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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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