Challenge

Week 1 Spur Gear Challenge

Aim : To perform static structural analysis on spur gear to find out equivalent stress, total deformation and stress intensity.

Objectives :

  • Objective 1 :
    • Case 1: Using Cast Iron(ductile) as a material, carry out a static structural analysis to find out Equivalent stress, Total deformation and Stress intensity. Find location for possible fracture.
    • Case 2: Using Cast Steel as material, perform the same analysis for the results as in Case 1.
    • Case 3: Using Cast Bronze as material, perform the same analysis for the results as in Case 1.
  • Objective 2 : Write a detailed report comparing the results of the three cases and conclude which material would you recommend based on these results.

Procedure :

Create materials and assign :

  • Open ansys workbench and set the case of structural analysis on workbench UI.
  • Now click on engineering data and it will open material window from there select ANSYS GRANTA materials and choose cast iron predefined material quality.
  • For second material create a second case by duplicating first case the from engineering data change the material. Duplicate for case-2.
  • Click on Engineering data in second case and then a new window will pop-up now from the engineering data sources we can select any predefined material or we can create a seperate material.
  • For that click on "click here to add a new material" and rename as cast steel and set all quality for that material as given for the challenge.
  • Set all material quality as given for this model.

Geometry :

  • Import cad file in Geometry and open it in space claim.
  • To check weather cad model is fine or not,if any error will found then fix it.

Model :

  • Check the geometry details and assign material from there as required.
  • Define contacts by selecting teeth contacting surface.

  • Joints :

For Right Gear

For Left Gear

Meshing :

  • Give element size of 0.5 mm.
  • Generate mesh.

Analysis Setting :

Joint Load :

For Right Gear

For Left Gear

Solution :

  • We will create output results for total deformation, equivalent stress, equivalent strain and stress intensity.
  • Click on Solve.

Results :

A) Objective 1 :

Case 1 : Using Cast Iron as a material.

Total Deformation :

Equivalent Stress :

Equivalent Strain :

Stress Intensity :

Case 2 : Using Cast Steel as a material.

Total Deformation :

Equivalent Stress :

Equivalent Strain :

Stress Intensity :

Case 3 : Using Cast Bronze as a material.

Total Deformation :

Equivalent Stress :

Equivalent Strain :

Stress Intensity :

Objective 2 : Comparing the results of the three cases.

Sr.No. Results Cast Iron Cast Steel Cast Bronze
1 Total deformation Maximum (mm) 30 30 30
2 Total deformation Minimum (mm) 18.5 18.5 18.5
3 Equivalent Stress Maximum (Mpa) 470.85 462.73 444.56
4 Equivalent Stress Minimum (Mpa) 3.3014 E-6 1.0041 E-5 6.3312 E-6
5 Stress Intensity Maximum (Mpa) 533.12 521.68 497.96
6 Stress Intensity Minimum (Mpa) 3.7951 E-6 1.0932 E-5 7.2917 E-6

Conclusion :

  • For all three materials, the maximum stress was observed in the root of the teeth. Similarly, the stress intensity measured was also found to be maximum in the same region.

  • The deformation is however found to be similar for all the cases.

  • Equivalent Stress : Cast Iron > Cast Steel > Cast Bronze.

  • Stress Intensity : Cast Iron > Cast Steel > Cast Bronze.

  • From the results, it is observed that the equivalent stress and the stress intensity is maximum for cast iron and minimum for the cast bronze.
  • The smaller values of stress intensity assist the part in achieving a longer life.
  • Hence, from the analysis, it can be considered that cast bronze is the a typical material for the spur gears due to the lower values of equivalent stress and stress intensity experienced by the material for the same boundary conditions.
Stress intensity factor (K) :

The stress intensity factor (K) is used in the field of fracture mechanics. It predicts stress intensity near the tip of a crack caused by a remote load or residual stresses.

The magnitude of K depends on:

  • Sample geometry
  • Size and location of the crack
  • Magnitude of load
  • Distribution of load

The following equation describes stress intensity:

Where ß describes the relationship between gross geometrical features to the stress intensity factors. ß can be a function of crack length as well as other geometrical features. K represents nominal stress that is remote from the crack.

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