Exam1solutions - m z = M d x-3 L 4 Therefore GDE becomes:...

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1 AE 221 – Spring 2004 – EXAM #1 Solutions 1. The is an extension and a bending problem in this case. Since the beam is of homogeneous and symmetric cross-section I yz = 0, thus uncoupling the y and z bending problems. Extension: GDE: EA ¢ u ( ) ¢ = - f x = 0 BCs: u 0 ( ) = 0 EA ¢ u L ( ) = P Bending: GDE: EA ¢ ¢ v ( ) ¢¢ = f y - ¢ m z Here f y = - 2 P 0 L x L ˆ ˜ stp L 2 - x ˆ ˜ - r gA and
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Unformatted text preview: m z = M d x-3 L 4 Therefore GDE becomes: EA v ( ) = -2 P L x L -gA-M d x-3 L 4 BCs: EI v ( ) = -M v L ( ) = EI v ( ) = -kv ( ) EI v L ( ) = -K v L ( ) 2. This is the same problem as question 0(f) in Chapter 2.1....
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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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