Week 3 Sheet metal Bending challenge

                              SHEET METAL BENDING CHALLENGE

Objective:-

To Simulate sheet metal bending challenge with three different materials
(Aluminium Alloy 1199, Copper Alloy NL & Magnesium Alloy NL) and find out Equivalent
Stress, Equivalent Elastic Strain & Total Deformation

Geometry:

Static structural:

selection of the step as follows 

1.  Material selection 
2. Geometry
3. Model
4. Setup
5. Solution 
6. Result

For Aluminium Alloy 1199:

First to import the given model and select the material one by one. Image of the 3D model
given below:

edit in space claim to check the or any correction in geometry :

After that select material from the Library for Sheet and for Punch and Die I select
Structural Steel. Properties of materials given below:

Aluminum Alloy NL as material selection:

1. Model:

update the model

after update the model 

image will be shown as :

after this edit the model :

change the name of the part through geometry :

after this apply connection :

connection :

information need to fill in the given condition :

1 ) punch to sheet 

2) sheet to die 

connection type	contact
type	frictional
frictional coefficient	0.1
behaviour	Auto  Assymetric
formulation	Augmented Lagrange

Mesh:

apply the finer mesh sizing for more accurate result :

insert sizing by using solid filter :

 element size: 4mm

select geometry for the punch and die :

insert sizing by using solid filter :

 element size: 1mm

select geometry for the sheet  :

insert sizing by using solid filter :

 element size: 1mm

select geometry for the faces that are in contact while analysis   :

insert method by using solid filter :

 element :tetrahedrons

select geometry for the sheet  :

 

now generaet the mesh :

ckeck for the number of node that it should not exceed limit :

static shructural :

analysis settings:

number of step	10
current step	10 (select all)
Large deflection	on
stabilisation	constant
energy desipation	0.1

all other displacement in static structural are :

 1) Displacement 

 2) Displacement 2

  3) Displacement 3

input for all the displacement given above is :

 1) Displacement 

analysis settings:

Scoping method	geometry selection
geometry	3 faces
Type	displacement
X component	tabular data
Y component	tabular data
Z component	tabular data

 for  2) Displacement 2

 

Scoping method	geometry selection
geometry	2 faces
Type	displacement
X component	free
Y component	free
Z component	tabular data

 

   3) Displacement 

 

Scoping method	geometry selection
geometry	5 faces
Type	displacement
X component	tabular data
Y component	tabular data
Z component	tabular data

To calculate the solution equation here are some input parameter are given which have to insert :

1. Equivalent stress
2. Equivalent elastic strain
3. Directional deformation
4. Equivalent stress 2
5. Contact tool-

          -Pressure 

         - Frictional stress 

      6.Contact tool-

1.           -Pressure 

            - Frictional stress 

                  

      7. Force reacton- 

Result for the Aluminium Alloy 1199:

1. Equivalent stress

1. Equivalent elastic strain

1. Directional deformation

1. Equivalent stress 2

1. Contact tool-

          -Pressure 

         - Frictional stress 

      6.Contact tool-

1.           -Pressure 

3. 

5.          - Frictional stress 

                  

      7. Force reacton- 

 case 1 result in tabular data

 compare the results for the three materials.

Material	Aluminium Alloy 1199	, Copper Alloy NL	Magnesium Alloy NL
Equivalent stress	4.829MPa	4.829MPa	3.5461MPa
Equivalent elastic strain	9.4292e-005	9.4292e-005	9.5327e-005
and Total Deformation	-3.1676e-002mm	-3.1676e-002mm	-3.1663e-002mm

NOTE:

for other material followed the same process as as done above i have attached the file for other material with this report 

 

 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

procedure for case 2

as to change the frictional coeffient 0.19

for both contact friction coefficient to 0.19

model edit> connection>contact> frictional >0.19>

1. punch to sheet
2. sheet to die

result for the case 2 study :

Material	Aluminium Alloy 1199 frictional coefficient = 0.1	Aluminium Alloy 1199 frictional coefficient = 0.19
Equivalent stress	4.829MPa	4.8246MPa
Equivalent elastic strain	9.4292e-005	9.4204e-005
and Total Deformation	-3.1676e-002mm	-3.168e-002mm

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.

prcedure as same which we followed for mesh above for change the el

1. body sizing
2.   body sizing
3. 2face sizing

give the proper value for the finer mesh and as shown in table result are shown :

Aluminium Alloy 1199

 mesh size 1 and 4 mm

statics : 6229

for finer mesh :

Result for the case three:

Material	Aluminium Alloy 1199  mesh size 1 and 4 mm	Aluminium Alloy 1199  mesh size 0.75 and 2 mm
Equivalent stress	4.829MPa	2.7722MPa
Equivalent elastic strain	9.4292e-005	4.7309e-005
and Total Deformation	-3.1676e-002mm	-3.2993e-002mm

  Conclusion :

Case 1:

Aluminium Alloy 1199, Copper Alloy NL, and Magnesium Alloy NL.  Equivalent stress, Equivalent elastic strain, and Total Deformation in the Y direction and comparision  for results for the three materials is done.

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 is done.

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 is done.