chap3_1 - 材材材材材材 Fundamental of Materials Prof...

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Unformatted text preview: 材材材材材材 Fundamental of Materials Prof: Tian Min Bo Prof: Tian Tel: 62795426 材 62772851 E-mail: [email protected] [email protected] Department of Material Science and Engineering Tsinghua University. Beijing 100084 Lesson eight Chapter Ⅲ Chapter The Structures of Alloys §3.1 Basic concepts of alloys §3.1 Ⅰ.Definition An alloy is the combination of metal(s) with other elements through chemical bonding. chemical Ⅱ.Terminology 1. Component (or constituent) 1. Component one component system two component system three component system four component system five component system binary system ternary system quarternary system quinary system multi-component multi-component system system 2. Composition 2. Composition It can be expressed either by atomic percentage (mol fraction) Xa or by mass percentage Xm X ai = Wmi M i n ∑(Xm j =1 j M j) X mi = X ai M i n ∑ ( X aj M j ) j =1 3. Phase 3. A phase is a homogeneous part of the material in which no abrupt change in composition, structure and properties occurs. An alloy may be single phase or multi-phase material. 4. Structure Structure is a general term for the combination of atom arrangement including the types amounts and distribution of all types of material as well as grain size, defect etc. Ⅲ. Classification of Alloy Phases 1. According to structure 1. structure Solid Solution Compound Compound In Solid Solution, atoms of different component share a common lattice in variable proportion. proportion. 2. According to position of the alloy in phase diagram diagram Terminal S.S. α β Intermediate S.S. Intermediate or secondary S.S. α+β A B §3.2 Factors Affecting the Structure of Alloy Phase Alloy 1. What is size factor Atomic radii metallic radii rA+rB =d ionic radii r++r - =d covalent radii single bond radius Van der Waals radii CN 12 → 8 12 → 6 12 → 4 BCC 3 a )+6(a) CN of BCC:8( CN 2 3 3 a α= × 0.290nm r α= 4 4 = 0.12557nm radii ∆ r = 3% − r ∆ r = 4% − r ∆ r = 12% − r FCC 2 a) CN of FCC:12( 2 2 2 aβ rβ= = 4 × 0.364nm 4 = 0.128674nm Goldschmidt atomic radii is the radii of atom in structures with CN=12 structures Size factor δ is defined as δ = d A − dB dA ×100% 2. What is Electrochemical factor —— Electronegativity X X represents the ability of an atom of an element in the compond to attract electrons to itself. Pauli’s empirical rule: n: valence n +1 X = 0.31 + 0.5 r(1) r(1): simple bond radius ( XA − XB) 2 EAA − EBB ∝( ) − EAB 2 EAA—— bonding energy between A-A atoms —— A-A atoms EBB—— bonding energy between B-B atoms —— B-B EAB—— bonding energy between A-B atoms —— A-B atoms 3. Electron concentration ( e/a ) e/a Electron concentration ( e/a ) is the number of valence Electron electrons per atom on the average. electrons i.g. for CuZn : e/a = 3/2 =1.5 e/a §3.3 Solid Solution §3.3 Ⅰ. Classification 1. According the position of solute atoms in the 1. lattice of the solvent lattice Substitutional S.S. Substitutional Interstitial S.S. Interstitial 2. According the regularity of the position occupied by solute atoms by Ordered S.S. Ordered Disordered S.S. Disordered Al : 4 Fe : 12 1 材 3 材 12×1/4 材 4 材 1 材 12 ∴ Fe12Al4 材 Fe3Al 3. According to solid solubility solid 0~100% continuous series of S.S 0~100% S.S with restricted solubility S.S summary substitutional S.S primary (terminal) interstitial S.S secondary (intermediate) ordered S.S continuous S.S disordered S.S S.S with restricted solubility Ex. Write out in full the coordinates of all Ex. cations and anions in nucleus, Wurtzite and CaF2 referred to a b c Wurtzite axes of the anions sublattice. axes Ⅱ. Determination of types of S.S nM = ρ exp V M ∴ n= Vρexp M is the average atomic weigh weighted by composition Compare n with no (atoms per unit cell of solvent) solvent) n=no : ideal substitutional S.S. n>no : interstitial S.S. n<no : vacant S.S. Ⅲ. Hume-Rothery Rule for primary Hume-Rothery solid solubility solid 1. Size factor: 1. Size factor 材 Size d0-dt d0 ×100 材 ×100 If size factor > 15 > solubility is very small. 15 solubility Example NiO can be added to MgO to produce a solid solution. What other ceramic systems are likely to exhibit 100% solid solubility with MgO? r(Å) rion − rMg + 2 ×100% rMg + 2 crystal structure Cd+2 in CdO 0.97 47 NaCl Ca+2 in CaO 0.99 50 NaCl Co+2 in CoO 0.72 9 NaCl Fe+2 in FeO 0.74 12 NaCl Sr+2 in SrO 1.12 70 NaCl Zn+2 in ZnO 0.74 12 NaCl FeO-MgO system will probably display unlimited solid solubility. CoO and ZnO systems also have appropriate radius ratios and crystal structures. d 1.15d0 d0 0.85d0 Z1 Z2 Z 2. Crystal structure 2. The materials must have the same crystal structure; otherwise there is some point at which a transition occurs from one phase to a second phase with a different structure. 3. Electrochemical factor 3. If the difference in X is great, the solubility is also very restricted. Formation of stable compound will restrict the solid solubility. Parameter: Electronegativity (x) Semiemperies formulas n′ + 1 X = 0.31 r + 0.5 1 Where: r1——single bond radius n ——valency x x0+0.4 x0 x0 > 0.4 0.85R0 R0 1.15R0 R Dorken-Gurry graphic 4. Electron concentration factor, e/a e/a e: the number of valence electrons a: the number of atoms e/a = average number of valence electrons per atom. Experimental findings: Experimental a. Zn, Ga, Ge, As in Cu (solute-solvent) If composition is expressed in terms of e/a rather than at%, the solid solubility of all elements in Cu will be roughly the same. b. b. Structure vs. e/a e/a for CuZn alloy system α(CuZn) —— e/a = 3/2 = 21/14 β(Cu5Zn8)—— e/a = 21/13 γ(CuZn3) —— e/a = 7/4 = 21/12 Ⅳ. Properties of S.S 1. lattice constants properties 1. Vagard’s law ass = ao + (a - ao)x for alloy S.S Δa = K(ZA-ZB)2 ZA, ZB are valences of solute and solvent. >>> a Au Cu Ni >>> 0 1.0 x 2. Mechanical properties σ0.2 is great, ductility is lower —— solid solution strengthening. strengthening. 3. Electrical properties IIn general ρS.S > ρele n S.S ρB L L+S ρA S Cu-Ni Examples and Discussions Examples Exercise Exercise Thank you ! Thank you ! 8 ...
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