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*mesh=stru et symax et viti et lili et slin;tass mesh;*opti sauv form 'bari10.msh';sauv form mesh;The input file is:BARI - 10*-----------------------------------------------------------------------ECHOCONV winCAST MESH*-----------------------------------------------------------Problem typeAXIS NONL ALE*-----------------------------------------------------------DimensioningDIMEPT2L 1000 Q41 360TABL 1 5ECRO 3960NALE 4 NBLE 354SLPC 1 SLPN 61TERM*---------------------------------------------------------------GeometryGEOM Q41 STRU TERM*--------------------------------------------------Geometric ComplementsCOMP EPAI 1 TOUS*------------------------------------------------------------Grid motionGRIL LAGR LECT P0 P1 TOP TERMALE TOUSLIGN BASE LECT P0 P1 TERM LIST LECT LILI TERMSLIP EQUI LECT SLIN TERMAUTO AUTR*----------------------------------------------------------Material dataMATE VM23 RO 8930. YOUN 1.17D11 NU 0.35D0 ELAS 4.D8TRAC 2 4.D8 3.418803D-03 1.004D11 1000.003418803TOUS*----------------------------------------------------Boundary conditionsLINK COUPBLOQ 2 LECT BASE TERMCONT SPLA NX 1 NY 0 LECT SYMAX TERM*-----------------------------------------------------Initial conditionsINIT VITE 2 -227. LECT VITI TERM*----------------------------------------------------------------OutputsECRI DEPL VITE TFRE 10.E-6POIN LECT P1 TOP TERMFICH K200 TFRE 10.E-6 POIN TOUSVARI DEPL VITE ECRO ECRC LECT 2 7 TERMfich alic temp FREQ 1poin lect P1 P2 termfich alic tfre 1.e-6*----------------------------------------------------------------OptionsOPTI NOTECSTA 0.5LOG 1*--------------------------------------------------Transient calculationCALC TINI 0. TEND 80.D-6*==============================================================ANIMATIONPLAYCAME 1 EYE 4.47289E-02 -3.66604E-02 6.13469E-02! Q 9.09576E-01 2.93578E-01 2.86788E-01 -6.50846E-02VIEW -4.83497E-01 5.71394E-01 -6.63129E-01RIGH 8.27033E-01 4.99905E-02 -5.59926E-01UP 2.86788E-01 8.19152E-01 4.96732E-01FOV 2.48819E+01SCEN GEOM NAVI FREELINE HEOU SSHAISO FILL FIEL ECRO 7 SCAL USER PROG 4.0D8 PAS 0.2D8 6.6D8 TERMTEXT ISCAcolo papeLIMA ONtitl tit1 'EUROPLEXUS (C) Animation'tit2 'Taylor Bar Impact'tit3 'Author: F. Casadei'2426 of 556
sler cam1 1 nfra 20trac offs fich avi nocl nfto 101 fps 10 kfre 10 comp -1AXIS 27 270 RENDsler cam1 1 nfra 1TRAC OFFS FICH AVI CONT NOCLAXIS 27 270 RENDFREQ 0 tfre 1.d-6GOTR LOOP 79 OFFS FICH AVI CONT NOCLAXIS 27 270 RENDGOTROFFS FICH AVI CONTAXIS 27 270 RENDENDPLAY*=========================================================POST-TREATMENTSUITPost-treatmentECHORESU ALIC TEMP GARD PSCRSORT GRAPAXTE 1000.0 'Time [ms]'*------------------------------------------------------Curve definitionsCOUR 1 'dx_P1' DEPL COMP 1 LECT P1 TERMCOUR 2 'dy_P2' DEPL COMP 2 LECT P2 TERM*------------------------------------------------------------------Plotstrac 1 2 axes 1.0 'DISPL. [M]' yzer*--------------------------------------------------Results qualificationQUAL DEPL COMP 1 lect P1 term REFE 3.90999E-3 TOLE 5.E-3DEPL COMP 2 lect P2 term REFE -1.09017E-2 TOLE 5.E-3*=======================================================================FINThe resulting final deformed mesh with superposed current yield stress (a measure ofplastic deformation) is:3427 of 556
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The mission of the Institute for th
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The present document contains the p
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6 of 556
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8 LOCALWWW.CONTEXTODEDURANGO.COM.MX
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Introductory Example (2)5Introducto
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Goals/Characteristics• Simulate f
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Mu f extB Tσ dVe eV= −∑ ∫Com
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Direct Time Integration (4)• The
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Geometric non-linearitiesA large-di
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Advantages of the Method• Transie
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Finding the Lagrange Multipliers (2
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Exercise 0 - Ideal ballistics (5)
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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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RESULTS:Results are in perfect agre
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VIEW 0.00000E+00 0.00000E+00 -1.000
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Trajectories:These results are in p
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848 of 556
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Hence:11 −8ES 2× 10 ⋅ 2.5×10
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The stress history is:Note that the
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AnalyticalTEST11Same as TEST01 but
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Comparison of natural vs. engineeri
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The total external forces at the tw
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1260 of 556
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TEST14Same as TEST13 but uses samee
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The resulting displacement is shown
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The load is reversed when the cable
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The (dynamic) equilibrium is expres
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Simple 1-D wave propagation problem
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Numerical SolutionsBARI01We discret
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BARI02We introduce some damping (10
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BARI09Use a 2-D geometry (elements
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or the velocities:or the Von Mises
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1480 of 556
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The system is initially at rest, an
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CAST MESHTRID NONL LAGR$$ Dimension
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Final deformation (with superposed
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Detailed Contents• Modeling the f
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Modeling the fluid domain• The fl
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Euler equations (3)• For a compre
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Time integrationEach time increment
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Time integration (5)10. Account for
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Euler Equations (FV)• They are wr
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Mesh rezoning - motivations• Rezo
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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 pseu
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Exercise 2 - Explosion inair-filled
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Exercise 3 - Bubble expansionin a T
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Exercise 4 - External blast on two
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Analytical solution (see e.g. Harlo
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3.05000E+01 5.00000E-01 3.10000E+01
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AnalyticalConclusion: the pseudo-vi
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Finally, we investigate the effect
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“Optimal” solution (for the pre
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Initial conditions: FE vs. FVThe tw
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Example of shock tube problem solve
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The pressure distribution at t = 0.
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TANK01Lagrangian solution: the whol
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The computed fluid pressures at the
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The computed pressure histories are
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8142 of 556
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TUBE06Lagrangian solution: the whol
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And the final velocity distribution
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The computed displacements are:To s
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43 30 30 43 44 31 31 44 45 3232 45
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10152 of 556
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p6 = 70 20 0;p7 = 70 40 0;p8 = 40 4
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$ high-pressure perfect gas (explos
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*r_cais = 965.00e-3 ;h_cais = 1560.
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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 ExampleElectric arc in
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FSI Motivation (2)Fully coupled ana
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Permanent FSI treatment (2)• Upon
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Geometrically complex cases• Bila
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The FSA algorithm (2)• Effect of
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The UP algorithmThe method simply r
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Application to Finite Volumes• Th
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Exercise 2 - Explosions insimple de
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Exercise/Example 3 - Wavepropagatio
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Exercise/Example 4 - CONT problemTh
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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(Comp
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Exercise/Example 8 : Steam Explosio
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Exercise/Example 9Explosion in Seco
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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 P5Exercise/Exampl
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Exercise/Example 12 - Sloshing (9)
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Exercise/Example 13 - Building Vuln
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Exercise 14 - Improving Initial Con
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DIMEPT3L 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 139121 120 139 140122 1
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and the final velocity field:The su
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The final velocities:The final liqu
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*----------------------------------
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TWIS07We study the phenomenon by a
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6236 of 556
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DIMEPT2L 293 PT3L 70 ZONE 3ED41 33
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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 TERMALE LECT fl
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The final bubble material mass frac
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Geometric data:MaterialsThe explosi
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Some results: global deformed mesh
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Detail of first diaphragm:Detail of
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BOUNDARY CONDITIONS:The step is ent
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The EUROPLEXUS input file reads:WOC
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WOCO3DThe mesh generation file (K20
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*RESU ALIC GARD PSCR*SORT GRAP*AXTE
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PT3L 16389PT6L 504 NBLE 16389NALE 1
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Final pressure distribution:4266 of
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BOUNDARY CONDITIONS:The vessel is e
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as01=daller c1 c2 c3 c4 plan;*c1=p1
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*trac oeil cach fluidstr;*opti donn
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cam3=cam31 et cam32;camb=cam1 et ca
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log 1dtmlREZO GAM0 0.5FLMP EPS1 5.0
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Final structure deformation:Final v
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concrete, and by the VMSF (von Mise
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flut=flui elem 'TRI3';fluq=flui ele
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COMPEPAI 1.15 LECT STR1 TERM0.61 LE
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Intermediate structure velocities:F
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10288 of 556
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LOADING:The event is initiated by t
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FLUT RO .242777373 EINT 6.865E5 GAM
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Final structure velocities:6294 of
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RESULTS:A paper by Bermudez and Rod
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The applied displacement and the co
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PROBLEM:A rigid tank with a deforma
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The EUROPLEXUS input file reads:GRA
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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 1112 13
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$$$91 92 93 94 95 96 97 98 99 100 T
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The computed displacements of the t
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TITLE:OILCUP2: uniformly accelerate
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2526 2527 2528 2529 2530 2531 25322
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Intermediate and final oil mass fra
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The QUAS STAT option is used to int
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t8 = (r4*c45) (r4*c45) h2;*Group a:
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t7t8 = cerc t7 (0. 0. h2) t8 dini l
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fill0 = fil7 et fil8 et fil9 et fil
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d13d3 = u13u3 proj coni a0 plan d0
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icolo0 et ibwal1 et iswal0 et ichoi
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sph0 = sph0 et (sph0 syme poin ec0)
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Some (qualitative) results are show
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Some selected snapshots:t = 2 msT =
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CALCULATIONS:Preliminary calculatio
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QUAS02Further preliminary calculati
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s1 = p2p d 5 p3p;s2 = p3p d 10 p4p;
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The velocities and pressures during
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Detailed contents• ALE descriptio
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Example 1 - Taylor bar impact7Examp
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Example 3 - CoiningBilateral,Plane
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Example 3 - Coining (5)Bilateral, 2
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Example 4 - Explosion in a 2D box19
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Example 5 - Explosion in a corridor
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Lagrangian contact• Non-permanent
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Example 6a - Deep drawing A (2)Velo
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Tube Crash (2)35Conventional contac
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The Basic Pinball Method[Belytschko
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Hierarchic Pinball Method (2)• Hi
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Example 7 - Cable impact (2)lLv 0MW
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Example 7b - Indentation (4)• 3D
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Example 7b - Indentation (8)• Com
Page 376 and 377: Example 7d - Falling Rock Catcher
Page 378 and 379: Smoothed Particles Hydrodynamics(Co
Page 380 and 381: SPH Formulation (2)• Basic idea:
Page 382 and 383: SPH impact [Courtesy of Samtech/Son
Page 384 and 385: PRGL01:Example 8 - SPH impacts(Cour
Page 386 and 387: Spectral Elements• Motivation: lo
Page 388 and 389: Spectral Elements (5)Convergence pr
Page 390 and 391: Example 9 - Closed FE/SE interface
Page 392 and 393: Example 11 - Cylindrical valley91Cy
Page 394 and 395: Performance Optimization• All exa
Page 396 and 397: Time Step Partitioning (4)• Intri
Page 398 and 399: Time Step Partitioning (8)• Activ
Page 400 and 401: Treatment of links in partitioning
Page 402 and 403: Example 12a - Taylor bar impact rev
Page 404 and 405: Coupling at the Interfaces• Two a
Page 406 and 407: Coupling at theInterfaces (5)MUT−
Page 408 and 409: Coupling at the Interfaces (9)Time
Page 410 and 411: Multiple scales in time (3)• Expr
Page 412 and 413: Multiple scales in space• Further
Page 414 and 415: Multiple scales in frequency• Som
Page 416 and 417: Example: power plant139Example: pow
Page 418 and 419: Example: aircraft (3)Thanks toCPU g
Page 420 and 421: Example 14 - Domains in 3D• Thick
Page 422 and 423: Example 15 - Bar Impact Revisited (
Page 424 and 425: 424 of 556
Page 428 and 429: The displacements are:4428 of 556
Page 430 and 431: c2=p1 d 8 p2;c3=p2 d 20 p3;c4=p3 d
Page 432 and 433: The resulting final deformed mesh w
Page 434 and 435: TERM*------------------------------
Page 436 and 437: 8436 of 556
Page 438 and 439: p4s=p4 plus (0 0);tol=0.001;c1=p1 d
Page 440 and 441: The deformed final mesh at 4 ms wit
Page 442 and 443: The input file:INFS - 02*----------
Page 444 and 445: The final mesh and pressures are:Th
Page 446 and 447: ZONE 2PMAT 2Q4GS 3904ECRO 858884BLO
Page 448 and 449: The final deformed mesh is:The fina
Page 450 and 451: 6450 of 556
Page 452 and 453: *----------------------------------
Page 454 and 455: The final deformed mesh is:The fina
Page 456 and 457: pc = 4.5 -1 -1.5;pd = 5.5 -1 -1.5;a
Page 458 and 459: The initial configuration (with par
Page 460 and 461: 6460 of 556
Page 462 and 463: LOADING:The indenter is pushed into
Page 466 and 467: The reaction force is:Approximatean
Page 468 and 469: The final velocities:8468 of 556
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is implicitly contained within the
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The input file is:plaque poinçon p
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The configurations at 0.5 ms (half-
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The first ruptured elements appear
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The same results, more detailed, ar
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The final view is:12486 of 556
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The two supports are clamped at the
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sauv form mesh;*fin;The input file
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UP -2.75643E-01 3.29707E-01 9.02948
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poin sphp!pinb cdescolo papelima on
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The final configuration (geometry o
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NETCF8Solution with a cables ruptur
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The strain in the cables at 53 ms i
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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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10510 of 556
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opti echo 0;opti donn 'D:\Users\Fol
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*opti dime 2 elem qua4;p0 = 0 0;p1
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The final X-displacement in the hyb
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*opti dime 2 elem qua4;opti titr 'V
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VALLPSThis calculation assumes a pu
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The receiver signal in the coupled
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8524 of 556
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VEC1=2. 0.;*P1=P11;R1=P13;S1=P13 PL
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S3=P33 PLUS VEC1;*N1=2;N2=3;N3=4;*P
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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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SUITPost-treatmentECHORESU ALIC TEM
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The final displacement field in the
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tol=0.01E-3;*base=p0 d 5 p1;stru=ba
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eli=stru elem i;si (ega j 0);sq42=e
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h3=8.1e-3;h4=16.2e-3;h5=24.3e-3;*n1
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The characteristic displacements in
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The final yield stress field in the
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554 of 556
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The mission of the Joint Research C
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