Phy351 ch 1


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Phy351 ch 1

  1. 1. PHY351 CHAPTER 1 Gas and Condensed Matter
  2. 2. What are Materials? Materials may be defined as substance of which something is composed or made. We obtain materials from earth crust and atmosphere. Examples : Silicon and Iron constitute 27.72 and 5.00 percentage of weight of earths crust respectively.  Nitrogen and Oxygen constitute 78.08 and 20.95 percentage of dry air by volume respectively.
  3. 3. Why the Study of Materials is Important? Production and processing of materials constitute a large part of our economy. Engineers choose materials to suite design. New materials might be needed for some new applications. Example :- High temperature resistant materials. - Space station and Mars Rovers should sustain conditions in space. (* High speed, low temperature, strong but light.) Modification of properties might be needed for some applications. Example :- Heat treatment to modify properties.
  4. 4. Materials Science and Engineering Materials science deals with basic knowledge about the internal structure, properties and processing of materials. Materials engineering deals with the application of knowledge gained by materials science to convert materials to products. Materials Science Basic Knowledge of Materials Materials Science and Engineering Resultant Knowledge of Structure and Properties Materials Engineering Applied Knowledge of Materials
  5. 5. Type of Material Most engineering materials are devided into three main or fundamental classes: Metallic material Polymeric material Ceramic In addition to the three main classes of materials, the other two processing or application classes are: Composite material Electronic material
  6. 6. Metallic Materials  Composed of one or more metallic elements. (Example:- Iron, Copper, Aluminum)  Metallic element may combine with nonmetallic elements. (Example:- Silicon Carbide, Iron Oxide)  Inorganic and have crystalline structure.  Good thermal and electric conductors. 6
  7. 7. Metallic Materials (cont..) Metals and Alloys Fig: The aircraft turbine engine is made principally of metal alloy. Ferrous Eg: Steel, Cast Iron Nonferrous Eg:Copper Aluminum 7
  8. 8. Polymeric (Plastic) Materials      Organic giant molecules and mostly noncrystalline. Some are mixtures of crystalline and noncrystalline regions. Poor conductors of electricity and hence used as insulators. Strength and ductility vary greatly. Low densities and decomposition temperatures. Examples :Poly vinyl Chloride (PVC), Polyester ( Applications :- DVDs, Fabrics etc. ) 8
  9. 9. Ceramic Materials  Metallic and nonmetallic elements are chemically bonded together.  Inorganic but can be either crystalline, noncrystalline or mixture of both.  High hardness, strength and wear resistance.  Very good insulator. Hence used for furnace lining for heat treating and melting metals.  Also used in space shuttle to insulate it during exit and reentry into atmosphere.  Other applications : Abrasives, construction materials, utensils etc. Example:Porcelain, Glass, Silicon nitride. 9
  10. 10. Composite Materials (cont..)     Mixture of two or more materials. Consists of a filler material and a binding material. Materials only bond, will not dissolve in each other. Mainly two types :o Fibrous: Fibers in a matrix o Particulate: Particles in a matrix (Matrix can be metals, ceramic or polymer)  Examples : Fiber Glass ( Reinforcing material in a polyester or epoxy matrix)  Concrete ( Gravels or steel rods reinforced in cement and sand)  Applications:- Aircraft wings and engine, construction. 10
  11. 11. Electronic Materials  Not major by volume but very important.  Silicon is a common electronic material.  Its electrical characteristics are changed by adding impurities.  Examples:Silicon chips, transistors  Applications :Computers, Integrated Circuits, Satellites etc. 11
  12. 12. Competition Among Materials • Materials compete with each other to exist in new market • Over a period of time usage of different materials changes depending on cost and performance. • New, cheaper or better materials replace the old materials when there is a breakthrough in technology. Aluminum Iron Plastic Steel 1600 1400 lb/Car 1200 1000 800 600 400 200 0 1985 1992 1997 Model Year Predictions and use of materials in US automobiles. 12
  13. 13. Recent Advances and Future Trends Smart Materials  React to environment Stimuli (temperature, strees, light, humidity and electric and magnetic field).  Change their properties by sensing external stimulus.  Examples: Shape memory alloys – used in the artery stents.  Microelectromechanical systems (MEMS) devices. 13
  14. 14. Nanomaterials  Smaller than 100 nm particle size.  Materials have special properties.  Very hard and strong characteristics.  Research in progress.  Example: Carbon nanofiber reinforced plastic: very light but stronger than metals. 14
  15. 15. Force between particles Nucleus of one ion attracts electron of another ion. The electron clouds of ion repulse each other when they are sufficiently close. These two forces will balance each other when the equilibrium interionic distance, a0, is reached and a bond is formed. Fig 2.16 The attraction repulsion forces developed during ionic bonding. Note that net force is zero when the bond is formed.
  16. 16. Net force Cl- Na+ a0 Fnet = Fattraction + Frepulsion 16
  17. 17. Attraction force Fattraction = -kZ1Z2 e2 a2 Where; Z1,Z2 = number of electrons removed or added during ion formation e = electron charge = 1.6 x 10-19C k = 8.99 x 109 Nm2/C2 a = interionic seperation distance 17
  18. 18. Example: Calculate the force of attraction between Na+ and Cl- ions. Given; Z1 = +1 for Na+ Z2 = -1 for Cle = 1.60 x 10-19 C ε0 = 8.85 x 10-12 C2/Nm2 a0 = sum of radii of Na+ and Cl- ions = 0.095 nm + 0.181 nm = 2.76 x 10-10 m (1)(1)(1.60 10 C )  Z Z e   4  a  4 (8.85 x 10 C /Nm2)(2.76 x 10 2 F 1 attraction 19 2 2 2 0 -12 2 -10 m)  3.02 109 N 18
  19. 19. Repulsion force Frepulsion = -nb an+1 Where n and b = constant a = interionic distance Example: The repulsion force between Na+(r = 0.095nm and Cl- (r = 0.181nm) ions at equilibrium is -3.02 x 10-9N. Assume n = 2. Calculate the value of constant b.
  20. 20. References  A.G. Guy (1972) Introduction to Material Science, McGraw Hill.  J.F. Shackelford (2000). Introduction to Material Science for Engineers, (5th Edition), Prentice Hall.  W.F. Smith (1996). Priciple to Material Science and Engineering, (3rd Edition), McGraw Hill.  W.D. Callister Jr. (1997) Material Science and Engineering: An Introduction, (4th Edition) John Wiley.