Week 7-Long Piston With Cam

OBJECTIVE :-

To perform a Transient Analysis on Piston & Cam Mechanism model. We have to run the simulation for 3 different cases.

1. Frictionless contact
2. Frictional coefficient= 0.1
3. Frictional coefficient = 0.2

Also find out the following results:

1. Directional Deformation
2. Equivalent stress
3. Equivalent Strain

& compare all the 3 results.

Procedure Followed:-

Create a new transient structural.

Import the geometry provided , Check for any errors in the geometry by opening it in space-claim.

Post assessment of geometry, open the model in mechanical workbench.

Rename the geometry, select both barrel and cam and change the flexible to RIGID.

Now its time for Contacts & joints.

Select the contact between barrel and cam follower and check for target faces & Contact faces for proper selection,

Select both contacts and for all the 3 cases as Frictionless , Frictional coefficient=0.1 & frictional coefficient=0.2 assisgn to the contacts,

After contacts insert a new joint

Body to ground & Fixed for Barrel

Body to ground & Revolute for Cam

Body to Body & Translational for Cam follower Check for the motion whether it is in alignment with the coordinate system.

Now analysis setting has to be done:

Number of steps = 9

Select 1^st step and assign the following

Initial time step = 0.1

Minimum time step = 5E-2

Maximum time step = 0.2

Weak springs = Program controlled

Large deflection = ON

Stabilization = constant

Energy dissipation ratio = 0.1

All output controls must be YES except the Contatc Miscellanous

Activation for first substep =YES

Now select steps 2 to 9 and assign the following

Initial time step = 0.1

Minimum time step = 5E-2

Maximum time step = 0.3

Weak springs = Program controlled

Large deflection = ON

Stabilization = constant

Energy dissipation ratio = 0.1

All output controls must be YES except the Contatc Miscellanous

Activation for first substep =YES

After the analysis setting, Insert a joint load Revolute ground to barrel cam

Type= Rotation

Magnitude has to be assigned using tabular data, and the data assigned must be 0,30,60,90,1320,150,180,210,240,270.

Now the output results requirement has to be given commanded. In this simulation we have to obtain output for Directional Deformation, Equivalent stress, Equivalent strain.

Now solve, and tabulate the results:

CASE 1:- For Frictionless contact

Directional Deformation:-

Equivalent Stress:-

Equivalent elastic strain:-

CASE 2:- Frictional contact = 0.1

Directional Deformation:-

Equivalent Stress:-

Equivalent elastic strain:-

CASE 3:- Frictional contact = 0.2

Directional Deformation:-

Equivalent Stress:-

Equivalent elastic strain:-

CONCLUSION:-

	Directional Deformation(mm)	Equivalent stress(MPa)	Equivalent elastic strain(mm/mm)
Frictionless Contact	20.373	1776.1	0.0089406
Frictional coefficient =0.1	20.512	1883.7	0.009787
Frictional coefficient =0.2	20.551	2353.7	0.012176

Above is the tabulated results for all the 3 cases.

As it can be concluded from the above result that all the 3 parameters, stress values, Equivalent elastic starin & Directional deformation increases with the increase in frictional coefficient value.

Google drive link is attached for reference files.

https://drive.google.com/file/d/1NuMQ1268SHOo0-9lfpfitFsCNSOvi1vX/view?usp=sharing