# Is dislocation plasticity governed by a power law?

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(2) Abstract Using discrete dislocation plasticity, the dependence of plastic stain rate on resolved shear stress has been studied in order to conﬁrm the power law proposed by Asaro and Needleman. A model has been made that contains sources, which nucleate two-dimensional edge dislocations, and obstacles, which can hinder the movement of dislocations. A limited strain rate is found, by assuming a constant nucleation time of dislocations. By making use of the collective behaviour of dislocations and a stress-dependent nucleation time, a power law dependence between strain rate and applied shear stress is found, at least in a limited range of stress..

(3) Contents 1 Introduction. 4. 2 Theory 2.1 Dislocations . . . . . . . . . . . . . . . . . . . . 2.1.1 Shear stress ﬁelds of edge dislocations 2.2 Discrete dislocation plasticity . . . . . . . . . . 2.2.1 Motion of dislocations . . . . . . . . . 2.2.2 Interactions between two dislocations 2.2.3 Nucleation of dislocations . . . . . . . 2.2.4 Annihilation of dislocations . . . . . . . 2.2.5 Obstacles . . . . . . . . . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. 5 5 6 8 9 10 11 13 13. 3 Model 3.1 Overview of the model . . . . . . . . . . . . . 3.1.1 Strain rate . . . . . . . . . . . . . . . . 3.1.2 Application of the theory to the model 3.2 Outline of the program . . . . . . . . . . . . . 3.2.1 Initialization . . . . . . . . . . . . . . . . 3.2.2 The computational loop . . . . . . . . 3.3 Results and discussion . . . . . . . . . . . . . . 3.3.1 One slip plane, one source . . . . . . . 3.3.2 One slip plane, three sources . . . . . 3.4 Bucket time . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . .. 14 14 15 16 18 18 18 19 19 25 28. 4 Improved model 4.1 Overview of the improved model 4.2 Outline of the program . . . . . . 4.3 Results and discussion . . . . . . . 4.3.1 Five slip planes . . . . . . . 4.3.2 Twenty-ﬁve slip planes . . 4.3.3 Fifty slip planes . . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. 31 31 31 32 32 36 38. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. 5 Conclusion. 45. References. 46. Appendices Appendix A Appendix B Appendix C. On the stress-dependent nucleation time . . . . . . . . . . . . . . . . . . . . . . . . . Fitting a power law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . List of symbols in order of ﬁrst appearance . . . . . . . . . . . . . . . . . . . . . . . .. 3. 47 47 49 52.

(4) 1 Introduction Plastic strain is the irreversible deformation of a material due to an applied load. It is irreversible in the sense that, opposite to elastic strain, the material will not return to its original state when it is unloaded, but stays in the same conﬁguration. Plastic behaviour is generally attributed to the movement of dislocations, which are line defects in an atomic lattice. The strain rate is the change in strain per unit time. An FCC metal contains twelve slip systems, which is more than the ﬁve independent components of plastic strain (rate). Therefore, the amount of slip per slip system cannot be determined from the strain tensor, which is a major barrier for the use of crystal plasticity. In order to circumvent this hurdle, Asaro and Needleman proposed a computational trick to relate the strain rate γ˙ (α) with the resolved shear stress τ on a slip system α in the form of an empirical power law by [1] γ˙. (α). =. (α) γ˙ 0.

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(6) m−1 τ

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(8) τ

(9)

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(12) (α)

(13) (α)

(14) τ τ 0. (α). (1.1). 0. (α). where γ˙ 0 and τ0 are normalizing constants and m is the rate sensitivity1.1 . The rate sensitivity is strongly material dependent and depends on e.g. the grain size of the material [2, 3]. Although the power law has been conﬁrmed experimentally for strain rates above 103 s-1 [4, 5], no computational simulations have been done so. Nevertheless, the discrete dislocation plasticity (DDP) framework of Van der Giessen and Needleman [6], has shown to be able to quantitatively describe small-grain size plasticity (see e.g. [7, 8]). This framework consists of a decomposition method to describe boundary value problems and the ingredients for two-dimensional dislocation dynamics. Hellinga tried to use this framework to conﬁrm the power law of Asaro and Needleman by prescribing a shear stress and getting the corresponding strain rate as output [9]. However, the model he used was simpliﬁed too much and the strain rate dependence he found was either limited at a certain strain rate or linear with resolved shear stress. This thesis is a follow-up study of Hellinga’s work. We try to extend his model and our goal is to see if DDP gives rise to the power law of Asaro and Needleman when collective dislocation motion is taken into account. Also, the aim is to ﬁnd the link between between the material parameters in Equation (1.1) and the parameters in the DDP model.. Outline of the thesis First, the theory behind dislocations and the DDP framework of Van der Giessen and Needleman is treated in Section 2. Section 3 covers the model, which is a continuation of Hellinga’s model, including results and discussion. An improvement of the model and its results are given in Section 4, together with a discussion. This thesis ends with the conclusion in Section 5.. 1.1 Note. that this rate sensitivity is not the same as ∂ log σy /∂ log ε,˙ which is used in macroscopic viscoplasticity.. 4.

(15) 2 Theory 2.1 Dislocations A dislocation is a line defect in an atomic lattice. It is a loop in which the atomic plane has shifted with respect to the lattice outside the loop, as shown in Figure 2.1. The shift inside the loop is uniform and is denoted by the Burgers vector b. At point A the dislocation line is perpendicular to b; this is called the edge part of the dislocation. At point B the dislocation is parallel with b; this is called the screw part of the dislocation. In between, the dislocation has a mixed character. Since a dislocation loop is a closed loop, every dislocation has at least two points where the dislocation is purely edge and two points where it is purely screw. Usually edge dislocations are denoted by ⊥ and screw dislocations by S or .. b b A. B. Figure 2.1: Schematic representation of a dislocation. From [10]. For simplicity, we make a cross-section perpendicular to the dislocation loop and through the opposite edge parts, in order to get only straight edge dislocations. From now on, each of these edge parts that become points in the cross-section will be referred to as a dislocation. In two dimensions, information about the original dislocation line is lost. By convention, a dislocation that has an extra half-plane of atoms above the dislocation, as shown Figure 2.2, has a positive Burgers vector and is denoted by ⊥. Consequently, a dislocation that has an extra half-plane of atoms under the dislocation has a negative Burgers vector and is denoted by ⊤. Note that the two dislocations from a cross-section are always oppositely signed, i.e. their Burgers vectors have opposite sign in two dimensions. The pair is called a dislocation dipole. Burgers vector b. Edge dislocation line. Figure 2.2: The atom positions around a positive edge dislocation. From [10]. Dislocations play a major role in slip. In a perfect crystal, i.e. without dislocations, a layer of atoms would have to slip over another layer all at once, which needs signiﬁcantly higher stresses, in the order of 10-2 µ, than 5.

(16) experimentally found values of the stress slip in real, non-perfect crystals, which are in the order of 10-4 – 10-8 µ, where µ is the shear modulus [11]. As shown in Figure 2.3, a dislocation can move by only breaking the bond between 1 and 3 and making a new bond between 1 and 2. When the dislocation keeps moving to the right, it produces slip by an amount of |b|, which needs lower stresses than in a perfect crystal. The plane at which a dislocation moves is the slip plane.. slip plane. 2 3 1. 2 3 1. 3 1. Figure 2.3: Movement of an edge dislocation: the arrows indicate the applied shear stress. From [11].. 2.1.1 Shear stress ﬁelds of edge dislocations Dislocations interact with each other due to their stress ﬁelds. The shear stress ﬁeld of a dislocation I in an inﬁnite medium is given by [12]. I (x1 , x2 ) = σ12. µbI x1 (x1 2 − x2 2 ) 2π(1 − ν) (x1 2 + x2 2 )2. (2.1). where µ is the shear modulus, ν is the Poisson’s ratio, bI is the Burger’s vector, x1 the distance to the dislocation parallel to the slip plane and x2 the distance to the dislocation perpendicular to the slip plane. The shear stress is singular at the dislocation itself and is long ranged, i.e. it goes with the distance r as 1/r. A contour plot is shown in Figure 2.4. The shear stress is zero at the thick black lines (x1 = 0 and x1 = x2 ) and is positive at one side and negative at the other side. x 12. 2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0. 0. -1 x -1 0 11 Figure 2.4: Contour plot of the the shear stress ﬁeld of an edge dislocation at (x1 , x2 ) = (0,0), normalized by µb/2π(1 − ν).. When one considers a periodic cell, one dislocation actually resembles a string of equal dislocation. The shear stress ﬁeld of an inﬁnitely long string of equal dislocations, i.e. with the same Burgers vector, on a slip plane, as 6.

(17) w. w x1 x2. j = -1. j=0. j=1. Figure 2.5: An inﬁnitely long string of dislocations with the same Burgers vector at mutual distance w. shown in Figure 2.5, is the sum of the shear stress ﬁelds of each individual dislocation and is therefore given by I σ12 (x1 , x2 ). ] [ +∞ ∑ (x1 − jw) (x1 − jw)2 − x2 2 µbI = 2 2π(1 − ν) j=−∞ [(x1 − jw)2 − x2 2 ]. (2.2). With the use of complex analysis this becomes [6] I σ12 (x1 , x2 ) =. [ ] µbI π sin(2πξ) sinh(2πη) 1 − 2πη 2π(1 − ν) w cosh(2πη) − cos(2πξ) cosh(2πη) − cos(2πξ). (2.3). where ξ ≡ x1 /w and η ≡ x2 /w. A contour plot is shown in Figure 2.6. Again, the shear stress is zero at the thick black lines (including ξ = 0 and ξ = 0.5) and is positive at one side and negative at the other side. η 1. 0.4 0.3 0.2 0.1 0.0 -0.1 -0.2 -0.3 -0.4. 0. -1 ξ -1 0 1 Figure 2.6: Contour plot of the the shear stress ﬁeld of an inﬁnitely long string of dislocations, with one dislocation at (ξ, η) = (0,0), normalized by µb/2(1 − ν)w.. 7.

(18) 2.2 Discrete dislocation plasticity The shear stress ﬁeld of a dislocation as given in Equation (2.1) is only valid in an inﬁnite medium. In order to ﬁnd solutions in a ﬁnite medium, Van der Giessen and Needleman proposed a solution based on superposition [6]. The displacement ﬁeld u, strain ﬁeld ε and stress ﬁeld σ are decomposed as ˜+u ˆ u=u ε = ε˜ + εˆ ˜ +σ ˆ σ=σ. (2.4a) (2.4b) (2.4c). where the (˜) ﬁelds are the superposition of the ﬁelds of the individual dislocations in an inﬁnite medium. The (ˆ) ﬁelds are calculated without the dislocations to account for the boundary conditions in terms of prescribed tractions T0 on part Sf of the the boundary and prescribed displacements u0 on part Su of the boundary. A schematic of this decomposition is given in Figure 2.7. The (ˆ) ﬁelds satisfy the elasticity equations, i.e. ˆ =0 equilibrium: div σ [ ] 1 T ˆ ˆ ˆ strains: ε = grad u + (grad u) 2 ˆ = L : εˆ elasticity: σ. T0 on Sf. (2.5a) (2.5b) (2.5c). ˜ Tˆ = T0 – T on Sf. T˜ on Sf. +. = u. u˜. uˆ. u˜ on Su. u0 on Su. ˜ on Su uˆ = u0 – u Figure 2.7: Decomposition of the problem for the dislocated body into the problem of interacting dislocations in the inﬁnite solid (the (˜) ﬁelds) and the complementary problem for the body without dislocations (the (ˆ) ﬁelds). From [6]. The tractions at the boundary Sf must satisfy T0 = σ · n ˜ ·n T˜ = σ ˆ ·n Tˆ = σ. (2.6a) (2.6b) (2.6c). where n is the outward normal to the boundary. The (ˆ) ﬁelds do not contain any dislocations and are therefore smooth and their boundary value problem can be solved using a ﬁnite element method (FEM).. 8.

(19) 2.2.1 Motion of dislocations Dislocations move due to the Peach–Koehler force f I , deﬁned by [6] ∑ I I J I ˆ+ f = t × σ σ ·b . (2.7). J̸=I. where tI is the unit vector in the direction of the dislocation line. The Peach–Koehler force does not depend on the stress ﬁeld of the dislocation itself; the dislocation only ’notices’ the other dislocations. Most important is the Peach–Koehler force component f I in the slip plane, which is ∑ ˆ+ f I = m I · σ σ J · bI. (2.8). J̸=I. where mI is the unit vector normal to the slip plane of dislocation I. This can also be written as f I = τ I bI where τ I is the resolved shear stress on dislocation I, deﬁned by ∑ bI ˆ+ τ I = m I · σ σJ · I b. (2.9). (2.10). J̸=I. When a dislocation moves, it generates phonons, which are elastic waves in the lattice. The dislocation feels these phonons as drag. Besides drag due to phonons, a dislocation can also experience drag due to electrons and impurity eﬀects. However, in metals phonon drag is dominating [6]. The sources for drag are combined in the drag coeﬃcient B, so that the net drag force on the dislocation is Bv I . For e.g. copper, the drag coeﬃcient is in the order of 10-4 Pa s [2]. We assume that drag is the only form of resistance and therefore the Peach– Koehler force should equal the drag force, so that the velocity of the dislocation is given by vI =. 9. fI B. (2.11).

(20) 2.2.2 Interactions between two dislocations For two dislocations on the same slip plane, i.e. x2 = 0, with the use of Equation (2.1) the Peach–Koehler force ˆ = 0) becomes on one of the dislocation due to the other in the absence of a (ˆ) ﬁeld (σ fI =. µbI bJ 1 2π(1 − ν) x1. (2.12). For two dislocations at mutual distance X, as shown in Figures 2.8(a) and (b), this yields fI = −. µbI bJ 1 2π(1 − ν) X. fJ =. (2.13a). µbJ bI 1 2π(1 − ν) X. (2.13b). Therefore two dislocations on the same slip plane repel each other when they are equally signed, i.e. their Burgers vectors have the same sign. On the other hand, two oppositely signed dislocations on the same slip plane attract each other. Interaction between two dislocations that are not on the same slip plane is not as trivial.. |fI|. I. J. |fJ|. I. |fI| |fJ|. J. X. X. (a) Interaction between two dislocations with a positive Burgers vector.. (b) Interaction between a dislocations with a positive Burgers vector and a dislocations with a negative Burgers vector.. Figure 2.8. 10.

(21) 2.2.3 Nucleation of dislocations One of the mechanisms to nucleate dislocations is by Frank–Read sources. Consider a dislocation line that is being pinned at points A and B, as shown in Figure 2.9(a). If a shear stress is applied on this segment, this dislocation line will bow out due to the Peach–Koehler force, as shown in Figure 2.9(b) (state 2). If the applied shear stress is high enough, points c and d come close to each other (state 4) and will eventually annihilate (state 5). After this process, there is a new closed loop around points A and B, which is the new dislocation, as well as a copy of the initial dislocation segment between A and B. This process is repeatable and therefore can be used as a source that nucleates new dislocations. The shear stress needed to nucleate a dislocation depends on lsrc , i.e. the length between A and B.. A. lsrc. τb. τb B. A. B (2). (1). slip plane A. τb. (3). τb. B. c. τb. d (4). (a) Schematic of a Frank–Read source. From [11].. (5). (b) Schematics of the dislocation line movement in a Frank–Read source. From [11].. Figure 2.9 A cross-section as indicated in Figure 2.10(a) gives a point source in two dimensions, as shown in Figure 2.10(b). A new dislocation line becomes two edge dislocations with opposite Burgers vector.. source. slip plane A. B. τ +b. -b lnuc. new dislocation line. lnuc. (b) The two-dimensional source with Burgers vector b nucleates a dislocation dipole at mutual distance of 1/2 lnuc . From [6].. (a) Cross-section of a Frank–Read source that has just nucleated a new dislocation.. Figure 2.10. 11.

(22) The Peach–Koehler force on source K is given by2.1 ( f. K. =m · K. ˆ+ σ. ∑. ) σ. J. · bK. (2.14). J. where mK is the unit vector normal to the slip plane of source K and bK the Burgers vector of the to be nucleated dislocations. This can also be written as f K = τ K bK where τ K is the resolved shear stress on source K, deﬁned by ( ) ∑ bK ˆ+ τ K = mK · σ σJ · K b. (2.15). (2.16). J. The

(23) K

(24) source is characterized by the critical nucleation stress τnuc . If the magnitude of the resolved shear stress

(25) τ

(26) exceeds τnuc for a period of time tnuc , called the nucleation time, the source nucleates a dislocation dipole, i.e. two dislocations with opposite Burgers vectors, with a certain distance lnuc between them. If τ K > 0, the right-hand side dislocation will have a Burgers vector of +b. If τ K < 0, i.e. the segment bowed out the other way, the left-hand side dislocation will have a Burgers vector of +b. Van der Giessen and Needleman proposed a practical criterion for the distance lnuc [6]. At the moment of nucleation, the left-hand side dislocation in Figure 2.10(b) feels a force to the right due to the right-hand side dislocation of (see Equation (2.13a)) f=. µb b 2π(1 − ν) lnuc. (2.17). Van der Giessen and Needleman proposed that the distance lnuc is chosen such that this force is canceled by τnuc b, i.e. the dipole is in static equilibrium if the resolved shear stress on the dipole is equal to the critical nucleation stress τnuc of the source, so that lnuc =. µb 2π(1 − ν)τnuc. (2.18). There are diﬀerent assumptions on the nucleation time tnuc . Van der Giessen and Needleman proposed a constant nucleation time [6], whereas Benzerga proposed a model for a source with critical nucleation stress [13] τnuc = β. µb lsrc. (2.19). where β is a line tension parameter. In the latter model, the nucleation time tnuc is given by the time needed to reach the critical shape of a Frank–Read source2.2 , which depends on the applied stress. Shishvan and Agnihotri reﬁned this model to account for the time it takes to go from the critical shape to a full loop [2,7,14]. Shishvan added a ﬁtting stress to account for the fact that the dislocation loop cannot always bow out freely, thus Equation (2.19) changes into [7] µb + τﬁtting lsrc. (2.20). Bηlsrc F (ζ) τnuc b. (2.21). τnuc = β The stress-dependent nucleation time is given by [7] tnuc =. where η is a character-dependent constant which is 1.5 for edge sources, ζ is the reduced resolved shear stress of source K deﬁned as ζ = τ K /τnuc and F is a fast decaying function with a singularity at ζ = 1. The exact form of F is stated in Appendix A, Equation (A.2). F has a minimum at ζ = 18.3. We will start out by using a constant nucleation time. If this assumption turns out to be too simple, we will start using the stress-dependent nucleation time. 2.1 Note 2.2 The. that the sum is over all present dislocations. critical shape of a Frank–Read source is the shape at which it starts to bow out (state 2 in Figure 2.9(b)).. 12.

(27) 2.2.4 Annihilation of dislocations Because two oppositely signed dislocations attract each other when they are on the same slip plane, they can come closely together. When their distance is less than the critical annihilation distance le , they will annihilate, leaving behind a normal crystal conﬁguration. In general two equally signed dislocations on the same slip plane will not meet, because they repel each other. However, if they would, their Burgers vectors are added up, becoming only one dislocation. Van der Giessen and Needleman proposed that le = 6 b.. 2.2.5 Obstacles When a dislocation moves along a slip plane, it can be hindered by diﬀerent kinds of obstacles, e.g. small precipitates, voids and dislocations on intersecting slip planes. These obstacles are modeled as point obstacles at which moving dislocations

(28) that

(29) pass an obstacle get pinned. A dislocation is released when the magnitude of the resolved shear stress

(30) τ I

(31) exceeds the obstacle strength τobs . Obstacles can lead to dislocation pile-ups. A pinned dislocation can prevent another dislocation from going any further. However, trailing dislocations can push a pinned dislocation over the obstacle. This process is illustrated in Figure 2.11.. τ. Source Source Obstacle Obstacle Dislocation Dislocation. τ. τ (a) Two dislocations are pinned at obstacles. τ. τ (b) A new dislocation dipole is nucleated. τ. (c) The dislocations at the obstacles prevent the other dislocations to move further. Two dislocation pile-ups are formed.. (d) A third dislocation dipole is nucleated and pushes the pinned dislocations over the obstacles.. Figure 2.11: Formation of pile-ups when a constant shear stress τ is applied.. 13.

(32) 3 Model 3.1 Overview of the model We model a two-dimensional cell with height h and width w that contains one slip system, i.e. all slip planes are parallel, that is parallel with the top and bottom of the cell. All the slip planes are near the middle of the cell and an uniform traction is imposed along the top and bottom. The height of the cell is chosen big enough in order that the (long ranged) stress ﬁeld of the dislocations at the top and bottom boundaries can be neglected compared to the stress ﬁeld due to the applied traction. The cell has periodic boundary conditions on the left-hand and right-hand side. These conditions imply that the displacement ﬁeld and stress ﬁeld are equal at both sides of the cell. Moreover, when a dislocation moves out of the cell at the left-hand side, it will appear at the right-hand side and vice versa. A schematic of this model is shown in Figure 3.1. Tapp. Tapp. Tapp. Tapp. Tapp. Tapp. Tapp. slip planes. h Tapp. Tapp. Tapp w. Figure 3.1: Schematic of a cell with periodic boundary conditions and uniform tractions at top and bottom. A traction Tapp is applied in the positive x1 -direction at the top boundary and in the negative x1 -direction at the bottom boundary. According to Equation (2.6a), the only non-zero of the stress ﬁeld σ at the boundaries is then σ12 . From now on we will call σ12 at the boundaries the applied shear stress τapp . The cell is discretized into a rectangular mesh with l1 elements in the x1 -direction and l2 elements in the x2 direction. Therefore, there are l1 + 1 nodes at both the top and bottom boundary. Since the most left and right node are actually the same because of the periodic boundary conditions, the traction Tapp applied to each node is given by {τ w app for the most left and right node 1 Tapp = τ2l (3.1) app w for the other nodes l1 To prevent the cell from moving, one node in the bottom boundary is being ﬁxed in the x1 -direction as well as the x2 -direction. Since the height of the cell is chosen big enough in order that the stress ﬁeld of the dislocations at the top and bottom boundaries can be neglected, σ ˆ12 ≈ τapp at the boundaries. The solution of this boundary value problem yields an almost uniform (ˆ) ﬁeld, i.e. σ ˆ12 is approximately τapp everywhere in the cell. Because the slip planes are parallel with the top and bottom of the cell, the applied shear stress τapp is also the resolved shear stress on every slip plane. The model is a numerical model and therefore it computes nsteps increments with a duration of tincr .. 14.

(33) 3.1.1 Strain rate The shear strain γ is deﬁned as γ=. u1 h. (3.2). where u1 is the average displacement in the x1 -direction of all the nodes at the top boundary. Therefore, the shear strain rate γ˙ is given by γ˙ =. ∆u1 htincr. (3.3). where ∆u1 is the diﬀerence in u1 between two increments. In this model the deformation is both elastic and plastic. However, since the elastic deformation at the boundary is only due to the applied shear stress, it is constant in time and therefore the shear strain rate is purely plastic. 3.1.1.1 Orowan strain rate Orowan proposed that the shear strain rate is given by [15] γ˙ ∗ = bρv. 3.1. (3.4). where ρ is the dislocation density and v is the mean dislocation velocity. In our model ρ = ndis /A, where ndis is the∑ number of present dislocations and A the area of the cell. By computing the mean dislocation velocity as v = J v J /ndis , Equation (3.4) can be written as ndis 1 ∑ J v A ndis J b ∑ J v = A. γ˙ ∗ = b. (3.5). J. 3.1.1.2 Dissipation strain rate Another deﬁnition of the strain∫ rate can be obtained when relating the macroscopic energy dissipation per unit of dislocation line length3.2 A τapp γ˙ † dA to the microscopic energy dissipation per unit of dislocation line ∑ length J f J v J .3.3 Combination with Equation (2.11) gives ∫ ∑ τapp γ˙ † dA = Bv J v J (3.6) A. J. In this way, when τapp is uniform, the dissipation strain rate γ˙ † can be deﬁned as γ˙ † =. B ∑ J J v v Aτapp. ∗ is added to avoid confusion with the strain rate of Equation (3.3). that the dislocation line is normal to the two-dimensional plane. 3.3 The † is added to avoid confusion with the strain rate of Equation (3.3). 3.1 The. 3.2 Note. 15. (3.7).

(34) 3.1.2 Application of the theory to the model Because we use only one slip system and σ ˆ12 ≈ τapp , the Peach–Koehler force on dislocation I in Equation (2.8) reduces to ∑ J I f I = σ ˆ12 + σ12 b J̸=I. . ∑. = τapp +. J I σ12 b. (3.8). J̸=I. Analogue, the Peach–Koehler force on source K in Equation (2.14) reduces to ) ( ∑ K J f = σ ˆ12 + σ12 bK J. ( =. τapp +. ∑. ) J σ12. bK. (3.9). J. The displacement ∆x1 I of dislocation I at an increment is given by ∆x1 I = v I tincr ) ( ∑ J bI τapp + J̸=I σ12 = tincr B. (3.10). The nucleation length as described in Equation (2.18) is based on a non-periodic cell. Because we have a periodic cell, we choose lnuc such that if the cell would contain only one source and the applied shear stress τapp is equal to the critical nucleation stress τnuc , at the moment of nucleation the dislocations would be in a static equilibrium, i.e. the Peach–Koehler force on either dislocations vanishes. The shear stress on dislocation I in Figure 3.2 due to the other dislocations with a positive Burgers vector cancel each other, see Equation (2.1). For the stresses on dislocation I due to the dislocations with a negative Burgers vector, Equation (2.3) reduces to J = σ12. µ(−b) π sin(2πξ) 2π(1 − ν) w 1 − cos(2πξ). (3.11). where ξ = X/w. We take X = lnuc and τapp = τnuc , so that the Peach–Koehler force from Equation (3.8) on dislocation I becomes ( ( nuc ) ) µb π sin 2π lw I f = τnuc − b (3.12) nuc 2π(1 − ν) w 1 − cos(2π lw ) With the aid of the identity, ( ) ( ) ( ) (x) 2 sin x2 cos x2 cos x2 sin(x) ( ) ( ) = = = cot 1 − cos(x) 2 sin x2 2 sin2 x2 The requirement f I = 0 can be reformulated as ( )) ( π lnuc µb cot π b 0 = τnuc − 2π(1 − ν) w w 16. (3.13).

(35) w. w. I x2. X. x1. Figure 3.2: Periodic cell with one Frank-Read source and one dislocation dipole. Hence, τnuc or. ( ) µb 1 lnuc = cot π 2(1 − ν) w w. ( ) 2(1 − ν)wτnuc w cot−1 π µb ( ) w µb = tan−1 π 2(1 − ν)wτnuc. lnuc =. (3.14). It can be shown that the Peach–Koehler force on the dislocations with a negative Burgers vector also vanishes when using the lnuc according to Equation (3.14).. 17.

(36) 3.2 Outline of the program The program is written in Fortran. Sections 3.2.1 and 3.2.2 will describe the outline of the program.. 3.2.1 Initialization From an input ﬁle the parameters as stated in Table 3.1 are imported. These parameters include material parameters, e.g. the elastic moduli, and model parameters, e.g. the applied shear stress. Next, the ﬁnite element mesh is created. Subsequently, information about the slip planes (positions), sources (slip plane, position, critical nucleation stress) and obstacles (slip plane, position, obstacle strength) is imported. h w l1 l2 E ν b. Height of the cell Width of the cell Number of elements in x1 -direction Number of elements in x2 -direction Young’s modulus Poisson’s ratio Magnitude of the Burgers vectors of the dislocations. B le tnuc τapp tincr nsteps. Drag coeﬃcient Critical annihilation distance Nucleation time Applied shear stress Duration of one increment Number of steps to be computed. Table 3.1: Parameters imported by the program The program goes through the loop described in Section 3.2.2 for a total of nsteps increments.. 3.2.2 The computational loop The program calculates the displacements of all present dislocations with the Peach–Koehler force at the present conﬁguration using Equation (3.10). After that, the positions of all dislocations are updated accordingly. If the dislocation leaves the cell at one side, it re-enters at the other side. A copy of this dislocation is put in the so called bucket. With this bucket the program can compute the displacements due to dislocations that have left the cell at some point in time. Next, the distances between neighbouring dislocations on the same slip plane are checked. When the dislocations have opposite sign and are less than the critical annihilation distance le apart, these dislocations are annihilated. After that, the program calculates the Peach–Koehler force f K on every source using Equation (3.9). If the Peach–Koehler force f K is larger or equal to the critical nucleation strength τnuc times bK , tincr will be added K K to tK elapsed . Otherwise, telapsed will be set to zero. When telapsed ≥ tnuc , a dislocation dipole will be created, each dislocation lnuc /2 from the nucleating source. In this model, the nucleation time tnuc is taken to be constant and 10 ns, as used in previous work by Hellinga [9]. Due to the generation of new dislocations, the distances between neighbouring dislocations on the same slip plane are checked once more. Again, when the dislocations are oppositely signed and less than the critical annihilation distance le apart, these dislocations are annihilated. Subsequently, the program calculates the displacements in the cell using the ﬁnite elements. The corresponding strain rate is calculated as in Equation (3.3), and written to a post-processing data ﬁle.. 18.

(37) 3.3 Results and discussion 3.3.1 One slip plane, one source We start out by considering a cell with one slip plane, a height h = 1 µm and width w = 4 µm, and Young’s modulus E = 70 GPA and Poisson’s ratio ν = 0.33, as used in previous work by Hellinga [9]. The slip plane is located halfway the height of the cell, as is shown in Figure 3.3. At the slip plane we take one source with a critical nucleation stress τnuc = 75 MPa and two obstacles, placed at 0.75 µm on either side of the source, with an obstacle strength τobs = 75 MPa. The source is placed just right of the center, to prevent dislocations reaching the sides of the cell at the same time.. h. 1.5 µm 4 µm. Figure 3.3: Cell with one source and two obstacles. 3.3.1.1 Strain rate cycle Until the applied shear stress reaches the so called yield stress, the strain rate is zero, because there is no dislocation motion. Figure 3.4 shows the strain rate versus time when a shear stress τapp = 75.4 MPa is applied, which is the yield stress of this speciﬁc cell. At state (1) there is a peak because a dipole is nucleated. At state (2) the dipole gets pinned, causing the strain rate to drop to zero, but since the Peach–Koehler force on the dipole is high enough to be unpinned, the strain rate returns to its value before the pinning. Between state (2) and (3) the strain rate increases, because the attractive force between the dislocations decreases as the dislocations move away from each other. When their distance is w/2 = 2 µm they even start to attract each other in the direction of the their movement. At state (4) the dislocations annihilate, causing the strain rate to drop to zero. After a while (approximately the nucleation time tnuc ), this process starts again. This behavior is periodic and we call one period, e.g. from state (1) to the next state (1), a strain rate cycle. To get an average strain rate that corresponds to an applied shear stress, we take the average of one strain rate cycle. 5 5 x 10 (1). (2). (3). 220 Time (ns). 230. Strain rate (1/s). 4 3 2 1 0 200. 210. Figure 3.4: Strain rate versus time at τapp = 75.4 MPa. 19. 240.

(38) Strain r. MJN. − τBQQ →τZJFME. 10. (γ) ˙ =. MJN. + τBQQ →τZJFME. (γ) ˙. 5IJT TUFQ GVODUJPO JT OPU JO BHSFFNFOU XJUI B QPXFS MBX XIJDI JT B DPOUJOVPVT GVODUJPO. 5 0. 80. 120 160 200 Applied shear stress (MPa). 240. 3.3.1.2 Strain rate versus shear stress 'JHVSF "WFSBHF TUSBJO SBUF WFSTVT BQQMJFE TIFBS TUSFTT 5IF TUSBJO SBUF JT MJNJUFE CZ b/tOVD h We compute the strain rates at various levels of the applied shear stress. Figure 3.5 shows the average strain 8F stress TFF UIBU rate versus applied shear τappUIF . TUSBJO SBUF DPOWFSHFT UP B NBYJNVN TUSBJO SBUF BOE UIFSFGPSF UIF DVSWF EPFT OPU SFTFNCMF B QPXFSMBX 8IFO B EJTMPDBUJPO EJQPMF IBT OVDMFBUFE BOE BOOJIJMBUFE JU IBT BMXBZT HJWFO B EJTQMBDFNFOU PG b 3 #FDBVTF UIF UJNF JT UBLFO UP CF DPOTUBOU UIF GSFRVFODZ PG OVDMFBUJPO JT MJNJUFE CZ 1/tOVD 5IFSFGPSF x 10OVDMFBUJPO GSPN 25 &RVBUJPO XF HFU UIBU UIF NBYJNVN TUSBJO SBUF JT. 20. 25.5. γ˙ NBY =. x 103. b. . tOVD h. Strain rate (1/s). Strain rate (1/s). *U BMTP MPPLT BT JG UIF TUSBJO SBUF JT EJTDSFUJ[FE 5IJT EJTDSFUJ[BUJPO WBOJTIFT XIFO UIF DIPTFO JODSFNFOU EVSBUJPO 25.3 tJODS JT15 SFEVDFE 8)"*+'*0&+0'&+ "U UIF ZJFME TUSFTT UIF TUSBJO SBUF JT B TUFQ GVODUJPO NFBOJOH UIBU. 10. 25.1. MJN. − τBQQ →τZJFME. (γ) ˙ =. MJN. + τBQQ →τZJFME. . (γ) ˙. 24.9. 5IJT TUFQ GVODUJPO JT OPU JO BHSFFNFOU XJUI B QPXFS MBX XIJDI JT B DPOUJOVPVT GVODUJPO 120. 5 0. 80. 140 160 Applied shear stress (MPa). 120 160 200 Applied shear stress (MPa). 240. Figure 3.5: Average strain rate versus applied shear stress. The strain rate is limited by b/tnuc h. The blue lines are the strain rates computed by Equation (3.16) for n = 201 to 206. We see that the strain rate converges to a maximum strain rate, and therefore the curve does not resemble a power-law. The origin of the asymptotic behaviour can be easily understood. Nucleation of a dipole (and subsequent annihilation) leads to a strain increment b/h. Because the nucleation time is taken to be constant, the frequency of nucleation is limited by 1/tnuc . Therefore, from Equation (3.3) we get that the maximum strain rate is γ˙ max =. b . tnuc h. (3.15). We also see that the strain rate is discretized. A closer look shows that the strain rate is discretized by γ˙ =. b ntincr h. (3.16). where n is a natural number. For this rather simple cell conﬁguration, n corresponds to the number of increments of one strain rate cycle. By reducing tincr , the curve becomes smoother, at the cost of longer computation times. At the yield stress, the strain rate is a step function, meaning that lim. − τapp →τyield. (γ) ˙ ̸=. lim. + τapp →τyield. (γ) ˙. (3.17). This step function is not in agreement with a power law, which is a continuous function. More on the yield stress is discussed in the next section.. 20.

(39) 3.3.1.3 Yield stress Liu et al. found that for a non-periodic cell with one source and two obstacles the yield stress is given by [16] if τ˜obs ≤ τ˜nuc − 1 (generating ﬁrst dipole) τ˜nuc if τ˜nuc − 1 ≤ τ˜obs ≤ τ˜nuc + 3 (pinning ﬁrst dipole) (3.18) τ˜yield = τ˜obs + 1 τ˜nuc + 4 if τ˜nuc + 3 ≤ τ˜obs ≲ 2˜ τnuc + 4 (generating second dipole)3.4 where dobs is the distance between the obstacles and τ˜ is a normalized stress, deﬁned by τ˜ = 2π(1 − ν). τ dobs µ b. (3.19). where τ is the unnormalized stress. Figure 3.6 shows the yield stress versus the obstacle strength at a normalized critical nucleation stress τ˜nuc = 23.2.. 35 τ~yield 30. τ~nuc=23.2. DD results Continuum pileup model. τ~yield=τ~nuc+4. Discrete pileup model 25. τ~yield=τ~obs+1. N=0 N=1 τ~yield=τ~nuc 40. N=2. N=3. 80 τ~obs. 120. Figure 3.6: Yield strength versus obstacle strength at τ˜nuc = 23.2. N is the number of dislocations in both pileups. From [16]. Since the results in Figures 3.4 and 3.5 are for τnuc = τobs , the normalized yield stress would be τ˜yield = τ˜obs + 1 in a non-periodic cell. For a periodic cell, this equation changes. If τapp = τyield , the Peach–Koehler force f I on pinned dislocation I should be equal to the force needed to be unpinned, which is τobs b, see Figure 3.7. w. w. I x2. x1. dobs. w–dobs. Figure 3.7: Periodic cell with one Frank-Read source and one dislocation dipole that is pinned at obstacles. 3.4 The. exact upper bound of this interval can be found in [16]. For τ˜nuc > 20, the approximation error is less than 0.2.. 21.

(40) Using Equations (2.3), (3.8) and (3.13), we ﬁnd τobs bI = f I = τyield +. ∑. J I σ12 b. J̸=I. or. (. µb 1 τobs b = τyield b − cot 2(1 − ν) w and hence τyield. µb 1 cot = τobs + 2(1 − ν) w. (. πdobs w. πdobs w. ) b. ) (3.20). or, in terms of the normalized stress from Equation (3.19), τ˜yield. πdobs cot = τ˜obs + w. (. πdobs w. ) (3.21). Note that Equation (3.21) is only valid for 0 < dobs ≤ w/2, because otherwise τyield < τnuc , which means that a dipole will not be nucleated in the ﬁrst place. Therefore, for dobs > w/2, the yield stress is given by τyield = τnuc when τnuc = τobs . 6TJOH &RVBUJPOT BOE XF mOE For dobs = 1.5 µm and w = 4 µm, Equation (3.21) gives τ˜y = τ˜obs + 0.49. To verify this result, we compute &RVBUJPOT shear stresses. BOE . XF mOE the applied the strain rate6TJOH for diﬀerent applied Figure 3.8 shows the average strain rate versus I I J I τZJFME + σ12 b τPCT b = f = shear stress. J=I J I bI =which f I = isthe + nucleation σ12 τPCTMPa, b stress of the source. τZJFMEcritical The ﬁrst time a dipole is nucleated is at τapp = 75.0 PS J=I 3.5 One would expect that the dipole would be pinned at τapp = 75.025 MPa , because lnuc is chosen such that the dislocations = τnuc . Nevertheless, does µb 1a dipoleπd PCT not PSare in a static equilibrium when the applied stressτ τappb = b τ b − DPU PCT is at τZJFME = 75.4 MPa. Therefore, the yield get pinned until τapp = 75.075 MPa. The ﬁrst time a dipole is released app w 2(1 −ν) w 1 should πdPCT stress τyield = 75.4 MPa. However, according toτEquation (3.21), the µb yield stress be atb75.52 MPa. DPU PCT b = τZJFME b − BOE IFODF w 2(1 − ν) w x 103 µb 1 πdPCT BOE IFODF τ = τ + DPU ZJFME PCT First w 2(1 −ν) w 15 µb 1 πdPCT dipole τ = τPCT + DPU PS JO UFSNTreleased PG UIFZJFME OPSNBMJ[FE TUSFTT 2(1 −GSPN ν) w&RVBUJPOw . . Strain rate (1/s). πdPCT πdPCT PS JO UFSNT PG UIF OPSNBMJ[FE TUSFTT GSPN &RVBUJPO . DPU τ˜BQQ = τ˜PCT + First w w 10 πdPCT πdPCT dipole τ ˜ DPU = τ ˜ + . BQQ PCT /PUF UIBU &RVBUJPO w GPS 0 <wdPCT ≤ w/2 CFDBVTF PUIFSXJTF τZJFME < τOVD XI nucleated JT POMZ WBMJE UIBU B EJQPMF XJMM OPU CF OVDMFBUFE JO UIF mSTU QMBDF 5IFSFGPSF GPS dPCT > w/2 UIF ZJFME TUSFTT /PUF UIBU &RVBUJPO . JT POMZ WBMJE < dPCT ≤ w/2 CFDBVTF PUIFSXJTF τZJFME < τOVD XIJDI NFBOT τZJFME = τOVD XIFO τOVDGPS = 0τPCT UIBU B EJQPMF XJMM OPU CF OVDMFBUFE JO UIF mSTU QMBDF 5IFSFGPSF GPS dPCT > w/2 UIF ZJFME TUSFTT JT HJWFO CZ First 5 'PS τZJFME = τOVD XIFO τOVD d= τ = dipole N BOE w = N &RVBUJPO HJWFT τ˜Z = τ˜PCT

(41) 5P WFSJGZ UIJT SFTVMU XF PCT PCT pinned UIF TUSBJO SBUF GPS EJóFSFOU BQQMJFE TIFBS TUSFTTFT 'JHVSF TIPXT UIF BWFSBHF TUSBJO SBUF WFSTVT U 'PS dPCT = NTIFBS BOE wTUSFTT = N &RVBUJPO HJWFT τ˜Z = τ˜PCT

(42) 5P WFSJGZ UIJT SFTVMU XF DPNQVUF UIF TUSBJO SBUF GPS EJóFSFOU BQQMJFE TIFBS TUSFTTFT 'JHVSF TIPXT UIF BWFSBHF TUSBJO SBUF WFSTVT UIF BQQMJFE TIFBS TUSFTT 5IF mSTU UJNF B EJQPMF JT OVDMFBUFE JT BU τBQQ = .1B XIJDI JT UIF DSJUJDBM OVDMFBUJPO TUSFTT PG UI 0 0OF XPVME FYQFDU UIBU UIF EJQPMF XPVME CF QJOOFE BU τ = .1B CFDBVTF l JT DIPTFO 75 75.2 75.4 75.6 75.8 76 BQQ OVD 5IF mSTU UJNF B EJQPMF JT OVDMFBUFE JT shear BU BτBQQ = FRVJMJCSJVN .1B XIJDI UIFBQQMJFE DSJUJDBM TUSFTT OVDMFBUJPO UIF TPVSDF B EJQPMF Applied stress (MPa) UIF EJTMPDBUJPOT BSF JO TUBUJD XIFOJTUIF τBQQ =TUSFTT τOVD PG /FWFSUIFMFTT 0OF XPVME UIBU UIF EJQPMF XPVME CFstress QJOOFE τBQQmSTU = UJNF SFMFBTFE CFDBVTF lJTOVD DIPTFO TVDI UIBU QJOOFE VOUJM τBQQ = .1B BU JTτBQQ = .1B 5IFSFGPSF Figure 3.8:FYQFDU StrainHFU rate versus applied shear for BU a5IF cell with heightB EJQPMF h.1B = 1 JTµm. UIF EJTMPDBUJPOT BSFTUSFTT JO B TUBUJD FRVJMJCSJVN XIFO UIF BQQMJFE TUSFTT τ = τ /FWFSUIFMFTT B EJQPMF EPFT BQQ OVD τZJFME = .1B )PXFWFS BDDPSEJOH UP &RVBUJPO UIF ZJFME TUSFTT TIPVME CFOPU BU .1 3.5 Actually at a applied HFU QJOOFE .1B mSTUbut UJNF B EJQPMF JT inSFMFBTFE JT BU τMPa. shear VOUJM stress τinﬁnitesimal greater than5IF 75 MPa, we applied τapp steps of 0.025 BQQ = BQQ = .1B 5IFSFGPSF UIF ZJFME 3 TUSFTT τZJFME = .1B )PXFWFS BDDPSEJOH UP &RVBUJPO . UIF ZJFME TUSFTT TIPVME CF BU .1B x 10. 15. 22. 15 First dipole released. 1/s). x 103. 10. First dipole released First. d. τ~app=. d.

(43) This discrepancy between theory and results can be explained when looking at the (ˆ) ﬁelds. Figure 3.9 shows a contour plot of σ ˆ12 at an applied shear stress τapp = 75 MPa, just as the dislocations are nucleated. Apparently, the stress ﬁeld of the dislocation are not negligible at the top and bottom boundary, because σ ˆ12 is not constant at the boundary. Consequently, σ ˆ12 is not uniform inside the cell, as is assumed in e.g. Equation (3.8). In other words, the height of the cell is not large enough to get an uniform σ ˆ12 inside the cell. Moreover, σ ˆ12 < τapp at the position of the dislocations, causing the dislocations to move to each other and eventually annihilate and not get pinned at the obstacles. 1. (MPa) 75.06 75.04 75.02 75.00 74.98 74.96 74.94 74.92 74.90. x2 (µm). 0.8 0.6 0.4 0.2 00. 0.5. 1. 1.5. 2 x1 (µm). 2.5. 3. 3.5. 4. Figure 3.9: Contour plot of σ ˆ12 in MPa at an applied shear stress τapp = 75 MPa with two dislocations nucleated from at source at x1 = 2.1 µm in a cell with height h = 1 µm. Figure 3.10 shows the yield stress versus the cell height.3.6 The yield stress converges to the theoretical yield stress found in Equation (3.21) when h ≥ 6 µm. But, as shown in the contour plot in Figure 3.11, σ ˆ12 still is not 6TJOHthrough &RVBUJPOT . . mOE uniform a cell of . heightBOE h= 6 µm.XF However, at the position of the dislocations, σ ˆ12 is approximately Computations with a cell height τapp and therefore the yield stress does match the theoretical yield stress. h = 12 µm show that σ ˆ12 deviates less thanI2 kPaIfromτapp , everywhere the cell. J inside σ12 τPCT b = f = τZJFME + bI J=I. PS. Yield stress (MPa). BOE IFODF. 75.8. 75.7. 75.6. πdPCT w. . πdPCT w. . µb 1 τPCT b = τZJFME b − DPU 2(1 − ν) w. . µb 1 DPU 2(1 − ν) w. . τZJFME = τPCT +. b. . PS JO UFSNT PG UIF OPSNBMJ[FE TUSFTT GSPN &RVBUJPO . 75.5 τ˜ZJFME. πdPCT DPU = τ˜PCT + w. . πdPCT w. . . 75.4JT POMZ WBMJE GPS 0 < dPCT ≤ w/2 CFDBVTF PUIFSXJTF τZJFME < τOVD XIJDI NFBOT /PUF UIBU &RVBUJPO . UIBU B EJQPMF XJMM OPU CF OVDMFBUFE JO UIF mSTU QMBDF 5IFSFGPSF GPS dPCT > w/2 UIF ZJFME TUSFTT JT HJWFO CZ 4 2 6 8 10 τZJFME = τOVD XIFO τOVD = τPCT0 6TJOH &RVBUJPOTHeight cell (µm) BOE XF mOE 'PS dPCT = N BOE3.10: w =Yield N &RVBUJPO HJWFThτ˜and τ˜PCTtheoretical

(44) 5P UIJT SFTVMU XF DPNQVUF Figure stress versus . cell height yieldWFSJGZ stress. Z = the &RVBUJPOT TIFBS TUSFTTFT BOE . XF mOE UIF BQQMJFE UIF TUSBJO SBUF6TJOH GPS EJóFSFOU BQQMJFE 'JHVSF TIPXT UIF BWFSBHF TUSBJO SBUF WFSTVT I I J I τZJFME + σ12 b τPCT b = f = TIFBS TUSFTT J=I I J I bcell. =XIJDI f I = JTUIF + OVDMFBUJPO σ12 τ the.1B b TUSFTT PG UIF TPVSDF τZJFMEDSJUJDBM 3.65IF UJNF EJQPMF JT OVDMFBUFE JT BU the τBQQheight = Note mSTU that the slip Bplane is always located halfway of PCT PS J = I 0OF XPVME FYQFDU UIBU UIF EJQPMF XPVME CF QJOOFE BU τBQQ = .1B CFDBVTF lOVD JT DIPTFO TVDI UIBU UIF EJTMPDBUJPOT = τOVD /FWFSUIFMFTT EPFT µb 1B EJQPMFπd PCT OPU PSBSF JO B TUBUJD FRVJMJCSJVN XIFO UIF23BQQMJFE TUSFTTτ τBQQb = b τ b − DPU PCT JT BU τZJFME = .1B 5IFSFGPSF UIF ZJFME HFU QJOOFE VOUJM τBQQ = .1B 5IF mSTU UJNF B EJQPMF JT SFMFBTFE BQQ w 2(1 −ν) w 1 TIPVME πdPCT TUSFTT τZJFME = .1B )PXFWFS BDDPSEJOH UPτ&RVBUJPO UIF µb ZJFME TUSFTT CF BUb .1B DPU PCT b = τZJFME b − BOE IFODF w 2(1 − ν) w x 103 µb 1 πdPCT BOE IFODF τ = τ + DPU ZJFME PCT First.

(45) x2 (µm). 6. (MPa) 75.52 75.49 75.46 75.43 74.40 74.37 74.34 74.31 74.28. 4. 2. 00. 0.5. 1. 1.5. 2 x1 (µm). 2.5. 3. 3.5. 4. Figure 3.11: Contour plot of σ ˆ12 in MPa at an applied shear stress τapp = 75.4 MPa with two dislocations pinned at obstacles in a cell with height h = 6 µm.. 24.

(46) 3.3.2 One slip plane, three sources In order to see the inﬂuence of diﬀerent obstacle strengths, we put the cell of Section 3.3.1 three times in a row, as shown in Figure 3.12. In contrast with the cell of Section 3.3.1, the obstacle strengths are drawn from a Gaussian distribution with mean ⟨τobs ⟩ = 1.5 τnuc and standard deviation SDτobs = 0.5 τnuc , where τnuc is still 75 MPa. The cell has a height h = 20 µm. 4 µm h 1.5 µm 12 µm. Figure 3.12: Cell with three sources and six obstacles. The height of the obstacles correlate with the obstacle strengths drawn from a Gaussian distribution. Figure 3.13 shows the average strain rate versus applied shear stress τapp . We see a higher yield stress, because at all times each obstacle has a dislocation pinned3.7 , as shown in Figure 3.14.. x 102 40 35 Strain rate (1/s). 30 25. 20 15. 10 5 0 78 78.5 79 79.5 80 80.5 81 81.5 82 82.5 Applied shear stress (MPa) Figure 3.13: Average strain rate versus applied shear stress.. 3.7 Corresponding. to the N = 1 regime of Figure 3.6.. 25.

(47) x2 (µm) Strain rate (1/s). 20 16 12 8 4 0 12. 0 2 3 x 10. 4. 6 x1 (µm). 8. 10. 12. 8 4 0. 32 34 36 38 40 42 44 46 48 50 Time (ns). Strain rate (1/s). x2 (µm). (a) Dislocations areTime pinned(ns) at all the obstacles. 20 16 12 8 4 0 12. 0 2 3 x 10. 4. 6 x1 (µm). 8. 10. 12. 8 4 0. 38 40 42 44 46 48 50 52 54 56 Time (ns). Strain rate (1/s). x2 (µm). (b) Nucleated dislocations push the pinned dislocations over the obstacles and become pinnedTime themselves. (ns) 20 16 12 8 4 0 12. 0 2 3 x 10. 4. 6 x1 (µm). 8. 10. 12. 8 4 0. 42 44 46 48 50 52 54 56 58 60 Time (ns). (c) The unpinned dislocations will annihilate with each other. Figure 3.14: Positions of the dislocations (top) and the strain rate at that moment (bottom). The shown process is periodical.. 26.

(48) 3.3.2.1 Orowan and dissipation strain rate When we compare the strain rate γ˙ computed by the program with the Orowan strain rate γ˙ ∗ and the dissipation strain rate γ˙ † , deﬁned in Equations (3.5) and (3.7) in Section 3.1.1, we see that the Orowan strain rate agrees nicely with the program’s strain rate, as shown in Figure 3.15. The area under the curve, and therefore the average strain rate, diﬀers 0.29 %. The dissipation strain rate does not follow the program’s strain rate quite as nice, but still the area under the curve diﬀers only 0.54 %.. x 102 25. Strain rate (1/s). 20 15 10 5 0 55. 60. 65. 70 75 Time (ns). 80. 85. 90. (a) Program’s strain rate γ˙ and Orowan strain rate γ˙ ∗ versus time. x 102. 25. Strain rate (1/s). 20 15 10 5 0 55. 60. 65. 70 75 Time (ns). 80. 85. 90. (b) Program’s strain rate γ˙ and dissipation strain rate γ˙ † versus time. Figure 3.15. 27.

(49) 3.4 Bucket time The FEM part of the model takes a signiﬁcant part of the computing time. The ﬁnite elements are used to calculate the (ˆ) ﬁelds and the displacements of the cell, and with that the strain rate. The (ˆ) ﬁelds don’t have to be computed, because it can be imposed that σ ˆ12 = τapp , provided that σ ˜12 can be neglected at the top and bottom boundary compared to the applied shear stress. If there is a way to calculate the strain rate without having to use the ﬁnite elements, computation times could be reduced drastically. When a dislocation reaches a side of the cell and after some time tbucket another dislocation reaches the same side of the cell, the total distance travelled by dislocations on that slip plane is the width of the cell, as shown in Figure 3.16. Therefore, this process can be seen as one dislocation that travelled a distance w in some time tbucket , so that the mean dislocation velocity v = w/tbucket . The dislocation density for one dislocation is given by ρ = 1/A, where A is the area of the cell. Since we consider only one slip plane, we deﬁne A = dslip w, where dslip is the spacing between slip planes. Therefore, for one slip plane, Equation (3.4) reduces to 1 w A tbucket b = dslip tbucket. γ˙ L = b. (3.22). where γ˙ L is the strain rate of slip plane L and tbucket is the bucket time, deﬁned as the time interval between two dislocations at slip plane L going into the bucket.3.8 In this way, the total strain rate of multiple slip planes is given by the mean of the strain rate of each individual slip plane, i.e. γ˙ =. 1 ∑ L γ˙ nslip. (3.23). L. where nslip is the number of slip planes.. w. w. (a) A positive dislocation reaches the side of the cell. w. (b) The left-hand source nucleates a dipole. w. (c) The two most left dislocations annihilate. w. (d) The two demeaning dislocations annihilate. w. (e) The right-hand source nucleates another dipole.. (f) The dislocation reaches the side of the cell. The total distance covered by dislocations is w.. Figure 3.16: Example of dislocations that reach the bucket. The red line indicates the distance travelled by the dislocations from the moment that the ﬁrst dislocation reaches the bucket.3.9 3.8 When a dislocation reaches the side of the cell, a copy goes into the bucket. When two dislocations of opposite sign reach the opposite sides of the cell at the same time, this counts as only one dislocation going into the bucket. 3.9 Note that the result (traveling a distance w) does not depend on which source did or did not nucleate. A negative dislocation that reaches the bucket instead of a positive dislocation would give the same result.. 28.

(50) To see whether we can use the bucket time and multiple slip planes, we consider a cell with ﬁve slip planes with a spacing dslip = 1 µm and a width w = 12 µm. Five sources are placed randomly at every slip plane. The critical nucleation stresses are drawn from a Gaussian distribution with mean ⟨τnuc ⟩ = 75 MPa and standard deviation SDτnuc = 0.1 ⟨τnuc ⟩. In addition, eight obstacles are placed randomly at every slip plane. Their obstacle strengths are drawn from a Gaussian distribution with mean ⟨τobs ⟩ = 100 MPa and standard deviation SDτobs = 0.1 ⟨τobs ⟩. We compute the strain rate by the program and compare it to the strain rate computed according to Equation (3.22). Figure 3.17 shows both strain rates versus time. Since we are only interested in the average strain rate, we do a 501-point moving average over the strain rate computed by the program to eliminate the strong ﬂuctuations. We see a nice agreement between the moving average and the strain rate computed using the bucket times.. Strain rate (1/s). x 103 35 30 25 20 100. 101. Strain rate (1/s). 3 40 x 10. 102 103 Time (µs). 104. 105. 30. 20 0. 25. Bucket time Program (its moving average) 50 75 100 125 150 175 200 225 250 Time (µs). Figure 3.17: Strain rate calculated by the program and its 501-point moving average and strain rate calculated using the bucket time versus time at an applied stress of 80 MPa. There is no clear strain rate cycle anymore and the behavior of the strain rate at the ﬁrst few steps is diﬀerent from the rest. Therefore, in order to get an average strain rate that corresponds to an applied shear stress, we average the strain rate per slip plane from the last computed step and go back. This method is called backward averaging and is demonstrated in Figure 3.18. When the average strain rate per slip plane converges, we know the average strain rate per slip plane and therefore the total average strain rate.. 29.

(51) x 102. Average strain rate (1/s). 220 216 212 208 204. 0. 0.5. 1. 1.5. 2. 2.5. Time (µs) Figure 3.18: Averaged strain rate per slip plane versus the time over which the strain rate is backward averaged at 80 MPa. Each colour corresponds to a slip plane.. 30.

(52) 4 Improved model 4.1 Overview of the improved model We found a way to compute the strain rate without having to use the ﬁnite elements by using the bucket times, as described in Section 3.4. Provided that σ ˜12 can be neglected at the top and bottom boundary compared to the applied shear stress, we impose that σ ˆ12 = τapp anywhere in the cell. Therefore, the ﬁnite elements are no longer needed and can be removed from the model.. 4.2 Outline of the program In general the program is the same as the program of the old model, as described in Section 3.2. From now on, we will call the old model the FEM model. In the improved model, during initialization there is no ﬁnite mesh created. In order to use Equation (2.21) to calculate the stress-dependent nucleation time, the program calculates the lsrc that corresponds to τnuc as described in Appendix A. In the computational loop the program no longer calculates the displacements in the cell using the ﬁnite elements. Instead, each slip plane has a timer. If a dislocation goes into the bucket, the value of the timer of the corresponding slip plane is written to a post-processing data ﬁle and that timer is set to zero.. 31.

(53) 4.3 Results and discussion 4.3.1 Five slip planes. . 4.3.1.1 Comparison with the FEM model. 3FTVMUT BOE EJTDVTTJPO. Figure 4.1 shows the average strain rate versus the applied shear stress for the cell as described in Section 3.4. We only see a small deviation with the strain rate of the FEM model at 70 MPa. The strain rate still converges $PNQBSJTPO UIF (3.15), '&. NPEFM to γ˙ max as inXJUI Equation but with h replaced with dslip . The computation with the FEM model takes 52 hours and 44 minutes on four CPUs, whereas the computation with the improved model takes only 1 hour TIPXT UIF and BWFSBHF TUSBJO SBUF UIF BQQMJFE TIFBS TUSFTT GPS UIF DFMM BT EFTDSJCFE JO 4FDUJPO 8F 2 minutes, whichWFSTVT is a factor 51. POMZ TFF B TNBMM EFWJBUJPO BU .1B 5IF TUSBJO SBUF TUJMM DPOWFSHFT UP γ˙ NBY BT JO &RVBUJPO CVU XJUI h 3 SFQMBDFE XJUI dTMJQ 5IF DPNQVUBUJPO XJUI UIF UBLFT IPVST BOE NJOVUFT PO GPVS $16T XIFSFBT x 10NPEFM UIF DPNQVUBUJPO XJUI UIF JNQSPWFE NPEFM UBLFT POMZ IPVS BOE NJOVUFT XIJDI JT B GBDUPS 25. Strain rate (1/s). 20. b tOVD dTMJQ. . 15. Strain rate (1/s). 10 FEM model 5 0 60. Improved model. 65 70 75 80 85 90 Applied shear stress (MPa) FEM model. 95. Figure 4.1: Average strain rate of the improved model and the FEM model versus applied shear stress. Improved model. 60 'JHVSF "WFSBHF TUSBJO SBUF PG UIF JNQSPWFE NPEFM BOE UIF '&. NPEFM WFSTVT BQQMJFE TIFBS TUSFTT. 3FEVDJOH UIF TQBDJOH CFUXFFO TMJQ QMBOFT *O QSFWJPVT %%1 TJNVMBUJPOT B TQBDJOH dTMJQ ∼ b o b BOE B TPVSDF EFOTJUZ ρOVD ∼ o N XFSF VTFE < > 5IFSFGPSF XF UBLF B DFMM XJUI dTMJQ = N w = N nOVD = BOE nPCT = XIFSF nOVD BOE nPCT BSF UIF OVNCFS PG TPVSDFT BOE PCTUBDMFT SFTQFDUJWFMZ 5IFSFGPSF ρOVD = N 5IF TPVSDFT BOE PCTUBDMFT BSF QVU SBOEPNMZ BU UIF TMJQ QMBOFT BOE IBWF UIF TBNF EJTUSJCVUJPO PG DSJUJDBM OVDMFBUJPO TUSFTT BOE PCTUBDMF TUSFOHUIT BT VTFE JO 4FDUJPO TIPXT UIF BWFSBHF TUSBJO SBUF QFS TMJQ QMBOF BOE UIF UPUBM TUSBJO SBUF WFSTVT UIF BQQMJFE TIFBS TUSFTT 5IF DVSWFT PG UIF TUSBJO SBUF BSF OPU BT TNPPUI BT JO FH CFDBVTF UIF NPNFOU PG EJTMPDBUJPOT PO B QBSUJDVMBS TMJQ QMBOF JT JOnVFODFE CZ EJTMPDBUJPOT PO PUIFS TMJQ QMBOFT 8F BMTP TFF B MBSHF EJQ PG UIF TUSBJO SBUF BU τBQQ = .1B JO TIPXT UIF TUSBJO SBUF WFSTVT TUFQT "GUFS BQQSPYJNBUFMZ NJMMJPO TUFQT UIF TUSBJO SBUF PG POF PG UIF TMJQ QMBOFT ESPQT UP [FSP "QQBSFOUMZ EJTMPDBUJPOT PO UIF PUIFS TMJQ QMBOF JOnVFODF UIF NPWFNFOU PG UIF EJTMPDBUJPOT PO UIJT TMJQ QMBOF DBVTJOH UIFN UP OPU SFBDI 32 UIF CVDLFU BOZNPSF.

(54) 4.3.1.2 Reducing the spacing between slip planes In previous DDP simulations, a spacing dslip ∼ 100 b – 200 b and a source density ρnuc ∼ 10 – 120 µm-2 were used [8, 17, 18]. Therefore, we take a cell with dslip = 0.05 µm, w = 8 µm, nnuc = 25 and nobs = 50, where nnuc and nobs are the number of sources and obstacles, respectively. Therefore, ρnuc = 12.5 µm-2 . The sources and obstacles are put randomly at the slip planes and have the same distribution of critical nucleation stresses and obstacle strengths as used in Section 4.3.1.1. Figure 4.2 shows the average strain rate per slip plane and the total strain rate versus the applied shear stress. The curves of the strain rate are not as smooth as in e.g. Figure 4.1, because the moment of dislocations on a particular slip plane is inﬂuenced by dislocations on other slip planes.. x 104 45. Total strain rate. 40 Strain rate (1/s). 35 30 25 20 15 10 5 0 65. 70. 75 80 85 90 95 Applied shear stress (MPa). 100. Figure 4.2: Average strain rate per slip plane (colour) and total strain rate (black) versus applied shear stress. We also see a large dip of the strain rate at τapp = 79 MPa in Figure 4.2. Figure 4.3 shows the strain rate versus steps. After approximately 18 µs, the strain rate of one of the slip planes drops to zero. Apparently dislocations on the other slip plane inﬂuence the movement of the dislocations on this slip plane, causing them to not reach the bucket anymore. Since the constant nucleation time limits the average strain rate and the curve still does not resemble a power law, we will use the stress-dependent nucleation time as stated in Equation (2.21) in the next section.. 33.

(55) 5 18 x 10. Strain rate (1/s). 15 12 9 6 3 0. 0. 5. 10. 15. 20 Time (µs). 25. 30. 35. 40. Figure 4.3: Strain rate per slip plane versus time at 79 MPa. Each colour corresponds to a slip plane. 4.3.1.3 Stress-dependent nucleation time Using the stress-dependent nucleation stress as stated in Equation (2.21) requires a ﬁtting stress τﬁtting . Agnihotri found that τﬁtting = 34.5 MPa for a material with Young’s modulus E = 110 GPa and Poisson’s ratio ν = 0.34. In order to use this ﬁtting stress, we have to use the same elastic moduli. However, the elastic moduli inﬂuence the strain rate, because the stress ﬁelds of the dislocations depend on the elastic moduli. Therefore, we compute the cell of Section 4.3.1.2 with the new elastic moduli using both a constant nucleation time and a stressdependent nucleation time. Figures 4.4(a) and (b) show the average strain rate per slip plane and the total strain rate versus the applied shear stress for a constant nucleation time and a stress-dependent nucleation time, respectively. The step function4.1 in the strain rate at the yield stress vanishes for the stress-dependent nucleation time and the behaviour is almost linear. Because dislocations inﬂuence the movement of dislocations on other slip planes, the slip rates depend on the order of the slip planes. Figure 4.5 shows the total strain rate versus the applied shear stress for three diﬀerent orders.. 4.1 As. deﬁned in Equation (3.17).. 34.

(56) 45. x 104. 40 Strain rate (1/s). 35 30 25 20 15 10. Total strain rate. 5 0 65. 70. 75 80 85 Applied shear stress (MPa). 90. (a) Contant nucleation time.. x 104. Strain rate (1/s). 20 15 10 5. Total strain rate. 0 65. 70. 75 80 85 Applied shear stress (MPa). 90. (b) Stress-dependent nucleation time. Figure 4.4: Average strain rate per slip plane (colour) and total strain rate (black) versus applied shear stress.. x 104. Strain rate (1/s). 20 15. 1-4-3-2-5. 10 4-2-3-1-5 5. 1-2-3-4-5. (black curve in Figure 4.4(b)). 0 65. 70. 75 80 85 Applied shear stress (MPa). 90. Figure 4.5: Total strain rate for conﬁgurations with a diﬀerent order of slip planes versus applied shear stress for a stress-dependent nucleation time.. 35.

(57) 4.3.2 Twenty-ﬁve slip planes We take a cell with dslip = 0.05 µm, w = 4 µm, nnuc = 54 and nobs = 104. Therefore, ρnuc = 10.8 µm-2 . The critical nucleation stresses are still drawn from a Gaussian distribution with mean ⟨τnuc ⟩ = 75 MPa and standard deviation SDτnuc = 0.1 ⟨τnuc ⟩ and the obstacle strengths with mean ⟨τobs ⟩ = 100 MPa and standard deviation SDτobs = 0.1 ⟨τobs ⟩. The strain rates for eight realizations with diﬀerent orders of slip planes versus the applied shear stress are plotted in Figure 4.6, along with the average of these strain rates. This curve might resemble a power law, but when looking at Figure 4.6(b) we see two domains with a diﬀerent slope, as indicated by the red dotted lines, for τapp < 67.5 MPa and τapp > 67.5 MPa.4.2 x 104 6. Strain rate (1/s). 5 4 3 2 1 0 60. 62. 64. 66 68 70 72 74 Applied shear stress (MPa). 76. 78. 80. (a) Normal plot. 105. Strain rate (1/s). 104. 103. 102 60. 62. 64. 66 68 70 72 74 Applied shear stress (MPa). 76. 78. 80. (b) Log-log plot. Figure 4.6: Strain rate versus applied shear stress for eight realizations with diﬀerent orders of slip planes (colour) and their average (black). The error bars denote the standard deviations. The red dotted lines indicate the two domains with a diﬀerent slope. 4.2 Note. that a straight line in a log-log plot is equivalent to a power law, where the slope is equivalent to the power.. 36.

(58) There are various ways to ﬁt a power law in the form of ( )m τ γ˙ = γ˙ 0 τ0. (4.1). as described in Appendix B. It is mathematically impossible to ﬁnd both γ˙ 0 and τ0 . Therefore, we use τ0 = ⟨τnuc ⟩.4.3 The strain rate is mostly determined by activated4.4 sources and the most sources get activated when their critical nucleation stress is ⟨τnuc ⟩ − SDτnuc ≤ τnuc ≤ ⟨τnuc ⟩. Therefore, we ﬁt only for ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩, which also makes it possible to compare ﬁts with diﬀerent SDτnuc . Figure 4.7 shows a minimum-variance unbiased (MVU) ﬁt. We ﬁnd for τ0 = 75 MPa that γ˙ 0 = 3.1 × 104 s-1 and m = 11.7. 7. x 104. Strain rate (1/s). 6 5 4 3 2 1 0 60. 62. 64. 66 68 70 72 74 Applied shear stress (MPa). 76. 78. 80. (a) Normal plot.. Strain rate (1/s). 104. 103. 60. 62. 64. 66 68 70 72 74 Applied shear stress (MPa). 76. 78. 80. (b) Log-log plot. Figure 4.7: MVU ﬁt of Figure 4.6 for the strain rate with ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩. 4.3 We. cannot use the yield stress for τ0 , because the experimental yield stress depends on the strain rate (see e.g. [4]). that nucleate dipoles.. 4.4 Sources. 37.

(59) 4.3.3 Fifty slip planes. Strain rate (1/s). We double the number of slip planes and keep the source and obstacle densities the same as in Section 4.3.2. The average strain rate of eight realizations with diﬀerent orders of slip planes versus the applied shear stress is plotted in Figure 4.8, along with a MVU ﬁt. We ﬁnd for τ0 = 75 MPa that γ˙ 0 = 3.4 × 104 s-1 and m = 16.9. x 104 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 60 62. 64. 66 68 70 72 74 Applied shear stress (MPa). 76. 78. (a) Normal plot.. Strain rate (1/s). 104. 103. 102. 101 60. 62. 64. 66 68 70 72 74 Applied shear stress (MPa). 76. 78. (b) Log-log plot. Figure 4.8: Average strain rate of eight realizations with diﬀerent orders of slip planes versus applied shear stress. The error bars denote the standard deviation. SDτnuc = 0.1 ⟨τnuc ⟩. The red line is a MVU ﬁt for the strain rate with ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩.. 38.

(60) 4.3.3.1 Other realizations of the random distributions We want to see whether the strain rate converges for diﬀerent realizations of the random distributions, i.e. the critical nucleation stresses, the obstacle strengths and the positions of sources and obstacles. Therefore, we average the strain rate of eight computations with diﬀerent orders of slip planes, for some realizations of the random distributions. Figure 4.9 shows the average strain rate for three diﬀerent realizations of the random distributions versus the applied shear stress, along MVU ﬁt. We ﬁnd for τ0 = 75 MPa that γ˙ 0 = 3.0 × 104 s-1 and m = 16.0. We see a small deviation between the diﬀerent curves. 3.5. x 104. Strain rate (1/s). 3 2.5 2 1.5 1 0.5 0. 68. 69. 70 71 72 73 Applied shear stress (MPa). 74. 75. 74. 75. (a) Normal plot.. Strain rate (1/s). 105. 104. 103. 68. 69. 70 71 72 73 Applied shear stress (MPa). (b) Log-log plot. Figure 4.9: Average strain rate versus applied shear stress. Each colour corresponds to the average strain rate of eight computations with diﬀerent orders of slip planes, where each colour has another realization of the random distributions. The green line is the black curve in Figure 4.8. The black line is the average of the colored curves. The error bars denote the standard deviation. SDτnuc = 0.1 ⟨τnuc ⟩. The red line is a MVU ﬁt for the strain rate with ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩.. 39.

(61) 4.3.3.2 Inﬂuence of SDτnuc. Strain rate (1/s). We take the same conﬁguration of Section 4.3.3 but change SDτnuc . Figure 4.10 shows the average strain rate of eight realizations with diﬀerent orders of slip planes versus the applied shear stress and a MVU ﬁt with SDτnuc = 0.2 ⟨τnuc ⟩. For τ0 = 75 MPa, we ﬁnd γ˙ 0 = 3.6 × 104 s-1 and m = 10.6. x 104 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 50. 55. 60 65 70 Applied shear stress (MPa). 75. (a) Normal plot.. Strain rate (1/s). 104. 103. 102. 101 50. 55. 60 65 70 Applied shear stress (MPa). 75. (b) Log-log plot. Figure 4.10: Average strain rate of eight realizations with diﬀerent orders of slip planes versus applied shear stress. The error bars denote the standard deviation. SDτnuc = 0.2 ⟨τnuc ⟩. The red line is a MVU ﬁt for the strain rate with ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩.. 40.

(62) Figure 4.11 shows the average strain rate of eight realizations with diﬀerent orders of slip planes versus the applied shear stress and a MVU ﬁt with SDτnuc = 0.05 ⟨τnuc ⟩. For τ0 = 75 MPa, we ﬁnd γ˙ 0 = 3.1 × 104 s-1 and m = 21.8. 3.5. x 104. Strain rate (1/s). 3 2.5 2 1.5 1 0.5 0. 71.5. 72. 72.5 73 73.5 74 Applied shear stress (MPa). 74.5. 75. 74.5. 75. (a) Normal plot.. Strain rate (1/s). 105. 104. 103. 71.5. 72. 72.5 73 73.5 74 Applied shear stress (MPa). (b) Log-log plot. Figure 4.11: Average strain rate of eight realizations with diﬀerent orders of slip planes versus applied shear stress. The error bars denote the standard deviation. SDτnuc = 0.05 ⟨τnuc ⟩. The red line is a MVU ﬁt for the strain rate with ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩.. 41.

(63) Figure 4.12 shows the average strain rate of eight realizations with diﬀerent orders of slip planes versus the applied shear stress and a MVU ﬁt with SDτnuc = 0.025 ⟨τnuc ⟩. For τ0 = 75 MPa, we ﬁnd γ˙ 0 = 2.8 × 104 s-1 and m = 15.8. 3.5. x 104. Strain rate (1/s). 3 2.5 2 1.5 1 0.5 0. 73.2 73.4 73.6 73.8 74 74.2 74.4 74.6 74.8 Applied shear stress (MPa). 75. (a) Normal plot. 104.6. Strain rate (1/s). 104.5 104.4 104.3 104.2 104.1. 73.2. 73.4. 73.6 73.8 74 74.2 74.4 74.6 74.8 Applied shear stress (MPa). 75. (b) Log-log plot. Figure 4.12: Average strain rate of eight realizations with diﬀerent orders of slip planes versus applied shear stress. The error bars denote the standard deviation. SDτnuc = 0.025 ⟨τnuc ⟩. The red line is a MVU ﬁt for the strain rate with ⟨τnuc ⟩ − SDτnuc ≤ τapp ≤ ⟨τnuc ⟩.. 42.

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