Solution5 - AE 321 Solutions to Practice Problems Chapter...

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AE 321 – Solutions to Practice Problems Chapter 5: Problem Formulation 1. The figure below shows both the Cartesian coordinates ( ) z y x , , and the cylindrical coordinates ( ) z r , , ! used to define the boundary conditions of the hollow circular cylinder. (i) Boundary conditions in Cartesian coordinates (a) Fixed end surface: 0 = z and b r a ! ! (remember ( ) 2 2 2 z y x r + + = ) 0 = = = z y x u u (b) Free end surface: L z = and b r a ! ! s Cauchy z e n ' ˆ ! = ! zz z yz y xz x P T T T = = = = = = 0 0 (c) Outer lateral surface: L z ! ! 0 and 2 2 2 b y x = +
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y x e b y e b x n ˆ ˆ + = ! Then T x = ! P 0 x b + q y b = " 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 (d) Inner lateral surface: L z ! ! 0 and 2 2 2 a y x = + . Following the same procedure as in part (c) ! 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 x a ! yy y a T z = 0 = ! xz x a ! yz y a (ii) Boundary conditions in cylindrical coordinates (a) Fixed end surface: 0 = z u r = u ! = u z = 0 (b) Free end surface: L z = and b r a ! ! z e n ˆ = ! ! T r = 0 = rz T = 0 = z T z = P = zz (c) Outer lateral surface: L z ! ! 0 and b r =
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r e n ˆ = ! ! T r = ! P 0 = " rr T # = ! q = r T z = 0 = rz (d) Inner lateral surface: L z ! ! 0 and
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Solution5 - AE 321 Solutions to Practice Problems Chapter...

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