SampleExam1_solution - AE 220 EXAM#1 Solutions 1 This...

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AE 220 – EXAM #1 Solutions 1. This question is the same as the last identity in Question #7 from Homework #1. The solution can be found there. 2. The figure below shows both the Cartesian coordinates ( x,y,z ) and cylindrical coordinates ( r, ± ,z ) used here. Traction in terms of stress components is obtained from Cauchy’s equation. (i) Boundary conditions in Cartesian coordinates (a) End surface: b r a z ± ± = and , 0 (remember 2 2 y x r + = ) r n = ± ) e z ± T x = 0 = xz T y = 0 = yz T z = P = zz (a) End surface: b r a L z ± ± = and , r n = ) e z ± T x = 0 = xz T y = 0 = yz T z = P = zz (c) Outer lateral surface: 2 2 2 and 0 b y x L z = + ± ±
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y x e b y e b x n ˆ ˆ + = r Then T x = ± P 0 x b + q y b = ² xx x b + xy y b T y = ± P 0 y b ± q x b = xy x b + yy y b T z = 0 = xz x b + yz y b Note: You can leave cos ± and sin in the expressions if you like. (d) Inner lateral surface 2 2 2 and 0 a y x L z = + ± ± . Following the same procedure as in part (c) r n = ± x a ˆ e x ± y a ˆ e y ± T x = P i x a = ± xx x a ± xy y a T y = P i y a = ± xy
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This note was uploaded on 03/07/2011 for the course AE 352 taught by Professor Namachchivaya during the Spring '08 term at University of Illinois, Urbana Champaign.

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SampleExam1_solution - AE 220 EXAM#1 Solutions 1 This...

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