only one straight slip planeis considered (see figure 2.1). We consider the stress that one dislocation withhorizontal coordinate x0 exerts on another dislocation with horizontal coordinate x1(see figure 2.1). The entire stress field of a single dislocation can be found in theliterature, for instance in Hull and Bacon (2001). Since in the previous section it waspointed out that only the shear component is relevant, only this component is givenhere. Furthermore, both dislocations have the same vertical coordinate (y = 0).The shear component of the stress tensor at x1 due to the dislocation at x0 then is:

σxy = σyx =Ḡb

x1 − x0(2.6)

with Ḡ =G

2π (1− ν)(2.7)

In this relation, b is the length of the Burgers vector, G is the shear modulus, andν is the Poisson ratio of the elastic medium which represents the lattice.

From equation (2.6) it is apparent that dislocations of the same sign repel eachother. To make this more clear, the stress field is plotted in figure 2.2. In thisplot x0 = 0 and x1 = x is taken as a variable. In addition, the diagram is non-dimensionalized by:

σ∗xy =σxyḠ

and x∗ =x

b(2.8)

From the graph it can be deduced that when the distance between the dislocationsgoes to zero, the stress that the one dislocation exerts on the other dislocation goesto infinity (thus as x1 → x0 the interaction stress σxy → ∞). This unboundedrepelling force prevents the dislocations from coalescing.

11

x

y

x0 x1

Figure 2.1: Two dislocations on a single slip plane

−5 0 5−10

−8

−6

−4

−2

0

2

4

6

8

10

x*

σ xy*

Figure 2.2: Interaction stress of two dislocation on a single slip plane, here: x0 = 0and x1 → x

12

2.2.2 Infinite vertical walls of dislocations on equidistant slip planes

The second problem is that of infinite vertical walls of dislocations on equidistantslip planes, as illustrated in figure 1.2. The driving force relevant for the motionof a wall is that acting on a single dislocation within the wall. Given the fact thatthe dislocations are distributed periodically within the walls and that the walls areinfinitely long, this driving force is identical for all dislocations within a wall. Itis however the result of interaction with all dislocations within all other walls. Wetherefore consider the stress acting at the position of an arbitrary dislocation withina wall at x1 due to all dislocations within a wall at x0.

What matters is again the shear component σxy of the stress tensor. Choosing,without loss of generality, the slip plane at y = 0, we have according to Roy et al.(2008)

σxy = Ḡb(πh

)2(x1 − x0) sinh−2

[πh

(x1 − x0)]. (2.9)

It is useful to make a comparison with the stress field for a single slip plane. In figure2.4 this comparison is made in the form of a graph of both stress fields, on a linearscale (2.4a) and a double logarithmic scale (2.4b). It comprises the stress field of asingle dislocation (the upper-right quadrant of the diagram in figure 2.2), and thestress field of an infinite vertical wall of dislocations on equidistant slip planes fordifferent vertical spacings h/b. This plot is non-dimensionalized using the definitionsof equation (2.8).

13

In analyzing figure 2.4 some important observations are made.

(i) It is observed that for small ratios h/b between the vertical spacing and theBurgers vector, the interaction stress field is occupying a very short range nearthe hard obstacle. The larger the ratio h/b is, the larger that range. Thisimplies that denser walls have a stress field of a short range nature.

(ii) In the limit that the vertical spacing is very large, the stress field of an infinitewall of dislocations on equidistant slip planes becomes identical to the stressfield of the single slip plane configuration. Mathematically, this can be shownas follows. For large h we have

σwallxy = Ḡb(πh

)2(x1 − x0) sinh−2

[πh

(x1 − x0)]

(2.10)

≈ Ḡb(πh

)2 x1 − x0[πh (x1 − x0)

]2 (2.11)=

Ḡb

x1 − x0= σplanexy (2.12)

Note that (2.10) reduces to (2.11) because sinh (x/h) ≈ x/h for finite x andlarge h.

(iii) On a double logarithmic scale, the stress field for a single slip plane is a straightline due to its 1/x dependence on x = x1. For an infinite wall of dislocations onequidistant slip planes this line is followed for small x, for which x/h

x

y

h

h

h

h

x0 x1

Figure 2.3: Infinite vertical wall of dislocations on equidistant slip planes

15

0 20 40 60 80 100 1200

0.02

0.04

0.06

0.08

0.1

0.12

x*

σ xy*

single slip planewalls of dislocations, h/b = 1walls of dislocations, h/b = 10walls of dislocations, h/b = 100walls of dislocations, h/b = 1000

(a) linear scale

10−1

100

101

102

10−3

10−2

10−1

100

101

102

x*

σ xy*

single slip planewalls of dislocations, h/b = 1walls of dislocations, h/b = 10walls of dislocations, h/b = 100walls of dislocations, h/b = 1000

(b) double logarithmic scale

Figure 2.4: Stress field comparison between a single slip plane of dislocations andvertical walls of dislocations on equidistant slip planes with different vertical spac-ing h

16

2.3 Model summary

In this chapter, some important equations have been introduced. Since the implica-tion of them might be difficult to capture, a short summary is given together withthe implications.

The shear component of the stress that a(n)

(i) single dislocation at x = xi exerts on another at x = xj , is found as

σxy =Ḡb

xj − xi. (2.6)

(ii) infinite vertical wall of dislocations on equidistant slip planes at x = xi exertson another at x = xj , is found as

σxy = Ḡb(πh

)2(xj − xi) sinh−2

[πh

(xj − xi)]. (2.9)

Superposition holds in calculating the net interaction stress due to interaction withmultiple dislocations (or walls of dislocations), together with the applied stress. Thismeans that:

σ(i)xy =n∑j 6=i

σ(j)xy,int + σxy,ext. (2.4)

The Peach-Koehler force is linked to the stress by

f (i) = b σ(i)xy . (2.2)

From this, the equilibrium positions could theoretically be calculated using forceequilibrium. However, this system of equations is to complex to solve. Therefore,equilibrium is searched using dislocation motion. The velocity of a dislocation isfound using a linear drag relation to the Peach-Koehler force, yielding

v(i) =1Bf (i). (2.1)

17

Chapter 3

Numerical implementation andverification

For the two problems of dislocation pile-up we are now applying a numerical al-gorithm to find the respective equilibrium positions of dislocations/walls as well asthe evolution in time towards these equilibrium positions. After an outline on thenumerical algorithm and a description of the method used to represent the discreteresults in terms of a continuous distribution, the results of some simple test casesare used to verify the method.

3.1 Numerical implementation

In chapter 1, it was sketched that this study focuses on the equilibrium state of dis-location pile-up. The equilibrium positions are found making use of the equationsgoverning dislocation motion, as presented in chapter 2. In accordance with theforward Euler method, the position of each dislocation is updated using the velocityof that dislocation based on the Peach-Koehler force acting on it. This process isrepeated - for all dislocations - until all dislocations have come to rest.

To give the reader a better understanding of the numerical implementation, anoutline of the program written is given below. For this study, it was implementedusing the C programming language.

(1) All dislocations are stored in a list; the order of this list is arbitrary. For simpli-city the dislocations are numbered according to their position (i.e. from lowestto highest x). One dislocation is used to model the hard obstacle, by imposinga fixed position of x = 0 on it.

19

(2) The interaction stress between the dislocations is calculated for all dislocations.Recall equation (2.4), which stipulates that the total interaction stress on adislocation i is found by summation of the stresses due to interaction witheach of the other dislocations. This process is repeated for the total number ofdislocations, i.e. for i = 1, 2, ..., n.This process can be thought of as a n×n matrix, the diagonal elements of whichare zero. Since the interaction stress that dislocation i exerts on dislocation jis opposite to the interaction stress that j exerts on i it is not necessary tocalculate both. This way, half of the evaluations can be saved. The evaluationsthat remain form a triangular matrix, depicted in figure 3.1. In terms of the“matrix” of figure 3.1, this implies that it is skew-symmetrical.

(3) The applied shear stress is added to all dislocations in the list.

(4) Using the net shear stress that is now known, the Peach-Koehler force is calcu-lated using equation (2.2).

(5) Using equation (2.1), the velocity v(i) is calculated for each dislocation.

(6) The position of each dislocation is updated, using the forward Euler method.This is done using the constant time step size dt:

x (t+ dt) = x (t) + vdt. (3.1)

(7) To formulate a relevant stopping criterion the velocity is squared and summedover all dislocations, forming an overall kinetic “energy”.

Ekin =n∑i=1

v2i (3.2)

(8) Steps (2) – (7) are repeated until the overall kinetic energy is lower than somecritical value Estop. This value is to be determined by trial and error. However,an effort could be made to relate it to the input parameters.

The initial positions of the dislocations are arbitrary. They have been selected suchthat the initial spacing equals the equilibrium distance of two dislocations on a singleslip plane under the same applied stress, i.e. the initial position of dislocation i is

x(i)

initial = i · xeq. (3.3)

20

ji 1 2 ... n

1 02 xx 0

... xx xx 0n xx xx xx 0

Figure 3.1: Evaluations done in the numerical model. Only the elements markedwith “xx” have to be evaluated.

One advantage of using the method described is that it is relatively straightforward.Another advantage is that the motion in time of the dislocations can be visualized.Especially during the development of the numerical implementation this capabilityproved useful.

An example of this visualization is depicted in figure 3.2. Input parameters forthis simulation are

Ḡ = 1.0 b = 1.0 B = 1.0 τ = −5 · 10−2. (3.4)The step size and the “critical kinetic energy”, which indicates that the dislocationshave come to rest, are

dt = 5.0 · 10−3 Estop = 1.0 · 10−9. (3.5)Both the step size dt and the “critical kinetic energy” Estop have been found by trialand error. Two remarks concerning the choice of them are given below.

• The step size is to be sufficiently small, such that the maximum distance dislo-cations travel during one time step is by all means smaller than the minimumdistance between two dislocations (vdt < ∆xmin). For more accurate timeevaluation the step size could be chosen smaller. However since the time eval-uation is mainly the means in finding the equilibrium positions, and not a goalon its own, a smaller step size is not pursued.

• The “critical kinetic energy” Estop is chosen with great margin for error, since– unlike for the step size – a value that is too big might result in incorrect re-sults. In that case, it is assumed that dislocations have found their equilibriumpositions, while in fact they are still in motion. Whether the value of Estop issmall enough can be verified visually. In case of equilibrium, dislocations willnot have been in motion for some time, corresponding to vertical lines in thetime evolution plot of figure 3.2. This illustrates the advantage of visualizingthe time evolution.

21

0 1000 2000 3000 4000 50000

0.5

1

1.5

2

2.5x 10

4

x/b

t |τ|

b/B

Figure 3.2: Time evaluation of a single slip plane, for n = 128 dislocations. Eachcurve represents a dislocation

3.2 Dislocation density

In this study the pile-up of dislocations is modeled in a discrete manner. This meansthat we consider each dislocation individually as a line defect moving through anelastic medium. Crystal plasticity theories consider a larger scale than is being con-sidered in this study. These theories describe the pile-up based on a continuumassumption. Therefore, a continuum-like quantity is needed to link our findings tothose of continuum based theories. The continuous quantity that is used to describethe distribution of dislocations in crystal plasticity is the dislocation density.

Using the discrete analyses, the positions of the dislocations in the pile-up are found.These positions are related to the dislocation (or wall) density using a finite differ-ence scheme:

F (xi) =2

xi+1 − xi−1(3.6)

Here is F (xi) the dislocation density at position xi (i = 2, 3, ..., n − 1 with n thenumber of dislocations considered). Recall that the hard obstacle is modeled as adislocation to which a fixed position is enforced; this “dislocation” is included in thetotal number of dislocations.

An example of the dislocation density of the pile-up is given in figure 3.3. Fromthis figure, it can be seen that the dislocation density increases rapidly towards theimpenetrable boundary at x = 0. Therefore, it proves useful to plot the density on

22

0 50 100 150 200 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/b

F(x

) b

128 dislocations

Figure 3.3: Dislocation density on a regular scale, for n = 128

a semi- or double-logarithmic scale.

3.3 Basic validation

When a problem is solved numerically, care should always be taken in validatingthe numerical implementation. A basic validation of the numerical implementation,put forward in the previous section, is attempted in this section. To this end, twosimple cases are formulated whose solution can be found analytically. This can thenbe compared to the numerically found solution. The two cases are discussed in thefollowing two subsections.

3.3.1 No dislocation interaction

The first validation case is that of a single dislocation with no interaction whatso-ever. This can be modeled by implementing only one dislocation, which is boundednowhere on the domain. The position this dislocation has, as a function of time,can easily be found analytically.

From equations (2.1) and (2.2) it follows that since the applied shear stress τ isthe only stress on the dislocation, its glide velocity equals:

v =τb

B(3.7)

23

−5 0 5 100

5

10

15

x/b

t |τ|

b/B

Simulated pathAnalytical path

Figure 3.4: Numerically and analytically found trajectory using parameters of equa-tion (3.9)

The position – as a function of time – can be found by integrating the equationabove, using the initial condition x(t = 0) = 10b. This gives

x(t) =τb

Bt+ 10b. (3.8)

If this relation is plotted in the same figure as the numerical result, both shouldyield the same curve. This is done in figure 3.4, using the parameters

Ḡ = 1.0 b = 1.0 B = 1.0 τ = −5 · 10−2 dt = 5.0 · 10−3. (3.9)

The two results indeed match perfectly.

3.3.2 Pile-up of one mobile dislocation

The second validation case is that of one mobile dislocation piling-up against a hardobstacle. In accordance with the first section of this chapter, the hard obstacle ismodeled as a dislocation to which a fixed position is enforced. In chapter 1, twoproblems are defined - one of a single slip plane, and the other of infinite verticalwalls on equidistant slip plane - both will be pursued in this subsection.

24

Equilibrium: single slip planeAnalytically the equilibrium position of one dislocation piling-up to a hard obstacleunder the influence of a shear stress τ can be found by force equilibrium. Therepulsive stress that the fixed dislocation exerts on the mobile dislocation must bein equilibrium with the applied shear stress. For the fixed dislocation at x0 = 0, thisresults in:

Ḡb

x+ τ = 0 → x = −Ḡb

τ(3.10)

Equilibrium: infinite vertical walls on equidistant slip planesFor the other problem introduced in chapter 1 – that of infinite vertical walls onequidistant slip planes - the equilibrium equation can be found in the same manneras for a single slip plane, and reads

Ḡb(πh

)2x sinh−2

(πhx)

+ τ = 0. (3.11)

This equation cannot be solved in closed form; therefore the secant method – anumerical iteration method for solving a non-linear equation – is employed, as elab-orated on in appendix A.

ResultsThe equilibrium position found using the numerical model should coincide with thosefound analytically. Furthermore, since the mobile dislocation is now repelled by thestationary dislocation at x = 0, dislocation glide should be slower than without dis-location interaction. Both observations are validated using the plots in figure 3.5.The parameters that are used for the single slip plane case (plotted in figure 3.5a)are:

Ḡ = 1.0 b = 1.0 B = 1.0 τ = −5 · 10−2

dt = 5.0 · 10−3 Estop = 1.0 · 10−9 (3.12)

For the infinite vertical wall of dislocations on equidistant slip planes (plotted infigure 3.5b), the parameters are the same as in equation (3.12), with the verticalspacing

h = 20b. (3.13)

Both diagrams show that the interaction with the immobile dislocation/wall indeedshows down the mobile dislocation/wall and that the equilibrium position it finallytakes coincides with the prediction. The same validation has been repeated fordifferent vertical spacing, the result of which is attached in appendix B

25

20 22 24 26 28 300

50

100

150

x/b

t |τ|

b/B

simulated pathanalytical equilibriumanalytically found path without interaction

(a) single slip plane

8 10 12 14 16 18 20 22 240

5

10

15

20

25

30

35

40

45

x/b

t |τ|

b/B


(b) walls of dislocations

Figure 3.5: Pile-up of one mobile dislocation(wall), together with the analyticallyfound equilibrium position and trajectory without interaction

26

Chapter 4

Results for a single slip plane

In this chapter the results for a single slip plane are presented, and in the nextchapter this is repeated for infinite vertical walls on equidistant slip planes.

4.1 Review of the literature

Concerning a single slip plane, a continuous formulation of the dislocation densitycan be found in the literature. In it, dislocations are smeared over an interval,obtaining a continuous formulation. According to Pande et al. (1972), the densityof the pile-up can be determined in closed form as described below. It has to satisfythe integral equation ∫ a

0

F (x)x− x0

dx =|τ |Ḡb

. (4.1)

Where a is the length of the pile-up. By solving this equation, the dislocation densitywas obtained as

F (x) =|τ |πḠ

√√√√∣∣∣2nḠbτ + x∣∣∣x

. (4.2)

In equation (4.2) the absolute signs appear because the applied shear stress τ istaken to be negative (the dislocations thus start at positive values of x and pile-upto a hard obstacle at x = 0). In the original work of Pande et al. (1972) it was takenpositive, meaning that the dislocations pile-up from negative values of x to a hardobstacle positioned at x = 0, and therefore the absolute signs were not necessary.

Applying the parameter values found in table 4.1, the dislocation density of thepile-up found using this continuum description is plotted in figure 4.1 for differentnumbers of dislocations. In analyzing this figure, we can conclude that the density

27

10−2

10−1

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b

128 dislocations64 dislocations32 dislocations16 dislocations8 dislocations4 dislocations

Figure 4.1: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 dislocations,found using the continuum model found in Pande et al. (1972), plotted on a doublelogarithmic scale

profiles appear to follow a straight line on a double logarithmic scale except nearthe end of the pile-up, near x = a. This can be understood from (4.2) by realizingthat for x

Table 4.1: Parameter values

Parameter Description Parameter valueτ Applied shear stress −5 · 10−2Ḡ Material parameter 1 · 100B Viscous drag coefficient 1 · 100b Burgers vector 1 · 100

dt Time step 5 · 10−3Estop Critical kinetic energy 1 · 10−9

is plotted in figure 4.2. As in the continuum model, the density shows a straightline on a double logarithmic scale, i.e. F (x) ∼ 1/

√x, for most of the pile-up. More

specifically, the density plot seems to be the same as found using the continuummodel formulated by Pande et al. (1972). This can be verified by plotting both thediscrete density and that produced by the continuum model in the same graph, asdone in figure 4.3. Notice that the dotted lines (representing the continuum model)and the solid lines (representing the discrete results) almost exactly coincide, espe-cially for a higher number of dislocations.

Furthermore, the conclusion can be drawn form the continuum description in equa-tion (4.2) that the dislocation scale with the square root of the number of dislocations(i.e.

√n). To verify whether this also holds for the discrete results, they are scaled

using:

Fn,scaled = Fn/√n where n = 4, 8, 16, 32, 64, 128. (4.3)

The result is plotted in figure 4.4. There, it can be seen that all scaled curves coin-cide, except at the end of the respective pile-ups. This means that – in accordancewith the continuous formulation by Pande et al. (1972) – the discretely found resultsalso scale with

√n.

29

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b


Figure 4.2: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 dislocations,using the discrete model. Plotted on a double logarithmic scale

30

10−2

10−1

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b


Figure 4.3: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 dislocations,using the continuum model (plotted dotted) and the discrete model (plotted solid).The dotted lines are hardly visible because they coincide with the solid lines; theyhave the same markers

31

100

101

102

103

104

10−4

10−3

10−2

10−1

100

x/b

F(x

) b

n−1/

2


Figure 4.4: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 dislocations,using the discrete model, scaled by

√n

32

Chapter 5

Results for infinite vertical walls

In the previous chapter the results found using the discrete model for the problem ofa single slip plane were discussed. This chapter presents the results found using thediscrete model applied to infinite vertical walls on equidistant slip planes. Again, alink will be made with the literature.

Similarly to the case of a single slip plane – discussed in the previous chapter –the dislocation density (defined in equation (3.6)) is used as a continuum-like de-scription of the results produced by the discrete model. Also, the same parametersas in the previous chapter are used throughout this chapter, which can be found intable 4.1 on page 29

5.1 Limit case: large vertical spacing

For walls of dislocations, the vertical spacing h is an important parameter. It isuseful to first consider the limit case where the vertical spacing h is sufficiently largefor the stress field of a wall of dislocations to be practically identical to that of asingle dislocation, see equation (2.10). For a vertical spacing of h/b = 10000, thedislocation density of the pile-up is plotted for various numbers of walls in figure5.1. Comparing this figure to figure 4.2 for a single slip plane, it appears that thegraphs are indeed identical. This means that the model does appear to obey thelimit as expected.

5.2 Results for different vertical spacing

The discrete model is now provided with different values of the vertical spacing h.Focussing on the case of n = 128 dislocations, the results are plotted in figure 5.2.From this picture the following conclusions can be drawn.

33

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b


Figure 5.1: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 walls ofdislocations with h/b = 10000, using the discrete model and plotted on a doublelogarithmic scale

(i) For large vertical spacing h/b, the density of the pile-up follows a straightline on a double logarithmic scale. This means that F (x) ∼ 1/

√x, which

is confirmed by the work of Roy et al. (2008). It should be remarked thattowards the end of the pile-up the density starts to deviate from the straightline. This can be considered to be a boundary effect, since there is no boundarycondition at that side of the domain. The deviation from the straight lines wasalso demonstrated in the continuum model of Pande et al. (1972).

(ii) for small vertical spacing h/b, the density shows a curved profile on a doublelogarithmic scale and no longer resembles F (x) ∼ 1/

√x. Review of the liter-

ature suggest this curve can be interpreted as F (x) ∼ exp−x, which can befound for instance in the work of Groma et al. (2003). For this to be true, thedata should be on a straight line on a semi-logarithmic scale, which is plottedin figure 5.3. From this graph, it can be seen that these lines are not straight.This deficiency is dealt with in the next section. Notice that the smallest ver-tical spacing is chosen as h = b/2. Although physically impossible, this valueis pursued to find out whether or not a F (x) ∼ exp−x dependence is found.

(iii) For intermediate vertical spacing h/b, the density profile appears to partiallyfollow the F (x) ∼ 1/

√x dependency close to the obstacle and starts to deviate

from this trend at some distance away from the obstacle. The distance from theobstacle at which this deviation occurs is smaller for smaller h/b. Interestingis that near the hard obstacle the density of the pile-up seems to obey the

34

continuum model for a single slip plane, presented in the previous chapter.This is supported by the observation, made in figure 5.4, that when the densityis scaled by

√n all lines seem to coincide for the part of the pile-up closest to

the hard obstacle. The scaling is done – in accordance with scaling done in theprevious chapter – by:

Fn,scaled = Fn/√n where n = 4, 8, 16, 32, 64, 128 walls of dislocations (5.1)

(iv) When the vertical spacing h/b is reduced, also the length of the pile-up de-creases. This can be explained by the stress field dependence of the verticalspacing. For smaller h/b, the stress field of a wall of dislocations becomesshorter range, and therefore the length of the pile-up decreases.

The complete set of graphs for each value of vertical spacing h/b, on both a doublelogarithmic scale and a semi-logarithmic scale, can be found in appendix C.1.

35

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b

h/b = 10000h/b = 1000h/b = 200h/b = 20h/b = 1/2

Figure 5.2: Dislocation density of the pile-up for n = 128 walls of dislocations, forseveral values of vertical spacing h/b and plotted on a double logarithmic scale

36

0 5 10 15 20 25 30 35 4010

0

101

x/b

F(x

) b


Figure 5.3: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 walls ofdislocations and h = 12b on a semi-logarithmic scale

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b


Figure 5.4: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 walls ofdislocations and h = 20b on a double logarithmic scale, scaled by

√n

37

5.3 Review of the literature

Review of the literature concerning the infinite vertical walls of dislocations onequidistant slip planes, leads to the work of Groma et al. (2003). In this work,it is found on the basis of a higher-order crystal plasticity model that the dislo-cation density of the pile-up obeys F (x) ∼ exp(−x). Observation (ii) – based onfigures 5.2 and 5.3 – shows however that the discrete analysis done in this studydoes not confirm this.

The basis of this difference lies in an assumption that was made by Groma et al.(2003), but is not made here. The assumption was made by Groma et al. (2003) thatonly the nearest neighbors of a dislocation should be accounted for in calculating thestress on a given wall of dislocations. No such assumption is made in our discretesimulations and all dislocation walls therefore interact with each other.

To verify that this is the explanation for the discrepancy between our results andthe density profile derived by Groma et al. (2003), simulations have been run withthe following additional assumption:

(m) In calculating the stress field on a wall of dislocations, only the interaction withthe two nearest neighboring wall is accounted for.

The result is plotted in figure 5.5 on both a double logarithmic and semi-logarithmicscale. It can be seen, firstly, that the length of the pile-up is smaller compared withthat for the fully interacting walls, which can be compared with figure 5.6. Secondly,that the dislocation density of the pile-up of infinite vertical walls of dislocations fol-lows straight lines on a semi-logarithmic scale, i.e. F (x) ∼ exp(−x). Again, theresults for different values of vertical spacing h/b can be found in the appendix C.2.The plots combined for the discrete analysis both with and without assumption (m)are included in appendix C.3. For comparison with figure 5.2, a similar plot is madeusing the assumption of only nearest-neighbor interaction, if figure 5.7

The results presented here give reason to believe that taking into account the in-teraction with all other walls of dislocation – not only the nearest neighbors – isimportant since it yields different results. This observation is important becausemost continuum descriptions – including most crystal plasticity models – use thenearest neighbor assumption.

38

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b


(a) double logarithmic scale

0 20 40 60 80 10010

−2

10−1

100

101

x/b

F(x

) b


(b) semi-logarithmic scale

Figure 5.5: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 walls ofdislocations, with h = 20b and the assumption of only nearest-neighbor interaction

39

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 50 100 150 200 250 30010

−2

10−1

100

101

x/b

F(x

) b



Figure 5.6: Dislocation density of the pile-up for n = 4, 8, 16, 32, 64, 128 walls ofdislocations, with h = 20b

40

100

101

102

103

10−2

10−1

100

101

x/b

F(x

) b

h/b = 1000h/b = 20h/b = 10h/b = 1h/b = 1/2

Figure 5.7: Dislocation density of the pile-up for n = 128 walls of dislocations, forseveral values of vertical spacing h/b and plotted on a double logarithmic scale andwith the assumption of only nearest-neighbor interaction.

41

Conclusion

In this study a discrete analysis of dislocation pile-up against a hard obstacle wasperformed. A distinction has been made between a single slip plane and infinitevertical walls of dislocations on equidistant slip planes. For both configurations thestress field of a single dislocation/wall has been taken from the literature (i.e. fromHull and Bacon (2001) and Roy et al. (2008)). From this stress field, the Peach-Koehler force was determined, which was then linked to dislocation motion via alinear drag relation.

A forward Euler method was applied to find the equilibrium positions for botha single slip plane and infinite vertical walls on equidistant slip planes, under the in-fluence of an applied shear stress. The equilibrium positions were used to constructa dislocation density plot of the pile-up. This was used to examine the validity ofresults found by Pande et al. (1972), Groma et al. (2003), and Roy et al. (2008).The density is the subject of study, because it links continuum models (for instancefrom Pande et al. (1972)) and discrete models (for instance from Groma et al. (2003)and Roy et al. (2008)).

For a single slip plane, the dislocation density of the pile-up found using the discreteanalysis coincides with the continuous distribution found in the work of Pande et al.(1972). Together with the fact that the discrete model is validated using simplecases, of which the analytical solution was known beforehand, this result gives rea-son to trust the results produced by the discrete numerical model constructed.

For infinite vertical walls of dislocations, the literature shows a different disloca-tion density distribution of the pile-up than produced in this study. The workof Roy et al. (2008) shows that the pile-up of these walls can be characterized byF (x) ∼ 1/

√x. The work of Groma et al. (2003) however, shows F (x) ∼ exp(−x). In

our study it is shown that for large vertical spacing, the first case seems to be valid.For a physically unrealistic small vertical spacing, the density seems to converge tothe latter case, however F (x) ∼ exp(x) is never exactly found. For intermediatevertical spacing, it was found that some combination of regimes seems to govern thedislocation density of the pile-up.

43

To investigate where the difference in the result of our study and the work of Gromaet al. (2003) originates, the additional assumption of only nearest neighbor interac-tion was implemented. Using this assumption the exponential density distributionfound by Groma et al. (2003) was reproduced. Because of this observation, the con-clusion can be drawn that the assumption of only nearest neighbor interaction maybe too strong.

Since most continuum models – exploited for instance in crystal plasticity mod-els – yield an exponential dislocation density distribution of the pile-up, our studyshows some adjustment may be needed to these continuum models.

44

Appendices

45

Appendix A

Secant method

For solving a nonlinear equation, several numerical iteration methods can be em-ployed, including Newton’s method and the secant method. First, both methodswill be discussed. Then, the choice for secant method is explained.

A.1 Numerical iteration

In Heath (2005) several numerical iteration methods for solving non-linear equationsare presented – including Newton’s and secant method. To employ these methods,the equation treated must first be converted to a root finding problem. For example,if we want to find the intersection between g(x) = x2 and h(x) = 1, we must convertthe problem to finding the root (i.e. intersection with the x-axis) of f(x) = x2 − 1.In other words, we solve f(x) = x2 − 1 = 0.

Newton’s method uses the derivative of the function for an approximate solutionto find a better approximation of the root of the function, and repeats this until theapproximation of the root is “good enough”. Each iteration uses the generic step,

xk+1 = xk −f(xk)f ′(xk)

. (A.1)

A sketch of this step is depicted in figure A.1a.

47

(a) Newton’s method (b) Secant method

Figure A.1: Numerical method’s for solving a nonlinear equation, taken from Heath(2005)

The secant method is similar to Newton’s method, but it uses an approximation forthe derivative of the function f(x) based on two previous estimates:

xk+1 = xk − f(xk)f ′(xk) (A.2)

f ′(xk) ≈ f(xk)−f(xk−1)xk−xk−1 (A.3)

Again, a sketch of this generic step is included, and can be found in figure A.1b.

48

A.2 Numerical iteration applied

Numerical iteration is employed in this study for solving equation (3.11), which forclarity is written as

f(x) = Ḡb(πh

)2x sinh−2

(πhx)

+ τ = 0. (A.4)

Note that for the sake of brevity x1 is written as x, and x0 = 0.

The derivative of the function f(x) is:

f ′(x) = Ḡb(πh

)2 {sinh−2

(πhx)− 2x cosh

(πhx)

sinh−3(πhx)}

(A.5)

When analyzing the derivative written above, it becomes apparent that the deriva-tive has O

(10−3

). Such an order poses the risk that the derivative goes to either

infinity or zero relatively easily. To overcome this problem, secant method is em-ployed.

As an example, the root of equation (A.4) is searched using the simulation param-eters used throughout this study – found in table 4.1 – additionally with verticalspacing h = 100b. To verify the correctness of this result, the function is sketchednear its root, together with the root found using the secant method. Judging fromthis sketch – depicted in figure A.2 – secant method produces the correct result.

A final remark is that the root found using the secant method was also searchedusing Newton’s method. As was expected – because of the order of the derivative– Newton’s method “overshot” already in three iteration steps. The choice for thesecant method is therefore justified.

49

4 4.2 4.4 4.6 4.8 5−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

x*

σ* xy

+ τ

f(x) = σ

xy + τ

g(x) = 0x

root

Figure A.2: Verification of secant method for h/b = 100

50

Appendix B

Validation

As described in chapter 3 the validation for infinite walls of dislocations on equidis-tant slip planes has been repeated for different vertical spacing. The results arepresented in figures B.1 and B.2 and the conclusion is identical to that for h/b = 20.

51

19 20 21 22 23 240

20

40

60

80

100

120

x/b

t |τ|

b/B


Figure B.1: Pile-up of one wall of dislocations with h/b = 1000, together with theanalytically found equilibrium position and trajectory without interaction

19 20 21 22 23 240

20

40

60

80

100

120

x/b

t |τ|

b/B


Figure B.2: Pile-up of one wall of dislocations with h/b = 10000, together with theanalytically found equilibrium position and trajectory without interaction

52

Appendix C

Additional results

In this appendix some additional figures are presented. Most of them were notincluded in the main text, but some are repeated for the sake of completeness. Allfigures concern infinite vertical walls of dislocations on equidistant slip planes. Inthe first section the complete discrete analysis is employed. In the second section thenumerical model is extended with assumption (m), i.e. that of only nearest-neighborinteraction. In the final section both the complete discrete analysis and the analysisemploying only nearest-neighbor interaction are combined.

C.1 Complete discrete analysis

In this section some additional results are presented, found using the complete dis-crete analysis.

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 5 10 15 20 25 30 35 4010

0

101

x/b

F(x

) b



Figure C.1: h = 12b

53

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 50 100 150 200 250 30010

−2

10−1

100

101

x/b

F(x

) b



Figure C.2: h = 20b

54

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 500 1000 1500 200010

−2

10−1

100

101

x/b

F(x

) b



Figure C.3: h = 1000b

100

101

102

103

104

10−3

10−2

10−1

100

101

x/b

F(x

) b



0 500 1000 1500 2000 2500 3000 3500 400010

−3

10−2

10−1

100

101

x/b

F(x

) b




55

C.2 Only nearest neighbor interaction

In this section some additional results are presented, found using the numericalmodel using additionally assumption (m), i.e. for only nearest neighbor interaction.

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 5 10 15 20 25 30 35 4010

0

101

x/b

F(x

) b



Figure C.5: h = 12b

57

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 20 40 60 80 10010

−2

10−1

100

101

x/b

F(x

) b



Figure C.6: h = 20b

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



0 20 40 60 80 10010

−2

10−1

100

101

x/b

F(x

) b




58

C.3 Comparison

In this section the results found using the numerical model both with and withoutassumption (m), that of only nearest neighbor interaction. In all plots the dashedcurves represent the results found using the assumption of only nearest neighborinteraction, the solid curves represent the complete discrete analysis.

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b


Figure C.8: Double logarithmic scale, h = 12b

59

100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



100

101

102

103

104

10−2

10−1

100

101

x/b

F(x

) b



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Bibliography

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