Week 3 Sheet metal Bending challenge
Aim : Performing sheet bending operation using structural analysis.
Objective :
1) Carry out a static structural analysis to find out Equivalent stress, Directional deformation and Stress intensity.
2) Using three different materials to perform sheet bending.
3) Comparing the three results on the basis of solution.
Procedure :
1) For sheet bending analysis, we use static structural tool in the analysis system tool box to define the geometry and other input parameters for the same.
2) In the engineering data we input the material and its properties. In this case we are using three types of non linear material and its properties are shown below.
Case 1:Aluminium Alloy 1199,Copper Alloy NL and Magnesium Alloy NL. Finding out the Equivalent stress, Equivalent elastic strain and Directional Deformation in Y direction and comparing the results for the three materials.
Case 2:With the material as Aluminium Alloy, changing the friction coefficient to 0.19 and run the analysis as mentioned in Case 1. Compare the results with that in case 1.
Case 3: Refine the mesh on the plate such that it doesn't cross the academic limit. With Aluminium alloy as material, run the analysis as in Case 1 and compare the results.
3) In connection, we have to define the contact faces of the geometry by selecting and deselecting as contact or target bodies. In contacts under connection tab we define the type of contact as frictional contact.
4) By opening the modelling the 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. Finer mesh is used for the sheet which is going to get deformed ann wherever sheet comes in contact.
5) We open analysis setting where we define the displacements
Displacement 1 : We select the top portion of the punch in negative Y axis and input the values in mm
Displacement 2 : Here we select the face of the plate and define displacement 0 in Z axis and keeping free in X and Y axis
Displacement 3 : After punching the sheet into the die, the die is pulled down a bit.
6) Setting up the solution for von mises equivalent stress, total deformation, equivalent stress and other factors.
Case 1 :
Aluminium alloy 1199 NL
a) Directional deformation Y axis
b) Equivalent Elastic strain
c) Equivalent stress
Copper alloy NL
a) Directional deformation Y axis
b) Equivalent Elastic strain
c) Equivalent stress
Magnesium alloy NL
a) Directional deformation Y axis
b) Equivalent Elastic strain
c) Equivalent stress
Case 2 :
Aluminium alloy with coefficient of friction 0.19
a) Directional deformation Y axis
b) Equivalent Elastic strain
c) Equivalent stress
Case 3 :
Aluminium alloy 1199 with finer mesh of 0.8 mm at sheet and 1.2 mm mesh at the contacts
a) Directional deformation Y axis
b) Equivalent Elastic strain
c) Equivalent stress
Results :
Case 1
| Materials/Solution | Directional Deformation | Equivalent strain | Equivalent stress |
| Aluminium alloy 1199 NL | 2.4242 | 0.33991 | 67869 |
| Copper alloy NL | 2.5141 | 0.51623 | 1.028e5 |
| Magnesium alloy NL | 3.1067 | 0.41486 | 82833 |
Case 2
| Materials/Solution | Directional Deformation | Equivalent strain | Equivalent stress |
| Aluminium alloy 1199 NL with cf of 0.1 | 2.4242 | 0.33991 | 67869 |
| Aluminium alloy 1199 NL with cf of 0.19 | 2.7443 | 0.3492 | 69620 |
Case 3
| Materials/Solution | Directional Deformation | Equivalent strain | Equivalent stress |
| Aluminium alloy 1199 NL(Refined mesh) | 2.2742 | 0.97217 | 1.9415e5 |
Conclusion :
1) In case 1, we can see that there is high stress value in copper alloy NL among the three materials given.
2) In case 2, Al alloy with 0.19 cofficient of fricton has higher stress than the with 0.1, which is obvious. Because when the punch forces the sheet into the die there is impact on it which is affected mostly by value of coefficient of friction.
3) In case 3 , finer the mesh greater chances of getting more accurate results of it but at the cost of computatonal time.