Week 3 Sheet metal Bending challenge
Objective: Carrying out a finite element simulation for the bending of a sheet using a punch (or ram) and a die and calculating stress and strain produced on sheets of different materials and in different simulation parameters.
Modelling:
In a sheet bending process, the sheet is placed on a die, which has a valley like opening. A punch applies a force on the sheet and plastically deforms it at the point of contact till the sheet starts touching the die opening sides. The main motive of this process is the plastic deformation of the sheet and according to the material of the sheet, appropriate force needs to be applied to the sheet by the punch in order to plastically deform it, i.e. the stresses generated in the sheet due to the punch should be beyond the yield strength of the sheet material.
In the model, the die is kept fixed at a place and the punch has a downward motion. Here, a downward displacement to the punch is more suitable boundary condition than a load, because the displacement specification means that the punch has to travel the specified distance and thus the sheet will deform that much. The force applied to the sheet gets adjusted on the basis of the amount of deformation the sheet has to go through for the specified punch movement, thus meaning that the stresses generated in the sheet will be enough for the plastic deformation (>yield strength) irrespective of the sheet material.
The maximum distance the punch can translate is 15mm as showcased in the figure. At 15mm displacement, the sheet will be completely touching the die opening and the punch will be on top of the sheet applying the load. After simulation of this model, it was found that due to this configuration at 15mm punch displacement, the load of the punch is transferred to the die through the sheet as the sheet is in complete contact with the die opening, and this can lead to very high stresses in the die, and thus the displacement of the punch is limited to 13.5mm. The die is kept fixed during the whole process.
Boundary Conditions:
The downward displacement of the punch is divided into 10 load steps, with the time step/increment for each load step being set according to the program itself (program controlled). Since the punch has to move downward only, its displacements in X and Z axis is constrained and equal to zero. For the die, which doesn’t move at all, all its displacements (X,Y,Z) are constrained and equated to zero for all the loadsteps in the simulation. For the sheet, only its Z displacement is constrained as we don’t want any sheet movement in the Z direction while it is free to displace in X and Y axes (since when it deforms, it will be displaced in those directions only).
The displacement of the punch across the loadsteps is illustrated below,
Analysis settings:
The number of loadsteps in the model are 10 and for each step, the time stepping is program controlled. The direct solver is used and since this problem includes a large amount of deformations and non-linearity, both material and geometric, large deflections has been activated and stabilization has been added, which helps in convergence of problems including large displacements by providing damping to the excessively deforming nodes. The stabilization is given by the energy dissipation ratio of the value 0.1.
Contacts:
All contacts in this model have been considered frictional. For all the contacts:
Behaviour = Auto-Asymmetric
Formulation: Augmented lagrange
Normal Stiffness/Contact stiffness = 0.1x(value of programmed controlled stiffness)
Interface Treatment: Add offset, Ramped Effects
For the contact between the punch and the sheet, the load is transferred from the punch to the sheet, thus making the punch the contact body and the sheet the target body.
For the contact between the sheet and the die, load is transferred from the sheet to the die and thus the sheet is the contact body and the die is the target body.
Mesh:
For this model, mesh refinement is required at the contact regions and the sheet, since majority of the loads are applied there. Rest of the model can have course mesh in order to reduce the computational times. The sheet has tetrahedron type elements and rest of the model has hexahedral type elements.
Element size at the contact regions and the sheet = 1mm (fine)
Element size in rest of the model = 4mm (course)
#Nodes = 17904
#Elements = 6262
Case 1:
The three materials to be considered for this analysis are:
1). Aluminium alloy 1199
Yield Strength = 140 MPa
Young’s Modulus = 62 GPa
Density = 2700 kg/m3
Tangent Modulus (slope of the linear curve in the plastic region of bilinear model) = 209.49 MPa
Hardening behaviour: Isotropic Hardening (both tensile and compressive yield limits increase after plastic deformation)
2). Magnesium Alloy
Yield Strength = 193 MPa
Young’s Modulus = 45 GPa
Density = 1800 kg/m3
Tangent Modulus = 920 MPa
Hardening behaviour: Isotropic Hardening
3). Copper Alloy
Yield Strength = 280 MPa
Young’s Modulus = 110 GPa
Density = 8300 kg/m3
Tangent Modulus = 1150 MPa
Hardening behaviour: Isotropic Hardening
Results for Aluminium Alloy 1199:
Stress:
As seen in the above plot, the stress levels easily bypass the yield strength of Aluminium alloy 1199 (140 MPa) and are thus able to plastically deform the sheet.
Sheet VM stress:
NOTE: The above picture shows the stress distribution in the sheet at t = 1.4 s (within 2nd loadstep), when the stress on the sheet is maximum.
Directional Deformation (Y):
Equivalent Elastic Strain:
The animations of the simulation is given below:
Aluminum alloy 1199 animation
Aluminum Alloy 1199 animation 2
Results for Magnesium Alloy:
Stress:
As seen in the above plot, the stress levels easily bypass the yield strength of Magnesium Alloy (193 MPa) and are thus able to plastically deform the sheet.
Sheet VM stress:
NOTE: The above picture shows the stress distribution in the sheet at t = 7 s (start of 8th loadstep), when the stress on the sheet is maximum.
Directional Deformation (Y):
Strain
Results for Copper Alloy
Stress:
As seen in the above plot, the stress levels easily bypass the yield strength of Copper Alloy (280 MPa) and are thus able to plastically deform the sheet.
Sheet VM stress:
NOTE: The above picture shows the stress distribution in the sheet at t = 7 s (start of 8th loadstep), when the stress on the sheet is maximum.
Directional deformation (Y):
Strain:
NOTE:
We can find out the force reaction due to the punch movement i.e the force which is being applied on the sheet. We observe the maximum downward forces applied for each of the materials are:
Aluminium Alloy 1199: 87.321 N
Magnesium Alloy: 129.01 N
Copper Alloy: 187.31 N
Out of these materials, Copper Alloy has the highest yield strength, thus a higher force is required to produce a plastic deformation in the sheet. Similarly, the aluminium alloy has the least yield strength and thus requires the lowest force for plastic deformation. This is the advantage of using displacements as boundary conditions as the load applied on the sheet adjusts according to the amount required to produce the specified deformation/displacement.
CASE 2 and CASE 3:
To see the sensitivity of the results to the changing frictional coefficient and refined mesh element size, we can carry out a parametric study, where the element size and the frictional coefficients are the input parameters and the maximum VM stresses, maximum strains and average Y deformations are the output parameters.
Design Point:
Initial design point (DP0): Refined mesh element size = 1mm and Frictional coefficient for both the contacts = 0.1
Case 2 design point (DP1): Refined mesh element size = 1mm and Frictional coefficient for both the contacts = 0.19
Case 3 design point (DP2): Refined mesh element size = 0.8mm and Frictional coefficient for both the contacts = 0.1
Results for 0.19 frictional coefficient (Case2/DP1)
Stress:
Sheet VM stress:
Directional deformation (Y):
Strain:
The animation of the sheet bending for Case 2 is given:
Case 2 animation
Comparison with case 1:
| Case | Max. VM stress (MPa) | Average Directional deformation (Y) (mm) | Max. VM stress (Sheet) (MPa) | Max. Elastic strain (mm/mm) |
| Case 1 (DP0) | 188.89 | -4.1612 | 188.89 | 0.00306 |
| Case 2 (DP1) | 188.97 | -4.1492 | 189.97 | 0.0030783 |
Result for refined mesh element size of 0.8mm on the sheet (Case 3/DP2):
For this case the body sized mesh on the plate is refined further from element size of 1mm to 0.8mm and it is ensured that the number of nodes and elements stay below the academic limit (32k)
#Nodes = 28022
#Elements = 10869
Stress:
Sheet VM stress:
Directional deformation (Y):
Strain:
Animation of the sheet bending for case 3 is given:
Case 3 animation
Comparison with case 1:
| Case | Max. VM stress (MPa) | Average Directional deformation (Y) (mm) | Max. VM stress (Sheet) (MPa) | Max. Elastic strain (mm/mm) |
| Case 1 (DP0) | 188.89 | -4.1612 | 188.89 | 0.00306 |
| Case 3 (DP2) | 188.53 | -4.6326 | 189.53 | 0.0030481 |
As seen above, the model has very little sensitivity to the mesh refinement in the plate.
The simulation files for the challenge are given below:
https://drive.google.com/file/d/1yNfvNRjwVJOFg4ytDCsgTIH40tEjrCLg/view?usp=sharing