726_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

726_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 09Ch09.qxd 9/27/08 720 1:31 PM Page 720 CHAPTER 9 Deflections of Beams EI B.C. L q0 x3 L x2 a + b + C 3x + C 4 24 6 2 (4) q0 L4 + C3 L + C4 72 (5) (L) 0 0 L B.C. ¿ L a b 2 L ¿R a b 2 1 q L3 + C1 96 0 1 q L3 + C3 64 0 1 q L4 1920 0 L For 0 … x … 2 q0 x (x) (200 x2 L2 5760LEI C4 B.C. La L b 2 Ra + 96 x4 (6) For L b 2 13 1 q0 L4 + C1 L 5760 2 5 1 qo L4 + C 3 L 1152 2 L q0 (40 x3 5760 EI + 83 L2 x 120 L x2 3 L3) ; 4 L ab 2 dC (7) ; L …x…L 2 (x) + C4 53 L4) 240 x3 L 3q0 L 1280 EI ; From (5)–(7) C1 53 q L3 5760 0 83 q L3 5760 0 C3 y Problem 9.3-17 The beam shown in the figure has a guided MA support at A and a roller support at B. The guided support permits vertical movement but no rotation. Derive the equation of the deflection curve and determine the deflection dA at end A and also dC at point C due to the uniform load of intensity q P/L applied over segment CB and load P at x L/3. (Note: Use the secondorder differential equation of the deflection curve.) L — 3 B A C L — 2 Solution 9.3-17 BENDING-MOMENT EQUATION For 0 … x … L 3 19 PL 24 EI – M(x) EI ¿ 19 PL x + C1 24 EI B.C. For ¿ (0) 0 L L …x… 3 2 C1 0 EI – 19 PL x2 + C 1 x + C2 48 19 EI EI ¿ PL x 24 M(x) 19 PL 24 P ax 19 PL x2 + C2 48 L b 3 P q=— L P L — 2 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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