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Tensile Test to Determine Material

Paper Type: Free Essay Subject: Engineering
Wordcount: 5063 words Published: 8th Feb 2020

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Aim & Objectives

Aim:

-          To determine the material of the different specimens by seeing how they respond to stress.

Objectives:

-          To determine the specimen’s modulus of elasticity, Ultimate Tensile Stress (UTS), Percentage of Elongation, Percentage of Reduction in Area, Stress, Strain,

 

Introduction

The usage of a material depends on its strengths and its properties. The Tensile Test can determine the mechanical properties of a sample such as Ultimate Tensile Strength (UTS), Percentage of Elongation, Percentage of Reduction in Area, Yield Point, and Fracture Point (Necking). These properties can then be used to determine the Stress Value, Strain Value, and Youngs Modulus. These properties are important as it can determine whether a material is brittle or ductile.

 

Stress= ForceArea      σ=FA

Where:

σ = Stress, in Newton/millimetre2 (Nmm-2) or converted into MNm2

F = Force, in Newtons (N)

A = Cross-sectional Area, in mm2

Strain= Change in lengthOriginal length      ε=LL0 

Where:

ε = Strain, no units

∆L = Extension or change in length, in millimetres (mm)

L0 = The original length, in millimetres (mm)

Youngs Modulus= StressStrain    E= σ ε 

Where:

E = Youngs Modulus, in Pascal (Newton metre2, Nm2)

σ = Stress, in Newton/millimetre2 (Nmm-2)  

ε = Strain, no units

Percentage of Elongation=LfLiLi×100%

Where:

Percentage of Elongation, no units

Li = Initial Gauge Length, mm

Lf = Final Gauge Length, mm

Percentage of Reduction in Area= AiAfAi×100%

 

Where:

Percentage of Reduction in Area, no units

Ai = Initial Area (mm2)

Ultimate Tensile Strength= Maximum LoadOriginal Cross Sectional Area=FA

Where:

UTS = Ultimate Tensile Strength, in Newton/millimetre2 (Nmm-2) or converted into MNm2

F = Force, in Newtons (N)

A = Cross-sectional Area, in mm2

Density= MassVolume p=mV

Where:

P = Density, in kilograms/Volume (kg.m-3)

m = Mass, in kg

V = Volume, in m3

 

Theory – Tensile Test of Materials

The Tensile Test works by having a hydraulic testing machine which applies a controlled load and displacement on the sample.

Method (Equipment & Procedures) & Materials

Equipment:

– Extensometer – A device which clamps onto the test piece at two locations. As the specimen extends, the extensometer also extends, and it measures the change in distance.

– Instron 3382 – A tensile testing machine which applies tension or compression for tests up to 100kN and is used with the software “BlueHill 3”. Used for the flat specimen

– Instron 1342 – Similar to the Instron 3382 except it’s used for circular specimens (specimens A, B, C)

– Safety Goggles

Material:

– A flat specimen with an initial length of 82.1mm and an initial thickness of 3.13mm.

– Circular specimens A, B and C with initial lengths of 87.49mm and initial diameters of 13.92mm. Has a gauge length of 50mm.

Procedure:

  • The specimen should have measurements and be prepared according to Data Sheet S191 [4].
  • Before placing the specimen into the machine, the following data should be recorded:

– the diameter of the gauge

– the length of the gauge

  • The specimen is placed in the tensile testing machine and should be clamped.
  • A small load is applied to ensure that the specimen is held in place.
  • The extensometer is fitted to where the marks/lines are according to Data Sheet S191 [4].
  • The machine is started but before the material fractures, the machine should be stopped to remove the extensometer.
  • The machine is then started again until fracture occurs which signals the stop of the procedure.
  • The specimen should be removed from the machine and the following data should be recorded:

– the diameter at the neck/fracture point

– Put the broken specimen together and measure the total length of the gauge

  • Test is repeated for other specimens A, B and C.

The data is sent to a software called “BlueHill 3” and the data is for the Load (kN) and Extension (mm). This data can be exported as an excel file where the stress and strain values can be calculated.

 

 

 

 

 

Results

Specimen A

Figure 1 – Load and Extension data of specimen A

 

Figure 2 – Stress and Strain graph from calculations from Figure 1.
 

UTS

Necking

Fracture

Plastic Region

         

Line Parallel to straight part of the graph.

Yield Point

Figure 3 – Stress Strain graph showing yield point
 

Elastic Region

Data:

Initial Length (mm)

87.49

Final Length (mm)

101.24

Initial Diameter (mm)

13.92

Final Diameter (mm)

10.44

Original Cross-Sectional Area (mm2) = 152.18

Final Cross-Sectional Area (mm2) = 85.6

Grip Diameter (m)

0.01912

Grip length (m)

0.0481

Mass (kg)

0.0369

 

Stress at Yield Point = 0.25

Strain at Yield Point = 0.0037

Maximum Load = 48.458752

 

Calculations

Youngs Modulus= StressStrain=0.250.0037=67.57

kNmm2 = 67.57GPa (GNm2)

 

Ultimate Tensile Strength= Maximum LoadOriginal Cross Sectional=48.458752152.18=0.318

kNmm2 = 318MPa

 

Percentage of Elongation=LfLiLi×100%= 101.2487.4987.49×100%=15.72%

 

Percentage of Reduction in Area= AiAfAi×100%= 152.1885.6152.18×100%=43.75%

Density= MassVolume=0.0369π0.0095620.0481=2671.9

kg/m3

 

 

Specimen B

Figure 4 – Load and Extension data of specimen B

Figure 5 – Parallel line which crosses extension axis at 0.1% of gauge length (50mm),0.1%= 0.05.

Yield Point

Elastic Region

 

 

 

0.05

UTS

Figure 4 – Load and Extension data of specimen B

Fracture and neck

Plastic Region

Data:

Initial Length (mm)

87.49

Final Length (mm)

87.94

Initial Diameter (mm)

13.92

Final Diameter (mm)

13.87

Original Cross-Sectional Area (mm2) = 152.18

Final Cross-Sectional Area (mm2) = 151.09

Grip Diameter (mm)

0.01912

Grip length (mm)

0.0481

Mass (kg)

0.0971

 

Load at Yield Point = 28.9kN   0.1% Proof Stress = 0.19

Extension at Yield Point = 0.47mm Strain at Yield Point = 0.0094

Maximum Load = 42.830109

 

Calculations

Youngs Modulus= StressStrain=0.190.0094=20.2

kNmm2 = 20.2GPa (GNm2)

 

Ultimate Tensile Strength= Maximum LoadOriginal Cross Sectional=42.830109152.18=0.281

kNmm2 = 281MPa

 

Percentage of Elongation=LfLiLi×100%= 87.9487.4987.49×100%=0.51%

 

Percentage of Reduction in Area= AiAfAi×100%= 152.18151.09152.18×100%=0.72%

Density= MassVolume=0.0971π0.0095620.0481=7030.86

kg/m3

 

Specimen C

UTS

Necking

Fracture

Plastic Region

 

Data:

Initial Length (mm)

87.49

Final Length (mm)

100.38

Initial Diameter (mm)

13.92

Final Diameter (mm)

10.2

Original Cross-Sectional Area (mm2) = 152.18

Final Cross-Sectional Area (mm2) = 81.71

Grip Diameter (mm)

0.01912

Grip length (mm)

0.0481

Mass (kg)

0.1046

 

Stress at Yield Point = 0.42

Strain at Yield Point = 0.0021

Maximum Load = 81.377167

 

Calculations

Youngs Modulus= StressStrain=0.420.0021=200

kNmm2 = 200GPa (GNm2)

 

Ultimate Tensile Strength= Maximum LoadOriginal Cross Sectional=81.377167152.18=0.535

kNmm2 = 535MPa

 

Percentage of Elongation=LfLiLi×100%= 100.3887.4987.49×100%=14.7%

 

Percentage of Reduction in Area= AiAfAi×100%= 152.1881.71152.18×100%=46.3%

Density= MassVolume=0.1046π0.0095620.0481=7574

kg/m3

 

Discussion

Specimen’s A and C are ductile materials as they have a plastic region on the stress strain graphs meaning that they will have permanent deform after the elastic region and extend in length. After reaching the UTS, the specimen begins to neck and then fractures.

Specimen B is a brittle material as it has no plastic region. When the UTS is reached, the specimen fractures immediately resulting in a minimal extension and minimal neck.

For Specimen A, the Young’s Modulus was calculated to be 67.57GPa. This suggests that the material is aluminium as the Young’s modulus for aluminium is 69GPa. This is further reinforced by the fact that the density was calculated to be 2671.9

kg/m3 and the density for aluminium is 2600-2720kg/m3 suggesting that it may be an aluminium alloy and not pure aluminium. [1][2]

For Specimen B, the Young’s Modulus was calculated to be 20.2GPa and the density was calculated to be 7030.86

kg/m3. It’s unsure which material it is but neodymium is the closest as it has a Young’s Modulus of 37.5 and a density of 7010kg/m3. [1][3] However, there are metals such as Zinc and Tin which are similar.

For Specimen C, the Young’s Modulus was calculated to be 200GPa and this suggests that the metal is steel or a variant of steel. The Density also matches steel as it was calculated to be 7574

kg/m3 and the density of steel is about 7700kg/m3. Iron also has very similar numbers however they are slightly higher than Steel.[1][2][5]

The results are valid however the results of Specimen B seem abnormal.

This may be due to parallel line of the 0.1% strain being done incorrectly and not being completely parallel.

The Load Extension graphs appear to begin from the origin point of 0 however due the tensile testing machine clamping onto the specimens, force must be exerted so there will be a minimal amount of force before the test begins.

The Software “Bluehill 3” exports the data from the tensile testing machine into Load and Extension. There are hundreds of values meaning that the use of a computer to convert the load extension values into stress strain values are required. However, if there are incorrect areas, we wouldn’t know, and it’d be near impossible to hand check every value.

Conclusion

Overall from our calculations and comparisons to other sources, we can conclude that the material of Specimen A is aluminium, Specimen B is Neodymium, and Specimen C is Steel.

 

Appendices

  1. https://www.engineersedge.com/materials/densities_of_metals_and_elements_table_13976.htm
  2. https://www.engineeringtoolbox.com/young-modulus-d_417.html 
  3. https://www.azom.com/properties.aspx?ArticleID=1595
  4. http://campusmoodle.rgu.ac.uk/pluginfile.php/4371395/mod_resource/content/1/STATICS-Lab_Sheets-Tensile_Test.pdf
  5. https://hypertextbook.com/facts/2004/KarenSutherland.shtml

 

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