Saturday, February 8, 2020

Lab report- materials Report Example | Topics and Well Written Essays - 750 words

- materials - Lab Report Example Steel is artificially produced and it exists in different forms of iron. Whereas steel has carbon content which is less than that of pig iron, its carbon content is more than that of wrought iron. Moreover, steel has such qualities as elasticity, hardness, as well as strength, which depends on the heat treatment and composition (Makelainen and Hassinen 488). Consequently, it is categorized as either having low, medium or high carbon content. On the other hand, aluminium is a silvery white metallic element that is ductile and has low density as well as high strength to weight ratio and is mainly bauxite. Owing to its good thermal and conductive properties, aluminium is usually used in forming hard light corrosion resistant alloys. Similarly, a polymer is defined as chemical compound that is formed through a process known as polymerization, and it consists of repeating structural units (Cheremisinoff 1). Inherently, polymers are normally characterized by their light weight and their ab ility to resist corrosion and reaction. Materials and methods The dumbbell specimens were put under tensile forces through a horizontal tensile testing machine. Consequently, the diameter of the specimen was obtained by use of vernier calliper and recorded before the test commenced. Moreover, during the application of the load, two cameras were used in determining the diameter of specimen. Subsequently, the cross-sectional area of the specimen was obtained and together with the load, both the engineering stress and the true stress were calculated in accordance to the following equations. The process was repeated for all the materials. Consequently, a horizontal tensile force was applied to the specimen. The elongation and the new diameter of the specimen, due to the applied load were similarly obtained through the use of VI monitor and the cameras, and they were recorded. Using the load (KN), the diameter (mm) and the elongation (m), engineering and true stress, engineering strain, yield stress *0.1 percent offset, Young modulus, ductility, ultimate tensile stress and work fracture were also obtained. Where where r = diameter/ 2 = Davis defines young elastic modulus as the measure of the resistance of a material to elastic deformation (32). It is equal to the slope of a stress/strain curve in elastic region. Therefore, where , and , are y axis points of the curve, and; , and are x axis points of the curve Moreover, according to Davis ductility of material describes the ability of a material to deform permanently before failure (37). It is actually the engineering strain at failure. = Davis claims that work fracture is equal to the area under stress-strain curve, and its unit is work per unit volume (Nm/m3) (45). More importantly, the ultimate tensile strength refers to the maximum amount of stress a material can bear. It is obtained from the engineering stress-strain curve as the highest point. Conclusion From the tensile test conducted on the materials A, B, and C, the young modulus, yield strength and ultimate tensile strength were obtained. Material A had the highest young modulus with a value of 190 GPa, and it was followed by material B and finally material C. Consequently, from the results it is clear that as the materials become brittle, the Young modulus tends to become higher. It was also noted that the polymer achieved the highest engineering strain, of 0.57, and was closely followed by steel with 0.35 and aluminium with 0.1.

No comments:

Post a Comment

Note: Only a member of this blog may post a comment.