1. ME221Lecture #291 ME 221 Statics Lecture #29 Sections 6.6 – 6.7

2. ME221Lecture #292 Homework #10 Chapter 7 problems: –2, 5, 6, 8, 19, 21, 24, 26 & 35 Due Today

3. ME221Lecture #293 Homework #11 Chapter 6 problems: –2, 3, 6 & 7 – Method of Joints –32, 36, 47 & 53 – Method of Sections –68 & 75 –Due Wednesday, November 19

4. ME221Lecture #294 Exam #3 Wednesday, November 12 Details on Monday

5. ME221Lecture #295 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.

6. ME221Lecture #296 Analysis of Trusses Using the Method of Joints AyAy AxAx P Ax P AY P By P Bx P Cy P Cx CyCy A B C  We need to solve for: (1) - Internal forces F AB, F AC, and F CA (2) - Reactions A x, A y and C y

7. ME221Lecture #297 Using Matrix Notation = Using manual calculations Look for joints with 2 unknowns

8. ME221Lecture #298 Example 1.2 m 1.2 m 6 kN 3 kN E B A CD 0.9 m

9. ME221Lecture #299 6 kN 3 kN E B A C D 44 16 4 4 9 9 15 5 5

10. ME221Lecture #2910 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

11. ME221Lecture #2911 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