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The total external forces at the two extremities have the form shown below. Note that the force applied to the moving end is the sum of the applied force (constant) and of the damping force caused by the quasi-static option, which is proportional to the velocity Damping force TEST08B Same as TEST08 but uses small-strain option OPTI EDSS. In that case the expected displacement is –0.2 and the expected stress is 4.E10. In this calculation we use the expected damping frequency for the linear case: f sys = 1.125 Hz. 10
TEST12 Same as TEST01 but uses a time increment 100 times larger, i.e. 10 ms instead of 0.1 ms. This value violates the Courant condition, as it is readily verified. In fact in this case the element length is L = 1 m while the sound speed in the cable material is: 11 c= E/ ρ = 2× 10 / 8000 = 5000 m/s Therefore the estimated critical time increment would be: crit ∆ t L/ c= 1/ 5000 = 0.2 ms Nevertheless, if one runs the program with the excessive value of time increment, the obtained results are quite similar to those obtained with the stability-compliant value (see graphs below). The reason is that this is a very special case. Having used just one element to “discretize” the problem geometry, there are no wave propagation phenomena in the numerical model: one node is fixed and the other one receives the applied load. Therefore, the stability condition does not apply in this case. Try out a finer discretization, involving more than one element, to see the disastrous effect of using a time increment beyond the stability limit on the numerical solution. 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
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 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 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
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
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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)
Page 309 and 310:
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
Page 329 and 330:
Example 7b - Indentation (2) • 2D
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Example 7b - Indentation (6) • 3D
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Smoothed Particles Hydrodynamics (C
Page 335 and 336:
SPH Formulation (2) • Basic idea:
Page 337 and 338:
SPH impact [Courtesy of Samtech/Son
Page 339 and 340:
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
Page 347 and 348:
Example 11 - Cylindrical valley 85
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Performance Optimization • All ex
Page 351 and 352:
Time Step Partitioning (4) • Intr
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Time Step Partitioning (8) • Acti
Page 355 and 356:
Treatment of links in partitioning
Page 357 and 358:
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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