750_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 750_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 09Ch09.qxd 9/27/08 744 1:33 PM Page 744 CHAPTER 9 Deflections of Beams dC 5qL4 384EI qb2 L2 a b 2 8EI qL2 (5L2 384EI 24b2) (downward is positive) (1) Rearrange and simplify the equation: 48b4 96b3L 24b2L2 16bL3 5L4 or b3 b2 b4 48 a b + 96 a b + 24 a b L L L BEAM AB: 0 (a) RATIO b 16 a b L 5 0 b L Solve the preceding equation numerically: b L Table G-1, Case 1: dA qb4 8EI uBb qb4 8EI qb (3b3 + 6b2L 24EI qL 2 (L 24EI 6b2)b dC 3 L) DEFLECTION dC EQUALS DEFLECTION dA 24b2) qb (3b3 + 6b2L 24EI b L Say, 0.4030 ; (b) DEFLECTION dC (EQ. 1) (downward is positive) qL2 (5L2 384EI 0.40301 L3) qL2 (5L2 24b2) 384EI qL2 24 (0.40301 L)2] [5L2 384EI qL4 0.002870 EI ; (downward deflection) Problem 9.5-24 A frame ABC is loaded at point C by a force P acting at an angle a to the horizontal (see figure). Both members of the frame have the same length and the same flexural rigidity. Determine the angle a so that the deflection of point C is in the same direction as the load. (Disregard the effects of axial deformations and consider only the effects of bending due to the load P.) Note: A direction of loading such that the resulting deflection is in the same direction as the load is called a principal direction. For a given load on a planar structure, there are two principal directions, perpendicular to each other. P L a B C L A Solution 9.5-24 Principal directions for a frame P1 and P2 are the components of the load P P1 P cos a P2 P sin a If P1 ACTS ALONE ¿ dH P1L3 3EI ¿ dv uBL (to the right) a P1L2 bL 2EI P1L3 2EI (downward) ...
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