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

2. Schedule Change Postponed Class Two Options
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. Example 26-01 (cont.) Draw FBD Due to symmetry TBC = TBD z TBD
F1= 3 kN
F2 = 4 kN Ax Ay Az

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

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

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

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

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

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

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

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

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

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 Az Cz

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 Az Cz

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

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

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

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

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

23. References Hibbeler (2007)