Presentation on theme: "GE 102 Manufacturing Technology. Workshop Technology W. A. Chapman Workshop Processes and Materials Bruce J. Black Materials and Processes in Manufacturing."— Presentation transcript:
Engineering Materials 2-1 Introduction Engineering materials are those the engineer uses in his work. Nearly all materials existing on and under the ground are used in engineering. Some of these materials are used directly as water, sand, etc., others need more or less treatment as iron ore, petroleum. Moreover, some materials are used alone in industry as wood, leather, etc., others are mixed together to produce other materials having specific properties as adding chromium to steel to improve its corrosion resistance.
2.3. Properties of Materials 2.3.1 Classification of properties a) Physical properties: As the shape, dimensions, porosity, etc. b) Chemical properties: As the chemical composition, acidity, etc. c) Thermal properties: As the expansion, thermal conductivity, specific heat, etc. d) Electrical and Magnetic properties: As the electrical resistivity and conductivity, magnetic permeability, etc. e) Optical properties: As the color, light reflection and absorption etc. f) Acoustical properties: As the acoustic reflection and absorption, etc. g) Mechanical properties: They are the properties, which determine the behavior of the material under loads.
2.3.2 Main Mechanical properties of Materials 1-Elasticity: Is the ability of the material to restore its original shape or volume at once when the load is released. 2- Plasticity: Is the ability of the material to change its shape and dimensions under load and to keep the new shape and dimensions after the load is released. 3- Ductility: Is the ability of the material to deform (elongate) in static tension without failure. 4- Malleability: Is the ability of the material to change its shape under pressure (compressive load) without failure. 5- Brittleness: Is the ability of the material to fail without a noticeable in its dimensions. 6- Hardness: Is the resistance of the material to penetration of another harder body?
7- Stiffness: Is the resistance of the material to any change of shape, it is measured by Young’s modulus. 8-Strength: Is the measure of the ability of materials to resist stresses (tensile, compressive, bending, shearing or torsion) under different conditions of loading (static and dynamic) and different temperatures. It is measured by the stress units ( = force/area) 9- Toughness: Is the ability of the material to resist the dynamic load (i.e., to resist shocks)
2.4 Main Mechanical Tests of Metals 2.4.1 Tensile Test Tensile test is of a static type, it is the easiest mechanical test to perform. It is carried out to determine the strength and plasticity of materials. Moreover, the result of the tensile test gives a clear idea about the other mechanical properties of the material under test, mainly its ductility. For the tensile test to be carried out, we use a test specimen and a tensile test machine. Test specimen: It is either round or flat shape cross-section. It ha s a standard shape and dimensions to be able to compare the obtained results. Fig 2.1 shows test specimens. The mechanical properties in tensions are determined on the gauge length l o of the specimen.
The elastic load P e : Is the maximum load that causes elastic deformation only, i.e., deformation that disappears when the load is removed. The corresponding stress is the elastic limit e e = P o /F o Pa or MPa. where F o : Initial cross-sectional area of the specimen Here is a line relation in the region of elastic deformation between stress and strain for metals and alloys. It confirms to the low of proportionality (Hook’s low): = E. Pa or MPa. E = / = tan Where (strain) = l/l o = (l 1 -l o )/l o The coefficient of proportionality E, called the nodules of elasticity or Young’s modulus, characterizes the rigidity of a material, i.e., its resistance to elastic deformation in tension.
Examples: 1- When testing a steel specimen of diameter D=10 mm., the maximum load P u is 28400 N. Calculate the ultimate strength u. Solution: F o = D 2 /4 = .10 2 /4 = 78.5 mm 2 = 78.5 x 10 -6 m 2 u = P u /F o = 28400/78.5 x 10 -6 = 361.8 MPa. 2- Determine the elongation of steel, if the specimen gauge lengths before and after teat lo and l1 are: 50 and 58 mm. respectively. Solution: = (l1 – lo)/lo x 100 = (58-50)/50 x 100 = 16%