ENGI 1313 Mechanics I Lecture 26 3 D

ENGI 1313 Mechanics I Lecture 26: 3 D Equilibrium of a Rigid Body Shawn Kenny, Ph. D. , P. Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland spkenny@engr. mun. ca

Schedule Change n Postponed Class Ø n Friday Nov. 9 Two Options Ø Use review class Wednesday Nov. 28 • Preferred option Ø n 2 Schedule time on Thursday Nov. 15 or 22 Please Advise Class Representative of Preference © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26

Lecture 26 Objective n 3 to illustrate application of scalar and vector analysis for 3 D rigid body equilibrium problems © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26

Example 26 -01 n 4 The pipe assembly supports the vertical loads shown. Determine the components of reaction at the balland-socket joint A and the tension in the supporting cables BC and BD. © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26

Example 26 -01 (cont. ) n Draw FBD z Due to symmetry TBC = TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 5 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n What are the First Steps? Define Cartesian coordinate system z Ø Resolve forces Ø • Scalar notation? • Vector notation? TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 6 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Cable Tension Forces Ø Position vectors z TBD Ø F 1= 3 k. N F 2 = 4 k. N TBC Az Unit vectors Ax Ay x 7 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Ball-and-Socket Reaction Forces Ø Unit vectors z TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 8 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n What Equilibrium Equation Should be Used? Ø Mo = 0 z • Why? Ø Find moment arm vectors TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 9 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Moment Equation z TBD Due to symmetry TBC = TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 10 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Moment Equation z TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 11 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Force Equilibrium z TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 12 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Force Equilibrium z TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 13 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -01 (cont. ) n Force Equilibrium z TBD F 1= 3 k. N F 2 = 4 k. N TBC Az Ax Ay x 14 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 y

Example 26 -02 n 15 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. © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26

Example 26 -02 (cont. ) Establish Cartesian Coordinate System n Draw FBD n W = 3500 lb F = 250 lb Az 16 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

Example 26 -02 (cont. ) n What Equilibrium Equation Should be Used? Ø Three equations to solve for three unknown vertical support reactions W = 3500 lb F = 250 lb Az 17 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

Example 26 -02 (cont. ) n Vertical Forces W = 3500 lb F = 250 lb Az 18 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

Example 26 -02 (cont. ) n Moment About x-axis W = 3500 lb F = 250 lb Az 19 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

Example 26 -02 (cont. ) n Moment About y-axis W = 3500 lb F = 250 lb Az 20 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

Example 26 -02 (cont. ) n System of Equations Ø Gaussian elimination W = 3500 lb F = 250 lb Az 21 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

Example 26 -02 (cont. ) n System of Equations Ø Gaussian elimination W = 3500 lb F = 250 lb Az 22 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26 Bz Cz

References Hibbeler (2007) n http: //wps. prenhall. com/esm_hibbeler_eng mech_1 n 23 © 2007 S. Kenny, Ph. D. , P. Eng. ENGI 1313 Statics I – Lecture 26