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Coupling at the Interfaces (7) ⎧ T ⎪MU 1 1,link −C1 Λ= 0 ⎪ T ⎨MU 2 2,link −C2Λ= 0 ⎪ ∆ t ∆ t ⎪− CU − CU = CU + CU ⎩ 2 2 1 1,link 2 2,link 1 1,free 2 2,free • By condensing on the Lagrange multipliers one obtains formally: 1 T U − 1,link = M1 C1 Λ 1 T U − 2,link = M2 C2 Λ ∆t ∆t − CM C Λ− CM 2 2 C Λ = CU + CU • By posing: −1 T −1 T 1 1 1 2 2 2 1 1,free 2 2,free ∆t −1 T −1 T H ( C1M1 C1 + C2M2 C2 ) 2 d CU + C U 1 1,free 2 2,free one obtains the linear system: HΛ= d (Steklov-Poincaré operator) which upon solution yields the Lagrange multipliers the link accelerations i,link Λ , then 115 U 116 Coupling at the Interfaces (8) Summarizing, the problem may be written in matrix form: • First, solve the problem without links (free). Symbolically: ⎡M1 0 0⎤⎡U ⎤ 1,free ⎡F1⎤ ⎢ 0 M 2 0 ⎥⎢ U ⎥ ⎢ 2,free F ⎥ ⎢ ⎥⎢ ⎥ = ⎢ 2 ⎥ ⎢⎣ 0 0 0⎥⎦ ⎢ 0 ⎥ ⎣ ⎦ ⎢⎣ 0 ⎥⎦ Standard problem! Use single-domain operators. • Then, solve the problem with links (link). Symbolically: T ⎡M1 0 −C ⎤⎡ 1 U ⎤ 1,link ⎡ 0 ⎤ ∆t ⎢ T ⎥⎢ ⎥ ⎢ ⎥ 0 M2 − C2 U 2,link 0 2 ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢−C1 −C2 0 ⎥⎢ Λ ⎥ ⎢CU ⎣ 1 1,free + CU ⎥ ⎣ ⎦⎣ ⎦ 2 2,free⎦ Thanks to diagonal mass matrix, the problem with links is confined only to the interfaces! 58
Coupling at the Interfaces (9) Time integration flowchart is as follows: • Let complete solution be known at time t n : U , U , U n n n i i i p n ∆t n • Compute mid-step velocities (i.e. predictors): U i U i + U i 2 n p • Compute new configurations: U = U +∆t U i i i • Solve free problem for accelerations: U = M F −1 i,free i i p t • Compute new free velocities: U ∆ i,free U i + U i,free 2 ∆t − • Compute: 1 T − 1 T H ( CM 1 1 C1 + CM 2 2 C2 ) and 2 d CU 1 1,free + CU 2 2,free 117 Coupling at the Interfaces (10) • Solve linear system for the Lagrange multipliers: −1 H Λ= d → Λ= H d H is diagonal for conforming interface! • Compute interaction forces: R i = C Λ T i −1 • Compute link accelerations: U M R i,link i i M is diagonal! • Compute total accelerations: U i = U i,free + U i,link t • Compute new velocities: U ∆ i = U i,free + U i,link 2 118 59
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European Laboratory for Structural
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Proceedings of a Course on: Numeric
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Programme of the Course 1. Introduc
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Credits & Acknowledgments • Struc
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Introductory Example (4) 7 Applicat
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x Computational Framework • Gover
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n+ 1 n ∆t n n+ 1 u = u + ( u + u
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Scheme start-up and marching (2)
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Stress Update • To solve the equi
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Geometric non-linearities (3) Set u
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Essential Boundary Conditions Essen
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Exercise 0 - Ideal ballistics • M
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Exercise 0 - Ideal ballistics (5)
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Exercise 1 - Suspended mass (3) •
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Exercise 1 - Suspended mass (7) •
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Exercise 2 - Wave propagation (4)
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Exercise 3 - Impact on Cooling Towe
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TITLE: PMAT04: motion of projectile
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sinon; tra2 = tra2 et ele; finsi; f
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Some results: horizontal and vertic
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PMAT01E This test is similar to PMA
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Hence: 11 −8 ES 2× 10 ⋅ 2.5×
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The stress history is: Note that th
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Analytical TEST11 Same as TEST01 bu
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Comparison of natural vs. engineeri
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The total external forces at the tw
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TEST13 Same as TEST01 but uses a 10
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The mass returns at the initial pos
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TEST22 To verify the independence o
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Linear dynamic analysis Consider th
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• The two waves f and g propagate
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Analytical solution For the bar imp
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Note that we have also put the Pois
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BARI08 We study the effect of the t
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ENDPLAY *==========================
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BARI10 Same as BARI09 but compares
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The system is initially at rest, an
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CAST MESH TRID NONL LAGR $ $ Dimens
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Final deformation (with superposed
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Detailed Contents • Modeling the
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Modeling the fluid domain • The f
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Euler equations (3) • For a compr
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Time integration Each time incremen
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Time integration (5) 10. Account fo
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Euler Equations (FV) • They are w
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Mesh rezoning - motivations • Rez
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Giuliani’s automatic rezoning •
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• Analytical solution (self-simil
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Shock tube (6) • Influence of pse
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Exercise 2 - Explosion in air-fille
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Exercise 3 - Bubble expansion in a
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Exercise 4 - External blast on two
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Geometric data: The calculation is
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SHOC01 Eulerian, “average” pseu
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COUR 7 'ro_a' ECRO COMP 2 ELEM LECT
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Analytical Conclusion: the pseudo-v
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SHOC13 Eulerian, very low pseudo-vi
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Analytical General conclusions: •
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The inverse path is not so straight
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The input files (mesh generation by
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Geometric data: The calculation is
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COUR 1 'dx_4' DEPL COMP 1 NOEU LECT
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TANK02 Eulerian solution: the whole
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The final pressure distribution is:
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TUBE06 Lagrangian solution: the who
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And the final velocity distribution
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The computed displacements are: To
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43 30 30 43 44 31 31 44 45 32 32 45
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Geometric data: External blast in a
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abs4 = dall (p4 d 28 p4u) (p4u d 40
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Geometric data: Internal blast in a
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air = air1 ET air2 ET air3 ET air4
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COURBE 8 'P e_p12' ECROU COMP 1 lec
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3D axisymmetric simulation: JWLS3G
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point lect 1 term elem lect 1 term
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Universitat Politècnica de Catalun
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• Motivation Detailed Contents
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FSI Classification FSI for compress
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The 2-D plane case (2) Use geometri
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Classification of FSI algorithms
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The FSA algorithm (4) The case of s
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The FSCR algorithm Combination of F
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Application to Finite Volumes (3) 2
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Exercise 2 - Explosions in simple d
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Exercise/Example 3 - Wave propagati
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Exercise/Example 4 - CONT problem (
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Exercise/Example 5 - Explosion in a
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Exercise/Example 6 : Woodward-Colel
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Geometry: Exercise/Example 8 Steam
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Exercise/Example 9 Explosion in Sec
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Exercise/Example 9 Explosion in Sec
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Boundary Condition Elements: CLxx (
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Exercise/Example 11 - Perforated Pl
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Exercise/Example 12 - Sloshing (3)
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DIME PT3L 23 PT2L 143 FL24 145 ED01
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The final pressure field in the flu
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ALE solution with FSA: it is not po
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120 119 138 139 121 120 139 140 122
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and the final velocity field: The s
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The final velocities: The final liq
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*----------------------------------
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TWIS07 We study the phenomenon by a
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This is a well-known reactor safety
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1.53330E+00 1.32310E+01 1.15000E+00
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200 199 239 240 200 200 200 240 241
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*----------------------------------
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The final fluid pressures: CONT02 T
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The final fluid velocities: The fin
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This example consists in a long 3D
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GEOM FL38 flui Q4GS stru TERM COMP
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Detail of first diaphragm: Detail o
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BOUNDARY CONDITIONS: The step is en
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The EUROPLEXUS input file reads: WO
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WOCO3D The mesh generation file (K2
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* RESU ALIC GARD PSCR * SORT GRAP *
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PT3L 16389 PT6L 504 NBLE 16389 NALE
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Final pressure distribution: 4
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BOUNDARY CONDITIONS: The vessel is
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as01=daller c1 c2 c3 c4 plan; * c1=
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*trac oeil cach fluidstr; *opti don
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cam3=cam31 et cam32; camb=cam1 et c
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log 1 dtml REZO GAM0 0.5 FLMP EPS1
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Final structure deformation: Final
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LOADING: The event is initiated by
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FLUT RO .242777373 EINT 6.865E5 GAM
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Final structure velocities: 6
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RESULTS: A paper by Bermudez and Ro
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The applied displacement and the co
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PROBLEM: A rigid tank with a deform
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The EUROPLEXUS input file reads: GR
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PROBLEM: A rigid tank with rigid or
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COMP EPAI 0.005 LECT 101 102 103 10
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Numerical Solution (flexible bottom
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! VARI DEPL VITE ECRO ! fich alic t
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The final pressures in the rigid an
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LOADING: A constant gravity in the
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Some results: intermediate and fina
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Universitat Politècnica de Catalun
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ALE description of structures (2)
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Example 1 - Taylor bar impact (3) B
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Page 313 and 314: • Tentative classification: Non-c
Page 315 and 316: Example 4 - Explosion in a 2D box (
Page 317 and 318: Example 6 - LOCA in the HDR (2) A -
Page 319 and 320: Examples of “Slow” Transient Co
Page 321 and 322: Examples of Fast Transient Impact
Page 323 and 324: Components of Contact-Impact Method
Page 325 and 326: Shortcomings • Slender or distort
Page 327 and 328: Example - Bar impact • (See Part
Page 329 and 330: Example 7b - Indentation (2) • 2D
Page 331 and 332: Example 7b - Indentation (6) • 3D
Page 333 and 334: Smoothed Particles Hydrodynamics (C
Page 335 and 336: SPH Formulation (2) • Basic idea:
Page 337 and 338: SPH impact [Courtesy of Samtech/Son
Page 339 and 340: PRGL01: Example 8 - SPH impacts (Co
Page 341 and 342: Spectral Elements • Motivation: l
Page 343 and 344: Spectral Elements (5) Convergence p
Page 345 and 346: Example 9 - Closed FE/SE interface
Page 347 and 348: Example 11 - Cylindrical valley 85
Page 349 and 350: Performance Optimization • All ex
Page 351 and 352: Time Step Partitioning (4) • Intr
Page 353 and 354: Time Step Partitioning (8) • Acti
Page 355 and 356: Treatment of links in partitioning
Page 357 and 358: Example 12a - Taylor bar impact rev
Page 359 and 360: Coupling at the Interfaces • Two
Page 361: Coupling at the Interfaces (5) MU T
Page 365 and 366: Multiple scales in time (3) • Exp
Page 367 and 368: Multiple scales in space • Furthe
Page 369 and 370: Multiple scales in frequency • So
Page 371 and 372: Example: power plant 133 Example: p
Page 373 and 374: Example: aircraft (3) Thanks to CPU
Page 375 and 376: Example 14 - Domains in 3D • Thic
Page 377: Example 15 - Bar Impact Revisited (
Page 382 and 383: * mesh=stru et symax et viti et lil
Page 384 and 385: The displacements are: 4
Page 386 and 387: c2=p1 d 8 p2; c3=p2 d 20 p3; c4=p3
Page 388 and 389: The resulting final deformed mesh w
Page 390 and 391: TERM *-----------------------------
Page 393 and 394: Geometric data and materials: The b
Page 395 and 396: BET0 1 KINT 0 AHGF 0 CL 0.5 CQ 2.56
Page 397 and 398: INFS02 We use a twice finer fluid m
Page 399 and 400: FREQ 1 GOTR LOOP 60 OFFS FICH AVI C
Page 401 and 402: Problem description: This example i
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Page 405: The final velocities: The final for
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Page 410 and 411: The final deformed mesh is: The fin
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pc = 4.5 -1 -1.5; pd = 5.5 -1 -1.5;
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The initial configuration (with par
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TITLE: Indentation problem. PROBLEM
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cp = p0 d 10 p13; elim tol (piece e
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The final velocities: The final dis
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The input file is: INDE - 13 ECHO !
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The final displacement norm: The fi
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elim tol (piece et cp); c_p = piece
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The final deformed shape is: 13
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Numerical Solutions PRGL01 Simplifi
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FOV 2.48819E+01 SCEN GEOM NAVI FREE
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ROMA01 Final plastic streen in the
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SONA01 7
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Final density in the projectile: Fi
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opti echo 0; opti donn 'D:\Users\Fo
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* opti dime 2 elem qua4; p0 = 0 0;
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The final X-displacement in the hyb
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* opti dime 2 elem qua4; opti titr
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VALLPS This calculation assumes a p
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The receiver signal in the coupled
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Problem description: This example r
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CALC TINI 0. TFIN 40.E-3 *=========
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REPETER BCL2 (2); REPETER BCL3 (2);
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*----------------------------------
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The tip displacement in the referen
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LOG 1 DPSD *-----------------------
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TRAC CACH OEIL2 ZONE1; TRAC CACH OE
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The tip displacement in the referen
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$ INIT VITE 2 -227. LECT VITI TERM
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* AXTE 1000.0 'Time [ms]' * COUR 1
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sq41=sq41 coul 'TURQ'; sq42=sq42 co
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The displacements in the case with
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European Commission DG Joint Resear
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