Exam3solutions - 3. There is an extension and a bending...

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1 AE 221 – Spring 2004 – EXAM #3 Solutions 1. (i) BVP approach: GDE: EI ¢ ¢ ¢ ¢ w = - f 0 EI ¢ ¢ ¢ w = - f 0 x + C 1 EI ¢ ¢ w = - f 0 x 2 2 + C 1 x + C 2 EI ¢ w = - f 0 x 3 6 + C 1 x 2 2 + C 2 x + C 3 EIw = - f 0 x 4 24 + C 1 x 3 6 + C 2 x 2 2 + C 3 x + C 4 BCs: w 0 ( ) = 0 C 4 = 0 ¢ w 0 ( ) = 0 C 3 = 0 ¢ ¢ ¢ w L ( ) = 0 - f 0 L + C 1 = 0 C 1 = f 0 L ¢ ¢ w L ( ) = 0 - f 0 L 2 2 + f 0 L 2 + C 2 = 0 C 2 = - 1 2 f 0 L 2 Tip deflection: w L ( ) = - f 0 L 4 8 EI (iii) Castigliano’s theorem: Find bending moment from FBD. Cut beam from right hand side, but also need to add a dummy point force at the tip: M P D x
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2 Bending moment: M x ( ) = P D L - x ( ) - f 0 2 L - x ( ) 2 Strain energy of beam: U = 1 2 EI P D L - x ( )- f 0 2 L - x ( ) 2 dx 0 L = 1 2 EI P D 2 L 3 3 - P D f 0 L 4 4 + f 0 2 4 L 5 5 Tip displacement: w L ( ) = lim P D 0 U P D = - f 0 L 4 8 EI , which agree with the previous approach, as it should! 2. Question 2 is exactly the same as question 10 from Section 2.2.
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Unformatted text preview: 3. There is an extension and a bending problem here. (Bending in z only since the section is symmetric). Extension: GDE EA ¢ ¢ u = -f x ( ) = -Q d x-3 L 4 ˆ ˜ BCs u ( ) = EA ¢ u L ( ) = -ku L ( ) Bending GDE EI ¢ ¢ ¢ ¢ w = f z x ( ) + ¢ m y x ( ) = -p stp x-L 2 ˆ ˜ -stp x-3 L 4 ˆ ˜ + M ¢ d x-L 4 ˆ ˜ BCs w ( ) = w L ( ) = EI ¢ ¢ w L ( ) = EI ¢ ¢ ¢ w ( ) = K ¢ w ( )...
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This note was uploaded on 01/18/2010 for the course AE 322 taught by Professor Lambros during the Spring '04 term at University of Illinois at Urbana–Champaign.

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Exam3solutions - 3. There is an extension and a bending...

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