Dytran Example Problem Manual

Dytran™ 2008 r1

Example Problem Manual

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C o n t e n t sDytran Example Problems Manual

Contents

1 Structural Dynamics

Overview 14

Impulsively Loaded Strip 15Problem Description 15Dytran Model 15Results Evaluation 16Files 17Abbreviated Dytran Input File 18

Impulsively Loaded Cylindrical Panel 20Problem Description 20Dytran Model 21Results 21Files 24Abbreviated Dytran Input File 24

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

Taylor Test with Euler 36Problem Description 36Dytran Model 36Dytran File 39

Hourglassing in Hexahedron Elements 41Problem Description 41Dytran Model 42

Dytran Example Problems Manual4

Results 43Files 45Abbreviated Dytran Input File 46

2 Structural Contact

Overview 50

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

Ball Penetrating a Steel Plate 58Problem Description 58Desired Results 60Dytran Modeling 61Results 61Files 63Abbreviated Dytran Input File 64

Tapered Beam Striking a Rigid Wall 66Problem Description 66Desired Results 66Dytran Modeling 66Results 67Files 70Abbreviated Dytran Input File 70

Impact Loading 73Problem Description 73Theoretical Results 73Dytran Model 74Results 74Files 75Abbreviated Dytran Input File 75

Pipe Whip 79Problem Description 79Dytran Model 80Results 81

5Contents

Files 82Abbreviated Dytran Input File 82

3 Fluid Dynamics

Overview 86

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

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

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

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


Modeling the JWL Explosion using 1-D Spherical Symmtery 125Problem Description 125Dytran Modeling 125Results 127Abbreviated Dytran Input File 134

Nonuniformity with MESH,BOX 137Problem Description 137Dytran Modeling 137Results 141Abbreviated Dytran Input File 143

4 Fluid-structure Interaction

Overview 147

Shock Formation 148Problem Description 148Theoretical Background 148Dytran Model 149Results 150Files 152Abbreviated Dytran Input File 152

Blast Containment in a Luggage Container 154Problem Description 154Dytran Modeling 155Results 155Files 157References 157Abbreviated Dytran Input File 157

Multiple Bird-strike on a Cylindrical Panel 161Problem Description 161Dytran Model 162Results 164Files 166Abbreviated Dytran Input File 166

Slanted Piston 169Problem Description 169Desired Results 170Dytran Modeling 171Results 171

7Contents

Files 173Abbreviated Dytran Input File 173

Sloshing using ALE Method 176Problem Description 176Dytran Model 177User Subroutine 177Results 178Files 179References 179Abbreviated Dytran Input File 180

Flow between Two Containers or Airbags 183Problem Description 183Theoretical Analysis of the 2-Vessel Flow Problem 183Dytran Model 188Results 190Abbreviated Dytran Input File 190

Blastwave Hitting a Bunker 193Problem Description 193Dytran Modeling 193Results 196

Mine Blast 197Problem Description 197Dytran Model 197Results 203Abbreviated Dytran Input File 204

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

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

Problem Description 216Dytran Model 218Results 221Abbreviated Dytran Input File 222


Fuel Tank Filling 224Problem Description 224Dytran Model 224Results 231Abbreviated Dytran Input File 233

Water Pouring into a Glass 237Problem Description 237Dytran Modeling 237Results 241Abbreviated Dytran Input File 242

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

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

Blast Wave with a Graded Mesh 273Problem Description 273Dytran Model 274Results 276Abbreviated Dytran Input File 281

Bubble Collapse with Hydrostatic Boundary Conditions 284Dytran Model 284General Setup 285Results 288Dytran Input files 290

Prestressed Concrete Beam 294Problem Description 294Analysis Scheme 295Dytran Model 297Results 298Abbreviated Dytran Input File 302

9Contents

Blast Simulation on Prestressed Concrete Beam 306Problem Description 306Analysis Scheme 306Dytran Model 307Results 309Abbreviated Dytran Input File 311

Vortex Shedding with Skin Friction 317Problem Description 317Dytran Model 318Results 320Abbreviated Dytran Input File 322

Geometric Eulerian Boundary Conditions 328Problem Description 328Dytran Model 328Results 332

Cohesive Friction 337Problem Description 337Dytran Modeling 337Results 340Dytran Input Deck 341

5 Forming

Overview 346

Square Cup Deep Drawing 347Problem Description 347Dytran Model 349Results 351Files 354Abbreviated Dytran Input File 355

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


Three-point Bending Test 368Problem Description 368Dytran Model 369Results 371Files 374Abbreviated Dytran Input File 375

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

6 Occupant Safety

Overview 390

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

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

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

11Contents

Appendix D: Properties and Materials of Hybrid III Dummy 455Appendix E: Groups created by Session File for Positioning within MD

Patran 458

Side Curtain Air Bag (Courtesy of Autoliv) 462Problem Description 462Dytran Model 464Results 469Input Deck 471

Hybrid III 5th%-tile Dummy 479Problem Description 479Model Description 479Conclusions and Recommendations 491Input Files 492Appendix A: Overview of the Hybrid III 5th%-tile Dummy 500Appendix B: SURFACE and SET1 Definition 504Appendix C: hyb305 Positioning 505Reference Database 508

Easy Postprocessing with Adaptive Meshing 509Problem Description 509Dytran Modeling 510Results 511Dytran Input File 514

7 Quasi-static Analysis

Overview 520

Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions 521

Problem Description 521Dytran Model 522Results 522MD Nastran Results 524Dytran Results 524Files 526Abbreviated Dytran Input File 526

Chapter 1: Structural Dynamics Dytran Example Problem Manual

1Structural Dynamics

Overview 14

Impulsively Loaded Strip 15

Impulsively Loaded Cylindrical Panel 20

Taylor Test (a rod impacted against a rigid wall) 26

Taylor Test with Euler 36

Hourglassing in Hexahedron Elements 41

Dytran Example Problem ManualOverview

14

OverviewIn 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.

15Chapter 1: Structural DynamicsImpulsively Loaded Strip

Impulsively Loaded Strip

Problem DescriptionAn 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:

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 ModelDue to the symmetry of the problem, only half of the strip needs to be modeled (see Figure 1-1). The left-hand 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.

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 υ = 0.3

yield stress σy = 2.854 108 N/m2

hardening modulus Eh = 0. N/m2

Dytran Example Problem ManualImpulsively Loaded Strip

16

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 EvaluationCalculations 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.]).


Figure 1-2 Vertical Z-displacement Time History of the (Free) Corner Grid Point (31) of the Half Strip

Files

Reference1. 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.

impulse_a.datimpulse_b.dat

Dytran input files

IMPULSE_A.OUTIMPULSE_B.OUT

Dytran output files

IMPULSE_A_ZDIS_0.THSIMPULSE_B_ZDIS_0.THS

Dytran time history files

Dytran Example Problem ManualImpulsively Loaded Strip

18

Abbreviated Dytran Input File

STARTCENDTITLE = IMPULSIVE LOADING OF A STRIP OF ALUMINUM (Units = m-Kg-N-sec)CHECK = NO ENDTIME = 0.001TIC = 1SPC = 1$$Data for Output Control Set 1$TYPE(zdis) = TIMEHISSAVE(zdis) = 99999GPOUT(zdis) = ZDISGRIDS(zdis) = 1SET 1 = 31TIMES(zdis) = 0tEndb1e-05$$------------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 $


$Material Properties $-------------------PSHELL1 1 1 Bely Gauss 5 Mid ++ .00318 $YLDVM 1 2.854+8 0.0 $DMATEP 1 2791. 7.17+10 .01 1 $$Initial Conditions$--------------------$TIC1 1 3 -132. 31 THRU 124 BY + + 31 30 THRU 123 BY 31 29 THRU + + 122 BY 31TIC1 1 3 -116. 28 THRU 121 BY + + 31 TIC1 1 3 -100. 25 THRU 118 BY + + 31 26 THRU 119 BY 31 27 THRU + + 120 BY 31ENDDATA

Dytran Example Problem ManualImpulsively Loaded Cylindrical Panel

20

Impulsively Loaded Cylindrical Panel

Problem DescriptionA 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 simply-supported 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


yield stress σy = 4.4 x 104 psi

21Chapter 1: Structural DynamicsImpulsively Loaded Cylindrical Panel

Size of the Panel

Initial Load

Dytran ModelBecause 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.

ResultsA 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.])

thickness t = 0.125 in

radius R = 2.94 in

length L = 12.56 in

velocity v0 = 5650 in/sec


22

.

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.


.

Figure 1-5 Deformation and Effective Plastic Strain of the Cylindrical Panel


24

Files

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.


$$ A 120 degree cylindrical panel is loaded impulsively. It is simply$ supported at the ends. The model consists of half the panel, with 12x32$ shell elements. The loading is generated by imposing initial velocities$ on the grid points.$STARTCENDTITLE = Cylindrical Panel (12x32 - B-T or Keyhoff Shell 5 G-Pts)CHECK = NOENDSTEP = 10000ENDTIME = 1.e-3TIC = 1SPC = 1$$$$ Data for output control Set 1$$$$ Time-History plot of the y-displacement of the midpoint along the crown$$ line of the cylinder (grid 209).$$TYPE(TH_G_24) = TIMEHISGPOUT(TH_G_24) = YDISSET 1 = 209GRIDS(TH_G_24) = 1TIMES(TH_G_24) = 0 THRU END BY 1.e-5

cylpan.datpanel_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


SAVE(TH_G_24) = 10000$$$$ Data for output control Set 15$$TYPE(PANEL) = ARCHIVEELEMENTS(PANEL) = 15SET 15 = 1 THRU 384ELOUT(PANEL) = EFFPL03,TXX03,TYY03,TXY03,TYZ03,TZX03ELOUT(PANEL) = EFFST03,EXX03,EYY03,EXY03TIMES(PANEL) = 0 THRU END BY 1e-4SAVE(PANEL) = 10000$BEGIN BULK$$ Definition of some parameters.$ -----------------------------PARAM INISTEP 1.e-6PARAM STEPFCT 0.9PARAM SHPLAST VECTPARAM HGCOEFF 0.1PARAM SHTHICK YES$$ Model geometry and boundary constrain.$ --------------------------------------INCLUDE panel_xl.dat$$ Shell Properties for the panel(B-T or Keyhoff).$ -----------------------------------------------$PSHELL1 1 1 keyhoff GAUSS 5 1.0 MID +CONTPSHELL1 1 1 bely GAUSS 5 1.0 MID +CONT+CONT .125$$ Material and yield model.$ -------------------------DMATEP 1 .00025 1.05e7 .33 10$YLDVM 10 44000 0$$ Initial condition for gridpoints.$ ---------------------------------CORD2C 1 0.0 0.0 0.0 0.0 0.0 -12.56 ++ 0.0 2.94 -6.28TICGP 1 1 CID1 1 XVEL -5650SET1 1 14 THRU 339 BY 13 15 THRU ++ 340 BY 13 16 THRU 341 BY 13 ++ 17 THRU 342 BY 13 18 THRU 343 ++ BY 13 19 THRU 344 BY 13 20 ++ THRU 345 BY 13$ENDDATA

Dytran Example Problem ManualTaylor Test (a rod impacted against a rigid wall)

26

Taylor Test (a rod impacted against a rigid wall)

Problem DescriptionThe 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 ModelThis 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 – TaylorTaylor developed a simple expression for the final length of the bar, , as function of the initial length

, the impact velocity , the yield strength , and the density .

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 Schematic Overview of the Taylor Impact Test

Lf

L0 V σy ρ

LfL0 ρV2

– 2σy⁄( )exp=

27Chapter 1: Structural DynamicsTaylor 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 ModelA 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 one-point Gauss integration scheme.

Figure 1-7 Dytran CHEXA Model – Impact of Cylinder - Quarter Model


28

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 Comparison of Dytran CHEXA Elements


Tetrahedral ElementsIn 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.

Figure 1-10 Dytran CTETRA Model – Impact of Cylinder - Quarter Model

Figure 1-11 Comparison of Different Element Types in Dytran


30

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 – TaylorIn Figure 1-9, Figure 1-11, and Figure 1-12, the theoretical reduction in length according to Taylor is plotted:

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

Lf L0⁄ 0.78=


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

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

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 timeFigure 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.


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

References5. 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


34


STARTCENDENDTIME=0.06E-3ENDSTEP=9999999CHECK=NOTITLE= Jobname is: Hexs-EPM-default-JCTLOAD=1TIC=1SPC=1$ Output result for request: Cylinder_ElemTYPE (Cylinder_Elem) = ARCHIVEELEMENTS (Cylinder_Elem) = 1SET 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-4SAVE (Cylinder_Elem) = 10000$ Output result for request: Cylinder_GridTYPE (Cylinder_Grid) = ARCHIVEGRIDS (Cylinder_Grid) = 2SET 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-4SAVE (Cylinder_Grid) = 10000$ Output result for request: GridTHSTYPE (GridTHS) = TIMEHISGRIDS (GridTHS) = 3SET 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-5SAVE (GridTHS) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,5E-9$------- BULK DATA SECTION -------BEGIN BULKINCLUDE 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 $


$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material Cylinder_VM id =2DMAT 28.96e-09 2 2 2 EOSPOL 2 143000SHREL 2 47700YLDVM 2 700$$ -------- Material Cylinder_JC id =3DMAT 38.96e-09 3 3 3 EOSPOL 3 143000SHREL 3 47700YLDJC 3 175 380 .32 .06 55 14.52e+08+A000002+A000002 1811 293$$ ======== Load Cases ========================$$$ ------- Initial Velocity BC InitialVelocity ----- SET1 4 1 THRU 1768TICGP 1 4 ZVEL -197000$$ ------- Rigid Plane BC RigidWall ----- WALL 9 0 0 0 0 0 1 5+A000004+A000004 PENALTYSET1 5 1 THRU 1768$$ENDDATA

Dytran Example Problem ManualTaylor Test with Euler

36

Taylor Test with EulerFor 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 DescriptionThe 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.

37Chapter 1: Structural DynamicsTaylor 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 mesh-size 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,4PEULER1,1,, MMSTREN,4

The material is defined by

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


38

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,,0.0,0.0,0.0,100.0BOX,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

39Chapter 1: Structural DynamicsTaylor 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

STARTMEMORY-SIZE = 6500000,6500000CENDENDTIME=7.e-5CHECK=NOTITLE= Jobname is: taylor-eulerTLOAD=1


40

TIC=1SPC=1$ Output result for request: eulerTYPE (euler) = ARCHIVEELEMENTS (euler) = 1SET 1 = ALLELEMENTS ELOUT (euler) = DENSITY TEMPTURE EFFSTS , VOID XVEL YVEL ZVEL MASS PRESSURE, FMAT EDIS EFFPLS TIMES (euler) = 0,THRU,END,BY,6.e-6SAVE (euler) = 10000$------- Parameter Section ------$PARAM,INISTEP,5e-9PARAM,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 =1DMAT,1,8.96e-9,1,1,1,,1PMINC,1,-1.e+20EOSPOL,1,143000SHREL,1,47700YLDJC ,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,,0.0,0.0,0.0,100.0BOX,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,1ENDDATA

41Chapter 1: Structural DynamicsHourglassing in Hexahedron Elements

Hourglassing in Hexahedron Elements

Problem DescriptionHourglass 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.

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.

Table 1-1 Summary of Analyses Settings

Case Element Integration MethodHourglass Control

MethodHourglass Control

Coefficient

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

Dytran Example Problem ManualHourglassing in Hexahedron Elements

42

Dytran ModelThe 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


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

ResultsThe 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 Case 1: 1-Point Integration Hexa, Hourglassing to a Small Extent in the Bottom and Top Rows

Note: Displacements are scaled up.


44

Figure 1-27 Case 2: 2x2x2 Fully Integrated Hexa, No Hourglassing

Figure 1-28 Case 3: 1-Point Integration Hexa with Lower Stiffness Damping Coefficient (0.01)




Figure 1-29 Case 4: 1-Point Integration Hexa with Viscous Hourglassing Suppression, Hourglassing very Clearly Present

Files

References7. Dytran reference manual, chapter 5, Bulk data description of Hourglass suppression modes

HGSUPPR

8. LS-Dyna version 970 Keyword's User Manual, *HOURGLASS


Case 1: EPM15-default.dat

Case 2: EPM15-2x2x2.dat

Case 3: EPM15-lowcoef.dat

Case 4: EPM15-hg-dyna.dat


46


STARTCENDENDTIME=0.1ENDSTEP=9999999CHECK=NOTITLE= Jobname is: Comparison-isoTLOAD=1TIC=1SPC=1$ Output result for request: elemsTYPE (elems) = ARCHIVEELEMENTS (elems) = 1SET 1 = 1 THRU 64 ELOUT (elems) = TXX TYY TZZ TXY TYZ TZX EFFSTS EFFPLS SIETIMES (elems) = 0 THRU END BY 0.05SAVE (elems) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1e-8$$ Case 1: no extra parameters$$ Case 2: no extra parameters$$ Case 3: $ PARAM,HGCSOL,0.01$$ Case 4:$ PARAM,HGSOLID,DYNA$$------- BULK DATA SECTION -------BEGIN BULK$ --- SPC-name = ClampSPC1 1 123 1 9 82 90$$ --- 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 10 82 83+A000001+A000001 92 91… .CHEXA 64 1 71 72 81 80 152 153+A000064+A000064 162 161$$ ========== PROPERTY SETS ========== $


$ 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 =1DMAT 1 .0012 1 1EOSPOL 1 23000SHREL 1 7500$$ ======== Load Cases ========================$$$ ------- Force BC Force ----- TLOAD1 1 5 0 1FORCE 5 77 0 1 0 -200 0FORCE 5 158 0 1 0 -200 0$$ ================ TABLES =================$$ ------- TABLE 1: Table -------TABLED1 1 +A000065+A000065 0 0 .1 1 ENDT$ENDDATA


48

Chapter 2: Structural ContactDytran Example Problem Manual

2Structural Contact

Overview 50

Three-plate Contact 51

Ball Penetrating a Steel Plate 58

Tapered Beam Striking a Rigid Wall 66

Impact Loading 73

Pipe Whip 79


50

OverviewIn 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.

51Chapter 2: Structural ContactThree-plate Contact

Three-plate Contact

Problem DescriptionPlates 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:

• Material of the plates (steel):

width in y-direction a = 2 m

width in z-direction b = 2 m



Young’s modulus E = 2.1 1011 N/mm2


Dytran Example Problem ManualThree-plate Contact

52

• Initial conditions:

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

Theoretical ResultThe 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 ModelThe 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:

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.

ResultsWhen 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

distance d = 0.005 m

velocity v0 = 15 m/sec

Master Slave Single Surface

CASE 1 Contact 1Contact 2

23

12

CASE 2 Contact 1Contact 2

2 12,3

CASE 3 Contact 1Contact 3Contact 2

21

12

2,3


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 Two Master-slave Contacts Result in a Symmetric Behavior


54

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


Files


TITLE = Three Plate ContactENDSTEP = 40TIC = 1$$ --------- Requesting the Position and Velocity of one $TYPE(grxpv) = TIMEHISSAVE(grxpv) = 9999GRIDS(grxpv) = 1SET 1 = 12t62b25STEPS(grxpv) = 1t100b1GPOUT(grxpv) = XPOS,XVEL$$ --------- Contact distance output vars$TYPE(cont_dis) = TIMEHISCONTS(cont_dis) = 2SET 2 = ALLCONTACTSCONTOUT(cont_dis) = DMIN,XFORCESTEPS(cont_dis) = 0,THRU,END,BY,1SAVE(cont_dis) = 9999$BEGIN BULKPARAM 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.

pl3contact1.datpl3contact2.datpl3contact3.dat

Dytran input file for the three cases

PL3CONTACT1.OUTPL3CONTACT2.OUTPL3CONTACT3.OUT

Dytran output files for the three cases

PL3CONTACT1_GRXPV_1.THSPL3CONTACT2_GRXPV_1.THSPL3CONTACT3_GRXPV_1.THS

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

PL3CONTACT1_CONT_DIS_0.THSPL3CONTACT2_CONT_DIS_0.THSPL3CONTACT3_CONT_DIS_0.THS

Dytran time history files for contact output for the three cases


56

.

.

.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 6CQUAD4 2 1 2 3 8 7CQUAD4 3 1 3 4 9 8CQUAD4 4 1 4 5 10 9.. CQUAD4 45 1 66 67 72 71CQUAD4 46 1 67 68 73 72CQUAD4 47 1 68 69 74 73CQUAD4 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 1CFACE 2 1 2 1CFACE 3 1 3 1CFACE 4 1 4 1...$PSHELL 1 1 .005$DMATEP 1 7800. 2.1+11 .3$SURFACE 1 SEG 1SURFACE 2 SEG 2SURFACE 3 SEG 3 $


CONTACT 1 SURF SURF 1 2 0.0 0.0 0.0 ++ Both Full ++ 0.0 .1 Distance1.+20 1.1 .1 Distance0.002 ++ 0.0 1.+20 On On 1.+20 CONTACT 2 SURF SURF 2 3 0.0 0.0 0.0 ++ Both Full ++ 0.0 .1 Distance1.+20 1.1 .1 Distance0.002 ++ 0.0 1.+20 On On 1.+20 ENDDATA

Dytran Example Problem ManualBall Penetrating a Steel Plate

58

Ball Penetrating a Steel Plate

Problem DescriptionConsider 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

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.

εp1 εp2 εp3 εp4+ + +( ) εp> 0.5=

59Chapter 2: Structural ContactBall Penetrating a Steel Plate

Plate

Figure 2-6 Ball Penetrating Plate using BJOINS

material steel

Young’s modulus E = 2. 1011 N/m2


yield stress σ = 4. 108 N/m2

percent of elongation 50%


length L = 0.3 m

width W = 0.3 m


60

Figure 2-7 Ball Penetrating Plate without using BJOINS

Ball

Desired ResultsThe 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.

material steel (rigid body)


mass m = 2.617 kg

radius R = 0.043 m

velocity v = 230 m/sec


Dytran ModelingThe 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.

ResultsFigure 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.


62

Figure 2-8 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


Files

Reference1. Gere, James M., and Timoshenko, Stephen P., Mechanics of Materials, 1985, Wadsworth

International.

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.OUTBALL_PLATE_ADAP.OUT

Dytran output files

BALL_PLATE_BALL_0.ARCBALL_PLATE_PLATE_0.ARC

Dytran archive files (BJOIN case)

BALL_PLATE_ADAP_BALL_0.ARCBALL_PLATE_ADAP_PLATE_0.ARC

Dytran archive files (eroding element)

BALL_PLATE_RIGID_BALL_0.THSBALL_PLATE_ENERGY_0.THS

Dytran time history files (BJOIN case)

BALL_PLATE_ADAP_RIGID_BALL_0.THSBALL_PLATE_ADAP_ENERGY_0.THS

Dytran time history files (eroding element)

exbrk.f User subroutine used for BJOIN case


64


STARTCENDTITLE = Ball penetrating a steel plateCHECK = NOENDSTEP = 1100TLOAD = 1TIC = 1SPC = 1$$$$ Data for output control Set 1$$TYPE (Ball) = ARCHIVEELEMENTS (Ball) = 8SET 8 = ALLDUMQUADELOUT (Ball) = ZUSERSTEPS (Ball) = 0,THRU,END,BY,100SAVE (Ball) = 10000$$$$ Data for output control Set 2$$TYPE (Plate) = ARCHIVEELEMENTS (Plate) = 7SET 7 = ALLSHQUADELOUT (Plate) = EFFPL-MID,EFFST-MIDSTEPS (Plate) = 0,THRU,END,BY,100SAVE (Plate) = 10000$$$$ Data for output control Set 3$$TYPE (Rigid_Ball) = TimeHisRIGIDS (Rigid_Ball) = 9SET 9 = 200RBOUT (Rigid_Ball) = XCG, YCG, ZCG, XVEL, YVEL, ZVELSTEPS (Rigid_Ball) = 0 THRU END BY 2SAVE(Rigid_Ball) = 10000$$$$ Data for output control Set 4$$TYPE (Energy) = TimeHisMATS (Energy) = 12SET 12 = 100MATOUT (Energy) = EKIN, EINT, EDIS STEPS (Energy) = 0 THRU END BY 2SAVE(Energy) = 10000$$BEGIN BULKSETTING,1,VERSION2PARAM,INISTEP,1.E-7PARAM,BULKL,0.06PARAM,BULKQ,1.44$


$ 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 $$ Contact surface on the plate.$ -----------------------------SURFACE,1,,ELEM,111SET1,111,1,THRU,900$$ 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

Dytran Example Problem ManualTapered Beam Striking a Rigid Wall

66

Tapered Beam Striking a Rigid Wall

Problem DescriptionIn 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

Desired ResultsThe 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 ModelingThe 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

material steel


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

67Chapter 2: Structural ContactTapered 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).

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

Figure 2-9 Moving Beam - Effective Stress Contour of the Deformed Beam after 1200 Cycles


68

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


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

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


STARTCENDTITLE = Moving beam strikes fixed wall.

tapered_beam1.dattapered_beam2.dattapered_beam_xl.dat

Dytran input files (case 1)

Dytran input files (case 2)

TAPERED_BEAM1.OUTTAPERED_BEAM2.OUT

Dytran output files

TAPERED_BEAM1_BEAM_0.ARCTAPERED_BEAM1_PLATE_0.ARC

Dytran archive files (BJOIN case)

TAPERED_BEAM2_BEAM_0.ARCTAPERED_BEAM2_PLATE_0.ARC

Dytran archive files (eroding element)

TAPERED_BEAM1_VEL_0.THSTAPERED_BEAM1_ENERGY_0.THS

Dytran time history files (BJOIN case)

TAPERED_BEAM2_VEL_0.THSTAPERED_BEAM2_ENERGY_0.THS

Dytran time history files (eroding element)


CHECK = NOENDSTEP = 1200TLOAD = 1TIC = 1SPC = 1$$$$ Data for output control Set 1$$TYPE (BEAM) = ARCHIVEELEMENTS (BEAM) = 8SET 8 = 1t80ELOUT (BEAM) = EFFSTS,PRESSURE,EFFPLSSTEPS (BEAM) = 0,THRU,END,BY,50SAVE (BEAM) = 10000$$$$ Data for output control Set 2$$TYPE (PLATE) = ARCHIVEELEMENTS (PLATE) = 17SET 17 = ALLDUMQUADELOUT (PLATE) = ZUSERSTEPS (PLATE) = 0,THRU,END,BY,50SAVE (PLATE) = 10000$$$$ Data for output control Set 3$$TYPE (VEL) = TIMEHISGRIDS (VEL) = 9SET 9 = 229GPOUT (VEL) = XVEL,YVEL,ZVEL,ZFORCESTEPS (VEL) = 0,THRU,END,BY,2SAVE (VEL) = 10000$$$$ Data for output control Set 4$$TYPE (ENERGY) = TIMEHISMATS (ENERGY) = 13SET 13 = 1MATOUT (ENERGY) = EKIN, EINT, EDIS,XMOM,YMOM,ZMOM STEPS (ENERGY) = 0,THRU,END,BY,2SAVE (ENERGY) = 10000$BEGIN BULKSETTING,1,VERSION2PARAM,INISTEP,1.E-7$$ Model geometry.$ All entities are measured in cm.$ --------------------------------INCLUDE tapered_beam_xl.dat$$ Define plate.$ -------------PSHELL1,2,,DUMMY


72

$$ A surface is needed to define the plate as a rigid body and$ to define a Lagrangian contact surface.$ ---------------------------------------SURFACE,2,,SEG,2$$ 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

73Chapter 2: Structural ContactImpact Loading

Impact Loading

Problem DescriptionA 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:

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 ResultsThe 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

The fixed velocity of the distortional pulse is given by

The time required by the pulse to traverse the entire length of the bar is given by

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

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.

Ppulse Aν0 Eρ 2,549 lb= =

c E ρ⁄ 2000,322 in/s= =

tLc--- 0.03 ms= =

PmaxEAm

L------------ xν0 7967 lb= =

Dytran Example Problem ManualImpact Loading

74

Dytran ModelDue 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

ResultsThe 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.


Figure 2-16 Plot for Axial Force at the Center of the Bar

Files

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).


STARTCENDENDTIME=0.0007ENDSTEP=999999CHECK=NOTITLE= Jobname is: barTLOAD=1TIC=1SPC=1$$Output request for the stress distribution in the elastic bar$TYPE (element) = ARCHIVE

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


76

ELEMENTS (element) = 1SET 1 = 1 THRU 40 ELOUT (element) = EFFSTSSTEPS (element) = 0 thru end by 20SAVE (element) = 10000$$Output request for the forces on the cross-sections of the bar$TYPE (sec) = TIMEHISCSECS (sec) = 5SET 5 = 1 2 3CSOUT (sec) = XFORCE YFORCE ZFORCE FMAGNSTEPS (sec) = 0 thru end by 1SAVE (sec) = 10000$$$------- Parameter Section ------$PARAM,INISTEP,1e-7PARAM,CONTACT,VERSION,V4PARAM,CONTACT,THICK,0.0PARAM,CONTACT,GAP,0.0$$------- BULK DATA SECTION -------$BEGIN BULK$SECTION 1 99 999SET1 99 161 THRU 164SET1 999 40$SECTION 2 88 888SET1 88 81 THRU 84SET1 888 20$SECTION 3 77 777SET1 77 1 THRU 4SET1 777 1 $$$ --- 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$$


$ -------- property set rigid ---------$CHEXA 41 1 165 166 168 167 169 170+A000049+A000049 172 171$$

$ -------- 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 =1DMATEL 2 0.000262 1.05e+007 0.33+A000090+A000090 $$ -------- Material rigid id =2MATRIG 1 .00311$$ ======== Load Cases ========================$$$ ------- Initial Velocity BC initial_velocity -----$SET1 1 165 THRU 172TICGP 1 1 ZVEL -240$$ -------- Contact : contact$$CONTACT 3 SURF SURF 1 2 +A000091+A000091 V4 BOTH SLAVE+A000092+A000092 +A000093+A000093 ON$$ Slave contact surface for contact$$


78

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

79Chapter 2: Structural ContactPipe Whip

Pipe Whip

Problem DescriptionA 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:

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.

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

Yield Stress σy = 3.1E11 Pa *

*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.

Dytran Example Problem ManualPipe Whip

80

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 ModelThe 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


ResultsFrom 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 Kinetic Energy of Rotating and Stationary Pipe


82

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

.

Figure 2-19 Deformed Shape at 3 msec.

Files


STARTCENDENDTIME=0.3E-2ENDSTEP=9999999CHECK=NOTITLE= Jobname is: Pipe_WhipTLOAD=1TIC=1SPC=1$ Output result for request: LagTYPE (Lag) = ARCHIVEELEMENTS (Lag) = 1

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


SET 1 = 1 THRU 7896 ELOUT (Lag) = EFFST-MID TIMES (Lag) = 0 THRU END BY 0.3E-3SAVE (Lag) = 10000$ Output result for request: MatTYPE (Mat) = TIMEHISMATS (Mat) = 2SET 2 = 1 2 MATOUT (Mat) = EKIN EINT EDIS STEPS (Mat) = 0 THRU END BY 10SAVE (Mat) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1.E-8PARAM,LIMCUB,5500PARAM,MINSTEP,1.E-15$------- BULK DATA SECTION -------BEGIN BULKINCLUDE 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 =1DMATEP 1 78272.07e+11 .3 1 YLDVM 1 3.1e+11$$ -------- Material m2 id =2DMATEP 2 78272.07e+11 .3 2 YLDVM 2 3.1e+11$$ ======== Load Cases ========================$$$ -------- Contact : Contact1$CONTACT 5 SURF SURF 1 2 +A000001+A000001 V4 TOP $


84

$ Slave contact surface for Contact1$SURFACE 1 SUB 1SUBSURF 1 1 ELEM 3SET1 3 3988 THRU 4011 4023 THRU 4046 4058+A000002+A000002 THRU 4081 4093 THRU 4116 4128 THRU 4151+A000003...$$ Master contact surface for Contact1$SURFACE 2 SUB 2SUBSURF 2 2 ELEM 4SET1 4 73 THRU 98 115 THRU 140 157+A000043+A000043 THRU 182 199 THRU 224 241 THRU 266+A000044...$$ -------- Contact : Contact2$CONTACT 6 SURF 3 +A000084+A000084 V4 BOTH$$ Self Contact Surface for Contact2$SURFACE 3 SUB 3SUBSURF 3 3 ELEM 5SET1 5 73 THRU 98 115 THRU 140 157+A000085+A000085 THRU 182 199 THRU 224 241 THRU 266+A000086...$$ ------- Initial Rotational Velocity: Inivel ----- TIC3 1 1 1 +A000126+A000126 0 0 0 0 200 0 +A000127...$$ENDDATA

Chapter 3: Fluid DynamicsDytran Example Problem Manual

3Fluid Dynamics

Overview 86

Shock Tube 87

Blast Wave 94

JWL Explosive Test 99

Modeling Blast Wave using 1-D Spherical Symmetry Method 107

Modeling the JWL Explosion using 1-D Spherical Symmtery 125

Nonuniformity with MESH,BOX


86

OverviewIn 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.

87Chapter 3: Fluid DynamicsShock Tube

Shock Tube

Problem DescriptionThe 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 Density and Velocity Profiles in a Shock Tube

Dytran Example Problem ManualShock Tube

88

Desired ResultsFor this example, the ideal gas equation of state is used. The initial conditions are:

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

Dytran ModelingTwo 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.)

High Pressure Low Pressure

ρ = 1 ρ = 0.125

u = 0 u = 0

e = 2.5 e = 2.0

P = 1 P = 0.1

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)


Coarse ModelThe 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 ModelThe shock tube is modeled using 5000 (5 x 5 x 200) CHEXA elements.

ResultsThe 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 ModelThe 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.


90

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


ModelThe 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).

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

Coarse Mesh

Fine Mesh

References1. 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.


STARTCENDTITLE = Shock Tube #1ENDTIME = 0.138ENDSTEP = 100TIC = 1$$$$ Data for Output Control Set 1$$TYPE(tube) = ARCHIVESAVE(tube) = 9999ELOUT(tube) = XVEL,YVEL,ZVEL,DENSITY,SIE,PRESSURE,SSPDSTEPS(tube) = 0,EndELEMENTS(tube) = 1SET 1 = 13t2488b25BEGIN BULKPARAM INISTEP 0.0005PARAM MINSTEP 0.0001$$ THIS SECTION CONTAINS BULK DATA$ Model Geometry: gridpoints and elements$------------------------------------------$

eul1_coarse.dat Dytran input files

EUL1_COARSE.OUT Dytran output files

EUL1_COARSE_TUBE_0.ARC Dytran archive files

eul1_fine.dat Dytran input file

EUL1_FINE.OUT Dytran output files

EUL1_FINE_TUBE_0.ARC Dytran archive files


GRID 1 .05 0.0 0.0....GRID 3636 -.05 -1.39-161.$CHEXA 1 1 1 37 38 2 7 43 ++ 44 8 ...CHEXA 2500 1 3593 3629 3630 3594 3599 3635 ++ 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 0.0 sie 2. TICVAL 12 density 1. zvel 0.0 sie 2.5 ENDDATA

Dytran Example Problem ManualBlast Wave

94

Blast Wave

Problem DescriptionThe 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)

Environment (Air at Room Temperature, r > R0)

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

Theoretical SolutionA 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:

(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

(3-2)

Inserting the initial conditions yields

specific internal energy

e1 = 1.29 1010 Joule/kg

density ρ1 = 7.74 kg/m3

specific internal energy e0 = 1.938 105 Joule/kg


ρ1

ρ0----- γ 1+

γ 1–------------=

ρ 0.155 E0r3–

=

95Chapter 3: Fluid DynamicsBlast Wave

(3-3)

and

(3-4)

Figure 3-8 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 ModelingThe 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:

E e1ρ143---πR0

31.6641E7 J= =

ρ1 R( ) 2.58E6 J R3–

=


96

(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 and Equation (3-5) yields for the time step

seconds. Therefore, the value 1.96E-7 sec is used for the INISTEP parameter.

ResultsFigure 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.

Δ t Sl

c0-----=

l 10R0 20⁄ 0.025 m= = c0 γ γ 1–( )e1= Δ t 1.96E-7=

97Chapter 3: Fluid DynamicsBlast Wave

Files

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


STARTCENDTITLE = Blast WaveCHECK = NO ENDSTEP = 60$$$$ Data for Output Control Set 1$$TYPE(profile) = ARCHIVESAVE(profile) = 99999ELEMENTS(profile) = 1SET 1 = 1t20STEPS(profile) = 0,5,10,15,20,30,40,50,60ELOUT(profile) = PRESSURE$$$$ Data for Output Control Set 2$$TYPE(contour) = ARCHIVESAVE(contour) = 999999ELEMENTS(contour) = 2SET 2 = 1t8000STEPS(contour) = 50ELOUT(contour) = PRESSUREBEGIN BULKPARAM 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 ....

blast.dat Dytran input file

BLAST.OUT Dytran output file

BLAST_PROFILE_0.ARCBLAST_CONTOUR_50.ARC

Dytran archive files


98

GRID 9258 .425 .5 .5 GRID 9259 .45 .5 .5 GRID 9260 .475 .5 .5 GRID 9261 .5 .5 .5 $CHEXA 1 1 1 442 443 2 22 463 ++ 464 23 CHEXA 2 1 2 443 444 3 23 464 ++ 465 24 CHEXA 3 1 3 444 445 4 24 465 ++ 466 25 CHEXA 4 1 4 445 446 5 25 466 ++ 467 26 ....CHEXA 7997 1 8795 9236 9237 8796 8816 9257 ++ 9258 8817 CHEXA 7998 1 8796 9237 9238 8797 8817 9258 ++ 9259 8818 CHEXA 7999 1 8797 9238 9239 8798 8818 9259 ++ 9260 8819 CHEXA 8000 1 8798 9239 9240 8799 8819 9260 ++ 9261 8820 $$PEULER1 1 Hydro 1 $DMAT 1 1. 1 $EOSGAM 1 1.4 $TICEUL 1 ++ SPHERE 1 1 1 10. + + ELEM 1 1 2 5. SET1 1 1 THRU 8000 $TICVAL 1 density 7.74 sie 1.29+10 TICVAL 2 density 1.29 sie 193800. $SPHERE 1 0.0 0.0 0.0 .05 ENDDATA

99Chapter 3: Fluid DynamicsJWL Explosive Test

JWL Explosive Test

Problem DescriptionA 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

• 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.

A = 5.24229 1011

B = 0.07678 1011

R1 = 4.2

R2 = 1.1

ω = 0.34

Dytran Example Problem ManualJWL Explosive Test

100

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

(3-6)

Conservation of energy

(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.)

p p0–D

2

V02

------ V0 V–( )=

e e0–12--- p p0+( ) V0 V–( ) q0+=


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 Rankine-Hugoniot relations). this process is shown in Figure 3-12.

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


102

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:

(3-8)

(3-9)

These relations are applicable also when using the JWL equation of state (with variable specific heat ratio) if measuring at the Chapman-Jouguet state 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 .

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 Arrival Times of Detonation Wave

Pcj 2 γ 1–( )q0ρ0=

Vcjγ

γ 1+------------V0=

γcj Pcj Vcj,( )

Dcj


The following input entries are required when modeling detonation:

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.

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.


104

ResultsThe 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

(3-10)

with γcj = 2.706 for TNT, this leads to

(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 Peak Pressure Profiles at Different Distances

Pcj 2 γcj 1–( )q0ρ0=

Pcj 2.9 * 1010

Pa=


Files


STARTCENDTITLE = HE DETONATION TESTCHECK = NO ENDTIME = 6e-05TIC = 1$$$$ Data for Output Control Set 1$$TYPE(p) = ARCHIVESAVE(p) = 99999ELEMENTS(p) = 1SET 1 = 1t200ELOUT(p) = PRESSURETIMES(p) = 0tEndb3e-06$---------------------------------BEGIN BULK$---------------------------------PARAM INISTEP 2e-07$$ THIS SECTION CONTAINS BULK DATA$$------------------mesh geometry$GRID 1 0.0 0.0 0.0GRID 2 0.0 .01 0.0GRID 3 0.0 0.0 .01GRID 4 0.0 .01 .01.... GRID 801 .5 0.0 0.0GRID 802 .5 .01 0.0GRID 803 .5 0.0 .01GRID 804 .5 .01 .01$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

jwl.dat Dytran input file

JWL.OUT Dytran output file

JWL_P_0.ARC Dytran archive file


106

CHEXA 4 1 13 17 18 14 15 19 ++ 20 16 ....CHEXA 197 1 785 789 790 786 787 791 ++ 792 788 CHEXA 198 1 789 793 794 790 791 795 ++ 796 792 CHEXA 199 1 793 797 798 794 795 799 ++ 800 796 CHEXA 200 1 797 801 802 798 799 803 ++ 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 200ENDDATA

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

Modeling Blast Wave using 1-D Spherical Symmetry Method

Problem DescriptionIn 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<R0)

Environment (r>R0)

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

Specific internal energy = 9.E+6 Joule/kg

Density = 10 kg/m3

Gamma = 1.4


Density = 1 kg/m3

Dytran Example Problem ManualModeling Blast Wave using 1-D Spherical Symmetry Method

108

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:


MESH,1,BOX,,,,,,,++,-1,-1,-1,2,2,2,,,++,60,60,60,,,,EULER,1

TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,2.0SPHERE,3,,0.0,0.0,0.0,1.0SPHERE,4,,0.0,0.0,0.0,20.0TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,++,R-VEL,400,density,200,SIE,300TICVAL,9,,DENSITY,1.0,SIE,3e+5$TABFILE,200,rho.xyd


TABFILE,300,sie.xydTABFILE,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.


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:


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


110

Results

Spherical Model

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

Mesh Density at Time = 0.11 ms

SIE at Time = 0.11 ms Radial Velocity at Time = 0.11 ms


The remap data are shown in the following figures.

Rho.xyd SIE.xyd

V el.xyd


112

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) Density at Cycle 1

SIE at Cycle 1 Velocity Magnitude at Cycle 1


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


114

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.


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.


116

Follow-up 3-D Coarse Model

3-D Coarse Mesh (30 x 30 x30)


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


118

Pressure distribution and radial pressure profile at Time = 0.877 ms

Comparison pressure profiles Run 2, 3, and 4


Spherical Symmetric Model

START


CENDENDSTEP = 100CHECK=NOTITLE= Jobname is: sphersymTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNCSTEPS(ALLEULER) = 0,thru,end,by,3SAVE (ALLEULER) = 10000$$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULK$ ------------------------------------------------------------------$PARAM,SPHERSYM,RECT,X,2.0$ * Euler.300 *$PARAM,MICRO,30PEULER1,1,,HYDRO,19$EOSGAM,2,1.4$DMAT 100 100 2 $$$ ========= MATERIAL DEFINITIONS ==========$$$ ======== Load Cases ========================TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,6.0SPHERE,3,,0.0,0.0,0,10000SPHERE,4,,0.0,0.0,0.0,0.1TICVAL,8,,density,1,sie,3e+5TICVAL,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


120

3-D Model with Mapping

MEMORY-SIZE=16000000,12000000STARTCENDENDSTEP = 100CHECK=NOTITLE= Jobname is: undex-2dTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNCSTEPS(ALLEULER) = 0,1,thru,end,by,15SAVE (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,4,100,9,2.0SPHERE,3,,0.0,0.0,0.0,1.0SPHERE,4,,0.0,0.0,0.0,20.0TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,++,R-VEL,400,density,200,SIE,300TICVAL,9,,DENSITY,1.0,SIE,3e+5


$TABFILE,200,rho.xydTABFILE,300,sie.xydTABFILE,400,vel.xydMESH,1,BOX,,,,,,,++,-1,-1,-1,2,2,2,,,++,60,60,60,,,,EULER,1$$ENDDATA

3-D Reference Model

MEMORY-SIZE=16000000,12000000STARTCENDENDSTEP = 100CHECK=NOTITLE= Jobname is: undex-2dTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNCSTEPS(ALLEULER) = 0,thru,end,by,6SAVE (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 ========================$


122

TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,6.0SPHERE,3,,0.0,0.0,0,10000SPHERE,4,,0.0,0.0,0.0,0.1TICVAL,8,,density,1,sie,3e+5TICVAL,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.xydMESH,1,BOX,,,,,,,++,-1,-1,-1,2,2,2,,,++,60,60,60,,,,EULER,1$$ENDDATA

Coarse Model with Remap

MEMORY-SIZE=16000000,12000000STARTCENDENDSTEP = 100CHECK=NOTITLE= Jobname is: undex-2dTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASS,DENSITY,FMATPLT, FVUNCSTEPS(ALLEULER) = 0,1,thru,end,by,15SAVE (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,4,100,9,2.0SPHERE,3,,0.0,0.0,0.0,1.0SPHERE,4,,0.0,0.0,0.0,20.0TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,++,R-VEL,400,density,200,SIE,300TICVAL,9,,DENSITY,1.0,SIE,3e+5$TABFILE,200,rho.xydTABFILE,300,sie.xydTABFILE,400,vel.xydMESH,1,BOX,,,,,,,++,-1,-1,-1,2,2,2,,,++,30,30,30,,,,EULER,1$$ENDDATA

Rho.xyd

.016666668 0.175999750.050000001 0.117557280.083333336 0.138206170.11666667 0.177306920.15000001 0.235730310.18333334 0.303575490.21666668 0.412329080.25 0.699942830.28333333 1.34443140.31666666 1.89834610.34999999 1.51421080.38333333 1.09903040.41666669 1.0090078


124

0.45000002 1.00038280.48333335 1.0000119

Sie.xyd

0.016666668 1644110.40.050000001 1521182.60.083333336 1650291.40.11666667 1871300.80.15000001 2126423.0.18333334 2316476.30.21666668 2376442.80.25 2337180.0.28333333 2127134.50.31666666 1557344.50.34999999 738252.380.38333333 328480.250.41666669 301103.970.45000002 300045.940.48333335 300001.44

Vel.xyd

0.016666668 551.551150.050000001 343.23770.083333336 417.905490.11666667 592.598940.15000001 727.586360.18333334 765.236820.21666668 778.49170.25 932.187380.28333333 1135.25390.31666666 1092.19250.34999999 602.565670.38333333 125.991880.41666669 9.38625050.45000002 0.33851680.48333335 0.010169516

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

Modeling the JWL Explosion using 1-D Spherical Symmtery

Problem DescriptionThe 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.


JWL model, refer to section 4 of Chapter 3: JWL Explosive Test.

Environment (r>R0)

The simulation will be run for 2.5 ms.

Dytran Modeling



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.

Ideal gas gamma 1.4


Density = 1 kg/m3

Dytran Example Problem ManualModeling the JWL Explosion using 1-D Spherical Symmtery

126

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


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,4,200,9,4.0SPHERE,3,,0.0,0.0,0.0,10.0SPHERE,4,,0.0,0.0,0.0,2.83244TICVAL,8,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,++,R-VEL,100,density,200,SIE,300TICVAL,9,RADIAL,X-CENTER,0,Y-CENTER,0,Z-CENTER,0,++,R-VEL,100,density,400,SIE,500$TABFILE,100,velj.xydTABFILE,200,rhoa.xydTABFILE,300,siea.xydTABFILE,400,rhoj.xydTABFILE,500,siej.xyd$

To allow outflow of air, a transmitting boundary condition is added:FLOWDIR,4,MMHYDRO,22,POSX,,,,,++,FLOW,OUTFLOWDIR,4,MMHYDRO,22,POSY,,,,,++,FLOW,OUTFLOWDIR,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.


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)


128

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

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


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

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


130

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)


Pressure at cycle 1 Comparison with 1-D

Comparison with 1-DDensity at cycle 1


132

SIE at cycle 1 Comparison with 1-D

Comparison with 1-DRadial Velocity at cycle 1


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).


134



STARTCENDENDSTEP = 4000CHECK=NOTITLE= Jobname is: sphersymTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = PRESSURE,XVEL,YVEL,ZVEL, FMAT100,DENSITY100,SIE100, FMAT200,DENSITY200,SIE200,FBURNSTEPS(ALLEULER) = 0,thru,end,by,50SAVE (ALLEULER) = 10000$$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULK$ ------------------------------------------------------------------$PARAM,SPHERSYM,RECT,X,1.0$ * Euler.300 *$PARAM,MICRO,30PEULER1,1,,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,,,,+


+,SPHERE,4,200,9,6.0SPHERE,3,,0.0,0.0,0,10000SPHERE,4,,0.0,0.0,0.0,0.25TICVAL,8,,density,1,sie,3e+5TICVAL,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,,MMHYDROFLOWDIR,4,MMHYDRO,22,POSX,,,,,++,FLOW,OUT$ENDDATA

3-D Model with Mapping

MEMORY-SIZE=40000000,40000000STARTCENDENDSTEP = 100CHECK=NOTITLE= Jobname is: undex-2dTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = PRESSURE,XVEL,YVEL,ZVEL,FBURN, FMAT100,FMAT200, DENSITY100,SIE100, DENSITY200,SIE200STEPS(ALLEULER) = 0,1,thru,end,by,15SAVE (ALLEULER) = 10000$$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULK$ ------------------------------------------------------------------$$ * Euler.300 *$PEULER1,1,,MMHYDRO,19$EOSGAM,100,1.4$DMAT 100 1.29 100$


136

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

137Chapter 3: Fluid DynamicsNonuniformity with MESH,BOX

Nonuniformity with MESH,BOXMESH,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 DescriptionIf 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


Environment (r>R0)

The simulation will be run for 9 ms.

Dytran ModelingFirst,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.


Density = 10 kg/m3

Gamma = 1.4


Density = 1 kg/m3

Dytran Example Problem ManualNonuniformity with MESH,BOX

138

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

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 X GROWTH ELMSIZE

1 -5.000E+00 1.000E+00 0.000E+00 2 -4.704E+00 9.481E-01 2.962E-01 3 -4.423E+00 9.481E-01 2.808E-01 4 -4.157E+00 9.481E-01 2.662E-01 5 -3.904E+00 9.481E-01 2.524E-01 6 -3.665E+00 9.481E-01 2.393E-01 7 -3.438E+00 9.481E-01 2.269E-01 8 -3.223E+00 9.481E-01 2.151E-01 9 -3.019E+00 9.481E-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.

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


For the y-direction:

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

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,19DMAT 3 2 2EOSGAM,2,1.4$TICEUL,19,,,,,,,,++,SPHERE,3,3,8,4.0,,,,++,SPHERE,4,3,9,6.0SPHERE,3,,0.0,0.0,0,10000SPHERE,4,,0.0,0.0,-3,0.2TICVAL,8,,density,1,sie,3e+5TICVAL,9,,density,10,sie,9e+6


1 -0.12 0.12 1 1


1 -5 -4 0.5 6

2 -4 -3 1 8

3 -3 -2 0.5 8

4 -2 5 2 20


140

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


ResultsPressures 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.


142



STARTCENDENDSTEP = 600CHECK=NOTITLE= Jobname is: non-uniformTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL,FVUNCSTEPS(ALLEULER) = 0,thru,end,by,20SAVE (ALLEULER) = 10000$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------


144

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

Chapter 4: Fluid-structural Interaction Dytran Example Problem Manual

4Fluid-structure Interaction

Overview 147

Shock Formation 148

Blast Containment in a Luggage Container 154

Multiple Bird-strike on a Cylindrical Panel 161

Slanted Piston 169

Sloshing using ALE Method 176

Flow between Two Containers or Airbags 183

Blastwave Hitting a Bunker 193

Mine Blast 197

Multiple Bird-strike on a Box Structure 209

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

Fuel Tank Filling 224

Water Pouring into a Glass 237

Fluid Flow through a Straight Pipe 246

Using Euler Archive Import in Blast Wave Analyses 263

Blast Wave with a Graded Mesh 273

Bubble Collapse with Hydrostatic Boundary Conditions 284

Dytran Example Problem Manual146

Prestressed Concrete Beam 294

Blast Simulation on Prestressed Concrete Beam 306

Vortex Shedding with Skin Friction 317

Geometric Eulerian Boundary Conditions 328

Cohesive Friction 337

147Chapter 4: Fluid-structural InteractionOverview

OverviewIn 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.

Dytran Example Problem ManualShock Formation

148

Shock Formation

Problem DescriptionGiven 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

.

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 BackgroundRelative 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

(4-1)

Second Hugoniot relation

(4-2)

Third Hugoniot relation

(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),

(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).


specific internal energy e0 = 2.5 106 Joule/kg

pressure p0 = 1000 N/m2

ratio of specific heats γ = 1.4

us

ρ ρ0–

ρ---------------⎝ ⎠⎛ ⎞ c0=

p p0– ρ0 us u⋅ ⋅=

e ep– 0.5 p p0+( )ρ ρ0–

ρρ0---------------⎝ ⎠⎛ ⎞=

p γ 1–( ) ρ e⋅ ⋅=

149Chapter 4: Fluid-structural InteractionShock Formation

Dytran ModelDue 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.


150

Figure 4-2 Numerical Problem Layout

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

Figure 4-3 Pressure Profile along the Tube at Different Times


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

(4-5)

(4-6)

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

(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):

(4-9)

p 3.49e3N m

2⁄=

ρ 2.25e3–kg m

3⁄=

p p0 ρ0 us u+ 3.46 e3N m

2⁄= =

ρ ρ0

us

us u–-------------- 2.3 e

3–kg m

3⁄= =

usγ 1+

4------------u

γ 1+4

------------⎝ ⎠⎛ ⎞ 2

u2

p01ρ0-----γ+⋅+ 2087 m sec⁄= =


152

Files


STARTCENDTITLE = SHOCK PROPAGATION (Units = m-Kg-N-sec)ENDSTEP = 80TIC = 1TLOAD = 1$$$$ Data for Output Control Set 1$$TYPE(GAS) = ARCHIVESAVE(GAS) = 99999ELOUT(GAS) = XVEL,YVEL,ZVEL,DENSITY,SIE,PRESSUREELEMENTS(GAS) = 1SET 1 = 1t50STEPS(GAS) = 1tEndb16BEGIN BULKPARAM INISTEP 1e-06PARAM,FASTCOUP$$ THIS SECTION CONTAINS BULK DATA$$GRID 1 0.0 0.0 0.0GRID 2 0.0 .01 0.0GRID 3 0.0 0.0 .01GRID 4 0.0 .01 .01....GRID 209 .015 -.005 -.005GRID 210 .015 .015 -.005GRID 211 .015 -.005 .015GRID 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 12CHEXA 4 1 13 17 18 14 15 19 ++ 20 16

shock.dat Dytran input file

SHOCK.OUT Dytran output file

SHOCK_GAS_1.ARC Dytran archive file


.

.

.

.CHEXA 48 1 189 193 194 190 191 195 ++ 196 192CHEXA 49 1 193 197 198 194 195 199 ++ 200 196CHEXA 50 1 197 201 202 198 199 203 ++ 204 200CHEXA 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 3CFACE 2 1 51 1CFACE 3 1 51 2CFACE 4 1 51 5CFACE 5 1 51 4CFACE 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

Dytran Example Problem ManualBlast Containment in a Luggage Container

154

Blast Containment in a Luggage Container

Problem DescriptionThe 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.

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

Dytran ModelingThe 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.

ResultsThe 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.”


156

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


Files

References1. 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.


STARTCENDTITLE = Container DemoCHECK = noENDSTEP = 4000ENDTIME = 0.003SPC = 1$$$$ Data for Output Control Set 1$$TYPE(eul) = ARCHIVESAVE(eul) = 1ELOUT(eul) = PRESSUREELEMENTS(eul) = 1SET 1 = 538t5907TIMES(eul) = 0tEndb0.0005$$$$ Data for Output Control Set 2$$TYPE(sh) = ARCHIVESAVE(sh) = 9999ELOUT(sh) = EFFPL2,EFFST2ELEMENTS(sh) = 2SET 2 = 1t537,5908t6787TIMES(sh) = 0tEndb0.0005$$$$ Data for Output Control Set 3

container.dat Dytran input file

CONTAINER.OUT Dytran output file

CONTAINER_EUL_XX.ARCCONTAINER_SH_0.ARC


CONTAINER_STRAIN_0.THS Dytran time history file


158

$$TYPE(strain) = TIMEHISSAVE(strain) = 9999ELOUT(strain) = EFFPL2ELEMENTS(strain) = 3SET 3 = 6402,6672,5953,6243,6503STEPS(strain) = 0tEndb50$--------------------------------------------------------BEGIN BULK$--------------------------------------------------------SETTING,1,VERSION2$PARAM STRNOUT YESPARAM INISTEP 2e-06PARAM MINSTEP 1e-06$$$ THIS SECTION CONTAINS BULK DATA$$GRID 1 0.0 .42 .8GRID 2 .07 .35 .8... GRID 8245 1.921 0.0 .96GRID 8246 1.921 0.0 .88$$$CQUAD4 1 801 6501 6502 6506 6505CQUAD4 2 801 6502 6503 6507 6506.... CQUAD4 6786 801 8245 8246 7366 7365CQUAD4 6787 801 8246 6556 6557 7366$$$CHEXA 538 1 1 2 6 5 681 682 ++ 683 684CHEXA 539 1 2 3 7 6 682 685 ++ 686 683 ...CHEXA 5906 1 5913 5915 5918 5916 6495 6497 ++ 6500 6498 CHEXA 5907 1 5915 5869 5872 5918 6497 6451 ++ 6454 6500 $$--------------------------------------------------------$


$ THIS SECTION CONTAINS THE LOADS, CONSTRAINTS, AND CONTROL BULK DATA$ ENTRIES$$SPC 1 7181 345SPC 1 7192 345....SPC 1 8226 345SPC 1 8237 345$$--------------------------------------------------------$$ THIS SECTION CONTAINS THE DEFINED FEFACES OF ELEMENTS$$CFACE 1418 8 1 1CFACE 1419 8 2 1....CFACE 2833 8 6786 1CFACE 2834 8 6787 1$$ CFACE 1 7 538 1CFACE 2 7 538 2...CFACE 1416 7 5905 5CFACE 1417 7 5906 5$$--------------------------------------------------------$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


160

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

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

Multiple Bird-strike on a Cylindrical Panel

Problem DescriptionLanding 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 Initial Situation

Dytran Example Problem ManualMultiple Bird-strike on a Cylindrical Panel

162

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

Plate

Bird 1

Bird 2

Dytran ModelThe 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

material titanium


bulk modulus K = 1.03 1011 N/m2


yield stress σij = 1.38 108 N/m2


radius R = 0.25 m

length L = 0.25 m

material jelly


speed of sound c = 1483 m/sec

mass m = 0.360 kg


material jelly


speed of sound c = 1483 m/sec

mass m = 0.285 kg



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


164

ResultsA theoretical check on the results is the peak pressure of the impact. For the maximum pressure of the impact can be written as

(4-10)

In this way, the maximum pressure of the impact of Bird 1 is

(4-11)

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)

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 Pressure Time History for the Impact of Bird 1

Pmax ρ c ΔV=

Pmax 207 MPa=

Pmax 117 MPa=


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


166

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


STARTCENDTITLE = Multiple bird strike using Lag-Eul ale couplingENDSTEP = 1000ENDTIME = .002TIC = 1SPC = 1$$$$ Data for output control Set 1$$TYPE(plate) = ARCHIVE

mbird.datgrid_xl.dat

Dytran input file

MBIRD.OUT Dytran output file

MBIRD_BIRD_XX.ARCMBIRD_PLATE_XX.ARC


MBIRD_PRESSURE_0.THS Dytran time history file


SAVE(plate) = 1STEPS(plate) = 0tEndb100ELEMENTS(plate) = 1SET 1 = 7393t7920ELOUT(plate) = EFFST01,EFFST02,EFFST03,EFFPL01,EFFPL02,EFFPL03$$$$ Data for output control Set 2$$TYPE(bird) = ARCHIVESAVE(bird) = 1STEPS(bird) = 0tEndb100ELEMENTS(bird) = 2SET 2 = 1t7392ELOUT(bird) = MASS,DENSITY,PRESSURE,FMAT,XVEL,YVEL,ZVEL$$$$ Data for output control Set 3$$TYPE(pressure) = TIMEHISSAVE(pressure) = 10000 ELEMENTS(pressure) = 3TIMES(pressure) = 0tEndb1e-6SET 3 = 7105,7119ELOUT(pressure) = PRESSURE,DENSITY,ENERGY$BEGIN BULKPARAM INISTEP 1e-6PARAM 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.$ --------------------------------------------


168

TICEUL 10 ++ CYLINDER1 3 1 3 ++ CYLINDER2 3 2 2 ++ ELEM 4 1$CYLINDER1 -.1381 .125 .26 -.2381 .125 .26 ++ .035CYLINDER2 .13 .125 .2252 .17 .125 .2944 ++ .035$SET1 4 1 THRU 7392$$ Initial material data.$ ----------------------TICVAL 1 XVEL 200TICVAL 2 XVEL -75 ZVEL -129.9$$ ========$ LAGRANGE$ ========$$ Property, material and yield model.$ -----------------------------------PSHELL1 2 2 Bely Gauss 3 .83333 Mid ++ .0015 $DMATEP 2 4527 .314 1.03e11 1$YLDVM 1 1.38e8 $$ Boundary constrain.$ --------------------CORD2C 1 0.0 0.0 0.0 0.0 0.25 0.0 ++ 0.0 0.125 0.25 $SPC3 1 1 23456 ++ 8671 THRU 8704 8705 THRU 9181 BY 34 ++ 8738 THRU 9214 BY 34 9215 THRU 9248$$ ============$ ALE COUPLING$ ============$SURFACE 1 SEG 1SURFACE 2 SEG 2$ALE 1 2 1$ENDDATA

169Chapter 4: Fluid-structural InteractionSlanted Piston

Slanted Piston

Problem DescriptionThis 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, two-dimensional 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 Two-dimensional Representation of the Problem

Dytran Example Problem ManualSlanted Piston

170

Figure 4-16 Three-dimensional Problem to be Modeled in Dytran

Desired ResultsFor this example, the initial conditions are:

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:

(4-13)

and combining this with the equation of state

(4-14)

(4-15)

Gas Piston

ρ = 0.1 g/cm3 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

de pdV–=

p γ 1–( )e V⁄=

dee

------ γ 1–( )dVV

-------–=


(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,

(4-17)

Using Equation (4-17), the following result is obtained for the kinetic energy:

(4-18)

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

and the process stops.

Dytran ModelingThe 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.

ResultsAs 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.

ee0-----

V0

V------⎝ ⎠⎛ ⎞

γ 1–( )=

KE e0 e– 1ee0-----–= =

KEe0--------⎝ ⎠⎛ ⎞ 1

V0

V------⎝ ⎠⎛ ⎞

γ 1–( )–=

V0

V------ 1

3---=


172

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 Energy Changes


Files


STARTCENDTITLE = Slanted PistonENDTIME = 250$$$$ Data for Output Control Set 1$$TYPE(gas) = ARCHIVESAVE(gas) = 9999ELOUT(gas) = XVEL,YVEL,ZVEL,PRESSURE,FMATELEMENTS(gas) = 1SET 1 = 2t1001TIMES(gas) = 0tEndb20$$$$ Data for Output Control Set 2$$TYPE(piston) = TIMEHISSAVE(piston) = 9999ELOUT(piston) = TXX,TYY,TZZELEMENTS(piston) = 2SET 2 = 1TIMES(piston) = 0tEndb5$$$$ Data for Output Control Set 3$$TYPE(grd) = ARCHIVESAVE(grd) = 9999GRIDS(grd) = 3SET 3 = 1t8GPOUT(grd) = XVEL,YVEL,ZVELTIMES(grd) = 0tEndb20$$$$ Data for Output Control Set 4$$TYPE(nrg) = TIMEHISSAVE(nrg) = 9999MATS(nrg) = 4SET 4 = 1,10MATOUT(nrg) = XMOM,YMOM,ZMOM,EKIN,EINT

slp.dat Dytran input file

SLP.OUT Dytran output file

SLP_GAS_0.ARCSLP_GRD_0.ARC


SLP_PISTON_0.THSSLP_NRG_0.THS



174

TIMES(nrg) = 0tEndb5BEGIN BULKPARAM INISTEP 0.1PARAM MINSTEP 0.005$$ THIS SECTION CONTAINS BULK DATA$$GRID 1 0.0 0.0 0.0GRID 2 0.0 0.0 1...GRID 1338 1.633333.7333333.7333333GRID 1339 1.666667.6666667.6666667$CHEXA 1 1 1 5 6 2 3 7 ++ 8 4CHEXA 2 10 9 130 131 10 20 141 ++ 142 21...CHEXA 1001 10 1206 1327 1328 1207 1217 1338 ++ 1339 1218$$$ THIS SECTION CONTAINS THE DEFINED FEFACES OF ELEMENTS$$CFACE 1 1 1 3CFACE 2 1 1 1CFACE 3 1 1 2CFACE 4 1 1 5CFACE 5 1 1 4CFACE 6 1 1 6$PSOLID 1 1$PEULER1 10 Hydro 10$DMAT 10 .1 1$DMATEL 1 7.85 210. .3$EOSGAM 1 1.4$TICEUL 10 ++ ELEM 1 10 1 1.SET1 1 2 THRU 1001$TICVAL 1 density .1 sie .025$SURFACE 5 SEG 1$


COUPLE 1 5 Inside On OnENDDATA

Dytran Example Problem ManualSloshing using ALE Method

176

Sloshing using ALE Method

Problem DescriptionMany 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 (a) Proposed Sloshsat System (b) Tank Dimensions

177Chapter 4: Fluid-structural InteractionSloshing using ALE Method

Dytran ModelA 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 SubroutineTo 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


178

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

ResultsThe 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.


Figure 4-22 Positions of Liquids in the Tank at Different Times in the Analysis

Files

References4. Simulation of Liquid Dynamics onboard Sloshsat FLEVO, NLR-TP-99236

sloshsat.datsloshsat.bdfeexout_com.f (user-subroutine)

Dytran input files

SLOSHSAT.OUTCOMEEUL.OUT (refer user-subroutine)

Dytran output files

SLOSHSAT_TANK_0.ARCSLOSHSAT_WATER_0.ARC


SLOSHSAT_MONITOR_0.THS Dytran time history file


180


STARTCENDENDTIME= 30.01ENDSTEP=99999999CHECK=NOTITLE= Jobname is: sloshTLOAD=1TIC=1SPC=1$$ Output result for request: monitor$TYPE (monitor) = TIMEHISGRIDS (monitor) = 1SET 1 = 14 40 44 83 117 181 GPOUT (monitor) = XVEL YVEL ZVEL XDIS YDIS ZDISTIMES (monitor) = 0 THRU END BY 1.SAVE (monitor) = 10000$$ Output result for request: tank$TYPE (tank) = ARCHIVEELEMENTS (tank) = 2SET 2 = 1 THRU 256 ELOUT (tank) = EFFPL EFFST TIMES (tank) = 0 THRU END BY 1.SAVE (tank) = 100000$$ Output result for request: water$TYPE (water) = ARCHIVEELEMENTS (water) = 3SET 3 = 1001 THRU 1736ELOUT (water) = XVEL YVEL ZVEL DENSITY PRESSURETIMES (water) = 0 THRU END BY 1.SAVE (water) = 100000$$ Output to user subroutine$ELEXOUT (ceegee)ELEMENTS (ceegee) = 99SET 99 = 1001 THRU 1736TIMES (ceegee) = 0 THRU END BY 1.SAVE (ceegee) = 10000$TYPE (step) = STEPSUMSTEPS (step) = 1 THRU END BY 250$TYPE (mat) = MATSUMSTEPS (mat) = 1 THRU END BY 250$$------- PARAMETER SECTION ------


$PARAM,INISTEP,1E-8PARAM,MINSTEP,1E-9PARAM,TOLCHK, 1.E-9PARAM, FMULTI, 1.0PARAM, ROMULTI, 1.E-15PARAM, VELMAX, 30., NOPARAM, VELCUT, 1.E-10$$------- BULK DATA SECTION -------BEGIN BULKINCLUDE slosh.bdf$$ ========== PROPERTY SETS ========== $$ * tank *$PSHELL, 1, 2, 0.002$$ * dummy *$PSHELL, 2, 3, 9999$ * water *$PEULER1, 3, , MMHYDRO, 1TICEUL, 1, , , , , , , , +AAAA02+AAAA02, ELEM, 7, 7, 7, 1.0, , , , +AAAA03+AAAA03, ELEM, 5, 1, 1, 2.0TICVAL, 7, , DENSITY, 1.29, SIE, 1.94E5TICVAL, 1, , DENSITY, 1000.$$$ ========= MATERIAL DEFINITIONS ==========$$ -------- Material water id =1$DMAT, 1, 1000, 1EOSPOL, 1, 2.2E6$$ -------- Material air (nitrogen) id =7$DMAT, 7, 1.29, 7EOSGAM, 7, 1.4$$ -------- Material aluminium id =2$DMATEP, 2, 36.5E3, 7.24E8, 0.33, , , 2,YLDVM, 2, 3.45E8$$$$ ======== LOAD CASES ========================$$ ------- Velocity BC tankvel -----


182

TLOAD1,1,6,,2,2$$FORCE,6,1,0,1,1,0,0.01FORCE,6,2,0,1,1,0,0.01::FORCE,6,257,0,1,1,0,0.01FORCE,6,258,0,1,1,0,0.01$$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, 6SURFACE, 2, , SEG, 2$ALE, 1, 1, 2$ALEGRID1, 1, 3$ENDDATA

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

Flow between Two Containers or Airbags

Problem DescriptionFlow 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 ProblemThis part contains the following sections:

A) Data Summary

B) Equations Summary

C) Analytical Estimates on Computed Results

Dytran Example Problem ManualFlow between Two Containers or Airbags

184

A) Data Summary

Let us briefly summarize the initial states of the two vessels:

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

Given these data, the following initial values are derived:

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.

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

P = (g-1) *ρ * E

with g = 1.517

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


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:

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

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.

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

Pc / P2 = 0.509

ρc / ρ2 = 0.641

cc / c2 = 0.8914

Tc / T2 = 0.7946


186

Figure 4-24 Function of Time

The second stage is the 'critical flow' stage. This can be considered as a gradual blowdown of the high-pressure 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.


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 thatdrhoc / 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 barP1 = 0.4 bardP / dt = 0.5 bar/msecPc = 0.51 * P2

We can now compute the following equality:

P1(t) = Pc(t), whereP1(t) = P1(0) + dP / dt * tcfPc(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


188

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 ModelThis 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


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,,,,,,++,,,,,,,,,++,,22COUPLE,20,50,OUTSIDE,,,11,,,++,,,,,,,,,++,,23SURFACE,25,,PROP,1SET1,1,1,3SURFACE,50,,PROP,2SET1,2,2,3

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,,1.0PORFLCPL,82,,,BOTH,10SUBSURF,31,50,ELEM,250SET1,250,32

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.


190

ResultsThe 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.


STARTCENDENDTIME = 0.01CHECK=NOTITLE= Jobname is: flowTLOAD=1TIC=1SPC=1PARAM,FASTCOUP PARAM,INISTEP,1.E-8$------- Parameter Section ------PARAM,CONTACT,THICK,0.0$------- BULK DATA SECTION -------$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLEULHYDROELOUT (ALLEULER) = PRESSURE,XVEL,YVEL,ZVELTIMES (ALLEULER) = ENDSAVE (ALLEULER) = 1$TYPE(Container1)=TIMEHISSURFACES(Container1)=155SET 155=25SURFOUT(Container1)=PRESSURE,MASS, MFL,MFLR,MFL-POR,MFLR-PORSTEPS(Container1)=0 THRU END BY 6SAVE(Container1)=9999999$TYPE(Container2)=TIMEHISSURFACES(Container2)=156SET 156=50SURFOUT(Container2)=PRESSURE,MASS, MFL,MFLR,MFL-POR,MFLR-PORSTEPS(Container2)=0 THRU END BY 6SAVE(Container2)=9999999$TYPE(HOLES) = TIMEHISSUBSOUT(HOLES) = MASS,MFLR-POR,MFL-POR,XVEL,YVEL,ZVEL, DENSITY,MFLR,MFL,PRESSURESUBSURFS(HOLES) = 88SET 88 = 31STEPS(HOLES) = 0 THRU END BY 5SAVE(HOLES) = 1000000


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


192

CQUAD4 18 2 46 42 58 68CQUAD4 25 2 71 72 58 57CQUAD4 27 2 72 78 68 58CQUAD4 29 2 78 84 61 68CQUAD4 31 2 84 71 57 61$ -------- property set Pset3.3 ---------CQUAD4 32 3 72 78 84 71$ -------- property set flowfaces1 ---------$ -------- property set flowfaces2 ---------$$ ========== PROPERTY SETS ========== $$ * Pset1.1 *$PSHELL1 1 DUMMY$$ * Pset2.2 *$PSHELL1 2 DUMMY$$ * Pset3.3 *$PSHELL1 3 DUMMY$ENDDATA

193Chapter 4: Fluid-structural InteractionBlastwave Hitting a Bunker

Blastwave Hitting a Bunker

Problem DescriptionThe 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 ModelingThe 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:

Dytran Example Problem ManualBlastwave Hitting a Bunker

194

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:


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

195Chapter 4: Fluid-structural InteractionBlastwave 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,111MESH,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,101MESH,1,BOX,,,,,,,++,-647.,0.,-1293.,1057.,447.,1293.,,,++,33,23,37,,,,EULER,201Euler archive output is generated using:ELEMENTS (AIR) = 1SET 1 = ALLEULHYDRO

Dytran Example Problem ManualBlastwave Hitting a Bunker

196

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

Figure 4-28 Deformed Effective Stress Plot of the Bunker

Figure 4-29 Isosurfaces Created using SIE Variable for the Two Euler Domains

197Chapter 4: Fluid-structural InteractionMine Blast

Mine Blast

Problem DescriptionThis 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 ModelAn 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 Outline of Basic Numerical Model

Dytran Example Problem ManualMine Blast

198

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:

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:

Density 7.85E-9 tonne/mm3

Modulus of elasticity 210000. tonne/mm/s2

Poison ratio 0.3

Yield stress 250. tonne/mm/s2

Density 1.29E-12 tonne/mm3

Gamma 1.4

Specific Internal Energy 1.9385E8 tonne-mm2/s2


In the input deck:

DMAT,230,1.29e-12,203,,,,,,++,,1.01TICVAL,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:

The input deck will show:

TICVAL,4,,DENSITY,107E-12,SIE,3.9e12SPHERE,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,3PEULER1,201,,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

Density 107E-12 tonne/mm3

Specific Internal Energy 4.9E12 tonne-mm2/s2


200

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,102TICEUL,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,,DUMMYSURFACE,999,,PROP,999SET1,999,999$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, ,12FORCE ,2000,999, ,0.0,1.0,1.0,1.0MOMENT,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.


Figure 4-32 Dummy Shell Elements Defined to Close the Volume

The input for dummy shell elements:

PSHELL1,900,,DUMMYPSHELL1,910,,DUMMYPSHELL1,920,,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,27SET1,27,60,61,62,110,135,150,900,++,910,920,999

The inner surface:

SURFACE,98,,PROP,28SET1,28,60,61,62,110,135,150,900,++,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


202

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,1.0SUBSURF,1,97,PROP,301SET1,301,900

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


ResultsThe 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.

Figure 4-33 Acceleration Plot

Figure 4-34 Stress Distribution Plot


204


STARTCENDENDTIME=0.005ENDSTEP=9999999CHECK=NOTITLE= Jobname is: test (mm/tonne/s/K)TLOAD=1TIC=1SPC=1$$ ========= OUTPUT REQUEST FOR AIR =========$TYPE (AIR1) = ARCHIVEELEMENTS (AIR1) = 1SET 1 = ALLEULHYDROELOUT (AIR1) = PRESSURE,XVEL,YVEL,ZVEL,SIE,DENSITYTIMES (AIR1) = 0,thru,end,by,.0002SAVE (AIR1) = 20000$CPLSURFS(BAGHT1)=300SET 300 = 97CPLSOUT(BAGHT1)=PRESSURE,XVEL,YVEL,ZVELTIMES (BAGHT1)= 0,thru,end,by,.0002TYPE (BAGHT1)=ARCHIVESAVE (BAGHT1)=1000000$CPLSURFS(BAGHT2)=301SET 301 = 98CPLSOUT(BAGHT2)=PRESSURE,XVEL,YVEL,ZVELTIMES (BAGHT2)= 0,thru,end,by,.0002TYPE (BAGHT2)=ARCHIVESAVE (BAGHT2)=1000000$$ ========= OUTPUT REQUEST FOR STRUCTURAL ELEMENTS =========$TYPE (VEHICLE) = ARCHIVEELEMENTS (VEHICLE) = 3SET 3 = 2, THRU, 105, 129, THRU, 1697, 1702, THRU, 1736, 1758, THRU, 1851, 2460, THRU, 2536ELOUT (VEHICLE) = EFFSTTIMES (VEHICLE) = 0,thru,end,by,.0002SAVE (VEHICLE) = 99999$TYPE (GP) = ARCHIVEGRIDS (GP) = 4GPOUT (GP) = RACC, RVELTIMES (GP) = 0,thru,end,by,.0002SAVE (GP) = 99999SET 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,


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$$------- Parameter Section ------$PARAM,RKSCHEME,3PARAM,FASTCOUPPARAM,STEPFCTL,0.9PARAM,INISTEP,.5E-7PARAM,MINSTEP,1.E-13$$------- 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$PEULER1,201,,2ndOrder,101TICEUL,101,,,,,,,,++,SPHERE,400,230,4,20.,,,,++,SPHERE,501,230,5,1.,SPHERE,400,,1797.5,0.,-450.,250.SPHERE,501,,0.,0.,-5000.,10000.$PEULER1,202,,2ndOrder,102TICEUL,102,,,,,,,,++,SPHERE,502,230,5,5.SPHERE,502,,0.,0.,-5000.,10000.$TICVAL,4,,DENSITY,107E-12,SIE,3.9e12TICVAL,5,,DENSITY,1.29E-12,SIE,1.938e11$DMAT,230,1.29e-12,203,,,,,,++,,1.01EOSGAM,203,1.4$FLOWDEF,202,,HYDRO,,,,,,++,FLOW,BOTH$COUPLE,1,97,INSIDE,ON,ON,11,,,+


206

+,,,,,,,,,++,,1SURFACE,97,,PROP,27$$ Define flow thru the holes$COUPOR,1,11,1,PORFLCPL,84,CONSTANT,1.0SUBSURF,1,97,PROP,301SET1,301,900$COUPOR,2,11,2,PORFLCPL,84,CONSTANT,1.0SUBSURF,2,97,PROP,302SET1,302,910$COUPOR,3,11,3,PORFLCPL,84,CONSTANT,1.0SUBSURF,3,97,PROP,303SET1,303,920$PORFLCPL,84,,,BOTH,2$COUPLE,2,98,OUTSIDE,ON,ON,,,,++,,,,,,,,,++,,2SURFACE,98,,PROP,28$SET1,27,60,61,62,110,135,150,900,++,910,920,999$SET1,28,60,61,62,110,135,150,900,++,910,920$$ ========== 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 *


$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 6PSHELL 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$ holePSHELL1,900,,DUMMY$ top cover


208

PSHELL1,910,,DUMMY$ side coverPSHELL1,920,,DUMMY$$ groundPSHELL1,999,,DUMMYSURFACE,999,,PROP,999SET1,999,999RIGID,999,999,1.0E10,,0.00,0.00,-800.,,++,,,,,,,,,++,,1.E10,,,1.E10,,1.E10$ constrain the motion of the groundTLOAD1,1,2000, ,12FORCE ,2000,999, ,0.0,1.0,1.0,1.0MOMENT,2000,999, ,0.0,1.0,1.0,1.0$$ * conm2 *$CONM2,5000,1145,,1.5 CONM2,5001,1146,,1.7$$ ========= MATERIAL DEFINITIONS ==========$DMATEP,111,7.85e-09,210000.,.3,,,111YLDVM,111,250E10$DMATEP,222,7.85e-09,210000.,.3 $INCLUDE model.bdfINCLUDE ground.dat$ENDDATA

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

Multiple Bird-strike on a Box Structure

Problem DescriptionBird 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 Bird-strike 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 Geometry of the Upper Plate and Position of the Birds Relative to the Plate

Dytran Example Problem ManualMultiple Bird-strike on a Box Structure

210

Dytran ModelingEach 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

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

Gamma 1.4

Thickness (m) 0.0015

Radius (m) 0.25

Length (m) 0.25

Mass (kg) 0.36 0.285

Initial velocity (m/s) 150 200

Fail (equiv. Plas. Strain) 0.1


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

ResultsIn 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


212

.

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


Files


STARTCENDTITLE = Multiple bird strike using Multi-Material-FVSurferENDSTEP = 1200ENDTIME = .0025TIC = 1SPC = 1$TYPE(plate) = ARCHIVESAVE(plate) = 1STEPS(plate) = 0,200,400,1200ELEMENTS(plate) = 1SET 1 = 7393 THRU 8448ELOUT(plate) = EFFST01,EFFST03,FAIL$$$$ Data for output control Set 2 : the 2 birdsTYPE(bird) = ARCHIVESAVE(bird) = 1STEPS(bird) = 0,200,400,1200ELEMENTS(bird) = 2SET 2 = ALLMULTIEULHYDROELOUT(bird) = XVEL,YVEL,ZVEL,FMAT3,FMAT5$TYPE(C1) = TIMEHISSAVE(C1) = 10000STEPS(C1) = 0tEndb10CPLSURFS(C1) = 3SET 3 = 1CPLSOUT(C1) = XFORCE,YFORCE,ZFORCE,RFORCE$type(out)=stepsumsteps(out)=0tendby10type(mat)=matsumsteps(mat)=0tendb10$BEGIN BULKPARAM INISTEP 1e-7PARAM MINSTEP 1e-8PARAM,FASTCOUP,INPLANE,FAILPARAM,FMULTI,1.0PARAM,NZEROVEL,YES

Ep4d9.dat

Ep4d9.bdf

Dytran input files

Ep4d9_C1_0.THS Dytran time history file

Ep4d9_BIRD_xx.ARC Dytran archive files


214

PARAM,FAILOUT,NOparam,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 1COUPLE 1 1 INSIDE ON ON +A000002+A000002 +A000002+A000002 1$SURFACE 1 ELEM 11SET1 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 2.2e9DMAT 5 930 5EOSPOL 5 2.2e9$$ Allocation of material to geometric regions.TICEUL 11 ++ CYLINDER1 3 1 3 ++ CYLINDER2 5 2 2 ++ SPHERE 4 4 5 1$CYLINDER1 .13 .125 .2252 .17 .125 .2944 ++ .035CYLINDER2 -.1381 .125 .26 -.2381 .125 .26 ++ .035SPHERE,4,,-.1381, .125, .26, 1000$$ 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 1YLDVM 1 1.38e8FAILMPS 1 0.1


$PSHELL1,3,,DUMMYPSHELL1,4,,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 =4DMAT 4 1.1848 4EOSGAM,4,1.4$$ Domain 2PEULER1,6,,MMHYDRO,12TICEUL,12,,,,,,,,++,SPHERE,7,4,5,1.0,,,,+SPHERE,7,,0.0,0.0,0.0,500.0TICVAL,5,,SIE,2.1388E5,DENSITY,1.1848$$===Coupling Surface 2$COUPLE,2,2,OUTSIDE,,,,,,++,,,,,,,,,++,,2$SURFACE 2 ELEM 12SET1 12 7393 THRU 8448 13729 THRU 14048 14577++ THRU 15236MESH,2,ADAPT,0.01,0.01,0.01,,,,++,-0.26,-0.015,-0.05,,,,,,++,,,,,,,EULER,6$$ coupling interactionCOUP1INT,2,2,1$ENDDATA

Dytran Example Problem ManualShaped Charge, using IG Model, Penetrating through Two Thick Plates

216

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

Problem DescriptionWhen 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.

217Chapter 4: Fluid-structural InteractionShaped 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 Dytran Model Setup


218

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 ModelThe 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:

a1 1.43E11 N/m2

a2 0.839E11 N/m2

a3 2.16E9 N/m2

b1 0.0

b2 0.0

b3 0.0


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

Other liner material properties of liner are as follows:

In the input deck:

DMAT 701 8960. 711 712 713 714EOSPOL, 711, 1.43+11, 0.839+11, 2.16+9SHREL,712,0.477E11$ Johnson-Cook$ A B n C m EPS0 CvYLDJC,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

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.,++,0.2939CYLINDER, 3,, 0.2, 2.0406, 0., 0.2047, 2.0406, 0.,++,2.0019

TICVAL,2,,DENSITY,8960.

A 1.2E8 N/m2

B 1.43E9 N/m2

C 0.0

n 0.5

m 1.0

1.0

Tmelt 1356.0 K

Troom 293.0 K

Cv 399.0 J/kg

Density 8960 Kg/m3

Constant shear model 0.477E11 N/m2

Constant spallation model -2.5E10 N/m2

ε· 0


220

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. 811 812 813 814EOSPOL, 811, 1.64E+11SHREL,812,0.818E11YLDVM,813,1.4E9PMINC,814,-3.8E9

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.,++,2.05CYLINDER, 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 1. ++ CYLINDER1 701 2 2. ++ CYLINDER2 3. ++ CYLINDER3 701 2 4. ++ CYLINDER4 801 3 5. ++ CYLINDER5 801 3 6.$SET1 1 1 THRU 15342TICVAL,1,,DENSITY,1630.,SIE,4.29E6


ResultsThe 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 Velocity Field of Explosive Gases, Liner, and Jet


222


STARTCENDTITLE = SHAPED CHARGES TEST (m/kg/s/K)CHECK = NO ENDTIME = 6.E-5 TIC = 1$$$$TYPE(EULER1) = ARCHIVEELEMENTS(EULER1) = 11SET 11 = 1,THRU,15342ELOUT(EULER1) = XVEL,YVEL,ZVEL,EFFSTS,EFFPLS, TXX, TYY, TZZ, TXY, TYZ, TZX, PRESSURE, SIE, FMAT100,FMAT701,DENSITY100,DENSITY701, FMAT801,DENSITY801,FMAT,DENSITYSAVE(EULER1) = 999999TIMES(EULER1) = 0.0, THRU, END, BY, 2.E-6$BEGIN BULKPARAM,INISTEP,1.E-11PARAM,MINSTEP,1.E-13PARAM,GEOCHECK,ONPARAM,VELMAX,20.0E+03$$ THIS SECTION CONTAINS BULK DATA$INCLUDE model.bdfINCLUDE wall.dat$$PEULER1, 1 ,, MMSTREN, 1$$ EXPLOSIVE$DMAT 100 1630. 100 101 102EOSIG,100,,,,,,,,++,,,,,,,,,++,,,,,99,,MCOMPB,SISHREL,101,3.E9YLDVM,102,2.E8$$$ COPPER$DMAT 701 8960. 711 712 713 714EOSPOL, 711, 1.43+11, 0.839+11, 2.16+9SHREL,712,0.477E11$ Johnson-Cook$ A B n C m EPS0 CvYLDJC,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$$ STEEL$DMAT 801 7830. 811 812 813 814EOSPOL, 811, 1.64E+11SHREL,812,0.818E11YLDVM,813,1.4E9PMINC,814,-3.8E9$TICEUL 1 ++ ELEM 1 100 1 1. ++ CYLINDER1 701 2 2. ++ CYLINDER2 3. ++ CYLINDER3 701 2 4. ++ CYLINDER4 801 3 5. ++ CYLINDER5 801 3 6.$SET1 1 1 THRU 15342CYLINDER, 1,, -0.5391, -0.56, 0., 2.0, 0.4147, 0.,++,0.2958 CYLINDER, 2,, -0.5391, -0.56, 0., 2.0, 0.4147, 0.,++,0.2939CYLINDER, 3,, 0.2, 2.0406, 0., 0.2047, 2.0406, 0.,++,2.0019CYLINDER, 4,, 0.22, 2.0406, 0., 0.223, 2.0406, 0.,++,2.05CYLINDER, 5,, 0.27, 2.0406, 0., 0.273, 2.0406, 0.,++,2.05$TICVAL,1,,DENSITY,1630.,SIE,4.29E6TICVAL,2,,DENSITY,8960.TICVAL,3,,DENSITY,7830.$$ DUMMY QUAD TO MODEL THE WALLET$PSHELL1,2,,,,,,,,++,9999.WALLET,1,2$$ENDDATA

Dytran Example Problem ManualFuel Tank Filling

224

Fuel Tank Filling

Problem DescriptionThe 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 ModelTank 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.

225Chapter 4: Fluid-structural InteractionFuel 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:

This is a reduced bulk modulus (1/100) to increase the time step and reduce CPU time.

In the input deck:

DMAT 2 8.5E-7 2EOSPOL 2 20000

The air properties are:

In the input deck:

DMAT 1 1.29E-9 1EOSGAM 1 1.4TICVAL,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

Density 8.5E-7 Kg/ mm3

Bulk modulus 2.0E+4 KPa

Density 1.29E-9 Kg/ mm3

Gamma 1.4

Specific internal energy 1.938E11 Kg-mm2/s2


226

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,0TICVAL,23,,DENSITY,8.5426E-7,SIE,0TICVAL,24,,DENSITY,8.54268E-7,SIE,0TICVAL,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$MESH,2,ADAPT,8.,8.,8.,,,,++,,,,,,,,,++,,,,,,,EULER,2TICEUL,200,,,,,,,,++,CYLINDER,51,2,41,1.0,,,,++,CYLINDER,52,1,21,2.0$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


• 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,3TICEUL,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,101SET1,101,7,8,12,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,201SET1,201,4,THRU,8


228


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,,1.0PORFLCPL,2,,,BOTH,1SUBSURF,2,201,PROP,250SET1,250,8

3) Vent pipe surface:

SURFACE,301,,PROP,301SET1,301,9,THRU,14



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,,1.0PORFLCPL,3,,,BOTH,1SUBSURF,3,301,PROP,350SET1,350,14

• 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).


230

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,,1.0PORFLOW,21,,XVEL,-1592.,DENSITY,8.5425E-7,FLOW,IN,++,YVEL,0.,ZVEL,0.,MATERIAL,2,,,++,SIE,0SUBSURF,21,201,PROP,251SET1,251,4

2) Outlet

The flow card of outlet is referenced from the third coupling surface.

COUPOR,31,33,31,PORFLOW,31,,1.0PORFLOW,31,,MATERIAL,1,DENSITY,1.29e-9,SIE,1.938e+11,++,PRESSURE,100.SUBSURF,31,301,PROP,351SET1,351,9

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.


• Miscellaneous

a) Fast coupling is to used:

PARAM,FASTCOUP

b) Gravity is applied to the whole model:

TLOAD1 1 444 0GRAV 444 -9800 1

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

PARAM,COSUBMAX,1000PARAM,COSUBCYC,100

d) To show the behavior of fuel and air, the following output request was added:

TYPE (euler) = ARCHIVEELEMENTS (euler) = 1SET 1 = ALLMULTIEULHYDROELOUT (euler) = XVEL,YVEL,ZVEL,PRESSURE, FMAT1,DENSITY1,MASS1,SIE1, FMAT2,DENSITY2,MASS2,SIE2, FMAT ,DENSITY ,MASS ,SIETIMES (euler) = 0 THRU END BY 10E-3SAVE (euler) = 10000

e) In order to output results of the flow boundaries, a history request is created:

TYPE (SUB) = TIMEHISSUBSURFS (SUB) = 3SET 3 = 21, 31SUBSOUT (SUB) = PRESSURE,AREA,XVEL,YVEL,ZVELSTEPS (SUB) = 0 THRU END BY 100SAVE (SUB) = 100000

ResultsFigure 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.

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.


232

Figure 4-49 Isosurfaces of FMAT


Figure 4-50 Velocities on the Flow Boundaries

Abbreviated Dytran Input File$ UNIT: mm/Kg/s/KSTARTTIME=99999CENDENDTIME=2000E-3ENDSTEP=9999999CHECK=NOTITLE= Jobname is: tank_fillingTLOAD=1TIC=1SPC=1$ Output result for request: eulerTYPE (euler) = ARCHIVEELEMENTS (euler) = 1SET 1 = ALLMULTIEULHYDROELOUT (euler) = XVEL,YVEL,ZVEL,PRESSURE, FMAT1,DENSITY1,MASS1,SIE1, FMAT2,DENSITY2,MASS2,SIE2, FMAT ,DENSITY ,MASS ,SIETIMES (euler) = 0 THRU END BY 10E-3SAVE (euler) = 10000$ Output result for request: Euler BoundaryTYPE (SUB) = TIMEHISSUBSURFS (SUB) = 3SET 3 = 21, 31SUBSOUT (SUB) = PRESSURE,AREA,XVEL,YVEL,ZVELSTEPS (SUB) = 0 THRU END BY 100SAVE (SUB) = 100000$TYPE (Shells) = ARCHIVEELEMENTS (Shells) = 8SET 8 = ALLDUMQUADELOUT (Shells) = ZUSER


234

TIMES (Shells) = 0 THRU END BY 10.E-3SAVE (Shells) = 100000$$------- Parameter Section ------PARAM,FASTCOUP,PARAM,INISTEP,1E-7PARAM,MINSTEP,1E-8PARAM,COSUBMAX,1000PARAM,COSUBCYC,100$------- BULK DATA SECTION -------BEGIN BULKINCLUDE tank.bdf$$1111111222222223333333344444444555555556666666677777777TLOAD1 1 444 0GRAV 444 -9800 1$ ========== PROPERTY SETS ==========$$ * Shell_1 *$PSHELL1 4 DUMMYPSHELL1 5 DUMMYPSHELL1 6 DUMMYPSHELL1 7 DUMMYPSHELL1 8 DUMMYPSHELL1 9 DUMMYPSHELL1 10 DUMMYPSHELL1 11 DUMMYPSHELL1 12 DUMMYPSHELL1 13 DUMMYPSHELL1 14 DUMMYPSHELL1 15 DUMMYPSHELL1 16 DUMMY$$$ -------- Material air_mat id =1$DMAT 1 1.29E-9 1EOSGAM 1 1.4$$ -------- Material oil_mat id =2$DMAT 2 8.5E-7 2EOSPOL 2 20000$$ ======== Load Cases ========================$RIGID,1,1,10.SURFACE,1,,PROP,1SET1,1,5,THRU,8,10,THRU,16$---------Tload1-tank----------------TLOAD1,1,1,,12FORCE,1,1,,0.,1.0,1.0,1.0MOMENT,1,1,,0.,1.0,1.0,1.0


$$-----------------------------Domain 1------------------------------PEULER1,1,,MMHYDRO,100$MESH,1,ADAPT,8.,8.,8.,,,,++,,,,,,,,,++,,,,,,,EULER,1$COUPLE,1,101,OUTSIDE,,,,,,++,,,,,,,,,++,,1$SURFACE,101,,PROP,101SET1,101,7,8,12,THRU,16$$---------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,1.938E11TICVAL,22,,DENSITY,8.54254E-7,SIE,0TICVAL,23,,DENSITY,8.5426E-7,SIE,0TICVAL,24,,DENSITY,8.54268E-7,SIE,0TICVAL,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$SURFACE,201,,PROP,201SET1,201,4,THRU,8$


236

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

237Chapter 4: Fluid-structural InteractionWater Pouring into a Glass

Water Pouring into a Glass

Problem DescriptionThis 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 ModelingThe 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 GPaPoisson's Ratio: 0.4

Water:

Density: 1E-6 kg/mm3

Bulk Modulus: 2.2 GPa

Dytran Example Problem ManualWater Pouring into a Glass

238

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,13MSEH,10,BOX,,,,,,,++,-63,-250,-65,147,300,130,,,++,21,42,19,,,,EULER,2


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


The couple card refers to a mesh number. The second Euler domain is an adaptive Euler domain created and initialized by

PEULER1,3,,MMHYDRO,12MSEH,11,ADAPT,7,7,7,,,,++, , , , , , ,,,++, , ,,,,,EULER,3,++,NONE


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,13MESH,12,ADAPT,7,7,7,,,,++, , , , , , ,,,++, , ,,,,,EULER,2,++,NONE


240


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,9COUPOR,15,30,46,PORFLCPL,95,,1.0SUBSURF,46,1,ELEM,59SET1,59,12,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,,1.0SUBSURF,47,3,ELEM,60SET1,60,3358,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) = ARCHIVEELEMENTS (euler) = 12SET 12 = ALLMULTIEULHYDROELOUT (euler) = DENSITY PRESSURE SIE FMAT2 FMAT3TIMES (euler) = 0.0,THRU,END,BY,10SAVE (euler) = 1


ResultsThe 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 .

Figure 4-54 Isosurface of FMAT at 0.0s



242



STARTCENDENDTIME=2000ENDSTEP=9999999CHECK=NOTITLE= Jobname is: water pouring in to a glassTLOAD=1TIC=1SPC=1TYPE(detail) = ARCHIVEELEMENTS(detail) = 2SET 2 = ALLSHQUAD TIMES(detail) = 0.0,THRU,END,BY,10ELOUT(detail) = MASSSAVE(detail) = 1000$$TYPE (euler) = ARCHIVEELEMENTS (euler) = 12SET 12 = ALLMULTIEULHYDROELOUT (euler) = DENSITY SIE PRESSURE FMAT FVUNC FMATPLT , VOLUME XVEL YVEL ZVEL XMOM YMOM ZMOM TEMPTURE FMAT2 FMAT3TIMES (euler) = 0.0,THRU,END,BY,10SAVE (euler) = 1$TYPE (restat) = RESTARTTIMES (restat) = 0 THRU END BY 2000SAVE (restat) = 1


$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,FASTCOUPPARAM,INISTEP,1e-3$------- BULK DATA SECTION -------BEGIN BULKINCLUDE water_pouring1.bdfINCLUDE glass.dat$ ------- GRAVITATION ----- TLOAD1 1 444 0GRAV 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 =1MATRIG 1 9e-07 1.4 .4 $$ -------- Material air id =2DMAT 21.28e-09 2EOSGAM 2 1.4 $$ -------- Material water id =3DMAT 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


244

$$ ------- General Coupling: couple_bottle_glass ----- $COUPLE 8 1 INSIDE ON ON 30 STANDARD+A000001+A000001 +A000002+A000002 10$SURFACE 1 ELEM 1SET1 1 2325 THRU 2704 1744 THRU 2324 1163+A000003+A000003 THRU 1743 582 THRU 1162 1 THRU 581++ 3000 THRU 3576$$ ------- General Coupling: couple_bottle----- $COUPLE 9 2 OUTSIDE ON ON STANDARD+A000004+A000004 +A000005+A000005 11$SURFACE 2 ELEM 2SET1 2 2526 THRU 2704 1945 THRU 2525 1364+A000006+A000006 THRU 1944 783 THRU 1363 202 THRU 782+A000007+A000007 1 THRU 201$$ ------- General Coupling: couple_glass ----- $COUPLE 10 3 OUTSIDE ON ON 31 STANDARD++A000004 ++A000005 12$SURFACE 3 ELEM 3SET1 3 3000 THRU 3576$$ ------- Mesh Box: outer_box_euler_mesh_for_bottle_glass$MESH 10 BOX +A000008+A000008 -63 -250 -65 147 300 130 +A000009+A000009 21 42 19 EULER 2$$ ------- Mesh Adap: inner_adapt_euler_mesh_for_bottle$MESH 11 ADAPT 7 7 7 +A000010+A000010 +A000011+A000011 EULER 3+A000012+A000012 NONE$$ ------- Mesh Adap: inner_adapt_euler_mesh_for_glass$MESH 12 ADAPT 7 7 7 +A000010+A000010 +A000011+A000011 EULER 2+A000012+A000012 NONE$$ ------- TICEUL BC inner_reg_def ----- TICEUL 12 +A000013


+A000013 SPHERE 3 2 5 1 +A000014+A000014 BOX 4 3 6 2 SPHERE 3 0 0 0 1000BOX 4 -80 -145 -60 270 133 120$$ ------- TICEUL BC outter_reg_def ----- TICEUL 13 +A000016+A000016 SPHERE 2 2 5 1 SPHERE 2 0 0 0 1000$$ ------- Rigid Body Object rbo ----- $ ---- No reference node is used.$ ---- enforced rotational velocity of the C.G of the bottleTLOAD1 1 16 12 FORCE 16 MR1 1 0 0 0TLOAD1 1 1016 12 6MOMENT 1016 MR1 1 0 0 1$$ ------- Rigid Body Object rbo1 ----- $ ---- No reference node is used.$ ---- fix the glass in space in all directionsTLOAD1 1 1 12FORCE 1 MR4 1 0 0 0TLOAD1 1 1001 12MOMENT 1001 MR4 1 0 0 0$ ================ TABLES =================$$ ------- TABLE 6: rotational_table -------TABLED1 6 +A000017+A000017 0 .0379 40 .0379 40.0001 0 2000 0+A000018+A000018 ENDT$$$ Porosity of the botte top$PORFLCPL,95,,,BOTH,9COUPOR,15,30,46,PORFLCPL,95,,1.0SUBSURF 46 1 ELEM 59SET1 59 12 THRU 63 317 THRU 368 501++ THRU 552 685 THRU 736$$ Porosity for the glass top$PORFLCPL,96,,,BOTH,8COUPOR,16,31,47,PORFLCPL,96,,1.0SUBSURF 47 3 ELEM 60SET1 60 3358 THRU 3490$ENDDATA

Dytran Example Problem ManualFluid Flow through a Straight Pipe

246

Fluid Flow through a Straight Pipe

Problem DescriptionThis 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.

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

Theoretical AnalysisTo 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

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

The flow rate is coupled to the pressure difference across the pipe by Poiseuille's law: ,

while the volumetric flow rate itself is defined by . Hence substitution yields

Here denoted dynamic viscosity, r the radius, l the length, u the inflow velocity. The shear stress is given by

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

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 1 12 1.SURFACE 12 ELEM 3TLOAD1 1 1 12FORCE 1 1 0. 1 1 1MOMENT 1 1 0. 1 1 1$

The viscous fluid is modeled as Eulerian material using EOSTAIT

DMAT 19.53e-05 1EOSTAIT 1 0 200000 19.529e-5 .001

Redensi ty*diameter*velocity

dydnamic vis i tycos–---------------------------------------------------------------------------- 9.52*10

5–*0.44*200

0.001-------------------------------------------------- 16= = =

1

Re----------- diameter

0.844

----------=

φ ΔP φ πr4

8μ l---------ΔP=

φ uπr2

=

ΔP8μ l

πr4

--------- φ 8μ l

r2

--------- u8*0.001*17.10.422*0.422

---------------------------------200 153.6psi= = = =

v

τyz τzxrΔP2 l

---------- 0.422*153.62*1.71

------------------------------ 1.896psi= = = =

F πr2ΔP 85.93 lbf= =


248

and the Euler mesh is created by MESH:

MESH 9 BOX +A000004+A000004 -.45 -.45 -.05 .9 .9 17.2 +A000005+A000005 11 11 80 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+A000006 SPHERE 12 1 10 1SPHERE 12 0 0 0 20$

To couple the Eulerian material to the pipe, the COUPLE card is used:

$COUPLE 8 1 OUTSIDE ON ON 8 STANDARD+A000001+A000001 +A000002+A000002 9$SURFACE 1 SUB 1 SUB 2 SUB 3$$ General Coupling Subsurface: SUB1$SUBSURF 1 1 ELEM 3SET1 3 1 THRU 456$SUBSURF 2 1 ELEM 4SET1 4 457 THR 514$$SUBSURF 3 1 ELEM 5SET1 5 515 THRU 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

$COUPOR 1 8 2 PORFLOW 1CONSTANT 1$PORFLOW 1 METHODVELOCITY FLOW IN DENSITY9.76e-05+A000003+A000003 XVEL 0 YVEL 0 ZVEL -200MATERIAL 1$SUBSURF 2 1 ELEM 4SET1 4 457 THRU 514


Note that COUPOR is referenced from the COUPLE card

$ General Coupling Subsurface: SUB3$COUPOR 2 8 3 PORFLOW 2CONSTANT 1$PORFLOW 2 METHODVELOCITY FLOW OUTPRESSURE 0. +A000009+A000009MATERIAL 1$SUBSURF 3 1 ELEM 5SET1 5 515 THRU 572

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

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

pipe. This yields the PORFLOWT model:

COUPOR 1 8 2PORFLOWT 1CONSTANT 1$PORFLOWT 1 IN ++ TABLE 100 ++ 1CONSTANT 9.76e-5CONSTANT 0.0TABLED1 100 ++ 0.0 0.0 0.0015 200 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 7 3 IN ++ TABLE 100 ++ 1CONSTANT 9.76e-5 TABLED1 100 ++ 0.0 0.0 0.0015 200 1.0 200.0

ResultsResults 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.

4*Length

soundspeed--------------------------------- 0.0015=


250


252


254


256


Figure 4-58 Model 1 with Time Dependent Flow and HYDRO SolverInflow and Outflow Results

Figure 4-59 Model 2 with Time Dependent Flow and HYDRO Solver Inflow and Outflow Results


258

Figure 4-60 Model 1 with Time Dependent Flow Model: Axial Force


Model 1

STARTCENDENDTIME=0.02ENDSTEP=9999999CHECK=NOTITLE= Jobname is: Cyl

TLOAD=1TIC=1SPC=1$ Output result for request: elemTYPE (elem) = ARCHIVEELEMENTS (elem) = 1SET 1 = ALLEULHYDRO

Table 4-2 Force on the Pipe, TYX is extrapolated to the Wall

Pressure Diff TYZ Z-Force

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%)


ELOUT (elem) = XVEL YVEL ZVEL DENSITY SIE PRESSURE Q TXY TYZ TZX TIMES (elem) = 0 THRU END BY 0.005SAVE (elem) = 10000$ Output result for request: thsTYPE (ths) = TIMEHISELEMENTS (ths) = 2SET 2 = 633,637,10192,10196ELOUT (ths) = XVEL YVEL ZVEL DENSITY SIE PRESSURE TZX TYZ TIMES (ths) = 0 THRU END BY 0.0001SAVE (ths) = 10000$TYPE (Coup) = TIMEHISRIGIDS(Coup) = 3SET 3 = 1RBOUT (Coup) = XFORCE YFORCE ZFORCE TIMES (Coup) = 0 THRU END BY 0.0001SAVE (Coup) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1.E-7PARAM,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 3TLOAD1 1 1 12FORCE 1 1 0. 1 1 1


260

MOMENT 1 1 0. 1 1 1$ $ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material euler id =1DMAT 19.53e-05 1EOSTAIT 1 0 200000 19.529e-5 .001$$ ======== 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 3SET1 3 1 THRU 456$$ General Coupling Subsurface: SUB2$COUPOR 1 8 2 PORFLOW 1CONSTANT 1$PORFLOW 1 METHODVELOCITY FLOW IN DENSITY9.76e-05+A000003+A000003 XVEL 0 YVEL 0 ZVEL -200MATERIAL 1$SUBSURF 2 1 ELEM 4SET1 4 457 THRU 514$$ General Coupling Subsurface: SUB3$COUPOR 2 8 3 PORFLOW 2CONSTANT 1$PORFLOW 2 METHODVELOCITY FLOW OUTPRESSURE 0. +A000009+A000009MATERIAL 1$SUBSURF 3 1 ELEM 5SET1 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$


$ ------- TICEL BC Simple ----- $$ ------- TICEUL BC Region ----- TICEUL 13 +A000006+A000006 SPHERE 12 1 10 1 SPHERE 12 0 0 0 20$$ENDDATA

Model2

STARTCENDENDTIME=2.E-2ENDSTEP=9999999CHECK=NOTITLE= Jobname is: cylsTLOAD=1TIC=1SPC=1$ Output result for request: elemTYPE (elem) = ARCHIVEELEMENTS (elem) = 1SET 1 = 1 THRU 2900 ELOUT (elem) = XVEL YVEL ZVEL DENSITY PRESSURE TYZ TZX TIMES (elem) = 0 THRU END BY 0.01 , 0.005SAVE (elem) = 10000$ Output result for request: thsTYPE (ths) = TIMEHISELEMENTS (ths) = 2SET 2 = 1 47 2843 2889 ELOUT (ths) = XVEL YVEL ZVEL DENSITY SIE PRESSURE TIMES (ths) = 0 THRU END BY 0.0001SAVE (ths) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1.E-7$------- BULK DATA SECTION -------BEGIN BULKINCLUDE cyls.bdf$$ ========== PROPERTY SETS ========== $$ * PEUL *$PEULER 1 1 HYDRO$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material eul id =1DMAT 19.53e-05 1


262

EOSTAIT 1 0 200000 19.529e-5 0.001$$ ======== Load Cases ========================$$$ ------- Flow BC INFLOW ----- TLOAD1 1 7 4FLOW 7 3MATERIAL 1 XVEL 0 YVEL 0+A000001+A000001 ZVEL -200 DENSITY9.76e-05CFACE 1 3 1 1....CFACE 58 3 58 1$$ ------- Flow BC OUTFLOW ----- TLOAD1 1 10 4FLOW 10 4 FLOW OUTMATERIAL 1PRESSURE 0+A000002+A000002 DENSITY9.53e-05CFACE 59 4 2900 4....CFACE 116 4 2843 4$$ ------- TICEL BC TIC ----- SET1 5 1 THRU 523 2790 THRU 2900 2209+A000003+A000003 THRU 2789 1628 THRU 2208 1047 THRU 1627+A000004+A000004 524 THRU 576 577 THRU 1046TICEL 1 5 DENSITY9.53e-05$$ENDDATA

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

Using Euler Archive Import in Blast Wave AnalysesIn 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 DescriptionA 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.


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 ModelThree 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

Dytran Example Problem ManualUsing Euler Archive Import in Blast Wave Analyses

264

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 SetupThe following entries apply to all three models.

The solver, material and initialization are given as

PEULER1,1,,MMHYDRO,19DMAT, 100, 1, 2 EOSGAM,2,1.4PARAM, BULKL,0.1$TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,6.0SPHERE,3,,5.0,0.0,5.0,10000SPHERE,4,,5.0,0.0,5.0,1TICVAL,8,,density,1,sie,3e+5TICVAL,9,,density,1,sie,9e+5PARAM,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,100PARAM,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.


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

STARTEULINIT,GAS2_ALLEULER_0.ARC,60CEND

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 5 2 4 8 6 3CQUAD4 6 2 3 6 5 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,ONSURFACE,200,,PROP,2PSHELL1 2 DUMMYSET1,2,2


266

Results

Model 1: Only the Blast wave and No Structure

The following three figures show the pressure distribution at cycles 3, 60, and 100.


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


268

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.


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.


270


$---Blast wave analysis------------------

STARTCENDENDSTEP = 100CHECK=NOTITLE= Blast Wave AnalysisTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASSSTEPS(ALLEULER) = 0,THRU,END,BY,3SAVE (ALLEULER) = 10000$TYPE (ARCMAT) = TIMEHISMATS (ARCMAT) = 15SET 15 = 100MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOMSTEPS (ARCMAT) = 0,THRU,END,BY,10SAVE (ARCMAT) = 99999$$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULKPARAM,BULKL,0.1PARAM,COSUBMAX,100PARAM,COSUBCYC,100$ ------------------------------------------------------------------$PARAM,MICRO,30PEULER1,1,,MMHYDRO,19$$DMAT, 100, 1, 2 EOSGAM,2,1.4$$TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,6.0SPHERE,3,,5.0,0.0,5.0,10000SPHERE,4,,5.0,0.0,5.0,1TICVAL,8,,density,1,sie,3e+5TICVAL,9,,density,1,sie,9e+5


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

$----Follow up run 2----------------------------------------------------------

STARTEULINIT,GAS2_ALLEULER_0.ARC,60CENDENDSTEP = 61CHECK=NOTITLE= Blast Wave AnalysisTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT,MASSSTEPS(ALLEULER) = 0,THRU,END,BY,3SAVE (ALLEULER) = 10000$TYPE (ARCMAT) = TIMEHISMATS (ARCMAT) = 15SET 15 = 100MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOMSTEPS (ARCMAT) = 0,THRU,END,BY,10SAVE (ARCMAT) = 99999$$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULKPARAM,BULKL,0.1PARAM,COSUBMAX,100PARAM,COSUBCYC,100$ ------------------------------------------------------------------$$PARAM,FASTCOUPPARAM,MICRO,30


272

PEULER1,1,,MMHYDRO,19$$$EOSGAM,2,1.4$DMAT 100 1 2 $$TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,6.0SPHERE,3,,5.0,0.0,5.0,10000SPHERE,4,,5.0,0.0,5.0,1TICVAL,8,,density,1,sie,3e+5TICVAL,9,,density,1,sie,9e+5$MESH,1,BOX,,,,,,,++,0,-0.12,0.0,10,0.24,10,,,++,30,1,30,,,,EULER,1$$COUPLE,100,200,INSIDE,ON,ON$SURFACE,200,,PROP,2$PSHELL1 2 DUMMYSET1,2,2$$GRID 1 7.50001-5.49900 7.4999 GRID 2 8.99999-5.49900 7.4999GRID 3 7.50001 5.49900 7.4999GRID 4 8.99999 5.49900 7.4999GRID 5 7.50001-5.49900 2.0001 GRID 6 7.50001 5.49900 2.0001 GRID 7 8.99999-5.49900 2.0001 GRID 8 8.99999 5.49900 2.0001 $$ --- Define 6 elements$CQUAD4 1 2 1 2 4 3CQUAD4 2 2 5 6 8 7CQUAD4 3 2 5 7 2 1CQUAD4 4 2 2 7 8 4CQUAD4 5 2 4 8 6 3CQUAD4 6 2 3 6 5 1 $ENDDATA

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

Blast Wave with a Graded MeshThe 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 DescriptionThe 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).


Explosive properties (r < R0)


Density = 1 kg/m3

Environment (r > R0)


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

Dytran Example Problem ManualBlast Wave with a Graded Mesh

274

Dytran ModelFor 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.


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, 6.0SPHERE, 3,, 0.0,0.0,0,10000SPHERE, 4, 1.125, 0.0, 7.5,1TICVAL,8,,density,1,sie,3e+5TICVAL,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


276

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,5SET1, 5,5PARAM,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,30PARAM,COSUBCYC,30

ResultsTo 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.


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 .


278

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.


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).


280

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 .



STARTCENDENDSTEP = 660CHECK=NOTITLE= Blast wave Graded MeshTIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMATPLTSTEPS(ALLEULER) = 0,THRU,END,BY,20SAVE (ALLEULER) = 10000$TYPE (BOX) = ARCHIVECPLSURFS (BOX) = 3SET 3 = 300CPLSOUT (BOX) = PRESSURE,XVELSTEPS (BOX) = 0,THRU,END,BY,20SAVE (BOX) = 10000$TYPE (LAG) = ARCHIVEELEMENTS (LAG) = 4SET 4 = ALLSHQUADELOUT (LAG) = EFFST-MID,EFFPL-MIDSTEPS (LAG) = 0,THRU,END,BY,20SAVE (LAG) = 10000$TYPE (ARCMAT) = TIMEHISMATS (ARCMAT) = 15SET 15 = 100MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOMSTEPS (ARCMAT) = 0,THRU,END,BY,10SAVE (ARCMAT) = 99999$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULK$ ============ Euler ==================================$PARAM,BULKL,0.1PARAM,COSUBMAX,100PARAM,COSUBCYC,100$PARAM,FASTCOUPPARAM,MICRO,30$PARAM,GRADED-MESH


282

PARAM,FLOW-METHOD,FACET$$ --- Material definitions ---$ PEULER1,1,,MMHYDRO,19$DMAT,100,1,2 $EOSGAM,2,1.4$$ --- Initial conditions ---$TICEUL,19,,,,,,,,++,SPHERE,3,100,8,4.0,,,,++,SPHERE,4,100,9,6.0SPHERE,3,,0.0,0.0,0,10000SPHERE,4,,1.125,0.0,7.5,1TICVAL,8,,density,1,sie,3e+5TICVAL,9,,density,1,sie,9e+5$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,5SET1,5,5$$=========== Structure ================$$ --- 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


GRID 12 11.8500 .190000 0.90000GRID 13 11.8500-.190000 1.28889 - - -CQUAD4 1018 5 145 146 606 605CQUAD4 1019 5 146 147 607 606CQUAD4 1020 5 147 148 608 607$ENDDATA

Dytran Example Problem ManualBubble Collapse with Hydrostatic Boundary Conditions

284

Bubble Collapse with Hydrostatic Boundary ConditionsA 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 ModelThe 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.

285Chapter 4: Fluid-structural InteractionBubble 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 SetupFirst, 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 1.e3 1$EOSPOL 1 2.2e9 $DMAT 3 2 2EOSGAM,2,1.4$DMAT 100 100 2


286

The initial conditions will be defined by using geometric regions:

TICEUL,19,,,,,,,,++,BOX,1,4,6,2.0,,,,++,SPHERE,3,100,8,4.0BOX,1,,-100,-100,0.0,200,200,20.0SPHERE,3,,0.0,0.0,4.652,0.25TICVAL,6,,SIE,0.,DENSITY,1000.0TICVAL,8,,density,100,sie,3e+5PARAM,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,,0GRAV,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 Pressure at Cycle 1: Hydrostatic Pressure Profile


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,120COUPOR,1,120,,PORHYDST,75PORHYDST 75SURFACE,200,,PROP,2PSHELL1 2 DUMMYSET1,2,2$GRID 1 -5.49990-5.49990 10.9999GRID 2 5.49990-5.49990 10.9999 GRID 3 -5.49990 5.49990 10.9999GRID 4 5.49990 5.49990 10.9999GRID 5 -5.49990-5.49990 0.0001 GRID 6 -5.49990 5.49990 0.0001 GRID 7 5.49990-5.49990 0.0001 GRID 8 5.49990 5.49990 0.0001 $$ --- Define 6 elements$ -------- property set pdum ---------CQUAD4 1 2 1 2 4 3CQUAD4 2 2 5 6 8 7CQUAD4 3 2 5 7 2 1CQUAD4 4 2 2 7 8 4CQUAD4 5 2 4 8 6 3CQUAD4 6 2 3 6 5 1

The boundary conditions are:

COUPOR,1,120,,PORHYDST,75PORHYDST 75

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


288

ResultsBoth 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.


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.


290

Dytran Input files

Model 1

STARTCENDENDTIME = 0.5CHECK=NOTITLE= Jobname is: undex-2dTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT4,MASS4,DENSITY4,FMATPLT4, FMATPLT,FMAT100,FLUXVOL, FVUNC FMAT100,MASS100,DENSITY100,SIE100,SIE4,HMATTIMES(ALLEULER) = 0,1E-8,thru,end,by,0.01SAVE (ALLEULER) = 10000$TYPE (ARCMAT) = TIMEHISMATS (ARCMAT) = 15SET 15 = 4MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOMSTEPS (ARCMAT) = 0,THRU,END,BY,10SAVE (ARCMAT) = 99999$$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULKPARAM,BULKL,0.1$ ------------------------------------------------------------------$$ * Euler.300 *$PARAM,MICRO,30PARAM,FMULTI,1.0PEULER1,1,,MMHYDRO,19$DMAT 4 1.e3 1 $EOSPOL 1 2.2e9 $DMAT 3 2 2EOSGAM,2,1.4$DMAT 100 100 2 $


$$ ========= MATERIAL DEFINITIONS ==========$$$ ======== Load Cases ========================$$TICEUL,19,,,,,,,,++,BOX,1,4,6,2.0,,,,++,SPHERE,3,100,8,4.0BOX,1,,-100,-100,0.0,200,200,20.0SPHERE,3,,0.0,0.0,4.652,0.25TICVAL,6,,SIE,0.,DENSITY,1000.0TICVAL,8,,density,100,sie,3e+5$MESH,1,BOX,,,,,,,++,-5.5,-0.12,0,11,0.24,11,,,++,41,1,41,,,,EULER,1$TLOAD1,1,444,,0GRAV,444,,9.8,,,-1$$HYDSTAT,123,4,,,0,0,13.00,104000$FLOWDEF,25,,MMHYDRO,,,,,,++,HYDSTAT,123$ENDDATAModel 2

STARTCENDENDSTEP = 5000CHECK=NOTITLE= Jobname is: undex-2dTLOAD=1TIC=1SPC=1$TYPE (ALLEULER) = ARCHIVEELEMENTS (ALLEULER) = 2SET 2 = ALLMULTIEULHYDROELOUT (ALLEULER) = DENSITY,SIE,PRESSURE,XVEL,YVEL,ZVEL, FMAT4,MASS4,DENSITY4,FMATPLT4, FMATPLT,FMAT100,HMAT, FVUNC FMAT100,MASS100,DENSITY100,SIE100,SIE4STEPS(ALLEULER) = 0,1,thru,end,by,100SAVE (ALLEULER) = 10000$TYPE (ARCMAT) = TIMEHISMATS (ARCMAT) = 15SET 15 = 4MATOUT (ARCMAT) = EKIN,MASS,ZMOM,XMOM,YMOM


292

STEPS (ARCMAT) = 0,THRU,END,BY,10SAVE (ARCMAT) = 99999$TYPE (SHELL) = ARCHIVEELEMENTS (SHELL) = 4SET 4 = 1,THRU,6ELOUT (SHELL) = ZUSERSTEPS(SHELL) = 0,1SAVE (SHELL) = 10000$$------- Parameter Section ------PARAM,INISTEP,1.E-8$$------- BULK DATA SECTION -------BEGIN BULKPARAM,BULKL,0.1PARAM,COSUBMAX,100PARAM,COSUBCYC,100$ ------------------------------------------------------------------$$ * Euler.300 *$PARAM,MICRO,30PARAM,FMULTI,1.0PEULER1,1,,MMHYDRO,19$DMAT 4 1.e3 1 $EOSPOL 1 2.2e9 $DMAT 3 2 2EOSGAM,2,1.4$DMAT 100 100 2 $$$ ========= MATERIAL DEFINITIONS ==========$$$ ======== Load Cases ========================$$TICEUL,19,,,,,,,,++,BOX,1,4,6,2.0,,,,++,SPHERE,3,100,8,4.0BOX,1,,-100,-100,0.0,200,200,20.0SPHERE,3,,0.0,0.0,4.652,0.25TICVAL,6,,SIE,0.,DENSITY,1000.0TICVAL,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,,0GRAV,444,,9.8,,,-1$$HYDSTAT,1,4,,,0,0,13.00,104000$COUPLE,100,200,OUTSIDE,ON,ON,120COUPOR,1,120,,PORHYDST,75PORHYDST 75SURFACE,200,,PROP,2$PSHELL1 2 DUMMYSET1,2,2$GRID 1 -5.49990-5.49990 10.9999GRID 2 5.49990-5.49990 10.9999 GRID 3 -5.49990 5.49990 10.9999GRID 4 5.49990 5.49990 10.9999GRID 5 -5.49990-5.49990 0.0001 GRID 6 -5.49990 5.49990 0.0001 GRID 7 5.49990-5.49990 0.0001 GRID 8 5.49990 5.49990 0.0001 $$ --- Define 6 elements$$ -------- property set pdum ---------CQUAD4 1 2 1 2 4 3CQUAD4 2 2 5 6 8 7CQUAD4 3 2 5 7 2 1CQUAD4 4 2 2 7 8 4CQUAD4 5 2 4 8 6 3CQUAD4 6 2 3 6 5 1 ENDDATA

Dytran Example Problem ManualPrestressed Concrete Beam

294

Prestressed Concrete Beam

Problem DescriptionThis 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.

295Chapter 4: Fluid-structural InteractionPrestressed Concrete Beam

Analysis SchemeThe 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 Analysis Scheme of Prestressed Beam


296

Figure 4-67 Analysis Models and Behavior at Each Step


Dytran ModelThere 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:

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

Concrete: Cross area: 40000 mm2

Young’s modulus = 21 GPa


Length = 2 m

Tendon: Cross area: 1390 mm2 (Using ten 15.2mm class 2 relaxation standard strands)

Young’s modulus = 200 GPa


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)


298

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.

STARTSOLINIT=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

ResultsAnalytical 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.

(Equilibrium: no external loading)

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

element)where

is the force on the concrete beam

is the force on the tendon

is the deformation of the concrete beam

is the deformation of the tendon

Pc Pt+ 0=

δc δt 0.012= =

Pc

Pt

δc

δt


where

Substituting the and into Equilibrium and Compatibility equations, the compressive stresses in

concrete can be obtained.

where

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.

is Young’s modulus of the concrete beam

is Young’s modulus of the tendon

is the cross area of the concrete beam

is the cross area of the tendon

is the original length of the concrete beam

is the original length of the tendon

is the compression Stress of the concrete beam

Pc

EcAc

Lc------------δc=

Pt

EtAt

Lt---------- δc=

Ec

Et

Ac

At

Lc

Lt

Pc Pt

σc

Ec

Lc-----

EtAt

Lt----------

EcAc

Lc------------

EtAt

Lt----------+

------------------------------ 0.012⋅ ⋅=

σc


300

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

Figure 4-68 Two Results of xx-stress at the Beam


Figure 4-69 Results at Each Step


302


Prestress Job Input

PRESTRESS$BULKOUT=GRID.DATNASTDISP=ZEROSOLUOUT=GDIL.SOLTIME=100000MEMORY-SIZE=50000000,40000000CENDENDTIME=0.1ENDSTEP=10000000TITLE= Jobname is: preTLOAD=1TIC=1SPC=1$ Output result for request: DISPTYPE (DISP) = TIMEHISGRIDS (DISP) = 1SET 1 = 1 6 11 16 21 626GPOUT (DISP) = XPOS XVEL XFORCE XDIS TIMES (DISP) = 0 THRU END BY 1e-5SAVE (DISP) = 10000$ Output result for request: LAGTYPE (LAG) = ARCHIVEELEMENTS (LAG) = 2SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 0.01SAVE (LAG) = 10000$ Output result for request: STRESSTYPE (STRESS) = ARCHIVEELEMENTS (STRESS) = 3SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 0.01SAVE (STRESS) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,0.0001PARAM,VDAMP,0.001PARAM,INITNAS,PATRANPARAM,INISTEP,1e-8PARAM,MINSTEP,1e-8$------- BULK DATA SECTION -------BEGIN BULK$INCLUDE pre.bdfNASINIT,100,yes,0.1,0.001$$ ========== PROPERTY SETS ========== $


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

Transient Job Input

STARTSOLINIT=GDIL.SOLTIME=100000MEMORY-SIZE=50000000,40000000CENDENDTIME=0.1ENDSTEP=10000


304

CHECK=NOTITLE= Jobname is: traTLOAD=1TIC=1SPC=1$ Output result for request: DISPTYPE (DISP) = TIMEHISGRIDS (DISP) = 1SET 1 = 1 6 11 16 21 626 153 258 468 573GPOUT (DISP) = XPOS XVEL XFORCE XDIS TIMES (DISP) = 0 THRU END BY 1e-4 SAVE (DISP) = 10000$ Output result for request: LAGTYPE (LAG) = ARCHIVEELEMENTS (LAG) = 2SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 1.e-3SAVE (LAG) = 10000$ Output result for request: STRESSTYPE (STRESS) = ARCHIVEELEMENTS (STRESS) = 3SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 1.e-3SAVE (STRESS) = 10000$ Output result for request: DISPTYPE (LOCAL) = TIMEHISELEMENTs (LOCAL) = 4SET 4 = 111 THRU 411 BY 20ELOUT (LOCAL) = TXX EFFSTS PRESSURETIMES (LOCAL) = 0 THRU END BY 1e-4SAVE (LOCAL) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1e-8$=================================damping======PARAM,VDAMP,0.01$==================================$------- BULK DATA SECTION -------BEGIN BULKINCLUDE pre.bdf$$ ========== PROPERTY SETS ========== $$ * GID *$PBEAM,1,1,0.00139,.1,.1$$ * Conc *$PSOLID,2,2$$


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

Dytran Example Problem ManualBlast Simulation on Prestressed Concrete Beam

306

Blast Simulation on Prestressed Concrete Beam

Problem DescriptionThis 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 SchemeThe 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 Analysis Scheme of Blast Simulation on Prestressed Beam

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

Dytran ModelThe 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 as air.

Figure 4-71 Simulation Model

Both prestressed and transient runs include the following Euler definitions:

DMAT* 5 1.2 5EOSGAM 5 1.4 TICVAL,18,,DENSITY,1.592+4,SIE,4.765E6TICVAL,19,,DENSITY,1.2,SIE,1.94E+05TICEUL,20,,,,,,,,+ + ,SPHERE,4,5,19,1,,,,+ + ,SPHERE,17,5,18,5,,,SPHERE,4,,0,0,0,5000SPHERE,17,,1,0,1.,0.1

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

Air: Density = 1.2 kg/m3

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

γ

γ 1.4=


308

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,100ACTIVE,2,INTERACT,COUPLE,,,,,,++,TABLE,100TABLED1,100,,,,,,,,++,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 61 2 INSIDE ON ON +A000012+A000012 +A000013+A000013 63

The Euler mesh is generated by MESH entry.

MESH 63 BOX +A000020+A000020 -1 -2 -1 4 4 4 +A000021+A000021 41 41 41 EULER 5


Results

Figure 4-72 Blast on Concrete Beam at Various Steps

Figure 4-73 Displacement and Effective Stress at the Center of the Beam


310

Figure 4-74 Deformation of Prestressed Concrete Beam



Prestress Job Input

PRESTRESSNASTDISP=ZEROSOLUOUT=GDIL.SOLTIME=100000MEMORY-SIZE=50000000,40000000CENDENDTIME=0.1ENDSTEP=10000000TITLE= Jobname is: preTLOAD=1TIC=1SPC=1$ Output result for request: DISPTYPE (DISP) = TIMEHISGRIDS (DISP) = 1SET 1 = 1 6 11 16 21 626GPOUT (DISP) = XPOS XVEL XFORCE XDIS TIMES (DISP) = 0 THRU END BY 1e-5SAVE (DISP) = 10000$ Output result for request: LAGTYPE (LAG) = ARCHIVEELEMENTS (LAG) = 2SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 0.01SAVE (LAG) = 10000$ Output result for request: STRESSTYPE (STRESS) = ARCHIVEELEMENTS (STRESS) = 3SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 0.01SAVE (STRESS) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,0.0001PARAM,VDAMP,0.001PARAM,INITNAS,PATRANPARAM,FASTCOUP,INPLANE,FAILPARAM,INISTEP,1e-8PARAM,MINSTEP,1e-8$PARAM,LIMITER,ROE$PARAM,RKSCHEME,3$PARAM,INITFILE,V1$------- BULK DATA SECTION -------BEGIN BULK$flow out of all faces of outer eulerFLOWDEF,1,,HYDRO,,,,,,++,FLOW,BOTH,MATERIAL,5


312

ACTIVE,1,ELEMENT,EULHYDRO,,,,,,++,TABLE,100ACTIVE,2,INTERACT,COUPLE,,,,,,++,TABLE,100TABLED1,100,,,,,,,,++, 0, -1, 11, -1$PARAM,FAILOUT,NOPARAM,NZEROVEL,YESINCLUDE pre.bdfNASINIT,100,yes,0.1,0.001$$ ========== PROPERTY SETS ========== $$ * GID *$PBEAM,1,1,.00139,0.1,0.1$$ * Conc *$PSOLID,2,2$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material Barzel id =1MAT1,1,200e+9,,0.3,7850.$$ -------- Material DEBESH id =2DMATEL,2,2400,21.e+9,0.2,,,,$$$ ------- Velocity BC VELO ----- TLOAD1,1,3,,2FORCE,3,21,0,1,1.2e-1,,$RBE2, 1, 2,23,1002RBE2, 2, 3,23,1003RBE2, 3, 4,23,1004RBE2, 4, 5,23,1005RBE2, 5, 6,23,1006RBE2, 6, 7,23,1007RBE2, 7, 8,23,1008RBE2, 8, 9,23,1009RBE2, 9,10,23,1010RBE2,10,11,23,1011RBE2,11,12,23,1012RBE2,12,13,23,1013RBE2,13,14,23,1014RBE2,14,15,23,1015RBE2,15,16,23,1016RBE2,16,17,23,1017RBE2,17,18,23,1018


RBE2,18,19,23,1019RBE2,19,20,23,1020RBE2,20,21,23, 626$$PEULER1 5 HYDRO 20$$DMAT* 5 1.186 5EOSGAM 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,,0,0,0,5000SPHERE,17,,1,0,1.,0.1$COUPLE 61 2 INSIDE ON ON +A000012+A000012 +A000013+A000013 63 $SURFACE 2 SEG 100$MESH 63 BOX +A000020+A000020 -1 -2 -1 4 4 4 +A000021+A000021 41 41 41 EULER 5$CFACE1,1011,100,101,102,122$...$ENDDATA

Transient Job Input

STARTSOLINIT=GDIL.SOLTIME=100000MEMORY-SIZE=50000000,40000000CENDENDTIME=0.03ENDSTEP=1000000


314

CHECK=NOTITLE= Jobname is: traTLOAD=1TIC=1SPC=1$ Output result for request: DISP TYPE (DISP) = TIMEHISGRIDS (DISP) = 1SET 1 = 1 6 11 16 21 626 111 THRU 342 BY 21GPOUT (DISP) = XPOS XVEL XFORCE XDIS ZPOSTIMES (DISP) = 0 THRU END BY 0.00001SAVE (DISP) = 10000$ Output result for request: LAGTYPE (LAG) = ARCHIVEELEMENTS (LAG) = 2SET 2 = ALLLAGSOLID ELOUT (LAG) = TXX EFFSTS PRESSURE TIMES (LAG) = 0 THRU END BY 0.001SAVE (LAG) = 10000$ Output result for request: STRESSTYPE (STRESS) = ARCHIVEELEMENTS (STRESS) = 3SET 3 = ALLELEM1D ELOUT (STRESS) = xFORCE TIMES (STRESS) = 0 THRU END BY 0.001SAVE (STRESS) = 10000$ Output result for request: DISPTYPE (LOCAL) = TIMEHISELEMENTs (LOCAL) = 4SET 4 = 111 THRU 411 BY 20ELOUT (LOCAL) = TXX EFFSTS TIMES (LOCAL) = 0 THRU END BY 0.00001SAVE (LOCAL) = 10000$TYPE (AIR) = ARCHIVEELEMENTS (AIR) = 5SET 5 = ALLEULHYDROELOUT (AIR) = DENSITY SIE PRESSURE FMAT FVUNC FMATPLT , VOLUME XVEL YVEL ZVEL TIMES (AIR) = 0.006 THRU 0.007 BY 0.00001SAVE (AIR) = 10000$ Output result for request: LAG1TYPE (LAG1) = ARCHIVEELEMENTS (LAG1) = 7SET 7 = ALLLAGSOLID ELOUT (LAG1) = EFFSTS TIMES (LAG1) = 0.006 THRU 0.007 BY 0.00001SAVE (LAG1) = 10000$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,FASTCOUP,INPLANE,FAILPARAM,INISTEP,1e-8PARAM,MINSTEP,1e-8PARAM,LIMITER,ROE


PARAM,RKSCHEME,3PARAM,VDAMP,0.01$------- BULK DATA SECTION -------BEGIN BULK$flow out of all faces of outer eulerFLOWDEF,1,,HYDRO,,,,,,++,FLOW,BOTHACTIVE,1,ELEMENT,EULHYDRO,,,,,,++,TABLE,100ACTIVE,2,INTERACT,COUPLE,,,,,,++,TABLE,100TABLED1,100,,,,,,,,++,0,-1,0.006,-1,0.006,1,0.007,1,++,0.007,-1,.5,-1$PARAM,FAILOUT,NOPARAM,NZEROVEL,YESINCLUDE pre.bdf$$ ========== PROPERTY SETS ========== $$ * GID *$PBEAM,1,1,.00139,0.1,0.1$$ * Conc *$PSOLID,2,2$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material Barzel id =1MAT1,1,200e+9,,0.3,7850.$$ -------- Material DEBESH id =2DMATEL,2,2400,21.e+9,0.2,,,,$RBE2, 1, 2,23,1002RBE2, 2, 3,23,1003RBE2, 3, 4,23,1004RBE2, 4, 5,23,1005RBE2, 5, 6,23,1006RBE2, 6, 7,23,1007RBE2, 7, 8,23,1008RBE2, 8, 9,23,1009RBE2, 9,10,23,1010RBE2,10,11,23,1011RBE2,11,12,23,1012RBE2,12,13,23,1013RBE2,13,14,23,1014RBE2,14,15,23,1015RBE2,15,16,23,1016


316

RBE2,16,17,23,1017RBE2,17,18,23,1018RBE2,18,19,23,1019RBE2,19,20,23,1020RBE2,20,21,123, 626$$PEULER1 5 HYDRO 20$$DMAT* 5 1.2 5EOSGAM 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,,0,0,0,5000SPHERE,17,,1,0,1.,0.1$COUPLE 61 2 INSIDE ON ON +A000012+A000012 +A000013+A000013 63 $SURFACE 2 SEG 100$MESH 63 BOX +A000020+A000020 -1 -2 -1 4 4 4 +A000021+A000021 41 41 41 EULER 5$$CFACE1,1011,100,101,102,122...$ENDDATA

317Chapter 4: Fluid-structural InteractionVortex 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 :

Here , is the local shear stress, 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 is about 1.2. The drag coefficient is defined as:

where is the 2-D drag force, the main stream velocity and 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 until the drag force

approximately reaches . After a few runs, it is found out that using results in

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 .

Cf

tw

12---ρu2-------------=

tw p

CD

CDD

12---ρU2d------------------=

D U d

Cf

CD 1.2= Cf 0.095=

CD 1.30=

+ x

Dytran Example Problem ManualVortex Shedding with Skin Friction

318

Dytran ModelFor 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.


The material properties of Air are as follows:

Density = 1.3 kg/m3Dynamic Viscosity = 18.E-6 Pa sGamma value = 1.4 -------- Material GAS id =2DMAT 2 1.3 2 2 EOSGAM 2 1.4 18.E-6PMINC 2 0.

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 of 0.095 is specified on the

COUPLEoption as:

COUPLE 1 1 INSIDE ON ON STANDARD+ + ++ .095 The flow boundary conditions are:$ ------- Flow BC flowin ----- TLOAD1 1 29 4FLOW 29 1 FLOW INMATERIAL 2 DENSITY 1.3+ + SIE 200000. XVEL 20.

$ ------- Flow BC flowout ----- TLOAD1 1 32 4FLOW 32 2 FLOW OUT

Cf


320

ResultsThe 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.


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 of

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

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.

CD

CDD

12--- pU

2A

------------------ 0.34 0.01⁄0.5 1.3 20 20 0.1⋅ ⋅ ⋅ ⋅------------------------------------------------------ 34

26------ 1.3====

StndU------ d

TU--------==


322

Abbreviated Dytran Input FileSTARTCENDENDTIME=1.0ENDSTEP=9999999CHECK=NOTITLE= Jobname is: skinTLOAD=1TIC=1SPC=1$ Output result for request: EULTYPE (EUL) = ARCHIVEELEMENTS (EUL) = 1SET 1 = ALLEULHYDRO ELOUT (EUL) = XVEL YVEL ZVEL DENSITY SIE PRESSURE, TXX TYY TXY EFFSTS TIMES (EUL) = 0 THRU END BY 0.005SAVE (EUL) = 10000$ Output result for request: COUPTYPE (COUP) = ARCHIVECPLSURFS (COUP) = 2SET 2 = 1 CPLSOUT (COUP) = XVEL YVEL ZVEL DENSITY PRESSURE, TXX TYY TXY EFFSTS TIMES (COUP) = 0 THRU END BY 0.005SAVE (COUP) = 10000$$ Output result for request: COUPTHSTYPE (COUPTHS) = TIMEHISCPLSURFS (COUPTHS) = 3SET 3 = 1 CPLSOUT (COUPTHS) = XFORCE YFORCE ZFORCE RFORCETIMES (COUPTHS) = 0 THRU END BY 0.001SAVE (COUPTHS) = 10000$TYPE (PROF) = ARCHIVEELEMENTS (PROF) = 4SET 4 = 4231 THRU 4320 ELOUT (PROF) = XVEL YVEL ZVEL DENSITY PRESSURE, TXX TYY TXY EFFSTSTIMES (PROF) = 0 THRU END BY 0.001SAVE (PROF) = 10000$TYPE (MAT) = TIMEHISMATS (MAT) = 5SET 5 = 2 MATOUT (MAT) = MASS, EKIN, EINTTIMES (MAT) = 0 THRU END BY 0.001SAVE (MAT) = 10000$$ Output result for request: COUP THS$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,EULTRAN,IMPULSE


PARAM,FASTCOUPPARAM,BULKL,0.2PARAM,INISTEP,1.E-7$------- BULK DATA SECTION -------BEGIN BULKINCLUDE skin.bdf$$ ========== PROPERTY SETS ========== $$ * p1 *$PSHELL 1 1 .001$$ * peul *$PEULER1 2 HYDRO 31$$ * p2 *$PSHELL1 3 DUMMY$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material matrig id =1MATRIG 1 7800 2.1e+11 .3 $$ -------- Material GAS id =2DMAT 2 1.3 2 2 EOSGAM 2 1.4 18.E-6PMINC 2 0.$$ ======== Load Cases ========================$$$ ------- General Coupling: coup ----- $COUPLE 1 1 INSIDE ON ON STANDARD+ + ++ .095 $SURFACE 1 ELEM 2SET1 2 1 THRU 180$$ ------- Rigid Body Object MR1 ----- $ ---- Reference Node for Rigid body is 122TLOAD1 1 20 12FORCE 20 MR1 0 1 1 1TLOAD1 1 1020 12MOMENT 1020 MR1 0 1 1 1$$ ------- TICVAL BC init ----- TICVAL 28 DENSITY 1.3 SIE 200000.


324

$$ ------- Flow BC flowin ----- TLOAD1 1 29 4FLOW 29 1 FLOW INMATERIAL 2 DENSITY 1.3+ + SIE 200000. XVEL 20.CFACE 1 1 181 3CFACE 2 1 271 3CFACE 3 1 361 3CFACE 4 1 451 3CFACE 5 1 541 3CFACE 6 1 631 3CFACE 7 1 721 3CFACE 8 1 811 3CFACE 9 1 901 3CFACE 10 1 991 3CFACE 11 1 1081 3CFACE 12 1 1171 3CFACE 13 1 1261 3CFACE 14 1 1351 3CFACE 15 1 1441 3CFACE 16 1 1531 3CFACE 17 1 1621 3CFACE 18 1 1711 3CFACE 19 1 1801 3CFACE 20 1 1891 3CFACE 21 1 1981 3CFACE 22 1 2071 3CFACE 23 1 2161 3CFACE 24 1 2251 3CFACE 25 1 2341 3CFACE 26 1 2431 3CFACE 27 1 2521 3CFACE 28 1 2611 3CFACE 29 1 2701 3CFACE 30 1 2791 3CFACE 31 1 2881 3CFACE 32 1 2971 3CFACE 33 1 3061 3CFACE 34 1 3151 3CFACE 35 1 3241 3CFACE 36 1 3331 3CFACE 37 1 3421 3CFACE 38 1 3511 3CFACE 39 1 3601 3CFACE 40 1 3691 3CFACE 41 1 3781 3CFACE 42 1 3871 3CFACE 43 1 3961 3CFACE 44 1 4051 3CFACE 45 1 4141 3CFACE 46 1 4231 3CFACE 47 1 4321 3CFACE 48 1 4411 3CFACE 49 1 4501 3


CFACE 50 1 4591 3CFACE 51 1 4681 3CFACE 52 1 4771 3CFACE 53 1 4861 3CFACE 54 1 4951 3CFACE 55 1 5041 3CFACE 56 1 5131 3CFACE 57 1 5221 3CFACE 58 1 5311 3CFACE 59 1 5401 3CFACE 60 1 5491 3CFACE 61 1 5581 3CFACE 62 1 5671 3CFACE 63 1 5761 3CFACE 64 1 5851 3CFACE 65 1 5941 3CFACE 66 1 6031 3CFACE 67 1 6121 3CFACE 68 1 6211 3CFACE 69 1 6301 3CFACE 70 1 6391 3CFACE 71 1 6481 3CFACE 72 1 6571 3CFACE 73 1 6661 3CFACE 74 1 6751 3CFACE 75 1 6841 3CFACE 76 1 6931 3CFACE 77 1 7021 3CFACE 78 1 7111 3CFACE 79 1 7201 3CFACE 80 1 7291 3CFACE 81 1 7381 3CFACE 82 1 7471 3CFACE 83 1 7561 3CFACE 84 1 7651 3CFACE 85 1 7741 3CFACE 86 1 7831 3CFACE 87 1 7921 3CFACE 88 1 8011 3CFACE 89 1 8101 3CFACE 90 1 8191 3$$ ------- TICEUL BC reg ----- TICEUL 31 + + SPHERE 2 2 28 5 SPHERE 2 0 0 0 10$$ ------- Flow BC flowout ----- TLOAD1 1 32 4FLOW 32 2 FLOW OUT$CFACE 91 2 270 6CFACE 92 2 360 6CFACE 93 2 450 6


326

CFACE 94 2 540 6CFACE 95 2 630 6CFACE 96 2 720 6CFACE 97 2 810 6CFACE 98 2 900 6CFACE 99 2 990 6CFACE 100 2 1080 6CFACE 101 2 1170 6CFACE 102 2 1260 6CFACE 103 2 1350 6CFACE 104 2 1440 6CFACE 105 2 1530 6CFACE 106 2 1620 6CFACE 107 2 1710 6CFACE 108 2 1800 6CFACE 109 2 1890 6CFACE 110 2 1980 6CFACE 111 2 2070 6CFACE 112 2 2160 6CFACE 113 2 2250 6CFACE 114 2 2340 6CFACE 115 2 2430 6CFACE 116 2 2520 6CFACE 117 2 2610 6CFACE 118 2 2700 6CFACE 119 2 2790 6CFACE 120 2 2880 6CFACE 121 2 2970 6CFACE 122 2 3060 6CFACE 123 2 3150 6CFACE 124 2 3240 6CFACE 125 2 3330 6CFACE 126 2 3420 6CFACE 127 2 3510 6CFACE 128 2 3600 6CFACE 129 2 3690 6CFACE 130 2 3780 6CFACE 131 2 3870 6CFACE 132 2 3960 6CFACE 133 2 4050 6CFACE 134 2 4140 6CFACE 135 2 4230 6CFACE 136 2 4320 6CFACE 137 2 4410 6CFACE 138 2 4500 6CFACE 139 2 4590 6CFACE 140 2 4680 6CFACE 141 2 4770 6CFACE 142 2 4860 6CFACE 143 2 4950 6CFACE 144 2 5040 6CFACE 145 2 5130 6CFACE 146 2 5220 6CFACE 147 2 5310 6


CFACE 148 2 5400 6CFACE 149 2 5490 6CFACE 150 2 5580 6CFACE 151 2 5670 6CFACE 152 2 5760 6CFACE 153 2 5850 6CFACE 154 2 5940 6CFACE 155 2 6030 6CFACE 156 2 6120 6CFACE 157 2 6210 6CFACE 158 2 6300 6CFACE 159 2 6390 6CFACE 160 2 6480 6CFACE 161 2 6570 6CFACE 162 2 6660 6CFACE 163 2 6750 6CFACE 164 2 6840 6CFACE 165 2 6930 6CFACE 166 2 7020 6CFACE 167 2 7110 6CFACE 168 2 7200 6CFACE 169 2 7290 6CFACE 170 2 7380 6CFACE 171 2 7470 6CFACE 172 2 7560 6CFACE 173 2 7650 6CFACE 174 2 7740 6CFACE 175 2 7830 6CFACE 176 2 7920 6CFACE 177 2 8010 6CFACE 178 2 8100 6CFACE 179 2 8190 6CFACE 180 2 8280 6$$ENDDATA

Dytran Example Problem ManualGeometric Eulerian Boundary Conditions

328

Geometric Eulerian Boundary Conditions

Problem DescriptionFor 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 ModelFor 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 = 1.3 kg/m3

329Chapter 4: Fluid-structural InteractionGeometric 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:

• 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.

The material properties of Air are as follows:

Density = 1.3 kg/m3Dynamic Viscosity = 18.E-6 Pa sGamma value = 1.4

$ -------- Material GAS id =2DMAT 2 1.3 2 2 EOSGAM 2 1.4 18.E-6PMINC 2 0.$

Specific internal energy = 2.E+5 Jkg

Density = 1.3 kg/m3

X-Velocity = 20 m/s

Y-Velocity = 0

Z-Velocity = 0

FLOW: INXVEL = 20 m/sYVEL = 0 m/sSIE = 2.E+5 J/kgDENSITY= 1.3 kg/m3


330

The Euler Mesh and the initial values for the Euler region are as follows:

$MESH 22 BOX + + -.40 -.40 0.0 .80 .80 .01 + + 90 90 1 EULER 2$$$ ------- TICEUL BC reg ----- TICEUL 31 + + SPHERE 2 2 28 5 SPHERE 2 0 0 0 10$$ ------- TICVAL BC init ----- TICVAL 28 DENSITY 1.3 SIE 200000. $

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 ON STANDARD+

+ ++ 22 .095 $SURFACE 1 ELEM 2SET1 2 1 THRU 180$

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 x-direction. 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

------------------------


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, NEGYWALLDIR, 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.


332

ResultsThe 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.

Results at the time around 0.495 seconds.

New model with Mesh and Geometric BC Old model with Grids and Chexa’s



Results at the time around 0.500 seconds.

The Time History plots of the FORCE and the Mass are shown in the following pictures:




334

The figures above show that this new modeling method gives exactly the same results as those of the old traditional method.


STARTCENDENDTIME=0.5ENDSTEP=9999999CHECK=NOTITLE= Jobname is: skinTLOAD=1TIC=1SPC=1$$$ Output result for request: EULTYPE (EUL) = ARCHIVEELEMENTS (EUL) = 1SET 1 = ALLEULHYDRO ELOUT (EUL) = XVEL YVEL ZVEL DENSITY SIE PRESSURE, TXX TYY TXY EFFSTS TIMES (EUL) = 0 THRU END BY 0.005SAVE (EUL) = 10000$ Output result for request: COUPTYPE (COUP) = ARCHIVECPLSURFS (COUP) = 2SET 2 = 1 CPLSOUT (COUP) = XVEL YVEL ZVEL DENSITY PRESSURE, TXX TYY TXY EFFSTS TIMES (COUP) = 0 THRU END BY 0.005SAVE (COUP) = 10000$$ Output result for request: COUPTHSTYPE (COUPTHS) = TIMEHISCPLSURFS (COUPTHS) = 3SET 3 = 1 CPLSOUT (COUPTHS) = XFORCE YFORCE ZFORCE RFORCE



TIMES (COUPTHS) = 0 THRU END BY 0.001SAVE (COUPTHS) = 10000$$TYPE (MAT) = TIMEHISMATS (MAT) = 5SET 5 = 2 MATOUT (MAT) = MASS, EKIN, EINTTIMES (MAT) = 0 THRU END BY 0.001SAVE (MAT) = 10000$$$ Output result for request: COUP THS$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,EULTRAN,IMPULSEPARAM,FASTCOUPPARAM,BULKL,0.2PARAM,INISTEP,1.E-7$------- BULK DATA SECTION -------BEGIN BULKINCLUDE skin.bdf$$ ========== PROPERTY SETS ========== $$ * p1 *$PSHELL 1 1 .001$$ * peul *$PEULER1 2 HYDRO 31$$ * p2 *$PSHELL1 3 DUMMY$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material matrig id =1MATRIG 1 7800 2.1e+11 .3 $$ -------- Material GAS id =2DMAT 2 1.3 2 2 EOSGAM 2 1.4 18.E-6PMINC 2 0.$$ ======== Load Cases ========================$$$ ------- General Coupling: coup ----- $


336

COUPLE 1 1 INSIDE ON ON STANDARD+

+ ++ 22 .095 $SURFACE 1 ELEM 2SET1 2 1 THRU 180$$ ------- Rigid Body Object MR1 ----- $ ---- Reference Node for Rigid body is 122TLOAD1 1 20 12FORCE 20 MR1 0 1 1 1TLOAD1 1 1020 12MOMENT 1020 MR1 0 1 1 1$MESH 22 BOX + + -.40 -.40 0.0 .80 .80 .01 + + 90 90 1 EULER 2$$$ ------- 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---------$WALLDIR,5,HYDRO,22,NEGYWALLDIR,6,HYDRO,22,POSYFLOWDIR,7,HYDRO,22,NEGX,,,,,++,FLOW,IN,MATERIAL,2,DENSITY,1.3,SIE,200000.,++,XVEL,20.FLOWDIR,8,HYDRO,22,POSX,,,,,++,FLOW,OUT$ENDDATA

337Chapter 4: Fluid-structural InteractionCohesive Friction

Cohesive Friction

Problem DescriptionWet 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 ModelingThe soil is modeled by:

$ -------- Material Soil id =2DMAT 2 1500 2 2 2 2EOSPOL 2 2e+07 SHREL 2 1e+07YLDMC 2 6.0e+07 2e+05 1.46PMINC 2 -5e+8

The elements are defined by MESH,BOX:

MESH 2 BOX +

+ -.2 -0.0101 0.0 0.4 0.0202 0.2 +

+ 160 8 80 EULER 1

Dytran Example Problem ManualCohesive Friction

338

and are initialized by

PEULER1 1 MMSTREN 6TICVAL 4 DENSITY 1500$$ ------- TICEUL BC PEuler_reg ----- TICEUL 6 +

+ CYLINDER 5 1 +

+ CYLINDER 3 2 4 2 CYLINDER 5 0 0 -2 0 0 2+

+ 1CYLINDER 3 0 0 -1 0 0 0.1+

+ 1

Here the multi-material solver will be used.

The wedge is defined by:

MATRIG 1 100 +

+ +

+ 0.00

GRID 1 .020000 -.03000 .120000GRID 2 -.020000 -.03000 .120000GRID 3 .000000 -.03000 .020000GRID 4 -.020000 .03000 .120000GRID 5 .000000 .03000 .020000GRID 9 .020000 .03000 .120000$$ --- Define 5 elements$$ -------- property set dum ---------CTRIA3 1 2 1 2 3CQUAD4 2 2 4 5 3 2CTRIA3 3 2 5 4 9CQUAD4 4 2 1 3 5 9CQUAD4 5 2 1 9 4 2

The interaction between wedge and Euler material is implemented by the COUPLE entry:

COUPLE 1 1 INSIDE ON ON STANDARD+

+ 0.8 0.8 +

+ 2$

Here a Coulomb friction model is used.


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 1 ELEM 6SET1 6 1 THRU 5

The motion of the wedge is given by:

$ ---- No reference node is used.TLOAD1 1 7 12 1FORCE 7 MR1 1 0 0 2.0TLOAD1 1 1007 12MOMENT 1007 MR1 1 0 0 0$tabled1,1,,,,,,,,++,0,0,0.2,1,endt

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.


340

ResultsThe pictures below show FMATPLT results for respectively 0 ms, 40 ms and 80 ms. The soils sticks to the wedge.


Dytran Input DeckSTARTCENDENDTIME=0.1ENDSTEP=9999999CHECK=NOTITLE= Jobname is: Wedge1910TLOAD=1TIC=1SPC=1$ Output result for request: CouplingTYPE (Coupling) = ARCHIVECPLSURFS (Coupling) = 1SET 1 = 1 CPLSOUT (Coupling) = PRESSURE TYY TZZ TXY TYZ TZX EFFSTS SXX TIMES (Coupling) = 0 THRU END BY 0.001SAVE (Coupling) = 10000$ Output result for request: Coupling_velTYPE (Coupling_vel) = TIMEHISCPLSURFS (Coupling_vel) = 2SET 2 = 1 CPLSOUT (Coupling_vel) = XFORCE YFORCE ZFORCE RFORCE STEPS (Coupling_vel) = 0 THRU END BY 1SAVE (Coupling_vel) = 10000$ Output result for request: PenTYPE (Pen) = ARCHIVEELEMENTS (Pen) = 3SET 3 = 1 THRU 5 ELOUT (Pen) = EXUSER1 TIMES (Pen) = 0 THRU END BY 0.001SAVE (Pen) = 10000$ Output result for request: Pen_velTYPE (Pen_vel) = TIMEHISRIGIDS (Pen_vel) = 4SET 4 = mr1 RBOUT (Pen_vel) = ZCG ZVEL ZACC ZFORCE STEPS (Pen_vel) = 0 THRU END BY 1SAVE (Pen_vel) = 10000$ Output result for request: SoilTYPE (Soil) = ARCHIVEELEMENTS (Soil) = 5SET 5 = ALLMULTIEULSTRENGTHELOUT (Soil) = XVEL YVEL ZVEL DENSITY PRESSURE FMAT2 FMATPLT, TXX TYY TZZ TXY TYZ TZX EFFSTS VOID FVUNC TIMES (Soil) = 0 THRU END BY 0.001SAVE (Soil) = 10000$------- Parameter Section ------PARAM,BULKL,1.0PARAM,CONTACT,THICK,0.0PARAM,FASTCOUPPARAM,INISTEP,1e-6PARAM,VELMAX,5.0,NO$------- BULK DATA SECTION -------BEGIN BULK


342

$$PARAM,COHESION,8e+10,8e+5,2.0$ --- Define 18 grid points --- $GRID 1 .020000 -.03000 .120000GRID 2 -.020000 -.03000 .120000GRID 3 .000000 -.03000 .020000GRID 4 -.020000 .03000 .120000GRID 5 .000000 .03000 .020000GRID 9 .020000 .03000 .120000$$ --- Define 5 elements$$ -------- property set dum ---------CTRIA3 1 2 1 2 3CQUAD4 2 2 4 5 3 2CTRIA3 3 2 5 4 9CQUAD4 4 2 1 3 5 9CQUAD4 5 2 1 9 4 2$ ========== PROPERTY SETS ========== $$ * PEul_MMStren *$PEULER1 1 MMSTREN 6$$ * Pen *$PSHELL 2 1 .001$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material Pen_rigid id =1MATRIG 1 100 + + + + 0.00 $$ -------- Material Soil id =2DMAT 2 1500 2 2 2 2EOSPOL 2 2e+07 SHREL 2 1e+07YLDMC 2 6.0e+07 2e+05 1.46PMINC 2 -5e+8 $$ ======== Load Cases ========================$$$ ------- General Coupling: Coup_pen ----- $COUPLE 1 1 INSIDE ON ON STANDARD+ + 0.8 0.8 + + 2


$SURFACE 1 ELEM 6SET1 6 1 THRU 5 $$ ------- Mesh Box: Mesh_box$MESH 2 BOX + + -.2 -0.0101 0.0 0.4 0.0202 0.2 + + 160 8 80 EULER 1$$ ------- TICVAL BC Soil_init ----- TICVAL 4 DENSITY 1500$$ ------- TICEUL BC PEuler_reg ----- TICEUL 6 + + CYLINDER 5 1 + + CYLINDER 3 2 4 2 CYLINDER 5 0 0 -2 0 0 2+ + 1CYLINDER 3 0 0 -1 0 0 0.1+ + 1$$ ------- Rigid Body Object RBO_Pen ----- $ ---- No reference node is used.TLOAD1 1 7 12 1FORCE 7 MR1 1 0 0 2.0TLOAD1 1 1007 12MOMENT 1007 MR1 1 0 0 0$tabled1,1,,,,,,,,++,0,0,0.2,1,endt$ENDDATA


344

Chapter 5: FormingDytran Example Problem Manual

5Forming

Overview 346

Square Cup Deep Drawing 347

Deep Drawing of a Cylindrical Cup 359

Three-point Bending Test 368

Sleeve Section Stamping 379


346

OverviewIn 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.

347Chapter 5: FormingSquare Cup Deep Drawing

Square Cup Deep Drawing

Problem DescriptionThis 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.

Dytran Example Problem ManualSquare Cup Deep Drawing

348

The dimensions of the plate, die, punch, and clamp are all given in Figure 5-1

.

Figure 5-1 Dimensions of Plate, Die, Punch, and Clamp (in Millimeters)


Dytran ModelThe 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


350

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

• 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

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.


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.

ResultsThe 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.


352

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)


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)


354

Figure 5-7 Contour Plot of the Forming Limit Parameter(Aluminum Plate at a Punch Depth of 20 mm)

Files

sq_cup_alual.dat

sq_cup_alual_xl.dat

Dytran input file

SQ_CUP_ALUAL.OUT Dytran output file

SQ_CUP_ALUAL_PLATE_0.ARCSQ_CUP_ALUAL_PLATE_0.ARC

Dytran archive file

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


References1. 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 FileSTART$TITLE = Deep Drawing of a Square Cup $CHECK = NOTLOAD= 1$ENDSTEP = 4700$$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS$TYPE(MYMAT) = MATSUMSTEPS(MYMAT) = 0 THRU END BY 1000$$ CHANGE THE OUTPUT FREQUENCY FOR MATRIG SUMMS$TYPE(MRMAT) = MRSUMSTEPS(MRMAT) = 0 THRU END BY 1000$$ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS$TYPE(MYCYC) = STEPSUMSTEPS(MYCYC) = 0 THRU END BY 250$$ Archive of Sheet Metal Output$ Note : Integration Layer 01 corresponds to the die side of the sheet$STEPS(PLATE) = 0,THRU,END,BY,470,3418TYPE(PLATE) = ARCHIVESAVE(PLATE) = 100000ELEMENTS(PLATE) = 60SET 60 = 1t1600ELOUT(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,470,3418TYPE(RIGID) = ARCHIVESAVE(RIGID) = 100000ELEMENTS(RIGID) = 63SET 63 = 2001t4468


356

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.$$TYPE(wall) = TIMEHISSAVE(wall) = 999999ELEMENTS(wall) = 1SET 1 = 468,492,509ELOUT(wall) = THICK,EPSMX01,EPSMN01,EZZ01,FLP01STEPS(wall) = 0,THRU,END,BY,47,3418$BEGIN BULK$$ Parameters :$ (i) Initial Time-stepPARAM,INISTEP,2.0-7$ (ii) Update of shell thicknessesPARAM,SHTHICK,YES$ (iii) Hourglass CoefficientPARAM,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$ die)$PSHELL1 61 2 Bely Gauss Mid ++ 1.-20PSHELL1 62 3 Bely Gauss Mid ++ 1.-20PSHELL1 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


$$ Planar Anisotropic Material Specification :$$ 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:$$ The effect of binder force is examined. Take F=19.6 kN.$TLOAD1 1 9999 13FORCE 9999 MR3 -19.6E6 1.$$ Prescribed motions:$$ (i) The punch velocity is prescribed as a constant downward value$ of 1000 mm/s (in Z-dir.).$TLOAD1 1 99 12FORCE 99 MR2 -1000. 1.$$ (ii) Rotational velocities of MATRIG 2, 3 and 4 (i.e. punch, binder$ and die) are equal to zero along X-, Y- and Z-axesTLOAD1 1 999 12MOMENT 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


358

CONTACT 1 SURF SURF 1 2 .162 .162 0.0 ++ Top Full 1.0 ++ 0.0 .1 Distance1.+20 1.1 .1 Factor 2. ++ 0.0 .04 On On 1.+20$ (ii) Contact between Blank and BinderCONTACT 2 SURF SURF 1 3 .162 .162 0.0 ++ Top Full 1.0 ++ 0.0 .1 Distance1.+20 1.1 .1 Factor 2. ++ 0.0 .04 On On 1.+20$ (iii) Contact between Blank and DieCONTACT 3 SURF SURF 1 4 .162 .162 0.0 ++ Bottom Full 1.0 ++ 0.0 .1 Distance1.+20 1.1 .1 Factor 2. ++ 0.0 .04 On On 1.+20$ENDDATA

359Chapter 5: FormingDeep Drawing of a Cylindrical Cup

Deep Drawing of a Cylindrical Cup

Problem DescriptionCylindrical 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:

70/30 Brass Material Properties

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

Young’s modulus E = 100 000 N/mm2


Density 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

Dytran Example Problem ManualDeep Drawing of a Cylindrical Cup

360

Dytran ModelSymmetry 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.


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.

ResultsThe 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.


362

Figure 5-10 Dytran and Experimental Punch Force

Figure 5-11 Dytran and Experimental Radial and Circumferential Strain Distribution


Files

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 FileSTART$TITLE = Cylindrical Cup Deep Drawing$TIME 99999CENDCHECK = NO ENDSTEP = 9868SPC = 1TLOAD = 1$$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS$TYPE (MYMAT) = MATSUMSTEPS (MYMAT) = 0 THRU END BY 1000$$ CHANGE THE OUTPUT FREQUENCY FOR THE MATRIG SUMMS$TYPE (MYMR) = MRSUMSTEPS (MYMR) = 0 THRU END BY 1000$$ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS$TYPE (MYCYC) = STEPSUMSTEPS (MYCYC) = 0 THRU END BY 250$$$$ Data for Output Control Set 1$$TYPE (blank) = ARCHIVESAVE (blank) = 999999ELEMENTS(blank) = 1

cyl_cup.dat

cyl_cup_xl.dat

Dytran input file

CYL_CUP.OUT Dytran output file

CYL_CUP_BLANK_0.ARCCYL_CUP_RIGID_0.ARC

Dytran archive file

CYL_CUP_CONTACT_0.THSCYL_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


364

SET 1 = 1t348ELOUT (blank) = THICK,EPSMX03,EPSMN03,EZZ03STEPS (blank) = 0,THRU,END,BY,1000,6323$$$$ Data for Output Control Set 2$$TYPE (rigid) = ARCHIVESAVE (rigid) = 999999ELEMENTS(rigid) = 2SET 2 = 1000t3035STEPS (rigid) = 0,THRU,END,BY,1000,6323ELOUT (rigid) = ZUSER$$ Time history of the Contact 1 (i.e. Punch)$CONTS (contact) = 3SET 3 = 1TYPE (contact) = TIMEHISCONTOUT (contact) = YFORCESAVE (contact) = 99999STEPS (contact) = 0,THRU,END,BY,100,6323$$ Time history of the Punch Motion (midpoint)$TYPE (punch_mot) = TIMEHISSAVE (punch_mot) = 99999GRIDS (punch_mot) = 4SET 4 = 1010GPOUT (punch_mot) = YDISSTEPS (punch_mot) = 0,THRU,END,BY,100,6323$BEGIN BULK$PARAM INISTEP 3e-07PARAM 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.-20PSHELL1 2222 3 Bely Gauss Mid +


+ 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


366

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 55FORCE 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+..$$ (ii) Rotational velocities of MATRIG 2, 3 and 4 (i.e. punch, die$ and binder) are equal to zero along X-, Y- and Z-axesTLOAD1 1 999 12MOMENT 999 MR2 0. 1. 1. 1.MOMENT 999 MR3 0. 1. 1. 1.MOMENT 999 MR4 0. 1. 1. 1.$$ (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$$ (i) Contact between Blank and PunchCONTACT 1 SURF SURF 1 2 .18 .18 0.0 ++ Both Full 1.0 ++ 0.0 .1 Distance1.+20 1.1 .1 Factor 2. ++ 0.0 1.+20 On On 1.+20$$ (ii) Contact between Blank and DieCONTACT 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


CONTACT 3 SURF SURF 1 4 .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.+20ENDDATA

Dytran Example Problem ManualThree-point Bending Test

368

Three-point Bending Test

Problem DescriptionThis 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:

Total punch displacement 92 mm

Total time for punch displacement 285 sec

Punch closure rate 0.3228 mm/sec

Assumed friction coefficient 0.05

369Chapter 5: FormingThree-point Bending Test

Aluminum 6111-T4 Material Properties

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 ModelDue 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.

Young’s modulus E = 69248 N/mm2


Density 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


370


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.


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.

ResultsFigure 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).


372

Figure 5-14 Deformed Configurations at various Stages throughout the Forming Process


Figure 5-15 Position of Grid Point Number 255

The springback angle φ is required as benchmark output and is compared with experimental results. In algebraic form:

in which XDIM_# denotes the X-dimension of grid point number #, and given by

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.

Note: 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).

φ 2 * arcXDIM_255-XDIM_225ZDIS_255-ZDIS_225

----------------------------------------------------------⎝ ⎠⎛ ⎞tan=

XDIM_# = XDIS_# + XPOS_#


374

Figure 5-16 Springback after Forming Process

Files

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.ARC3PBT_UNDAMP_ROLL_0.ARC

Dytran archive file (undamped)

3PBT_DMP_BLANK_0.ARC3PBT_DMP_PUNCH_0.ARC3PBT_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


Abbreviated Dytran Input FileSTART$$USERCODE=exvdmp.f( used for the damped problem )$TITLE = 3-Point Bending Test on Al6111-T4 $-----------------------------------------------------------------------Global system damping is applied:( used for the damped problem )$ The user-subroutine exvdmp.f is employed in which the parameter $ VDAMP is switched on at the end of the forming stage. VDAMP $ corresponds with the critical damping coefficient :$VDAMP = omega * timestep = 2*pi*4.9665E-7/0.02343= 1.33186E-4$-----------------------------------------------------------------------TIME 99999CENDCHECK = NOENDSTEP = 744000SPC = 1TLOAD = 1$$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS$TYPE (MYMAT) = MATSUMSTEPS (MYMAT) = 0 THRU END BY 5000$$ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS$TYPE (MYCYC) = STEPSUMSTEPS (MYCYC) = 0 THRU END BY 2500$ CHANGE THE OUTPUT FREQUENCY FOR THE MATRIG-RBE2 SUMMS$TYPE (MYMR) = MRSUMSTEPS (MYMR) = 0 THRU END BY 5000$$$ Data for Output Control Set 1$$TYPE (blank) = ARCHIVESAVE (blank) = 999999ELEMENTS(blank) = 1SET 1 = 1t200ELOUT (blank) = ZUSERSTEPS (blank) = 0,THRU,744000,BY,74400,573739$$$$ Data for Output Control Set 2$$TYPE (punch) = ARCHIVESAVE (punch) = 999999ELEMENTS(punch) = 2SET 2 = 601t780STEPS (punch) = 0,THRU,744000,BY,74400,573739ELOUT (punch) = ZUSER$$$$ Data for Output Control Set 3TYPE (roll) = ARCHIVESAVE (roll) = 999999ELEMENTS(roll) = 3SET 3 = 781t1050TIMES (roll) = 0ELOUT (roll) = ZUSER$$------------------------------------------------------------$$ Case Control which refers to User-subroutine( used for the damped$ problem)$$ELEXOUT (userout)


376

$ELEMENTS(userout) = 4 $SET 4 = 1$STEPS (userout) = 0tendb1$$------------------------------------------------------------$ Time history of the Position of Grid #255$TYPE (grid_mot) = TIMEHISSAVE (grid_mot) = 99999GRIDS (grid_mot) = 5SET 5 = 255STEPS (grid_mot) = 0,THRU,744000,BY,744,573739 GPOUT (grid_mot) = XDIS$$$BEGIN BULK$PARAM INISTEP 2e-7PARAM SHTHICK YES$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 1SET1 200 1111SET1 300 2222


$$ Contact specification for which the blank (i.e. surface 1) is $ in all cases the slave.$ (i) Contact between Blank and PunchCONTACT 1 SURF SURF 1 2 .05 .05 0.0 ++ Both Full 1.0 ++ 0.0 .1 Distance1.+20 1.1 .1 Factor 2. ++ 0.0 0.285 On On 1.+20$$ (ii) Contact between Blank and RollCONTACT 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$$ Prescribed motions:$ (i) The punch velocity is described by a sine function such that$ both the velocity and acceleration are equal to zero at the$ start and end of the stroke. The maximum downward velocity$ amounts to 645.6 mm/s (in Z-dir.) and mean velocity is 322.8 mm/sTLOAD1 1 99 12 55FORCE 99 MR2 -322.8 1.TABLED1 55 ++ .000E+000.000E+0.594E-020.856E-2.119E-010.341E-1.178E-010.761E-1++ .237E-010.134E+0.297E-010.207E+0.356E-010.293E+0.416E-010.391E+0+...$$ (ii) Rotational velocities of MATRIG 2 and 3 (i.e. punch and roll) $ are equal to zero around X-, Y- and Z-axesTLOAD1 1 999 12


378

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

379Chapter 5: FormingSleeve Section Stamping

Sleeve Section Stamping

Problem DescriptionThis 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:

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

Dytran Example Problem ManualSleeve Section Stamping

380

Aluminum 6111-T4 Material Properties

Workhardening data are provided in tabulated form.

Dytran ModelDue 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.


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.

Young’s modulus E = 181302 N/mm2


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


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:

in which T denotes the natural period of free vibration and 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.

VDAMP2 * π * δt

T-------------------------=

δt


382

Figure 5-19 Deformed Configurations at various Stages throughout the Stamping Process

Figure 5-20 Position of Grid Point Number 10

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:

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 = 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 Experiment

Chord Length [mm] 62.54 61.58

Depth [mm] 13.77 14.38

Total Force [kN] 84.90 40 - 225


384

Files

Reference5. Slagter, W. “Simulation of Sheet Metal Forming Processes using Dytran.”

Abbreviated Dytran Input FileSTART$$USERCODE=exvdmp.f( used for the damped problem )$TITLE = Sleeve Section Stamping $$-------------------------------------------------------------------$ 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 99999CENDCHECK = NOENDTIME = 0.025SPC = 1TLOAD = 1$$ CHANGE THE OUTPUT FREQUENCY FOR MATERIAL SUMMS$TYPE (MYMAT) = MATSUMSTEPS (MYMAT) = 0 THRU END BY 1000$$ CHANGE THE OUTPUT FREQUENCY FOR RIGID SUMMS$TYPE (MRMAT) = MRSUM

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.ARCSLEEVE_UNDAMP_PUNCH_0.ARCSLEEVE_UNDAMP_DIE_0.ARC

Dytran archive file (undamped)

SLEEVE_DMP_BLANK_0.ARCSLEEVE_DMP_PUNCH_0.ARCSLEEVE_DMP_DIE_0.ARC

Dytran archive file (damped)

SLEEVE_UNDAMP_CONTACT_0.THSSLEEVE_UNDAMP_GRID_MOT_0.THS

Dytran time history file (undamped)

SLEEVE_DMP_CONTACT_0.THSSLEEVE_DMP_GRID_MOT_0.THS

Dytran time history file (damped)


STEPS (MRMAT) = 0 THRU END BY 1000$$ CHANGE THE OUTPUT FREQUENCY FOR THE CYCLE SUMMS$TYPE (MYCYC) = STEPSUMSTEPS (MYCYC) = 0 THRU END BY 250$$$$ Data for Output Control Set 1$$TYPE (blank) = ARCHIVESAVE (blank) = 999999ELEMENTS(blank) = 1SET 1 = 1t600ELOUT (blank) = EFFSTS,EFFPLSTIMES (blank) = 0,0.003,0.004,0.007,0.025$$$$ Data for Output Control Set 2$$TYPE (punch) = ARCHIVESAVE (punch) = 999999ELEMENTS(punch) = 2SET 2 = 12000t12419TIMES (punch) = 0,0.003,0.004,0.007,0.025ELOUT (punch) = ZUSER$$$$ Data for Output Control Set 3$$TYPE (die) = ARCHIVESAVE (die) = 999999ELEMENTS(die) = 3SET 3 = 13000t13559TIMES (die) = 0ELOUT (die) = ZUSER$$ Time history of the MATRIG #2 (i.e. Punch)$RIGIDS (contact) = 4SET 4 = MR2TYPE (contact) = TIMEHISRBOUT (contact) = ZFORCESAVE (contact) = 99999TIMES (contact) = 0tendb0.000025,0.007$$ Time history of the Position of Grid #10 (at YZ-plane)$TYPE (grid_mot) = TIMEHISSAVE (grid_mot) = 99999GRIDS (grid_mot) = 5SET 5 = 10TIMES (grid_mot) = 0tendb0.000025GPOUT (grid_mot) = ZPOS$$-------------------------------------------------------------------$ Case Control which refers to User-subroutine( used for the damped problem )$$ELEXOUT (userout)$ELEMENTS(userout) = 6$SET 6 = 1


386

$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.-20PSHELL1 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$$ Geometry description of blank, punch and die in single surface$ contact. SET1 11 1111SET1 22 2222$SURFACE 1 SEG 1 PROP 11 PROP 22 $CONTACT 1 SURF 1 .30 .30 0.0 ++ V4 0.0 0.0 ++ ++ 0.007$$ 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


$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$$ Prescribed motions:$ (i) The punch velocity is described by a sine function such that$ both the velocity and acceleration are equal to zero at the$ start and end of the stroke. The maximum downward velocity$ amounts of 5000 mm/s (in Z-dir.) and mean velocity is 2.5e3 mm/s.TLOAD1 1 99 12 66FORCE 99 MR2 -2500. 1.TABLED1 66 ++ .000E+00 .000E+0.146E-03 .856E-2.292E-03 .341E-1.438E-03 .761E-1++ .583E-03 .134E+0.729E-03 .207E+0.875E-03 .293E+0.102E-02 .391E+0+....$$ (ii) Rotational velocities of MATRIG 2 and 3 (i.e. punch and$ die) are equal to zero around X-, Y- and Z-axesTLOAD1 1 999 12MOMENT 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


388

Chapter 6: Occupant Safety Dytran Example Problem Manual

6Occupant Safety

Overview 390

Flat Unfolded Air Bag Inflation (GBAG) 391

Rigid Ellipsoid Dummy Hitting a Passenger Air Bag 400

Sled Test Verification of the Enhanced Hybrid III Dummy (50%) 416

Side Curtain Air Bag (Courtesy of Autoliv) 462

Hybrid III 5th%-tile Dummy 479

Easy Postprocessing with Adaptive Meshing 509


390

OverviewIn 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.

391Chapter 6: Occupant SafetyFlat Unfolded Air Bag Inflation (GBAG)

Flat Unfolded Air Bag Inflation (GBAG)

Problem DescriptionThis 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 Air Bag Model

Dytran Example Problem ManualFlat Unfolded Air Bag Inflation (GBAG)

392

The material behavior of the air bag fabric is modeled with a linear elastic isotropic material model with the following properties:

The following input entries are required when modeling a gas bag:

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 Initial Position of the Rigid Ellipsoid

density 600 kg/m3

Young’s modulus 6 107 N/mm2

Poisson’s ratio 0.3

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 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.

ResultsAs 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 Inflation of the Air Bag


394

The time history of pressure, volume, and mass of the gas inside the air bag is plotted in Figure 6-4.

Figure 6-4 Time Histories of Pressure, Mass, and Temperature of Gas inside the Air Bag


Files


Abbreviated Input file

$ file: demo1.dat$ ---------------$$ - inflation of a flat, unfolded demo passenger-side airbag $ - uniform pressure assumption $ - contact with a rigid elllipsoid$STARTCENDCHECK=NOENDSTEP=100000ENDTIME=125.E-3TITLE=INFLATION OF DEMO PASSENGER-BAG WITH UNIFORM PRESSURE MODELTIC=1SPC=1TLOAD=1$ --------------------------------------------------------------$ define the output of the gbag variables:$GBAGS (GB) = 5SET 5 = 1GBOUT (GB) = VOLUME,PRESSURE,FLGAS,TGAS,TEMP,MASSTIMES (GB) = 0. THRU END BY 1.E-3TYPE (GB) = TIMEHISSAVE (GB) = 1000000$$ --------------------------------------------------------------$ define the output for the membrane elements:$ELEMENTS(AIRBAG) = 996SET 996 = ALLMEMTRIAELOUT (AIRBAG) = THICK,SMDFER

demo1.datdemo1_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.THSDEMO1_TELLIPS_0.THS



396

TIMES (AIRBAG) = 0. THRU END BY 20.E-3TYPE (AIRBAG) = ARCHIVESAVE (AIRBAG) = 1000000$$ --------------------------------------------------------------$ define the output for the ellipsoid:$RELS (ELLIPS) = 600$ nb: setc,600 defined in demo1_ellips_contact.datRELOUT (ELLIPS) = GEOMETRYTIMES (ELLIPS) = 0. THRU END BY 20.E-3TYPE (ELLIPS) = ARCHIVESAVE (ELLIPS) = 1000000$$ --------------------------------------------------------------$ define more output for the ellipsoid:$RELS (TELLIPS) = 600$ nb: setc,600 defined in demo1_ellips_contact.datRELOUT (TELLIPS) = XVEL,YVEL,ZVEL,XAVEL,YAVEL,ZAVELTIMES (TELLIPS) = 0. THRU END BY 1.E-3TYPE (TELLIPS) = TIMEHISSAVE (TELLIPS) = 1000000$BEGIN BULK$ -------------------------------------------------------------$ Initial timestep & minimum timestep$PARAM,INISTEP,0.7E-05PARAM,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


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


398

$ ^$ |$ ---------------------------------------------------------------$ | | \ 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,,,,,,+


+,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$SURFACE,25,,PROP,199SET1,199,501,THRU,513SET1,199,551,THRU,563SET1,199,601,THRU,604,606,THRU,613SET1,199,651,THRU,663$ -------------------------------------------------------------ENDDATA

Dytran Example Problem ManualRigid Ellipsoid Dummy Hitting a Passenger Air Bag

400

Rigid Ellipsoid Dummy Hitting a Passenger Air Bag

Problem DescriptionAs 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 ModelThe 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.

401Chapter 6: Occupant SafetyRigid 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.


402

ResultsFigure 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.


.

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 Comparison of the Resultant Head Acceleration with MADYMO


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Files


Abbreviated Main Input File$$ - inflation of a flat, unfolded demo passenger-side airbag $ - inflated by uniformi pressure $ - contact with a rigid elllipsoid$$MEMORY-SIZE=4000000,2000000STARTCENDCHECK=NOENDSTEP=100000ENDTIME=.1TITLE=DEMO PASSENGER-BAG WITH ATB DUMMYTIC=1SPC=1TLOAD=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

demo4.datdemo4_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.ARCDEMO4_AIRBAG_0.ARCDEMO4_DUMQ_0.ARCDEMO4_DUMT_0.ARCDEMO4_ELLIPS_0.ARC

Dytran archive file

DEMO4_GB_0.THSDEMO4_TELLIPS_0.THS



$ - write 1000000 times to the same file, which will be named:$$ DEMO2_GB_0.ARC$GBAGS (GB) = 5SET 5 = 1GBOUT (GB) = VOLUME,PRESSURE,FLGAS,TGAS,TEMP,MASSTIMES (GB) = 0. THRU END BY 1.E-3TYPE (GB) = TIMEHISSAVE (GB) = 1000000$$ --------------------------------------------------------------$ 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) = 995SET 995 = ALLEULHYDROELOUT(EUL) = PRESSURE,DENSITY,FMAT,XVEL,YVEL,ZVEL,SIETIMES(EUL) = 0. THRU END BY 1.E-2TYPE(EUL) = ARCHIVESAVE(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:$$ DEMO4_AIRBAG_0.ARC$ELEMENTS(AIRBAG) = 996SET 996 = ALLMEMTRIAELOUT(AIRBAG) = THICK,SMDFER,EXUSER1TIMES(AIRBAG) = 0. THRU END BY 1.E-2TYPE(AIRBAG) = ARCHIVESAVE(AIRBAG) = 1000000$$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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$ 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$ELEMENTS(DUMQ) = 997SET 997 = ALLDUMQUADELOUT(DUMQ) = ZUSERTIMES(DUMQ) = 0. THRU END BY 1.E-2TYPE(DUMQ) = ARCHIVESAVE(DUMQ) = 1000000$ELEMENTS(DUMT) = 998SET 998 = ALLDUMTRIAELOUT(DUMT) = ZUSERTIMES(DUMT) = 0. THRU END BY 1.E-2TYPE(DUMT) = ARCHIVESAVE(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:$$ DEMO4_ELLIPS_0.ARC$RELS (ELLIPS) = 600$ nb: setc,600 defined in demo4_ellips_contact.datRELOUT (ELLIPS) = GEOMETRYTIMES (ELLIPS) = 0. THRU END BY 1.E-2TYPE (ELLIPS) = ARCHIVESAVE (ELLIPS) = 1000000$$ --------------------------------------------------------------$ 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$


RELS (TELLIPS) = 600$ nb: setc,600 defined in demo4_ellips_contact.datRELOUT (TELLIPS) = XVEL,YVEL,ZVEL,XAVEL,YAVEL,ZAVELTIMES (TELLIPS) = 0. THRU END BY 1.E-3TYPE (TELLIPS) = TIMEHISSAVE (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$$ACTIVE,1,ELEMENT,EULHYDRO,,,,,,+$+,TABLE,9999$ACTIVE,1,INTERACT,COUPLE,,,,,,+$+,TABLE,9999$TABLED1,9999,,,,,,,,+$+,0.0,0.0,1.0,0.0$ -------------------------------------------------------------$ Define the switch from coupling to gasbag, start checking$ after 100 msecs$GBAGCOU,1,10,1,100.0E-3$ -------------------------------------------------------------$ Initial timestep & minimum timestep$PARAM,INISTEP,0.7E-05PARAM,MINSTEP,.5E-8$$ Make the contact default version V2 for this problemPARAM,CONTACT,VERSION,V2$$ -------------------------------------------------------------$ Apply gravity to the modelGRAV,2, ,-9.81,0.,0.,1.$TLOAD1 1 2 0 0$$ -------------------------------------------------------------$ 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.datINCLUDE demo4_bag_contact.dat


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$$ -------------------------------------------------------------$ Define the euler$INCLUDE demo4_euler.dat$$ -------------------------------------------------------------$ 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$INCLUDE xl_demo4.dat$ -------------------------------------------------------------ENDDATAAbbreviated Input File with ATB-related Definitions$ file:demo4_atb.datBEGIN BULK$-------------------------------------------------------------$ Define a setc containing all the atb ellipsoids$ ---> used for visualization$SETC,600,LT,CT,UT,N,HSETC,600,RUL,RLL,RF,LUL,LLLSETC,600,LF,RUA,RLA,LUA,LLA$-------------------------------------------------------------$ 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.


$ 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 SPCCONTACT,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$PSHELL1,801,,DUMMYPSHELL1,802,,DUMMY...PSHELL1,817,,DUMMY$PSHELL1,818,,DUMMY$PSHELL1,857,,DUMMYPSHELL1,859,,DUMMY$SURFACE,801,,PROP,801SET1,801,801...SURFACE,817,,PROP,817


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SET1,817,817$SURFACE,857,,PROP,857SET1,857,857SURFACE,859,,PROP,859SET1,859,859$$ Give them all a “contact-mass” of 2. kg$RIGID,801,801,2.0...RIGID,817,817,2.0$RIGID,857,857,2.0RIGID,859,859,2.0$$ connect the rigids to the atb ellipsoids$RCONREL,1,801,RIGID,401SETC,801,LTSET1,401,801..RCONREL,6,806,RIGID,406SETC,806,HSET1,406,806$$ also connect the spheres for ellbow & shoulder to uparml$RCONREL,7,807,RIGID,407SETC,807,LUASET1,407,807,857RCONREL,8,808,RIGID,408SETC,808,LLASET1,408,808RCONREL,9,809,RIGID,409$$ also connect the spheres for ellbow & shoulder to uparmr$SETC,809,RUASET1,409,809,859...RCONREL,17,817,RIGID,417SETC,817,RFSET1,417,817$ -------------------------------------------------------------$ Define contact of all the rigid surfaces connected to the $ ellipsoids with the car-parts$SURFACE,890,,PROP,890


$ HEAD,NECK,KNEES AND SHOULDER ARE ROMOVEDSET1,890,801,802,803,807SET1,890,808,809,810,811,812,813SET1,890,815,816,817SET1,890,857,859$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$$OCT. 27 1994 0 0 0.0 CARD A1ADEMO4: ACCELERATED DUMMY HITS AIRBAG (UNITS: SI) CARD A1BDR TREEP, MSC BV, HOLLAND CARD A1C$$ 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:$


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$ 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 B6ODUMMY 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


-0.040 0.25 0.149 CARD D2B -0.305 0.25 0.689 CARD D2C -0.305 -0.25 0.689 CARD D2D 3 FLOOR CARD D2A 0.89 0.35 0.017 CARD D2B -1.0 0.35 0.017 CARD D2C -1.0 -1.05 0.017 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 E4B0.07710 2822.50765 0.08390 3963.27115 0.08850 340.32926CARD E4B0.09090 2493.21509 0.09740 -446.89704 0.10300 305.77166CARD E4B0.11050 -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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$ 11 FORCE DEFLECTION CARD E.1 0. -0.11 0. 0. 0. CARD E2 4 CARD E4A 0. 0. 0.02 1500. 0.06 9500.CARD E4B 0.11 14500. CARD E4B... 16 FORCE DEFLECTION CARD E.1 0. -0.10 0. 0. 0. CARD E2 2 CARD E4A 0. 0. 0.10 2000. CARD E4B9999$$ Contact definition between segments and ATB-planes and$ between segments and segements$ CARD E.1 1 2 2 CARD F1A 1 17 1 1 11 0 0 0 14 0 CARD F1B 2 17 1 1 12 0 0 0 14 -1 CARD F1B... 6 6 13 13 16 0 0 0 15 CARD F3B 9 9 15 15 16 0 0 0 15 CARD F3B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 CARD F.4$$ 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 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..


. CARD H11

Dytran Example Problem ManualSled Test Verification of the Enhanced Hybrid III Dummy (50%)

416

Sled Test Verification of the Enhanced Hybrid III Dummy (50%)

Problem DescriptionThe 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:

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

417Chapter 6: Occupant SafetySled 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 IIIThe 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)


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2) ADULT HUMAN FEMALE

3) ADULT HUMAN MALE

4) USER-SUPPLIED BODY DIMENSIONS5) SITTING HYBRID III DUMMY (50%)6) SITTING HYBRID III DUMMY (50%)7) HYBRID II DUMMY (50%) ENTER NUMBER CORRESPONDING TO DESIRED SUBJECT TYPE 5 SELECT UNITS FOR OUTPUT1) ENGLISH2) METRIC1 IS ATB MODEL FORMATTED OUTPUT DESIRED (Y/N) ?YWHAT IS THE NAME OF THE OUTPUT FILE TO CONTAIN THE ATB INPUT DATA ? USE THE FILE EXTENSION “.AIN”.EXAMPLE.AIN IS IT DESIRED TO PRODUCE ANOTHER BODY DESCRIPTION DATA SET (Y/N) ?NSTOP FROM BRESULTS

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 FileThis 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.


How to Create FEM Entities for ATB DummyIt 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.$ -----------------------------------------------------------------------$CHECK=YESPARAM,INISTEP,1.E-3$BEGIN BULK$$ Define the segment contact ellipsoids$ -> card B.2$RELEX,LT,ATBRELEX,MT,ATBRELEX,UT,ATBRELEX,N,ATBRELEX,H,ATBRELEX,RUL,ATBRELEX,RLL,ATBRELEX,RF,ATBRELEX,LUL,ATBRELEX,LLL,ATBRELEX,LF,ATBRELEX,RUA,ATBRELEX,RLA,ATBRELEX,LUA,ATBRELEX,LLA,ATBRELEX,RHD,ATB


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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 )


$ - 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$ATBJNT ,1 ,P ,,,,,,,++ ,20201 ,20202 ,20203 ,20204 ,20201 ,20202 ,20203 ,,++ ,20205 ,20206 ,20207 ,20208 ,20204 ,20205 ,20206$ENDDATA

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.


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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 FEM Entities generated by Dytran for Hybrid III Dummy


Attaching Dytran Finite Element to ATB SegmentsIt 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,20001SETC,20001,LTSET1,20001,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 Motion of the Digitized Dummy during Crash


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Positioning the Dummy with PatranIn 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$


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 Dytran-MD Patran Preference Interface Guide.

1. Start the dummy positioner:

AnalysesAction: Special FeaturesObject: Dummy PositionerApply

2. Read in the dummy data set:

Import DummySelect hybridIII-50p.bdfApply

3. Create the groups of the dummy:

Build DummyDefine Dummy Prefix NameOK

4. Position the dummy:

Manipulate Dummy

Translation of the Dummy:

Action: TransformObject: Full Dummy

5. Define Translation Vector:

Select the Dummy for ManipulationApply


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6. Rotation of a part of the dummy:

Apply

7. Write out the new positional dummy:

Export DummySelect Original Dummy NameSelect Output Dummy’s File NameOK

Integrating the Dummy into other FEM ModelsThe dummy model as delivered is using the following numbers

: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 BeltsThe 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$ -----------------------

Action: Transform

Object: Part Dummy

Method: Rotate

Parts Select group

Power User Tools Choose axis

Relation Parameters Define axis and angle

Entity Minimum Maximum

Grids 20001 24138

Elements 20001 24299

Material IDs 20001 20017

Property IDs 20001 20404


$BEGIN BULK$ ==========================================================$belt with the dummy $All gridpoints within a distance of 2.54 mm will be repositioned$$ belt gridpointsSET1,20100,70000t70058$SURFACE,20401,,PROP,20401SET1,20401,20101,20303,20103,20302,20104,20301CONTACT,101,GRID,SURF,20100,20401,1.0,1.0,1.0,++,BELT1,TOP,,,,,,SLAVE,++,,,,,,,,,++,,,,,,,2.54E-03$$ Check for contact with the digitized upper-legs:$ Upper legs (dummy shells)$SURFACE,20402,,PROP,20402SET1,20402,20304,20307CONTACT,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


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$RCONREL,2,70001,GRID,70001SETC,70001,SLEDSET1,70001,70000,70001,70004,70005$

Applying an Acceleration Field to the DummyStandard 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 ExperimentAn 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


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 16 SITTING HYBRID III CARD B.1 LT 544.4602.45751.29691.2080 5.000 7.185 4.800-1.000 0.000 0.000 1 CARD B.2 0.00-37.32 180.0 MT 4 4.8900.06120.05930.0205 4.775 6.500 4.000 1.000 0.000-1.000 1 CARD B.2 0.00 4.16 180.0 UT 339.2162.62032.05171.7336 4.825 6.500 7.785 0.000 0.000 0.000 1 CARD B.2 0.00 4.99 180.0 N 2 2.6680.02540.02570.0084 1.675 1.675 3.000 0.000 0.000 0.000 1 CARD B.2


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0.00 0.00 180.0 H 1 9.9210.14080.21280.1956 4.250 2.875 4.000 0.000 0.000 0.000 1 CARD B.2 0.00-26.58 180.0 RUL 613.7130.60920.59340.1068 2.950 3.050 7.285 0.000 0.000 0.000 1 CARD B.2 0.00 4.13 180.0 RLL 7 7.2370.67080.67450.0397 2.165 2.050 9.750 0.000 0.000 2.000 1 CARD B.2 0.00 -1.90 180.0 RF 8 2.7560.00670.05240.0491 4.900 1.675 1.675 0.000 0.000 0.000 1 CARD B.2 0.96 -9.57 158.2 LUL 913.7130.60920.59340.1068 2.950 3.050 7.285 0.000 0.000 0.000 1 CARD B.2 0.00 4.13 180.0 LLL 0 7.2370.67080.67450.0397 2.165 2.050 9.750 0.000 0.000 2.000 1 CARD B.2 0.00 -1.90 180.0 LF 1 2.7560.00670.05240.0491 4.900 1.675 1.675 0.000 0.000 0.000 1 CARD B.2 -0.96 -9.57-158.2 RUA 2 4.5970.10240.09970.0109 1.900 1.800 6.000 0.000 0.000-1.000 1 CARD B.2 0.00 -1.31 180.0 RLA 3 3.8000.11910.11280.0069 1.775 1.775 5.800 0.000 0.000 0.000 1 CARD B.2 0.00 -1.31 180.0 LUA 4 4.5970.10240.09970.0109 1.900 1.800 6.000 0.000 0.000-1.000 1 CARD B.2 0.00 -1.31 180.0 LLA 5 3.8000.11910.11280.0069 1.775 1.775 5.800 0.000 0.000 0.000 1 CARD B.2 0.00 -1.31 180.0 RHD 6 1.2900.01140.00930.0036 1.000 1.870 3.650 0.000 0.000 0.000 1 CARD B.2 -5.35 30.75 173.5 LHD 7 1.2900.01140.00930.0036 1.000 1.870 3.650 0.000 0.000 0.000 1 CARD B.2 5.35 30.75-173.5 P M 1 0 -2.15 0.00 -1.66 -0.35 0.00 2.56 0 0.0 0.0 CARD B.3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 W N 2 0 -0.35 0.00 -2.56 -0.89 0.00 5.85 0 0.0 0.0 CARD B.3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 NP O 3 0 0.00 0.00 -5.96 0.00 0.00 2.76 0 0.0 0.0 CARD B.3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 HP P 4 0 0.00 0.00 -2.84 -0.55 0.00 2.00 0 0.0 0.0 CARD B.3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 RH Q 1 0 -0.11 3.15 1.24 0.00 0.00 -9.96 0 0.0 0.0 CARD B.3 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 RK R 6 1 0.00 0.00 6.56 -0.20 0.00 -6.74 0 0.0 0.0 CARD B.3 0.00 0.00 0.00 0.00 55.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 RA S 7 -8 -0.20 0.00 9.65 -2.12 0.00 -1.54 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 90.00 0.00 0.00 0.00-10.00 0.00 0 0 0 0 0 0 LH T 1 0 -0.11 -3.15 1.24 0.00 0.00 -9.96 0 0.0 0.0 CARD B.3 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 LK U 9 1 0.00 0.00 6.56 -0.20 0.00 -6.74 0 0.0 0.0 CARD B.3 0.00 0.00 0.00 0.00 55.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 LA V 10 -8 -0.20 0.00 9.65 -2.12 0.00 -1.54 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 90.00 0.00 0.00 0.00-10.00 0.00 0 0 0 0 0 0 RS W 3 0 -0.88 7.38 -2.66 0.00 0.00 -5.43 0 0.0 0.0 CARD B.3 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 RE X 12 -8 0.00 0.00 4.94 0.00 0.00 -3.67 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 90.00 0.00 0.00 0.00 62.50 0.00 0 0 0 0 0 0 LS Y 3 0 -0.88 -7.38 -2.66 0.00 0.00 -5.43 0 0.0 0.0 CARD B.3 0.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0 0 0 0 0 0 LE Z 14 -8 0.00 0.00 4.94 0.00 0.00 -3.67 0 0.0 0.0 CARD B.3


90.00 0.00 0.00 90.00 0.00 0.00 0.00 62.50 0.00 0 0 0 0 0 0 RW 13 -8 0.00 0.00 6.07 -0.30 0.00 -2.13 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 90.00 0.00 0.00 0.00 15.00 0.00 0 0 0 0 0 0 LW 15 -8 0.00 0.00 6.07 -0.30 0.00 -2.13 0 0.0 0.0 CARD B.3 90.00 0.00 0.00 90.00 0.00 0.00 0.00 15.00 0.00 0 0 0 0 0 0-45.00 0.00 0.00 1.00 0.00 34.38 0.00 0.00 1.00 0.00CARD B.4-45.00 0.00 0.00 1.00 0.00 34.38 0.00 0.00 1.00 0.00CARD B.4-46.00 0.00 0.00 1.00 0.00 15.00 0.00 0.00 1.00 0.00CARD B.4-47.00 0.00 0.00 1.00 0.00 15.00 0.00 0.00 1.00 0.00CARD B.4-43.00 0.00 0.00 1.00 0.00 7.50 75.00 75.00 1.00 55.00CARD B.4 0.00 2.32 0.00 1.00 48.90 0.00 0.00 0.00 1.00 0.00CARD B.4 1.00 10.00 10.00 1.00 27.00 0.00 6.69 0.00 1.00 30.00CARD B.4 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00-44.00 0.00 0.00 1.00 0.00 7.50 75.00 75.00 1.00 55.00CARD B.4 0.00 2.32 0.00 1.00 48.90 0.00 0.00 0.00 1.00 0.00CARD B.4 1.00 10.00 10.00 1.00 27.00 0.00 6.69 0.00 1.00 30.00CARD B.4 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00-41.00 0.00 0.00 1.00 0.00 0.00100.00100.00 1.00 125.00CARD B.4 0.00 20.00 20.00 1.00 52.00 0.00 4.65 0.00 1.00 65.30CARD B.4 0.00 0.00 0.00 1.00 0.00 0.00 90.00 0.00-42.00 0.00 0.00 1.00 0.00 0.00100.00100.00 1.00 125.00CARD B.4 0.00 20.00 20.00 1.00 52.00 0.00 4.65 0.00 1.00 65.30CARD B.4 0.00 0.00 0.00 1.00 0.00 0.00 90.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 4.32 0.14 1.00 55.50CARD B.4 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 4.32 0.14 1.00 55.50CARD B.4 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.200 0. 30. 0. 0. 0. 0. CARD B.5 1.200 0. 30. 0. 0. 0. 0. CARD B.5 0.350 0. 30. 0. 0. 0. 0. CARD B.5 0.192 0. 30. 0. 0. 0. 0. CARD B.5 1.000 0. 30. 0. 0. 0. 0. CARD B.5 1.000 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.000 0. 30. 0. 0. 0. 0. CARD B.5 1.000 0. 30. 0. 0. 0. 0. CARD B.5 1.000 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.000 0. 30. 0. 0. 0. 0. CARD B.5 0.100 0. 30. 0. 0. 0. 0. CARD B.5 0.100 0. 30. 0. 0. 0. 0. CARD B.5 0.100 0. 30. 0. 0. 0. 0. CARD B.5 0.000 0. 30. 0. 0. 0. 0. CARD B.5 0.100 0. 30. 0. 0. 0. 0. CARD B.5 0.100 0. 30. 0. 0. 0. 0. CARD B.5 0.100 0. 30. 0. 0. 0. 0. CARD B.5 0.000 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.000 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5 0.500 0. 30. 0. 0. 0. 0. CARD B.5


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0.000 0. 30. 0. 0. 0. 0. CARD B.5 0.01 0.01 0.01 0.01 0.01 0.01 0.10 0.10 0.10 0.10 0.10 0.01CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 0.01 0.01 0.01 0.00 0.00 0.00 0.10 0.10 0.10 0.00 0.00 0.00CARD B.6 1 HEAD SURFACE CARD E.1 0.00 0.71 -0.80 0.00 0.00 CARD E.2 2.57 -175.98 4495.67 -8953.11 6029.59 0.00CARD E.3 3 CARD E.4 0.71 358.00 0.75 3580.00 0.80 35800.00CARD E.4 2 BACK OF SHOULDER CARD E.1 0.00 1.62 -1.70 0.00 0.00 CARD E.2 0.01 -26.59 380.91 -678.22 456.53 0.00CARD E.3 3 CARD E.4 1.62 48.30 1.65 483.00 1.70 4830.00CARD E.4 3 CHEST CARD E.1 0.00 1.31 -1.40 0.00 0.00 CARD E.2 0.55 5.46 73.49 -15.01 0.00 0.00CARD E.3 3 CARD E.4 1.31 158.60 1.35 1586.00 1.40 15860.00CARD E.4 4 ANTERIOR PELVIS CARD E.1 0.00 2.02 -2.10 0.00 0.00 CARD E.2 -0.14 1.58 24.39 -29.33 13.76 0.00CARD E.3 3 CARD E.4 2.02 89.70 2.05 897.00 2.10 8970.00CARD E.4 5 POSTERIOR PELVIS CARD E.1 0.00 1.96 -2.10 0.00 0.00 CARD E.2 -0.76 35.10 40.04 -12.49 0.00 0.00CARD E.3 3 CARD E.4 1.96 127.80 2.03 1278.00 2.10 12780.00CARD E.4 6 UPPER ARM CARD E.1 0.00 0.89 -1.00 0.00 0.00 CARD E.2 1.54 -30.63 627.09 -1197.70 917.23 0.00CARD E.3 3 CARD E.4 0.89 203.10 0.95 2031.00 1.00 20310.00CARD E.4 7 FOREARM CARD E.1 0.00 1.31 -1.40 0.00 0.00 CARD E.2 -2.40 107.37 -313.36 504.61 -196.37 0.00CARD E.3 3 CARD E.4 1.31 156.20 1.35 1562.00 1.40 15620.00CARD E.4 8 HAND CARD E.1


0.00 0.38 -0.45 0.00 0.00 CARD E.2 -0.04 -31.04 2384.47 -10193.90 19172.50 0.00CARD E.3 3 CARD E.4 0.38 139.40 0.41 1394.00 0.45 13940.00CARD E.4 9 UPPER LEG CARD E.1 0.00 1.31 -1.40 0.00 0.00 CARD E.2 -0.28 14.19 81.94 -122.62 98.73 -29.09CARD E.3 3 CARD E.4 1.31 61.60 1.35 616.00 1.40 6160.00CARD E.4 10 KNEE CARD E.1 0.00 0.54 -0.58 0.00 0.00 CARD E.2 -0.04 2.22 867.34 -401.61 0.00 0.00CARD E.3 3 CARD E.4 0.54 187.80 0.56 1878.00 0.58 18780.00CARD E.4 11 FRONT OF LOWER LEG CARD E.1 0.00 1.78 -1.90 0.00 0.00 CARD E.2 -0.74 26.79 3.92 29.05 0.00 0.00CARD E.3 3 CARD E.4 1.78 223.60 1.84 2236.00 1.90 22360.00CARD E.4 12 BACK OF LOWER LEG CARD E.1 0.00 1.66 -1.75 0.00 0.00 CARD E.2 -0.08 29.42 -2.08 0.00 0.00 0.00CARD E.3 3 CARD E.4 1.66 43.20 1.70 432.00 1.75 4320.00CARD E.4 13 FOOT CARD E.1 0.00 0.83 -0.95 0.00 0.00 CARD E.2 -0.68 82.75 -704.17 3367.10 -5570.69 3188.30CARD E.3 3 CARD E.4 0.83 121.20 0.89 1212.00 0.95 12120.00CARD E.4 999 CARD E.1 41 RIGHT SHOULDER JOINT CARD E.7 -4 8 CARD E.7 151.000 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 CARD E.7 151.000 0.000 87900.000 0.000 153.000 0.000 13000.000 0.000 111.000 0.000 2470.000 17600.000 89.000 0.000 22000.000 0.000 68.000 0.000 22200.000 376000.000 66.000 0.000 7780.000 3010000.000 0.000 0.000 -312.000 11800.000 75.000 0.000 13700.000 0.000 43 SEATED RIGHT HIP CARD E.7


434

19 4 CARD E.7 0.000 7.000 22.000 52.000 112.000 232.000CARD E.7 472.000 952.000 1910.000 3830.000 7670.000 15400.000CARD E.7 30700.000 61400.000 123000.000 246000.000 492000.000 983000.000CARD E.7 1970000.000 0.000 17.000 51.000 119.000 255.000 527.000CARD E.7 1070.000 2160.000 4340.000 8690.000 17400.000 34900.000CARD E.7 69700.000 139000.000 279000.000 557000.000 1110000.000 2230000.000CARD E.7 4460000.000 0.000 8.000 17.000 28.000 50.000 94.000CARD E.7 182.000 358.000 710.000 1410.000 2820.000 5640.000CARD E.7 11300.000 22500.000 45100.000 90100.000 180000.000 360000.000CARD E.7 721000.000 0.000 7.000 13.000 18.000 23.000 33.000CARD E.7 53.000 93.000 173.000 333.000 653.000 1290.000CARD E.7 2570.000 5130.000 10300.000 20500.000 41000.000 81900.000CARD E.7 164000.000 44 SEATED LEFT HIP CARD E.7 19 4 CARD E.7 0.000 7.000 22.000 52.000 112.000 232.000CARD E.7 472.000 952.000 1910.000 3830.000 7670.000 15400.000CARD E.7 30700.000 61400.000 123000.000 246000.000 492000.000 983000.000CARD E.7 1970000.000 0.000 7.000 13.000 18.000 23.000 33.000CARD E.7 53.000 93.000 173.000 333.000 653.000 1290.000CARD E.7 2570.000 5130.000 10300.000 20500.000 41000.000 81900.000CARD E.7 164000.000 0.000 8.000 17.000 28.000 50.000 94.000CARD E.7 182.000 358.000 710.000 1410.000 2820.000 5640.000CARD E.7 11300.000 22500.000 45100.000 90100.000 180000.000 360000.000CARD E.7 721000.000 0.000 17.000 51.000 119.000 255.000 527.000CARD E.7 1070.000 2160.000 4340.000 8690.000 17400.000 34900.000CARD E.7 69700.000 139000.000 279000.000 557000.000 1110000.000 2230000.000CARD E.7 4460000.000 45 SEATED LUMBAR SPINE CARD E.7 19 4 CARD E.7 0.000 4600.000 9200.000 13800.000 18400.000 23000.000CARD E.7 27600.000 32200.000 36800.000 41400.000 50600.000 69000.000CARD E.7 105800.000 179400.000 326600.000 621000.000 1209800.000 2387400.000CARD E.7 4742600.000 0.000 6800.000 13600.000 20400.000 27200.000 34000.000CARD E.7 40800.000 47600.000 54400.000 61200.000 74800.000 102000.000CARD E.7 156400.000 265200.000 482800.000 918000.000 1788400.000 3529200.000CARD E.7 7010800.000 0.000 3000.000 6000.000 9000.000 12000.000 15000.000CARD E.7 18000.000 21000.000 24000.000 27000.000 33000.000 45000.000CARD E.7 69000.000 117000.000 213000.000 405000.000 789000.000 1557000.000CARD E.7 3093000.000 0.000 6800.000 13600.000 20400.000 27200.000 34000.000CARD E.7 40800.000 47600.000 54400.000 61200.000 74800.000 102000.000CARD E.7 156400.000 265200.000 482800.000 918000.000 1788400.000 3529200.000CARD E.7


7010800.000 46 NECK PIVOT CARD E.7 19 4 CARD E.7 0.000 700.000 1400.000 2100.000 2800.000 3500.000CARD E.7 4200.000 4900.000 5600.000 6300.000 7000.000 7700.000CARD E.7 8400.000 9800.000 12600.000 18200.000 29400.000 51800.000CARD E.7 96600.000 0.000 600.000 1200.000 1800.000 2400.000 3000.000CARD E.7 3600.000 4200.000 4800.000 5400.000 6000.000 6600.000CARD E.7 7200.000 8400.000 10800.000 15600.000 25200.000 44400.000CARD E.7 82800.000 0.000 300.000 600.000 900.000 1200.000 1500.000CARD E.7 1800.000 2100.000 2400.000 2700.000 3000.000 3300.000CARD E.7 3600.000 4200.000 5400.000 7800.000 12600.000 22200.000CARD E.7 41400.000 0.000 600.000 1200.000 1800.000 2400.000 3000.000CARD E.7 3600.000 4200.000 4800.000 5400.000 6000.000 6600.000CARD E.7 7200.000 8400.000 10800.000 15600.000 25200.000 44400.000CARD E.7 82800.000 47 HEAD PIVOT CARD E.7 19 4 CARD E.7 0.000 490.000 980.000 1680.000 2380.000 3080.000CARD E.7 3780.000 4480.000 5180.000 5880.000 6580.000 7280.000CARD E.7 7980.000 9380.000 12180.000 17780.000 28980.000 51380.000CARD E.7 96180.000 0.000 600.000 1200.000 1800.000 2400.000 3000.000CARD E.7 3600.000 4200.000 4800.000 5400.000 6000.000 6600.000CARD E.7 7200.000 8400.000 10800.000 15600.000 25200.000 44400.000CARD E.7 82800.000 0.000 250.000 500.000 800.000 1100.000 1400.000CARD E.7 1700.000 2000.000 2300.000 2600.000 2900.000 3200.000CARD E.7 3500.000 4100.000 5300.000 7700.000 12500.000 22100.000CARD E.7 41300.000 0.000 600.000 1200.000 1800.000 2400.000 3000.000CARD E.7 3600.000 4200.000 4800.000 5400.000 6000.000 6600.000CARD E.7 7200.000 8400.000 10800.000 15600.000 25200.000 44400.000CARD E.7 82800.000 999 CARD E.7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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:


436

$$ program atb_clean$*$* clean up comments in an atb input file$*$* 1) compile and link$* f77 -o atb_clean.exe atb_clean.f$* 2) clean up the input file with :$* atb_clean.exe < run50_commented.ain > run50.ain$*$ character*(80) line$ 1 continue$ read(*,’(a)’,end=9900) line$ if (line(1:1).ne.’$’) then$ write(*,’(a)’) line$ endif$ goto 1$ 9900 continue$ end$$---------------------------------------------------------------------------$ Date & Restarts - Format (3A4,2I4,F8.0)$$234A234A234I234I234F2345678FEB. 13.1995 0 0 0.0 CARD A1A$---------------------------------------------------------------------------$$ Comment - Format (20A4 / 20A4)$SITTING HYBRID III DUMMY (50%) GENERATED WITH GEBOD CARD A1BFORD SLED TEST 112 MS ; 23.7 G PEAK ACCELERATION CARD A1C$---------------------------------------------------------------------------$$ Units & Gravity definition - Format (3A4, 4F12.0)$$ 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)


$$23456I23456A2345678A234A234A234A234A234 18 17 SITTING HYBRID III CARD B.1$-------------------------------------------------------------------------------$$ 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


438

RHD 6 5.74 .0013 .0011 .0004 .025 .047 .093 .000 .000 .000 1 CARD B.2 -2.35-31.09-175.6$17 LHD 7 5.74 .0013 .0011 .0004 .025 .047 .093 .000 .000 .000 1 CARD B.2 2.35-31.09 175.6$$ a segment connected to a vehicle -> used for connecting MSC/DYTRAN FEM model$$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


90.00 .00 .00 90.00 .00 .00 .00 62.50 .00 0 0 0 0 0 0$13: UT-LUA LS Y 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$14: LUA-LLA LE Z 14 -8 .00 .00 .13 .00 .00 -.09 0 .00 .00 CARD B.3 90.00 .00 .00 90.00 .00 .00 .00 62.50 .00 0 0 0 0 0 0$15: RLA-RHD RW 13 -8 .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$16: LLA-LHD LW 15 -8 .00 .00 .15 -.01 .00 -.05 0 .00 .00 CARD B.3 90.00 .00 .00 90.00 .00 .00 .00 15.00 .00 0 0 0 0 0 0NJ1 1 0 0 0.00 0.00 0.0 0.00 0.00 0.0 CARD B.3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00$-------------------------------------------------------------------------------$$ Joints spring characteristics - Format (2( 4F6.0, F12.0) )$ Format (2( 4F6.0, F12.0) )$$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


440

.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)


$ 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)$$234567890F234567890F234567890F234567890F234567890F234567890F234567890$XXXXXXXX|-Xlinear-|-Ylinear-|-Zlinear-|-Xangulr-|-Yangulr-|-Zangulr-|0.000E+00 0.000E+001.000E-02 0.0$ZERO ACCELERATION FOR VEHICLE - WE APPLY ACCELERATION ON DYMMY CARD C1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -2 .000.01000 0CARD C2A 0 0.0 0.0 0.0 CARD C2B 0.000E+00 0.000E+001.000E-02 0.0$---------------------------------------------------------------------------$$ 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


442

$$ Cards D.2 -> a) FORMAT (I4, 4X, 5A4)$ b) FORMAT (3F12.0)$ 1 BACK SEAT CARD D2A$23456789012$23456789012$23456789012 -3.09131 -.27140 -.57808 -3.36387 -.27140 -1.1516 -3.09131 .27140 -.57808 2 BOTTOM SEAT CARD D2A$23456789012$23456789012$23456789012 -3.09131 -.27140 -.57808 -3.09131 .27140 -.57808 -2.66830 -.27140 -.57808 3 FLOOR$23456789012$23456789012$23456789012 -2.45313 -.27140 -.4251959 -2.45313 .27140 -.4251959 -2.21513 -.27140 -.4251959 4 TOE BOARD$23456789012$23456789012$23456789012 -2.21513 -.27140 -.4251959 -2.21513 .27140 -.4251959 -2.02641 -.27140 -.7561072 5 FIRE WALL$23456789012$23456789012$23456789012 -2.02641 -.27140 -.7561072 -2.02641 .27140 -.7561072 -2.02641 -.27140 -1.137110$------------------------------------------------------------------------------$$ 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


$ - 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 E.1 .00 .02 -.02 .00 .00 CARD E.2 .11432E+02 -.30820E+05 .30998E+08 -.24304E+10 .64440E+11 .00000E+00CARD E.3 3 CARD E.4 .02 9.09 .02 90.93 .02 909.32CARD E.4 2 BACK OF SHOULDER CARD E.1 .00 .04 -.04 .00 .00 CARD E.2 .44484E-01 -.46568E+04 .26264E+07 -.18411E+09 .48790E+10 .00000E+00CARD E.3 3 CARD E.4 .04 1.23 .04 12.27 .04 122.68CARD E.4 3 CHEST CARD E.1 .00 .03 -.04 .00 .00 CARD E.2 .24466E+01 .95623E+03 .50671E+06 -.40746E+07 .00000E+00 .00000E+00CARD E.3 3 CARD E.4 .03 4.03 .03 40.28 .04 402.84CARD E.4 4 ANTERIOR PELVIS CARD E.1 .00 .05 -.05 .00 .00 CARD E.2 -.62278E+00 .27671E+03 .16817E+06 -.79618E+07 .14706E+09 .00000E+00CARD E.3 3 CARD E.4 .05 2.28 .05 22.78 .05 227.84CARD E.4 5 POSTERIOR PELVIS CARD E.1 .00 .05 -.05 .00 .00 CARD E.2 -.33808E+01 .61472E+04 .27608E+06 -.33905E+07 .00000E+00 .00000E+00CARD E.3 3 CARD E.4 .05 3.25 .05 32.46 .05 324.61CARD E.4 6 UPPER ARM CARD E.1 .00 .02 -.03 .00 .00 CARD E.2 .68505E+01 -.53643E+04 .43238E+07 -.32512E+09 .98026E+10 .00000E+00CARD E.3 3 CARD E.4 .02 5.16 .02 51.59 .03 515.87CARD E.4 7 FOREARM CARD E.1 .00 .03 -.04 .00 .00 CARD E.2 -.10676E+02 .18804E+05 -.21606E+07 .13698E+09 -.20987E+10 .00000E+00CARD E.3 3 CARD E.4 .03 3.97 .03 39.67 .04 396.75CARD E.4 8 HAND CARD E.1 .00 .01 -.01 .00 .00 CARD E.2 -.17794E+00 -.54361E+04 .16441E+08 -.27672E+10 .20490E+12 .00000E+00CARD E.3 3 CARD E.4 .01 3.54 .01 35.41 .01 354.08CARD E.4 9 UPPER LEG CARD E.1 .00 .03 -.04 .00 .00 CARD E.2 -.12456E+01 .24851E+04 .56498E+06 -.33286E+08 .10552E+10 -.12240E+11CARD E.3 3 CARD E.4 .03 1.56 .03 15.65 .04 156.46CARD E.4 10 KNEE CARD E.1 .00 .01 -.01 .00 .00 CARD E.2


444

-.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 E29999 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 CARD E.7 151.000 .000 9931.734 .000 75.000 .000 1547.949 .000 .000 .000 -35.253 1333.270 66.000 .000 879.054 340096.906 68.000 .000 2508.356 42483.871 89.000 .000 2485.758 .000 111.000 .000 279.083 1988.606 153.000 .000 1468.857 .000 42 LEFT SHOULDER JOINT CARD E.7 -4 8 CARD E.7 151.000 .000 9931.734 .000 153.000 .000 1468.857 .000 111.000 .000 279.083 1988.606 89.000 .000 2485.758 .000 68.000 .000 2508.356 42483.871 66.000 .000 879.054 340096.906 .000 .000 -35.253 1333.270


75.000 .000 1547.949 .000 43 SEATED RIGHT HIP CARD E.7 19 4 CARD E.7 .000 .791 2.486 5.875 12.655 26.213CARD E.7 53.331 107.566 215.809 432.748 866.626 1740.031CARD E.7 3468.762 6937.523 13897.650 27795.301 55590.590 111068.203CARD E.7 222588.406 .000 1.921 5.762 13.446 28.812 59.545CARD E.7 120.898 244.056 490.372 981.875 1966.009 3943.316CARD E.7 7875.332 15705.470 31523.930 62934.879 125417.797 251965.500CARD E.7 503931.000 .000 .904 1.921 3.164 5.649 10.621CARD E.7 20.564 40.450 80.222 159.314 318.629 637.258CARD E.7 1276.776 2542.253 5095.805 10180.310 20338.020 40676.039CARD E.7 81465.063 .000 .791 1.469 2.034 2.599 3.729CARD E.7 5.988 10.508 19.547 37.625 73.782 145.756CARD E.7 290.382 579.634 1163.787 2316.275 4632.551 9253.801CARD E.7 18530.199 44 SEATED LEFT HIP CARD E.7 19 4 CARD E.7 .000 .791 2.486 5.875 12.655 26.213CARD E.7 53.331 107.566 215.809 432.748 866.626 1740.031CARD E.7 3468.762 6937.523 13897.650 27795.301 55590.590 111068.203CARD E.7 222588.406 .000 .791 1.469 2.034 2.599 3.729CARD E.7 5.988 10.508 19.547 37.625 73.782 145.756CARD E.7 290.382 579.634 1163.787 2316.275 4632.551 9253.801CARD E.7 18530.199 .000 .904 1.921 3.164 5.649 10.621CARD E.7 20.564 40.450 80.222 159.314 318.629 637.258CARD E.7 1276.776 2542.253 5095.805 10180.310 20338.020 40676.039CARD E.7 81465.063 .000 1.921 5.762 13.446 28.812 59.545CARD E.7 120.898 244.056 490.372 981.875 1966.009 3943.316CARD E.7 7875.332 15705.470 31523.930 62934.879 125417.797 251965.500CARD E.7 503931.000 45 SEATED LUMBAR SPINE CARD E.7 19 4 CARD E.7 .000 519.750 1039.499 1559.248 2078.998 2598.747CARD E.7 3118.497 3638.246 4157.996 4677.746 5717.242 7796.242CARD E.7 11954.240 20270.230 36902.211 70166.188 136694.094 269749.906CARD E.7 535861.688 .000 768.325 1536.651 2304.976 3073.301 3841.626CARD E.7 4609.953 5378.277 6146.602 6914.926 8451.578 11524.880CARD E.7 17671.480 29964.680 54551.090 103723.898 202069.594 398760.813CARD E.7 792143.313 .000 338.967 677.934 1016.901 1355.868 1694.835CARD E.7 2033.802 2372.769 2711.736 3050.703 3728.637 5084.504CARD E.7 7796.242 13219.710 24066.660 45760.551 89148.313 175923.906CARD E.7 349475.000


446

.000 768.325 1536.651 2304.976 3073.301 3841.626CARD E.7 4609.953 5378.277 6146.602 6914.926 8451.578 11524.880CARD E.7 17671.480 29964.680 54551.090 103723.898 202069.594 398760.813CARD E.7 792143.313 46 NECK PIVOT CARD E.7 19 4 CARD E.7 .000 79.092 158.185 237.277 316.369 395.461CARD E.7 474.554 553.646 632.739 711.831 790.923 870.015CARD E.7 949.108 1107.292 1423.662 2056.400 3321.877 5852.832CARD E.7 10914.740 .000 67.793 135.587 203.380 271.174 338.967CARD E.7 406.760 474.554 542.347 610.141 677.934 745.728CARD E.7 813.521 949.108 1220.281 1762.628 2847.323 5016.711CARD E.7 9355.488 .000 33.897 67.793 101.690 135.587 169.484CARD E.7 203.380 237.277 271.174 305.070 338.967 372.864CARD E.7 406.760 474.554 610.141 881.314 1423.662 2508.356CARD E.7 4677.746 .000 67.793 135.587 203.380 271.174 338.967CARD E.7 406.760 474.554 542.347 610.141 677.934 745.728CARD E.7 813.521 949.108 1220.281 1762.628 2847.323 5016.711CARD E.7 9355.488 47 HEAD PIVOT CARD E.7 19 4 CARD E.7 .000 55.365 110.729 189.822 268.914 348.006CARD E.7 427.098 506.191 585.283 664.375 743.468 822.560CARD E.7 901.652 1059.837 1376.206 2008.945 3274.421 5805.375CARD E.7 10867.280 .000 67.793 135.587 203.380 271.174 338.967CARD E.7 406.760 474.554 542.347 610.141 677.934 745.728CARD E.7 813.521 949.108 1220.281 1762.628 2847.323 5016.711CARD E.7 9355.488 .000 28.247 56.495 90.391 124.288 158.185CARD E.7 192.081 225.978 259.875 293.771 327.668 361.565CARD E.7 395.461 463.255 598.842 870.015 1412.363 2497.057CARD E.7 4666.445 .000 67.793 135.587 203.380 271.174 338.967CARD E.7 406.760 474.554 542.347 610.141 677.934 745.728CARD E.7 813.521 949.108 1220.281 1762.628 2847.323 5016.711CARD E.7 9355.488$$-----------------------------------------------------------------------------$$ 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


$ 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


448

$-------------------------------------------------------------------------------$$ Card F.5.a: What joint is using what joint-restoring function - Format (18I4)$ ****************************************************************$ ***** is allways required if NJNT>0 (card B.1) and NJNTF>0 (Card D.1) ******$ ****************************************************************$$ 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.


$ 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)


450

$$23456I23I23F23456789012F23456789012F23456789012$ Lower Torso 3 0 -1 0.000 0.000 0.000 CARD H2$ Upper Torso 0 -3 0.000 0.000 0.000 CARD H2$ Head 0 -5 0.000 0.000 0.000 CARD H2$-------------------------------------------------------------------------------$$ 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.$ -----------------------------------------------------------------------$CHECK=YESPARAM,INISTEP,1.E-3$BEGIN BULK$$ Define the segment contact ellipsoids


$ -> card B.2$RELEX,LT,ATBRELEX,MT,ATBRELEX,UT,ATBRELEX,N,ATBRELEX,H,ATBRELEX,RUL,ATBRELEX,RLL,ATBRELEX,RF,ATBRELEX,LUL,ATBRELEX,LLL,ATBRELEX,LF,ATBRELEX,RUA,ATBRELEX,RLA,ATBRELEX,LUA,ATBRELEX,LLA,ATBRELEX,RHD,ATBRELEX,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 )$


452

$ 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$$ 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,+


+,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,,+


454

+,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


Appendix D: Properties and Materials of Hybrid III Dummy

Bar elements representing ATB segment and joint coordinate systems.

Digitized segments

Covered ATB ellipsoids

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

Name pid mid

lower torso 101 1

middle torso 2 Modeled as dummy shells

upper torso 103 3

neck 104 4

head 105 5

right upper leg 106 6 Leg itself is modeled as dummy shells

right lower leg 107 7

right foot 108 8

left upper leg 109 9 Leg itself is modeled as dummy shells


456

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 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

Name pid mid


Digitized dummy shells (mid = 0)

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 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

Name 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

Appendix E: Groups created by Session File for Positioning within MD Patran

Complete Dummy

Ankle

Knee

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

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

Group Name Contents

H3ROT–KNEE–BLW–LEFT All items in this group use the same logic as is used for Ankle

H3ROT–KNEE–BLW–RIGHT

H3ROT–KNEE–BLW–SYMM

H3ROT–KNEE–ABV–LEFT

H3ROT–KNEE–ABV–RIGHT

H3ROT–KNEE–ABV–SYMM


Hip

Lumbar Spine

Thoracic Spine

Upper Neck

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

Group Name

H3ROT–LSPINE–BLW

H3ROT–LSPINE–ABV

Group Name

H3ROT–TSPINE–BLW

H3ROT–TSPINE–ABV

Group Name

H3ROT–UPNECK–ABV

H3ROT–UPNECK–BLW


460

Lower Neck

Wrist

Elbow

Shoulder

References1. Prasad, P, “CoR:<n=5>-<n=1>.\tmparative Evaluation of the Dynamic Responses of the Hybrid

II and Hybrid III Dummies” SAE paper 902318 (Also published in 2).

Group Name

H3ROT–LWNECK–ABV

H3ROT–LWNECK–BLW

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

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

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


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.

Dytran Example Problem ManualSide Curtain Air Bag (Courtesy of Autoliv)

462

Side Curtain Air Bag (Courtesy of Autoliv)

Problem DescriptionA 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.

Figure 6-14 Side Curtain Air Bag (folded pattern)

Figure 6-15 Side Curtain Air Bag (deployed pattern)

463Chapter 6: Occupant SafetySide Curtain Air Bag (Courtesy of Autoliv)

The properties of the bag membrane are as follows:

Air properties:

Initial conditions of air inside the bags:

Conditions environment:

Properties of the gas from the inflator:

Young's Modulus 2.6e8 N/m2

Poison’s Ratio .3

Damp Factor .1

Membrane Thickness .0007 m

Density 1.527 kg/m3

Gamma 1.517

Gas Constant R 226.45 m2/sec2*K

Density 1.527 kg/m3

Specific Internal Energy 1.2815e5 m2/sec2

Pressure 101325.2222e2 N/m2

Pressure 101325. N/m^2

Gamma 111.557

Gas Constant R 243. m2/sec2*K

Temperature 400. K (assumed constant)

Mass Flow Rate:


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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,,,,,,,,++,22SURFACE,25,,PROP,125SET1,125,3,10,14,200,201,301,302,++,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,++,,,,,,,,,++,23SURFACE,35,,PROP,135SET1,135,4,15,301


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,1.0PORFLCPL,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,1001SET1,1001,200COUINFL,1,10,1001,INFLATR1,85,CONSTANT,0.7INFLATR1,85,1,,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,.00035MATRIG,1,783.TLOAD1,1,1,,12FORCE,1,MR1,,1.0,0.0,0.0,0.0MOMENT,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


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The gas itself is represented by the Gamma Law equation of state:

DMAT,3,1.527,3EOSGAM,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,1,3,5,1.0SPHERE,1,,0.0,0.0,0.0,500.0TICVAL,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.


In summary, the following contact options are used

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.

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


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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 sub-cycle, 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) = TIMEHISSURFACES (Surface) = 3SET 3 = 25SURFOUT (Surface) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (Surface) = 0,thru,end,by,0.0001SAVE (Surface) = 10000

For each compartment, another set of these is defined.


4. To visualize the gas flow in the Euler mesh, the following output request is used:

TYPE (euler) = ARCHIVEELEMENTS (euler) = 1000SET 1000 = ALLEULHYDROELOUT (euler) = XVEL YVEL ZVEL PRESSURETIMES (euler) = 0,thru,end,by,1.E-3SAVE (euler) = 1

There are six separate Euler domains, and different files are created for each Euler domain, using the naming convention:

FOLDED_EULER_FV1_#.ARCFOLDED_EULER_FV2_#.ARCFOLDED_EULER_FV3_#.ARCFOLDED_EULER_FV4_#.ARCFOLDED_EULER_FV5_#.ARCFOLDED_EULER_FV6_#.ARC

A new set of files is written at each output step.

ResultsThe 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 Average Pressures in Vertical Compartments


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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 Euler Domains at End of Simulation


Input DeckMain Input file folded.dat:

MEMORY-SIZE = 6000000,6000000IMMFILE=flat_imm.datSTARTCENDENDTIME=0.02CHECK=NOTITLE= Jobname is: foldedTLOAD=1TIC=1SPC=1$$ Output result for request: elementsTYPE (elements) = ARCHIVEELEMENTS (elements) = 1SET 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 17004ELOUT (elements) = EFFSTSTIMES (elements) = 0,THRU,END,BY,4e-4SAVE (elements) = 1$INCLUDE output_for_euler.ccf$------- Parameter Section ------PARAM,INISTEP,1.0e-7PARAM,MINSTEP,1.0e-8PARAM,STEPFCTL,0.9PARAM,IMM,,ON,5e-3,5E-3PARAM,FASTCOUPPARAM,COSUBMAX,10$------- BULK DATA SECTION -------BEGIN BULKINCLUDE bag.bdfINCLUDE couple_settings.bdfINCLUDE euler_mesh.bdfINCLUDE fix_inflator.bdfINCLUDE pressure_gauge.bdfINCLUDE folded.bdf$------- 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


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$$ ========== PROPERTY SETS ==========$$ * P_SET.1 *$PSHELL1 1 2 MEMB +A000001+A000001 .0007$$ * P_SET.2 *$PSHELL1 2 2 MEMB +A000002+A000002 .0007$$ * P_SET.3 *$PSHELL1 3 2 MEMB +A000003+A000003 .00035$$ * P_SET.4 *$PSHELL1 4 2 MEMB +A000004+A000004 .00035$$ * P_SET.5 *$PSHELL1 5 2 MEMB +A000005+A000005 .00035$$ * P_SET.6 *$PSHELL1 6 2 MEMB +A000006+A000006 .00035$$ * P_SET.7 *$PSHELL1 7 2 MEMB +A000007+A000007 .00035$$ * P_SET.8 *$PSHELL1 8 2 MEMB +A000008+A000008 .00035$$ * P_SET.14 *$PSHELL1 14 2 MEMB +A000010+A000010 .00035$$ * P_SET.15 *$PSHELL1 15 2 MEMB +A000011+A000011 .00035$$ * P_SET.16 *


$PSHELL1 16 2 MEMB +A000012+A000012 .00035$$ * P_SET.17 *$PSHELL1 17 2 MEMB +A000013+A000013 .00035$$ * P_SET.18 *$PSHELL1 18 2 MEMB +A000014+A000014 .00035$$ * P_SET.19 *$PSHELL1 19 2 MEMB +A000015+A000015 .00035$$ * inlet.200 *$PSHELL 200 1 .00035$$ * Euler.300 *$PEULER1,300,,HYDRO,19$$ * face.201 *$PSHELL 201 1 .00035$$ * diffusor.10 *$PSHELL1 10 2 MEMB +A000016+A000016 .00035$PSHELL1 301 DUMMYPSHELL1 302 DUMMYPSHELL1 303 DUMMYPSHELL1 304 DUMMYPSHELL1 305 DUMMY$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- Material Rigid id =1MATRIG 1 783 +A000017+A000017$$ -------- Material Fabric id =2DMATEL 2 783 2.6e+08 .3 ++ .01$


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$ -------- Material Ideal_Gas id =3DMAT 3 1.527 3$ |$ -> density$EOSGAM,3,1.517,226.45$ | |$ | -> R$ -------> Cp/Cv$$ -------- Material Webbing id =4DMATEL 4 670 6.5e+09 .3 ++ .1$$ Transient Dynamic LoadTLOAD1,1,1,,12$ translational velocityFORCE,1,MR1,,1.0,0.0,0.0,0.0$ rotational velocityMOMENT,1,MR1,,1.0,0.0,0.0,0.0$ENDDATAThe bulkdata file folded.bdf:

$ --- SPC-name = AttachmentsSPC1 1 123456 555 THRU 563 566 THRU 574++ 611 THRU 636$$ --- Define 6722 grid points ---$GRID 1 .0184701.0143125 .459878GRID 2 .0125105.0143125 .459878GRID 61 1.89e-08.0005000 .247850GRID 62 1.45e-08.0005000 .172850...$$ --- Define 13727 elements$$ -------- property set P_SET.1 ---------CTRIA3 59 1 183 252 61CTRIA3 60 1 253 270 252CTRIA3 61 1 252 270 61CTRIA3 62 1 248 189 62CTRIA3 63 1 213 247 62CTRIA3 64 1 248 62 247...The IMM reference file flat_imm.dat:

$$ --- Define 6871 grid points ---$GRID 1 .0009375-.626128 .230000GRID 2 .0009375-.626128 .220000GRID 61 .0005000-.414100 .450000


GRID 62 .0005000-.339100 .450000...$$ --- Define 13727 elements$$ -------- property set P_SET.1 ---------CTRIA3 59 1 183 252 61CTRIA3 60 1 253 270 252CTRIA3 61 1 252 270 61CTRIA3 62 1 248 189 62CTRIA3 63 1 213 247 62CTRIA3 64 1 248 62 247The Surface definitions file bag.bdf:

$$ -------------------------------------------------------------$ bag 1, connected to the inflator$ this surface is used for gbag/couple definition$ main bagSURFACE,25,,PROP,125SET1,125,3,10,14,200,201,301,302,++,303,304,305$SET1,125,3,THRU,8,10,14,THRU,19,+$+,200,201$$ compartment 1SURFACE,35,,PROP,135SET1,135,4,15,301$$ compartment 2SURFACE,45,,PROP,145SET1,145,5,16,302$$ compartment 3SURFACE,55,,PROP,155SET1,155,6,17,303$$ compartment 4SURFACE,65,,PROP,165SET1,165,7,18,304$$ compartment 5SURFACE,75,,PROP,175SET1,175,8,19,305$$ this surface is used for contact (not 10, 200,201)SURFACE,26,,PROP,126SET1,126,3,THRU,8,14,THRU,19$SUBSURF,301,25,PROP,1301SET1,1301,301$SUBSURF,302,25,PROP,1302SET1,1302,302


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$SUBSURF,303,25,PROP,1303SET1,1303,303$SUBSURF,304,25,PROP,1304SET1,1304,304$SUBSURF,305,25,PROP,1305SET1,1305,305$$ define gasbag$ See section 2.1.1 in user manual for units$$ define euler coupled airbag$COUPLE,1,25,OUTSIDE,ON,ON,9,11,AIRBAG,++,10,,,,,,,,++,,22COUPLE,2,35,OUTSIDE,ON,ON,,12,AIRBAG,++,,,,,,,,,++,,23COUPLE,3,45,OUTSIDE,ON,ON,,13,AIRBAG,++,,,,,,,,,++,,24COUPLE,4,55,OUTSIDE,ON,ON,,14,AIRBAG,++,,,,,,,,,++,,25COUPLE,5,65,OUTSIDE,ON,ON,,15,AIRBAG,++,,,,,,,,,++,,26COUPLE,6,75,OUTSIDE,ON,ON,,16,AIRBAG,++,,,,,,,,,++,,27$ ..............................................................$ inflator$SUBSURF,1001,25,PROP,1001SET1,1001,200$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,1.0COUPOR,2,9,302,PORFLCPL,3002,constant,1.0


COUPOR,3,9,303,PORFLCPL,3003,constant,1.0COUPOR,4,9,304,PORFLCPL,3004,constant,1.0COUPOR,5,9,305,PORFLCPL,3005,constant,1.0$PORFLCPL,3001,,,,2PORFLCPL,3002,,,,3PORFLCPL,3003,,,,4PORFLCPL,3004,,,,5PORFLCPL,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.$MESH,22,ADAPT,0.009,0.009,0.009,,,,++,,,,,,,,,++,,,,,,,EULER,300MESH,23,ADAPT,0.011,0.011,0.011,,,,++,,,,,,,,,++,,,,,,,EULER,300MESH,24,ADAPT,0.011,0.011,0.011,,,,++,,,,,,,,,++,,,,,,,EULER,300MESH,25,ADAPT,0.011,0.011,0.011,,,,++,,,,,,,,,++,,,,,,,EULER,300MESH,26,ADAPT,0.011,0.011,0.011,,,,++,,,,,,,,,++,,,,,,,EULER,300MESH,27,ADAPT,0.011,0.011,0.011,,,,++,,,,,,,,,++,,,,,,,EULER,300$TICEUL,19,,,,,,,,++,SPHERE,1,3,5,1.0SPHERE,1,,0.0,0.0,0.0,500.0TICVAL,5,,DENSITY,1.527,SIE,1.2815E5The output request file output_for_euler.ccf:

$ Output result for request: eulerTYPE (euler) = ARCHIVE


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ELEMENTS (euler) = 1000SET 1000 = ALLEULHYDROELOUT (euler) = XVEL YVEL ZVEL PRESSURE$ SIE XMOM YMOM ZMOM DENSITYTIMES (euler) = 0,thru,end,by,1.E-3SAVE (euler) = 1$TYPE (Surface) = TIMEHISSURFACES (Surface) = 3SET 3 = 25SURFOUT (Surface) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (Surface) = 0,thru,end,by,0.0001SAVE (Surface) = 10000$TYPE (part1) = TIMEHISSURFACES (part1) = 4SET 4 = 35SURFOUT (part1) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (part1) = 0,thru,end,by,0.0001SAVE (part1) = 10000$TYPE (part2) = TIMEHISSURFACES (part2) = 5SET 5 = 45SURFOUT (part2) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (part2) = 0,thru,end,by,0.0001SAVE (part2) = 10000$TYPE (part3) = TIMEHISSURFACES (part3) = 6SET 6 = 55SURFOUT (part3) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (part3) = 0,thru,end,by,0.0001SAVE (part3) = 10000$TYPE (part4) = TIMEHISSURFACES (part4) = 7SET 7 = 65SURFOUT (part4) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (part4) = 0,thru,end,by,0.0001SAVE (part4) = 10000$TYPE (part5) = TIMEHISSURFACES (part5) = 8SET 8 = 75SURFOUT (part5) = AREA VOLUME MASS PRESSURE TEMPTURETIMES (part5) = 0,thru,end,by,0.0001SAVE (part5) = 10000

479Chapter 6: Occupant SafetyHybrid III 5th%-tile Dummy

Hybrid III 5th%-tile Dummy

Problem DescriptionIn 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-Of-Position (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 DescriptionThe 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).

Dytran Example Problem ManualHybrid III 5th%-tile Dummy

480

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.


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


482

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.


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 Results of Neck Extension for the Pivot andbber Neck Models for every 20 Mseconds


484

Figure 6-24 Time History of D-Plane Extension

Figure 6-25 Time History of Neck Extension Moment


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.


486

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 Results of Flexion for the Pivot and Rubber Neck ls for every 20 mseconds


Figure 6-28 Time History of D-Plane Flexion

Figure 6-29 Time History of Neck Flexion Moment


488

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.


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 Results of Thorax Impact of Dummy Model


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Figure 6-32 Time History of Thorax Displacement

Figure 6-33 Time History of thorax Impact


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.


492

Input Files

Neck Extension

CENDENDTIME=0.120ENDSTEP=99999999CHECK=NOTITLE= Jobname is: neck_extensionTLOAD=1TIC=1SPC=1$TYPE(MR)=MRSUMSTEPS(MR)= 0 THRU END BY 1000$TYPE(MAT)=MATSUMSTEPS(MAT)= 0 THRU END BY 1000$TYPE(STEP)=STEPSUMSTEPS(STEP)= 0 THRU END BY 25$$ Output result for request: neck_extensionTYPE (elem) = ARCHIVEELEMENTS (elem) = 1SET 1 = ALLELEMENTSELOUT (elem) = MASSTIMES (elem) = 0 THRU END BY 0.005SAVE (elem) = 10000$TYPE (dplane) = TIMEHISGRIDS (dplane) = 2SET 2 = 1080020 1080023GPOUT (dplane) = XPOS YPOS ZPOSTIMES (dplane) = 0 THRU END BY 0.0001SAVE (dplane) = 10000$TYPE(loadcells) = TIMEHISELEMENTS(loadcells) = 1000003SET 1000003 = 1080011, 1080012, 1080021, 1080022, 1080031, 1080032, 1080041, 1080042, 1080051, 1080052, 1080061, 1080062ELOUT(loadcells) = XFORCETIMES(loadcells) = 0.000 THRU END BY 0.0001SAVE(loadcells) = 10000$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$BEGIN BULK$PARAM, INISTEP, 1.0e-6PARAM, MINSTEP, 1.0e-9PARAM, GEOCHECK, ONPARAM, RBE2INFO, GRIDONPARAM, MATRMERG, FR1, MR1040000, MR2000000


$INCLUDE hyb305_pivot_neck.dat$$--------------------------------------------------------------------------$ BEGIN PENDULUM COMPONENT$--------------------------------------------------------------------------GRID 2000000 -.098729 .0381 .296886GRID 2000001 -.098729 3.725-9 .296886GRID 2000002 -.098729-.0381 .296886-

$CQUAD4 2000000 2000000 2000000 2000001 2000004 2000003CQUAD4 2000001 2000000 2000001 2000002 2000005 2000004-

$ pid mid form quad nip shfact ref contPSHELL1, 2000000,2000000, , , , , , ,+$ thick+, 0.005$ mid rho E nu mass cogX cogY cogZMATRIG, 2000000, , , , 29.57,-.047929, 3.027-9 ,2.13209$TLOAD1 1 1001 12 2000001MOMENT,1001,FR1,,1,0,1,0FORCE,1001,FR1,,0,0,0,0$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


494

Neck Flexion

STARTCENDENDTIME=0.120ENDSTEP=99999999CHECK=NOTITLE= Jobname is: neck_extensionTLOAD=1TIC=1SPC=1$TYPE(MR)=MRSUMSTEPS(MR)= 0 THRU END BY 1000$TYPE(MAT)=MATSUMSTEPS(MAT)= 0 THRU END BY 1000$TYPE(STEP)=STEPSUMSTEPS(STEP)= 0 THRU END BY 25$$ Output result for request: neck_extensionTYPE (elem) = ARCHIVEELEMENTS (elem) = 1SET 1 = ALLELEMENTSELOUT (elem) = MASSTIMES (elem) = 0 THRU END BY 0.005SAVE (elem) = 10000$TYPE (dplane) = TIMEHISGRIDS (dplane) = 2SET 2 = 1080020 1080023GPOUT (dplane) = XPOS YPOS ZPOSTIMES (dplane) = 0 THRU END BY 0.0001SAVE (dplane) = 10000$TYPE(loadcells) = TIMEHISELEMENTS(loadcells) = 1000003SET 1000003 = 1080011, 1080012, 1080021, 1080022, 1080031, 1080032, 1080041, 1080042, 1080051, 1080052, 1080061, 1080062ELOUT(loadcells) = XFORCETIMES(loadcells) = 0.000, THRU, END, BY, 0.0001SAVE(loadcells) = 10000$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$BEGIN BULK$PARAM, INISTEP, 1.0e-6PARAM, MINSTEP, 1.0e-9PARAM, GEOCHECK, ONPARAM, RBE2INFO, GRIDONPARAM, MATRMERG,FR1,MR1040000,MR2000000$INCLUDE hyb305_pivot_neck.dat


$$--------------------------------------------------------------------------$ BEGIN PENDULUM COMPONENT$--------------------------------------------------------------------------GRID 2000000 -.098729 .0381 .296886GRID 2000001 -.098729 3.725-9 .296886-

$CQUAD4 2000000 2000000 2000000 2000001 2000004 2000003CQUAD4 2000001 2000000 2000001 2000002 2000005 2000004-$ pid mid form quad nip shfact ref contPSHELL1, 2000000,2000000, , , , , , ,+$ thick+, 0.005$ mid rho E nu mass cogX cogY cogZMATRIG, 2000000, , , , 29.57,-.047929, 3.027-9 ,2.13209$TLOAD1 1 1001 12 2000001MOMENT,1001,FR1,,1,0,1,0FORCE,1001,FR1,,0,0,0,0$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


496

Thorax ImpactSTARTCENDENDTIME=0.050CHECK=NOTITLE= Jobname is: testTLOAD=1TIC=1SPC=1$TYPE(MR)=MRSUMSTEPS(MR)= 0 THRU END BY 1000$TYPE(MAT)=MATSUMSTEPS(MAT)= 0 THRU END BY 1000$TYPE(STEP)=STEPSUMSTEPS(STEP)= 0 THRU END BY 25$TYPE (HIII) = ARCHIVEELEMENTS (HIII) = 1SET 1 = ALLELEMENTSELOUT (HIII) = MASSTIMES (HIII) = 0 THRU END BY 0.005SAVE (HIII) = 100000$TYPE (grid_CG) = TIMEHISGRIDS (grid_CG) = 2SET 2 = 1040100 1050000GPOUT (grid_CG) = XPOS YPOS ZPOSTIMES (grid_CG) = 0 THRU END BY 0.0001SAVE (grid_CG) = 10000$$ Output result for request: contactTYPE (contact) = TIMEHISCONTS (contact) = 4SET 4 = 3 4CONTOUT (contact) = XFORCE YFORCE ZFORCE FMAGN DMINTIMES (contact) = 0 THRU END BY 0.001SAVE (contact) = 100000$TYPE (loadcells) = TIMEHISELEMENTS (loadcells) = 5SET 5 = 1080011, 1080012, 1080021, 1080022, 1080031, 1080032, 1080041, 1080042, 1080051, 1080052, 1080061, 1080062, 1050011, 1050012, 1050021, 1050022, 1050031, 1050032, 1050041, 1050042, 1050051, 1050052, 1050061, 1050062ELOUT (loadcells) = XFORCETIMES (loadcells) = 0.000 THRU END BY 0.0001SAVE (loadcells) = 10000$$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1e-7


PARAM,STEPFCT,0.9$$------- BULK DATA SECTION -------BEGIN BULK$INCLUDE hyb305_bulk.datINCLUDE floor.datINCLUDE impactbar.dat$$ Rigid Impactor can move only in global X dir$TLOAD1 1 1001 12FORCE, 1001,MR1,,0.0, ,1.0,1.0MOMENT,1001,MR1,,0.0,1.0,1.0,1.0$$ Rigid floor can not move$TLOAD1 1 1002 12FORCE, 1002,MR2,,0.0,1.0,1.0,1.0MOMENT,1002,MR2,,0.0,1.0,1.0,1.0$$ ------- Initial Velocity BC vel -----SET1 1 1281154 THRU 1282050TICGP 1 1 XVEL -6.47$$ -------- Contact between dummy and floor$CONTACT 3 SURF SURF 2 1 ++ V4 TOP FULL 0.0 NONE++ ++ ++ 33CONTFORC 33 3 3 33$TABLED1,3,,,,,,,,++,0.0,0.0,0.050,1000.0$TABLED1,33,,,,,,,,++,-1.000,-200.0,1.000,200.0$$ Slave contact surface: Dummy$SURFACE 2 PROP 1011002 PROP 1091001 PROP 1101001++ PROP 1111001 PROP 1121001 PROP 1151001 PROP 1161001++ PROP 1191001 PROP 1201001$$ Master contact surface: Floor$SURFACE 1 PROP 2SET1 2 2$$ -------- Contact between chest and impactbar$CONTACT 4 GRID SURF 4 1051001 +


498

+ V4 TOP FULL 0.0 NONE++ ++ ++ 44CONTFORC 44 4 4 5$TABLED1,4,,,,,,,,++,0.0,0.0,0.050,1000.0$TABLED1,5,,,,,,,,++,-1.000,-200.0,1.000,200.0$$ Slave contact gridpoints: ImpactbarSET1 4 1281154 THRU 1282050$$$ ========== PROPERTY SETS ==========$$ * impactbar *$PSOLID 1 1$$ * Floor *$PSHELL 2 2 .003$$$ ========= MATERIAL DEFINITIONS ==========$$$ -------- ImpactbarMATRIG 1 2590 7e+10 .3$$ -------- FloorMATRIG 2 7850 7e+10 .3$ENDDATAfloor.dat:CQUAD4 1289794 2 1282051 1282052 1282073 1282072CQUAD4 1289795 2 1282052 1282053 1282074 1282073CQUAD4 1289796 2 1282053 1282054 1282075 1282074CQUAD4 1289797 2 1282054 1282055 1282076 1282075CQUAD4 1289798 2 1282055 1282056 1282077 1282076...

GRID 1282051 -.553700-.250000-.091000GRID 1282052 -.478700-.250000-.091000GRID 1282053 -.403700-.250000-.091000GRID 1282054 -.328700-.250000-.091000...impact_bar.dat:$GRID 1281154 .176400-.006968 .267826GRID 1281155 .176400-.013936 .289273


GRID 1281156 .211960-.013936 .289273GRID 1281157 .211960-.006968 .267826GRID 1281158 .247520-.013936 .289273GRID 1281159 .247520-.006968 .267826-

$ -------- property set impactbar_prop ---------CHEXA 1289014 1 1282001 1282002 1281991 1281990 1281154 1281157+A000001+A000001 1281156 1281155CHEXA 1289015 1 1282002 1282003 1281992 1281991 1281157 1281159+A000002+A000002 1281158 1281156CHEXA 1289016 1 1282003 1282004 1281993 1281992 1281159 1281161+A000003+A000003 1281160 1281158CHEXA 1289017 1 1282004 1282005 1281994 1281993 1281161 1281163+A000004+A000004 1281162 1281160-

CPENTA 1289474 1 1282001 1281154 1282012 1282002 1281157 1282013CPENTA 1289475 1 1282002 1281157 1282013 1282003 1281159 1282014CPENTA 1289476 1 1282003 1281159 1282014 1282004 1281161 1282015CPENTA 1289477 1 1282004 1281161 1282015 1282005 1281163 1282016CPENTA 1289478 1 1282005 1281163 1282016 1282006 1281165 1282017-


500

Appendix A: Overview of the Hybrid III 5th%-tile Dummy

Schematic Setup


Joint Information


502

General Numbering Information


Component Numbering Information


504

Joint Numbering Information

Appendix B: SURFACE and SET1 Definition

SurfaceSURFACE 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


Appendix C: hyb305 PositioningThe 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

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

$ 5th Percentile Hybrid III Dummy Database Definition Model$$ @(#) HYB305 : 1.0 2003/05/01 :$$ Developed For : Dytran 2003$ Unit System : kg, m, s, K$ Release Status : NOT RELEASED$ Performance : Functional Model Only - NOT VALIDATED$$ IMPORTANT NOTICE:$

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

SurfaceSURFACE and SET1

Identification Number Association Typical Usage


506

$ 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 componentsSELECT, NeckModel, 2$$ Note: Angles in Degrees$$ nameID dim dof1 dof2 dof3INIVAR, NumberingBase, 1, 1000000$$ translate in direction : forward left verticalINIVAR, HPointPosition, 3, 0.000, 0.00, 0.000$$ rotate around direction: forward left verticalINIVAR, HPointAngle, 3, 0.0, 0.0, 0.0$$ rotate around direction: forward left verticalINIVAR, LowerTorsoAngle, 3, 0.0, 0.0, 0.0$$ rotate around direction: leftINIVAR, NeckBracketAngle, 1, 0.0$$ rotate around direction: left forward verticalINIVAR, LeftHipAngle, 3, 0.0, 0.0, 0.0$$ rotate around direction: left forward verticalINIVAR, RightHipAngle, 3, 0.0, 0.0, 0.0$$ rotate around direction: leftINIVAR, LeftKneeAngle, 1, -90.0$$ rotate around direction: leftINIVAR, RightKneeAngle, 1, -90.0$$ rotate around direction: left forward verticalINIVAR, LeftAnkleAngle, 3, 0.0, 0.0, 0.0$$ rotate around direction: left forward verticalINIVAR, RightAnkleAngle, 3, 0.0, 0.0, 0.0$


$ rotate around direction: left vertical forwardINIVAR, LeftShoulderAngle, 3, -90.0, 0.0, 0.0$$ rotate around direction: left vertical forwardINIVAR, RightShoulderAngle, 3, -90.0, 0.0, 0.0$$ rotate around direction: left verticalINIVAR, LeftElbowAngle, 2, 0.0, 0.0$$ rotate around direction: left verticalINIVAR, RightElbowAngle, 2, 0.0, 0.0$$ rotate around direction: left forwardINIVAR, LeftWristAngle, 2, 0.0, 0.0, 0.0$$ rotate around direction: left forwardINIVAR, RightWristAngle, 2, 0.0, 0.0, 0.0


508

Reference Database

509Chapter 6: Occupant SafetyEasy Postprocessing with Adaptive Meshing

Easy Postprocessing with Adaptive MeshingDuring 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 DescriptionTo illustrate static output for adaptive meshing, example problem 6.1 will be run with the Euler solver.

Dytran Example Problem ManualEasy Postprocessing with Adaptive Meshing

510

Airbag Model

Dytran ModelingFor 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.


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,.01INFLATR1,82,1,2,,1.4,,286.

The Euler is initialized by:

PEULER1,1,,HYDRO, 1,,,,,+SPHERE, 1,, 0., 0. , 0., 1.E10TICEUL,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.

ResultsAll 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.


512

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.


514

Velocities on the fmat=0.5 iso-surface for 20 ms and 60 ms.

Dytran Input FileNIF = FORMSTART

Dytran input deck

CENDCHECK=NOENDSTEP=100000ENDTIME=60.E-3


TITLE=INFLATION OF DEMO PASSENGER-BAG USING EULERTIC=1SPC=1TLOAD=1$ --------------------------------------------------------------$ define the output for the euler elements:$ELEMENTS(EUL) = 995SET 995 = ALLEULHYDROELOUT(EUL) = PRESSURE,DENSITY,FMAT,XVEL,YVEL,ZVEL,SIE,VOLUMEELOUT(EUL) = XMOM,YMOM,ZMOMTIMES(EUL) = 0. THRU END BY 4.E-3TYPE(EUL) = ARCHIVESAVE(EUL) = 1000$$ --------------------------------------------------------------$ define the output for the membrane elements:$ELEMENTS(AIRBAG) = 996SET 996 = ALLMEMTRIAELOUT(AIRBAG) = THICK,SMDFERTIMES(AIRBAG) = 0. THRU END BY 4.E-3TYPE(AIRBAG) = ARCHIVESAVE(AIRBAG) = 100000$$$BEGIN BULK$ Initial timestep & minimum timestep$PARAM,FASTCOUPPARAM,INISTEP,0.7E-05PARAM,MINSTEP,.5E-8$CTRIA3 1 501 1 12 13CTRIA3 2 501 1 13 2----------------CTRIA3 3514 505 1248 1242 2053$GRID 1 0.0 0.0 0.0GRID 2 -.03 0.0 0.0----------------GRID 2145 .15 .0496 -.04$$ First quarter$PSHELL1,501,1,MEMB,,,,,,++,5.E-4PSHELL1,502,1,MEMB,,,,,,++,5.E-4------------


516

PSHELL1,513,1,MEMB,,,,,,++,5.E-4$$ Second quarter (PID first + 50)$PSHELL1,551,1,MEMB,,,,,,++,5.E-4PSHELL1,552,1,MEMB,,,,,,++,5.E-4------PSHELL1,563,1,MEMB,,,,,,++,5.E-4$$ Third quarter (PID first + 100)$PSHELL1,601,1,MEMB,,,,,,++,5.E-4PSHELL1,602,1,MEMB,,,,,,++,5.E-4------PSHELL1,613,1,MEMB,,,,,,++,5.E-4$$ fourth quarter (PID first + 150)$PSHELL1,651,1,MEMB,,,,,,++,5.E-4PSHELL1,652,1,MEMB,,,,,,++,5.E-4------------PSHELL1,663,1,MEMB,,,,,,++,5.E-4$$ 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 !$PSHELL1,510, ,DUMMYPSHELL1,560, ,DUMMYPSHELL1,610, ,DUMMYPSHELL1,660, ,DUMMY$

DMATEL,1,600.,6.E7,0.3$DMAT 5 .681864 1EOSGAM 1 1.4 297.


$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.E10TICEUL,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,.01INFLATR1,82,1,2,,1.4,,286.$TABLED1 1 + + 0.00E+00 0.0 0.50E-02 .15 0.10E-01 3.05 0.15E-01 3.2 + + 0.20E-01 3.25 0.25E-01 3.25 0.50E-01 2.8 0.55E-01 1.75 + + 0.60E-01 1.0 0.65E-01 .5 0.70E-01 0.0 0.10E+00 0.0+ + ENDT TABLED1 2 + + 0.00E+00 500. 0.50E-02 500. 0.10E-01 600. 0.10E+006.000E+0+ + ENDT SPC 1 253 123456SPC 1 254 123456------------SPC 1 2145 123456$SURFACE,25,,SUB,42$SUBSURF,42,25,PROP,42SET1,42,501,THRU,513,551,THRU,563,601,++,THRU,604,606,THRU,613,651,THRU,663$SUBSURF,81,25,PROP,81SET1,81,510,560,610,660$SUBSURF,82,25,PROP,82SET1,82,505


518

$ENDDATA

Chapter 7: Quasi-static Analysis Dytran Example Problem Manual

7Quasi-static Analysis

Overview 520

Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions 521


520

OverviewIn 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:

521Chapter 7: Quasi-static AnalysisComparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

Comparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

Problem DescriptionA 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)

Initial Load

Total Force = 76.5 N

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

Dytran Example Problem ManualComparison of Quasi-static Dytran to MD Nastran Shell Element Solutions

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Dytran ModelA 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:

In which T denotes the natural period of free vibration and 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.

ResultsThe 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.

VDAMP2∗π∗δt

T-----------------=

δt


Figure 7-3 shows the displacement of the nodes at the free end of the plate for the KEYHOFF model, with and without damping.

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)


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MD Nastran ResultsMD Natran’s linear static solution (SOL 101) and the nonlinear static solution (SOL 106) give very similar results. See the table below.

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 ResultsWhen 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 Maximum Stress

SOL 101 (linear statics) 0.0203 m 3.42E8 Pa

SOL 106 (nonlinear statics) 0.0202 m 3.43E8 Pa

Maximum Displacement (*) Maximum Stress (*)

KEYHOFF without RBRCOR 0.0199 m (1.9%) 3.11E8 Pa (9.0%)

KEYHOFF with RBRCOR 0.0202 m. (0.0%) 3.55E8 Pa (3.4%)

(*) = difference with MD Nastran SOL 106 solution


Figure 7-4 Fringe Plots from MD Nastran Results

Figure 7-5 Fringe Plot for Dytran Results


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Files


STARTCENDENDTIME=0.070ENDSTEP=999999CHECK=NOTITLE= Jobname is: dytranjobTLOAD=1TIC=1SPC=1$ Output result for request: elementsTYPE (elem) = ARCHIVEELEMENTS (elem) = 1SET 1 = 1 THRU 800 ELOUT (elem) = EFFSTS TIMES (elem) = 0 thru end by 0.0005SAVE (elem) = 10000$TYPE (grid) = TIMEHISGRIDS (grid) = 10SET 10 = 977 713 449GPOUT (grid) = XDIS YDIS ZDIS XFORCE RVEL RDIS RFORCE TIMES (grid) = 0 thru end by 0.0005SAVE (grid) = 10000$$------- Parameter Section ------PARAM,CONTACT,THICK,0.0PARAM,INISTEP,1e-7PARAM,VDAMP,0.0001$------- BULK DATA SECTION -------BEGIN BULK$ --- SPC-name = nodal_constraintsSPC1 1 123456 11 14 15 16 42 43+A000001$$ --- Define 889 grid points --- $GRID 1 .100000.0750000 .00000GRID 2 .100000 .100000 .00000:GRID 1008 .106455 .00000.0005027$$ --- Define 800 elements

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


$$ -------- property set prop ---------CQUAD4 1 1 7 19 5 2CQUAD4 2 1 8 20 19 7:CQUAD4 800 1 975 229 226 1008$$ ========== 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+A000007 .3 1 .05$$ ========= MATERIAL DEFINITIONS ==========$$ -------- Material steel1 id =1DMATEP 1 8027 2e+11 .3 1YLDVM 11.86e+09 4e+09 +A000008+A000008$$ ======== Load Cases ========================$ ------- Force BC nodal_force ----- TLOAD1 1 9 0 2FORCE 9 449 0 1 1.7913 0 -4.1281:FORCE 9 977 0 1 1.7913 0 -4.1281$$ ================ TABLES =================$ ------- TABLE 2: force_table -------TABLED1 2 +A000009+A000009 0 0 .003 1 .07 1 ENDTENDDATA


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Dytran Example Problem Manual

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