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Detailed contents• ALE description for structures• Non-conforming FSI• Lagrangian contact:‣ Classical methods‣ Pinballs‣ SPH• Spectral elements• Space partitioning and domain decomposition3ALE description of structuresDonea, Huerta, Casadei (early 1990s)• Extend ALE formulation to materials with memory,namely, non-linear path-dependent materialsyf ( x)yxxxxNo memoryMemory4348 of 5562
ALE description of structures (2)• Major extra difficulty w/r to Newtonian fluids: necessity to transportstresses and stress-related quantities across inter-element boundaries.Fields are usually discontinuous and evaluated only at Gauss Points• Previous attempts had used implicit interpolation techniques: tooexpensive in the present explicit fast transient context• Two distinct strategies, borrowed from the fluid dynamic experience:a) a Lax-Wendroff scheme based on nodal averaging and smoothingof the stress gradients, and b) a Godunov scheme inspired bymethods often used for conservation laws in finite volumes• Both techniques implemented in 2D quadrilateral finite elements usingsingle-point as well as multiple-point spatial numerical integration5ALE description of structures (3)• When ALE is applied to nonlinear path-dependent materials, threeconservation equations basically similar to Euler equations areobtained and therefore the same techniques (time integrationstrategy, rezoning algorithms, treatment of boundary conditions, etc.)developed for the fluids can be directly applied, with the importantexception, however, of the stress transport algorithm• Time integration is achieved by same fractional step strategy seenabove for fluids: 1) Lagrangian phase, in which it is assumed thatthe mesh follows the material particles and transport terms vanish;2) Convective flux calculation (transport terms only)• The time integration procedure still remains completely explicit andonly an extra loop over the elements to deal with transport is requiredat each time step, compared with the Lagrangian case6349 of 5563
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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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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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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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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
Page 298 and 299: The applied displacement and the co
Page 300 and 301: PROBLEM:A rigid tank with a deforma
Page 302 and 303: The EUROPLEXUS input file reads:GRA
Page 304 and 305: The velocity at 200 ms is presented
Page 306 and 307: compute a reasonable initial distri
Page 308 and 309: FSA LECT 2 3 4 5 6 7 8 9 10 1112 13
Page 310 and 311: $$$91 92 93 94 95 96 97 98 99 100 T
Page 312 and 313: The computed displacements of the t
Page 314 and 315: TITLE:OILCUP2: uniformly accelerate
Page 316 and 317: 2526 2527 2528 2529 2530 2531 25322
Page 318 and 319: Intermediate and final oil mass fra
Page 320 and 321: The QUAS STAT option is used to int
Page 322 and 323: t8 = (r4*c45) (r4*c45) h2;*Group a:
Page 324 and 325: t7t8 = cerc t7 (0. 0. h2) t8 dini l
Page 326 and 327: fill0 = fil7 et fil8 et fil9 et fil
Page 328 and 329: d13d3 = u13u3 proj coni a0 plan d0
Page 330 and 331: icolo0 et ibwal1 et iswal0 et ichoi
Page 332 and 333: sph0 = sph0 et (sph0 syme poin ec0)
Page 334 and 335: Some (qualitative) results are show
Page 336 and 337: Some selected snapshots:t = 2 msT =
Page 338 and 339: CALCULATIONS:Preliminary calculatio
Page 340 and 341: QUAS02Further preliminary calculati
Page 342 and 343: s1 = p2p d 5 p3p;s2 = p3p d 10 p4p;
Page 344 and 345: The velocities and pressures during
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Page 350 and 351: Example 1 - Taylor bar impact7Examp
Page 352 and 353: Example 3 - CoiningBilateral,Plane
Page 354 and 355: Example 3 - Coining (5)Bilateral, 2
Page 356 and 357: Example 4 - Explosion in a 2D box19
Page 358 and 359: Example 5 - Explosion in a corridor
Page 360 and 361: Lagrangian contact• Non-permanent
Page 362 and 363: Example 6a - Deep drawing A (2)Velo
Page 364 and 365: Tube Crash (2)35Conventional contac
Page 366 and 367: The Basic Pinball Method[Belytschko
Page 368 and 369: Hierarchic Pinball Method (2)• Hi
Page 370 and 371: Example 7 - Cable impact (2)lLv 0MW
Page 372 and 373: Example 7b - Indentation (4)• 3D
Page 374 and 375: 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
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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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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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*----------------------------------
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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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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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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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The mission of the Joint Research C
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