Week 3 Sheet metal Bending challenge

SHEET METAL BENDING SIMULATION

In Sheet metal bending simulation the Punch is forced down on sheet which is fixed on a die of certain mould to get the desired output from the sheet

• In this Scenario We will be Discussing The Following three Cases as CASEI, CASEII and CASEIII
• i) 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.
• ii) 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.
• iii) 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 I :

In the following case scenario the sheet on which the punch force is going to be applied wil be assigned three different materials and their respective results will be tested accordingly.

These material cases will be named as a,b,c respectively.

GEOMETRY FOR CASE I:

Fig. Geometry for Sheet metal Bending Simulation

In this CASE just the material property of the sheet is going to be adhered rest of the mesh, contacts and boundary condition remains unchanged.'

In case a the material assigned is Aluminium alloy 1199

CONTACT ANALYSIS:

Fig. Contact Analysis

In the above simulation to be carried out the Contact is Between

i) Punch face and Sheet which is frictional contact with co-efficient of 0.1

ii) Sheet and Die Face contact with frictional Co-efficient of 0.1

MESH:

Fig.Mesh Generation

Fig.Element Quality

From the above grid generation it is clear that the mesh generated is the combinaion of the Tetra mesh and the hexa hedral mesh as to get a pitch perfect deformation the plate has been assigned tetrahedrons with the body sizing of 1mm and its corresponding top face sizing of 2mm respectively.

So basically the punch and die are given a element lengt of 4mm respectively.And the avg element quality lies about 0.78 with 12520 nodes and 5250 elements respectively.

BOUNDARY CONDITIONS:

In the Case I the Boundary Conditions asre as follows:

Fig.Boundary Conditions

In the above Case the sheet is constrained in X and Z direction respectively and the punch is alowed a maximun displacement of 13.5 mm in negative y directon in the following 10 Sub Steps respectivey. And while reverting back the Die is given a displcement of 3 mm max in y direction.

SOLUTION:

Fig.Directional Deformation For CASE I(a)

Fig. Von Mises Stress For CASE I(a)

Fig.Elastic Strain For CASE I (a)

CASE I (b):

In this Case the material to be used is Copper Alloy

Fig.Deformation for CASE I(b)

Fig.Von Mises Stress for CASE I(b)

Fig. Equivalent Strain For CASE I(c)

SOLUTION For CASE I(c): Material is Magnesium Alloy

Fig.Deformation For CASE I(c)

Fig.Von Mises Stress For CASE I(c)

Fig.Equivalent Strain For CASE I (c)

CASE II:

In this CASE the frictional co-efficient is set to 0.19 basically no either chnage is made

Fig. Deformation CASEII

Fig.Equivalent Stress For CASE II

Fig.Strain For CASE II

CASE III:

Fig.Deformation For CASE III

Fig.Stress For CASE III

Fig.Strain For CASE III

CONCLUSION: From the above table it is Concluded as Follows:

CASE I SOLUTION

Sr. No	CASE	Material	Total Deformation (mm)	Von Mises Stress (MPa)	Equivalent Strain
1	a	Aluminium1199	1.89	71.45	0.0015
2	b	Copper	2.32	150.46	0.0018
3	c	Magnesium	2.64	99.08	0.0021

From the above CASE it is Clear that max directional deformation is of Magnesium aoy anf max Stress generated of Copper respectively.

CASE II SOLUTION:

In this Solution the CASES will be compared on the basis of co-efficient of friction for Aluminium alloy.

Sr. No	CASE	Co-Efficient of Friction	Deformation(mm)	Von Mises Stress (MPa)	Equivalent Strain
1	I(a)	0.1	1.89	71.45	0.0015
2	II	0.19	1.84	72.32	0.0016

From the above table it is clear that the higher the co-efficient of friction higher the magnitude, but there is not much difference in the values due to less difference in frictional values and mesh

CASE III SOLUTION:

In this CASE the Mesh is varied as 0.7mm and the results are compared respectively for CASE I and CASEIII

Sr.No	CASE	Mesh Size	Deformation(mm)	Von Mises Stress(MPa)	Equivalent Strain
1	I	1 mm	1.89	71.45	0.0015
2	III	0.7 mm	1.96	82.09	0.0016

From the Above it is Clear that the higher the mesh high accurate vaues we get , as due to limitation ins mesh size , of the mesh size is further improved more accurate we eill get the result until the equillibrium condition is achieved.