...Tensile test report (full length) Abstract The Maximum load, breaking load, percentage elongation and minimum diameter at fracture of Mild (low carbon) Steel, Duralmin and Copper were calculated by applying a tensile load to the respective specimen until it fractured. Copper and Steel were the most ductile materials because their percentage elongation and their percentage reduction in area were the highest. The Young modulus of Steel revealed that it is also a stiff material. The Duralmin was found to be the strongest material because of its high tensile and breaking strength. The accuracy of the results were satisfactory but they could have been improved by repeating the tensile test several times and taking an average. Introduction The tensile test is conducted to determine the basic stress-strain properties of a metal. The test is also used to measure the resistance of a material to a static or slowly applied force. The tensile test was performed on mild Steel, Duralmin and Copper. The purpose of the experiment was to investigate the properties of mild steel in detail by performing tensile tests on a specimen rod of constant cross section. Theory The maximum load is the greatest load that the specimen can withstand without breaking. The breaking load is the load at which the specimen breaks. Elongation is the amount that the sample has increased in length after it has fractured compared to its original length. The percentage elongation is extent...
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...Universal Interface | UI-5000 | 1 | PASCO Capstone Software | UI-5400 | Introduction The objective of this lab is to find the relationship between tensile stress and strain for various materials. The Stress-Strain Apparatus stretches (and in some cases breaks) a test coupon while it measures the amount of stretch and force experienced by the test coupon. Software is used to generate a plot of stress versus strain, which allows Young's Modulus, the elastic region, the plastic region, the yield point, and the break point to be ascertained. Theory The ratio of the force (F) applied to the cross-sectional area (A) of a material is called the stress: Stress=FA (1) The ratio of the change in length (L) to the original length (Lo) of a material is called the strain: Strain=∆LLo (2) Stress (Pa) Strain Elastic Region Plastic Region Yield Point Stress (Pa) Strain Elastic Region Plastic Region Yield Point In the elastic region, the stress is proportional to the strain and the proportionality constant is called Young’s Modulus, Y. Y=StressStrain (3) Set-Up Figure 1. Stress/Strain Apparatus Assembly 1. Open the PASCO Capstone software on the computer. 2. Connect the Stress/Strain apparatus to the computer using USB cable and the PASCO Capstone software will automatically recognize the device. 3. Make a calculation for the stretch. For every 360o turn of the Rotary Motion Sensor, the screw moves 1 mm, so the stretch, x, in mm is given...
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...Name: Kevin Lundstrom Class: Monday Group: 3 Lab Partners: Rusty Flocken Oliver Beres Torsion Introduction The purpose of this experiment was to ultimately determine the shear modulus, 0.2% offset shear yield point, and ultimate shear strength of two different materials: aluminum and brass. This was accomplished using a torsion test machine connected to a computer that recorded the torsion data. After determining these three material properties, the measured experimental values could be compared to published data to determine the accuracy of the test. Experimental Procedure In this experiment the specimens underwent torsion by means of a torsion test machine. This machine consisted of two drill chucks: one was attached to a rotating chain-driven wheel, and the other was held stationary. A potentiometer was used to determine the number of revolutions the rotating wheel had undergone. Strain gauges were attached to the stationary drill chuck. LabVIEW was used to measure and record the data from the potentiometer as well as the strain gauges. Each sample was measured until it failed. The length of each specimen as well as the diameter was recorded in order to perform calculations. Also, the diameter of the potentiometer as well as the drill chuck, as well as the minimum and maximum resistance of the potentiometer was recorded for calculations. Results/Discussion The recorded measurements are shown in the table below. These were necessary to perform the...
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...Carbon fibre reinforced polymer Name Institutional analysis Introduction Carbon Fibre reinforced polymer commonly known as CFRP is a very strong composite material that is majorly preferred for many automotive and aeronautical application due to the fact that despite their strength that still light that the traditional metallic materials. The polymers often used to make this composite material is epoxy, polyester, vinyl and even nylon. Thus in making CFRP carbon fibre is often combined with polymers such as epoxy to result in a material that has a very high strength-to-weight ratio. Carbon fibre reinforced polymers despite being relative expensive as compared to other traditional metallic materials; they have been preferred for many applications such as in the automotive industry and aeronautical industry where high strength-to-weight ratio is demanded. (Image 1: carbon fibre reinforced polymer) Advantages of CFRP over Traditional metallic material Carbon fibre reinforced polymers are very strong which has made them of interest in the field of engineering. According to Gesellschaft (1981), CFRPs is important because they can offer high strength value at a relatively low value of weight. In the construction of body parts of automobiles, it is important that the designs are very low in fuel consumption. In order to have the minimal fuel consumption then, the first issue that the automobile fields have to address is the weight of the designs. CFRPs are of interest because...
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...the volume fraction Vf of fiber. Use the properties from attached Tables. 3. (5 points) Plot the longitudinal tensile strength of an E-glass/epoxy unidirectionallyreinforced composite, as a function of the volume fraction of fiber, assuming tensile strength follows a volume rule of additivity. Use the properties from attached Tables. 4. (5 points) What is the maximum volume fraction of spherical fillers that could be obtained in the particle-reinforced composite assuming cubic packing? Please show the step. 5. (4 points) Using the Takayanagi model and assuming uniform strain in the matrix, derive a relationship for the transverse, tensile compliance of a unidirectionally-reinforced composite. 6. (6 points) Two test specimens have the same Young’s modulus in tension. However, one of the specimens is homogeneous while the other one has two layers with one layer stiffer than other layer. How will the flexural modulus of the two specimens differ? Explain. (Hint, use Parallel and Series model to calculate the modulus to compare them. Assume the modulus of stiff material is 1X109MPa as and the modulus of soft material is 1X108 , and the volume fraction of each one is 0.5) Properties will be used for solving above problems: The modulus of S-glass is 85.5 GPa, the modulus of E-glass is 72.4 GPa, the modulus of Epoxy is 3.5 GPa and the modulus of polyamide (nylon) is 3.0 GPa. The tensile break strength of E-glass is 2.4 GPa and that of epoxy is 45 MPa. The break strain for E-glass and epoxy...
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...Title: How different springs behave under the influence of same mass depending on the spring constant (k) and connection types (parallel or series)? Summary: This experiment will be carried out to investigate spring constants of different springs by measuring the length of springs under the influence of same mass and different masses, and observe how overall spring constant changes depending on connection types of springs by measuring amount of extension when springs are connected in series and parallel. Different springs with different constant, different masses, weighing machine and a measuring ruler will be used. Research: The course textbook and different textbooks (S.Beichner, Fizik1, Ankara: Palme Yayınevi, 2002.), some laboratory experiments, which retrieved from the Internet, some lectures and videos comprise sources for the theoretical analysis of the experiment. -Prof. Walter Lewin (1999), Hooke's Law, Simple Harmonic Oscillator (online) (http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-10), -Combination of springs (online) (http://engineeronadisk.com/V2/book_modelling/engineeronadisk-10.html) -Sal Khan (2008), Intro to springs and Hooke's Law (online) (http://www.youtube.com/watch?v=ZzwuHS9ldbY) -Forces and elasticity (online) (http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/forces/forceselasticityrev2.shtml) -Deformation of materials (online) (http://qatemplates.everythingscience.co.z...
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...M E S H C O N V E R G E N C E A N D H E AT F L U X A NA LY S I S ( A B A Q U S ) Ricardo Loretz - A01182657 Mesh Controls / Global Seeds For this analysis I used Hex as Element Shape with a sweep technique in order to get the best mesh finish. I changed from 1 to 0.15 of approximate global size to get the different number of elements that I used in the mesh convergence analysis. Boundary Conditions Global Size: 1 – Elements: 60 Max. Stress and Loc. Global Size: 0.5 – Elements: 3840 15% Displacement Global Size: 0.3 – Elements: 28556 Mesh Convergence on 15% displacement analysis. For this analysis we need to determine a displacement of 15% in the initial height for a cylinder. MATERIAL PROPERTIES Material Titanium, Ti Height (h) 3 mm Diameter 6 mm Young Modulus (E) 116 GPa Poisson Ratio (V) 0.34 BC’s CSYS- Encastred on the bottom of the scaffold and 15% displacement in U3 direction. MESH CONVERGENCE ANALYSIS Mesh Type Element Max. Stress Mesh 1 60 2.07E+10 Mesh 2 792 2.43E+10 Mesh 3 3840 2.86E+10 Mesh 4 13254 Mesh 5 28560 3.16E+10 3.47E+10 Results As we can see in the mesh sensitivity study, with more elements we get better results but we increase the computational time so we don’t need too much elements for this analysis According to Fourier´s Law, the heat flux is from highest to lowest temperature; therefore, our analysis is correct...
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... ELASTICITY The ability of a strained body to recover its size and shape after deformation. stress – characterizes the strength of forces causing the deformation. F stress A in 2 or pascal (Pa) strain – the resulting deformation. Hooke’s Law: stress elastic mod ulus strain 3 Types of Stress tensile or compressive stress – changes the length of a strained body. tensile stress bulk stress – changes the volume of a strained body. bulk stress shear stress – opposite forces act at different points in the body. shear stress Tensile and Compressive Stress Consider a body of initial length lo acted upon by a tensile or compressive force of F. F stress A l strain lo Young’s Modulus (Y) F stress A Y strain l l o Flo Y Al Example 1 Bulk Stress When uniform pressure p acts on a body, the volume of the body decreases. F pressure stress A V strain Vo F stress A B strain V V o Bulk Modulus (B) p B V V o B p V Vo Example 2 Shear Stress and Strain When equal and opposite forces act tangent to the opposite sides of a body, a shear strain is produced. F// stress A x strain h x h x tan h Shear Modulus (S) F// h S Ax F// S A tan Example 3 F// = 9.0 x 105 N A = 5.0 x 1-4 m2 F Ssteel...
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...Recording and presenting your data 1 Calculate the cross-sectional area A of the wire using the equation A = ¼πd2 2 Record your values for m and x in a table of results. Include columns for F and for the stress and the strain, where F = mg Stress = F/A Strain = x / L Where F is the applied force, x is the extension, A is the cross-sectional area and L is the original length and g = 9.81 m.s-2. 3 Plot a graph of stress (y-axis) against strain (x-axis). Analyzing your data 1 Calculate the gradient of your graph. 2 The Young modulus E of the material of the wire is given by the equation E = stress / strain. Use your answer to 1 to determine the Young modulus of the material of the wire. Evaluation The procedure you carried out including what steps were taken to reduce error. 1 Estimate the actual error in each measurement taken. 2 Describe any limitations or problems with the method used to determine the Young modulus. 3 Suggest why a large value of L was used in this experiment. 4 Suggest two ways in which the accuracy of the measurements taken could be improved. 5 Look up the ‘true’ value for the Young modulus of the material of your wire. Comment on the difference between this ‘true’ value and the value you obtained from your experiment. Key learning objectives • To experimentally determine a value of the Young modulus of a material. • To illustrate the importance of making some variables large in order to reduce errors in small readings. Expected Results material...
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...RULES OF MIXTURES FOR ELASTIC PROPERTIES The paper is about the rules of mixtures which are used to express the dependencies of the physical properties and mechanical properties which depend on type, form, quality and arrangement of its constituents, but they are based on various assumptions so one should with caution, especially if they are used anything more than preliminary design. The paper mainly concentrates on expressions for elastic properties which are as follows: Unidirectional Ply- longitudinal modulus: The Figure 1 clearly shows the orthogonal axes and fiber direction, the fiber directions. The first approximation made is E3= E4. And also for deriving the rules of mixtures the following assumptions are made: Fibers are uniform, parallel and continuous. Perfect bonding exists between fibers and matrix. A longitudinal load produces equal strain in fiber and matrix. Using the above assumptions and approximations two rules of mixtures are derived which are E1 = EfVf + EmVm = EfVf + Em (1- Vf) V12= vfVf+vm+Vm These two rules of mixtures are generally accepted as it goes well with experimental data. Unidirectional Ply- Transverse modulus: In this the rules of mixtures are less reliable than those for longitudinal properties as they are based on assumptions of stress distribution. In this the poisson’s contraction is ignored and the stress is assumed to be the same...
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...| Civ Eng 4G04 | Resilient Modulus Lab | | Procedure: 1. Soil mould was assembled with rubber membrane lining inside of mould. 2. The mould was filled with soil so the approximate compacted height was equal to 1/6th the height of the mould (approx. 50mm). 3. Soil was compacted using hammer drill equipped with tamping attachment 4. Steps 2 and 3 were repeated 5 additional times until the mould was filled. 5. The mould was removed and the sample was placed into the testing apparatus. 6. Rubber membrane was secured using rubber rings placed at the top and bottom of the sample, ensuring contact with the sample did not occur. 7. Pressure cell was placed over sample and secured into place. 8. Sample was brought to cell pressure of 3 kPa. 9. Sample was conditioned for shakedown 10. Cyclic loading was applied to sample, with a loading duration of 0.1 second and a rest period of 0.9 seconds, displacements and forces during this loading were recorded using supplied software. 11. This loading was applied in groups of five with a 95 second rest period between sets of load application. A total of three sets of load applications were applied to the sample. 12. Steps 8 and 10 were repeated at cell pressures of 5, 10, 15 and 20 kPa. 13. At the end of testing soil samples from the top, middle and bottom of the sample were taken in order to determine moisture content. Results: Tabulated results can be seen in Table 2 in the...
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...California State University, Long Beach Mechanical and Aerospace Engineering Department MAE 361: Materials and Properties Laboratory Torsion Test of Metals and Polymers Date: October 29, 2013 Submitted To: Dr. Parvin Shariat Submitted By: Ryan Kim D. Lim - 008142015 Steven Reid - 007777066 Thomas Leininger- 008403601 Marco Evangelista – 007774076 Objective: The objective of this experiment is to determine the shear stress, shear strain, and shear modulus of elasticity of solid cylindrical specimens of steel, brass, aluminum, and a polymer. Procedure: 1. Plug the cord for the machine into the outlet hanging from the ceiling. 2. Zero the drum angle using start, forward/reverse, and speed on machine control. When start is triggered, you must specify the direction of movement, either forward or reverse. There are values on the drum which will indicate when you zero it out. 3. Measure diameter of the sample using a caliper and make sure to measure three different areas, preferably the two ends and the middle for an accurate measurement. Record the average of the three measurements. 4. Mount the specimen in the self tightening chucks of the machine: a. Slide weighing chuck horizontally of the machine and open jaws of the chuck to permit insertion of the specimen. b. Insert the specimen into the loading chuck and then slide the weighing chuck assembly to the right until it is stopped by the specimen ends. Move the weighing chuck around ½ inch away from...
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...III Międzynarodowa Konferencja Naukowo-Techniczna „Nowoczesne technologie w budownictwie drogowym” Poznań, 8–9 września 2005 3rd International Conference „Modern Technologies in Highway Engineering” Poznań, 8–9 September, 2005 Lysandros Pantelidis Aristotle University of Thessaloniki, Greece DETERMINATION OF SOIL STRENGTH CHARACTERISTICS PERFORMING THE PLATE BEARING TEST Summary: In the international literature there is often confusion between the terms Modulus of Elasticity EYoung and Modulus of Deformation EDef for homogenous, isotropic masses, which, sometimes, plainly referred as Modulus E, without any further explanation. In this paper a distinction between the EYoung and the EDef is being done and the interest is concentrated in the Plate Bearing Test, which is commonly used in highway earthworks. Moreover, a soil classification diagram is given, in which the soils are classified by their shear strength parameters and their behavior, as totally elastic or elastoplastic, under specific loading applied by rigid circular plate. Finally, a method for the estimation of the soil parameters is proposed, using only the in-situ Plate Bearing Test. The study and the documentation were supported by a presentation of theoretical examples, based on the Finite Element Method. 1. INTRODUCTION In highway earthworks the knowledge of the Modulus of Elasticity of the soils is very important, whether the projects are in the stage of design, construction or compaction check and this...
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...Representation of riveted joints in finite element modelling Modelling of riveted joints in Finite Element Analysis (FEA) programs can be a complicated task. Previously beam elements have been used to represent fasteners where the shear stiffness of the beam has been approximated to that of the shear stiffness of the rivet. Errors result in this method since the beam can also support bending. Test results of rivets joining sheet metal components have shown that beams are not ideal at simulating a rivet in FEA. An acceptable method for simulating riveted joints in FEA is using a spring element. The spring element has the benefit of providing the appropriate shear stiffness whilst not being able to support any bending. Riveted joints that attached thin sheet metal components are capable of transferring shear loads but due to the close proximity of the bending planes, they do not transfer any, or very little, bending loads. The spring element connects two nodes between the components as shown in Figure 1 below. DOUBLER / SPLICE PLATE KSHEAR KAXIAL SKIN Figure 1 – Spring Element Connection A rivet provides an axial stiffness which is a function of the material properties and the geometry of the rivet. The axial stiffness can be calculated as follows: K AXIAL AE L Where A = Rivet Cross Sectional Area E = Rivet Modulus of Elasticity L = Rivet Grip Length The shear stiffness of a riveted joint is calculated by the following equation for rivets installed in...
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...PST 522E – Synthesis and Characterization of Macromolecules CHAPTER 4 DYNAMIC MECHANICAL ANALYSIS OF EPOXY-CARBON FIBER COMPOSITES 1. THEORETICAL PART Composite materials have been used widely for daily life applications for many years. Polymer composite systems have a large scale for industry or research area due to their light weight, design flexibility, and processability [1,2]. Using polymer composite is widespread to use for special engineering materials such as aerospace industry, automotive and civil engineering structures because of their outstanding mechanical properties. Polymer composite theories were established on properties of composite constituents, volume fraction, shape, matrix-inclusion interface [3,4,1,2]. Mechanical properties of polymer composites are significant when they use for a building or an instrument which has critical points about temperature and strength properties [5-7]. Composite materials properties have been developed such as chemical, physical and mechanical, so their application area comes much bigger. Because of improvements on composites materials, their industry are growing on automotive, aerospace, electronics and biotechnology [8,9]. The important point of using nanocomposites is large scale, low cost, easy applicable [10]. In composites, two materials are mixed to make improvement in the mechanical properties. Matrix material is dominant and including material. The second material is particle, filler or fiber. Composites are manmade...
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