Week 5 Bending of iPhone

Aim : To simulate the bending of iPhone.

Objectives :

• Case 1: Simulate the model as it is.

• Case 2: Move the bottom fingers from their defined position to the given position X= 22.5mm & Z= 10mm and obtain the results for the simulation.

• Obtain the directional deformation for the model.

• Factor of safety of Aluminium Alloy NL.

• Life of the Aluminium alloy for the case

• Compare the results and present with the conclusion.

Procedure :

Engineering data and material :

• Open engineering data.
• Go to general non linear material from engineering data sources.
• Add Aluminium Alloy NL to library.
• Go to general materials.
• Add Polyethylene, Structural steel.

• Duplicate Polyethylene name it as Glass and give following parameter values.

• Go to Engineering Data and select the Aluminium Alloy NL.
• Create the S-N curve as following for the Aluminium alloy NL.

Geometry :

• Import geometry.
• Check for abnormalities.

• For Case 2, go to spaceclaim geometry use the move option.
• Now the position of the pushing fingers is changed in the spaceclaim geometry to the given position X=22.5mm, Z=10mm.

Model Setup :

• Open geometry and rename the components as following,

• Assign materials to each components.

Display = Glass

Frame = Alumunium Alloy NL

Inside = Polyethylene

Pushing fingers and fingers = Structural steel

• Select all fingers and one pushing finger and make stiffness behavior as Rigid.

Contacts :

• Delete unwanted contacts.
• Select all contacts and rename based on defination.
• Select contact as bonded for Display to Frame.
• Define Other contacts as frictionless.

Joints :

• Define fixed joint for Fingers.

• Define translational joint for Push fingers.
• Reference coordinate system for both should be in same direction i.e. X-axis.

Meshing :

• Insert sizing for frame and push finger contact surface.
• Give element size as 4 mm.
• Generate mesh.

Analysis Setting :

Fixed support :

• Insert fixed support for Inside body.

Displacement :

• Insert displacement for frame to constrain the motion in X and Z direction.

Joint load :

• Insert joint load and give displacement for both Push fingers.

Solution :

Create output results for :

• Directional deformation for model.
• Equivalent stress on Frame.
• Equivalent strain on Frame.
• Fatigue tool for Life and Safety factor of Frame.
• Equivalent stress for Model.
• Equivalent strain for Model.

Results :

Case 1 : Simulate the model as it is without changing push fingers position.

Directional deformation for model :

Equivalent stress for Model :

Equivalent strain for Model :

Equivalent stress on Frame :

Equivalent strain on Frame :

Life of Aluminium Alloy NL :

Safety factor of Aluminium Alloy NL :

Case 2 : Move the bottom fingers from their defined position to the given position X= 22.5 mm & Z= 10 mm.

Directional deformation for model :

Equivalent stress for Model :

Equivalent strain for Model :

Equivalent stress on Frame :

Equivalent strain on Frame :

Life of Aluminium Alloy NL :

Safety factor of Aluminium Alloy NL :

Comparison of Results :

Sr. No.	Results	Case 1	Case 2
1	Maximum Directional deformation for model (mm)	5.5643	4.3723
2	Minimum Directional deformation for model (mm)	-0.9282	-0.2094
3	Maximum Equivalent stress for Model (MPa)	856.16	1005
4	Minimum Equivalent stress for Model (MPa)	6.3808 E-7	6.3811 E-7
5	Maximum Equivalent strain for Model	0.1227	0.1322
6	Minimum Equivalent strain for Model	7.0767 E-12	7.077 E-12
7	Maximum Equivalent stress on Frame (MPa)	605.83	396.68
8	Minimum Equivalent stress on Frame (MPa)	2.2513	5.3425
9	Maximum Equivalent strain on Frame	0.008614	0.00572
10	Minimum Equivalent strain on Frame	0.000103	0.00030
11	Maximum Life of Aluminium Alloy NL	1 E+08	1 E+08
12	Minimum Life of Aluminium Alloy NL	0	0
13	Maximum Safety factor of Aluminium Alloy NL	15	14.974
14	Minimum Safety factor of Aluminium Alloy NL	0.1320	0.2016

Conclusion :

• By observing the above results Case 2 gives the best results of directional deformation and equivalent elastic strain.
• We can conclude that Case 2 can make the model survive comparing to the Case 1 as the stress and strain developed were comparitively low.