PowerPoint PresentationHomework #10Homework #11Exam #3TrussesSlide 6Slide 7Slide 8Slide 9Slide 10Slide 11ME221 Lecture #29 1ME 221 StaticsLecture #29Sections 6.6 – 6.7ME221 Lecture #29 2Homework #10Chapter 7 problems:–2, 5, 6, 8, 19, 21, 24, 26 & 35•Due TodayME221 Lecture #29 3Homework #11Chapter 6 problems:–2, 3, 6 & 7 – Method of Joints–32, 36, 47 & 53 – Method of Sections–68 & 75 –Due Wednesday, November 19ME221 Lecture #29 4Exam #3Wednesday, November 12Details on MondayME221 Lecture #29 5TrussesComposed of slender straight pieces connected together by frictionless pins where all the loads (no moments) are applied.Each member will act as a two-force member (either in tension or compression). All the forces acting on a truss member are axial.ME221 Lecture #29 6Analysis of Trusses Using the Method of JointsAyAxPAxPAYPByPBxPCyPCxCyABC We need to solve for:(1) - Internal forces FAB, FAC, and FCA(2) - ReactionsAx, Ay and CyME221 Lecture #29 7Using Matrix Notation100sin00000cos10000sin0sin000cos0cos01000sin00101cosCyCxByBxAyAxPPPPPPyyxBCACABCAAFFF=-1Using manual calculationsLook for joints with 2 unknownsME221 Lecture #29 8 Example 1.2 m 1.2 m6 kN3 kNEBAC D0.9 mME221 Lecture #29 96 kN3 kNEBACD4416164499151555ME221 Lecture #29 101.8 m 12 kN 12 kN4 @ 2.4 m=9.6 mB D F H J A C E G IMethod of SectioningIf the question is to find internal forces in selected members of the truss, then one can alternatively use the method of sectioning.Example: Determine the force in members FG and FHME221 Lecture #29 111.8 m 12 kN 12 kNB D F H J A C E G IFGEFGFFHF1.8 m 12 kN 12 kNB D F H J A C E G IFEGFFG FFH
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