Lecture 29 sect 6.6

# Lecture 29 sect 6.6 - ------Cy Cx By Bx Ay Ax P P P P P P y...

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ME221 Lecture #29 1 ME 221 Statics Lecture #29 Sections 6.6 – 6.7

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ME221 Lecture #29 2 Homework #10 Chapter 7 problems: Due Today
ME221 Lecture #29 3 Homework #11 Chapter 6 problems: 68 & 75 Due Wednesday, November 19

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ME221 Lecture #29 4 Exam #3 Wednesday, November 12 Details on Monday
ME221 Lecture #29 5 Trusses Composed 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.

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ME221 Lecture #29 6 Analysis of Trusses Using the Method of Joints A y A x C y A B C α β We need to solve for: (1) - Internal forces F AB , F AC , and F CA (2) - Reactions A x , A y and C y
ME221 Lecture #29 7 Using Matrix Notation - - - - - 1 0 0 sin 0 0 0 0 0 cos 1 0 0 0 0 sin 0 sin 0 0 0 cos 0 cos 0 1 0 0 0 sin 0 0 1 0 1 cos β α

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Unformatted text preview: ------Cy Cx By Bx Ay Ax P P P P P P y y x BC AC AB C A A F F F =-1 Using manual calculations Look for joints with 2 unknowns ME221 Lecture #29 8 Example 1.2 m 1.2 m 6 kN 3 kN E B A C D 0.9 m ME221 Lecture #29 9 6 kN 3 kN E B A C D 4 4 16 16 4 4 9 9 15 15 5 5 ME221 Lecture #29 10 1.8 m 12 kN 12 kN 4 @ 2.4 m=9.6 m B D F H J A C E G I Method of Sectioning If 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 FH ME221 Lecture #29 11 1.8 m 12 kN 12 kN B D F H J A C E G I F GE F GF F HF 1.8 m 12 kN 12 kN B D F H J A C E G I F EG F FG F FH = = = int Anypo y x M F F...
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## This note was uploaded on 07/25/2008 for the course ME 221 taught by Professor Buch during the Summer '08 term at Michigan State University.

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Lecture 29 sect 6.6 - ------Cy Cx By Bx Ay Ax P P P P P P y...

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