{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4020-F2011-HW9Solutions

4020-F2011-HW9Solutions - MSE 4020 HW9-SOLUTION Problem 1 1...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
1 MSE 4020 HW9-SOLUTION 10/26/2011 Problem 1 1. For two parallel edge dislocations with antiparallel Burger’s vectors (shown in the figure above), the force F(Glide force F x and Climing force F y ) can be written as: F x = ! Gb 2 2 " (1 ! # ) x(x 2 ! y 2 ) (x 2 + y 2 ) 2 (1) F y = ! Gb 2 2 " (1 ! # ) y(3x 2 + y 2 ) (x 2 + y 2 ) 2 (2) (a) x = -2, y = 1 From Eq. 1 F x = 2.28 ! 10 " 3 dym / cm (b) x = -y F x = 0 dym / cm (c) x = 0 F x = 0 dym / cm (d) x y F x F y r θ A B
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 x = y F x = 0 dym / cm (e) dF x dx = ! 9.502 " 10 ! 7 ( ) " (x 2 + y 2 ) 2 " (3x 2 ! y 2 ) ! 4x 2 " (x 2 ! y 2 ) " (x 2 + y 2 ) (x 2 + y 2 ) 4 # $ % & ' ( = 0 ) (x 2 + y 2 ) 2 " (3x 2 ! y 2 ) ! 4x 2 " (x 2 ! y 2 ) " (x 2 + y 2 ) = 0 ) (x 2 + y 2 ) " (3x 2 ! y 2 ) ! 4x 2 " (x 2 ! y 2 ) = 0 ) x 4 ! 6x 2 y 2 + y 4 = 0 ) x 2 = ! ! 6y 2 ± 36y 4 ! 4y 4 2 = 3 ± 2 2 ( ) y 2 y = 1 μ m ) x 2 ( ) 1 = 5.83 μ m ) x 1 = 2.414 μ m ) F x = 2.4 " 10 ! 3 dyn / cm x 2 = ! 2.414 μ m ) F x = ! 2.4 " 10 ! 3 dyn / cm * + , - , x 2 ( ) 2 = 0.41 μ m ) x 1 = 0.41 μ m ) F x = ! 2.4 " 10 ! 3 dyn / cm x 2 = ! 0.41 μ m ) F x = 2.4 " 10 ! 3 dyn / cm * + , - , * + , , , - , , , ) F x max = 2.4 " 10 ! 3 dyn / cm
Background image of page 2
3 Problem 2 ! SF = μ b 2 8 " d 2 # v 1 # v 1 # 2vcos2 $ 2 # v % & ' ( ) * where b is the Burger’s vector of a partial dislocation, and angle β is the angle between Burger’s vector (before dissociation) and corresponding dislocation line. For screw dislocation, β =0 ! " SF = Gb 2 8 # d $ 2 % 3v 1 % v b = a 0 6 ! " SF = Ga 0 2 288 # d $ 2 % 3v 1 % v In SIunit ! d = Ga 0 2 288 #" SF $ 2 % 3v 1 % v = 10 11 $ 4 $ 10 % 10 ( ) 2 288 $ 3.142 $ 30 $ 10 % 3 2 % 3 $ 0.33 1 % 0.33 = 8.9 $ 10 % 10 m
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4 Problem 3 Dislocation line sense vector: ! = 1 2 1 01 " # $ % Burger’s vector: ! = a 2 1 01 " # $ % Stress state: ! ! = 0 0 0 0 0 0 0 0 100 " # $ $ $ % & ' ' ' psi In general, ! F = G ! " where G = # $ b = 0 0 0 0 0 0 0 0 100 % & ' ' ' ( ) * * * $ a 2 1 0 1 % & ' ' ' ( ) * * * = 0 0 50a % & ' ' ' ( ) * * * + ! F = G ! " =
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}