Container Blast Test Report

Simulation of a Standard ISO Steel Container Subjected to Blast Loading

Torbjoern Dyngeland PUBSY JRC56386 -2010

The mission of the JRC-IPSC is to provide research results and to support EU policy-makers in their effort towards global security and towards protection of European citizens from accidents deliberate attacks fraud and illegal actions against EU policies European Commission Joint Research Centre Institute for the Protection and Security of the Citizen Contact information Torbjoern Dyngeland Address European Commission - JRC TP 480 Via Fermi 2749 I-21027 Ispra E-mail torbjoerndyngelandjrcit Tel +390332786197 Fax +390332789049 httpipscjrceceuropaeu httpwwwjrceceuropaeu Legal Notice Neither the European Commission nor any person acting on behalf of the Commission is responsible for the use which might be made of this publication

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TABLE OF CONTENTS 1 Introduction 1 11 Background 1 12 Collaboration framework 1 13 Organisation of the report 1 2 Blast test of a 20 ft ISO container 3 21 General 3 22 Test set-up 4 23 Test results 5 3 The finite element model of the container 8 31 General 8 32 FE-model 10 4 Numerical simulations 13 41 General 13 42 Initial calculations 13 43 Refined calculations 18 5 Discussions and conclusions 27 6 References 29 7 Appendix 30 71 Cast3M (file type dgibi) and EUROPLEXUS (file type epx)

input files for the numerical calculations 30

i

1

1 Introduction 11 Background The use of the finite element method (FEM) to simulate structural responses to extreme loadings is a well established practise in design of buildings and structures today Modern FEM software codes in structural engineering are verified and validated against a vast number of experimental results ranging from minute testing of engineering materials for the verification of constitutive models via high precision testing of structural elements for investigating local failure modes to full scale tests of real structures to verify the global response under realistic loading scenarios Full scale blast tests of real structures are rare because of their size complexity and above all the costs required for their execution Hence high precision tests of full scale structures are always met with great interest and enthusiasm in the scientific society since such tests represent an excellent opportunity to check a complex FEM-model with all its assumptions against a verified outcome The present report presents the outcome of a numerical study of a full scale blast test of an unprotected 20 ft standard ISO steel container The blast test was part of a comprehensive research programme executed by the SIMLab group at the Department of Structural Engineering The Norwegian University of Science and Technology (NTNU) in Trondheim Norway The tests aimed at verifying the behaviour of standard 20 ft ISO containers used as shelters in international operations such as peace-keeping operations rescue and reconstruction operations in international conflict areas 1 The blast tests were carried out in the Large Blast Simulator (LBS) at the Bundeswehr Technical Center for Protective and Special Technologies (WTD 52) in Oberjettenberg Germany 2

12 Collaboration framework The present study was carried out under the Physical Vulnerability Assessment of Critical Structures (PVACS) action of the JRC ELSA unit The numerical simulations were performed by use of the explicit finite element (FE) code EUROPLEXUS 3 while the specific detailing and build-up of the FE-model of the container was carried out by use of the general purpose finite element code Cast3M 4 Access to the detailed test results from the blast tests and the experimental set-up has been granted to the JRC under a collaboration agreement titled Structural Safety and Security between the JRC and the Department of Structural Engineering NTNU in 2007 5 Valuable additional information related to the test site in Germany has been provided by direct input from the WTD 52 in Germany

13 Organisation of the report The numerical simulations in this study include two phases In the first phase the container was subjected to an applied pressure-time loading on the longitudinal wall of the container that was facing the explosive charge in the tunnel The applied pressure-time loading was derived by a simple pressure-time routine in EUROPLEXUS called AIRB 6 which calculates the pressure acting on an object

2

based on the explosive charge and the stand-of distance between the explosive charge and the object In the second phase the container was loaded with the set of pressure-time curves actually recorded during the experiment and reported in 1 The experimental set up is briefly presented in Chapter 2 together with some of the main results from the blast test Due to ownership issues and restricted information related to the NTNU project the interested reader is referred to reports issued from NTNU eg 6 7 and 8 for further details Chapter 3 presents the detailed build-up of the FE model of the container while the load conditions the main results from the simulations and comparison between the test results and the numerical findings are presented in Chapter 4 Discussions and conclusions are presented in Chapter 5 followed by a list of relevant publications in Chapter 6

3

2 Blast test of a 20 ft ISO container 21 General A 20 ft standard ISO steel container was subjected to a blast test in the Large Blast Simulator (LBS) at WTD 52 in Germany The blast simulator is basically a dead-end tunnel of approximately 100 m length and a rectangular-semicircular 1284 m wide cross-section with a height of about 700 m The tunnel is slightly inclined towards the opening with a slope of around 4 See the sketch in figure 1 below The tunnel is equipped with a battery of 100 high pressure steel vessels located at the dead-end of the tunnel each of them with the capacity of delivering 384 litres of compressed air up to 200 bars When the compressed air in the pressure vessels is simultaneously released by explosive devices cutting of steel diaphragms at the end of each vessel the air expands and forms a plane blast wave travelling down towards the opening of the tunnel By manipulating the pressure level in the vessels as well as the release time of the diaphragms this blast facility can simulate pressure waves with side-on pressure in the range of 5 to 100 kPa and with durations in the order of seconds The tunnel is further equipped with baffle sections three internal walls with rectangular openings gradually reduced down to a final rectangular opening of 6x5 m forcing an improved plane pressure wave to arrive at the test specimen location 40-100 m down the tunnel Figure 1 A schematic view of the Large Blast Simulator (LBS) at Bundeswehr

Technical Center for Protective and Special Technologies (WTD 52) in Oberjettenberg Germany 2 Courtesy WTD 52

10000 m

Test zone with test specimen

Battery of 100 steel bottles filled with compressed air

Cross-section of the tunnel with baffle walls

Cross-section of the tunnel

700

1284

4

22 Test set-up A standard 20ft ISO steel container was positioned at the mouth of the tunnel 95 m away from the pressured steel vessels delivering the pressure wave A series of blasts against an instrumented rigid concrete wall of 3x3 m were conducted in order to calibrate the pressure-time behaviour of the blast to a given design blast load representing 4000 kg TNT at a standoff distance of 120 m 1 The container was mounted with one of the longitudinal walls facing the blast front and fixed to the ground by a total of 7 clamps two on each short wall sides and 3 along the rear longitudinal wall These claps were made of angular steel brackets welded to the bottom frame of the container and bolted to the concrete floor of the blast tunnel The container was fully instrumented by means of a set of pressure gauges for recording the pressure at various locations of the container during the blast A laser gauge was set up to measure deflections and 3 high-speed cameras were used for visualization of the blast load response Though some of the instruments failed during the test a fairly comprehensive and consistent set of data was recorded by the data acquisition system For the unprotected container a total of 5 pressure gauges were mounted on the container at the mid-point of the longitudinal front wall and the longitudinal rear wall at the mid-point of the short wall on the roof and finally one gauge was mounted on the inside of the doors of the container A laser gauge registered the deflections of the mid-point of the longitudinal front wall The locations of the gauges are shown in figure 2 Additional details about the test set-up can be found in 1

Figure 2 Location of gauges on unprotected container Container mounted in the LBS Courtesy SIMLab NTNU 1

5

23 Test results A complete discussion of the test results for the blast load on the container is given in 1 Here only the main findings will be listed for the purpose of the comparison between the experimental results and the present numerical simulations of the container subjected to the same blast load as in the experiments Unfortunately the pressure gauges mounted on the longitudinal front wall and the longitudinal rear wall failed during the test However a later test performed on a similar container but with a protective wall mounted in front of the container facing the blast source gave a complete set of pressure-time readings throughout The authors of 1 opined that those results would suffice as a substitute for the lacking readings from the blast test of the unprotected container That position has been adopted also in the present study The pressure-time curves registered during the test are shown in figure 3 below It shall be noted that these are the overpressure values that is relative to the atmospheric pressure Hence the zero value in the pressure time curves represents the atmospheric pressure level (1 atm asymp 1 bar = 100 kPa) The peak value of the pressure recorded for the roof reached 38 kPa after approximately 380 ms and then gradually turned into negative pressure levels around 410 ms with peak values from -15 to - 28 kPa in the time interval of 440 - 500 ms See figure 3 top graph This rather pronounced peak in the negative pressure-time evolution was possibly due to reflection from the roof of the tunnel The registered pressure on the longitudinal front wall see figure 3 bottom graph exhibited a somewhat similar build-up of the pressure as for the roof The peak pressure level reached about 35 kPa at approximately 375 ms and only minor negative pressure levels around -5 kPa were registered in the final part of the pressure-time curves A significant internal pressure did build up in the container during the blast reaching a peak value of 22 kPa at 395 ms then to gradually turn into a negative pressure phase at 440 ms with a peak value in the order of -5 kPa around 450 ms It is interesting to compare the pressure levels on the container from both the outside and inside pressure gauges with the time level of the outward deflection of the roof as seen in figure 4 The deformation of the container was registered on the mid-point of the longitudinal front wall by use of a laser gauge The total deflection went beyond the working area of the laser gauge hence the maximum deflection of the wall was not properly registered However 1 reported that based upon post-assessment of the container the total plastic and elastic deflection of the mid-point was estimated to reach ndash 485 mm (inward deflection) The detailed deflection-time curve is given in figure 4 1 It was further reported in 1 that the top beam of the container had a permanent downward deformation of about 110 mm and a permanent inward displacement of 60 mm The roof exhibited an outward plastic deflection of about 300 mm No fracture of the container was observed

6

Figure 3 Pressure ndash time curves for unprotected (top) and protected (bottom)

container Courtesy SIMLab 1

a) pressure time curves for the roof the inside the longitudinal front and rear walls and the side walls of the container

b) pressure time curves for the longitudinal front wall of the protected container

7

a) b) Figure 4 a) Deformed container at various stages of the blast incident b)

recorded deflection of mid-point on front wall Courtesy SIMLab 1

8

3 The finite element model of the container 31 General The finite element (FE) models of the container were established by use of the general purpose finite element software Cast3M 4 Due to the rather complicated geometry of the various structural components of the container with corrugated wall panels and roof elements open-shaped columns and beams as well as the assemblage of the structural components the geometry of the finite element (FE) model of the container was established using a special algorithm developed at JRC ELSA for handling node numbering and mesh resolution issues originally developed for masonry structures 9 The algorithm allows for the meshing of structures made of blocks or separate regions connected together by joint interfaces By starting from the definition of the corner nodes of a given region the algorithm is able to produce the mesh of each region in such a way that the faces that are in contact are topologically identical This approach allows also an easy generation of joint elements if those are to be included in the analyses eg welds However for the present FE-models of the container complete material connections between the various structural elements were assumed hence no joint elements were defined The generation of compatible regions is based on an algorithm that inspects each region and if a master node of another region geometrically lies on this segment the node is duplicated and added to the description of the current region As a result the master nodes lying on the contact lines between two regions are two-by-two placed at the same location Hence the meshing strategy allows automatic meshing of the regions in such a way that the contour lines of two regions in contact are discretized in a similar way The principle is illustrated in figure 5 below while a complete description of this approach both for 2D and 3D examples is given in 9 Figure 5 Additional master nodes in non-compatible regions 9

Initial master nodes for the blockregion generation

additional master nodes for compatibility

9

This meshing method was very efficient and useful for merging together in a topologically consistent way the rather complicated intersections between the corrugated wall panels and the flanges of the bottom and top beams with their U-shaped cross-sections As illustrated in figure 7 these intersections would have been very difficult to discretize by a more traditional method Together with the objected oriented nature of Cast3M which allows for duplicating mirroring and translating geometrical objects the FE-model of the container could be defined in a consistent way by an input file that when executed by use of Cast3M generated a FE-model of the container that was later imported into Europlexus for the final numerical blast load simulations A highly useful outcome from this approach to the build-up of the FE-model of the container was that in order to change the mesh resolution of the FE-model only the density parameter of the meshes of the longitudinal walls had to be changed in the input file This approach proved also useful for the mesh size dependency runs carried out in the present study in that it saved the operator much tedious work and the gain in time was significant The complete input file for the FE-model of the container is given in Appendix A Figure 6 Standard 20 ft ISO container 1

L = 6058 mm W = 2438 mm H = 2591 mm

H

W

L

10

32 FE-model The FE-models of the container were detailed based upon drawings and information received from SIMLab 1 The global dimensions of the 20 ft ISO container were 2438x2591x6058 mm and the container was made up by a frame structure of cold formed channel-shaped or hollow beams and the walls roof and doors were made of corrugated steel panels of various shapes The base structure was constructed of two longitudinal beams of 48x158x30 mm 45 mm thick welded together with a front beam of 40x166x40 mm and 40 mm thick and a rear beam with the cross-sectional dimensions 40x150x50x70 and a thickness of 40 mm 16 equally spaced cross-over beams with dimensions 45x122x45 40 mm thick were welded between the longitudinal side beams of the base frame The floor of the container was made of plywood plates fixed to the base structure by self-tapping screws A somewhat similar layout shaped up the top frame however hollow squared beams of 60x60 mm 30 mm thickness were used there for the longitudinal beams and the front beam while a channel shaped beam of 132x113x132 mm with a thickness of 40 mm was used for the rear beam The top frame was supported by four corner columns welded to the top and bottom frames The front corner columns had multi-faceted cross sections of 50x50x154x170x50x50 mm with a thickness of 60 mm while the rear corner columns were made up of two profiles a 50x40x166x50 mm 60 thick profile and a 50x113x50 100 mm thick profile respectively joined together by continuous welds Corrugated steel panels of 20 mm thickness welded continuously to the frame structures were used for the side walls the front wall and the roof respectively The shape of the corrugated steel panels varied a bit as can be seen from the detailed drawing in figure 9 The rear part of the container was made up by two doors hinged to the rear corner columns and closed to the rear frame of the container with 4 vertical bar locks Each door consisted of a closed steel frame of rectangular hollow steel beams with the dimensions 100x50 mm and 32 mm thick framing a 20 mm thick corrugated steel panel fixed to the door frame by continuous welds See 1 for further details The main structural parts of the container such as the corrugated panels the corner columns and most of the beams were made of anti-corrosive steel named Corten A SPA-H B480 or equivalent with a yield stress around 345 MPa and a tensile strength of about 520 MPa Though some minor parts of the container were made of steel with a somewhat lower yield stress and tensile strength limit the above values have been used for all parts of the container for the calculations carried out in the present study

11

Figure 7 Structural details of the FE-model of the 20 ft ISO container

Rear corner column 50 ndash 36 ndash 232 ndash 46 t = 60

Longitudinal roof beam 600 ndash 600 ndash 600 ndash 600 t = 30

Transverse rear floor beam 400 ndash 200 ndash 1210 ndash 1400 ndash 600 t = 45

Rear door frames 500 ndash 1000 ndash 500 ndash 1000 t = 32

Transverse rear top beam 1020 ndash 1130 ndash1320 t = 40

Front corner column 500 ndash 450 ndash 1540 ndash 1740 ndash 360 ndash 500 t = 60

Sidewall panel 700 ndash 680 ndash 720 ndash 680 ndash 700 Indentation = 350 t = 20

Front wall panel 1040 ndash 180 ndash 1080 ndash 180 Indentation = 450 t = 20

Roof panel main part 910 ndash 135 ndash 910 ndash 135 ndash 910 Indentation = 200 t = 30

Roof panel front and rear part 955 ndash 200 ndash 4365 Indentation = 300 t = 30

Longitudinal floor beam 500 ndash 1580 ndash 30 0 t = 45

Transverse front roof beam 600 ndash 600 ndash 600 ndash 600 t = 30

Floor beam 450 ndash 1220 ndash 450 t = 40

Transverse front floor beam 400 ndash 500 ndash 1660 ndash 400 t = 40

Rear door panels - ndash 180 ndash 1100 ndash 180 - - Indentation = 450 t = 20

Rear door frames 500 ndash 1500 ndash 50 t = 30

12

The constitutive model applied for the Corten steel was a Von Mises material model with elasto-plastic behaviour implemented via a radial return algorithm Only isotropic hardening was treated and neither temperature nor strain rate dependency were introduced in the calculations The key parameters for the constitutive law used in the calculations were the yield stress 0 2 345 MPaσ = the tensile strength limit

520failure MPaσ = density 37850 kg mρ = Youngrsquos modulus 210000E MPa= and the Poissonrsquos ratio 0 3υ = The container geometry was discretized by use of 4-node shell elements (Batoz) with 4 integration points in the plane and 5 integration points over the thickness for plasticity combined with 3-node shell elements (Discrete Kirchhoff Triangle) based on the thick shell element theory (Mindlin) The various structural elements of the FE-model of the container are presented in figure 7 above identifying in a simplistic form the cross-sectional dimensions and thicknesses of the various elements

13

4 Numerical simulations 41 General The numerical simulations were grouped in 2 different sets of calculations The first initial set of calculations were based on imposed pressure-time curves on only the longitudinal front side of the container derived from the findings in the tests 1 These pressure-time curves had been calibrated against rigid wall blasts in order to aim towards prescribed design loads similar to the design pressure loads calculated by the ConWep 10 procedures 1 Based on the blast tests towards a rigid wall 1 the explosive charge and the stand-off distance were set to 4000 kg and 120 m respectively These initial set of calculations were used to study the mesh size dependency of the numerical models of the container in that two different mesh resolutions were compared against each other The second set of calculations was a more elaborate study of the behaviour of the numerical model of the container In particular a more complete pressure-time loading of the container was performed including also the pressure history recorded for the roof of the container The pressure-time histories were taken directly from the registered values in the blast tests 1 In the following these two sets of calculations are discussed in more detail 42 Initial calculations Two different mesh resolutions were used in these initial calculations one with a typical mesh size of 100 mm relative to the global dimensions of the container and one with a typical mesh size of 50 mm respectively This yielded a total number of 16332 finite elements for the coarser meshed model named cont250 in the following and a total of 43034 finite elements for the finer meshed model named cont500 Some data for the models are listed in Table 1 The imposed pressure-time curve on the longitudinal front wall of the container was calculated automatically by use of the AIRB-routine in EUROPLEXUS code The AIRB-routine was developed by M Larcher 11 and the routine is based on the same underlying equations as for the ConWep 10 formulae and gives similar pressure-time curves for identical inputs The AIRB-routine calculates an imposed pressure-time sequence on a given object in this case the longitudinal front wall towards the source of the explosion The input parameters of the routine are the explosive charge in kilograms the scaled distance between the explosive charge and the object (the steel container in this case) and the nature of the explosion that is whether it takes place on the ground above the ground etc Further details can be found in 11 The AIRB-routine allows to load the structures without having to model the fluid domain It does not take into account multiple wave reflections on structural walls but optionally allows to take into account in a very simplified way the first wave reflection at a wall It is clear that for the steel container tested in the blast tunnel in

14

2 this approach only serves as a rough first assessment of the numerical model of the container The following input has been used in the calculations of the blast wave Explosive charge in kilograms = 4000 kg Stand-off distance from object = 120 m Nature of the explosion = hemispherical charge no reflective blast considered The pressure-time curve generated by the AIRB-routine and applied as pressure-time loading of the longitudinal front wall of the steel container is shown in figure 8 below Figure 8 Pressure versus time function applied on the container The two FE-models cont250 and cont500 respectively were subjected to the above pressure-time function applied to the longitudinal front wall of the container The number of FE-elements the total cpu time and the duration of the pressure-time loading for the two models are summarised in Table 1 below

Table 1 FE-model No of elements Total cpu-time Final time step Cont250 16332 26271 s 0100 s Cont500 43034 103728 s 0100 s

The overall behaviours of the two models were rather equal the coarser meshed model cont250 acting somewhat stiffer than the model cont500 with the finer mesh as seen in figure 9 Both models exhibited similar stress levels during the loading as

15

can be seen in the von Mises stress levels versus time plots in figure 10 below however the response of the cont250 model was clearly stiffer than that of the cont500 model as illustrated in the x-displacement versus time plots in figure 11 Both plots relate to nodes at the mid-level of the longitudinal front wall of the container model Detailed plots of the deformed configurations after loading for the models cont250 and cont500 are compared in figure 12 below From these preliminary calculations it was decided to continue the simulations with the finer mesh resolution although the differences between the two models were quite small

16

Figure 9 Global behaviour of the cont250 (left column) and cont500 (right

column) container model respectively The deformed figures from top down represent the situation at 0025s 0050 s 0075 s and 0100 s for the applied pressure-time history

17

Figure 10 Stress-time curves at the mid-point of the longitudinal front wall of

container models cont250 and cont500 respectively Figure 11 Displacement-time curves at the mid-point of the longitudinal front

wall of container models cont250 and cont500 respectively

18

Figure 12 Deformed configurations after loading of the cont250 and cont500

models respectively 43 Refined calculations The initial calculations were based on a simplified pressure-time function for the explosive impact on the container It is clear that such an approach may well fall short in describing the real behaviour of the rather complex loading situation the container underwent in the blast test tunnel in Germany In particular the container in the blast tunnel test experienced a pressure loading quite different from that of an imposed pressure-time loading on the longitudinal wall only As can be seen from the recorded pressure-time development in figure 3 the container was engulfed by pressure waves as the initial pressure wave travelled through the blast tunnel Both pressure as well as suction took place on all sides of the container and internal pressure built up during the deformation and successive collapse of the container In particular the roof of the container experienced a significant increased suction loading after the initial pressure wave hit the roof indicating a possible reflection from the blast tunnel roof 1 A more realistic representation of the pressure loading of the container would hence yield better results compared to the test results from 1 It was therefore decided to

a) Deformed configuration after loading of the cont250 model

b) Deformed configuration after loading of the cont500 model

19

apply two pressure-time curves on the container one on the longitudinal wall and one on the roof no sides no back These pressure-time loadings were derived directly from the recorded results during the tests in the blast tunnel in Germany 1 and 2 The resulting pressure-time curves for the longitudinal wall and the roof were derived from the recordings in figure 3 in that the net resulting pressure of the wall was calculated as the external pressure minus the internal pressure over the duration of the test and the roof pressure as the external pressure registered on the roof minus the internal pressure registered in the container The two simplified pressure-time curves applied to the container in these somewhat refined calculations are shown in figure 13 The refined calculations were run with the same mesh resolution as in the cont500 FE-model The model was named cont1000 to distinguish it from the models used in the initial calculations The total number of finite elements the total cpu time and the duration of the pressure-time loading are in Table 2 below Note that the final duration is twice that used for the initial calculations Further the imposed pressure-time loading was applied directly to the structure and not calculated by the AIRB function used in the calculations of the cont250 and cont500 FE-models respectively

Table 2

FE-model No of elements Total cpu-time Final time step

Cont1000 43034 108842 s 0200 s The deformed shape of the container is presented in figure 14 below for 8 various time steps of the applied pressure-time histories A fairly good correlation between the global deformation pattern of the numerical model cont1000 and the real behaviour of the container during the blast tests can be observed in figure 16 where the permanent deformed configuration of the numerical container model after loading is compared with the final image of the container during the explosion test in the WTD 52 blast tunnel Both the distinct yield line failure pattern of the longitudinal front wall as well as the outward deflection of the container roof were well captured by the numerical FE-model of the container It shall be stressed that this is to be expected to some extent in that the applied pressure-time history is the same as registered during the test However it is an important result in itself in particular from an engineering design point of view that the numerical model if sufficiently discretized in terms of mesh resolution and precisely reproduced in terms of geometry and material parameters is able to grasp the global behaviour of the rather geometrically complex container as precisely as seen in figure 14 and figure 15

20

Figure 13 Applied pressure-time curves for the longitudinal wall and the roof of

the container respectively

21

Figure 14 Global behaviour of the cont1000 container model The deformed

figures from top down left column then right column represent the situation at 0025s 0050 s 0075 s 0100 s 0125 s 0150 s 0175 s and 0200 s into the applied pressure-time history

22

Figure 15 a) Global behaviour of the cont1000 container model at the final load

step permanent plastic deformation pattern b) Deformed container during the blast tests in the WTD 52 blast tunnel 1

a) Global behaviour of the cont1000 container model at the final load step Permanent plastic deformation pattern

b) Deformed container during the blast tests in the WTD 52 blast tunnel

23

The mid-wall deflection and stress history during the imposed pressure-time loading were also fairly close to those observed during the tests A maximum stress level in the mid-point of the wall arrived at approximately 360 MPa pushing the material into permanent deformations The total displacement of the mid-point of the longitudinal front wall reached around 270 mm while the remaining permanent displacement after unloading was about 220 mm For the roof structure the maximum stress level reached 360 MPa (mid-point of the roof) during the inward deflection of the roof then up to 380 MPa when the roof bent outwards snapping back due to the negative pressure pulse hitting the roof later in the pressure-time loading history Hence also the roof exhibited large permanent deformations as seen in the above figures The maximum displacement of the roof reached -500 mm during the inward deflection and a displacement of 500 mm during the outward deflection in the final phase of the loading The permanent outward deflection of the roof was about 380 mm The top wall-roof beam experienced a maximum inward displacement (x-direction) of 180 mm and a final permanent displacement of 100 mm The displacement in the vertical direction (z-direction) arrived at about ndash 50 mm and a permanent downward deformation in the order of -25 mm worth recalling test values Stress-time curves for the longitudinal front wall and the roof are shown in figure 16 below while the maximum displacement-time curves for the same points are shown in figure 17 below Figure 16 Deformed container model The dots represent the locations of the

various nodes

24

Figure 17 Von Mises stress-time curves for selected elements at the mid-point of

the longitudinal front wall the upper wall-roof beam and the roof respectively

Figure 18 Hydrostatic pressure versus time curves for selected elements at the

mid-point of the longitudinal front wall the upper wall-roof beam and the roof respectively

25

Figure 19 Displacement in the x-direction and the z-the direction respectively

for various nodes of the cont1000 model

26

The only direct displacement measurements carried out during the blast test were at the mid-point of the longitudinal front wall presented in figure 4 above 1 Direct comparison between these recordings and the numerical results for the cont1000 model is given in figure 20 below Although the numerical simulations failed to some extent in replicating the maximum displacement value reported in 1 the overall displacements for the cont1000 model showed good accordance with the experimental results

Figure 20 Displacement of the mid-point of the longitudinal wall of the container

Experimental results (red curve) compared with numerical results for the cont1000 model (green curve)

27

5 Discussions and conclusions The current numerical calculations results presented above for the 20 ft steel container subjected to blast loadings showed relatively good agreement with the reported results from the blast tunnel tests executed by the SIMLab team 1 The two numerical models cont250 and cont500 that were subjected to a simplified pressure-time loading fell somewhat short in describing the global failure mode of the container This was mainly due to the fact that only the longitudinal front wall of the container was loaded in the numerical calculations for cont250 and cont500 while the container during the blast tests was engulfed by a rather complex pressure wave loading which included a significant negative pressure on the roof of the container during the final part of the blast Hence the simplified numerical calculations did not include the collapse of the roof structure and consequently the significant weakening of the support of the longitudinal front wall due to the deflection of the roof However both the cont250 and cont500 calculations were able to replicate the failure mode and failure level of the longitudinal front wall with fair accuracy Both the distinct yield line pattern of the longitudinal front wall as well as the permanent plastic deformations seen in the blast tests were clearly identified in the numerical calculations The displacement of the longitudinal front wall was less for the cont250 model than for the cont500 model due to a stiffer overall response caused by the coarser finite element mesh resolution used for the former model The numerical calculations for the cont1000 model differed from the others by the applied pressure-time loading The same finite element mesh resolution as for the cont500 model was used but the applied pressure-time loading was derived directly from the recorded pressure values from the blast test performed by SIMLab 1 The longitudinal front wall and the roof were loaded by two separate pressure-time loadings over a duration of 0200 s The cont1000 calculations replicated the global failure mode of the container very well The yield failure pattern of the longitudinal front wall was in close agreement with the observed failure of the container in the blast test and the inward deflection of the roof followed by an outward deflection caused by the negative pressure in the final phase of the blast were precisely captured by the cont1000 model Although the final permanent deformations of the container generally were lower than those observed during the blast test of the container the overall behaviour of the cont1000 model was generally in close agreement with the observations from the blast test The maximum deformations of selected regions of the container are compared in Table3 The permanent deformations of the same regions of the container are listed in Table 4 below The results from the current study demonstrate that a sufficiently discretized finite element model with well described material parameters and realistic representation of the applied blast loadings can replicate the global behaviour of a structure to a very high degree Both local behaviour of critical regions in terms of stress-levels and deformations were well captured by the numerical models and the overall global failure modes were closely reproduced when compared to the experimental blast test results

28

Table 3 Maximum deformations of the container

Displacement of

wall (x-direction)

Displacement of top wall-roof beam

Displacement of roof

(z-direction) x-direction z-direction cont250 270 mm - - - cont500 335 mm - - - cont1000 270 mm 180 mm -50 mm 500-500 mm SIMLab container 485 mm - - -

estimated 1 Table 4 Maximum permanent deformations of the container

Displacement of

wall (x-direction)

Displacement of top wall-roof beam

Displacement of roof

(z-direction) x-direction z-direction cont250 220 mm - - - cont500 270 mm - - - cont1000 220 mm 100 mm -25 mm 380 mm SIMLab container 400 mm 60 mm -100 mm 300 mm

29

6 References 1 T Boslashrvik Burbach A Langberg H Langseth ldquoOn the ballistic and blast load

response of a 20 ft ISO container protected with aluminium panels filled with local mass ndash Phase II Validation of protective systemrdquo Engineering Structures Volume 30 Issue 6 June 2008 pp 1621-1631

2 Large Blast Simulator (LBS) The Bundeswehr Technical Center for

Protective and Special Technologies (WTD 52) in Oberjettenberg Germany 3 The EUROPLEXUS code ndash an explicit finite element code for fast dynamic

fluid-structure interaction calculations Developed in collaboration between the French Commissariat agrave lrsquoEnergie Atomique (CEA Saclay) and the Joint Research Centre of the European Commission (JRC Ispra)

4 Cast3M - General purpose finite element code Jointly developed by the Joint

Research Centre of the European Commission (JRC Ispra) and the French Commissariat agrave lrsquoEnergie Atomique

5 Safe Structures ndash A collaboration agreement between the NTNU SIMLab and

the JRC Ispra 6 M Larcher ldquoSimulation of the Effects of an Air Blast Waverdquo JRC Technical

Note JRC Ispra 2007 7 T Boslashrvik A G Hanssen S Dey H Langberg M Langseth ldquoOn the ballistic

and blast load response of a 20 ft ISO container protected with aluminum panels filled with local mass ndash Phase I Design of protective systemrdquo Engineering Structures Volume 30 Issue 6 June 2008 pp 1605-1620

8 BoslashrvikT A G Hanssen M Langseth L Olovsson ldquoResponse of structures to

planar blast loads - A finite element engineering approachrdquo Computers and Structures Volume 87 Issue 9-10 May 2009 pp 507-520

9 P Pegon ldquoAutomatic generation of blocks connected with jointsrdquo JRC-Special

Publication No I99101 JRC Ispra 1999 10 ConWep-Conventional Weapons Effects Protective Design Center United

States Army Corps of Engineers httpspdcusacearmymilsoftwareconwep

30

7 Appendix 71 Cast3M (file type dgibi) and EUROPLEXUS (file type epx)

input files for the numerical calculations cont250dgibi WARNING in meshface REGU used for meshing the 4th face of the top longitudinal bar (very distorted elements) debproc meshface m1MAILLAGE ct1ENTIER repe lab1 (nbel m1) e1=m1 elem amplab1 c1=(e1 poin 1) d (e1 poin 2) si (amplab1 ega 1) c2=c1 sinon c2=c2 et c1 finsi fin lab1 si (exis ct1) m2=surf PLAN c2 REGU ct1 sinon m2=surf PLAN c2 finsi finproc m2 opti echo 1 opti lang angl opti titr Container Blast Test - NTNU WTD52 density for the computation in europlexus dens1= 100 dens dens1 density for testing the mesh generation dens1=100 dens dens1 tol1=1d-5 opti dime 3 elem cub8 p0=0 0 0 we will use (with care) the operator CBLO to manage all the possible overlaping tab1 = table LISTE_DE_BLOCS ttb1 = table NAME_OF_MESH Longitudinal bottom beam - lbb thickness = 45 mm total length = 605800 mm plbb1 = 2000 000 000 mdum=(plbb1 et plbb1) elem 1 plbb2 = 4100 000 000 plbb3 = 4100 000 15350 plbb4 = 000 000 15350 clbb1 = plbb1 droit 1 plbb2 droit 1 plbb3 droit 1 plbb4 llbb1 = 605800 vlbb1 = 0 llbb1 0 slbb1 = clbb1 tran 1 vlbb1 slbb1=slbb1 coul roug tab2 = table LISTE_DE_FACES repe lab1 (nbel slbb1) tab2 (dime tab2) = conto (slbb1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Longitudinal bottom beam Longitudinal top beam - ltb

thickness = 30 mm total length = 605800 mm pltb1 = 000 000 251400 pltb2 = 4100 000 251400 pltb3 = 4100 000 257100 pltb4 = 000 000 257100 cltb1 = pltb1 droit 1 pltb2 droit 1 pltb3 droit 1 pltb4 droit 1 pltb1 sltb1 = cltb1 tran 1 vlbb1 sltb1=sltb1 coul roug tab2 = table LISTE_DE_FACES repe lab1 (nbel sltb1) tab2 (dime tab2) = conto (sltb1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Longitudinal top beam Transverse bottom beam - tbb thickness = 40 mm total length = 236000 mm Warning we add as a first face an additional rectangle ptbb1 = 4100 000 000 ptbb2 = 4100 4300 000 ptbb3 = 4100 4300 12000 ptbb4 = 4100 000 12000 ctbb1 = ptbb1 droit 1 ptbb2 droit 1 ptbb3 droit 1 ptbb4 vtbb1 = (2360002) 0 0 stbb1 = ctbb1 tran 1 vtbb1 stbb1=stbb1 coul vert ctbb0 = ctbb1 et (ptbb4 d 1 ptbb1) llbb2 = 5000 ntbb1 = 18 atbb1 = plbb3 coor 1 dtbb1 = ((llbb1-llbb2) - (ntbb1atbb1))(-1+ntbb1) stbb1 = depl stbb1 plus (0 (llbb2+dtbb1+atbb1) 0) stbb0 = stbb1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ctbb0 repe lab1 (nbel stbb0) tab2 (dime tab2) = conto (stbb0 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Transverse bottom beam number 1 repe lab1 (-3+ntbb1) ctbb0 stbb0=ctbb0 stbb0 plus (0 (dtbb1+atbb1) 0) stbb1=stbb1 et stbb0 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ctbb0 repe lab2 (nbel stbb0) tab2 (dime tab2) = conto (stbb0 elem amplab2) fin lab2 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Transverse bottom beam number (1+amplab1) fin lab1 Transverse bottom back beam - tbb

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thickness = 45 mm total length = 236000 mm ptbbb1 = 4100 17200 12000 ptbbb2 = 4100 11700 12000 ptbbb3 = 4100 11700 15350 ptbbb4 = 4100 000 15350 ptbbb5 = 4100 000 000 ptbbb6 = 4100 5600 000 ctbbb1 = ptbbb1 d 1 ptbbb2 d 1 ptbbb3 d 1 ptbbb4 d 1 ptbbb5 d 1 ptbbb6 depl ctbbb1 plus (0 llbb2 0) vtbbb1 = (2360002) 0 0 vtbbb3 = 42000 0 0 vtbbb4 = (340002) 0 0 vtbbb2 = vtbbb1 moin vtbbb3 moin vtbbb4 ptbbbbc1 = ptbbb4 plus vtbbb2 ptbbbbc2 = ptbbbbc1 plus vtbbb3 stbbb1 = ctbbb1 tran 1 vtbbb1 stbbb1=stbbb1 coul vert to take into consideration the door closure points stbbb1 = ctbbb1 tran 1 vtbbb2 tran 1 vtbbb3 tran 1 vtbbb4 elim (stbbb1 et ptbbbbc1 et ptbbbbc2) stbbb1=stbbb1 coul vert ctbbb0 = ctbbb1 et (ptbbb6 d 1 ptbbb1) tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ctbbb0 repe lab1 (nbel stbbb1) tab2 (dime tab2) = conto (stbbb1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Transverse bottom back beam Transverse bottom front beam - tbb thickness = 40 mm total length = 236000 mm ptbfb1 = 4100 -11200 12000 ptbfb2 = 4100 -5000 12000 ptbfb3 = 4100 -5000 15350 ptbfb4 = 4100 000 15350 ptbfb5 = 4100 000 000 ptbfb6 = 4100 -5000 000 ctbfb1 = ptbfb1 d 1 ptbfb2 d 1 ptbfb3 d 1 ptbfb4 d 1 ptbfb5 d 1 ptbfb6 vtbfb1 = (2360002) 0 0 stbfb1 = ctbfb1 tran 1 vtbfb1 stbfb1=stbfb1 coul vert ctbfb0 = ctbfb1 et (ptbfb6 d 1 ptbfb1) depl stbfb1 plus vlbb1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ctbfb0 repe lab1 (nbel stbfb1) tab2 (dime tab2) = conto (stbfb1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Transverse bottom front beam Transverse top beam back and front - ttb thickness = 30 mm total length = 236000 mm2 Warning we add as a first face an additional rectangle back with the door closure points

pttb1 = 4100 11350 255100 pttb2 = 4100 11350 243000 pttb3 = 4100 000 243000 pttb4 = 4100 000 257100 pttb5 = 4100 5000 257100 pttb6 = 4100 9550 257100 pttb7 = 4100 000 251400 pttb8 = 4100 11350 251400 cttb1 = pttb1 d 1 pttb8 d 1 pttb2 d 1 pttb3 d 1 pttb7 d 1 pttb4 d 1 pttb5 cttb2 = (pttb6 d 1 pttb1) et cttb1 cttb3 = (pttb5 d 1 pttb6) et cttb2 cttb4 = pttb8 d 1 pttb2 d 1 pttb3 d 1 pttb7 cttb5 = (pttb7 d 1 pttb8) et cttb4 depl cttb3 plus (0 llbb2 0) ptbbb4 = 4100 000 15350 ptbbbbc1 = ptbbb4 plus vtbbb2 ptbbbbc2 = ptbbbbc1 plus vtbbb3 pttbc1 pttbc2 = ptbbbbc1 ptbbbbc2 plus (pttb3 moins ptbbb4) sttbb1= cttb1 tran 1 vtbfb1 sttbb1 = cttb1 tran 1 vtbbb2 tran 1 vtbbb3 tran 1 vtbbb4 sttbb2 = cttb2 tran 1 vtbbb2 tran 1 vtbbb3 tran 1 vtbbb4 sttbb4 = cttb4 tran 1 (-4100 0 0) sttbb5 = cttb5 tran 1 (-4100 0 0) cttb5 = cttb5 plus (-4100 0 0) elim (sttbb1 et sttbb2) tol1 elim (sttbb4 et sttbb5) tol1 elim (sttbb1 et pttbc1 et pttbc2) tol1 sttbb1=(sttbb1 et sttbb4) coul blan mincl1=mdum repe lab1 ((nbel sttbb2)(nbel cttb2)) mincl1=mincl1 et (sttbb2 elem (1 + ((-1+amplab1)(nbel cttb2)))) fin lab1 mincl1=mincl1 diff mdum sttbb2=(mincl1 coul roug) et (sttbb2 diff mincl1) sttbb3=(sttbb5 elem 1) coul roug sttbb2=sttbb3 et sttbb2 et (sttbb5 diff sttbb3) tab2 = table LISTE_DE_FACES tab2 (dime tab2) = cttb5 repe lab1 (nbel sttbb2) tab2 (dime tab2) = conto (sttbb2 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Transverse top back beam front pttf1 = 4100 000 251400 pttf2 = 4100 5000 251400 pttf3 = 4100 5000 257100 pttf4 = 4100 000 257100 cttf1 = pttf1 droit 1 pttf2 droit 1 pttf3 droit 1 pttf4 droit 1 pttf1 depl cttf1 plus (vlbb1 moin (0 5000 0)) sttfb1= cttf1 tran 1 vtbfb1 sttfb1=sttfb1 coul blan sttfb1 = sttfb1 coul blan tab2 = table LISTE_DE_FACES tab2 (dime tab2) = cttf1 repe lab1 (nbel sttfb1) tab2 (dime tab2) = conto (sttfb1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Transverse top front beam Column back side of container - cbs

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thickness = 60 mm total height = 23605 mm pcbs1 = 4100 000 15350 pcbs2 = 000 000 15350 pcbs3 = 000 22600 15350 pcbs4 = 4100 22600 15350 pcbs5 = 4100 27900 15350 pcbs6 = 000 27900 15350 ccbs1 = pcbs1 droit 1 pcbs2 droit 1 pcbs3 droit 1 pcbs4 droit 1 pcbs5 vcbs1 = 0 0 23605 to take into consideration the hinge supports pchv0 = 4100 5000 15350 v1 = 0 0 (-15350+2100+300) z1=coor 3 pchv0 z2=coor 3 pttb3 z3=coor 3 v1 z3=(-180+z2-z1-(2z3))3 v2 = 0 0 (z3+600) v3 = v2 v4 = v2 v5 = vcbs1 moin v1 moin v2 moin v3 moin v4 we have to care for the contact points scbs1 = ccbs1 tran 1 vcbs1 scbs1=scbs1 coul rose scbs1 = ccbs1 tran 1 v1 tran 1 v2 tran 1 v3 tran 1 v4 tran 1 v5 scbs1=scbs1 coul rose ccbs2 = pcbs1 d 1 pcbs2 d 1 pcbs3 d 1 pcbs4 d 1 pchv0 d 1 pcbs1 ccbs3 = pcbs3 droit 1 pcbs4 droit 1 pcbs5 d 1 pcbs6 d 1 pcbs3 ccbs4 ccbs5=ccbs2 ccbs3 plus vcbs1 elim (scbs1 et ccbs4 et ccbs5) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ccbs2 tab2 (dime tab2) = ccbs3 tab2 (dime tab2) = ccbs4 tab2 (dime tab2) = ccbs5 repe lab1 (nbel scbs1) tab2 (dime tab2) = conto (scbs1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Column back side Hinge support pchv1 hing1 = pchv0 ccbs2 plus v1 pchv2 hing2 = pchv1 hing1 plus v2 pchv3 hing3 = pchv2 hing2 plus v3 pchv4 hing4 = pchv3 hing3 plus v4 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = hing1 tab2 (dime tab2) = hing2 tab2 (dime tab2) = hing3 tab2 (dime tab2) = hing4 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Hinge support hing1 = (hing1 et hing2 et hing3 et hing4) coul rouge door columns close to the hinges (32mmm) phdoo1 = 4100 5000 15350 phdoo2 = 14100 5000 15350 phdoo3 = 14100 10000 15350

phdoo4 = 4100 10000 15350 v6 = 0 0 (z2-z1) chdoo1 = phdoo1 d 1 phdoo2 d 1 phdoo3 d 1 phdoo4 d 1 phdoo1 shdoo1 = chdoo1 tran 1 v1 tran 1 v2 tran 1 v3 tran 1 v4 tran 1 v1 shdoo1 = shdoo1 coul JAUN chdoo2 = chdoo1 plus v6 elim (shdoo1 et chdoo1) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = chdoo1 tab2 (dime tab2) = chdoo2 repe lab1 (nbel shdoo1) tab2 (dime tab2) = conto (shdoo1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Door hinge column door central columns (32mmm) u1=10000 0 0 u2=vtbbb1 moin u1 cmdoo1 = chdoo1 plus u2 smdoo1 = cmdoo1 tran 1 (v62) tran 1 (v62) smdoo1 = smdoo1 coul JAUN cmdoo2 = cmdoo1 plus v6 elim (smdoo1 et cmdoo1) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = cmdoo1 tab2 (dime tab2) = cmdoo2 repe lab1 (nbel smdoo1) tab2 (dime tab2) = conto (smdoo1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Door central column door bottom beam (3mmm) pbdoo1 = 14100 10000 15350 pbdoo2 = 14100 5000 15350 pbdoo3 = 14100 5000 30350 pbdoo4 = 14100 10000 30350 cbdoo1 = pbdoo1 d 1 pbdoo2 d 1 pbdoo3 d 1 pbdoo4 cbdoo2 = cbdoo1 et (pbdoo4 d 1 pbdoo1) cbdoo3 = cbdoo2 plus u2 sbdoo1 = cbdoo1 tran 1 (vtbbb2 moin u1) tran 1 vtbbb3 tran 1 (vtbbb4 moin u1) sbdoo1 = sbdoo1 coul turq elim (sbdoo1 et cbdoo3) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = cbdoo2 tab2 (dime tab2) = cbdoo3 repe lab1 (nbel sbdoo1) tab2 (dime tab2) = conto (sbdoo1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Door bottom beam door top beam (3mmm) ctdoo2 ctdoo3 stdoo1= cbdoo2 cbdoo3 sbdoo1 plus (v6 moin (pbdoo3 moin pbdoo2)) tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ctdoo2 tab2 (dime tab2) = ctdoo3 repe lab1 (nbel stdoo1) tab2 (dime tab2) = conto (stdoo1 elem amplab1) fin lab1

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tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Door top beam door central part z5 = (pbdoo3 coor 3)-(pbdoo2 coor 3) pcdoo1 = 14100 5000 30350 pcdoo2 = 14100 10000 32150 pcdoo3 = 14100 10000 43150 pcdoo4 = 14100 5000 44950 ccdoo0 = pcdoo1 d 1 pcdoo2 d 1 pcdoo3 d 1 pcdoo4 z6 = (pcdoo4 coor 3)-(pcdoo1 coor 3) z7 = (z2-z1-(2z5)-(3z6))4 v7 = 00 00 z7 v8 = 00 00 (z7+z6) ccdoo0 = ccdoo0 plus v7 ccdoo1 = pcdoo1 d 1 ccdoo0 ccdoo0 = ccdoo0 plus v8 ccdoo1 = ccdoo1 d 1 ccdoo0 ccdoo0 = ccdoo0 plus v8 ccdoo1 = ccdoo1 d 1 ccdoo0 ccdoo1 = ccdoo1 d 1 ((ccdoo1 poin FINAL) plus v7) scdoo1 = (ccdoo1 tran 1 (vtbbb1 moin (2u1))) coul rose v9=0 100 0 ccdoo2 = (pcdoo1 moin (0 100 0)) d 1 ccdoo1 d 1 ((ccdoo1 poin FINAL) moin v9) d 1 ccdoo3 = ccdoo2 plus (vtbbb1 moin (2u1)) elim (ccdoo3 et scdoo1) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = ccdoo2 tab2 (dime tab2) = ccdoo3 repe lab1 (nbel scdoo1) tab2 (dime tab2) = conto (scdoo1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Door central part Column front side of container - cfs thickness = 60 mm total height = 23605 mm pcfs0 = 000 583700 15350 pcfs1 = 4100 583700 15350 pcfs2 = 4100 589000 15350 pcfs3 = 000 589000 15350 pcfs4 = 000 605800 15350 pcfs5 = 16800 605800 15350 pcfs6 = 16800 600800 15350 pcfs7 = 22100 600800 15350 pcfs8 = 22100 605800 15350 pcfs9 = 16800 589000 15350 ccfs1 = pcfs1 droit 1 pcfs2 droit 1 pcfs3 droit 1 pcfs4 droit 1 pcfs5 droit 1 pcfs6 droit 1 pcfs7 vcfs1 = 0 0 23605 scfs1 = ccfs1 tran 1 vcfs1 scfs1=scfs1 coul rose scfs2 = pcfs0 droit 1 pcfs1 droit 1 pcfs2 droit 1 pcfs3 droit 1 pcfs0 scfs3 = pcfs5 droit 1 pcfs6 droit 1 pcfs7 droit 1 pcfs8 droit 1 pcfs5 scfs6 = pcfs3 d 1 pcfs4 d 1 pcfs5 d 1 pcfs9 d 1 pcfs3 scfs4 scfs5 scfs7 = scfs2 scfs3 scfs6 plus vcfs1 elim (scfs1 et scfs4 et scfs5 et scfs7) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = scfs2 tab2 (dime tab2) = scfs3 tab2 (dime tab2) = scfs4

tab2 (dime tab2) = scfs5 tab2 (dime tab2) = scfs6 tab2 (dime tab2) = scfs7 repe lab1 (nbel scfs1) tab2 (dime tab2) = conto (scfs1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Column front side Sidewall section - sws thickness = 20 mm total height = 23605 mm Warning 2 faces that will be split have been added psws0 = -5000 27900 15350 psws1 = 4100 27900 15350 psws2 = 4100 31400 15350 psws3 = 000 38200 15350 psws4 = 000 45400 15350 psws5 = 4100 52200 15350 psws6 = 4100 55690 15350 psws7 = -5000 55690 15350 csws1 = psws1 droit 1 psws2 droit 1 psws3 droit 1 psws4 droit 1 psws5 droit 1 psws6 vsws1 = 0 0 23605 vsws2 = (psws6 moin psws1) nsws0 = (pcfs1 moin pcbs5) coor 2 nsws0 = nsws0(coor vsws2 2) nsws0=enti (nsws0+tol1) csws0 = csws1 repe lab1 (-1+nsws0) csws0=csws0 plus vsws2 depl psws7 plus vsws2 csws1=csws1 et csws0 fin lab1 elim csws1 tol1 csws0=csws1 el0=csws0 elem 1 csws1=el0 repe lab1 (-1 + (nbel csws0)) el1=csws0 elem (1+amplab1) p1=el0 poin 1 p2=el0 poin 2 p3=el1 poin 2 aa1=(coor 1 (bary (p1 et p2 et p3))) - (coor 1 p1) si ((abs aa1) lt tol1) csws1=csws1 diff el0 csws1=csws1 et (p1 d 1 p3) sinon csws1=csws1 et el1 finsi el0=el1 fin lab1 ssws1 = csws1 tran 1 vsws1 ssws1=ssws1 coul bleu csws2=(psws0 d 1 (csws1 poin INITIAL)) et csws1 et ((csws1 poin FINAL) d 1 psws7 d 1 psws0) csws3=csws2 plus vsws1 elim (ssws1 et csws3) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = csws2 tab2 (dime tab2) = csws3 repe lab1 (nbel ssws1) tab2 (dime tab2) = conto (ssws1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Sidewall Frontwall section - fws thickness = 20 mm total height = 23605 mm

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Warning 2 faces that will be split have been added pfws0 = 22100 610800 15350 pfws1 = 22100 600800 15350 pfws2 = 27300 600800 15350 pfws3 = 29100 605800 15350 pfws4 = 40100 605800 15350 pfws5 = 41900 600800 15350 pfws6 = 47100 600800 15350 pfws7 = 47100 610800 15350 cfws1 = pfws1 droit 1 pfws2 droit 1 pfws3 droit 1 pfws4 droit 1 pfws5 droit 1 pfws6 vfws1 = 0 0 23605 nfws0=((ptbbb1 plus vtbbb1) coor 1) - ((pcfs7 moin pcfs4) coor 1) nfws0=nfws0((pfws6 moins pfws1) coor 1) nfws0=enti (nfws0 + tol1) vfws2=pfws6 moin pfws1 cfws0 = cfws1 repe lab1 (-1+nfws0) cfws0=cfws0 plus vfws2 depl pfws7 plus vfws2 cfws1=cfws1 et cfws0 fin lab1 elim cfws1 tol1 cfws0=cfws1 el0=cfws0 elem 1 cfws1=el0 repe lab1 (-1 + (nbel cfws0)) el1=cfws0 elem (1+amplab1) p1=el0 poin 1 p2=el0 poin 2 p3=el1 poin 2 aa1=(coor 2 (bary (p1 et p2 et p3))) - (coor 2 p1) si ((abs aa1) lt tol1) cfws1=cfws1 diff el0 cfws1=cfws1 et (p1 d 1 p3) sinon cfws1=cfws1 et el1 finsi el0=el1 fin lab1 sfws1 = cfws1 tran 1 vfws1 sfws1=sfws1 coul bleu cfws2=(pfws0 d 1 (cfws1 poin INITIAL)) et cfws1 et ((cfws1 poin FINAL) d 1 pfws7 d 1 pfws0) cfws3=cfws2 plus vfws1 elim (sfws1 et cfws3) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = cfws2 tab2 (dime tab2) = cfws3 repe lab1 (nbel sfws1) tab2 (dime tab2) = conto (sfws1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain Frontwall floor thickness = mm pflo1 pflo2=ptbbb2 ptbfb2 plus p0 sflo1= (pflo1 d 1 pflo2) tran 1 vtbfb1 sflo1=sflo1 coul jaun tab2 = table LISTE_DE_FACES repe lab1 (nbel sflo1) tab2 (dime tab2) = conto (sflo1 elem amplab1) fin lab1

tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain floor side part of the roof thickness = 30 mm profb1 = 000 000 257100 profb2 = 000 9550 257100 profb3 = 000 11350 255100 profb4 = 000 41650 255100 profb5 = 000 43650 255100 vrofb1 = (vtbfb1 plus ptbbb5) moin (0 llbb2 0) depl (profb1 et profb2 et profb3) plus (0 llbb2 0) depl (profb4 et profb5) plus (0 (llbb22) 0) this enticipate on the central part vrofm3 = (coor 1 pltb2) 0 0 vrofm2 = vrofm3 plus (2700 0 0) crofb1 = profb1 d 1 profb2 d 1 profb3 d 1 profb4 d 1 profb5 crofb1 = profb1 d 1 profb2 d 1 profb3 d 1 profb4 lrofb1 = coor 2 (profb4 moin profb1) lrofb2 = coor 2 (profb5 moin profb1) crofb2 = inve (crofb1 syme DROIT profb4 (profb4 plus vcbs1)) depl crofb2 PLUS (0 (llbb1-(2lrofb2)) 0) depl crofb2 PLUS (0 (llbb1-llbb2-(2lrofb1)) 0) crofb1 = crofb1 et crofb2 depl crofb1 plus vrofm2 srofb1 = crofb1 tran 1 (vrofb1 moin vrofm2) srofb1=srofb1 coul jaun we add ad-hoc faces defined by hand vadd1=2700 0 0 padd0=000 llbb2 257100 padd1=padd0 plus (0 955 0) padd2=padd0 plus (0 ((profb4 moin profb1) coor 2) 0) padd3=padd2 plus (4100 0 0) padd4=padd0 plus (4100 955 0) crofbb2=profb1 d 1 profb2 d 1 padd4 d 1 padd3 d 1 padd2 d 1 padd1 d 1 padd0 d 1 padd5=profb3 moin (0 0 (-257100+257100)) padd6=padd5 plus (profb4 moin profb3) crofbb3=padd5 d 1 padd6 d 1 padd3 d 1 padd4 d 1 crofbb4=padd5 d 1 padd4 d 1 profb2 d 1 crofbb5 crofbb6 crofbb7 = crofbb2 crofbb3 crofbb4 syme PLAN profb4 (profb4 plus vcbs1) (profb4 plus vadd1) depl (crofbb5 et crofbb6 et crofbb7) plus (0 (llbb1-llbb2-(2lrofb1)) 0) elim (crofbb5 et crofbb6 et crofbb7 et crofb2) tol1 tab2 = table LISTE_DE_FACES tab2 (dime tab2) = crofbb2 tab2 (dime tab2) = crofbb3 tab2 (dime tab2) = crofbb4 tab2 (dime tab2) = inve crofbb5 tab2 (dime tab2) = inve crofbb6 tab2 (dime tab2) = inve crofbb7 repe lab1 (nbel srofb1) tab2 (dime tab2) = conto (srofb1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain front and back parts of the roof

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central part of the roof thickness = 20 mm profm1 = 000 41650 255100 profm2 = 000 46200 255100 profm3 = 000 47550 257100 profm4 = 000 56650 257100 profm5 = 000 58000 255100 profm6 = 000 62550 255100 crofm1 = profm1 d 1 profm2 d 1 profm3 d 1 profm4 d 1 profm5 d 1 profm6 depl crofm1 plus (0 (llbb22) 0) drofm0 = llbb1 - llbb2 - (2 lrofb1) vrofm1 = profm6 moin profm1 drofm1 = coor 2 vrofm1 nrofm0 = drofm0drofm1 nrofm0=enti (nrofm0 + tol1) crofm0 = crofm1 repe lab1 (-1+nrofm0) crofm0=crofm0 plus vrofm1 crofm1=crofm1 et crofm0 fin lab1 elim crofm1 tol1 crofm0=crofm1 el0=crofm0 elem 1 crofm1=el0 repe lab1 (-1 + (nbel crofm0)) el1=crofm0 elem (1+amplab1) p1=el0 poin 1 p2=el0 poin 2 p3=el1 poin 2 aa1=(coor 3 (bary (p1 et p2 et p3))) - (coor 3 p1) si ((abs aa1) lt tol1) crofm1=crofm1 diff el0 crofm1=crofm1 et (p1 d 1 p3) sinon crofm1=crofm1 et el1 finsi el0=el1 fin lab1 zrofm0 = (coor 3 crofm1) exco SCAL UZ zrofm1 = (coor 3 profb1) (zrofm0 masq SUPERIEUR 0) zrofm1 = zrofm1 - zrofm0 crofm3 = crofm1 plus zrofm1 vrofm3 = (coor 1 pltb2) 0 0 crofm2 = crofm3 plus vrofm3 vrofm2 = vrofm3 plus (2700 0 0) depl crofm1 plus vrofm2 srofm3 = dall crofm3 ((crofm3 poin FINAL) d 1 (crofm2 poin FINAL)) (inve crofm2) ((crofm2 poin INITIAL) d 1 (crofm3 poin INITIAL)) QUELCONQUE srofm2 = dall crofm2 ((crofm2 poin FINAL) d 1 (crofm1 poin FINAL)) (inve crofm1) ((crofm1 poin INITIAL) d 1 (crofm2 poin INITIAL)) QUELCONQUE srofm1 = crofm1 tran 1 (vrofb1 moin vrofm2) vrofm4 = profm5 moin profm4 vrofm4 = 0 0 (coor 3 vrofm4) vrofm4 = (vrofm2 moin vrofm3) plus vrofm4 srofm4 = (crofm2 elem 1) tran 1 vrofm4 tran 1 (vrofb1 moin vrofm2) srofm5 = (crofm2 elem (nbel crofm2)) tran 1 vrofm4 tran 1 (vrofb1 moin vrofm2) srofm1 = srofm3 et srofm2 et srofm1 srofm1 = srofm1 coul vert

elim (srofm1 et srofm4 et srofm5) tol1 exte1=srofm3 et srofm4 et srofm5 inte1=srofm1 diff exte1 only the external part is put on the faces tab2 = table LISTE_DE_FACES repe lab1 (nbel exte1) tab2 (dime tab2) = conto (exte1 elem amplab1) fin lab1 tab1 (dime tab1) = tab2 ttb1 (dime ttb1) = chain central part of the roof meshto1=(ssws1 et stbb1 et slbb1 et scbs1 et scfs1 et stbbb1 et stbfb1 et sfws1 et sltb1 et sttbb1 et sttfb1 et sflo1 et srofb1 et srofm1) trak meshto1 trak (shdoo1 et smdoo1 et sbdoo1 et stdoo1) trak (shdoo1 et smdoo1 et sbdoo1 et stdoo1 et scdoo1) opti donn 5 automatic treatment of the various overlapings tbb1 = cblo tab1 tol1 opti donn 5 verif lverif=faux si lverif repe lab1 (-1+(dime tbb1)) mess amplab1 tab2 = tbb1 amplab1 titre ttb1 amplab1 repe lab2 (-1+(dime tab2)) si (amplab2 ega 1) meshbi = tab2 1 sinon meshbi = meshbi et tab2 amplab2 finsi fin lab2 si (amplab1 gt 20) trak meshbi mess input a character obte aaaMOT finsi fin lab1 finsi lverif=faux si lverif amplab1=23 tab2 = tbb1 amplab1 titre ttb1 amplab1 repe lab2 (-1+(dime tab2)) si (amplab2 ega 1) meshbi = tab2 1 sinon meshbi = meshbi et tab2 amplab2 finsi fin lab2 trak meshbi finsi sect1 = clbb1 et cltb1 et ctbb1 et cttb1 et ccbs1 et ccfs1 et csws1 sect1 = sect1 et cfws1 trac sect1 si lverif trak (slbb1 et sltb1 et ssws1) finsi trak (ssws1 et stbb1 et slbb1 et scbs1 et scfs1 et stbbb1 et stbfb1 et sfws1 et sltb1 et sttbb1 et sttfb1 et sflo1) trak (sfws1 et stbfb1 et scfs1 et sttfb1) opti donn 5 we generate now the real mesh