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tol=0.01E-3; * base=p0 d 5 p1; stru=base TRAN 50 p2; elim tol (stru et p2); * symax=stru poin droi p0 p2 tol; top=stru poin droi p2 (p1 PLUS p2) tol; * str1=chan 'POI1' stru; bas1=chan 'POI1' base; viti=diff str1 bas1; * nstr=nbel stru; nten=(nstr + 1) / 8; nten=enti nten; j=0; * i=0 - 3;repe loop nten;i=i+8; si (
$ INIT VITE 2 -227. LECT VITI TERM $ ECRI DEPL VITE TFRE 10.E-6 POIN LECT P1 TOP TERM fich alic tfre 80.e-8 $ OPTI NOTE csta 0.5 log 1 $ CALC TINI 0. TEND 80.D-6 *====================================================================== SUIT Post-treatment of bar impact problem ECHO * RESU alic GARD PSCR * SORT GRAP * AXTE 1000.0 'Time [ms]' * COUR 1 'dr_p1' DEPL COMP 1 NOEU LECT p1 TERM COUR 2 'dz_p2' DEPL COMP 2 NOEU LECT p2 TERM COUR 3 'vr_p1' VITE COMP 1 NOEU LECT p1 TERM COUR 4 'vz_p2' VITE COMP 2 NOEU LECT p2 TERM * TRAC 1 2 AXES 1.0 'DISPL. [M]' yzer TRAC 3 4 AXES 1.0 'VELOC. [M/S]' yzer * QUAL DEPL COMP 1 LECT p1 TERM REFE 3.88425E-3 TOLE 1.E-3 DEPL COMP 2 LECT p2 TERM REFE -1.09666E-2 TOLE 1.E-3 *====================================================================== FIN BAMD03 This calculation uses the seame mesh as the previous one (no domain decomposition). The time step partitioning technique is activated by specifying OPTI PART. The mesh generation file is: *%siz 40 * opti echo 1; * opti titr 'BAMD - 03'; opti dime 2 elem qua4; * p0=0 0; p1=3.2E-3 0; p2=0 32.4E-3; tol=0.01E-3; * base=p0 d 5 p1; stru=base TRAN 50 p2; elim tol (stru et p2); * symax=stru poin droi p0 p2 tol; top=stru poin droi p2 (p1 PLUS p2) tol; * str1=chan 'POI1' stru; bas1=chan 'POI1' base; viti=diff str1 bas1; * nstr=nbel stru; nten=(nstr + 1) / 8; nten=enti nten; j=0; * i=0 - 3;repe loop nten;i=i+8; si (
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European Laboratory for Structural
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Proceedings of a Course on: Numeric
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Programme of the Course 1. Introduc
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Credits & Acknowledgments • Struc
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Introductory Example (4) 7 Applicat
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x Computational Framework • Gover
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n+ 1 n ∆t n n+ 1 u = u + ( u + u
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Scheme start-up and marching (2)
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Stress Update • To solve the equi
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Geometric non-linearities (3) Set u
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Essential Boundary Conditions Essen
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Exercise 0 - Ideal ballistics • M
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Exercise 0 - Ideal ballistics (5)
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Exercise 1 - Suspended mass (3) •
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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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TITLE: PMAT04: motion of projectile
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sinon; tra2 = tra2 et ele; finsi; f
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Some results: horizontal and vertic
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PMAT01E This test is similar to PMA
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Hence: 11 −8 ES 2× 10 ⋅ 2.5×
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The stress history is: Note that th
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Analytical TEST11 Same as TEST01 bu
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Comparison of natural vs. engineeri
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The total external forces at the tw
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TEST13 Same as TEST01 but uses a 10
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The mass returns at the initial pos
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TEST22 To verify the independence o
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Linear dynamic analysis Consider th
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• The two waves f and g propagate
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Analytical solution For the bar imp
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Note that we have also put the Pois
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BARI08 We study the effect of the t
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ENDPLAY *==========================
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BARI10 Same as BARI09 but compares
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The system is initially at rest, an
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CAST MESH TRID NONL LAGR $ $ Dimens
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Final deformation (with superposed
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Detailed Contents • Modeling the
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Modeling the fluid domain • The f
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Euler equations (3) • For a compr
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Time integration Each time incremen
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Time integration (5) 10. Account fo
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Euler Equations (FV) • They are w
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Mesh rezoning - motivations • Rez
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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 pse
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Exercise 2 - Explosion in air-fille
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Exercise 3 - Bubble expansion in a
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Exercise 4 - External blast on two
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Geometric data: The calculation is
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SHOC01 Eulerian, “average” pseu
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COUR 7 'ro_a' ECRO COMP 2 ELEM LECT
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Analytical Conclusion: the pseudo-v
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SHOC13 Eulerian, very low pseudo-vi
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Analytical General conclusions: •
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The inverse path is not so straight
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The input files (mesh generation by
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Geometric data: The calculation is
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COUR 1 'dx_4' DEPL COMP 1 NOEU LECT
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TANK02 Eulerian solution: the whole
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The final pressure distribution is:
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TUBE06 Lagrangian solution: the who
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And the final velocity distribution
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The computed displacements are: To
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43 30 30 43 44 31 31 44 45 32 32 45
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Geometric data: External blast in a
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abs4 = dall (p4 d 28 p4u) (p4u d 40
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Geometric data: Internal blast in a
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air = air1 ET air2 ET air3 ET air4
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COURBE 8 'P e_p12' ECROU COMP 1 lec
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3D axisymmetric simulation: JWLS3G
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point lect 1 term elem lect 1 term
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Universitat Politècnica de Catalun
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• Motivation Detailed Contents
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FSI Classification FSI for compress
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The 2-D plane case (2) Use geometri
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Classification of FSI algorithms
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The FSA algorithm (4) The case of s
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The FSCR algorithm Combination of F
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Application to Finite Volumes (3) 2
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Exercise 2 - Explosions in simple d
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Exercise/Example 3 - Wave propagati
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Exercise/Example 4 - CONT problem (
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Exercise/Example 5 - Explosion in a
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Exercise/Example 6 : Woodward-Colel
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Geometry: Exercise/Example 8 Steam
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Exercise/Example 9 Explosion in Sec
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Exercise/Example 9 Explosion in Sec
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Boundary Condition Elements: CLxx (
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Exercise/Example 11 - Perforated Pl
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Exercise/Example 12 - Sloshing (3)
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DIME PT3L 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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120 119 138 139 121 120 139 140 122
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and the final velocity field: The s
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The final velocities: The final liq
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*----------------------------------
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TWIS07 We study the phenomenon by a
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This is a well-known reactor safety
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1.53330E+00 1.32310E+01 1.15000E+00
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200 199 239 240 200 200 200 240 241
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*----------------------------------
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The final fluid pressures: CONT02 T
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The final fluid velocities: The fin
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This example consists in a long 3D
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GEOM FL38 flui Q4GS stru TERM COMP
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Detail of first diaphragm: Detail o
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BOUNDARY CONDITIONS: The step is en
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The EUROPLEXUS input file reads: WO
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WOCO3D The mesh generation file (K2
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* RESU ALIC GARD PSCR * SORT GRAP *
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PT3L 16389 PT6L 504 NBLE 16389 NALE
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Final pressure distribution: 4
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BOUNDARY CONDITIONS: The vessel is
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as01=daller c1 c2 c3 c4 plan; * c1=
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*trac oeil cach fluidstr; *opti don
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cam3=cam31 et cam32; camb=cam1 et c
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log 1 dtml REZO GAM0 0.5 FLMP EPS1
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Final structure deformation: Final
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LOADING: The event is initiated by
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FLUT RO .242777373 EINT 6.865E5 GAM
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Final structure velocities: 6
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RESULTS: A paper by Bermudez and Ro
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The applied displacement and the co
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PROBLEM: A rigid tank with a deform
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The EUROPLEXUS input file reads: GR
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PROBLEM: A rigid tank with rigid or
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COMP EPAI 0.005 LECT 101 102 103 10
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Numerical Solution (flexible bottom
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! VARI DEPL VITE ECRO ! fich alic t
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The final pressures in the rigid an
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LOADING: A constant gravity in the
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Some results: intermediate and fina
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Universitat Politècnica de Catalun
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ALE description of structures (2)
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Example 1 - Taylor bar impact (3) B
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Example 3 - Coining (3) Influence o
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• Tentative classification: Non-c
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Example 4 - Explosion in a 2D box (
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Example 6 - LOCA in the HDR (2) A -
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Examples of “Slow” Transient Co
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Examples of Fast Transient Impact
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Components of Contact-Impact Method
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Shortcomings • Slender or distort
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Example - Bar impact • (See Part
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Example 7b - Indentation (2) • 2D
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Example 7b - Indentation (6) • 3D
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Smoothed Particles Hydrodynamics (C
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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 (Co
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Spectral Elements • Motivation: l
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Spectral Elements (5) Convergence p
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Example 9 - Closed FE/SE interface
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Example 11 - Cylindrical valley 85
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Performance Optimization • All ex
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Time Step Partitioning (4) • Intr
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Time Step Partitioning (8) • Acti
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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
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Coupling at the Interfaces (5) MU T
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Coupling at the Interfaces (9) Time
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Multiple scales in time (3) • Exp
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Multiple scales in space • Furthe
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Multiple scales in frequency • So
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Example: power plant 133 Example: p
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Example: aircraft (3) Thanks to CPU
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Example 14 - Domains in 3D • Thic
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Example 15 - Bar Impact Revisited (
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* mesh=stru et symax et viti et lil
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The displacements are: 4
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c2=p1 d 8 p2; c3=p2 d 20 p3; c4=p3
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The resulting final deformed mesh w
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TERM *-----------------------------
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Geometric data and materials: The b
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BET0 1 KINT 0 AHGF 0 CL 0.5 CQ 2.56
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INFS02 We use a twice finer fluid m
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FREQ 1 GOTR LOOP 60 OFFS FICH AVI C
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Problem description: This example i
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*----------------------------------
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The final velocities: The final for
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*----------------------------------
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The final deformed mesh is: The fin
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pc = 4.5 -1 -1.5; pd = 5.5 -1 -1.5;
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The initial configuration (with par
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TITLE: Indentation problem. PROBLEM
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cp = p0 d 10 p13; elim tol (piece e
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The final velocities: The final dis
Page 423 and 424: The input file is: INDE - 13 ECHO !
Page 425 and 426: The final displacement norm: The fi
Page 427 and 428: elim tol (piece et cp); c_p = piece
Page 429 and 430: The final deformed shape is: 13
Page 431 and 432: Numerical Solutions PRGL01 Simplifi
Page 433 and 434: FOV 2.48819E+01 SCEN GEOM NAVI FREE
Page 435 and 436: ROMA01 Final plastic streen in the
Page 437 and 438: SONA01 7
Page 439: Final density in the projectile: Fi
Page 442 and 443: opti echo 0; opti donn 'D:\Users\Fo
Page 444 and 445: * opti dime 2 elem qua4; p0 = 0 0;
Page 446 and 447: The final X-displacement in the hyb
Page 448 and 449: * opti dime 2 elem qua4; opti titr
Page 450 and 451: VALLPS This calculation assumes a p
Page 452 and 453: The receiver signal in the coupled
Page 455 and 456: Problem description: This example r
Page 457 and 458: CALC TINI 0. TFIN 40.E-3 *=========
Page 459 and 460: REPETER BCL2 (2); REPETER BCL3 (2);
Page 461 and 462: *----------------------------------
Page 463 and 464: The tip displacement in the referen
Page 465 and 466: Problem description: This example r
Page 467 and 468: LOG 1 DPSD *-----------------------
Page 469 and 470: TRAC CACH OEIL2 ZONE1; TRAC CACH OE
Page 471 and 472: The tip displacement in the referen
Page 473: Problem description: This example r
Page 477 and 478: * AXTE 1000.0 'Time [ms]' * COUR 1
Page 479 and 480: sq41=sq41 coul 'TURQ'; sq42=sq42 co
Page 481 and 482: The displacements in the case with
Page 485 and 486: European Commission DG Joint Resear
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