Week 11 Car Crash simulation

Aim: To perform a parametric study using 3 different values of thickness for the car body.

Objective:

Analyse the total deformation and Equivalent stresses acting on the model.

Procedure:

1) To verify welds in the setup, we use the Explicit Dynamics Tool in the analysis system toolbox to define the geometry and other input parameters for the same.

First, we open SpaceClaim to identify and rectify the errors in the geometry. In the below geometry we can see the Car Crash Model.

2) In the engineering data(explicit materials) we input the material and its properties. In this case, we are using four different types of material i.e., Structural Steel and Steel NL and their properties are shown below.

3) In the geometry of the model, the Wall is made up of Structural Steel.

Case 1: The Car Body is made up of Steel NL with a thickness of 3 mm.

Case 2: The Car Body is made up of Steel NL with a thickness of 5 mm.

Case 3: The Car Body is made up of Steel NL with a thickness of 7 mm.

4) In Symmetry, creating the symmetry for the face of the wall and edges of the car body.

5) In Connection, defining the Body Interaction and making Type Frictionless

6) By opening the modelling tab, we mesh the geometry with element size as default and finer mesh size where contacts are present. Generating the mesh thereafter by right-clicking on the Mesh icon. Using the method for creating the mesh. Using the Refined mesh using the face sizing.

7) We open the Explicit Dynamics in that choosing Analysis Setting where stimulation is run for 1 step with End Time to 1e-3 sec for Displacement and Fixed-Joint

Fixed Support is given for the face of the car body.

Displacement is given for the Wall 500 mm

8) Setting up the solution:

Case 1: Thickness of body 3 mm

a) Equivalent Stress:

b) Total Deformation:

Case 2: Thickness of body 5 mm

a) Equivalent Stress:

b) Total Deformation:

Case 3: Thickness of body 7 mm

a) Equivalent Stress:

b) Total Deformation:

Parameters Values for all three cases:

Result:

Material	Case 1 (3 mm)	Case 2 (5 mm)	Case 3 (7 mm)
Equivalent Stress in MPa (Maximum)	7217.6	7067.7	7061.9
Equivalent Stress in MPa (Minimum)	0	0	0
Total Deformation in mm (Maximum)	1020.2	1016.5	994.37
Total Deformation in mm (Minimum)	0	0	0

Conclusion:

• From the above result, we can conclude that the equivalent stress is maximum in Case 1 to other Cases. This means 3 mm thickness car body is more likely to get facture compared to other too.
• Total deformation is maximum in Case 1. This means lesser the thickness more deformation to take place.

Note: google drive link is provided below in a text file. Please check the link for the model.

Thank You...