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MT 2 Tensile Hardness Testing of MetalsName: SHI Tai Student Number: 528882271. IntroductionPurposes: (1). To master the operation of tensile and hardness testing on different materials.(2) . Knowing the basic tensile properties and hardness of a certain metal.(3) . To illustrate the strain-stress curve of a metal and understand the phenomenon of yield.Instruments: Tensile testing instrument (Zwick Z030)Hardness testing machineVernier caliperMicrometerAluminium tensile sampleCopper tensile sampleSteel tensile sampleBackground: Tensile properties, describes how the materials react to tension, represents the maximun stress of a material being stretched or pulled before cracking. Totally, it is a important reference factor when choose a material for the ability to bear high loads with no fraction or unrecoverable deformation. So a tensile test becomes necessary to measure the effects of forces putting on the material. In generally, tensile tests are used to illustrate the modulus of elasticity, elastic limit, yield point, maximum load, proportional limit, tensile strength, yield strength, reduction in area and many other properties.It is possible to obtain a stress-strain curve of a material using the tensile testing instrument, the sample is then put on an increasing tensile and the length increases slightly, all these data are recorded by a computer software. After the whole operation, we get a curve shown in Fig.1, which indicates some basic properties of the sample.Fig.1 A typical tensile stress-strain curveIt is clear to find the relationship between load and extensile from figure 1. There are total six points in the graph, which represent A, B, C, D, E and F. It is worth to point out that the sample indicates a direct proportional relationship between the applied load and extensile at the beginning of the testing. Point A stands for the elastic limit of the sample, in which the material exerts the linear strain-stress behavior.1 The whole process should obey Hookes law as the stress is directly proportional to strain and the linearity region is called elastic region. During this region, if the load is removed from the sample, the material will recover to its original shape. The slop of this line in the region is called modulus of elasticity or Youngs modulus,2 the modulus of elasticity (E) is described as the ratio of the elastic stress and the elastic strain of the sample, as shown in the formular below: Where: F = the load in one point of elastic stagel = the corresponding extension degreel0 = the initial length of sampleS0 = the original cross-sectional areaAs can be found the F/l represents the slope of the straight curve, thus we have Point B, the yield point, suffers from a sudden increase without the corresponding increase in stress. The curve between AB is not a linear line and the rate of decrease of stress is quite low. The stress put on B point is named as yield stress and it is determined by Where S0 = the original cross-sectional area of the sample.Beyond the point B are points C and D, the region between B and C is called plastic deformation area. The cross-sectional area begins to decrease and a slight decrease of stress appears between the point C and D. Point E shows the maximum load to produce extension, the portion DE indicates a slightly increase in stress and the strain changes obviously. The extension leads to a decrease in the diameter at somewhere of the sample. A concrete diagram is shown below in Fig.2.Fig.2 extension of tensile test sampleThe tensile stress relates to the stress corresponds to the maximum point E and it is determined by From point E onwards, the deformation of the sample becomes harder, thus need more stress until point F, which considered to be the fracture point or ultimate point. A material is considered to break down under the ultimate stress. The percentage elongation after cracking is given by In which lu is the final measuring length of the cracking sample, the percentage reduction of the cross-section area is determined by Where Su represents the fracture area of the cross-section of the specimen.3 The ductility of a traditional material is described by the percentage elongation. With the measurement of the cross-section area, the deformation length and the applied compressive force on the sample, it is available to draw a true stress-strain curve.Hardness is a important property of a material, which gives it the power to resist abrasion and indentation. Hardness testing vary form macro-, micro- to nano- scale due to the force applied and deformation obtained. Many mechanisms and methods are put into use in which a hard steel ball or some other hard indenter is pressed into the surface of the sample by a normal pressure. The hardness data then obtained form the depth of the indentation or the surface area by the hardness testing machine. One standard method for making a comparative tests on those metals with significantly hard surface. To put it simply, a pyramid-shaped diamond, made of a polishing square-based pyramid diamond with the intersection angle of 136 as shown in Fig.3, is introduced to acquire a standard pressure put on the surface of the sample. Fig.3 a traditional diamond pyramid indenterAnother advantage of the pyramid shape is the ease of calculate accurate diagonal, d, of the square impression. It is possible to count the hardness by the formula below: While it is feasible now to direct read HV from the hardness testing machine. The well know machine is designed by Vickers-Armstrong Ltd. in the United Kingdom, which permits a load varying from 1 to approximately 120 kg and the diagonal size (usually on more than 0.5 mm) is survey with the help of a calibrated microscope.4 The Vickers test is suitable for not only the metal hardness testing, but also applied on ceramics. Though the considerable precise for measuring the hardness of different materials under varying loads and the ability to adapt to a wide range, the price of the Vickers machine is quite expensive than the Rockwell (shown in Fig.4) or Brinill instruments (shown in Fig.5).5 Fig.4 the schematic of Fig.5 Schematic diagram for Rockwell hardness test the Brinell hardness testProcedures:(1) . Measuring the original length, width and thickness of the three samples-aluminum, copper and steel, take 80 mm marks on the surface on each sample.(2) . Checking the tensile testing machine and open the data processing software on computer.(3) . Setting the extensile rate at 5 mm/second and the start position of the grips at 770 mm.(4) . plugging the copper sample in the tensile test machine.(5) . Starting to apply the load on the sample and the computer will record the data and plot the load-extension curve.(6) . After the fracture appears, taking down the samples and measuring the final length, width and thickness. The maximum force applied on the sample can be read from the computer directly.(7) . Testing the other two samples follow the same operation as the copper sample.(8) . After measuring the tensile properties, test the hardness of the three samples by the hardness tester. Choosing three scalar points to reduce the errors of the testing on the edge of the fractured side and the intact area. Details shows in Fig.6.Fig.6 Vickers hardness test on metals2. Results and discussions (1). After the tensile and hardness tests, the data are recorded in the following tables 1 and 2, it is clearly to see the hardness of sample increase after fracturing, the hardness increased percentage of stress is the highest while the hardness of aluminium change slightly:Table 1 tensile test datasamplesLength(mm)Width(mm)Thickness(mm)Maximum loadFracture loadoriginalfinaloriginalfinaloriginalfinalForce(N)Length(mm)Force(N)Length(mm)Al80849.989.790.990.791278772.47721.14.2Cu801099.988.281.020.792590790.76203831.34Fe801159.976.151.030.782824792.96284.537.48Table 2 the hardness test datasampleHardness(HV)Fracture sideIntact side123average123averageAl47.347.047.447.2344.144.145.444.53Cu132.8135.7134.1134.290.390.789.190.03Fe148.2163.5145.9152.53105.1104.0107.6105.57Let us take an overview of the fractured samples after tensile testing described in Fig.7, it is obvious that the deformation of steel and copper is lager than that of aluminium, which is mainly due to the density of the metal and its specific ductility. According to the results, it is possible to draw the load-extension curves of the three samples as shown below in Fig.8, 9 and 10. and a composite graph of Fig.11.Fig.7 the fractured samplesFig.8 the strain-stress curve of AluminiumFig.9 the strain-stress curve of CopperFig.10 the strain-stress curve of SteelFig.11 the composite curves of Al, Cu and FeIllustrated from the strain-stress curves above, it is easy to draw the conclusion that the aluminium is the most brittleness one of the three and the steel has the largest fracture point and the ultimate stress is higher than the rest two.Inquiring from reference books,6 it is available to get the formula It is feasible to calculate the Youngs modulus E using formula or , the yield stress through formula , the tensile stress from equation ,the elongation by equation and the area reduction by formula precisely, describing those results in table 3 below.Table 3 the engineering dataProperties AlCuFeOriginal area S0 (mm2)9.8810.1810.27Final area Su (mm2)7.736.544.79Young modulus E (Gpa)25.47645.7962.64Yield stress y (Mpa)118.62212.97244.52Tensile stress T (Mpa)129.34254.43274.99Percentage elongation (%)4.0528.4934.08Reduction in area (%)21.7235.7453.293. Answers to the questions(1). Explain the reason why the yield point is not obvious in aluminium sample.The yield point is known as the turning point changing from the elastic deformation region to plastic deformation area. Before the yield point, the materials will return to its former shape if the stress is removed from it. Once pass the yield point, the deformation will result to a permanent transformation and can not recovery to the initial state. The main reason for the absence of the distinct yield point in tensile testing of aluminium is that the density of aluminium is relative smaller than some other metals and the transformation happens very fast from elastic state to plastic condition, which may lead to the overlap of the two deformation regions and it is hardly to have a clearly observation of the yield point.(2) . Why the hardness of the samples increase due to the deformation?The deformation contributes to a internal structure changing of the material. To put it simply, the grain slipping from each other and the dislocation motion occurs, which lead to some properties act differently. When a external stress put on the material, there exits another inter force trying to reduce the deformation caused by the external stress. This action makes the cross-section area reduction to meet the volume balance, just as a tighten stress put on the material and thus result to a increase in the hardness of the samples.(3) . Calculating the work done in deforming stress with the help of the tensile data.Work is the integral of the dot product of force and dislocation, which represents a change of energy.7 The steel work area is marked by green colour in Fig.12 and the calculating results of integral is illustrated in Fig.13.Fig.12 the steel work area Fig.13 the integral resultAccording to the calculating result shown in Fig.13, the area equal to 77338.917 Nmm. So it is distinct that the work done is 77338.917 J.(4) . Draw a hardness-yield strength curve and analysis the relationship between them.Yield strength represents the stress at yield point where the curve lever off the the plastic deformation happens, we have the hardness and and yield stress of these three samples list in the table below:Samples AlCuFeHardness (HV)47.23134.2152.53Yield strength (Mpa)118.62212.97244.52The hardness-yield strength curve is as follow:Fig.14 hardness- yield strength curveIt can seen from the graph that the yield strength is proportional to the hardness, that is to say, the the yield strength increase with the increase of hardness.(5) . Plot a hardness- ultimate tensile strength and analysis the relationship between them.Ultimate tensile strength describes the stress under the maximum load, so we get the data below:Samples AlCuFeHardness (HV)47.23134.2152.53Ultimate tensile strength (Mpa)129.34254.43274.99The relationship curve is following:Fig.15 the hardness-ultimate tensile strength curveIt is show in the graph that the ultimate tensile
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