807_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

807_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 10.3 Differential Equations of the Deflection Curve 801 (4) (5) B . C . 1 (6) B . C . 2 B . C . 3 B . C . 4 (7) Solve Eqs. (6) and (7): S HEAR FORCE ( EQ . 2) V ± q 0 2 L p cos a p x 2 L b ² 6 q 0 L p 2 ² 4 p + 8 p 4 C 2 ± 2 q 0 L 2 p 2 ² 12 p + 24 p 4 C 1 ±² 6 q 0 L p 2 ² 4 p + 8 p 4 + q 0 a 2 L p b 4 6 L 2 ² q 0 a 2 L p b 3 L 6 L 2 C 1 L + 3 C 2 ± ³ ( L ) ± 0 ³ (0) ± 0 C 4 ± 0 ³ ¿ (0) ± 0 C 3 ± q 0 a 2 L p b 3 C 1 L + C 2 ±² q 0 4 L 2 p 2 ³ ( L ) ± 0 + C 3 x + C 4 EI ³± ² q 0 a 2 L p b 4 sin a p x 2 L b + C 1 x 3 6 + C 2 x 2 2 + C 2 x + C 3 EI ³ ¿ ±² q 0 a 2 L p b 3 cos a p x 2 L b + C 1 x 2 2 R EACTIONS From equilibrium D EFLECTION CURVE ( EQ . 5) ´ q 0 a 2 L p b 3 x d ; ´ 2 q 0 L 2 # p 2 ² 12 p + 24 p 4 x 2 2 ² 6 q 0 L p 2 ² 4 p + 8 p 4 x 3 6 ³± 1 EI c ² q 0 a 2 L p b 4 sin a p x 2 L b + C 1 x 3 6 + C 2 x 2 2 + C 3 x + C 4 , or EI ³± ² q 0 a 2 L p b 4 sin a p x 2 L b M A ±² C 2 ±² 2 q 0 L 2 p 2 ² 12 p + 24 p 4 ; ± a 6 p 2 ² 4 p + 8 p 4 b q 0 L ; R B ±² V ( L ) ± 0.327 q 0 L ± a 2 p ² 6 p 2 ² 4 p + 8 p 4 b q 0 L ; R A ± V (0) ± 0.31 q 0 L Problem 10.3-7 A fixed-end beam of length L is loaded by distributed load in the form of a cosine curve with maximum intensity
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Unformatted text preview: q at A . (a) Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve. (b) Repeat (a) using the distributed load q sin( x / L ). p L 2 L 2 A y p x L q 0 cos q x B M A M B R A R B ( ) 10Ch10.qxd 9/27/08 7:29 AM Page 801...
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