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OBJECTIVE: To perform a transient analysis on a piston and cam mechanism using ANSYS Workbench. The following objectives have to be satisfied. Case 1: To run the analysis with frictionless contact Case 2: To run the analysis with frictional contact with 0.1 Case 3: To run the analysis with frictional contact with 0.2 PROCEDURE…
Ashwen Venkatesh
updated on 29 Dec 2020
OBJECTIVE:
To perform a transient analysis on a piston and cam mechanism using ANSYS Workbench. The following objectives have to be satisfied.
Case 1: To run the analysis with frictionless contact
Case 2: To run the analysis with frictional contact with 0.1
Case 3: To run the analysis with frictional contact with 0.2
PROCEDURE FOR CASE SETUP:
1. Open ANSYS >> Drag and drop static structural in the project schematic window.
2. Go to engineering data for defining the materials given in the problem. For this simulation, the default material structural steel is used.
3. Select the model tab to establish the meshing, contact definitions, and analysis settings definition. Rename the parts according to convenience.
4. Go to contact and rename the contact based on definitions. Now, create a frictionless contact between the cam follower and barrel cam and bar and cam follower.
The contact definition is shown in the image below.
For the next cases, change the type to frictional and input the given frictional coefficient values.
5. The joint definitions are shown in the figure below.
6. For meshing, face sizing is given for the faces which are in contact with a mesh size of 3mm.
The final meshed model is shown in the figure below.
The mesh metrics are shown in the figure below.
7. Go to analysis settings. The number of steps defined for this analysis is 9. For the first time step, the definition is shown in the figure below.
For the rest of the time steps the definition is shown in the figure below.
7. The boundary conditions for all the cases are shown below.
9. The output requests for equivalent stress, equivalent elastic strain, and directional deformation are placed.
10. From the analysis settings, hit on solve to start the simulation.
RESULTS AND DISCUSSION:
1. The equivalent stress obtained is shown in the figure below.
2. The directional deformation is shown in the figure below.
3. The equivalent elastic strain obtained is shown in the figure below.
ANIMATION FILES:
1. The equivalent stress obtained is shown in the figure below.
2. The directional deformation is shown in the figure below.
3. The equivalent elastic strain obtained is shown in the figure below.
CONCLUSION:
From the simulation, it can be seen that for all the cases the solution converged without any errors.
The output parameters are tabulated below.
Case Setup | Directional Deformation (in mm) | Equivalent Stress (in MPa) | Equivalent Elastic Strain |
Frictionless Contact | 20.35 | 2008.3 | 0.010109 |
Frictional Coefficient - 0.1 | 20.454 | 2199.4 | 0.011059 |
Frictional Coefficient - 0.2 | 20.499 | 2395.8 | 0.012039 |
From the above table, it can be seen that magnitude of directional deformation is almost the same for all the cases with a value of 20.4 mm. There is a slight increase in the value of directional deformation with increasing frictional coefficient values.
The equivalent stress is highest for frictional contact with a frictional coefficient of 0.2. The value obtained is 2395.8 MPa. The value obtained for frictionless contact is the least with a value of 2008.3 MPa.
The equivalent elastic strain is highest for frictional contact with a frictional coefficient of 0.2. The value obtained is 0.012. For the frictional coefficient, the value obtained is 0.011. For frictionless contact, the value obtained is 0.0101.
Therefore, it can be concluded that the directional deformation slightly increases in the three cases. The equivalent stress is higher for frictional contact with a frictional coefficient of 0.2. This is because the higher the frictional coefficient, the more resistance to the movement of cam over the follower, and due to this higher stress is developed at the follwer. The equivalent elastic strain is higher for frictional contact with a frictional coefficient of 0.2. This is due to more resistance because of the higher frictional coefficient. Hence, all the objectives are satisfied.
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