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Unformatted text preview: x y B A C D P 10 ’ 10 ’ P B C A D E 25’ 25’ 2I I I 2I A 25’ 20’ 4.1 Determine of the structure shown in Fig. P4 1 for the given parameters: Fig. P41 , , Case 1: Case 2: Case 3: Compare the solutions by the Energy method and the slopedeflection equations and provide comments. Solution Slopedeflection equation method 1 Let (1) @ b 2 (2) Symbolically, The smallest values of from the solution of the characteristic equation resulting from the stability condition determinant are Case 1: =2,730.4 kips 3 Case 2: =1,944.7 kips Case 3: =1,966.3 kips Energy method Total potential energy functional Assume For Case 1, From Maple ® Case 1: =2,736.1 kips For Case 2, 4 Case 2: =1,971.1 kips For Case 3, Case 3: =1,992.7 kips Try whether these solutions could be improved by assuming more terms. Proceeding exactly the same manner as that used for oneterm function, Maple ® yields Case 1: =2,731.9 kips Case 2: =1,945.1 kips Case 3: =1,966.7 kips In view of the small improvement of the solutions, just oneterm assumed function is...
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This note was uploaded on 09/24/2011 for the course CIVL 3110 taught by Professor Melville during the Spring '08 term at Auburn University.
 Spring '08
 MELVILLE

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