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4020-F2011-PrelimSolutions

# 4020-F2011-PrelimSolutions - MSE 4020 Prelim#1-SOLUTION...

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1 MSE 4020 Prelim #1-SOLUTION 10/05/2011 Problem 1 Det ! xx " ! N ! xy ! xz ! yx ! yy " ! N ! yz ! zx ! zy ! zz " ! N = 0 # Det ! xx " ! N ! xy ! xz ! xy ! yy " ! N ! yz ! xz ! yz ! zz " ! N = 0 # ! N 3 " ! xx + ! yy + ! zz ( ) ! N 2 + ! xx ! yy + ! yy ! zz + ! xx ! zz " ! xy 2 " ! xz 2 " ! yz 2 ( ) ! N " "! xz 2 ! yy + 2 ! xy ! xz ! yz " ! xy 2 ! zz + ! xx "! yz 2 + ! yy ! zz ( ) ( ) Thus, I 1 = ! xx + ! yy + ! zz I 2 = ! xx ! yy + ! yy ! zz + ! xx ! zz " ! xy 2 " ! xz 2 " ! yz 2

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2 Problem 2 Fail if the following conditions are satisfied, ! 1 N + ! 2 N + ! 3 N 3 = ! 1 N + ! 2 N 3 " P cave # 3p cave 3p cave ! 2 N ! 1 N SAFE UNSAFE
3 Problem 3 (a) For homogeneous strain, in general we can write down following displacement field for 2D problem: u x = r ! x + s ! y + t u y = m ! x + n ! y + k where, r, s, t, m, n, k are constants to be determined from given geometry. For Point A(0,0), after deformation, its position is unchanged, thus, u x = r ! 0 + s ! 0 + t u y = m ! 0 + n ! 0 + k " # \$ % \$ & t = 0 k = 0 ' ( \$ ) \$ For Point B(2,3), after deformation, its position becomes (1.9,3.05), thus, 1.9 ! 2 = r " 2 + s " 2 + 0 = ! 0.1 1.9 ! 2 = m " 2 + n " 2 + 0 = ! 0.1 # \$ % & 2r + 2s = ! 0.1 2m + 2n = ! 0.1 ' ( ) * ) For Point C(5,3), after deformation, its position becomes (4.8,3), thus, 4.8 ! 5 = r " 5 + s " 2 + 0 = ! 0.2 1.8 ! 2 = m " 5 + n " 2 + 0 = ! 0.2 # \$ % & 5r + 2s = ! 0.2 5m + 2n = ! 0.2 ' ( ) * ) Solve the above four equations for the four unknowns, giving that r = ! 0.03333 s = ! 0.01667 m = ! 0.03333 n = ! 0.01667 " # \$ \$ % \$ \$

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