Supplementary Technical Document

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1 Supplementary Technical Document Qi Guo, Xuchen Han, Chuyuan Fu, Theodore Gast, Rasmus Tamstorf, Joseph Teran 1 FEM Force computation We compute forces on the control points x p by fp KL = ΨS F KL,Etr x KL x KL p = V 0 ψf KL,Etr q x q q x KL q p = q V 0 q ψ F FKL,Etr KL q x q : FKL,Etr q x KL x q, p where x q s are positions of the quadrature points. We give expressions for each FKL q x KL p k x q with fixed p, q and k, where k represents the x, y, or z direction. For simplicity of notation, we omit the subscripts p, q and superscript KL for now. Recall from the paper that we have where F = 3 g i ḡ i, with g α = a α + ξ 3 a 3,α, g 3 = a 3, i=1 a α = j Nj SD x j ξ 1, ξ 2, α = 1, 2 ξ α a 3 = a 1 a 2 a 3,α = I a 3 a 3 a 1,α a 2 + a 1 a 2,α = ã a 3 a 3 ã 1

2 in which we define ã to be Now we compute F. ã = a 1,α a 2 + a 1 a 2,α. F = 3 i=1 g i ḡ i, and where g α = a α + ξ 3 a 3,α g 3 = a 3 a α = N SD k ξ 1, ξ 2 ξ α e k summation convention does not apply here 1 and Finally, a 3 = a 1 a 2 + a 1 a 2 a 1 a 2 a 3 = a 3 a 1 a 2 + a 1 a 2, a 3,α = ã a 3 a 3 ã ã + a 3 a 3 a 3 ã, where ã = a 1,α a 2 + a 1,α a 2 + a 1 a 2,α + a 1 a 2,α a 1,α a 2 + a 1 a 2,α 2, in which a α,β = N SD k ξ 1, ξ 2 e k summation convention does not apply here. ξ β ξ α 2

3 2 Grid force computation The force on the MPM grid fi iii x computes as follows: f iii i x = p I iii = p I iii + p I iii χa pα ā pα + a E p3 ā p3 Vp 0 χa pα ā pα + a E p3 ā p3 : apα ā pα + a E p3 ā p3 F E a pβ χa pα ā pα + a E p3 ā p3 F E : apα ā pα + a E p3 ā p3 a E p3 Then, omitting the subscript p, we compute each term in the contraction: χa α ā α + a E 3 ā 3 F E = τ S a α ā α + a E 3 ā 3 T = τ S ã α ā α + ã 3 ā 3 where τ S is the Kirchhoff stress and ã α and ã 3 are the contravariant counterparts of a α and a E 3 respectively. And using index notation, we see that a α ā α + a E 3 ā 3 = a α i ā αj a β a βk = δ αβ δ ik ā αj = δ ik ā βj : a pβ Vp 0 x i : ae p3 x i V 0 p. Similarly, a α ā α + a E 3 ā 3 a E = δ ik ā 3j 3 Hence, contracting the first two terms in the summation, each term in the summation becomes τ S ã α ā α + ã 3 ā 3 ā β : a β + τ S ã α ā α + ã 3 ā 3 ā 3 : ae 3 = τ S ã β : a β + τ S ã 3 : ae 3 3

4 Note that a β = a β x p x p = a β x p w n ip, and the expression for a β x p is given equation 1. Ignoring further plastic flow, we have a E 3 x = j x j wn jp a E,n 3, and thus, a E 3 = w n ip ae,n 3 Therefore, we arrive at the final expression for the force of type iii: f iii i x = 3 Laminate Stress p I iii τ S p ãβ p : In this section we derive the expression for τ KL = τ αβ q KL,E α a pβ wip n x + τ S p ã3 p : wip n ae,n p3 p q KL,E β, τ KL αβ = 2µɛL αβ + λɛl γγδ αβ. 2 First notice that we may replace the right Hencky strain with left Hencky strain in the definition of energy because of the isotropic nature of the energy function. We now give the drivation of Equation 2 with index free notation assuming all variables are in 2D. ψf = ψuσv T because the energy is isotropic. Hence, PF = PUΣV T = UPΣV T PF = UPΣV T = U ψ Σ VT = U 2µ logσσ 1 + λtrlog ΣΣ 1 V T. 4

5 Therefore, τ KL = U 2µ logσσ 1 + λtrlog ΣΣ 1 V T F T = U 2µ logσ + λtrlog Σ U T = 2µɛ L + λtrɛ L 4 QR and Elastic Potential We can use QR orthogonalization of deformed material directions to define q i r ij = Fā j, F = r ij q i ā j, r ij = 0 for i > j Change of basis tensor Define the change of basis tensor Q = Q ij ā i ā j 4 with Q ij = q j ā i. With this convention we see that Qā i = q i and Q T Q = I. Furthermore, defining R = r ij ā i ā j we have F = QR. 4.2 Differentials The QR differential satisfies q k δq i r ij + δr kj = q k δfā j, δf = δr ij q i ā j + r ij δq i ā j 5 where q k δq i = q i δq k from orthogonality of the q i. And δf = δqr + QδR 6 where δq T Q = Q T δq from Q T Q = I. Furthermore, and the δr ij = 0 for i > j. δq = δq ij ā i ā j, δq ij = δq j ā i, δq i = δqā i 7 δr = δr ij ā i ā j 8 5

6 5 Elastic potential and stresses Define the hyperelastic potential as ψf = ˆψ[R] 9 where [R] = r 11 r 12 r 13 r 22 r The differential satisfies δψf = ψ F : δf = P : δf = [R]δr ij 11 F where P = ψ F F. Therefore Similarly, δr ij q i Pā j + r ij δq i Pā j = [R]δr ij. 12 P : δf = P : δqr + P : QδR = [R]δr ij 13 Choosing δf = δr ij q i ā j i.e. δq i = 0, we can conclude that q i Pā j δr ij = [R]δr ij 14 for arbitrary δr ij with i j. Therefore the q i Pā j = [R] for i j. Similarly, P : QδR = Q T P : δr = δr ij ā i Q T Pā j = δrij q i Pā j = [R]δr ij. Choosing δf = r ij δq i ā j i.e. δr ij = 0, we can conclude that 15 0 = r ij δq i Pā j. 16 6

7 Similarly, 0 = P : δqr = PR T : δq = PR T : δqq T Q = PR T Q T : δqq T = PF T : δqq T 17 In other words, the Kirchhoff stress τ = PF T is symmetric since δqq T is arbitrary skew. Furthermore, and we know P ij = τ 11 τ 12 τ 13 τ 21 τ 22 τ 23 = τ 31 τ 32 τ 33 P = P ij q i ā j, τ = P ij r kj q i q k = τ ik q i q k 18 = for i j from Equation 14. Thus P 11 P 12 P 13 P 21 P 22 P 23 P 31 P 32 P 33 r 11 r 12 r 22 r 13 r P 11 r 11 + P 12 r 12 + P 13 r 13 P 12 r 22 + P 13 r 32 P 13 P 21 r 11 + P 22 r 12 + P 23 r 13 P 22 r 22 + P 23 r 32 P 23 P 31 r 11 + P 32 r 12 + P 33 r 13 P 32 r 22 + P 33 r 32 P 33 20, and since τ = τ T and P ij = for i j, τ 11 τ 12 τ 13 τ 21 τ 22 τ 23 τ 31 τ 32 τ 33 = r 11 r 11 + r 12 r 12 + r 12 r 22 + r 13 r 13 r 32 r 13 r 13 r 12 r 22 + r 13 r 32 r 22 r 22 + r 23 r 23 r 32 r 13 r In particular, the matrix representation of τ S reads τ 11 τ 12 τ 13 τ 21 τ 22 τ 23 τ 31 τ 32 τ 33 = = 0 0 γs γs f s 3 γs 2 1 γs 1 s 2 γs 1 s 3 γs 1 s 2 γs 2 2 γs 2 s 3 γs 1 s 3 γs 2 s 3 f s s 1 s 2 s

8 6 Frictional Contact Yield Condition Coulomb friction places a constraint on the stress as t S c F σ n 24 where σ n = a KL 3 σa KL 3. Recall that a KL 3 = q 3 and thus σ n = q 3 σq 3. On the other hand, t S is the tangential component of the force density and has the form t S = cq 1 + sq 2 σq 3 for some c and s such that c 2 + s 2 = 1. Hence, we may rewrite the constraint on stress as cq 1 + sq 2 σq 3 + c F q 3 σq Using the fact that σ = detfτ, we rewrite the constraint as cq 1 + sq 2 τ q 3 + c F q 3 τ q Substituting in the expression for τ from equation 23, we find that the maximum on the left-hand-side is ±γs 3 s s2 2 + c F f s 3 We apply the particular form of f in the paper where fx = 1 3 kc 1 x 3 for x 1 and 0 otherwise. When s 3 > 1, the maximum is γs 3 s s 2 2. In this case the return mapping set s 1 and s 2 to 0. If 0 < s 3 1, the maximum is and thus we need γs 3 s s2 2 c F k c s s 3, s s2 2 c F k c γ 1 s 3 2. In this case we uniformly scale back s 1 and s 2 to satisfy the constraint. 7 Denting Yield Condition and Return Mapping We apply the von Mises yield condition to the Kirchhoff-Stress in Equation 2 This condition states that the deviatoric component of the stress is less than a threshold value c vm f vm τ = τ trτ 3 I F c vm. 27 8

9 This condition defines a cylindrical region of feasible states in the principal stress space since 2 f vm τ = 3 τ τ 22 + τ 32 τ 1 τ 2 + τ 2 τ 3 + τ 1 τ 3 28 where τ = i τ iu i u i with principal stresses τ i. The plane stress nature of τ KL = α τ α KL u α u α means that feasible stresses are those where the principal stresses are in the ellipsoidal intersection of the cylinder and the τα KL plane. The yield condition is satisfied via associative projection or return mapping of the stress to the feasible region. The elastic and plastic strains are then computed to be consistent with the projected stress. We use F KL,Etr, F KL,P tr to denote the trial state of elastoplastic strains with associated trial stress τ KLtr. We use F KL,E, F KL,P, τ KL to denote their projected counterparts. F KL,Etr, F KL,P tr, τ KLtr F KL,E, F KL,P, τ KL. 29 The deformation gradient constraint must be equal to the product of trial and projected elastic and plastic deformation gradients, creating the constraint on the projection F KL = F KL,Etr F KL,P tr = F KL,E F KL,P. 30 The projection is completed by first computing the trial state of stress τ KLtr from F KL,Etr using Equation 2. This is done by computing the QR decomposition of the trial elastic deformation gradient F KL,Etr = r KL,E tr αβ q KL,E α ā β + q KL,E 3 ā 3. Then we compute the SVD of matrix [r KL,Etr ] R 2 2 and the trial strain [ɛ Ltr ] [r KL,Etr ] = [U E ] [ɛ Ltr ] = [U E ] σ1 E tr logσ E 1 tr σ2 E tr [V E ] T 31 logσ2 E tr [U E ] T 32 From Equation 2 we see that the two non-zero principal stresses τ KLtr α of τ KLtr are equal to the eigenvalues of the matrix [τ KLtr ] [τ KLtr ] = 2µ[ɛ Ltr ] + λtr[ɛ Ltr ]I = [U E ] τ1 KL tr τ2 KL tr [U E ] T. 33 9

10 We therefore project the eigenvalues τ KLtr α τ KL α into the ellipsoidal intersection the von Mises yield surface and the τ 1, τ 2 plane in the direction that maximizes energy dissipation. We approximate this region by the diamond shaped region whose boundaries have slopes of ±1 to simplify the return mapping. Note that the direction of the return that maximizes energy dissipation is a function of the Cauchy-Green strain derivative of the Kirchhoff stress and thus is non-trivial to find in general. Fortunately, the quadratic Hencky strain model has the favorable property that the return direction is perpendicular to the yield surface [1] which greatly simplifies the return mapping. We illustrate this property in Figure 1. After projection, we rebuild the matrix without changing the eigenvectors and rebuild τ KL from the matrix [τ KL ] = [U E τ KL ] 1 τ KL 2 [U E ] T, τ KL = ταβ KL qkl,e α q KL,E β 34 where ταβ KL are the entries in the projected matrix [τ KL ] R 2 2. The projected strain [ɛ L ] is computed from the projected principal stresses from [ɛ L ] = [U E logσ E ] 1 logσ2 E [U E ] T 35 logσ E 1 1 2µ + λ λ τ KL logσ2 E = 1 36 λ 2µ + λ τ KL 2 and the projected elastic deformation gradient is F KL,E = F KL,E αβ q KL,E α ā β + q KL,E 3 ā 3 where [ˆF KL,E ] = [U E σ E ] 1 [V E ] T. 37 The projected plastic deformation gradient is computed from F KL,P = F KL,E 1 F KL in order to maintain the constraint in Equation 30. References [1] C. Mast. Modeling landslide-induced flow interactions with structures using the Material Point Method. PhD thesis, σ E 2 10

11 a b Figure 1: Return Mapping. In general in the return mapping direction is non trivial left. Quadratic Hencky strain energy density simplifies the return mapping right. 11