Week 2 Railwheel and Track

Aim : To perform static structural analysis on the Railwheel and Track setup.

Conclusion :

1) Carry out a static structural analysis to find out Equivalent stress, Total deformation and Life.

2) Comparing results of 100000N and 500000N load for analysis.

3) Implementing a user defined result and calculate the total deformation from the results and comparing it with results of previous analysis on the basis of solution.

Introduction :

Railway transportation system, as one of the notable means of commuting systems, has served for human societies andhas pursued its improvements as other promoted aspects of life. In recent years, the capacity of carrying axial loads for world railways as well as their velocities has been enhanced which results in increasing the amount of strains and stresses on lines and digression of rails. We are performing the railwheel and track analysis carrying load.

Procedure :

1) For both cases, we use static structural tool in the analysis system tool box to define the geometry and other input parameters for the same.

2) In the engineering data we input the material and its properties.

3) Input the given rail wheel and track geometry from the geometry tab. Open in model manager to check for any errors if visible.

4) In connection, we have to define the contact faces of the rail wheel and track geometry by selecting and deselecting as contact or target bodies. In contacts under connection tab we define the type of contact as frictional contact with coefficient of friction as 0.3

Contact Region 1 :

Contact Region 2 :

5) To define type of joint we use contact tab and specifying the body to ground connection type. Also scoping the geometry by selecting the faces of the geometry for fixed, planar and translational motion of the bodies.

6) By opening the modelling the tab, we mesh the geometry with element size of 50 mm. Generating the mesh thereafter by right clicking on the Mesh icon. Face sizing under mesh tab is used to refine the mesh where contacts are present.

Finer mesh on the contact regions and 50 mm on the rest of the geometry.

7) After defining the contacts and joints, we open analysis setting where we define step control settings and input the parameters as needed.

For 1st step :

For 2nd to 5th step :

8) Defining displacement of 100 mm by incrementing for each step in X direction about Z axis.

Input the value of bearing load

9) Setting up the solutions

For Case 1 :

a) 100000N of bearing load

Total deformation :

Equivalent stress :

Life :

Equivalent elastic strain :

b) For 500000N of bearing load

Total deformation :

Equivalent stress :

Life :

Equivalent elastic strain :

For Case 2 :

User defined Total deformation :

Results :

For Case 1

For bearing load of 100000N

Total deformation	Equivalent stress	Life	Equivalent elastic strain
999.06 mm	86.052 MPa	1e6	0.00057013 mm/mm

For bearing load of 500000N

Total deformation	Equivalent stress	Life	Equivalent elastic strain
999.14 mm	349.09 MPa	1e6	0.0022547 mm/mm

For Case 2

Total deformation	User defined Total deformation
999.06 mm	999.06 mm

Conclusion :

1) In Case 1 it is visible that the equivalent stress with 100000N is low 86.052 MPa where as with 500000N is higher 349.09 MPa. It can bear or say withstand the load for both cases if we observe total deformation.

2) For Case 2, if we compare it with previous result of total deformation it is same as user defined total deformation.