Week 9 Tension and Torsion test challenge

Objective: To perform the tension and torsion test on the specimen and finding out the total deformation, equivalent stress and temperature of the specimen.

Setting up analysis: The model is impoerted and Steel 1006 is given as the material to the specimen.

Assign 2 local coordinate systems at both sides of the specimen

Case 1: Tension Test

Mesh:

Entire specimen is selected and tetrahydron method is used for meshing

Next sphere of influence is used at botj coordinate systems and meshed

Boundary conditions: One side is fixed and other side is subjected to displacement of 18mm

Analysis settings:

Analysis settings are same for both the cases

Solution:

Total deformation

Equivalent stress

Temperature

Case 2: Torsion test

In this case, a cylindrical coordinate system is used to give the torsion load

Mesh: Face sizing is carried out at the selected parts with the size of 1.5mm and sphere of influence is used for other selected part with the element size of 1mm

Boundary conditions:

One side is fixed and displcement of 1200 degress is given with the cylindrical coordinate system

Solution:

Total deformation:

Equivalent stress:

Temperature:

Results:

Case	Total deformation(mm)	Equivalent stress(MPa)	Temperature ('c)
Tension	18.2	631	273
Torsion	11	538	520

Conclusion:

As we can observe, the specimen encountered fracture in second case and subjected to large deformation in the first which proves that the ductile material is strong in tension and poor in torsion.