Week 10 Bullet penetrating a Bucket Challenge

Aim: To perform the tension and torsion test on the specimen provided below the video corresponding to the challenge.

Objective:

For the tension test, displace one end of the specimen to 18mm while keeping the other fixed. You will need to find out the total deformation, equivalent stress and temperature of the specimen and attach appropriate images to show your results.

For the torsion test, you will need to displace one end of the specimen to an angle of 1200 degrees. Find out the total deformation, equivalent stress and temperature of the specimen and attach images to show your results.

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 Bullet and Bucket 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., Tantalum, Aluminium Alloy NL, Copper Alloy NL and Steel NL and their properties are shown below.

3) In the geometry of the model, the Bullet is made up of Tantalum and made rigid behaviour.

Case 1: The Bucket is made up of Aluminium Alloy NL.

Case 2: The Bucket is made up of Copper Alloy NL.

Case 3: The Bucket is made up of Steel NL.

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

5) 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 Refined mesh using the face sizing.

6) We open the Explicit Dynamics in that choosing Analysis Setting where stimulation is run for 1 step with End Time to 1e-2 sec for Velocity and Fixed-Joint

Fixed Support is given for the base of the bucket.

Velocity is given for Bullet at the rate of 139000 mm/sec.

7) Setting up the solution:

Case 1: Bucket of Aluminium Alloy NL

a) Equivalent Stress:

b) Equivalent Elastic Strain:

c) Total Deformation:

Case 2: Bucket of Copper Alloy NL

a) Equivalent Stress:

b) Equivalent Elastic Strain:

c) Total Deformation:

Case 3: Bucket of Steel NL

a) Equivalent Stress:

b) Equivalent Elastic Strain:

c) Total Deformation:

Result:

Material	Case 1	Case 2	Case 3
Equivalent Stress in MPa (Maximum)	2262.7	2093.9	2058.8
Equivalent Stress in MPa (Minimum)	0	0	0
Equivalent Elastic Strain (Maximum)	0.03187	0.022567	0.013251
Equivalent Elastic Strain (Minimum)	0	0	0
Total Deformation in mm (Maximum)	123.63	757.61	359.42
Total Deformation in mm (Minimum)	0	0	0

Conclusion:

• From the above result, we can conclude that the equivalent stress is maximum in Case 2 Copper Alloy NL compared to other Cases.
• Equivalent Elastic Strain is more in Case 2 than in the other two cases.
• Total deformation is maximum in Case 2 i.e., Copper Alloy NL due to flexible material.

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

Thank You...