484_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

484_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 05Ch05.qxd 9/25/08 478 2:29 PM Page 478 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.12-4 Rigid frame AXIAL FORCE: N P sin a 2 RA sin a BENDING MOMENT: M PL 2 RAL TENSILE STRESS Load P at midpoint B REACTIONS: RA RC sin a 1/ 12 d SUBSTITUTE NUMERICAL VALUES P 2 BAR AB: tan a H L 1H2 + L2 P 8.0 kN L A 1.4 m a 46.37 * 106 mm4 2(11.31 * 103 mm2) (8.0 kN)(1.4 m)(200 mm) + 4(46.37 * 106 mm4) d/2 0.250 MPa + 12.08 MPa ; 11.83 MPa (tension) sc N A Mc I 0.250 MPa 12.08 MPa ; 12.33 MPa (compression) Problem 5.12-5 A palm tree weighing 1000 lb is inclined at an angle of 60° (see figure). The weight of the tree may be resolved into two resultant forces, a force P1 900 lb acting at a point 12 ft from the base and a force P2 100 lb acting at the top of the tree, which is 30 ft long. The diameter at the base of the tree is 14 in. Calculate the maximum tensile and compressive stresses st and sc, respectively, at the base of the tree due to its weight. 45° 200 mm 11.31 * 103 mm2 I st diameter c H (8.0 kN)(1/ 12) sina H d P sin a PLd + 2A 4I N Mc + A I st P2 = 100 lb 30 ft 12 ft P1 = 900 lb 60° ...
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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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