Engineering Materials and Their ApplicationsThis edition of the classic text/reference book has been updated and revised to provide balanced coverage of metals, ceramics, polymers and composites. The first five chapters assess the different structures of metals, ceramics and polymers and how stress and temperature affect them. Demonstrates how to optimize a material's structure by using equilibrium data (phase diagrams) and nonequilibrium conditions, especially precipitation hardening. Discusses the structures, characteristics and applications of the important materials in each field. Considers topics common to all materials--corrosion and oxidation, failure analysis, processing of electrical and magnetic materials, materials selection and specification. Contains special chapters on advanced and large volume engineering materials plus abundant examples and problems. |
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Page 455
... cm / in . ) 2 x ( π / 4 ) 107.9 x 10-6 ohm - cm 1 = 8,600 cm = 86 m It is more positive and simpler in the text ahead to think of the material as conducting rather than resisting the passage of current , so we use the well - known ...
... cm / in . ) 2 x ( π / 4 ) 107.9 x 10-6 ohm - cm 1 = 8,600 cm = 86 m It is more positive and simpler in the text ahead to think of the material as conducting rather than resisting the passage of current , so we use the well - known ...
Page 523
... ( cm / sec - K ) or cal / cm - sec - K . The third is the change in length per unit of length per K , that is , cm / cm - K or simply ( K ) -1 . [ The interrelation of these units with British thermal units ( BTU ) and with the joule ...
... ( cm / sec - K ) or cal / cm - sec - K . The third is the change in length per unit of length per K , that is , cm / cm - K or simply ( K ) -1 . [ The interrelation of these units with British thermal units ( BTU ) and with the joule ...
Page 532
... cm - sec - K . Thermal expansion The change in unit length per unit of temperature change ; AL / AT cm / cm - K = ( K ) -1 . = PROBLEMS 15.1 The most valuable diamonds have a bluish tint . Given that the wave- length of blue light is ...
... cm - sec - K . Thermal expansion The change in unit length per unit of temperature change ; AL / AT cm / cm - K = ( K ) -1 . = PROBLEMS 15.1 The most valuable diamonds have a bluish tint . Given that the wave- length of blue light is ...
Contents
A General View of the Problems | 3 |
Summary | 14 |
Summary | 45 |
Copyright | |
19 other sections not shown
Other editions - View all
Engineering Materials and Their Applications Richard Aloysius Flinn,Paul K. Trojan Snippet view - 1986 |
Engineering Materials and Their Applications Richard Aloysius Flinn,Paul K. Trojan Snippet view - 1975 |
Engineering Materials and Their Applications Richard Aloysius Flinn,Paul K. Trojan Snippet view - 1975 |
Common terms and phrases
0.8 percent carbon 10-3 to obtain alloys aluminum Annealed atoms austenite bainite bonds brittle Calculate carbon content cast iron ceramics Chap chemical chromium cold-worked composition cooling rate copper corrosion covalent crystal density diffusion discussed ductile iron effect electrons elements engineering eutectoid example Fe2+ ferrite fibers fracture glass grain graphite gray iron H H H hardening hardness heat treatment important ions iron carbide liquid load magnesium martensite material matrix melt metal microstructure mold molecules Multiply psi nickel obtain kg/mm² obtain MN/m² oxide oxygen pearlite Percent Elongation percent silicon phase diagram plane plastic polyethylene polymer precipitate produce properties quenched reaction resistance room temperature shown in Fig silica single-phase slip solid solution specimen strain stress structure surface Table tempered tensile strength thermoplastic thermosetting titanium transformation two-phase typical unit cell Weight percentage yield strength zinc