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TEST14 Same as TEST13 but uses same expected. ∆ t as TEST01. This calculation is unstable, as TEST21 To conclude this exercise, let us consider a similar problem whereby we replace the gravity g by an initial velocity v 0 directed downwards. The initial kinetic energy of the system is: 1 2 EK 0 = Mv0 2 The mass will stop when all this energy has been transformed into elastic energy in the cable or bar. Assuming linear elastic behaviour and Poisson’s coefficient ν = 0 the resistance force of the bar is: ES R= σS = EεS = λ L0 where λ is the elongation: λ L− L 0 The elastic energy is: 2 ∆L ES ES ( ∆L) EE = ∫ λdλ = 0 L0 L0 2 By posing EE = EK0 one obtains from the above expressions: LM 0 ∆ L= v0 ES In order to obtain an elongation of, say, 20 % of the initial length: ∆ L= 0.2L0 = 0.2 m the initial velocity should be: 11 −8 ES 2.0× 10 ⋅ 2.5× 10 v0 = ∆ L= = 1.414 m/s LM 0 1⋅100 Let us now compute the time τ to reach the maximum elongation. The equation of motion of the system may be written as: ES ES M λ =− L λ → λ =− 0 ML λ 0 The solution of this equation, for the particular initial conditions considered in this problem, is the following harmonic motion: λ() t = sin( ωt) with the pulsation given by: ES −1 ω = = 7.071 s ML0 i.e. the same value as in the case considered previously, with gravity and zero initial velocity. The desired time τ is clearly ¼ of an oscillation period T , and therefore is given by: T 2π τ = = = 0.222 s 4 4ω 14
The mass returns at the initial position after a time 2τ = 0.444 s . Note that these times do not depend upon the initial velocity of the mass. To check these analytical results, a calculation (TEST21) is performed, similar to case TEST11 i.e. by the small strain option EDSS, but without gravity and by an initial velocity of –1.414 m/s in the vertical direction. The input file is as follows: TEST - 21 *----------------------------------------------------------------------- ECHO *CONV win *-----------------------------------------------------------Problem type CPLA NONL LAGR *-----------------------------------------------------------Dimensioning DIME PT2L 2 FUN2 1 PMAT 1 ZONE 2 TABL 1 2 FORC 1 TERM *---------------------------------------------------------------Geometry GEOM LIBR POIN 2 FUN2 1 PMAT 1 TERM 0 0 0 -1 1 2 2 *------------------------------------------------Geometrical complements COMP EPAI 2.5E-8 LECT 1 TERM *----------------------------------------------------------Material data MATE VM23 RO 8000. YOUN 2.0E11 NU 0.0 ELAS 2.0E11 TRAC 1 2.0E11 1.E0 LECT 1 TERM MASS 100.0 LECT 2 TERM *----------------------------------------------------Boundary conditions LINK COUP BLOQ 2 LECT 1 TERM *-----------------------------------------------------Initial conditions INIT VITE 2 -1.4142 LECT 2 TERM *----------------------------------------------------------------Outputs ECRI DEPL VITE CONT ECRO TFREQ 0.5 FICH ALIC TEMP FREQ 20 POIN LECT 1 2 TERM ELEM LECT 1 TERM *----------------------------------------------------------------Options OPTI PAS UTIL NOTEST EDSS *--------------------------------------------------Transient calculation CALCUL TINI 0. TEND 1.5 PASF 0.1E-3 *=========================================================POST-TREATMENT SUIT Post-treatment ECHO RESU ALIC TEMP GARD PSCR SORT GRAP AXTE 1.0 'Time [s]' *------------------------------------------------------Curve definitions COUR 1 'dy_2' DEPL COMP 2 NOEU LECT 2 TERM COUR 2 'sg_1' CONT COMP 1 ELEM LECT 1 TERM COUR 3 'fe_2' FORC COMP 2 NOEU LECT 2 TERM COUR 4 'fe_1' FORC COMP 2 NOEU LECT 1 TERM DCOU 5 'tau' 2 0.222 -0.20 0.222 -0.25 DCOU 6 '2tau' 2 0.444 0.00 0.444 -0.25 DCOU 7 'd_min' 2 0. -0.20 1.5 -0.20 DCOU 8 'd_max' 2 0. 0.20 1.5 0.20 *------------------------------------------------------------------Plots trac 1 5 6 7 8 axes 1.0 'DISPL. [M]' yzer colo noir roug roug vert vert dash 0 2 2 2 2 noyl 0 1 1 1 1 trac 2 axes 1.0 'CONTR. [PA]' yzer trac 3 4 axes 1.0 'FORCE [N]' yzer *--------------------------------------------------Results qualification QUAL DEPL COMP 2 LECT 2 TERM REFE 1.85058E-01 TOLE 1.E-2 CONT COMP 1 LECT 1 TERM REFE -3.70116E+10 TOLE 1.E-2 *======================================================================= FIN 15
Page 1 and 2:
European Laboratory for Structural
Page 3 and 4:
Proceedings of a Course on: Numeric
Page 5:
Programme of the Course 1. Introduc
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Credits & Acknowledgments • Struc
Page 12 and 13: Introductory Example (4) 7 Applicat
Page 14 and 15: x Computational Framework • Gover
Page 16 and 17: n+ 1 n ∆t n n+ 1 u = u + ( u + u
Page 18 and 19: Scheme start-up and marching (2)
Page 20 and 21: Stress Update • To solve the equi
Page 22 and 23: Geometric non-linearities (3) Set u
Page 24 and 25: Essential Boundary Conditions Essen
Page 26 and 27: Exercise 0 - Ideal ballistics • M
Page 28 and 29: Exercise 0 - Ideal ballistics (5)
Page 30 and 31: Exercise 1 - Suspended mass (3) •
Page 32 and 33: Exercise 1 - Suspended mass (7) •
Page 34 and 35: 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: TEST13 Same as TEST01 but uses a 10
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
Page 77 and 78: ENDPLAY *==========================
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 86: Final deformation (with superposed
Page 90 and 91: Detailed Contents • Modeling the
Page 92 and 93: Modeling the fluid domain • The f
Page 94 and 95: Euler equations (3) • For a compr
Page 96 and 97: Time integration Each time incremen
Page 98 and 99: Time integration (5) 10. Account fo
Page 100 and 101: Euler Equations (FV) • They are w
Page 102 and 103: Mesh rezoning - motivations • Rez
Page 104 and 105: Giuliani’s automatic rezoning •
Page 106 and 107: • Analytical solution (self-simil
Page 108 and 109: Shock tube (6) • Influence of pse
Page 110 and 111: Exercise 2 - Explosion in air-fille
Page 112 and 113:
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
Page 137 and 138:
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
Page 301 and 302:
Some results: intermediate and fina
Page 305 and 306:
Universitat Politècnica de Catalun
Page 307 and 308:
ALE description of structures (2)
Page 309 and 310:
Example 1 - Taylor bar impact (3) B
Page 311 and 312:
Example 3 - Coining (3) Influence o
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
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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 and 362:
Coupling at the Interfaces (5) MU T
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Coupling at the Interfaces (9) Time
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
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Example 15 - Bar Impact Revisited (
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* mesh=stru et symax et viti et lil
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The displacements are: 4
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c2=p1 d 8 p2; c3=p2 d 20 p3; c4=p3
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The resulting final deformed mesh w
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TERM *-----------------------------
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Geometric data and materials: The b
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BET0 1 KINT 0 AHGF 0 CL 0.5 CQ 2.56
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INFS02 We use a twice finer fluid m
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FREQ 1 GOTR LOOP 60 OFFS FICH AVI C
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Problem description: This example i
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*----------------------------------
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The final velocities: The final for
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*----------------------------------
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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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