4020-F2011-HW8Solutions

# 4020-F2011-HW8Solutions - C 12 Therefore the shear modulus...

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1 MSE 4020 HW8-SOLUTION 10/18/2011 Problem 1 ! xy = "# b # < x > " : dislocation density " = 1cm 1cm 3 = 1cm \$ 2 = 1 # 10 4 m \$ 2 b :Burger's vector b = 2 ! A = 2 # 10 \$ 10 m < x > :Dislocation displacement < x >= 1cm = 1 # 10 \$ 2 m % ! xy = 1 # 10 4 # 2 # 10 \$ 10 # 1 # 10 \$ 2 = 2 # 10 \$ 8 % x ! xy ! xy y ' ( ) ) * + , , = 0 2 # 10 \$ 8 2 # 10 \$ 8 0 ' ( ) ) * + , ,

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2 Problem 2 ! = G " b 2 4 # ln R r 0 G: Shear modulus b :Burger's vector r 0 : dislocation core radius R : cutoff distance of dislocation strain field Let, r 0 = 5b; R = 20b; \$ ! = G " b 2 4 # ln4 = 5 " 10 10 " 2.4 " 10 % 10 ( ) 2 4 " 3.14159 = 3.1772 " 10 % 10 J m = 1.98575 " 10 7 eV cm 1cm = N " a nickel \$ N = 1cm a nickel = 1 2.4 " 10 % 8 = 4.1667 " 10 7 \$ ! = 1.98575 " 10 7 eV N " a nickel = 1.98575 " 10 7 4.1667 " 10 7 " eV a nickel = 0.4766 eV a nickel
3 Problem 3 Force by external shear stress is equal to, ! " b = # y 2 " b Force by the image dislocation is equal to, G ! b 2 4 "! x # \$ y 2 ! b = G ! b 2 4 "! x # x = G ! b 2 "!\$ y (a) G = C 44 = 7.54 ! 10 10 Pa # x = 7.54 ! 10 10 ! 0.255 ! 10 % 9 2 ! 3.14159 ! 70 ! 10 6 = 4.3715 ! 10 % 8 m (b) G = C 11 % C 12 2 = 2.35 ! 10 10 Pa # x = 2.35 ! 10 10 ! 0.255 ! 10 % 9 2 ! 3.14159 ! 70 ! 10 6 = 1.3625 ! 10 % 8 m S X X S

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4 (c) Shear modulus depends on crystal orientation. For Cu, the Burger’s vector of screw dislocation is, a 2 110 ! " # \$ which is 45° rotation with respect to x or y axis Referring back to HW5, we can see under rotation, G = 1 2 C 11 !

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Unformatted text preview: C 12 ( ) Therefore, the shear modulus above should apply to this problem. 5 Problem 4 Similar to problem 1, ! xy = "# b # < x > " : dislocation density " = 1cm 1cm 3 = 1cm \$ 2 = 1 # 10 4 m \$ 2 b :Burger's vector b = 2 ! A = 2 # 10 \$ 10 m < x > :Dislocation displacement < x >= 1 2 cm = 0.5 # 10 \$ 2 m % ! xy = 1 # 10 4 # 2 # 10 \$ 10 # 0.5 # 10 \$ 2 = 1 # 10 \$ 8 rad 6 Problem 5 dislocation sense vector: along z direction (b) Burger’s vector along x direction -> i.e., perpendicular to the sense vector -> edge dislocation (c) Burger’s vector along y direction -> i.e., perpendicular to the sense vector -> edge dislocation (d) Burger’s vector along z direction -> i.e., parallel to the sense vector -> screw dislocation...
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4020-F2011-HW8Solutions - C 12 Therefore the shear modulus...

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