4020-F2011-HW2Solutions - MSE 4020 HW2-SOLUTION 09/05/2011...

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1 MSE 4020 HW2-SOLUTION 09/05/2011 Problem 1 (a) ! (a) = 0 0 0 F 0 A 0 (b) F = ! 2 2 F 0 ! 2 2 F 0 (c) Force along x direction F x = F 0 cos θ , and F y = F 0 sin θ F x y F x F y x y θ F
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2 ! xx = F x A x = F 0 cos " A 0 / cos " = F 0 A 0 cos 2 " ! yy = F y A y = F 0 sin " A 0 / sin " = F 0 A 0 sin 2 " ! xy = ! yx = F x A y = F 0 cos " A 0 / sin " = F 0 A 0 sin " cos " Thus ! (c) = ! xx ! xy ! yx ! yy = F 0 A 0 cos 2 " F 0 A 0 sin " cos " F 0 A 0 sin " cos " F 0 A 0 sin 2 " The applied force is antiparallel to y axis, thus, Fx = 0, Fy = - F 0 , i.e., F = 0 ! F 0 (d) Compare σ (a) and σ (c) from above, Tr( σ (a)) = Tr( σ (c)) = F 0 /A 0 Det( σ (a)) = Det( σ (c)) = 0
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3 Problem 2 (a) Det ! x "! N # # ! y "! N = 0 $ ! N 2 " ! x + ! y ( ) ! N + ! x ! y " # 2 = 0 $ ! N1 = ! x + ! y ( ) + ! x + ! y ( ) 2 " 4 ! x ! y " # 2 ( ) 2 = ! x + ! y ( ) + ! x "! y ( ) 2 + 4 # 2 2 ! N2 = ! x + ! y ( ) + ! x + ! y ( ) 2 " 4 ! x ! y " # 2 ( ) 2 = ! x + ! y ( ) " ! x "! y
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4020-F2011-HW2Solutions - MSE 4020 HW2-SOLUTION 09/05/2011...

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