Rail Wheel and Track Simulation with User Defined Results

Rail wheel and Track

The objective in this challenge is to perform the simulation by multiplying the applied load by five times and, also learn about the user defined results and check the Total Deformation and compare it with the inbuilt results in Ansys.

Case 1

To perform this simulation, following setup is used and material applied is structural with following properties.

	Density	Young’s Modulus	Poisson’s ratio	Yield Strength	Ultimate Tensile Strength
Structural Steel	7850 Kg m^-3	200 GPa	0.3	250 MPa	460 MPa

A frictional contact is applied between the track and the wheel and a frictionless contact is applied between the contacting surfaces of Wheel and the Shaft.

For joints, a fixed joint is applied on the track to prevent them from moving and all degrees of freedom are constrained.

A translational joint is applied on the shaft in the X direction, as the wheel will roll on the track in that direction. The shaft only allowed to translate along X axis and all other degrees of freedom are constrained.

A planar joint is applied on the wheel as shown in the picture below. The degrees of freedom constrained in the planar joints are- UZ, ROTX and ROTY. The wheel is allowed to move in X and Y direction as well as rotate along Z axis.

For the mesh, element size of 50 mm is applied, and a face sizing of 35 mm is applied on the surfaces between the wheel and the track. The grid dependency is not performed on this occasion as Student version is being used and therefore, restricted element numbers are available.

A bearing load of 1e5 N is applied in the negative Y direction on the surface of the wheel which is in contact with the shaft. This load is transferred on the wheel by the shaft.

Also, a Displacement joint is applied on the surface of the shaft for total distance of 500mm. In the analysis settings, 5 steps are used and from the step 2 to 5, Min Time Step applied is 1e-3s and Max time step is 0.1s.

The solver type is Direct, and the Large deflection is turned On. All the output controls are switched to ‘yes’.

Results for load of 1e5 N

Total Deformation= 1000.6 mm (Max)

Equivalent Stress= 146.34 MPa (Max)

Equivalent Stress on the wheel= 143.87

Fatigue Life

Results for the load of 5e5 N

Total Deformation= 1000.7mm (Max)

Equivalent Stress= 564.42 MPa (Max)

Equivalent Stress on the wheel=517.52 MPa (Max)

Fatigue Life

	Total Deformation	Equivalent Stress	Equivalent Stress (wheel)	Fatigue Life
Load=1e5 N	1000.6mm (Max)	146.34 MPa (Max) 1.76e-10 MPa (Min)	143.87 MPa (Max)	1e6 cycles (Max) 5.73e5 cycles (Min)
Load=5e5 N	1000.7 mm (Max)	564.42 MPa (Max) 5.65e-12 MPa (Min)	517.52 MPa (Max)	1e6 cycles (Max) 2494.9 cycles (Min)

Comparing the results for both the loads, it is observed that no change occurs in Total deformation. It is obvious that Equivalent Stress will increase with the increased load and the maximum Equivalent stress for the load of 5e5 N is 564.42 MPa which is larger than 146.34 MPa for the load of 1e5 N.

The equivalent stress on the wheel also increases to 517.52 MPa when the applied load increases, and it is slightly lower than the equivalent stress on the track. But the stress values on wheel in both cases is close to the stress value occurring on the track.

The Max Equivalent stress occurs on the point where both the surfaces come in contact.

The min fatigue life in case of the load of 1e5 N is 5.73e5 cycles. A drastic change is observed in the min fatigue life value when the load is increased to 5e5 N. The min fatigue life is reduced to 2494.9 cycles.

This means that the material will fail in Low cycle fatigue (LCF) when the load in increased by 5 times. The material almost has infinite life if the operating load remains at 1e5 N.

The track and wheel design can be modified to increase their fatigue strength and some manufacturing processes can be applied to relive the stresses in the material.

Case 2

Lets first understand what Directional Deformation and Total Deformation is.

When an analysis in performed, displacement occurs in 3 directions, viz., X,Y and Z. If the user can check the deformation in the desired direction.

Total Deformation is the overall deformation of the system and it is the square root of the summation of the square of x-direction, y-direction and z-direction.

This is the expression for the Total deformation-

Total deformation=SQRT(X^2+Y^2+Z^2) (X,Y,Z are directional deformations)

To simply check how the user defined results works in Ansys, this expression is used as shown below.

The obtained user defined results are same as the obtained inbuild results in Ansys. The Total Deformation from UDR is 1000.6 mm (Max) same as above in inbuilt Total Deformation for the loading of 1e5 N.

Animations are attached.