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
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
5 Problem 4. Force by external shear stress is equal to, ! " b = # y 2 " b Force by the image dislocation is equal to, F img = G ! b 2 4 "! d # \$ y 2 ! b = G ! b 2 4 "! d # d = G ! b 2 "!\$ y = 3.9 ! 10 6 ! 2.86 2 "! 40 ! 10 3 = 44.38 Angstrom Note: If using the units of MPa, the final answer will be different(much larger than the value above) because of inconsistent transformation between Ksi and MPa. Anyway, full credits are given as long as you showed the equation for ‘d’.

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6 Problem 5.
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