Frontal Crash Simulation of Chrysler Neon BIW

Objective:

In this project, I will run a frontal crash simulation of a Chrysler Neon BIW model. A rigid wall is placed in front of the BIW and the intial veloctity is set to be 35 mph = 15.56 m/s.

In this simulation, the pre-processing is carried out by HyperMesh and HyperCrash. After that, the nonlinear explicit dynamic model is solved by using Radioss. Finally, the results are assessed by HyperView and HyperGraph.

Pre-Processing Main Points:

1. The unit system used in this simulation is kg/mm/ms.

2. Element properties used in this simulation are Ishell = 24 for shell elements and Isolid = 24 for solid elements. Choosing type 24 properties with 1 Gauss point for shell and solid elements help reduce the hourglass energy using the physical hourglass control.

3. Create Type 7 Self-contact Interface for the whole model (Master segments and Slave nodes) to prevent self penetration with the following parameters:

- Istf = 4

- Igap = 2

- Idel = 2

- Gapmin = 1

- Inacti = 6

- Friction = 0.2

- Iform = 2

4. Create Type 7 Interface between bumper (Master segments) and front rails (Slave nodes) to calculate the force transduced from the bumper to the rails. The parameters are the same as above.

5. Create a rigid wall right in front of the BIW (normal to x axis) with friction 0.1, Dsearch = 1000.

6. Use Mesh Editing with HyperMesh transfer to optimize the mesh and fix any penetration and interaction in HyperMesh.

7. Use added mass to distribute lumped mass to nodes on the floor panel in order to reach the target mass 700 kg at the CG position which is equivalent to the full model of 300000 nodes.

8. Initial velocity is created for the BIW with a value Vx = 15.56 m/s.

9. Use model checker to ensure the quality of the model: delete floating nodes, delete unnecessary spring joints,...

10. Timestep is set to 0.0005 ms and Tscale = 0.67 and Termination time is 80 ms.

11. Time history TH for the following:

- Rigid wall to get Rigid wall reaction force.

- Sections at left and right Rails to get rail sectional forces.

- Sections at left and right Shotguns to get shotgun sectional forces.

- Sections at left and right A-Pillars to get A-pillar sectional forces.

- Node acceleration on left and right B-Pillars at the bottom rocker panel.

Results:

1. Energy Conservation:

From the graph above, we can see that the total energy is conserved during the crash event. The kinetic energy due to the movement of the BIW is transformed into internal energy in forms of deformation. The energy error (contributed by contact energy and hourglass energy) in this simulation is -2.9% and the mass error DM/M = 0.16% which are acceptable.

It takes 2432.75 seconds and 160057 cycles to reach the normal termination time.

2. Rigid Wall reaction force:

The rigid wall reaction force is the force the wall received from the whole BIW during the crash event (which is also the force transferred to the whole BIW model according to Newton\'s third law of motion). Depending on the load path distribution and the material property of each individual part, their deformation will differ.

The first peak force of 90 kN occurs at around 2.5 ms. This is the time around which the bumper first hit the the rigid wall and decelerated from initial velocity of 15.56 m/s to 0 m/s.

Everytime there is a next spike in force in the graph, there is a component receiving the force from the impact:

- At the spike force around 15 ms is the time when the rails received the load path (see figures in 3. and 4. below). At this time, the rails crumpled and after that the force reduced because of the reduction in rail stiffness.

- At the spike force around 33 ms is the time when the shotguns received the load path (see figures in 5. below). At this time, the shotguns crumpled and after that the force reduced because of the reduction in shotgun stiffness.

- At the spike force around 80 ms is the time when the rails and the bumper are stacked together due to the impact from the components behind them (see figures in 3. below).

The mean reaction force of the rigid wall is estimated by dividing the area under the load curve by the value 80 (maximum time), which is 38 kN.

3. Force received by the front rails from the bumper:

The first peak force occurs at around 5 ms to 15 ms during which the rails received the load path from the rigid wall. At this time, the rails crumpled and after that the force reduced because of the reduction in rail stiffness.

The mean force received from the bumper is estimated by dividing the area under the load curve by the value 80 (maximum time), which is 4 kN for the left rail and 6 kN for the right rail.

We can see that the right rail received more load than the left rail. This is because of the difference in cross sections along the length of the rails. The right rail seems to have a harder-to-break geometry along its length, so its absorbing force is larger.

4. Sectional force at left and right Rails:

The first peak force occurs at around 5 ms to 15 ms during which the rails received the load path from the rigid wall. At this time, the rails crumpled and after that the force reduced because of the reduction in rail stiffness.

The mean sectional force is estimated by dividing the area under the load curve by the value 80 (maximum time), which is 5 kN for the left rail and 9.7 kN for the right rail.

We can see that the right rail received more load than the left rail. This is because of the difference in cross sections along the length of the rails. The right rail seems to have a harder-to-break geometry along its length (the right rail is tilted inwards from the rail section to the back), so its absorbing force is larger.

The sections of the rails are defined at following positions:

The inclined structures in front the rail sections have created more strength for the rails. That is why the loads received on the rail sections are higher from the loads at the interface between the bumper and the rails.

5. Sectional force at left and right Shotguns:

The first peak force occurs at around 33 ms at which the shotguns received the load path from the rigid wall. At this time, the shotguns crumpled and after that the force reduced because of the reduction in shotgun stiffness.

The mean sectional force is estimated by dividing the area under the load curve by the value 80 (maximum time), which is 3.15 kN for the left shotgun and 2 kN for the right shotgun.

There is a spike force of 12.5 kN at around 60 ms on the left shotgun section. This is because of an asymmetry in the impact distribution and the angle of impact of the left shotgun section. The back components seem to pile up more on the left shotgun in addition to the straighter orientation along the x axis which receives more load.

6. Sectional force at left and right A-Pillars:

The spikes in force occur at around 33 ms and 60 ms.

The mean sectional force is estimated by dividing the area under the load curve by the value 80 (maximum time), which is 1.3 kN for the left A-pillar and 1 kN for the right A-pillar.

We can see that the forces on these sections are smaller than those from the rails and the shotguns thanks to the load path distribution across the front components.

7. Acceleration at left and right B-Pillar rockers:

The deceleration at the B-Pillar rocker has a peak value at around 3.5 mm/ms^2.

8. As observed from the deformation, the node 61585 at the center rib of the floor panel suffers the least deformation in the x direction. Therefore, I will pick this node as a reference to measure the intrusion of dashboard and foot pedal in the x direction.

- Intrusion of dashboard at node 66695:

The intrusion of the dashboard is then the difference of x distances at the beginning and at the end of the crash event = 897.18 - 665.802 = 231.4 mm.

- Intrusion of foot pedal at node 66244:

The intrusion of the foot pedal is then the difference of x distances at the beginning and at the end of the crash event = 917.499 - 786.506 = 131 mm.

We can see that the dashboard suffers more intrusion into the driver direction (-x direction).

• Some comments on the load distribution:

We can see that in this frontal crash simulation of a BIW model, the load path is transferred from the rigid wall and distributed from the bumper into the components behind it. If we use the mean force as a reference, the frontal components that take up most of the impact force are the rails and the shotguns. They take up for about (5 + 9.7 + 3.15 + 2)/38 = 52.2% the total impact force received from the rigid wall.