809_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

809_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 10.3 Differential Equations of the Deflection Curve 803 (5) B . C . 1 From symmetry, B . C . 2 B . C . 3 B . C . 4 S HEAR FORCE ( EQ . 2) R B ± R A ± q 0 L p ; V ± q 0 L p cos p x L R A ± V (0) ± q 0 L p ; v (0) ± 0 C 4 ± 0 ² ¿ ( L ) ± 0 C 2 ±³ 2 q 0 L 2 p 3 ² ¿ (0) ± 0 C 3 ± q 0 a L p b 3 V a L 2 b ± 0 C 1 ± 0 + C 2 x 2 2 + C 3 x + C 4 EI ²±³ q 0 L 4 p 4 sin p x L + C 1 x 3 6 B ENDING MOMENT ( EQ . 3) D EFLECTION CURVE ( FROM EQ . 5) or ²± ³ q 0 L 2 p 4 EI a L 2 sin p x L + p x 2 ³ p Lx b ; EI ²±³ q 0 L 2 p 4 sin p x L ³ q 0 L 2 x 2 p 3 + q 0 L 3 x p 3 M B ± M A ± 2 q 0 L 2 p 3 ; M A ±³ M (0) ± 2 q 0 L 2 p 3 M ± q 0 L 2 p 3 a p sin p x L ³ 2 b Problem 10.3-8 A fixed-end beam of length L is loaded by a distributed load in the form of a cosine curve with maximum intensity q 0 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) if the distributed load is now
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Unformatted text preview: . q (1 x 2 / L 2 ) A L y x B M A M B R A R B p x 2 L q 0 cos q ( ) Solution 10.3-8 (a) Loading D IFFERENTIAL EQUATION (1) (2) (3) + C 1 x + C 2 EI M q 0 a 2 L p b 2 cos a p x 2 L b EI q 2 L p sin a p x 2 L b + C 1 EI q q 0 cos a p x 2 L b q q cos a p x 2 L b (4) (5) + C 1 x 3 6 + C 2 x 2 2 + C 3 x + C 4 EI q a 2 L p b 4 cos a p x 2 L b + C 1 x 2 2 + C 2 x + C 3 EI q a 2 L p b 3 sin a p x 2 L b 10Ch10.qxd 9/27/08 7:29 AM Page 803...
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