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Unformatted text preview: AE322 HW 7 Solutions Soln 10.7 Given that: (i) EA and EI are constant. (iii) No axial forces. Governing Di erential Equations are: Figure 1: Problem 10.7 EI yy w 0000 ( x ) + EI yz v 0000 ( x ) = f z = f 1 x L 2 EI yz w 0000 ( x ) + EI zz v 0000 ( x ) = 0 BC's : v (0) = v (0) = v ( L ) = v 00 ( L ) = 0 w (0) = w (0) = 0 , w ( L ) = d and w 00 ( L ) = 0 Since there is no axial force, the GDE can be decoupled as, ER zz w 0000 ( x ) = f 1 x L 2 ER yz v 0000 ( x ) = f 1 x L 2 1 Where R yy = I yy I zz ( I yz ) 2 I yy R zz = I yy I zz ( I yz ) 2 I zz The decoupled GDE's are similar except for a sign change on the RHS, hence integration on one GDE is su cient. Integrating zdirection di erential equation, ER zz w 0000 ( x ) = f f x 2 L 2 ER zz w 000 ( x ) = f x f x 3 3 L 2 + C 1 ER zz w 00 ( x ) = f x 2 2 f x 4 12 L 2 + C 1 x + C 2 ER zz w ( x ) = f x 3 6 f x 5 60 L 2 + C 1 x 2 2 + C 2 x + C 3 ER zz w ( x ) = f x 4 24 f x 6 360 L 2 + C 1 x 3 6 + C 2 x 2 2 + C 3 x + C 4 Imposing BC's, w (0) = w (0) = 0 yield C 3 = C 4 = 0 ER zz w 00 ( L ) = 0 ⇒ = f L 2 2 f L 4 12 L 2 + C 1 L + C 2 = 0 ER zz w ( L ) = d ⇒ = f L 4 24 f L 6 360 L 2 + C 1 L 3 6 + C 2 L 2 2 = d Simplifying, C 1 L + C 2 = 5 12 f L 2 C 1 L 6 + C 2 1 2 = 14 360 f L 2 + ER zz d L 2 C 1 and C 2 can be solved for in above equations as, C 1 = 61 f L 2 120 3 ER zz d L 3 C 2 = 11 f L 2 120 + 3 ER zz d L 2 2 Substituting for C 1 and C 2 in the expression for w ( x ) , the slop at the end L can be computed as, w ( L ) = 3 d 2 L f L 3 80 ER zz Soln 10.8 Figure 2: Problem 10.8 The decoupled governing di erential equations are,...
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This note was uploaded on 04/12/2010 for the course AE 353 taught by Professor Voulgaris,p during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Voulgaris,P

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