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Exercise 0 – Ideal ballistics • Motion in vacuum is analytical: v = v = v cosφ x 0x 0 0 v = v − gt = v sinφ −gt y 0y 0 0 • The trajectory is a parabola: φ = φ and v = v 2 0 2 0 • Time to reach highest point is when v = 0 , i.e.: v0 y v0 t1 = = sinφ0 g g y • By symmetry, time to impact is twice as long: v0 y v0 t2 = 2t1 = 2 = 2 sinφ0 g g 35 Exercise 0 – Ideal ballistics (2) • The rangeis therefore: 2 2 v0 y v0 v0 X = vxt2 = 2v0x = 2 sinφ0cosφ0 = sin(2 φ0) g g g v = v = v cosφ x 0x 0 0 v = v − gt = v sinφ −gt y 0y 0 0 v0 y v0 t1 = = sinφ0 g g v0 y v0 t2 = 2t1 = 2 = 2 sinφ0 g g • Max range is when shooting at 45º: 2 0 X max ( φ0) = v for sin(2 φ0) = 1 ⇔ φ0 = π 4 g • The position at the generic time is: xt () = v t= vtcosφ t t yt () = v0yt− g = vt 0 sinφ0 −g 2 2 0x 0 0 2 2 • Max elevation depends only on v 0y : 2 2 2 2 v0y g v0y v0y v0 2 Y = ymax = y( t1) = − = = sin φ 2 0 g 2 g 2g 2g 36 18
Exercise 0 – Ideal ballistics (3) • Study motion of projectile as a function of the shooting angle: v 0 = 100 φ 0 = 30 ° ,45 ° ,60° • Computed vs. analytical positions: 37 Exercise 0 – Ideal ballistics (4) • Computed trajectories: 38 19
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 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 and 74: Note that we have also put the Pois
Page 75 and 76: 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
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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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