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The mission of the Institute for the Protection and Security of the Citizen is to provids research- based, systemsorientedsupport to EU policies so as to protect the citizen against economic and technological risk. The Institutemaintains and develops its expertise and networks in information, communication, space and engineeringtechnologies in support of its mission. The strong cross-fertilisation between its nuclear and non-nuclear activitiesstrengthens the expertise it can bring to the benefit of customers in both domains.European CommissionDirectorate-General Joint Research CentreInstitute the Protection and Security of the CitizenContact informationAddress: Folco CasadeiE-mail: folco.casadei@jrc.itTelephone : +39 033278 9563Fax : +39 033278 9049http://ipsc.jrc.ec.europa.eu/home.phphttp://www.jrc.cec.eu.intLegal NoticeNeither the European Commission nor any person acting on behalf of theCommission is responsible for the use which might be made of this publication.A great deal of additional information on the European Union is available on the Internet.It can be accessed through the Europa serverhttp://europa.eu.int© European Communities, 2007Reproduction is authorised provided the source is acknowledgedPrinted in Italy2 of 556
Proceedings of a Course on:Numerical Simulationof Fast Transient Dynamic Phenomenain Fluid-Structure Systems(Presentation slides and trace of proposed exercises/examples)A Short Course given by:F. CasadeiatUniversitat Politècnica de Catalunya, Barcelona, 14-17 May 20073 of 556
Page 4 and 5: The present document contains the p
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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 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
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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
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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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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 =
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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;
Page 344 and 345:
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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Page 472 and 473:
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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
Page 520 and 521:
VALLPSThis calculation assumes a pu
Page 522 and 523:
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)'*-------------
Page 534 and 535:
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=
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
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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
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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