454_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

454_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 05Ch05.qxd 9/25/08 448 2:29 PM Page 448 CHAPTER 5 Stresses in Beams (Basic Topics) (b) FIND M BASED ON ALLOWABLE BENDING STRESS AT h/2 FROM NA AT LOCATION (xm) OF MAX. BENDING MOMENT, Mm qx2 2 qL M a + bx 2 L M(x) d M(x) dx use to find location of zero shear where max. moment occurs qx2 d 2 M d qL ca + bx dx 2 L M 1 qL + 2 L xm qx Mm 0 simplifying 2 2 1 1qL + 2 M2 8q L2 sa S Mm Mm sa a bh2 b 6 Equating both Mm expressions & solving for M where sa 8 MPa A MAX. MOMENT Mm qL M a + b xm 2 L Mm L M2 + b 2 qL 2 also L M + 2 qL qL M L M + ba + b 2 L 2 qL qa Mm 0 a qxm 2 2 M Mmax Problem 5.8-9 A wood beam AB on simple supports with span length equal to 10 ft is subjected to a uniform load of intensity 125 lb/ft acting along the entire length of the beam, a concentrated load of magnitude 7500 lb acting at a point 3 ft from the right-hand support, and a moment at A of 18,500 ft-lb (see figure). The allowable stresses in bending and shear, respectively, are 2250 psi and 160 psi. sa a bh2 b a 8 qL2 b 6 qL2 2 9.01 N # m ; 7500 lb 18,500 ft-lb 125 lb/ft 3 ft A B (a) From the table in Appendix F, select the lightest beam that will support the loads (disregard the weight of the beam). (b) Taking into account the weight of the beam (weight density 5 35 lb/ft3), verify that the selected beam is satisfactory, or if it is not, select a new beam. 10 ft Solution 5.8-9 (a) q L 125 1b ft 10 ft sAllow P 75001b M d 3 ft RB 7.725 * 103 1b 7.725 * 103 1b Vmax t allow 2250 psi 18500 ft-b 160 psi RB Mmax RB d Mmax 2.261 * 104 1b-ft Vmax qd2 2 RA qL d +P 2 L RA 1.025 * 103 1b tmax 3V 2A RB qL L d M +P + 2 L L Areq 72.422 in.2 M L Areq 3Vmax 2tallow ...
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