Week 3 Sheet metal Bending challenge
Objective:
To perform a sheet metal bending analysis using three different materials ( Al Alloy 1199, Cu Alloy NL, Mg Alloy NL) and compare their results for Equivalent Stress, Elastic strain and total deformation in Y direction and also to perform a grid dependency test using Al Alloy NL as a material for the sheet metal.
Case Setup and procedure:
- Firstly we have assigned different material for this analysis from the engineering data source of the static structural analysis.
- Secondly we have imported the geometry of the Sheet metal for the bending analysis.
- In the Geometry section of the model tab we have assigned Al alloy 1199 to the sheet and we have assigned structural steel to the punch and die. For the other cases we have assigned Cu, Mg and Al alloy to the sheet.
- For the contact we have manually assigned the contact to the geometry as shown below
- For the meshing we have assigned a body sizing to the die and the punch with a element size of 4mm and for the sheet plate we have assigned element size of 1 mm. The faces for the contact and the target has also been assigned an element size of 1mm.
- The analysis setting for the simulation has been shown below:
- In the loading region we have assigned three different displacement to the three different parts punch, sheet and die. The Displacement assigned are shown below.
Results:
Case 1
Material (Al 1199 Alloy)
Total deformation (Y direction):
Equivalent stress:
Elastic strain:
Material (Cu Alloy NL)
Total deformation (Y direction):
Equivalent stress:
Elastic strain:
Material ( Mg Alloy NL)
Total deformation (Y direction)
Equivalent Stress:
Elastic strain:
Case 2 ( Material Al Alloy NL)
The contact and the target between the two bodies has been assigned as frictional with coefficient of frictional as 0.19
Total deformation (Y direction)
Equivalent Stress:
Elastic strain:
Case 3 (Material Al Alloy NL, MESH Refinement)
The mesh refinement is done for the sheet plate with body sizing with element size of 0.8mm.
Total deformation (Y direction):
Equivalent Stress:
Elastic strain:
Discussion & Conclusion:
Case 1
| Sheet materials | Total Deformations (Y direction) mm | Equivalent Stress MPa | Elastic strain mm/mm |
| Al Alloy 1199 | 2.4495 mm | 2.426e5 MPa | 1.2146 mm/mm |
| Cu Alloy NL | 2.7882 mm | 2.019e5 MPa | 1.0106 mm/mm |
| Mg Alloy NL | 1.9292 mm | 1.68e5 MPa | 0.844 mm/mm |
- From the above table it can be justified that total deformation in the direction of deformation is maximum in case of Cu Alloy and minimum in case of Mg Alloy.
- In the section of the Equivalent Stress it can be observed that the maximum is in the case of Al Alloy 1199 and minimum in case of Mg Alloy NL.
- Elastic Strain is maximum in case of Al Alloy 1199.
- Thus copper being more ductile can undergo more deformation.
Case 2
| Sheet materials | Total Deformations (Y direction) mm | Equivalent Stress MPa | Elastic Strain mm/mm |
| Al Alloy 1199 | 2.4495 mm | 2.426e5 MPa | 1.2146 mm/mm |
| Al Alloy NL | 2.5269 mm | 1.2837e5 MPa | 0.6455 mm/mm |
- From the above table it can be observed that equivalent stress is maximum in case of sheet material of Al Alloy 1199 and because of this resistive force the deformation is also less.
- Thus with increase in the frictional coefficient the elastic strain decreases.
Case 3
| Sheet materials | Total Deformations (Y direction) mm | Equivalent Stress MPa | Elastic Strain mm/mm |
| Al Alloy 1199 | 2.4495 mm | 2.426e5 MPa | 1.2146 mm/mm |
| Al Alloy NL | 2.5269 mm | 1.2837e5 MPa | 0.6455 mm/mm |
| Al Alloy NL (mesh refinement) | 2.4512 mm | 1.818e5 MPa | 0.910 mm/mm |
- Here we can observe that due to the mesh refinement the deformation has decreased and so does the equivalent stress.
- Thus reducing the mesh size the overall computational time increases but we get more accurate result.
https://drive.google.com/file/d/1MXTJuL1lXzHuxB5o4oZBwT3YSLQ2wVkT/view?usp=sharing