Static Structural Analysis of Sheet Metal Bending

OBJECTIVE:

To perform Sheet metal bending for 3 different materials and perform variation of certain settings.

Case 1: Aluminium Alloy 1199, Copper Alloy NL and Magnesium Alloy NL. To find out the Equivalent stress, Equivalent elastic strain and Total Deformation in Y direction and compare the results for the three materials.

Case 2: With the material as Aluminium Alloy and friction coefficient as 0.19 and run the analysis as mentioned in Case 1 and compare the results with that in case 1.

Case 3: Refine the mesh on the plate to the academic version limit. With Aluminium alloy as material, run the analysis as in case 1 and compare the results.

To present the findings as a report and give explanation as to which property/properties of the materials affect the outcome.

SOLUTION:

Step 1: Design

The model has 3 bodies namely, die, punch and the sheet.

Step 2: Material

The punch and die were made using structural steel and the sheet material was taken as aluminium alloy 1199, copper alloy NL and magnesium alloy NL and the results for each material was evaluated and compared. The properties of the material is as given below:

Step 3: Connections

The default contacts between the bodies were deleted and new contact were added. The first contact was defined between the punch and the sheet and the second contact between the sheet and the die.

The contacts were defined as frictional type and the coeffcient of friction was set as 0.1. The behaviour of the contact is defined as auto asymmetric and formulation as augmented lagrange. A normal stiffness factor of 0.1 was given and the interface treatment was defined as add offset with ramped effects.

Step 4: Meshing

The models was given a default mesh of 4mm size. The sheet and the contacts was given a finer mesh size of 1. Further, the sheet was given a tertragonal mesh using the patch conforming method.

Step 5: Analysis settings

In this step, the movement of the bodies were defined. The punch has to move down such that it bends the sheet into the die cavity and was given a displacement along the y direction. This displacement was defined over 10 steps. Since the sheet should move along the z axis, the displacement along this axis was constrained. Similary, the movement of the die body was also constrained along all directions. The direct solver type was chosen for the simulation, the large deflection was turned on and the energy dissipation factor was set as 0.1.

Step 6: Results

Case 1:

The equivalent stress, strain and total deformation in y direction was evaluated for the above cases.

1. Equivalent stress

2. Directional deformation (y direction)

3. Equivalent elastic strain

Material	Equivalent stress (MPa)	Equivalent strain	Directional deformation (mm) Max Min
Al alloy 1199	57.75	0.0015926	2.6258 -12.145
Cu alloy NL	117.87	0.0017575	2.8103 -11.646
Mg alloy NL	76.78	0.0028534	3.217 -10.592

Case 2:

In case 2, the coefficient of friction was increased to 0.19 and with aluminium alloy 1199 as the material results were evaluated and compared with case 1.

Coefficient of friction (µ)	Equivalent stress (MPa)	Equivalent strain	Directional deformation (mm) Max Min
0.1	57.75	0.0015926	2.6258 -12.145
0.19	58.744	0.001574	2.7718 -12.078

Case 3:

For case 3, the mesh was refined to the academic version limit. This was done maintaining the default mesh as 4mm and the refining the mesh size for the plate and the contact faces to 0.8mm. The results obtained are compared with case 1.

Case	Equivalent stress (MPa)	Equivalent strain	Directional deformation (mm) Max Min
1	57.75	0.0015926	2.6258 -12.145
3	60.02	0.0016842	2.5379 -12.074

CONCLUSIONS:

1. The variation in the properties with respect to change in materials is seen as below.

Equivalent stress: Cu alloy NL > Mg alloy NL > Al alloy 1199.

Equivalent strain: Mg alloy NL > Cu alloy NL > Al alloy 1199.

Deformation (max negative): Al alloy 1199 > Cu alloy NL > Mg alloy NL.

The yield strength and tensile strength of the material plays a very important role in sheet metal bending. In order for the material to permanently deform, the load has to exceed the yield strength of the material. Also, to avoid failure the load has to be well below the tensile strength of the material.

Here, the highest stress is experienced by Cu alloy and this behaviour is consistent with the above theory. Among the three alloys, Cu alloy has the highest yield strength. Hence a higher load is required to deform the metal sheet, which in turn generates a higher stress.

2. When the coefficient of friction was increased, the equivalent stress increased. Further, the strain and deformation values saw a decrease.
3. In case 3, as the mesh is made finer, the stress and strain value increases while the deformation value shows a small decrease.