725_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 725_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 09Ch09.qxd 9/27/08 1:31 PM Page 719 719 SECTION 9.3 Deflections by Integration of the Bending-Moment Equation EI – q0 ( 3 L2 x + L3 3L M(x) + 3 L x2 EI ¿ B.C. x4 b + C3 4 5q0 L3 96 EI L ¿a b 2 L ab 2 x3) q0 3 22 a L x + L3x 3L 2 + L x3 EI B.C. q0 L4 64 EI q0 ( 960LEI (x) + 3 L5) 15 x b + C3 x + C4 20 dB (L) 1 q L4 320 0 C4 160 L2 x3 + 160 L3 x2 + 80 L x4 q0 1 1 1 23 a L x + L3 x2 + L x4 3L 2 2 4 5 q L3 192 0 C3 16 x5 25 L4 x ; 7q0L4 160EI Problem 9.3-16 Derive the equations of the deflection curve for a simple beam ; y AB with a distributed load of peak intensity q0 acting over the left-hand half of the span (see figure). Also, determine the deflection dC at the midpoint of the beam. (Note: Use the second-order differential equation of the deflection curve.) q0 A L /2 C B L /2 RA RB Solution 9.3-16 BENDING-MOMENT EQUATION For 0 … x … EI – M(x) L 2 2 x b 2 L a 2 q0 (5 L2 x 24L 22 EI ¿ q0 5 L x a 24L 2 + C1 x4 L + 2 x5 b 5 + C 1x + C 2 5q0 L x 24 a EI – q0 5 L2 x3 a 24L 6 EI 2 q0 L a L2 1 cq 20 xb B.C. n(0) 0 C2 0 23 q0 5 L x a 24L 6 EI 2 q0 L x4 L + 2 x5 b 5 (2) + C 1x 2 xb d x x 3 For L …x…L 2 12 x2 L + 8 x3) 1L q ax 2 02 EI – 5q0 L x 24 EI – 4 x3 L + 2 x4 b M(x) L q0 ( 24 x + L) EI ¿ L q0 x2 a + L x b + C3 24 2 L b 6 (1) (3) x ...
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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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