1 / 23

ENGI 1313 Mechanics I

ENGI 1313 Mechanics I . Lecture 26: 3D Equilibrium of a Rigid Body. Schedule Change. Postponed Class Friday Nov. 9 Two Options Use review class Wednesday Nov. 28 Preferred option Schedule time on Thursday Nov.15 or 22 Please Advise Class Representative of Preference. Lecture 26 Objective.

payton
Download Presentation

ENGI 1313 Mechanics I

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. ENGI 1313 Mechanics I Lecture 26: 3D Equilibrium of a Rigid Body

  2. Schedule Change • Postponed Class • Friday Nov. 9 • Two Options • Use review class Wednesday Nov. 28 • Preferred option • Schedule time on Thursday Nov.15 or 22 • Please Advise Class Representative of Preference

  3. Lecture 26 Objective • to illustrate application of scalar and vector analysis for 3D rigid body equilibrium problems

  4. Example 26-01 • The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD.

  5. z TBD F1= 3 kN TBC F2 = 4 kN Az Ay Ax y x Example 26-01 (cont.) • Draw FBD Due to symmetry TBC = TBD

  6. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • What are the First Steps? • Define Cartesian coordinate system • Resolve forces • Scalar notation? • Vector notation? F1= 3 kN F2 = 4 kN

  7. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Cable Tension Forces • Position vectors • Unit vectors F1= 3 kN F2 = 4 kN

  8. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Ball-and-Socket Reaction Forces • Unit vectors F1= 3 kN F2 = 4 kN

  9. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • What Equilibrium Equation Should be Used? • Mo = 0 • Why? • Find moment arm vectors F1= 3 kN F2 = 4 kN

  10. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Moment Equation Due to symmetry TBC = TBD F1= 3 kN F2 = 4 kN

  11. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Moment Equation F1= 3 kN F2 = 4 kN

  12. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Force Equilibrium F1= 3 kN F2 = 4 kN

  13. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Force Equilibrium F1= 3 kN F2 = 4 kN

  14. z TBD TBC Az Ay Ax y x Example 26-01 (cont.) • Force Equilibrium F1= 3 kN F2 = 4 kN

  15. Example 26-02 • The silo has a weight of 3500 lb and a center of gravity at G. Determine the vertical component of force that each of the three struts at A, B, and C exerts on the silo if it is subjected to a resultant wind loading of 250 lb which acts in the direction shown.

  16. Example 26-02 (cont.) • Establish Cartesian Coordinate System • Draw FBD W = 3500 lb F = 250lb Bz Cz Az

  17. Example 26-02 (cont.) • What Equilibrium Equation Should be Used? • Three equations to solve for three unknown vertical support reactions W = 3500 lb F = 250lb Bz Cz Az

  18. Example 26-02 (cont.) • Vertical Forces W = 3500 lb F = 250lb Bz Cz Az

  19. Example 26-02 (cont.) • Moment About x-axis W = 3500 lb F = 250lb Bz Cz Az

  20. Example 26-02 (cont.) • Moment About y-axis W = 3500 lb F = 250lb Bz Cz Az

  21. Example 26-02 (cont.) • System of Equations • Gaussian elimination W = 3500 lb F = 250lb Bz Cz Az

  22. Example 26-02 (cont.) • System of Equations • Gaussian elimination W = 3500 lb F = 250lb Bz Cz Az

  23. References • Hibbeler (2007) • http://wps.prenhall.com/esm_hibbeler_engmech_1

More Related