Posted: November 15th, 2022

The tensile properties of mechanical materials

Date of Submission

Contents
Abstract 3
Introduction 3
Objectives of the experiment 5
Materials and methods 5
Results 9
Definitions 9
Material A 11
Material B 12
Material C 13
Discussion 17
Conclusion 17
References 18

Abstract
The experiment seeks to find out the tensile properties of mechanical materials. Some of the fundamental aspects of materials are expounded by determining its properties when subjected to various forces such as strain and stress. The properties of these materials are obtained by carrying out a testing process for elastic materials. In essence, laboratory experiments are normally conducted in a careful manner in order to determine the mechanical properties of different materials when they reach breakage limit. Behavior of various mechanical materials such as stress are obtained and a subsequent curves plotted. Primarily, the plotted curves will help to understand the stress-strain of the material in respect to yield and fracture aspects. Finally, fracture stress, ultimate tensile strength, ductility as well as work to fracture of the material will be obtained hence helping to draw conclusion based on the relationship developed between the Vickers hardness and yield strength. Resultantly, the information will help in Assessment of the fracture process.

Introduction
Contextually, engineering materials are technically defined based on plastic and tensile properties. These properties are unique and distinctive in every mechanical material hence used to specify its properties. Ideally, determination of any material involves a lab test process to specifically measure and identify the properties that defines them.
Profoundly, the experiment involves a lab test to enable better understanding of various materials based on their elastic properties. Evidently, properties of mechanical materials are identified by considering a number of fundamental aspects such as the Young’s modulus, elasticity, strain and the stress applied on the material. Ostensibly, the young modulus refers to a measure of toughness in plastic materials. It is used to determine elasticity of various materials such as wires when a force is exerted on it. Young’s modulus is defined by the following formula;
Young^’ smodulus (E)=(stress (F/A))/(strain(dL/L))
Tensile modulus is normally a ratio of stress of force exerted per unit area to the strain force along a given axis. It is used to obtain the elongation or compression present in a material if and only if the magnitude of the stress is lower in comparison to the yield strength of the object when at rest.
The strain is the deformation occurring on a solid due to stress force. It is normally expressed as;

ε=dL/L
Where;
ε-strain
L -the length of the object
On the other hand, stress is defined as the force that is exerted on the object per unit area. It is expressed as follows;
σ=F/A
Where;
σ – stress,
F -force
A –area
Conclusively, the yield strength property is used to define all mechanical materials. Yield strength refers to the amount of stress needed for an object to transform from elastic state deformation to plastic state deformation. Definitely, ultimate tensile strength is another important aspect in the study of material properties. Ideally, it refers to the allowable threshold limit stress to which the material will break due to exceeding the elastic limit resulting to the release of potential elastic energy.
Objectives of the experiment
The experiment is aimed at finding the ultimate tensile strength, fracture stress as well as the work to fracture.
To establish the correlation between Vickers hardness and yield strength.
To carry out experimental tensile tests for three materials; A, B, C and to determine their mechanical properties.
To find the Young’s Modulus (E), 0.1% proof stress and strain yield.

Materials and methods
Basically, the following outlined are the required materials;
Calipers
Software
Measuring gauge
Strain gauge
Clamp

Methods
The functioning of the tensile test software should be well understood after which it is turned on and the materials A, B, and C tested.

The diameter and gauge length of the specimen to be tested is then measured by use of calipers instrument. The thinnest measurement of the specimen was measured as illustrated on the in the figure below.

Figure 1: measurement of the gauge diameter

The specimen sample is then mounted on the instrument wherebythe collets are removed, first, the sample was then placed on the test machine. The collets around the specimen were then closed and a ring inserted around it.
The screw was then firmly tightened by use of Allen key which was provided.
The second ring was then placed around the specimen while unsecured. The cross head is then brought by use of black switch to enable the specimen to fit in between the collets. The top collets around the specimen was then closed while a second ring is placed around the collets. the Allen key is used to tighten the screw of the ring
Figure 2: placed specimen on a ring

The load output reading is assessed by keeping a close eye, a load of about 100N is placed on the sample and the gauge length reset.
Calibration of the gauge length is initiated. It was done by clicking on strain 1 settings on the top of the right page. The strain gauge was picked gently. The two sides of the cup-and –cones are gently pressed in order to close the gauge in that position. The procedure sets the gauge strain to 0% to allow the calibration.
Figure 3: calibrating the gauge length

“Calibrate” is clicked on the screen. This was to start the calibration process. The procedure takes the less time. The strain page 1 is then shut.
The sample is placed on the strain gauge. The procedure is done by holding the strain gauge in closed position and the cup-and-cone pressed together in a gently manner as shown in the figure below.
Figure 4: Placed strain gauge around the sample

The door of the machine is then closed. The demonstrator is then used to check the instrument. The demonstrator permits the procedure of starting the test. The layout was described as follows;
Specified load of specimen was loaded and the machine automatically ceases. 30kN for specimen A, 3.5kN for B and 3kN for C. The strain gauge is then removed after five minutes. The strain gauge was removed from the specimen and placed on the machine side, the door was closed and “End Hold” clicked. When the machine stops, the door was opened and diameter of the thinnest region measured. After completion of measurements of each specimen, the door was closed “End Hold” clicked again.

The specimen is then loaded at intervals of extension and will stop regularly. When the machine stops, open the door, measure the diameter of the specimen at the thinnest region and input this value into the b. b. the machine will load the specimen until failure after all the measurements have been taken.
The machine stops automatically after the specimen reaches failure and at this point the specimen was removed from the machine. Calipers were then used to measure the final length and diameter of all the specimens at necking region.

When the procedure is done, next option on the screen is clicked. A directive of the final length aad diameter of the specimen will be required. After entering the values, it is then completed by click of next again.

Figure 5: measurement of the final length and diameter

The finish option is clicked to complete the task thus ending the test. The data is then saved.
The procedure for other two remaining specimens was repeated.

Results
Definitions
Engineering stress is the force exerted divided by the original cross-sectional area of the material.
Engineering stress(Pa)=(Load(kN))/(original area(M2))
Engineering strain- it is the threshold limit of a material which results to deformation when exceeded.
Engineering strain=(L-Lo)/Lo
True stress- load per unit cross sectional area of the specimen load.

Stress=Force/Area
True strain- is the deformation occurring on a solid due to stress force.
strain ε=in(dL/L )

Young’s modulus
Young^’ smodulus (E)=(stress (F/A))/(strain(dL/L))
Vicker hardness-

Hv=3xσyield/g

Yield strength- the value of stress at yield point, usually plotted at specific young’s modulus percent of offset (offset=0.1%)
Work to fracture- it is the area in the force against deformation curve.
Ductility- ability of a material to completely deform. It is also engineering strain at failure.
Elongation- this is a change in gauge length based on the original gauge length.

Material A
Calculations

Original Area= (D/2)2π= (4×10-3)2 xπ=5.026×10-5
Area after deformation= (3.4×10-3) xπ=3.631×10-5
Engineering stress;
σ=(Load(kN))/(area(M2))=30000/0.00005026 596.9Mpa
Engineering strain
62.49-58.25= 4.24 mm therefore stress = 4.24/58.25=0.07279

True stress
True stress(Pa)=(Load(kN))/(area (m□(2)))=30000/0.00003631 826.2Mpa

True strain
strain ε=in(dL/L)=in(4.24/58.25)=0.3176 compressive

Young’s modulus
Young modulus=stress/strain=596.9Mpa/0.3176=1879Mpa
Vicker hardness
Hv=3xσyield/g=(3x 596.9MPa))/9.81=179Mpa
Work to fracture
εmax=σL/Lo=596.9Mpa/58.25=10247Mpa
Ductility
εf=(lf-Lo)/Lo X100=(62.49-58.25)/58.25=7.278%
Where Lo is the gauge length
Elongation
Lf-Lo=62.49-58.25=4.24mm
Material B

Original Area= (D/2)2π= (4×10-3)2 xπ=5.026×10-5
Area after stress= (3.4×10-3) xπ=1.56×10-5
Engineering stress;
σ=(Load(kN))/(area(M2))=3500/0.00005026=69.6Mpa
Engineering strain
68.55-57.56= 10.99 mm therefore stress = 10.99/57.56=0.191

True stress
True stress(Pa)=(Load(kN))/(area (m□(2)))=3500/0.0000156=224.4Mpa

True strain
strain ε=in(dL/L)=in(10.99/57.56)=1.656 compressive

Young’s modulus
Young modulus=stress/strain=224.4Mpa/1.656=135.5Mpa
Vicker hardness
Hv=3xσyield/g=(3x 224.4MPa))/9.81=68.6Mpa
Work to fracture
εmax=σL/Lo=224.4MpaX68.55/57.56=267.24Mpa
Ductility
εf=(lf-Lo)/Lo X 100%=(68.55-57.56)/57.56=19.1%
Where Lo is the gauge length
Elongation
Lf-Lo=68.55-57.56=10.99mm

Material C

Original Area= (D/2)2π= (4×10-3)2 xπ=5.026×10-5
Area after stress= (3.0×10-3) xπ=2.87 x10-5
Engineering stress;
σ=(Load(kN))/(area(M2))=3000/0.00005026 59.7Mpa
Engineering strain
60-58.28= 1.72 mm therefore stress =1.72/58.28=0.0295

True stress
Force for material A was 30kN. Area = 0.002011M2
Stress=30000/0.0020011=149Mpa.
True stress(Pa)=(Load(kN))/(area (m□(2)))=3000/0.0000287 104.5MPa

True strain
strain ε=in(dL/L)=in(1.72/58.28)=3.5229compressive

Young’s modulus
Young modulus=stress/strain=104.5Mpa/3.5229=29.7Mpa
Vicker hardness
Hv=3xσyield/g=(3x 104.5MPa))/9.81=31.96Mpa
Work to fracture
εmax=σL/Lo=(104.5Mpa X 60)/58.25=107.6MPa
Ductility
εf=(lf-Lo)/Lo X 100%=(60-58.28)/58.28=0.02951=2.95%
Where Lo is the gauge length
Elongation
Lf-Lo=60-58.28=1.72mm

Figure 6: Graph of load versus extension for specimen for 1

Figure 7: Graph of load versus extension for specimen 2

Figure 8: Graph of load versus extension for specimen for material A and C

Discussion
The young’s modulus for materials A, B and C were obtained as 249.9, 35.16 and 29.15 in that order. The ultimate strength was obtained to be 8000 from the graph.
The toughness of the sample specimen was computed from the area under the curve. Various length elongation were noted from the material specimens. A material which was subjected to a great change was that of 10.99mm length while the other experience a significant change.
Typically, graph shown on figure 6 indicates a true relationship of stress-strain engineering curve for the material A. The young’s modulus was obtained from the slope of the straight line of the curve. Ideally, the area shown under the curve is the resilience whereas toughness is the total area under the curve.
Deformation occurs when the material is subjected to a great load values. Material A started to deform when a load of 37kN was subjected to it. However, determination of yield point was hard to define.
Materials A, B and C were basically identified by use of the previous calculated properties. CES Edupack 2010 was used to identify these materials. Mechanical material properties such as hardness, elongation percentage, young’s modulus as well as the ultimate strength values were placed into CES. Based on the calculated moduli of material A, B and C being 249.9, 35.16 and 29.15 respectively, material A could be Aluminum, B steel while C could be polymer.
Conclusively, hardness is a fundamental property to identify engineering materials. Strong bonding in a material results to higher Vickers hardness. Therefore, Higher Vickers hardness lead to a higher young modulus of a material.

Conclusion
Remarkably, the laboratory test experiment was a successful process for determining the various properties of mechanical materials. Several curves of stress against strain were plotted. In addition, different young’s modulus, stress and strain of various materials were determined. The values obtained from the experiment is somehow different from the exact values given from the handout. This is of course the experiment may have been subjected to different kinds of errors thus reducing the probability of accuracy.

References
Roylance, D. (2008), Mechanical properties of materials

Olsen, T (2010), Interpretation of Stress-Strain Curves and Mechanical Properties of Materials. Testing machine co.

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