Brutal partial damage a case study for the interaction between - PowerPoint PPT Presentation

– Brutal partial damage – a case study for the interaction between evolution and relaxation Gilles A. Francfort Universit´ e Paris 13 & Institut Universitaire de France Analyse Variationnelle et Microstructuration, Quiberon, September 2014 Comes from various works from G. Allaire, me, A. Garroni, C. Larsen, J.-J. Marigo

1/ Brutal damage – a mechanical model – 1 Quasi-static evolution Rate independence no kinetic energy no viscous type behavior Energy density: W ( " | d ) kinematic variable: internal variable: ✏ ( u ) = " = 1 / 2( Du + Du T ) d u : Ω ⇢ R 3 ! R 3 + loads: f ( t ): volume or surface forces g ( t ): imposed displacements Principles: instantaneous equilibrium: positivity of dissipation : � D d W ( ✏ ( u )( t ) , d ( t )) 2 @ D ( ˙ � div D ε W ( ✏ ( u )( t ) , d ( t )) = f ( t ) d ( t )) u ( t ) = g ( t ) on @ Ω D convex, � 0 , D (0) = 0

2/ Brutal damage – a mechanical model – 2 • Assumptions: I on D : I on W : d 2 [0 , 1] W ( " | d ) := 1 2 A ( d ) " · " A ( d ) & with d isotropy: A ( d ) = � ( d ) i ⌦ i + 2 µ ( d ) I • 8 � div ( A ( d ( t )) ✏ ( u )( t )) = f ( t ) > > > > > u ( t ) = 0 on @ Ω > > > > < � 1 / 2 A 0 ( d ( t )) ✏ ( u )( t ) · ✏ ( u )( t )  k > � 1 / 2 A 0 ( d ( t )) ✏ ( u )( t ) · ✏ ( u )( t ) = k , if ˙ > d ( t ) > 0 > > > > > > t > : d ( t ) %

3/ Brutal damage and rate independence – 1 Unilateral Stationarity of: R 1 Ω A ( ⌘ ) ✏ ( v ) · ✏ ( v ) dx 2 R � div ( ... ) = f ( t ) � Ω f ( t ) · vdx R $ • + Ω D ( ⌘ � d ( t )) dx u ( t ) = 0 on @ Ω � 1 / 2 A 0 ( d ( t )) ✏ ( u )( t ) · ✏ ( u )( t )  k v = 0 on @ Ω • ˙ d ( t ) = 0 if strict ineq. $ Energy Balance: Z d E ˙ dt = � f ( t ) · u ( t ) dx Ω with Z E ( t ) := 1 / 2 A ( d ( t )) ✏ ( u )( t ) · ✏ ( u )( t ) dx Ω Z t Z Z ˙ � f ( t ) · u ( t ) dx + k d ( s ) dxds Ω 0 Ω In what follows d ⌘ � 2 { 0 , 1 } , A ( � ); = � A w + (1 � � ) A s , A w  A s , and � (0) ⌘ 0.

4/ Brutal damage and rate independence – 2 • First departure: replace Stationarity by Global Minimality • Initial time step: u 0 , � 0 minimizes Z 1 ( � A w + (1 � � ) A s ) ✏ ( v ) · ✏ ( v ) � f 0 · v + k � � dx 2 Ω + get rid of � u 0 minimizer of R Ω ( W 0 ( ✏ ( u )) � f 0 · u ) dx where W 0 ( ✏ ) = min χ 2 { 0 , 1 } { 1 2 ( � A w ✏ + (1 � � ) A s ✏ ) + k � } W 0 ( " ) = min { 1 2 A s " · " + k , 1 or still 2 A w " · " } No minimizers! = ) Relaxation through quasiconvexification u 0 minimizes R � QW 0 ( ✏ ( u )) � f 0 · u dx Ω where ⇢Z � QW 0 ( " ) := min W 0 ( " + ✏ ( ' )) dy : ' 2 H 1 # ( Y ; R 3 ) Y periodic instead of " H 1 0

5/ Relaxation first time step – 1 • General homogenization formula for two-phase periodic mixtures: ⇢ " or R A 0 " · " = Y ( � B + (1 � � ) C )( " + ✏ ( w ε )) · + ✏ ( w ε )) dy , ( " with w ε periodic solution of div A ( y )( ✏ ( w ε ) + " ) = 0 with 0-mean. • Define for any ✓ 2 [0 , 1] and any B , C : G θ ( B ; C ) = set of (periodic) homogenized tensors asstd. with R Y � ( y ) dy = ✓ (Also define G ( B ; C ) = [ θ { G θ ( B ; C ) } ) + n o ⇥ 1 ⇤ QW 0 ( " ) := min 0  θ  1 2 A 0 " · " min A 0 2 G θ ( A w ; A s ) + k ✓ • How does one compute QW 0 ? Problem: G θ ( A w ; A s ) is unknown as of yet!!! • Only need to compute G θ ( A w ; A s ) " · " . The minimum inside the brackets is attained for a finite rank laminate. Indeed,......

6/ Energy bounds – 1 – Lamination R • Assume � ( y ) = � ( y · ⇠ ) with Y � ( y ) dy = ✓ : then (1 � ✓ )( A lam � A w ) � 1 = ( A s � A w ) � 1 + ✓ G ( ⇠ ) , where 1 λ w + µ w G ( ⇠ ) " := µ w "⇠ � ⇠ � µ w ( λ w +2 µ w ) ( "⇠ · ⇠ ) ⇠ � ⇠ . Proof: Seek ✏ ( w ε ) = " w � + " s (1 � � ) , " w , " s constant. ⇧ " w � + " s (1 � � ) sym. grad. = ) " w � " s = ⌧ � ⇠ for some ⌧ ⇧ A ( y )( " + ✏ ( w ε )) = A w ( " + " w ) � + A s ( " + " s )(1 � � ) has 0 div. = ) [ A w ( " + " w ) � A s ( " + " s )] ⇠ = 0 = ) [( A s � A w )( " + " s )] ⇠ = µ w ⌧ + ( � w + µ w )( ⌧ · ⇠ ) ⇠ λ w + µ w 1 ⇧ Set h := ( A s � A w )( " + " s ) ) ⌧ = µ w h ⇠ � µ w ( λ w +2 µ w ) ( h ⇠ · ⇠ ) ⇠ λ w + µ w 1 = ) ⌧ � ⇠ = µ w h ⇠ � ⇠ � µ w ( λ w +2 µ w ) ( h ⇠ · ⇠ ) ⇠ � ⇠ ( ✓" w +(1 � ✓ ) " s = 0 ) " s = � ✓⌧ � ⇠ def . of h " =( A s � A w ) � 1 h + ✓⌧ � ⇠ ) ⇧ A lam " = ✓ A w ( " + " w ) + (1 � ✓ ) A s ( " + " s ) ( A lam � A w ) " = ✓ A w ( " + " w )+(1 � ✓ ) A s ( " + " s ) � A w ( " + ✓" w +(1 � ✓ ) " s ) = (1 � ✓ )( A s � A w )( " + " s ) = (1 � ✓ ) h ) " = (1 � ✓ )( A lam � A w ) � 1 h also = ( A s � A w ) � 1 h + ✓⌧ � ⇠ . =

7/ Energy bounds – 2 – Lamination • Formula iterates (finite rank laminates): (1 � ✓ )( A lam � A w ) � 1 = ⇢ m i � 0 ( A s � A w ) � 1 + ✓ P p 1 m i G ( e i ) , P p 1 m i = 1 . + • In general any finite rank laminate is given by (1 � ✓ )( A lam � A w ) � 1 = ( A s � A w ) � 1 + ✓ R S N � 1 G ( e ) d ⌫ ( e ) , with ⌫ probability measure on the sphere ( since extreme points of the set of such measures are Dirac masses). • Multi-ranks ( > 1) laminates are not periodic! A subtle point....

8/ Energy bounds – 3 – Hashin-Shtrikman bounds • Thm: Any A 0 in G θ ( A w ; A s ) is such that there exist two finite rank laminates A � and A + with A �  A 0  A + . Proof: A 0 " · " = �R R Y A w idem dy : v 2 H 1 inf Y (1 � � )( A s � A w )( " + ✏ ( v )) · ( " + ✏ ( v )) dy + per R Y (1 � � )(2 ⌘ · ( " + ✏ ( v )) � ( A s � A w ) � 1 ⌘ · ⌘ ) dy + idem } � = inf v { sup η { same with constant ⌘ = η cst . { A w " · " + (1 � ✓ )[2 ⌘ · " � ( A s � A w ) � 1 ⌘ · ⌘ ] sup Z + inf v { A w ✏ ( v ) · ✏ ( v ) � 2 �⌘ · ✏ ( v )) dy }} Y • the inf. in v is computed with Fourier series...... = ) n A 0 " · " � sup A w " · " + (1 � ✓ )[2 ⌘ · " � ( A s � A w ) � 1 ⌘ · ⌘ ] η cst . � k | 2 G ( k o X � | ˆ | k | ) ⌘ · ⌘ k 6 =0 R P � k | 2 = Y ( � � ✓ ) 2 dy = ✓ (1 � ✓ ) k 6 =0 | ˆ

9/ Energy bounds – 4 – Hashin-Shtrikman bounds – part II P 1 � k | 2 � k Set ⌫ := k 6 =0 | ˆ θ (1 � θ ) | k | + A 0 " · " � A w " · " + Z h ⇣ ⌘ i ( A s � A w ) � 1 + ✓ (1 � ✓ ) sup 2 ⌘ · " � S N � 1 G ( e ) d ⌫ ( e ) ⌘ · ⌘ η cst . | {z } = (1 � ✓ )( A lam � A w ) � 1 for some laminate A 0 " · " � A lam " · " Thus: ⇤ + ⇥ 1 ⇤ 2 A 0 " · " min A 0 2 G θ ( A w ; A s ) = A w " · " + (1 � ✓ ) min ν { sup η .... } + [1 2 A 0 " · " ] = min in 2d A 0 2 G θ ( A w ; A s ) η [2 ⌘ · " � ( A s � A w ) � 1 ⌘ · ⌘ � ✓ max A w " · " + (1 � ✓ ) sup e 2 S N � 1 G ( e ) ⌘ · ⌘ ]  2( ⌘ 1 " 1 + ⌘ 2 " 2 ) � ( ⌘ 1 � ⌘ 2 ) 2 4( K s � K w ) � ( ⌘ 1 + ⌘ 2 ) 2 = A w " · " +(1 � ✓ ) sup 4( µ s � µ w ) η 1 , η 2 ✓ ◆ � � ✓ 2 (1 � ⌫ ) � � w + µ w ⌘ 2 1 ⌫ + ⌘ 2 ( ⌫⌘ 1 + (1 � ⌫ ) ⌘ 2 ) 2 max µ w � w + 2 µ w 0  ν  1

10/ Relaxation first time step – 2 – 2d case • We have to compute min A 0 2 G θ ( A w ; A s ) [ 1 2 A 0 " · " ] using the previous expression. Not a simple task. • Explicit result is unimportant: There are three regimes ⇧ regime 1: ( K s � K w )( ✓ µ s +(1 � ✓ ) µ w ) | tr " | < p ( µ s � µ w )( ✓ K s +(1 � ✓ ) K w ) 2 k " d k = ) rank-one layering ⇧ regime 2: p ✓ ( K s � K w ) | tr " | � ( µ w + ✓ K s + (1 � ✓ ) K w ) 2 k " d k = ) rank-one layering ⇧ regime 3: the rest = ) rank-2 layering Still have to minimize in ✓ , so as to obtain QW 0 ( " )! In the end: we get a pair ( ✓ 0 ( " ) , A 0 ( " )) with QW 0 ( " ) = 1 / 2 A 0 ( " ) " · " + k ✓ 0 ( " )

11/ Relaxation next time steps t n i , 0 = t n 0  ...  t n k ( n ) = T • Say we perform time stepping. At time step 1, it seems that we should impose ✓ 1 � ✓ 0 . Bad: At next time step mix weak material with result of previous step, paying at maximum in terms of dissipated energy the vol. frac. of remaining strong material at previous step at time t n + Θ n i − 1 v.f. strong mat. at t n i with i − 1 � 1 i � 1 , 1 W ( t n 2 A w " · " + k Θ n 2 A n i , " ) := min i � 1 " · " + h i QW ( t n i � 1 ) { 1 2 A " . " } + k Θ n i , " ) = min 0  θ  1 min A 2 G θ ( A w , A n i � 1 ✓