Anisotropic elastoplasticity for cloth, knit and hair frictional contact supplementary technical document

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1 Anisotroic elastolasticity for cloth, knit and hair frictional contact sulementary technical document Chenfanfu Jiang, Theodore Gast, Joseh Teran 1 QR differentiation Here we discuss how to comute δq and δr from F and δf. This is used for comuting the linearized force in imlicit time integration. If F = QR where Q is orthonormal with Q T Q = I, and r 11 r 12 r 13 R = r 22 r 23 is uer triangular, and we define the function Ψ(F) = ˆΨ(R) then we have r 33 δf = QδR + δqr δψ(f) = δ ˆΨ(R) δq T Q + Q T δq = 0 δr is uer triangular Q T δf = δr + Q T δqr 0 ω 3 ω 2 Ω = Q T δq = ω 3 0 ω 1 ω 2 ω ω 3 ω 2 Q T δf = δr + ω 3 0 ω 1 R ω 2 ω 1 0 a b c 0 ω 3 ω 2 r 11 r 12 r 13 d e f = 0 + ω 3 0 ω 1 0 r 22 r 23 g h i 0 0 ω 2 ω r 33 This imlies d = ω 3 r 11 g = ω 2 r 11 h = ω 2 r 12 + ω 1 r 22 We can use these three equations to solve for ω 1, ω 2, ω 3. After that, we can construct δq = QΩ and then δr = Q T δf Q T δqr. 1

2 The corresonding 2D result is a = (QT δf) 21 δq = aq r 11 ( 0 ) δr = Q T δf Q T δqr 2 Comuting stress δψ = δ ˆΨ : δf = F R : δr : (QδR + δqr) = F R : δr : (QδR) + : (δqr) = : δr (1) F F R It can be shown that the above holds for any δq and δr that satisfy δq T Q + Q T δq = 0 and δr being uer triangular. Secifically, it can be shown that given arbitrary δq, δr with δq T Q + Q T δq = 0 and δr uer triangular, δf such that δr = R Q (F) : δf, and δq = (F) : δf. F F Using null[ Q F (F)] = san {N 1, N 2, N 3, N 4, N 5, N 6 }, for all uer triangular δr, δf Q null[ Q F (F)] such that δq = Q F (F) : δf Q = Q F (F) : δ(f Q + F R ) F R null[ Q F (F)]. Further, it can be shown that QδR + δqr δf Q null[ Q F (F)] and with δf R = QδR + δqr δf Q, δf = δf Q + δf R roduces If we choose δq = 0 in Equation (1), then δr = R Q (F) : δf, and δq = (F) : δf. F F : (QδR) = F R : δr (Q T ) : δr = F R : δr ) and R have the same uer triangular art. is lower triangular, it is easy to show (entry wise rovable) that Q T F RT and Recall δr is any uer triangular matrix, therefore (Q T F Further more, since R T R RT have the same uer triangular art. 2

3 If we choose δr = 0 in Equation (1), then F : (δqr) = 0 ( F RT ) : δq = 0 ( F RT ) : (δqq T Q) = 0 ( F RT Q T ) : (δqq T ) = 0 ( F FT ) : (δqq T ) = 0 Since δqq T is an arbitrary skew symmetric matrix (due to δq T Q + Q T δq = 0), for the above equation to hold, we know F FT has to be symmetric. This also roves that Kirchoff stress τ = F FT is symmetric without needing to use conservation of angular momentum. Now we have τ = τ T F FT = F( F )T F RT Q T = QR( F )T Q T F RT = R( F )T Q Q T F RT is symmetric. In summary, Q T F RT and R RT have the same uer triangular art. Q T F RT is symmetric. We further denote this tensor with A := Q T F RT = Q T F FT Q = Q T τq. Therefore A is just τ written in the Q basis. 3 A curve in 3D We can construct the uer triangular art of A with R RT, then fill the rest using symmetry of A. Assuming then ˆΨ = f(r 11 ) g(r r 2 13) + h(r 22, r 23, r 33 ), R = [ f g r T 0 ˆP [ where r = (r 12, r 13 ) T r11 r T ], R =, 0 ˆR ˆP = h ˆR. We can show f A = r 11 + g r T r g r T ˆRT g r T ˆRT ˆP ˆRT Here, we choose ˆP ˆR T to be the dilational art of the Kirchoff stress from the 2 2 Stvk Hencky Drucker- Prager. assuming we get some ˆτ after the return maing of dry sand from an inut F = ˆR, we relace the bottom right corner of A with I, where = tr(ˆτ)/2. ], 3

4 Under such a choice, f A = r 11 + g r T r g r T ˆRT g r T. ˆRT I Mohr friction criteria leads to the following maximization roblem: Maximize d T τn + c F n T τn over all ossible n that is erendicular to the fiber direction, with all d erendicular to n and has unit length. Maximize d T QAQ T n + c F n T QAQ T n over all ossible n that is erendicular to the fiber direction, with all d erendicular to n and has unit length. Maximize d T Añ + c F ñ T Añ over all ossible n that is erendicular to the fiber direction, with all d erendicular to n and has unit length, where ñ = Q T n and d = Q T d. Since in our discretization, the fiber direction is q 1, therefore q T 1 n = 0, then we know ñ = (0, c, s) for some θ. Maximize d T Añ + c F ñ T Añ over all ossible n that is (0, c, s) over all θ, with all d erendicular to ñ and has unit length. Lagrangian multilier can solve this roblem and gives the maximum ˆRr g + c F. If we choose g(x) = x, then we have our yield surface ˆRr + c F < 0. The return maing is then simly a scaling on r so that the yield criteria is satisfied. 4 A surface in 3D Surface is very similar to curve in our framework. With our codimensional discretization, we ma the triangle back to x-y lane. If the inut 3D triangle at rest is Â, ˆB, Ĉ, we can define ˆD 1 = ˆB Â and ˆD 2 = Ĉ Â. This forms an imaginary ˆD s, whose QR decomosition gives us the rotation from x-y lane of this triangle ˆQ, as well as the to art of ˆR 3 2 being the 2 2 version D m. The third column of ˆQ is the rotated D 3 where D 3 = e 3. With these recomutations, for any triangle d 1, d 2 in world sace, we can construct the full F as D 1 m F = [d 1, d 2, d 3 ]. 1 In MPM, we have d n+1 3 = (I+ t v)d n 3, with d 0 3 = ˆq 3. Kee in mind that D m is uer triangular, therefore D 1 m is uer triangular. Now do the thin QR decomosition [d 1, d 2 ] = [q 1, q 2 ] R where R is 2 2 uer. If we construct q 3 = q 1 q 2, then R (hx, h [d 1, d 2, d 3 ] = [q 1, q 2, q 3 ] y ) T 0 T h z where Qh := d 3. Now we have constructed the (unique) QR decomosition of F as R (hx, h F = [q 1, q 2, q 3 ] y ) T D 1 0 T m := QR = Q r 11 r 12 r 13 r h z 1 22 r 23 r 33 We can see h = r 3 = Q T d 3. 4

5 The revious lemma in curve still holds: have the same uer triangular art. Q T F RT is symmetric. We further denote this tensor with A := Q T F RT = Q T F FT Q = Q T τq. Therefore A is just τ written in the Q basis. We can construct the uer triangular art of A with R RT, then fill the rest using symmetry of A. 4.1 Surface elastolasticity Recall D 3 = e 3, Q T F RT and R RT r 11 r 12 r 13 F = Q r 22 r 23 Physically, the to left is in lane (x-y) deformation of the triangle, since r 11 r 12 r 13 r 22 r 23 e 3 = r 3, r 33 we know r 3 reresents the deformation of D 3. From h = r 3 = Q T d 3, we can say r 3 is the length change, r r 2 23 is the shearing (or the deviation from being erendicular to the x-y triangle lane). Note that shearing also ernalizes length change. From these, we can define then r 33 ˆΨ = f(r 33 ) g(r r 2 23) + h(r 11, r 12, r 21, r 22 ), R = ˆP g r 0 T f, ˆR where r = (r 13, r 23 ) T r, R = 0 T, r ˆP = h 33 ˆR. We can show ˆP A = ˆRT + g rr T g r 33 r g r 33 r T f r 33 Mohr friction criteria leads to the following maximization roblem: Maximize d T τn + c F n T τn over all ossible n that is erendicular to the manifold lane, with all d erendicular to n and has unit length. Maximize d T QAQ T n + c F n T QAQ T n over all ossible n that is erendicular to the manifold lane, with all d erendicular to n and has unit length. Maximize d T Añ + c F ñ T Añ over all ossible n that is erendicular to the manifold lane, with all d erendicular to n and has unit length, where ñ = Q T n and d = Q T d. n erendicular to manifold lane means n = k(d 1 d 2 ) for some k (unit length constraint is extra). Recall [d 1, d 2 ] = [q 1, q 2 ] R, where R is uer triangular 2 2, this means k(d 1 d 2 ) = z(q 1 q 2 ) for some z. Therefore, n = ±q 3. Therefore ñ = Q T n = ±e 3. Maximize d T Añ + c F ñ T Añ over ñ = ±e 3, with all d = (c, s, 0) for some θ 5

6 The maximum is ±g r 33 r + c F f r 33 Assume f(x) = 1 3 k(1 x)3 for x 1, 0 otherwise. g(x) = γx. When r 33 > 1, f = 0, the maximum is γr 33 r. In this case the return maing is making r to be zero. When r 33 < 1, the maximum is ±γr 33 r c F k(r 33 1) 2 r 33. The yield surface is therefore max(± γ k r 33 r c F (r 33 1) 2 r 33 ) 0 If r 33 is negative (corresonding to inverted collision), the max should choose γ k. The return maing is setting r to be 0. Otherwise, try to scale r (when necessary) to satisfy 5 A curve in 2D γ k r c F (r 33 1) 2 0 2D derivation follows 3D curve derivation: Q T F RT and R RT have the same uer triangular art. Q T F RT is symmetric. We further denote this tensor with A := Q T F RT = Q T F FT Q = Q T τq. Therefore A is just τ written in the Q basis. The energy choice is then r11 r where R = 12, 0 r ˆP = h 22 ˆR. We can show ˆΨ = f(r 11 ) g(r2 12) + h(r 22 ), R = f g r 12 0 h, f A = r 11 + g r12 2 g r 12 r 22 g r 12 r 22 h r 22 Mohr friction criteria leads to the following maximization roblem: Maximize d T τn + c F n T τn over all ossible n that is erendicular to the fiber direction, with all d erendicular to n and has unit length. Maximize d T QAQ T n + c F n T QAQ T n over all ossible n that is erendicular to the fiber direction, with all d erendicular to n and has unit length. Maximize d T Añ + c F ñ T Añ over all ossible n that is erendicular to the fiber direction, with all d erendicular to n and has unit length, where ñ = Q T n and d = Q T d. Since in our discretization, the fiber direction is q 1, therefore q T 1 n = 0, then we know ñ = ±(0, 1) Maximize d T Añ + c F ñ T Añ over ñ = (0, ±1) and d = (±1, 0) 6

7 Maximize d T Añ + c F ñ T Añ over ñ = (0, ±1) and d = (±1, 0) The maximum is ±g r 12 r 22 + c F h r 22 Assume g(x) = γx, h(x) = 1 3 s(1 x)3 for x 1, 0 otherwise. When r 22 > 1, h = 0, the maximum is γr 12 r 22, the return maing is setting r 12 = 0. When r 22 < 1, the maximum is The yield surface is therefore ±γ r 12 r 22 c F s(1 r 22 ) 2 r 22 max(± γ s r 12 r 22 c F (1 r 22 ) 2 r 22 ) 0 If r 22 < 0, the max should choose γ s. The return aming is setting r 12 = 0. Otherwise, r 22 [0, 1], try to scale r 12 to satisfy γ s r 12 c F (1 r 22 ) Derivative of ˆF E In the aer we mention that ˆF E x i is a third order tensor, and does not deend on ˆx because ˆF E is linear in ˆx i. Here we give the derivation of comuting this derivative. We have by definition Plugging in ˆF E,α = i Differentiating we have ˆF E,α x = ˆd E,β(ˆx) = ˆx mesh(,β) ˆx mesh(,0) (2) (wi,mesh(,β) n wn i,mesh(,0) )x i,αd 1 (wi,mesh(,β) n wn i,mesh(,0) )δ αζd 1 ˆd E,β(ˆx) = ( ˆx),β (3) ˆx q = ˆx i wiq n i (4) ˆF E (ˆx) = ˆd E D 1 (5),β +,β + w n i x κ,κβ x i,αd 1,β w n i x κ,κβ δ αζd 1,β which doesn t deend on ˆx. Note we can think of the undeformed segment or triangle as being aligned with the x-axis or xy-lane resectively. This will mean that the initial F is no longer I, but rather the rotation which mas the axis aligned element to its initial osition in sace. This is not an issue as our elastic energy density ψ is world sace rotation invariant. However this allows us to assume that D is block diagonal of the form D D = for segments and D 11 D 12 0 D = 0 D for triangles. This saves memory for D, and simlifies comuting D 1. 7

8 7 Comutation of Force on the Grid Define Then i (ˆx) = V 0 (ˆx) = (ˆx) = (ˆx) = F : ˆF E (6) x i α f (ii) qζ (ˆx) = = =1 F ζ F ζ ˆF,α E (7) F α x ( (wi,mesh(,β) n wn i,mesh(,0) )D 1,β + (8) wi n ),κβ x D 1,β κ ( (wi,mesh(,β) n wn i,mesh(,0) )D 1,β + (10) F ζ F ζ (ˆx) = f (ii) qζ (ˆx)wn i I (ii) (ˆx) = f (ii) qζ (ˆx)wn i I (ii) wi n ),κβ x δ β κ (δ q,mesh(,β) δ q,mesh(,0) )D 1 (δ q,mesh(,β) δ q,mesh(,0) )D 1 =γ+1 F ζ F ζ κ=1 (9) (11),β (12),β (13) w n i x κ,κβ δ β (14) w n i x κ,κ (15) 8

9 8 Pseudocode Algorithm 1 Simulate 1: rocedure Time Ste 2: Transfer to grid 3: Grid ste 4: Transfer to articles 5: Udate article state 6: Plasticity 1: rocedure Transfer to grid 2: for all grid nodes i do 3: m n i wn i m 4: vi n 1 m n wn i m ( v n i + C n (x i x n ) ) 1: rocedure Grid ste 2: vi vn i + Force increment( FE,n ) 3: v n+1 i Grid collisions( vi ) 4: ṽ n+1 i Friction( v n+1 i, v n+1 i vi ) 1: rocedure Transfer to articles 2: for all articles of tye (i) and (ii) do 3: v n+1 i β=0 w n iṽn+1 i 4: for all articles of tye (iii) do 5: v n+1 1 γ vn+1 mesh(,β) 6: for all articles do 7: C n+1 i w n iṽn+1 i (x i x n ) T 1: rocedure Udate article state 2: for all articles of tye (i) do 3: x n+1 wi(x n n i + tv n+1 i ) i 4: v i v n+1 i ( wi) n T 5: ˆFE,n+1 (I + t v )F E,n 6: for all articles of tye (ii) do 7: x n+1 i w n i(x n i + tv n+1 i ) 8: for all articles of tye (iii) do 9: x n+1 1 γ xn+1 mesh(,β) β=0 10: v v n+1 i ( wi) n T i 11: for β = 1 to γ do 12: ˆdE,n+1,β x n+1 mesh(,β) xn+1 mesh(,0) 13: for β =γ + 1 to 3 do 14: ˆdE,n+1,β 15: ˆFE,n+1 (I + t v ) E,n+1 ˆd D 1 1: rocedure Plasticity 2: for all articles of tye (i) and (iii) do 3: F E,n+1 Return maing(ˆf E,n+1 ) 9