4020-F2011-PrelimSolutions

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

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Unformatted text preview: 1 MSE 4020 Prelim #1-SOLUTION 10/05/2011 Problem 1 Det ! xx "! N ! xy ! xz ! yx ! yy "! N ! yz ! zx ! zy ! zz "! N = # Det ! xx "! N ! xy ! xz ! xy ! yy "! N ! yz ! xz ! yz ! zz "! N = # ! 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 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 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 ! + s ! + t u y = m ! + n ! + k " # $ % $ & t = k = ' ( $ ) $ For Point B(2,3), after deformation, its position becomes (1.9,3.05), thus, For Point B(2,3), after deformation, its position becomes (1....
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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-PrelimSolutions - 1 MSE 4020 Prelim #1-SOLUTION...

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