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BARI02 We introduce some damping (10% of the critical value: OPTI AMOR LINE 0.1) to reduce the oscillations in the numerical solution. However, this has also an effect on wavefronts, which become less steep and thus deviate more from the analytical solution. The resulting stress at the bar mid-point is: 8
BARI08 We study the effect of the time increment. The critical time step for the chosen problem is: crit L 1/100 −6 ∆ t = = 2× 10 s E / ρ 5000 −6 crit In the previous runs we have assumed a fixed time increment ∆ t = 1× 10 = 0.5⋅∆ t . Let us see what happens by taking a time increment larger than the estimated stability limit. −6 crit By assuming ∆ t = 4× 10 = 2.0⋅∆ t , the numerical solution “explodes” ( ∞ velocities) after just 10 time steps (try out to see). −6 crit By assuming ∆ t = 2× 10 = 1.0⋅∆ t , the numerical solution is carried out until the end and the results are very close to the analytical solution: Note that this calculation is done without any added damping. The fact that an almost analytical result is obtained is remarkable, but it should be noted that in practice using a time step equal to the critical value is impossible for real calculations where the mesh size and/or the material properties change in space and in time. It is always advisable to use a time increment as close as possible to the stability limit, but by bearing in mind that the latter is often just an estimation. BARI08_B Same as BARI08 but twice larger ∆ t : the calculation is unstable as expected. 9
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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
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 and 46: Some results: horizontal and vertic
Page 47: PMAT01E This test is similar to PMA
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: Note that we have also put the Pois
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
Page 96 and 97: Time integration Each time incremen
Page 98 and 99: Time integration (5) 10. Account fo
Page 100 and 101: Euler Equations (FV) • They are w
Page 102 and 103: Mesh rezoning - motivations • Rez
Page 104 and 105: Giuliani’s automatic rezoning •
Page 106 and 107: • Analytical solution (self-simil
Page 108 and 109: Shock tube (6) • Influence of pse
Page 110 and 111: Exercise 2 - Explosion in air-fille
Page 112 and 113: Exercise 3 - Bubble expansion in a
Page 114 and 115: Exercise 4 - External blast on two
Page 119 and 120: Geometric data: The calculation is
Page 121 and 122: SHOC01 Eulerian, “average” pseu
Page 123 and 124: 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
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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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