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Trajectories: These results are in perfect agreement with the analytical results. An animation of the motion allows to visually compare the computed positions with the analytical ones: 6
PMAT01E This test is similar to PMAT04 but uses only one projectile (the one at 30°) and just one time step to compute the final position of the projectile as it hits the ground. The solution is in perfect agreement with the analytical one. This should not be surprising, since: • In this problem there are no deformable structures, and therefore no wave propagation phenomena. Consequently, the usual stability requirements (Courant) of explicit time integration methods do not hold in the present case and one may use an arbitrarily large time increment (as far as stability is concerned). • As concerns accuracy, the central difference scheme is second-order accurate and therefore it reproduces exactly the analytical solution which, in this case, is a second-degree polynomial (parabola). A summary of solutions computed with different time increments is given in the following Table: 7
Page 1 and 2: European Laboratory for Structural
Page 3 and 4: Proceedings of a Course on: Numeric
Page 5: Programme of the Course 1. Introduc
Page 10 and 11: Credits & Acknowledgments • Struc
Page 12 and 13: Introductory Example (4) 7 Applicat
Page 14 and 15: x Computational Framework • Gover
Page 16 and 17: n+ 1 n ∆t n n+ 1 u = u + ( u + u
Page 18 and 19: Scheme start-up and marching (2)
Page 20 and 21: Stress Update • To solve the equi
Page 22 and 23: Geometric non-linearities (3) Set u
Page 24 and 25: Essential Boundary Conditions Essen
Page 26 and 27: Exercise 0 - Ideal ballistics • M
Page 28 and 29: Exercise 0 - Ideal ballistics (5)
Page 30 and 31: Exercise 1 - Suspended mass (3) •
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
Page 41 and 42: TITLE: PMAT04: motion of projectile
Page 43 and 44: sinon; tra2 = tra2 et ele; finsi; f
Page 45: Some results: horizontal and vertic
Page 50 and 51: Hence: 11 −8 ES 2× 10 ⋅ 2.5×
Page 52 and 53: The stress history is: Note that th
Page 54 and 55: Analytical TEST11 Same as TEST01 bu
Page 56 and 57: Comparison of natural vs. engineeri
Page 58 and 59: The total external forces at the tw
Page 61 and 62: TEST13 Same as TEST01 but uses a 10
Page 63 and 64: The mass returns at the initial pos
Page 65 and 66: TEST22 To verify the independence o
Page 67 and 68: Linear dynamic analysis Consider th
Page 69 and 70: • The two waves f and g propagate
Page 71 and 72: Analytical solution For the bar imp
Page 73 and 74: Note that we have also put the Pois
Page 75 and 76: BARI08 We study the effect of the t
Page 77 and 78: ENDPLAY *==========================
Page 79: BARI10 Same as BARI09 but compares
Page 82 and 83: The system is initially at rest, an
Page 84 and 85: CAST MESH TRID NONL LAGR $ $ Dimens
Page 86: Final deformation (with superposed
Page 90 and 91: Detailed Contents • Modeling the
Page 92 and 93: Modeling the fluid domain • The f
Page 94 and 95: 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
Page 183 and 184:
Application to Finite Volumes (3) 2
Page 185 and 186:
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
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The input file is: INDE - 13 ECHO !
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The final displacement norm: The fi
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elim tol (piece et cp); c_p = piece
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The final deformed shape is: 13
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Numerical Solutions PRGL01 Simplifi
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FOV 2.48819E+01 SCEN GEOM NAVI FREE
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ROMA01 Final plastic streen in the
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SONA01 7
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Final density in the projectile: Fi
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opti echo 0; opti donn 'D:\Users\Fo
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* opti dime 2 elem qua4; p0 = 0 0;
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The final X-displacement in the hyb
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* opti dime 2 elem qua4; opti titr
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VALLPS This calculation assumes a p
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The receiver signal in the coupled
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Problem description: This example r
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CALC TINI 0. TFIN 40.E-3 *=========
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REPETER BCL2 (2); REPETER BCL3 (2);
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*----------------------------------
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The tip displacement in the referen
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LOG 1 DPSD *-----------------------
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TRAC CACH OEIL2 ZONE1; TRAC CACH OE
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The tip displacement in the referen
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$ INIT VITE 2 -227. LECT VITI TERM
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* AXTE 1000.0 'Time [ms]' * COUR 1
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sq41=sq41 coul 'TURQ'; sq42=sq42 co
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The displacements in the case with
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European Commission DG Joint Resear
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