Week 9 Tension and Torsion test challenge

objective:For this challenge, you will have to perform the tension and torsion test on the specimen provided and then simulate the both the tests and then compare the results.

procedure:

case1:Tensile test

It is used to find out how strong a material is and also how much it can be stretched before it breaks. This test method is used to determine yield strength, ultimate tensile strength, ductility, strain hardening characteristics, Young's modulus and Poisson's ratio.

1.Edit the material in the engineering data sources add the material as the steel 1008 to the engineering data library and then take the explicit dynamics and import the geometry into the spaceclaim.

2.Assign the steel 1008 material to the given geomtery and then create the coordinate systems for both the sides of the geometry .

Fig shows the coordinate system of the geometry with both the sides.

3.Mesh the geometry with the automatic method of tetra hedrons to the hole geometry and then face sizing to the both the sides of the geometry with sphere of radius 8mm and with the sizing of 1.5mm.

Fig shows the mesh of the geometry with face sizing and the tetrahedrons.

4.Insert the displacement in the one side of the geometry with the displacement of 18mm the geometry in x direction.

Fig shows the insert of the displacement in x direction because of the tension will be cause in the particular direction.

5.Insert the fixed support to the another side of the geometry so that only one side of the geomtry will move another side will be fixed.

Fig shows the fixed support to the one side of geometry.

6.Analysis settings give the end time as the 1e-03sec and max no of cycles be 1e+07 as the default.

7.Add the solution of total deformation equivalent stress and temperature of the specimen and then compare the results of both tension and torsion tests and then add the user defined result from the work sheet after the problem is solved and that is taken for the temperature as output units.

Fig shows the total deformation of min of 0mm and max of 18mm.

Fig shows the equivalent elastic strain of min of 4.045e-07mm/mm and max of 0.0030446mm/mm.

Fig shows the equivalent stress of min of 0mpa and max of 637.58mpa.

Fig shows the temperature of the geometry when undergo tensile test will be min of 22degree and max of 267.24degree.

case2:

Torsion test

Torsion testing is a type of mechanical testing that evaluates the properties of materials or devices while under stress from angular displacement. ... The most common mechanical properties measured by torsion testing are: Modulus of elasticity in shear. Yield shear strength.

• Repeat the same procedure as in case1 but insert the coordinate system for the centre of the geometry and then insert the mesh sizing of 1.5mm and the sphere of influence on both the sides of the geometry with a radius of 8mm and the mesh size of 1mm  and fixed the geometry at one end and give the displacement of the geometry in y direction of 1200degree.

Fig shows the coordinate system of the geometry for the 3 faces with the rotation of the one face in the y direction.

Fig shows the mesh of the geometry with the mesh size of 1mm to the centre and remaining will be tetrahedrons on both sides.

Fig shows the displacement of the geometry with the angle of 1200degree.

Fig shows the fixed support to the geometry.

Fig shows the total deformation of min of 0mm and max of 17.877mm.

Fig shows the equivalent elastic strain of min of 0mm/mm and max of 0.0031886mm/mm.

Fig shows the equivalent stress of min of 0mpa and max of 660.24mpa.

Fig shows the temperature of min of 21.993 degree and max of 399.38 degree.

Results and comparision

• The total deformation is of max of 18mm for the tensile test due to the elogation of the bar.
• The total deformation is min of 17.877 compare to tensile test due to the twisting moment of the bar.
• The equivalent stress is max of 660.24mpa in case of torsion test due to the angle of rotation of the work piece .
• The equivalent stress is min of 637.58mpa in case of tensile test due to elongation of the bar makes that reduce the crosssection but volume remains constant.
• The equivalent elastic strain is max of 0.0031886mm/mm in case of torsion test due to the change in deformation of the bar.

conclusion

In this challenge learnt about the tensile test and torsion test for the given geometry and then comparision of the results are made for both the cases.

Note : In this both the challenges the solution and results section will be cleared due to large file size.