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The FSA algorithm (2) • Effect of progressive “sharpening” of interface corner: n v n γ v n γ v v = 0 n 2 γ y n 1 x γ > γ → 2 normals! S k • Face vectors are subdivided into one or more groups. Vectors in same group form angle < than given value . • Each group is used to generate one normal. • If the number of independent normals equals the space dimension, the node is set Lagrangian and “tied” to the structural node: vF = vS γ 19 The FSA algorithm (3) • Thus, in 3D cases we may have 1 or 2 normals in an ALE node: A B C v = 0 z (Lagr.) y x A: 1 normal (blocked), 2 free B: 2 normals (blocked), 1 free C: 3 normals (blocked), node becomes Lagrangian, tied to structure 20 10
The FSA algorithm (4) The case of submerged structural edges: • If the structural element has no topological thickness, the node at the tip is a singular point as concerns the normal • FSA strategy outlined so far would lead to two mutually opposite normals: one is redundant and should be rejected • Aligned flow is undisturbed, transversal flow “sticks” at tip 21 Shortcomings of the FSA algorithm • Onset of spurious velocities in 3-D models with warped element faces. Example: patch test • Reason: slight but non-negligible accuracy mismatch in calculation of internal pressure forces and of reactions (direction of the normal) • FSA method is purely geometrical and uses no information about internal fluid element formulation (e.g., spatial integration rules) • Investigate alternative methods based on equilibrium 22 11
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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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Page 129 and 130: Analytical General conclusions: •
Page 131 and 132: The inverse path is not so straight
Page 133 and 134: The input files (mesh generation by
Page 135 and 136: Geometric data: The calculation is
Page 137 and 138: COUR 1 'dx_4' DEPL COMP 1 NOEU LECT
Page 139 and 140: TANK02 Eulerian solution: the whole
Page 141: The final pressure distribution is:
Page 144 and 145: TUBE06 Lagrangian solution: the who
Page 146 and 147: And the final velocity distribution
Page 148 and 149: The computed displacements are: To
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Page 153 and 154: Geometric data: External blast in a
Page 155 and 156: abs4 = dall (p4 d 28 p4u) (p4u d 40
Page 157 and 158: Geometric data: Internal blast in a
Page 159 and 160: air = air1 ET air2 ET air3 ET air4
Page 161 and 162: COURBE 8 'P e_p12' ECROU COMP 1 lec
Page 163 and 164: 3D axisymmetric simulation: JWLS3G
Page 165 and 166: point lect 1 term elem lect 1 term
Page 169 and 170: Universitat Politècnica de Catalun
Page 171 and 172: • Motivation Detailed Contents
Page 173 and 174: FSI Classification FSI for compress
Page 175 and 176: The 2-D plane case (2) Use geometri
Page 177: Classification of FSI algorithms
Page 181 and 182: 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
Page 187 and 188: Exercise/Example 3 - Wave propagati
Page 189 and 190: Exercise/Example 4 - CONT problem (
Page 191 and 192: Exercise/Example 5 - Explosion in a
Page 193 and 194: Exercise/Example 6 : Woodward-Colel
Page 195 and 196: Geometry: Exercise/Example 8 Steam
Page 197 and 198: Exercise/Example 9 Explosion in Sec
Page 199 and 200: Exercise/Example 9 Explosion in Sec
Page 201 and 202: Boundary Condition Elements: CLxx (
Page 203 and 204: Exercise/Example 11 - Perforated Pl
Page 205 and 206: Exercise/Example 12 - Sloshing (3)
Page 207 and 208: Exercise/Example 12 - Sloshing (7)
Page 209: Exercise/Example 12 - Sloshing (11)
Page 214 and 215: DIME PT3L 23 PT2L 143 FL24 145 ED01
Page 216 and 217: The final pressure field in the flu
Page 218 and 219: ALE solution with FSA: it is not po
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Page 222 and 223: and the final velocity field: The s
Page 224 and 225: The final velocities: The final liq
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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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