Week 9 Tension and Torsion test 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 Torsion and Tension Test Model.
2) In the engineering data(explicit materials) we input the material and its properties. In this case, we are using Steel 1006 material i.e., and its properties are shown below.
3) In the geometry of the model, the model is made up of Steel 1006.
4) In Coordinate System, defining Global Coordinate System and Coordinate System.
Case 1: Tension Test
Case 2: Torsion Test
Cylindrical Type
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, Method of meshing and sizing.
Case 1: Tension Test
Case 2: Torsion Test
6) We open the analysis setting where stimulation is run for 1 step for Displacement and Fixed-Joint
Case 1: Tension test
Displacement for 18 mm
Case 2: Torsion test
Displacement for 1200 mm
Fixed-Joint is given as shown below
7) Setting up the solution:
Case 1: Tension Test
a) Equivalent Stress:
b) Temperature:
c) Total Deformation:
Case 2: Torsion Test
a) Equivalent Stress:
b) Temperature:
c) Total Deformation:
Result:
| Material | Case 1 | Case 2 |
| Equivalent Stress in MPa (Maximum) | 626.92 | 641.44 |
| Equivalent Stress in MPa (Minimum) | 0 | 0 |
| Temperature in °C (Maximum) | 281.92 | 425.32 |
| Temperature in °C (Minimum) | 22 | 21.999 |
| Total Deformation in mm (Maximum) | 18.007 | 13.814 |
| Total Deformation in mm (Minimum) | 0 | 0 |
Conclusion:
- From the above result, we can conclude that the equivalent stress is maximum in Case 2 has deformation happens in a circular motion where stress generation will more.
- Temperature is maximum in Case 2 due to rotation the particle remains there place without much deformation leading to high temperature.
- Total deformation is maximum in Case 1 as in Case 2, in the torsion test, it is the circular motion we can't observe much deformation but in the tension test, it is an expansion of length where we observe more deformation.
Note: google drive link is provided below in a text file. Please check the link for the model.
Thank You...