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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)
- Page 36 and 37: Exercise 3 - Impact on Cooling Towe
- Page 41 and 42: TITLE: PMAT04: motion of projectile
- Page 43 and 44: sinon; tra2 = tra2 et ele; finsi; f
- Page 45 and 46: Some results: horizontal and vertic
- Page 47: PMAT01E This test is similar to PMA
- Page 50 and 51: Hence: 11 −8 ES 2× 10 ⋅ 2.5×
- Page 52 and 53: The stress history is: Note that th
- Page 54 and 55: Analytical TEST11 Same as TEST01 bu
- Page 56 and 57: Comparison of natural vs. engineeri
- Page 58 and 59: The total external forces at the tw
- Page 61 and 62: TEST13 Same as TEST01 but uses a 10
- Page 63 and 64: The mass returns at the initial pos
- Page 65 and 66: TEST22 To verify the independence o
- Page 67 and 68: Linear dynamic analysis Consider th
- Page 69 and 70: • The two waves f and g propagate
- Page 71 and 72: Analytical solution For the bar imp
- Page 73 and 74: Note that we have also put the Pois
- Page 75 and 76: BARI08 We study the effect of the t
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- Page 79: BARI10 Same as BARI09 but compares
- Page 82 and 83: The system is initially at rest, an
- Page 84 and 85: CAST MESH TRID NONL LAGR $ $ Dimens
- Page 89 and 90: Universitat Politècnica de Catalun
- Page 91 and 92: Building Vulnerability (2) P max co
- Page 93 and 94: Euler equations • These equations
- Page 95 and 96: Treatment of momentum equation •
- Page 97 and 98: Time integration (3) Some pseudo-vi
- Page 99 and 100: ALE description The ALE description
- Page 101 and 102: Initial Conditions FE / FV • Equi
- Page 103 and 104: Mean-based rezoning • Simple: opt
- Page 105 and 106: Free surface modeling • Lagrangia
- Page 107 and 108: Shock tube (4) • Assumed paramete
- Page 109 and 110: Exercise 2 - Explosion in air-fille
- Page 111 and 112: Exercise 3 - Bubble expansion in a
- Page 113 and 114: Exercise 4 - External blast on two
- Page 115: Exercise 5 - Confined detonation of
- Page 120 and 121: Analytical solution (see e.g. Harlo
- Page 122 and 123: 3.05000E+01 5.00000E-01 3.10000E+01
- Page 124 and 125: Analytical Conclusion: the pseudo-v
- Page 126 and 127: Finally, we investigate the effect
- Page 128 and 129: “Optimal” solution (for the pre
- Page 130 and 131: Initial conditions: FE vs. FV The t
- Page 132 and 133: Example of shock tube problem solve
- Page 134 and 135: The pressure distribution at t = 0.
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TANK01 Lagrangian solution: the who
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The computed fluid pressures at the
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The computed pressure histories are
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Geometric data: The calculation is
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The computed vertical displacements
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TUBE04 ALE solution with a multi-ph
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The input file for the test TUBE04
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gotr loop 249 offs fich avi cont no
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p6 = 70 20 0; p7 = 70 40 0; p8 = 40
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$ high-pressure perfect gas (explos
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* r_cais = 965.00e-3 ; h_cais = 156
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flut ro 1.3 eint 0.21978e6 gamm 1.3
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The fluid pressures and velocities
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mess 'NB_POIN =' (NBNO tout) ; mess
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The fluid pressures and velocities
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Further FSI Example Electric arc in
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FSI Motivation (2) Fully coupled an
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Permanent FSI treatment (2) • Upo
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Geometrically complex cases • Bil
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The FSA algorithm (2) • Effect of
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The UP algorithm The method simply
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Application to Finite Volumes • T
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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 T
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Exercise/Example 4 - CONT problem (
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Exercise/Example 6 : Woodward-Colel
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Exercise/Example 7 : Exfs Test (Com
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Exercise/Example 8 : Steam Explosio
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Exercise/Example 9 Explosion in Sec
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Boundary Condition Elements: CLxx
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Boundary Condition Elements: CLxx (
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Exercise/Example 12 - Sloshing (Cou
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Spectrum at point P5 Exercise/Examp
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Exercise/Example 12 - Sloshing (9)
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Geometric data: For the fluid domai
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ECHO RESU ALIC TEMP GARD PSCR SORT
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The displacements of the horizontal
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0.00000E+00 7.11050E+00 1.06860E+00
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scen geom navi free !vect scco scal
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Alternative 2 Use multi-phase multi
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Geometric data: Complex shell-like
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At 0.5 ms, spurious (non-physical)
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Here is an example of the computed
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DIME PT2L 293 PT3L 70 ZONE 3 ED41 3
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2.10000E+01 8.33330E+00 2.10000E+01
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FLUT RO 2.4278E3 EINT 0. GAMM 0.75D
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The final mesh (colors indicate mat
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GRIL LAGR LECT stru TERM ALE LECT f
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The final bubble material mass frac
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Geometric data: Materials The explo
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Some results: global deformed mesh
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The well-known Woodward-Colella tes
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Numerical Solutions WOCO2D The mesh
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ECHO * RESU ALIC GARD PSCR * SORT G
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The EUROPLEXUS input file reads: WO
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This example is a patch of four sho
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CONT SPLA NX 1 NY 0 NZ 0 LECT symx
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TITLE: Cavi51: steam explosion in a
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p10p=p10 plus p0; p11=0.75 0 1.7; p
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*opti donn 5; * vol3=vol2 syme plan
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si (nn2 ega 1); str2=ei; sinon; str
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The EUROPLEXUS input file reads: CA
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Some results: final pressure distri
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TITLE: PPLA04: unsteady flow throug
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liq2=daller c1 c2 c3 c4 plan; lag=l
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Some results: final fluid pressure:
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PROBLEM: A deformable tank complete
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s1 = p5 d 4 p9; s2 = p9 d 16 p12; s
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The final pressure distribution in
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RESULTS: Linear theory predicts a 1
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The velocity at 200 ms is presented
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compute a reasonable initial distri
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FSA LECT 2 3 4 5 6 7 8 9 10 11 12 1
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$ $ $ 91 92 93 94 95 96 97 98 99 10
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The computed displacements of the t
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TITLE: OILCUP2: uniformly accelerat
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2526 2527 2528 2529 2530 2531 2532
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Intermediate and final oil mass fra
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Detailed contents • ALE descripti
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Example 1 - Taylor bar impact 7 Exa
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Example 3 - Coining Bilateral, Plan
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Example 3 - Coining (5) Bilateral,
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Example 4 - Explosion in a 2D box 1
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Example 5 - Explosion in a corridor
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Lagrangian contact • Non-permanen
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Example 6a - Deep drawing A (2) Vel
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Tube Crash (2) 35 Conventional cont
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The Basic Pinball Method [Belytschk
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Hierarchic Pinball Method (2) • H
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Example 7 - Cable impact (2) l L v
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Example 7b - Indentation (4) • 3D
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Example 7b - Indentation (8) • Co
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Why a Particle-Based Method? (2)
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SPH Formulation (4) • To write th
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Bird Strike [Courtesy of Snecma/CEA
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Example 8 - SPH impacts (2) SONA01:
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Spectral Elements (3) • To obtain
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Spectral Elements (7) Coupling betw
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Example 10 - Sediment valley • Ve
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Example 12 - Matsuzaki valley • M
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Time Step Partitioning (2) • Buil
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Time Step Partitioning (6) • This
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Treatment of links in partitioning
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Example 12a - Taylor bar impact rev
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Example 12a - Taylor bar impact rev
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⎧ M1U 1 = F1+ R1 ⎪ ⎨M U = F +
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Coupling at the Interfaces (7) ⎧
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Multiple scales in time • Use app
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Multiple scales in time (5) • The
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Multiple scales in space (3) •
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Example: simplified engine S 1 S 2
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Example: aircraft Material is linea
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Example 13 - Domains in 2D (2) •
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Example 14 - Domains in 3D (3) •
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Geometric data and materials: See s
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sler cam1 1 nfra 20 trac offs fich
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Geometric data and materials: See s
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POIN TOUS CHAMELEM FICH TPLO FREQ 1
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* p6=1.2E-2 0; p7=1.2E-2 1.0E-2; *
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COUR 1 'dx_P1' DEPL COMP 1 NOEU LEC
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p4s=p4 plus (0 0); tol=0.001; c1=p1
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The deformed final mesh at 4 ms wit
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The input file: INFS - 02 *--------
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The final mesh and pressures are: T
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ZONE 2 PMAT 2 Q4GS 3904 ECRO 858884
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The final deformed mesh is: The fin
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Problem description: This example i
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LECT die punch holder TERM MASS 0.1
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Geometric data and materials: See s
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geom !navi free line heou poin sphp
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An example of intermediate velociti
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LOADING: The indenter is pushed int
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The initial configuration (with par
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The reaction force is: Approximate
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The final velocities: 8
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The reaction force is: Approximate
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The initial configuration (with par
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The reaction force is: Approximate
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-5.10000E-01 1.33333E+00 2.00000E+0
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Some hardening quantity in the targ
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Final plastic streen in the target
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Final plastic strain in the target:
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Problem description: This example r
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The input file is: HOLE - 06 $ ECHO
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trac 2 1 AXES 1.0 'DISPL. [M]' yzer
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Problem description: This example r
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* fin lab3; nodf=c1p; * tpln=0.0 40
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The input file is: VALL - PS $ ECHO
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The final displacement norm in the
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VEC1=2. 0.; * P1=P11; R1=P13; S1=P1
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S3=P33 PLUS VEC1; * N1=2; N2=3; N3=
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CHPO2=CHPO2 / SCAL1; TSTAT2 . &BCL9
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AXTE 1E3 'Temps (ms)' *------------
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The final displacement field in the
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P9=40. 5. 5.; P14=20. 5. 5.; * BASE
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CLIM1=(P5 ET P6 ET P7 ET P8); CLIM1
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SUIT Post-treatment ECHO RESU ALIC
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The final displacement field in the
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tol=0.01E-3; * base=p0 d 5 p1; stru
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eli=stru elem i; si (ega j 0); sq42
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h3=8.1e-3; h4=16.2e-3; h5=24.3e-3;
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The characteristic displacements in
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The final yield stress field in the
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The mission of the Joint Research C