1 PROBLEM STATEMENT

CONDUCT TENSILE TEST & TORSION TEST ON THE SPECIMEN.

Refer following table for boundary condition

NATURE OF TEST	B.C. ON END-1	B.C. ON END-2
TENSILE	FIXED SUPPORT	18mm displacement
TORSION	FIXED SUPPORT	1200 degree of rotation

USE STEEL 1006 AS A MATTERIAL

Find out the total deformation, equivalent stress and temperature of the specimen and attach appropriate images to show your results.

1 INTRODUCTION

Material testing is plays very important role in our life. Testing of material gives accurate data about life span of material under different loading conditions. Testing helps in designing of machine component such that it can perform well under applied load. We can obtain different material property using different material testing technology. These properties define strength of material.

Some of material testing technology are Tensile test, Torsion test, Bending test, Hardness test, Compression test, Impact test, Fracture-Toughness test, Creep test, Fatigue test and the list is very long. These testing gives different material strength in that specific loading condition. While using those material as a machine component we can’t exceed the loading value from limiting material value obtained from material testing in any condition. If we do so, our material we fail.

In present work we will focus on tensile testing & torsion test of material. Tensile test gives us important properties of material like ultimate tensile strength, maximum elongation, fracture stress of material directly. There are some other material property which we can measure from above obtained properties are Young’s modulus, Poisson’s ratio, yield strength, & strain hardening characteristics. Similarly, torsion test helps in determining shear modulus, yield shear strength, torsion fatigue life, ductility, ultimate shear strength, modulus of rupture in shear.

2 AIM

The aim of project is to conduct tensile & torsion test on specimen & determine mechanical properties of material. We will also compute total deformation, equivalent stress and temperature of the specimen at the end of test.

3 GEOMETRY & MATTERIAL PROPERTY

A cylindrical cross section was used to conduct both tests. The total length of specimen was 106mm & the gauge length is 15mm. The diameter of cross-section at grip is 12mm & diameter at gauge part is 6.5mm.

Following diagram shows the detailed geometry of specimen.

STEEL 1006 (Explicit material) has been taken for performing tensile & torsion test.

Following table shows the material property.

PARAMETER	VALUE
Density	7896 Kg/m3
Initial yield stress	3.5E+08 Pa
Shear modulus	8.18E+10 Pa

4 MESHING

Total number of nodes is 673 & element is 2407. Tetrahedron method of meshing has been done. Sphere of influence was chosen for sizing. Radius of sphere is 22mm & element size is 3mm for grip section.

In torsion test, additionally gauge part was meshed with element size of 1 mm. Reducing cross-section was also meshed separately using sphere of influence whose radius is 8mm & element size with 1mm. Total number of nodes is 4559 & element is 20908.

Above figure shows meshing for tension test & below figure shows meshing for torsion test

5 BOUNDARY CONDITION & LOADING

Following figure shows the loading conditions. Left side of geometry has been given a fixed support & right side has been given as a displacement of 18mm for tension test. In torsion test, instead of displacement rotation of 1200 degree has been applied.

Following figure shows the applied boundary condition for tensile testing

Following figure shows the applied boundary condition for torsion testing

6 RESULT & DISSCUSION

6.1 TENSILE TESING

We have already discussed the importance of tensile testing in introduction section. In this test load is applied is along longitudinal axis & perpendicular to the cross section. We will now discuss the nature of failure in tensile testing. The nature of failure depends on type of material also. Ductile material is found to be strongest in tension loading & weakest in shear loading. Similarly, brittle material is strongest in compression & weakest in tension. The strength of brittle material in shear lies somewhere in between of tension & compression.

If a ductile material is loaded in tension then it will keep elongating. A point will come where fracture will take place & material will break in 2 separate part. The cup-cone failure phenomenon can be seen after fracture.

6.1.1  TOTAL DEFORMATION

Following figure shows the deformation specimen

The blue colour tag shows the minimum deformation in specimen that occurs on fixed side & the red colour tag shows the maximum deformation that occurs on side where displacement is given. The value of minimum & maximum deformation is 0mm & 18.003mm respectively.

Fracture of material has not been seen at this displacement of 18mm. For fracture, a minimum displacement of 45mm has to be given for this specimen.

At the same time, we can observe that the middle part of specimen is getting elongated due to applied tensile load & the section is getting thinner with respect of time. This phenomenon leads to necking (area of cross section is minimum). When fracture will occur, it will break this specimen into 2 separate parts forming cup-cone type structure.

6.1.2  EQUIVALENT STRESS

The blue colour tag shows the minimum stress in specimen that occurs on fixed side & the red colour tag shows the maximum stress that occurs on thinner section. The value of minimum & maximum stress is 3.3809e^-002 MPa & 609.13MPa respectively.

6.1.3  TEMPERATURE

The blue colour tag shows the minimum temperature in specimen that occurs on fixed side & the red colour tag shows the maximum temperature that occurs on middle of the section. The value of minimum & maximum temperature is 22 & 267.44 degree centigrade respectively.

6.2 TORSION TEST

In torsion test moment is applied along the longitudinal axis on the cross-section. This moment tries to twist the specimen. The maximum rotation of component has been found at free end & the minimum rotation is found on entire fixed part. We have already discussed the hierarchy of   strength of material in shear, tension & compression in section 6.1.

Now we will discuss mode of failure due to torsion test. If ductile material is loaded in torsion it will break very easily because it is weakest in shear. Fracture will occur along vertical plane or normal plane at an angle of 90 degree. If brittle material is loaded in shear it will fail along shear plane at an angle of 45 degree with vertical plane.

6.2.1  TOTAL DEFORMATION

Following figure shows the deformation specimen

The blue colour tag shows the minimum deformation in specimen that occurs on fixed side & the red colour tag shows the maximum deformation that occurs on fractured part of material that is no more part of specimen. The value of minimum & maximum deformation is 0mm & 19.39mm respectively.

Fracture of material has been seen at this displacement of 1200 degree of rotation. Fracture is happening at vertical place due to shear so we can conclude that this material is ductile material

6.2.2  EQUIVALENT STRESS

The blue colour tag shows the minimum stress in specimen that occurs on fixed side & the red colour tag shows the maximum stress that occurs on fracture plane. The value of minimum & maximum stress is 0.284 MPa & 565.99 MPa respectively.

6.2.3 TEMPERATURE

The blue colour tag shows the minimum temperature in specimen that occurs on fixed side & the red colour tag shows the maximum temperature that occurs on middle of the section. The value of minimum & maximum temperature is 22 & 466.42 degree centigrade respectively.

7 CONCLUSION

7.1  COMPARISION OF PARAMETER

Following table shows the comparison of different parameter. The comparison has been done for maximum value only. The minimum value for stress & deformation is 0(or tending to 0) in all cases & minimum temperature is 22 degree centigrade.

PARAMETER	DEFORMATION In mm	STRESS In MPa	TEMPERATURE In degree centigrade
TENSION	18.003	609.13	267.44
TORSION	19.39	565.99	466.42

7.2  FRACTURE MODE

The following figure shows the fracture surface in torsion test

It can be clearly seen fracture due to torsion is along normal plane or vertical plane & material get separated.  Following figure shows fracture due to tensile load

Tensile test has been performed again in order to see the fracture of surface. This time displacement load has been kept as 45mm. Cup-cone fracture surface can be seen clearly in tensile fracture.

TENSILE AS WELL AS TORSION TEST, BOTH INDICATES THAT MATERIAL IS DUCTILE.