chap2_3

# chap2_3 - 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 four §2.3 Crystal Structure and Complex Lattice Ⅰ.Crystal structure is the real .Crystal arrangement of atom in crystals arrangement Crystal structure 材 Space lattice + Basis Space or structure unit or + = The difference between space lattice and crystal structure Fe Fe : Al = 1 : 1 Fe : Al = 1 : 1 Al 2×3 atoms / cell Ⅱ.Typical crystal structures of metals 1. BCC Example: α-Fe , V, Nb, Ta, Cr, Mo, W, alkali metals n = 2 atoms/cell CN=8 The number of nearest neighbours around each atom is called —— Coordination Number. Packing fraction ξ＝ Volume of atoms / cell Volume of unit cell Volume To determineξ, The atom is looked as a hard sphere, and the nearest neighbours touch each other. ∴ For BCC, 4 33 2× π ( a) 3 4 ξ= = 0.68 3 a 2. FCC 2. Example: γ-Fe , Al , Ni , Pb , Cu , Ag , Au , stainless steal n= 8×1/8+6×1/2=4 atoms/cell CN=12 4 23 4× π ( ) 3 4 ξ= = 0.74 3 a 3. HCP 3. Example: Be, Mg, Zn, Cd, Zr, Hf Ti( low temperature) 1 1 n ＝ 6 × 12 + 2 × 2 + 3 = 6 •• CN ＝ 12 ξ ＝ 0.74 c 2 2 3 a =( ) + × a 2 3 2 • • 2 2 c 8 ∴= = 1.633 a 3 CN = 12 ξ = 0.74 •• • • • • • •• • • •• 4. Summary Structure Structure a0 vs. r Atoms per cell SC a0 = 2 r 1 BCC 4 a0 = r 3 FCC 4 a0 = r 2 HCP a0 = 2 r c0 ≈ 1.633a0 2 4 2 Coordination Number Packing factor Examples 6 0.52 Polonium (Po),α-Mn 0.68 Fe,Ti,W,Mo, Nb,Ta,K,Na, V,Zr,Cr 0.74 Fe,Cu,Au,Pt ,Ag,Pb,Ni 0.74 Ti,Mg,Zn,Be ,Co,Zr,Cd 8 12 12 §2.4 Interstices in typical crystals of metals §2.4 Definition: In any of the crystal structures, there are small holes between the usual atoms into which smaller atoms may be placed. These locations are called interstitial sites. Ⅰ.Two types of Interstitials in .Two typical crystals typical Octahedral interstitial Tetrahedral interstitial 1. Octahedral interstitial 1. BCC 3 a 2 a 2 a 2 FCC FCC a 2 a 2 HCP HCP a 2 2. Tetrahedral interstitial 2. BCC 3 a 2 5 a 4 a FCC FCC 3 a 4 a 2 HCP HCP 7 c 8 1 c 8 3 c 8 5 c 8 Ⅱ.Determination of the sizes of .Determination interstitials Definition: By size of an interstitial we mean diameter of the maximum hard sphere which can be accommodated in the interstitial without distorting the lattice. di ＝ da diameter of interstitial atom diameter of atom in lattice point 1. Octahedral interstitial 1. condition for touching di + d a = a di a = −1 da da For BCC For 3 da = a 2 di a = −1 = 0.15 da 3 2 For FCC 2 da = a 2 di a = −1 = 0.41 da 2 2 2. Tetrahedral interstitial 2. B L A interstitial d a + di 2 H D C host atom d a + di 2 L2 H2 ∴( ) =( ) +( ) 2 2 2 d i + d a = L2 + H 2 di ∴= da L2 + H 2 −1 da For BCC For L=a 3 da = a 2 a H= 2 di ∴ = 0.29 da For FCC 2 L= a 2 2 da = a 2 H= a 2 di = 0.22 da 3. Summary 3. n CN BCC 2 8 ξ interstices oct. tete. di/da oct. tete. 0.68 6 6/2=3 12 12/2=6 0.15 0.29 FCC 4 12 0.74 4 4/4=1 8 8/4=2 0.41 0.22 HCP 6 12 0.74 6 6/6=1 12 12/6=2 0.41 0.22 Examples and Discussions Examples 1. Both FCC and BCC are close-packed Both structures while BCC is more open? structures 2. The interstitial atoms most likely occupy the The oct. interstitial position in FCC and HCP, while in BCC two types of interstitial can be occupied equally. equally. 3. The solid solubility in BCC is much lower than 3. in FCC. in 4. Diffusion of interstitial atoms in BCC diffusion is much faster than in FCC or HCP at same temperature. temperature. 5. Determine the relationship between the atomic radius and the lattice parameter in SC, BCC, and FCC structures when one atom is located at each lattice point. at 6. Determine the density of BCC iron, which has a 6. lattice parameter of 0.2866nm. lattice Solution: For a BCC cell, Atoms/cell = 2 a0 = 0.2866nm = 2.866×10 材 8cm Atomic mass = 55.847g/mol Volume of unit cell = a03 = 23.54×10 Density cm3/cell 材 24 (number of atoms / cell)(atomic mass of iron ) ( volume of unit cell)(Avogadro's number) (2)(55.847) = = 7.882g / cm3 (23.54 × 10-24 )(6.02 × 10 23 ) ρ= Exercise Exercise 1. Determine the coordinates of centers of both Determine the octahedral and the tetrahedral interstitials in HCP referred to a, b and c. c b a 120o Thank you ! Thank you ! 4 ...
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