Assignment 7-Side Pole Crash Simulation Challenge
Assignment 7-Side Pole Crash Simulation Challenge
1. Aim
- Perform a side pole crash simulation with the mentioned model checks and procedures for the given model and find the answers for specified output requests.
2. Objective
Model Checks And Procedures for the Neon Side Crash -BIW
- Check the unit system and either follow [Mg mm s] or [Kg mm ms].
- Create an appropriate interface, friction 0.2, and recommended parameters.
- Make sure of no penetration and intersections.
- Create a rigid pole with a friction of 0.1 as per the FMVSS 214 standards.
- Compare the model weight with the reference model and use added masses to reach the target weight of 700kg while getting CG about the required range.
- Initial velocity - 20mph, as per the standards.
- Use a model checker to ensure good quality.
- Timestep: 0.5 to 0.1 microseconds.
- Run 80ms.
The Output Requests
- Sectional force in the cross member.
- Intrusion at B pillar, hinge pillar and fuel tank region. Provide recommendations on what can help to reduce Fuel tank intrusion.
- Peak velocity of an inner node of the door.
Fig 1: Isometric View of Neon Side Crash -BIW
3. Procedure
Step 1: Importing of Model and Checking of Unit System
Initially import the model 'neon_front_0000.rad' to hypermesh using the solver deck. After importing we have to check the unit system specified on the model. It can be done by using a solver browser. From the solver browser, by clicking the 'Begin Card' we can see the unit system for the model because we have to use the same unit system for the entire crash test (for hyper-crash also). Otherwise, it will affect the results.
Fig 2: Checking of Unit System
So the unit system for the model is,
- Mass = Kg
- Length = mm
- Time = ms
Step 2: Penetration and Intersection Checking
Before the simulation, we have to make sure that there are no intersections (elements passing completely through one another) or penetrations (Overlapping of the material thickness of shell elements). The checking can be done by following,
- Tools (Menu Bar) >> Penetration Check >> Entity Type= Componets >> Selection= Select All Components >> Check.
Fig 3: Penetration Checking Using Hypermesh
So from the penetration check, we can see any kind of penetrations or intersections available on the model. So we can move on to the next step.
Step 03: Checking For Free Parts.
We have to make sure that the model does not have any free parts. There needs to be a proper connection between the components. To check the connectivity,
- Tools >> Find Connectivity >> Select All Components >> Check.
Fig 4: Connectivity Checking
Fr3 components haveonents that have no connectivity. We need to connect them with proper rigid connections. We can use 1D rigid connections for the connectivity. After all the connections are generated, check the connectivity once again and make sure that there are no free parts.
Fig 5: Rigid Connections
Step 04: Creating an Appropriate Interface With Recommended Values
A general node-to-surface contact or type 7 interface can be used for any application. By analysing the model, pre-existing interfaces are available on the model. To ensure the proper interface connections, we have to delete all of those interfaces and create a new interface Type 7. The recommended parameters for type 7 include
| Sl No. | Parameter | Values |
| 1 | Istiff | 4 |
| 2 | Igap | 2 |
| 3 | Idel | 2 |
| 4 | Iform | 2 |
| 5 | Inacti | 6 |
| 6 | Stmin | 1 |
| 7 | GAPmin | 0.5 |
| 8 | Fric | 0.2 |
Table 1: Type 7 Parameters and Their Values
To assign a type 7 contact, go to the solver browser and right-click on,
- Create >> INTER >> TYPE7 >> Select Master and Slave >> Assign Recommended Values
Fig 6: Assigning Interface Type 7
Step 4: Achieving Mass And CoG Of The Full-Scale Model/Mass Balancing
Because it is one-third of the full-scale model (~30K Nodes), we have to achieve the same mass and CoG condition as this model. The CoG coordinates of the current model before the mass addition,
Cogx = 2426.4347
Cogy = 251.798
Cogz = 589.5146
The coordinates of the CoG point of a full-scale model is
Cogx = 3124.1047
Cogy = 5.7746
Cogz = 531.1952
Fig 7: Mass And Cog Of The Model (Hypermesh)
The total mass = 700 Kg, and the mass of the current model is 166 Kg, We have to add an extra 535 Kg in the proper place to make the mass and location of CoG consistent. For that, we use hypercrash. Import the RADIOSS file to the hyper crash that was saved from hypermesh. The point of the cog should lie inside the model.
We have to isolate a part to add the additional mass. We use TYPE 1 to add mass.
Fig 8: Mass And Cog Of The Model (Hypercrash)
Then add mass to the nodes of these components by,
- LoadCase >> Added Masses >> Create New Object >> Select Type= 1(Mass is distributed to each node of the node group) Give Title >> Enter Required Mass Value >> Grnod_Id(Support)= Right Click >> Select In Graphics >> Add/Remove By Picking Selection >> Select Nodes >> Yes >> Save.
Fig 9: Adding Mass To The Model
Check the updated Cog coordinates by,
- Mass >> Balancing
The Cog coordinates are updated and the values of updated coordinates are,
Cogx = 3071.3514
Cogy = 5.639
Cogz = 393.11
Step 5: Appling Initial Velocity To The Model
As per the objective, we need to define the velocity of 35 mph for the model. We have to convert the unit to mm/ms. That gives
35 mph = 15.65 mm/ms.
Using the hypercrash again, we can assign the velocity to the whole model. If we consider the global coordinates as a reference we need only the translational velocity in the y direction because the simulation is a side pole crash.
- LoadCase >> Initial Velocity >> Create New Object >> Give Title >> Select Type= 0: Translational>> Vx=0, Vy=15.65, Vz=0 >> Gnod_Id(Support)= Right Click >> Select In Graphics >> Include Picked Parts >> Select Whole Parts >> Yes >> Save
Fig 10: Assigning Initial Velocity
Step 6: Creating a Rigid Side Pole
For creating the rigid side pole, we can use hypermesh.
- Right Click On Solver Browser >> Create >> RWALL >> CYL
After that, we have to assign proper values to the side pole (as per the FMVSS 214 standards.). For the coordinates of the base node, we can define using a temp node. The normal should be along the '-z axis'. Other assigned values are
- Slide = 2: Sliding With Friction
- Dsearch = 1500
- FRIC =0.1
- Diameter = 254
Assign the TH_Rwall file for the rigid wall. Make sure that, you select the required variables so that it improves the efficiency of the simulation
Fig 11: Creating Rigid Side Pole
Step 7: Sections
Per the output requests, we must create sections across both cross-members. Before this, we have to make frames at these mentioned locations. These are used in sections as a Frame ID. For that, go to the solver browser and right-click on,
- Create >> FRAME >> MOV >> Select Origin Node >> Select Node To Define Axis >> Select Node To Define Plane.
Fig 12: MOV Frames
The construct sections with these frames. From the solver browser,
- Create >> SECT >> SECT >> Select N1, N2, N3 for the Normal Direction>> Frame_ID =Select System corresponding MOV frames ID >> grShel_ID >> Right Click >> Select required nodes from part >> delta T=0.1.
Fig 13: Sections
Create a TH file for Sections and assign values. Make sure that, you select the required variables so that it improves the efficiency of the simulation
Step 7: Intrusions on the B pillar, hinge pillar and fuel tank region
Here we have to find the displacement of these mentioned regions B pillar, hinge pillar and fuel tank region. For that, we can define springs as having very small stiffness. And we can find the displacement of those springs with the TH file. To create spring, go to the solver browser and right-click on,
- Create >> Elements >> Spring >> Spring2N >> Define one node by node ID and the other node in the cross member.
Fig 14: Spring Element Settings
Create a spring property card having the values of K=0.001 and M=0.001. Assign these properties to the newly created springs. Create a TH file for a Springs and assign values. Make sure that, you select the required variables so that it improves the efficiency of the simulation
Step 8: Peak Velocity Of An Inner Node Of Door (337773)
Initially, we have to define a skew as a reference. To create a skew
- Create >> SKEW >> MOV >> Select Origin Node >> Select Node To Define Axis >> Select Node To Define Plane.
Fig 15: MOV Skews
After that, Create a TH file for an output block (Node) and assign values. Select Skew System as the Skew that is created now. Make sure that, you select the required variables so that it improves the efficiency of the simulation
Step 09: Model Checking
Before the simulation, we have to make sure that, the model is error-free and ready to simulate. So initiate a model check by,
- Tools >> Model Checker >> RadiossBlock
Fig 16: Model Checking
After the checking, there are some unused sets and other minute errors. The model checker automatically resolves those. Apart from a warning, the model has no other issues and is ready for the simulation. Check the minimum time step and total run time.
Step 10: Simulate the Model and Analysis of the Results
The simulation is done by following,
- Analysis >> Radioss >> Browse Input File Location >> Option = -nt 4 (No. Of Threads) >> Radioss
Fig 17: Loading Simulation Using Hyper Mesh
4. Observations And Conclusions
After the simulation, we have to check the job summary to analyse the results. It decides whether the results are acceptable or not.
From the output file,
- Energy Error = -0.9%
- Mass Error = 0.0004422%
- Mass Addition = 0.03097
- Engine Run Time = 3597.04 s
- Total No. Of Cycles = 160001
The errors are within the acceptable range. So we can continue with the model.
Fig 18: Animation Of Simulation
1. Von Mises Stresses
To get Von Mises Stress distribution, load the '.h3d file' in hypergraph and select the contour type as a Von Mises Stress.
Fig 19: Von Mises Stress Distribution
By analysing the animation,
- Von Mises Stress (Max) = 3.144E-01 KN/mm2 at node 23242.
= 314.4 N/mm2
= 314.4 Mpa
From the simulation, initially, the side doors and pillars made contact with the side pole. So the number gets maximum deformation from the collision and then the forces are reduced throughout the simulation because of the energy absorption of each component. And we can see that most stresses are concentrated where the pole made contact with the car.
2. Energy Graphs Of The Simulation
By loading the time history file (T01) to Hypergraph 2D, we can plot the necessary graphs and analyse the results.
Plot 1: Time Vs Internal Energy / Kinetic Energy
| Sl No | Time (ms) | Internal Energy (Kgmm2/ms2) (Joules) | Kinetic Energy (Kgmm2/ms2) (Joules) |
| 1 | 0 | 0.00 (Min) | 85780.039 (Max) |
| 2 | 6 | 7.1762 | 85768.30 |
| 3 | 6.5 | 84.399 | 85679.835 |
| 4 | 30 | 6604.715 | 78690.226 |
| 5 | 79.5 | 13012.8525 (Max) | 71593.32 (Min) |
Table 1: Time Vs Internal Energy / Kinetic Energy
There are no contact take places for the first 6ms. So the internal energy and kinetic energy remains at their minimum and maximum position. After the 6ms, the car crashes the pole at 35Mph of speed resulting in an increase of internal energy as well as a decrease of Kinetic energy.
Plot 2: Time vs. Total Energy
Plot 3: Time vs. Hourglass Energy
| Sl No | Time (ms) | Total Energy (Kgmm2/ms2) (Joules) | HourGlass Energy (Kgmm2/ms2) (Joules) |
| 1 | 0 | 85780.039 (Max) | 0.00 (Min) |
| 2 | 6 | 85775.48 | 6.3520e-5 |
| 3 | 6.5 | 85764.23 | 0.0611 |
| 4 | 10 | 85461.75 | 9.09385 |
| 5 | 25 | 85378.48 | 32.281 |
| 6 | 79.5 | 84710.507 (Min) | 65.955 (Max) |
Table 2: Time vs. Total Energy / Hourglass Energy
Take account of the hourglass energy and total energy plots. The total energy is slightly reduced as the simulation runs, it is because of the presence of minute hourglass energy. Because of the Improved under-integrated element formulation method (QEPH), the majority of hourglass energy is controlled by physical stabilisation. So the hourglass energy is negligible compared to total energy.
3. Peak Velocity Of Node 337773
Plot 4: Time vs. Peak Velocity
| l No | Time (ms) | Peak Velocity (mm/ms) |
| 1 | 0 | 15.694 |
| 2 | 8 | 14.706 |
| 3 | 8.5 | 16.038 (Max) |
| 4 | 10 | 15.818 |
| 5 | 15 | 3.727 |
| 6 | 41 | 4.639 |
| 7 | 45.5 | 1.136 (Min) |
| 8 | 54 | 3.361 |
| 9 | 79.5 | 1.871 |
Table 3: Time vs. Peak Velocity
Before the collision with the pole, the node followed the constant velocity of 15.394 mm/ms. When the pole is made contact with the door panel, the velocity of a node is rapidly decreased to a value. Then with small ups and downs the velocity of the specified node. So the
Peak Velocity Of Node 337773 = 16.038 at t=8.5ms
4. Sectional force in the cross-members
Plot 5: Time vs. Total Resultant Forces On Cross Members
| Sl No | Time (ms) | Cross Member 1 (N) | Cross Section 2 (N) |
| 1 | 0 | 0.00 (Min) (~5.398e-19) | 0.00 (Min) (~2.396e-19) |
| 2 | 7 | 0.522 | 1.112 |
| 3 | 10 | 82.977 | 172.244 |
| 4 | 16 | 1236.16 | 1535.524 |
| 38.5 | 1527.387 | 1768.076 | |
| 40 | 2156 | 4418.488 | |
| 40.5 | 1654.094 | 4685.11 (Max) | |
| 52 | 1452.928 | 1059.536 | |
| 59 | 2192.46 | 377.203 | |
| 72 | 2629.65 (Max) | 577.499 | |
| 79.5 | 1661.682 | 804.009 |
Table 4: Time vs. Total Resultant Force On Cross Members
By analysing the forces on each cross-section, cross member 2 exerts more resultant forces than cross member 1. The reason behind this is that cross member 2 lies directly behind the B-pillar where the pole directly makes contact. Cross member 1 lies just left to the B pillar. After the collision, the forces of cross member 1. rapidly rise and follow a zig-zag graph throughout the simulation and finally, it starts to decrease. But in the case of cross member 2, After the small leap occurs at the beginning of the collision after some time it immediately rises to its maximum value and decreases with a rapid fall. Then gradually reduced to its minimum value.
The max total resultant force experienced at Cross Member_1= 2629.65 N
= 2.629 KN at t=72ms
The max total resultant force experienced at Cross Member_2 = 4685.11 N
= 4.685 KN at t=40.5ms
5. Intrusions
Plot 6: Time vs. Total Elongation On Springs
| Sl No | Time (ms) | Spring 14526 (A-Pillar) Total Elongation (mm) | Spring 14456 (B-Pillar) Total Elongation (mm) | Spring 14545 (Fuel Tank) Total Elongation (mm) |
| 1 | 0 | 0 | 0 | 0 |
| 2 | 10 | 0.22 | 0.807 | 0.248 |
| 3 | 11 | 0.742 | 7.244 | 0.239 |
| 4 | 40 | 27.656 | 349.482 | 218.220 |
| 5 | 50 | 58.713 | 427.003 | 295.527 |
| 6 | 79.5 | 185.1 | 668.942 | 438.23 |
Table 4: Time vs. Total Elongation On Springs
- The max elongation of Spring 14526 (A-Pillar) = 185.1 mm at t=79.5 ms
- The max elongation of Spring 14456 (B-Pillar) = 668.942 mm at t=79.5 ms
- The max elongation of Spring 14526 (Fuel Tank) = 438.23 mm at t=79.5 ms
The elongation of the spring situated on the B-Pillar has higher elongation leading to higher intrusion. Because it is located near to crash point compared to other springs and leads to higher elongation and higher intrusion.
The higher intrusion on fuel tanks may lead to explosions and hazards. So we have to reduce those intrusions. We can add a cross member where the fuel tank is located. The cross member reduces the intrusions. Also, we can use certain composite materials to increase the strength and rigidity of rails and it reduces intrusions.
5. Results
- For the given model, simulate the crash scenario with the mentioned parameters and objectives and with the help of necessary plots, point out the outcomes and their values in the mentioned output requests.
Simulation File::
https://drive.google.com/drive/folders/1oeeMPOmrCAtG8_6ptJaGl2RWKowoFnkH?usp=drive_link