chap2_7

chap2_7 - 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 six §2.7 Some Important Crystallographic Formulas Formulas 1. Zone and zone equation All Planes parellel to common axis [u v w] constitute a (crystal) zone, the [u v w] zone. If (h k l) belongs to zone [u v w], then hu 材 kv 材 lw 材 0 2. If hu 材 kv 材 lw 材 0 材 then [u v w] lies in (h k l) 2. hu 3. For cubic crystals, [h k l] ⊥(h k l) 3. cubic 4. The zone [u v w] contains two planes (h1 k1 l1) 4. and (h2 k2 l2),then and u :v:w = (k1l2 − k 2l1 ) : (l1h2 − l2 h1 ) : (h1k 2 − h2 k1 ) 5. The plane (h k l) belongs to two zones [u1 v1 w1] and [u2 v2 w2] if and h:k :l = (v1w2 − v2 w1 ) : ( w1u2 − w2u1 ) : (u1v2 − u2 v1 ) 6. The distance between adjacent plane 6. ( interplanar distance ) interplanar d(hkl) = f(a,b,c α,β,γ h,k,l) f(a,b,c For cubic: d= For orthorhombic: a h2 + k 2 + l 2 1 h2 k 2 l 2 = 2+ 2+ 2 2 d a b c For hexagonal crystals: 2 2 2 1 4 h + hk + k l = +2 2 2 3 d a c 7. The length of [u v w] 7. L[uvw] = (ua ) + (vb ) + ( wc ) + 2vwbc cos α 2 2 2 + 2uwac cos β + 2uvab cos γ L[uvw] = a = u 2 + v 2 + w2 For cubic: 8. The angle Ф between (h1 k1 l1) and (h2 k2 l2) 8. For cubic: cos ϕ = h1h2 + k1k 2 + l1l2 2 2 2 2 2 2 (h1 + k1 + l1 )(h2 + k 2 + l2 ) For orthorhombic: For h1h2 k1k 2 l1l2 + 2+2 2 a b c cos ϕ = h1 2 k1 2 l1 2 h2 2 k 2 2 l2 2 [( ) + ( ) + ( ) ] × [( ) + ( ) + ( ) ] a b c a b c For hexagonal: For cos ϕ = 3a2 1 h1h2 + k1k 2 + ( 2 ) l1l2 + (h1k 2 + h2 k1 ) 4c 2 3a2 2 3a2 2 2 2 2 2 h1 + k1 + ( ) l1 + h1k1 h2 + k 2 + ( ) l2 + h2 k 2 4c 4c 9. The angle ϕbetween [u1 v1 w1] and 9. and [u2 v2 w2] For cubic: cos ϕ = u1u2 + v1v2 + w1w2 2 2 u1 + v1 + w1 2 ⋅ 2 2 u2 + v2 + w2 2 For orthorhombic: For cos ϕ = u1u2 a 2 + v1v2b 2 + w1w2c 2 (u1a) 2 + (v1b) 2 + ( w1c) 2 ⋅ (u2 a) 2 + (v2b) 2 + ( w2c) 2 For hexagonal: cos ϕ = For c2 1 u1u2 + v1v2 + w1w2 ( ) − (u1v2 + u2 v1 ) a 2 2 2 2c2 2 2 2c2 u1 + v1 + w1 ( ) − u1v1 ⋅ u2 + v2 + w2 ( ) − u2 v2 a a 10. The volume of unit cells V 10. V = abc 1 − cos 2 α − cos 2 β − cos 2 γ + 2 cos α cos β cos γ §2.8 Stacking Mode of Crystals §2.8 Ⅰ.A crystal can be considered as the result .A of stacking the atomic layers, say (h k l ), layers, one over another in a specific sequence. another For simple cubic (001) aaaa…… (110) abab…… 1 shift = [ 1 10] , 2 along [ 1 10] This sequence is called the stacking order Ⅱ.Comparison of stacking mode of .Comparison HCP and FCC HCP 1. HCP HCP stacking order: ABABABAB…… 2. FCC 2. Stacking order of (111) 材 ABCABCABC…… A B A C A B B C C Ⅲ.Stacking fault • For HCP: normal order: fault order: • ABABAB…… ABCABAB…… For FCC 材 normal order: ABCABCABC…… fault order: ABCACABCA…… ABCACBCABC …… ABCABABAB …… ABCACBA …… Ⅳ.Transformation of hexagonal to .Transformation rhombohedral indices and vice versa rhombohedral cH aR aH cR bR bH 1 1 1 aR = aH − bH + cH 3 3 3 1 2 1 bR = aH + bH + cH 3 3 3 2 1 1 cR = − aH − bH + cH 3 3 3 or a R 1 − 1 1 a H b = 1 1 b 2 1 H R 3 cR − 2 − 1 1 cH Examples and Discussions Examples Exercise Exercise Thank you ! Thank you ! 6 ...
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This note was uploaded on 10/15/2011 for the course ENG 242 taught by Professor Mingjiezhao during the Spring '11 term at BYU.

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