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Lecture 15

ENGR-1100 Introduction to Engineering Analysis. Lecture 15. Today Lecture Outline. Equilibrium of a rigid body in 3D. F ree B ody D iagrams in 3D Equilibrium equations (6 independent equations in 3D). Resultant of a general 3D system of non-parallel non-coplanar forces. z. z. y. y.

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Lecture 15

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 15

  2. Today Lecture Outline • Equilibrium of a rigid body in 3D. • Free Body Diagrams in 3D • Equilibrium equations (6 independent equations in 3D)

  3. Resultant of a general 3D system of non-parallel non-coplanar forces z z y y x1 xR x y1 x yR The procedure 1) Choose a point O (arbitrary). 2) Replace each force by a force at O and a couple perpendicular to it. 3) Combine the forces at O (vectorially) to obtain the resultant force. 4) Combine the couples at O (vectorially) to obtain the resultant couple

  4. Particle equilibrium Rigid body equilibrium Necessary and sufficient conditions of equilibrium of a rigid body The necessary AND sufficient condition for a rigid body to be in equilibrium is that theresultant forceandcoupleacting on the rigid body must be zero. Generally, a system of forces in 3-D can be reduced to aforce-couple system.

  5. 6 Independent Equations in 3-D

  6. 1. Ball and socket 2. Hinge Three-dimensional reaction at supports and connections

  7. 3. Ball bearing 4. Journal bearing

  8. 6. Smooth pin and bracket 5. Thrust bearing 7. Fixed support

  9. Example P6-80 Bar AC of Fig. P6-80 rests against a smooth surface at end C and is supported at end A with a ball-and-socket joint. The cable at B is attached midway between the ends of the bar. Determine the reactions at supports A and C and the tension in thecable at B.

  10. Z Az Ay Ax TBD FBE x y C =388.1 i - 388.1 j - 582.1 k N =TBD(-0.3714 i +0.7428 j - 0.5571 k N

  11. Z Az Ay Ax TBD FBE x y C From a free-body diagram on the bar: The moment equilibrium: = (0.8 i +0.4 j-1.2k)x(388.1i -388.1 j -582.1 k ) +(1.6 j - 1.2 k ) x (-0.3714 TBDi + 0.7428 TBDj -0.5571 TBDk ) +(0.8 i + 1.6 j - 1.2 k ) x (Czk) = ( 1.6 Cz - 698.56) i + (0.4457TBD - 0.8 Cz) j + (0.5942TBD - 465.72) k = 0

  12. Z Az Ay Ax TBD FBE x y C For which: Cz = 436.6 N C = 437 k N TBD= 783.7 N TBD= -291.1 i + 582.2 j -436.7 k N The force equilibrium: = (Axi + Ayj + Azk )+ (388.1i -388.1 j -582.1 k ) - (291.1 i +582.2 j –436.7k ) +436.6 k =0 (Ax +97)i + (Ay +194.1)j + (Az-582.2)k =0

  13. (Ax +97)i + (Ay +194.1)j + (Az-582.2)k =0 Ax =-97 N Ay =-194.1 N Az=582.2 Or: Question: What does it mean to have 5 unknown forces vs, 6 equations?

  14. Class Assignment: Exercise set 6-79 please submit to TA at the end of the lecture Bar AB is used to support an 850-lb load as shown in Fig. P6-79. End A of the bar is supported with a ball-and-socket joint. End B of the bar is supported with two cables. Determine the components of the reaction at support A and the tensions in the two cables. Answer: TBC= -1046 j lb; TBD= -523 i lb; A= 523 i + 1046 j + 850 k lb

  15. Class Assignment: Exercise set 6-87 please submit to TA at the end of the lecture The plate shown in Fig. P6-87 weighs 150 lb and is supported in a horizontal position by two hinges and a cable, The hinges have been properly aligned; therefore, they exert only force reactions on the plate. Assume that the hinge at B resists any force along the axis of the hinge pins. Determine the reactions at supports A and B and the tension in the cable. Answer: TC= 83i –91.67 j + 75 k lb; A= 107j +50 k lb; B=- 83 i –15.31j + 25 k lb

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