...SCHOOL OF ENGINEERING &TECHNOLOGY MECHANICAL & AUTOMOBILE ENGINEERING DEPARTMENT III TERM SECOND YEAR 1 Course number MEC211 2 Course Title STRENGTH OF MATERIALS 3 Credits 5 4 Contact Hours (LT- P) 3-1-2 5 Course Objective To understand the relationship between stress and strain in solids. 6 Course Outcomes On successful completion of this module students will be able to 1. Understand the concept of strain and stress, stress- strain diagram, Elastic constants and constitutive relations.. 2. Determine principal stresses and strain and locate principal planes. 3. Apply the theory of simple bending to compute stresses in beams of homogenous and composite sections of different shapes. 4. Calculate slope and deflection in beams.Use Double integration method, Macaulay’s method, moment area method methods to calculate slope and deflection for the following : a) Cantilevers b) Simply supported beams with or without overhang Under concentrated loads, uniformly distributed loads or combination of concentrated and uniformly distributed loads. 5. Apply different formulae to analyze stresses in struts and columns subjected to axial loads. 7 Outline syllabus 7.01 MEC211.A Unit A Simple stresses and strains 7.02 MEC211.A1 Unit A Topic 1 Concept of stress and strain, St. Venant’s principle, Stress and strain diagram, Hooke’s law, Young’s modulus (E), Modulus of Rigidity(G), Bulk modulus(K), Poisson ratio. 7.03 MEC211.A2 Unit A Topic 2 Stress and elongation...
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...American Society of Civil Engineers (ASCE) Pipelines Conference 2005, Houston, TX Longitudinal Mechanics of Buried Thermoplastic Pipe: Analysis of PVC Pipes of Various Joint Types Shah Rahman1, Reynold K. Watkins2 ABSTRACT The analysis of longitudinal deformations compared to performance limits of deformation in buried pipes is referred to as longitudinal mechanics. Principal causes of longitudinal stress and strain within a pipe system include changes in temperature, internal pressure or vacuum, and beam bending. The widespread use of thermoplastic pipes, namely polyvinyl chloride (PVC), in municipal applications throughout North America in the past four decades has made it necessary to re-visit the topic of longitudinal mechanics for pipes that incorporate various types of joining systems. Typical North American PVC pipe joints are either bell-and-spigot gasket-joint or welded (heat-fused Fusible PVC and solvent-cement joints). Analysis herein focuses on the three main causes of longitudinal stresses and strains in bell-and-spigot gasket joints and welded joints, and includes discussion of theoretical concepts such as the Poisson effect and the Reissner effect. Topics which have raised issues in the field such as ponding due to sags in a PVC gravity line and the occasional cracking of PVC pipe bells during or after installation are also discussed. Current industry and manufacturer recommendations of various design parameters are provided in conjunction with the analysis...
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... Relationship between Kmax and σf Kannan et al. [23] had examined the applicability of a modified two-parameter fracture criterion MTPFC which introduced by Christopher et al. [24, 25] while assessing the fracture strength of structure components. They utilized a relation between the stress intensity factor (Kmax) and the corresponding stress at failure (σf) as (2) where σf is the hoop stress at the failure pressure of the flawed pipe. σu is the hoop stress at the failure pressure of the unflawed pipe. KF, m and p are fracture parameters. The failure stress, σf of a pipe decreases with increasing crack size. If σf is less than the yield strength (σys) of the material, then there exists a linear relationship between σf and Kmax. For small sizes of cracks where σf is more than σys and less than σu, the relationship between σf and Kmax is expected to be non-linear. The idea of expressing Kmax as a function of σf in Eq. 2 is mainly for estimation of failure strength of a cracked body whether it contains through-thickness or part-through cracks which are small or large in size. In the absence of cracks, σf is equal to σu and Kmax= 0. The exponential form of the third term in Eq. 2 represents the nonlinear variation of Kmax with σf, when σf is more than σys. The fracture parameter can be determine from the test data of cracked configurations. KF has the units of the fracture toughness...
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...Stress-Strain Equipment | Included: | | 1 | Stress-Strain Apparatus | AP-8214A | 1 | Force Sensor | PS-2104 | 1 | Rotary Motion Sensor | PS-2120 | 1 | Calipers | SF-8711 | | Required but Not Included: | | 1 | 850 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...
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...MATS1101 Materials Strand/MATS9520 INTRODUCTION TO MATERIALS ENGINEERING Dislocations Ashby & Jones, Volume 1, Chapter 9 Ideal Strength of Materials Last week we saw that the slope of the interatomic force-distance curve at equilibrium spacing is proportional to Young's Modulus, E. Interatomic forces become negligible for r > 2r0. F ~2Fmax Fmax Attraction 0 Repulsion r0 2r0 r0 r0 ~1.25r0 2r0 r Figure 8 Interatomic force-distance curve The maximum in the force-distance curve occurs at ~1.25r0, (where F = Fmax). If applied stress is greater than Fmax per bond, bonds between atoms are broken and fracture occurs. Ideal strength, , corresponds to bond rupture at Fmax. Calculation of ideal strength: Slope E 0.25r0 r0 2r0 r0 Figure 9 Ideal Strength ~ From the force-distance curve, where r = 1.25r0, = 0.25 and 2 is ideal strength). Copyright School of Materials Science and Engineering, UNSW, 2012. E ~ 2 0.25 ~ E 8 ~ E 15 E 15 A better estimate using interatomic potential gives Glasses and some ceramics have a yield strength of For other ceramics and polymers, For metals, y y 10 1 E 15 E 15 y 10 1 10 5 Actual vs Ideal Strength 10-1 10-2 10-3 Cement (nonreinforced Ceramics Silica glass Diamond Soda glass SiC Al2O3, Si3N4 MgO, ice Alkali halides Metals Polymers Low density PE Epoxies PP, PMMA High density PE Nylons...
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...------------------------------------------------- TENSILE TESTING OF MATERIALS Lecturer: Dr. Fadi Ghaith Submitted by___________________________________________ Sminoy Mohandas - 101668769 CONTENTS 1. Summary........................................................................................................3 2. Introduction ..................................................................................................3 3. Applications * Aerospace Industry ...............................................................................4 * Automotive Industry .............................................................................4 4. Theory ...........................................................................................................5 5. Experimental Method....................................................................................6 6. Results ...........................................................................................................6 7. Conclusion…...................................................................................................9 8. References ....................................................................................................9 Summary: Tensile testing is one of the simplest and most widely used mechanical tests. By measuring the force required to elongate a specimen to breaking point, material properties can be determined that will allow designers and quality...
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...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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...Chapter 1 – Materials Engineering Concepts * 1.1 Economic Factors * Availability and cost of raw materials * Manufacturing costs * Transportation * Placing * Maintenance * 1.2 Mechanical Properties * Loading conditions * Static load – A sustained loading of the structure over a period of time * Dynamic load – A load that generates a shock or vibration in the structure (periodic – repeats itself with time, random- load pattern never repeats, or transient- impulse load applied over short interval) * Dead load- Static load, always applied (weight of the structure) * Live load- Dynamic loads, (people in a building or movable objects) * Stress-Strain Relations * Stress : σ=FA (psi, ksi, kPa, MPa, GPa) * Strain: ε=Change in LengthOriginal Length (%, in/in, mm/mm) * Elastic behavior * Elasticity- instantaneous response (deformation to load) and must return to its original shape when the load is removed. Stretches the bonds between atoms-doesn’t change arrangement of atoms * Young’s Modulus (Modulus of Elasticity)- Slope of the linear portion of the stress-strain curve E=σε * Poisson’s ratio- Ratio of the lateral strain to the axial strain (must be between 0.0(compressible material)-0.5(does not change volume when load applied)) Usually between 0.1 and 0.45 ν=-εlεa * Generalized Hooke’s Law * εx=σx-ν(σy+σz)E ...
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...Thin Cylinder Summary/ Abstract Thin-walled pressure vessel provides an important application of the analysis of plane stress. Since their walls offer litter resistance to bending, it may be assumed that the internal forces exerted on a given portion of wall are tangent to the surface of the vessel. The resulting stresses on an element of the wall will thus be contained in a plane tangent to the surface of the vessel. In this experiment, we are going to determine the circumferential stress under open condition, and analysis of combined axial and circumferential stress. We propose to determine the stresses exerted on a small element of wall with sides respectively parallel and perpendicular to the axis of the cylinder. Because of axis symmetry of the vessel and its contents, it is clear that no shearing stress is exerted on the element. It can be seen that all strains are used to calculate stress. As it is not possible to measure all strains, these have to be computed on the basis of marginal conditions. With this experiment it is not possible to measure in particular longitudinal strain in the body and radian strain perpendicular to the surface. Initial stress at the surface must be zero, longitudinal stress is constant over the radius are the marginal conditions to obtain the solution. The average E value we obtain from graph was 64 GN/m2. with the help of the computer, the Poisson’s ratio, v =0.35 and the principal strains for the thin cylinder were also calculated using...
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...1.0 INTRODUCTION 1.1 Beam Deflections 1.2 Theory - Calculations DeflectionF formula for the load given above: A determination of flexural stress yields: When rectangular it is Where; δ = Deflection (mm) E = Coefficient of Elasticity L = Span (mm) I = Inertia Factor Mb = Moment of flexure (Nmm) F1 = Load occasioned by weight Wb = Resistance to flexure (mm3) of Load Device (N) σb = Flexural Stress (N/mm2) F = Load of occasioned by additional weight (N) 1.3 Objectives * To investigate the relationship between load, span, width, height and deflection of a beam placed on two bearers and affected by a concentrated load at the center. * To ascertain the coefficient of elasticity for steel brass and aluminium 2.0 METHODOLOGY 2.1 Procedure - Experiement 1A * Investigate the relationship between load and deflection. 1) Set the bearers so that a span of 600 mm is obtained. The interval between each groove on the shafts of the apparatus is 100 mm. 2) Place a test specimen with dimensions of 4 x 25 mm, on the bearers and mount the load device in the center of the test specimen. 3) Set the testing device so that the top of the gauge is centered on the upper plane of the load device. Lower the gauge so that its small hand is at about 10 and set the gauge to zero by twisting its outer ring. 4) Load the weights as shown in the table below and read off the deflection...
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...Theory: Forming process are solid state manufacturing processes involving minimum amount of material wastage and faster production. It does have been one of the common methods in formation of product in multi shape and size. In fact some of casting product will undergoes forming process to achieve more precise shape. In metal forming, transformation to desire shape is done at solid state and usually required high forces. If the temperature is below crystallization temperature, it is called ‘cold work’. Large force is applied such that the material flows and takes the desired shape. These processes are normally used for large scale production rates. It is also generally economical and many cases improve the mechanical properties. Some of the metals forming process are rolling, forging, extrusion, drawing, sheet metal forming and bending. In fabrication of metal dustbin, below process forming process are involved. a. Sheet Metal Cutting -Sheet metal cutting can be done by shearing operation. Shearing is the mechanical cutting of material without the formation of chips or the use of burning or melting. When sheets of metal are to be sheared along a straight line, squaring shears are frequently used. As the upper ram descends, a clamping bar or set of clamping fingers presses the sheet of metal against the machine table to hold firmly in formation. A moving blade then comes down across a fixed blade and shears the metal. On larger shears, the moving blade is often set...
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...The Title of the experiment: Tensile Properties of Aluminum using Lloyds Testing Machine The author`s name: Jr., E. Russell Johnston , Ferdinand P. Beer The date of the experiment: 15/11/2012 Lab Group – 3 Report by: 1-Turky Abdullah AL-Bussairi 2-Yasser Habib AL-Mutairi 3-Abdulrahamn Sa`ad AL-Huaifi 4-Abdulrahman hemdy AL-Harbi Abstract: We determined the elastic modulus, yield strength, tensile strength, modulus of toughens, elongation, reduction of area, as well as true stress and strain at rupture point for one specimen, aluminum,. We accomplished this by placing our specimen at a time into a universal testing machine (UTM), which, under computer control, slowly increased the tension force on each specimen, stretching each until failure. Purpose: The purpose of this experiment is to extract data on the material properties one specimen (aluminum), using a mechanically driven universal testing machine (UTM). The material properties include the following: the elastic modulus, 0.2% offset yield strength, ultimate tensile strength, modulus of rupture, modulus of resilience, as well as true strain and true stress at the point of rupture. Theory: Certain materials (those that are linear, homogeneous, elastic, and isotropic) can be described by their material properties. These properties include the modulus of elasticity, modulus of toughness, modulus of resilience...
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...Laboratory Experiment: Uniaxial Testing of Civil Engineering Materials Table of Contents Introduction 1 Procedure 1 Experimental 1 Analysis 1 Results 2 Conclusion 4 References 5 Introduction In this set of laboratory experiments, uniaxial tests were performed on two specimens: 1) a steel bar in tension, and 2) an aluminum bar in tension. The specific types of materials used in this experiment are grade 50 for the steel, and type 2017-T351 for the aluminum. The experimental procedure and analysis give a better insight to the behavior of different ductile materials under normal stresses. Through the analysis of the experimental data, various properties of the material can be found and these properties are used in the design of engineering structures which use these materials. Through the use of plots of the stress versus strain for each material, and tables providing material properties, the report will include the following objectives for all materials: • Describe the behavior in the elastic range by determining the moduli of elasticity, and yielding stresses. • Describe the behavior of the materials beyond yielding by determining plastic stresses, ultimate stresses, rupture stresses, and ductility as well as indicating the range of strain hardening and necking. • Compare computed values obtained from the test results to expected values found for steel and aluminum. Further...
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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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...ceramic. And also learn what the degradation process is. Assessors feedback Action plan Assessors signature Rishi Mali Date Learners signature Nagesh Date Name: Nagesh Kom. Assignment No.: 4 Unit Name: Properties and Applications of Engineering Material Unit No.: 19 Task: 1 Describe the principles of the modes of failure known as ductile/brittle fracture, fatigue and creep. Brittle fracture: In the brittle fracture extensive plastic deformation and heat absorption before a fracture. In the brittle fracture crystalline materials, fracture can be occurring by cleavage as the result of tensile stress, acting normal to crystallographic planes with low bonding. In the amorphous solid by contrast, the lack of a crystalline structure results in as fracture, with cracks proceeding normal to the applied tension. http://hycalin.nl/line-types/chain/ Ductile fracture: In the ductile fracture, plastic deformation takes place before fracture. The terms rupture or ductile rupture describes the ultimate failure of the ductile materials loaded in tension. Many ductile metals, especially materials with high purity, can sustain very large deformation of 50 to 100% or more strain before fracture under loading condition and environmental condition. The strain at which the fracture happens is controlled by the purity of the materials. At room temperature, pure iron can undergo deformation up to 100% strain before breaking, while...
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