Tensile Test to Determine Material
✅ Paper Type: Free Essay  ✅ Subject: Engineering 
✅ Wordcount: 5063 words  ✅ Published: 8th Feb 2020 
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.
$\mathit{Stress}=\frac{\mathit{Force}}{\mathit{Area}}\mathit{}\textcolor[rgb]{}{\sigma}=\frac{F}{A}$
Where:
σ = Stress, in Newton/millimetre^{2 }(Nmm^{2}) or converted into MNm^{2}
F = Force, in Newtons (N)
A = Crosssectional Area, in mm^{2 }
$\textcolor[rgb]{}{\mathit{Strain}}\textcolor[rgb]{}{=}\textcolor[rgb]{}{}\frac{\textcolor[rgb]{}{\mathit{Change\; in\; length}}}{\textcolor[rgb]{}{\mathit{Original\; length}}}\textcolor[rgb]{}{\mathit{\epsilon}}\textcolor[rgb]{}{=}\frac{\textcolor[rgb]{}{\u2206}\textcolor[rgb]{}{L}}{{\textcolor[rgb]{}{L}}_{\textcolor[rgb]{}{0}}}\textcolor[rgb]{}{}$
Where:
ε = Strain, no units
∆L = Extension or change in length, in millimetres (mm)
L_{0} = The original length, in millimetres (mm)
$\mathit{Youngs\; Modulus}=\frac{\mathit{Stress}}{\mathit{Strain}}\mathit{E}=\mathrm{}\frac{\textcolor[rgb]{}{\sigma}}{\textcolor[rgb]{}{\epsilon}}$
Where:
E = Youngs Modulus, in Pascal (Newton metre^{2}, Nm^{2})
σ = Stress, in Newton/millimetre^{2 }(Nmm^{2})
ε = Strain, no units
$\mathit{Percentage\; of\; Elongation}=\frac{{L}_{f}\u2013{L}_{i}}{{L}_{i}}\times 100\%$
Where:
Percentage of Elongation, no units
L_{i} = Initial Gauge Length, mm
L_{f} = Final Gauge Length, mm
$\mathit{Percentage\; of\; Reduction\; in\; Area}=\frac{{A}_{i}\u2013{A}_{f}}{{A}_{i}}\times 100\%$
Where:
Percentage of Reduction in Area, no units
A_{i} = Initial Area (mm^{2})
$\mathit{Ultimate\; Tensile\; Strength}=\frac{\mathit{Maximum\; Load}}{\mathit{Original\; Cross\; Sectional\; Area}}=\frac{F}{A}$
Where:
UTS = Ultimate Tensile Strength, in Newton/millimetre^{2 }(Nmm^{2}) or converted into MNm^{2}
F = Force, in Newtons (N)
A = Crosssectional Area, in mm^{2 }
$\mathit{Density}=\frac{\mathit{Mass}}{\mathit{Volume}}\mathit{p}=\frac{m}{V}$
Where:
P = Density, in kilograms/Volume (kg.m^{3})
m = Mass, in kg
V = Volume, in m^{3}
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 CrossSectional Area (mm^{2}) = 152.18 
Final CrossSectional Area (mm^{2}) = 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
$\mathit{Youngs\; Modulus}=\frac{\mathit{Stress}}{\mathit{Strain}}=\frac{0.25}{0.0037}=67.57$
kNmm^{2} = 67.57GPa (GNm^{2})
$\mathit{Ultimate\; Tensile\; Strength}=\frac{\mathit{Maximum\; Load}}{\mathit{Original\; Cross\; Sectional}}=\frac{48.458752}{152.18}=0.318$
kNmm^{2} = 318MPa
$\mathit{Percentage\; of\; Elongation}=\frac{{L}_{f}\u2013{L}_{i}}{{L}_{i}}\times 100\%=\frac{101.24\u201387.49}{87.49}\times 100\%=15.72\%$
$\mathit{Percentage\; of\; Reduction\; in\; Area}=\frac{{A}_{i}\u2013{A}_{f}}{{A}_{i}}\times 100\%=\frac{152.18\u201385.6}{152.18}\times 100\%=43.75\%$
$\mathit{Density}=\frac{\mathit{Mass}}{\mathit{Volume}}=\frac{0.0369}{\pi {\left(0.00956\right)}^{2}0.0481}=2671.9$
kg/m^{3}
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 CrossSectional Area (mm^{2}) = 152.18 
Final CrossSectional Area (mm^{2}) = 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
$\mathit{Youngs\; Modulus}=\frac{\mathit{Stress}}{\mathit{Strain}}=\frac{0.19}{0.0094}=20.2$
kNmm^{2} = 20.2GPa (GNm^{2})
$\mathit{Ultimate\; Tensile\; Strength}=\frac{\mathit{Maximum\; Load}}{\mathit{Original\; Cross\; Sectional}}=\frac{42.830109}{152.18}=0.281$
kNmm^{2} = 281MPa
$\mathit{Percentage\; of\; Elongation}=\frac{{L}_{f}\u2013{L}_{i}}{{L}_{i}}\times 100\%=\frac{87.94\u201387.49}{87.49}\times 100\%=0.51\%$
$\mathit{Percentage\; of\; Reduction\; in\; Area}=\frac{{A}_{i}\u2013{A}_{f}}{{A}_{i}}\times 100\%=\frac{152.18\u2013151.09}{152.18}\times 100\%=0.72\%$
$\mathit{Density}=\frac{\mathit{Mass}}{\mathit{Volume}}=\frac{0.0971}{\pi {\left(0.00956\right)}^{2}0.0481}=7030.86$
kg/m^{3}
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 CrossSectional Area (mm^{2}) = 152.18 
Final CrossSectional Area (mm^{2}) = 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
$\mathit{Youngs\; Modulus}=\frac{\mathit{Stress}}{\mathit{Strain}}=\frac{0.42}{0.0021}=200$
kNmm^{2} = 200GPa (GNm^{2})
$\mathit{Ultimate\; Tensile\; Strength}=\frac{\mathit{Maximum\; Load}}{\mathit{Original\; Cross\; Sectional}}=\frac{81.377167}{152.18}=0.535$
kNmm^{2} = 535MPa
$\mathit{Percentage\; of\; Elongation}=\frac{{L}_{f}\u2013{L}_{i}}{{L}_{i}}\times 100\%=\frac{100.38\u201387.49}{87.49}\times 100\%=14.7\%$
$\mathit{Percentage\; of\; Reduction\; in\; Area}=\frac{{A}_{i}\u2013{A}_{f}}{{A}_{i}}\times 100\%=\frac{152.18\u201381.71}{152.18}\times 100\%=46.3\%$
$\mathit{Density}=\frac{\mathit{Mass}}{\mathit{Volume}}=\frac{0.1046}{\pi {\left(0.00956\right)}^{2}0.0481}=7574$
kg/m^{3}
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/m^{3 }and the density for aluminium is 26002720kg/m^{3 }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/m^{3}. 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/m^{3}. [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/m^{3 }and the density of steel is about 7700kg/m^{3}. 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
 https://www.engineersedge.com/materials/densities_of_metals_and_elements_table_13976.htm
 https://www.engineeringtoolbox.com/youngmodulusd_417.html
 https://www.azom.com/properties.aspx?ArticleID=1595
 http://campusmoodle.rgu.ac.uk/pluginfile.php/4371395/mod_resource/content/1/STATICSLab_SheetsTensile_Test.pdf
 https://hypertextbook.com/facts/2004/KarenSutherland.shtml
Cite This Work
To export a reference to this article please select a referencing stye below:
Related Services
View allDMCA / Removal Request
If you are the original writer of this essay and no longer wish to have your work published on UKEssays.com then please: