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h3=8.1e-3;h4=16.2e-3;h5=24.3e-3;*n1=0;n2=0;n3=0;n4=0;n5=0;n6=0;i=0;repe loop2 (nbel stru);i=i+1;ei = stru elem i;bi = bary ei;xb yb = coor bi;si (
sq41=sq41 coul 'TURQ';sq42=sq42 coul 'ROUG';trac face (sq41 et sq42);trac qual mesh;fin;The input file is:BAMD - 05$ECHO$VERI*CONV winCAST MESHAXIS NONL$DIMEPT2L 306 Q41L 219 Q42L 31 ZONE 2mtpo 7TERM$GEOM Q41L SQ41 Q42L SQ42 TERM$COMP EPAI 1 TOUS$MATE VM23 RO 8930. YOUN 1.17D11 NU 0.35D0 ELAS 4.D8TRAC 2 4.D8 3.418803D-03 1.004D11 1000.003418803TOUS$LINK COUPBLOQ 2 LECT BASE TERMBLOQ 1 LECT SYMAX TERM$INIT VITE 2 -227. LECT VITI TERM$ECRI DEPL VITE TFRE 10.E-6POIN LECT P1 TOP TERMfich alic tfre 80.e-8$OPTI NOTEcsta 0.5log 1*STRU 6DOMA LECT s1 TERMDOMA LECT s2 TERMDOMA LECT s3 TERMDOMA LECT s4 TERMDOMA LECT s5 TERMDOMA LECT s6 TERM$CALC TINI 0. TEND 80.D-6*======================================================================SUITPost-treatment of bar impact problemECHO*RESU alic GARD PSCR*SORT GRAP*AXTE 1000.0 'Time [ms]'*COUR 1 'dr_p1' DEPL COMP 1 NOEU LECT p1 TERMCOUR 2 'dz_p2' DEPL COMP 2 NOEU LECT p2 TERMCOUR 3 'vr_p1' VITE COMP 1 NOEU LECT p1 TERMCOUR 4 'vz_p2' VITE COMP 2 NOEU LECT p2 TERM*TRAC 1 2 AXES 1.0 'DISPL. [M]' yzerTRAC 3 4 AXES 1.0 'VELOC. [M/S]' yzer*QUAL DEPL COMP 1 LECT p1 TERM REFE 3.88425E-3 TOLE 1.E-2DEPL COMP 2 LECT p2 TERM REFE -1.09666E-2 TOLE 1.E-2*======================================================================FIN7549 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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88 of 556
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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
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Example 7d - Falling Rock Catcher
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Smoothed Particles Hydrodynamics(Co
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SPH Formulation (2)• Basic idea:
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SPH impact [Courtesy of Samtech/Son
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PRGL01:Example 8 - SPH impacts(Cour
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Spectral Elements• Motivation: lo
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Spectral Elements (5)Convergence pr
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Example 9 - Closed FE/SE interface
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Example 11 - Cylindrical valley91Cy
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Performance Optimization• All exa
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Time Step Partitioning (4)• Intri
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Time Step Partitioning (8)• Activ
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Treatment of links in partitioning
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Example 12a - Taylor bar impact rev
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Coupling at the Interfaces• Two a
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Coupling at theInterfaces (5)MUT−
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Coupling at the Interfaces (9)Time
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Multiple scales in time (3)• Expr
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Multiple scales in space• Further
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Multiple scales in frequency• Som
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Example: power plant139Example: pow
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Example: aircraft (3)Thanks toCPU g
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Example 14 - Domains in 3D• Thick
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Example 15 - Bar Impact Revisited (
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*mesh=stru et symax et viti et lili
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The displacements are:4428 of 556
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c2=p1 d 8 p2;c3=p2 d 20 p3;c4=p3 d
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The resulting final deformed mesh w
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TERM*------------------------------
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8436 of 556
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p4s=p4 plus (0 0);tol=0.001;c1=p1 d
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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:Th
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ZONE 2PMAT 2Q4GS 3904ECRO 858884BLO
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The final deformed mesh is:The fina
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6450 of 556
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*----------------------------------
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The final deformed mesh is:The fina
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pc = 4.5 -1 -1.5;pd = 5.5 -1 -1.5;a
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The initial configuration (with par
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6460 of 556
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LOADING:The indenter is pushed into
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The reaction force is:Approximatean
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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
Page 498 and 499: NETCF8Solution with a cables ruptur
Page 500 and 501: The strain in the cables at 53 ms i
Page 502 and 503: -5.10000E-01 1.33333E+00 2.00000E+0
Page 504 and 505: Some hardening quantity in the targ
Page 506 and 507: Final plastic streen in the target
Page 508 and 509: Final plastic strain in the target:
Page 510 and 511: 10510 of 556
Page 512 and 513: opti echo 0;opti donn 'D:\Users\Fol
Page 514 and 515: *opti dime 2 elem qua4;p0 = 0 0;p1
Page 516 and 517: The final X-displacement in the hyb
Page 518 and 519: *opti dime 2 elem qua4;opti titr 'V
Page 520 and 521: VALLPSThis calculation assumes a pu
Page 522 and 523: The receiver signal in the coupled
Page 524 and 525: 8524 of 556
Page 526 and 527: VEC1=2. 0.;*P1=P11;R1=P13;S1=P13 PL
Page 528 and 529: S3=P33 PLUS VEC1;*N1=2;N2=3;N3=4;*P
Page 530 and 531: CHPO2=CHPO2 / SCAL1;TSTAT2 . &BCL9=
Page 532 and 533: AXTE 1E3 'Temps (ms)'*-------------
Page 534 and 535: The final displacement field in the
Page 536 and 537: P9=40. 5. 5.;P14=20. 5. 5.;*BASE =
Page 538 and 539: CLIM1=(P5 ET P6 ET P7 ET P8);CLIM1=
Page 540 and 541: SUITPost-treatmentECHORESU ALIC TEM
Page 542 and 543: The final displacement field in the
Page 544 and 545: tol=0.01E-3;*base=p0 d 5 p1;stru=ba
Page 546 and 547: eli=stru elem i;si (ega j 0);sq42=e
Page 550 and 551: The characteristic displacements in
Page 552 and 553: The final yield stress field in the
Page 554 and 555: 554 of 556
Page 556: The mission of the Joint Research C
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