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清华大学:《材料科学基础》课程教学资源(PPT课件讲稿)Chapter 4.8 TwinnIng

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4. Rule for elongation or reduction of R D All R.D. in the region where K1 and K2 makes an obtuse(acute) angle will increase (reduce) its length after twinning
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34.8 Twinning Macroscop ic Aspect .L=l+r(l·n)b I Rules governing the change in length 1. Change in shape of a hemisphere with unit radius before twinning: x2+y+z=1 x=xy=y (2) substitute Into x'+(y'-rz)+z=1 equation for ellipsoid

§4.8 Twinning (I) ——Macroscopic Aspect ◆ Ⅰ. ◆ Ⅱ.Rules governing the change in length. 1. Change in shape of a hemisphere with unit radius. before twinning: L l r l n b      = + (  ) z 1 (1) 2 2 2 x + y + = x  = x y  = y z  = z (2) substitute (2) into (1) ( ) z 1 equation for ellipsoid 2 2 2 x  + y − rz  +  =

2. Undistorted planes and directions Ki--T. P 7h1--T.D IK In2--the intersecti on of K2 and shear plane Twin elements-K1、mh,K2、m

2. Undistorted planes and directions.    − − − − T.D. T.P. 1 1  K    − −the intersecti on of and shear plane 2 2 2 K K  Twin elements —— K1、1,K2、2

K K FCC(111) 112 11)112 BCC(12)[1(12)11 HCP(012)01(02)oi 1012)[1011 (1012)101 K1,7 K2, n2 K1,7 FCC BCC HCP

K1 1 K2 2 FCC BCC HCP (111) (112) (1 11) (1012) (1012) (1 12) [1 12] [11 1] [1011] [112] [111] [1011] FCC K1 , 1 K2 , 2 BCC K1 , 1 K2 , 2 c a1 a2 a3 HCP (1012)[1011] (1012)[1011]

3.y=AA=2ctg(20) 72K2 72K2 无法显示该图片 B 2b2 K 7 4. Rule for elongation or reduction of RD All R.D. in the region where K, and K, makes an obtuse(acute)angle will increase(reduce) its length after twinning

3. 4. Rule for elongation or reduction of R.D. All R.D. in the region where K1 and K2 makes an obtuse (acute) angle will increase (reduce) its length after twinning.  = AA = 2ctg(2) C D

◆Ⅲ. Appl ication: take hcp for examp le Suppose:K1=(1012),K2=(1012) crystal Zn(-=1.86) (10121011 When=√320=90 (10121011 √3 √3a/2√3 tgp /2c/a /2 a> acute angle a<√3 obtuse angle Zn(=1.86)→2小=86 HCP

◆ Ⅲ.Application: take HCP for example Suppose: , (1012) K1 = (1012) K2 = crystal ( =1.86) a c Zn c c a a / 3 / 2 3 / 2 tg = = 2 3a c/2 (1012) c a1 a2 a3 HCP (1012)[1011] [1011] c a  3 acute angle c a  3 obtuse angle Zn( =1.86)  2 = 86 a c  When = 3 2 = 90 a c

(1102)(0112 K12K2 I,ID(IILIVI(VVD) 29 (1012) zone A ⅥI 1210 B (1012) B D D 110 1010

C I II III VI IV V 12 1 0 1120 10 1 0 21 10 (1012) (0112) (0112) (1 120) (1102) (1012) (I,II) (III,IV) (V,VI) 2 A - - - B + - - C + - + D + + + zone1 2 K ,K

Examples and Discussions

Examples and Discussions

Exercise

Exercise

Thank you 15

Thank you ! 15

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