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Exercise 1 – Suspended mass (7)• TEST12 : time increment 50 times the critical value47Exercise 2 – Wave propagation• Obtain 1D analytical solution• Discuss numerical solutionE, ρ,νv 0L• Why was cross-section not specified?• Study effect of time increment• Study effect of Poisson’s ratio …L = 1 mv = 100 m/s0E = ×112 10 Paρ = 8000 kg/mν = 0.334832 of 55624
Exercise 2 – Wave propagation (2)• BARI01 : 100 elements of type FUN2 (cable), VM23 material(traction/compression), small-strain option (OPTI EDSS)Analytical49Exercise 2 – Wave propagation (3)• EUROPLEXUS input file:BARI - 01*-----------------------------------------------------------------------ECHO*CONV winCAST mesh*-----------------------------------------------------------Problem typeCPLA NONL LAGR LAGC*-----------------------------------------------------------DimensioningDIMEPT2L 102 FUN2 100 PMAT 1 ZONE 2BLOQ 2TABL 1 2FORC 1IMPA 1 PSIM 1TERM*---------------------------------------------------------------GeometryGEOM FUN2 bar PMAT obst TERM*------------------------------------------------Geometrical complementsCOMP EPAI 2.5E-8 LECT bar TERM*----------------------------------------------------------Material dataMATE VM23 RO 8000. YOUN 2.0E11 NU 0.0 ELAS 2.0E11TRAC 1 2.0E11 1.E0LECT bar TERMMASS 1.0 LECT obst TERM*----------------------------------------------------Boundary conditionsLIAI freq 1BLOQ 12 LECT obst TERMIMPA DDL 1 COTE -1PROJ LECT obst TERMCIBL LECT p2 TERM*-----------------------------------------------------Initial conditionsINIT VITE 1 100.0 LECT bar TERM*----------------------------------------------------------------OutputsECRI DEPL VITE CONT ECRO TFREQ 0.1E-3FICH ALIC TEMP FREQ 1POIN LECT p3 p2 TERMELEM LECT e1 e2 TERM*----------------------------------------------------------------OptionsOPTI PAS UTIL NOTESTLOG 1EDSS*--------------------------------------------------Transient calculationCALCUL TINI 0. TEND 0.5E-3 PASF 0.1E-5*=========================================================POST-TREATMENTSUITPost-treatmentECHORESU ALIC TEMP GARD PSCRSORT GRAPAXTE 1000.0 'Time [ms]'*------------------------------------------------------Curve definitionsCOUR 1 'dx_2' DEPL COMP 1 NOEU LECT p2 TERMCOUR 2 'dx_3' DEPL COMP 1 NOEU LECT p3 TERMCOUR 3 'sg_1' CONT COMP 1 ELEM LECT e1 TERMCOUR 4 'sg_2' CONT COMP 1 ELEM LECT e2 TERMDCOU 5 'Analytical' 60.0 0.00.1E-3 0.00.1E-3 -4.E90.3E-3 -4.E90.3E-3 0.00.5E-3 0.0*------------------------------------------------------------------Plotstrac 1 axes 1.0 'DISPL. [M]'trac 2 axes 1.0 'DISPL. [M]'trac 3 axes 1.0 'CONTR. [PA]'trac 4 axes 1.0 'CONTR. [PA]'trac 5 3 4 axes 1.0 'CONTR. [PA]' yzercolo roug noir noirdash 2 0 0*--------------------------------------------------Results qualificationQUAL DEPL COMP 1 LECT p2 TERM REFE -9.76759E-3 TOLE 1.E-2DEPL COMP 1 LECT p3 TERM REFE -9.86359E-3 TOLE 1.E-2CONT COMP 1 LECT e1 TERM REFE 1.00128E+9 TOLE 1.E-2CONT COMP 1 LECT e2 TERM REFE 1.31288E+9 TOLE 1.E-2*=======================================================================FIN5033 of 55625
Page 2 and 3: The mission of the Institute for th
Page 4 and 5: The present document contains the p
Page 6 and 7: 6 of 556
Page 8: 8 LOCALWWW.CONTEXTODEDURANGO.COM.MX
Page 11 and 12: Introductory Example (2)5Introducto
Page 13 and 14: Goals/Characteristics• Simulate f
Page 15 and 16: Mu f extB Tσ dVe eV= −∑ ∫Com
Page 17: Direct Time Integration (4)• The
Page 21 and 22: Geometric non-linearitiesA large-di
Page 23 and 24: Advantages of the Method• Transie
Page 25 and 26: Finding the Lagrange Multipliers (2
Page 28: Exercise 0 - Ideal ballistics (5)
Page 34 and 35: Exercise 2 - Wave propagation (4)
Page 36 and 37: Exercise 3 - Impact on Cooling Towe
Page 38 and 39: 38 of 556
Page 40 and 41: 40 of 556
Page 42 and 43: RESULTS:Results are in perfect agre
Page 44 and 45: VIEW 0.00000E+00 0.00000E+00 -1.000
Page 46 and 47: Trajectories:These results are in p
Page 48 and 49: 848 of 556
Page 50 and 51: Hence:11 −8ES 2× 10 ⋅ 2.5×10
Page 52 and 53: The stress history is:Note that the
Page 54 and 55: AnalyticalTEST11Same as TEST01 but
Page 56 and 57: Comparison of natural vs. engineeri
Page 58 and 59: The total external forces at the tw
Page 60 and 61: 1260 of 556
Page 62 and 63: TEST14Same as TEST13 but uses samee
Page 64 and 65: The resulting displacement is shown
Page 66 and 67: The load is reversed when the cable
Page 68 and 69: The (dynamic) equilibrium is expres
Page 70 and 71: Simple 1-D wave propagation problem
Page 72 and 73: Numerical SolutionsBARI01We discret
Page 74 and 75: BARI02We introduce some damping (10
Page 76 and 77: BARI09Use a 2-D geometry (elements
Page 78 and 79: or the velocities:or the Von Mises
Page 80 and 81: 1480 of 556
Page 82 and 83:
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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116 of 556
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118 of 556
Page 120 and 121:
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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168 of 556
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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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Page 214 and 215:
Page 216 and 217:
Exercise 14 - Improving Initial Con
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218 of 556
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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 =
Page 338 and 339:
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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346 of 556
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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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424 of 556
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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
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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
Page 552 and 553:
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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