4020-F2011-Prelim2Solutions

4020-F2011-Prelim2Solutions - MSE 4020 Prelim #2-SOLUTION...

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1 MSE 4020 Prelim #2-SOLUTION 11/27/2011 Problem 1 F SN = ! D 2 4 2K ( ) = ! D 2 K 2 ! D "#" K = F SN = ! D 2 K 2 $ # = D 2 $ # = # / 1 n = D 2 / 1 n = D " n 2 = 1 " 16 2 = 8
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2 Problem 2 (a) ! " z = D #" 4L = D # 3 $ 4L = D # 3 $ 4L = 6 # 3 $ 4 # 30 = 3 $ 20 % ! f = 2 3 #! " z = 2 3 # 3 $ 20 = 3 $ 10 f = 30000 + 20000 # ! f ( ) 1/2 = 30000 + 20000 # 3 $ 10 ' ( ) ) * + , , 1/2 = 44753 psi f = 3 " z " z = f 3 = 25838 psi Twist moment M = $ D # t ( ) area !"# " z # D / 2 = $ D # t ( ) # f 3 # D / 2 = $ D 2 # t 2 ' ( ) * + ,# f 3 = $ # 6 2 # 3 8 2 ' ( ) ) ) ) * + , , , , # 44753 3 = 54792 in # lb = 45660 ft # lb
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3 (b) Before twisting: ! f = 30000 + 20000 " # f ( ) 1/2 = 30000 psi $ ! y = ! f 3 = 17321psi After twisting: ! f = 30000 + 20000 " # f ( ) 1/2 = 44753 psi $ ! y = ! f 3 = 25838 psi
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4 Problem 3 1 -> No dislocation 2 -> No dislocation 3 -> No dislocation 4 -> Screw dislocation 5 -> Frank dislocation 6 -> Edge dislocation 7 -> No dislocation
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5 Problem 4. Force by external shear stress is equal to,
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This note was uploaded on 12/05/2011 for the course MSE 4020 at Cornell.

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4020-F2011-Prelim2Solutions - MSE 4020 Prelim #2-SOLUTION...

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