1 / 13

ME 221 Statics Lecture #32 Section 6.9

ME 221 Statics Lecture #32 Section 6.9. Homework #11. Chapter 6 problems: 2, 3, 6 & 7 – Method of Joints 32, 36, 47 & 53 – Method of Sections 68 & 75 Due Friday, November 21. Quiz #7. Friday, November 21 Analysis of Structures Method of Joints or Method of Sections. P By. P Bx. B.

georgio-betagh
Download Presentation

ME 221 Statics Lecture #32 Section 6.9

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ME 221 StaticsLecture #32Section 6.9 Lecture #32

  2. Homework #11 Chapter 6 problems: • 2, 3, 6 & 7 – Method of Joints • 32, 36, 47 & 53 – Method of Sections • 68 & 75 • Due Friday, November 21 Lecture #32

  3. Quiz #7 Friday, November 21 Analysis of Structures Method of Joints or Method of Sections Lecture #32

  4. PBy PBx B PAY PCy PAx a b PCx Ax C A Ay Cy Analysis of Trusses Using the Method of Joints We need to solve for: (1) - Internal forces FAB, FAC, and FBC (2) - Reactions Ax, Ay and Cy Lecture #32

  5. 6 kN A 4 B 4 5 9 15 3 kN 9 15 5 C 16 16 4 4 E D Lecture #32

  6. Method of Sectioning 12 kN 12 kN A C E G I B D F H J 4 @ 2.4 m=9.6 m 1.8 m 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 Lecture #32

  7. 12 kN 12 kN A C E G I FGE 1.8 m FGF FHF B D F H J 12 kN 12 kN A C E G I FEG FFG FEH 1.8 m B D F H J Lecture #32

  8. Frames Are designed to support, prevent or transmit loads (forces, no moments). At least one member is not a two-force member. Machines Are designed to transmit force, motion or energy (forces and moments). Will always have at least one member with multiple forces. Lecture #32

  9. Frames and Machines • Frames and machines have at least one multi-force member • Frames are designed to support loads • Machines are designed to transmit and modify loads • Just like trusses, frames and machines need to be disassembled to determine member forces • Need to draw a FBD for each member Lecture #32

  10. D E C B W A D Cx Cy E Cx FBE Cy W C T T FBE FBE W B A Ax Ax FBE Ay Ay Disassembly Exposes Internal Forces • First, reactions of entire structure • Second, disassemble and solve for internal forces FBE is a two force member Lecture #32

  11. Notes on Previous Problem • Without recognizing BE is a two force member, the problem is not tractable. • MUST draw a FBD for entire structure and each disassembled part • Each FBD may have three unknown forces which are found from SFx = 0; SFy = 0; SMz = 0 Lecture #32

  12. General Procedure • Draw FBD for entire structure and solve for reaction forces SFx = 0; SFy = 0; SMz = 0 • Disassemble into component parts • Recognize two force members • Draw FBD and solve for reaction forces - using SFx = 0 ; SFy = 0 ; SMz = 0 Lecture #32

  13. Frame Example Lecture #32

More Related