Week - 4 - crash box simulation

OBJECTIVE:

• To create a simulation of crash box impact against a rigid wall at 50kmph using a LS-Dyna using elastic and plastic material and shell thickness of 1.2 mm and 1.5 mm. Based on simulation observe and compare the behaviour of the result
• Different case that are studied are
  1) elastic with 1.2mm 2) elastic with 1.5 mm 3) plastic with 1.2mm 4) plastic with 1.5 mm

Note: Here we will follow the unit system of g-mm-ms

PROCEDURE:

Import the geometry

• Import the given keyword file and prepare the simulation file using following step.

Rigidwall

• We require rigidwall to have an impact simulation based on our objective. Rigidwall is a an arbitrary plane like structure which acts as a rigid member without any elements or section
• To create rigidwall follow the procedure
  Create Entity -> Rigidwall -> create -> Planar -> N1-N2 -> select two nodes (in the direct normal to which you need rigidwall) -> translate to some distance-> done
• This rigid wall acts as a master and thus we may skip the contact steps.

Section

• Keyword Manager -> Section -> Shell -> given section id = 1 -> thickness = 1.2 mm -> Accept -> Done
• Similarly create shell with 1.5 mm thickness and sec id = 2

Material

• Here we are using the steel as crash box material
• Keyword Manager -> Mat -> 001 elastic -> Mid = 1 -> give properties -> Accept -> Done
• Similarly create the material using 024 piecewise linear plasticity with Mid = 2

Part

• Assign the material and section to the crash box with Pid = 1
• For 1^st simulation we will use elastic material with 1.2 mm thickness
• Repeat the procedure with different section and material for different simulation

Boundary Condition

• Create entity -> Initial -> Velocity -> select part -> 14 mm/ms (approx. 50 kmph) -> Accept -> Done

Control

• In order to determine how long the simulation will run, we need to give termination time.
• Keyword Manager -> Control -> Termination -> time = 5

Midsection

• Since we require the force along the length at mid section of the crash box, we need to create a section set to define at which elements or node we require the forces to be computed.
• For this we need to create a node set using all the nodes at midsection and shell set using pair of elements surrounding those nodes.
• This can be done using Create Entity -> set option

Database

• To get the output of the simulation we need to define different Binary and ASCI outputs with appropriate time steps.
• Binary -> d3plot
• ASCI -> select GLSTAT, MATSUM, SLEOUT, RCFORC, RWFORC, SECFORC.
• Each keyword yields different output which can be referred in manual
• Here timestep i.e. dt = 0.01 for all outputs results.
• Also since we require section force, we need to define cross section set for output.

RESULTS

We will discuss the results for 1.2 mm vs 1.5mm with following parameters for Elastic and Plastic Material

ELASTIC MATERIAL

Stress

	Min	Max
1.2 mm	0	717.106
1.5 mm	0	704.803

From the table it can be seen that, for same input condition, the stress induced in crash box with 1.2 mm is more than that of box with 1.5 mm which means for elastic material more thickness yield better stress capabilities.

Energy

From the fig, it can be seen that energy is conserved as total energy is constant and thus our simulation is correct.

Cross Section Force

Normal Force at Rigid Wall

Cross sectional Force at midsection along length

Acceleration at Mid section

Plastic Material

Stress

Von Mises Stress

X Stress

Y Stress

Z Stress

	V-M stress	X stress	Y stress	Z stress
1.2 mm	432.6	308.7	455.4	307
1.5 mm	427.7	339	352.3	252.3

From the table and the fig above, it can be seen that stress in the specimen with 1.2 mm thickness are comparatively more than that of 1.5 mm thickness which shows the better resistance of 1.5 mm thickness specimen against same sets of input.

Strain

Lower Ipt Plastic Strain

Upper Ipt Plastic Strain

	Lower Plastic Strain	Upper Plastic Strain
1.2 mm	0.008239	0.007849
1.5 mm	0.007841	0.00805

From the table it can be seen that strain in 1.2 mm thickness specimen is more than that of 1.5 mm thickness specimen. This shows that the amount of change in dimension in 1.2 mm thickness is more than that in 1.5 mm thickness specimen and this is caused due to larger resistance offered by more thickness of the shell

Energy

Total, Internal, Kinetic and hourglass energies are extracted.

From the above figures we can see that total energy in all scenarios remain nearly constant which depicts the success of our simulation. Also whatever drop we see in Kinetic energy is the compensated with the rise in Internal energy which is the second parameter which validates our simulation. Hourglass energy is the unwanted distortion of element caused due to FEA solver. Since it is negligible, it can be ignored

Cross Sectional Force

Normal Force at Rigid Wall

Cross sectional force at mid section along length

Here we have calculated two sets of force. One is normal force on the rigid wall and other is the cross sectional force at mid-section along the length

From the figure, it can be inferred that force produced by 1.5 mm thickness section is more than that of 1.2 mm thickness which is acceptable due to higher cross sectional area. Similar is the case for mid cross section along the length as well. The forces comes into play in both the cases at same time showing the impact at that instance.

Acceleration

The above figure shows the acceleration at mid node 1328. It shows acceleration in 1.5 mm specimen is little more than that of 1.2 mm thickness.

CONCLUSION

The study helped us understand different steps involved in impact analysis. Also it helps us understand the procedure to create the arbitrary rigidwall and its related data along with the requirement of different output database cards defined.

LINK

https://drive.google.com/drive/folders/1AJsFgxvIz2ue1qv17ji1KnFClqPnGCpR?usp=sharing