732_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

732_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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726 CHAPTER 9 Deflections of Beams B . C .4 n (0) ± 0 ± C 4 ± 0 d B ±²³ ( L ) ± 19 q 0 L 4 360 EI ; ³±² q 0 x 2 360 L 2 EI (45 L 4 ² 40 L 3 x + 15 L 2 x 2 ² x 4 ) ; u B ¿ ( L ) ± q 0 L 3 15 EI ; ³ ¿ ±² q 0 x 60 L 2 EI (15 L 4 ² 20 L 3 x + 10 L 2 x 2 ² x 4 ) Problem 9.4-7 A beam on simple supports is subjected to a parabolically distributed load of intensity q ± 4 q 0 x ( L ² x )/ L 2 , where q 0 is the max- imum intensity of the load (see figure). Derive the equation of the deflection curve, and then determine the maximum deflection d max . Use the fourthorder differential equation of the deflection curve (the load equation). A y x 4 q 0 x L 2 B q = ( L ² x ) L Solution 9.4-7 Simple beam (parabolic load) L OAD EQUATION (E Q . 9-12c) B . C .1 EI n ´´ ± ME I n ´´ (0) ± 0 ± C 2 ± 0 B . C .2 EI n ( L ) ± 0 EI ³ ¿ q 0 30 L 2 ( ² 5 L 3 x 2 + 5 Lx 4 ² 2 x 5 ) + C 3 C 1 ± q 0 L 3 EI ³ q 0 3 L 2 (2 Lx 3 ² x 4 ) + C 1 x + C 2 EI ³ –¿ 2 q 0 3 L 2 (3 Lx 2 ² 2 x 3 ) + C 1 EI ³ –– q 4 q 0 x L 2 ( L ² x ) 4 q 0 L 2 ( Lx ² x 2 ) B . C . 3 (Symmetry) B . C n (0) ± 0 ± C 4 ± 0 d max a L 2 b ± 61 q 0 L 4 5760 EI ; q 0 x 90 L 2 EI (3 L 5 ² 5 L 3 x 2 + 3 Lx 4 ² x 5 ) ; EI q 0 30 L 2 a L 5 x ² 5 L 3 x 3 3 + Lx 5 ²
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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