Week 3 Sheet metal Bending challenge

OBJECTIVE :

Sheet metal bending is to be performed for 3 different materials mentioned below. Variation of certain settings is also to be performed and your conclusions are to be added to the report

Run the analysis for the material :

Case 1:Aluminium Alloy 1199( mentioned in the course video),Copper Alloy NL and Magnesium Alloy NL. 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, change 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.

CASE SETUP :

Case set up is as mentioned below, with only changes being:

Case 1 : material is changed

Case 2 : friction coefficient is changed.

Case 3 : mesh count is changed.

1. Import the geometry.

2. Assign appropriate material as mentioned in the objective.

3. Assign the contact property as shown below for case 1 and 3 . For case 2 change friction coefficient to 0.19.

4. Mesh setting.

Select punch and die and give of size of 4 mm

Select sheet and assign size of 1mm . (case 1 and 2) for case 3 enter size of 0.7

Select all contact faces and enter a size of 1 mm( case 1& 2) for case 3 enter size of 0.75 mm.

Also for sheet enter type of elements as tetrahedron and Generate the mesh.

	case 1 & 2	Case 3
No of elements	18058	31804

5. Analysis Setting

6. Boundary Condition:

a) Create a displacement for the punch.

b) constrain the movement of sheet in z-direction.

c) To remove the sheet from die after bending , displacement to the die to be given.

7. Run the solution.

RESULT AND CONCLUSION:

CASE 1:

Equivalent stress plot

Max equivalent stress is found in Copper alloy and minimum in Aluminum Alloy 1199. The Maximum value of equivalent stress is due to higher moment of resistance.

Stress is propotional moment of resistance which is directly related to (E*I)

Equivalent strain Plot

Total deformation in Y direction

Thought the sheet is bend by 13.5 mm from initial position, yet the final deformation is less than 13.5 mm because of spring back effect. which is highest is magnesium alloy NL.

Case 2:

Equivalent stress plot

Increasing the friction coefficient increased the stress value.

Equivalent strain plot

Total deformation in Y-direction

Increasing the friction have no effect on the deformation.

Case 3:

Equivalent stress:

Total deformation in Y-direction:

From above plots of equivalent stress and deformation , it is observed that increasing the mesh count resulted in better result at the expence of the time.