Dytran Example Problem Manual

Dytran 2008 r1 ™

Example Problem Manual

Main Index

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Main Index

Contents Dytran Example Problems Manual

Contents

1

Structural Dynamics Overview

14

Impulsively Loaded Strip 15 Problem Description 15 Dytran Model 15 Results Evaluation 16 Files 17 Abbreviated Dytran Input File 18 Impulsively Loaded Cylindrical Panel Problem Description 20 Dytran Model 21 Results 21 Files 24 Abbreviated Dytran Input File 24

20

Taylor Test (a rod impacted against a rigid wall) Problem Description 26 Johnson-Cook Material Model 26 Theoretical Approach – Taylor 26 Dytran Model 27 Dytran Results 27 Tetrahedral Elements 29 Theoretical – Taylor 30 Comparison of Shape 31 Comparison of CPU time 32 Abbreviated Dytran Input File 34 Taylor Test with Euler 36 Problem Description 36 Dytran Model 36 Dytran File 39 Hourglassing in Hexahedron Elements Problem Description 41 Dytran Model 42

Main Index

41

26

4 Dytran Example Problems Manual

Results 43 Files 45 Abbreviated Dytran Input File

2

46

Structural Contact Overview

50

Three-plate Contact 51 Problem Description 51 Theoretical Result 52 Dytran Model 52 Results 52 Files 55 Abbreviated Dytran Input File

55

Ball Penetrating a Steel Plate 58 Problem Description 58 Desired Results 60 Dytran Modeling 61 Results 61 Files 63 Abbreviated Dytran Input File 64 Tapered Beam Striking a Rigid Wall Problem Description 66 Desired Results 66 Dytran Modeling 66 Results 67 Files 70 Abbreviated Dytran Input File 70 Impact Loading 73 Problem Description 73 Theoretical Results 73 Dytran Model 74 Results 74 Files 75 Abbreviated Dytran Input File Pipe Whip 79 Problem Description Dytran Model 80 Results 81

Main Index

79

75

66

Contents 5

Files 82 Abbreviated Dytran Input File

3

82

Fluid Dynamics Overview

86

Shock Tube 87 Problem Description 87 Desired Results 88 Dytran Modeling 88 Coarse Model 89 Fine Model 89 Results 89 Coarse Model 89 Model 91 Coarse Mesh 92 Fine Mesh 92 Abbreviated Dytran Input File

92

Blast Wave 94 Problem Description 94 Theoretical Solution 94 Dytran Modeling 95 Results 96 Files 97 Abbreviated Dytran Input File

97

JWL Explosive Test 99 Problem Description 99 Theoretical Background 99 Dytran Model 102 Results 104 Files 105 Abbreviated Dytran Input File

105

Modeling Blast Wave using 1-D Spherical Symmetry Method Problem Description 107 Dytran Modeling 107 Results 110 Abbreviated Dytran Input File 118

Main Index

107

6 Dytran Example Problems Manual

Modeling the JWL Explosion using 1-D Spherical Symmtery Problem Description 125 Dytran Modeling 125 Results 127 Abbreviated Dytran Input File 134 Nonuniformity with MESH,BOX 137 Problem Description 137 Dytran Modeling 137 Results 141 Abbreviated Dytran Input File 143

4

Fluid-structure Interaction Overview

147

Shock Formation 148 Problem Description 148 Theoretical Background 148 Dytran Model 149 Results 150 Files 152 Abbreviated Dytran Input File 152 Blast Containment in a Luggage Container Problem Description 154 Dytran Modeling 155 Results 155 Files 157 References 157 Abbreviated Dytran Input File 157 Multiple Bird-strike on a Cylindrical Panel Problem Description 161 Dytran Model 162 Results 164 Files 166 Abbreviated Dytran Input File 166 Slanted Piston 169 Problem Description 169 Desired Results 170 Dytran Modeling 171 Results 171

Main Index

154

161

125

Contents 7

Files 173 Abbreviated Dytran Input File Sloshing using ALE Method Problem Description 176 Dytran Model 177 User Subroutine 177 Results 178 Files 179 References 179 Abbreviated Dytran Input File

173 176

180

Flow between Two Containers or Airbags 183 Problem Description 183 Theoretical Analysis of the 2-Vessel Flow Problem 183 Dytran Model 188 Results 190 Abbreviated Dytran Input File 190 Blastwave Hitting a Bunker Problem Description 193 Dytran Modeling 193 Results 196 Mine Blast 197 Problem Description 197 Dytran Model 197 Results 203 Abbreviated Dytran Input File

193

204

Multiple Bird-strike on a Box Structure Problem Description 209 Dytran Modeling 210 Results 211 Files 213 Abbreviated Dytran Input File 213

209

Shaped Charge, using IG Model, Penetrating through Two Thick Plates 216 Problem Description 216 Dytran Model 218 Results 221 Abbreviated Dytran Input File 222

Main Index

8 Dytran Example Problems Manual

Fuel Tank Filling 224 Problem Description 224 Dytran Model 224 Results 231 Abbreviated Dytran Input File Water Pouring into a Glass Problem Description 237 Dytran Modeling 237 Results 241 Abbreviated Dytran Input File

233 237

242

Fluid Flow through a Straight Pipe Problem Description 246 Theoretical Analysis 247 Dytran Model 247 Results 249 Abbreviated Dytran Input File 258

246

Using Euler Archive Import in Blast Wave Analyses Problem Description 263 Dytran Model 263 General Setup 264 Results 266 Abbreviated Dytran Input File 270

263

Blast Wave with a Graded Mesh 273 Problem Description 273 Dytran Model 274 Results 276 Abbreviated Dytran Input File 281 Bubble Collapse with Hydrostatic Boundary Conditions Dytran Model 284 General Setup 285 Results 288 Dytran Input files 290 Prestressed Concrete Beam Problem Description 294 Analysis Scheme 295 Dytran Model 297 Results 298 Abbreviated Dytran Input File

Main Index

294

302

284

Contents 9

Blast Simulation on Prestressed Concrete Beam Problem Description 306 Analysis Scheme 306 Dytran Model 307 Results 309 Abbreviated Dytran Input File 311 Vortex Shedding with Skin Friction Problem Description 317 Dytran Model 318 Results 320 Abbreviated Dytran Input File 322

317

Geometric Eulerian Boundary Conditions Problem Description 328 Dytran Model 328 Results 332 Cohesive Friction 337 Problem Description 337 Dytran Modeling 337 Results 340 Dytran Input Deck 341

5

Forming Overview

346

Square Cup Deep Drawing Problem Description 347 Dytran Model 349 Results 351 Files 354 Abbreviated Dytran Input File

347

355

Deep Drawing of a Cylindrical Cup Problem Description 359 Dytran Model 360 Results 361 Files 363 Abbreviated Dytran Input File 363

Main Index

359

328

306

10 Dytran Example Problems Manual

Three-point Bending Test Problem Description 368 Dytran Model 369 Results 371 Files 374 Abbreviated Dytran Input File

368

375

Sleeve Section Stamping 379 Problem Description 379 Dytran Model 380 Files 384 Abbreviated Dytran Input File 384

6

Occupant Safety Overview

390

Flat Unfolded Air Bag Inflation (GBAG) Problem Description 391 Results 393 Files 395 Abbreviated Dytran Input File 395

391

Rigid Ellipsoid Dummy Hitting a Passenger Air Bag Problem Description 400 Dytran Model 400 Results 402 Files 404 Abbreviated Dytran Input File 404

400

Sled Test Verification of the Enhanced Hybrid III Dummy (50%) 416 Problem Description 416 Usage of GEBOD to get the Basic Setup of a 50% HYBRID III 417 Completion of the ATB Input File 418 How to Create FEM Entities for ATB Dummy 419 Attaching Dytran Finite Element to ATB Segments 423 Positioning the Dummy with Patran 424 Integrating the Dummy into other FEM Models 426 Defining Lap and Shoulder Belts 426 Applying an Acceleration Field to the Dummy 428 Comparison of the Results with Experiment 428 Appendix A: File EXAMPLE.AIN as generated by GEBOD 429 Appendix B: Complete ATB Input File for Sled Test Calculation 435 Appendix C: File Create_fem_dummy.dat 450

Main Index

Contents 11

Appendix D: Properties and Materials of Hybrid III Dummy 455 Appendix E: Groups created by Session File for Positioning within MD Patran 458 Side Curtain Air Bag (Courtesy of Autoliv) Problem Description 462 Dytran Model 464 Results 469 Input Deck 471

462

Hybrid III 5th%-tile Dummy 479 Problem Description 479 Model Description 479 Conclusions and Recommendations 491 Input Files 492 Appendix A: Overview of the Hybrid III 5th%-tile Dummy Appendix B: SURFACE and SET1 Definition 504 Appendix C: hyb305 Positioning 505 Reference Database 508 Easy Postprocessing with Adaptive Meshing Problem Description 509 Dytran Modeling 510 Results 511 Dytran Input File 514

7

500

509

Quasi-static Analysis Overview

520

Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions 521 Problem Description 521 Dytran Model 522 Results 522 MD Nastran Results 524 Dytran Results 524 Files 526 Abbreviated Dytran Input File 526

Main Index

12 Dytran Example Problems Manual

Main Index

Chapter 1: Structural Dynamics Dytran Example Problem Manual

1

Main Index

Structural Dynamics J

Overview

J

Impulsively Loaded Strip

J

Impulsively Loaded Cylindrical Panel

J

Taylor Test (a rod impacted against a rigid wall)

J

Taylor Test with Euler

J

Hourglassing in Hexahedron Elements

14 15 20

36 41

26

14 Dytran Example Problem Manual Overview

Overview In this chapter, a number of example problems are presented that show the structural capabilities of Dytran. The user can find in these examples how to model dynamic structural problems, what material models to use, how to apply loads and constraints.

Main Index

Chapter 1: Structural Dynamics 15 Impulsively Loaded Strip

Impulsively Loaded Strip Problem Description An aluminum strip clamped at its left and right edges is subjected to an impulsive initial velocity (vz = -132 m/sec) over its central portion (one-fourth of the length). The overall dimensions and material properties are shown below: length

L = 0.254 m

width

W = 0.0305 m

thickness

t = 0.00318 m

density

ρ = 2791 kg/m3

Young’s modulus

E = 7.17 1010 N/m2

Poisson’s ratio yield stress

υ = 0.3 σy = 2.854 108 N/m2

hardening modulus

Eh = 0. N/m2

The purpose is to investigate the sensitivity of the response to Poisson’s ratio (ν =0.01, ν = 0.3) and to check results against data available from experiments performed at the Air Freight Flight Dynamics Laboratory.

Dytran Model Due to the symmetry of the problem, only half of the strip needs to be modeled (see Figure 1-1). The lefthand half strip is discretized by a regular mesh of (30x3) quadrilateral elements. The DMATEP and YLDVM entries are used to input the aluminum elastic-plastic material data. Since plastic deformations occur, the PSHELL1 entry is used to specify five integration points across the thickness. The PSHELL entry defaults to three integration points. The SPC entry is used to impose the zero displacement/rotations of the damped left-hand edge and zero out-of-plane displacement/rotations of the right-hand edge (symmetry plane). To better represent the actual experimental conditions, the initial velocity distribution is slightly modified so as to have a smooth transition to the remainder of the strip. The TIC1 entry is used to input initial velocities. Using the PARAM, INISTEP entry, the initial time step is set to 0.1e-6 sec according to the COURANT criterion. The Case Control entry ENDTIME is used to follow the dynamic behavior in the range up to 0.001 sec.

Main Index

16 Dytran Example Problem Manual Impulsively Loaded Strip

The Case Control entries TYPE, SAVE, GPOUT, GRIDS, and TIMES are used to build the z-displacement time history of Grid Point 31 (in the plane of symmetry) by saving results every 0.01 msec. The experimental results are available for every 0.1 msec.

Figure 1-1

Numerical Model Layout

Results Evaluation Calculations have been carried out for two values of the Poisson’s ratio (ν = 0.01, ν = 0.3). The numerical and experimental results are shown in Figure 1-2. The results demonstrate that the behavior is not sensitive to the value of the Poisson’s ratio. Furthermore, the numerical and experimental results agree satisfactorily ([Ref 1.]).

Main Index

Chapter 1: Structural Dynamics 17 Impulsively Loaded Strip

Figure 1-2

Vertical Z-displacement Time History of the (Free) Corner Grid Point (31) of the Half Strip

Files impulse_a.dat impulse_b.dat

Dytran input files

IMPULSE_A.OUT IMPULSE_B.OUT

Dytran output files

IMPULSE_A_ZDIS_0.THS IMPULSE_B_ZDIS_0.THS

Dytran time history files

Reference 1. Balmer, H. A. and Witmer, E. A. “Theoretical-experimental Correlation of Large Dynamic and Permanent Deformation of Impulsively Loaded Simple Structure,” 1964, Air Force Flight Dynamics Laboratory report FDRTDR-64-108.

Main Index

18 Dytran Example Problem Manual Impulsively Loaded Strip

Abbreviated Dytran Input File START CEND TITLE = IMPULSIVE LOADING OF A STRIP OF ALUMINUM (Units = m-Kg-N-sec) CHECK = NO ENDTIME = 0.001 TIC = 1 SPC = 1 $ $Data for Output Control Set 1 $ TYPE(zdis) = TIMEHIS SAVE(zdis) = 99999 GPOUT(zdis) = ZDIS GRIDS(zdis) = 1 SET 1 = 31 TIMES(zdis) = 0tEndb1e-05 $ $ -----------BEGIN BULK -----------SETTING,1,VERSION2 $ PARAM INISTEP 1e-07 $ $ THIS SECTION CONTAINS BULK DATA $Geometry definition $--------------------$ $ GRID 1 0.0 0.0 0.0 GRID 2 .00423330.0 0.0 . . . $ CQUAD4 1 1 1 2 33 32 CQUAD4 2 1 2 3 34 33 $ $ THIS SECTION CONTAINS THE LOADS, CONSTRAINTS, AND CONTROL BULK DATA ENTRIES $Boundary condition $--------------------$ SPC 1 1 123456 SPC 1 32 123456 SPC 1 63 123456 SPC 1 94 123456 SPC 1 31 156 SPC 1 62 156 SPC 1 93 156 SPC 1 124 156 $

Main Index

Chapter 1: Structural Dynamics 19 Impulsively Loaded Strip

$Material Properties $------------------PSHELL1 1 1 + .00318 $ YLDVM 1 2.854+8 $ DMATEP 1 2791. $ $Initial Conditions $-------------------$ TIC1 1 3 + 31 30 + 122 BY TIC1 1 3 + 31 TIC1 1 3 + 31 26 + 120 BY ENDDATA

Main Index

Bely

Gauss

5

Mid

+

0.0 7.17+10 .01

THRU 31

THRU 31

1

-132. 123

31 BY

THRU 31

124 29

BY THRU

+ +

-116.

28

THRU

121

BY

+

-100. 119

25 BY

THRU 31

118 27

BY THRU

+ +

20 Dytran Example Problem Manual Impulsively Loaded Cylindrical Panel

Impulsively Loaded Cylindrical Panel Problem Description A 120° cylindrical panel (see Figure 1-3) is loaded impulsively by imposing a initial velocity normal to the surface over a region along a part of the crown line of the cylinder. The ends of the panel are simplysupported boundaries, while the sides of the panel are fixed.

Figure 1-3

Initial State

The properties and initial conditions are listed below. Material of the Panel (6061 – T6 aluminum) density

ρ = 2.5 x 10-4 lb sec2/in4

Young’s modulus

E = 1.05 x 107 psi

Poisson’s ratio

υ = 0.33 σy = 4.4 x 104 psi

yield stress

Main Index

Chapter 1: Structural Dynamics 21 Impulsively Loaded Cylindrical Panel

Size of the Panel thickness

t = 0.125 in

radius

R = 2.94 in

length

L = 12.56 in

Initial Load velocity

v0 = 5650 in/sec

Dytran Model Because the panel is symmetric, only one half of the panel is modeled using 12 x 32 (348) shell elements (CQUAD4). Through the thickness, a five-point Gauss integration is applied to produce sufficient accuracy. Two types of shell formulations are used, Belytschko-Tsay and Key-Hoff ([Ref. 2.]). The initial velocity is defined in a cylindrical coordinate system using the TICGP entry. The cylindrical system is defined by the CORD2C entry.

Results A time history is used to give the y-displacement at the midpoint of the crown line of the cylinder for both the Belytschko-Tsay and Key-Hoff shell element. The results are shown in (Figure 1-4) and can be compared with the experimental results of Balmer and Witmer (see [Ref. 3.]). The differences between the results of the two shell formulations are small because Key-Hoff only performs better when a significant part of the element is warped. In comparison with the experiments, the results are acceptable. If one takes into account that the experimental edges were not ideally damped ([Ref. 4.])

Main Index

22 Dytran Example Problem Manual Impulsively Loaded Cylindrical Panel

.

Figure 1-4

Displacement of the Midpoint of the Crown Line

Figure 1-5 shows the deformation and effective plastic strain of the cylindrical panel for the

Belytschko-Tsay shell elements at different time steps.

Main Index

Chapter 1: Structural Dynamics 23 Impulsively Loaded Cylindrical Panel

.

Figure 1-5

Main Index

Deformation and Effective Plastic Strain of the Cylindrical Panel

24 Dytran Example Problem Manual Impulsively Loaded Cylindrical Panel

Files cylpan.dat panel_xl.dat

Dytran input files

CYLPAN.OUT

Dytran Output File

CYLPAN_PANEL_0.ARC

Dytran Archive File

CYLPAN_TH_G_24_0.THS

Dytran Time History File

References (Continued) 2. Belytschko, T., Wong, B. L., and Chiang, H.-Y. “Advances in One-point Quadrature Shell Elements,” Computer Methods in Applied Mechanics and Engineering 96, 1992, pp. 93–107. 3. Balmer, H. A. and Witmer, E. A. “Theoretical-experimental Correlation of Large Dynamic and Permanent Deformation of Impulsively Loaded Simple Structure,” 1964, Air Force Flight Dynamics Laboratory report FDRTDR-64-108. 4. Morino, L., Leach, J. W. and Witmer, E. A. “An Improved Numerical Calculation Technique for Large Elastic-Plastic Transient Deformations of Thin Shell = Part 2 – Evaluation and Applications,” Journal of Applied Mechanics, June 1971, pp. 429–436.

Abbreviated Dytran Input File $ $ A 120 degree cylindrical panel is loaded impulsively. It is simply $ supported at the ends. The model consists of half the panel, with 12x32 $ shell elements. The loading is generated by imposing initial velocities $ on the grid points. $ START CEND TITLE = Cylindrical Panel (12x32 - B-T or Keyhoff Shell 5 G-Pts) CHECK = NO ENDSTEP = 10000 ENDTIME = 1.e-3 TIC = 1 SPC = 1 $$ $$ Data for output control Set 1 $$ $$ Time-History plot of the y-displacement of the midpoint along the crown $$ line of the cylinder (grid 209). $$ TYPE(TH_G_24) = TIMEHIS GPOUT(TH_G_24) = YDIS SET 1 = 209 GRIDS(TH_G_24) = 1 TIMES(TH_G_24) = 0 THRU END BY 1.e-5

Main Index

Chapter 1: Structural Dynamics 25 Impulsively Loaded Cylindrical Panel

SAVE(TH_G_24) = 10000 $$ $$ Data for output control Set 15 $$ TYPE(PANEL) = ARCHIVE ELEMENTS(PANEL) = 15 SET 15 = 1 THRU 384 ELOUT(PANEL) = EFFPL03,TXX03,TYY03,TXY03,TYZ03,TZX03 ELOUT(PANEL) = EFFST03,EXX03,EYY03,EXY03 TIMES(PANEL) = 0 THRU END BY 1e-4 SAVE(PANEL) = 10000 $ BEGIN BULK $ $ Definition of some parameters. $ ----------------------------PARAM INISTEP 1.e-6 PARAM STEPFCT 0.9 PARAM SHPLAST VECT PARAM HGCOEFF 0.1 PARAM SHTHICK YES $ $ Model geometry and boundary constrain. $ -------------------------------------INCLUDE panel_xl.dat $ $ Shell Properties for the panel(B-T or Keyhoff). $ ----------------------------------------------$PSHELL1 1 1 keyhoff GAUSS 5 1.0 PSHELL1 1 1 bely GAUSS 5 1.0 +CONT .125 $ $ Material and yield model. $ ------------------------DMATEP 1 .00025 1.05e7 .33 $ YLDVM 10 44000 0 $ $ Initial condition for gridpoints. $ --------------------------------CORD2C 1 0.0 0.0 0.0 0.0 + 0.0 2.94 -6.28 TICGP 1 1 CID1 1 XVEL -5650 SET1 1 14 THRU 339 BY 13 + 340 BY 13 16 THRU 341 + 17 THRU 342 BY 13 18 + BY 13 19 THRU 344 BY + THRU 345 BY 13 $ ENDDATA

Main Index

MID MID

+CONT +CONT

10

0.0

-12.56

+

15 BY THRU 13

THRU 13 343 20

+ + + +

26 Dytran Example Problem Manual Taylor Test (a rod impacted against a rigid wall)

Taylor Test (a rod impacted against a rigid wall) Problem Description The Taylor "bar test" is an important laboratory test in the science of ballistics. It enables to determine an average value of the dynamic yield stress of a material. It consists of accelerating a cylindrical bar (the velocity being parallel to the axis of symmetry), and then let the bar hit a rigid target. As a result, the bar shortens and the impact side expands radially acquiring a mushroom like shape. Much research work has been done on these impact tests. In this example, the experimental work done by Johnson/Cook is validated with a number of simulations in Dytran with different element types. On top of this work, the results are compared against the theoretical solution developed by Taylor. Johnson and Cook developed a material model that represents a constitutive model for materials subjected to large strains, high strain rates and high temperatures. The Dytran implementation is validated for a set of constitutive constants presented in ([Ref. 5.]).

Johnson-Cook Material Model This material model is described in the Dytran Reference and User manuals. For more detailed information and data sets for various materials, see ([Ref. 5.]).

Theoretical Approach – Taylor Taylor developed a simple expression for the final length of the bar, L0 ,

the impact velocity

V,

the yield strength

σy ,

and the density ρ .

2

L f L 0 = exp ( – ρ V ⁄ 2σ y )

The assumptions made in this expression are: • There is a stationary yield front near the wall • A quasi-steady process is assumed

Figure 1-6

Main Index

Schematic Overview of the Taylor Impact Test

Lf ,

as function of the initial length

Chapter 1: Structural Dynamics 27 Taylor Test (a rod impacted against a rigid wall)

In this Taylor test, the process stops when all the kinetic energy has converted to plastic work. It is interesting to note that the governing parameters of the process give a non-dimensional quantity, therefore, the result depends on this combination and not on the specific value of each parameter.

Dytran Model A cylindrical rod with a length of 25.4 mm and a diameter of 3.82 impacts a rigid wall with a velocity of 190 mm/s. The material data: • rho=8.96e-9 tonne/mm3 • Bulk modulus K=143e3 MPa • Shear modulus G=47.7e3 MPa • Two different material models have been used: • Johnson/Cook • Constant von Mises: σy = 600 MPa (elastic-rigid-plastic)

The interface between the wall and the rod are assumed frictionless.

Dytran Results Hexa Elements As a first step, a comparison is made between an experimental result in ([Ref. 5.]) and a simulation that uses Johnson-Cooks constitutive model that is presented in the same reference. The simulation performed with Dytran is setup with CHEXA elements. The simulation is done with the widely used onepoint Gauss integration scheme.

Figure 1-7

Main Index

Dytran CHEXA Model – Impact of Cylinder - Quarter Model

28 Dytran Example Problem Manual Taylor Test (a rod impacted against a rigid wall)

To be able to compare this result with the other element solvers in Dytran, an equivalent von Mises yield stress is chosen such that a simulation with the von Mises yield criterion gives the same answers. The von Mises yield stress is found to be ~ 600 MPa in order to give the same results as with the Johnson Cook material model. This 600 MPa seems realistic, see Figure 1-8.

Figure 1-8

Armco Iron Material Data – Johnson Cook

Next, a simulation is done with the new 2x2x2 CHEXA element in Dytran 2004. This element has a larger reduction in length. The results of the simulations with Dytran are shown in Figure 1-9. In summary: the Hexa – one point (von Mises) result is tuned to be on top of the Hexa – one point (Johnson Cook) result. The CHEXA – 2x2x2 result has the largest length reduction.

Figure 1-9

Main Index

Comparison of Dytran CHEXA Elements

Chapter 1: Structural Dynamics 29 Taylor Test (a rod impacted against a rigid wall)

Tetrahedral Elements In order to evaluate the new and old tetrahedral elements in Dytran, the model is meshed with CTETRA elements. The old tetrahedral implementation is a 8-noded hexa element with 1 gauss point, degenerated into a tet shape. This element is known to give imprecise answers. This is clearly shown in Figure 1-11 and Figure 1-12. The new 4 nodal tetrahedral element gives answers much closer to the Hexahedral one point element.

Main Index

Figure 1-10

Dytran CTETRA Model – Impact of Cylinder - Quarter Model

Figure 1-11

Comparison of Different Element Types in Dytran

30 Dytran Example Problem Manual Taylor Test (a rod impacted against a rigid wall)

In Figure 1-12, a detailed section of Figure 1-11 is shown. The results all show an oscillating behavior that is common in all transient dynamic events and is physical. Furthermore, the new 2x2x2 CHEXA element is deviating from the experimental result. and further developments are targeted to improve the behavior of the new CHEXA element in impact events involving metal plasticity. The 2x2x2 CHEXA element has specifically been implemented and tested to reduce the hour-glassing phenomenon in rubber and foam modeling

Figure 1-12

Detailed Comparison of Different Element Types in Dytran

Theoretical – Taylor In Figure 1-9, Figure 1-11, and Figure 1-12, the theoretical reduction in length according to Taylor is plotted: L f ⁄ L 0 = 0.78

This theoretical reduction is based on an assumption of constant yield of 700 MPa.

Main Index

Chapter 1: Structural Dynamics 31 Taylor Test (a rod impacted against a rigid wall)

Comparison of Shape The shape is compared in Figure 1-13 through Figure 1-17.

Main Index

Figure 1-13

Dytran Simulation - Johnson Cook Material Model with CHEXA One Point Gauss

Figure 1-14

Dytran Simulation - von Mises Material Model with CHEXA - One Point Gauss

Figure 1-15

Dytran Simulation - von Mises Material Model with CHEXA - 2x2x2

32 Dytran Example Problem Manual Taylor Test (a rod impacted against a rigid wall)

Figure 1-16

Dytran Simulation - von Mises Material Model with CTETRA - Old Element

Figure 1-17

Dytran Simulation - von Mises Material Model with CTETRA - New Element

Comparison of CPU time Figure 1-18 shows the total CPU time for the various simulations. This doesn't show the performance of each element very clearly, because the total timings for the various runs are still heavily dependent on the physics and modeling used: the deformation and the size of the elements. The number of elements in all models is approximately the same, but the model with the Tet elements contain a few tiny elements which determine the timestep. At the end of the simulation, the time-step of the hex-models is 1.53e-5 msec, while the time-step of the tet-models is 1.28e-5 msec. Therefore, Figure 1-19 shows the timings weighted per element-cycle. This allows a fair comparison of the element's performance. The new tetrahedral element shows a dramatic speed up compared to the old implementation, and is in fact the fastest element available.

Main Index

Chapter 1: Structural Dynamics 33 Taylor Test (a rod impacted against a rigid wall)

Figure 1-18

Comparison of Total CPU Time for Several Dytran Elements

Figure 1-19

Comparison of Timings for the New CHEXA and CTETRA Elements in Dytran

References 5. Johnson, G.R. and Hook, W.H. "A constitutive model and data for metals subjected to large strains, high strain rates and high temperatures", April 1983, 7th Ballistic Symposium, The Hague, The Netherlands 6. Wilkins, L.M and Guinan M.W. " Impact of cylinders on a rigid boundary", August 1972, Lawrence Livermore Laboratory

Main Index

34 Dytran Example Problem Manual Taylor Test (a rod impacted against a rigid wall)

Abbreviated Dytran Input File START CEND ENDTIME=0.06E-3 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: Hexs-EPM-default-JC TLOAD=1 TIC=1 SPC=1 $ Output result for request: Cylinder_Elem TYPE (Cylinder_Elem) = ARCHIVE ELEMENTS (Cylinder_Elem) = 1 SET 1 = 1 THRU 1224 ELOUT (Cylinder_Elem) = TXX TYY TZZ TXY TYZ TZX EFFSTS PRESSURE EFFPLS SIE TIMES (Cylinder_Elem) = 0 THRU END BY 0.03e-4 SAVE (Cylinder_Elem) = 10000 $ Output result for request: Cylinder_Grid TYPE (Cylinder_Grid) = ARCHIVE GRIDS (Cylinder_Grid) = 2 SET 2 = 1 THRU 1768 GPOUT (Cylinder_Grid) = XVEL YVEL ZVEL RVEL XACC YACC ZACC RACC , XFORCE YFORCE ZFORCE RFORCE TIMES (Cylinder_Grid) = 0 THRU END BY 0.03e-4 SAVE (Cylinder_Grid) = 10000 $ Output result for request: GridTHS TYPE (GridTHS) = TIMEHIS GRIDS (GridTHS) = 3 SET 3 = 1 1735 GPOUT (GridTHS) = XPOS YPOS ZPOS RPOS XVEL YVEL ZVEL RVEL XACC YACC ZACC , RACC TIMES (GridTHS) = 0 THRU END BY 0.06e-5 SAVE (GridTHS) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,5E-9 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE Hexs-EPM-default-JC.bdf $ $ ========== PROPERTY SETS ========== $ $ * Prop_cylinder 2x2x2 * $ PSOLID 1 2 TWO FULL $ $ * Prop_cylinder_VM * $ PSOLID 2 3 $ CORD2R 1 0 0 0 0 0 0 1+A000001 +A000001 1 0 0 $

Main Index

Chapter 1: Structural Dynamics 35 Taylor Test (a rod impacted against a rigid wall)

$ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Cylinder_VM id =2 DMAT 28.96e-09 2 2 2 EOSPOL 2 143000 SHREL 2 47700 YLDVM 2 700 $ $ -------- Material Cylinder_JC id =3 DMAT 38.96e-09 3 3 3 EOSPOL 3 143000 SHREL 3 47700 YLDJC 3 175 380 .32 .06 55 +A000002 1811 293 $ $ ======== Load Cases ======================== $ $ $ ------- Initial Velocity BC InitialVelocity ----SET1 4 1 THRU 1768 TICGP 1 4 ZVEL -197000 $ $ ------- Rigid Plane BC RigidWall ----WALL 9 0 0 0 0 0 +A000004 PENALTY SET1 5 1 THRU 1768 $ $ ENDDATA

Main Index

14.52e+08+A000002

1

5+A000004

36 Dytran Example Problem Manual Taylor Test with Euler

Taylor Test with Euler For most structural problems, the Lagrange approach is the preferred solution. But if deformations become large, the Euler approach can become more accurate. As deformations are increased, the Lagrangian elements are severely distorted affecting the accuracy of the solution. In the Euler approach, severe distortion is irrelevant and does not exist. Eulerian elements are stationary and have a fixed volume; only the material inside the elements can move and is transported over the faces of the stationary Euler elements. In addition, the Euler approach does not need any contact definition when an impactor hits a plate. This example shows how the model from Example1-4 is handled by the multi-material Euler solver with strength. It also illustrates the use of PARAM, AXIALSYM.

Problem Description The problem is the same as in Example 1-4. A bar impacts a rigid wall. This example will compare the results of a Lagrangian approach using the Johnson-Cook material model as shown in Example 1-4 and an Euler simulation.

Dytran Model Setup of the Euler Domain To construct the Euler mesh, a pie shape consisting of one layer of elements will be created as shown in Figure 1-20. Because the simulation will assume axi-symmetric behaviour the following requirements are imposed on the Euler mesh domain: • For accuracy the angle of this pie should be small, say 5°. • When creating a pie mesh with MD Patran, the grid points are written out in single precision,

resulting in very small errors. The small errors are large enough to cause errors in the normals of the faces of Euler faces. A normal that is supposed to point in the circumferential direction gets a very small component in the axial and radial direction. For a problem involving strength, this leads to small dynamics in the circumferential direction. To keep the dynamics small and bounded, the stable time step has also to be based on the circumferential direction. The pie model has a small angle which results in a small mesh size and therefore a small time step. With PARAM, AXIALSYM, these grid points can be slightly corrected resulting in aligned normals. This technique results in a much larger time step. In addition, PARAM, AXIALSYM can be used to create a pie mesh directly from a rectangular block slab of elements. This approach has been followed in this example. First of all, a rectangular Euler mesh consisting of one layer has been defined in the input file. Secondly, PARAM,AXIALSYM transforms this mesh automatically into a pie shape mesh in the Dytran solver. Please refer to the reference manual for further information on this option.

Main Index

Chapter 1: Structural Dynamics 37 Taylor Test with Euler

The Euler elements are defined by MESH,1,BOX,,,,,,,+ +,0.0,0.0,0.0,26.0,0.2,3.82,,,+ +,128,1,44,,,,EULER,1

Figure 1-20

Pie Model

To make this rectangular mesh into a pie-shaped mesh PARAM,AXIALSYM will be used: PARAM,AXIALSYM,RECT,X,ZX,2.5

The pie-shaped mesh will also be used in the initialization. The time step will only be based on the meshsize in x and z-direction. The mesh size in the y-direction would require a much smaller time step. Since there is no dynamics in the y direction, the mesh size in y-direction can be left out. This is also defined by PARAM,AXIALSYM. Setup of Material and Initialization Data The multi-material Euler with strength solver will be used: First model: PEULER1,1,, STRENGTH,4 PEULER1,1,, MMSTREN,4

The material is defined by DMAT,1,8.96e-9,1,1,1,,1 PMINC,1,-1.e+20 EOSPOL,1,143000 SHREL,1,47700 YLDJC ,1,175,380,0.32,0.06,0.55,1,4.52e+008,+ + ,1811,293

Main Index

38 Dytran Example Problem Manual Taylor Test with Euler

The Euler elements are initialized using geometric regions: TICVAL,3,,XVEL,-197000,DENSITY,8.96e-9 $ $ ------- TICEUL BC init-euler ----TICEUL,4,,,,,,,,+ + ,SPHERE,1,,,1,,,,+ + ,BOX,2,1,3,2,,,,+ SPHERE,1,,0.0,0.0,0.0,100.0 BOX,2,,0.0,-0.3,0.0,25.4,0.6,1.91

Results Results at time 4.8 e-5 sec are shown in Figure 1-21 and Figure 1-22 .

Figure 1-21

FMAT of the metal

Figure 1-22

Plastic strain, Range 0 to 1.08

As in Example 4.1, the rod elongates slightly after 0.48e-5 sec as shown in Figure 1-23. The total length reached at the end is 19.9. This translates into a reduction of length of 0.78. These results compare well with the results presented in Example 1.4

Main Index

Chapter 1: Structural Dynamics 39 Taylor Test with Euler

Figure 1-23

Elongation as Function of Time

If the compression is more severe or when damage is used it is preferred to add PARAM, EULSTRESS,MASS. With this PARAM stresses are transported by using mass instead of volume. The mass approach can be more accurate, but for this simulation it is not needed. Running the same problem with the single material strength solver also gives reasonable results, although the material is not as smooth. This is explained by the fact that the transport algorithms of the multi-mat strength solver are currently more sophisticated.

Figure 1-24

FMAT at Time 4.8e-5 seconds with the Strength Solver

Dytran File START MEMORY-SIZE = 6500000,6500000 CEND ENDTIME=7.e-5 CHECK=NO TITLE= Jobname is: taylor-euler TLOAD=1

Main Index

40 Dytran Example Problem Manual Taylor Test with Euler

TIC=1 SPC=1 $ Output result for request: euler TYPE (euler) = ARCHIVE ELEMENTS (euler) = 1 SET 1 = ALLELEMENTS ELOUT (euler) = DENSITY TEMPTURE EFFSTS , VOID XVEL YVEL ZVEL MASS PRESSURE, FMAT EDIS EFFPLS TIMES (euler) = 0,THRU,END,BY,6.e-6 SAVE (euler) = 10000 $------- Parameter Section -----$ PARAM,INISTEP,5e-9 PARAM,MINSTEP,1e-10 $------- BULK DATA SECTION ------BEGIN BULK $ $ ========== PROPERTY SETS ========== $ $ * peuler1 * PARAM,AXIALSYM,RECT,X,ZX,2.5 $ PEULER1,1,, MMSTREN,4 $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material JC id =1 DMAT,1,8.96e-9,1,1,1,,1 PMINC,1,-1.e+20 EOSPOL,1,143000 SHREL,1,47700 YLDJC ,1,175,380,0.32,0.06,0.55,1,4.52e+008,+ + ,1811,293 $ $ ======== Load Cases ======================== $ $ $ ------- TICVAL BC init-bar-val ----TICVAL,3,,XVEL,-197000,DENSITY,8.96e-9 $ $ ------- TICEUL BC init-euler ----TICEUL,4,,,,,,,,+ + ,SPHERE,1,,,1,,,,+ + ,BOX,2,1,3,2,,,,+ SPHERE,1,,0.0,0.0,0.0,100.0 BOX,2,,0.0,-0.3,0.0,25.4,0.6,1.91 $ MESH,1,BOX,,,,,,,+ +,0.0,0.0,0.0,26.0,0.2,3.82,,,+ +,128,1,44,,,,EULER,1 ENDDATA

Main Index

Chapter 1: Structural Dynamics 41 Hourglassing in Hexahedron Elements

Hourglassing in Hexahedron Elements Problem Description Hourglass modes in simulations with 1-point integration hexahedron and quadrilateral elements are always present. In the past, a lot of research in suppressing these modes have been done in Dytran and other explicit codes. Therefore, nowadays, the hourglass modes are often suppressed to a sufficient level by default. However, in certain occasions, the modes are still perseverant and can easily trouble the solution. A rigorous solution for this is the use of fully integrated quadrilateral and/or hexahedron elements. These elements do not exhibit hourglassing modes, and therefore, do not need hourglass suppressing terms in the algorithm. To illustrate the above, a load is applied to a plate modeled with one layer of hexahedron elements. The model has a load (linearly) applied in the top mid center and is supported in the lower corners. It will be shown that the new fully integrated hexahedron element in Dytran does not suffer from hourglassing. Theoretically, this can be proven because no hourglassing modes exist in the solution, contrary to the 1-point integration hexahedron element, which has several hourglassing modes. The model will be run with 4 different settings, as summarized in Table 1-1. Table 1-1

Summary of Analyses Settings Hourglass Control Method

Element Integration Method

1

1-point – constant stress

Stiffness Based (FBS)

0.1

2

2x2x2 points – fully integrated

–

-

3

1-point – constant stress

Stiffness Based (FBS)

0.01

4

1-point – constant stress

Viscosity Based (DYNA)

0.1

The model has following dimensions: LxB = 5x5 Thickness = 1 The material, linear elastic: G = 7500 K = 16250 (as a result: ν = 0.3) r = 1.2e-3 Loads and BC's: F = 400, linear in time up to 0.1 [sec] Clamped at bottom corner nodes.

Main Index

Hourglass Control Coefficient

Case

42 Dytran Example Problem Manual Hourglassing in Hexahedron Elements

Dytran Model The hexahedron mesh is shown in Figure 1-25. The model has one element over the thickness, and 64 elements in total. The force is applied on two nodes in the top center and the bottom corner nodes are clamped using a SPC. The force is ramped up linearly in 0.1 seconds to minimize oscillations. The ENDTIME is 0.1 seconds.

Figure 1-25

Model

Case 1 This case is using default settings. Case 2 Starting with the model of Case 1, choosing the following options on the PSOLID entry activates the 2x2x2 fully integrated hexa element: IN = 2 ISOP = FULL Case 3 Starting with the model of Case 1, choosing the following option on the HGSUPPR entry activates the lower hourglass coefficient: HGCSOL = 0.01

Main Index

Chapter 1: Structural Dynamics 43 Hourglassing in Hexahedron Elements

Case 4 Starting with the model of Case 1, the overall hourglass control method for solids is modified to the viscous method, by entering the parameter: PARAM, HGCSOLID, DYNA

Results The results for the four cases are shown in Figure 1-26 through Figure 1-29 below. Figure 1-26 shows that applying point loads as done in this example introduce hourglass modes when using 1-point integration. Figure 1-27 shows that the hourglass modes do not occur when using the 2x2x2 integration method. Figure 1-28 shows that by lowering the hourglass control coefficient more severe hourglass modes occur. Figure 1-29 shows that the viscous hourglass mode suppression method is not working well in this example. This is understandable, because in general a viscous hourglass mode suppression method is only suited for problems with high velocities.

Note that in all the figures, the displacements are all scaled up in order to show the hourglassing modes more clearly.

Figure 1-26

Note:

Main Index

Case 1: 1-Point Integration Hexa, Hourglassing to a Small Extent in the Bottom and Top Rows Displacements are scaled up.

44 Dytran Example Problem Manual Hourglassing in Hexahedron Elements

Figure 1-27 Note:

Figure 1-28

Note:

Main Index

Case 2: 2x2x2 Fully Integrated Hexa, No Hourglassing Displacements are scaled up.

Case 3: 1-Point Integration Hexa with Lower Stiffness Damping Coefficient (0.01) Displacements are scaled up.

Chapter 1: Structural Dynamics 45 Hourglassing in Hexahedron Elements

Figure 1-29

Note:

Case 4: 1-Point Integration Hexa with Viscous Hourglassing Suppression, Hourglassing very Clearly Present Displacements are scaled up.

Files Case 1:

EPM15-default.dat

Case 2:

EPM15-2x2x2.dat

Case 3:

EPM15-lowcoef.dat

Case 4:

EPM15-hg-dyna.dat

References 7. Dytran reference manual, chapter 5, Bulk data description of Hourglass suppression modes HGSUPPR 8. LS-Dyna version 970 Keyword's User Manual, *HOURGLASS

Main Index

46 Dytran Example Problem Manual Hourglassing in Hexahedron Elements

Abbreviated Dytran Input File START CEND ENDTIME=0.1 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: Comparison-iso TLOAD=1 TIC=1 SPC=1 $ Output result for request: elems TYPE (elems) = ARCHIVE ELEMENTS (elems) = 1 SET 1 = 1 THRU 64 ELOUT (elems) = TXX TYY TZZ TXY TYZ TZX EFFSTS EFFPLS TIMES (elems) = 0 THRU END BY 0.05 SAVE (elems) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1e-8 $ $ Case 1: no extra parameters $ $ Case 2: no extra parameters $ $ Case 3: $ PARAM,HGCSOL,0.01 $ $ Case 4: $ PARAM,HGSOLID,DYNA $ $------- BULK DATA SECTION ------BEGIN BULK $ --- SPC-name = Clamp SPC1 1 123 1 9 82 $ $ --- Define 162 grid points --$ GRID 1 .00000 .00000 .00000 … . GRID 162 5.00000 5.00000 1.00000 $ $ --- Define 64 elements $ $ -------- property set Property --------CHEXA 1 1 1 2 11 +A000001 92 91 … . CHEXA 64 1 71 72 81 +A000064 162 161 $ $ ========== PROPERTY SETS ========== $

Main Index

SIE

90

10

82

83+A000001

80

152

153+A000064

Chapter 1: Structural Dynamics 47 Hourglassing in Hexahedron Elements

$ Case 1,3 and 4: $ * Property * $ PSOLID 1 1 $ $ Case 2: $ * Property 2x2x2 Hexa fully integrated * $ $ PSOLID 1 1 2 1 $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Material-iso id =1 DMAT 1 .0012 1 1 EOSPOL 1 23000 SHREL 1 7500 $ $ ======== Load Cases ======================== $ $ $ ------- Force BC Force ----TLOAD1 1 5 0 1 FORCE 5 77 0 1 0 -200 0 FORCE 5 158 0 1 0 -200 0 $ $ ================ TABLES ================= $ $ ------- TABLE 1: Table ------TABLED1 1 +A000065 +A000065 0 0 .1 1 ENDT $ ENDDATA

Main Index

48 Dytran Example Problem Manual Hourglassing in Hexahedron Elements

Main Index

Chapter 2: Structural Contact Dytran Example Problem Manual

2

Main Index

Structural Contact J

Overview

J

Three-plate Contact

J

Ball Penetrating a Steel Plate

J

Tapered Beam Striking a Rigid Wall

J

Impact Loading

J

Pipe Whip

50

79

73

51 58 66

50 Dytran Example Problem Manual Overview

Overview In this chapter, a number of example problems are presented that show the capabilities of Dytran to model structural interaction. Through these examples, the user can learn how to model the interaction between structural parts using the different types of contact available in Dytran.

Main Index

Chapter 2: Structural Contact 51 Three-plate Contact

Three-plate Contact Problem Description Plates 1 and 3 are moving towards a third, resting Plate 2, which is located between them. The initial situation is given in Figure 2-1 below:

Figure 2-1

Initial Situation

The required data is shown below. • Size of all three plates:

width in y-direction

a=2m

width in z-direction

b=2m

thickness

t = 0.005 m

• Material of the plates (steel):

Main Index

density

ρ = 7800 kg/m3

Young’s modulus

E = 2.1 1011 N/mm2

Poisson’s ratio

υ = 0.3

52 Dytran Example Problem Manual Three-plate Contact

• Initial conditions:

distance

d = 0.005 m

velocity

v0 = 15 m/sec

The purpose of this example is to investigate the treatment of contact between structures in Dytran.

Theoretical Result The conservation of momentum and energy will only be achieved if the plates are reflected with a velocity equal to the initial velocity but with reversed direction. For symmetry reasons, the central plate stays at rest. As shell elements (which are infinitely stiff in their normal direction) are used, the duration of the contact is theoretically approaching zero.

Dytran Model The plates are modeled by 4 x 4 shell elements. The initial grid point velocities are prescribed using TIC entries. The CONTACT entry refers to SURFACE entries, which define the surfaces by sets of element faces or by property IDs of shell elements. In this example, a CFACE is defined for each shell element. The CFACEs of the plates 1, 2, and 3 are grouped in the face sets IDs 1, 2, and 3, respectively. Three different ways to model the contact are compared. The following table shows how the surfaces and contacts are defined in three cases: Master

Slave

CASE 1

Contact 1 Contact 2

2 3

1 2

CASE 2

Contact 1 Contact 2

2

1

Contact 1 Contact 3 Contact 2

2 1

CASE 3

Single Surface

2,3 1 2 2,3

Time histories for the position and velocities of grid points on each plate are requested for output. In addition, two Contact variables are requested for output. One is the magnitude of Contact force in the x-direction and the other is the smallest distance between slave and master face.

Results When a contact between two surfaces is defined, the code applies a constraint force pushing the corresponding parts away from each other, as soon as penetration occurs. In the case of the master-slave contact, the grid points of the slave surface are checked for penetration of the faces of the master surface

Main Index

Chapter 2: Structural Contact 53 Three-plate Contact

and the force is related to the depth of that penetration. The grid points of the master surface are not checked against the slave surface. In the case of the single surface contact, the algorithm checks all grid points for the penetration of any face of the surface. As the stiffness of the contact (which is a numerical and not a physical entity) is related to the number of penetrating grid points, the single surface contact is stiffer than the master-slave contact. In this example, it should behave exactly as two master-slave contacts with interchanged surfaces. The time histories of the central plate grid points velocities in x-direction agree with the expectations. The plots of Case 1 (Figure 2-2) and Case 3 (Figure 2-3) show that the energy and momentum are conserved and that symmetry is achieved. The acceleration of the plates during the contact has a finite value according to the finite numerical stiffness of the contact. Comparison of the plots show the stiffer behavior of the single surface contact. As a result of this, in Case 2 (Figure 2-4), there is a delayed exchange of momentum in the master-slave contact causing a part of the momentum to be transferred to the Plate 2.

Figure 2-2

Main Index

Two Master-slave Contacts Result in a Symmetric Behavior

54 Dytran Example Problem Manual Three-plate Contact

Main Index

Figure 2-3

A Single-surface Contact behaves as Stiff as Two Master-slave Contacts Interchanged Surfaces

Figure 2-4

The Combination of a Single Surface Contact with a Single Master-slave Contact produces an Unsymmetric Result

Chapter 2: Structural Contact 55 Three-plate Contact

Files pl3contact1.dat pl3contact2.dat pl3contact3.dat

Dytran input file for the three cases

PL3CONTACT1.OUT PL3CONTACT2.OUT PL3CONTACT3.OUT

Dytran output files for the three cases

PL3CONTACT1_GRXPV_1.THS PL3CONTACT2_GRXPV_1.THS PL3CONTACT3_GRXPV_1.THS

Dytran time history files for grid point output for the three cases

PL3CONTACT1_CONT_DIS_0.THS PL3CONTACT2_CONT_DIS_0.THS PL3CONTACT3_CONT_DIS_0.THS

Dytran time history files for contact output for the three cases

Abbreviated Dytran Input File TITLE = Three Plate Contact ENDSTEP = 40 TIC = 1 $ $ --------- Requesting the Position and Velocity of one $ TYPE(grxpv) = TIMEHIS SAVE(grxpv) = 9999 GRIDS(grxpv) = 1 SET 1 = 12t62b25 STEPS(grxpv) = 1t100b1 GPOUT(grxpv) = XPOS,XVEL $ $ --------- Contact distance output vars $ TYPE(cont_dis) = TIMEHIS CONTS(cont_dis) = 2 SET 2 = ALLCONTACTS CONTOUT(cont_dis) = DMIN,XFORCE STEPS(cont_dis) = 0,THRU,END,BY,1 SAVE(cont_dis) = 9999 $ BEGIN BULK PARAM INISTEP 1e-05 $ $ THIS SECTION CONTAINS BULK DATA $ $ GRID 1 -.005 -1. -1. GRID 2 -.005 -.5 -1. GRID 3 -.005 0.0 -1. GRID 4 -.005 .5 -1.

Main Index

56 Dytran Example Problem Manual Three-plate Contact

. . . GRID 72 .005 -.5 1. GRID 73 .005 -4.44-161. GRID 74 .005 .5 1. GRID 75 .005 1. 1. $ CQUAD4 1 1 1 2 7 6 CQUAD4 2 1 2 3 8 7 CQUAD4 3 1 3 4 9 8 CQUAD4 4 1 4 5 10 9 . . CQUAD4 45 1 66 67 72 71 CQUAD4 46 1 67 68 73 72 CQUAD4 47 1 68 69 74 73 CQUAD4 48 1 69 70 75 74 $ $ THIS SECTION CONTAINS THE LOADS, CONSTRAINTS, AND CONTROL BULK DATA $ ENTRIES $ $ TIC 1 1 1 15. TIC 1 2 1 15. TIC 1 3 1 15. TIC 1 4 1 15. . . TIC 1 51 1 -15. TIC 1 52 1 -15. TIC 1 53 1 -15. TIC 1 54 1 -15. $ $ $ THIS SECTION CONTAINS THE DEFINED FEFACES OF ELEMENTS $ $ CFACE 1 1 1 1 CFACE 2 1 2 1 CFACE 3 1 3 1 CFACE 4 1 4 1 . . . $ PSHELL 1 1 .005 $ DMATEP 1 7800. 2.1+11 .3 $ SURFACE 1 SEG 1 SURFACE 2 SEG 2 SURFACE 3 SEG 3 $

Main Index

Chapter 2: Structural Contact 57 Three-plate Contact

CONTACT + + + CONTACT + + + ENDDATA

Main Index

1 0.0 0.0 2 0.0 0.0

SURF Both .1 1.+20 SURF Both .1 1.+20

SURF 1 Full Distance1.+20 On On SURF 2 Full Distance1.+20 On On

2

0.0

0.0

0.0

1.1 1.+20 3

.1

Distance0.002

0.0

0.0

1.1 1.+20

.1

Distance0.002

0.0

+ + + + + +

58 Dytran Example Problem Manual Ball Penetrating a Steel Plate

Ball Penetrating a Steel Plate Problem Description Consider the situation where a rigid steel ball with a weight of 2.6 kg strikes a square steel plate with a thickness of 0.005 m at a velocity of 230 m/sec. The ball hits the center of the plate perpendicularly. Two alternative models of the plate have been computed ([Ref. 1.]). In the first model, the plate is composed of individual shell elements without any grid points in common. the connection between adjacent elements is defined by constraints (BJOINS), which are eliminated when the average plastic strain of the connected elements exceeds a specified value (see Figure 2-5). Additionally, when the ball penetrates the plate, the elements that break off do not go to pieces but remain in the Dytran model, as is shown in Figure 2-6. This approach might be used to simulate e.g. welding lines.

Figure 2-5

BJOINS: Connection Fails if 1/4

( ε p1 + ε p 2 + ε p 3 + ε p4 ) > ε p = 0.5

The second model concerns a plate composed of shell elements with a failure criterion (eroding elements). Input data are equal as for the first model. As is shown in Figure 2-7, the failed elements vanish if they lose their stiffness. The mass and momentum at the grid points is retained. An advantage of this approach is that the CPU time is about five times smaller than for the first model. The properties of the plate and ball are listed below.

Main Index

Chapter 2: Structural Contact 59 Ball Penetrating a Steel Plate

Plate material

steel

Young’s modulus

E = 2. 1011 N/m2

Poisson’s ratio

υ = 0.3

yield stress

σ = 4. 108 N/m2

percent of elongation

50%

thickness

t = 0.005 m

length

L = 0.3 m

width

W = 0.3 m

Figure 2-6

Main Index

Ball Penetrating Plate using BJOINS

60 Dytran Example Problem Manual Ball Penetrating a Steel Plate

Figure 2-7

Ball Penetrating Plate without using BJOINS

Ball material

steel (rigid body)

density

ρ = 7850 kg/m3

mass

m = 2.617 kg

radius

R = 0.043 m

velocity

v = 230 m/sec

Desired Results The two models will be compared with respect to the failure of the plate. In the BJOIN model, the elements should break off without failing themselves and they should remain visible in the Dytran model. In the eroding element model, the failed elements are only used in the calculation but should be invisible to the user.

Main Index

Chapter 2: Structural Contact 61 Ball Penetrating a Steel Plate

Dytran Modeling The ball is modeled as a rigid body using the RIGID entry. The outer surface of the ball consists of 1000 CQUAD4 dummy elements, which are defined as such by the PSHELL1 entry. The elements are covered by CFACEs so that by using the SURFACE entry, the outer surface can be both defined as a rigid surface that is needed in the RIGID entry as well as a Lagrangian interface in the contact with the plate. There are no constraints imposed on the ball, but it is given an initial velocity of 230 m/sec in the RIGID entry. The plate is modeled using 30 x 30 (900) Belytschko-Tsay shell elements (CQUAD4). Furthermore, a nonuniform mesh with smaller elements is used near the center, where the impact takes place. By using the SPC entry, the boundary of the plate is fixed in all directions. In the BJOIN model, there are 4 x 900 (3600) different grid points defined, since the 900 elements have no grid points in common. Constraints are imposed on the grid points, using the BJOIN entry to tie the elements together, and a user-written subroutine exbrk.f to define the failure mechanism (see Figure 2-5). For the BJOIN entry, you must define an input set that contains all the joined grid point pairs, while the subroutine exbrk.f causes the connection of a set of tied nodes to break when the average plastic strain of the four adjacent elements surrounding this set of nodes exceeds 50%. The entire plate is defined as a Lagranagian interface by the SURFACE entry. Finally, the Lagrangian interaction between ball and plate is established by the CONTACT entry, where the ball is defined as “master” and the plate as “slave”. Note that the plate must be a slave surface since the elements (segments of the SURFACE) are separate during the contact, which will cause problems with sliding. In the eroding element case (normal failure criterion), there are again 900 elements but only 961 grid points for defining the plate. Instead of the BJOIN entry, the FAILMPS entry is used and the maximum plastic strain is set to 0.5. The results are shown in Figure 2-7.

Results Figure 2-6 and Figure 2-7 show the penetration of the ball through a plate modeled with BJOIN and

modeled with eroding elements, respectively. In the BJOIN case, the elements that break off remain visible, whereas in the eroding element case, the failed elements disappear as was expected. In Figure 2-8, the velocity of the ball, the kinetic energy, and the internal energy of the plate are compared for both models. In the BJOIN case, the ball loses more kinetic energy during the impact than in the eroding element case. Finally, the BJOIN case consumes about five times more CPU time than the eroding element case.

Main Index

62 Dytran Example Problem Manual Ball Penetrating a Steel Plate

Figure 2-8

Main Index

Difference in the Velocity of the Ball, and the Kinetic and Internal Energy Plate between both Cases. The Thick (Upper) Curve is made using BJOIN

Chapter 2: Structural Contact 63 Ball Penetrating a Steel Plate

Files ball_plate.dat

Dytran input file for BJOIN case

ball_plate_adap.dat

Dytran input file for the eroding element case

ball_xl.dat

Geometry input file for ball

non_equi_plate.dat

Geometry input file for BJOIN_plate

bjoin_grid_set.dat

BJOIN grid_pairs file

fail_plate_xl.dat

Geometry input file for plate for eroding element case. Dytran Input Files

BALL_PLATE.OUT BALL_PLATE_ADAP.OUT

Dytran output files

BALL_PLATE_BALL_0.ARC BALL_PLATE_PLATE_0.ARC

Dytran archive files (BJOIN case)

BALL_PLATE_ADAP_BALL_0.ARC BALL_PLATE_ADAP_PLATE_0.ARC

Dytran archive files (eroding element)

BALL_PLATE_RIGID_BALL_0.THS BALL_PLATE_ENERGY_0.THS

Dytran time history files (BJOIN case)

BALL_PLATE_ADAP_RIGID_BALL_0.THS Dytran time history files (eroding element) BALL_PLATE_ADAP_ENERGY_0.THS exbrk.f

User subroutine used for BJOIN case

Reference 1. Gere, James M., and Timoshenko, Stephen P., Mechanics of Materials, 1985, Wadsworth International.

Main Index

64 Dytran Example Problem Manual Ball Penetrating a Steel Plate

Abbreviated Dytran Input File START CEND TITLE = Ball penetrating a steel plate CHECK = NO ENDSTEP = 1100 TLOAD = 1 TIC = 1 SPC = 1 $$ $$ Data for output control Set 1 $$ TYPE (Ball) = ARCHIVE ELEMENTS (Ball) = 8 SET 8 = ALLDUMQUAD ELOUT (Ball) = ZUSER STEPS (Ball) = 0,THRU,END,BY,100 SAVE (Ball) = 10000 $$ $$ Data for output control Set 2 $$ TYPE (Plate) = ARCHIVE ELEMENTS (Plate) = 7 SET 7 = ALLSHQUAD ELOUT (Plate) = EFFPL-MID,EFFST-MID STEPS (Plate) = 0,THRU,END,BY,100 SAVE (Plate) = 10000 $$ $$ Data for output control Set 3 $$ TYPE (Rigid_Ball) = TimeHis RIGIDS (Rigid_Ball) = 9 SET 9 = 200 RBOUT (Rigid_Ball) = XCG, YCG, ZCG, XVEL, YVEL, ZVEL STEPS (Rigid_Ball) = 0 THRU END BY 2 SAVE(Rigid_Ball) = 10000 $$ $$ Data for output control Set 4 $$ TYPE (Energy) = TimeHis MATS (Energy) = 12 SET 12 = 100 MATOUT (Energy) = EKIN, EINT, EDIS STEPS (Energy) = 0 THRU END BY 2 SAVE(Energy) = 10000 $ $ BEGIN BULK SETTING,1,VERSION2 PARAM,INISTEP,1.E-7 PARAM,BULKL,0.06 PARAM,BULKQ,1.44 $

Main Index

Chapter 2: Structural Contact 65 Ball Penetrating a Steel Plate

$ Model geometry. $ --------------INCLUDE ball_xl.dat $ INCLUDE non_equi_plate.dat $ $ Definition of joined grid points. $ --------------------------------BJOIN,1,229,1.E-5,USER,,EXBRK,,,+ +,,,,,YES $ INCLUDE bjoin_grid_set.dat $ $ Dummy elements to build the ball. $ --------------------------------PSHELL1,200, ,DUMMY $ $ Define the ball as a rigid body. $ -------------------------------RIGID,200,2,2.617,,.15,.15,.044,,+ +,,0.0,0.0,-230.,,,,,+ +,,.124767,0.0,0.0,.124767,0.0,.124767 $ $ Structural elements to build the steel plate. $ --------------------------------------------PSHELL,100,100,.005 YLDVM,222,4.+8,0.0 DMATEP,100,7850.,2.+11,.3,,,222 $ $ Contact surface on the plate. $ ----------------------------SURFACE,1,,ELEM,111 SET1,111,1,THRU,900 $ $ Surface for defining the ball as a rigid body, $ which is also used as contact surface. $ ---------------------------------------SURFACE,2,,SEG,1 $ $ Contact surface for master (ball) - slave (plate). $ -------------------------------------------------CONTACT,1,SURF,SURF,1,2 $ ENDDATA

Main Index

66 Dytran Example Problem Manual Tapered Beam Striking a Rigid Wall

Tapered Beam Striking a Rigid Wall Problem Description In the case of an impact between a moving object (given an initial velocity v0) and a rigid stationary wall, it is common for the user to request forces, stresses, and deformations resulting from this impact. In some cases, it might be easier to consider an equivalent problem where the object has no initial velocity and where the rigid wall is given a constant velocity v(t) = –v0. The resulting forces, stresses, and deformations are exactly equal for both cases since the first problem is identical to the second problem if the frame of reference is moving with the same velocity v0. To verify that Dytran is consistent with this theoretical result, two alternative analyses are used to model the impact of a tapered beam (made of steel with a length of 1 m and a weight of 40.71 kg) with a rigid plate. In the first case, the beam is given an initial velocity of 100m/sec while the plate is fixed. In the second case, the plate is given a constant velocity of –100 m/sec while the beam has no initial velocity but is free to move. The properties of the beam are as follows: Beam material

steel

density

ρ = 7830 kg/m3

mass

m = 40.71 kg

bulk modulus

K = 1.64 1013 N/m2

shear modulus

G = 8.18 1012 N/m2

yield stress

σ = 1.4 1011 N/m2

spall pressure

ps = –3.8 109 N/m2

Desired Results The two alternative analyses should yield the same results with respect to forces, stresses, and deformations. Furthermore, the resulting velocities and momentum of the tapered beam for both cases should differ by exactly 100 m/sec and 100 kg m/sec for every time step, respectively.

Dytran Modeling The plate is modeled using 5 x 5 (25) Belytschko-Tsay shell elements (CQUAD4) covered by CFACEs. Using the PSHELL1 (with the DUMMY option), the SURFACE, and the RIGID entries, the plate is defined as a rigid body as well as a Lagrangian contact surface. Using the TLOAD1 entry (TYPE = 12) and the FORCE entry, the plate is given a constant velocity of 0 m/sec and 100 m/sec in the first and second case, respectively. The tapered beam is built from 4 x 20 (80) nonuniform CHEXA elements that get smaller towards the one end. The PSOLID entry together with the DMAT, EOSPOL, SHREL, YLDVM, and PMINC entries

Main Index

Chapter 2: Structural Contact 67 Tapered Beam Striking a Rigid Wall

define a steel material property for the beam. The front part of the beam, which is in contact with the plate, is covered by CFACEs to define a Lagrangian contact surface. In the first case, all the grid points of the beam are given an initial velocity of 100 m/sec by using the TICGP entry. Finally, the CONTACT entry defines the contact between the beam (slave) and the plate (master).

Results As shown in Figure 2-9 through Figure 2-14, the results are identical for both cases, as expected.

Figure 2-9

Main Index

Moving Beam - Effective Stress Contour of the Deformed Beam after 1200 Cycles

68 Dytran Example Problem Manual Tapered Beam Striking a Rigid Wall

Main Index

Figure 2-10

Moving Plate – Effective Stress Contour of the Deformed Beam after 1200 Cycles

Figure 2-11

Z-Forces for a Grid Point on the Front of the Beam – The Curves are identical for both Cases

Chapter 2: Structural Contact 69 Tapered Beam Striking a Rigid Wall

Main Index

Figure 2-12

Z-velocities for Grid Point on the Front of the Beam – The Upper Curve is derived from the Moving Beam Case

Figure 2-13

Z-momentum averaged over all the Grid Points of the Beam – The Upper Curve is derived from the Moving Beam Case

70 Dytran Example Problem Manual Tapered Beam Striking a Rigid Wall

Figure 2-14

Average Kinetic Energy of all the Grid Points of the Beam – The Thick Curve is derived from the Moving Beam Case

Files tapered_beam1.dat tapered_beam2.dat tapered_beam_xl.dat

Dytran input files (case 1)

TAPERED_BEAM1.OUT TAPERED_BEAM2.OUT

Dytran output files

TAPERED_BEAM1_BEAM_0.ARC TAPERED_BEAM1_PLATE_0.ARC

Dytran archive files (BJOIN case)

TAPERED_BEAM2_BEAM_0.ARC TAPERED_BEAM2_PLATE_0.ARC

Dytran archive files (eroding element)

TAPERED_BEAM1_VEL_0.THS TAPERED_BEAM1_ENERGY_0.THS

Dytran time history files (BJOIN case)

TAPERED_BEAM2_VEL_0.THS TAPERED_BEAM2_ENERGY_0.THS

Dytran time history files (eroding element)

Dytran input files (case 2)

Abbreviated Dytran Input File START CEND TITLE = Moving beam strikes fixed wall.

Main Index

Chapter 2: Structural Contact 71 Tapered Beam Striking a Rigid Wall

CHECK = NO ENDSTEP = 1200 TLOAD = 1 TIC = 1 SPC = 1 $$ $$ Data for output control Set 1 $$ TYPE (BEAM) = ARCHIVE ELEMENTS (BEAM) = 8 SET 8 = 1t80 ELOUT (BEAM) = EFFSTS,PRESSURE,EFFPLS STEPS (BEAM) = 0,THRU,END,BY,50 SAVE (BEAM) = 10000 $$ $$ Data for output control Set 2 $$ TYPE (PLATE) = ARCHIVE ELEMENTS (PLATE) = 17 SET 17 = ALLDUMQUAD ELOUT (PLATE) = ZUSER STEPS (PLATE) = 0,THRU,END,BY,50 SAVE (PLATE) = 10000 $$ $$ Data for output control Set 3 $$ TYPE (VEL) = TIMEHIS GRIDS (VEL) = 9 SET 9 = 229 GPOUT (VEL) = XVEL,YVEL,ZVEL,ZFORCE STEPS (VEL) = 0,THRU,END,BY,2 SAVE (VEL) = 10000 $$ $$ Data for output control Set 4 $$ TYPE (ENERGY) = TIMEHIS MATS (ENERGY) = 13 SET 13 = 1 MATOUT (ENERGY) = EKIN, EINT, EDIS,XMOM,YMOM,ZMOM STEPS (ENERGY) = 0,THRU,END,BY,2 SAVE (ENERGY) = 10000 $ BEGIN BULK SETTING,1,VERSION2 PARAM,INISTEP,1.E-7 $ $ Model geometry. $ All entities are measured in cm. $ -------------------------------INCLUDE tapered_beam_xl.dat $ $ Define plate. $ ------------PSHELL1,2,,DUMMY

Main Index

72 Dytran Example Problem Manual Tapered Beam Striking a Rigid Wall

$ $ A surface is needed to define the plate as a rigid body and $ to define a Lagrangian contact surface. $ --------------------------------------SURFACE,2,,SEG,2 $ $ Plate is rigid. In this case v=(0,0,0). $ --------------------------------------RIGID,2,2,50.,,5.,5.,101.,,+ +,,0.,0.,0.,,,,,+ +,,1.e20,0.,0.,1.e20,0.,1.e20 $ $ Plate is fixed in space for this case. $ By changing the data in the Force-entry, the plate can be $ given a constant velocity. $ -------------------------TLOAD1,1,2,,12 $ FORCE,2,2,,0.,0.,0.,1. $ $ Initial velocity beam is 10000 cm/s for this case. $ The TICGP-entry is skipped when the plate is given a velocity. $ -------------------------------------------------------------TICGP,1,5,ZVEL,10000. $ SET1,5,45,THRU,233 $ $ Properties beam: $ --------------PSOLID,1,1 $ $ Material beam is steel. $ ----------------------DMAT,1,0.007830,1,1,1,,1 $ EOSPOL,1,1.64e9 $ SHREL,1,8.18e8 $ YLDVM,1,1.4e7 $ PMINC,1,-3.8e7 $ $ A contact surface is needed for the beam. $ ----------------------------------------SURFACE,1,,SEG,1 $ $ Contact between beam (slave) and plate (master): $ -----------------------------------------------CONTACT,1,SURF,SURF,1,2 $ ENDDATA

Main Index

Chapter 2: Structural Contact 73 Impact Loading

Impact Loading Problem Description A simple elastic bar is subjected to an impact load on one end and supported on the other end. The impact load on the elastic bar is applied by a rigid body with mass m = 0.00311 lbs2/in and an initial velocity v0= 240 in/s. The elastic bar has the following properties: Cross sectional area A = 0.2025 in2 Elastic modulus

E = 10.5 x 106 psi

Material density

ρ = 0.000262 lb.s2/in4

Bar length

L = 6.00 in.

The purpose of this model is to study the response of the simple elastic bar to impact loading and to check the results against theoretical data.

Theoretical Results The contacted end of the elastic bar experiences an impulse, which causes an internal force locally in the bar. This pulse starts traveling down the length of the bar and the force in the bar at any point away from the impacted end remains zero until the distortional pulse reaches that point. The distortional pulse (stress wave) travels with at a fixed velocity equal to the speed of sound in solids, which is a function of material density and stiffness. The magnitude of the initial pulse is given by P pul se = A ν 0 Eρ = 2,549 lb

The fixed velocity of the distortional pulse is given by c =

E ⁄ ρ = 2000,322 in/s

The time required by the pulse to traverse the entire length of the bar is given by L t = --- = 0.03 ms c

From the conservation of energy, the magnitude of peak load in the bar is P max =

Main Index

E A m x ν = 7967 lb -----------0 L

74 Dytran Example Problem Manual Impact Loading

Dytran Model Due to one-dimensional problem, a simple mesh is set up (see Figure 2-15). The rigid body is modeled as a single Lagrangian solid CHEXA element of dimension 0.5 in × 0.5 in × 0.5 in. The elastic bar is modeled as 1 × 40(40) equidistant Lagrangian solid CHEXA elements for the entire length of 6 inches and all the nodes at one end of the bar is constrained in all directions. All the nodes of the rigid body are given an initial velocity of 240 in/s. A cross-section is defined at the center of the bar using SECTION card to monitor the axial force at the center of the bar.

Figure 2-15

Numerical Problem Layout showing CHEXA Lagrangian Solid Elements

Results The plot for the axial cross-sectional force at the center of the bar is shown in Figure 2-16. It can be seen from the plot that the time taken for the stress wave to travel the entire length of the bar is t = 0.03ms. After the stress wave reaches the fixed ends of the bar, it gets reflected as a compressive wave traveling back towards the impactor. Because the initial pulse did not impart the entire impact energy to the bar, the impactor is still moving forward. Therefore, a secondary compressive pulse is initiated with a slightly lower magnitude than the initial pulse because the velocity of the impactor is slightly reduced. This secondary pulse adds directly to the reflected initial pulse. This process continues with the addition of ever diminishing pulse magnitudes until all the impact energy has been absorbed. The correlation between the theoretical values and the numerical values obtained from theDytran result plots are very good.

Main Index

Chapter 2: Structural Contact 75 Impact Loading

Figure 2-16

Plot for Axial Force at the Center of the Bar

Files bar.dat

Dytran input file

BAR.OUT

Dytran output file

BAR_ELEMENT_0.ARC

Dytran archive file

BAR_SEC_0.THS

Dytran time history file

References (Continued) 2. Jones, Rodney H., Hethcock, J. Donn, and Mullins, B. R., “Analysis and Design of Dynamically Loaded Fittings” American Helicopter Society 57th Annual Forum (2001).

Abbreviated Dytran Input File START CEND ENDTIME=0.0007 ENDSTEP=999999 CHECK=NO TITLE= Jobname is: bar TLOAD=1 TIC=1 SPC=1 $ $Output request for the stress distribution in the elastic bar $ TYPE (element) = ARCHIVE

Main Index

76 Dytran Example Problem Manual Impact Loading

ELEMENTS (element) = 1 SET 1 = 1 THRU 40 ELOUT (element) = EFFSTS STEPS (element) = 0 thru end by 20 SAVE (element) = 10000 $ $Output request for the forces on the cross-sections of the bar $ TYPE (sec) = TIMEHIS CSECS (sec) = 5 SET 5 = 1 2 3 CSOUT (sec) = XFORCE YFORCE ZFORCE FMAGN STEPS (sec) = 0 thru end by 1 SAVE (sec) = 10000 $ $ $------- Parameter Section -----$ PARAM,INISTEP,1e-7 PARAM,CONTACT,VERSION,V4 PARAM,CONTACT,THICK,0.0 PARAM,CONTACT,GAP,0.0 $ $------- BULK DATA SECTION ------$ BEGIN BULK $ SECTION 1 99 999 SET1 99 161 THRU 164 SET1 999 40 $ SECTION 2 88 888 SET1 88 81 THRU 84 SET1 888 20 $ SECTION 3 77 777 SET1 77 1 THRU 4 SET1 777 1 $ $ $ --- SPC-name = fixed_ends $ SPC1 1 123 161 THRU 164 $ $ --- Define 172 grid points --$ GRID 1 .0000000.0000000.0000000 : : : GRID 172 .4750000.4750000.0250000 $ $ --- Define 41 elements$ $

Main Index

Chapter 2: Structural Contact 77 Impact Loading

$ -------- property set rigid --------$ CHEXA 41 1 165 166 168 167 +A000049 172 171 $ $

169

170+A000049

$ -------- property set bar --------$ CHEXA 1 2 1 2 4 3 5 6+A000050 +A000050 8 7 . .CHEXA 40 2 157 158 160 159 161 162+A000089 +A000089 164 163 $ ========== PROPERTY SETS ========== $ $ * rigid * $ PSOLID 1 1 $ $ * bar * $ PSOLID 2 2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ -------- Material bar_mat id =1 DMATEL 2 0.000262 1.05e+007 0.33+A000090 +A000090 $ $ -------- Material rigid id =2 MATRIG 1 .00311 $ $ ======== Load Cases ======================== $ $ $ ------- Initial Velocity BC initial_velocity ----$ SET1 1 165 THRU 172 TICGP 1 1 ZVEL -240 $ $ -------- Contact : contact $ $ CONTACT 3 SURF SURF 1 2 +A000091 +A000091 V4 BOTH SLAVE+A000092 +A000092 +A000093 +A000093 ON $ $ Slave contact surface for contact $ $

Main Index

78 Dytran Example Problem Manual Impact Loading

SURFACE 1 SEG 1 CFACE 1 1 1 2 . . CFACE 162 1 40 3 $ $ Master contact surface for contact $ SURFACE 2 SEG 2 CFACE 163 2 41 2 CFACE 164 2 41 5 CFACE 165 2 41 1 CFACE 166 2 41 6 CFACE 167 2 41 4 CFACE 168 2 41 3 $ $ ENDDATA

Main Index

Chapter 2: Structural Contact 79 Pipe Whip

Pipe Whip Problem Description A tubular pipe that rotates with an initial velocity of 200 radians/s around one end hits another tubular pipe that is clamped on both ends:

Figure 2-17

Initial Configuration of Pipe Whip Model

The overall dimensions and material properties of both pipes are the same, and are shown below: Radius

R = 0.1778 m

Length

L = 1.524 m

Thickness

t = 0.01097 m

Density

ρ = 7827 kg/m3

Young's modulus

E = 2.07E11 Pa

Poisson’s Ratio

ν = 0.3 σy = 3.1E11 Pa *

Yield Stress

*A very high Yield Stress has been defined to mimic almost elastic behavior. This will result in a quick rebound of the rotating pipe, and is meant for demonstration purposes only. The center of rotation of the rotating pipe is equal to the center point of its one rear end. The impact point is at the middle of the stationary pipe and for the rotation pipe, it is at a distance of 0.9144 m from the rotation center. The initial rotational velocity is 200 radians/s.

Main Index

80 Dytran Example Problem Manual Pipe Whip

The calculation is carried out from the time just before the impact until the time that the rotating pipe begins to reverse. The Problem Time is 3 msec.

Dytran Model The pipes are modeled with quadrilateral shell elements, and a contact is defined between the rotating pipe and the stationary pipe. An additional self-contact is defined for the rotating pipe. A DMATEP with YLDVM is used to define the material. In order to easily apply the rotational boundary condition, a plate closes off the pipe, and an SPC is applied to the center node. This plate is given a thickness 10 times higher than the thickness of the pipe itself. This mimics a rigid connection between all the nodes in the rotation plane. The entire analysis from modeling to postprocessing is carried out by using the Dytran Preference of MD Patran. The complete description for this step-by-step Patran based Analysis performance is reported in the workshop 12. For quick reproduction of this model in Patran, two session files are available: Pipe_Whip_1.ses: From modeling to submission of the Dytran Job Pipe_Whip_2.ses: From reading the result files to postprocessing

Main Index

Chapter 2: Structural Contact 81 Pipe Whip

Results From the Time History of the Kinetic Energy of Material_1, it can be seen that the rotating pipe has bounced back around 2.5 mseconds.

Figure 2-18

Main Index

Kinetic Energy of Rotating and Stationary Pipe

82 Dytran Example Problem Manual Pipe Whip

The deformation of the pipes at 3 msec is shown in the Figure 2-19 .

Figure 2-19

Deformed Shape at 3 msec.

Files Pipe_Whip.dat

Dytran input file

Pipe_Whip.bdf

Dytran input file

PIPE_WHIP.OUT

Dytran output file

PIPE_WHIP_LAG_0.ARC

Dytran ARC file

PIPE_WHIP_MAT_0.THS

Dytran THS file

Pipe_Whip_1.ses

MD Patran session file: Modeling & JobSubmission

Pipe_Whip_2.ses

MD Patran session file: Postprocessing

Abbreviated Dytran Input File START CEND ENDTIME=0.3E-2 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: Pipe_Whip TLOAD=1 TIC=1 SPC=1 $ Output result for request: Lag TYPE (Lag) = ARCHIVE ELEMENTS (Lag) = 1

Main Index

Chapter 2: Structural Contact 83 Pipe Whip

SET 1 = 1 THRU 7896 ELOUT (Lag) = EFFST-MID TIMES (Lag) = 0 THRU END BY 0.3E-3 SAVE (Lag) = 10000 $ Output result for request: Mat TYPE (Mat) = TIMEHIS MATS (Mat) = 2 SET 2 = 1 2 MATOUT (Mat) = EKIN EINT EDIS STEPS (Mat) = 0 THRU END BY 10 SAVE (Mat) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1.E-8 PARAM,LIMCUB,5500 PARAM,MINSTEP,1.E-15 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE Pipe_Whip.bdf $ $ ========== PROPERTY SETS ========== $ $ * p.11 * $ PSHELL 11 1 .1097 $ $ * p.1 * $ PSHELL 1 1 .01097 $ $ * p.2 * $ PSHELL 2 2 .01097 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material m1 id =1 DMATEP 1 78272.07e+11 .3 YLDVM 1 3.1e+11 $ $ -------- Material m2 id =2 DMATEP 2 78272.07e+11 .3 YLDVM 2 3.1e+11 $ $ ======== Load Cases ======================== $ $ $ -------- Contact : Contact1 $ CONTACT 5 SURF SURF 1 2 +A000001 V4 TOP $

Main Index

1

2

+A000001

84 Dytran Example Problem Manual Pipe Whip

$ Slave contact surface for Contact1 $ SURFACE 1 SUB 1 SUBSURF 1 1 ELEM 3 SET1 3 3988 THRU 4011 4023 +A000002 THRU 4081 4093 THRU 4116 . . . $ $ Master contact surface for Contact1 $ SURFACE 2 SUB 2 SUBSURF 2 2 ELEM 4 SET1 4 73 THRU 98 115 +A000043 THRU 182 199 THRU 224 . . . $ $ -------- Contact : Contact2 $ CONTACT 6 SURF 3 +A000084 V4 BOTH $ $ Self Contact Surface for Contact2 $ SURFACE 3 SUB 3 SUBSURF 3 3 ELEM 5 SET1 5 73 THRU 98 115 +A000085 THRU 182 199 THRU 224 . . . $ $ ------- Initial Rotational Velocity: Inivel ----TIC3 1 1 1 +A000126 0 0 0 0 200 . . . $ $ ENDDATA

Main Index

THRU 4128

4046 THRU

4058+A000002 4151+A000003

THRU 241

140 THRU

157+A000043 266+A000044

+A000084

THRU 241

0

140 THRU

157+A000085 266+A000086

+A000126 +A000127

Chapter 3: Fluid Dynamics Dytran Example Problem Manual

3

Main Index

Fluid Dynamics J

Overview

J

Shock Tube

87

J

Blast Wave

94

J

JWL Explosive Test

J

Modeling Blast Wave using 1-D Spherical Symmetry Method 107

J

Modeling the JWL Explosion using 1-D Spherical Symmtery

J

Nonuniformity with MESH,BOX

86

99

125

86 Dytran Example Problem Manual Overview

Overview In this chapter, a number of example problems are presented that show the fluid dynamics capabilities of Dytran. The user can find in these examples how to model the dynamic behavior of fluids and gasses using Eulerian technology in Dytran, what material models to use, and how to apply loads and constraints. The use of materials includes high energy explosive materials.

Main Index

Chapter 3: Fluid Dynamics 87 Shock Tube

Shock Tube Problem Description The propagation of shock waves is an important consideration for many Dytran application areas. Shock tube gas dynamics constitute a relatively simple process and as such a closed form solution can be found by analytical means. For one-dimensional plane shock, two adjacent regions of gas are initialized at time = 0 and there is a single discontinuity at the boundary between them .

Figure 3-1

One-Dimensional Plane Shock

At time > 0, the situation is much more complex. At point b, the expansion front of the compressed material has a velocity Ub, the sound speed of the compressed material. Point c is the contact front between the compressed and uncompressed materials and has a velocity Uc, and point s is the shock front with velocity Us. In the region between c and s, the material variables are constant (ρ, P, Uc) with a local sound speed. This time region is therefore useful for the study of hypervelocity events.

Figure 3-2

Main Index

Density and Velocity Profiles in a Shock Tube

88 Dytran Example Problem Manual Shock Tube

Desired Results For this example, the ideal gas equation of state is used. The initial conditions are: High Pressure

Low Pressure

ρ=1

ρ = 0.125

u=0

u=0

e = 2.5

e = 2.0

P=1

P = 0.1

A time > 0, the analytical technique produces ([Ref. 1.] to [Ref. 3.]), the following results: P

=

3.03130

Pc

=

0.303130

ρR

=

0.265574

ρL

=

0.426319

eR

=

2.85354

eL

=

1.77760

Uc

=

0.927453 (peak flow velocity: at point c)

Dytran Modeling Two models were analyzed with different mesh densities. In both models, hydrodynamic Euler elements are used, and the gas regions are initialized using a TICEUL entry. (See Chapter 2, Eulerian Elements of the Dytran User’s Guide.)

Main Index

Chapter 3: Fluid Dynamics 89 Shock Tube

Coarse Model The shock tube is modeled using 2500 (5 x 5 x 100) CHEXA elements. The numerical mesh is given in Figure 3-3.

Figure 3-3

Coarse Finite Element Mesh

Fine Model The shock tube is modeled using 5000 (5 x 5 x 200) CHEXA elements.

Results The sharpness of the steps in the graph as produced by a Dytran analysis is dependent on how fine a mesh is used. This effect can be seen by comparing the graphs produced from the two analyses.

Coarse Model The Results are plotted in Figure 3-4 and Figure 3-5 show that the variation of material variables along the length of the shock tube. A selective number of elements along the axis of the shock tube were selected for this purpose.

Main Index

90 Dytran Example Problem Manual Shock Tube

Main Index

Figure 3-4

Length along the Shock Tube plotted against Material Density

Figure 3-5

Length along the Shock Tube plotted against Material Sound Speed and Material Velocity

Chapter 3: Fluid Dynamics 91 Shock Tube

Model The fine model shows the effect of a finer mesh on the resolution of the results graphs (see the graphs shown in Figure 3-6 and Figure 3-7).

Main Index

Figure 3-6

Length Along the Shock Tube Plotted against Material Density

Figure 3-7

Length along the Shock Tube plotted against Material Sound Speed and Material Velocity

92 Dytran Example Problem Manual Shock Tube

Coarse Mesh eul1_coarse.dat

Dytran input files

EUL1_COARSE.OUT

Dytran output files

EUL1_COARSE_TUBE_0.ARC

Dytran archive files

Fine Mesh eul1_fine.dat

Dytran input file

EUL1_FINE.OUT

Dytran output files

EUL1_FINE_TUBE_0.ARC

Dytran archive files

References 1. Buis, J. P., “Analysis of shock Tube Calculations Performed by the PISCES-2DELK/V4 Code,” 1967, PISCES International B.V. 2. Harlow, F. H., and Amsden, A. A., “Fluid Dynamics,” 1971, Los Alamos Scientific Laboratory Monograph. 3. Jardin, S. C., and Hoffman, R., “Simulation of Explosive Processes in PISCES-1DL,” 1972, Physics International Company.

Abbreviated Dytran Input File START CEND TITLE = Shock Tube #1 ENDTIME = 0.138 ENDSTEP = 100 TIC = 1 $$ $$ Data for Output Control Set 1 $$ TYPE(tube) = ARCHIVE SAVE(tube) = 9999 ELOUT(tube) = XVEL,YVEL,ZVEL,DENSITY,SIE,PRESSURE,SSPD STEPS(tube) = 0,End ELEMENTS(tube) = 1 SET 1 = 13t2488b25 BEGIN BULK PARAM INISTEP 0.0005 PARAM MINSTEP 0.0001 $ $ THIS SECTION CONTAINS BULK DATA $ Model Geometry: gridpoints and elements $-----------------------------------------$

Main Index

Chapter 3: Fluid Dynamics 93 Shock Tube

GRID 1 .05 0.0 0.0 . . . . GRID 3636 -.05 -1.39-161. $ CHEXA 1 1 1 37 38 + 44 8 . . . CHEXA 2500 1 3593 3629 3630 + 3636 3600 $ $Property,material and equation of state data $-------------------------------------------PEULER1 1 Hydro 1 $ DMAT 1 1. 1 $ EOSGAM 1 1.4 $ $Allocation of material to geometric regions. $-------------------------------------------TICEUL 1 + ELEM 1 1 11 1. + ELEM 2 1 12 2. SET1 1 1251 THRU 2500 SET1 2 1 THRU 1250 $ $Initial material data $----------------------TICVAL 11 density .125 zvel TICVAL 12 density 1. zvel ENDDATA

Main Index

2

7

43

+

3594

3599

3635

+

+ +

0.0 0.0

sie sie

2. 2.5

94 Dytran Example Problem Manual Blast Wave

Blast Wave Problem Description The effect of a detonation on the environment can be simulated by assuming that the detonated material can be idealized by a sphere of hot gas with a homogeneous density and specific internal energy. This approach is suited for problems in which the processes inside of the explosive material are not to be investigated. In this example, the propagation of a blast wave will be simulated starting from the initial shock front radius R0 = 0.05 m at the time t = 0 sec until it reaches a radius of R = 10 R0. Both, the gas in the sphere and the surrounding environment behave as an ideal gas (γ = 1.4). The initial conditions are: Blast Products (r < R0) specific internal energy

e1 = 1.29 1010 Joule/kg

density

ρ1 = 7.74 kg/m3

Environment (Air at Room Temperature, r > R0) specific internal energy

e0 = 1.938 105 Joule/kg

density

ρ0 = 1.29 kg/m3

The index “1” denotes values at the shock front and the index “0” values in the ambient atmosphere.

Theoretical Solution A theoretical solution is available for a spherical blast wave originating from a point source (Taylor’s similarity solution; see [Ref. 3.]). Assuming ρ1 >> ρ0, the evaluation of the Rankine-Hugoniot relations deliver in the case of a blast wave: ρ1 γ+1 ----- = -----------γ–1 ρ0

(3-1)

This relation has been considered in the given problem setup. Assuming a point source and internal energy E, the pressure ρ1 at the shock front (r = R) should be ρ = 0.155 E 0 r

–3

Inserting the initial conditions yields

Main Index

(3-2)

Chapter 3: Fluid Dynamics 95 Blast Wave

4 3 E = e 1 ρ 1 --- π R 0 = 1.6641 E7 J 3

(3-3)

and ρ 1 ( R ) = 2.58E6 J R

Figure 3-8

–3

(3-4)

Radial Pressure Distribution for several Time Steps and the Theoretical Maximal Pressure

The results of the simulation should converge to this solution as the radius of the shock front increases. The analysis should generate an x-y plot of the pressure distribution along the x-axis and compare it with the theoretical solution. Also, a contour plot should be produced to check the spherical shape of the expanding pressure front.

Dytran Modeling The motion of the gas is radial. Therefore, only a part of the area of interest need to be analyzed. Here, a mesh of 20 x 20 x 20 hexahedral Eulerian elements has been used for investigating the volume (0 < x < 0.3 m, 0 < y < 0.5 m, 0 < z < 0.5 m). The center of the blast wave is located at the origin of the global coordinate system. As discussed in Chapter 2, Eulerian Elements of the Dytran User’s Guide, TICEUL entry can be used for describing the initial conditions. A higher level has to be used to indicate the higher priority of the spherical condition. The value for the parameter INISTEP has to be below the minimal time step that follows from the Courant Criterion:

Main Index

96 Dytran Example Problem Manual Blast Wave

l Δ t = S ----c0

(3-5)

where l denotes the smallest element dimension, c0 the initial speed of sound, and S is a safety factor Dytran’s default value is 2/3). With l = 10 R 0 ⁄ 20 = 0.025 m and c 0 = γ ( γ – 1 ) e 1 Equation (3-5) yields for the time step seconds. Therefore, the value 1.96E-7 sec is used for the INISTEP parameter.

Δ t = 1.96 E-7

Results Figure 3-9 shows pressure profiles in the elements 1 to 20, which are located along the edge y = z = 0 m

of the control volume. The theoretical value of the shock front pressure is also included in this plot. Though this mesh is very coarse, the analysis results is a fairly good approximation of the theoretical values. Note that at t = 0, element 2 is not completely inside of the sphere of high energy gas. Therefore, its pressure is below that of element 1 at t = 0 sec.

Figure 3-9

Pressure Contour Plot for Step 50

The contour plot of the pressure in Figure 3-9 shows an almost spherical shape of the shock front. Deviations are due to the fact that mass transport takes place along element faces only. Thus, the relief of pressure is hampered for the elements on the diagonal.

Main Index

Chapter 3: Fluid Dynamics 97 Blast Wave

Files blast.dat

Dytran input file

BLAST.OUT

Dytran output file

BLAST_PROFILE_0.ARC BLAST_CONTOUR_50.ARC

Dytran archive files

Reference (Continued) 4. Baker, W. E., Explosions in Air, University of Texas Press, 1973, Austin and London.

Abbreviated Dytran Input File START CEND TITLE = Blast Wave CHECK = NO ENDSTEP = 60 $$ $$ Data for Output Control Set 1 $$ TYPE(profile) = ARCHIVE SAVE(profile) = 99999 ELEMENTS(profile) = 1 SET 1 = 1t20 STEPS(profile) = 0,5,10,15,20,30,40,50,60 ELOUT(profile) = PRESSURE $$ $$ Data for Output Control Set 2 $$ TYPE(contour) = ARCHIVE SAVE(contour) = 999999 ELEMENTS(contour) = 2 SET 2 = 1t8000 STEPS(contour) = 50 ELOUT(contour) = PRESSURE BEGIN BULK PARAM INISTEP 1.5e-07 $ $ THIS SECTION CONTAINS BULK DATA $ $ GRID 1 0.0 0.0 0.0 GRID 2 .025 0.0 0.0 GRID 3 .05 0.0 0.0 GRID 4 .075 0.0 0.0 . . . .

Main Index

98 Dytran Example Problem Manual Blast Wave

GRID GRID GRID GRID $ CHEXA + CHEXA + CHEXA + CHEXA + . . . . CHEXA + CHEXA + CHEXA + CHEXA + $ $ PEULER1 $ DMAT $ EOSGAM $ TICEUL + + SET1 $ TICVAL TICVAL $ SPHERE ENDDATA

Main Index

9258 9259 9260 9261

.425 .45 .475 .5

.5 .5 .5 .5

.5 .5 .5 .5

1 464 2 465 3 466 4 467

1 23 1 24 1 25 1 26

1

442

443

2

22

463

+

2

443

444

3

23

464

+

3

444

445

4

24

465

+

4

445

446

5

25

466

+

7997 9258 7998 9259 7999 9260 8000 9261

1 8817 1 8818 1 8819 1 8820

8795

9236

9237

8796

8816

9257

+

8796

9237

9238

8797

8817

9258

+

8797

9238

9239

8798

8818

9259

+

8798

9239

9240

8799

8819

9260

+

Hydro

1

1 1

1.

1

1.4

1 SPHERE ELEM 1

1 1 1

1

1 1 THRU

+ +

1 2 8000

10. 5.

1 2

density 7.74 density 1.29

sie sie

1.29+10 193800.

1

0.0

0.0

.05

0.0

Chapter 3: Fluid Dynamics 99 JWL Explosive Test

JWL Explosive Test Problem Description A slab of explosive, COMPOSITION B, 1 cm x 50 cm, is detonated at one end. The explosive material data is shown below: • JWL equation of state parameters

A

=

5.24229 1011

B

=

0.07678 1011

R1

=

4.2

R2

=

1.1

ω

=

0.34

• density ρ0 = 1717 kg/m3 • specific chemical energy q0 = 4.95 106 Joule/kg • detonation velocity D = 7980 m/sec

The purpose of this example is to model the detonation of the explosive and to check the pressure behind the detonation front (peak pressure) against the theoretical Chapman-Jouguet value.

Theoretical Background Detonation of High Explosives (HE) Detonation is the mechanism by which the “HIGH EXPLOSIVE” materials release their chemical energy: • The chemical reaction, causing the energy release, takes place in a narrow zone (reaction zone)

which propagates at high speed through the explosive that transforms the solid explosive into hot compressed gasses. The reaction zone is then a form of discontinuous wave, like a shock wave, with physical behavior that is governed solely by the properties of the unreached and completely reacted material on either side of the wave. This means that a hydrodynamic approach of detonation can be used.

Main Index

100 Dytran Example Problem Manual JWL Explosive Test

Hydrodynamic Theory of Steady-state Plane Detonation The model of the plane, steady-state reaction zone propagating at constant speed D through the explosive is depicted in Figure 3-10.

Figure 3-10

Plane Reaction Zone Propagating at Constant Speed

The Rankine-Hugoniot relations, which express the conservation of mass, momentum, and energy in the material stream flowing through the reaction zone, are used to relate the hydrodynamic variable across the reaction zone. Conservation of mass and momentum 2

D p – p 0 = ------2 ( V 0 – V ) V0

(3-6)

Conservation of energy 1 e – e 0 = --- ( p + p 0 ) ( V 0 – V ) + q 0 2

(3-7)

Equation (3-6) describes a straight line (Rayleigh line) defining the locus of all possible final states (p,V) attainable by a discontinuous transition from the initial state (p0, V0) consistent with conservation of mass and momentum. Equation (3-7) is purely thermodynamic from which, with a given equation of state p = p (V, e) for the detonation products, the energy term may be eliminated, resulting in the Hugoniot curve of the explosive. The Hugoniot curve defines a concave downward curve locus of all possible final states (p, V) attainable by a discontinuous transition from the initial state (p0, V0) consistent with conservation of energy. (See Figure 3-11.)

Main Index

Chapter 3: Fluid Dynamics 101 JWL Explosive Test

Figure 3-11

Hugoniat Curve and Rayleigh Line

The forms of the Rayleigh line and Hugoniot curve are such that their interaction permits the existence of any detonation speed D above a “minimum value” and each value of D is consistent with two possible final states for the detonation products. One further condition is therefore required. Chapman and Jouquet added the following condition to conservation of mass, momentum, and energy: • The detonation speed D is such that the Rayleigh line is tangent to the Hugoniot curve of the

explosive (or the detonation speed is the minimum velocity consistent with the RankineHugoniot relations). this process is shown in Figure 3-12.

Figure 3-12 Hugoniot Curve and Rayleigh Line for Detonation Process

Main Index

102 Dytran Example Problem Manual JWL Explosive Test

According to the above considerations if the ideal gas equation of state (with constant specific heat ratio γ) is used to model the detonation products, the following formulas result: P cj = 2 ( γ – 1 )q 0 ρ 0

(3-8)

γ V cj = ------------ V 0 γ+1

(3-9)

These relations are applicable also when using the JWL equation of state (with variable specific heat ratio) if measuring γ cj at the Chapman-Jouguet state ( P cj, V cj ) behind the detonation front.

Dytran Model Steady-state Detonation Modeling withDytran Dytran use the “programmed burn” technique to model the detonation of high explosives (HE). The basic assumption of this technique is that the reaction zone propagates in all directions at a constant speed equal to the Chapman-Jouguet detonation velocity D cj . As the reaction zone reaches and proceeds into an element, the chemical energy is proportionally released into that element over its “burn time.” The arrival and burn times of each element are computed according to Figure 3-13 below:

Figure 3-13

Main Index

Arrival Times of Detonation Wave

Chapter 3: Fluid Dynamics 103 JWL Explosive Test

The following input entries are required when modeling detonation: EOSJWL

Defines equation of state of detonation products.

TICEL

Is used to assign the specific chemical energy as initial specific internal energy of each element.

DETSPH

Defines the ignition point, ignition time and speed of a “spherical” detonation front.

Preparing Input Due to symmetry, the reaction zone (detonation front) is a plane traveling along the length of the slab. Therefore, it is sufficient to model only a portion of the slab with all boundary faces closed to transport. A mesh of 200 elements along the 50 cm of slab length is used (element thickness of 0.25 cm).

Figure 3-14

Geometrical Layout of Explosive and Detonation Point

At the start time, all the elements are filled with explosive material. Therefore, they all reference a EOSJWL equation of state defined by a DMAT entry. The specific chemical energy q0 is assigned as the initial specific internal energy of the explosive by using the TICEL entry. The Chapman-Jouguet detonation velocity Dcj, the ignition point, and the ignition time are specified by the DETSPH entry. The ignition time is taken as the start time of the analysis and the ignition point is the center of the left face of the mesh. (See Figure 3-14.) The duration of the analysis is set to 60 μsec necessary to burn the slab (50 cm) at a detonation speed of 7980 m/sec. Edits of pressure profiles are requested every 10 μsec.

Main Index

104 Dytran Example Problem Manual JWL Explosive Test

Results The solution of the detonation analysis is shown in Figure 3-15. The peak pressure of each profile corresponding to the pressure behind the detonation front is depicted in Figure 3-15 in terms of the Chapman-Jouguet pressure fraction. From the hydrodynamic theory of steady-state plane detonation for a γ-law gas, the CJ pressure is P cj = 2 ( γ cj – 1 )q 0 ρ 0

(3-10)

with γcj = 2.706 for TNT, this leads to 10

P cj = 2.9 * 10 P a

(3-11)

The code needs about 60 elements to build up the detonation front at which time the pressure reaches approximately 0.85 of Pcj. Subsequently, the front propagates with only a small increase of pressure. These results are acceptable if the fact is taken into account that Dytran is a first-order code that smears the shock front over a number of elements (always conserving momentum and energy). The result of that will be a reduction in peak pressure .

Figure 3-15

Main Index

Peak Pressure Profiles at Different Distances

Chapter 3: Fluid Dynamics 105 JWL Explosive Test

Files jwl.dat

Dytran input file

JWL.OUT

Dytran output file

JWL_P_0.ARC

Dytran archive file

Abbreviated Dytran Input File START CEND TITLE = HE DETONATION TEST CHECK = NO ENDTIME = 6e-05 TIC = 1 $$ $$ Data for Output Control Set 1 $$ TYPE(p) = ARCHIVE SAVE(p) = 99999 ELEMENTS(p) = 1 SET 1 = 1t200 ELOUT(p) = PRESSURE TIMES(p) = 0tEndb3e-06 $--------------------------------BEGIN BULK $--------------------------------PARAM INISTEP 2e-07 $ $ THIS SECTION CONTAINS BULK DATA $ $------------------mesh geometry $ GRID 1 0.0 0.0 GRID 2 0.0 .01 GRID 3 0.0 0.0 GRID 4 0.0 .01 . . . . GRID 801 .5 0.0 GRID 802 .5 .01 GRID 803 .5 0.0 GRID 804 .5 .01 $ CHEXA 1 1 1 5 + 8 4 CHEXA 2 1 5 9 + 12 8 CHEXA 3 1 9 13 + 16 12

Main Index

0.0 0.0 .01 .01

0.0 0.0 .01 .01 6

2

3

7

+

10

6

7

11

+

14

10

11

15

+

106 Dytran Example Problem Manual JWL Explosive Test

CHEXA 4 1 13 17 18 14 15 + 20 16 . . . . CHEXA 197 1 785 789 790 786 787 + 792 788 CHEXA 198 1 789 793 794 790 791 + 796 792 CHEXA 199 1 793 797 798 794 795 + 800 796 CHEXA 200 1 797 801 802 798 799 + 804 800 $ $------------------all elements filled with HE material $ $ PEULER 1 100 Hydro $ DMAT 100 1717. 100 $ EOSJWL 100 5.242+117.6783+94.2 1.1 .34 $ $------------------initial state of HE material $ DETSPH 1 100 0.0 .01 .01 7980. 0.0 $ $------------------detonation characteristics $ TICEL 1 1 density 1717. sie 4950000. SET1 1 1 THRU 200 ENDDATA

Main Index

19

+

791

+

795

+

799

+

803

+

Chapter 3: Fluid Dynamics 107 Modeling Blast Wave using 1-D Spherical Symmetry Method

Modeling Blast Wave using 1-D Spherical Symmetry Method Problem Description In many blast applications, the explosive is spherically symmetric and small compared to the distance from ignition point to target. An efficient method is developed to reduce the computation time significantly by initializing the spherical blast wave by constructing a 1-D model first, and then as the wave progresses and expands, the pressure wave is mapped to a full 3-D model just before the blast wave reaches the structure. The pressure in the wave front should be adequately maintained during the mapping process by avoiding excessive coarse mesh. This limits the amount of mesh coarsening one can apply. To validate mapping, the 1-D mesh and the 3-D mesh will have comparable mesh-sizes. In addition, mapping on a coarser mesh is also considered. To model the blast, ideal gas is used. The initial radius of the sphere will be R0=0.1 m. Ideal gas model: Blast Products (r
= 10 kg/m3

Gamma

= 1.4

Environment (r>R0) Specific internal energy = 3.E+5 Joule/kg Density

= 1 kg/m3

The simulation will be run for 0.843 ms. Remapped results will be compared with results that do not use remapping.

Dytran Modeling The Spherical model The Euler region is defined and initialized as follows: MESH,1,BOX,,,,,,,+ +,0,-0.001,-0.001,0.5,0.002,0.002,,,+ +,15,1,1,,,,EULER,1

Main Index

108 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

This defines one row of Euler elements. To make the mesh suitable for spherical symmetric analyses PARAM SPHERSYM is added PARAM,SPHERSYM,RECT,X,2.0 This transforms the rectangular 1-D mesh into a wedge-shaped mesh. In addition, only the radial mesh size will be taken into account in the time step computation. In the 1-D spherical modeling technique, the analysis has to be terminated as soon as the material starts leaving the 1-D mesh model. In this example, the last cycle (cycle 48, t=0.11 ms,) where the material is still inside the domain, is considered. To import the 1-D result into the 3D simulation, the following steps can be followed. • Read in the Euler archives in Patran and select a cycle in which the material has not left the

domain yet. • In Patran Click Results->Graph • Set as target entries the centers of the Euler elements and select variables. After apply, a graph

is shown • Converting the graph to a text file (.xyd file):

a. Click XYPOT. b. Select Create->XYwindow and create a new window. c. Click POST> Curve. Select a curve and Apply. d. Click Modify->Curve. e. Select the Curve and Apply. f. Select Data from Keyboard . g. Check Write XY Data to file and apply. This writes out a text file. For the remap, text files have to be created for density, specific internal energy (SIE), and radial velocity. These are called rho.xyd, sie.xyd, and vel.xyd. The follow-up 3-D model: The Euler region is defined and initialized as follows: MESH,1,BOX,,,,,,,+ +,-1,-1,-1,2,2,2,,,+ +,60,60,60,,,,EULER,1 TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,2.0 SPHERE,3,,0.0,0.0,0.0,1.0 SPHERE,4,,0.0,0.0,0.0,20.0 TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,400,density,200,SIE,300 TICVAL,9,,DENSITY,1.0,SIE,3e+5 $ TABFILE,200,rho.xyd

Main Index

Chapter 3: Fluid Dynamics 109 Modeling Blast Wave using 1-D Spherical Symmetry Method

TABFILE,300,sie.xyd TABFILE,400,vel.xyd TABFILE allows defining a table from a text file.

The full 3-D model: To check the integrity of the remapped results, a full 3-D model will also be simulated and results will be compared to the previous method. The Euler region is defined and initialized as follows: MESH,1,BOX,,,,,,,+ +,-1,-1,-1,2,2,2,,,+ +,60,60,60,,,,EULER,1 The material initialization is identical to the spherical 1-D model. The follow-up 3-D model coarse model: The Euler region is defined and initialized as follows: MESH,1,BOX,,,,,,,+ +,-1,-1,-1,2,2,2,,,+ +,30,30,30,,,,EULER,1 In this example, the following calculations have been performed. 1. Spherical Model 2. Follow-up 3-D Model 3. Full 3-D Model 4. Follow-up 3-D Coarse model

Main Index

110 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

Results Spherical Model The mesh and the results of the spherical symmetric model are shown in the following figures.

Main Index

Mesh

Density at Time = 0.11 ms

SIE at Time = 0.11 ms

Radial Velocity at Time = 0.11 ms

Chapter 3: Fluid Dynamics 111 Modeling Blast Wave using 1-D Spherical Symmetry Method

The remap data are shown in the following figures.

Rho.xyd

SIE.xyd

V el.xyd

Main Index

112 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

Follow-up 3-D Model The initial conditions of the follow-up 3-D model at cycle 1 closely resemble the spherical symmetric solution results at cycle 48 as shown in the following figures.

3-D Mesh (60 x 60 x 60)

SIE at Cycle 1

Main Index

Density at Cycle 1

Velocity Magnitude at Cycle 1

Chapter 3: Fluid Dynamics 113 Modeling Blast Wave using 1-D Spherical Symmetry Method

The validation of the 1-D to 3-D remap is shown in the following figures.

Main Index

114 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

The following figure shows the pressure at time = 0.733 ms (cycle 76). This time represents the total problem time of 0.843 ms (1D + 3D=0.773+0.11).

Full 3-D Model The results of the full 3-D model at Time = 0.11 ms (cycle 48) is shown in the following figure. This time is equivalent to the time of cycle 1 of the 3-D remap run.

Main Index

Chapter 3: Fluid Dynamics 115 Modeling Blast Wave using 1-D Spherical Symmetry Method

The use of an orthogonal mesh is clearly visible in the square shape. The following figure shows the pressure profile comparison with the 3-D remap run at cycle 1.

The following figure shows the pressure distribution at time = 0.836 ms. This compares well with the 3-D results using remap. The pressures in the wave front are very similar.

Main Index

116 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

Follow-up 3-D Coarse Model

3-D Coarse Mesh (30 x 30 x30)

Main Index

Chapter 3: Fluid Dynamics 117 Modeling Blast Wave using 1-D Spherical Symmetry Method

Comparing initial conditions at cycle 1 with 1-D result.

Main Index

118 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

Pressure distribution and radial pressure profile at Time = 0.877 ms

Comparison pressure profiles Run 2, 3, and 4

Abbreviated Dytran Input File Spherical Symmetric Model START

Main Index

Chapter 3: Fluid Dynamics 119 Modeling Blast Wave using 1-D Spherical Symmetry Method

CEND ENDSTEP = 100 CHECK=NO TITLE= Jobname is: sphersym TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNC STEPS(ALLEULER) = 0,thru,end,by,3 SAVE (ALLEULER) = 10000 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK $ -----------------------------------------------------------------$ PARAM,SPHERSYM,RECT,X,2.0 $ * Euler.300 * $ PARAM,MICRO,30 PEULER1,1,,HYDRO,19 $ EOSGAM,2,1.4 $ DMAT 100 100 2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,6.0 SPHERE,3,,0.0,0.0,0,10000 SPHERE,4,,0.0,0.0,0.0,0.1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,10,sie,9e+6 $ MESH,1,BOX,,,,,,,+ +,0,-0.001,-0.001,0.5,0.002,0.002,,,+ +,15,1,1,,,,EULER,1 $ $ ENDDATA

Main Index

120 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

3-D Model with Mapping MEMORY-SIZE=16000000,12000000 START CEND ENDSTEP = 100 CHECK=NO TITLE= Jobname is: undex-2d TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNC STEPS(ALLEULER) = 0,1,thru,end,by,15 SAVE (ALLEULER) = 10000 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK $ -----------------------------------------------------------------$ $ * Euler.300 * $ PEULER1,1,,HYDRO,19 $ EOSGAM,2,1.4 $ DMAT 100 100 2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,2.0 SPHERE,3,,0.0,0.0,0.0,1.0 SPHERE,4,,0.0,0.0,0.0,20.0 TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,400,density,200,SIE,300 TICVAL,9,,DENSITY,1.0,SIE,3e+5

Main Index

Chapter 3: Fluid Dynamics 121 Modeling Blast Wave using 1-D Spherical Symmetry Method

$ TABFILE,200,rho.xyd TABFILE,300,sie.xyd TABFILE,400,vel.xyd MESH,1,BOX,,,,,,,+ +,-1,-1,-1,2,2,2,,,+ +,60,60,60,,,,EULER,1 $ $ ENDDATA 3-D Reference Model MEMORY-SIZE=16000000,12000000 START CEND ENDSTEP = 100 CHECK=NO TITLE= Jobname is: undex-2d TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNC STEPS(ALLEULER) = 0,thru,end,by,6 SAVE (ALLEULER) = 10000 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK $ -----------------------------------------------------------------$ $ * Euler.300 * $ PEULER1,1,,HYDRO,19 $ EOSGAM,2,1.4 $ DMAT 100 100 2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== $

Main Index

122 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,6.0 SPHERE,3,,0.0,0.0,0,10000 SPHERE,4,,0.0,0.0,0.0,0.1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,10,sie,9e+6 $ $TICEUL,19,,,,,,,,+ $+,SPHERE,3,100,8,4.0,,,,+ $+,SPHERE,4,100,9,2.0 $SPHERE,3,,0.0,0.0,0.0,1.0 $SPHERE,4,,0.0,0.0,0.0,20.0 $TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ $+,R-VEL,400,density,200,SIE,300 $TICVAL,9,,DENSITY,1.0,SIE,3e+5 $$ $TABFILE,200,rho.xyd $TABFILE,300,sie.xyd $TABFILE,400,vel.xyd MESH,1,BOX,,,,,,,+ +,-1,-1,-1,2,2,2,,,+ +,60,60,60,,,,EULER,1 $ $ ENDDATA Coarse Model with Remap MEMORY-SIZE=16000000,12000000 START CEND ENDSTEP = 100 CHECK=NO TITLE= Jobname is: undex-2d TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNC STEPS(ALLEULER) = 0,1,thru,end,by,15 SAVE (ALLEULER) = 10000 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK

Main Index

Chapter 3: Fluid Dynamics 123 Modeling Blast Wave using 1-D Spherical Symmetry Method

$ -----------------------------------------------------------------$ $ * Euler.300 * $ PEULER1,1,,HYDRO,19 $ EOSGAM,2,1.4 $ DMAT 100 100 2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,2.0 SPHERE,3,,0.0,0.0,0.0,1.0 SPHERE,4,,0.0,0.0,0.0,20.0 TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,400,density,200,SIE,300 TICVAL,9,,DENSITY,1.0,SIE,3e+5 $ TABFILE,200,rho.xyd TABFILE,300,sie.xyd TABFILE,400,vel.xyd MESH,1,BOX,,,,,,,+ +,-1,-1,-1,2,2,2,,,+ +,30,30,30,,,,EULER,1 $ $ ENDDATA

Rho.xyd .016666668 0.17599975 0.050000001 0.11755728 0.083333336 0.13820617 0.11666667 0.17730692 0.15000001 0.23573031 0.18333334 0.30357549 0.21666668 0.41232908 0.25 0.69994283 0.28333333 1.3444314 0.31666666 1.8983461 0.34999999 1.5142108 0.38333333 1.0990304 0.41666669 1.0090078

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124 Dytran Example Problem Manual Modeling Blast Wave using 1-D Spherical Symmetry Method

0.45000002 1.0003828 0.48333335 1.0000119 Sie.xyd 0.016666668 1644110.4 0.050000001 1521182.6 0.083333336 1650291.4 0.11666667 1871300.8 0.15000001 2126423. 0.18333334 2316476.3 0.21666668 2376442.8 0.25 2337180. 0.28333333 2127134.5 0.31666666 1557344.5 0.34999999 738252.38 0.38333333 328480.25 0.41666669 301103.97 0.45000002 300045.94 0.48333335 300001.44 Vel.xyd 0.016666668 551.55115 0.050000001 343.2377 0.083333336 417.90549 0.11666667 592.59894 0.15000001 727.58636 0.18333334 765.23682 0.21666668 778.4917 0.25 932.18738 0.28333333 1135.2539 0.31666666 1092.1925 0.34999999 602.56567 0.38333333 125.99188 0.41666669 9.3862505 0.45000002 0.3385168 0.48333335 0.010169516

Main Index

Chapter 3: Fluid Dynamics 125 Modeling the JWL Explosion using 1-D Spherical Symmtery

Modeling the JWL Explosion using 1-D Spherical Symmtery Problem Description The JWL model requires very fine mesh to reach the correct peak pressure. After sufficient expansion of the blast wave, fine elements are no longer useful and it would be very computationally efficient to replace the fine mesh by a coarse mesh. Furthermore, the initial stages of the explosion are spherical symmetric and allows the use of a 1D model. As in the previous example, the spherical initialization is done by constructing a 1-D model and results are remapped on a first quadrant 3-D model. The air outside the explosive will also be modeled. The multi-material Eulerian solver is used. The initial radius of the sphere will be R0=0.25 m. Blast Products (rR0) Ideal gas

gamma 1.4

Specific internal energy

= 3.E+5 Joule/kg

Density

= 1 kg/m3

The simulation will be run for 2.5 ms.

Dytran Modeling Spherical Symmetric Model The Euler region is defined and initialized as follows: MESH,1,BOX,,,,,,,+ +,0,-0.001,-0.001,4,0.002,0.002,,,+ +,1600,1,1,,,,EULER,1 This defines one row of Euler elements. To make the mesh suitable for spherical symmetric analyses PARAM SPHERSYM is added PARAM,SPHERSYM,RECT,X,1.0 This transforms the rectangular 1-D mesh into a wedge shaped mesh. Also, in the time step computation, only the radial mesh-size will be taken into account.

Main Index

126 Dytran Example Problem Manual Modeling the JWL Explosion using 1-D Spherical Symmtery

To allow outflow of air, a transmitting boundary condition is added: FLOWDIR,4,MMHYDRO,22,POSX,,,,,+ +,FLOW,OUT As soon as material is starting to leave the 1-D mesh, the analysis has to be terminated. In this example cycle 3000 and time 1.1 ms will be taken for the remap. For details as how to import the 1-D results into the 3-D simulation refer to the previous example. For the remap text files have to be created for air density, JWL density, air specific internal energy (SIE), JWL SIE and radial velocity. These are called rhoa.xyd, rhoj.xyd, siea.xyd, siej.xyd, and velj.xyd. In the files containing density information the entries with zero densities have to be removed. Follow-up 3-D Model for the First Quadrant The Euler region is defined and initialized as follows: MESH,1,BOX,,,,,,,+ +,0,0,0,5,5,5,,,+ +,75,75,75,,,,EULER,1 Using the material fraction the position of the interface between JWL material and air in the 1-D run can be computed. This interface is at 2.83244 m. This is used in SPHERE to define a sphere of explosive products. TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,2.0,,,,+ +,SPHERE,4,200,9,4.0 SPHERE,3,,0.0,0.0,0.0,10.0 SPHERE,4,,0.0,0.0,0.0,2.83244 TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,100,density,200,SIE,300 TICVAL,9,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,100,density,400,SIE,500 $ TABFILE,100,velj.xyd TABFILE,200,rhoa.xyd TABFILE,300,siea.xyd TABFILE,400,rhoj.xyd TABFILE,500,siej.xyd $

To allow outflow of air, a transmitting boundary condition is added: FLOWDIR,4,MMHYDRO,22,POSX,,,,,+ +,FLOW,OUT FLOWDIR,4,MMHYDRO,22,POSY,,,,,+ +,FLOW,OUT FLOWDIR,4,MMHYDRO,22,POSZ,,,,,+ +,FLOW,OUT $ Here, a geometric boundary condition has been used. Otherwise a CHEXA mesh and CFACES would have to be made with Patran.

Main Index

Chapter 3: Fluid Dynamics 127 Modeling the JWL Explosion using 1-D Spherical Symmtery

All JWL material should have a burn factor of 1 at cycle 1. To enforce this, the VEL field of the DETSPH entry is set to a sufficiently large value: DETSPH

1

200

.0

.0

.0

1E+12

0.0

Results Spherical Model The mesh and the results of the spherical symmetric model are shown in the following figures.

Mesh (1600 elements)

Main Index

128 Dytran Example Problem Manual Modeling the JWL Explosion using 1-D Spherical Symmtery

Density Explosive Material at Time = 1.0951 ms (cycle 3000)

SIE Explosive Material at Time = 1.0951 ms (cycle 3000)

Main Index

Chapter 3: Fluid Dynamics 129 Modeling the JWL Explosion using 1-D Spherical Symmtery

Radial Velocity Explosive Material at Time = 1.0951 ms (cycle 3000)

Pressure Explosive Material at Time = 1.0951 ms (cycle 3000)

Main Index

130 Dytran Example Problem Manual Modeling the JWL Explosion using 1-D Spherical Symmtery

Follow-up 3-D Model The initial conditions of the follow-up 3-D model are equal to the remapping result at cycle1 as shown in the following figures.

Mesh (75 x 75 x75)

Main Index

Chapter 3: Fluid Dynamics 131 Modeling the JWL Explosion using 1-D Spherical Symmtery

Main Index

Pressure at cycle 1

Comparison with 1-D

Density at cycle 1

Comparison with 1-D

132 Dytran Example Problem Manual Modeling the JWL Explosion using 1-D Spherical Symmtery

Main Index

SIE at cycle 1

Comparison with 1-D

Radial Velocity at cycle 1

Comparison with 1-D

Chapter 3: Fluid Dynamics 133 Modeling the JWL Explosion using 1-D Spherical Symmtery

The following figure shows the pressure at time = 0.799 ms (cycle 76). This time represents the total problem time of 1.894 ms (1D + 3D).

The following figure shows the pressure at time = 1.431 ms (cycle 100). This time represents the total problem time of 2.526 ms (1D + 3D).

Main Index

134 Dytran Example Problem Manual Modeling the JWL Explosion using 1-D Spherical Symmtery

Abbreviated Dytran Input File Spherical Symmetric Model START CEND ENDSTEP = 4000 CHECK=NO TITLE= Jobname is: sphersym TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = PRESSURE,XVEL,YVEL,ZVEL, FMAT100,DENSITY100,SIE100, FMAT200,DENSITY200,SIE200,FBURN STEPS(ALLEULER) = 0,thru,end,by,50 SAVE (ALLEULER) = 10000 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK $ -----------------------------------------------------------------$ PARAM,SPHERSYM,RECT,X,1.0 $ * Euler.300 * $ PARAM,MICRO,30 PEULER1,1,,MMHYDRO,19 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ EOSGAM,100,1.4 $ DMAT 100 1.29 100 $ DMAT 200 1717. 200 $ EOSJWL 200 5.242+117.6783+94.2 1.1 .34 $ DETSPH 1 200 .0 .0 .0 7980. 0.0 $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+

Main Index

Chapter 3: Fluid Dynamics 135 Modeling the JWL Explosion using 1-D Spherical Symmtery

+,SPHERE,4,200,9,6.0 SPHERE,3,,0.0,0.0,0,10000 SPHERE,4,,0.0,0.0,0.0,0.25 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,1717.,sie,4950000. $ MESH,1,BOX,,,,,,,+ +,0,-0.001,-0.001,4,0.002,0.002,,,+ +,1600,1,1,,,,EULER,1 $ $FLOWDEF,25,,MMHYDRO FLOWDIR,4,MMHYDRO,22,POSX,,,,,+ +,FLOW,OUT $ ENDDATA 3-D Model with Mapping MEMORY-SIZE=40000000,40000000 START CEND ENDSTEP = 100 CHECK=NO TITLE= Jobname is: undex-2d TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = PRESSURE,XVEL,YVEL,ZVEL,FBURN, FMAT100,FMAT200, DENSITY100,SIE100, DENSITY200,SIE200 STEPS(ALLEULER) = 0,1,thru,end,by,15 SAVE (ALLEULER) = 10000 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK $ -----------------------------------------------------------------$ $ * Euler.300 * $ PEULER1,1,,MMHYDRO,19 $ EOSGAM,100,1.4 $ DMAT 100 1.29 100 $

Main Index

136 Dytran Example Problem Manual Modeling the JWL Explosion using 1-D Spherical Symmtery

DMAT 200 1717. 200 $ EOSJWL 200 5.242+117.6783+94.2 1.1 .34 $ DETSPH 1 200 .0 .0 .0 1E+12 $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,2.0,,,,+ +,SPHERE,4,200,9,4.0 SPHERE,3,,0.0,0.0,0.0,10.0 SPHERE,4,,0.0,0.0,0.0,2.83244 TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,100,density,200,SIE,300 TICVAL,9,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,+ +,R-VEL,100,density,400,SIE,500 $ TABFILE,100,velj.xyd TABFILE,200,rhoa.xyd TABFILE,300,siea.xyd TABFILE,400,rhoj.xyd TABFILE,500,siej.xyd $ MESH,1,BOX,,,,,,,+ +,0,0,0,5,5,5,,,+ +,75,75,75,,,,EULER,1 $ FLOWDIR,4,MMHYDRO,22,POSX,,,,,+ +,FLOW,OUT FLOWDIR,4,MMHYDRO,22,POSY,,,,,+ +,FLOW,OUT FLOWDIR,4,MMHYDRO,22,POSZ,,,,,+ +,FLOW,OUT $ ENDDATA

Main Index

0.0

Chapter 3: Fluid Dynamics 137 Nonuniformity with MESH,BOX

Nonuniformity with MESH,BOX MESH,BOX allows the user to create nonuniform block meshes. These meshes were traditionally constructed only by preprocessors such as Patran. A uniform Euler block consists of a number of planes in each direction. These are at fixed distances. MESH, BOX has a functionality that allows full control over the locations of the planes. This functionality is activated by defining BIAS entries and using them on the MESH entry. The BIAS entries specify the locations of the planes.

Problem Description If the explosive is small and the object far away, it is often necessary to use a non-uniform mesh that is fine near the explosive and coarse further away. This example problem illustrates the use of nonuniform meshes created by MESH,BOX. To model the blast, ideal gas will be considered. The initial radius of the sphere will be R0=0.2 m. Ideal gas model Blast Products (r
= 9.E+6 Joule/kg

Density

= 10 kg/m3

Gamma

= 1.4

Environment (r>R0) Specific internal energy

= 3.E+5 Joule/kg

Density

= 1 kg/m3

The simulation will be run for 9 ms.

Dytran Modeling First,consider an Euler mesh without biasing that is defined by MESH,1,BOX,,,,0,0,7.5,+ +,-5,-0.12,-5,10,0.24,10,,,+ +,30,1,30,,123000,345000,EULER,1,+ By defining three BIAS entries ,this MESH entry will be extended to create a nonuniform mesh. After the extension, the number of elements like 30 and 1 will be overruled by the BIAS definition.

Main Index

138 Dytran Example Problem Manual Nonuniformity with MESH,BOX

The locations of planes in x-direction are specified by giving growth factors and number of elements for a series of subsequent intervals. These interval make up the x-range of [-5, 5] that is specified on the MESH,BOX. The biasing per interval reads INTERVAL

BEGIN

END

GROWTH

N

1

-5

-2

0.5

14

2

-2

-1

1

7

3

-2

0

0.5

10

4

0

1

2

10

5

1

2

1

7

6

2

5

2

14

Here BEGIN and END are, respectively, the begin coordinate and end coordinate of each interval. N is the number of elements inside the interval and GROWTH is the growth factor between two subsequent elements within the interval. In translating this to the BIAS entry, only the begin points are put in the BIAS entry. The end points of an interval follow from the begin point of the next interval. For the x-direction the bias definition reads: BIAS,100,,,,,,,,+ +,-5,0.5 ,14,,,,,,+ +,-2 ,1 ,7,,,,,,+ +,-1 ,0.5 ,10,,,,,,+ +, 0 ,2 ,10,,,,,,+ +, 1 ,1 ,7,,,,,,+ +, 2 ,2 ,14 The OUT file list these planes as: PLANE 1 2 3 4 5 6 7 8 9

X

GROWTH -5.000E+00 -4.704E+00 -4.423E+00 -4.157E+00 -3.904E+00 -3.665E+00 -3.438E+00 -3.223E+00 -3.019E+00

ELMSIZE

1.000E+00 9.481E-01 9.481E-01 9.481E-01 9.481E-01 9.481E-01 9.481E-01 9.481E-01 9.481E-01

0.000E+00 2.962E-01 2.808E-01 2.662E-01 2.524E-01 2.393E-01 2.269E-01 2.151E-01 2.039E-01

The third column shows the growth of element size in between planes. This growth is well within the [0.7,1.3] range. Growth factors outside this range should be avoided. By slightly changing the growth or N this can easily be achieved. The fourth column lists the element size of the elements that are to the immediate left of a plane. The first plane has no elements to the left and ELMSIZE = 0.

Main Index

Chapter 3: Fluid Dynamics 139 Nonuniformity with MESH,BOX

For the y-direction: INTERVAL

BEGIN

END

GROWTH

N

1

-0.12

0.12

1

1

For the y-direction, there is no need to specify a bias because the definition on the mesh entry already contains all information. Defining a bias for the y-direction is BIAS,200,,,,,,,,+ +,-0.12,,1 For the z-direction, planes are given by INTERVAL

BEGIN

END

GROWTH

N

1

-5

-4

0.5

6

2

-4

-3

1

8

3

-3

-2

0.5

8

4

-2

5

2

20

and the bias definition reads BIAS,300,,,,,,,,+ +,-5, 1 ,6,,,,,,+ +,-4, 0.5,8,,,,,,+ +,-3, 2.0,8,,,,,,+ +,-2, 3.0,20 Using these three BIAS definitions gives the MESH entry MESH,1,BOX,,,,0,0,7.5,+ +,-5,-0.12,-5,10,0.24,10,,,+ +,30,1,30,,123000,345000,EULER,1,+ +,,,,,,,,,+ +,,,,,100,200,300 Material and initial condition are defined by PEULER1,1,,HYDRO,19 DMAT 3 2 2 EOSGAM,2,1.4 $ TICEUL,19,,,,,,,,+ +,SPHERE,3,3,8,4.0,,,,+ +,SPHERE,4,3,9,6.0 SPHERE,3,,0.0,0.0,0,10000 SPHERE,4,,0.0,0.0,-3,0.2 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,10,sie,9e+6

Main Index

140 Dytran Example Problem Manual Nonuniformity with MESH,BOX

At the bottom of the Euler domain, a wall will be specified. This is done by using a geometric boundary conditions WALLDIR,120,HYDRO,,NEGZ All boundary faces pointing in the negative z-direction will get the WALLET definition. The remaining boundaries should be transmitting and a FLOWDEF is added: FLOWDEF,25,,HYDRO

Main Index

Chapter 3: Fluid Dynamics 141 Nonuniformity with MESH,BOX

Results Pressures are shown for cycle numbers 60, 100,160 and 300. The mesh is finest at the blast and coarsens away from the blast and wall. The cycle 100 result shows the reflection at the wall. The cycle 300 result show that the lateral boundaries are transmitting.

Main Index

142 Dytran Example Problem Manual Nonuniformity with MESH,BOX

Main Index

Chapter 3: Fluid Dynamics 143 Nonuniformity with MESH,BOX

Abbreviated Dytran Input File START CEND ENDSTEP = 600 CHECK=NO TITLE= Jobname is: non-uniform TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL,FVUNC STEPS(ALLEULER) = 0,thru,end,by,20 SAVE (ALLEULER) = 10000 $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION -------

Main Index

144 Dytran Example Problem Manual Nonuniformity with MESH,BOX

BEGIN BULK PARAM,BULKL,0.1 $ -----------------------------------------------------------------$ $ * Euler.300 * $ PARAM,MICRO,30 PEULER1,1,,HYDRO,19 $ DMAT 3 2 2 EOSGAM,2,1.4 $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,3,8,4.0,,,,+ +,SPHERE,4,3,9,6.0 SPHERE,3,,0.0,0.0,0,10000 SPHERE,4,,0.0,0.0,-3,0.2 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,10,sie,9e+6 $ BIAS,100,,,,,,,,+ +,-5,0.5 ,14,,,,,,+ +,-2 ,1 ,7,,,,,,+ +,-1 ,0.5 ,10,,,,,,+ +, 0 ,2 ,10,,,,,,+ +, 1 ,1 ,7,,,,,,+ +, 2 ,2 ,14 BIAS,200,,,,,,,,+ +,-0.12,,1 BIAS,300,,,,,,,,+ +,-5, 1 ,6,,,,,,+ +,-4, 0.5,8,,,,,,+ +,-3, 2.0,8,,,,,,+ +,-2, 3.0,20 $ MESH,1,BOX,,,,0,0,7.5,+ +,-5,-0.12,-5,10,0.24,10,,,+ +,30,1,30,,123000,345000,EULER,1,+ +,,,,,,,,,+ +,,,,,100,200,300 $ FLOWDEF,25,,HYDRO WALLDIR,120,HYDRO,,NEGZ $ ENDDATA

Main Index

Chapter 4: Fluid-structural Interaction Dytran Example Problem Manual

4

Main Index

Fluid-structure Interaction J

Overview

J

Shock Formation

J

Blast Containment in a Luggage Container

J

Multiple Bird-strike on a Cylindrical Panel

J

Slanted Piston

J

Sloshing using ALE Method

J

Flow between Two Containers or Airbags

J

Blastwave Hitting a Bunker

J

Mine Blast

J

Multiple Bird-strike on a Box Structure

J

Shaped Charge, using IG Model, Penetrating through Two Thick Plates 216

J

Fuel Tank Filling

J

Water Pouring into a Glass

J

Fluid Flow through a Straight Pipe

J

Using Euler Archive Import in Blast Wave Analyses

J

Blast Wave with a Graded Mesh

J

Bubble Collapse with Hydrostatic Boundary Conditions

147 148 154 161

169 176 183

193

197 209

224 237 246 263

273 284

146 Dytran Example Problem Manual

Main Index

J

Prestressed Concrete Beam

J

Blast Simulation on Prestressed Concrete Beam

J

Vortex Shedding with Skin Friction

J

Geometric Eulerian Boundary Conditions

J

Cohesive Friction

337

294

317 328

306

Chapter 4: Fluid-structural Interaction 147 Overview

Overview In this chapter, a number of example problems are presented that show the fluid-structure interaction capabilities of Dytran. The user can find in these examples how to model the interaction of fluids and gasses with structural parts using the general coupling and ALE-Coupling techniques of Dytran. These examples include the modeling of an explosion inside a structure and a birdstrike on a panel.

Main Index

148 Dytran Example Problem Manual Shock Formation

Shock Formation Problem Description Given a column of ideal gas at rest, up = 0 m/sec. A rigid piston at one end of the gas is instantaneously accelerated to the velocity up = 1180 m/sec that is subsequently maintained .

density

ρ0 = 0.001 kg/m3

specific internal energy

e0 = 2.5 106 Joule/kg

pressure

p0 = 1000 N/m2

ratio of specific heats

γ = 1.4

The purpose of this example is to model the propagation of the (shock) perturbation inside the gas and to check results against one-dimensional theoretical data.

Theoretical Background Relative to one-dimensional dynamics, where the shock perturbation travels in only one direction in a medium (plane shock front). the following steady-state Hugoniot relations can be applied: First Hugoniot relation ρ–ρ u s ⎛ --------------0-⎞ = c 0 ⎝ ρ ⎠

(4-1)

Second Hugoniot relation p – p0 = ρ0 ⋅ us ⋅ u

(4-2)

Third Hugoniot relation ρ – ρ0 e – e p = 0.5 ( p + p 0 ) ⎛ ---------------⎞ ⎝ ρ ρ0 ⎠

(4-3)

where ρ, e, p, and u, respectively, represent the shocked values of material density, specific internal energy, pressure, and velocity, c0 is the sound speed, and us is the speed of the shock front. Together with the equation of state for the medium (which in this case is treated as an ideal gas), p = (γ – 1) ⋅ ρ ⋅ e

(4-4)

The four equations define uniquely the values of the four variables of the dynamic, namely ρ, e, p, and us (the value of u is known and represents the excitation in the current problem).

Main Index

Chapter 4: Fluid-structural Interaction 149 Shock Formation

Dytran Model Due to the one-dimensional behavior of the problem, a simple mesh is set up (see Figure 4-1 and Figure 4-2): • The piston is modeled as a single Lagrangian solid CHEXA element of 0.02 m x 0.02 m. • A row of Eulerian cells, each of 0.01 m x 0.01 m x 0.01 m, is used to model the one-dimensional

flow of the gas.

Figure 4-1 Geometrical Problem Layout To simulate the steady-state shock excitation of the gas from rest to the velocity u = 1180 m/sec, the piston modeled as RIGID is COUPLED with the Eulerian mesh and a prescribed constant velocity of 1180 m/sec is enforced. • The RIGID entry of Dytran references a SURFACE entry that defines a surface wrapped around

the part to be made rigid. All grid points on this surface are treated as being part of the rigid body. • The COUPLE entry references a SURFACE entry that defines a surface wrapped around

the Lagrangian mesh. This surface acts as a constraint (moving boundary condition) to the material flow in the Eulerian mesh and at the same time is itself loaded by the pressure in the Eulerian mesh. • The TLOAD1 entry is used to enforce a prescribed constant velocity on the rigid body

center of gravity.

Main Index

150 Dytran Example Problem Manual Shock Formation

Figure 4-2 Numerical Problem Layout

Results The pressure and density profiles of the column of gas are shown in Figure 4-3 and Figure 4-4.

Figure 4-3

Main Index

Pressure Profile along the Tube at Different Times

Chapter 4: Fluid-structural Interaction 151 Shock Formation

Figure 4-4

Density Profile along the Tube at Different Times

Note the steady-state values of the pressure and density behind the shock front 3

p = 3.49 e N ⁄ m –3

2

ρ = 2.25e k g ⁄ m

(4-5) 3

(4-6)

From the Hugoniot Equations (4-1) to (4-3), the theoretical values of these variables can be derived 3

p = p 0 + ρ 0 u s u = 3.46 e N ⁄ m

2

us –3 3 ρ = ρ 0 -------------- = 2.3 e k g ⁄ m us – u

(4-7) (4-8)

There is agreement between the theoretical and computed values. The left vertical part of each graph represents the position of the piston while the right quasi-vertical part represents the position of the shock front. Note that their relative distance during the dynamic is growing due to the difference in velocity of the piston (1180 m/sec) and the shock front (2087 m/sec). The theoretical value of the shock front speed can be derived from the Hugoniot Equations (4-1) to (4-4): γ+1 u s = ------------ u + 4

Main Index

+ 1-⎞ 2 2 1 ⎛ γ----------⋅ u + p 0 ----- γ = 2087 m ⁄ sec ⎝ 4 ⎠ ρ0

(4-9)

152 Dytran Example Problem Manual Shock Formation

Files shock.dat

Dytran input file

SHOCK.OUT

Dytran output file

SHOCK_GAS_1.ARC

Dytran archive file

Abbreviated Dytran Input File START CEND TITLE = SHOCK PROPAGATION (Units = m-Kg-N-sec) ENDSTEP = 80 TIC = 1 TLOAD = 1 $$ $$ Data for Output Control Set 1 $$ TYPE(GAS) = ARCHIVE SAVE(GAS) = 99999 ELOUT(GAS) = XVEL,YVEL,ZVEL,DENSITY,SIE,PRESSURE ELEMENTS(GAS) = 1 SET 1 = 1t50 STEPS(GAS) = 1tEndb16 BEGIN BULK PARAM INISTEP 1e-06 PARAM,FASTCOUP $ $ THIS SECTION CONTAINS BULK DATA $ $ GRID 1 0.0 0.0 0.0 GRID 2 0.0 .01 0.0 GRID 3 0.0 0.0 .01 GRID 4 0.0 .01 .01 . . . . GRID 209 .015 -.005 -.005 GRID 210 .015 .015 -.005 GRID 211 .015 -.005 .015 GRID 212 .015 .015 .015 $ CHEXA 1 1 1 5 6 2 3 7 + 8 4 CHEXA 2 1 5 9 10 6 7 11 + 12 8 CHEXA 3 1 9 13 14 10 11 15 + 16 12 CHEXA 4 1 13 17 18 14 15 19 + 20 16

Main Index

+ + + +

Chapter 4: Fluid-structural Interaction 153 Shock Formation

. . . . CHEXA 48 1 189 193 194 190 191 195 + 196 192 CHEXA 49 1 193 197 198 194 195 199 + 200 196 CHEXA 50 1 197 201 202 198 199 203 + 204 200 CHEXA 51 2 205 209 210 206 207 211 + 212 208 $ $ THIS SECTION CONTAINS THE LOADS, CONSTRAINTS, AND CONTROL BULK DATA $ ENTRIES $ $ FORCE 1 1 0 1180. 1. 0.0 0.0 $ $ $ THIS SECTION CONTAINS THE DEFINED FEFACES OF ELEMENTS $ $ CFACE 1 1 51 3 CFACE 2 1 51 1 CFACE 3 1 51 2 CFACE 4 1 51 5 CFACE 5 1 51 4 CFACE 6 1 51 6 $ PSOLID 2 2 $ PEULER 1 1 Hydro $ DMAT 1 .001 1 $ DMATEL 2 1. 1. 0.0 $ EOSGAM 1 1.4 $ TLOAD1 1 1 12 $ TICEL 1 1 DENSITY .001 SIE 2500000. SET1 1 1 THRU 50 $ SURFACE 1 SEG 1 $ COUPLE 1 1 Inside On On $ RIGID 1 1 1. 0.0 .005 .005 + 1180. 0.0 0.0 0 0.0 0.0 0.0 + 10000. 0.0 0.0 10000. 0.0 10000. ENDDATA

Main Index

+ + + +

+ +

154 Dytran Example Problem Manual Blast Containment in a Luggage Container

Blast Containment in a Luggage Container Problem Description The purpose of this example is to demonstrate the application of ALE Euler and ALE coupling to a practical problem, in this case, an explosion in the baggage hold of an aircraft. The problem consists of a generic luggage container as would typically be found in a commercial passenger aircraft (see Figure 4-5).

Figure 4-5

Aircraft Baggage Container (Dimensions in Meters)

The analysis makes use of a symmetry line and therefore half the container and its surrounding environment is modeled. The Dytran model consists of an ALE Eulerian region for the expanding explosive gasses, which is coupled to a shell structure, and represents the aluminum container. The structural and fluid parts of the analysis are connected using ALE coupling. The explosive event is modeled using the blast wave approach ([Ref. 1.] to [Ref. 3.]); a sphere of high density, high energy gas is used to represent the detonation products of the explosive. The consequence of this is that the single material hydrodynamic formulation can be used for the Eulerian region. This reduces computation times compared to other more complex Eulerian formulations. This approach is discussed in Chapter 3: Fluid Dynamics in the “Blast Wave” problem.

Main Index

Chapter 4: Fluid-structural Interaction 155 Blast Containment in a Luggage Container

Dytran Modeling The explosive/air region is modeled using 5370 Eulerian elements and the EOSGAM equation of state. The container is modeled with 1417 CQUAD4 shell elements. The numerical model is shown in Figure 4-6.

Figure 4-6

Lagrangian Shell Element Mesh

The grid points of the Lagrangian and Eulerian elements at the interface are coincident but distinct. ALE coupling is used to define the interaction between these two surfaces. Each Eulerian grid point follows the Lagrangian grid point with which it is coincident at time = 0. The remaining Eulerian grid points (excluding those constrained on the symmetry axis) have their position updated each time step. In simple terms, each grid point is moved to the midpoint of all its neighbors. The use of the ALE approach, in this case, is appropriate due to the smooth nature of the deformation that results from the pressure loading on the structure and is very efficient in terms of CPU usage.

Results The results of the analysis at an event time of 3 msec are given in Figure 4-7. body 4-7 shows the deformed Eulerian mesh and the pressure of the expanding gas. This is a good example of the type of deformation that the ALE coupling can be used for. In this case, the original box-like structure expands, becoming more rounded as the analysis proceeds. The Dytran User’s Guide states that if ALE coupling is to be used, then the deformation must be “smooth.” This type of transition from undeformed to deformed shape is typical of what is meant by “smooth deformation.”

Main Index

156 Dytran Example Problem Manual Blast Containment in a Luggage Container

Figure 4-7

Deformed Pressure Contour Plot of Eulerian Mesh at Time = 3 msec

Figure 4-8 shows the deformed Lagrangian mesh at the same time, 3 msec. Superimposed on the deformed geometry is a contour of the effective plastic strain in the structure.

Figure 4-8

Main Index

Deformed Plastic Strain Contour Plot of Lagrangian Mesh at Time = 3 msec

Chapter 4: Fluid-structural Interaction 157 Blast Containment in a Luggage Container

Files container.dat

Dytran input file

CONTAINER.OUT

Dytran output file

CONTAINER_EUL_XX.ARC CONTAINER_SH_0.ARC

Dytran archive files

CONTAINER_STRAIN_0.THS

Dytran time history file

References 1. Kivity, Y. and Feller, S., “Blast Venting from a Cubicle,” 1986, 22nd DoD Explosives Safety Seminar. 2. Kivity, Y., Florie, C., and Lenselink, H., “The Plastic Response of a Cylindrical Shell Subjected to an Internal Blast Wave with a Finite Width Shock Front,” 1993, 33rd Israel Conference on Aviation and Aeronautics. 3. Kivity, Y., Florie, C., and Lenselink, H., “Response of Protective Structures to Internal Explosions with Blast Venting,” 1993, MSC World Users’ Conference.

Abbreviated Dytran Input File START CEND TITLE = Container Demo CHECK = no ENDSTEP = 4000 ENDTIME = 0.003 SPC = 1 $$ $$ Data for Output Control Set 1 $$ TYPE(eul) = ARCHIVE SAVE(eul) = 1 ELOUT(eul) = PRESSURE ELEMENTS(eul) = 1 SET 1 = 538t5907 TIMES(eul) = 0tEndb0.0005 $$ $$ Data for Output Control Set 2 $$ TYPE(sh) = ARCHIVE SAVE(sh) = 9999 ELOUT(sh) = EFFPL2,EFFST2 ELEMENTS(sh) = 2 SET 2 = 1t537,5908t6787 TIMES(sh) = 0tEndb0.0005 $$ $$ Data for Output Control Set 3

Main Index

158 Dytran Example Problem Manual Blast Containment in a Luggage Container

$$ TYPE(strain) = TIMEHIS SAVE(strain) = 9999 ELOUT(strain) = EFFPL2 ELEMENTS(strain) = 3 SET 3 = 6402,6672,5953,6243,6503 STEPS(strain) = 0tEndb50 $-------------------------------------------------------BEGIN BULK $-------------------------------------------------------SETTING,1,VERSION2 $ PARAM STRNOUT YES PARAM INISTEP 2e-06 PARAM MINSTEP 1e-06 $ $ $ THIS SECTION CONTAINS BULK DATA $ $ GRID 1 0.0 .42 .8 GRID 2 .07 .35 .8 . . . GRID 8245 1.921 0.0 .96 GRID 8246 1.921 0.0 .88 $ $ $ CQUAD4 1 801 6501 6502 6506 6505 CQUAD4 2 801 6502 6503 6507 6506 . . . . CQUAD4 6786 801 8245 8246 7366 7365 CQUAD4 6787 801 8246 6556 6557 7366 $ $ $ CHEXA 538 1 1 2 6 5 681 + 683 684 CHEXA 539 1 2 3 7 6 682 + 686 683 . . . CHEXA 5906 1 5913 5915 5918 5916 6495 + 6500 6498 CHEXA 5907 1 5915 5869 5872 5918 6497 + 6454 6500 $$-------------------------------------------------------$

Main Index

682

+

685

+

6497

+

6451

+

Chapter 4: Fluid-structural Interaction 159 Blast Containment in a Luggage Container

$ THIS SECTION CONTAINS THE LOADS, CONSTRAINTS, AND CONTROL BULK DATA $ ENTRIES $ $ SPC 1 7181 345 SPC 1 7192 345 . . . . SPC 1 8226 345 SPC 1 8237 345 $ $-------------------------------------------------------$ $ THIS SECTION CONTAINS THE DEFINED FEFACES OF ELEMENTS $ $ CFACE 1418 8 1 1 CFACE 1419 8 2 1 . . . . CFACE 2833 8 6786 1 CFACE 2834 8 6787 1 $ $ CFACE 1 7 538 1 CFACE 2 7 538 2 . . . CFACE 1416 7 5905 5 CFACE 1417 7 5906 5 $ $-------------------------------------------------------$ ALEGRID 1 0.0 1. SPECIAL COMPUTED + 1 THRU 16 18 THRU 20 22 THRU + 24 26 THRU 28 30 THRU 32 38 + THRU 40 42 THRU 44 46 THRU 48 . . . + THRU 679 681 THRU 6500 $ $-------------------------------------------------------$ PSHELL 801 801 .001 $ PEULER1 1 Hydro 11 $ YLDVM 801 1.2+8 1.+9

Main Index

+ + + +

160 Dytran Example Problem Manual Blast Containment in a Luggage Container

$ FAILMPS 801 9 $ $-------------------------------------------------------$ DMAT 2 1.25 2 $ DMATEP 801 10800. 4.+10 .33 801 $ EOSGAM 2 1.408 $ TICEUL 11 + SPHERE 1 2 10 10. + ELEM 1 2 1 1. SET1 1 538 THRU 5907 $ TICVAL 1 DENSITY 1.25 SIE 200000. TICVAL 10 DENSITY 40. SIE 5625000. $ SPHERE 1 1.0125 .774 1.6 .109124 $ SURFACE 77 SEG 7 SURFACE 88 SEG 8 $ ALE 1 88 77 $ $ ENDDATA

Main Index

801

+ + +

Chapter 4: Fluid-structural Interaction 161 Multiple Bird-strike on a Cylindrical Panel

Multiple Bird-strike on a Cylindrical Panel Problem Description Landing and climbing aircraft sometimes encounter difficulties with bird swarms, since an impact of several birds at high velocity can cause severe damage to the aircraft structure. Therefore, this problem considers a situation where two birds strike a curved titanium plate at arbitrary time (see Figure 4-9). Bird 1 hits the plate perpendicularly, while Bird 2 hits the plate with the lower side at an angle of 25°. The birds are modeled as cylindrical slugs of jelly. The plate is constrained in such a way that the edges can only move in radial direction.

Figure 4-9

Main Index

Initial Situation

162 Dytran Example Problem Manual Multiple Bird-strike on a Cylindrical Panel

The properties and initial conditions of the plate and birds are as follows: Plate material

titanium

density

ρ = 4527 kg/m3

bulk modulus

K = 1.03 1011 N/m2

Poisson’s ratio

υ = 0.314

yield stress

σij = 1.38 108 N/m2

thickness

t = 0.0015 m

radius

R = 0.25 m

length

L = 0.25 m

Bird 1 material

jelly

density

ρ = 930 kg/m3

speed of sound

c = 1483 m/sec

mass

m = 0.360 kg

velocity

v = 150 m/sec

Bird 2 material

jelly

density

ρ = 930 kg/m3

speed of sound

c = 1483 m/sec

mass

m = 0.285 kg

velocity

v = 200 m/sec

Dytran Model The curved plate is modeled using 33 x 16 (528) Belytschko-Tsay shell elements (CQUAD4). The boundary conditions applied at the edges of the plate are defined within a cylindrical coordinate system, where the local z-axis is aligned with the length axis of the plate. The cylindrical system is defined by a CORD2C entry. The two birds are modeled in an Eulerian frame of reference, which is built from 33 x 16 x 14 (7392) CHEXA elements. The birds are modeled as cylinders using the TICEUL entry where the remaining part of the mesh initially is void. The material is allowed to flow out of the Eulerian mesh by defining an outflow boundary condition to all free faces of the mesh by means of a FLOWDEF entry. From this

Main Index

Chapter 4: Fluid-structural Interaction 163 Multiple Bird-strike on a Cylindrical Panel

description, it is clear that the user easily can define any number of arbitrarily shaped birds in an Eulerian mesh with each having its own initial conditions. The structure and the fluid (birds) are allowed to interact at the fluid-structure interface. For birdstrikes, the Arbitrary-Lagrange-Euler interaction is the most efficient FSI to use. To define the FSI, an ALE interface is defined, consisting of a Lagrangian and an Eulerian surface. The structural plate serves as an interface by defining CFACEs on the elements of the plate. The CFACEs then are used to define a Lagrangian SURFACE. The Eulerian mesh coincides with the plate and in the plane of coincidence the faces of the Eulerian mesh are covered with CFACEs that define an Eulerian SURFACE. Both SURFACEs are defined to be used as the ALE interface. To propagate the interface motion into the Eulerian mesh, the Eulerian grid points are defined as ALEGRIDs. The type option on the ALEGRID entry is set to SPECIAL in which case Dytran will automatically use corrections on the Eulerian grid point motion depending on the boundary condition in which the point is included. Figure 4-10 shows the model with the fluid (birds) inside the mesh. The birds are shown by plotting the material fraction (FMAT) for elements.

Figure 4-10

Main Index

Cross Section of the Eulerian Mesh with the Fluid (Birds) in the Initial Situation

164 Dytran Example Problem Manual Multiple Bird-strike on a Cylindrical Panel

Results A theoretical check on the results is the peak pressure of the impact. For the maximum pressure of the impact can be written as P max = ρ c Δ V

(4-10)

In this way, the maximum pressure of the impact of Bird 1 is (4-11)

P max = 207 MP a

and the maximum pressure of the impact of the lower side of Bird 2, (where one has to take the angle of 25° into account) is given as: (4-12)

P max = 117 MP a

In the Figure 4-11 and Figure 4-12, both peak pressures are presented in a time-history format. The maximum pressures are lower than the theoretical values. This is because of two reasons. First, the Dytran Euler processor is first order and therefore smears out the peak pressures. This gives lower pressure peaks, but the momentum is the same. Second, the impact causes a dent in the plate and, because of the interaction, the Eulerian elements near the plate will change in volume, thus causing cavitation in which case the pressure drops. This second aspect shows stronger for the impact of Bird 1, because the plate deformation is larger.

Figure 4-11

Main Index

Pressure Time History for the Impact of Bird 1

Chapter 4: Fluid-structural Interaction 165 Multiple Bird-strike on a Cylindrical Panel

Figure 4-12

Pressure Time History for the Impact of Bird 2

A picture of the cross section of the deformed Eulerian mesh with the remainders of the birds at Time Step 500 (0.8345 msec) is shown in Figure 4-13.

Figure 4-13

Main Index

Cross Section of the Deformed Eulerian Mesh with the Fluid on Time Step 500 (0.8345 msec)

166 Dytran Example Problem Manual Multiple Bird-strike on a Cylindrical Panel

Figure 4-14 shows the deformed shape of the plate at Time Step 500 with the effective stress at integration Layer 2 superimposed on it.

Figure 4-14

Effect Stress at Integration Layer 2 of the Plate on Time Step 500

Files mbird.dat grid_xl.dat

Dytran input file

MBIRD.OUT

Dytran output file

MBIRD_BIRD_XX.ARC MBIRD_PLATE_XX.ARC

Dytran archive files

MBIRD_PRESSURE_0.THS

Dytran time history file

Abbreviated Dytran Input File START CEND TITLE = Multiple bird strike using Lag-Eul ale coupling ENDSTEP = 1000 ENDTIME = .002 TIC = 1 SPC = 1 $$ $$ Data for output control Set 1 $$ TYPE(plate) = ARCHIVE

Main Index

Chapter 4: Fluid-structural Interaction 167 Multiple Bird-strike on a Cylindrical Panel

SAVE(plate) = 1 STEPS(plate) = 0tEndb100 ELEMENTS(plate) = 1 SET 1 = 7393t7920 ELOUT(plate) = EFFST01,EFFST02,EFFST03,EFFPL01,EFFPL02,EFFPL03 $$ $$ Data for output control Set 2 $$ TYPE(bird) = ARCHIVE SAVE(bird) = 1 STEPS(bird) = 0tEndb100 ELEMENTS(bird) = 2 SET 2 = 1t7392 ELOUT(bird) = MASS,DENSITY,PRESSURE,FMAT,XVEL,YVEL,ZVEL $$ $$ Data for output control Set 3 $$ TYPE(pressure) = TIMEHIS SAVE(pressure) = 10000 ELEMENTS(pressure) = 3 TIMES(pressure) = 0tEndb1e-6 SET 3 = 7105,7119 ELOUT(pressure) = PRESSURE,DENSITY,ENERGY $ BEGIN BULK PARAM INISTEP 1e-6 PARAM MINSTEP 1e-7 $ $ Model geometry. $ --------------INCLUDE grid_xl.dat $ $ ===== $ EULER $ ===== $ $ Definition of ALE motion. $ ------------------------ALEGRID 1 .0 1e20 SPECIAL + 1 THRU 8670 $ $ Flow boundary, property, material and equation of state data. $ ------------------------------------------------------------FLOWDEF 1 HYDRO + MATERIAL3 FLOW OUT $ PEULER1 1 HYDRO 10 $ DMAT 3 930 3 $ EOSPOL 3 2.2e9 $ $ Allocation of material to geometric regions. $ --------------------------------------------

Main Index

+

+

168 Dytran Example Problem Manual Multiple Bird-strike on a Cylindrical Panel

TICEUL 10 + CYLINDER1 3 1 + CYLINDER2 3 2 + ELEM 4 $ CYLINDER1 -.1381 .125 + .035 CYLINDER2 .13 .125 + .035 $ SET1 4 1 THRU 7392 $ $ Initial material data. $ ---------------------TICVAL 1 XVEL 200 TICVAL 2 XVEL -75 $ $ ======== $ LAGRANGE $ ======== $ $ Property, material and yield model. $ ----------------------------------PSHELL1 2 2 Bely Gauss + .0015 $ DMATEP 2 4527 .314 $ YLDVM 1 1.38e8 $ $ Boundary constrain. $ -------------------CORD2C 1 0.0 0.0 + 0.0 0.125 0.25 $ SPC3 1 1 23456 + 8671 THRU 8704 8705 + 8738 THRU 9214 BY $ $ ============ $ ALE COUPLING $ ============ $ SURFACE 1 SEG 1 SURFACE 2 SEG 2 $ ALE 1 2 1 $ ENDDATA

Main Index

+ + +

3 2 1 .26

-.2381

.125

.26

+

.2252

.17

.125

.2944

+

ZVEL

-129.9

3

.83333

Mid

+

1.03e11 1

0.0

0.0

0.25

0.0

THRU 34

9181 9215

BY THRU

34 9248

+

+ +

Chapter 4: Fluid-structural Interaction 169 Slanted Piston

Slanted Piston Problem Description This is a simple coupled Euler/Lagrange problem that uses the general coupling algorithm. A box of gas of 1 cm x 1 cm x 1 cm, modeled with Eulerian elements, is intersected by a piston of the same dimension that is modeled with Lagrangian elements. The piston is initially at rest but will start to accelerate because of the pressure exerted on it by the gas. Eventually, the piston will leave the box. However, no gas will leave the box because the piston uncovers wall boundaries as it recedes. In the following figures, twodimensional representation of the problem (Figure 4-15) and a three-dimensional representation of the problem (Figure 4-16) illustrate how the pressure of the gas will accelerate the piston.

Figure 4-15

Main Index

Two-dimensional Representation of the Problem

170 Dytran Example Problem Manual Slanted Piston

Figure 4-16

Three-dimensional Problem to be Modeled in Dytran

Desired Results For this example, the initial conditions are: Gas ρ = 0.1

g/cm3

Piston ρ = 7.85 g/cm3

e = 0.025 108 J

e = 21.109 N/m2

ρ = 0.001 Mbar

μ = 0.3

γ = 1.4 The symmetry of the problem can be checked by comparing the velocity profiles of the piston in the three component directions. These profiles should be identical. Also, the gas will expand approximately adiabatically and, therefore, if e is the gas internal energy and V is the volume of the gas, the following result is obtained: d e = – pd V

(4-13)

and combining this with the equation of state

Main Index

p = (γ – 1)e ⁄ V

(4-14)

de dV ------ = – ( γ – 1 ) ------e V

(4-15)

Chapter 4: Fluid-structural Interaction 171 Slanted Piston

V ( γ – 1) e ----- = ⎛ -----0-⎞ ⎝ V⎠ e0

(4-16)

The gas internal energy will be converted into piston kinetic energy, KE, (gas kinetic energy and piston internal energy are not included), therefore, e K E = e 0 – e = 1 – ----e0

(4-17)

Using Equation (4-17), the following result is obtained for the kinetic energy: V K E⎞ ⎛ ------- = 1 – ⎛ -----0-⎞ ⎝ V⎠ ⎝ e0 ⎠

( γ – 1)

(4-18)

The initial volume of gas is one-third covered by the piston. When the piston has completely left the box, V0 1 ------ = --3 V

and the process stops.

Dytran Modeling The gas is modeled using 1000 (10 x 10 x 10) Eulerian elements and the EOSGAM equation of state. A single Lagrangian element is used to model the piston, and a high elastic modulus restricts any deformation.

Results As this is a symmetrical problem, the results can be checked for symmetry in the three component directions of the basic coordinate system. Also, the decrease in the internal energy of the gas should be equal to the increase in the kinetic energy of the piston. The velocity profile of the piston in the x, y, and z directions is shown in Figure 4-17 and confirms the symmetry of the analysis.

Main Index

172 Dytran Example Problem Manual Slanted Piston

Figure 4-17

Velocity Components of Piston

The change in internal energy of the gas and in kinetic energy of the piston is shown in Figure 4-18. The results confirm that the energy lost by the gas is equal to the energy gained by the piston.

Figure 4-18

Main Index

Energy Changes

Chapter 4: Fluid-structural Interaction 173 Slanted Piston

Files slp.dat

Dytran input file

SLP.OUT

Dytran output file

SLP_GAS_0.ARC SLP_GRD_0.ARC

Dytran archive files

SLP_PISTON_0.THS SLP_NRG_0.THS

Dytran time history files

Abbreviated Dytran Input File START CEND TITLE = Slanted Piston ENDTIME = 250 $$ $$ Data for Output Control Set 1 $$ TYPE(gas) = ARCHIVE SAVE(gas) = 9999 ELOUT(gas) = XVEL,YVEL,ZVEL,PRESSURE,FMAT ELEMENTS(gas) = 1 SET 1 = 2t1001 TIMES(gas) = 0tEndb20 $$ $$ Data for Output Control Set 2 $$ TYPE(piston) = TIMEHIS SAVE(piston) = 9999 ELOUT(piston) = TXX,TYY,TZZ ELEMENTS(piston) = 2 SET 2 = 1 TIMES(piston) = 0tEndb5 $$ $$ Data for Output Control Set 3 $$ TYPE(grd) = ARCHIVE SAVE(grd) = 9999 GRIDS(grd) = 3 SET 3 = 1t8 GPOUT(grd) = XVEL,YVEL,ZVEL TIMES(grd) = 0tEndb20 $$ $$ Data for Output Control Set 4 $$ TYPE(nrg) = TIMEHIS SAVE(nrg) = 9999 MATS(nrg) = 4 SET 4 = 1,10 MATOUT(nrg) = XMOM,YMOM,ZMOM,EKIN,EINT

Main Index

174 Dytran Example Problem Manual Slanted Piston

TIMES(nrg) = 0tEndb5 BEGIN BULK PARAM INISTEP 0.1 PARAM MINSTEP 0.005 $ $ THIS SECTION CONTAINS $ $ GRID 1 GRID 2 . . GRID 1338 GRID 1339 $ CHEXA 1 1 + 8 4 CHEXA 2 10 + 142 21 . . . CHEXA 1001 10 + 1339 1218 $ $ $ THIS SECTION CONTAINS $ $ CFACE 1 1 CFACE 2 1 CFACE 3 1 CFACE 4 1 CFACE 5 1 CFACE 6 1 $ PSOLID 1 1 $ PEULER1 10 $ DMAT 10 .1 $ DMATEL 1 7.85 $ EOSGAM 1 1.4 $ TICEUL 10 + ELEM 1 SET1 1 2 $ TICVAL 1 $ SURFACE 5 $

Main Index

BULK DATA

0.0 0.0

0.0 0.0

0.0 1.

1.633333.7333333.7333333 1.666667.6666667.6666667 1

5

6

2

3

7

+

9

130

131

10

20

141

+

1206

1327

1328

1207

1217

1338

+

THE DEFINED FEFACES OF ELEMENTS

1 1 1 1 1 1

3 1 2 5 4 6

Hydro

10

1 210.

.3

10 THRU

1 1001

+

density .1 SEG

1

1.

sie

.025

Chapter 4: Fluid-structural Interaction 175 Slanted Piston

COUPLE 1 ENDDATA

Main Index

5

Inside

On

On

176 Dytran Example Problem Manual Sloshing using ALE Method

Sloshing using ALE Method Problem Description Many spacecraft, such as satellites, contain liquid fuel containers. Sloshing in these partially filled containers can have a large influence on spacecraft stability and control. An experimental setup to investigate sloshing phenomena in space is called Sloshsat. This example is a schematic representation of the liquid motion in the Sloshsat tank. The tank itself consists of two spherical endcaps with a cylindrical middle part. In this example it is composed of aluminium. The tank is partially filled with a fluid (water); the rest of the tank is filled with nitrogen gas. For simplicity only the tank itself is modeled, not the surrounding spacecraft structure. The tank mass is adjusted so the tank still has a weight representative of the whole satellite. This way, not only the fluid motion but also the motion of the satellite as a whole can be calculated.

Figure 4-19

Main Index

(a) Proposed Sloshsat System (b) Tank Dimensions

Chapter 4: Fluid-structural Interaction 177 Sloshing using ALE Method

Dytran Model A full three-dimensional model is set up, with the x-axis pointing along the length of the tank. The tank itself is modeled using CQUAD4 shell elements. The inside of the tank, that must contain the fluid and gas, is defined by 736 CHEXA solid elements. To be able to define a coupling surface, additional dummy shell elements are defined on the surface of the solids. Arbitrary Lagrange-Euler (ALE) coupling is used to connect the Euler and Lagrange meshes. Because the deformation of the tank will be smooth this is allowable. ALE coupling means the nodes on the coupling surface move with the Lagrange mesh. For this coupling, it is necessary that all nodes of the Euler coupling surface have a corresponding node on the Lagrange surface. In essence, this means the mesh on both surfaces must be the same. To ensure the interior Euler nodes also follow the motion of the tank (so the Euler mesh stays inside) the ALEGRID option is used. The 736 Euler elements use a multi-material PEULER1 formulation so they can contain both water and gas. Results have shown major discrepancies in case the gas is not taken into consideration. Different regions as indicated on the TICEUL specify the fluid and gas location. TICVAL entries give each region different initial conditions. The 256 shell elements that make up the tank use the default PSHELL formulation. The elements of the ALE coupling surface also use PSHELL, but with a thickness set to 9999. These elements will be converted to surface segments by Dytran automatically. In order to obtain a faster analysis, a technique known as bulk scaling is used. By reducing the elastic moduli of the tank and the fluid, Dytran uses a smaller timestep for the problem, reducing computational costs. The fluid bulk modulus is therefore reduced by a factor of 1000 to 2.2 MPa, the tank modulus of elasticity is reduced by a factor of 100 to 0.724 GPa. A TLOAD1 entry with subsequent FORCE descriptions is used to prescribe a velocity profile to the tank. See Figure 4-21

Figure 4-20

(a) Euler Mesh (b) Surface Mesh

User Subroutine To obtain results for the center of mass of the Eulerian fluid, a user subroutine EEXOUT was used. This subroutine can be used to request user-specified output. In this case, the mass and location of each

Main Index

178 Dytran Example Problem Manual Sloshing using ALE Method

Euler element are obtained. Knowing this, the location of the center of mass of the fluid as a whole can be computed.

Figure 4-21

COMFLO and Dytran Calculations of the Liquid Center of Mass Trajectory

Results The fluid is initially at rest in one end of the tank. The tank is subjected to a constant deceleration to bring the fluid from one end of the tank to the other. This deceleration along the X-axis is 0.002G, or 0.0196 m/s2. A small lateral acceleration is also applied along the Z-axis; the magnitude of this is 0.01 times the longitudinal acceleration, or 0.00002G. The end time for the simulation is taken as 30 s. Fluid motion is checked against predictions made with the CFD package COMFLO. Figure 4-21 shows the X- and Z-coordinate of the water c.o.m. for both calculation methods. The results can be seen to be in good agreement, especially in the x-direction.

Main Index

Chapter 4: Fluid-structural Interaction 179 Sloshing using ALE Method

Figure 4-22

Positions of Liquids in the Tank at Different Times in the Analysis

Files sloshsat.dat sloshsat.bdf eexout_com.f (user-subroutine)

Dytran input files

SLOSHSAT.OUT COMEEUL.OUT (refer user-subroutine)

Dytran output files

SLOSHSAT_TANK_0.ARC SLOSHSAT_WATER_0.ARC

Dytran archive files

SLOSHSAT_MONITOR_0.THS

Dytran time history file

References 4. Simulation of Liquid Dynamics onboard Sloshsat FLEVO, NLR-TP-99236

Main Index

180 Dytran Example Problem Manual Sloshing using ALE Method

Abbreviated Dytran Input File START CEND ENDTIME= 30.01 ENDSTEP=99999999 CHECK=NO TITLE= Jobname is: slosh TLOAD=1 TIC=1 SPC=1 $ $ Output result for request: monitor $ TYPE (monitor) = TIMEHIS GRIDS (monitor) = 1 SET 1 = 14 40 44 83 117 181 GPOUT (monitor) = XVEL YVEL ZVEL XDIS YDIS ZDIS TIMES (monitor) = 0 THRU END BY 1. SAVE (monitor) = 10000 $ $ Output result for request: tank $ TYPE (tank) = ARCHIVE ELEMENTS (tank) = 2 SET 2 = 1 THRU 256 ELOUT (tank) = EFFPL EFFST TIMES (tank) = 0 THRU END BY 1. SAVE (tank) = 100000 $ $ Output result for request: water $ TYPE (water) = ARCHIVE ELEMENTS (water) = 3 SET 3 = 1001 THRU 1736 ELOUT (water) = XVEL YVEL ZVEL DENSITY PRESSURE TIMES (water) = 0 THRU END BY 1. SAVE (water) = 100000 $ $ Output to user subroutine $ ELEXOUT (ceegee) ELEMENTS (ceegee) = 99 SET 99 = 1001 THRU 1736 TIMES (ceegee) = 0 THRU END BY 1. SAVE (ceegee) = 10000 $ TYPE (step) = STEPSUM STEPS (step) = 1 THRU END BY 250 $ TYPE (mat) = MATSUM STEPS (mat) = 1 THRU END BY 250 $ $------- PARAMETER SECTION ------

Main Index

Chapter 4: Fluid-structural Interaction 181 Sloshing using ALE Method

$ PARAM,INISTEP,1E-8 PARAM,MINSTEP,1E-9 PARAM,TOLCHK, 1.E-9 PARAM, FMULTI, 1.0 PARAM, ROMULTI, 1.E-15 PARAM, VELMAX, 30., NO PARAM, VELCUT, 1.E-10 $ $------- BULK DATA SECTION ------BEGIN BULK INCLUDE slosh.bdf $ $ ========== PROPERTY SETS ========== $ $ * tank * $ PSHELL, 1, 2, 0.002 $ $ * dummy * $ PSHELL, 2, 3, 9999 $ * water * $ PEULER1, 3, , MMHYDRO, 1 TICEUL, 1, , , , , , , , +AAAA02 +AAAA02, ELEM, 7, 7, 7, 1.0, , , , +AAAA03 +AAAA03, ELEM, 5, 1, 1, 2.0 TICVAL, 7, , DENSITY, 1.29, SIE, 1.94E5 TICVAL, 1, , DENSITY, 1000. $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ -------- Material water id =1 $ DMAT, 1, 1000, 1 EOSPOL, 1, 2.2E6 $ $ -------- Material air (nitrogen) id =7 $ DMAT, 7, 1.29, 7 EOSGAM, 7, 1.4 $ $ -------- Material aluminium id =2 $ DMATEP, 2, 36.5E3, 7.24E8, 0.33, , , 2, YLDVM, 2, 3.45E8 $ $ $ $ ======== LOAD CASES ======================== $ $ ------- Velocity BC tankvel -----

Main Index

182 Dytran Example Problem Manual Sloshing using ALE Method

TLOAD1,1,6,,2,2 $ $ FORCE,6,1,0,1,1,0,0.01 FORCE,6,2,0,1,1,0,0.01 : : FORCE,6,257,0,1,1,0,0.01 FORCE,6,258,0,1,1,0,0.01 $ $ TABLED1, 2, , , , , , , , +XXX01 +XXX01, 0., 0., 30., -0.5886, 30.00001, FREE, ENDT $ 0.002G accel, 0-30s $----------- SETS -----------$ $Setname: alegrid with Set ID: 3 contains Nodes $ SET1, 3, 1001, THRU, 1885 $ $Setname: euler_init with Set ID: 5 contains Elements $ SET1,5,1001,THRU,1068,1093,1097,1101,1105,+A000002 +A000002,1109,1113,1117,1121,1185,THRU,1252,1277,+A000003 +A000003,1281,1285,1289,1293,1297,1301,1305,1369,+A000004 +A000004,THRU,1436,1461,1465,1469,1473,1477,1481,+A000005 +A000005,1485,1489,1553,THRU,1620,1645,1649,1653,+A000006 +A000006,1657,1661,1665,1669,1673 $ $ $Setname: shell_surf with Set ID: 6 contains Properties $ SET1, 6, 1 $ $Setname: euler with Set ID: 7 contains Elements $ SET1, 7, 1001, THRU, 1736 $ SURFACE, 1, , PROP, 6 SURFACE, 2, , SEG, 2 $ ALE, 1, 1, 2 $ ALEGRID1, 1, 3 $ ENDDATA

Main Index

Chapter 4: Fluid-structural Interaction 183 Flow between Two Containers or Airbags

Flow between Two Containers or Airbags Problem Description Flow between containers or airbags occurs in • Multi-compartment airbags, including airbags with strips • Explosive blast wave containment

For simplicity, this problem examines the flow between two rigid, stationary containers, without failure. The problem and Euler element meshing is shown in Figure 4-23. Both containers consists of a box of dimensions of 0.1 m by 0.1 m by 0.1 m and one half of a hose. This hose has dimension 0.02 m by 0.02 m by 0.04 m. Both containers are filled with gas. The left container is initialized with density 0.2 kg/m3 and specific energy 400000 J/kg and the container on the right is initialized with a density of 1.9 kg/m3 and with the same specific energy as the left one. This corresponds to a pressure of 0.04136 Mpa in container 1, and 0.39292 Mpa in container 2.

Figure 4-23

Dytran Model

In the analysis, gas will flow across the hole.

Theoretical Analysis of the 2-Vessel Flow Problem This part contains the following sections: A) Data Summary B) Equations Summary C) Analytical Estimates on Computed Results

Main Index

184 Dytran Example Problem Manual Flow between Two Containers or Airbags

A) Data Summary Let us briefly summarize the initial states of the two vessels: V1

= 1.E-3 m3

volume

ρ1

= 0.2 kg/m3

density

E1

= 4.E5 J/kg

specific internal energyE2

V2

= 1.E-3 m3

volume

ρ2

= 1.9 kg/m3

density

E2

=4.E5 J/kg

specific internal energy

The Euler fluid is modeled by means of an ideal gas equation of state: P

= (g-1) *ρ * E

with g

= 1.517

Given these data, the following initial values are derived: P1

= 0.4136E5 N/m2

pressure

c1

= 560.10 m/sec

sound speed

M1

= 2.E-4 kg

mass

P2

= 3.9292E5 N/m2

pressure

c2

= 560.10 m/sec

sound speed

M2

= 1.9E-3 kg

mass

The connection between the two vessels has a cross-section area of 4.E-4 m2 and a length of 4.E-2 m. For the purpose of the analysis, the length of this 'hose' connecting the vessels will be neglected.

Main Index

Chapter 4: Fluid-structural Interaction 185 Flow between Two Containers or Airbags

B) Equations Summary During the initial flow developing from V2 into V1, a critical flow will develop. Below is a brief summary of equations governing critical flow, using the 'upstream' high-pressure data as the reference: Pc

= P2 * [2/(g+1)] ** [g/(g-1)]

critical pressure

ρc

= ρ2 * [2/(g+1)] ** [1/(g-1)]

critical density

cc

= c2 * sqrt[2/(g+1)]

critical sound speed

Tc

= T2 * [2/(g+1)]

critical temperature

Using the defined value for g = 1.517, then the critical ratios can be computed as: Pc / P2

= 0.509

ρc / ρ2

= 0.641

cc / c2

= 0.8914

Tc / T2

= 0.7946

We now have sufficient data to analytically estimate the correctness of the calculated transient results. C) Analytical Estimates on Computed Results The blowdown transient from the high-pressure volume V2 through the hose into the low-pressure volume V1 has three stages. The initial stage is a 'shock-tube' stage, which exists only for a very short duration inside the hose, where the high and low pressure regions are adjacent and an imaginary wall is assumed to be instantaneously removed at time t=0. The duration of this stage is very short, typically of duration of half the hose length divided by the sound speed. Beyond that time, pressure reflections at both larger volumes will cause a step-wise increase of the flow through the hose, until critical flow conditions are reached. The duration is estimated as: tau_st = 0.02 / 500 = 4.E-5 sec, or 40 microsec. The pressure reflections in each volume have a typical run time of the vessel length divided by the sound speed, or: tau_v = 0.1 / 500 = 2.E-4 sec, or 0.2 msec. Oscillations with this period can be observed in the flowrate plot Figure 4-24b during the first few millisecs in the transient. Over the first 2 msec some 10 oscillations can be seen, giving the average oscillation period of 0.2 msec.

Main Index

186 Dytran Example Problem Manual Flow between Two Containers or Airbags

Figure 4-24

Function of Time

The second stage is the 'critical flow' stage. This can be considered as a gradual blowdown of the highpressure volume, where the flowrate is restricted to the critical flow in the hose. Most high-frequency oscillations following from the initial shock formation and reflections, have meanwhile been dissipated. The critical flow is estimated as follows: dM / dt = rhoc * uc * A, where:

A is the hose cross-section, uc =cc, for critical flow the velocity at critical flow is the velocity of sound.

This would be the theoretical maximum flowrate if the critical flow conditions would exist ideally at time=0. The actual run in Figure 4-24b shows a max peak of approximate 0.2 kg/sec. This is slightly lower due to the following effects: • neglecting energy effects in the flow, and pressure/energy coupling • neglecting flowrate during the initial 'shock-tube' stage

Both these effects are included in the calculation but are not taken into account in the above analytical estimate.

Main Index

Chapter 4: Fluid-structural Interaction 187 Flow between Two Containers or Airbags

When neglecting the pressure/energy dependency for simplicity's sake, we can also estimate the rate of pressure drop: dP / dt = (g-1) * E2 * drho2/dt Rewriting drho2 / dt as (1/V2) * dM / dt, we obtain: dP / dt = (1/V2) * (g-1) * E2 * dM / dt = 5.0293E7 N/m2 per second. This means, the pressure P2 drops at a rate of 5.0293E4 N/m2 per msec. Figure 4-24a shows the pressure P2 to drop from 4.E5 down to 3.5E5 N/m2 during the first millesec, so this is in good agreement with the estimated rate of pressure drop. Simultaneously, the lower pressure P1 will rise with the same rate, from 0.4E5 up to 0.9E5 N/m2 during the first millisec, as shown in Figure 4-24a. The flowrate itself reduces during the critical flow, not so much by a change in the flow velocity but primarily because the upstream density reduces when the upstream pressure drops. Note that the critical sound speed, equaling the critical flow velocity, only depends on the gamma value and is dependent on upstream pressure only as a square-root of (slowly varying) upstream energy. Estimating the reduction of critical density change then as: drhoc / dt = 1./[(g-1)*E2] * dP2 / dt, it follows that drhoc / dt = 1./(2.068E5) * 5.0293E7 = -234.2 kg/[m3.sec] Since the change in rhoc is directly proportional to the critical flowrate (discounting the small change in cc), it follows that the flowrate reduces by approximate 0.234 kg/sec. Observation of the flowrate Figure 4-24b shows that during the first 2 msec, the flowrate drops from approximate -0.2 down to -0.14 kg/sec, which is in good agreement with the estimate. Let us now estimate when the critical flow stage is completed. This is the case when the downstream pressure P1 has risen up to the level of critical pressure Pc which, in turn, drops proportional to the drop in upstream pressure P2. Using slightly rounded numbers for simplicity, we recapitulate: P2 = 4 bar P1 = 0.4 bar dP / dt = 0.5 bar/msec Pc = 0.51 * P2 We can now compute the following equality: P1(t) = Pc(t), where P1(t) = P1(0) + dP / dt * tcf Pc(t) = 0.51 * [ P2(0) - dP / dt * tcf ] and solve for time tcf in millisec. It follows that: tcf = 2.17 msec, which is the time at which the critical flow stage ends. In the flowrate plot of Figure 4-24b, this point in time can be recognized by the change in slope of the flowrate. When the critical flow ends, then the flow

Main Index

188 Dytran Example Problem Manual Flow between Two Containers or Airbags

velocity is no longer a more or less constant value but starts to become dependent on pressure differences across the hose. The transient has now reached the 'sub-critical flow' stage. In this stage, the flow solution is no longer determined mainly by the upstream values, but by both the upstream and downstream values, requiring e.g., a Riemann solution, to determine pressure and flowrate for the remaining part of the blowdown transient. A detailed analytical solution of the subcritical stage will not be presented here, but the following observations are interesting. Since both density and flow velocity are now reducing in proportion to the pressure difference, the slope in mass flowrate reduction is now twice as large as during the critical flow stage. Final pressure reduction is now also seen more clearly to follow a quadratic curve. The final oscillations in flowrate and pressures are physical rather than numerical in nature. Upon reaching pressure equality, there is still some fluid motion. This motion causes pressure overshoot and reverse flow, much like a pendulum. Of course, due to dissipation this oscillatory flow quickly vanishes, after 6 msec. In reality, dissipation originates from friction and viscous effects, while in the calculation residual numerical viscosity causes a similar dissipative effect.

Dytran Model This model could have been modeled as one coupling surface but to illustrate the flow between coupling surfaces, the two containers will be modeled as two separate coupling surfaces. These two surfaces are connected by a hole that is located half way down the hose. This hole is modeled as a surface consisting of either quads or trias that are fully porous. In this case, the hole is modeled by one quad. The elements in the surface of the hole connect the two coupling surface and are included in the definition of both coupling surfaces. The first coupling surface consists of: • The cube on the left • The left half of the hose • The surface modeling the hole, this is a square

Taking only the first two surfaces does not give a closed surface. The missing part is exactly the third surface. These three objects form a closed surface. The second coupling surface similarly is defined as: • The cube on the right • The right half of the hose • The surface modeling the hole

Main Index

Chapter 4: Fluid-structural Interaction 189 Flow between Two Containers or Airbags

In the input deck listed, the container to the left and the left half of the hole make up property set 1. The container on the right and the other half of the hole make up property set 2. The hole is property set 3. So property set 3 has to be used by both coupling surfaces as follows: COUPLE,10,25,OUTSIDE,,,,,,+ +,,,,,,,,,+ +,,22 COUPLE,20,50,OUTSIDE,,,11,,,+ +,,,,,,,,,+ +,,23 SURFACE,25,,PROP,1 SET1,1,1,3 SURFACE,50,,PROP,2 SET1,2,2,3

There are two options to create the Euler mesh: 1. MESH with TYPE=BOX This method should be used when the coupling surface does not deform, and there is no need for the Euler Domain to adapt itself. When using this method to create the Euler domains, one needs to ensure that the two Euler domains have at least one element overlapping at the hole. Also, one needs to ensure that holes are not precisely on the Euler element faces. 2. MESH with TYPE=ADAPT This method should be used when the coupling surface moves and deforms, and there is a need for the Euler Domain to adapt itself. Since the coupling surfaces are stationary in this model, the option TYPE=BOX has been used. To activate the hole between the coupling surfaces, the following needs to be entered: • The MESHID must be referenced from the COUPLE entry • Define a suitable flow model for the hole. This can be either the PORFLCPL entry or the

PORFCPL entry. Note that latter one is only meant for small holes. In this case, we used a PORFLCPL entry that is suited for large holes. This entry with the corresponding COUPOR entry reads COUPOR,2,11,31,PORFLCPL,82,,1.0 PORFLCPL,82,,,BOTH,10 SUBSURF,31,50,ELEM,250 SET1,250,32

Here the SUBSURF consists of the element in property set 3, that is element 32. In defining the coupling surface, the elements constituting the hole were formatted as a property set but it can also be formatted in all ways supported by SURFACE.

Main Index

190 Dytran Example Problem Manual Flow between Two Containers or Airbags

Results The problem has been run with a problem time of 0.01s. Figure 4-24a shows the pressure in both containers as function of time and Figure 4-24b shows the mass flow rate across the hole as function of time.

Abbreviated Dytran Input File START CEND ENDTIME = 0.01 CHECK=NO TITLE= Jobname is: flow TLOAD=1 TIC=1 SPC=1 PARAM,FASTCOUP PARAM,INISTEP,1.E-8 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 $------- BULK DATA SECTION ------$ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLEULHYDRO ELOUT (ALLEULER) = PRESSURE,XVEL,YVEL,ZVEL TIMES (ALLEULER) = END SAVE (ALLEULER) = 1 $ TYPE(Container1)=TIMEHIS SURFACES(Container1)=155 SET 155=25 SURFOUT(Container1)=PRESSURE,MASS, MFL,MFLR,MFL-POR,MFLR-POR STEPS(Container1)=0 THRU END BY 6 SAVE(Container1)=9999999 $ TYPE(Container2)=TIMEHIS SURFACES(Container2)=156 SET 156=50 SURFOUT(Container2)=PRESSURE,MASS, MFL,MFLR,MFL-POR,MFLR-POR STEPS(Container2)=0 THRU END BY 6 SAVE(Container2)=9999999 $ TYPE(HOLES) = TIMEHIS SUBSOUT(HOLES) = MASS,MFLR-POR,MFL-POR,XVEL,YVEL,ZVEL, DENSITY,MFLR,MFL,PRESSURE SUBSURFS(HOLES) = 88 SET 88 = 31 STEPS(HOLES) = 0 THRU END BY 5 SAVE(HOLES) = 1000000

Main Index

Chapter 4: Fluid-structural Interaction 191 Flow between Two Containers or Airbags

$ BEGIN BULK $ --- Define 24 grid points --$ $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material id =0 $ $ ======== Load Cases ======================== $ $ -------- Material Ideal_Gas id =3 DMAT 3 1.167 3 $ | $ -> density $ $ ========= MATERIAL DEFINITIONS ========== $ GRID 57 .140000.0600000.0400000 GRID 58 .140000.0600000.0600000 GRID 61 .140000.0400000.0400000 GRID 68 .140000.0400000.0600000 GRID 71 .120000.0600000.0400000 GRID 72 .120000.0600000.0600000 GRID 78 .120000.0400000.0600000 GRID 84 .120000.0400000.0400000 $ $ --- Define 28 elements $ $ -------- property set Pset1.1 --------CQUAD4 1 1 1 2 4 CQUAD4 2 1 3 4 8 CQUAD4 3 1 1 3 7 CQUAD4 4 1 2 1 11 CQUAD4 5 1 7 18 20 CQUAD4 6 1 21 18 7 CQUAD4 7 1 21 8 15 CQUAD4 8 1 8 4 2 CQUAD4 9 1 20 27 15 CQUAD4 24 1 18 21 72 CQUAD4 26 1 21 27 78 CQUAD4 28 1 27 20 84 CQUAD4 30 1 20 18 71 $ -------- property set Pset2.2 --------CQUAD4 10 2 37 38 40 CQUAD4 11 2 38 42 44 CQUAD4 12 2 42 46 48 CQUAD4 13 2 44 48 39 CQUAD4 14 2 46 37 39 CQUAD4 15 2 57 58 42 CQUAD4 16 2 61 57 38 CQUAD4 17 2 37 46 68

Main Index

3 7 11 15 11 8 27 15 11 71 72 78 84 39 40 44 40 48 38 37 61

192 Dytran Example Problem Manual Flow between Two Containers or Airbags

CQUAD4 18 2 46 42 CQUAD4 25 2 71 72 CQUAD4 27 2 72 78 CQUAD4 29 2 78 84 CQUAD4 31 2 84 71 $ -------- property set Pset3.3 --------CQUAD4 32 3 72 78 $ -------- property set flowfaces1 --------$ -------- property set flowfaces2 --------$ $ ========== PROPERTY SETS ========== $ $ * Pset1.1 * $ PSHELL1 1 DUMMY $ $ * Pset2.2 * $ PSHELL1 2 DUMMY $ $ * Pset3.3 * $ PSHELL1 3 DUMMY $ ENDDATA

Main Index

58 58 68 61 57

68 57 58 68 61

84

71

Chapter 4: Fluid-structural Interaction 193 Blastwave Hitting a Bunker

Blastwave Hitting a Bunker Problem Description The purpose is to demonstrate application of multi-Euler domains to failing coupling surfaces. The problem simulates a bunker, located on the ground, that is open at the sides and is surrounded by air. Gas can flow freely through the sides of the bunker. A blast wave is ignited close to the bunker and expands into the air. When by the impact of the blast wave, the bunker surface fails gas will flow trough the bunker surface .

Figure 4-25

Bunker Model

Dytran Modeling The bunker and the ground consist of cquad4 shell elements. The elements of the bunker are Lagrangian deformable shells and the ground is modeled as rigid, using a MATRIG. The explosive/air region is modeled by two Euler meshes. The first domain models the inside of the bunker and the second one models the outside of the bunker. For the interaction between the bunker and an Euler domain, a unique coupling surface has to be used, therefore, two coupling surfaces are needed. The first coupling surface, for modeling the inside of the bunker, consists of the following facets: • The 180 degrees cylindrical surface and the two open sides of the bunker. The two open sides are

represented by dummy shell elements. These are elements 1 to 2240. • The top of the ground that lies within the bunker. This is a square and is formed by elements

2241 to 3280. These facets make up a closed coupling surface, as shown in Figure 4-26:

Main Index

194 Dytran Example Problem Manual Blastwave Hitting a Bunker

Figure 4-26

Coupling Surface 1

This coupling surface contains gas inside, and therefore Euler elements outside the coupling surface should not be processed and so the COVER is OUTSIDE. The second coupling surface consists of the following facets: • The 180 degrees cylindrical surface and the two open sides of the bunker. These are elements 1

to 2240. The top of the ground inside the bunker is not part of the second COUPLE. • The top of the ground that is outside the bunker and 5 dummy surfaces of the ground that are

used to close the coupling surfaces. These are formed by the elements 3413 to 4012, 4095 to 4340, 4505 to 4709, 4894 to 7904. These facets make up a closed coupling surface, as shown in the Figure 4-27:

Figure 4-27

Coupling Surface 2

This coupling surface is used for simulating the gas outside the coupling surface. So Euler elements inside the coupling surface should not be processed and the COVER has to be set to INSIDE. The second

Main Index

Chapter 4: Fluid-structural Interaction 195 Blastwave Hitting a Bunker

coupling surface uses the second Euler mesh and serves as inner boundary surface for this Euler mesh. The outside boundary of this mesh is where the Euler domains ends and boundary conditions for this boundary are provided by a FLOWDEF. The FLOWDEF is chosen as non-reflecting. Waves exit the Euler domain with only little reflection. To get an accurate expansion of the blast wave, the diffusion should be kept at a minimum, and therefore the Roe solver with second-order is used. Interactive failure will be used for the bunker structure, while porosity will be used for the open sides: • The bunker elements itself can fail and gas will flow through the failed elements from outside

the bunker into the bunker. All elements of the bunker itself are assigned to a SUBSURF, and occur in both coupling surfaces. They are able to fail interactively, using the COUP1FL entry. These parts are formed by elements 1 to 1600. The nodes of the failed elements are constrained in space by using PARAM, NZEROVEL, YES, to preserve the geometry of the coupling surfaces from severe distortion. • Since gas can flow through the two sides without any obstruction, these two areas are modeled

with 'SUBSURF' entries, and are opened by using a PORFLCPL entry. These sides are modeled with dummy shell elements and consist of elements 1601 to 2400. The couple cards refer to mesh-number. The first mesh for the Euler elements inside the bunker is created and initialized by: PEULER1,301,,2ndOrder,111 MESH,2,BOX,,,,,,,+ +,-430.,0.,-1287.,837.,480.,1296.,,,+ +,24,16,30,,,,EULER,301

The value "2ndOrder" activates the Roe solver with second-order accuracy. The property id is the link between the TICEUL card 101 and the mesh card. The second Euler mesh for the Euler elements outside the bunker is created and initialized by: PEULER1,201,,2ndOrder,101 MESH,1,BOX,,,,,,,+ +,-647.,0.,-1293.,1057.,447.,1293.,,,+ +,33,23,37,,,,EULER,201 Euler archive output is generated using: ELEMENTS (AIR) = 1 SET 1 = ALLEULHYDRO

Main Index

196 Dytran Example Problem Manual Blastwave Hitting a Bunker

Results Figures Figure 4-28 and Figure 4-29 show a fringe plot and an isosurface. Figure 4-29 has been created by Ensight.

Main Index

Figure 4-28

Deformed Effective Stress Plot of the Bunker

Figure 4-29

Isosurfaces Created using SIE Variable for the Two Euler Domains

Chapter 4: Fluid-structural Interaction 197 Mine Blast

Mine Blast Problem Description This is a simulation of an explosion under a vehicle. The vehicle has triggered a mine that is exploding underneath the bottom shield. In this example, the actual explosion of the mine will not be modeled. Instead, the simulation will be started moments after the mine has exploded. This is called the blast wave approach. At the location of the mine, a high density and high specific energy is assumed in the shape of a small sphere. During the simulation, this region of high density, energy and therefore also high pressure, will expand rapidly. The blast wave will interact with the bottom shield and cause an acceleration of parts of the flexible body. The intent of this simulation is to find the location and the value of the maximum acceleration.

Dytran Model An outline of the basic numerical model is shown in the Figure 4-30 below. It is composed of the following main components: A. Vehicle Structure B. Euler Domain 1 - air outside vehicle and compressed air (explosive) C. Euler Domain 2 - air inside vehicle D. Ground E. Fluid Structural Coupling

Figure 4-30

Main Index

Outline of Basic Numerical Model

198 Dytran Example Problem Manual Mine Blast

The Vehicle: Vehicle structure is modeled by QUAD, TRIA shell elements and some BAR elements.

Figure 4-31

Vehicle Structure

Material properties are taken as follows: Density

7.85E-9

tonne/mm3

Modulus of elasticity

210000.

tonne/mm/s2

Poison ratio

0.3

Yield stress

250.

tonne/mm/s2

Assumed that there will be no failure of the structure. In a part of the structure, there is a hole through which air and pressure waves can freely flow. This hole will be modeled with dummy shell elements. Euler Domain 1: The first Euler domain is the air on the outside of the vehicle. The properties of air at rest are:

Main Index

Density

1.29E-12

Gamma

1.4

Specific Internal Energy

1.9385E8

tonne/mm3 tonne-mm2/s2

Chapter 4: Fluid-structural Interaction 199 Mine Blast

In the input deck: DMAT,230,1.29e-12,203,,,,,,+ +,,1.01 TICVAL,5,,DENSITY,1.29E-12,SIE,1.938e11

At the location of the mine, a small region will be modeled with high density and specific internal energy equivalent to TNT of 7kg when the sphere has a radius of .25 meter: Density

107E-12

tonne/mm3

Specific Internal Energy

4.9E12

tonne-mm2/s2

The input deck will show: TICVAL,4,,DENSITY,107E-12,SIE,3.9e12 SPHERE,400,,1797.5,0.,-450.,250.

The Euler region will be modeled by using the MESH card. The region will have to be large enough to contain the entire vehicle, including when the vehicle is in motion: MESH,1,BOX,,,,,,,+ +,-2623.,-1403.,-903.,6100.,2800.,2150.,,,+ +,30,10,10,,,,EULER,201

For the most accurate blastwave simulations, it is advised to use the Second-order Euler solver of Dytran. This is activated by specifying the second-order option on the Euler property card and specifying the parameter to use the second-order Range Kutta integration method: PARAM,RKSCHEME,3 PEULER1,201,,2ndOrder,101

To initialize the whole first Euler mesh, a TICEUL card will be defined. To initialize the Euler domain, other than within the sphere of the explosion, a 2nd large sphere is used. Because it has lower priority, the Euler elements within the mine blast sphere will still be initialized with high density and energy: TICEUL,101,,,,,,,,+ +,SPHERE,400,230,4,20.,,,,+ +,SPHERE,501,230,5,1. SPHERE,501,,0.,0.,-5000.,10000.

The Euler domain has infinite boundaries. This can be achieved by defining a zero gradient flow boundary on the outside of the Euler mesh. Use an empty FLOWDEF card: FLOWDEF,202,,HYDRO,,,,,,+ +,FLOW,BOTH

Main Index

200 Dytran Example Problem Manual Mine Blast

Euler Domain 2: The second Euler region represents the air inside the vehicle. Also for the second Euler region, a MESH card is used. The air is at rest again, so the same properties apply: PEULER1,202,,2ndOrder,102 TICEUL,102,,,,,,,,+ +,SPHERE,502,230,5,5. SPHERE,502,,0.,0.,-5000.,10000.

Many of the previous cards will be used to initialize the density and energy (TICVAL) and material (DMAT/EOSGAM) in this Euler region. The Ground: The ground is modeled as rigid body using dummy QUAD elements. It is used to close the Euler boundary under the vehicle so the blast wave will reflect on this boundary: PSHELL1,999,,DUMMY SURFACE,999,,PROP,999 SET1,999,999 $ RIGID,999,999,1.0E10,,0.00,0.00,-800.,,+ +,,,,,,,,,+ +,,1.E10,,,1.E10,,1.E10

The motion of the ground is constrained by specifying zero velocities: TLOAD1,1,2000, ,12 FORCE ,2000,999, ,0.0,1.0,1.0,1.0 MOMENT,2000,999, ,0.0,1.0,1.0,1.0

Fluid Structure Interaction: In order to make fluid structure interaction possible, a closed volume needs to be defined. The car model itself is not closed, so a dummy boundary will be defined to close the volume. This extra surface consists of three parts: Part 1 resides on the back, Part 2 is the top cover, and Part 3 is the vent on the bottom of the vehicle. For all these parts, dummy shell elements are defined and hole definitions will be defined.

Main Index

Chapter 4: Fluid-structural Interaction 201 Mine Blast

Figure 4-32

Dummy Shell Elements Defined to Close the Volume

The input for dummy shell elements: PSHELL1,900,,DUMMY PSHELL1,910,,DUMMY PSHELL1,920,,DUMMY

With this closed volume, the coupling surface can be defined. For each Euler domain, a separate surface is required. However, in this model, the interaction surface consists of the same elements, except for the extra ground elements (pid=999) for the outer Euler domain region 1. The surface definition will make use of the properties of the elements. The outer surface: SURFACE,97,,PROP,27 SET1,27,60,61,62,110,135,150,900,+ +,910,920,999

The inner surface: SURFACE,98,,PROP,28 SET1,28,60,61,62,110,135,150,900,+ +,910,920

Now the coupling surfaces can be defined. For the outer region, all elements inside the volume are not active. The covered option will, therefore, be set to INSIDE. Attached to this surface will be the first Euler MESH: COUPLE,1,97,INSIDE,ON,ON,11,,,+ +,,,,,,,,,+ +,,1

The inner Euler domain is constrained by surface 2. For this volume, the outer Euler elements will be covered: COUPLE,2,98,OUTSIDE,ON,ON,,,,+ +,,,,,,,,,+ +,,2

Main Index

202 Dytran Example Problem Manual Mine Blast

As discussed before, there are holes in the coupling surface. To this end, a flow definition is required for one of the coupling surfaces. In this example, the flow cards are referenced from the first coupling surface. The input to define flow between the regions is: COUPOR,1,11,1,PORFLCPL,84,CONSTANT,1.0 SUBSURF,1,97,PROP,301 SET1,301,900

Also, for each of the other two flow surfaces, these set of cards are repeated. Finally, the flow definition itself prescribes that the Euler region from coupling surface 1 is interacting with the Euler region from coupling surface 2: PORFLCPL,84,,,BOTH,2

Miscellaneous: a. Because this model uses the coupling surface interface, the time step safety factor for Eulerian elements has to be .6. However, the Lagrangian elements (the quadratic and triangular elements) determine the time-step, and it is beneficial to use a higher time step safety factor for the Lagrangian elements: PARAM,STEPFCTL,0.9

b. To show the stress on the structure, the following output request was added: TYPE (Vehicle) = ARCHIVE ELEMENTS (Vehicle) = 3 SET 3 = {list of the element numbers of the vehicle} ELOUT (Vehicle) = EFFST TIMES (Vehicle) = 0,thru,end,by,0.0002 SAVE (Vehicle) = 10000

c. In order to find the location of the maximum acceleration a gridpoint archive output request will be created. In this case, location and also the approximate value of the acceleration can be visualized in the postprocessor: TYPE (Surface) = ARCHIVE GRIDS (Surface) = 4 SET 4 = {list of the gridpoints of the vehicle} GPOUT (Surface) = RACC, RVEL TIMES (Surface) = 0,thru,end,by,0.0002 SAVE (Surface) = 10000

Main Index

Chapter 4: Fluid-structural Interaction 203 Mine Blast

Results The Figure 4-33 below shows the location, value, and time of the maximum acceleration. The stress distribution at this time is also in Figure 4-34.

Main Index

Figure 4-33

Acceleration Plot

Figure 4-34

Stress Distribution Plot

204 Dytran Example Problem Manual Mine Blast

Abbreviated Dytran Input File START CEND ENDTIME=0.005 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: test (mm/tonne/s/K) TLOAD=1 TIC=1 SPC=1 $ $ ========= OUTPUT REQUEST FOR AIR ========= $ TYPE (AIR1) = ARCHIVE ELEMENTS (AIR1) = 1 SET 1 = ALLEULHYDRO ELOUT (AIR1) = PRESSURE,XVEL,YVEL,ZVEL,SIE,DENSITY TIMES (AIR1) = 0,thru,end,by,.0002 SAVE (AIR1) = 20000 $ CPLSURFS(BAGHT1)=300 SET 300 = 97 CPLSOUT(BAGHT1)=PRESSURE,XVEL,YVEL,ZVEL TIMES (BAGHT1)= 0,thru,end,by,.0002 TYPE (BAGHT1)=ARCHIVE SAVE (BAGHT1)=1000000 $ CPLSURFS(BAGHT2)=301 SET 301 = 98 CPLSOUT(BAGHT2)=PRESSURE,XVEL,YVEL,ZVEL TIMES (BAGHT2)= 0,thru,end,by,.0002 TYPE (BAGHT2)=ARCHIVE SAVE (BAGHT2)=1000000 $ $ ========= OUTPUT REQUEST FOR STRUCTURAL ELEMENTS ========= $ TYPE (VEHICLE) = ARCHIVE ELEMENTS (VEHICLE) = 3 SET 3 = 2, THRU, 105, 129, THRU, 1697, 1702, THRU, 1736, 1758, THRU, 1851, 2460, THRU, 2536 ELOUT (VEHICLE) = EFFST TIMES (VEHICLE) = 0,thru,end,by,.0002 SAVE (VEHICLE) = 99999 $ TYPE (GP) = ARCHIVE GRIDS (GP) = 4 GPOUT (GP) = RACC, RVEL TIMES (GP) = 0,thru,end,by,.0002 SAVE (GP) = 99999 SET 4 = 1,thru,176 189,thru,202 218,thru,225, 253,thru,259 272,thru,285 294,thru,389, 414,thru,421 471,thru,489 500,thru,543, 545,thru,563 575,thru,578 595,thru,643,

Main Index

Chapter 4: Fluid-structural Interaction 205 Mine Blast

776,thru,1137 1149,thru,1284 1301,thru,1308, 1343,thru,1361 1364,thru,1479 1492,thru,1593, 1630,thru,1687 1908 1909 2038 2124, 2227 2304 2343 2448 2526 2702,thru,2727, 2755,thru,2762 2807,thru,2825 2838,thru,2878, 2907,thru,2910 2944,thru,2962 2969,thru,2972, 2988,thru,3007 3016,thru,3030 3037,thru,3044, 3049,thru,3052 3056,thru,3060 3064,thru,3077, 3087,thru,3090 3094,thru,3097 3101,thru,3104, 3107,thru,3150 3313,thru,3374 3651,thru,3705, 3707,thru,3714 3718,thru,3729 $ $------- Parameter Section -----$ PARAM,RKSCHEME,3 PARAM,FASTCOUP PARAM,STEPFCTL,0.9 PARAM,INISTEP,.5E-7 PARAM,MINSTEP,1.E-13 $ $------- BULK DATA SECTION ------BEGIN BULK $ MESH,1,BOX,,,,,,,+ +,-2623.,-1403.,-903.,6100.,2800.,2150.,,,+ +,30,10,10,,,,EULER,201 $ MESH,2,BOX,,,,,,,+ +,-2621.,-1201.,-251.,5900.,2400.,1250.,,,+ +,30,10,10,,,,EULER,202 $ PEULER1,201,,2ndOrder,101 TICEUL,101,,,,,,,,+ +,SPHERE,400,230,4,20.,,,,+ +,SPHERE,501,230,5,1., SPHERE,400,,1797.5,0.,-450.,250. SPHERE,501,,0.,0.,-5000.,10000. $ PEULER1,202,,2ndOrder,102 TICEUL,102,,,,,,,,+ +,SPHERE,502,230,5,5. SPHERE,502,,0.,0.,-5000.,10000. $ TICVAL,4,,DENSITY,107E-12,SIE,3.9e12 TICVAL,5,,DENSITY,1.29E-12,SIE,1.938e11 $ DMAT,230,1.29e-12,203,,,,,,+ +,,1.01 EOSGAM,203,1.4 $ FLOWDEF,202,,HYDRO,,,,,,+ +,FLOW,BOTH $ COUPLE,1,97,INSIDE,ON,ON,11,,,+

Main Index

206 Dytran Example Problem Manual Mine Blast

+,,,,,,,,,+ +,,1 SURFACE,97,,PROP,27 $ $ Define flow thru the holes $ COUPOR,1,11,1,PORFLCPL,84,CONSTANT,1.0 SUBSURF,1,97,PROP,301 SET1,301,900 $ COUPOR,2,11,2,PORFLCPL,84,CONSTANT,1.0 SUBSURF,2,97,PROP,302 SET1,302,910 $ COUPOR,3,11,3,PORFLCPL,84,CONSTANT,1.0 SUBSURF,3,97,PROP,303 SET1,303,920 $ PORFLCPL,84,,,BOTH,2 $ COUPLE,2,98,OUTSIDE,ON,ON,,,,+ +,,,,,,,,,+ +,,2 SURFACE,98,,PROP,28 $ SET1,27,60,61,62,110,135,150,900,+ +,910,920,999 $ SET1,28,60,61,62,110,135,150,900,+ +,910,920 $ $ ========== PROPERTY SETS ========== $ $ * pbar.9988 * $ PBAR 9988 222 3600.1000000.1000000.2000000. $ $ * pbar.9989 * $ PBAR 9989 222 100000. 3.E+8 3.E+8 6.E+8 $ $ * pbar.9990 * $ PBAR 9990 222 3000. 200000.2500000.3000000. $ $ * pbar.9993 * $ PBAR,9993,111,459.96,25066.,55282.,16543. $ $ * pbar.9996 * $ PBAR,9996,111,895.52,309450.,55349.,48782. $ $ * pbar.9999 *

Main Index

Chapter 4: Fluid-structural Interaction 207 Mine Blast

$ PBAR,9999,111,736.,490275.,827555.,2095137. $ $ * pshell.30 * $ PSHELL 30 111 3 $ $ * pshell.40 * $ PSHELL 40 111 4 $ $ * pshell.50 * $ PSHELL 50 111 5 $ $ * pshell.60 * $ PSHELL 60 111 6 PSHELL 61 111 6 PSHELL 62 111 6 $ * pshell.80 * $ PSHELL 80 111 8 $ $ * pshell.110 * $ PSHELL 110 111 11 $ $ * pshell.120 * $ PSHELL 120 111 12 $ $ * pshell.135 * $ PSHELL 135 111 13.5 $ $ * pshell.150 * $ PSHELL 150 111 15 PSHELL 151 111 15 $ $ * pshell.200 * $ PSHELL 200 111 20 $ $ * pshell.450 * $ PSHELL 450 111 45 $ $ dummy elements for coupling surface $ fail immediately $ hole PSHELL1,900,,DUMMY $ top cover

Main Index

208 Dytran Example Problem Manual Mine Blast

PSHELL1,910,,DUMMY $ side cover PSHELL1,920,,DUMMY $ $ ground PSHELL1,999,,DUMMY SURFACE,999,,PROP,999 SET1,999,999 RIGID,999,999,1.0E10,,0.00,0.00,-800.,,+ +,,,,,,,,,+ +,,1.E10,,,1.E10,,1.E10 $ constrain the motion of the ground TLOAD1,1,2000, ,12 FORCE ,2000,999, ,0.0,1.0,1.0,1.0 MOMENT,2000,999, ,0.0,1.0,1.0,1.0 $ $ * conm2 * $ CONM2,5000,1145,,1.5 CONM2,5001,1146,,1.7 $ $ ========= MATERIAL DEFINITIONS ========== $ DMATEP,111,7.85e-09,210000.,.3,,,111 YLDVM,111,250E10 $ DMATEP,222,7.85e-09,210000.,.3 $ INCLUDE model.bdf INCLUDE ground.dat $ ENDDATA

Main Index

Chapter 4: Fluid-structural Interaction 209 Multiple Bird-strike on a Box Structure

Multiple Bird-strike on a Box Structure Problem Description Bird strike on a box structure is a typical problem in aircraft industries. The box structure simulates the leading edge of lifting surfaces, e.g. wing, vertical, and horizontal stabilizers. The box can be simplified to consist of a curve leading edge panel and a front spar. The acceptable design criteria for bird strike is that the leading edge panel may fail but the front spar strength may not degrade to a certain level. In this example, two cylindrical panels are put in parallel. Two birds strike the upper panel. One bird strikes in horizontal direction and the second one vertically. The second bird will perforate the first panel and impact the second one. The problem is based on the model described in the example Multiple Birdstrike on a Cylindrical Panelwhere an ALE feature of Dytran is used. Due to the introduction of failure in the structure (coupling surface), the standard Euler solver will be used. In this example, the emphasis is given on the modeling technique. The relative positions of the birds to the upper plate and their initial velocities are shown in Figure 4-35. The data for analysis are given in Table 4-1.

Figure 4-35

Main Index

Geometry of the Upper Plate and Position of the Birds Relative to the Plate

210 Dytran Example Problem Manual Multiple Bird-strike on a Box Structure

Table 4-1

Material Data and Initial Conditions Plate

Ambience

Bird 1

Bird 2

Material

Titanium

Air

Jelly

Jelly

Density (kg/m3)

4527

1.1848

930

930

Bulk modulus (Pa)

1.03e11

2.2e9

2.2e9

Poisson’s ratio

0.314

Yield stress (Pa)

1.38e8

Mass (kg)

0.36

0.285

Initial velocity (m/s)

150

200

Gamma

1.4

Thickness (m)

0.0015

Radius (m)

0.25

Length (m)

0.25

Fail (equiv. Plas. Strain)

0.1

Dytran Modeling Each curved plate is modeled using 33x16 BLT-shells. The boundary conditions applied at the edges of the plate are defined within a cylindrical coordinate system, where the local z-axis is aligned with the length axis of the plate. The cylindrical system is defined using a CORD2C entry. To create a closed surface, required by COUPLING option, the two plates are connected with dummy quad elements. The two birds and air are modeled using Multi Material Eulerian (FV) elements, also known as MMHYDRO. The location of the bird in the Euler domain is defined using TICEUL option. The material for the birds and air are modeled using EOSPOL and EOSGAM, respectively. To allow the bird perforating the first plate and impact the second one, several modeling techniques can be used. One of them is using two Eulerian domains and two coupling surfaces. Both the Eulerian domains and the coupling surfaces have to be logically different. Each coupling surface associates with one Eulerian domain. In this model the two coupling surfaces share the same physical space. By specifying that one domain is covered outside and the other inside, the Eulerian domain represents the correct physical space. The two Eulerian domains cannot interact with each other except through coupling surfaces. When coupling surfaces share the same shell elements, and some or all shells fail, then material can flow from one Eulerian domain into another one. The interaction between the Eulerian domains is activated using COUP1INT option and PARAM, FASTCOUP, INPLANE, FAIL. The rest of the Euler domain is filled with Air. Please notice that when the effect of air is neglected then the rest of the Eulerian domain should be filled with void. It will speed up the analysis. The first domain is associated with a coupling surface that is INSIDE covered. Therefore, it cannot be adaptive and is defined using MESH,,BOX card. The second domain is adaptive and defined using

Main Index

Chapter 4: Fluid-structural Interaction 211 Multiple Bird-strike on a Box Structure

MESH,,ADAPT. The ADAPT option will let Dytran create and update the Eulerian domain to minimize memory allocation and consequently lowered CPU time. The default Eulerian boundary condition is set to that only outflow is allowed using FLOWDEF option. In this case, a bird that reaches the free face boundary will flow out of the domain. The initial velocity of the birds is defined using TICVAL option. The finite element model of the upper and lower plates, the Eulerian domains and the initialization of the birds are shown in Figure 4-36. The dummy quad elements used to create closed coupling surfaces are not shown in this figure .

Figure 4-36

Finite Element Model and Finite Volume Domains

Results In this simulation, the time history of total z-force on the coupling surface is requested as shown in Figure 4-37. This force is the sum of all z-forces on the nodes that belong to both the upper and the lower plate. From this figure, it is obvious that there are three large impact forces occurring on the plate. The first one is when the first bird impacts the upper plate, which is subject to a significant damage. The second one is when the second bird impacts the upper plate. The last peak is caused by the first bird impacting the lower plate. Snapshots of the motion of the two birds and the deformation of the plates are shown in Figure 4-38 at various time steps of the simulation. Figure 4-38a is the initial condition. Figure 4-38b is at the moment when the first bird penetrates the upper plate and second bird touches the plate. This corresponds with the first peak in the time history plot shown in Figure 4-37. Figure 4-38c is at the moment when the second bird penetrates the upper plate. It corresponds with the second peak of the time history plot. Figure 4-38d is at the moment when the second bird has left the plate and the first bird penetrates the lower plate. This corresponds with the third peak in the time history plot

Main Index

212 Dytran Example Problem Manual Multiple Bird-strike on a Box Structure

.

Main Index

Figure 4-37

Time History of the Total Z-force on the Plates

Figure 4-38

Snapshots of Simulation Results Showing the Motion of the Birds and Plates

Chapter 4: Fluid-structural Interaction 213 Multiple Bird-strike on a Box Structure

Files Ep4d9.dat

Dytran input files

Ep4d9.bdf Ep4d9_C1_0.THS

Dytran time history file

Ep4d9_BIRD_xx.ARC

Dytran archive files

Abbreviated Dytran Input File START CEND TITLE = Multiple bird strike using Multi-Material-FVSurfer ENDSTEP = 1200 ENDTIME = .0025 TIC = 1 SPC = 1 $ TYPE(plate) = ARCHIVE SAVE(plate) = 1 STEPS(plate) = 0,200,400,1200 ELEMENTS(plate) = 1 SET 1 = 7393 THRU 8448 ELOUT(plate) = EFFST01,EFFST03,FAIL $$ $$ Data for output control Set 2 : the 2 birds TYPE(bird) = ARCHIVE SAVE(bird) = 1 STEPS(bird) = 0,200,400,1200 ELEMENTS(bird) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT(bird) = XVEL,YVEL,ZVEL,FMAT3,FMAT5 $ TYPE(C1) = TIMEHIS SAVE(C1) = 10000 STEPS(C1) = 0tEndb10 CPLSURFS(C1) = 3 SET 3 = 1 CPLSOUT(C1) = XFORCE,YFORCE,ZFORCE,RFORCE $ type(out)=stepsum steps(out)=0tendby10 type(mat)=matsum steps(mat)=0tendb10 $ BEGIN BULK PARAM INISTEP 1e-7 PARAM MINSTEP 1e-8 PARAM,FASTCOUP,INPLANE,FAIL PARAM,FMULTI,1.0 PARAM,NZEROVEL,YES

Main Index

214 Dytran Example Problem Manual Multiple Bird-strike on a Box Structure

PARAM,FAILOUT,NO param,stepfctl,0.9 $ INCLUDE ep4d9.bdf $ $ domain 1 $ MESH,1,BOX,,,,,,,+ +,-0.26,-0.015,-0.05,0.50,0.28,0.44,,,+ +,50,28,44,,,,EULER,1 $ $ COUPLING SURFACE 1 COUPLE 1 1 INSIDE ON ON +A000002 +A000002 +A000002 +A000002 1 $ SURFACE 1 ELEM 11 SET1 11 7393 THRU 8448 13729 THRU 14048 14577+ + THRU 15236 $ $ Flow boundary, property, material and equation of state data. FLOWDEF 1 MMHYDRO + + FLOW OUT $ PEULER1 1 MMHYDRO 11 $ $--------Material Bird -----------------------------------DMAT 3 930 3 EOSPOL 3 2.2e9 DMAT 5 930 5 EOSPOL 5 2.2e9 $ $ Allocation of material to geometric regions. TICEUL 11 + + CYLINDER1 3 1 3 + + CYLINDER2 5 2 2 + + SPHERE 4 4 5 1 $ CYLINDER1 .13 .125 .2252 .17 .125 .2944 + + .035 CYLINDER2 -.1381 .125 .26 -.2381 .125 .26 + + .035 SPHERE,4,,-.1381, .125, .26, 1000 $ $ Initial material data. TICVAL 1 XVEL -75 ZVEL -129.9 TICVAL 2 XVEL 200 $ $ Property, material and yield model. PSHELL1 2 2 Blt Gauss 3 .83333 Mid + + .0015 DMATEP 2 4527 .314 1.03e11 1 1 YLDVM 1 1.38e8 FAILMPS 1 0.1

Main Index

Chapter 4: Fluid-structural Interaction 215 Multiple Bird-strike on a Box Structure

$ PSHELL1,3,,DUMMY PSHELL1,4,,DUMMY $ $ Boundary constrain. $ -------------------CORD2C 1 0.0 0.0 0.0 0.0 0.25 0.0 + + 0.0 0.125 0.25 $ $ -------- Material Air id =4 DMAT 4 1.1848 4 EOSGAM,4,1.4 $ $ Domain 2 PEULER1,6,,MMHYDRO,12 TICEUL,12,,,,,,,,+ +,SPHERE,7,4,5,1.0,,,,+ SPHERE,7,,0.0,0.0,0.0,500.0 TICVAL,5,,SIE,2.1388E5,DENSITY,1.1848 $ $===Coupling Surface 2 $ COUPLE,2,2,OUTSIDE,,,,,,+ +,,,,,,,,,+ +,,2 $ SURFACE 2 ELEM 12 SET1 12 7393 THRU 8448 13729 THRU 14048 14577+ + THRU 15236 MESH,2,ADAPT,0.01,0.01,0.01,,,,+ +,-0.26,-0.015,-0.05,,,,,,+ +,,,,,,,EULER,6 $ $ coupling interaction COUP1INT,2,2,1 $ ENDDATA

Main Index

216 Dytran Example Problem Manual Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Problem Description When a metal cone is explosively collapsed onto its axis, a high-velocity rod of molten metal, the jet, is ejected out of the open end of the cone. The cone is called a liner and is typically made of copper. The jet has a mass approximately 20 percent of the cone mass, and elongates rapidly due to its high velocity gradient. This molten rod is followed by the rest of the mass of the collapsed cone, the slug. Typical shaped charges have liner slope angles of less than 42 degrees ensuring the development of a jet; with jet velocities ranging from 3000 to 8000 m/s. A typical construction of a shaped charge is shown in Figure 4-39.

Main Index

Chapter 4: Fluid-structural Interaction 217 Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Figure 4-39

A Typical Construction of a Shaped Charge

An example simulation of shaped charge formation is carried out to demonstrate the ability of Dytran to perform such a simulation. A simplified axisymmetric model of explosives and a copper liner is created in a finite volume Euler mesh. Explosive are detonated starting from a point on the axis of symmetry at the end of the explosives. The simulation is carried out for 60 μs after detonation of the explosives. The jet is formed and penetrates two thick plates. Please see Figure 4-40 for the model layout .

Figure 4-40

Main Index

Dytran Model Setup

218 Dytran Example Problem Manual Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Typical shaped charges are axisymmetric. However, aiming at higher velocity, 3-D designs are targeted. 3-D simulation of shaped charge formation would be necessary to avoid excessive experimental work. Dytran has full abilities to perform such a 3-D simulation.

Dytran Model The model is simplified as shown in Figure 4-40. The aluminum casting is replaced with a rigid body. Detonation is assumed to start at a point on the axis at the rear end of the explosives. The liner shape is slightly simplified as shown in the figure. The retaining ring is assumed rigid and is modeled as a wall boundary for the Euler Mesh (WALLET). SI units are used in this example. Euler Mesh and Liner: A triangular prismatic Finite Volume Euler mesh is used with head angle of 5 degrees as shown in Figure 4-41. A very fine mesh is used to accurately simulate the behavior of the extremely thin liner. The

liner is placed in this Euler mesh. Symmetry conditions (closed boundary, default Euler boundary condition) are imposed on the two rectangular faces of the prism to create an axisymmetric behavior.

Figure 4-41

Euler Mesh

The liner material pressure – density relationship is modeled with EOSPOL model. The liner is made of copper and the constants are taken as follows:

Main Index

a1

1.43E11

N/m2

a2

0.839E11

N/m2

a3

2.16E9

N/m2

b1

0.0

b2

0.0

b3

0.0

Chapter 4: Fluid-structural Interaction 219 Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Material yield strength is modeled with a Johnson-Cook yield model. The constants are taken as follows: A

1.2E8

N/m2

B

1.43E9

N/m2

C

0.0

n

0.5

m

1.0

· ε0

1.0

Tmelt

1356.0

K

Troom

293.0

K

Cv

399.0

J/kg

Other liner material properties of liner are as follows: Density

8960

Kg/m3

Constant shear model

0.477E11

N/m2

Constant spallation model

-2.5E10

N/m2

In the input deck: DMAT 701 8960. 711 712 713 EOSPOL, 711, 1.43+11, 0.839+11, 2.16+9 SHREL,712,0.477E11 $ Johnson-Cook $ A B n C m EPS0 Cv YLDJC,713, 1.2E8, 1.43E9, 0.5, 0.0, 1.0, 1.0, 399.0,+ $ TMELT TROOM +, 1356.0, 293.0 $ PMINC,714,-2.5E10

714

It is very easy to define the shape and position of the liner by using the method of geometrical regions when creating the initial conditions of the liner material. CYLINDER, 1,, -0.5391, -0.56, 0., 2.0, 0.4147, 0.,+ +,0.2958 CYLINDER, 2,, -0.5391, -0.56, 0., 2.0, 0.4147, 0.,+ +,0.2939 CYLINDER, 3,, 0.2, 2.0406, 0., 0.2047, 2.0406, 0.,+ +,2.0019 TICVAL,2,,DENSITY,8960.

Main Index

220 Dytran Example Problem Manual Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Casting and Retaining Ring: The casting is assumed to be rigid. It is modeled by the default Eulerian boundary condition (closed boundary). The retaining ring is also assumed to be rigid and is modeled by a wallet. Plates: Two thick plates are placed in this Euler mesh. Plate material is defined as steel: DMAT 801 7830. EOSPOL, 811, 1.64E+11 SHREL,812,0.818E11 YLDVM,813,1.4E9 PMINC,814,-3.8E9

811

812

813

814

The shapes and positions of the plates are defined by using the method of geometrical regions. CYLINDER, 4,, 0.22, 2.0406, 0., 0.223, 2.0406, 0.,+ +,2.05 CYLINDER, 5,, 0.27, 2.0406, 0., 0.273, 2.0406, 0.,+ +,2.05 TICVAL,3,,DENSITY,7830.

Explosive: The explosive is modeled by ignition and growth equation of state. The explosive is placed in this Euler mesh. EOSIG,100,,,,,,,,+ +,,,,,,,,,+ +,,,,,99,,MCOMPB,SI

The explosive material is taken from the database that is build into Dytran. To initialize the whole Euler mesh, a TICEUL card will be defined. TICEUL 1 + ELEM 1 100 1 + CYLINDER1 701 2 + CYLINDER2 + CYLINDER3 701 2 + CYLINDER4 801 3 + CYLINDER5 801 3 $ SET1 1 1 THRU 15342 TICVAL,1,,DENSITY,1630.,SIE,4.29E6

Main Index

1. 2. 3. 4. 5. 6.

+ + + + + +

Chapter 4: Fluid-structural Interaction 221 Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Results The figure below shows the initial position of the copper liner and two thick plates at 0μs, snap shots of liner collapse, jet formation and plates penetrated at 10 μs, 20 μs, 30 μs, 40 μs, 50 μs and 60 μs.

Figure 4-42

Initial Position of the Copper Liner and Two Thick Plates, Snap Shots of Liner Collapse, Jet Formation and Plates Penetrated (Courtesy – Postprocessing by CEI Ensight)

Figure 4-43 shows the velocity field of explosive gases, liner, and jet at 20 μs. A jet velocity of about 6000 m/s is achieved.

Figure 4-43

Main Index

Velocity Field of Explosive Gases, Liner, and Jet

222 Dytran Example Problem Manual Shaped Charge, using IG Model, Penetrating through Two Thick Plates

Abbreviated Dytran Input File START CEND TITLE = SHAPED CHARGES TEST (m/kg/s/K) CHECK = NO ENDTIME = 6.E-5 TIC = 1 $$ $$ TYPE(EULER1) = ARCHIVE ELEMENTS(EULER1) = 11 SET 11 = 1,THRU,15342 ELOUT(EULER1) = XVEL,YVEL,ZVEL,EFFSTS,EFFPLS, TXX, TYY, TZZ, TXY, TYZ, TZX, PRESSURE, SIE, FMAT100,FMAT701,DENSITY100,DENSITY701, FMAT801,DENSITY801,FMAT,DENSITY SAVE(EULER1) = 999999 TIMES(EULER1) = 0.0, THRU, END, BY, 2.E-6 $ BEGIN BULK PARAM,INISTEP,1.E-11 PARAM,MINSTEP,1.E-13 PARAM,GEOCHECK,ON PARAM,VELMAX,20.0E+03 $ $ THIS SECTION CONTAINS BULK DATA $ INCLUDE model.bdf INCLUDE wall.dat $ $ PEULER1, 1 ,, MMSTREN, 1 $ $ EXPLOSIVE $ DMAT 100 1630. 100 101 102 EOSIG,100,,,,,,,,+ +,,,,,,,,,+ +,,,,,99,,MCOMPB,SI SHREL,101,3.E9 YLDVM,102,2.E8 $ $ $ COPPER $ DMAT 701 8960. 711 712 713 EOSPOL, 711, 1.43+11, 0.839+11, 2.16+9 SHREL,712,0.477E11 $ Johnson-Cook $ A B n C m EPS0 Cv YLDJC,713, 1.2E8, 1.43E9, 0.5, 0.0, 1.0, 1.0, 399.0,+ $ TMELT TROOM

Main Index

714

Chapter 4: Fluid-structural Interaction 223 Shaped Charge, using IG Model, Penetrating through Two Thick Plates

+, 1356.0, 293.0 $ PMINC,714,-2.5E10 $ $ STEEL $ DMAT 801 7830. 811 812 EOSPOL, 811, 1.64E+11 SHREL,812,0.818E11 YLDVM,813,1.4E9 PMINC,814,-3.8E9 $ TICEUL 1 + ELEM 1 100 1 + CYLINDER1 701 2 + CYLINDER2 + CYLINDER3 701 2 + CYLINDER4 801 3 + CYLINDER5 801 3 $ SET1 1 1 THRU 15342 CYLINDER, 1,, -0.5391, -0.56, 0., 2.0, +,0.2958 CYLINDER, 2,, -0.5391, -0.56, 0., 2.0, +,0.2939 CYLINDER, 3,, 0.2, 2.0406, 0., 0.2047, +,2.0019 CYLINDER, 4,, 0.22, 2.0406, 0., 0.223, +,2.05 CYLINDER, 5,, 0.27, 2.0406, 0., 0.273, +,2.05 $ TICVAL,1,,DENSITY,1630.,SIE,4.29E6 TICVAL,2,,DENSITY,8960. TICVAL,3,,DENSITY,7830. $ $ DUMMY QUAD TO MODEL THE WALLET $ PSHELL1,2,,,,,,,,+ +,9999. WALLET,1,2 $ $ ENDDATA

Main Index

813

1. 2. 3. 4. 5. 6.

0.4147, 0.,+ 0.4147, 0.,+ 2.0406, 0.,+ 2.0406, 0.,+ 2.0406, 0.,+

814

+ + + + + +

224 Dytran Example Problem Manual Fuel Tank Filling

Fuel Tank Filling

Problem Description The process of filling up an automobile fuel tank must be easy and comfortable for the customer. Effects like the gasoline pump prematurely switching off or the back splashing of fuel must be avoided. Furthermore, the legal requirements must be met. The space available for the whole system is constantly being minimized, leading to additional difficulties in fulfilling the above criteria. Usually costly and time-consuming experiments are necessary for this optimization. Numerical simulation is a desirable tool to avoid excessive experimental work. The purpose is to demonstrate application of Multiple Adaptive Euler Domains for Multiple Material to fuel tank filling process. The problem simulates a fuel tank that contains a filling pipe and a vent pipe. Tank is full with fuel up to 80 mm from the bottom. The rest is full with air. In the simulation, the fuel is made to flow into the tank through the inlet of the filling pipe. The air and the fuel escape out of the tank through the outlet of the vent pipe.

Dytran Model Tank and pipes are modeled as rigid bodies. The fuel/air region is modeled by three Euler meshes. The first domain models the inside of the tank, the second domain models the inside of the filling pipe, and the third domain models the inside of vent pipe. For the interaction between the structure and Euler domains, three coupling surfaces are needed. • Units

Length = mm, Mass = Kg and Time = second • Tank and Pipes Figure 4-44 shows the structure mesh. All elements are defined as dummy shell elements. A surface is created and defined as a rigid body. The tank is fixed in position by defining zero velocity in all directions and zero rotation in all directions.

Main Index

Chapter 4: Fluid-structural Interaction 225 Fuel Tank Filling

Figure 4-44 Tank and Piping Structure Mesh • Euler domain 1

The first Euler domain has the fuel and air inside of the tank. The properties of fuel are: Density

8.5E-7

Kg/ mm3

Bulk modulus

2.0E+4

KPa

This is a reduced bulk modulus (1/100) to increase the time step and reduce CPU time. In the input deck: DMAT EOSPOL

2 2

8.5E-7 20000

2

The air properties are: Density

1.29E-9

Gamma

1.4

Specific internal energy

1.938E11

Kg/ mm3 Kg-mm2/s2

In the input deck: DMAT 1 1.29E-9 1 EOSGAM 1 1.4 TICVAL,21,,DENSITY,1.29E-9,SIE,1.938E11

The Euler region is modeled by using the MESH card. The ADAPT option is used: PEULER1,1,,MMHYDRO,100 $ MESH,1,ADAPT,8.,8.,8.,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,1

Main Index

226 Dytran Example Problem Manual Fuel Tank Filling

To initialize the whole first Euler mesh, a TICEUL card is defined. Tank is full with fuel up to 80 mms from the bottom. The rest is full with air. Initial air pressure is set to 100 KPa. Fuel hydrostatic pressure is defined starting from 100 KPa at the surface and increasing going down. The four layers with different pressures are defined: TICEUL,100,,,,,,,,+ +,CYLINDER,31,1,21,1.0,,,,+ +,CYLINDER,32,2,22,2.0,,,,+ +,CYLINDER,33,2,23,3.0,,,,+ +,CYLINDER,34,2,24,4.0,,,,+ +,CYLINDER,35,2,25,5.0 $ CYLINDER,31,,-350.,150.,-10000.,50.,150.,-10000.,+ +,20000. CYLINDER,32,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10020. CYLINDER,33,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10040. CYLINDER,34,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10060. CYLINDER,35,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10080. $ TICVAL,22,,DENSITY,8.54254E-7,SIE,0 TICVAL,23,,DENSITY,8.5426E-7,SIE,0 TICVAL,24,,DENSITY,8.54268E-7,SIE,0 TICVAL,25,,DENSITY,8.54275E-7,SIE,0

• Euler domain 2

The second Euler region represents the fuel and air inside the filling pipe. For smooth start of the simulation, the part near the inlet of the filling pipe is initially filled with fuel. The rest is full with air. For the second Euler region a MESH card is used: PEULER1,2,,MMHYDRO,200 $ MESH,2,ADAPT,8.,8.,8.,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,2 TICEUL,200,,,,,,,,+ +,CYLINDER,51,2,41,1.0,,,,+ +,CYLINDER,52,1,21,2.0 $ CYLINDER,51,,-150.,150.,-10000.,250.,150.,-10000.,+ +,10260. CYLINDER,52,,-150.,150.,-10000.,250.,150.,-10000.,+ +,10205. $ TICVAL,41,,DENSITY,8.5425E-7,SIE,0

Main Index

Chapter 4: Fluid-structural Interaction 227 Fuel Tank Filling

• Euler domain 3

The third Euler region represents the fuel and air inside the vent pipe. The vent pipe is initially full with air. For the third Euler region, a MESH card is used: PEULER1,3,,MMHYDRO,300 $ MESH,3,ADAPT,8.,8.,8.,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,3 TICEUL,300,,,,,,,,+ +,CYLINDER,51,1,21,1.0

• Fluid structure interaction

For each Euler domain, a separate surface is required. The surface definition makes use of the properties of the elements. 1) Tank surface: SURFACE,101,,PROP,101 SET1,101,7,8,12,THRU,16

The surface has been closed to constitute valid coupling surface (Figure 4-45).

Figure 4-45 Coupling Surface of Euler Domain 1 The Euler domain 1 is constrained by surface 101. All elements outside the volume are not active. The covered option is, therefore, set to OUTSIDE. Attached to this surface is the first Euler MESH: COUPLE,1,101,OUTSIDE,,,,,,+ +,,,,,,,,,+ +,,1

2) Filling pipe surface: SURFACE,201,,PROP,201 SET1,201,4,THRU,8

Main Index

228 Dytran Example Problem Manual Fuel Tank Filling

The surface has been closed to constitute valid coupling surface (Figure 4-46).

Figure 4-46 Coupling Surface of Euler Domain 2 (filling pipe) The Euler domain 2 is constrained by surface 201. For this volume, the outer Euler elements are covered: COUPLE,2,201,OUTSIDE,,,22,,,+ +,,,,,,,,,+ +,,2

Surface 101 and surface 201 share some elements (with property number 7, 8). A hole is modeled as a sub-surface consisting of quads (with property number 8) that are fully porous. The elements in this sub-surface connect the two coupling surface and are included in the definition of both coupling surfaces. A flow definition is required for one of the coupling surfaces. The flow card is referenced from the second coupling surface. The input to define flow between the two regions: COUPOR,2,22,2,PORFLCPL,2,,1.0 PORFLCPL,2,,,BOTH,1 SUBSURF,2,201,PROP,250 SET1,250,8

3) Vent pipe surface: SURFACE,301,,PROP,301 SET1,301,9,THRU,14

The surface has been closed to constitute valid coupling surface (Figure 4-47).

Main Index

Chapter 4: Fluid-structural Interaction 229 Fuel Tank Filling

Figure 4-47 Coupling Surface of Euler Domain 3 (vent pipe) The Euler domain 3 is constrained by surface 301. For this volume, the outer Euler elements are covered: COUPLE,3,301,OUTSIDE,,,33,,,+ +,,,,,,,,,+ +,,3

Surface 101 and surface 301 share some elements (with property number 12, 13, 14). A hole is modeled as a sub-surface consisting of quads (with property number 14) that are fully porous. The elements in this sub-surface connect the two coupling surface and are included in the definition of both coupling surfaces. A flow definition is required for one of the coupling surfaces. The flow card is referenced from the third coupling surface. The input to define flow between the two regions: COUPOR,3,33,3,PORFLCPL,3,,1.0 PORFLCPL,3,,,BOTH,1 SUBSURF,3,301,PROP,350 SET1,350,14

• Inlet and outlet

Two flow boundaries are defined to the coupling surface (pipe ends, see Figure 4-48). The first is to define fuel flow into the tank at a predefined flow rate (velocity × area). The second is to allow air (or fuel) to escape out of the tank. Pressure at the second boundary is defined as 1.0 at (100 KPa).

Main Index

230 Dytran Example Problem Manual Fuel Tank Filling

Figure 4-48 Flow Boundaries 1) Inlet The flow card of inlet is referenced from the second coupling surface. As input, the velocity is defined such that the flow rate keeps 2.0 liter/second over 2 seconds. Since the area of the inlet hole is 1256 mm2, the velocity is 1592 mm/s. COUPOR,21,22,21,PORFLOW,21,,1.0 PORFLOW,21,,XVEL,-1592.,DENSITY,8.5425E-7,FLOW,IN,+ +,YVEL,0.,ZVEL,0.,MATERIAL,2,,,+ +,SIE,0 SUBSURF,21,201,PROP,251 SET1,251, 4

Note:

1. In the case of material flow into a multi-material Euler mesh, the density and specific energy have to be set. 2. Prescribing both pressure and velocity may lead to the instabilities.

2) Outlet The flow card of outlet is referenced from the third coupling surface. COUPOR,31,33,31,PORFLOW,31,,1.0 PORFLOW,31,,MATERIAL,1,DENSITY,1.29e-9,SIE,1.938e+11,+ +,PRESSURE,100. SUBSURF,31,301,PROP,351 SET1,351, 9

Main Index

Chapter 4: Fluid-structural Interaction 231 Fuel Tank Filling

Note:

a) Since tank flow is in general subsonic, a prescribed pressure condition to the flow condition is necessary. The boundary condition without the prescribed pressure actually assumes that flow is supersonic. b) When material flows out of a multi-material Euler mesh, it is assumed that each of the materials present in the outflow Euler element contributes to the out flow of mass. The materials are transported in proportion to their relative volume fractions.

• Miscellaneous

a) Fast coupling is to used: PARAM,FASTCOUP

b) Gravity is applied to the whole model: TLOAD1 GRAV

1 444

444

0 -9800

1

c) Since the tank is stationary and rigid, large subcycling intervals are used to offer significant CPU savings: PARAM,COSUBMAX,1000 PARAM,COSUBCYC,100

d) To show the behavior of fuel and air, the following output request was added: TYPE (euler) = ARCHIVE ELEMENTS (euler) = 1 SET 1 = ALLMULTIEULHYDRO ELOUT (euler) = XVEL,YVEL,ZVEL,PRESSURE, FMAT1,DENSITY1,MASS1,SIE1, FMAT2,DENSITY2,MASS2,SIE2, FMAT ,DENSITY ,MASS ,SIE TIMES (euler) = 0 THRU END BY 10E-3 SAVE (euler) = 10000

e) In order to output results of the flow boundaries, a history request is created: TYPE SUBSURFS SET 3 SUBSOUT STEPS SAVE

(SUB) = TIMEHIS (SUB) = 3 = 21, 31 (SUB) = PRESSURE,AREA,XVEL,YVEL,ZVEL (SUB) = 0 THRU END BY 100 (SUB) = 100000

Results Figure 4-49 show isosurfaces of the fuel and air. The images are created with CEI.Ensight. Figure 4-50 show time history curves of the velocities on the flow boundaries. XVEL-SUB21 is the X-Velocity of the inlet and XVEL-SUB31 is the X-Velocity of the outlet. The outflow velocity is much higher, because the outlet vent is small. At 1.45 seconds fuel starts to vent out.

Main Index

232 Dytran Example Problem Manual Fuel Tank Filling

Figure 4-49

Main Index

Isosurfaces of FMAT

Chapter 4: Fluid-structural Interaction 233 Fuel Tank Filling

Figure 4-50

Velocities on the Flow Boundaries

Abbreviated Dytran Input File $ UNIT: mm/Kg/s/K START TIME=99999 CEND ENDTIME=2000E-3 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: tank_filling TLOAD=1 TIC=1 SPC=1 $ Output result for request: euler TYPE (euler) = ARCHIVE ELEMENTS (euler) = 1 SET 1 = ALLMULTIEULHYDRO ELOUT (euler) = XVEL,YVEL,ZVEL,PRESSURE, FMAT1,DENSITY1,MASS1,SIE1, FMAT2,DENSITY2,MASS2,SIE2, FMAT ,DENSITY ,MASS ,SIE TIMES (euler) = 0 THRU END BY 10E-3 SAVE (euler) = 10000 $ Output result for request: Euler Boundary TYPE (SUB) = TIMEHIS SUBSURFS (SUB) = 3 SET 3 = 21, 31 SUBSOUT (SUB) = PRESSURE,AREA,XVEL,YVEL,ZVEL STEPS (SUB) = 0 THRU END BY 100 SAVE (SUB) = 100000 $ TYPE (Shells) = ARCHIVE ELEMENTS (Shells) = 8 SET 8 = ALLDUMQUAD ELOUT (Shells) = ZUSER

Main Index

234 Dytran Example Problem Manual Fuel Tank Filling

TIMES (Shells) = 0 THRU END BY 10.E-3 SAVE (Shells) = 100000 $ $------- Parameter Section -----PARAM,FASTCOUP, PARAM,INISTEP,1E-7 PARAM,MINSTEP,1E-8 PARAM,COSUBMAX,1000 PARAM,COSUBCYC,100 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE tank.bdf $ $1111111222222223333333344444444555555556666666677777777 TLOAD1 1 444 0 GRAV 444 -9800 1 $ ========== PROPERTY SETS ========== $ $ * Shell_1 * $ PSHELL1 4 DUMMY PSHELL1 5 DUMMY PSHELL1 6 DUMMY PSHELL1 7 DUMMY PSHELL1 8 DUMMY PSHELL1 9 DUMMY PSHELL1 10 DUMMY PSHELL1 11 DUMMY PSHELL1 12 DUMMY PSHELL1 13 DUMMY PSHELL1 14 DUMMY PSHELL1 15 DUMMY PSHELL1 16 DUMMY $ $ $ -------- Material air_mat id =1 $ DMAT 1 1.29E-9 1 EOSGAM 1 1.4 $ $ -------- Material oil_mat id =2 $ DMAT 2 8.5E-7 2 EOSPOL 2 20000 $ $ ======== Load Cases ======================== $ RIGID,1,1,10. SURFACE,1,,PROP,1 SET1,1,5,THRU,8,10,THRU,16 $---------Tload1-tank---------------TLOAD1,1,1,,12 FORCE,1,1,,0.,1.0,1.0,1.0 MOMENT,1,1,,0.,1.0,1.0,1.0

Main Index

Chapter 4: Fluid-structural Interaction 235 Fuel Tank Filling

$ $-----------------------------Domain 1-----------------------------PEULER1,1,,MMHYDRO,100 $ MESH,1,ADAPT,8.,8.,8.,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,1 $ COUPLE,1,101,OUTSIDE,,,,,,+ +,,,,,,,,,+ +,,1 $ SURFACE,101,,PROP,101 SET1,101,7,8,12,THRU,16 $ $---------Euler initial condition--------------TICEUL,100,,,,,,,,+ +,CYLINDER,31,1,21,1.0,,,,+ +,CYLINDER,32,2,22,2.0,,,,+ +,CYLINDER,33,2,23,3.0,,,,+ +,CYLINDER,34,2,24,4.0,,,,+ +,CYLINDER,35,2,25,5.0 $ CYLINDER,31,,-350.,150.,-10000.,50.,150.,-10000.,+ +,20000. CYLINDER,32,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10020. CYLINDER,33,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10040. CYLINDER,34,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10060. CYLINDER,35,,-350.,150.,-10000.,50.,150.,-10000.,+ +,10080. $ TICVAL,21,,DENSITY,1.29E-9,SIE,1.938E11 TICVAL,22,,DENSITY,8.54254E-7,SIE,0 TICVAL,23,,DENSITY,8.5426E-7,SIE,0 TICVAL,24,,DENSITY,8.54268E-7,SIE,0 TICVAL,25,,DENSITY,8.54275E-7,SIE,0 $ $-----------------------------Domain 2-----------------------------PEULER1,2,,MMHYDRO,200 $ MESH,2,ADAPT,8.,8.,8.,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,2 $ COUPLE,2,201,OUTSIDE,,,22,,,+ +,,,,,,,,,+ +,,2 $ SURFACE,201,,PROP,201 SET1,201,4,THRU,8 $

Main Index

236 Dytran Example Problem Manual Fuel Tank Filling

COUPOR,2,22,2,PORFLCPL,2,,1.0 PORFLCPL,2,,,BOTH,1 SUBSURF,2,201,PROP,250 SET1,250,8 $ COUPOR,21,22,21,PORFLOW,21,,1.0 PORFLOW,21,,XVEL,-1592.,DENSITY,8.5425E-7,FLOW,IN,+ +,YVEL,0.,ZVEL,0.,MATERIAL,2,,,+ +,SIE,0 SUBSURF,21,201,PROP,251 SET1,251,4 $ $---------Euler initial condition--------------TICEUL,200,,,,,,,,+ +,CYLINDER,51,2,41,1.0,,,,+ +,CYLINDER,52,1,21,2.0 $ CYLINDER,51,,-150.,150.,-10000.,250.,150.,-10000.,+ +,10260. CYLINDER,52,,-150.,150.,-10000.,250.,150.,-10000.,+ +,10205. $ TICVAL,41,,DENSITY,8.5425E-7,SIE,0 $-----------------------------Domain 3-----------------------------PEULER1,3,,MMHYDRO,300 $ MESH,3,ADAPT,8.,8.,8.,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,3 $ COUPLE,3,301,OUTSIDE,,,33,,,+ +,,,,,,,,,+ +,,3 $ SURFACE,301,,PROP,301 SET1,301,9,THRU,14 $ COUPOR,3,33,3,PORFLCPL,3,,1.0 PORFLCPL,3,,,BOTH,1 SUBSURF,3,301,PROP,350 SET1,350,14 $ COUPOR,31,33,31,PORFLOW,31,,1.0 PORFLOW,31,,MATERIAL,1,DENSITY,1.29e-9,SIE,1.938e+11,+ +,PRESSURE,100. SUBSURF,31,301,PROP,351 SET1,351,9 $ $---------Euler initial condition--------------TICEUL,300,,,,,,,,+ +,CYLINDER,51,1,21,1.0 $ $ ENDDATA

Main Index

Chapter 4: Fluid-structural Interaction 237 Water Pouring into a Glass

Water Pouring into a Glass

Problem Description This problem demonstrates the use of multiple Euler domains that can interact with each other when porosity is defined to the coupling surfaces associated with them. This problem simulates the water pouring into a glass. A bottle that is partially filled with water is rotated allowing water to flow out into a glass. The Euler domains in this model handle multiple hydrodynamic materials (air and water).

Dytran Modeling The bottle and the glass are modeled with cquad4 elements. The elements of the bottle and the glass are modeled as rigid using MATRIG material model. The bottle is partially filled with water and air. The C.G of the bottle is constrained in global X, Y, Z translations, and X, Y rotations. An enforced rotation about global Z direction is defined. The glass is fixed in space in all directions and is initialized with air. A gravitational force defined in the negative Y direction applies to the entire model. The material properties used in the model are listed below. Glass and Bottle: Density: 0.9E-6 kg/mm3 Young's Modulus: 1.4 GPa Poisson's Ratio: 0.4 Water: Density: 1E-6 kg/mm3 Bulk Modulus: 2.2 GPa

Main Index

238 Dytran Example Problem Manual Water Pouring into a Glass

Air: Density: 1.28E-9 kg/mm3 Gas Constant: 1.4 The bulk stiffness scaling technique is used to speed up the calculation. By reducing the water bulk modulus by a factor of 1000, Dytran uses a larger time-step for the problem reducing computational cost. There are three coupling surfaces associated with three Euler domains in the model. The first coupling surface is shown in Figure 4-51. This coupling surfaces is used to simulate fluid outside the bottle and glass so Euler elements inside the coupling surface should not be processed and the COVER in the COUPLE definition is set to INSIDE. The couple card refers to a mesh number. The first mesh for the Euler elements is created and initialized by PEULER1,2,,MMHYDRO,13 MSEH,10,BOX,,,,,,,+ +,-63,-250,-65,147,300,130,,,+ +,21,42,19,,,,EULER,2

Figure 4-51

Coupling Surface 1

The second coupling surface is shown in Figure 4-52. This coupling surface is used to model the fluid inside the bottle. So all elements outside the coupling surface should not be processed and the COVER in the COUPLE definition is set to OUTSIDE

Main Index

Chapter 4: Fluid-structural Interaction 239 Water Pouring into a Glass

The couple card refers to a mesh number. The second Euler domain is an adaptive Euler domain created and initialized by PEULER1,3,,MMHYDRO,12 MSEH,11,ADAPT,7,7,7,,,,+ +, , , , , , ,,,+ +, , ,,,,,EULER,3,+ +,NONE

Figure 4-52

Coupling Surface 2

The third coupling surface and its Euler domain are shown in Figure 4-53. This coupling surface is used to model the fluid inside the glass. So all elements outside the coupling surface should not be processed and the COVER in the COUPLE definition is set to OUTSIDE. The couple card refers to a mesh number. The third Euler domain is an adaptive Euler domain created and initialized by PEULER1,2,,MMHYDRO,13 MESH,12,ADAPT,7,7,7,,,,+ +, , , , , , ,,,+ +, , ,,,,,EULER,2,+ +,NONE

Main Index

240 Dytran Example Problem Manual Water Pouring into a Glass

Figure 4-53

Coupling Surface 3

Coupling Surface 1 interacts with Coupling surface 2 through porosity defined to the bottle top through COUPOR/PORFLCPL/SUBSURF entries below. The elements of the coupling surface that define porosity are defined in the SET1 entry below. PORFLCPL,95,,,BOTH,9 COUPOR,15,30,46,PORFLCPL,95,,1.0 SUBSURF,46,1,ELEM,59 SET1,59,12,THRU,63,317,THRU,368,501,+ +,THRU,552,685,THRU,736

Coupling Surface 3 interacts with Coupling surface 1 through porosity defined to the glass top through COUPOR/PORFLCPL/SUBSURF entries below. The elements of the coupling surface that define porosity are defined in the SET1 entry below. PORFLCPL, 96,,,BOTH,8 COUPOR,16,31,47,PORFLCPL,96,,1.0 SUBSURF,47,3,ELEM,60 SET1,60,3358,THRU,3490

The Euler element output is requested using ALLMULTIEULHYDRO. Also since adaptive Euler mesh is used, SAVE=1 is used as shown below. TYPE (euler) = ARCHIVE ELEMENTS (euler) = 12 SET 12 = ALLMULTIEULHYDRO ELOUT (euler) = DENSITY PRESSURE SIE FMAT2 FMAT3 TIMES (euler) = 0.0,THRU,END,BY,10 SAVE (euler) = 1

Main Index

Chapter 4: Fluid-structural Interaction 241 Water Pouring into a Glass

Results The simulations results at 0, 1.0 and 2.0 seconds are shown in the below figures. FMAT of water was used to create the isosurface .

Main Index

Figure 4-54

Isosurface of FMAT at 0.0s

Figure 4-55

Isosurface of FMAT at 1.0s

242 Dytran Example Problem Manual Water Pouring into a Glass

Figure 4-56

Isosurface of FMAT at 2.0s

Abbreviated Dytran Input File START CEND ENDTIME=2000 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: water pouring in to a glass TLOAD=1 TIC=1 SPC=1 TYPE(detail) = ARCHIVE ELEMENTS(detail) = 2 SET 2 = ALLSHQUAD TIMES(detail) = 0.0,THRU,END,BY,10 ELOUT(detail) = MASS SAVE(detail) = 1000 $ $TYPE (euler) = ARCHIVE ELEMENTS (euler) = 12 SET 12 = ALLMULTIEULHYDRO ELOUT (euler) = DENSITY SIE PRESSURE FMAT FVUNC FMATPLT , VOLUME XVEL YVEL ZVEL XMOM YMOM ZMOM TEMPTURE FMAT2 FMAT3 TIMES (euler) = 0.0,THRU,END,BY,10 SAVE (euler) = 1 $ TYPE (restat) = RESTART TIMES (restat) = 0 THRU END BY 2000 SAVE (restat) = 1

Main Index

Chapter 4: Fluid-structural Interaction 243 Water Pouring into a Glass

$------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,FASTCOUP PARAM,INISTEP,1e-3 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE water_pouring1.bdf INCLUDE glass.dat $ ------- GRAVITATION ----TLOAD1 1 444 0 GRAV 444 -0.0098 1 $ $ ========== PROPERTY SETS ========== $ $ * prop.1 * $ PSHELL 1 1 1.5 $ $ * outer_euler * $ PEULER1 2 MMHYDRO 13 $ $ * inner_euler * $ PEULER1 3 MMHYDRO 12 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material DMATEP.1 id =1 MATRIG 1 9e-07 1.4 .4 $ $ -------- Material air id =2 DMAT 21.28e-09 2 EOSGAM 2 1.4 $ $ -------- Material water id =3 DMAT 3 1e-06 3 $ $ Due to bulk scaling bulk modulus is reduced by a factor of 1000. $ This results in larger time step and lower computational cost. $ EOSPOL 3 2.2e-03 $EOSPOL 3 2.2 $ $ ======== Load Cases ======================== $ $ $ ------- TICVAL BC air_ini ----TICVAL 5 DENSITY1.28e-09 SIE 194000 $ $ ------- TICVAL BC water_ini ----TICVAL 6 DENSITY 1e-06

Main Index

244 Dytran Example Problem Manual Water Pouring into a Glass

$ $ ------$ COUPLE +A000001 +A000002 $ SURFACE SET1 +A000003 + $ $ ------$ COUPLE +A000004 +A000005 $ SURFACE SET1 +A000006 +A000007 $ $ ------$ COUPLE +A000004 +A000005 $ SURFACE SET1 $ $ ------$ MESH +A000008 +A000009 $ $ ------$ MESH +A000010 +A000011 +A000012 $ $ ------$ MESH +A000010 +A000011 +A000012 $ $ ------TICEUL

Main Index

General Coupling: couple_bottle_glass ----8

1

INSIDE

ON

ON

30

ELEM THRU 582 3576

1 2704 THRU

1744 1162

THRU 1

STANDARD+A000001 +A000002

10 1 1 THRU 3000

2325 1743 THRU

2324 THRU

1163+A000003 581+

General Coupling: couple_bottle----9

2 OUTSIDE

ON

ON

2 2704 THRU

1945 1363

STANDARD+A000004 +A000005

11 2 2 THRU 1

2526 1944 THRU

ELEM THRU 783 201

THRU 202

2525 THRU

1364+A000006 782+A000007

General Coupling: couple_glass ----10

3 OUTSIDE

ON

ON

31

STANDARD+ +

12 3 3

3000

ELEM THRU

3 3576

Mesh Box: outer_box_euler_mesh_for_bottle_glass 10 -63 21

BOX -250 42

-65 19

147

300

+A000008 +A000009

130 EULER

2

EULER

+A000010 +A000011 3+A000012

EULER

+A000010 +A000011 2+A000012

Mesh Adap: inner_adapt_euler_mesh_for_bottle 11

ADAPT

7

7

7

NONE Mesh Adap: inner_adapt_euler_mesh_for_glass 12

ADAPT

7

7

7

NONE TICEUL BC inner_reg_def ----12

+A000013

Chapter 4: Fluid-structural Interaction 245 Water Pouring into a Glass

+A000013 SPHERE 3 2 5 1 +A000014 BOX 4 3 6 2 SPHERE 3 0 0 0 1000 BOX 4 -80 -145 -60 270 133 $ $ ------- TICEUL BC outter_reg_def ----TICEUL 13 +A000016 SPHERE 2 2 5 1 SPHERE 2 0 0 0 1000 $ $ ------- Rigid Body Object rbo ----$ ---- No reference node is used. $ ---- enforced rotational velocity of the C.G of the bottle TLOAD1 1 16 12 FORCE 16 MR1 1 0 0 0 TLOAD1 1 1016 12 6 MOMENT 1016 MR1 1 0 0 1 $ $ ------- Rigid Body Object rbo1 ----$ ---- No reference node is used. $ ---- fix the glass in space in all directions TLOAD1 1 1 12 FORCE 1 MR4 1 0 0 0 TLOAD1 1 1001 12 MOMENT 1001 MR4 1 0 0 0 $ ================ TABLES ================= $ $ ------- TABLE 6: rotational_table ------TABLED1 6 +A000017 0 .0379 40 .0379 40.0001 0 2000 +A000018 ENDT $ $ $ Porosity of the botte top $ PORFLCPL,95,,,BOTH,9 COUPOR,15,30,46,PORFLCPL,95,,1.0 SUBSURF 46 1 ELEM 59 SET1 59 12 THRU 63 317 THRU 368 + THRU 552 685 THRU 736 $ $ Porosity for the glass top $ PORFLCPL,96,,,BOTH,8 COUPOR,16,31,47,PORFLCPL,96,,1.0 SUBSURF 47 3 ELEM 60 SET1 60 3358 THRU 3490 $ ENDDATA

Main Index

+A000014

120

+A000016

+A000017 0+A000018

501+

246 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

Fluid Flow through a Straight Pipe Problem Description This problem demonstrates a viscous fluid flow through a pipe with a prescribed inlet flow and with prescribed pressure at the end of the pipe. The pipe is modeled by two approaches, see Figure 4-57. • Model 1: Rectangular Euler mesh with coupling surface • Model 2: A general Euler mesh. The boundary of the Euler mesh forms the pipe.

Figure 4-57 Model Setup Fluid properties are: 1. Reference density: 9.53E-5 lbf/inch3 2. EOSTAIT with A0=0, A1=2E+5, G=1 3. Initial pressure 0 psi 4. Dynamic viscosity of 0.001 lb.s/inch2 The pipe has a diameter of 0.844 inch and a length of 17.1 inches. At the inflow boundary, a constant velocity of 200 inch/s is prescribed. The inflow material has a density of 9.76E-5 lbf/inch3. At the outflow boundary, a pressure of 0 psi is used.

Main Index

Chapter 4: Fluid-structural Interaction 247 Fluid Flow through a Straight Pipe

Theoretical Analysis To obtain a good representation of the viscous boundary layer, there should be at least a few elements across its thickness. To estimate this size, the Reynolds number is computed. This is given by –5

de nsi t y *di am et e r* v e l oc it y 9.52*10 *0.44*200 R e = ---------------------------------------------------------------------------- = -------------------------------------------------- = 16 d yd na mi c – vi s cos i ty 0.001

This means that the size of the boundary layer is in the order of

10.84 ---------di am et e r = ---------4 Re

Inch=0.2 inch.

Therefore, it is acceptable to use about 10 elements across the diameter. If the Reynolds number had been larger, then more elements would have been needed to capture the viscous boundary layer. note that in flow with high Reynolds numbers, it is not always necessary to capture the viscous boundary layer. Model 1 will be meshed with 11*11*80 Euler elements. Model 2 will be meshed by 2900 Euler elements. 4

The flow rate

φ

is coupled to the pressure difference across the pipe

while the volumetric flow rate itself is defined by

φ = uπ r

2

ΔP

by Poiseuille's law:

πr φ = --------- Δ P , 8 μl

. Hence substitution yields

8 μl 8 μl 8*0.001*17.1 - u = --------------------------------- 200 = 153.6 p si Δ P = --------4- φ = -------2 0.422*0.422 πr r

Here v denoted dynamic viscosity, r the radius, l the length, u the inflow velocity. The shear stress is given by rΔ P 0.422*153.6 τ yz = τ zx = ---------- = ------------------------------ = 1.896p si 2*1.71 2l

The resulting axial wall force exerted by the fluid on the pipe is: 2

F = π r Δ P = 85.93l bf

Dytran Model Model 1: The pipe is modeled as a coupling surface and the viscous fluid by a block Eulerian mesh. In addition, the pipe surface is modeled as rigid: RIGID SURFACE TLOAD1 FORCE MOMENT $

1 12 1 1 1

12

1. ELEM

1 1 1

3 12 0. 0.

1 1

1 1

The viscous fluid is modeled as Eulerian material using EOSTAIT DMAT EOSTAIT

Main Index

19.53e-05 1 0

1 200000

19.529e-5

.001

1 1

248 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

and the Euler mesh is created by MESH: MESH +A000004 +A000005 $

9 -.45 11

BOX -.45 11

-.05 80

.9

.9

+A000004 +A000005

17.2 EULER

2

Initialization of Eulerian element is done via TICEUL and TICVAL TICVAL 10 DENSITY9.53e-05 $ $ ------- TICEL BC Simple ----$ $ ------- TICEUL BC Region ----TICEUL 13 +A000006 SPHERE 12 1 10 SPHERE 12 0 0 $

+A000006 1 0

20

To couple the Eulerian material to the pipe, the COUPLE card is used: $ COUPLE 8 1 OUTSIDE +A000001 +A000002 9 $ SURFACE 1 SUB $ $ General Coupling Subsurface: SUB1 $ SUBSURF 1 1 ELEM SET1 3 1 THRU $ SUBSURF 2 1 ELEM SET1 4 457 THR $ $ SUBSURF 3 1 ELEM SET1 5 515 THRU

ON

ON

8

1

SUB

2

STANDARD+A000001 +A000002

SUB

3

3 456 4 514

5 572

Since the Eulerian material is inside the pipe, the COVER field on the COUPLE card is OUTSIDE. Boundary conditions are imposed by the PORFLOW entry $ COUPOR $ PORFLOW +A000003 $ SUBSURF SET1

Main Index

1

8

2 PORFLOW

1 XVEL

0

METHODVELOCITY YVEL 0

2 4

1 457

ELEM THRU

4 514

1CONSTANT FLOW ZVEL

1

IN DENSITY9.76e-05+A000003 -200MATERIAL 1

Chapter 4: Fluid-structural Interaction 249 Fluid Flow through a Straight Pipe

Note that COUPOR is referenced from the COUPLE card $ General Coupling Subsurface: SUB3 $ COUPOR 2 8 3 PORFLOW $ PORFLOW 2 METHODVELOCITY +A000009MATERIAL 1 $ SUBSURF 3 1 ELEM 5 SET1 5 515 THRU 572

2CONSTANT FLOW

1

OUTPRESSURE 0.

+A000009

The time histories of inflow pressure and velocity show significant oscillations. They occur because the fluid inside the pipe is initially at rest and is subjected to a step-change of inflowing material. To avoid oscillatory transients, the inflow can be made time dependent. Inflow starts at zero and grows linearly until it reaches the stationary target value. The time at which the switch from linear to constant takes place should be sufficiently large to enable flow to settle down. This time is generally given by Le ng th 4 * --------------------------------- = 0.0015 . sou nd spe e d

This is the time it takes for acoustic waves to travel four times the length of the

pipe. This yields the PORFLOWT model: COUPOR $ PORFLOWT + + TABLED1 +

1

8

2PORFLOWT

1 IN TABLE 100 1CONSTANT 9.76e-5CONSTANT 100 0.0 0.0 0.0015 200

1CONSTANT

1 + +

0.0 + 1.0

200.0

Model 2: Euler elements are defined by CHEXAs, and boundary conditions are imposed by use of FLOW. This model has also been run with a time depended flow model by replacing the inflow FLOW by a FLOWT model: FLOWT + + TABLED1 +

7 3 IN TABLE 100 1CONSTANT 9.76e-5 100 0.0 0.0 0.0015

+ + + 200

1.0

200.0

Results Results and comparisons to theory are shown for the standard HYDRO solver, MMHYDRO solver, and the Roe solver. Also shown is that by using time dependent flow models oscillations in pressure and velocity are significantly reduced. As a result, the steady state flow condition is reached within a much shorter problem time. See example on the following page.

Main Index

250 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

Main Index

Chapter 4: Fluid-structural Interaction 251 Fluid Flow through a Straight Pipe

Main Index

252 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

Main Index

Chapter 4: Fluid-structural Interaction 253 Fluid Flow through a Straight Pipe

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254 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

Main Index

Chapter 4: Fluid-structural Interaction 255 Fluid Flow through a Straight Pipe

Main Index

256 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

Main Index

Chapter 4: Fluid-structural Interaction 257 Fluid Flow through a Straight Pipe

Main Index

Figure 4-58

Model 1 with Time Dependent Flow and HYDRO Solver Inflow and Outflow Results

Figure 4-59

Model 2 with Time Dependent Flow and HYDRO Solver Inflow and Outflow Results

258 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

Figure 4-60

Model 1 with Time Dependent Flow Model: Axial Force

Table 4-2

Force on the Pipe, TYX is extrapolated to the Wall Pressure Diff

TYZ

THEORY

153.6

1.896

85.93

HYDRO

159.2 (+3.6%)

1.98 (+4.4%)

N.A.

HYDRO (FSI)

151.1 (-1.6%)

1.84 (-3.0%)

83.38 (-3.0%)

MMHYDRO

159.1 (+3.6%)

1.98 (+4.4%)

N.A.

MMHYDRO (FSI)

151.1 (-1.6%)

1.84 (-3.0%)

83.38 (-3.0%)

HYDRO – ROE

170.2 (+10.8%)

1.91 (+0.7%)

N.A.

HYDRO (FSI) – ROE

165.9 (+8.0%)

1.887 (-1.3%)

85.46 (-0.5%)

Abbreviated Dytran Input File Model 1 START CEND ENDTIME=0.02 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: Cyl TLOAD=1 TIC=1 SPC=1 $ Output result for request: elem TYPE (elem) = ARCHIVE ELEMENTS (elem) = 1 SET 1 = ALLEULHYDRO

Main Index

Z-Force

Chapter 4: Fluid-structural Interaction 259 Fluid Flow through a Straight Pipe

ELOUT (elem) = XVEL YVEL ZVEL DENSITY SIE PRESSURE Q TXY TYZ TZX TIMES (elem) = 0 THRU END BY 0.005 SAVE (elem) = 10000 $ Output result for request: ths TYPE (ths) = TIMEHIS ELEMENTS (ths) = 2 SET 2 = 633,637,10192,10196 ELOUT (ths) = XVEL YVEL ZVEL DENSITY SIE PRESSURE TZX TYZ TIMES (ths) = 0 THRU END BY 0.0001 SAVE (ths) = 10000 $ TYPE (Coup) = TIMEHIS RIGIDS(Coup) = 3 SET 3 = 1 RBOUT (Coup) = XFORCE YFORCE ZFORCE TIMES (Coup) = 0 THRU END BY 0.0001 SAVE (Coup) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1.E-7 PARAM,FASTCOUP,INPLANE $------- BULK DATA SECTION ------BEGIN BULK $ $ --- Define 574 grid points --$ GRID 1 .422000 .00000 .00000 .. .. GRID 641 .319874-.177903 .00000 $ $ --- Define 572 elements $ $ -------- property set pipe --------CQUAD4 1 1 1 2 27 26 .. .. CQUAD4 572 1 617 618 641 616 $ $ ========== PROPERTY SETS ========== $ $ * pipe * $ PSHELL1 1 DUMMY $ $ * Peuler * $ PEULER1 2 HYDRO 13 $ $ RIGID 1 12 1. SURFACE 12 ELEM 3 TLOAD1 1 1 12 FORCE 1 1 0. 1 1 1

Main Index

260 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

MOMENT 1 1 0. 1 1 1 $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material euler id =1 DMAT 19.53e-05 1 EOSTAIT 1 0 200000 19.529e-5 .001 $ $ ======== Load Cases ======================== $ $ $ ------- General Coupling: Couple ----$ COUPLE 8 1 OUTSIDE ON ON 8 STANDARD+A000001 +A000001 +A000002 +A000002 9 $ SURFACE 1 SUB 1 SUB 2 SUB 3 $ $ General Coupling Subsurface: SUB1 $ SUBSURF 1 1 ELEM 3 SET1 3 1 THRU 456 $ $ General Coupling Subsurface: SUB2 $ COUPOR 1 8 2 PORFLOW 1CONSTANT 1 $ PORFLOW 1 METHODVELOCITY FLOW IN DENSITY9.76e-05+A000003 +A000003 XVEL 0 YVEL 0 ZVEL -200MATERIAL 1 $ SUBSURF 2 1 ELEM 4 SET1 4 457 THRU 514 $ $ General Coupling Subsurface: SUB3 $ COUPOR 2 8 3 PORFLOW 2CONSTANT 1 $ PORFLOW 2 METHODVELOCITY FLOW OUTPRESSURE 0. +A000009 +A000009MATERIAL 1 $ SUBSURF 3 1 ELEM 5 SET1 5 515 THRU 572 $ $ ------- Mesh Box: MESH1 $ MESH 9 BOX +A000004 +A000004 -.45 -.45 -.05 .9 .9 17.2 +A000005 +A000005 11 11 80 EULER 2 $ $ ------- TICVAL BC INIT ----TICVAL 10 DENSITY9.53e-05 $

Main Index

Chapter 4: Fluid-structural Interaction 261 Fluid Flow through a Straight Pipe

$ ------- TICEL BC Simple ----$ $ ------- TICEUL BC Region ----TICEUL 13 +A000006 SPHERE 12 1 SPHERE 12 0 $ $ ENDDATA

+A000006 10 0

1 0

Model2 START CEND ENDTIME=2.E-2 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: cyls TLOAD=1 TIC=1 SPC=1 $ Output result for request: elem TYPE (elem) = ARCHIVE ELEMENTS (elem) = 1 SET 1 = 1 THRU 2900 ELOUT (elem) = XVEL YVEL ZVEL DENSITY PRESSURE TYZ TZX TIMES (elem) = 0 THRU END BY 0.01 , 0.005 SAVE (elem) = 10000 $ Output result for request: ths TYPE (ths) = TIMEHIS ELEMENTS (ths) = 2 SET 2 = 1 47 2843 2889 ELOUT (ths) = XVEL YVEL ZVEL DENSITY SIE PRESSURE TIMES (ths) = 0 THRU END BY 0.0001 SAVE (ths) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1.E-7 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE cyls.bdf $ $ ========== PROPERTY SETS ========== $ $ * PEUL * $ PEULER 1 1 HYDRO $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material eul id =1 DMAT 19.53e-05 1

Main Index

20

262 Dytran Example Problem Manual Fluid Flow through a Straight Pipe

EOSTAIT 1 0 200000 19.529e-5 0.001 $ $ ======== Load Cases ======================== $ $ $ ------- Flow BC INFLOW ----TLOAD1 1 7 4 FLOW 7 3MATERIAL 1 XVEL 0 YVEL +A000001 ZVEL -200 DENSITY9.76e-05 CFACE 1 3 1 1 .. .. CFACE 58 3 58 1 $ $ ------- Flow BC OUTFLOW ----TLOAD1 1 10 4 FLOW 10 4 FLOW OUTMATERIAL 1PRESSURE +A000002 DENSITY9.53e-05 CFACE 59 4 2900 4 .. .. CFACE 116 4 2843 4 $ $ ------- TICEL BC TIC ----SET1 5 1 THRU 523 2790 THRU 2900 +A000003 THRU 2789 1628 THRU 2208 1047 THRU +A000004 524 THRU 576 577 THRU 1046 TICEL 1 5 DENSITY9.53e-05 $ $ ENDDATA

Main Index

0+A000001

0+A000002

2209+A000003 1627+A000004

Chapter 4: Fluid-structural Interaction 263 Using Euler Archive Import in Blast Wave Analyses

Using Euler Archive Import in Blast Wave Analyses In blast wave analyses, the distance between the structure and the blast is often large resulting in excessive CPU time for the blast wave to reach the structure. In addition, the simulation may have to be repeated several times for different structures or for different positions of the structure. To reduce the CPU time, the simulation is split up in two parts. In the first simulation, the structure is omitted. In the subsequent runs, the result can be imported into a simulation which includes the structure. The blast wave will almost immediately hit the structure at the start of the run, without using a lot of cpu time. In addition, importing the Euler archive also allows mapping the fine mesh to a coarser mesh. This can also save much time. Importing an Euler Archive is done by means of the File Management Section (FMS) entry EULINIT, which is similar to the SOLINIT entry for Prestress Analysis. An import archive should include several required variables and can also include several cycles. On the EULINIT entry, the user specifies which cycles need to be imported. This example demonstrates the method of importing the results of an Euler archive in a blast wave analysis. Three simulations will be shown. The first run simulates the propagation of a blast wave only. In the two follow up runs, the structure is added and the simulation is started at the time when the blast wave starts to hit the structure. The first follow up run uses the same Euler mesh and the second run uses a coarser mesh.

Problem Description A blast wave hits a box shaped shell structure. A 2-D model is used with an Euler mesh of 10 x 10 m and an element thickness of 0.24 m. The shock front is located at the center of the model and it has an initial radius of R0 = 1 m at time t = 0 seconds. The initial conditions are: Explosive Properties (r < R0) Specific Internal Energy = 9.E+5 Joule/kg Density = 1 kg/m3 Environment (r > R0) Specific Internal Energy = 3.E+5 Joule/kg Density = 1 kg/m3

Dytran Model Three models will be used. The first two models will use an Euler mesh that is created by MESH,1,BOX,,,,,,,+ +,0,-0.12,0.0,10,0.24,10,,,+ +,40,1,40,,,,EULER,1

Main Index

264 Dytran Example Problem Manual Using Euler Archive Import in Blast Wave Analyses

The Euler mesh of the third model is created by MESH,1,BOX,,,,,,,+ +,0,-0.12,0.0,10,0.24,10,,,+ +,30,1,30,,,,EULER,1

The models are summarized as follows: 1. An initial run without structure 2. A second run using results from the first run by means of the EULER import functionality. The structure is included. 3. The same as model 2 but now the results of the initial run (without structure) will be imported and mapped onto a coarser mesh.

General Setup The following entries apply to all three models. The solver, material and initialization are given as PEULER1,1,,MMHYDRO,19 DMAT, 100, 1, 2 EOSGAM,2,1.4 PARAM, BULKL,0.1 $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,6.0 SPHERE,3,,5.0,0.0,5.0,10000 SPHERE,4,,5.0,0.0,5.0,1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,1,sie,9e+5 PARAM,MICRO,30

Non-reflecting boundary conditions will be applied FLOWDEF,24,,MMHYDRO

To speed up the coupling surface computation the following PARAM’s are added PARAM, COSUBMAX,100 PARAM,COSUBCYC,100

Specific Setup for Model 1 The first model is stopped when the blast wave front reaches the model boundary. This occurs after about 100 cycles. In the output request the variables DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL,FMAT,MASS

have to be included to assure a full continuation in the follow up run.

Main Index

Chapter 4: Fluid-structural Interaction 265 Using Euler Archive Import in Blast Wave Analyses

Specific Setup for Model 2 and Model 3 For the follow up runs the input file of model 1 is used with the following additions: To import the Euler archive GAS2_ALLEULER_0.ARC that was created by the initial model 1 run, an EULINIT entry is added to the File Management Section START EULINIT,GAS2_ALLEULER_0.ARC,60 CEND

As a result Cycle 60 will be imported. The follow-up runs are only run for one additional cycle to check if the import of the result variables is done correctly. ENDSTEP = 61

The structure which is hit by the blast wave is modeled with dummy CQUAD4 elements. GRID

1

7.50001-5.49900 7.4999

CQUAD4 CQUAD4

5 6

2 2

4 3

8 6

6 5

3 1

To enable interaction between blast wave and structure, a closed coupling surface is needed. This coupling surface serves as a transmitter of data between the Euler solver and the Lagrange solver. The shell element of the structure will be taken for that COUPLE,100,200,INSIDE,ON,ON SURFACE,200,,PROP,2 PSHELL1 2 DUMMY SET1,2,2

Main Index

266 Dytran Example Problem Manual Using Euler Archive Import in Blast Wave Analyses

Results Model 1: Only the Blast wave and No Structure The following three figures show the pressure distribution at cycles 3, 60, and 100.

Main Index

Chapter 4: Fluid-structural Interaction 267 Using Euler Archive Import in Blast Wave Analyses

The pressure profiles along the X-axis across the center of the model is shown in the following picture.

Main Index

268 Dytran Example Problem Manual Using Euler Archive Import in Blast Wave Analyses

Model 2: Follow-up Run with Structure and Using the Same Mesh The pressure distribution at cycle 60 (the first cycle in this run) is shown in the following picture.

The pressure profile along the X-axis across the center point of the model is shown in the following picture.

Main Index

Chapter 4: Fluid-structural Interaction 269 Using Euler Archive Import in Blast Wave Analyses

Model 3: Follow Run-up with Structure and Using a Coarser Mesh The pressure distribution at cycle 60 (the first cycle in this run) is shown in the following picture .

The pressure profile along the X-axis across the center point of the model is shown in the following picture.

Main Index

270 Dytran Example Problem Manual Using Euler Archive Import in Blast Wave Analyses

Abbreviated Dytran Input File $---Blast wave analysis-----------------START CEND ENDSTEP = 100 CHECK=NO TITLE= Blast Wave Analysis TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS STEPS(ALLEULER) = 0,THRU,END,BY,3 SAVE (ALLEULER) = 10000 $ TYPE (ARCMAT) = TIMEHIS MATS (ARCMAT) = 15 SET 15 = 100 MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOM STEPS (ARCMAT) = 0,THRU,END,BY,10 SAVE (ARCMAT) = 99999 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK PARAM,BULKL,0.1 PARAM,COSUBMAX,100 PARAM,COSUBCYC,100 $ -----------------------------------------------------------------$ PARAM,MICRO,30 PEULER1,1,,MMHYDRO,19 $ $ DMAT, 100, 1, 2 EOSGAM,2,1.4 $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,6.0 SPHERE,3,,5.0,0.0,5.0,10000 SPHERE,4,,5.0,0.0,5.0,1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,1,sie,9e+5

Main Index

Chapter 4: Fluid-structural Interaction 271 Using Euler Archive Import in Blast Wave Analyses

$ MESH,1,BOX,,,,,,,+ +,0,-0.12,0.0,10,0.24,10,,,+ +,40,1,40,,,,EULER,1 $ FLOWDEF,24,,MMHYDRO $ $ ENDDATA

$----Follow up run 2---------------------------------------------------------START EULINIT,GAS2_ALLEULER_0.ARC,60 CEND ENDSTEP = 61 CHECK=NO TITLE= Blast Wave Analysis TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS STEPS(ALLEULER) = 0,THRU,END,BY,3 SAVE (ALLEULER) = 10000 $ TYPE (ARCMAT) = TIMEHIS MATS (ARCMAT) = 15 SET 15 = 100 MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOM STEPS (ARCMAT) = 0,THRU,END,BY,10 SAVE (ARCMAT) = 99999 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK PARAM,BULKL,0.1 PARAM,COSUBMAX,100 PARAM,COSUBCYC,100 $ -----------------------------------------------------------------$ $ PARAM,FASTCOUP PARAM,MICRO,30

Main Index

272 Dytran Example Problem Manual Using Euler Archive Import in Blast Wave Analyses

PEULER1,1,,MMHYDRO,19 $ $ $ EOSGAM,2,1.4 $ DMAT 100 1 2 $ $ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,6.0 SPHERE,3,,5.0,0.0,5.0,10000 SPHERE,4,,5.0,0.0,5.0,1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,1,sie,9e+5 $ MESH,1,BOX,,,,,,,+ +,0,-0.12,0.0,10,0.24,10,,,+ +,30,1,30,,,,EULER,1 $ $ COUPLE,100,200,INSIDE,ON,ON $ SURFACE,200,,PROP,2 $ PSHELL1 2 DUMMY SET1,2,2 $ $ GRID 1 7.50001-5.49900 GRID 2 8.99999-5.49900 GRID 3 7.50001 5.49900 GRID 4 8.99999 5.49900 GRID 5 7.50001-5.49900 GRID 6 7.50001 5.49900 GRID 7 8.99999-5.49900 GRID 8 8.99999 5.49900 $ $ --- Define 6 elements $ CQUAD4 1 2 1 2 CQUAD4 2 2 5 6 CQUAD4 3 2 5 7 CQUAD4 4 2 2 7 CQUAD4 5 2 4 8 CQUAD4 6 2 3 6 $ ENDDATA

Main Index

7.4999 7.4999 7.4999 7.4999 2.0001 2.0001 2.0001 2.0001

4 8 2 8 6 5

3 7 1 4 3 1

Chapter 4: Fluid-structural Interaction 273 Blast Wave with a Graded Mesh

Blast Wave with a Graded Mesh The effect of a blast wave on a structure will be simulated using the graded mesh technique. To get accurate results for the initial expansion of the blast wave, small mesh-sizes are needed. This is especially true if EOSJWL or EOSIG are used. These Equations of states simulate the actual detonation of the explosive in detail, but they do require a small mesh size in the vicinity of the explosive. This means that this approach is only efficient if the target is close. If the distance between the explosive and target is large, then the simulation takes a lot of CPU time. Moreover, after the initial expansion, the blast wave becomes larger in radius and less steep as it propagates; thus making the finer elements in the mesh less effective to predict the physics of problem. To run the simulation more efficiently, part of the fine mesh can be replaced by a coarser mesh. The initial stage of the expansion takes place in a small fine mesh. This mesh is then glued to a larger mesh with coarser elements by using the graded mesh capability.

Problem Description The effect of a detonation on the environment can be simulated by assuming that the detonated material can be idealized by a sphere of hot gas with a homogenous density and specific internal energy. This approach is suited for problems in which the processes inside the explosive material are not investigated. The technique is called the “Blast Wave” approach. In this example, the propagation of the blast wave will be simulated starting from the initial shock front radius R0 = 1 m at the time t = 0 second until it reaches a radius of about R = 10 R0. During the expansion the blast wave will hit a box structure at a distance of 8.475 m from the center point of the explosion. Both the gas in the sphere and the surrounding environment behave as an ideal gas (Gamma = 1.4). The initial conditions are: Explosive properties (r < R0) Specific internal energy = 9.E+5 Joule/kg Density

= 1 kg/m3

Environment (r > R0) Specific internal energy

= 3.E+5 Joule/kg

Density

= 1 kg/m3

The material of the structure (box) is steel: Density = 7800 kg/m3 Young’s modulus = 2.1E+11 Pa Poison’s ratio = 0.3 Yield Stress = 2.E+8 Pa

Main Index

274 Dytran Example Problem Manual Blast Wave with a Graded Mesh

Dytran Model For the purpose of illustrating the graded mesh technique, a 2-D mesh will be used. A coarse and a fine mesh are created and then glued/connected together. The connecting process requires that one mesh fits nicely into the other. Connecting a coarse mesh with a fine mesh results in a number of nodes that are only part of the fine mesh but not part of the coarse mesh. These free “hanging” nodes are allowed in the model. In Figure 4-61, the node directly below the top right marked node is a hanging node. For meshes created by MESH, BOX, the only requirement for connecting coarse and fine meshes is that the eight corner points of the smaller MESH-box coincide with nodes of the largest mesh.

Figure 4-61

Coarse mesh with Fine Mesh and the Target

The coarse and fine meshes are given by: Finer mesh covering a small region for the initialization of the blast sphere (18, 1, 18): ( -1.5 < X < 3.75 ; -0.18 < Y < 0.18 ; 5 < Z < 10) Coarse mesh covering the entire model (48, 1, 30): ( -4.5 < X < 13.5 ; -0.18 < Y < 0.18 ; 0 < Z < 15) By gluing these meshes, a new mesh will be created that is shown in Figure 4-61. Four locations have been marked. For gluing to work, both meshes need to have a grid point at each of these four locations. The four grid points of the coarse mesh do not need to be exactly on top of the grid points of the fine mesh. A tolerance is used to “equivalence” these grid points in the Dytran solver.

Main Index

Chapter 4: Fluid-structural Interaction 275 Blast Wave with a Graded Mesh

The two meshes are defined as follows: MESH,1,BOX,,,,,,,+ +,-4.5,-0.18,0.0,18,0.36,15,,,+ +,48,1,30,,,,EULER,1 MESH,2,BOX,,,,,,,+ +,-1.5,-0.18,5,5.25,0.36,5,,,+ +,18,1,18,,,,EULER,1

To activate the Graded Mesh method the following parameter is added to the input file: PARAM,GRADED-MESH

All coarse elements that are completely covered by fine elements will be made inactive. In addition, special faces will be made that connect elements of the fine mesh to elements of the coarse mesh. The multi-material Euler solver will be used. PEULER1,1,,MMHYDRO,19

The initialization is given by TICEUL,19,,,,,,,,+ +, SPHERE, 3, 100, 8, 4.0,,,,+ +, SPHERE, 4,100, 9, 6.0 SPHERE, 3,, 0.0,0.0,0,10000 SPHERE, 4, 1.125, 0.0, 7.5,1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,1,sie,9e+5 $ DMAT, 100, 1, 2 EOSGAM, 2, 1.4 $ PARAM,MICRO,30

When the blast wave reaches the boundary of the Euler mesh no reflections should occur. Reflections can be avoided by defining a zero gradient flow boundary on the outside of the Euler mesh. This is defined by means of a FLOWDEF entry: FLOWDEF,1,,MMHYDRO,,,,,,+ +,FLOW,OUT

The target structure as shown in Figure 4-61 is modeled by QUAD shell elements. PSHELL1, 5, 5, GAUSS,,,,,,+ +, .003 $ DYMAT24, 5, 7800, 2.1e+11, .3,,,,,+ +, 2e+08

To hold the structure stationary, constraints are given to the four corner points of the back side of the box: SPC1, 1, 123456, 11, 12, 101, 102

Main Index

276 Dytran Example Problem Manual Blast Wave with a Graded Mesh

To enable interaction of the blast wave with the structure, a coupling surface needs to be defined. Since the structure itself is closed (box), all the elements can serve as the coupling surface. COUPLE,200,300,INSIDE,ON,ON, SURFACE,300,,PROP,5 SET1, 5,5 PARAM,FASTCOUP

To get optimal performance, a set of switches is added to redo the coupling surface computation only after significant movement of the structure. PARAM,COSUBMAX,30 PARAM,COSUBCYC,30

Results To assess the accuracy of the graded mesh simulation, results of the fine mesh will also be examined. The figure below shows the pressure distribution at the beginning of the analysis.

Main Index

Chapter 4: Fluid-structural Interaction 277 Blast Wave with a Graded Mesh

For comparison, the result with a fine mesh is shown in the following picture

The pressure distribution at the time when the blast wave is crossing the interface between the coarse and fine mesh is shown in the following picture .

Main Index

278 Dytran Example Problem Manual Blast Wave with a Graded Mesh

For comparison, the results for the fine mesh are shown below.

The result at the time that the blast wave hits the structure is shown in the figure below.

For comparison, the result for the case with a fine mesh is shown in the figure below.

Main Index

Chapter 4: Fluid-structural Interaction 279 Blast Wave with a Graded Mesh

The figure below shows the pressure profiles along the model at different times. It also shows that the pressure profiles propagate smoothly from the fine to the coarse mesh (at X = -1.5 and X = 3.75).

Main Index

280 Dytran Example Problem Manual Blast Wave with a Graded Mesh

For comparison, the results with the fine mesh are shown in the following figure.

The figure below compares the time history of the deflection of the structure at a point in the center area for the two cases .

Main Index

Chapter 4: Fluid-structural Interaction 281 Blast Wave with a Graded Mesh

Abbreviated Dytran Input File START CEND ENDSTEP = 660 CHECK=NO TITLE= Blast wave Graded Mesh TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMATPLT STEPS(ALLEULER) = 0,THRU,END,BY,20 SAVE (ALLEULER) = 10000 $ TYPE (BOX) = ARCHIVE CPLSURFS (BOX) = 3 SET 3 = 300 CPLSOUT (BOX) = PRESSURE,XVEL STEPS (BOX) = 0,THRU,END,BY,20 SAVE (BOX) = 10000 $ TYPE (LAG) = ARCHIVE ELEMENTS (LAG) = 4 SET 4 = ALLSHQUAD ELOUT (LAG) = EFFST-MID,EFFPL-MID STEPS (LAG) = 0,THRU,END,BY,20 SAVE (LAG) = 10000 $ TYPE (ARCMAT) = TIMEHIS MATS (ARCMAT) = 15 SET 15 = 100 MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOM STEPS (ARCMAT) = 0,THRU,END,BY,10 SAVE (ARCMAT) = 99999 $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK $ ============ Euler ================================== $ PARAM,BULKL,0.1 PARAM,COSUBMAX,100 PARAM,COSUBCYC,100 $ PARAM,FASTCOUP PARAM,MICRO,30 $ PARAM,GRADED-MESH

Main Index

282 Dytran Example Problem Manual Blast Wave with a Graded Mesh

PARAM,FLOW-METHOD,FACET $ $ --- Material definitions --$ PEULER1,1,,MMHYDRO,19 $ DMAT,100,1,2 $ EOSGAM,2,1.4 $ $ --- Initial conditions --$ TICEUL,19,,,,,,,,+ +,SPHERE,3,100,8,4.0,,,,+ +,SPHERE,4,100,9,6.0 SPHERE,3,,0.0,0.0,0,10000 SPHERE,4,,1.125,0.0,7.5,1 TICVAL,8,,density,1,sie,3e+5 TICVAL,9,,density,1,sie,9e+5 $ MESH,1,BOX,,,,,,,+ +,-4.5,-0.18,0.0,18,0.36,15,,,+ +,48,1,30,,,,EULER,1 $ MESH,2,BOX,,,,,,,+ +,-1.5,-0.18,5,5.25,0.36,5,,,+ +,18,1,18,,,,EULER,1 $ $ --- Boundary conditions --------$ FLOWDEF,1,,MMHYDRO,,,,,,+ +,FLOW,OUT $ $ COUPLE,200,300,INSIDE,ON,ON, SURFACE,300,,PROP,5 SET1,5,5 $ $=========== Structure ================ $ $ --- Material properties --$ PSHELL1, 5, 5,,GAUSS,,,,,+ +,.003$ DYMAT24, 5, 7800, 2.1e+11 ,.3,,,,,+ +, 2e+08 $ $ --- Boundary condition --$ SPC1, 1, 123456, 11, 12, 101, 102 $ $ --- Define 1012 grid points --$ GRID 11 11.8500-.190000 0.90000

Main Index

Chapter 4: Fluid-structural Interaction 283 Blast Wave with a Graded Mesh

GRID GRID

12 13

11.8500 .190000 0.90000 11.8500-.190000 1.28889

CQUAD4 CQUAD4 CQUAD4 $ ENDDATA

Main Index

1018 1019 1020

5 5 5

145 146 147

146 147 148

606 607 608

605 606 607

284 Dytran Example Problem Manual Bubble Collapse with Hydrostatic Boundary Conditions

Bubble Collapse with Hydrostatic Boundary Conditions A weak explosion is ignited in the ocean. Initially the bubble, formed due to chemical explosion, expands. During this expansion water is pushed away towards the boundaries. This will result in higher pressure levels in the water until the equilibrium state, where the pressure inside the bubble and water pressure outside the bubble are the same. During the equilibrium, the bubble expansion stops and then reverses itself because the displaced water, now moves back again resulting in severe compression of the bubble. Due to momentum of the water the compression causes a large pressure build up in the bubble. The bubble pressure becomes so large that a second blast wave is ignited. During the entire expansion and contraction event, the bubble does not reach the water surface. The explosion will be modeled by a blast wave approach. The initial conditions of the hot gas inside the initial region of the blast wave are: Specific energy = 3.E+5 Joule/KgK Density

= 100kg/m3

The water properties are: Reference density = 1000 kg/mm3 Bulk modulus K = 2.2e+9 Pa

Dytran Model The ocean will be modeled by a container of water. The top of the container is two meters below the surface and the bottom is 13 meters below the surface. The water will be initialized by a hydrostatic pressure profile. Using a wall as boundary would give undesirable reflections. A transmitting boundary condition allows water to flow out of the mesh but prevents the inflow of water back into the container. Therefore, a special boundary condition is defined on the side walls of the container that enables back flow of water into the container. This special boundary condition prescribes a hydrostatic pressure profile on the boundary of the container. The density of material flowing back is computed from the hydrostatic pressures. A container of water can be modeled by a block of Euler elements. On the boundary of the block of elements, a hydrostatic pressure profile is imposed. This approach suffices in general. But, if the effect of the blast wave on a structure is to be studied in more detail, it may be necessary to allow for water to enter the structure in case of ruptures. In this case, multiple Euler domains with coupling surfaces have to be used. This means that the block of Euler is wrapped with a structural surface that is fully porous. The porosity model will be of the hydrostatic type. This structural surface has to consist of dummy shell elements and is used as coupling surface for the Euler domain.

Main Index

Chapter 4: Fluid-structural Interaction 285 Bubble Collapse with Hydrostatic Boundary Conditions

Therefore, there are two models to consider: 1. The general approach using a flow definition to impose a boundary condition. 2. A coupling surface with a porosity model.

General Setup First, the entries are discussed that apply to both models .

Figure 4-62

Models

The Euler mesh is created by using a MESH,BOX entry: MESH,1,BOX,,,,,,,+ +,-5.5,-0.12,0,11,0.24,11,,,+ +,41,1,41,,,,EULER,1

and the multi-material Euler solver will be used: PEULER1,1,,MMHYDRO,19

Two materials representing air and water are defined as: DMAT 4 $ EOSPOL 1 $ DMAT 3 EOSGAM,2,1.4 $ DMAT 100

Main Index

1.e3

1

2.2e9 2

100

2

2

286 Dytran Example Problem Manual Bubble Collapse with Hydrostatic Boundary Conditions

The initial conditions will be defined by using geometric regions: TICEUL,19,,,,,,,,+ +,BOX,1,4,6,2.0,,,,+ +,SPHERE,3,100,8,4.0 BOX,1,,-100,-100,0.0,200,200,20.0 SPHERE,3,,0.0,0.0,4.652,0.25 TICVAL,6,,SIE,0.,DENSITY,1000.0 TICVAL,8,,density,100,sie,3e+5 PARAM,MICRO,30

Figure 4-62a shows the initial conditions. The two elements that are colored red are initialized as gas, the

other elements are water. The blast is ignited at a depth of 13 m-4.6 m = 8.4 m. Gravity is applied on water and air regions as follow: TLOAD1,1,444,,0 GRAV,444,,9.8,,,-1

Because of the gravity, a hydrostatic pressure profile will develop. To effectively simulate a blast wave in water, the hydrostatic pressure profile should be present at cycle 1. To do this, the HYDSTAT option will be used. HYDSTAT,123,4,,,0,0,13.00,104000

The water level will be at a height of 13 meters. The HYDSTAT functionality changes the density initialization as specified by the previous TICVAL entry such that the element pressure equals the hydrostatic pressure. Figure 4-63 shows the pressure at cycle 1. At the center, several elements with large pressure have been hidden to make the hydrostatic pressure initialization visible.

Figure 4-63

Main Index

Pressure at Cycle 1: Hydrostatic Pressure Profile

Chapter 4: Fluid-structural Interaction 287 Bubble Collapse with Hydrostatic Boundary Conditions

There are two methods to set up the model: Model 1 The boundary condition will be given by FLOWDEF,25,,MMHYDRO,,,,,,,, ,HYDSTAT,23

The boundary conditions will be taken from HYDSTAT entry 123. FLOWDEF will also work on the front and back faces, but this will not make any contributions to the elements, since there is no transport over these front and back faces. Model 2 The water and fluid are enclosed by a coupling surface as shown in Figure 4-62b. COUPLE,100,200,OUTSIDE,ON,ON,120 COUPOR,1,120,,PORHYDST,75 PORHYDST 75 SURFACE,200,,PROP,2 PSHELL1 2 DUMMY SET1,2,2 $ GRID 1 -5.49990-5.49990 GRID 2 5.49990-5.49990 GRID 3 -5.49990 5.49990 GRID 4 5.49990 5.49990 GRID 5 -5.49990-5.49990 GRID 6 -5.49990 5.49990 GRID 7 5.49990-5.49990 GRID 8 5.49990 5.49990 $ $ --- Define 6 elements $ -------- property set pdum --------CQUAD4 1 2 1 2 CQUAD4 2 2 5 6 CQUAD4 3 2 5 7 CQUAD4 4 2 2 7 CQUAD4 5 2 4 8 CQUAD4 6 2 3 6

10.9999 10.9999 10.9999 10.9999 0.0001 0.0001 0.0001 0.0001

4 8 2 8 6 5

3 7 1 4 3 1

The boundary conditions are: COUPOR,1,120,,PORHYDST,75 PORHYDST 75

For the boundary condition, the HYDSTAT will be taken that is referenced by the COUPLE entry.

Main Index

288 Dytran Example Problem Manual Bubble Collapse with Hydrostatic Boundary Conditions

Results Both models yield almost identical results as shown in Figure 4-64. Detailed results are only shown for Model 1.Figure 4-64 also shows that after 0.2 seconds, a considerable amount of mass has left the domain. When the bubble starts to collapse (from 0.2 to 0.4 seconds), the water flows back into the domain through the hydrostatic boundary.

Figure 4-64

Total Mass of Water in the Container for Model 1 (Red) and Model 2 (Blue)

Figure 4-65 shows the bubble rise and collapse. After 0.23 sec the expansions stops and the bubble is compressed. Also it starts to rise.

Main Index

Chapter 4: Fluid-structural Interaction 289 Bubble Collapse with Hydrostatic Boundary Conditions

Figure 4-65

Bubble Rise and Collapse

After 0.44 seconds, the bubble has collapsed. Due to the momentum of the water, the gas has been severely compressed and a second blast occurs results in the subsequent phase.

Main Index

290 Dytran Example Problem Manual Bubble Collapse with Hydrostatic Boundary Conditions

Dytran Input files Model 1 START CEND ENDTIME = 0.5 CHECK=NO TITLE= Jobname is: undex-2d TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT4,MASS4,DENSITY4,FMATPLT4, FMATPLT,FMAT100,FLUXVOL, FVUNC FMAT100,MASS100,DENSITY100,SIE100,SIE4,HMAT TIMES(ALLEULER) = 0,1E-8,thru,end,by,0.01 SAVE (ALLEULER) = 10000 $ TYPE (ARCMAT) = TIMEHIS MATS (ARCMAT) = 15 SET 15 = 4 MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOM STEPS (ARCMAT) = 0,THRU,END,BY,10 SAVE (ARCMAT) = 99999 $ $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK PARAM,BULKL,0.1 $ -----------------------------------------------------------------$ $ * Euler.300 * $ PARAM,MICRO,30 PARAM,FMULTI,1.0 PEULER1,1,,MMHYDRO,19 $ DMAT 4 1.e3 1 $ EOSPOL 1 2.2e9 $ DMAT 3 2 2 EOSGAM,2,1.4 $ DMAT 100 100 2 $

Main Index

Chapter 4: Fluid-structural Interaction 291 Bubble Collapse with Hydrostatic Boundary Conditions

$ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== $ $ TICEUL,19,,,,,,,,+ +,BOX,1,4,6,2.0,,,,+ +,SPHERE,3,100,8,4.0 BOX,1,,-100,-100,0.0,200,200,20.0 SPHERE,3,,0.0,0.0,4.652,0.25 TICVAL,6,,SIE,0.,DENSITY,1000.0 TICVAL,8,,density,100,sie,3e+5 $ MESH,1,BOX,,,,,,,+ +,-5.5,-0.12,0,11,0.24,11,,,+ +,41,1,41,,,,EULER,1 $ TLOAD1,1,444,,0 GRAV,444,,9.8,,,-1 $ $ HYDSTAT,123,4,,,0,0,13.00,104000 $ FLOWDEF,25,,MMHYDRO,,,,,,+ +,HYDSTAT,123 $ ENDDATA Model 2 START CEND ENDSTEP = 5000 CHECK=NO TITLE= Jobname is: undex-2d TLOAD=1 TIC=1 SPC=1 $ TYPE (ALLEULER) = ARCHIVE ELEMENTS (ALLEULER) = 2 SET 2 = ALLMULTIEULHYDRO ELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT4,MASS4,DENSITY4,FMATPLT4, FMATPLT,FMAT100,HMAT, FVUNC FMAT100,MASS100,DENSITY100,SIE100,SIE4 STEPS(ALLEULER) = 0,1,thru,end,by,100 SAVE (ALLEULER) = 10000 $ TYPE (ARCMAT) = TIMEHIS MATS (ARCMAT) = 15 SET 15 = 4 MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOM

Main Index

292 Dytran Example Problem Manual Bubble Collapse with Hydrostatic Boundary Conditions

STEPS (ARCMAT) = 0,THRU,END,BY,10 SAVE (ARCMAT) = 99999 $ TYPE (SHELL) = ARCHIVE ELEMENTS (SHELL) = 4 SET 4 = 1,THRU,6 ELOUT (SHELL) = ZUSER STEPS(SHELL) = 0,1 SAVE (SHELL) = 10000 $ $------- Parameter Section -----PARAM,INISTEP,1.E-8 $ $------- BULK DATA SECTION ------BEGIN BULK PARAM,BULKL,0.1 PARAM,COSUBMAX,100 PARAM,COSUBCYC,100 $ -----------------------------------------------------------------$ $ * Euler.300 * $ PARAM,MICRO,30 PARAM,FMULTI,1.0 PEULER1,1,,MMHYDRO,19 $ DMAT 4 1.e3 1 $ EOSPOL 1 2.2e9 $ DMAT 3 2 2 EOSGAM,2,1.4 $ DMAT 100 100 2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ ======== Load Cases ======================== $ $ TICEUL,19,,,,,,,,+ +,BOX,1,4,6,2.0,,,,+ +,SPHERE,3,100,8,4.0 BOX,1,,-100,-100,0.0,200,200,20.0 SPHERE,3,,0.0,0.0,4.652,0.25 TICVAL,6,,SIE,0.,DENSITY,1000.0 TICVAL,8,,density,100,sie,3e+5 $ MESH,1,BOX,,,,,,,+ +,-5.5,-0.12,0,11,0.24,11,,,+ +,41,1,41,,,,EULER,1 $

Main Index

Chapter 4: Fluid-structural Interaction 293 Bubble Collapse with Hydrostatic Boundary Conditions

TLOAD1,1,444,,0 GRAV,444,,9.8,,,-1 $ $ HYDSTAT,1,4,,,0,0,13.00,104000 $ COUPLE,100,200,OUTSIDE,ON,ON,120 COUPOR,1,120,,PORHYDST,75 PORHYDST 75 SURFACE,200,,PROP,2 $ PSHELL1 2 DUMMY SET1,2,2 $ GRID 1 -5.49990-5.49990 GRID 2 5.49990-5.49990 GRID 3 -5.49990 5.49990 GRID 4 5.49990 5.49990 GRID 5 -5.49990-5.49990 GRID 6 -5.49990 5.49990 GRID 7 5.49990-5.49990 GRID 8 5.49990 5.49990 $ $ --- Define 6 elements $ $ -------- property set pdum --------CQUAD4 1 2 1 2 CQUAD4 2 2 5 6 CQUAD4 3 2 5 7 CQUAD4 4 2 2 7 CQUAD4 5 2 4 8 CQUAD4 6 2 3 6 ENDDATA

Main Index

10.9999 10.9999 10.9999 10.9999 0.0001 0.0001 0.0001 0.0001

4 8 2 8 6 5

3 7 1 4 3 1

294 Dytran Example Problem Manual Prestressed Concrete Beam

Prestressed Concrete Beam Problem Description This problem demonstrates how to prestress a concrete beam prior to transient loading condition. Prestressed concrete is an architectural and structural member providing great strength. Generally, concrete has a big advantage in compression, but is naturally weak in tension. In typical reinforced concrete, the concrete’s great compressive strength is combined with the high tensile strength of steel to create a structural member that is strong in both compression and tension. The concept in prestress concrete is that compressive stresses induced by high strength steel tendons in the concrete structure prior to actual loading will offset and balance the tensile stresses that are subjected to the member during service.

Main Index

Chapter 4: Fluid-structural Interaction 295 Prestressed Concrete Beam

Analysis Scheme The example consists of two jobs: Prestress and Transient. The analysis scheme is explained by a flow chart as Figures Figure 4-66 and Figure 4-67.

Figure 4-66

Main Index

Analysis Scheme of Prestressed Beam

296 Dytran Example Problem Manual Prestressed Concrete Beam

Figure 4-67

Main Index

Analysis Models and Behavior at Each Step

Chapter 4: Fluid-structural Interaction 297 Prestressed Concrete Beam

Dytran Model There are two different material properties defined for concrete and the tendon. The details of two materials and applied loads are explained below. Each input deck will show: Concrete:

Cross area: 40000 mm2 Young’s modulus = 21 GPa Density = 2400 kg/m3 Length = 2 m

Tendon:

Cross area: 1390 mm2 (Using ten 15.2mm class 2 relaxation standard strands) Young’s modulus = 200 GPa Density = 7850 kg/m3 Length = 2 m

Applied velocity

= 0.12 m/seconds

End time of simulation

= 0.1 second

Final applied displacement

= 0.12×0.1 = 0.012 m (equivalent to 1670 kN as force)

DMATEL,2,2400,21.e+9,0.2,, PSOLID,2,2 MAT1,1,200e+9,,0.3,7850. PBEAM,1,1,.00139,0.1,0.1 TLOAD1,1,3,,2 FORCE,3,21,0,1,1.2e-1,,

(a) Prestress Model PRESTRESS entry in File Management Section (FMS) shows that this job is the prestress one. PRESTRESS

SOLUOUT entry in File Management Section (FMS) specifies an output file to which the solution data is written at the end of a prestress analysis. Output file of this example is GDIL.SOL. SOLUOUT=GDIL.SOL

PARAM, VDAMP is for control on the global damping in the dynamic. PARAM,VDAMP,0.001

A dynamic relaxation is defined to dampen the solution and to prevent high frequency oscillations. For this purpose, NASINIT entry is used. Relaxation phase lasts 0.1 second. NASINIT,100,yes,.1,0.001

To connect the tendon and concrete, RBE2 is used as below.

Main Index

298 Dytran Example Problem Manual Prestressed Concrete Beam

RBE2, 1, 2,23,1002 ...

(b) Transient Model The transient analysis starts with START entry. The results from the prestress analysis which are stored in GDIL.SOL are used as a pre-condition for transient run by using the SOLINIT entry. START SOLINIT=GDIL.SOL

PARAM, VDAMP is for control on the global damping in the dynamic. PARAM,VDAMP,0.001

The RBE2 definition at the end is now changed by defining additional restraint in the z-direction for ensuring a common behavior for tendon and concrete. RBE2,20,21,123, 626

Results Analytical results are used to match the simulation results. If there is no local deformation at the end, the theoretical results can be obtained by following scheme. Pc + Pt = 0

(Equilibrium: no external loading)

δ c = δ t = 0.012

(Compatibility: total deformation is the same as the sum of deformations of each

element) where is the force on the concrete beam Pc

Main Index

Pt

is the force on the tendon

δc

is the deformation of the concrete beam

δt

is the deformation of the tendon

Chapter 4: Fluid-structural Interaction 299 Prestressed Concrete Beam

Ec Ac P c = ------------ δ c Lc Et At P t = ---------- δ c Lt

where Ec

is Young’s modulus of the concrete beam

Et

is Young’s modulus of the tendon

Ac

is the cross area of the concrete beam

At

is the cross area of the tendon

Lc

is the original length of the concrete beam

Lt

is the original length of the tendon

Substituting the P c and P t into Equilibrium and Compatibility equations, the compressive stresses in concrete can be obtained. E t At ---------Ec Lt σ c = ----- ⋅ ------------------------------ ⋅ 0.012 Lc E c Ac E t A t ------------ + ---------Lc Lt

where σc

is the compression Stress of the concrete beam

Substituting the material properties of tendon and concrete in the equation above, the computed concrete stress is 3.14 × 107. This value is the same as the xx-stress result in Figure 4-68b. The value of Figure 4-68a is little different from the theoretical value due to the local deformation at the end where the loading is applied and stress is concentrated.

Main Index

300 Dytran Example Problem Manual Prestressed Concrete Beam

Other results with local deformation allowed at each step are shown in Figure 4-69.

Figure 4-68

Main Index

Two Results of xx-stress at the Beam

Chapter 4: Fluid-structural Interaction 301 Prestressed Concrete Beam

Figure 4-69

Main Index

Results at Each Step

302 Dytran Example Problem Manual Prestressed Concrete Beam

Abbreviated Dytran Input File Prestress Job Input PRESTRESS $BULKOUT=GRID.DAT NASTDISP=ZERO SOLUOUT=GDIL.SOL TIME=100000 MEMORY-SIZE=50000000,40000000 CEND ENDTIME=0.1 ENDSTEP=10000000 TITLE= Jobname is: pre TLOAD=1 TIC=1 SPC=1 $ Output result for request: DISP TYPE (DISP) = TIMEHIS GRIDS (DISP) = 1 SET 1 = 1 6 11 16 21 626 GPOUT (DISP) = XPOS XVEL XFORCE XDIS TIMES (DISP) = 0 THRU END BY 1e-5 SAVE (DISP) = 10000 $ Output result for request: LAG TYPE (LAG) = ARCHIVE ELEMENTS (LAG) = 2 SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 0.01 SAVE (LAG) = 10000 $ Output result for request: STRESS TYPE (STRESS) = ARCHIVE ELEMENTS (STRESS) = 3 SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 0.01 SAVE (STRESS) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,0.0001 PARAM,VDAMP,0.001 PARAM,INITNAS,PATRAN PARAM,INISTEP,1e-8 PARAM,MINSTEP,1e-8 $------- BULK DATA SECTION ------BEGIN BULK $ INCLUDE pre.bdf NASINIT,100,yes,0.1,0.001 $ $ ========== PROPERTY SETS ========== $

Main Index

Chapter 4: Fluid-structural Interaction 303 Prestressed Concrete Beam

$ * GID * $ PBEAM,1,1,.00139,0.1,0.1 $ $ * Conc * $ PSOLID,2,2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Barzel id =1 MAT1,1,200e+9,,0.3,7850. $ $ -------- Material DEBESH id =2 DMATEL,2,2400,21.e+9,0.2,,,, $ $ ------- Velocity BC VELO ----TLOAD1,1,3,,2 FORCE,3,21,0,1,1.2e-1,, $ RBE2, 1, 2,23,1002 RBE2, 2, 3,23,1003 RBE2, 3, 4,23,1004 RBE2, 4, 5,23,1005 RBE2, 5, 6,23,1006 RBE2, 6, 7,23,1007 RBE2, 7, 8,23,1008 RBE2, 8, 9,23,1009 RBE2, 9,10,23,1010 RBE2,10,11,23,1011 RBE2,11,12,23,1012 RBE2,12,13,23,1013 RBE2,13,14,23,1014 RBE2,14,15,23,1015 RBE2,15,16,23,1016 RBE2,16,17,23,1017 RBE2,17,18,23,1018 RBE2,18,19,23,1019 RBE2,19,20,23,1020 RBE2,20,21,23, 626 $ ENDDATA

Transient Job Input START SOLINIT=GDIL.SOL TIME=100000 MEMORY-SIZE=50000000,40000000 CEND ENDTIME=0.1 ENDSTEP=10000

Main Index

304 Dytran Example Problem Manual Prestressed Concrete Beam

CHECK=NO TITLE= Jobname is: tra TLOAD=1 TIC=1 SPC=1 $ Output result for request: DISP TYPE (DISP) = TIMEHIS GRIDS (DISP) = 1 SET 1 = 1 6 11 16 21 626 153 258 468 573 GPOUT (DISP) = XPOS XVEL XFORCE XDIS TIMES (DISP) = 0 THRU END BY 1e-4 SAVE (DISP) = 10000 $ Output result for request: LAG TYPE (LAG) = ARCHIVE ELEMENTS (LAG) = 2 SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 1.e-3 SAVE (LAG) = 10000 $ Output result for request: STRESS TYPE (STRESS) = ARCHIVE ELEMENTS (STRESS) = 3 SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 1.e-3 SAVE (STRESS) = 10000 $ Output result for request: DISP TYPE (LOCAL) = TIMEHIS ELEMENTs (LOCAL) = 4 SET 4 = 111 THRU 411 BY 20 ELOUT (LOCAL) = TXX EFFSTS PRESSURE TIMES (LOCAL) = 0 THRU END BY 1e-4 SAVE (LOCAL) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1e-8 $=================================damping====== PARAM,VDAMP,0.01 $================================== $------- BULK DATA SECTION ------BEGIN BULK INCLUDE pre.bdf $ $ ========== PROPERTY SETS ========== $ $ * GID * $ PBEAM,1,1,0.00139,.1,.1 $ $ * Conc * $ PSOLID,2,2 $ $

Main Index

Chapter 4: Fluid-structural Interaction 305 Prestressed Concrete Beam

$ ========= MATERIAL DEFINITIONS ========== $ $ -------- Material Barzel id =1 MAT1,1,200e+9,,0.3,7850. $ $ -------- Material DEBESH id =2 DMATEL,2,2400,21.e+9,0.2,, $ RBE2, 1, 2,23,1002 RBE2, 2, 3,23,1003 RBE2, 3, 4,23,1004 RBE2, 4, 5,23,1005 RBE2, 5, 6,23,1006 RBE2, 6, 7,23,1007 RBE2, 7, 8,23,1008 RBE2, 8, 9,23,1009 RBE2, 9,10,23,1010 RBE2,10,11,23,1011 RBE2,11,12,23,1012 RBE2,12,13,23,1013 RBE2,13,14,23,1014 RBE2,14,15,23,1015 RBE2,15,16,23,1016 RBE2,16,17,23,1017 RBE2,17,18,23,1018 RBE2,18,19,23,1019 RBE2,19,20,23,1020 RBE2,20,21,123, 626 $ ENDDATA

Main Index

306 Dytran Example Problem Manual Blast Simulation on Prestressed Concrete Beam

Blast Simulation on Prestressed Concrete Beam Problem Description This problem is a continuation of the Prestressed Concrete Beam example and demonstrates how to apply a blast load on the beam. To model the explosive, blast technique is used where the Eulerian elements inside the explosive region are given high density and internal energy values. In this analysis, the explosive properties are defined by ideal gas model.

Analysis Scheme The example consists of two jobs: Prestress and Transient. The analysis scheme is explained by a flow chart as shown in Figure 4-70. The additional steps to prestress run are highlighted in bold italics.

Figure 4-70

Main Index

Analysis Scheme of Blast Simulation on Prestressed Beam

Chapter 4: Fluid-structural Interaction 307 Blast Simulation on Prestressed Concrete Beam

Dytran Model The material properties of concrete and tendon and the applied load in tendon are the same as Prestressed Concrete Beam example. The material models of explosive and air are explained below. In this simulation the explosive is defined as a compressed hot gas with the same γ as air. Explosive (sphere shape with 0.1 radious:

Explosive center distance from the bear: 1 m Total mass = 500 kg Density = 1.592 x 104 kg/m3 Specific internal energy = 4.765 x 106 Kg-m2/s2 Density = 1.2 kg/m3

Air:

Specific internal energy = 1.94 x 105 Kg-m2/s2 γ = 1.4

Figure 4-71

Simulation Model

Both prestressed and transient runs include the following Euler definitions: DMAT* 5 1.2 EOSGAM 5 1.4 TICVAL,18,,DENSITY,1.592+4,SIE,4.765E6 TICVAL,19,,DENSITY,1.2,SIE,1.94E+05 TICEUL,20,,,,,,,,+ + ,SPHERE,4,5,19,1,,,,+ + ,SPHERE,17,5,18,5,,, SPHERE,4,,0,0,0,5000 SPHERE,17,,1,0,1.,0.1

Main Index

5

308 Dytran Example Problem Manual Blast Simulation on Prestressed Concrete Beam

Prestress Model The Prestress model is identical with the Prestress Concrete Beam example except the definition of Eulerian parts. The Eulerian part is explained in Transient model. Transient Model The Lagrangian part is the same as the Prestress Concrete Beam example. Here, the explanation of Eulerian parts is added. To decrease simulation time, Euler elements and coupling are active only for 0.001 seconds after they become activated in 0.006 seconds. In Prestressed Concrete Beam example, the vibration from the Prestress Run is diminished after 0.006 seconds. ACTIVE,1,ELEMENT,EULHYDRO,,,,,,+ +,TABLE,100 ACTIVE,2,INTERACT,COUPLE,,,,,,+ +,TABLE,100 TABLED1,100,,,,,,,,+ +,0,-1,0.006,-1,0.006,1,0.007,1,+ +,0.007,-1,.5,-1

To couple the Euler material to the concrete beam, the COUPLE entry is used COUPLE +A000012 +A000013

61

2

INSIDE

ON

ON

4

4

+A000012 +A000013

63

The Euler mesh is generated by MESH entry. MESH +A000020 +A000021

Main Index

63 -1 41

BOX -2 41

-1 41

+A000020 +A000021

4 EULER

5

Chapter 4: Fluid-structural Interaction 309 Blast Simulation on Prestressed Concrete Beam

Results

Main Index

Figure 4-72

Blast on Concrete Beam at Various Steps

Figure 4-73

Displacement and Effective Stress at the Center of the Beam

310 Dytran Example Problem Manual Blast Simulation on Prestressed Concrete Beam

Figure 4-74

Main Index

Deformation of Prestressed Concrete Beam

Chapter 4: Fluid-structural Interaction 311 Blast Simulation on Prestressed Concrete Beam

Abbreviated Dytran Input File Prestress Job Input PRESTRESS NASTDISP=ZERO SOLUOUT=GDIL.SOL TIME=100000 MEMORY-SIZE=50000000,40000000 CEND ENDTIME=0.1 ENDSTEP=10000000 TITLE= Jobname is: pre TLOAD=1 TIC=1 SPC=1 $ Output result for request: DISP TYPE (DISP) = TIMEHIS GRIDS (DISP) = 1 SET 1 = 1 6 11 16 21 626 GPOUT (DISP) = XPOS XVEL XFORCE XDIS TIMES (DISP) = 0 THRU END BY 1e-5 SAVE (DISP) = 10000 $ Output result for request: LAG TYPE (LAG) = ARCHIVE ELEMENTS (LAG) = 2 SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 0.01 SAVE (LAG) = 10000 $ Output result for request: STRESS TYPE (STRESS) = ARCHIVE ELEMENTS (STRESS) = 3 SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 0.01 SAVE (STRESS) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,0.0001 PARAM,VDAMP,0.001 PARAM,INITNAS,PATRAN PARAM,FASTCOUP,INPLANE,FAIL PARAM,INISTEP,1e-8 PARAM,MINSTEP,1e-8 $PARAM,LIMITER,ROE $PARAM,RKSCHEME,3 $PARAM,INITFILE,V1 $------- BULK DATA SECTION ------BEGIN BULK $flow out of all faces of outer euler FLOWDEF,1,,HYDRO,,,,,,+ +,FLOW,BOTH,MATERIAL,5

Main Index

312 Dytran Example Problem Manual Blast Simulation on Prestressed Concrete Beam

ACTIVE,1,ELEMENT,EULHYDRO,,,,,,+ +,TABLE,100 ACTIVE,2,INTERACT,COUPLE,,,,,,+ +,TABLE,100 TABLED1,100,,,,,,,,+ +, 0, -1, 11, -1 $ PARAM,FAILOUT,NO PARAM,NZEROVEL,YES INCLUDE pre.bdf NASINIT,100,yes,0.1,0.001 $ $ ========== PROPERTY SETS ========== $ $ * GID * $ PBEAM,1,1,.00139,0.1,0.1 $ $ * Conc * $ PSOLID,2,2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Barzel id =1 MAT1,1,200e+9,,0.3,7850. $ $ -------- Material DEBESH id =2 DMATEL,2,2400,21.e+9,0.2,,,, $ $ $ ------- Velocity BC VELO ----TLOAD1,1,3,,2 FORCE,3,21,0,1,1.2e-1,, $ RBE2, 1, 2,23,1002 RBE2, 2, 3,23,1003 RBE2, 3, 4,23,1004 RBE2, 4, 5,23,1005 RBE2, 5, 6,23,1006 RBE2, 6, 7,23,1007 RBE2, 7, 8,23,1008 RBE2, 8, 9,23,1009 RBE2, 9,10,23,1010 RBE2,10,11,23,1011 RBE2,11,12,23,1012 RBE2,12,13,23,1013 RBE2,13,14,23,1014 RBE2,14,15,23,1015 RBE2,15,16,23,1016 RBE2,16,17,23,1017 RBE2,17,18,23,1018

Main Index

Chapter 4: Fluid-structural Interaction 313 Blast Simulation on Prestressed Concrete Beam

RBE2,18,19,23,1019 RBE2,19,20,23,1020 RBE2,20,21,23, 626 $ $ PEULER1 5 HYDRO 20 $ $ DMAT* 5 1.186 EOSGAM 5 1.4 $ $ $ ======== Load Cases ======================== $ $ $ ------- TICVAL BC initblast ----TICVAL,18,,DENSITY,1.61e+1,SIE,3.88E+05 $ $ ------- TICVAL BC air_steady ----TICVAL,19,,DENSITY,1.2,SIE,1.94E+05 $ $ ------- TICEUL BC out_euler ----TICEUL,20,,,,,,,,+ + ,SPHERE,4,5,19,1,,,,+ + ,SPHERE,17,5,18,5,,, SPHERE,4,,0,0,0,5000 SPHERE,17,,1,0,1.,0.1 $ COUPLE 61 2 INSIDE ON ON +A000012 +A000013 63 $ SURFACE 2 SEG 100 $ MESH 63 BOX +A000020 -1 -2 -1 4 4 +A000021 41 41 41 $ CFACE1,1011,100,101,102,122 $ ... $ ENDDATA

Transient Job Input START SOLINIT=GDIL.SOL TIME=100000 MEMORY-SIZE=50000000,40000000 CEND ENDTIME=0.03 ENDSTEP=1000000

Main Index

5

+A000012 +A000013

+A000020 +A000021

4 EULER

5

314 Dytran Example Problem Manual Blast Simulation on Prestressed Concrete Beam

CHECK=NO TITLE= Jobname is: tra TLOAD=1 TIC=1 SPC=1 $ Output result for request: DISP TYPE (DISP) = TIMEHIS GRIDS (DISP) = 1 SET 1 = 1 6 11 16 21 626 111 THRU 342 BY 21 GPOUT (DISP) = XPOS XVEL XFORCE XDIS ZPOS TIMES (DISP) = 0 THRU END BY 0.00001 SAVE (DISP) = 10000 $ Output result for request: LAG TYPE (LAG) = ARCHIVE ELEMENTS (LAG) = 2 SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 0.001 SAVE (LAG) = 10000 $ Output result for request: STRESS TYPE (STRESS) = ARCHIVE ELEMENTS (STRESS) = 3 SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 0.001 SAVE (STRESS) = 10000 $ Output result for request: DISP TYPE (LOCAL) = TIMEHIS ELEMENTs (LOCAL) = 4 SET 4 = 111 THRU 411 BY 20 ELOUT (LOCAL) = TXX EFFSTS TIMES (LOCAL) = 0 THRU END BY 0.00001 SAVE (LOCAL) = 10000 $ TYPE (AIR) = ARCHIVE ELEMENTS (AIR) = 5 SET 5 = ALLEULHYDRO ELOUT (AIR) = DENSITY SIE PRESSURE FMAT FVUNC FMATPLT , VOLUME XVEL YVEL ZVEL TIMES (AIR) = 0.006 THRU 0.007 BY 0.00001 SAVE (AIR) = 10000 $ Output result for request: LAG1 TYPE (LAG1) = ARCHIVE ELEMENTS (LAG1) = 7 SET 7 = ALLLAGSOLID ELOUT (LAG1) = EFFSTS TIMES (LAG1) = 0.006 THRU 0.007 BY 0.00001 SAVE (LAG1) = 10000 $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,FASTCOUP,INPLANE,FAIL PARAM,INISTEP,1e-8 PARAM,MINSTEP,1e-8 PARAM,LIMITER,ROE

Main Index

Chapter 4: Fluid-structural Interaction 315 Blast Simulation on Prestressed Concrete Beam

PARAM,RKSCHEME,3 PARAM,VDAMP,0.01 $------- BULK DATA SECTION ------BEGIN BULK $flow out of all faces of outer euler FLOWDEF,1,,HYDRO,,,,,,+ +,FLOW,BOTH ACTIVE,1,ELEMENT,EULHYDRO,,,,,,+ +,TABLE,100 ACTIVE,2,INTERACT,COUPLE,,,,,,+ +,TABLE,100 TABLED1,100,,,,,,,,+ +,0,-1,0.006,-1,0.006,1,0.007,1,+ +,0.007,-1,.5,-1 $ PARAM,FAILOUT,NO PARAM,NZEROVEL,YES INCLUDE pre.bdf $ $ ========== PROPERTY SETS ========== $ $ * GID * $ PBEAM,1,1,.00139,0.1,0.1 $ $ * Conc * $ PSOLID,2,2 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Barzel id =1 MAT1,1,200e+9,,0.3,7850. $ $ -------- Material DEBESH id =2 DMATEL,2,2400,21.e+9,0.2,,,, $ RBE2, 1, 2,23,1002 RBE2, 2, 3,23,1003 RBE2, 3, 4,23,1004 RBE2, 4, 5,23,1005 RBE2, 5, 6,23,1006 RBE2, 6, 7,23,1007 RBE2, 7, 8,23,1008 RBE2, 8, 9,23,1009 RBE2, 9,10,23,1010 RBE2,10,11,23,1011 RBE2,11,12,23,1012 RBE2,12,13,23,1013 RBE2,13,14,23,1014 RBE2,14,15,23,1015 RBE2,15,16,23,1016

Main Index

316 Dytran Example Problem Manual Blast Simulation on Prestressed Concrete Beam

RBE2,16,17,23,1017 RBE2,17,18,23,1018 RBE2,18,19,23,1019 RBE2,19,20,23,1020 RBE2,20,21,123, 626 $ $ PEULER1 5 HYDRO 20 $ $ DMAT* 5 1.2 EOSGAM 5 1.4 $ $ $ ======== Load Cases ======================== $ $ $ ------- TICVAL BC initblast ----TICVAL,18,,DENSITY,1.592+4,SIE,4.765E6 $ $ ------- TICVAL BC air_steady ----TICVAL,19,,DENSITY,1.2,SIE,1.94E+05 $ $ ------- TICEUL BC out_euler ----TICEUL,20,,,,,,,,+ + ,SPHERE,4,5,19,1,,,,+ + ,SPHERE,17,5,18,5,,, SPHERE,4,,0,0,0,5000 SPHERE,17,,1,0,1.,0.1 $ COUPLE 61 2 INSIDE ON ON +A000012 +A000013 63 $ SURFACE 2 SEG 100 $ MESH 63 BOX +A000020 -1 -2 -1 4 4 +A000021 41 41 41 $ $ CFACE1,1011,100,101,102,122 ... $ ENDDATA

Main Index

5

+A000012 +A000013

+A000020 +A000021

4 EULER

5

Chapter 4: Fluid-structural Interaction 317 Vortex Shedding with Skin Friction

Vortex Shedding with Skin Friction To illustrate the use of skin friction, the flow of gas around a 2-D cylinder will be simulated. For information on boundary layers and the Reynolds number, see the Dytran User’s Guide, Chapter 3: Constraints and Loading, Viscosity and Skin Friction in Euler . By default shear stresses at walls are derived from the difference between tangential Euler element velocity and wall velocity. If the characteristic element size is larger than the boundary layer then the wall shear stress is underestimated. An estimate of the size of this layer can be obtained using the Reynolds number as described in chapter 3. If the boundary layer is too small it is preferred to base the shear stress on the skin friction coefficient as given by : tw C f = ------------1 --- ρu 2 2

Here , t w is the local shear stress, p and u are, respectively, the element density and the relative tangential velocity in the Euler element adjacent to the wall.

Problem Description For this example flow conditions are chosen such that the Reynolds number is about 144444. For such a high Reynolds number the boundary layer is much smaller than the size of an Euler element and therefore skin friction is required to get a realistic shear stress at the surface of the cylinder. Friction coefficients have to be taken from literature, experiment or fine tuning. Here the Friction coefficient will be taken from literature. At the Reynolds number of 144444 a realistic value for the drag coefficient C D is about 1.2. The drag coefficient is defined as: D C D = -----------------1 --- ρ U 2 d 2

where D is the 2-D drag force, U the main stream velocity and d is the diameter of the cylinder. It is not possible to specify the drag coefficient directly in Dytran. Instead, the skin friction can be specified. To determine the skin friction coefficient, a few trial runs are made to adjust the C f until the drag force approximately reaches C D = 1.2 . After a few runs, it is found out that using C f = 0.095 results in C D = 1.30 which is approximate enough for the purpose of this example. The shedding of the vortices becomes visible when the flow has reached steady state. Vortices are then alternatively shed from the top and bottom half of the cylinder. This gives rise to a periodically varying force operating in a diection perpendicular to the flow direction. The flow direction is taken as + x .

Main Index

318 Dytran Example Problem Manual Vortex Shedding with Skin Friction

Dytran Model For this example, a 2-D model is used with an Euler Mesh of 0.8 x 0.8 m and element thickness of 0.01 m. The number of elements in X and Y directions is 90. The diameter of the cylinder is 0.1 m. The cylinder is modeled as a rigid coupling surface that can’t move. The initial conditions in the Euler region are: Density = 1.3 kg/m3 Specific Internal Energy = 2.E+5 J/kg The boundary conditions at the borders of the Euler mesh are as follows: • At the left side, an Inflow boundary with the following data: • Density = 1.3 kg/m3, Specific Internal Energy = 2.E+5 J/kg, X-Velocity = 20 m/s

Y-Velocity

= 0 and Z-Velocity =0 • At the right side, a transmitting Outflow boundary • At the top and the bottom side, no boundaries are defined which means that these sides are

modeled as walls.

Main Index

Chapter 4: Fluid-structural Interaction 319 Vortex Shedding with Skin Friction

The material properties of Air are as follows: Density = 1.3 kg/m3 Dynamic Viscosity = 18.E-6 Pa s Gamma value = 1.4 -------- Material GAS id =2 DMAT 2 1.3 2 EOSGAM 2 1.4 PMINC 2 0.

2 18.E-6

The initial values for the Euler region are as follows: $ ------- TICVAL BC init ----TICVAL 28 DENSITY $

1.3

SIE

200000.

The coupling surface is similar to a wall and for coupling surface segments shear stresses can be computed using a skin friction coefficient The skin friction factor C f of 0.095 is specified on the COUPLEoption as: COUPLE 1 1 INSIDE + + + The flow boundary conditions are: $ ------- Flow BC flowin ----TLOAD1 1 29 FLOW 29 1 FLOW + SIE 200000. XVEL

$ ------- Flow BC flowout ----TLOAD1 1 32 FLOW 32 2 FLOW

Main Index

ON

ON

STANDARD+

.095

4 INMATERIAL 20.

4 OUT

2 DENSITY

1.3+

320 Dytran Example Problem Manual Vortex Shedding with Skin Friction

Results The figures below show the velocity plots at two different times around 0.5 second. In these plots, the shedding of vortices is visible.

Similar plots at the time of about 1 second.

Main Index

Chapter 4: Fluid-structural Interaction 321 Vortex Shedding with Skin Friction

The Time History plots of the FORCE is shown in the following picture

The X-force slowly increases. A representative value is 0.34. This given a

CD

of

D 0.34 ⁄ 0.01 34 C D = ------------------ = ------------------------------------------------------ = ------ = 1.3 0.5 ⋅ 1.3 ⋅ 20 ⋅ 20 ⋅ 0.1 26 1 --- pU 2 A 2

The Y-Force oscillates due to the alternate shedding of vortices. The time period of this oscillation is 0.021 s. To compare this with experiment the Strouhal number given by nd d S t = ------ = -------U TU

is computed. Here n is the frequency of the vortex shedding, d the diameter, U the main flow velocity and T the time period of the vortex shedding. For this simulation the value is 0.23. This is close to the value of 0.21 found in literature.

Main Index

322 Dytran Example Problem Manual Vortex Shedding with Skin Friction

Abbreviated Dytran Input File START CEND ENDTIME=1.0 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: skin TLOAD=1 TIC=1 SPC=1 $ Output result for request: EUL TYPE (EUL) = ARCHIVE ELEMENTS (EUL) = 1 SET 1 = ALLEULHYDRO ELOUT (EUL) = XVEL YVEL ZVEL DENSITY SIE PRESSURE, TXX TYY TXY EFFSTS TIMES (EUL) = 0 THRU END BY 0.005 SAVE (EUL) = 10000 $ Output result for request: COUP TYPE (COUP) = ARCHIVE CPLSURFS (COUP) = 2 SET 2 = 1 CPLSOUT (COUP) = XVEL YVEL ZVEL DENSITY PRESSURE, TXX TYY TXY EFFSTS TIMES (COUP) = 0 THRU END BY 0.005 SAVE (COUP) = 10000 $ $ Output result for request: COUPTHS TYPE (COUPTHS) = TIMEHIS CPLSURFS (COUPTHS) = 3 SET 3 = 1 CPLSOUT (COUPTHS) = XFORCE YFORCE ZFORCE RFORCE TIMES (COUPTHS) = 0 THRU END BY 0.001 SAVE (COUPTHS) = 10000 $ TYPE (PROF) = ARCHIVE ELEMENTS (PROF) = 4 SET 4 = 4231 THRU 4320 ELOUT (PROF) = XVEL YVEL ZVEL DENSITY PRESSURE, TXX TYY TXY EFFSTS TIMES (PROF) = 0 THRU END BY 0.001 SAVE (PROF) = 10000 $ TYPE (MAT) = TIMEHIS MATS (MAT) = 5 SET 5 = 2 MATOUT (MAT) = MASS, EKIN, EINT TIMES (MAT) = 0 THRU END BY 0.001 SAVE (MAT) = 10000 $ $ Output result for request: COUP THS $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,EULTRAN,IMPULSE

Main Index

Chapter 4: Fluid-structural Interaction 323 Vortex Shedding with Skin Friction

PARAM,FASTCOUP PARAM,BULKL,0.2 PARAM,INISTEP,1.E-7 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE skin.bdf $ $ ========== PROPERTY SETS ========== $ $ * p1 * $ PSHELL 1 1 .001 $ $ * peul * $ PEULER1 2 HYDRO 31 $ $ * p2 * $ PSHELL1 3 DUMMY $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material matrig id =1 MATRIG 1 7800 2.1e+11 .3 $ $ -------- Material GAS id =2 DMAT 2 1.3 2 EOSGAM 2 1.4 18.E-6 PMINC 2 0. $ $ ======== Load Cases ======================== $ $ $ ------- General Coupling: coup ----$ COUPLE 1 1 INSIDE ON ON + + .095 $ SURFACE 1 ELEM 2 SET1 2 1 THRU 180 $ $ ------- Rigid Body Object MR1 ----$ ---- Reference Node for Rigid body is 122 TLOAD1 1 20 12 FORCE 20 MR1 0 1 1 TLOAD1 1 1020 12 MOMENT 1020 MR1 0 1 1 $ $ ------- TICVAL BC init ----TICVAL 28 DENSITY 1.3 SIE 200000.

Main Index

2

STANDARD+ +

1 1

324 Dytran Example Problem Manual Vortex Shedding with Skin Friction

$ $ ------- Flow BC flowin ----TLOAD1 1 29 FLOW 29 1 FLOW + SIE 200000. XVEL CFACE 1 1 181 CFACE 2 1 271 CFACE 3 1 361 CFACE 4 1 451 CFACE 5 1 541 CFACE 6 1 631 CFACE 7 1 721 CFACE 8 1 811 CFACE 9 1 901 CFACE 10 1 991 CFACE 11 1 1081 CFACE 12 1 1171 CFACE 13 1 1261 CFACE 14 1 1351 CFACE 15 1 1441 CFACE 16 1 1531 CFACE 17 1 1621 CFACE 18 1 1711 CFACE 19 1 1801 CFACE 20 1 1891 CFACE 21 1 1981 CFACE 22 1 2071 CFACE 23 1 2161 CFACE 24 1 2251 CFACE 25 1 2341 CFACE 26 1 2431 CFACE 27 1 2521 CFACE 28 1 2611 CFACE 29 1 2701 CFACE 30 1 2791 CFACE 31 1 2881 CFACE 32 1 2971 CFACE 33 1 3061 CFACE 34 1 3151 CFACE 35 1 3241 CFACE 36 1 3331 CFACE 37 1 3421 CFACE 38 1 3511 CFACE 39 1 3601 CFACE 40 1 3691 CFACE 41 1 3781 CFACE 42 1 3871 CFACE 43 1 3961 CFACE 44 1 4051 CFACE 45 1 4141 CFACE 46 1 4231 CFACE 47 1 4321 CFACE 48 1 4411 CFACE 49 1 4501

Main Index

4 INMATERIAL 20. 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

2 DENSITY

1.3+

Chapter 4: Fluid-structural Interaction 325 Vortex Shedding with Skin Friction

CFACE 50 1 4591 CFACE 51 1 4681 CFACE 52 1 4771 CFACE 53 1 4861 CFACE 54 1 4951 CFACE 55 1 5041 CFACE 56 1 5131 CFACE 57 1 5221 CFACE 58 1 5311 CFACE 59 1 5401 CFACE 60 1 5491 CFACE 61 1 5581 CFACE 62 1 5671 CFACE 63 1 5761 CFACE 64 1 5851 CFACE 65 1 5941 CFACE 66 1 6031 CFACE 67 1 6121 CFACE 68 1 6211 CFACE 69 1 6301 CFACE 70 1 6391 CFACE 71 1 6481 CFACE 72 1 6571 CFACE 73 1 6661 CFACE 74 1 6751 CFACE 75 1 6841 CFACE 76 1 6931 CFACE 77 1 7021 CFACE 78 1 7111 CFACE 79 1 7201 CFACE 80 1 7291 CFACE 81 1 7381 CFACE 82 1 7471 CFACE 83 1 7561 CFACE 84 1 7651 CFACE 85 1 7741 CFACE 86 1 7831 CFACE 87 1 7921 CFACE 88 1 8011 CFACE 89 1 8101 CFACE 90 1 8191 $ $ ------- TICEUL BC reg ----TICEUL 31 + SPHERE 2 2 SPHERE 2 0 $ $ ------- Flow BC flowout ----TLOAD1 1 32 FLOW 32 2 FLOW $ CFACE 91 2 270 CFACE 92 2 360 CFACE 93 2 450

Main Index

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

+ 28 0

4 OUT 6 6 6

5 0

10

326 Dytran Example Problem Manual Vortex Shedding with Skin Friction

CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE

Main Index

94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

540 630 720 810 900 990 1080 1170 1260 1350 1440 1530 1620 1710 1800 1890 1980 2070 2160 2250 2340 2430 2520 2610 2700 2790 2880 2970 3060 3150 3240 3330 3420 3510 3600 3690 3780 3870 3960 4050 4140 4230 4320 4410 4500 4590 4680 4770 4860 4950 5040 5130 5220 5310

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Chapter 4: Fluid-structural Interaction 327 Vortex Shedding with Skin Friction

CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE CFACE $ $ ENDDATA

Main Index

148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

5400 5490 5580 5670 5760 5850 5940 6030 6120 6210 6300 6390 6480 6570 6660 6750 6840 6930 7020 7110 7200 7290 7380 7470 7560 7650 7740 7830 7920 8010 8100 8190 8280

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

328 Dytran Example Problem Manual Geometric Eulerian Boundary Conditions

Geometric Eulerian Boundary Conditions Problem Description For many applications, the Euler modeling can be simplified by using the MESH Generator entry. However this method has some shortcomings with regard to the definition of the boundary conditions at the boundaries of the Euler meshes. In the past if flow boundaries were required one had to create CFACES with Patran. Since CFACES only support CHEXA’s, the only way to define flow boundaries for MESH,BOX was to define a PORFLOW entry and a COUPLE entry. Geometric Euler Boundary Condition entries allow to directly prescribing boundary conditions on the boundaries of the Euler domain. This functionality also supports Euler domains consisting of CHEXA’s. Running the simulation with different mesh-sizes is much easier, since the creation of CHEXA’s and CFACES by Patran can be avoided. The geometric boundary entries are FLOWDIR, WALLDIR, FLOWSQ, and FLOWTSQ. The first two allow the assignment of a boundary condition to all Eulerian boundary faces pointing in a certain direction. FLOWDIR allows a general flow boundary condition, whereas WALLDIR applies a WALLET boundary condition. The FLOWSQ and FLOWTSQ assign flow and time dependent flow conditions to all parts of Eulerian boundary faces that are within a given square. Geometric boundary condition can only apply to the boundary of the domain. Internal WALLETs cannot be defined by geometric boundary conditions. FLOW, FLOWDIR, and FLOWDEF can be used in one input deck. If a face is referred to by multiple boundary conditions, then the FLOWSQ overrules all others, If it not present, FLOWDIR and WALLDIR overrule the FLOW, WALLET, FLOWDEF definitions for the face under consideration. Also the example Modeling Blast Wave using 1-D Spherical Symmetry Method in Chapter 3 illustrates the use of FLOWDIR.

This example illustrates use of Geometric Euler Boundary condition in the Vortex shedding with skin friction calculation. For a description of the problem refer to Example Problem 4.17.

Dytran Model For this example, a 2-D model is used with an Euler Mesh of 0.8 x 0.8 m and element thickness of 0.01 m. The number of elements in X and Y directions is 90. The diameter of the cylinder is 0.1 m. The cylinder is modeled as a rigid coupling surface that can’t move. The initial conditions in the Euler region are: Specific internal energy = 2.E+5 Jkg Density

Main Index

= 1.3 kg/m3

Chapter 4: Fluid-structural Interaction 329 Geometric Eulerian Boundary Conditions

The boundary conditions at the borders of the Euler mesh are as follows: • At the left side, an Inflow boundary with the following data:

Specific internal energy = 2.E+5 Jkg Density

= 1.3 kg/m3

X-Velocity

= 20 m/s

Y-Velocity

= 0

Z-Velocity

= 0

• At the right side, a transmitting Outflow boundary. • At the top and the bottom side, no boundaries are defined which means that these sides are

modeled as walls.

FLOW: IN XVEL = 20 m/s YVEL = 0 m/s SIE = 2.E+5 J/kg DENSITY= 1.3 kg/m3

The material properties of Air are as follows: Density = 1.3 kg/m3 Dynamic Viscosity = 18.E-6 Pa s Gamma value = 1.4 $ -------- Material GAS id =2 DMAT 2 1.3 2 EOSGAM 2 1.4 PMINC 2 0. $

Main Index

2 18.E-6

330 Dytran Example Problem Manual Geometric Eulerian Boundary Conditions

The Euler Mesh and the initial values for the Euler region are as follows: $ MESH 22 BOX + -.40 -.40 0.0 + 90 90 1 $ $$ ------- TICEUL BC reg ----TICEUL 31 + SPHERE 2 2 SPHERE 2 0 $ $ ------- TICVAL BC init ----TICVAL 28 DENSITY $

.80

.80

.01 EULER

+ + 2 +

28

5 0

0

1.3

SIE

10 200000.

The coupling surface is similar to a wall and for coupling surface segments shear stresses can be computed using a skin friction coefficient. The skin friction factor of 0.095 is specified on the COUPLE option as: $ ------- General Coupling: coup ----$ COUPLE 1 1 INSIDE ON + + $ SURFACE SET1 $

ON

STANDARD+ +

22 1 2

1

.095 ELEM THRU

2 180

The Geometric boundary conditions are defined as followed: The inflow boundary condition has to be put on all Euler boundary faces that point in the negative xdirection. So one has to go over all Euler boundary faces. Compute the normal and if it points in the negative x direction than the face gets the inflow boundary condition. The following geometric boundary condition just does this. FLOWDIR, 7, HYDRO, 22, NEGX,,,,,+ +,FLOW,IN,MATERIAL,2,DENSITY,1.3,SIE,200000.,+ +, XVEL, 20. Here NEGX means in the negative x-direction. Other allowed values on this field are POSX, NEGY, POSY, NEGZ, and POSZ. The area of boundary faces pointing in the negative x-direction is 0.01*0.8 = 0.008. This can be checked with the OUT file that gives as message PRESCRIBED FLOW BOUNDARY – FLOWDR7 ------------------------

Main Index

Chapter 4: Fluid-structural Interaction 331 Geometric Eulerian Boundary Conditions

ALL FACES IN THE NEGATIVE X DIRECTION TOTAL AREA BOUNDARY FACES:

8.0000E-03

The OUTFLOW boundary condition has to be put on all boundary faces that point in the positive x-direction. This is done by FLOWDIR, 8, HYDRO, 22, POSX,,,,,+ +, FLOW, OUT The boundary faces pointing in the negative and positive y-direction are assigned a WALLET condition by: WALLDIR, 5, HYDRO, 22, NEGY WALLDIR, 6, HYDRO, 22, POSY To illustrate the use of FLOWSQ entry, the FLOWDIR entry used to define the inflow could be replaced by: FLOWSQ,6,MMHYDRO,,,,,,,+ +,-0.4,,-0.4,0.4,0,0.1,,,+ +, FLOW,IN,MATERIAL,2,DENSITY,1.3,SIE,200000.,+ +, XVEL, 20. This specifies a flow condition for the square (x=-0.4, y=-0.4 to y=0.4 and z=0 to 0.01). The FLOWSQ condition is imposed by going over all boundary faces. Compute the intersection of each Euler face with the square and assign to any non empty intersection the condition. Also, here the total area assigned to the boundary condition is written out in the OUT file.

Main Index

332 Dytran Example Problem Manual Geometric Eulerian Boundary Conditions

Results The figures below show the velocity plots at time around 0.25 and 0.50 seconds. For comparison, at the right side the reference result plots from the Example Problem 4.17 have been added. New model with Mesh and Geometric BC

Old model with Grids and Chexa’s

Results at the time around 0.495 seconds. New model with Mesh and Geometric BC

Main Index

Old model with Grids and Chexa’s

Chapter 4: Fluid-structural Interaction 333 Geometric Eulerian Boundary Conditions

Results at the time around 0.500 seconds. New model with Mesh and Geometric BC

Old model with Grids and Chexa’s

The Time History plots of the FORCE and the Mass are shown in the following pictures: New model with Mesh and Geometric BC

Main Index

Old model with Grids and Chexa’s

334 Dytran Example Problem Manual Geometric Eulerian Boundary Conditions

New model with Mesh and Geometric BC

Old model with Grids and Chexa’s

The figures above show that this new modeling method gives exactly the same results as those of the old traditional method.

Abbreviated Dytran Input File START CEND ENDTIME=0.5 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: skin TLOAD=1 TIC=1 SPC=1 $ $ $ Output result for request: EUL TYPE (EUL) = ARCHIVE ELEMENTS (EUL) = 1 SET 1 = ALLEULHYDRO ELOUT (EUL) = XVEL YVEL ZVEL DENSITY SIE PRESSURE, TXX TYY TXY EFFSTS TIMES (EUL) = 0 THRU END BY 0.005 SAVE (EUL) = 10000 $ Output result for request: COUP TYPE (COUP) = ARCHIVE CPLSURFS (COUP) = 2 SET 2 = 1 CPLSOUT (COUP) = XVEL YVEL ZVEL DENSITY PRESSURE, TXX TYY TXY EFFSTS TIMES (COUP) = 0 THRU END BY 0.005 SAVE (COUP) = 10000 $ $ Output result for request: COUPTHS TYPE (COUPTHS) = TIMEHIS CPLSURFS (COUPTHS) = 3 SET 3 = 1 CPLSOUT (COUPTHS) = XFORCE YFORCE ZFORCE RFORCE

Main Index

Chapter 4: Fluid-structural Interaction 335 Geometric Eulerian Boundary Conditions

TIMES (COUPTHS) = 0 THRU END BY 0.001 SAVE (COUPTHS) = 10000 $ $ TYPE (MAT) = TIMEHIS MATS (MAT) = 5 SET 5 = 2 MATOUT (MAT) = MASS, EKIN, EINT TIMES (MAT) = 0 THRU END BY 0.001 SAVE (MAT) = 10000 $ $ $ Output result for request: COUP THS $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,EULTRAN,IMPULSE PARAM,FASTCOUP PARAM,BULKL,0.2 PARAM,INISTEP,1.E-7 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE skin.bdf $ $ ========== PROPERTY SETS ========== $ $ * p1 * $ PSHELL 1 1 .001 $ $ * peul * $ PEULER1 2 HYDRO 31 $ $ * p2 * $ PSHELL1 3 DUMMY $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material matrig id =1 MATRIG 1 7800 2.1e+11 .3 $ $ -------- Material GAS id =2 DMAT 2 1.3 2 EOSGAM 2 1.4 PMINC 2 0. $ $ ======== Load Cases ======================== $ $ $ ------- General Coupling: coup ----$

Main Index

2 18.E-6

336 Dytran Example Problem Manual Geometric Eulerian Boundary Conditions

COUPLE

1

1

INSIDE

ON

ON

+ + 22 .095 $ SURFACE 1 ELEM 2 SET1 2 1 THRU 180 $ $ ------- Rigid Body Object MR1 ----$ ---- Reference Node for Rigid body is 122 TLOAD1 1 20 12 FORCE 20 MR1 0 1 1 TLOAD1 1 1020 12 MOMENT 1020 MR1 0 1 1 $ MESH 22 BOX + -.40 -.40 0.0 .80 .80 .01 + 90 90 1 $ $$ ------- TICEUL BC reg ----TICEUL 31 + SPHERE 2 2 28 5 SPHERE 2 0 0 0 10 $ $ ------- TICVAL BC init ----TICVAL 28 DENSITY 1.3 SIE 200000. $ $--Geometric Boundary Condition--------$ WALLDIR,5,HYDRO,22,NEGY WALLDIR,6,HYDRO,22,POSY FLOWDIR,7,HYDRO,22,NEGX,,,,,+ +,FLOW,IN,MATERIAL,2,DENSITY,1.3,SIE,200000.,+ +,XVEL,20. FLOWDIR,8,HYDRO,22,POSX,,,,,+ +,FLOW,OUT $ ENDDATA

Main Index

STANDARD+ +

1 1

EULER

+ + 2 +

Chapter 4: Fluid-structural Interaction 337 Cohesive Friction

Cohesive Friction Problem Description Wet soil that is in contact with a tire can stick to it and show cohesive behavior. The friction model at the coupling surface modeling the tire should predict this viscous behavior. This example illustrates the use of the cohesive friction model. A rigid wedge is pulled out of wet soil. Initially the wedge is 80% immerged in the soil. The simulation will be run for 80 ms.

Dytran Modeling The soil is modeled by: $ -------- Material Soil id =2 DMAT 2 1500 2 EOSPOL 2 2e+07 SHREL 2 1e+07 YLDMC 2 6.0e+07 2e+05 PMINC 2 -5e+8

2

2

2

1.46

The elements are defined by MESH,BOX: MESH + +

Main Index

2

BOX

-.2 -0.0101 160

8

+ 0.0

0.4 0.0202 80

0.2

+ EULER

1

338 Dytran Example Problem Manual Cohesive Friction

and are initialized by PEULER1 1 MMSTREN 6 TICVAL 4 DENSITY 1500 $ $ ------- TICEUL BC PEuler_reg ----TICEUL 6 +

CYLINDER

5

+ CYLINDER CYLINDER 5 + CYLINDER

+ 1

3

2

4

+ 2

0

0

-2

0

0

2+

0

0

-1

0

0

0.1+

1 3

+

1

Here the multi-material solver will be used. The wedge is defined by: MATRIG

1

100

+

+

+

+

0.00

GRID 1 .020000 -.03000 GRID 2 -.020000 -.03000 GRID 3 .000000 -.03000 GRID 4 -.020000 .03000 GRID 5 .000000 .03000 GRID 9 .020000 .03000 $ $ --- Define 5 elements $ $ -------- property set dum --------CTRIA3 1 2 1 2 CQUAD4 2 2 4 5 CTRIA3 3 2 5 4 CQUAD4 4 2 1 3 CQUAD4 5 2 1 9

.120000 .120000 .020000 .120000 .020000 .120000

3 3 9 5 4

2 9 2

The interaction between wedge and Euler material is implemented by the COUPLE entry: COUPLE

1

1 INSIDE

+

0.8

+ $

2

Here a Coulomb friction model is used.

Main Index

ON 0.8

ON

STANDARD+ +

Chapter 4: Fluid-structural Interaction 339 Cohesive Friction

The COUPLE entry requires a surface that is wrapped around the structure. Since the structure consists of shell element, the wedge elements themselves can be used to define the wrapping surface. SURFACE SET1

1 6

1

ELEM THRU

6 5

The motion of the wedge is given by: $ ---- No reference node is used. TLOAD1 1 7 FORCE 7 MR1 TLOAD1 1 1007 MOMENT 1007 MR1 $ tabled1,1,,,,,,,,+ +,0,0,0.2,1,endt

12 1 12 1

1 0

0

2.0

0

0

0

and is shown in the figure below.

To extend the Coulomb friction model to account for cohesive conditions: PARAM,COHESION,8e+10,8e+5,2.0 is added. This allows soil to stick to some extend to the wedge that is pulled up. The 8e+10 is the maximum tensile stress. The 8e+5 and 2.0 describe a viscous-like friction law. This law specifies a friction stress that is linear in the relative tangential velocity between material and wedge. Under compressive conditions the Coulomb friction model will be used as specified on the COUPLE entry. Only for tensile conditions will the model be used as specified on the COHESION entry.

Main Index

340 Dytran Example Problem Manual Cohesive Friction

Results The pictures below show FMATPLT results for respectively 0 ms, 40 ms and 80 ms. The soils sticks to the wedge.

Main Index

Chapter 4: Fluid-structural Interaction 341 Cohesive Friction

Dytran Input Deck START CEND ENDTIME=0.1 ENDSTEP=9999999 CHECK=NO TITLE= Jobname is: Wedge1910 TLOAD=1 TIC=1 SPC=1 $ Output result for request: Coupling TYPE (Coupling) = ARCHIVE CPLSURFS (Coupling) = 1 SET 1 = 1 CPLSOUT (Coupling) = PRESSURE TYY TZZ TXY TYZ TZX EFFSTS SXX TIMES (Coupling) = 0 THRU END BY 0.001 SAVE (Coupling) = 10000 $ Output result for request: Coupling_vel TYPE (Coupling_vel) = TIMEHIS CPLSURFS (Coupling_vel) = 2 SET 2 = 1 CPLSOUT (Coupling_vel) = XFORCE YFORCE ZFORCE RFORCE STEPS (Coupling_vel) = 0 THRU END BY 1 SAVE (Coupling_vel) = 10000 $ Output result for request: Pen TYPE (Pen) = ARCHIVE ELEMENTS (Pen) = 3 SET 3 = 1 THRU 5 ELOUT (Pen) = EXUSER1 TIMES (Pen) = 0 THRU END BY 0.001 SAVE (Pen) = 10000 $ Output result for request: Pen_vel TYPE (Pen_vel) = TIMEHIS RIGIDS (Pen_vel) = 4 SET 4 = mr1 RBOUT (Pen_vel) = ZCG ZVEL ZACC ZFORCE STEPS (Pen_vel) = 0 THRU END BY 1 SAVE (Pen_vel) = 10000 $ Output result for request: Soil TYPE (Soil) = ARCHIVE ELEMENTS (Soil) = 5 SET 5 = ALLMULTIEULSTRENGTH ELOUT (Soil) = XVEL YVEL ZVEL DENSITY PRESSURE FMAT2 FMATPLT, TXX TYY TZZ TXY TYZ TZX EFFSTS VOID FVUNC TIMES (Soil) = 0 THRU END BY 0.001 SAVE (Soil) = 10000 $------- Parameter Section -----PARAM,BULKL,1.0 PARAM,CONTACT,THICK,0.0 PARAM,FASTCOUP PARAM,INISTEP,1e-6 PARAM,VELMAX,5.0,NO $------- BULK DATA SECTION ------BEGIN BULK

Main Index

342 Dytran Example Problem Manual Cohesive Friction

$ $ PARAM,COHESION,8e+10,8e+5,2.0 $ --- Define 18 grid points --$ GRID 1 .020000 -.03000 .120000 GRID 2 -.020000 -.03000 .120000 GRID 3 .000000 -.03000 .020000 GRID 4 -.020000 .03000 .120000 GRID 5 .000000 .03000 .020000 GRID 9 .020000 .03000 .120000 $ $ --- Define 5 elements $ $ -------- property set dum --------CTRIA3 1 2 1 2 3 CQUAD4 2 2 4 5 3 CTRIA3 3 2 5 4 9 CQUAD4 4 2 1 3 5 CQUAD4 5 2 1 9 4 $ ========== PROPERTY SETS ========== $ $ * PEul_MMStren * $ PEULER1 1 MMSTREN 6 $ $ * Pen * $ PSHELL 2 1 .001 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Pen_rigid id =1 MATRIG 1 100 + + 0.00 $ $ -------- Material Soil id =2 DMAT 2 1500 2 2 2 EOSPOL 2 2e+07 SHREL 2 1e+07 YLDMC 2 6.0e+07 2e+05 1.46 PMINC 2 -5e+8 $ $ ======== Load Cases ======================== $ $ $ ------- General Coupling: Coup_pen ----$ COUPLE 1 1 INSIDE ON ON + 0.8 0.8 + 2

Main Index

2 9 2

+ +

2

STANDARD+ +

Chapter 4: Fluid-structural Interaction 343 Cohesive Friction

$ SURFACE 1 ELEM 6 SET1 6 1 THRU 5 $ $ ------- Mesh Box: Mesh_box $ MESH 2 BOX + -.2 -0.0101 0.0 0.4 0.0202 + 160 8 80 $ $ ------- TICVAL BC Soil_init ----TICVAL 4 DENSITY 1500 $ $ ------- TICEUL BC PEuler_reg ----TICEUL 6 + CYLINDER 5 1 + CYLINDER 3 2 4 2 CYLINDER 5 0 0 -2 + 1 CYLINDER 3 0 0 -1 + 1 $ $ ------- Rigid Body Object RBO_Pen ----$ ---- No reference node is used. TLOAD1 1 7 12 1 FORCE 7 MR1 1 0 TLOAD1 1 1007 12 MOMENT 1007 MR1 1 0 $ tabled1,1,,,,,,,,+ +,0,0,0.2,1,endt $ ENDDATA

Main Index

+ + 1

0.2 EULER

+ + 0

0

2+

0

0

0.1+

0

2.0

0

0

344 Dytran Example Problem Manual Cohesive Friction

Main Index

Chapter 5: Forming Dytran Example Problem Manual

5

Main Index

Forming J

Overview

J

Square Cup Deep Drawing

J

Deep Drawing of a Cylindrical Cup

J

Three-point Bending Test

368

J

Sleeve Section Stamping

379

346 347 359

346 Dytran Example Problem Manual Overview

Overview In this special chapter, a number of example problems are presented that highlight the capabilities of Dytran in the area of sheet-metal forming. The user can find in these examples guidelines how to model sheet-metal forming problems, how to use the special anisotropic material model for sheet-metal forming, how to model contact between dies and sheet and how to apply loads and constraints. All examples show a correlation with experiments.

Main Index

Chapter 5: Forming 347 Square Cup Deep Drawing

Square Cup Deep Drawing Problem Description This case, considers a typical metal-forming example, that of a plate of anisotropic sheet metal being drawn through a square hole by means of a punch. This particular example has experimental results as it was provided as a verification problem for participants of the 1993 NUMISHEET Conference (see [Ref. 1.]) held in Japan. The analysis involved results obtained at single punch depth (20 mm punch travel) for an aluminum alloy plate. The material is seen to be planarly anisotropic, i.e., the material behavior is different in all directions in the plane of the sheet metal as well as in the out of plane direction. The state-of-the-art material model in Dytran, developed originally by Dr. Raymond Krieg of the University of Tennessee (see [Ref. 2.]), is capable of capturing such behavior of the sheet metal, which can be vital for the modeling of the actual behavior exhibits by sheet metal. The data obtained from the NUMISHEET Conference were as follows: Aluminum Alloy • Thickness = 0.81 mm • Young’s modulus = 71 GPa • Poisson’s ratio = 0.33 • Density = 2700 kg/m3 • Yield stress = 135.3 MPa • Stress = 576.79 * (0.01658 + εp)0.3593 MPa • Lankford parameters: R0 = 0.71, R45 = 0.58, R90 = 0.70 • Friction coefficient = 0.162

Size of the plate modeled was 0.15 x 0.15 (in meters). No strain-rate dependency effects were included in the material data, so the metal sheet was analyzed without these effects.

Main Index

348 Dytran Example Problem Manual Square Cup Deep Drawing

The dimensions of the plate, die, punch, and clamp are all given in Figure 5-1 .

Figure 5-1

Main Index

Dimensions of Plate, Die, Punch, and Clamp (in Millimeters)

Chapter 5: Forming 349 Square Cup Deep Drawing

Dytran Model The essential components in this finite element model are: • the sheet metal • the punch • the die • the clamp

The Dytran model of the above components is given in Figure 5-2, followed by descriptions of each component:

Figure 5-2

Dytran Model (Exploded View)

The Sheet Metal The Dytran material model for sheet metals is a highly sophisticated model and includes full anisotropic behavior, strain-rate effects, and customized output options that are dependent on material choice. Since not all of the materials can be derived from the simplified set given by the NUMISHEET organization most participants in the conference used an isotropic material model. In reality, the process is definitely anisotropic and effect due to these differences can be seen in the transverse direction. For materials displaying in-plane anisotropic behavior, the effect would be even more noticeable. The parameters on the SHEETMAT (refer to the Dytran Reference Manual, [Ref. 3.) entry specify planar anisotropic behavior and are as follows (for the aluminum sheet): • SHEETMAT elastic material properties entry. • Isotropic behavior was assumed in the elastic range:

Exx = 71.0 GPa

υ = 0.33

Main Index

350 Dytran Example Problem Manual Square Cup Deep Drawing

The material “rolling” direction equals the global x-direction so the material vector XMAT,YMAT,ZMAT = {1,0,0}. • Planar anisotropic yielding and isotropic hardening were assumed in the plastic range:

A = Stress constant = 0.0 MPa B = Hardening modulus = 576.79 MPa C = Strain offset = 0.01658 n = Exponent for power-law hardening = 0.3593 • Lankford parameters:

R0 = 0.71 R45 = 0.58 R90 = 0.70 Note:

The strain ratio RΘ is found by carrying tensile test in the corresponding “rolling” direction, Θ, of the metal sheet and is equal to the ratio of strain in the in-plane “cross rolling” direction to the out-of-plane direction. As such, it is an indication of the orthotropic behavior in the transverse direction.

• Material output

If sufficient material data is θ supplied, the user can specify the Forming Limit Parameter as output. This is an extremely useful variable for metal-forming engineers and is normally difficult to calculate from normal output. For this reason, Dytran automatically calculates this parameters for each integration layer allowing the user to make contour plots that will immediately reveal where the so-called “forming-limit” has been exceeded. To enable users to get a feel for this parameter, material data based on [4] were specified in this case: C1 = 0.24421, C2 = –0.195, C3 = 0.857187, C4 = 3.43924, C5 = –11.9292 D2 = –0.41688, D8 = –1.5667, D4 = –4.8485, D5 = –6.0606

Main Index

Chapter 5: Forming 351 Square Cup Deep Drawing

Punch, Die, and Clamp These three components provide the constraints and driving displacement for the analysis and are modeled as rigid bodies. DUMMY formulation quadrilateral elements are used to define the three geometric shapes; CFACEs are defined on each quadrilateral; these CFACEs are referred to from a SURFACE entry; this SURFACE entry is then referred to from a RIGID entry defining the rigid body properties for each of the three bodies. Contact is then specified with the metal sheet using the friction coefficient values provided. The three CONTACT entry specify the following: • Contact between the punch and the sheet • Contact between the die and sheet • Contact between the clamp and sheet

Lastly, the punch is given a scaled downward velocity providing the driving displacement for the analysis.

Results The results requested by the NUMISHEET organization were as follows: • Strain distributions along line OB on the die side of the metal sheet as shown in Figure 5-3 through Figure 5-6. The strain measures requested were:

– In-plane major principal strain – In-plane minor principal strain – Out-of-plane thickness strain • A contour plot of the thickness strain plotted upon the deformed shape of the sheet metal. • The amount of “draw-in” DX, DY, and DD measured from the undeformed shape edges to the

edges after deformation along the three lines OA, OB, and OC. Some example plots are given in Figure 5-3 through Figure 5-6. Note that the last result listed above is not included as it involves three simple values indicating the degree of draw-in. Dytran gave a solution well within the spread of experimental values. Figure 5-7 shows a contour plot of the forming limit parameter.

Main Index

352 Dytran Example Problem Manual Square Cup Deep Drawing

Figure 5-3

Contour Plot of Thickness Strain Superimposed on the Deformed Shape (Aluminum Plate at a Punch Depth of 15 mm)

Figure 5-4

Major Principal Strain Distribution along Line OB (Aluminum Plate at a Punch Depth of 15 mm)

Main Index

Chapter 5: Forming 353 Square Cup Deep Drawing

Figure 5-5

Minor Principal Strain Distribution along Line OB (Aluminum Plate at a Punch Depth of 15 mm)

Figure 5-6

Thickness-strain Distribution along Line OB (Aluminum Plate at a Punch Depth of 15 mm)

Main Index

354 Dytran Example Problem Manual Square Cup Deep Drawing

Figure 5-7

Contour Plot of the Forming Limit Parameter (Aluminum Plate at a Punch Depth of 20 mm)

Files sq_cup_alual.dat

Dytran input file

sq_cup_alual_xl.dat SQ_CUP_ALUAL.OUT

Dytran output file

SQ_CUP_ALUAL_PLATE_0.ARC Dytran archive file SQ_CUP_ALUAL_PLATE_0.ARC

Main Index

SQ_CUP_ALUAL_WALL_0.THS

Dytran time history file

alual_plate_15_mm.ext

Dytran translated results file for the aluminium alloy metal sheet corresponding to a punch travel of 15 mm

alual_15_experiment_##.ext

Experimental data obtained for the aluminium alloy corresponding to a punch travel of 15 mm from NUMISHEET ’93 participants

Chapter 5: Forming 355 Square Cup Deep Drawing

References 1. Makinouchi, A., Nakamachi, E., Onate, E., and Wagoner, R. H., “Numerical Simulation of 3-D Sheet Metal Forming Processes, Verification of Simulation with Experiment,” NUMISHEET ‘93 2nd International Conference. 2. Krieg, Prof. Raymond Dl, “Constitutive Model for Sheet Metal Forming,” Part 2, Engineering Science and Mechanics, The University of Tennessee, Knoxville. 3. Dytran Reference Manual, Version 2004, MSC.Software Corporation.

Abbreviated Dytran Input File START $TITLE = Deep Drawing of a Square Cup $ CHECK = NO TLOAD= 1 $ ENDSTEP = 4700 $ $ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS $ TYPE(MYMAT) = MATSUM STEPS(MYMAT) = 0 THRU END BY 1000 $ $ CHANGE THE OUTPUT FREQUENCY FOR MATRIG SUMMS $ TYPE(MRMAT) = MRSUM STEPS(MRMAT) = 0 THRU END BY 1000 $ $ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS $ TYPE(MYCYC) = STEPSUM STEPS(MYCYC) = 0 THRU END BY 250 $ $ Archive of Sheet Metal Output $ Note : Integration Layer 01 corresponds to the die side of the sheet $STEPS(PLATE) = 0,THRU,END,BY,470,3418 TYPE(PLATE) = ARCHIVE SAVE(PLATE) = 100000 ELEMENTS(PLATE) = 60 SET 60 = 1t1600 ELOUT(PLATE) = THICK,EPSMX01,EPSMN01,EZZ01,FLP01 $ $Archive of Rigid Body Output $Note : The dummy user variable ZUSER is used so as to obtain the grid $ point displacements. This enables the user to visualise the $ motion and position of the Rigid Bodies. $STEPS(RIGID) = 0,THRU,END,BY,470,3418 TYPE(RIGID) = ARCHIVE SAVE(RIGID) = 100000 ELEMENTS(RIGID) = 63 SET 63 = 2001t4468

Main Index

356 Dytran Example Problem Manual Square Cup Deep Drawing

ELOUT(RIGID) = ZUSER $$ $$ Time history of specific (i.e. highly strained) elements in the cup $$ wall in order to determine the strain path in a FLD plot. $$ TYPE(wall) = TIMEHIS SAVE(wall) = 999999 ELEMENTS(wall) = 1 SET 1 = 468,492,509 ELOUT(wall) = THICK,EPSMX01,EPSMN01,EZZ01,FLP01 STEPS(wall) = 0,THRU,END,BY,47,3418 $ BEGIN BULK $$ Parameters : $ (i) Initial Time-step PARAM,INISTEP,2.0-7 $ (ii) Update of shell thicknesses PARAM,SHTHICK,YES $ (iii) Hourglass Coefficient PARAM,HGCOEFF,0.05 $THIS SECTION CONTAINS BULK DATA $Geometry Data is input from a seperate file. $INCLUDE sq_cup_alual_xl.dat $$ Element formulation is Belytschko-Lin-Tsay for the blank $ which thickness amounts to 0.81 mm $PSHELL1 1 1 BLT Gauss Mid + .81 $$ Bely element formulations for MATRIG bodies (i.e. punch, binder and$ $PSHELL1 61 2 Bely Gauss Mid + 1.-20 PSHELL1 62 3 Bely Gauss Mid + 1.-20 PSHELL1 63 4 Bely Gauss Mid + 1.-20 $$ Punch $MATRIG 2 210.e9 0.3 1. 0.0 0.0 0.0 + 1.+2 0.0 0.0 1.+2 0.0 1.+2 + 0.0 0.0 0.0 0.0 0.0 0.0 $ $ Binder $ MATRIG 3 210.e9 0.3 1. 0.0 0.0 0.0 + 1.+2 0.0 0.0 1.+2 0.0 1.+2 + 0.0 0.0 0.0 0.0 0.0 0.0 $ $ Die $MATRIG 4 210.e9 0.3 1. 0.0 0.0 0.0 + 1.+2 0.0 0.0 1.+2 0.0 1.+2 + 0.0 0.0 0.0 0.0 0.0 0.0 $$ The steel blank is modelled with the Sheetmat yield model, $ in which the strain rate effect is assumed to be negligible. $ Mass-Scaling by a factor 100 (additional to speed-up of punch). $$ Material Properties : $$ Young’s modulus E = 71.0 GPa, Nu = 0.33, Density = 2.700E-6 kg/mm**3

Main Index

+ die) + + +

+ +

+ +

+ +

Chapter 5: Forming 357 Square Cup Deep Drawing

$$ Planar Anisotropic Material Specification : $$ Power Law Stress Constant, a = 0.0 Pa $ Hardening Modulus, b = 576.79e3 kg/(mm*s**2) $ Hardening exponent, n = 0.3593 $ Strain offset, c = 0.01658 $ Lankford Parameter : R(0) = 0.71, R(45) = 0.58, R(90) = 0.70 $$ The coefficients of the Forming Limit Diagram correspond to engineering values !! $SHEETMAT1 2.7E-4 7.1E7 + + 0.33 ISO 1.0 0.0 0.0 + + 0.0 576.79E3.01658 .3593 + + PLANANI .71 .58 .70 + + ISO + + .24421 -.195 .857187 3.43924 -11.9292 + + -.41688 -1.5667 -4.8485 -6.0606 $$ Prescribed Binder Force: $$ The effect of binder force is examined. Take F=19.6 kN. $TLOAD1 1 9999 13 FORCE 9999 MR3 -19.6E6 1. $$ Prescribed motions: $$ (i) The punch velocity is prescribed as a constant downward value $ of 1000 mm/s (in Z-dir.). $TLOAD1 1 99 12 FORCE 99 MR2 -1000. 1. $ $ (ii) Rotational velocities of MATRIG 2, 3 and 4 (i.e. punch, binder $ and die) are equal to zero along X-, Y- and Z-axes TLOAD1 1 999 12 MOMENT 999 MR2 0. 1. 1. 1. MOMENT 999 MR3 0. 1. 1. 1. MOMENT 999 MR4 0. 1. 1. 1. $ $ (iii) Die has a translational restraint in X-, Y- and Z-direction! FORCE 999 MR4 0. 1. 1. 1. $ $ (iv) Punch and Binder have only a translational restraint in $ X- and Y-direction! FORCE 999 MR2 0. 1. 1. FORCE 999 MR3 0. 1. 1. $$ Geometry description of blank(1), punch(2), binder(3) and die(4) in$ order to model the contact. These surfaces refer to sets of segments$ defined using the CFACE entries for set number 1, 61, 62 and 63,$ respectively. Surface 1: blank $ 2: punch $ 3: binder $ 4: die $ SURFACE 1 SEG 1 SURFACE 2 SEG 61 SURFACE 3 SEG 62 SURFACE 4 SEG 63 $$ Contact specification for which the blank is in all cases the slave: $$ The thickness of the blank is taken into account in the contact$ problem by using THICK = 1.0 $$ (i) Contact between Blank and Punch

Main Index

358 Dytran Example Problem Manual Square Cup Deep Drawing

CONTACT + + + $ (ii) CONTACT + + + $ (iii) CONTACT + + + $ ENDDATA

Main Index

1

SURF Top 0.0 .1 0.0 .04 Contact between 2 SURF Top 0.0 .1 0.0 .04 Contact between 3 SURF Bottom 0.0 .1 0.0 .04

SURF 1 2 Full 1.0 Distance1.+20 1.1 On On 1.+20 Blank and Binder SURF 1 3 Full 1.0 Distance1.+20 1.1 On On 1.+20 Blank and Die SURF 1 4 Full 1.0 Distance1.+20 1.1 On On 1.+20

.162

.162

0.0

.1

Factor

2.

.162

.162

0.0

.1

Factor

2.

.162

.162

0.0

.1

Factor

2.

+ + +

+ + +

+ + +

Chapter 5: Forming 359 Deep Drawing of a Cylindrical Cup

Deep Drawing of a Cylindrical Cup Problem Description Cylindrical cup-drawing tests are generally the best means of studying the effect of process conditions on drawing behavior. Simulation by using Dytran can help the stamping engineer to determine optimum process conditions. For example, analysis can determine that a cylindrical cup is most successfully drawn under variable (instead of constant) blank-holder force conditions. The present problem is a cylindrical cup, which is punched out of 70/30 brass sheet material while the sheet is clamped by a blank holder against a die. The results obtained by using Dytran can be compared with experimental results presented in [Ref.4.]. Drawing Process Data: Blank

100 mm

Blank thickness

0.7 mm

Punch radius

50.0 mm

Punch corner radius

13.0 mm

Die inner radius

51.25 mm

Die corner radius

5 mm

Total punch stroke

40 mm

Punch velocity

1000 mm/s

Blank-holder inner radius

56.25 mm

Blank-holder force

80 kN

Assumed friction coefficient

0.18/0.04

70/30 Brass Material Properties

Main Index

Young’s modulus

E = 100 000 N/mm2

Poisson’s ratio Density

υ = 0.33 π = 8.413 106 kg/mm3

Lankford parameter

R0 = R45 = R90 = 0.85

Hardening modulus

b = 895 N/mm2

Strain offset

c = 2.94 10-4

Strain hardening exponent

n = 0.42

360 Dytran Example Problem Manual Deep Drawing of a Cylindrical Cup

Dytran Model Symmetry allows a one-quarter model to be constructed although a mesh for the whole geometry is shown in Figure 5-8. It is well known that the mesh size of the blank is determined by the radii of the tools. The element side length recommended is at most half the smallest corner radius over which the blank is sliding. Consequently, the blank can be modeled with 348 shell elements (CQUAD4).

Figure 5-8

Dytran Model (Exploded View)

The blank holder is considered as a flat plate. The punch, die, and blank holder are modeled as rigid bodies using the material definition MATRIG entry. All surfaces including the blank are covered with CFACEs. In the three contact operations (i.e. blank punch, blank holder, blank die), the surface of the blank is chosen to be the slave surface because of its mesh fineness. The coefficient of friction between the blank and punch is 0.18, and that between the die and the blank holder is 0.04. The latter value implies a higher degree of lubrication between the surfaces. Since springback is not considered in this example, the entire analysis is carried out in a single step representing the actual drawing stage. The applied force on the blank holder is 80 kN and is kept constant during the drawing stage. This is realized by using the FORCE entry and setting the TYPE field of the TLOAD1 entry equal to 13. The G field in the FORCE entry refers to the property number of the MATRIG. Since all tools are not allowed to have a rotational velocity, the MOMENT entry is used in conjunction with setting the SCALE factor for the moment to zero in all three directions. Similarly, the translational velocities of the tools are kept zero in X- and Y-direction by using the FORCE entry. Because of the velocity prescription, the TYPE field of the TLOAD1 entry is equal to 12.

Main Index

Chapter 5: Forming 361 Deep Drawing of a Cylindrical Cup

In order to permit larger stable time steps, the mass density is increased two orders of magnitude. The punch is moving with a mean velocity of 1 m/s to a depth of 40 mm according to a kind of sine function against time. Similar to the Example Problem (describing Square Cup Deep Drawing), the material behavior of this blank is represented by the SHEETMAT material model. Therefore, no further description on this model will be given in this section. During the drawing process, the membrane behavior of the blank must be modeled accurately. Thickness changes due to membrane stretching are accounted for by putting the parameter SCHTHICK equals YES into the Bulk Data Section.

Results The deformed shape predicted at the end of the stamping process is shown in Figure 5-9.

Figure 5-9

Final Shape of the Cup at 40 mm Punch Stroke

In a drawing process such as this problem, the magnitude of blank-holder force and accompanying friction dictates the extent of metal flow into the die cavity. This is reflected by the magnitude of drawing force as a function of punch stroke. Therefore, a convenient way to validate the FE calculations is to compare the drawing force measured in tests with Dytran results (Figure 5-10). Some oscillating behavior can be observed that is likely to be dependent on process time and coarseness of the mesh used. The experimental results were taken from [Ref. 4.]. The distribution of radial and circumferential strain is shown in Figure 5-9. It can be clearly seen that the flat bottom of the cup is biaxially stretched while drawing prevails in the wall and flange region.

Main Index

362 Dytran Example Problem Manual Deep Drawing of a Cylindrical Cup

Main Index

Figure 5-10

Dytran and Experimental Punch Force

Figure 5-11

Dytran and Experimental Radial and Circumferential Strain Distribution

Chapter 5: Forming 363 Deep Drawing of a Cylindrical Cup

Files cyl_cup.dat

Dytran input file

cyl_cup_xl.dat CYL_CUP.OUT

Dytran output file

CYL_CUP_BLANK_0.ARC CYL_CUP_RIGID_0.ARC

Dytran archive file

CYL_CUP_CONTACT_0.THS CYL_CUP_PUNCH_MOT_0.THS

Dytran time history file

cyl_cup_blank_0_6323

Dytran translated results file corresponding to punch travel of 30mm

experiment_cir.ext

Experimental data obtained from [Ref. 4.] for circumferential strains corresponding to punch travel of 30 mm

experiment_rad.ext

Experimental data obtained from [Ref.4.] for the radial strains corresponding to punch travel of 30 mm

Reference (Continued) 4. Saran, M., Schedin, El, Samuelsson, Al, Melander, A., and Gustafsson, C. “Numerical and Experimental Investigations of Deep Drawing of Metal Sheets”. ASME J. Vol. 112, 272-277 (1990).

Abbreviated Dytran Input File START $TITLE = Cylindrical Cup Deep Drawing $TIME 99999 CEND CHECK = NO ENDSTEP = 9868 SPC = 1 TLOAD = 1 $$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS $TYPE (MYMAT) = MATSUM STEPS (MYMAT) = 0 THRU END BY 1000 $$ CHANGE THE OUTPUT FREQUENCY FOR THE MATRIG SUMMS $TYPE (MYMR) = MRSUM STEPS (MYMR) = 0 THRU END BY 1000 $$ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS $TYPE (MYCYC) = STEPSUM STEPS (MYCYC) = 0 THRU END BY 250 $$ $$ Data for Output Control Set 1 $$ TYPE (blank) = ARCHIVE SAVE (blank) = 999999 ELEMENTS(blank) = 1

Main Index

364 Dytran Example Problem Manual Deep Drawing of a Cylindrical Cup

SET 1 = 1t348 ELOUT (blank) = THICK,EPSMX03,EPSMN03,EZZ03 STEPS (blank) = 0,THRU,END,BY,1000,6323 $$ $$ Data for Output Control Set 2 $$ TYPE (rigid) = ARCHIVE SAVE (rigid) = 999999 ELEMENTS(rigid) = 2 SET 2 = 1000t3035 STEPS (rigid) = 0,THRU,END,BY,1000,6323 ELOUT (rigid) = ZUSER $ $ Time history of the Contact 1 (i.e. Punch) $ CONTS (contact) = 3 SET 3 = 1 TYPE (contact) = TIMEHIS CONTOUT (contact) = YFORCE SAVE (contact) = 99999 STEPS (contact) = 0,THRU,END,BY,100,6323 $ $ Time history of the Punch Motion (midpoint) $ TYPE (punch_mot) = TIMEHIS SAVE (punch_mot) = 99999 GRIDS (punch_mot) = 4 SET 4 = 1010 GPOUT (punch_mot) = YDIS STEPS (punch_mot) = 0,THRU,END,BY,100,6323 $ BEGIN BULK $ PARAM INISTEP 3e-07 PARAM SHTHICK YES $ $ MODEL GEOMETRY $ ============== INCLUDE cyl_cup_xl.dat $ $ Element formulation is Belytschko-Lin-Tsay for the blank $ which thickness amounts to 0.70 mm $ PSHELL1 1 1 BLT Gauss Mid + .70 $ $ Bely element formulations for MATRIG bodies (i.e. punch, die $ and blankholder) $ PSHELL1 1111 2 Bely Gauss Mid + 1.-20 PSHELL1 1112 2 Gauss Mid + 1.-20 PSHELL1 2222 3 Bely Gauss Mid

Main Index

+

+ + +

Chapter 5: Forming 365 Deep Drawing of a Cylindrical Cup

+ 1.-20 PSHELL1 3333 4 Bely Gauss Mid + + 1.-20 $ $ Punch $ ===== MATRIG 2 210.e9 0.3 1. 0.0 0.0 5.0 + 1.+2 0.0 0.0 1.+2 0.0 1.+2 + + 0.0 0.0 0.0 0.0 0.0 0.0 $ $ Die $ === MATRIG 3 210.e9 0.3 1. 0.0 0.0 -20. + + 1.+2 0.0 0.0 1.+2 0.0 1.+2 + + 0.0 0.0 0.0 0.0 0.0 0.0 $ $ Blankholder $ =========== MATRIG 4 210.e9 0.3 1. 0.0 0.0 0.0 + + 1.+2 0.0 0.0 1.+2 0.0 1.+2 + + 0.0 0.0 0.0 0.0 0.0 0.0 $ $ The BR brass blank is modelled with the Sheetmat material model, $ in which the strain rate effect is assumed to be negligible. $ Mass-scaling is applied by a factor of 100. $ $ Material Properties : $ $ Youngs modulus E = 100.0 GPa, Nu = 0.33, Density=8.413e-6 Kg/mm3 $ $ Transversely Isotropic Material Specification : $ $ Power Law Stress Constant, a = 0.0 Pa $ Hardening Modulus, b = 895.0E+3 kg/(mm*s**2) $ Hardening exponent, n = .42 $ Strain offset, c = 2.94E-4 $ Lankford Parameter : R(0) = 0.85, R(45) = 0.85, R(90) = 0.85 +$ $ NOTE : c is chosen such that the initial yield stress is 29 MPa. $ SHEETMAT1 8.413E-41.0E8 + + 0.33 ISO 1.0 0.0 0.0 + + 0.0 895.E3 2.94E-4 0.42 + + NORMANI 0.85 0.85 0.85 + + NORMANI $ $ ----------------------------------------------------------------------$ $ Prescribed Binder Force: $ $ The total binder force is 80 kN which implies 20 kN to $ the quarter of the model. $ TLOAD1 1 9999 13

Main Index

366 Dytran Example Problem Manual Deep Drawing of a Cylindrical Cup

FORCE 9999 MR4 -2.0E7 1. $ $ Prescribed motions: $ $ (i) The punch velocity is described by a sine function such that $ both the velocity and acceleration are equal to zero at the $ start and end of the stroke. The maximum downward velocity $ amounts of 2000 mm/s (in Y-dir.) TLOAD1 1 99 12 55 FORCE 99 MR2 -1000. 1. TABLED1 55 + + .000E+00.000E+00.833E-03.856E-02.167E-02.341E-01.250E-02.761E-01+ + .333E-02.134E+00.417E-02.207E+00.500E-02.293E+00.583E-02.391E+00+ . . $ $ (ii) Rotational velocities of MATRIG 2, 3 and 4 (i.e. punch, die $ and binder) are equal to zero along X-, Y- and Z-axes TLOAD1 1 999 12 MOMENT 999 MR2 0. 1. 1. 1. MOMENT 999 MR3 0. 1. 1. 1. MOMENT 999 MR4 0. 1. 1. 1. $ $ (iii) Die has a translational restraint in X-, Y- and Z-direction! FORCE 999 MR3 0. 1. 1. 1. $ $ (iv) Punch and Blankholder have a translational restraint $ in X- and Z-direction! FORCE 999 MR2 0. 1. 1. FORCE 999 MR4 0. 1. 1. $ $ Geometry description of blank(1), punch(2), die(3) $ and binder(4) in order to model the contact. $ SURFACE 1 SEG 1 SURFACE 2 SEG 1000 SURFACE 3 SEG 2000 SURFACE 4 SEG 3000 $$ Contact specification for which the blank is in all cases the slave: $$ The thickness of the blank is taken into account in the contact $ problem by using THICK = 1.0 $$ (i) Contact between Blank and Punch CONTACT 1 SURF SURF 1 2 .18 .18 0.0 + + Both Full 1.0 + + 0.0 .1 Distance1.+20 1.1 .1 Factor 2. + + 0.0 1.+20 On On 1.+20 $ $ (ii) Contact between Blank and Die CONTACT 2 SURF SURF 1 3 .04 .04 0.0 + + Both Full 1.0 + + 0.0 .1 Distance1.+20 1.1 .1 Factor 2. + + 0.0 1.+20 On On 1.+20 $ $ (iii) Contact between Blank and Blankholder

Main Index

Chapter 5: Forming 367 Deep Drawing of a Cylindrical Cup

CONTACT 3 + + 0.0 + 0.0 ENDDATA

Main Index

SURF Both .1 1.+20

SURF 1 Full Distance1.+20 On On

4 1.0 1.1 1.+20

.04

.04

0.0

.1

Factor

2.

+ + +

368 Dytran Example Problem Manual Three-point Bending Test

Three-point Bending Test Problem Description This example is one of the benchmark problems examined in the TEAM-Virtual Manufacturing project and established at the Oak Ridge National Laboratory. The test setup is shown schematically in Figure 5-12. The test set-up consists of two outer rolls, a punch, and a blank initially located between the punch and both outer rolls. In the test, both outer rolls are moving upwards and are punching the blank around the fixed punch. Both the punch and the outer rolls have cylindrical shapes. Two materials were considered in the experimental program. In the present problem, the final shape upon springback is only sought for aluminum.

Figure 5-12

Three-point Bending Test Setup

Forming Process Data:

Main Index

Total punch displacement

92 mm

Total time for punch displacement

285 sec

Punch closure rate

0.3228 mm/sec

Assumed friction coefficient

0.05

Chapter 5: Forming 369 Three-point Bending Test

Aluminum 6111-T4 Material Properties Young’s modulus

E = 69248 N/mm2

Poisson’s ratio Density

υ = 0.33 ρ = 2.7 * 10-6 kg/mm3

Lankford parameter

R0 = R45 = R90 = 0.64

Proportional limit stress

a = 158.5 N/mm2

Hardening modulus

b = 350.48 N/mm2

Strain offset

c = 0.0

Strain hardening exponent

n = 0.4434

The values for a, b, c, and n could be easily derived from the workhardening data delivered in tabulated form. Experimental data were also obtained for two and four times the above closure rate. In view of the quasi-static nature of this process, it is demonstrated that the explicit dynamics based Dytran can predict the final shape after springback rather accurately.

Dytran Model Due to the symmetry of the geometry and loading conditions, only a one-quarter model is constructed. The mesh of the blank is regular and consists of 200 shell elements (CQUAD4). In order to cover both yielding and springback correctly, five through-the-thickness integration points are defined. The punch and outer roll are modeled as rigid bodies using the material definition MATRIG entry. The surfaces of punch and roll are built up of segments indirectly defined by the property number of the shell elements. In the two contact operations (i.e. blank-punch, blank-roll), the surface of the blank is chosen to be the slave surface because of its mesh fineness.

Main Index

370 Dytran Example Problem Manual Three-point Bending Test

Figure 5-13

Dytran Model

Since symmetry enables a one-quarter model to be constructed, single-point constraints are applied on grid points along the X- and Y-axis of symmetry. All tools are not allowed to have a rotational velocity whereas the punch is only allowed to have a translation velocity in Z-direction. To model these constraints, the FORCE and MOMENT entries can be employed by defining the corresponding enforced zero-motions. Because of the velocity prescription, the TYPE field of the TLOAD1 entry is equal to 12. The explicit dynamics based procedure of Dytran uses very small time steps dictated by the shortest natural period of the mesh. In conjunction with the quasi-static nature of this problem (taking at least more than 70 sec testing time), it would be computationally very expensive to model the real process velocity. Instead, we have modeled the process by using an increased punch velocity and some mass scaling. These are well known, speed-up techniques which are widely accepted in explicit dynamics methods. However, the user should be aware of possible inertia effects produced in the simulation but not visible in the test itself. Due to kinematic constraints on the tools, most sheet-metal forming simulations are not very sensitive for inertia effects when using artificial speed-up techniques. Because there are no blank holders in this problem, there is a risk that the blank slaps around the punch at the moment the punch impacts the blank with too high a velocity. Trial runs were carried out at different closure rates and show that inertia effects start to dominate at simulation rates higher than 1000 times the (fastest) testing rate. The possibility of significant inertia effects also decreases by using a velocity which is described by a kind of sine function such that both the velocity and acceleration are equal to zero at the beginning and end of the punch stroke. In conjunction with a four-fold increase in mass scaling, the mean velocity can now be increased up to 322.8 mm/s without inertia effects affecting the results in an unacceptable way. Similar to the Example Problem (describing Square Cup Deep Drawing, )the material blank is represented by the SHEETMAT material model. Therefore, no further description on this model will be given in this section. Note that the parameter SHTHICK equals YES is put in the Bulk Data Section to take shell thickness changes due to membrane stretching into account.

Main Index

Chapter 5: Forming 371 Three-point Bending Test

The approach of modeling a springback analysis is described in Example Problem (Sleeve Section Stamping) and therefore should be read in advance. For the sake of brevity, the springback analysis is carried out in two sequential steps. The first run is needed to assess the global damping parameter VDAMP. This parameter is used in the succeeding run to obtain the steady state of the blank after springback.

Results Figure 5-14 shows deformed shapes predicted at various stages of the forming process up to end of stroke (Step 1). The position of an edge grid point is plotted against time to evaluate the natural period Figure 5-15. In conjunction with DLTH obtained from the .OUT file, the parameter VDAMP can be determined, and finally applied for the succeeding run (Step 2).

Main Index

372 Dytran Example Problem Manual Three-point Bending Test

Figure 5-14

Main Index

Deformed Configurations at various Stages throughout the Forming Process

Chapter 5: Forming 373 Three-point Bending Test

Figure 5-15 Note:

Position of Grid Point Number 255

The forming stage ending at 0.285 sec is succeeded by an undamped free vibration analysis (Step 1) and a springback analysis in which the structure is critically damped (Step 2).

The springback angle φ is required as benchmark output and is compared with experimental results. In algebraic form: XDIM _255-XDIM_225 φ = 2 * a rc tan ⎛ ----------------------------------------------------------⎞ ⎝ ZDIS _255-ZDIS_225 ⎠

in which XDIM_# denotes the X-dimension of grid point number #, and given by XDIM_# = XDIS_# + XPOS_#

Substitution of displacements/positions predicted by Dytran in the above equation will give us a springback angle of 37.71 degrees. The mean experimental value is 41.88 degrees with a standard deviation of 5.86 degrees. The deformed shape after springback is shown in Figure 5-16.

Main Index

374 Dytran Example Problem Manual Three-point Bending Test

Figure 5-16

Springback after Forming Process

Files

Main Index

3pbt_undmp.dat (undamped) 3pbt_dmp.dat (damped) exvdmp.f (user-subroutine) 3pbt_xl.dat

Dytran input file

3PBT_UNDAMP.OUT (undamped) 3PBT_DMP.OUT (damped)

Dytran output file

3PBT_UNDAMP_BLANK_0.ARC 3PBT_UNDAMP_PUNCH_0.ARC 3PBT_UNDAMP_ROLL_0.ARC

Dytran archive file (undamped)

3PBT_DMP_BLANK_0.ARC 3PBT_DMP_PUNCH_0.ARC 3PBT_DMP_ROLL_0.ARC

Dytran archive file (damped)

3PBT_UNDAMP_GRID_MOT_0.THS (Undamped) 3PBT_DMP_GRID_MOT_0.THS (damped)

Dytran time history file

Chapter 5: Forming 375 Three-point Bending Test

Abbreviated Dytran Input File START $$USERCODE=exvdmp.f( used for the damped problem ) $TITLE = 3-Point Bending Test on Al6111-T4 $----------------------------------------------------------------------Global system damping is applied:( used for the damped problem ) $ The user-subroutine exvdmp.f is employed in which the parameter $ VDAMP is switched on at the end of the forming stage. VDAMP $ corresponds with the critical damping coefficient : $VDAMP = omega * timestep = 2*pi*4.9665E-7/0.02343= 1.33186E-4 $----------------------------------------------------------------------TIME 99999 CEND CHECK = NO ENDSTEP = 744000 SPC = 1 TLOAD = 1 $$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS $TYPE (MYMAT) = MATSUM STEPS (MYMAT) = 0 THRU END BY 5000 $$ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS $TYPE (MYCYC) = STEPSUM STEPS (MYCYC) = 0 THRU END BY 2500 $ CHANGE THE OUTPUT FREQUENCY FOR THE MATRIG-RBE2 SUMMS $TYPE (MYMR) = MRSUM STEPS (MYMR) = 0 THRU END BY 5000 $$$ Data for Output Control Set 1 $$TYPE (blank) = ARCHIVE SAVE (blank) = 999999 ELEMENTS(blank) = 1 SET 1 = 1t200 ELOUT (blank) = ZUSER STEPS (blank) = 0,THRU,744000,BY,74400,573739 $$$$ Data for Output Control Set 2 $$TYPE (punch) = ARCHIVE SAVE (punch) = 999999 ELEMENTS(punch) = 2 SET 2 = 601t780 STEPS (punch) = 0,THRU,744000,BY,74400,573739 ELOUT (punch) = ZUSER $$$$ Data for Output Control Set 3 TYPE (roll) = ARCHIVE SAVE (roll) = 999999 ELEMENTS(roll) = 3 SET 3 = 781t1050 TIMES (roll) = 0 ELOUT (roll) = ZUSER $ $-----------------------------------------------------------$ $ Case Control which refers to User-subroutine( used for the damped $ problem) $ $ELEXOUT (userout)

Main Index

376 Dytran Example Problem Manual Three-point Bending Test

$ELEMENTS(userout) = 4 $SET 4 = 1 $STEPS (userout) = 0tendb1 $ $-----------------------------------------------------------$ Time history of the Position of Grid #255 $ TYPE (grid_mot) = TIMEHIS SAVE (grid_mot) = 99999 GRIDS (grid_mot) = 5 SET 5 = 255 STEPS (grid_mot) = 0,THRU,744000,BY,744,573739 GPOUT (grid_mot) = XDIS $$ $ BEGIN BULK $ PARAM INISTEP 2e-7 PARAM SHTHICK YES $PARAM VDAMP 1.3319-4( used for the damped problem ) $ $ MODEL GEOMETRY $ ============== INCLUDE 3pbt_xl.dat $ $ Five integration points through the thickness of the blank. PSHELL1 1 1 BLT Gauss 5 Mid + .91 $ $ Bely element formulations for MATRIG bodies (i.e. punch and roll) PSHELL1 1111 2 Bely Gauss Mid + 1.-20 PSHELL1 2222 3 Bely Gauss Mid + 1.-20 $ $ Punch $ ===== MATRIG 2 210.e9 0.3 1. 0.0 0.0 0.0 + 1.+2 0.0 0.0 1.+2 0.0 1.+2 $ $ Roll $ ==== MATRIG 3 210.e9 0.3 1. 0.0 0.0 0.0 + 1.+2 0.0 0.0 1.+2 0.0 1.+2 $ $ Geometry description of blank(1), punch(2) and roll(3) in $ order to model the contact. SURFACE 1 PROP 100 SURFACE 2 PROP 200 SURFACE 3 PROP 300 $ SET1 100 1 SET1 200 1111 SET1 300 2222

Main Index

+

+ +

+

+

Chapter 5: Forming 377 Three-point Bending Test

$ $ Contact specification for which the blank (i.e. surface 1) is $ in all cases the slave. $ (i) Contact between Blank and Punch CONTACT 1 SURF SURF 1 2 .05 .05 0.0 + + Both Full 1.0 + + 0.0 .1 Distance1.+20 1.1 .1 Factor 2. + + 0.0 0.285 On On 1.+20 $ $ (ii) Contact between Blank and Roll CONTACT 2 SURF SURF 1 3 .05 .05 0.0 + + Both Full 1.0 + + 0.0 .1 Distance1.+20 1.1 .1 Factor 2. + + 0.0 0.285 On On 1.+20 $ $ The aluminium blank is modelled with the Sheetmat material model. $ Strain rate effect is assumed to be negligible in this example. $ Mass-scaling is applied by a factor of 4 x density. $ $ Material Properties of the aluminium 6111-T4 alloy: $ $ Youngs modulus E = 69.248 GPa, Nu = 0.33, Density = 2.7E-6 Kg/mm3 $ $ Transversely Isotropic Material Specification : $ (i.e. normal anisotropy assumed) $ $ Power Law Stress Constant, a = 158.5E+3 Kg/mm*s2 $ Hardening Modulus, b = 350.48E+3 Kg/mm*s2 $ Strain offset, c= 0.0 $ Hardening exponent, n = 0.4434 $ Lankford Parameters : R(0) = R(45) = R(90) = 0.64 $ SHEETMAT1 1.08E-5 6.9248E7 + + 0.33 ISO 1.0 0.0 0.0 + + 158.5E3 350.48E30.0 0.4434 + + NORMANI 0.64 0.64 0.64 + + NORMANI $$ Prescribed motions: $ (i) The punch velocity is described by a sine function such that $ both the velocity and acceleration are equal to zero at the $ start and end of the stroke. The maximum downward velocity $ amounts to 645.6 mm/s (in Z-dir.) and mean velocity is 322.8 mm/s TLOAD1 1 99 12 55 FORCE 99 MR2 -322.8 1. TABLED1 55 + + .000E+000.000E+0.594E-020.856E-2.119E-010.341E-1.178E-010.761E-1+ + .237E-010.134E+0.297E-010.207E+0.356E-010.293E+0.416E-010.391E+0+ . . . $ $ (ii) Rotational velocities of MATRIG 2 and 3 (i.e. punch and roll) $ are equal to zero around X-, Y- and Z-axes TLOAD1 1 999 12

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378 Dytran Example Problem Manual Three-point Bending Test

MOMENT 999 MR2 0. 1. 1. 1. MOMENT 999 MR3 0. 1. 1. 1. $ $ (iii) Roll has a translational restraint in X-, Y- and Z-direction! FORCE 999 MR3 0. 1. 1. 1. $ $ (iv) Punch has a translational restraint in X- and Y-direction! FORCE 999 MR2 0. 1. 1. $ENDDATA

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Chapter 5: Forming 379 Sleeve Section Stamping

Sleeve Section Stamping Problem Description This example is one of the benchmark problems examined in the TEAM Virtual Manufacturing project, established at the Oak Ridge National Laboratory and reported in [Ref. 5.]. The test setup is shown schematically in Figure 5-17. Both punch and die are double-curved surfaces, and there is lack of symmetry. The objective is to predict the deformed shape of the sheet after springback and to compare this with experimental results.

Figure 5-17

Schematic Diagram of Sleeve Section Stamping Test Setup

Forming Process Data:

Main Index

Minimum top die total punch displacement (just touching)

17.34 mm

Maximum nominal top die total displacement

17.66 mm

Nominal die separation at closure

2.03 mm

Nominal closure time

0.7 s

Nominal closure rate

25 mm/s

Assumed friction coefficient

0.3

380 Dytran Example Problem Manual Sleeve Section Stamping

Aluminum 6111-T4 Material Properties Young’s modulus

E = 181302 N/mm2

Poisson’s ratio

υ = 0.16

Density

ρ = 7.916 * 10-6 kg/mm3

Proportional limit stress

σ0 = 637.76 N/mm2

0.2% yield stress

σ0.2 = 892.00 N/mm2

Workhardening data are provided in tabulated form.

Dytran Model Due to lack of symmetry, the whole model is constructed. The mesh is delivered by the organizing committee of the TEAM project and contains two solid elements through the thickness (Figure 5-18). Solid elements (CHEXA8) are used because the punch stroke leads to some overclosure of the die. Therefore, thickness changes as a result of pressure imposed by the punch must be modeled accurately. This cannot be realized by using shell elements because of their incompressibility in thickness direction.

Figure 5-18

Dytran Model

The punch and die are modeled as rigid bodies using the material definition MATRIG entry. The blank surface is covered with CFACEs whereas the surfaces of punch and die are built up of segments indirectly defined by the property number of the specific shell elements. Due to a very fine mesh in the corner region of the die (i.e., mesh of blank is even coarser), the contact operation is taken to be single surface. Since the slave surface (commonly the blank) is only checked for penetration of the master surface (die/punch) in master-slave contact, the single-surface contact works correctly regardless of the higher mesh density of the die in the corner region.

Main Index

Chapter 5: Forming 381 Sleeve Section Stamping

A combination of kinematic of kinematic boundary conditions is defined for which both translational and rotational velocity of the RIGIDs are restrained. For example, since all tools are not allowed to have a rotational velocity, the MOMENT entry is used in conjunction with setting the SCALE factor for the moment to zero in all three directions. Because of the velocity prescription, the TYPE field of the TLOAD1 entry is equal to 12. The analysis is carried out with a 100-fold mass density. The punch velocity is chosen to be described by a sine function such that both the velocity and acceleration are equal to zero at the start and end of the stroke. The mean velocity can now be increased up to 2.5 m/s without inertia effects affecting the stress/strain results of the blank in an unacceptable way. The material behavior of this blank satisfies the von Mises yield model in combination with isotropic hardening. The workhardening curve is described in accordance with a table of true stress versus true plastic strain data. To treat springback in this example, Dytran offers the so-called Dynamic Relaxation method that uses the global system damping parameter (VDAMP). The theory is discussed in the Dytran User’s Guide. The springback analysis is performed in two steps. Step 1: In order to assess VDAMP, the first run is meant to capture undamped free vibrations upon instantaneous removal of the tools at the end of the stroke. Removal of the tools can be realized by setting the ENDTIME field in the CONTACT entry to the time that corresponds to the end of the forming stage. On the basis of the Dytran User’s Guide, the parameter VDAMP can be given in the following algebraic form: 2 * π * δt V DAM P = ------------------------T

in which T denotes the natural period of free vibration and δt is the time increment needed for the free vibration analysis. The latter represents DLTH in .OUT file and can be found to be almost constant throughout the free vibration analysis. The period T can be determined by plotting a relevant grid point displacement/position versus time. Step 2: In the second final run, the free vibration mode is damped out by defining the parameter VDAMP assessed in Step 1. The user-subroutine exvdmp.f (Three-point Bending Test) is used to activate this parameter only for solid elements at the end of the forming analysis. To call this user subroutine, the ELEXOUT entry is required in Case Control of the input file. The File Management Section contains the USERCODE entry. Figure 5-19 shows deformed shapes predicted at various stages of the stamping process up to full die

closure (Step 1). As stated above, the position of an edge grid point is plotted against time to evaluate the natural period (Figure 5-20). In conjunction with DLTH obtained from the .OUT file, the parameter VDAMP can be determined and finally applied to the succeeding run (Step 2) in an appropriate way as described above.

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382 Dytran Example Problem Manual Sleeve Section Stamping

Figure 5-19

Deformed Configurations at various Stages throughout the Stamping Process

Figure 5-20

Position of Grid Point Number 10

Note:

The forming stage ending at 0.7E – 2 sec is succeeded by an undamped free vibration analysis (Step 1) and a springback analysis in which the structure is critically damped (Step 2)

Chord length and depth of the deformed shape of the sheet after springback, as well as the total applied load at full die closure are considered to be major benchmark output results. In algebraic form:

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Chapter 5: Forming 383 Sleeve Section Stamping

Chord Length = XDIM_10 – XDIM_1060 Depth = ZDIM_535 – (ZDIM_12 + zdim_1062)/2 Total Force = ZFORCE_MATRIG_2 in which XDIM_# and ZDIM_# denote the X- and Z-dimension of grid point number #, calculated by using XDIM_# = XCIS_# + XPOS_# Substitution of displacements/positions predicted by Dytran in the above equation will provide the results shown in the following table. The deformed shape after springback is shown in Figure 5-21.

Figure 5-21

Springback after Forming Process

The original mesh delivered for the blank contains only two elements through the thickness and therefore, can be expected to do a relatively poor job of representing pure bending deformation. Refined mesh would provide better correlation with experiment ([Ref. 5.]). This would also increase the cost of the analysis and consequently, is beyond the scope of this Dytran Example Problem Manual. Comparison between Dytran and Experimental Results Dytran

Main Index

Experiment

Chord Length [mm]

62.54

61.58

Depth [mm]

13.77

14.38

Total Force [kN]

84.90

40 - 225

384 Dytran Example Problem Manual Sleeve Section Stamping

Files sleeve_undmp.dat (undamped) sleeve_dmp.dat (damped) exvdmp.f (user-subroutine) sleeve_xl.dat

Dytran input file

SLEEVE_UNDAMP.OUT (undamped) SLEEVE_DMP.OUT (damped)

Dytran output file

SLEEVE_UNDAMP_BLANK_0.ARC SLEEVE_UNDAMP_PUNCH_0.ARC SLEEVE_UNDAMP_DIE_0.ARC

Dytran archive file (undamped)

SLEEVE_DMP_BLANK_0.ARC SLEEVE_DMP_PUNCH_0.ARC SLEEVE_DMP_DIE_0.ARC

Dytran archive file (damped)

SLEEVE_UNDAMP_CONTACT_0.THS SLEEVE_UNDAMP_GRID_MOT_0.THS

Dytran time history file (undamped)

SLEEVE_DMP_CONTACT_0.THS SLEEVE_DMP_GRID_MOT_0.THS

Dytran time history file (damped)

Reference 5. Slagter, W. “Simulation of Sheet Metal Forming Processes using Dytran.”

Abbreviated Dytran Input File START $$USERCODE=exvdmp.f( used for the damped problem ) $TITLE = Sleeve Section Stamping $$------------------------------------------------------------------$ Note : Global system damping is applied:( used for the damped problem ) $ The user-subroutine exvdmp.f is employed in which the $ parameter VDAMP is switched on at the end of the formingstage. VDAMP corresponds with the critical damping$ coefficient: $VDAMP = omega * timestep = 2*pi*1.324E-6/5.6E-3 = = 1.5E-3 $------------------------------------------------------------------$ TIME 99999 CEND CHECK = NO ENDTIME = 0.025 SPC = 1 TLOAD = 1 $$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS $TYPE (MYMAT) = MATSUM STEPS (MYMAT) = 0 THRU END BY 1000 $$ CHANGE THE OUTPUT FREQUENCY FOR RIGID SUMMS $ TYPE (MRMAT) = MRSUM

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Chapter 5: Forming 385 Sleeve Section Stamping

STEPS (MRMAT) = 0 THRU END BY 1000 $ $ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS $ TYPE (MYCYC) = STEPSUM STEPS (MYCYC) = 0 THRU END BY 250 $$ $$ Data for Output Control Set 1

$$ TYPE (blank) = ARCHIVE SAVE (blank) = 999999 ELEMENTS(blank) = 1 SET 1 = 1t600 ELOUT (blank) = EFFSTS,EFFPLS TIMES (blank) = 0,0.003,0.004,0.007,0.025 $$ $$ Data for Output Control Set 2 $$ TYPE (punch) = ARCHIVE SAVE (punch) = 999999 ELEMENTS(punch) = 2 SET 2 = 12000t12419 TIMES (punch) = 0,0.003,0.004,0.007,0.025 ELOUT (punch) = ZUSER $$ $$ Data for Output Control Set 3 $$ TYPE (die) = ARCHIVE SAVE (die) = 999999 ELEMENTS(die) = 3 SET 3 = 13000t13559 TIMES (die) = 0 ELOUT (die) = ZUSER $ $ Time history of the MATRIG #2 (i.e. Punch) $ RIGIDS (contact) = 4 SET 4 = MR2 TYPE (contact) = TIMEHIS RBOUT (contact) = ZFORCE SAVE (contact) = 99999 TIMES (contact) = 0tendb0.000025,0.007 $$ Time history of the Position of Grid #10 (at YZ-plane) $TYPE (grid_mot) = TIMEHIS SAVE (grid_mot) = 99999 GRIDS (grid_mot) = 5 SET 5 = 10 TIMES (grid_mot) = 0tendb0.000025 GPOUT (grid_mot) = ZPOS $$------------------------------------------------------------------$ Case Control which refers to User-subroutine( used for the damped problem ) $$ELEXOUT (userout) $ELEMENTS(userout) = 6 $SET 6 = 1

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386 Dytran Example Problem Manual Sleeve Section Stamping

$STEPS (userout) = 0tendb1 $ $------------------------------------------------------------------$ BEGIN BULK $ PARAM INISTEP 1e-07 $PARAM VDAMP 0.0015 ( used for the damped problem ) PARAM GEOCHECKON $ $ MODEL GEOMETRY $ ============== INCLUDE sleeve_xl.dat $ $ Lagrangian solid elements for blank. $ Two elements through the thickness of the blank. PSOLID 1 1 $ $ Bely element formulations for MATRIG bodies (i.e. punch and die) PSHELL1 1111 2 Bely Gauss Mid + + 1.-20 PSHELL1 2222 3 Bely Gauss Mid + + 1.-20 $ $ Punch $ ===== MATRIG 2 210.e9 0.3 1. 0.0 0.0 0.0 + + 1.+20 0.0 0.0 1.+20 0.0 1.+20 $ $ Die $ === MATRIG 3 210.e9 0.3 1. 0.0 0.0 0.0 + + 1.+20 0.0 0.0 1.+20 0.0 1.+20 $ $ Geometry description of blank, punch and die in single surface $ contact. SET1 11 1111 SET1 22 2222 $ SURFACE 1 SEG 1 PROP 11 PROP 22 $ CONTACT 1 SURF 1 .30 .30 0.0 + + V4 0.0 0.0 + + + + 0.007 $ $ Material Properties of 301 Half-Hard Stainless Steel: $ Mass-scaling is applied by a factor of 100 (original density=7.916e-6) $ DYMAT24 1 7.916-4 1.813+8 .16 55 PLAST + + + + 1.0 0.0 $ HGSUPPR 1 SOLID 1 FBS

Main Index

Chapter 5: Forming 387 Sleeve Section Stamping

$ TABLED1 55 + + .000E+00.638E+06.229E-03.840E+06.485E-03.869E+06.672E-03.906E+06+ + .338E-02.986E+06.548E-02.102E+07.812E-02.103E+07.102E-01.104E+07+ + .139E-01.105E+07.241E-01.108E+07.329E-01.110E+07.415E-01.113E+07+ + .562E-01.116E+07.856E-01.124E+07.100E+00.128E+07.195E+00.150E+07+ + ENDT $ $ Prescribed motions: $ (i) The punch velocity is described by a sine function such that $ both the velocity and acceleration are equal to zero at the $ start and end of the stroke. The maximum downward velocity $ amounts of 5000 mm/s (in Z-dir.) and mean velocity is 2.5e3 mm/s. TLOAD1 1 99 12 66 FORCE 99 MR2 -2500. 1. TABLED1 66 + + .000E+00 .000E+0.146E-03 .856E-2.292E-03 .341E-1.438E-03 .761E-1+ + .583E-03 .134E+0.729E-03 .207E+0.875E-03 .293E+0.102E-02 .391E+0+ . . . . $ $ (ii) Rotational velocities of MATRIG 2 and 3 (i.e. punch and $ die) are equal to zero around X-, Y- and Z-axes TLOAD1 1 999 12 MOMENT 999 MR2 0. 1. 1. 1. MOMENT 999 MR3 0. 1. 1. 1. $ $ (iii) Die has a translational restraint in X-, Y- and Z-direction! FORCE 999 MR3 0. 1. 1. 1. $ $ (iv) Punch has a translational restraint in X- and Y-direction! FORCE 999 MR2 0. 1. 1. $ ENDDATA

Main Index

388 Dytran Example Problem Manual Sleeve Section Stamping

Main Index

Chapter 6: Occupant Safety Dytran Example Problem Manual

6

Main Index

Occupant Safety J

Overview

J

Flat Unfolded Air Bag Inflation (GBAG)

J

Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

J

Sled Test Verification of the Enhanced Hybrid III Dummy (50%) 416

J

Side Curtain Air Bag (Courtesy of Autoliv)

J

Hybrid III 5th%-tile Dummy

J

Easy Postprocessing with Adaptive Meshing

390 391

462

479 509

400

390 Dytran Example Problem Manual Overview

Overview In this chapter, a number of example problems are presented that highlight the capabilities of Dytran in the area of Occupant Safety. In these examples, the user can find the guidelines to model Occupant Safety problems that include air bags and occupant dummies. In the area of air bag modeling, the constant pressure method and full gas dynamics approach are shown. The examples include the use of rigid ellipsoid dummies as well as the use to the enhanced HYBRIDIII rigid body dummy. The following example problems are described in this chapter.

Main Index

Chapter 6: Occupant Safety 391 Flat Unfolded Air Bag Inflation (GBAG)

Flat Unfolded Air Bag Inflation (GBAG) Problem Description This problem demonstrates the interaction between a passenger-side, unfolded air bag and a rigid body. The air bag is inflated by means of a uniform pressure law (GBAG) and impacts a spherical ellipsoid during the inflation. Dytran uses a closed surface to define the gas bag. The surface must be built from faces connected to the finite elements. The elements can be either membranes, shells, solids, dummy shells, or a combination of those elements. To model the flat rectangular air bag, two parallel planes are defined by using triangular constant strain membrane elements. The surface is closed by coupling the nodes located on the four edges of the two planes. The inflator is located on the center of the back plane and is modeled by membrane elements with a SPC constraint to keep its shape rigid and in place (see Figure 6-1).

Figure 6-1

Main Index

Air Bag Model

392 Dytran Example Problem Manual Flat Unfolded Air Bag Inflation (GBAG)

The material behavior of the air bag fabric is modeled with a linear elastic isotropic material model with the following properties: density

600 kg/m3

Young’s modulus

6 107 N/mm2

Poisson’s ratio

0.3

The following input entries are required when modeling a gas bag: SURFACE

Defines the element faces that are part of the gas bag. The surface must be closed.

GBAG

Defines the gas bag properties such as: • Mass inflow rate • Trigger time • Area and discharge coefficient of holes • Area and discharge coefficient of porous fabric • Volumetric porosity • Initial state of the gas inside • Material properties of the gas inside

The rigid ellipsoid is modeled by defining a 0.1 m radius sphere with a mass of 0.5 kg. The sphere is located .15 m above the air bags at the center (see Figure 6-2).

Figure 6-2

Main Index

Initial Position of the Rigid Ellipsoid

Chapter 6: Occupant Safety 393 Flat Unfolded Air Bag Inflation (GBAG)

The contact surface between the rigid ellipsoid and the air bag is defined by using the CONTREL entry. This option is used to define the grid points on the Lagrangian (air bag) surface that can come into contact with the rigid ellipsoid.

Results As shown in the animation plots (see Figure 6-3), the air bag is inflated from time 0 sec and will hit the rigid ellipsoid at 20 msec. From that moment, the rigid ellipsoid is loaded by the air bag and will be pushed away from the air bag. After 45 msec, there is no contact anymore between rigid ellipsoid and the air bag, and the rigid ellipsoid will move away from the air bag at a constant speed.

Figure 6-3

Main Index

Inflation of the Air Bag

394 Dytran Example Problem Manual Flat Unfolded Air Bag Inflation (GBAG)

The time history of pressure, volume, and mass of the gas inside the air bag is plotted in Figure 6-4.

Figure 6-4

Main Index

Time Histories of Pressure, Mass, and Temperature of Gas inside the Air Bag

Chapter 6: Occupant Safety 395 Flat Unfolded Air Bag Inflation (GBAG)

Files demo1.dat demo1_bag.dat (property input file) demo1_gbag.dat (gas bag and mass flow input definition) demo1_ellips_contact.dat (contact input definition) xl_demo1.dat (geometry file)

Dytran input file

DEMO1.OUT

Dytran output file

DEMO1_AIRBAG_0.ARC DEMO1_ELLIPS_0.ARC

Dytran archive file

DEMO1_GB_0.THS DEMO1_TELLIPS_0.THS

Dytran time history files

Abbreviated Dytran Input File Abbreviated Input file $ file: demo1.dat $ --------------$ $ - inflation of a flat, unfolded demo passenger-side airbag $ - uniform pressure assumption $ - contact with a rigid elllipsoid $ START CEND CHECK=NO ENDSTEP=100000 ENDTIME=125.E-3 TITLE=INFLATION OF DEMO PASSENGER-BAG WITH UNIFORM PRESSURE MODEL TIC=1 SPC=1 TLOAD=1 $ -------------------------------------------------------------$ define the output of the gbag variables: $ GBAGS (GB) = 5 SET 5 = 1 GBOUT (GB) = VOLUME,PRESSURE,FLGAS,TGAS,TEMP,MASS TIMES (GB) = 0. THRU END BY 1.E-3 TYPE (GB) = TIMEHIS SAVE (GB) = 1000000 $ $ -------------------------------------------------------------$ define the output for the membrane elements: $ ELEMENTS(AIRBAG) = 996 SET 996 = ALLMEMTRIA ELOUT (AIRBAG) = THICK,SMDFER

Main Index

396 Dytran Example Problem Manual Flat Unfolded Air Bag Inflation (GBAG)

TIMES (AIRBAG) = 0. THRU END BY 20.E-3 TYPE (AIRBAG) = ARCHIVE SAVE (AIRBAG) = 1000000 $ $ -------------------------------------------------------------$ define the output for the ellipsoid: $ RELS (ELLIPS) = 600 $ nb: setc,600 defined in demo1_ellips_contact.dat RELOUT (ELLIPS) = GEOMETRY TIMES (ELLIPS) = 0. THRU END BY 20.E-3 TYPE (ELLIPS) = ARCHIVE SAVE (ELLIPS) = 1000000 $ $ -------------------------------------------------------------$ define more output for the ellipsoid: $ RELS (TELLIPS) = 600 $ nb: setc,600 defined in demo1_ellips_contact.dat RELOUT (TELLIPS) = XVEL,YVEL,ZVEL,XAVEL,YAVEL,ZAVEL TIMES (TELLIPS) = 0. THRU END BY 1.E-3 TYPE (TELLIPS) = TIMEHIS SAVE (TELLIPS) = 1000000 $ BEGIN BULK $ ------------------------------------------------------------$ Initial timestep & minimum timestep $ PARAM,INISTEP,0.7E-05 PARAM,MINSTEP,.5E-8 $ $ ------------------------------------------------------------$ Define the airbag $ INCLUDE demo1_bag.dat $ $ ------------------------------------------------------------$ Define the gbag ( uniform pressure parameters ) $ INCLUDE demo1_gbag.dat $ $ ------------------------------------------------------------$ Define the rigid ellipsoid and the contact with it $ INCLUDE demo1_ellips_contact.dat $ $ ------------------------------------------------------------$ $ Rest is defined with xl $ INCLUDE xl_demo1.dat $ ------------------------------------------------------------ENDDATA

Main Index

Chapter 6: Occupant Safety 397 Flat Unfolded Air Bag Inflation (GBAG)

Properties Input Definition $ file: demo1_bag.dat $ ------------------$BEGIN BULK $ ------------------------------------------------------------$ Define the membrane properties & material $ During modelling we took care of the folding lines, and four $ holes at the back of the airbag $ The use of the folding lines will be clear in demo2 (folding) $ The use of the holes will be clear in demo3 (porosity with euler) $$ Every part of the airbag has his own PID $ The use of this will be clear in demo2 (folding) $$ The model with the PID’s looks like this: $$ front: $$ y $ ^ $ | $ --------------------------------------------------------------$ | | | | | $ | | | | | $ | | | | | $ | 504 | 503 | 603 | 604 | $ | | | | | $ | | | | | $ | | | | | $ |---------------------------------------------------------------| $ | | | | | $ | | | | | $ | | | | | $ | 502 | 501 | 601 | 602 | $ | | | | | $ | | | | | $ | | | | | $ |---------------------------------------------------------------|--> x $ | | | | | $ | | | | | $ | | | | | $ | 552 | 551 | 651 | 652 | $ | | | | | $ | | | | | $ | | | | | $ |---------------------------------------------------------------| $ | | | | | $ | | | | | $ | | | | | $ | 554 | 553 | 653 | 654 | $ | | | | | $ | | | | | $ | | | | | $ --------------------------------------------------------------$ $ back: $ $ y

Main Index

398 Dytran Example Problem Manual Flat Unfolded Air Bag Inflation (GBAG)

$ ^ $ | $ --------------------------------------------------------------$ | | \ 512 /| \ 612 /| | $ | | \ / | \ / | | $ | | --| --| | $ | 513 | 511 |510| 509 | 609 |610| 611 | 613 | $ | | --| --| | $ | | / \ | / \ | | $ | | / 508 \ | / 608 \ | | $ |---------------------------------------------------------------| $ | | | | | | $ | | | | | | $ | | | | | | $ | 507 | 506 | 505 | 606 | 607 | $ | | | | | | $ | | | | | | $ | | | | | | $ |---------------------------------------------------------------|--> x $ | | | | | | | $ | | | | | | | $ | | | | | | $ | 557 | 556 | 655 | 555 | 656 | 657 | $ | | | | | | | $ | | | | | | | $ | | | | | | $ |---------------------------------------------------------------| $ | | \ 558 /| \ 658 /| | $ | | \ / | \ / | | $ | | --| --| | $ | 563 | 561 |560| 559 | 659 |660| 661 | 663 | $ | | --| --| | $ | | / \ | / \ | | $ | | / 562 \ | / 662 \ | | $ --------------------------------------------------------------$ $ PID = 505 ---> inflator (see xl-pictures) $$ PID = 510,560,610,660 ---> holes $$ First quarter $PSHELL1,501,1,MEMB,,,,,,+ +,5.E-4 .. . PSHELL1,513,1,MEMB,,,,,,+ +,5.E-4 $ $Second quarter (PID first + 50) $ PSHELL1,551,1,MEMB,,,,,,+ +,5.E-4 . . . PSHELL1,563,1,MEMB,,,,,,+

Main Index

Chapter 6: Occupant Safety 399 Flat Unfolded Air Bag Inflation (GBAG)

+,5.E-4 $ $ Third quarter (PID first + 100) $ PSHELL1,601,1,MEMB,,,,,,+ +,5.E-4 . . . PSHELL1,613,1,MEMB,,,,,,+ +,5.E-4 $ $ fourth quarter (PID first + 150) $ PSHELL1,651,1,MEMB,,,,,,+ +,5.E-4 . . . PSHELL1,663,1,MEMB,,,,,,+ +,5.E-4 $ $ Define the inflator as membranes with a SPC to keep it at its $ place. $ PSHELL1,505,1,MEMB,,,,,,+ +,5.E-4 $ $ ------------------------------------------------------------$ $ definition of material, elastic with: $ - density = 600 kg/m**3 $ - young’s = 6.E7 N/m**2 $ - poisson = 0.3 $ DMATEL,1,600.,6.E7,0.3 $ ------------------------------------------------------------$ $ Define the closed gbag-surface $ $ surface 25 : - airbag + inflator $ $ nb. pid=605 does not exist, since the inflator is completely $ modeled by pid=505 $ SURFACE,25,,PROP,199 SET1,199,501,THRU,513 SET1,199,551,THRU,563 SET1,199,601,THRU,604,606,THRU,613 SET1,199,651,THRU,663 $ ------------------------------------------------------------ENDDATA

Main Index

400 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

Rigid Ellipsoid Dummy Hitting a Passenger Air Bag Problem Description As an example for occupant modeling, this problem demonstrates the interaction between an inflated air bag and a rigid ellipsoid dummy model, which behavior is simulated by the Crash Victim Simulation (CVS) program ATB, which has been integrated into Dytran. This example problem will not describe how to model an inflated air bag, but will focus on the simulation of a rigid dummy by ATB and its interaction with structural parts like an air bag. It will also not describe the advanced technique of positioning of an ATB dummy through Dytran.

Dytran Model The model consists of a flat, folded passenger-side air bag, which is inflated (using full gas dynamics) and interacts with an accelerated (Hybrid III-like) ATB dummy. The rigid ATB dummy model consists of 15 rigid ellipsoids, which are connected by 14 rigid body joints. Characteristics and initial position of the dummy are completely defined in the ATB input file. Each ATB segment contact ellipsoid must be defined in the Dytran input file with the RELEX entry. The names of these ellipsoids are the same as the segment names as defined in the ATB input file on the B.2 entries. For contact purposes, 15 rigid bodies were created, which were rigidly connected to the ATB ellipsoids using RCONREL entries. In this case, the shape of the rigid bodies coincide with the ATB ellipsoids, but their shapes can be arbitrary. These rigid bodies are only used as contact surfaces in the contact definitions between the air bag and other car parts. Two additional rigid bodies, created to model elbows, are rigidly connected with the upper arms. Besides contact with the air bag, contact is defined between other rigid car parts (not shown in Figure 6-5) and the ATB dummy, except for head and neck.

Main Index

Chapter 6: Occupant Safety 401 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

Figure 6-5

Air-bag Model and ATB Dummy

The specification of GEOMETRY on the RELOUT entry causes Dytran to cover the ATB ellipsoids with unique dummy shell elements. Primarily, the created dummy geometry is used for the visualization of the ellipsoids. It can also be used to create the above-mentioned rigid bodies. To obtain the shape of the ATB segments, a short separate Dytran pre-run with the positioned ATB dummy is necessary. The rigid bodies in this example problem (except for the elbows) have been created this way.

Main Index

402 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

Results Figure 6-6 shows air bag and ATB dummy halfway (at 0.05 seconds) and at the end of the simulation (at

0.1 seconds).

Figure 6-6

Deformed Geometry of Air Bag and ATB Dummy

Figure 6-7 is a time history of the velocity of the head and lower torso of the dummy. As can be seen from this graph, the entire dummy is equally accelerated at the start of the simulation. The deceleration of the lower torso is caused by contact of the dummy feet and legs with some rigid car parts, which were not shown in the previous figures. The head is decelerated by the unfolding air bag.

Main Index

Chapter 6: Occupant Safety 403 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

.

Figure 6-7

Time History of the Forward Velocity of Head and Lower Torso

To verify the ATB coupling with Dytran, an analogous simulation using MADYMO has been done.Figure 6-8 shows the resultant Head acceleration both from ATB and MADYMO coupling. Deviations between the results are mainly due to differences between the joint models in ATB and MADYMO dummy.

Figure 6-8

Main Index

Comparison of the Resultant Head Acceleration with MADYMO

404 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

Files demo4.dat demo4_bag.dat (airbag property definition) demo4_bag_contact.dat (definition of airbag contacts) demo4_euler.dat (Eulerian definition for gas dynamics of airbag) demo4_euler_coupling.dat (definition of coupling) demo4_gbag.dat (definition of gas bag parameters) demo4_car.dat (definition of rigid car parts) demo4_atb.dat (definition of ATB interactions) xl_demo4.dat (geometry file) atb_demo4.ain (ATB input file with dummy definition)

Dytran input file

DEMO4.OUT

Dytran output file

DEMo4_EUL_0.ARC DEMO4_AIRBAG_0.ARC DEMO4_DUMQ_0.ARC DEMO4_DUMT_0.ARC DEMO4_ELLIPS_0.ARC

Dytran archive file

DEMO4_GB_0.THS DEMO4_TELLIPS_0.THS

Dytran time history files

Abbreviated Dytran Input File Abbreviated Main Input File $ $ - inflation of a flat, unfolded demo passenger-side airbag $ - inflated by uniformi pressure $ - contact with a rigid elllipsoid $ $ MEMORY-SIZE=4000000,2000000 START CEND CHECK=NO ENDSTEP=100000 ENDTIME=.1 TITLE=DEMO PASSENGER-BAG WITH ATB DUMMY TIC=1 SPC=1 TLOAD=1 $ -------------------------------------------------------------$ define the output of the gbag variables: $ $ - data asked for: GBAG,1 $ - ask for: volume,pressure,flgas,tgas,temp,mass (see manual) $ - ask for output every 1 msec $ - create file for time-history plots

Main Index

Chapter 6: Occupant Safety 405 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ - write 1000000 times to the same file, which will be named: $ $ DEMO2_GB_0.ARC $ GBAGS (GB) = 5 SET 5 = 1 GBOUT (GB) = VOLUME,PRESSURE,FLGAS,TGAS,TEMP,MASS TIMES (GB) = 0. THRU END BY 1.E-3 TYPE (GB) = TIMEHIS SAVE (GB) = 1000000 $ $ -------------------------------------------------------------$ define the output for the euler elements: $ $ - data asked for: all euler elements $ - ask for: pressure,density,fmat,xvel,yvel,zvel,sie (see manual) $ - ask for output every 4 msec $ - create file for plots of contours $ - write 1000000 times to the same file, which will be named: $ $ DEMO4_EUL_0.ARC $ ELEMENTS(EUL) = 995 SET 995 = ALLEULHYDRO ELOUT(EUL) = PRESSURE,DENSITY,FMAT,XVEL,YVEL,ZVEL,SIE TIMES(EUL) = 0. THRU END BY 1.E-2 TYPE(EUL) = ARCHIVE SAVE(EUL) = 1000000 $ $ -------------------------------------------------------------$ define the output for the membrane elements: $ $ - data asked for: all membrane elements $ - ask for: thick,smdfer (see manual) $ - ask for output every 4 msec $ - create file for plots of deformed geometry and contours $ - write 1000000 times to the same file, which will be named: $ $ DEMO4_AIRBAG_0.ARC $ ELEMENTS(AIRBAG) = 996 SET 996 = ALLMEMTRIA ELOUT(AIRBAG) = THICK,SMDFER,EXUSER1 TIMES(AIRBAG) = 0. THRU END BY 1.E-2 TYPE(AIRBAG) = ARCHIVE SAVE(AIRBAG) = 1000000 $ $ELEMENTS(USEROUT) = 1001 $SET 1001 = 1,THRU,654,657,THRU,1460,1463,THRU,2266,2269, $ THRU,3072,3075,THRU,3514 $ELEXOUT(USEROUT) $STEPS(USEROUT) = 0 THRU END BY 1 $ $ --------------------------------------------------------------

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406 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ define the output for the dummy shell elements: $ $ - data asked for: all dummy shell elements $ - ask for: zuser (see manual) $ - ask for output every 4 msec $ - create file for plots of deformed geometry and contours $ - write 1000000 times to the same file, which will be named: $ $ DEMO4_DUMQ_0.ARC $ DEMO4_DUMT_0.ARC $ ELEMENTS(DUMQ) = 997 SET 997 = ALLDUMQUAD ELOUT(DUMQ) = ZUSER TIMES(DUMQ) = 0. THRU END BY 1.E-2 TYPE(DUMQ) = ARCHIVE SAVE(DUMQ) = 1000000 $ ELEMENTS(DUMT) = 998 SET 998 = ALLDUMTRIA ELOUT(DUMT) = ZUSER TIMES(DUMT) = 0. THRU END BY 1.E-2 TYPE(DUMT) = ARCHIVE SAVE(DUMT) = 1000000 $ $ -------------------------------------------------------------$ define the output for the ellipsoids: $ $ - data asked for all rigid ellipsoids $ - ask for: geometry (see manual) $ - ask for output every 4 msec $ - create file for plots of displaced geometry $ - write 1000000 times to the same file, which will be named: $ $ DEMO4_ELLIPS_0.ARC $ RELS (ELLIPS) = 600 $ nb: setc,600 defined in demo4_ellips_contact.dat RELOUT (ELLIPS) = GEOMETRY TIMES (ELLIPS) = 0. THRU END BY 1.E-2 TYPE (ELLIPS) = ARCHIVE SAVE (ELLIPS) = 1000000 $ $ -------------------------------------------------------------$ define more output for the ellipsoid: $ $ - data asked for rigid ellipsoid with name SPHERE $ - ask for: xvel,yvel,zvel,xavel,yavel,zavel (see manual) $ - ask for output every 4 msec $ - create file for time-history plots $ - write 1000000 times to the same file, which will be named: $ $ DEMO4_TELLIPS_0.ARC $

Main Index

Chapter 6: Occupant Safety 407 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

RELS (TELLIPS) = 600 $ nb: setc,600 defined in demo4_ellips_contact.dat RELOUT (TELLIPS) = XVEL,YVEL,ZVEL,XAVEL,YAVEL,ZAVEL TIMES (TELLIPS) = 0. THRU END BY 1.E-3 TYPE (TELLIPS) = TIMEHIS SAVE (TELLIPS) = 1000000 $ BEGIN BULK $ ------------------------------------------------------------$ Deactivate Euler & Euler-coupling with an offset in time $ This allows the user to use a trigger-time in the massflow-tables $ $ACTIVE,1,ELEMENT,EULHYDRO,,,,,,+ $+,TABLE,9999 $ACTIVE,1,INTERACT,COUPLE,,,,,,+ $+,TABLE,9999 $TABLED1,9999,,,,,,,,+ $+,0.0,0.0,1.0,0.0 $ ------------------------------------------------------------$ Define the switch from coupling to gasbag, start checking $ after 100 msecs $ GBAGCOU,1,10,1,100.0E-3 $ ------------------------------------------------------------$ Initial timestep & minimum timestep $ PARAM,INISTEP,0.7E-05 PARAM,MINSTEP,.5E-8 $ $ Make the contact default version V2 for this problem PARAM,CONTACT,VERSION,V2 $ $ ------------------------------------------------------------$ Apply gravity to the model GRAV,2, ,-9.81,0.,0.,1. $ TLOAD1 1 2 0 0 $ $ ------------------------------------------------------------$ Nodal velocity damping for the membrane elements is activated $ after some time, to take care of hysteresis in material when $ it comes in tension. $ When more advanced material models become available this will $ not be necessary any more. $ $VISCDMP,,,,,,,,,+ $+,,,,,,,,,+ $+,60.E-3,,,0.05 $ $ ------------------------------------------------------------$ Define the airbag & the contacts for unfolding $ INCLUDE demo4_bag.dat INCLUDE demo4_bag_contact.dat

Main Index

408 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ $ ------------------------------------------------------------$ Define the euler $ INCLUDE demo4_euler.dat $ $ ------------------------------------------------------------$ Define the euler_coupling $ INCLUDE demo4_euler_coupling.dat $ $ ------------------------------------------------------------$ Define the gasbag parameters $ INCLUDE demo4_gbag.dat $ ------------------------------------------------------------$ Define parts of the car the contact with it $ INCLUDE demo4_car.dat $ $ ------------------------------------------------------------$ Define atb-related stuff $ INCLUDE demo4_atb.dat $ $ ------------------------------------------------------------$ $ Rest is defined with xl $ INCLUDE xl_demo4.dat $ ------------------------------------------------------------ENDDATA Abbreviated Input File with ATB-related Definitions $ file:demo4_atb.dat BEGIN BULK $------------------------------------------------------------$ Define a setc containing all the atb ellipsoids $ ---> used for visualization $ SETC,600,LT,CT,UT,N,H SETC,600,RUL,RLL,RF,LUL,LLL SETC,600,LF,RUA,RLA,LUA,LLA $------------------------------------------------------------$ Define ellipsoids that are used for dummy in ATB $ RELEX,LT,ATB . . . RELEX,LLA,ATB $ $ ------------------------------------------------------------$ Define contact of all the membrane grid-points at the front with $ a subset of the atb ellipsoids.

Main Index

Chapter 6: Occupant Safety 409 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ The grid-point list can easily be retrieved from grouping with $ xl and listing the group. $ The points with SPC will be overruled by the SPC CONTACT,801,SURF,SURF,25,801,,,,+ +,,TOP,,,,,,,+ +,0.,,,,,0.4,,,+ +,,,,,0. . . . $ $ Define contact with spheres at the ellbow & shoulder $ CONTACT,857,SURF,SURF,25,857,,,,+ +,,TOP,,,,,,,+ +,0.,,,,,0.4,,,+ +,,,,,0. $ CONTACT,859,SURF,SURF,25,859,,,,+ +,,TOP,,,,,,,+ +,0.,,,,,0.4,,,+ +,,,,,0. $ ------------------------------------------------------------$ Define the rigid surfaces connected to atb-ellipsoids $ $ lotorso pid=801 $ spine pid=802 . . . . $ footrig pid=817 $ kneerig pid=818 $ $ spheres at ellbow & shoulder on left side pid=857 $ spheres at ellbow & shoulder on right side pid=859 $ PSHELL1,801,,DUMMY PSHELL1,802,,DUMMY . . . PSHELL1,817,,DUMMY $PSHELL1,818,,DUMMY $ PSHELL1,857,,DUMMY PSHELL1,859,,DUMMY $ SURFACE,801,,PROP,801 SET1,801,801 . . . SURFACE,817,,PROP,817

Main Index

410 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

SET1,817,817 $ SURFACE,857,,PROP,857 SET1,857,857 SURFACE,859,,PROP,859 SET1,859,859 $ $ Give them all a “contact-mass” of 2. kg $ RIGID,801,801,2.0 . . . RIGID,817,817,2.0 $ RIGID,857,857,2.0 RIGID,859,859,2.0 $ $ connect the rigids to the atb ellipsoids $ RCONREL,1,801,RIGID,401 SETC,801,LT SET1,401,801 . . RCONREL,6,806,RIGID,406 SETC,806,H SET1,406,806 $ $ also connect the spheres for ellbow & shoulder to uparml $ RCONREL,7,807,RIGID,407 SETC,807,LUA SET1,407,807,857 RCONREL,8,808,RIGID,408 SETC,808,LLA SET1,408,808 RCONREL,9,809,RIGID,409 $ $ also connect the spheres for ellbow & shoulder to uparmr $ SETC,809,RUA SET1,409,809,859 . . . RCONREL,17,817,RIGID,417 SETC,817,RF SET1,417,817 $ ------------------------------------------------------------$ Define contact of all the rigid surfaces connected to the $ ellipsoids with the car-parts $ SURFACE,890,,PROP,890

Main Index

Chapter 6: Occupant Safety 411 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ HEAD,NECK,KNEES AND SHOULDER ARE ROMOVED SET1,890,801,802,803,807 SET1,890,808,809,810,811,812,813 SET1,890,815,816,817 SET1,890,857,859 $ CONTACT,890,SURF,SURF,890,2000,,,,+ +,,TOP,,,,,,,+ +,0.,,,,,0.4,,,+ +,,,,,0. $ $ENDDATA

Abbreviated ATB Input File $ file:atb_demo4.ain $ ------------------$ $ ATB-input-file, containing definition and initial position of the dummy $ $ For the exact format of this file, see the ATB User’s Guide $ $ OCT. 27 1994 0 0 0.0 CARD A1A DEMO4: ACCELERATED DUMMY HITS AIRBAG (UNITS: SI) CARD A1B DR TREEP, MSC BV, HOLLAND CARD A1C $ $ Units of length,force and time, gravity vector,gravitational constant: $ M N SEC 0.0 0.0 0.00 9.81 CARD A3 $ $ Initial , maximum and minimum integration step (resp. on field 4, 5 and 6) : $ 4 999 0.002 0.0005 0.001 .000063 CARD A4 $ $ ATB-output specification: $ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0CARD A5 $ $Definition of the 15 segments and 14 joints: $ 15 14 MADYMO DEMO4 DUMMY CARD B.1 $ Definition of the 15 segments: $ Name, plot-symbol, weight (not mass!), principal moments of inertia LT 1115.37 .1297 .0817 .1393 0.115 0.165 0.115 .019 .000 .034 CARD B.2 CT 226.389 .0140 .0159 .0186 0.110 0.150 0.110 -.009 .000 -.006 CARD B.2 . . . LUA E21.778 .0161 .0156 .01 0.047 0.042 0.141 .000 .000-.0085 CARD B.2 LLA F21.092 .0311 .0301 .01 0.040 0.040 0.235 .000 .000 -.018 CARD B.2 $ $ Definition of the 14 joints: $

Main Index

412 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ Joint j connects segment j=1 and the lower segment number in 3rd field: P M 1 0-0.026 0.000 0.079-0.033 0.000-0.072 CARD B.3 0.00 0.00 180. 0.00 -5.00 180. 0 0 0 . . . $ $ Ankles are made rigud, by locking them and supplying huge forces to unlock: $ RA S 7 -4 0.00 0.00 -0.17 -.084 0.00 0.048 0-99999999999 CARD B.3 0.00-90.00 180. 0.00 -100. 180. 0 0 0 . LE Z 14 1 0.000 0.000-0.139 0.000 0.000 0.167 CARD B.3 0.00 0.00 180. 0.00 70.00 180. 0 0 0 $ $ Spring characteristics of the joints: $ .4067 0.000 0.000 1.000 5.0000 .5640 0.000 0.000 1.000 5.0000CARD B4A .4067 0.000 0.000 1.000 35.0000 .5640 0.000 0.000 1.000 35.0000CARD B4B . . . .8896 0.000 0.000 0.700 122.5000 .0000 0.000 0.000 0.700 80.0000CARD B4M .8896 0.000 0.000 0.700 65.3000 .0000 0.000 0.000 0.000 0.0000CARD B4N $ $ Viscous charateristics of the joints $ .03016 0.000 30. 0. 0. 0. 0. CARD B5A .00302 0.000 30. 0. 0. 0. 0. CARD B5B . . . .00115 1.037 30. 0. 0.0. 0. CARD B5M .00115 1.671 30. 0. 0. 0. 0. CARD B5N $ $ Convergence test parameters: $ .01 .01 .01 .01 .01 .01 .10 .10 .10 .10 .10 .01CARD B6A . . . .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .0 .00CARD B6O DUMMY VEHICLE CARD C1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3 0.0 0.010 0CARD C2A 0.0 0.0 0.0 CARD C3A 3 0 0 0 0 0 0 0 0 15 CARD D1 $ $ Definition of some ATB contact planes: $ 1 SEAT DR CARD D2A 0.365 0.25 0.262 CARD D2B -0.015 0.25 0.145 CARD D2C -0.015 -0.25 0.145 CARD D2D 2 SEAT BACK DR CARD D2A

Main Index

Chapter 6: Occupant Safety 413 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

3

-0.040 -0.305 -0.305 FLOOR 0.89 -1.0 -1.0

0.25 0.25 -0.25

0.149 0.689 0.689

0.35 0.35 -1.05

0.017 0.017 0.017

CARD D2B CARD D2C CARD D2D CARD D2A CARD D2B CARD D2C CARD D2D CARD D7

$ $ Force definitions, used to apply time-dependent forward acceleration: $ $ The neagtive value for NFVNT on cards D.9 force ATB to define the force$ coordinate sysytem with respect to the global reference system, instead of the local reference system of the segment. This option was added to ATB$ to define a time dependent gravity field in a fixed global direction. $ $ Segement number, function-number, local point of application and direction: $ 1 -1 0. 0. 0. 0.0 0.0 0.0CARD D9 2 -2 0. 0. 0. 0.0 0.0 0.0CARD D9 . . . 14 -9 0. 0. 0. 0.0 0.0 0.0CARD D9 15 -10 0. 0. 0. 0.0 0.0 0.0CARD D9 $ $ Fuction definitions,used to define time-dependent force on each segment: $ (Dependent on mass of segment !) $ 1 DECEL FORCE LT CARD E1 0. -1.E-1 0. 0. 0. CARD E2 24 CARD E4A 0.00000 0.00000 0.00300 35.28135 0.00790 1646.46279CARD E4B 0.01400 834.99185 0.01810 1528.85831 0.02530 1434.77472CARD E4B 0.02850 882.03364 0.03450 1658.22324 0.03930 705.62691CARD E4B 0.04660 2963.63303 0.05000 1999.27625 0.05430 2105.12029CARD E4B 0.06500 3704.54128 0.06900 2493.21509 0.07400 3669.25994CARD E4B 0.07710 2822.50765 0.08390 3963.27115 0.08850 340.32926CARD E4B 0.09090 2493.21509 0.09740 -446.89704 0.10300 305.77166CARD E4B 0.11050 -764.42915 0.11500 0.00000 1.16000 0.00000CARD E4B . . . 10 DECEL FORCE RLA/LLA CARD E1 0. -1.E-1 0. 0. 0. CARD E2 24 CARD E4A 0.00000 0.00000 0.00300 6.45015 0.00790 301.00714CARD E4B 0.01400 152.65362 0.01810 279.50663 0.02530 262.30622CARD E4B 0.02850 161.25382 0.03450 303.15719 0.03930 129.00306CARD E4B 0.04660 541.81284 0.05000 365.50866 0.05430 384.85912CARD E4B 0.06500 677.26606 0.06900 455.81081 0.07400 670.81590CARD E4B 0.07710 516.01223 0.08390 724.56718 0.08850 427.86014CARD E4B 0.09090 455.81081 0.09740 -81.70194 0.10300 55.90133CARD E4B 0.11050 -139.75331 0.11500 0.00000 1.16000 0.00000CARD E4B $ $ Definition of contact forces between segments and ATB-planes:

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414 Dytran Example Problem Manual Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

$ 11

FORCE DEFLECTION 0. -0.11

0.

0.

0.

4 0. 0.11

0. 14500.

0.02

1500.

0.06

CARD E.1 CARD E2 CARD E4A 9500.CARD E4B CARD E4B

. . . 16

FORCE DEFLECTION 0. -0.10 2

0.

0.

0.

CARD E.1 CARD E2 CARD E4A CARD E4B

0. 0. 0.10 2000. 9999 $ $ Contact definition between segments and ATB-planes and $ between segments and segements $ CARD 1 2 2 CARD 1 17 1 1 11 0 0 0 14 0 CARD 2 17 1 1 12 0 0 0 14 -1 CARD . . . 6 6 13 13 16 0 0 0 15 CARD 9 9 15 15 16 0 0 0 15 CARD 0 0 0 0 0 0 0 0 0 0 0 0 0 0 CARD $ $ Initialisation of dummy: $ $ The 9th field forces an initial equilibrium by substraction of initial $ joint-forces. Care should be taken using this added option ! $ 0.0 0.0 0.0 0 0 0 0 1 1 CARD

E.1 F1A F1B F1B

F3B F3B F.4

G1A

$ Initial coordinates and velocity of reference segment in inertial reference: $ .027711637 0.000 .24136108 0.0 0.0 0.0 CARD G2A $ $ Initial rotation and angular velocity of the segments in local reference: $ 0.0-22.918312 0.0 0.0 0.0 0.0 CARD G3A 0.0-22.918312 0.0 0.0 0.0 0.0 CARD G3B . . . 0.0-22.918312 0.0 0.0 0.0 0.0 CARD G3N 0.0-74.484513 0.0 0.0 0.0 0.0 CARD G3O $ $ ATB timehistory output specification: $ CARD H1A . .

Main Index

Chapter 6: Occupant Safety 415 Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

. CARD H11

Main Index

416 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Sled Test Verification of the Enhanced Hybrid III Dummy (50%) Problem Description The analysis presented in this example problem is a validation of the integrated ATB occupant model against a sled test. The sled test is described in [Ref. 1.]. The model setup is shown in Figure 6-9. It shows the digitized Hybrid III dummy, the planes representing the sled and the belt elements. Not shown are the contact ellipsoids of the ATB dummy.

Figure 6-9

Model Setup

The calculation and the modeling techniques can be subdivided into:

Main Index

Occupant kinematics

ATB

Sled model

ATB rigid planes

Occupant/sled interaction

ATB contact

Occupant positioning

MD Patran

Lap & shoulder belts

Dytran

Belt pretensioning

Dytran

Occupant/belt interaction

Dytran contact

Chapter 6: Occupant Safety 417 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

This example problem will show: • How to use GEBOD to get the basic setup of a 50% HYBRID III. • How to create an ATB input file for a sled test, using the output from GEBOD as the

starting point. • How to create FEM entities to visualize and position the ATB dummy, by using: • Shell elements coated on the ATB contact ellipsoids • Beams representing the local coordinate system of the ATB segments • Beams representing the local coordinate system of the ATB joints • How to attach Dytran finite elements to ATB segments, by creating: • A FEM representation of the ATB planes • An geometric representation of the ATB contact ellipsoids • An accurate geometric representation of the dummy • How to position the dummy with MD Patran. • Integrating the dummy into other FEM models using MD Patran. • How to define lap and shoulder belts and how to prestress the belts. • How to apply an acceleration field to the dummy. • Comparison of the results with the experiment.

Usage of GEBOD to get the Basic Setup of a 50% HYBRID III The Generator of Body Data (GEBOD) program is supplied on the Dytran installation tape. Like ATB, this program is developed and maintained by Wright Patterson Air Force Base. The objective of the GEBOD program is to automatically generate data sets for human and dummy rigid body dynamics modeling. The data generated is a partial input file for the ATB program. The input entries for ATB are described in [Ref. 3.]. The data set of the HYBRID III dummy in ATB corresponds with the definition of the HYBRID III as given in [Ref.2.]. #gebod PROGRAM GEBOD GEBOD GENERATES BODY DESCRIPTION DATA SUITABLE FOR INPUT TO THE ATB MODEL PLEASE ENTER A DESCRIPTION OF THE SUBJECT (<60 CHARS.) EXAMPLE PROBLEM - HYBRID III WHAT IS THE NAME OF THE OUTPUT FILE FOR THE TABLE FORM AND RESULTS ? USE THE FILE EXTENSION ".ATB". EXAMPLE.TAB EXAMPLE PROBLEM - HYBRID III 1) CHILD (2-19 YEARS)

Main Index

418 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

2) ADULT HUMAN FEMALE 3) ADULT HUMAN MALE 4) USER-SUPPLIED BODY DIMENSIONS 5) SITTING HYBRID III DUMMY (50%) 6) SITTING HYBRID III DUMMY (50%) 7) HYBRID II DUMMY (50%) ENTER NUMBER CORRESPONDING TO DESIRED SELECT UNITS FOR OUTPUT 1) ENGLISH 2) METRIC 1 IS ATB MODEL FORMATTED OUTPUT DESIRED Y WHAT IS THE NAME OF THE OUTPUT FILE TO USE THE FILE EXTENSION “.AIN”. EXAMPLE.AIN IS IT DESIRED TO PRODUCE ANOTHER BODY N STOP FROM BRESULTS

SUBJECT TYPE 5

(Y/N) ? CONTAIN THE ATB INPUT DATA ?

DESCRIPTION DATA SET (Y/N) ?

This session generates the file EXAMPLE.AIN as shown in Appendix A: File E XAMPLE.AIN as generated by GEBOD. The Metric system is using leg-m-s for the mass-length-time units, while the English system is using slug-in-s.

Completion of the ATB Input File This file needs some more input entries to create a complete input file. The complete input file as used for the calculation is given in Appendix B: Complete ATB Input File for Sled Test Calculation. The file shown contains comments to clarify the meaning of the input cards. The input file used for the calculation doesn’t contain the comment lines. Most important additions to the input file are the input entries: • Entries defining an extra vehicle, named SLED • D.2 entries defining the planes: • BACK SEAT • BOTTOM SEAT • FLOOR • TOEBOARD • FIRE WALL • G.2 entry defining the initial position of the Lower Torso • G.3 entries giving the dummy an initial sitting position, such that it is sitting correctly in the sled.

Main Index

Chapter 6: Occupant Safety 419 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

How to Create FEM Entities for ATB Dummy It is possible to ask for a Bulk Data file written out by Dytran containing the following: • Shell elements coated on the ATB contact ellipsoids • Beams representing the local coordinate system of the ATB segments • Beams representing the local coordinate system of the ATB joints

The input entities needed are: • ATBJNT • ATBSEG • PARAM, ATBSEGCREATE

An input file to create the FEM representation for the lower torso segments LT and the middle torso MT is as follows: $ SI Units: kg - meter - seconds $ -----------------------------$ $ file: create_fem_lt_mt.dat $ ========================== $ $ ----------------------------------------------------------------------$ This example shows how to create an fem representation of the atb $ segments LT & MT and the joint P that connects the two. $ ----------------------------------------------------------------------$ CHECK=YES PARAM,INISTEP,1.E-3 $ BEGIN BULK $ $ Define the segment contact ellipsoids $ -> card B.2 $ RELEX,LT,ATB RELEX,MT,ATB RELEX,UT,ATB RELEX,N,ATB RELEX,H,ATB RELEX,RUL,ATB RELEX,RLL,ATB RELEX,RF,ATB RELEX,LUL,ATB RELEX,LLL,ATB RELEX,LF,ATB RELEX,RUA,ATB RELEX,RLA,ATB RELEX,LUA,ATB RELEX,LLA,ATB RELEX,RHD,ATB

Main Index

420 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

RELEX,LHD,ATB $ $ Tell DYTRAN to create the ATBSEGs & ATBJNTs with length of the $ bars/beams equal to: segment coordinate systems -> LENGTH=.025 m $ joint coordinate systems -> LENGTH=.05 m $ PARAM,ATBSEGCREATE,YES,LT+MT,.025,.05 $ $ atbseg 1 : name = LT $ will be covered with 72 (NUMELM) shell elements with: $ - Gridpoint ids start at 11000 (GSTART) $ - element ids start at 11000 (ESTART) $ - material id is 20001 (MID ) $ - property id is 20201 (PIDCOV) $ $ if PARAM,ATBSEGCREATE has been specified, beam elements $ representing the local coordinate system will be generated $ with: $ - Grid id of grid located at the origin = 20001 (G0 ) $ - Grid id of grid located on local x-axis = 20002 (G1 ) $ - Grid id of grid located on local y-axis = 20003 (G2 ) $ - Grid id of grid located on local z-axis = 20004 (G3 ) $ - Cbar id of element representing local x-axis = 20001 (EID1 ) $ - Cbar id of element representing local y-axis = 20002 (EID2 ) $ - Cbar id of element representing local z-axis = 20003 (EID3 ) $ - Property id of cbar elements = 20001 (PIDCG) $ - Material id of cbar elements = 20001 (MID ) $ $ if PARAM,ATBSEGCREATE has not been specified the position and $ orientation of the ATB segments as spcified on the G.2 and G.3 $ entries in the ATB input file will be overruled by the definitions $ given here. The local coordinate system is defined by: $ - Grid id of grid located at the origin = 20001 (G0 ) $ - Grid id of grid located on local x-axis = 20002 (G1 ) $ - Grid id of grid located on local y-axis = 20003 (G2 ) $ - Grid id of grid located on local z-axis = 20004 (G3 ) $ $ ATBSEG ID NAME COVER NUMELM GSTART ESTART MID PIDCOV + $ + G0 G1 G2 G3 EID1 EID2 EID3 PIDCG $ ATBSEG ,1 ,LT ,YES ,72 ,11000 ,11000 ,20001 ,20201 ,+ + ,20001 ,20002 ,20003 ,20004 ,20001 ,20002 ,20003 ,20001 $ ATBSEG,2,MT,YES,72 ,11250,11250,20002,20202,+ +,20005,20006,20007,20008,20004,20005,20006,20002 $ $ $ atbjnt 1 with name P connects atb segments 1 (LT) with 2 (MT) $ (see card B.3 in ATB input deck) $ if PARAM,ATBSEGCREATE has been specified, beam elements $ representing the joint coordinate systems will be generated $ with: $ -> coordinate system connected to atb segment 1 (LT) $ - Grid id of grid located at the origin = 20201 (G0 )

Main Index

Chapter 6: Occupant Safety 421 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ $ $ $ $ $ $ $ $ $ $ $ -> $ $ $ $ $ $ $ $ $ $ $ $ $ $ ATBJNT ID $ + G0 $ + G4 $ ATBJNT ,1 + ,20201 + ,20205 $ ENDDATA

-

Grid id of grid located on local x-axis = 20202 (G1 ) Grid id of grid located on local y-axis = 20203 (G2 ) Grid id of grid located on local z-axis = 20204 (G3 ) Cbar id of element representing local x-axis = 20201 (EID1 ) Cbar id of element representing local y-axis = 20202 (EID2 ) Cbar id of element representing local z-axis = 20203 (EID3 ) Property id of cbar elements = 20001 (PIDCG-LT) Material id of cbar elements = 20001 (MID -LT) where: PIDCG-LT is the PIDCG as specified on ATBSEG,,LT MID -LT is the MID as specified on ATBSEG,,LT

coordinate system connected to atb segment 2 (MT) - Grid id of grid located at the origin = 20205 (G4 ) - Grid id of grid located on local x-axis = 20206 (G5 ) - Grid id of grid located on local y-axis = 20207 (G6 ) - Grid id of grid located on local z-axis = 20208 (G7 ) - Cbar id of element representing local x-axis = 20204 (EID4 ) - Cbar id of element representing local y-axis = 20205 (EID5 ) - Cbar id of element representing local z-axis = 20206 (EID6 ) - Property id of cbar elements = 20002 (PIDCG-MT) - Material id of cbar elements = 20002 (MID -MT) where: PIDCG-MT is the PIDCG as specified on ATBSEG,,MT MID -MT is the MID as specified on ATBSEG,,MT

NAME G1 G5

G2 G6

G3 G7

EID1 EID4

,P ,20202 ,20203 ,20204 ,20201 ,20206 ,20207 ,20208 ,20204

EID2 EID5

EID3 EID6

,20202 ,20205

,20203 ,20206

+ +

,,,,,,,+ ,,+

Running this input file with the command: dytran jid=create_fem_lt_mt atb=run

will generate a Bulk Data file named: CREATE_FEM_LT+MT_ATBSEGAT. Figure 6-10 shows a plot of the FEM entries written to the Bulk Data file. It can be seen that both the ellipsoids are covered with shell elements. The smaller bars represent the local coordinate systems for the segments LT and MT, respectively. The larger bars represent the joint coordinate system s for the joint P connecting the two segments together. In this particular case, the joint coordinate systems are coincident.

Main Index

422 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Figure 6-10

FEM Entities generated by Dytran for LT and MT

Appendix C: File Create_fem_dummy.dat gives the input file to generate FEM entities for the

complete dummy. Figure 6-11 shows a plot of the generated entities.

Figure 6-11

Main Index

FEM Entities generated by Dytran for Hybrid III Dummy

Chapter 6: Occupant Safety 423 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Attaching Dytran Finite Element to ATB Segments It is possible to attach Dytran finite elements to ATB segments by means of the RCONREL entry. For example, to attach elements to ATB segment 1 (LT), give the elements the rigid material (MATRIG) 20001. Then insert the following input entries in the input file: RCONREL,1,20001,RIGID,20001 SETC,20001,LT SET1,20001,MR20001

The finite elements that are generated in How to Create FEM Entities for ATB Dummy all have a MATRIG; therefore, these elements are easily attached to the corresponding ATB segment. To visualize the sled, some shell elements were defined within MD Patran. These shell elements are attached to the ATB segment named SLED by giving these shell elements a rigid material and using a correct RCONREL entry. Similarly, the shell elements generated by digitizing an actual HYBRID III dummy were connected to the corresponding ATB segments. The resulting model is as depicted in Figure 6-9. The digitized dummy contains some elements that are dummy shell elements. Dummy shell elements are shell elements without structural behavior (strength). These dummy shell elements have the purpose to create a closed surface between the different rigid parts of the dummy and to maintain a realistic shape during the motion of the dummy. These dummy shell elements themselves are not attached to any ATB segment. The grid points on one side are attached to one ATB segment while the grid points on the other side are attached to the other ATB segment. Figure 6-12 shows the motion of the digitized dummy during a crash analyses, where the effect of the dummy shell elements is most clear for the neck, elbow, middle torso, and upper legs.

Figure 6-12

Main Index

Motion of the Digitized Dummy during Crash

424 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Positioning the Dummy with Patran In order to position the dummy with MD Patran, the ATBSEG entry has been introduced in Dytran. The ATBSEG entry references four grid points that span a coordinate system. This coordinate system becomes the local coordinate system of the referenced ATB segment. The local coordinate system as given on an ATBSEG entry in the Dytran input file will overrule the definitions in the ATB input file. For the HYBRID III dummy, the ATBSEG entries are: ATBSEG,1,LT,,,,,,,+ +,20001,20002,20003,20004 $ ATBSEG,2,MT,,,,,,,+ +,20005,20006,20007,20008 $ ATBSEG,3,UT,,,,,,,+ +,20009,20010,20011,20012 $ ATBSEG,4,N ,,,,,,,+ +,20013,20014,20015,20016 $ ATBSEG,5,H ,,,,,,,+ +,20017,20018,20019,20020 $ ATBSEG,6,RUL,,,,,,,+ +,20021,20022,20023,20024 $ ATBSEG,7,RLL,,,,,,,+ +,20025,20026,20027,20028 $ ATBSEG,8,RF,,,,,,,+ +,20029,20030,20031,20032 $ ATBSEG,9,LUL,,,,,,,+ +,20033,20034,20035,20036 $ ATBSEG,10,LLL,,,,,,,+ +,20037,20038,20039,20040 $ ATBSEG,11,LF,,,,,,,+ +,20041,20042,20043,20044 $ ATBSEG,12,RUA,,,,,,,+ +,20045,20046,20047,20048 $ ATBSEG,13,RLA,,,,,,,+ +,20049,20050,20051,20052 $ ATBSEG,14,LUA,,,,,,,+ +,20053,20054,20055,20056 $ ATBSEG,15,LLA,,,,,,,+ +,20057,20058,20059,20060 $

Main Index

Chapter 6: Occupant Safety 425 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

ATBSEG,16,RHD,,,,,,,+ +,20061,20062,20063,20064 $ ATBSEG,17,LHD,,,,,,,+ +,20065,20066,20067,20068 $

The locations of grid points 20001 to 20068 are defining the position and orientation of the dummy. In order to easily position the dummy, the dummy positioner functionality inside the Dytran preference for MD Patran can be used. To start positioning, do the following within MD Patran (see also the DytranMD Patran Preference Interface Guide. 1. Start the dummy positioner: Analyses Action: Special Features Object: Dummy Positioner Apply 2. Read in the dummy data set: Import Dummy Select hybridIII-50p.bdf Apply 3. Create the groups of the dummy: Build Dummy Define Dummy Prefix Name OK 4. Position the dummy: Manipulate Dummy Translation of the Dummy: Action: Transform Object: Full Dummy 5. Define Translation Vector: Select the Dummy for Manipulation Apply

Main Index

426 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

6. Rotation of a part of the dummy: Action:

Transform

Object:

Part Dummy

Method:

Rotate

Parts

Select group

Power User Tools

Choose axis

Relation Parameters

Define axis and angle

Apply 7. Write out the new positional dummy: Export Dummy Select Original Dummy Name Select Output Dummy’s File Name OK

Integrating the Dummy into other FEM Models The dummy model as delivered is using the following numbers Entity

Minimum

Maximum

Grids

20001

24138

Elements

20001

24299

Material IDs

20001

20017

Property IDs

20001

20404

:The dummy date file hybridIII-50p.bdf as described above is set up such that the dummy can be easily integrated into an existing FEM model. The numbering of the grids in the dummy data set is consistent with the grid numbering in the ATBSEG entries and, whenever the grids of the dummy data set are renumbered, the grid numbering in the ATBSEG entries must be changed also in the same manner.

Defining Lap and Shoulder Belts The belts are defined by means of CROD elements, referencing a PBELT property. The mass per unit length is defined and the load/unloading curves are defined on the PBELT entry. The load/unload curves are in force per unit strain.The initial position of the grid points is allowed to be a little bit inside or outside the contact surface. If the contact version BELT1 is chosen, the grids of the belt are automatically positioned on top of the master surface. The contact entries are: $ file: belt_contacts.dat $ -----------------------

Main Index

Chapter 6: Occupant Safety 427 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$BEGIN BULK $ ========================================================== $belt with the dummy $All gridpoints within a distance of 2.54 mm will be repositioned $$ belt gridpoints SET1,20100,70000t70058 $ SURFACE,20401,,PROP,20401 SET1,20401,20101,20303,20103,20302,20104,20301 CONTACT,101,GRID,SURF,20100,20401,1.0,1.0,1.0,+ +,BELT1,TOP,,,,,,SLAVE,+ +,,,,,,,,,+ +,,,,,,,2.54E-03 $ $ Check for contact with the digitized upper-legs: $ Upper legs (dummy shells) $ SURFACE,20402,,PROP,20402 SET1,20402,20304,20307 CONTACT,102,GRID,SURF,20100,20402,1.0,1.0,1.0,+ +,BELT1,TOP,,,,,,SLAVE,+ +,,,,,,,,,+ +,,,,,,,2.54E-03 $ $ ========================================================== $ ENDDATA

The first 10 msecs of the calculaton are being used to prestress the belt. In the experiment, the belt was pulled snug. This is simulated by introducing a small prestress strain during the first 5 msecs, and releasing the prestress strain during the following 5 msecs. The end points of the belts are connected to the sled by means of an RCONREL entry. The complete input for the belt elements is: $$ file: sled.dat $ -------------$PBELT,20100,101,101,.1860205,0.1,0.1,,201 $ $ Load/unload curve $ TABLED1,101,,,,,,,,+ +,0.0,0.0,.065,1779.288,.25,11120.55,XSMALL,ENDVAL,+ +,XLARGE,EXTRAP $ $ $ prestress table: $ -> first position the belt by giving a small prestress strain $ -> simulate pulling the belt snug, by releasing the prestress $ TABLED1,201,,,,,,,,+ +,.0,.0,25.E-3,0.02,50.E-3,0.0 $ $ Rconrel entry to connect the belt to the sled ellipsoid

Main Index

428 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ RCONREL,2,70001,GRID,70001 SETC,70001,SLED SET1,70001,70000,70001,70004,70005 $

Applying an Acceleration Field to the Dummy Standard ATB is not able to apply an acceleration field to the segments. This capability has been introduced into the version integrated into Dytran by means of the ATBACC entry. To apply an acceleration pulse to the complete dummy, the following Bulk Data entries are specified. $$ file: acc_low.dat $ ================= $BEGIN BULK $Apply an x-acceleration to the dummy $ TLOAD1,1,201,,,202 $ $Offset with time to: $- allow the belt to prestress itself $ TABLED1,202,,,,,,,,+ +,0.0E+00, 0.0,1.0E-02, 5.0,2.0E-02,12.0,3.0E-02,17.0,+ +,4.0E-02,22.0,5.0E-02,23.7,6.0E-02,23.0,7.0E-02,19.0,+ +,8.0E-02,15.0,9.0E-02,12.5,1.0E-01, 7.5,1.1E-01, 1.0,+ +,1.2E-01, 0.0,1.3E-01, 0.0,XOFFSET,50.0E-3 $ $ Table is acceleration in G $ Use SCALE to convert it into m/s2 : 1 G = 9.8 m/s2 $ ATBACC,201,,9.8,1.0,0.0,0.0,,,+ +,LT,MT,UT,N,H,RUL,RLL,RF,+ +,LUL,LLL,LF,RUA,RLA,LUA,LLA,RHD,+ +,LHD $ ENDDATA

Comparison of the Results with Experiment An animation of the simulation is available as an mpg file: Sledtest.mpg Figure 6-13 shows the comparison with experimental data. The acceleration of the head, chest, and lower torso is requested in the ATB input file on Cards A.5 and Cards H.1. The acceleration curves are output in Dytran time history format because the following parameter was set: PARAM,ATB-H-OUTPUT,YES

Main Index

Chapter 6: Occupant Safety 429 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Dytran generates a file name: SLEDTEST_ATB_H1.THS containing the accelerations in time. The accelerations are output in g.

Figure 6-13

Acceleration Time Histories for Low Severity Level

Appendix A: File EXAMPLE.AIN as generated by GEBOD 17 LT MT UT N

Main Index

16 SITTING HYBRID III 544.4602.45751.29691.2080 5.000 0.00-37.32 180.0 4 4.8900.06120.05930.0205 4.775 0.00 4.16 180.0 339.2162.62032.05171.7336 4.825 0.00 4.99 180.0 2 2.6680.02540.02570.0084 1.675

7.185 4.800-1.000 0.000 0.000

CARD B.1 1 CARD B.2

6.500 4.000 1.000 0.000-1.000

1

CARD B.2

6.500 7.785 0.000 0.000 0.000

1

CARD B.2

1.675 3.000 0.000 0.000 0.000

1

CARD B.2

430 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

H RUL RLL RF LUL LLL LF RUA RLA LUA LLA RHD LHD P W NP HP RH RK RA LH LK LA RS RE LS LE

Main Index

0.00 0.00 180.0 1 9.9210.14080.21280.1956 4.250 2.875 4.000 0.000 0.000 0.000 0.00-26.58 180.0 613.7130.60920.59340.1068 2.950 3.050 7.285 0.000 0.000 0.000 0.00 4.13 180.0 7 7.2370.67080.67450.0397 2.165 2.050 9.750 0.000 0.000 2.000 0.00 -1.90 180.0 8 2.7560.00670.05240.0491 4.900 1.675 1.675 0.000 0.000 0.000 0.96 -9.57 158.2 913.7130.60920.59340.1068 2.950 3.050 7.285 0.000 0.000 0.000 0.00 4.13 180.0 0 7.2370.67080.67450.0397 2.165 2.050 9.750 0.000 0.000 2.000 0.00 -1.90 180.0 1 2.7560.00670.05240.0491 4.900 1.675 1.675 0.000 0.000 0.000 -0.96 -9.57-158.2 2 4.5970.10240.09970.0109 1.900 1.800 6.000 0.000 0.000-1.000 0.00 -1.31 180.0 3 3.8000.11910.11280.0069 1.775 1.775 5.800 0.000 0.000 0.000 0.00 -1.31 180.0 4 4.5970.10240.09970.0109 1.900 1.800 6.000 0.000 0.000-1.000 0.00 -1.31 180.0 5 3.8000.11910.11280.0069 1.775 1.775 5.800 0.000 0.000 0.000 0.00 -1.31 180.0 6 1.2900.01140.00930.0036 1.000 1.870 3.650 0.000 0.000 0.000 -5.35 30.75 173.5 7 1.2900.01140.00930.0036 1.000 1.870 3.650 0.000 0.000 0.000 5.35 30.75-173.5 M 1 0 -2.15 0.00 -1.66 -0.35 0.00 2.56 0 0.0 0.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 N 2 0 -0.35 0.00 -2.56 -0.89 0.00 5.85 0 0.0 0.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 O 3 0 0.00 0.00 -5.96 0.00 0.00 2.76 0 0.0 0.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 P 4 0 0.00 0.00 -2.84 -0.55 0.00 2.00 0 0.0 0.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Q 1 0 -0.11 3.15 1.24 0.00 0.00 -9.96 0 0.0 0.0 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 R 6 1 0.00 0.00 6.56 -0.20 0.00 -6.74 0 0.0 0.0 0.00 0.00 0.00 0.00 55.00 0.00 0.00 0.00 0.00 S 7 -8 -0.20 0.00 9.65 -2.12 0.00 -1.54 0 0.0 0.0 90.00 0.00 0.00 90.00 0.00 0.00 0.00-10.00 0.00 T 1 0 -0.11 -3.15 1.24 0.00 0.00 -9.96 0 0.0 0.0 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 U 9 1 0.00 0.00 6.56 -0.20 0.00 -6.74 0 0.0 0.0 0.00 0.00 0.00 0.00 55.00 0.00 0.00 0.00 0.00 V 10 -8 -0.20 0.00 9.65 -2.12 0.00 -1.54 0 0.0 0.0 90.00 0.00 0.00 90.00 0.00 0.00 0.00-10.00 0.00 W 3 0 -0.88 7.38 -2.66 0.00 0.00 -5.43 0 0.0 0.0 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 X 12 -8 0.00 0.00 4.94 0.00 0.00 -3.67 0 0.0 0.0 90.00 0.00 0.00 90.00 0.00 0.00 0.00 62.50 0.00 Y 3 0 -0.88 -7.38 -2.66 0.00 0.00 -5.43 0 0.0 0.0 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Z 14 -8 0.00 0.00 4.94 0.00 0.00 -3.67 0 0.0 0.0

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

1

CARD B.2

CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD 0 0 0 0 CARD

B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3 0 0 B.3

Chapter 6: Occupant Safety 431 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

RW LW -45.00 -45.00 -46.00 -47.00 -43.00 0.00 1.00 0.00 -44.00 0.00 1.00 0.00 -41.00 0.00 0.00 -42.00 0.00 0.00 0.00 0.00 0.00 0.00 1.200 1.200 0.350 0.192 1.000 1.000 0.500 0.500 0.000 1.000 1.000 0.500 0.500 0.000 0.100 0.100 0.100 0.000 0.100 0.100 0.100 0.000 0.500 0.500 0.000 0.500 0.500

Main Index

90.00 0.00 0.00 0.00 90.00 0.00 15 -8 0.00 0.00 90.00 0.00 0.00 0.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 2.32 0.00 1.00 10.00 10.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 2.32 0.00 1.00 10.00 10.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 20.00 20.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 20.00 20.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 0.00 0.00 1.00 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 0. 30. 0. 13

-8

0.00 6.07 0.00 6.07 0.00

0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.

90.00 0.00 0.00 0.00 62.50 0.00 0 0 0 0 0 0 -0.30 0.00 -2.13 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 0.00 15.00 0.00 0 0 0 0 0 0 -0.30 0.00 -2.13 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 0.00 15.00 0.00 0 0 0 0 0 0 0.00 34.38 0.00 0.00 1.00 0.00CARD B.4 0.00 34.38 0.00 0.00 1.00 0.00CARD B.4 0.00 15.00 0.00 0.00 1.00 0.00CARD B.4 0.00 15.00 0.00 0.00 1.00 0.00CARD B.4 0.00 7.50 75.00 75.00 1.00 55.00CARD B.4 48.90 0.00 0.00 0.00 1.00 0.00CARD B.4 27.00 0.00 6.69 0.00 1.00 30.00CARD B.4 0.00 0.00 0.00 0.00 0.00 7.50 75.00 75.00 1.00 55.00CARD B.4 48.90 0.00 0.00 0.00 1.00 0.00CARD B.4 27.00 0.00 6.69 0.00 1.00 30.00CARD B.4 0.00 0.00 0.00 0.00 0.00 0.00100.00100.00 1.00 125.00CARD B.4 52.00 0.00 4.65 0.00 1.00 65.30CARD B.4 0.00 0.00 90.00 0.00 0.00 0.00100.00100.00 1.00 125.00CARD B.4 52.00 0.00 4.65 0.00 1.00 65.30CARD B.4 0.00 0.00 90.00 0.00 0.00 0.00 4.32 0.14 1.00 55.50CARD B.4 0.00 0.00 0.00 0.00 0.00 0.00 4.32 0.14 1.00 55.50CARD B.4 0.00 0.00 0.00 0.00 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5 0. 0. CARD B.5

432 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

0.000 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 1

0. 30. 0. 0.01 0.01 0.01 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 0.01 0.01 0.00 HEAD SURFACE 0.00 0.71 2.57 -175.98

0. 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0. 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0. 0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

-0.80 4495.67

0.00 -8953.11

0.00 6029.59

0.75

3580.00

0.80

-1.70 380.91

0.00 -678.22

0.00 456.53

48.30

1.65

483.00

1.70

1.31 5.46

-1.40 73.49

0.00 -15.01

0.00 0.00

1.31 158.60 ANTERIOR PELVIS 0.00 2.02 -0.14 1.58

1.35

1586.00

1.40

-2.10 24.39

0.00 -29.33

0.00 13.76

2.05

897.00

2.10

-2.10 40.04

0.00 -12.49

0.00 0.00

2.03

1278.00

2.10

-1.00 627.09

0.00 -1197.70

0.00 917.23

203.10

0.95

2031.00

1.00

1.31 107.37

-1.40 -313.36

0.00 504.61

0.00 -196.37

156.20

1.35

1562.00

1.40

3 0.71 358.00 BACK OF SHOULDER 0.00 1.62 0.01 -26.59

2

3 1.62 CHEST 0.00 0.55

3

3 4

3 2.02 89.70 POSTERIOR PELVIS 0.00 1.96 -0.76 35.10

5

3 1.96 127.80 UPPER ARM 0.00 0.89 1.54 -30.63

6

3 0.89 FOREARM 0.00 -2.40

7

3 8

Main Index

1.31 HAND

CARD 0.01CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD 0.00CARD CARD CARD 0.00CARD CARD 35800.00CARD CARD CARD 0.00CARD CARD 4830.00CARD CARD CARD 0.00CARD CARD 15860.00CARD CARD CARD 0.00CARD CARD 8970.00CARD CARD CARD 0.00CARD CARD 12780.00CARD CARD CARD 0.00CARD CARD 20310.00CARD CARD CARD 0.00CARD CARD 15620.00CARD CARD

0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

B.5 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 B.6 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1

Chapter 6: Occupant Safety 433 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

0.00 -0.04

0.38 -31.04

-0.45 2384.47

0.00 -10193.90

0.00 19172.50

0.38 139.40 UPPER LEG 0.00 1.31 -0.28 14.19

0.41

1394.00

0.45

-1.40 81.94

0.00 -122.62

0.00 98.73

61.60

1.35

616.00

1.40

0.54 2.22

-0.58 867.34

0.00 -401.61

0.00 0.00

0.56

1878.00

0.58

-1.90 3.92

0.00 29.05

0.00 0.00

1.84

2236.00

1.90

-1.75 -2.08

0.00 0.00

0.00 0.00

3 9

3 1.31 KNEE 0.00 -0.04

10

3 0.54 187.80 FRONT OF LOWER LEG 0.00 1.78 -0.74 26.79

11

3 1.78 223.60 BACK OF LOWER LEG 0.00 1.66 -0.08 29.42

12

3 13

1.66 FOOT 0.00 -0.68

43.20

1.70

432.00

1.75

0.83 82.75

-0.95 -704.17

0.00 3367.10

0.00 -5570.69

0.83

121.20

0.89

1212.00

0.95

3 999 41

Main Index

RIGHT SHOULDER JOINT

CARD 0.00CARD CARD 13940.00CARD CARD CARD -29.09CARD CARD 6160.00CARD CARD CARD 0.00CARD CARD 18780.00CARD CARD CARD 0.00CARD CARD 22360.00CARD CARD CARD 0.00CARD CARD 4320.00CARD CARD CARD 3188.30CARD CARD 12120.00CARD CARD CARD

E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.7

-4 8 151.000 0.000 87900.000 0.000 75.000 0.000 13700.000 0.000 0.000 0.000 -312.000 11800.000 66.000 0.000 7780.000 3010000.000 68.000 0.000 22200.000 376000.000 89.000 0.000 22000.000 0.000 111.000 0.000 2470.000 17600.000 153.000 0.000 13000.000 0.000 42 LEFT SHOULDER JOINT

CARD E.7

-4 8 151.000 0.000 153.000 0.000 111.000 0.000 89.000 0.000 68.000 0.000 66.000 0.000 0.000 0.000 75.000 0.000 43 SEATED RIGHT HIP

CARD E.7

CARD E.7

87900.000 0.000 13000.000 0.000 2470.000 17600.000 22000.000 0.000 22200.000 376000.000 7780.000 3010000.000 -312.000 11800.000 13700.000 0.000 CARD E.7

434 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

19

4 0.000 7.000 472.000 952.000 30700.000 61400.000 1970000.000 0.000 17.000 1070.000 2160.000 69700.000 139000.000 4460000.000 0.000 8.000 182.000 358.000 11300.000 22500.000 721000.000 0.000 7.000 53.000 93.000 2570.000 5130.000 164000.000 44 SEATED LEFT HIP

112.000 7670.000 492000.000

51.000 4340.000 279000.000

119.000 255.000 527.000CARD E.7 8690.000 17400.000 34900.000CARD E.7 557000.000 1110000.000 2230000.000CARD E.7

17.000 710.000 45100.000

28.000 1410.000 90100.000

50.000 2820.000 180000.000

94.000CARD E.7 5640.000CARD E.7 360000.000CARD E.7

13.000 173.000 10300.000

18.000 333.000 20500.000

23.000 653.000 41000.000

33.000CARD E.7 1290.000CARD E.7 81900.000CARD E.7 CARD E.7

4 0.000 7.000 22.000 472.000 952.000 1910.000 30700.000 61400.000 123000.000 1970000.000 0.000 7.000 13.000 53.000 93.000 173.000 2570.000 5130.000 10300.000 164000.000 0.000 8.000 17.000 182.000 358.000 710.000 11300.000 22500.000 45100.000 721000.000 0.000 17.000 51.000 1070.000 2160.000 4340.000 69700.000 139000.000 279000.000 4460000.000 45 SEATED LUMBAR SPINE 19

Main Index

E.7 E.7 E.7 E.7

52.000 3830.000 246000.000

19

4 0.000 27600.000 105800.000 4742600.000 0.000 40800.000 156400.000 7010800.000 0.000 18000.000 69000.000 3093000.000 0.000 40800.000 156400.000

CARD 232.000CARD 15400.000CARD 983000.000CARD

22.000 1910.000 123000.000

CARD 232.000CARD 15400.000CARD 983000.000CARD

E.7 E.7 E.7 E.7

52.000 3830.000 246000.000

112.000 7670.000 492000.000

18.000 333.000 20500.000

23.000 653.000 41000.000

33.000CARD E.7 1290.000CARD E.7 81900.000CARD E.7

28.000 1410.000 90100.000

50.000 2820.000 180000.000

94.000CARD E.7 5640.000CARD E.7 360000.000CARD E.7

119.000 255.000 527.000CARD E.7 8690.000 17400.000 34900.000CARD E.7 557000.000 1110000.000 2230000.000CARD E.7 CARD E.7

4600.000 32200.000 179400.000

9200.000 36800.000 326600.000

CARD 13800.000 18400.000 23000.000CARD 41400.000 50600.000 69000.000CARD 621000.000 1209800.000 2387400.000CARD

E.7 E.7 E.7 E.7

6800.000 47600.000 265200.000

13600.000 54400.000 482800.000

20400.000 27200.000 34000.000CARD E.7 61200.000 74800.000 102000.000CARD E.7 918000.000 1788400.000 3529200.000CARD E.7

3000.000 21000.000 117000.000

6000.000 24000.000 213000.000

9000.000 27000.000 405000.000

6800.000 47600.000 265200.000

13600.000 54400.000 482800.000

20400.000 27200.000 34000.000CARD E.7 61200.000 74800.000 102000.000CARD E.7 918000.000 1788400.000 3529200.000CARD E.7

12000.000 15000.000CARD E.7 33000.000 45000.000CARD E.7 789000.000 1557000.000CARD E.7

Chapter 6: Occupant Safety 435 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

7010800.000 46 NECK PIVOT

CARD E.7

19

4 0.000 700.000 4200.000 4900.000 8400.000 9800.000 96600.000 0.000 600.000 3600.000 4200.000 7200.000 8400.000 82800.000 0.000 300.000 1800.000 2100.000 3600.000 4200.000 41400.000 0.000 600.000 3600.000 4200.000 7200.000 8400.000 82800.000 47 HEAD PIVOT 4 0.000 3780.000 7980.000 96180.000 0.000 3600.000 7200.000 82800.000 0.000 1700.000 3500.000 41300.000 0.000 3600.000 7200.000 82800.000 999 0 0 0

1400.000 5600.000 12600.000

2100.000 6300.000 18200.000

2800.000 7000.000 29400.000

CARD 3500.000CARD 7700.000CARD 51800.000CARD

1200.000 4800.000 10800.000

1800.000 5400.000 15600.000

2400.000 6000.000 25200.000

3000.000CARD E.7 6600.000CARD E.7 44400.000CARD E.7

600.000 2400.000 5400.000

900.000 2700.000 7800.000

1200.000 3000.000 12600.000

1500.000CARD E.7 3300.000CARD E.7 22200.000CARD E.7

1200.000 4800.000 10800.000

1800.000 5400.000 15600.000

2400.000 6000.000 25200.000

3000.000CARD E.7 6600.000CARD E.7 44400.000CARD E.7 CARD E.7

19

0

E.7 E.7 E.7 E.7

490.000 4480.000 9380.000

980.000 5180.000 12180.000

1680.000 5880.000 17780.000

2380.000 6580.000 28980.000

CARD 3080.000CARD 7280.000CARD 51380.000CARD

600.000 4200.000 8400.000

1200.000 4800.000 10800.000

1800.000 5400.000 15600.000

2400.000 6000.000 25200.000

3000.000CARD E.7 6600.000CARD E.7 44400.000CARD E.7

250.000 2000.000 4100.000

500.000 2300.000 5300.000

800.000 2600.000 7700.000

1100.000 2900.000 12500.000

1400.000CARD E.7 3200.000CARD E.7 22100.000CARD E.7

600.000 4200.000 8400.000

1200.000 4800.000 10800.000

1800.000 5400.000 15600.000

2400.000 6000.000 25200.000

3000.000CARD E.7 6600.000CARD E.7 44400.000CARD E.7

0

0

0

0

0

0

0

0

0

0

0

0

0

E.7 E.7 E.7 E.7

CARD E.7 0CARD F.4

Appendix B: Complete ATB Input File for Sled Test Calculation $ ========================= $ file: run50_commented.ain $ ========================= $--------------------------------------------------------------------------$ Model of a Sitting, 50 % Hybrid III $ $ The model contains comment lines, that start with a $ $ The atb code is not able to handle these kind of comment lines, so they need $ to be stripped out. This can be done easily by the following fortran code:

Main Index

436 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ $ program atb_clean $* $* clean up comments in an atb input file $* $* 1) compile and link $* f77 -o atb_clean.exe atb_clean.f $* 2) clean up the input file with : $* atb_clean.exe < run50_commented.ain > run50.ain $* $ character*(80) line $ 1 continue $ read(*,’(a)’,end=9900) line $ if (line(1:1).ne.’$’) then $ write(*,’(a)’) line $ endif $ goto 1 $ 9900 continue $ end $ $--------------------------------------------------------------------------$ Date & Restarts - Format (3A4,2I4,F8.0) $ $234A234A234I234I234F2345678 FEB. 13.1995 0 0 0.0 CARD A1A $--------------------------------------------------------------------------$ $ Comment - Format (20A4 / 20A4) $ SITTING HYBRID III DUMMY (50%) GENERATED WITH GEBOD CARD A1B FORD SLED TEST 112 MS ; 23.7 G PEAK ACCELERATION CARD A1C $--------------------------------------------------------------------------$ $ Units & Gravity definition - Format (3A4, 4F12.0) $ $ The influence of gravity is assumed to be negligible $ $234A234A234F23456789012F23456789012F23456789012F23456789012 M. N. SEC. 0.0 0.0 9.8 9.8 CARD A3 $------------------------------------------------------------------------------$ $ Integrator variables - Format (2I4, 4F8.0) $ $234I234F2345678F2345678F2345678F2345678 6 55 0.003 .000125 .001 .000125 CARD A4 $------------------------------------------------------------------------------$ $ Output and program control features - Format (36I2) $ $10203040506070809101112131415161718192021222324252627282930313233343536 0 0-1 2 0 0 0 0 0 0 0 0 1 0 0 0 0 9 0 0 0 0 0 0 0-1 0 0 0 1 0 0 0 0 0 1CARD A5 $-----------------------------------------------------------------------------$ $ Number of segments & joints + name - Format (2I6, 8X, 5A4)

Main Index

Chapter 6: Occupant Safety 437 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ $23456I23456A2345678A234A234A234A234A234 18 17 SITTING HYBRID III CARD B.1 $------------------------------------------------------------------------------$ $ Definition of the segments - Format (A4, 1X, A1, 10F6.0, I4) $ Format (12X, 3F6.0) $ $234XAF23456F23456F23456F23456F23456F23456F23456F23456F23456F23456I234 $1 LT 5197.78 .2777 .1465 .1365 .127 .182 .122 -.025 .000 .000 1 CARD B.2 .00 35.00 180.0 $2 MT 4 21.75 .0069 .0067 .0023 .121 .165 .102 .025 .000 -.025 1 CARD B.2 .00 .00 180.0 $3 UT 3174.45 .2961 .2318 .1959 .123 .165 .198 .000 .000 .000 1 CARD B.2 .00 4.99 180.0 $4 N 2 11.87 .0029 .0029 .0009 .043 .043 .076 .000 .000 .000 1 CARD B.2 .00 .00 180.0 $5 H 1 44.13 .0159 .0240 .0221 .108 .073 .102 .000 .000 .000 1 CARD B.2 .00-26.58 180.0 $6 RUL 6 61.00 .0688 .0670 .0121 .075 .077 .185 .000 .000 .000 1 CARD B.2 .00 4.13 180.0 $7 RLL 7 32.19 .0758 .0762 .0045 .055 .052 .248 .000 .000 .051 1 CARD B.2 .00 -1.90 180.0 $8 RF 8 12.26 .0008 .0059 .0055 .124 .043 .043 .000 .000 .000 1 CARD B.2 -2.69 -9.23-158.0 $9 LUL 9 61.00 .0688 .0670 .0121 .075 .077 .185 .000 .000 .000 1 CARD B.2 .00 4.13 180.0 $10 LLL 0 32.19 .0758 .0762 .0045 .055 .052 .248 .000 .000 .051 1 CARD B.2 .00 -1.90 180.0 $11 LF 1 12.26 .0008 .0059 .0055 .124 .043 .043 .000 .000 .000 1 CARD B.2 2.69 -9.23 158.0 $12 RUA 2 20.45 .0116 .0113 .0012 .048 .046 .152 .000 .000 -.025 1 CARD B.2 .00 -1.31 180.0 $13 RLA 3 16.90 .0135 .0127 .0008 .045 .045 .147 .000 .000 .000 1 CARD B.2 .00 -1.31 180.0 $14 LUA 4 20.45 .0116 .0113 .0012 .048 .046 .152 .000 .000 -.025 1 CARD B.2 .00 -1.31 180.0 $15 LLA 5 16.90 .0135 .0127 .0008 .045 .045 .147 .000 .000 .000 1 CARD B.2 .00 -1.31 180.0 $16

Main Index

438 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

RHD 6 $17 LHD 7

5.74 .0013 .0011 .0004 -2.35-31.09-175.6

.025

.047

.093

.000

.000

.000

1

CARD B.2

5.74 .0013 .0011 .0004 2.35-31.09 175.6

.025

.047

.093

.000

.000

.000

1

CARD B.2

$ $ a segment connected to a vehicle -> used for connecting MSC/DYTRAN FEM model $ $234XAF23456F23456F23456F23456F23456F23456F23456F23456F23456F23456I234 $18 SLED 8 1.000 .0100 .0100 .0100 0.500 0.500 0.500 0.000 0.000 0.000 CARD B.2 $------------------------------------------------------------------------------$ $ Definition of Joints - Format (A4, 1X, A1, 2I4, 6F6.0, I4, 2F6.0 ) $ Format (14X, 9F6.0, 6I2) $ $234XAI234I234F23456F23456F23456F23456F23456F23456I234F23456F23456 $2345678901234F23456F23456F23456F23456F23456F23456F23456F23456F23456I2I2I2I2I2 $I2 $1: LT-MT P M 1 0 -.05 .00 -.04 -.01 .00 .07 0 .00 .00 CARD B.3 .00 .00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $2: MT-UT W N 2 0 -.01 .00 -.07 -.02 .00 .15 0 .00 .00 CARD B.3 .00 .00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $3: UT-N NP O 3 0 .00 .00 -.15 .00 .00 .07 0 .00 .00 CARD B.3 .00 .00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $4: N-H HP P 4 0 .00 .00 -.07 -.01 .00 .05 0 .00 .00 CARD B.3 .00 .00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $5: LT-RUL RH Q 1 0 .00 .08 .03 .00 .00 -.25 0 .00 .00 CARD B.3 .00 90.00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $6: RUL-RLL RK R 6 1 .00 .00 .17 -.01 .00 -.17 0 .00 .00 CARD B.3 .00 .00 .00 .00 55.00 .00 .00 .00 .00 0 0 0 0 0 0 $7: RF-RLL RA S 7 -8 -.01 .00 .25 -.05 .00 -.04 0 .00 .00 CARD B.3 90.00 .00 .00 90.00 .00 .00 .00-10.00 .00 0 0 0 0 0 0 $8: LT-LUL LH T 1 0 .00 -.08 .03 .00 .00 -.25 0 .00 .00 CARD B.3 .00 90.00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $9: LUL-LLL LK U 9 1 .00 .00 .17 -.01 .00 -.17 0 .00 .00 CARD B.3 .00 .00 .00 .00 55.00 .00 .00 .00 .00 0 0 0 0 0 0 $10: LLL-LF LA V 10 -8 -.01 .00 .25 -.05 .00 -.04 0 .00 .00 CARD B.3 90.00 .00 .00 90.00 .00 .00 .00-10.00 .00 0 0 0 0 0 0 $11: UT-RUA RS W 3 0 -.02 .19 -.07 .00 .00 -.14 0 .00 .00 CARD B.3 .00 90.00 .00 .00 .00 .00 .00 .00 .00 0 0 0 0 0 0 $12: RUA-RLA RE X 12 -8 .00 .00 .13 .00 .00 -.09 0 .00 .00 CARD B.3

Main Index

Chapter 6: Occupant Safety 439 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

90.00 $13: UT-LUA LS Y 3

0

$14: LUA-LLA LE Z 14 -8 $15: RLA-RHD RW 13 -8 $16: LLA-LHD LW 15 -8

.00

-.02 -.19 .00 90.00

.00 90.00 -.07 .00

.00

.00

.00 62.50

.00 0 0 0 0 0 0

.00 .00

.00 .00

-.14 .00

0 .00 .00 CARD B.3 .00 .00 .00 0 0 0 0 0 0

.00 90.00

.00 .00

.13 .00 .00 90.00

.00 .00

-.09 .00

0 .00 .00 CARD B.3 .00 62.50 .00 0 0 0 0 0 0

.00 90.00

.00 .00

.15 -.01 .00 90.00

.00 .00

-.05 .00

0 .00 .00 CARD B.3 .00 15.00 .00 0 0 0 0 0 0

.00 .00 .15 -.01 .00 -.05 0 .00 .00 CARD B.3 90.00 .00 .00 90.00 .00 .00 .00 15.00 .00 0 0 0 0 0 0 NJ1 1 0 0 0.00 0.00 0.0 0.00 0.00 0.0 CARD B.3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 $------------------------------------------------------------------------------$ $ Joints spring characteristics - Format (2( 4F6.0, F12.0) ) $ Format (2( 4F6.0, F12.0) ) $ $23456F23456F23456F23456F23456789012$23456F23456F23456F23456F23456789012 $ .00 .00 .00 1.00 .00 3.88 .00 .00 1.00 .00CARD B.4 .00 .00 .00 1.00 .00 3.88 .00 .00 1.00 .00CARD B.4 .00 .00 .00 1.00 .00 1.69 .00 .00 1.00 .00CARD B.4 .00 .00 .00 1.00 .00 1.69 .00 .00 1.00 .00CARD B.4 .00 .00 .00 1.00 .00 .85 8.47 8.47 1.00 55.00CARD B.4 .00 .26 .00 1.00 48.90 .00 .00 .00 1.00 .00CARD B.4 .11 1.13 1.13 1.00 27.00 .00 .76 .00 1.00 30.00CARD B.4 .00 .00 .00 1.00 .00 .00 .00 .00 .00 .00 .00 1.00 .00 .85 8.47 8.47 1.00 55.00CARD B.4 .00 .26 .00 1.00 48.90 .00 .00 .00 1.00 .00CARD B.4 .11 1.13 1.13 1.00 27.00 .00 .76 .00 1.00 30.00CARD B.4 .00 .00 .00 1.00 .00 .00 .00 .00 .00 .00 .00 1.00 .00 .00 11.30 11.30 1.00 125.00CARD B.4 .00 2.26 2.26 1.00 52.00 .00 .53 .00 1.00 65.30CARD B.4 .00 .00 .00 1.00 .00 .00 90.00 .00 .00 .00 .00 1.00 .00 .00 11.30 11.30 1.00 125.00CARD B.4 .00 2.26 2.26 1.00 52.00 .00 .53 .00 1.00 65.30CARD B.4 .00 .00 .00 1.00 .00 .00 90.00 .00 .00 .00 .00 1.00 .00 .00 .49 .02 1.00 55.50CARD B.4 .00 .00 .00 1.00 .00 .00 .00 .00 .00 .00 .00 1.00 .00 .00 .49 .02 1.00 55.50CARD B.4 .00 .00 .00 1.00 .00 .00 .00 .00 0.00 NJ1 $---------------------------------------------------------------------------$ $ Joints viscous characteristics - Format (5F6.0, 18X , 2F6.0) $ Format (5F6.0, 18X , 2F6.0) $ $23456F23456F23456F23456F23456X23456789012345678F23456F23456 $ .136 0. 30. 0. 0. 0. 0. CARD B.5 .136 0. 30. 0. 0. 0. 0. CARD B.5

Main Index

440 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

.040 0. 30. 0. 0. 0. 0. CARD B.5 .022 0. 30. 0. 0. 0. 0. CARD B.5 .113 0. 30. 0. 0. 0. 0. CARD B.5 .113 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .000 0. 30. 0. 0. 0. 0. CARD B.5 .113 0. 30. 0. 0. 0. 0. CARD B.5 .113 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .000 0. 30. 0. 0. 0. 0. CARD B.5 .011 0. 30. 0. 0. 0. 0. CARD B.5 .011 0. 30. 0. 0. 0. 0. CARD B.5 .011 0. 30. 0. 0. 0. 0. CARD B.5 .000 0. 30. 0. 0. 0. 0. CARD B.5 .011 0. 30. 0. 0. 0. 0. CARD B.5 .011 0. 30. 0. 0. 0. 0. CARD B.5 .011 0. 30. 0. 0. 0. 0. CARD B.5 .000 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .000 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .056 0. 30. 0. 0. 0. 0. CARD B.5 .000 0. 30. 0. 0. 0. 0. CARD B.5 0.00 NJ1 $---------------------------------------------------------------------------$ $ Convergence tests settings - Format (12F6.0) $ $23456F23456F23456F23456F23456F23456F23456F23456F23456F23456F23456F23456 $ .01 .01 .012.5E-42.5E-4 .01 .10 .10 .102.5E-32.5E-3 .01CARD B6A .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6B .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6C .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6D .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6E .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6F .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6G .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6H .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6I .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6J .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6K .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6L .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6M .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6N .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6O .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6P .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00CARD B6Q .01 .01 .01 .00 .00 .00 .10 .10 .10 .00 .00 .00 SLED $---------------------------------------------------------------------------$ $ Description of the crash vehicle deceleration - Format (20A4)

Main Index

Chapter 6: Occupant Safety 441 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ ZERO ACCELERATION FOR SLED - WE APPLY ACCELERATION ON DYMMY CARD C1 $------------------------------------------------------------------------------$ $ - Angles of decelaration pulse vector from the inertial coordinate system $ - Initial velocity of the prescribed motion segment $ - Shape of the decelartion pulse $ $ Format (8F6.0, I6, 2F6.0, I6) $ $.....ANGLES.....|-VIPS|VTIME|--------X0-------|NATAB|-AT0-|-ATD-|-MSEG$23456F23456F23456F23456F23456F23456F23456F23456I23456F23456F23456I23456 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -2 .000.01000 18CARD C2A 0 0.0 0.0 0.0 CARD C2B $ $ - The x,y, and z components (g’s) of the linear deceleration of the $ vehicle origin at time T(I) $ $ - The angular accelerations (deg/sec**2) about the local x,y and z axes $ of the vehicle at time T(I) $ $ Cards C.4 -> Format (10X, 6F10.0) $ $234567890F234567890F234567890F234567890F234567890F234567890F234567890 $XXXXXXXX|-Xlinear-|-Ylinear-|-Zlinear-|-Xangulr-|-Yangulr-|-Zangulr-| 0.000E+00 0.000E+00 1.000E-02 0.0 $ ZERO ACCELERATION FOR VEHICLE - WE APPLY ACCELERATION ON DYMMY CARD C1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -2 .000.01000 0CARD C2A 0 0.0 0.0 0.0 CARD C2B 0.000E+00 0.000E+00 1.000E-02 0.0 $--------------------------------------------------------------------------$ $ Number of: - planes describing contact panels NPL $ - belts used to restrain the crash victim NBLT $ - airbags used to restrain the crash victim NBAG $ - contact ellipsoids or hyper-ellipsoids on Cards D.5 NELP $ - constraints on Cards D.6 NQ $ - spring-dampers on Cards D.8 NSD $ - harness-belt systems on Cards F.8 NHRNSS $ - wind force and drag coefficient functions on Cards E.6 NWINDF $ - joint restoring force functions on cards E.7 NJNTF $ - force and/or torque functions on Cards D.9 NFORCE $ $ Format (10I6) $ $23456I23456I23456I23456I23456I23456I23456I23456I23456I23456 $-NPL-|-NBLT|-NBAG|-NELP|-NQ--|-NSD-|NHRNS|NWIND|NJNTF|NFORCE 5 0 0 0 0 0 0 0 7 0 CARD D.1 $------------------------------------------------------------------------------$ $ Contact planes

Main Index

442 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ $ Cards D.2 -> a) FORMAT (I4, 4X, 5A4) $ b) FORMAT (3F12.0) $ 1 BACK SEAT CARD D2A $23456789012$23456789012$23456789012 -3.09131 -.27140 -.57808 -3.36387 -.27140 -1.1516 -3.09131 .27140 -.57808 2 BOTTOM SEAT CARD D2A $23456789012$23456789012$23456789012 -3.09131 -.27140 -.57808 -3.09131 .27140 -.57808 -2.66830 -.27140 -.57808 3 FLOOR $23456789012$23456789012$23456789012 -2.45313 -.27140 -.4251959 -2.45313 .27140 -.4251959 -2.21513 -.27140 -.4251959 4 TOE BOARD $23456789012$23456789012$23456789012 -2.21513 -.27140 -.4251959 -2.21513 .27140 -.4251959 -2.02641 -.27140 -.7561072 5 FIRE WALL $23456789012$23456789012$23456789012 -2.02641 -.27140 -.7561072 -2.02641 .27140 -.7561072 -2.02641 -.27140 -1.137110 $-----------------------------------------------------------------------------$ $ Contact (hyper)-ellipsoids $ $ Cards D.5 $ $-----------------------------------------------------------------------------$ $ Symmetry options => ALLWAYS REQUIRED $ => supply blank card for normal 3D motion $ $ Card D.7 - Format (18I4) $ CARD D7 $-----------------------------------------------------------------------------$ $ CARDs E.1 - E.5 ; Functions referred to by: $ $ - Card F.1.b : Segment-Segment planes contact $ - Card F.2.b : Belt-Ellipsoid contact $ - Card F.3.b : Segment-Segment ellipsoid contact $ - Card F.4.b : Globalgraphic joint $ - Card F.8.c : Harness-Belt system $ - Card F.8.d1 : Harness-Belt system $ - Card D.8 : Spring-dampers

Main Index

Chapter 6: Occupant Safety 443 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ - Card D.9 : Force or Torque applied to a segment $ $ $ CARD E.6 ; Functions referred to by: $ - Card F.7.b : Windforce $ $ 1 HEAD SURFACE CARD .00 .02 -.02 .00 .00 CARD .11432E+02 -.30820E+05 .30998E+08 -.24304E+10 .64440E+11 .00000E+00CARD 3 CARD .02 9.09 .02 90.93 .02 909.32CARD 2 BACK OF SHOULDER CARD .00 .04 -.04 .00 .00 CARD .44484E-01 -.46568E+04 .26264E+07 -.18411E+09 .48790E+10 .00000E+00CARD 3 CARD .04 1.23 .04 12.27 .04 122.68CARD 3 CHEST CARD .00 .03 -.04 .00 .00 CARD .24466E+01 .95623E+03 .50671E+06 -.40746E+07 .00000E+00 .00000E+00CARD 3 CARD .03 4.03 .03 40.28 .04 402.84CARD 4 ANTERIOR PELVIS CARD .00 .05 -.05 .00 .00 CARD -.62278E+00 .27671E+03 .16817E+06 -.79618E+07 .14706E+09 .00000E+00CARD 3 CARD .05 2.28 .05 22.78 .05 227.84CARD 5 POSTERIOR PELVIS CARD .00 .05 -.05 .00 .00 CARD -.33808E+01 .61472E+04 .27608E+06 -.33905E+07 .00000E+00 .00000E+00CARD 3 CARD .05 3.25 .05 32.46 .05 324.61CARD 6 UPPER ARM CARD .00 .02 -.03 .00 .00 CARD .68505E+01 -.53643E+04 .43238E+07 -.32512E+09 .98026E+10 .00000E+00CARD 3 CARD .02 5.16 .02 51.59 .03 515.87CARD 7 FOREARM CARD .00 .03 -.04 .00 .00 CARD -.10676E+02 .18804E+05 -.21606E+07 .13698E+09 -.20987E+10 .00000E+00CARD 3 CARD .03 3.97 .03 39.67 .04 396.75CARD 8 HAND CARD .00 .01 -.01 .00 .00 CARD -.17794E+00 -.54361E+04 .16441E+08 -.27672E+10 .20490E+12 .00000E+00CARD 3 CARD .01 3.54 .01 35.41 .01 354.08CARD 9 UPPER LEG CARD .00 .03 -.04 .00 .00 CARD -.12456E+01 .24851E+04 .56498E+06 -.33286E+08 .10552E+10 -.12240E+11CARD 3 CARD .03 1.56 .03 15.65 .04 156.46CARD 10 KNEE CARD .00 .01 -.01 .00 .00 CARD

Main Index

E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2 E.3 E.4 E.4 E.1 E.2

444 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

-.17794E+00 .38880E+03 .59803E+07 -.10902E+09 .00000E+00 .00000E+00CARD E.3 3 CARD E.4 .01 4.77 .01 47.70 .01 477.01CARD E.4 11 FRONT OF LOWER LEG CARD E.1 .00 .05 -.05 .00 .00 CARD E.2 -.32918E+01 .46918E+04 .27028E+05 .78858E+07 .00000E+00 .00000E+00CARD E.3 3 CARD E.4 .05 5.68 .05 56.79 .05 567.94CARD E.4 12 BACK OF LOWER LEG CARD E.1 .00 .04 -.04 .00 .00 CARD E.2 -.35587E+00 .51524E+04 -.14342E+05 .00000E+00 .00000E+00 .00000E+00CARD E.3 3 CARD E.4 .04 1.10 .04 10.97 .04 109.73CARD E.4 13 FOOT CARD E.1 .00 .02 -.02 .00 .00 CARD E.2 -.30249E+01 .14492E+05 -.48553E+07 .91402E+09 -.59535E+11 .13415E+13CARD E.3 3 CARD E.4 .02 3.08 .02 30.78 .02 307.85CARD E.4 14 CONTACT STIFFNESS CARD E.1 0.0 -.1016 0.0 0.0 1.0 CARD E2 2 CARD E4A 0.0 0.0 2.54E-02 17792.9 15 FRICTION FUNC 1 CARD E1 0.0 0.0 0.62 0.0 1.0 CARD E2 16 FRICTION FUNC 2 CARD E1 0.0 0.0 1.00 0.0 1.0 CARD E2 9999 CARD E.1 $----------------------------------------------------------------------------$ $ Joint restoring force functions : Cards E.7A - E.7D $ $ Referred to by Card F.5 : specifies what function is used by what joint $ 41 RIGHT SHOULDER JOINT CARD E.7 -4 8 151.000 .000 75.000 .000 .000 .000 66.000 .000 68.000 .000 89.000 .000 111.000 .000 153.000 .000 42 LEFT SHOULDER JOINT -4 8 151.000 153.000 111.000 89.000 68.000 66.000 .000

Main Index

CARD E.7 9931.734 1547.949 -35.253 879.054 2508.356 2485.758 279.083 1468.857

.000 .000 1333.270 340096.906 42483.871 .000 1988.606 .000 CARD E.7 CARD E.7

.000 .000 .000 .000 .000 .000 .000

9931.734 1468.857 279.083 2485.758 2508.356 879.054 -35.253

.000 .000 1988.606 .000 42483.871 340096.906 1333.270

Chapter 6: Occupant Safety 445 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

75.000 .000 SEATED RIGHT HIP

43

4 .000 .791 53.331 107.566 3468.762 6937.523 222588.406 .000 1.921 120.898 244.056 7875.332 15705.470 503931.000 .000 .904 20.564 40.450 1276.776 2542.253 81465.063 .000 .791 5.988 10.508 290.382 579.634 18530.199 44 SEATED LEFT HIP

1547.949

.000 CARD E.7

19

2.486 215.809 13897.650 5.762 490.372 31523.930

13.446 981.875 62934.879

28.812 1966.009 125417.797

59.545CARD E.7 3943.316CARD E.7 251965.500CARD E.7

1.921 80.222 5095.805

3.164 159.314 10180.310

5.649 318.629 20338.020

10.621CARD E.7 637.258CARD E.7 40676.039CARD E.7

1.469 19.547 1163.787

2.034 37.625 2316.275

2.599 73.782 4632.551

3.729CARD E.7 145.756CARD E.7 9253.801CARD E.7 CARD E.7

4 .000 .791 2.486 53.331 107.566 215.809 3468.762 6937.523 13897.650 222588.406 .000 .791 1.469 5.988 10.508 19.547 290.382 579.634 1163.787 18530.199 .000 .904 1.921 20.564 40.450 80.222 1276.776 2542.253 5095.805 81465.063 .000 1.921 5.762 120.898 244.056 490.372 7875.332 15705.470 31523.930 503931.000 45 SEATED LUMBAR SPINE

CARD 26.213CARD 1740.031CARD 111068.203CARD

E.7 E.7 E.7 E.7

5.875 432.748 27795.301

12.655 866.626 55590.590

2.034 37.625 2316.275

2.599 73.782 4632.551

3.729CARD E.7 145.756CARD E.7 9253.801CARD E.7

3.164 159.314 10180.310

5.649 318.629 20338.020

10.621CARD E.7 637.258CARD E.7 40676.039CARD E.7

13.446 981.875 62934.879

28.812 1966.009 125417.797

59.545CARD E.7 3943.316CARD E.7 251965.500CARD E.7 CARD E.7

19

Main Index

E.7 E.7 E.7 E.7

12.655 866.626 55590.590

19

4 .000 3118.497 11954.240 535861.688 .000 4609.953 17671.480 792143.313 .000 2033.802 7796.242 349475.000

CARD 26.213CARD 1740.031CARD 111068.203CARD

5.875 432.748 27795.301

519.750 3638.246 20270.230

1039.499 4157.996 36902.211

1559.248 4677.746 70166.188

2078.998 5717.242 136694.094

CARD 2598.747CARD 7796.242CARD 269749.906CARD

E.7 E.7 E.7 E.7

768.325 5378.277 29964.680

1536.651 6146.602 54551.090

2304.976 6914.926 103723.898

3073.301 8451.578 202069.594

3841.626CARD E.7 11524.880CARD E.7 398760.813CARD E.7

338.967 2372.769 13219.710

677.934 2711.736 24066.660

1016.901 3050.703 45760.551

1355.868 3728.637 89148.313

1694.835CARD E.7 5084.504CARD E.7 175923.906CARD E.7

446 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

.000 768.325 4609.953 5378.277 17671.480 29964.680 792143.313 46 NECK PIVOT 4 .000 79.092 474.554 553.646 949.108 1107.292 10914.740 .000 67.793 406.760 474.554 813.521 949.108 9355.488 .000 33.897 203.380 237.277 406.760 474.554 4677.746 .000 67.793 406.760 474.554 813.521 949.108 9355.488 47 HEAD PIVOT

1536.651 6146.602 54551.090

2304.976 6914.926 103723.898

3073.301 8451.578 202069.594

CARD E.7

19

4 .000 427.098 901.652 10867.280 .000 406.760 813.521 9355.488 .000 192.081 395.461 4666.445 .000 406.760 813.521 9355.488

3841.626CARD E.7 11524.880CARD E.7 398760.813CARD E.7

158.185 632.739 1423.662

237.277 711.831 2056.400

316.369 790.923 3321.877

CARD 395.461CARD 870.015CARD 5852.832CARD

E.7 E.7 E.7 E.7

135.587 542.347 1220.281

203.380 610.141 1762.628

271.174 677.934 2847.323

338.967CARD E.7 745.728CARD E.7 5016.711CARD E.7

67.793 271.174 610.141

101.690 305.070 881.314

135.587 338.967 1423.662

169.484CARD E.7 372.864CARD E.7 2508.356CARD E.7

135.587 542.347 1220.281

203.380 610.141 1762.628

271.174 677.934 2847.323

338.967CARD E.7 745.728CARD E.7 5016.711CARD E.7 CARD E.7

19

55.365 506.191 1059.837

110.729 585.283 1376.206

189.822 664.375 2008.945

268.914 743.468 3274.421

CARD 348.006CARD 822.560CARD 5805.375CARD

E.7 E.7 E.7 E.7

67.793 474.554 949.108

135.587 542.347 1220.281

203.380 610.141 1762.628

271.174 677.934 2847.323

338.967CARD E.7 745.728CARD E.7 5016.711CARD E.7

28.247 225.978 463.255

56.495 259.875 598.842

90.391 293.771 870.015

124.288 327.668 1412.363

158.185CARD E.7 361.565CARD E.7 2497.057CARD E.7

67.793 474.554 949.108

135.587 542.347 1220.281

203.380 610.141 1762.628

271.174 677.934 2847.323

338.967CARD E.7 745.728CARD E.7 5016.711CARD E.7

$ $----------------------------------------------------------------------------$$ Allowed Contacts $$ Plane-segment contact (if NPL on Card D.1 > 0) $ $234$234$234$234$234$234$234$234$234$234$234$234$234$234$234$234$234$234 5 1 2 2 2 CARD F1A $-NJ-NS1-NS2-NS3-NF1-NF2-NF3-NF4-NF5-NX $ $ NJ = plane $ NS1 = segment to which plane is connected $ NS2 = segment that checks for contact

Main Index

Chapter 6: Occupant Safety 447 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ NS3 = $ NF1 = function number of force deflection curve $ NF5 = function number of friction curve $ NX = ege option for contact $ $234$234$234$234$234$234$234$234$234$234$234$234$234$234$234$234$234$234 $ Back-seat with Lower Torso $ Middle Torso $ Upper Torso $ neck $ head 1 18 1 1 14 0 0 0 15 1 1 18 2 2 14 0 0 0 15 1 1 18 3 3 14 0 0 0 15 1 1 18 4 4 14 0 0 0 15 1 1 18 5 5 14 0 0 0 15 1 $ Bottom-seat with Lower Torso 2 18 1 1 14 0 0 0 15 1 $ Floor with Right Foot $ Left Foot 3 18 8 8 14 0 0 0 16 1 3 18 11 11 14 0 0 0 16 1 $ Toe-board with Right Foot $ Left Foot 4 18 8 8 14 0 0 0 16 1 4 18 11 11 14 0 0 0 16 1 $ Fire-wall with Right Foot $ Left Foot 5 18 8 8 14 0 0 0 16 1 5 18 11 11 14 0 0 0 16 1 $----------------------------------------------------------------------------$ $ Card F.3.a: segment-segment contact - Format (18I4) $ ******************** $ ***** is allways required ****** $ ******************** $ ==> blank card will specify that no segment-segment forces are to $ be computed by the program $ $ 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11| 12| 13| 14| 15| 16| 17| 18| $234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234 CARD F3A $------------------------------------------------------------------------------$ $ Card F.4.a: Globalgraphic joints - Format (18I4) $ *************************************************** $ ***** is allways required if NJNT is nonzero on card B.1 ****** $ *************************************************** $ ==> IGLOB=1 for globalgraphic joint, $ blank or zero if otherwise $ $ 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11| 12| 13| 14| 15| 16| 17| 18| $234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0CARD F.4

Main Index

448 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$------------------------------------------------------------------------------$ $ Card F.5.a: What joint is using what joint-restoring function - Format (18I4) $ **************************************************************** $ ***** is allways required if NJNT>0 (card B.1) and NJNTF>0 (Card D.1) ****** $ **************************************************************** $ $ 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11| 12| 13| 14| 15| 16| 17| 18| $234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234I234 45 45 46 47 43 0 0 44 0 0 41 0 42 0 0 0 0 0CARD F.5 $------------------------------------------------------------------------------$ $ Card F.8: Harness-belt systems. $ $------------------------------------------------------------------------------$ $ Card G.1.a: - Plot info - Format (3F10.0, 6I4) $ - call to EQUILB or not $ - segment linear and angular velocities supplied or not $ - MSC-update: NSJF=0 -> use initial joint forces $ 1 -> subtract initial joint forces $ $ ******************* $ ***** is allways required ****** $ ******************* $ $234567890F234567890F234567890I234I234I234I234I234I234 1 CARD G1 $------------------------------------------------------------------------------$ $ Card G.2: Initial position/velocity of reference segments - Format (6F10.0) $ ******************* $ ***** is allways required ****** $ ******************* $ $234567890F234567890F234567890F234567890F234567890F234567890 -2.98508 0.00 -.710202 CARD G2 0.00 0.0 0.00 SLED $------------------------------------------------------------------------------$ $ Card G.3: Initial position/velocity of other segments - Format (6F10.0, 4I3) $ ******************* $ ***** is allways required ****** $ ******************* $ $234567890F234567890F234567890F234567890F234567890F234567890I234I234I234 $ LT 0.0000 21.9970 0.0000 3 2 1 CARD 1. $ MT 0.0000 22.0130 0.0000 3 2 1 CARD 2. $ UT 0.0000 22.0320 0.0000 3 2 1 CARD 3. $ N 0.0000 -4.4170 0.0000 3 2 1 CARD 4.

Main Index

Chapter 6: Occupant Safety 449 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ H 0.0000 -4.4170 0.0000 3 2 1 CARD 8. $ RUL 0.0000 106.1720 0.0000 3 2 1 CARD 9. $ RLL 0.0000 49.6150 0.0000 3 2 1 CARD 10 $ RF 0.0000 57.7190 0.0000 3 2 1 CARD 11 $ LUL 0.0000 106.1720 0.0000 3 2 1 CARD 12 $ LLL 0.0000 49.6150 0.0000 3 2 1 CARD 13. $ LF 0.0000 57.7190 0.0000 3 2 1 CARD 14. $ RUA 0.0000 53.4650 0.0000 3 2 1 CARD 15. $ RLA 00.000 90.1750 0.0000 3 2 1 CARD 16. $ LUA 0.0000 53.4650 0.0000 3 2 1 CARD 17. $ LLA 00.000 90.1750 0.0000 3 2 1 CARD 18 $ RHD 0.0 90.614 0.0 0.0 0.0 0.0 CARD G3P $ LHD 0.0 90.614 0.0 0.0 0.0 0.0 CARD G3Q 0.0 0.0 0.0 0.0 0.0 0.0 SLED $------------------------------------------------------------------------------$ $ Card H.1: Linear acceleration time history output $ ******************* $ ***** is allways required ****** $ ******************* $ $ Card H.1.a: Format (I6, 2I3, 3F12.6) $ Card H.1.b: Format (I9, I3, 3F12.6 ) $ $23456I23I23F23456789012F23456789012F23456789012 $23456789I23F23456789012F23456789012F23456789012 $ Lower Torso 3 0 -1 0.000 0.000 0.000 CARD H1 $ Upper Torso 0 -3 0.000 0.000 0.000 CARD H1 $ Head 0 -5 0.000 0.000 0.000 CARD H1 $------------------------------------------------------------------------------$ $ Card H.2: Relative velocity time history output $ ******************* $ ***** is allways required ****** $ ******************* $ $ Card H.2.a: Format (I6, 2I3, 3F12.6) $ Card H.2.b: Format (I9, 2I3, 3F12.6)

Main Index

450 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ $23456I23I23F23456789012F23456789012F23456789012 $ Lower Torso 3 0 -1 0.000 0.000 0.000 CARD H2 $ Upper Torso 0 -3 0.000 0.000 0.000 CARD H2 $ Head 0 -5 0.000 0.000 0.000 CARD H2 $------------------------------------------------------------------------------$ $ Card H.3: Relative displacement time history output $ ******************* $ ***** is allways required ****** $ ******************* $ $ Card H.3.a: Format (I6, 2I3, 3F12.6) $ Card H.3.b: Format (I9, 2I3, 3F12.6) $ $23456I23I23F23456789012F23456789012F23456789012 $ Lower Torso 3 0 -1 0.000 0.000 0.000 CARD H2 $ Upper Torso 0 -3 0.000 0.000 0.000 CARD H2 $ Head 0 -5 0.000 0.000 0.000 CARD H2 3 16 1 16 3 16 5 CARD H4 CARD H5 CARD H6 2 3 4 CARD H7 CARD H8 2 4 3 4 4 CARD H9 0 CARD H10A 3 2 CARD H11

Appendix C: File Create_fem_dummy.dat $ SI Units: kg - meter - seconds $ -----------------------------$ $ file: create_fem_dummy.dat $ ========================== $ $ ----------------------------------------------------------------------$ This example shows how to create the necessary bulk data entries for $ easy positioning. $ ----------------------------------------------------------------------$ CHECK=YES PARAM,INISTEP,1.E-3 $ BEGIN BULK $ $ Define the segment contact ellipsoids

Main Index

Chapter 6: Occupant Safety 451 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ -> card B.2 $ RELEX,LT,ATB RELEX,MT,ATB RELEX,UT,ATB RELEX,N,ATB RELEX,H,ATB RELEX,RUL,ATB RELEX,RLL,ATB RELEX,RF,ATB RELEX,LUL,ATB RELEX,LLL,ATB RELEX,LF,ATB RELEX,RUA,ATB RELEX,RLA,ATB RELEX,LUA,ATB RELEX,LLA,ATB RELEX,RHD,ATB RELEX,LHD,ATB $ $ Tell DYTRAN to create the ATBSEGs & ATBJNTs with length of the $ bars/beams equal to: segment coordinate systems -> LENGTH=.025 m $ joint coordinate systems -> LENGTH=.05 m $ PARAM,ATBSEGCREATE,YES,HYBRID-III,.025,.05 $ $ atbseg 1 : name = LT $ will be covered with 72 (NUMELM) shell elements with: $ - Gridpoint ids start at 11000 (GSTART) $ - element ids start at 11000 (ESTART) $ - material id is 20001 (MID ) $ - property id is 20201 (PIDCOV) $ $ if PARAM,ATBSEGCREATE has been specified, beam elements$ representing the local coordinate system will be generated with: $ - Grid id of grid located at the origin = 20001 (G0 ) $ - Grid id of grid located on local x-axis = 20002 (G1 ) $ - Grid id of grid located on local y-axis = 20003 (G2 ) $ - Grid id of grid located on local z-axis = 20004 (G3 ) $ - Cbar id of element representing local x-axis = 20001 (EID1 ) $ - Cbar id of element representing local y-axis = 20002 (EID2 ) $ - Cbar id of element representing local z-axis = 20003 (EID3 ) $ - Property id of cbar elements = 20001 (PIDCG) $ - Material id of cbar elements = 20001 (MID ) $ $ if PARAM,ATBSEGCREATE has not been specified the position and $ orientation of the ATB segments as spcified on the G.2 and G.3 $ entries in the ATB input file will be overruled by the definitions $ given here. The local coordinate system is defined by: $ - Grid id of grid located at the origin = 20001 (G0 ) $ - Grid id of grid located on local x-axis = 20002 (G1 ) $ - Grid id of grid located on local y-axis = 20003 (G2 ) $ - Grid id of grid located on local z-axis = 20004 (G3 ) $

Main Index

452 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

$ ATBSEG ID NAME COVER NUMELM GSTART ESTART $ + G0 G1 G2 G3 EID1 EID2 $ ATBSEG ,1 ,LT ,YES ,72 ,11000 ,11000 + ,20001 ,20002 ,20003 ,20004 ,20001 ,20002 $ $ similar explanation for all the other atbsegs $ ATBSEG,2,MT,YES,72 ,11250,11250,20002,20202,+ +,20005,20006,20007,20008,20004,20005,20006,20002 $ ATBSEG,3,UT,YES,72 ,11500,11500,20003,20203,+ +,20009,20010,20011,20012,20007,20008,20009,20003 $ ATBSEG,4,N ,YES,72 ,11750,11750,20004,20204,+ +,20013,20014,20015,20016,20010,20011,20012,20004 $ ATBSEG,5,H ,YES,72 ,12000,12000,20005,20205,+ +,20017,20018,20019,20020,20013,20014,20015,20005 $ ATBSEG,6,RUL,YES,72 ,12250,12250,20006,20206,+ +,20021,20022,20023,20024,20016,20017,20018,20006 $ ATBSEG,7,RLL,YES,72 ,12500,12500,20007,20207,+ +,20025,20026,20027,20028,20019,20020,20021,20007 $ ATBSEG,8,RF,YES,72 ,12750,12750,20008,20208,+ +,20029,20030,20031,20032,20022,20023,20024,20008 $ ATBSEG,9,LUL,YES,72 ,13000,13000,20009,20209,+ +,20033,20034,20035,20036,20025,20026,20027,20009 $ ATBSEG,10,LLL,YES,72 ,13250,13250,20010,20210,+ +,20037,20038,20039,20040,20028,20029,20030,20010 $ ATBSEG,11,LF,YES,72 ,13500,13500,20011,20211,+ +,20041,20042,20043,20044,20031,20032,20033,20011 $ ATBSEG,12,RUA,YES,72 ,13750,13750,20012,20212,+ +,20045,20046,20047,20048,20034,20035,20036,20012 $ ATBSEG,13,RLA,YES,72 ,14000,14000,20013,20213,+ +,20049,20050,20051,20052,20037,20038,20039,20013 $ ATBSEG,14,LUA,YES,72 ,14250,14250,20014,20214,+ +,20053,20054,20055,20056,20040,20041,20042,20014 $ ATBSEG,15,LLA,YES,72 ,14500,14500,20015,20215,+ +,20057,20058,20059,20060,20043,20044,20045,20015 $ ATBSEG,16,RHD,YES,72 ,14750,14750,20016,20216,+ +,20061,20062,20063,20064,20046,20047,20048,20016 $ ATBSEG,17,LHD,YES,72 ,15000,15000,20017,20217,+

Main Index

MID EID3 ,20001 ,20003

PIDCOV PIDCG ,20201 ,20001

+

,+

Chapter 6: Occupant Safety 453 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

+,20065,20066,20067,20068,20049,20050,20051,20017 $ $ atbjnt 1 with name P connects atb segments 1 (LT) with 2 (MT)(see card B.3 in ATB input deck)if PARAM,ATBSEGCREATE has been specified, beam elements$ representing the joint coordinate systems will be generated with: $ -> coordinate system connected to atb segment 1 (LT) $ - Grid id of grid located at the origin = 20201 (G0 ) $ - Grid id of grid located on local x-axis = 20202 (G1 ) $ - Grid id of grid located on local y-axis = 20203 (G2 ) $ - Grid id of grid located on local z-axis = 20204 (G3 ) $ - Cbar id of element representing local x-axis = 20201 (EID1 ) $ - Cbar id of element representing local y-axis = 20202 (EID2 ) $ - Cbar id of element representing local z-axis = 20203 (EID3 ) $ - Property id of cbar elements = 20001 (PIDCG-LT) $ - Material id of cbar elements = 20001 (MID -LT) $ where: PIDCG-LT is the PIDCG as specified on ATBSEG,,LT $ MID -LT is the MID as specified on ATBSEG,,LT $ $ -> coordinate system connected to atb segment 2 (MT) $ - Grid id of grid located at the origin = 20205 (G4 ) $ - Grid id of grid located on local x-axis = 20206 (G5 ) $ - Grid id of grid located on local y-axis = 20207 (G6 ) $ - Grid id of grid located on local z-axis = 20208 (G7 ) $ - Cbar id of element representing local x-axis = 20204 (EID4 ) $ - Cbar id of element representing local y-axis = 20205 (EID5 ) $ - Cbar id of element representing local z-axis = 20206 (EID6 ) $ - Property id of cbar elements = 20002 (PIDCG-MT) $ - Material id of cbar elements = 20002 (MID -MT) $ where: PIDCG-MT is the PIDCG as specified on ATBSEG,,MT $ MID -MT is the MID as specified on ATBSEG,,MT $ $ $ ATBJNT ID NAME + $ + G0 G1 G2 G3 EID1 EID2 EID3 + $ + G4 G5 G6 G7 EID4 EID5 EID6 $ ATBJNT ,1 ,P ,,,,,,,+ + ,20201 ,20202 ,20203 ,20204 ,20201 ,20202 ,20203 ,,+ + ,20205 ,20206 ,20207 ,20208 ,20204 ,20205 ,20206 $ ATBJNT,2,W,,,,,,,+ +,20209,20210,20211,20212,20207,20208,20209,,+ +,20213,20214,20215,20216,20210,20211,20212 $ ATBJNT,3,NP,,,,,,,+ +,20217,20218,20219,20220,20213,20214,20215,,+ +,20221,20222,20223,20224,20216,20217,20218 $ ATBJNT,4,HP,,,,,,,+ +,20225,20226,20227,20228,20219,20220,20221,,+ +,20229,20230,20231,20232,20222,20223,20224 $ ATBJNT,5,RH,,,,,,,+ +,20233,20234,20235,20236,20225,20226,20227,,+

Main Index

454 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

+,20237,20238,20239,20240,20228,20229,20230 $ ATBJNT,6,RK,,,,,,,+ +,20241,20242,20243,20244,20231,20232,20233,,+ +,20245,20246,20247,20248,20234,20235,20236 $ ATBJNT,7,RA,,,,,,,+ +,20249,20250,20251,20252,20237,20238,20239,,+ +,20253,20254,20255,20256,20240,20241,20242 $ ATBJNT,8,LH,,,,,,,+ +,20257,20258,20259,20260,20243,20244,20245,,+ +,20261,20262,20263,20264,20246,20247,20248 $ ATBJNT,9,LK,,,,,,,+ +,20265,20266,20267,20268,20249,20250,20251,,+ +,20269,20270,20271,20272,20252,20253,20254 $ ATBJNT,10,LA,,,,,,,+ +,20273,20274,20275,20276,20255,20256,20257,,+ +,20277,20278,20279,20280,20258,20259,20260 $ ATBJNT,11,RS,,,,,,,+ +,20281,20282,20283,20284,20261,20262,20263,,+ +,20285,20286,20287,20288,20264,20265,20266 $ ATBJNT,12,RE,,,,,,,+ +,20289,20290,20291,20292,20267,20268,20269,,+ +,20293,20294,20295,20296,20270,20271,20272 $ ATBJNT,13,LS,,,,,,,+ +,20297,20298,20299,20300,20273,20274,20275,,+ +,20301,20302,20303,20304,20276,20277,20278 $ ATBJNT,14,LE,,,,,,,+ +,20305,20306,20307,20308,20279,20280,20281,,+ +,20309,20310,20311,20312,20282,20283,20284 $ ATBJNT,15,RW,,,,,,,+ +,20313,20314,20315,20316,20285,20286,20287,,+ +,20317,20318,20319,20320,20288,20289,20290 $ ATBJNT,16,LW,,,,,,,+ +,20321,20322,20323,20324,20291,20292,20293,,+ +,20325,20326,20327,20328,20294,20295,20296 $ $ ---------------------------------------------------------------$ ENDDATA

Main Index

Chapter 6: Occupant Safety 455 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Appendix D: Properties and Materials of Hybrid III Dummy Bar elements representing ATB segment and joint coordinate systems. Digitized segments Name

ATB-name

pid

mid

lower torso

LT(1)

1

1

middle torso

MT(2)

2

2

upper torso

UT(3)

3

3

neck

N(4)

4

4

head

H(5)

5

5

right upper leg

RUL(6)

6

6

right lower leg

RLL(7)

7

7

right foot

RF(8)

8

8

left upper leg

LUL(9)

9

9

left lower leg

LLL(10)

10

10

left foot

LF(11)

11

11

right upper arm

RUA(12)

12

12

right lower arm

RLA(13)

13

13

left upper arm

LUA(14)

14

14

left lower arm

LLA(15)

15

15

right hand

RHD(16)

16

16

left hand

LHD(17)

17

17

Covered ATB ellipsoids Name lower torso

pid 101

middle torso

Main Index

mid 1 2

upper torso

103

3

neck

104

4

head

105

5

right upper leg

106

6

right lower leg

107

7

right foot

108

8

left upper leg

109

9

Modeled as dummy shells

Leg itself is modeled as dummy shells

Leg itself is modeled as dummy shells

456 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Name

mid

left lower leg

110

10

left foot

111

11

right upper arm

112

12

right lower arm

113

13

left upper arm

114

14

left lower arm

115

15

right hand

116

16

left hand

117

17

Name

Main Index

pid

ATB-name

pid

mid

lower torso

LT(1)

201

1

middle torso

MT(2)

202

2

upper torso

UT(3)

203

3

neck

N(4)

204

4

head

H(5)

205

5

right upper leg

RUL(6)

206

6

right lower leg

RLL(7)

207

7

right foot

RF(8)

208

8

left upper leg

LUL(9)

209

9

left lower leg

LLL(10)

210

10

left foot

LF(11)

211

11

right upper arm

RUA(12)

212

12

right lower arm

RLA(13)

213

13

left upper arm

LUA(14)

214

14

left lower arm

LLA(15)

215

15

right hand

RHD(16)

216

16

left hand

LHD(17)

217

17

Chapter 6: Occupant Safety 457 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Digitized dummy shells (mid = 0) Name

pid

CONNECTING

upper neck

301

head

neck

lower neck

302

neck

upper torso

middle torso

303

upper torso

lower torso

right upper leg

304

lower torso

front of right upper leg

right knee

305

front of right upper leg

right lower leg

right ankle

306

right lower leg

right foot

left upper leg

307

lower torso

front of left upper leg

left knee

308

front of left upper leg

left lower leg

left ankle

309

left lower leg

left foot

right shoulder

310

upper torso

right upper arm

right elbow

311

right upper arm

right lower arm

right wrist

312

right lower arm

right hand

shoulder

313

upper torso

left upper arm

left elbow

314

left upper arm

left lower arm

left wrist

315

left lower arm

left hand

Bar elements inside torso These bars are used to generate an overlap in the contact surface where the legs and arms connect to the torso. In these regions there are sharp corners, and the belt might slip through. By using bar elements inside the torso, this problem is solved. Name

Main Index

pid

mid

right shoulder

401

3

mid of upper torso

right leg

402

1

mid of lower torso

left leg

403

1

mid of lower torso

left shoulder

404

3

mid of upper torso

458 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Appendix E: Groups created by Session File for Positioning within MD Patran Complete Dummy Group Name

Contents

H3ALL

All FEM entities defining the dummy

H3F–ATB

Covered ATB ellipsoids (shell elements)

H3F–DIGIT

Digitized HIII surface (shell element)

H3F–JOINTS

CBARS+grids representing the joint coordinate systems

H3F–SEGMENT–COORD

CBARS+grids representing the segment coordinate systems

H3F–GRIDS –ATB

Covered ATB ellipsoids (grids only)

H3F–GRIDS–DIGIT

Digitized HIII surface (grids only)

H3F–OVERLAP–BARS

Bar elements inside torso

Ankle Group Name

Contents

H3ROT–ANKLE–BLW–LEFT

Elements of left foot

H3ROT–ANKLE–BLW–RIGHT

Elements of right foot

H3ROT–ANKLE–BLW–SYMM

Elements of left+right feet

H3ROT–ANKLE–ABV–LEFT

Elements of complete dummy except left foot and dummy shells connecting lower left leg to left foot

H3ROT–ANKLE–ABV–RIGHT

Elements of complete dummy except right foot and dummy shells connecting lower right leg to right foot

H3ROT–ANKLE–ABV–SYMM

Elements of complete dummy except left+right foot and dummy shells connecting lower left leg to left tool and dummy shells connecting lower right leg to right foot

Knee Group Name H3ROT–KNEE–BLW–LEFT H3ROT–KNEE–BLW–RIGHT H3ROT–KNEE–BLW–SYMM H3ROT–KNEE–ABV–LEFT H3ROT–KNEE–ABV–RIGHT H3ROT–KNEE–ABV–SYMM

Main Index

Contents All items in this group use the same logic as is used for Ankle

Chapter 6: Occupant Safety 459 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Hip Group Name H3ROT–HIP–BLW–LEFT H3ROT–HIP–BLW–RIGHT H3ROT–HIP–BLW–SYMM H3ROT–HIP–ABV–LEFT H3ROT–HIP–ABV–RIGHT H3ROT–HIP–ABV–SYMM Lumbar Spine Group Name H3ROT–LSPINE–BLW H3ROT–LSPINE–ABV Thoracic Spine Group Name H3ROT–TSPINE–BLW H3ROT–TSPINE–ABV Upper Neck Group Name H3ROT–UPNECK–ABV H3ROT–UPNECK–BLW

Main Index

460 Dytran Example Problem Manual Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Lower Neck Group Name H3ROT–LWNECK–ABV H3ROT–LWNECK–BLW Wrist Group Name H3ROT–WRIST–BLW–LEFT H3ROT–WRIST–BLW–RIGHT H3ROT–WRIST–BLW–SYMM H3ROT–WRIST–ABV–LEFT H3ROT–WRIST–ABV–RIGHT H3ROT–WRIST–ABV–SYMM Elbow Group Name H3ROT–ELBOW–BLW–LEFT H3ROT–ELBOW–BLW–RIGHT H3ROT–ELBOW–BLW–SYMM H3ROT–ELBOW–ABV–LEFT H3ROT–ELBOW–ABV–RIGHT H3ROT–ELBOW–ABV–SYMM Shoulder Group Name H3ROT–SHOULD–BLW–LEFT H3ROT–SHOULD–BLW–RIGHT H3ROT–SHOULD–BLW–SYMM H3ROT–SHOULD–ABV–LEFT H3ROT–SHOULD–ABV–RIGHT H3ROT–SHOULD–ABV–SYMM References 1. Prasad, P, “CoR:-.\tmparative Evaluation of the Dynamic Responses of the Hybrid II and Hybrid III Dummies” SAE paper 902318 (Also published in 2).

Main Index

Chapter 6: Occupant Safety 461 Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

2. Backaitis, Stanley H., and Mertz, Harold J. eds., “Hybrid III: The First Human-Like Crash Test Dummy”. SAE PT-44 3. Obergefell, Louise A., Gardner, Thomas R., Kaleps, Ints, and Fleck, John T., Articulated Total Body Model Enhancements, Volume 2: User’s Guide. January 1988.

Main Index

462 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

Side Curtain Air Bag (Courtesy of Autoliv) Problem Description A folded side curtain air bag consists of an inflator, a horizontal distribution tube, and five vertical compartments. The five compartments are connected to the distribution tube with five holes. The purpose of this model is to predict the sequence of filling of the five vertical compartments.

Main Index

Figure 6-14

Side Curtain Air Bag (folded pattern)

Figure 6-15

Side Curtain Air Bag (deployed pattern)

Chapter 6: Occupant Safety 463 Side Curtain Air Bag (Courtesy of Autoliv)

The properties of the bag membrane are as follows: Young's Modulus

2.6e8

Poison’s Ratio

.3

Damp Factor

.1

Membrane Thickness

.0007

N/m2

m

Air properties: Density

1.527

Gamma

1.517

Gas Constant R

226.45

kg/m3 m2/sec2*K

Initial conditions of air inside the bags: Density

1.527

kg/m3

Specific Internal Energy

1.2815e5

m2/sec2

Pressure

101325.2222e2

N/m2

Conditions environment: Pressure

101325.

N/m^2

Properties of the gas from the inflator: Gamma

111.557

Gas Constant R

243.

m2/sec2*K

Temperature

400.

K (assumed constant)

Mass Flow Rate:

Main Index

464 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

Figure 6-16

Mass Flow Rate Curve

Dytran Model The Distribution Tube and the Vertical Compartments: For the distribution tube and for each vertical compartment, a separate coupling surface is defined. At locations where the holes are located, a subsurface is defined with dummy triangular shell elements. These dummy shell elements do not carry any stiffness and are only needed to obtain a closed surface and to define the flow connections between the distribution tube and each of the vertical compartments. The distribution tube is supported with two flanges that are fixed in space using a SPC. The coupling surface for the distribution tube is defined as: COUPLE,1,25,OUTSIDE,ON,ON,9,11,AIRBAG,+ +,10,,,,,,,,+ +,22 SURFACE,25,,PROP,125 SET1,125,3,10,14,200,201,301,302,+ +,303,304,305

The property id's 301 through 305 represent the holes to the compartments. The coupling surface for the first compartment is defined with: COUPLE,2,35,OUTSIDE,ON,ON,,12,AIRBAG,+ +,,,,,,,,,+ +,23 SURFACE,35,,PROP,135 SET1,135,4,15,301

Main Index

Chapter 6: Occupant Safety 465 Side Curtain Air Bag (Courtesy of Autoliv)

As can be seen in the coupling surface definitions, the dummy shell elements with PID=301 are common to the distribution tube and the first compartment. To allow gas flow between the distribution tube and the first compartment, these elements must be defined as a hole, using: COUPOR,1,9,301,PORFLCPL,3001,CONSTANT,1.0 PORFLCPL,3001,,,,2

The external pressure on the coupling surface of the distribution tube is defined as: COUOPT,1,11,,,,,,,+ +,CONSTANT,101325.

Similar COUPLE, COUPOR, and COUOPT cards are defined for each vertical compartment. The vertical compartments are not connected to each other and there are no external holes The Inflator: The inflator is defined by an inflow boundary condition on a subsurface of the distribution tube: SUBSURF,1001,25,PROP,1001 SET1,1001,200 COUINFL,1,10,1001,INFLATR1,85,CONSTANT,0.7 INFLATR1,85,1,,350.,1.557,,243. TABLED1,1,,,,,,,,+ +,0.0,0.0,0.001,3.0,0.015,0.0

The inflator is assumed rigid and is rigidly constrained in all 6 directions: PSHELL,200,1,.00035 MATRIG,1,783. TLOAD1,1,1,,12 FORCE,1,MR1,,1.0,0.0,0.0,0.0 MOMENT,1,MR1,,1.0,0.0,0.0,0.0

The Eulerian Domains: The Eulerian domain for the distribution tube is defined by: MESH,22,ADAPT,0.009,0.009,0.009,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300

The Eulerian domain is defined with the "adaptive" mesh functionality. This results in a fully automatic generation of the Euler domain, and the Euler domain will adapt itself according to the deploying coupling surface. A separate MESH card is defined for each vertical compartment. The element size for the vertical compartments is 11.0 mm, while the element size for the distribution tube is 9.0 mm. The distribution tube requires a smaller Euler element size to get an accurate calculation of the gas flow. As the entire bag model is filled with the same inflator gas, only one property for the gas is required. All MESH cards refer to the same Eulerian property: PEULER1,300,,HYDRO,19

Main Index

466 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

The gas itself is represented by the Gamma Law equation of state: DMAT,3,1.527,3 EOSGAM,3,1.517,226.45

The gas inside the tube and the compartments is initialized with one set of cards, using a very large sphere that contains to complete side-curtain airbag: TICEUL,19,,,,,,,,+ +,SPHERE,1,3,5,1.0 SPHERE,1,,0.0,0.0,0.0,500.0 TICVAL,5,,DENSITY,1.527,SIE,1.2815E5

Note that the pressure is not initialized directly, but indirectly using the density and the specific internal energy. By using the Gamma Law equation of state, the pressure is calculated by Dytran and is equal to 101325.0 Pascal. This initial pressure is equal to the external pressure defined on the COUOPT entries, to avoid motion at the beginning of the calculation. Contact: The contact between all layers is defined with one CONTACT card. Because we have one continuous surface, the single surface contact definition can be used. This airbag model is folded using offsets between the folds. For offset folded air bags, a special contact option is available. It is called the BPFULL contact option, which consists of two basic parts. One part looks for new nodal penetrations and the second part takes special care of contacts near sharp corners. The first part of the BPFULL contact is quite expensive but it is only necessary during the first few milliseconds of the calculation. This logic is switched off at 3 milliseconds, using the TENDNEW option. The GAP value is set equal to the thickness of the membranes (.0007 m). A rule of thumb for airbag simulations is to set the monitoring distance double the GAP value, the maximum penetration four times the GAP value, and the initial monitoring distance to a large number, so all potential future contacts are verified. Using the BPFULL option also activates the checking on crossing membrane layers that might have been introduced during the folding process. The default projection tolerances are used. For air bags, a value of 1.e-3 is typical.

Main Index

Chapter 6: Occupant Safety 467 Side Curtain Air Bag (Courtesy of Autoliv)

In summary, the following contact options are used SEARCH

BPFULL

GAP

7.e-3

m

MONDISV

14.e-3

m

PENV

28.e-3

m

INITMON

.1

m

TENDNEW

3.e-3

seconds

TOLPROJ1

1e-3

TOLPROJ2

1e-3

IMM – The Reference Air Bag model: The air bag model is folded using offsets between all layers. Due to the nature of offset folding, the original element shapes are distorted, and it is necessary to use the Initial Metric Method (IMM). The bag model in unfolded shape (reference shape) is used to initialize the IMM strains in the folded model (initial model). The IMM strains are re-initialized every 5 milliseconds, using the following parameter: PARAM,IMM,,ON,5e-3,5E-3

The reference shape of the airbag is shown below. To activate the Initial Metric Method, the reference shape must be stored in a separate file, which is referenced using the IMMFILE option in the main input file. For example: IMMFILE=flat_imm.dat

For more information about the IMM method, please see the User's Guide.

Main Index

468 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

Figure 6-17

Reference Shape of Air Bag

Miscellaneous: 1. Because this model uses the coupling surface interface, the time step safety factor for Eulerian elements has to be .6. However, the Lagrangian elements (the bag membrane elements) determine the time-step, and it is beneficial to use a higher time step safety factor for the Lagrangian elements: PARAM,STEPFCTL,0.9

2. To speedup the simulation, sub-cycling is activated for the coupling calculations. During a subcycle, it is assumed that the coupling surface location remains constant, and therefore, the cover sections for each Euler element remains constant. The software will use a safety mechanism to determine the number of sub-cycles based upon the velocities of the coupling surface. The maximum number of sub-cycles is set by: PARAM,COSUBMAX,10

3. The pressure in each compartment will be monitored. For this, a surface output request is used: TYPE (Surface) = TIMEHIS SURFACES (Surface) = 3 SET 3 = 25 SURFOUT (Surface) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (Surface) = 0,thru,end,by,0.0001 SAVE (Surface) = 10000

For each compartment, another set of these is defined.

Main Index

Chapter 6: Occupant Safety 469 Side Curtain Air Bag (Courtesy of Autoliv)

4. To visualize the gas flow in the Euler mesh, the following output request is used: TYPE (euler) = ARCHIVE ELEMENTS (euler) = 1000 SET 1000 = ALLEULHYDRO ELOUT (euler) = XVEL YVEL ZVEL PRESSURE TIMES (euler) = 0,thru,end,by,1.E-3 SAVE (euler) = 1

There are six separate Euler domains, and different files are created for each Euler domain, using the naming convention: FOLDED_EULER_FV1_#.ARC FOLDED_EULER_FV2_#.ARC FOLDED_EULER_FV3_#.ARC FOLDED_EULER_FV4_#.ARC FOLDED_EULER_FV5_#.ARC FOLDED_EULER_FV6_#.ARC

A new set of files is written at each output step.

Results The model is set to run for 20 milliseconds. The Figure 6-18 below shows the average pressure in each of the 5 vertical compartments.

Figure 6-18

Main Index

Average Pressures in Vertical Compartments

470 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

The air bag at the end of the calculation is shown below (Figure 6-19 )

Figure 6-19

Air Bag at End of Simulation

The six Euler domains at the end of the simulation are shown in Figure 6-20:

Figure 6-20

Main Index

Euler Domains at End of Simulation

Chapter 6: Occupant Safety 471 Side Curtain Air Bag (Courtesy of Autoliv)

Input Deck Main Input file folded.dat: MEMORY-SIZE = 6000000,6000000 IMMFILE=flat_imm.dat START CEND ENDTIME=0.02 CHECK=NO TITLE= Jobname is: folded TLOAD=1 TIC=1 SPC=1 $ $ Output result for request: elements TYPE (elements) = ARCHIVE ELEMENTS (elements) = 1 SET 1 = 16805 THRU 16875 16877 THRU 16947 16949 THRU 18080 , 59 THRU 945 1067 1068 1069 1070 1181 THRU 1412 , 1423 THRU 2965 3076 THRU 3930 4041 THRU 4895 , 5006 THRU 5860 5971 THRU 7744 10855 THRU 12025 , 12138 THRU 12990 13103 THRU 13955 14068 THRU 14920 , 15033 THRU 15885 15998 THRU 16875 16877 THRU 16947 , 16949 THRU 18080 16805 THRU 16875 16877 THRU 16947 , 17005 THRU 18080 16949 THRU 17004 ELOUT (elements) = EFFSTS TIMES (elements) = 0,THRU,END,BY,4e-4 SAVE (elements) = 1 $ INCLUDE output_for_euler.ccf $------- Parameter Section -----PARAM,INISTEP,1.0e-7 PARAM,MINSTEP,1.0e-8 PARAM,STEPFCTL,0.9 PARAM,IMM,,ON,5e-3,5E-3 PARAM,FASTCOUP PARAM,COSUBMAX,10 $------- BULK DATA SECTION ------BEGIN BULK INCLUDE bag.bdf INCLUDE couple_settings.bdf INCLUDE euler_mesh.bdf INCLUDE fix_inflator.bdf INCLUDE pressure_gauge.bdf INCLUDE folded.bdf $------- Contact Definition ----CONTACT,25,SURF,,26,,,,,+ +,V4,BOTH,BPFULL,,0.0,.0007,,,+ +,,,DISTANCE,.0028,,0.45,DISTANCE,.0014,+ +,,10.E-3,,,,,.1,,+ +,,,1.E-3,1.E-3,,,,,+ +,3.e-3

Main Index

472 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

$ $ ========== PROPERTY SETS ========== $ $ * P_SET.1 * $ PSHELL1 1 2 MEMB +A000001 .0007 $ $ * P_SET.2 * $ PSHELL1 2 2 MEMB +A000002 .0007 $ $ * P_SET.3 * $ PSHELL1 3 2 MEMB +A000003 .00035 $ $ * P_SET.4 * $ PSHELL1 4 2 MEMB +A000004 .00035 $ $ * P_SET.5 * $ PSHELL1 5 2 MEMB +A000005 .00035 $ $ * P_SET.6 * $ PSHELL1 6 2 MEMB +A000006 .00035 $ $ * P_SET.7 * $ PSHELL1 7 2 MEMB +A000007 .00035 $ $ * P_SET.8 * $ PSHELL1 8 2 MEMB +A000008 .00035 $ $ * P_SET.14 * $ PSHELL1 14 2 MEMB +A000010 .00035 $ $ * P_SET.15 * $ PSHELL1 15 2 MEMB +A000011 .00035 $ $ * P_SET.16 *

Main Index

+A000001

+A000002

+A000003

+A000004

+A000005

+A000006

+A000007

+A000008

+A000010

+A000011

Chapter 6: Occupant Safety 473 Side Curtain Air Bag (Courtesy of Autoliv)

$ PSHELL1 16 2 MEMB +A000012 .00035 $ $ * P_SET.17 * $ PSHELL1 17 2 MEMB +A000013 .00035 $ $ * P_SET.18 * $ PSHELL1 18 2 MEMB +A000014 .00035 $ $ * P_SET.19 * $ PSHELL1 19 2 MEMB +A000015 .00035 $ $ * inlet.200 * $ PSHELL 200 1 .00035 $ $ * Euler.300 * $ PEULER1,300,,HYDRO,19 $ $ * face.201 * $ PSHELL 201 1 .00035 $ $ * diffusor.10 * $ PSHELL1 10 2 MEMB +A000016 .00035 $ PSHELL1 301 DUMMY PSHELL1 302 DUMMY PSHELL1 303 DUMMY PSHELL1 304 DUMMY PSHELL1 305 DUMMY $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Material Rigid id =1 MATRIG 1 783 +A000017 $ $ -------- Material Fabric id =2 DMATEL 2 783 2.6e+08 .3 + .01 $

Main Index

+A000012

+A000013

+A000014

+A000015

+A000016

+A000017

+

474 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

$ -------- Material Ideal_Gas id =3 DMAT 3 1.527 3 $ | $ -> density $ EOSGAM,3,1.517,226.45 $ | | $ | -> R $ -------> Cp/Cv $ $ -------- Material Webbing id =4 DMATEL 4 670 6.5e+09 + .1 $ $ Transient Dynamic Load TLOAD1,1,1,,12 $ translational velocity FORCE,1,MR1,,1.0,0.0,0.0,0.0 $ rotational velocity MOMENT,1,MR1,,1.0,0.0,0.0,0.0 $ ENDDATA The bulkdata file folded.bdf:

.3

$ --- SPC-name = Attachments SPC1 1 123456 555 THRU 563 + 611 THRU 636 $ $ --- Define 6722 grid points --$ GRID 1 .0184701.0143125 .459878 GRID 2 .0125105.0143125 .459878 GRID 61 1.89e-08.0005000 .247850 GRID 62 1.45e-08.0005000 .172850 ... $ $ --- Define 13727 elements $ $ -------- property set P_SET.1 --------CTRIA3 59 1 183 252 61 CTRIA3 60 1 253 270 252 CTRIA3 61 1 252 270 61 CTRIA3 62 1 248 189 62 CTRIA3 63 1 213 247 62 CTRIA3 64 1 248 62 247 ... The IMM reference file flat_imm.dat: $ $ --- Define 6871 grid points --$ GRID 1 .0009375-.626128 .230000 GRID 2 .0009375-.626128 .220000 GRID 61 .0005000-.414100 .450000

Main Index

+

566

THRU

574+

Chapter 6: Occupant Safety 475 Side Curtain Air Bag (Courtesy of Autoliv)

GRID 62 .0005000-.339100 .450000 ... $ $ --- Define 13727 elements $ $ -------- property set P_SET.1 --------CTRIA3 59 1 183 252 61 CTRIA3 60 1 253 270 252 CTRIA3 61 1 252 270 61 CTRIA3 62 1 248 189 62 CTRIA3 63 1 213 247 62 CTRIA3 64 1 248 62 247 The Surface definitions file bag.bdf: $ $ ------------------------------------------------------------$ bag 1, connected to the inflator $ this surface is used for gbag/couple definition $ main bag SURFACE,25,,PROP,125 SET1,125,3,10,14,200,201,301,302,+ +,303,304,305 $SET1,125,3,THRU,8,10,14,THRU,19,+ $+,200,201 $ $ compartment 1 SURFACE,35,,PROP,135 SET1,135,4,15,301 $ $ compartment 2 SURFACE,45,,PROP,145 SET1,145,5,16,302 $ $ compartment 3 SURFACE,55,,PROP,155 SET1,155,6,17,303 $ $ compartment 4 SURFACE,65,,PROP,165 SET1,165,7,18,304 $ $ compartment 5 SURFACE,75,,PROP,175 SET1,175,8,19,305 $ $ this surface is used for contact (not 10, 200,201) SURFACE,26,,PROP,126 SET1,126,3,THRU,8,14,THRU,19 $ SUBSURF,301,25,PROP,1301 SET1,1301,301 $ SUBSURF,302,25,PROP,1302 SET1,1302,302

Main Index

476 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

$ SUBSURF,303,25,PROP,1303 SET1,1303,303 $ SUBSURF,304,25,PROP,1304 SET1,1304,304 $ SUBSURF,305,25,PROP,1305 SET1,1305,305 $ $ define gasbag $ See section 2.1.1 in user manual for units $ $ define euler coupled airbag $ COUPLE,1,25,OUTSIDE,ON,ON,9,11,AIRBAG,+ +,10,,,,,,,,+ +,,22 COUPLE,2,35,OUTSIDE,ON,ON,,12,AIRBAG,+ +,,,,,,,,,+ +,,23 COUPLE,3,45,OUTSIDE,ON,ON,,13,AIRBAG,+ +,,,,,,,,,+ +,,24 COUPLE,4,55,OUTSIDE,ON,ON,,14,AIRBAG,+ +,,,,,,,,,+ +,,25 COUPLE,5,65,OUTSIDE,ON,ON,,15,AIRBAG,+ +,,,,,,,,,+ +,,26 COUPLE,6,75,OUTSIDE,ON,ON,,16,AIRBAG,+ +,,,,,,,,,+ +,,27 $ .............................................................. $ inflator $ SUBSURF,1001,25,PROP,1001 SET1,1001,200 $ COUINFL,1,10,1001,INFLATR1,85,constant,0.7 $ | $ +- referenced by COUPLE entry $ INFLATR1,85,1,,350.,1.557,,243. $ | $ | $ +-> mass flow rate defined in TABLED1,1 $ TABLED1,1,,,,,,,,+ +,0.0,0.0,0.001,3.0,0.015,0.0 $ $ COUPOR,1,9,301,PORFLCPL,3001,constant,1.0 COUPOR,2,9,302,PORFLCPL,3002,constant,1.0

Main Index

Chapter 6: Occupant Safety 477 Side Curtain Air Bag (Courtesy of Autoliv)

COUPOR,3,9,303,PORFLCPL,3003,constant,1.0 COUPOR,4,9,304,PORFLCPL,3004,constant,1.0 COUPOR,5,9,305,PORFLCPL,3005,constant,1.0 $ PORFLCPL,3001,,,,2 PORFLCPL,3002,,,,3 PORFLCPL,3003,,,,4 PORFLCPL,3004,,,,5 PORFLCPL,3005,,,,6 $ $ Model atmospheric pressure by PLCOVER on the COUOPT card $ $ -> referenced from couple card $ | COUOPT,1,11,,,,,,,+ +,CONSTANT,101325. COUOPT,2,12,,,,,,,+ +,CONSTANT,101325. COUOPT,3,13,,,,,,,+ +,CONSTANT,101325. COUOPT,4,14,,,,,,,+ +,CONSTANT,101325. COUOPT,5,15,,,,,,,+ +,CONSTANT,101325. COUOPT,6,16,,,,,,,+ +,CONSTANT,101325. $ MESH,22,ADAPT,0.009,0.009,0.009,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300 MESH,23,ADAPT,0.011,0.011,0.011,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300 MESH,24,ADAPT,0.011,0.011,0.011,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300 MESH,25,ADAPT,0.011,0.011,0.011,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300 MESH,26,ADAPT,0.011,0.011,0.011,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300 MESH,27,ADAPT,0.011,0.011,0.011,,,,+ +,,,,,,,,,+ +,,,,,,,EULER,300 $ TICEUL,19,,,,,,,,+ +,SPHERE,1,3,5,1.0 SPHERE,1,,0.0,0.0,0.0,500.0 TICVAL,5,,DENSITY,1.527,SIE,1.2815E5 The output request file output_for_euler.ccf: $ Output result for request: euler TYPE (euler) = ARCHIVE

Main Index

478 Dytran Example Problem Manual Side Curtain Air Bag (Courtesy of Autoliv)

ELEMENTS (euler) = 1000 SET 1000 = ALLEULHYDRO ELOUT (euler) = XVEL YVEL ZVEL PRESSURE $ SIE XMOM YMOM ZMOM DENSITY TIMES (euler) = 0,thru,end,by,1.E-3 SAVE (euler) = 1 $ TYPE (Surface) = TIMEHIS SURFACES (Surface) = 3 SET 3 = 25 SURFOUT (Surface) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (Surface) = 0,thru,end,by,0.0001 SAVE (Surface) = 10000 $ TYPE (part1) = TIMEHIS SURFACES (part1) = 4 SET 4 = 35 SURFOUT (part1) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (part1) = 0,thru,end,by,0.0001 SAVE (part1) = 10000 $ TYPE (part2) = TIMEHIS SURFACES (part2) = 5 SET 5 = 45 SURFOUT (part2) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (part2) = 0,thru,end,by,0.0001 SAVE (part2) = 10000 $ TYPE (part3) = TIMEHIS SURFACES (part3) = 6 SET 6 = 55 SURFOUT (part3) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (part3) = 0,thru,end,by,0.0001 SAVE (part3) = 10000 $ TYPE (part4) = TIMEHIS SURFACES (part4) = 7 SET 7 = 65 SURFOUT (part4) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (part4) = 0,thru,end,by,0.0001 SAVE (part4) = 10000 $ TYPE (part5) = TIMEHIS SURFACES (part5) = 8 SET 8 = 75 SURFOUT (part5) = AREA VOLUME MASS PRESSURE TEMPTURE TIMES (part5) = 0,thru,end,by,0.0001 SAVE (part5) = 10000

Main Index

Chapter 6: Occupant Safety 479 Hybrid III 5th%-tile Dummy

Hybrid III 5th%-tile Dummy Problem Description In this part of the Example Problem Manual, a Finite Element model for the Hybrid III 5th%-tile small female dummy will be introduced. This dummy has been especially calibrated and validated for Out-OfPosition (OOP) situations as described in regulation FMVSS 208, revision 2001 and beyond set forward by the National Highway Traffic Safety Administration (NHTSA). In this chapter, the complete dummy is described first. This is followed by the description of the calibration tests that have been performed.

Figure 6-21

Finite Element Model for Hybrid III 5th%-tile

Model Description The dummy model comprises 28 rigid body segments, linked together with multi-axis joints. Each body segment has predefined inertia properties, and surface contour. The inertia properties are defined through the MATRIG card, whilst rigid shell elements define the surface contour. Between the individual body segments, “null” elements may be defined to obtain a closed surface for smooth contact behavior. The kinematic joints, connecting the body segments, are modeled with directional springs (CELAST1) and dampers (CDAMP1).

Main Index

480 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

The rigid bodies are connected with directional springs and dampers. For the neck area, the user may choose between two types of neck models. One is the Pivot Neck Model and the second is the Rubber Neck Model: 1. Pivot Neck Model: This model uses a series of rigid bodies connected with directional springs and dampers. There are solid elements between the rigid bodies, but they are only modeled to capture the inertia effects during the simulation. The stiffness is set to be very low. 2. Rubber Neck Model: The rigid parts of the neck are modeled with solid rubber elements. Contact is defined between the slits to avoid penetration during flexion of the neck. The spring and damper elements are completely deactivated when this model is chosen. The rubber elements have validated material properties. Tetra elements are used to model the slits to improve the quality and of the robustness of the behavior of the neck during flexion. The rubber is modeled with a linear viscoelastic material model.

Figure 6-22 Pivot and Rubber Neck Models The structure of linked rigid bodies is generally known as a tree structure. Starting at the Lower Torso, which is at the root of the structure, the tree branches out to the Legs, and Arms and Head, through the Upper Torso. Several coordinate systems are associated with each rigid body to define its properties and the linkage with adjoining bodies. Every rigid body has a coordinate system that defines the body center of mass and the three principal inertia axes. Each body, except the Lower Torso, which is at the root of the body structure, has a coordinate system which defines the location and orientation of the joint that connects it to its preceding, or Parent body. This coordinate system is referred to as the Joint Child Coordinate System. Body segments that have a connection to an outer branch of the tree structure, further have a coordinate system that locates the origin of that branch and its orientation relative to the parenting body. This coordinate system is referred to as the Joint Parent Coordinate System. Note that each body may have several children, but has only one parent. The relative position and orientation of the Joint Parent and Child coordinate systems determines the forces and moments that act between them. Depending on the body, two more types of coordinate systems may have been defined. For the Lower Torso, Upper Torso, and Head, a coordinate system is defined in the location and with the orientation of the accelerometers. Joint Parent coordinate systems in Femurs, Tibias and Neck double as coordinate frames for the load transducers that are present at these locations.

Main Index

Chapter 6: Occupant Safety 481 Hybrid III 5th%-tile Dummy

To help visualize the various coordinate systems, rod elements have been defined in their locations. For example of the Right Femur, see the following page:

Coord A: Right Femur Principal Inertia Frame Coord B: Right Femur Load Cell Parent Frame Coord C: Right Hip Joint Child Frame A complete schematic overview of the dummy is attached in Appendix A: File E XAMPLE.AIN as generated by GEBODFor each of the body component, a set of SURFACE's and property SET1's are predefined that can be used when contact is required between the dummy and external objects such as air bags or the instrument panel. The detailed list of property and surface definitions can be found in Appendix B: Complete ATB Input File for Sled Test CalculationThe structure of the Dytran input deck necessitates that the dummy model data is split into two files. The file hyb305_case.mod contains the CASE data, comprising cards for output of the dummy kinematics (into ARC files) and instrumentation responses (into THS files). The file hyb305_bulk.mod comprises the bulk data cards defining the dummy model. The model defined in this file represents the dummy in so-called “reference position”. For a typical simulation analysis, the dummy will be used in a position and with an orientation that differs from this reference position. A utility program is provided to change the position and orientation of the dummy as a whole and of the limbs individually, to accommodate this. The user provides the desired location of the H-point, orientation of the Lower Torso, and rotation angles of arm and leg segments in a position definition file. The utility program hyb305 will then assemble a dummy model in the correct position and orientation from the hyb305_bulk.mod file and the user provided position definition file. The resulting bulk data files hyb305_bulk.dat and hyb305_case.dat can be included in a simulation model for analysis. Positioning or changing the orientation of dummy parts through other means may seriously affect the integrity of the dummy model and is therefore discouraged Details of the structure of the position definition file and the execution of the positioning program are contained in Appendix C: File Create_fem_dummy.dat Calibration Tests

Main Index

482 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

In this section the calibration simulations will be described. The simulation performed follows the guidelines of Ref 1. [Technical Report – Development and Evaluation of the Hybrid III fifth Percentile Female Crash Test Dummy (H-III5F), NHTSA, August 1998]. However, the focus is on the dummy behavior during the impact of an air bag. To this end, only in the area of the neck and chest calibration tests have been performed. The head impact and knee impact are skipped at this point in time. Each one of the following sections will describe the details of the test and show the results. Neck Extension Calibration Test Experiment In this test, the neck and head of the dummy are attached to a pendulum. The structure is then lifted to a specific height and released. The pendulum impacts a honeycomb block which will give it a certain deceleration profile. Due to the deceleration, the head of the dummy starts to show extension. In this test, the moment and the head D-plane rotation are measured as a function of time. Simulation The neck portion of the dummy was extracted from the main hybrid305_bulk.dat file and was attached to the pendulum by merging the MATRIG definitions using the parameter MATRMERG. In order to save CPU time, the simulation will start just before the pendulum contacts the honeycomb block. The contact with the foam will not be modeled, but instead the pendulum will be given a predefined rotational velocity profile. All the rest of the neck and head nodes will be given an initial rotational and translational velocity by using the TIC3 input card. At the moment of impact, the radial velocity of the system is 3.5523 Rad/sec. The force in rotational spring 1080051 and damper 1080052 will be monitored. This represents the moment at the occipital condyle. For the D-Plane, the locations of nodes 1080020 and 1080023 are monitored. These nodes are part of the coordinate system that describes the location and orientation of the head accelerometer. From the relative location of these two nodes, the orientation of the head D-Plane is calculated.

Main Index

Chapter 6: Occupant Safety 483 Hybrid III 5th%-tile Dummy

Results Figure 6-23 shows the results for the pivot neck and the rubber neck models for every 20 mseconds

followed by the time history graphics of the neck moment and the orientation of the head D-Plane.

Figure 6-23

Main Index

Results of Neck Extension for the Pivot andbber Neck Models for every 20 Mseconds

484 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Main Index

Figure 6-24

Time History of D-Plane Extension

Figure 6-25

Time History of Neck Extension Moment

Chapter 6: Occupant Safety 485 Hybrid III 5th%-tile Dummy

Figure 6-26

Time History of D-Plane Moment Extension

Neck Flexion Calibration Test Experiment In this test, the same pendulum is used but the neck bracket will be mounted in the opposite direction.Also in this test, the neck moment and the head D-Plane rotation will be measured as a function of time. Simulation In this simulation, the same model will be used as the neck extension model, but the enforced rotational and initial velocity will be in the negative direction. Also, the absolute value of the radial velocity on the TIC3 card was increased to 4.14435 as the deceleration curve is different according to the NHTSA specifications.

Main Index

486 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Results On the next page, the results for the pivot neck model and the rubber neck model for every 20 mseconds followed by the time history graphics of the neck moment and orientation of the head D-Plane.

Figure 6-27

Main Index

Results of Flexion for the Pivot and Rubber Neck ls for every 20 mseconds

Chapter 6: Occupant Safety 487 Hybrid III 5th%-tile Dummy

Main Index

Figure 6-28

Time History of D-Plane Flexion

Figure 6-29

Time History of Neck Flexion Moment

488 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Figure 6-30

Time History of D-Plane Moment Flexion

Thorax Impact Experiment In this calibration test, an impactor with a weight of 14 kg will contact the chest at a velocity of 6.47 m/s. Simulation First, the dummy was positioned using the hyb305 positioner, the arms and legs were straightened, then the floor and impactor were added to the test input file. A contact was defined between the impactor and the dummy chest. During the simulation, the force in the spring 1050031 was monitored and is the impact force. Also, the location of nodes 1040100 and 1050000 was monitored. The difference in x-position is a measure for the chest deflection.

Main Index

Chapter 6: Occupant Safety 489 Hybrid III 5th%-tile Dummy

Results On the next page, the results for the thorax impact are given for every 10 mseconds followed by the graphs for thorax displacement in time and thorax force deflection curve.

Figure 6-31

Main Index

Results of Thorax Impact of Dummy Model

490 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Main Index

Figure 6-32

Time History of Thorax Displacement

Figure 6-33

Time History of thorax Impact

Chapter 6: Occupant Safety 491 Hybrid III 5th%-tile Dummy

Conclusions and Recommendations Rubber Neck Model During extension, the rubber neck model fits the experiment very well, up to the time that the neck is bend backwards to approximately 90 degrees. At that moment, the rubber layers on the disks contact each other, and the peak moment that occurs at that moment is not captured completely. Some more attention to this contact modeling during extension will be given in a future release. In flexion, the rubber neck model initially follows the experimental data well, but then starts to show a too soft a behavior. Given that the same material properties in flexion are used as in extension, and the extension results were accurate, it can only be that the slit contact, the behavior of the tet elements during compression, or the material definition during compression are causing the softening. In a future release, more attention will be given to this slit contact and to a more detailed modeling of the rubber in the slit area. Pivot Neck Model During extension, the pivot neck model initially follows the experimental data well, but it appears that a stiffer contact between the disks is required. The extension side could also do with a little less damping, but it remains a concern this can be accomplished without affecting the flexion response. Perhaps, by lowering the damping little by little might help, yet care must be taken that a complete change in eigen mode can occur when the damping becomes too small. These items will be investigated for a future release. In flexion, the pivot neck behaves excellent. Care must be taken that this behavior remains at this high level while further fine-tuning the behavior during extension. Thorax Impact The characteristics of the spring 1050031 were taken directly from the thorax experiment. This implies that the damper characteristics of the system are already included in the spring data. Therefore, the damper part of the chest deflection was deactivated for this release of the dummy. As a result, the chest deflection matches the experiment well, but since the damper was deactivated, the deflection returned to zero quicker than the experiment. For a future release, the chest model will be adapted to be modeled with both a spring and a damper. The chest deflection also needs to be validated against different velocities of the impactor.

Main Index

492 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Input Files Neck Extension CEND ENDTIME=0.120 ENDSTEP=99999999 CHECK=NO TITLE= Jobname is: neck_extension TLOAD=1 TIC=1 SPC=1 $ TYPE(MR)=MRSUM STEPS(MR)= 0 THRU END BY 1000 $ TYPE(MAT)=MATSUM STEPS(MAT)= 0 THRU END BY 1000 $ TYPE(STEP)=STEPSUM STEPS(STEP)= 0 THRU END BY 25 $ $ Output result for request: neck_extension TYPE (elem) = ARCHIVE ELEMENTS (elem) = 1 SET 1 = ALLELEMENTS ELOUT (elem) = MASS TIMES (elem) = 0 THRU END BY 0.005 SAVE (elem) = 10000 $ TYPE (dplane) = TIMEHIS GRIDS (dplane) = 2 SET 2 = 1080020 1080023 GPOUT (dplane) = XPOS YPOS ZPOS TIMES (dplane) = 0 THRU END BY 0.0001 SAVE (dplane) = 10000 $ TYPE(loadcells) = TIMEHIS ELEMENTS(loadcells) = 1000003 SET 1000003 = 1080011, 1080012, 1080021, 1080022, 1080031, 1080032, 1080041, 1080042, 1080051, 1080052, 1080061, 1080062 ELOUT(loadcells) = XFORCE TIMES(loadcells) = 0.000 THRU END BY 0.0001 SAVE(loadcells) = 10000 $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ BEGIN BULK $ PARAM, INISTEP, 1.0e-6 PARAM, MINSTEP, 1.0e-9 PARAM, GEOCHECK, ON PARAM, RBE2INFO, GRIDON PARAM, MATRMERG, FR1, MR1040000, MR2000000

Main Index

Chapter 6: Occupant Safety 493 Hybrid III 5th%-tile Dummy

$ INCLUDE hyb305_pivot_neck.dat $ $-------------------------------------------------------------------------$ BEGIN PENDULUM COMPONENT $-------------------------------------------------------------------------GRID 2000000 -.098729 .0381 .296886 GRID 2000001 -.098729 3.725-9 .296886 GRID 2000002 -.098729-.0381 .296886 $ CQUAD4 CQUAD4 -

2000000 2000000 2000000 2000001 2000004 2000003 2000001 2000000 2000001 2000002 2000005 2000004

$ pid mid form quad nip shfact ref cont PSHELL1, 2000000,2000000, , , , , , ,+ $ thick +, 0.005 $ mid rho E nu mass cogX cogY cogZ MATRIG, 2000000, , , , 29.57,-.047929, 3.027-9 ,2.13209 $ TLOAD1 1 1001 12 2000001 MOMENT,1001,FR1,,1,0,1,0 FORCE,1001,FR1,,0,0,0,0 $ TABLED1,2000001,,,,,,,,+ +,0,-3.5523,0.01,-2.664,0.02,-1.7170,0.03,-0.8288,+ +,0.04,0,ENDT $-------------------------------------------------------------------------$ END PENDULUM $-------------------------------------------------------------------------$ TIC3,1,2000040,,1.0,,,,,+ +, , , , ,-3.5523, ,,,+ + 1040010 THRU 1040013 1040020 THRU 1040023 1040200 THRU+ + 1040203 1041601 THRU 1041675 1060000 THRU 1060003 1060010+ + THRU 1060013 1060100 THRU 1060103 1061001 THRU 1061139+ + 1070100 THRU 1070103 1070110 THRU 1070113 1070190 THRU+ + 1070193 1070200 THRU 1070203 1070210 THRU 1070213 1070290+ + THRU 1070293 1070300 THRU 1070303 1070310 THRU 1070313+ + 1070390 THRU 1070393 1070400 THRU 1070403 1070410 THRU+ + 1070413 1070490 THRU 1070493 1070500 THRU 1070503 1070510+ + THRU 1070513 1070590 THRU 1070593 1070600 THRU 1070603+ + 1071001 THRU 1074344 1080000 THRU 1080003 1080010 THRU+ + 1080013 1080020 THRU 1080023 1081001 THRU 1081380 $ ENDDATA

Main Index

494 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Neck Flexion START CEND ENDTIME=0.120 ENDSTEP=99999999 CHECK=NO TITLE= Jobname is: neck_extension TLOAD=1 TIC=1 SPC=1 $ TYPE(MR)=MRSUM STEPS(MR)= 0 THRU END BY 1000 $ TYPE(MAT)=MATSUM STEPS(MAT)= 0 THRU END BY 1000 $ TYPE(STEP)=STEPSUM STEPS(STEP)= 0 THRU END BY 25 $ $ Output result for request: neck_extension TYPE (elem) = ARCHIVE ELEMENTS (elem) = 1 SET 1 = ALLELEMENTS ELOUT (elem) = MASS TIMES (elem) = 0 THRU END BY 0.005 SAVE (elem) = 10000 $ TYPE (dplane) = TIMEHIS GRIDS (dplane) = 2 SET 2 = 1080020 1080023 GPOUT (dplane) = XPOS YPOS ZPOS TIMES (dplane) = 0 THRU END BY 0.0001 SAVE (dplane) = 10000 $ TYPE(loadcells) = TIMEHIS ELEMENTS(loadcells) = 1000003 SET 1000003 = 1080011, 1080012, 1080021, 1080022, 1080031, 1080032, 1080041, 1080042, 1080051, 1080052, 1080061, 1080062 ELOUT(loadcells) = XFORCE TIMES(loadcells) = 0.000, THRU, END, BY, 0.0001 SAVE(loadcells) = 10000 $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ BEGIN BULK $ PARAM, INISTEP, 1.0e-6 PARAM, MINSTEP, 1.0e-9 PARAM, GEOCHECK, ON PARAM, RBE2INFO, GRIDON PARAM, MATRMERG,FR1,MR1040000,MR2000000 $ INCLUDE hyb305_pivot_neck.dat

Main Index

Chapter 6: Occupant Safety 495 Hybrid III 5th%-tile Dummy

$ $-------------------------------------------------------------------------$ BEGIN PENDULUM COMPONENT $-------------------------------------------------------------------------GRID 2000000 -.098729 .0381 .296886 GRID 2000001 -.098729 3.725-9 .296886 $ CQUAD4 2000000 2000000 2000000 2000001 2000004 2000003 CQUAD4 2000001 2000000 2000001 2000002 2000005 2000004 $ pid mid form quad nip shfact ref cont PSHELL1, 2000000,2000000, , , , , , ,+ $ thick +, 0.005 $ mid rho E nu mass cogX cogY cogZ MATRIG, 2000000, , , , 29.57,-.047929, 3.027-9 ,2.13209 $ TLOAD1 1 1001 12 2000001 MOMENT,1001,FR1,,1,0,1,0 FORCE,1001,FR1,,0,0,0,0 $ TABLED1,2000001,,,,,,,,+ +,0,4.14435, 0.01,2.0465, 0.02,1.253, 0.03,0.501,+ +,0.04,0,ENDT $ $-------------------------------------------------------------------------$ END PENDULUM $-------------------------------------------------------------------------$ TIC3,1,2000040,,1.0,,,,,+ +, , , , ,4.14435, ,,,+ + 1040010 THRU 1040013 1040020 THRU 1040023 1040200 THRU+ + 1040203 1041601 THRU 1041675 1060000 THRU 1060003 1060010+ + THRU 1060013 1060100 THRU 1060103 1061001 THRU 1061139+ + 1070100 THRU 1070103 1070110 THRU 1070113 1070190 THRU+ + 1070193 1070200 THRU 1070203 1070210 THRU 1070213 1070290+ + THRU 1070293 1070300 THRU 1070303 1070310 THRU 1070313+ + 1070390 THRU 1070393 1070400 THRU 1070403 1070410 THRU+ + 1070413 1070490 THRU 1070493 1070500 THRU 1070503 1070510+ + THRU 1070513 1070590 THRU 1070593 1070600 THRU 1070603+ + 1071001 THRU 1074344 1080000 THRU 1080003 1080010 THRU+ + 1080013 1080020 THRU 1080023 1081001 THRU 1081380 $ ENDDATA

Main Index

496 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Thorax Impact START CEND ENDTIME=0.050 CHECK=NO TITLE= Jobname is: test TLOAD=1 TIC=1 SPC=1 $ TYPE(MR)=MRSUM STEPS(MR)= 0 THRU END BY 1000 $ TYPE(MAT)=MATSUM STEPS(MAT)= 0 THRU END BY 1000 $ TYPE(STEP)=STEPSUM STEPS(STEP)= 0 THRU END BY 25 $ TYPE (HIII) = ARCHIVE ELEMENTS (HIII) = 1 SET 1 = ALLELEMENTS ELOUT (HIII) = MASS TIMES (HIII) = 0 THRU END BY 0.005 SAVE (HIII) = 100000 $ TYPE (grid_CG) = TIMEHIS GRIDS (grid_CG) = 2 SET 2 = 1040100 1050000 GPOUT (grid_CG) = XPOS YPOS ZPOS TIMES (grid_CG) = 0 THRU END BY 0.0001 SAVE (grid_CG) = 10000 $ $ Output result for request: contact TYPE (contact) = TIMEHIS CONTS (contact) = 4 SET 4 = 3 4 CONTOUT (contact) = XFORCE YFORCE ZFORCE FMAGN TIMES (contact) = 0 THRU END BY 0.001 SAVE (contact) = 100000 $ TYPE (loadcells) = TIMEHIS ELEMENTS (loadcells) = 5 SET 5 = 1080011, 1080012, 1080021, 1080041, 1080042, 1080051, 1050011, 1050012, 1050021, 1050041, 1050042, 1050051, ELOUT (loadcells) = XFORCE TIMES (loadcells) = 0.000 THRU END BY 0.0001 SAVE (loadcells) = 10000 $ $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1e-7

Main Index

DMIN

1080022, 1080052, 1050022, 1050052,

1080031, 1080061, 1050031, 1050061,

1080032, 1080062, 1050032, 1050062

Chapter 6: Occupant Safety 497 Hybrid III 5th%-tile Dummy

PARAM,STEPFCT,0.9 $ $------- BULK DATA SECTION ------BEGIN BULK $ INCLUDE hyb305_bulk.dat INCLUDE floor.dat INCLUDE impactbar.dat $ $ Rigid Impactor can move only in global X dir $ TLOAD1 1 1001 12 FORCE, 1001,MR1,,0.0, ,1.0,1.0 MOMENT,1001,MR1,,0.0,1.0,1.0,1.0 $ $ Rigid floor can not move $ TLOAD1 1 1002 12 FORCE, 1002,MR2,,0.0,1.0,1.0,1.0 MOMENT,1002,MR2,,0.0,1.0,1.0,1.0 $ $ ------- Initial Velocity BC vel ----SET1 1 1281154 THRU 1282050 TICGP 1 1 XVEL -6.47 $ $ -------- Contact between dummy and floor $ CONTACT 3 SURF SURF 2 1 + V4 TOP FULL 0.0 + + + 33 CONTFORC 33 3 3 33 $ TABLED1,3,,,,,,,,+ +,0.0,0.0,0.050,1000.0 $ TABLED1,33,,,,,,,,+ +,-1.000,-200.0,1.000,200.0 $ $ Slave contact surface: Dummy $ SURFACE 2 PROP 1011002 PROP 1091001 + PROP 1111001 PROP 1121001 PROP 1151001 + PROP 1191001 PROP 1201001 $ $ Master contact surface: Floor $ SURFACE 1 PROP 2 SET1 2 2 $ $ -------- Contact between chest and impactbar $ CONTACT 4 GRID SURF 4 1051001

Main Index

+ NONE+ + +

PROP 1101001+ PROP 1161001+

+

498 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

+ V4 TOP FULL 0.0 + + + 44 CONTFORC 44 4 4 $ TABLED1,4,,,,,,,,+ +,0.0,0.0,0.050,1000.0 $ TABLED1,5,,,,,,,,+ +,-1.000,-200.0,1.000,200.0 $ $ Slave contact gridpoints: Impactbar SET1 4 1281154 THRU 1282050 $ $ $ ========== PROPERTY SETS ========== $ $ * impactbar * $ PSOLID 1 1 $ $ * Floor * $ PSHELL 2 2 .003 $ $ $ ========= MATERIAL DEFINITIONS ========== $ $ $ -------- Impactbar MATRIG 1 2590 7e+10 .3 $ $ -------- Floor MATRIG 2 7850 7e+10 .3 $ ENDDATA floor.dat: CQUAD4 1289794 2 1282051 1282052 1282073 CQUAD4 1289795 2 1282052 1282053 1282074 CQUAD4 1289796 2 1282053 1282054 1282075 CQUAD4 1289797 2 1282054 1282055 1282076 CQUAD4 1289798 2 1282055 1282056 1282077 ... GRID 1282051 GRID 1282052 GRID 1282053 GRID 1282054 ... impact_bar.dat: $ GRID 1281154 GRID 1281155

Main Index

-.553700-.250000-.091000 -.478700-.250000-.091000 -.403700-.250000-.091000 -.328700-.250000-.091000

.176400-.006968 .267826 .176400-.013936 .289273

NONE+ + + 5

1282072 1282073 1282074 1282075 1282076

Chapter 6: Occupant Safety 499 Hybrid III 5th%-tile Dummy

GRID GRID GRID GRID -

1281156 1281157 1281158 1281159

.211960-.013936 .211960-.006968 .247520-.013936 .247520-.006968

.289273 .267826 .289273 .267826

$ -------- property set impactbar_prop --------CHEXA 1289014 1 1282001 1282002 1281991 +A000001 1281156 1281155 CHEXA 1289015 1 1282002 1282003 1281992 +A000002 1281158 1281156 CHEXA 1289016 1 1282003 1282004 1281993 +A000003 1281160 1281158 CHEXA 1289017 1 1282004 1282005 1281994 +A000004 1281162 1281160 CPENTA CPENTA CPENTA CPENTA CPENTA -

Main Index

1289474 1289475 1289476 1289477 1289478

1 1 1 1 1

1282001 1282002 1282003 1282004 1282005

1281154 1281157 1281159 1281161 1281163

1282012 1282013 1282014 1282015 1282016

1281990 1281154 1281157+A000001 1281991 1281157 1281159+A000002 1281992 1281159 1281161+A000003 1281993 1281161 1281163+A000004

1282002 1282003 1282004 1282005 1282006

1281157 1281159 1281161 1281163 1281165

1282013 1282014 1282015 1282016 1282017

500 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Appendix A: Overview of the Hybrid III 5th%-tile Dummy Schematic Setup

Main Index

Chapter 6: Occupant Safety 501 Hybrid III 5th%-tile Dummy

Joint Information

Main Index

502 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

General Numbering Information

Main Index

Chapter 6: Occupant Safety 503 Hybrid III 5th%-tile Dummy

Component Numbering Information

Main Index

504 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Joint Numbering Information

Appendix B: SURFACE and SET1 Definition Surface

Main Index

SURFACE and SET1 Identification Number

Association

Typical Usage

Buttocks

1011002

Lower Torso

Seat Contact

Lap

1011003

Lower Torso, Abdomen, Femurs

Lap Belt Contact

ChestRib

1041001

Upper Torso

Chin to Chest Contact

Jacket

1041002

Upper Torso, Chest

Belt and Airbag Contact

Back

1041003

Upper Torso

Seat Contact

Chapter 6: Occupant Safety 505 Hybrid III 5th%-tile Dummy

SURFACE and SET1 Identification Number

Surface

Association

Typical Usage

Chest

1051001

Chest

Steering Wheel Contact

Neck

1071001

Neck

Belt and Airbag Contact

Head

1081001

Head

Airbag Contact

Chin

1081002

Head

Chin to Chest Contact

Left Upper Leg

1091001

Left Femur

Seat Contact

Right Upper Leg

1101001

Right Femur

Seat Contact

Left Knee

1111001

Left Knee

Bolster Contact

Right Knee

1121001

Right Knee

Bolster Contact

Left Lower Leg

1151001

Left Tibia

Bolster Contact

Right Lower Leg

1161001

Right Tibia

Bolster Contact

Left Shoe

1191001

Left Foot

Floor Contact

Right Shoe

1201001

Right Shoe

Floor Contact

Left Upper Arm

1231001

Left Upper Arm

Airbag Contact

Right Upper Arm

1241001

Right Upper Arm

Airbag Contact

Left Lower Arm

1251001

Left Lower Arm

Airbag Contact

Right Lower Arm

1261001

Right Lower Arm

Airbag Contact

Left Hand

1271001

Left Hand

IP Contact

Right Hand

1281001

Right Hand

IP Contact

Appendix C: hyb305 Positioning The positioning of the dummy can be done by means of a little fortran program, called hyb305.f. In the remainder of this text, it is assumed that the user has access to the executable form: hyb305.exe Definition file: hyb305.def $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $ $ $ $ $ $ $ $ $ $

Main Index

5th Percentile Hybrid III Dummy Database Definition Model @(#) HYB305

: 1.0

Developed For Unit System Release Status Performance

: : : :

IMPORTANT NOTICE:

2003/05/01 :

Dytran 2003 kg, m, s, K NOT RELEASED Functional Model Only - NOT VALIDATED

506 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

$ This Database is distributed under the GNU General Public License. $ This database can be freely used, redistributed and/or modified under $ the terms of the GNU General Public License as published by the Free $ Software Foundation; either version 2 of the License, or any later $ version, provided that this message is retained. $ $ This database is distributed in the hope that it will be useful, but $ WITHOUT ANY WARRANTY or FITNESS FOR A PARTICULAR PURPOSE. $ See the GNU General Public License for more details $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

$ $ NOTE: Interpretation of this file is case sensitive $ $ $ 1 = pivot model for Rubber Neck components $ 2 = FEM model for Rubber Neck components SELECT, NeckModel, 2 $ $ Note: Angles in Degrees $ $ nameID dim dof1 dof2 INIVAR, NumberingBase, 1, 1000000 $ $ translate in direction : forward left INIVAR, HPointPosition, 3, 0.000, 0.00, $ $ rotate around direction: forward left INIVAR, HPointAngle, 3, 0.0, 0.0, $ $ rotate around direction: forward left INIVAR, LowerTorsoAngle, 3, 0.0, 0.0, $ $ rotate around direction: left INIVAR, NeckBracketAngle, 1, 0.0 $ $ rotate around direction: left forward INIVAR, LeftHipAngle, 3, 0.0, 0.0, $ $ rotate around direction: left forward INIVAR, RightHipAngle, 3, 0.0, 0.0, $ $ rotate around direction: left INIVAR, LeftKneeAngle, 1, -90.0 $ $ rotate around direction: left INIVAR, RightKneeAngle, 1, -90.0 $ $ rotate around direction: left forward INIVAR, LeftAnkleAngle, 3, 0.0, 0.0, $ $ rotate around direction: left forward INIVAR, RightAnkleAngle, 3, 0.0, 0.0, $

Main Index

dof3

vertical 0.000 vertical 0.0 vertical 0.0

vertical 0.0 vertical 0.0

vertical 0.0 vertical 0.0

Chapter 6: Occupant Safety 507 Hybrid III 5th%-tile Dummy

$ INIVAR, $ $ INIVAR, $ $ INIVAR, $ $ INIVAR, $ $ INIVAR, $ $ INIVAR,

Main Index

rotate around direction: left LeftShoulderAngle, 3, -90.0,

vertical 0.0,

forward 0.0

rotate around direction: left RightShoulderAngle, 3, -90.0,

vertical 0.0,

forward 0.0

rotate around direction: LeftElbowAngle, 2,

left 0.0,

vertical 0.0

rotate around direction: RightElbowAngle, 2,

left 0.0,

vertical 0.0

rotate around direction: LeftWristAngle, 2,

left 0.0,

forward 0.0,

0.0

rotate around direction: RightWristAngle, 2,

left 0.0,

forward 0.0,

0.0

508 Dytran Example Problem Manual Hybrid III 5th%-tile Dummy

Reference Database

Main Index

Chapter 6: Occupant Safety 509 Easy Postprocessing with Adaptive Meshing

Easy Postprocessing with Adaptive Meshing During a simulation with adaptive meshing, the Euler mesh changes continuously and “adapts itself” to follow the coupling surface moves and deformations. The Euler mesh adapts to the coupling surface by adding and removing elements. The adaptive algorithm ensures that the coupling surface is contained inside the Euler mesh at all times with the minimum amount of elements. This can save a lot of computational time but postprocessing can be a little tricky. Since the geometry can change at any cycle, Euler archives contain only one cycle. This makes postprocessing with Patran cumbersome, especially if one needs to postprocess several cycles. Usually, all one needs to do is to have Dytran output results of all cycles in one big archive and read that into Patran. Not so with adaptive meshing. Here, one has to read in a separate archive for each cycle which makes animation a difficult task. For this purpose, a postprocessor must have multi-model capability to postprocess the multiple archives each having a different geometry simultaneously. A new functionally now allows easy postprocessing with Patran. The user defines a box. This box consists of virtual elements and all adaptive elements completely inside the box will be mapped onto matching virtual elements, making virtual element real. Virtual elements have grid points that are compatible with real elements. Therefore, virtual and the real element have the same geometry and user number, but only the adaptive elements have nonzero element values. When a virtual element is written to output and if it has a matching real element the values of the real element are written out. The virtual elements that have no matching real element are written out with zeroes. Since this box is fixed during the simulation, the Euler element geometry does not change. Therefore, archives written with this option can contain multiple cycles and can be postprocessed by Patran. Then, also with Patran, one can make animations with adaptive mesh results. This new functionality is activated by specifying the STATBOX entries on the MESH entry. The size of the box can be chosen freely, but then, adaptive elements may be created that are outside of the box. The run will continue with a warning and no output will be written for elements that are outside. In general the box should be large enough to contain all adaptive elements that are created during the simulation. For an adaptive run the adaptive meshing summary in the OUT file lists the largest box that surrounds all adaptive elements. This box can be used to define the static output box. This option does not require a significant amount of CPU-time.

Problem Description To illustrate static output for adaptive meshing, example problem 6.1 will be run with the Euler solver.

Main Index

510 Dytran Example Problem Manual Easy Postprocessing with Adaptive Meshing

Airbag Model

Dytran Modeling For details, see Example 6 1. The GBAG is replaced by a MESH entry and COUPLE entry. The MESH entry will use adaptive meshing. MESH,1,ADAPT,0.02,0.02,0.02,0.002,0.002,-0.002,+ +,,,,,,,,,+ +,,,,,,,EULER,1,+ +,,,,,,,,,+ +,,,,,,,,,+ +,-0.538,-0.438,-0.242,1.060,0.960,0.880 The dimensions of the static output box are set by the STATBOX fields on the sixth line of the MESH entry. To get appropriate dimensions for the static output box the simulation is first run without the STATBOX option. In this OUT file, there are several messages about adaptive meshing that monitor the largest box that surrounds all adaptive elements so far. The last of this message reads: * *

SMALLEST BOX CONTAINING ALL ELEMENTS: *

* *

NUMBER OF ELEMENTS IN X,Y,Z DIRECTION:

53

44

21

* * POINT OF ORIGIN -0.5380E+00 -0.4380E+00 -0.2420E+00 * * WIDTH OF BOX

0.1060E+01 0.9000E+00 0.4200E+00

* This is used on the STABOX fields. Also the simulation with STATBOX will show this message at the end. Therefore, the box is large enough to contain the complete deployment process.

Main Index

Chapter 6: Occupant Safety 511 Easy Postprocessing with Adaptive Meshing

The coupling surface and inflator are defined by: COUOPT, 1, 10,,,,,,,+ +,CONSTANT, 101325. $ COUPLE,10,25,OUTSIDE,ON,ON, ,10,AIRBAG,+ +,10,,,,,,,,+ +, ,1 $ COUINFL,10,10,82,INFLATR1,82,CONSTANT,.01 INFLATR1,82,1,2,,1.4,,286. The Euler is initialized by: PEULER1,1,,HYDRO, 1,,,,,+ SPHERE, 1,, 0., 0. , 0., 1.E10 TICEUL,1,,,,,,,,+ +,SPHERE,1, 5, 1, 1. TICVAL,1, , DENSITY, 1.500, SIE , 371500. To keep this example straightforward, no resizing has been applied. In this type of simulation, resizing is worthwhile and static output supports it. The mapping between statbox element and adaptive elements is one to one. Therefore, after each resize, the current Euler archive is closed and a new one is opened.

Results All figures below where obtained by post processing just one Euler archive with Patran. The variable VOLUME indicates whether an element is really existing as an adaptive element or as nonexisting (VOLUME=0). The figures with Eulerian output show the first layer of element after x=0.

Main Index

512 Dytran Example Problem Manual Easy Postprocessing with Adaptive Meshing

The figure at cycle 0 of Volume shows that initially a lot of elements have been created by adaptive meshing. Many elements have been deleted as shown in the figure at cycle 219.

Main Index

Chapter 6: Occupant Safety 513 Easy Postprocessing with Adaptive Meshing

Main Index

514 Dytran Example Problem Manual Easy Postprocessing with Adaptive Meshing

Velocities on the fmat=0.5 iso-surface for 20 ms and 60 ms.

Dytran Input File NIF = FORM START Dytran input deck CEND CHECK=NO ENDSTEP=100000 ENDTIME=60.E-3

Main Index

Chapter 6: Occupant Safety 515 Easy Postprocessing with Adaptive Meshing

TITLE=INFLATION OF DEMO PASSENGER-BAG USING EULER TIC=1 SPC=1 TLOAD=1 $ -------------------------------------------------------------$ define the output for the euler elements: $ ELEMENTS(EUL) = 995 SET 995 = ALLEULHYDRO ELOUT(EUL) = PRESSURE,DENSITY,FMAT,XVEL,YVEL,ZVEL,SIE,VOLUME ELOUT(EUL) = XMOM,YMOM,ZMOM TIMES(EUL) = 0. THRU END BY 4.E-3 TYPE(EUL) = ARCHIVE SAVE(EUL) = 1000 $ $ -------------------------------------------------------------$ define the output for the membrane elements: $ ELEMENTS(AIRBAG) = 996 SET 996 = ALLMEMTRIA ELOUT(AIRBAG) = THICK,SMDFER TIMES(AIRBAG) = 0. THRU END BY 4.E-3 TYPE(AIRBAG) = ARCHIVE SAVE(AIRBAG) = 100000 $ $ $ BEGIN BULK $ Initial timestep & minimum timestep $ PARAM,FASTCOUP PARAM,INISTEP,0.7E-05 PARAM,MINSTEP,.5E-8 $ CTRIA3 1 501 1 12 13 CTRIA3 2 501 1 13 2 --------------CTRIA3 3514 505 1248 1242 2053 $ GRID 1 0.0 0.0 0.0 GRID 2 -.03 0.0 0.0 --------------GRID 2145 .15 .0496 -.04 $ $ First quarter $ PSHELL1,501,1,MEMB,,,,,,+ +,5.E-4 PSHELL1,502,1,MEMB,,,,,,+ +,5.E-4 -----------

Main Index

516 Dytran Example Problem Manual Easy Postprocessing with Adaptive Meshing

PSHELL1,513,1,MEMB,,,,,,+ +,5.E-4 $ $ Second quarter (PID first + 50) $ PSHELL1,551,1,MEMB,,,,,,+ +,5.E-4 PSHELL1,552,1,MEMB,,,,,,+ +,5.E-4 -----PSHELL1,563,1,MEMB,,,,,,+ +,5.E-4 $ $ Third quarter (PID first + 100) $ PSHELL1,601,1,MEMB,,,,,,+ +,5.E-4 PSHELL1,602,1,MEMB,,,,,,+ +,5.E-4 -----PSHELL1,613,1,MEMB,,,,,,+ +,5.E-4 $ $ fourth quarter (PID first + 150) $ PSHELL1,651,1,MEMB,,,,,,+ +,5.E-4 PSHELL1,652,1,MEMB,,,,,,+ +,5.E-4 ----------PSHELL1,663,1,MEMB,,,,,,+ +,5.E-4 $ $ Define the inflator as membranes with a SPC to keep it at its $ place. $ PSHELL1,505,1,MEMB,,,,,,+ +,5.E-4 $ $ ------------------------------------------------------------$ define the holes as dummy elements, so they have no strength $ --> No material associated with them either ! $ PSHELL1,510, ,DUMMY PSHELL1,560, ,DUMMY PSHELL1,610, ,DUMMY PSHELL1,660, ,DUMMY $ DMATEL,1,600.,6.E7,0.3 $ DMAT 5 .681864 1 EOSGAM 1 1.4 297.

Main Index

Chapter 6: Occupant Safety 517 Easy Postprocessing with Adaptive Meshing

$ MESH,1,ADAPT,0.02,0.02,0.02,0.002,0.002,-0.002,+ +,,,,,,,,,+ +,,,,,,,EULER,1,+ +,,,,,,,,,+ +,,,,,,,,,+ +,-0.538,-0.438,-0.242,1.060,0.960,0.880 PEULER1,1,,HYDRO, 1,,,,,+ SPHERE, 1,, 0., 0. , 0., 1.E10 TICEUL,1,,,,,,,,+ +,SPHERE,1, 5, 1, 1. TICVAL,1, , DENSITY, 1.500, SIE , 371500. $ $ COUOPT, 1, 10,,,,,,,+ +,CONSTANT, 101325. $ COUPLE,10,25,OUTSIDE,ON,ON, ,10,AIRBAG,+ +,10,,,,,,,,+ +, ,1 $ $ $ COUINFL,10,10,82,INFLATR1,82,CONSTANT,.01 INFLATR1,82,1,2,,1.4,,286. $ TABLED1 1 + + 0.00E+00 0.0 0.50E-02 .15 0.10E-01 3.05 0.15E-01 3.2 + + 0.20E-01 3.25 0.25E-01 3.25 0.50E-01 2.8 0.55E-01 1.75 + + 0.60E-01 1.0 0.65E-01 .5 0.70E-01 0.0 0.10E+00 0.0+ + ENDT TABLED1 2 + + 0.00E+00 500. 0.50E-02 500. 0.10E-01 600. 0.10E+006.000E+0+ + ENDT SPC 1 253 123456 SPC 1 254 123456 ----------SPC 1 2145 123456 $ SURFACE,25,,SUB,42 $ SUBSURF,42,25,PROP,42 SET1,42,501,THRU,513,551,THRU,563,601,+ +,THRU,604,606,THRU,613,651,THRU,663 $ SUBSURF,81,25,PROP,81 SET1,81,510,560,610,660 $ SUBSURF,82,25,PROP,82 SET1,82,505

Main Index

518 Dytran Example Problem Manual Easy Postprocessing with Adaptive Meshing

$ ENDDATA

Main Index

Chapter 7: Quasi-static Analysis Dytran Example Problem Manual

7

Main Index

Quasi-static Analysis J

Overview

J

Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions 521

520

520 Dytran Example Problem Manual Overview

Overview In this chapter, a number of example problems are presented that highlight the capabilities of Dytran in the area of Quasi-static Analysis. The user can find in these examples guidelines to model Quasi-static problems that include comparison of Dytran results with that of MD Nastran. The following example problem is described:

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Chapter 7: Quasi-static Analysis 521 Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions Problem Description A simple curved plate is subjected to constant load on one end and supported on the other end. The motion of the nodes is damped by means of dynamic relaxation (VDAMP) to reach a static deformed state. The purpose of this model is to study the different options of the Dytran shell element formulations and compare it to the solution of MD Nastran.

Figure 7-1

Curved Plate with CQUAD4 Elements

Material of the Plate (M300 Steel) Density

ρ = 8027 Kg/m3

Young’s modulus

E = 2E+11 Pa

Poisson’s ratio

ν = 0.3

Yield stress

σy= 1.86E+9 Pa

Hardening modulus

Eh= 4E+9 Pa

Initial Load Total Force = 76.5 N

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522 Dytran Example Problem Manual Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

Dytran Model A simple mesh is set up (see Figure 7-1). The curved plate, with overall length L = 0.3 m and width W = 0.1 m, is modeled with 800 CQUAD4 elements with a thickness of 2mm. The nodes near the holes of the plate are constrained in all direction. The nodes on the end of the plate are loaded by a constant force, using the combination of TLOAD1, FORCE, and TABLED1 cards. In order to be able to compare the results with MD Natran, the LOBATTO quadrature is used on the PSHELL1 card. This option will give the effective stress output on the face of the shell element. The model is run using both the KEYHOFF and BLT formulation. The analysis is performed in two steps: Step 1: In order to find the most optimal dynamic relaxation value (VDAMP), the first run is meant to capture the oscillation frequency of the nodes, which are subjected to load. The parameter VDAMP can be given in the following algebraic form: 2∗ π∗ δ t V DA MP = ----------------T

In which T denotes the natural period of free vibration and δ t is the time step used during the analysis. The latter represents DLTH in .OUT file and can be found constant throughout the analysis. The period T can be determined from the displacement versus time. Step 2: In the second, by defining the parameter VDAMP found using the first step, damp out the oscillations in the model.

Results The correct solution of the undamped model is an oscillation with constant amplitude and frequency, around an equilibrium state. Figure 7-2 shows the displacement of the nodes at the free end of the plate for both the BLT and KEYHOFF shell formulations. The KEYHOFF shell element shows the correct behavior, while the BLT shell diverges over time. The reason for this is that the BLT element formulation does not capture a twisting deformation mode.

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Chapter 7: Quasi-static Analysis 523 Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

Figure 7-3 shows the displacement of the nodes at the free end of the plate for the KEYHOFF model, with

and without damping.

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Figure 7-2

Resultant Displacement Time History of the Free GridPoint using BLT and KEYHOFF Shell Formulation

Figure 7-3

Resultant Displacement Time History of the Free Grid Point using KEYHOFF Shell Formulation (Damped and Undamped Models)

524 Dytran Example Problem Manual Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

MD Nastran Results MD Natran’s linear static solution (SOL 101) and the nonlinear static solution (SOL 106) give very similar results. See the table below. Maximum Displacement

Maximum Stress

SOL 101 (linear statics)

0.0203 m

3.42E8 Pa

SOL 106 (nonlinear statics)

0.0202 m

3.43E8 Pa

It is important to note that MD Nastran by default computes the stress results on the face of the elements. The load increment is done in steps of five for the nonlinear static analysis.

Dytran Results When shell elements undergo a large rigid body rotation, it is important to activate the rigid body rotation correction in the hourglass control (RBRCOR on the HGSUPPR card). The table below shows the results obtained for the damped model, using the KEYHOFF shell formulation, with and without the rigid body rotation correction. It can be seen that the result of Dytran compares well with MD Nastran. When the rigid body rotation correction is active, the displacements are identical, while the Maximum Stress differs by only 3.4% . Maximum Displacement (*) KEYHOFF without RBRCOR 0.0199 m (1.9%)

3.11E8 Pa (9.0%)

KEYHOFF with RBRCOR

3.55E8 Pa (3.4%)

0.0202 m. (0.0%)

(*) = difference with MD Nastran SOL 106 solution

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Maximum Stress (*)

Chapter 7: Quasi-static Analysis 525 Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

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Figure 7-4

Fringe Plots from MD Nastran Results

Figure 7-5

Fringe Plot for Dytran Results

526 Dytran Example Problem Manual Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

Files plate.dat

Dytran input file

PLATE.OUT

Dytran output file

PLATE_ELEM_0.ARC

Dytran archive file

PLATE_GRID_0.THS

Dytran time history file

Abbreviated Dytran Input File START CEND ENDTIME=0.070 ENDSTEP=999999 CHECK=NO TITLE= Jobname is: dytranjob TLOAD=1 TIC=1 SPC=1 $ Output result for request: elements TYPE (elem) = ARCHIVE ELEMENTS (elem) = 1 SET 1 = 1 THRU 800 ELOUT (elem) = EFFSTS TIMES (elem) = 0 thru end by 0.0005 SAVE (elem) = 10000 $ TYPE (grid) = TIMEHIS GRIDS (grid) = 10 SET 10 = 977 713 449 GPOUT (grid) = XDIS YDIS ZDIS XFORCE RVEL RDIS RFORCE TIMES (grid) = 0 thru end by 0.0005 SAVE (grid) = 10000 $ $------- Parameter Section -----PARAM,CONTACT,THICK,0.0 PARAM,INISTEP,1e-7 PARAM,VDAMP,0.0001 $------- BULK DATA SECTION ------BEGIN BULK $ --- SPC-name = nodal_constraints SPC1 1 123456 11 14 15 16 42 43+A000001 $ $ --- Define 889 grid points --$ GRID 1 .100000.0750000 .00000 GRID 2 .100000 .100000 .00000 : GRID 1008 .106455 .00000.0005027 $ $ --- Define 800 elements

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Chapter 7: Quasi-static Analysis 527 Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

$ $ -------- property set prop --------CQUAD4 1 1 7 19 5 2 CQUAD4 2 1 8 20 19 7 : CQUAD4 800 1 975 229 226 1008 $ $ ========== PROPERTY SETS ========== $ * prop * $ PSHELL1 1 1 KEYHOFF LOBATTO +A000001 .0020 $ HGSUPPR,1,SHELL,1,,,,,,+ +,YES $ CORD2R 1 0 .3 0 .05-.617356 0-.348067 +A000007 .3 1 .05 $ $ ========= MATERIAL DEFINITIONS ========== $ $ -------- Material steel1 id =1 DMATEP 1 8027 2e+11 .3 1 YLDVM 11.86e+09 4e+09 +A000008 $ $ ======== Load Cases ======================== $ ------- Force BC nodal_force ----TLOAD1 1 9 0 2 FORCE 9 449 0 1 1.7913 0 -4.1281 : FORCE 9 977 0 1 1.7913 0 -4.1281 $ $ ================ TABLES ================= $ ------- TABLE 2: force_table ------TABLED1 2 +A000009 +A000009 0 0 .003 1 .07 1 ENDT ENDDATA

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+

+A000007

+A000008

528 Dytran Example Problem Manual Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

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