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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.

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ENGI 1313 Mechanics I

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  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

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