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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 ¿ ( L ) 0 C 2 2 q 0 L 2 p 3 ¿ (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 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 q 0 L 2 p 4 EI a L 2 sin p x L + p x 2 p Lx b ; EI q 0 L 2 p 4 sin p x L 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 M (0) 2 q 0 L 2 p 3 M q 0 L 2 p 3 a p sin p x L 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 ³ 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 ² – ± M ± q 0 a 2 L p b 2 cos a p x 2 L b EI ² –¿ ± ³ q 2 L p sin a p x 2 L b + C 1 EI ² –– ± ³ q ± ³ 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 ² ± ³ 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 ² ¿ ± 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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