Date of Submission<\/p>\n
Contents
\nAbstract\t3
\nIntroduction\t3
\nObjectives of the experiment\t5
\nMaterials and methods\t5
\nResults\t9
\nDefinitions\t9
\nMaterial A\t11
\nMaterial B\t12
\nMaterial C\t13
\nDiscussion\t17
\nConclusion\t17
\nReferences\t18<\/p>\n
Abstract
\nThe 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.<\/p>\n
Introduction
\nContextually, 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.
\nProfoundly, 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\u2019s 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\u2019s modulus is defined by the following formula;
\nYoung^’ smodulus (E)=(stress (F\/A))\/(strain(dL\/L))
\nTensile 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.
\nThe strain is the deformation occurring on a solid due to stress force. It is normally expressed as;<\/p>\n
\u03b5=dL\/L
\nWhere;
\n\u03b5-strain
\nL -the length of the object
\nOn the other hand, stress is defined as the force that is exerted on the object per unit area. It is expressed as follows;
\n\u03c3=F\/A
\nWhere;
\n\u03c3 – stress,
\nF -force
\nA \u2013area
\nConclusively, 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.
\nObjectives of the experiment
\nThe experiment is aimed at finding the ultimate tensile strength, fracture stress as well as the work to fracture.
\nTo establish the correlation between Vickers hardness and yield strength.
\nTo carry out experimental tensile tests for three materials; A, B, C and to determine their mechanical properties.
\nTo find the Young\u2019s Modulus (E), 0.1% proof stress and strain yield.<\/p>\n
Materials and methods
\nBasically, the following outlined are the required materials;
\nCalipers
\nSoftware
\nMeasuring gauge
\nStrain gauge
\nClamp<\/p>\n
Methods
\nThe functioning of the tensile test software should be well understood after which it is turned on and the materials A, B, and C tested.<\/p>\n
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.<\/p>\n
Figure 1: measurement of the gauge diameter<\/p>\n
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.
\nThe screw was then firmly tightened by use of Allen key which was provided.
\nThe 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
\nFigure 2: placed specimen on a ring<\/p>\n
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.
\nCalibration 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 \u2013cones are gently pressed in order to close the gauge in that position. The procedure sets the gauge strain to 0% to allow the calibration.
\nFigure 3: calibrating the gauge length<\/p>\n
\u201cCalibrate\u201d 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.
\nThe 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.
\nFigure 4: Placed strain gauge around the sample<\/p>\n
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;
\nSpecified 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 \u201cEnd Hold\u201d 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 \u201cEnd Hold\u201d clicked again.<\/p>\n
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.
\nThe 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.<\/p>\n
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.<\/p>\n
Figure 5: measurement of the final length and diameter<\/p>\n
The finish option is clicked to complete the task thus ending the test. The data is then saved.
\nThe procedure for other two remaining specimens was repeated.<\/p>\n
Results
\nDefinitions
\nEngineering stress is the force exerted divided by the original cross-sectional area of the material.
\nEngineering stress(Pa)=(Load(kN))\/(original area(M2))
\nEngineering strain- it is the threshold limit of a material which results to deformation when exceeded.
\nEngineering strain=(L-Lo)\/Lo
\nTrue stress- load per unit cross sectional area of the specimen load.<\/p>\n
Stress=Force\/Area
\nTrue strain- is the deformation occurring on a solid due to stress force.
\nstrain \u03b5=in(dL\/L )<\/p>\n
Young\u2019s modulus
\nYoung^’ smodulus (E)=(stress (F\/A))\/(strain(dL\/L))
\nVicker hardness-<\/p>\n
Hv=3x\u03c3yield\/g<\/p>\n
Yield strength- the value of stress at yield point, usually plotted at specific young\u2019s modulus percent of offset (offset=0.1%)
\nWork to fracture- it is the area in the force against deformation curve.
\nDuctility- ability of a material to completely deform. It is also engineering strain at failure.
\nElongation- this is a change in gauge length based on the original gauge length.<\/p>\n
Material A
\nCalculations<\/p>\n
Original Area= (D\/2)2\u03c0= (4×10-3)2 x\u03c0=5.026×10-5
\nArea after deformation= (3.4×10-3) x\u03c0=3.631×10-5
\nEngineering stress;
\n\u03c3=(Load(kN))\/(area(M2))=30000\/0.00005026 596.9Mpa
\nEngineering strain
\n62.49-58.25= 4.24 mm therefore stress = 4.24\/58.25=0.07279<\/p>\n
True stress
\nTrue stress(Pa)=(Load(kN))\/(area (m\u25a1(2)))=30000\/0.00003631 826.2Mpa<\/p>\n
True strain
\nstrain \u03b5=in(dL\/L)=in(4.24\/58.25)=0.3176 compressive<\/p>\n
Young\u2019s modulus
\nYoung modulus=stress\/strain=596.9Mpa\/0.3176=1879Mpa
\nVicker hardness
\nHv=3x\u03c3yield\/g=(3x 596.9MPa))\/9.81=179Mpa
\nWork to fracture
\n\u03b5max=\u03c3L\/Lo=596.9Mpa\/58.25=10247Mpa
\nDuctility
\n\u03b5f=(lf-Lo)\/Lo X100=(62.49-58.25)\/58.25=7.278%
\nWhere Lo is the gauge length
\nElongation
\nLf-Lo=62.49-58.25=4.24mm
\nMaterial B<\/p>\n
Original Area= (D\/2)2\u03c0= (4×10-3)2 x\u03c0=5.026×10-5
\nArea after stress= (3.4×10-3) x\u03c0=1.56×10-5
\nEngineering stress;
\n\u03c3=(Load(kN))\/(area(M2))=3500\/0.00005026=69.6Mpa
\nEngineering strain
\n68.55-57.56= 10.99 mm therefore stress = 10.99\/57.56=0.191<\/p>\n
True stress
\nTrue stress(Pa)=(Load(kN))\/(area (m\u25a1(2)))=3500\/0.0000156=224.4Mpa<\/p>\n
True strain
\nstrain \u03b5=in(dL\/L)=in(10.99\/57.56)=1.656 compressive<\/p>\n
Young\u2019s modulus
\nYoung modulus=stress\/strain=224.4Mpa\/1.656=135.5Mpa
\nVicker hardness
\nHv=3x\u03c3yield\/g=(3x 224.4MPa))\/9.81=68.6Mpa
\nWork to fracture
\n\u03b5max=\u03c3L\/Lo=224.4MpaX68.55\/57.56=267.24Mpa
\nDuctility
\n\u03b5f=(lf-Lo)\/Lo X 100%=(68.55-57.56)\/57.56=19.1%
\nWhere Lo is the gauge length
\nElongation
\nLf-Lo=68.55-57.56=10.99mm<\/p>\n
Material C<\/p>\n
Original Area= (D\/2)2\u03c0= (4×10-3)2 x\u03c0=5.026×10-5
\nArea after stress= (3.0×10-3) x\u03c0=2.87 x10-5
\nEngineering stress;
\n\u03c3=(Load(kN))\/(area(M2))=3000\/0.00005026 59.7Mpa
\nEngineering strain
\n60-58.28= 1.72 mm therefore stress =1.72\/58.28=0.0295<\/p>\n
True stress
\nForce for material A was 30kN. Area = 0.002011M2
\nStress=30000\/0.0020011=149Mpa.
\nTrue stress(Pa)=(Load(kN))\/(area (m\u25a1(2)))=3000\/0.0000287 104.5MPa<\/p>\n
True strain
\nstrain \u03b5=in(dL\/L)=in(1.72\/58.28)=3.5229compressive<\/p>\n
Young\u2019s modulus
\nYoung modulus=stress\/strain=104.5Mpa\/3.5229=29.7Mpa
\nVicker hardness
\nHv=3x\u03c3yield\/g=(3x 104.5MPa))\/9.81=31.96Mpa
\nWork to fracture
\n\u03b5max=\u03c3L\/Lo=(104.5Mpa X 60)\/58.25=107.6MPa
\nDuctility
\n\u03b5f=(lf-Lo)\/Lo X 100%=(60-58.28)\/58.28=0.02951=2.95%
\nWhere Lo is the gauge length
\nElongation
\nLf-Lo=60-58.28=1.72mm<\/p>\n
Figure 6: Graph of load versus extension for specimen for 1<\/p>\n
Figure 7: Graph of load versus extension for specimen 2<\/p>\n
Figure 8: Graph of load versus extension for specimen for material A and C<\/p>\n
Discussion
\nThe young\u2019s 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.
\nThe 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.
\nTypically, graph shown on figure 6 indicates a true relationship of stress-strain engineering curve for the material A. The young\u2019s 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.
\nDeformation 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.
\nMaterials 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\u2019s 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.
\nConclusively, 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.<\/p>\n
Conclusion
\nRemarkably, 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\u2019s 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.<\/p>\n
References
\nRoylance, D. (2008), Mechanical properties of materials<\/p>\n
Olsen, T (2010), Interpretation of Stress-Strain Curves and Mechanical Properties of Materials. Testing machine co.<\/p>\n","protected":false},"excerpt":{"rendered":"
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 […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8197],"tags":[8278],"class_list":["post-89173","post","type-post","status-publish","format-standard","hentry","category-assessment-assignment-help-uk","tag-the-tensile-properties-of-mechanical-materials"],"_links":{"self":[{"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/posts\/89173","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/comments?post=89173"}],"version-history":[{"count":0,"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/posts\/89173\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/media?parent=89173"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/categories?post=89173"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.colapapers.com\/assessments\/wp-json\/wp\/v2\/tags?post=89173"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}