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CHAP 5 Equilibrium of a Rigid body

CHAP 5 Equilibrium of a Rigid body. rigid body. Consider a rigid body which is at rest or moving with x y z reference at constant velocity. 5.1 Conditions for Rigid body equilibrium. Free body diagram of i th particle of the body. E xternal force ( 外力 )

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CHAP 5 Equilibrium of a Rigid body

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  1. CHAP 5 Equilibrium of a Rigid body

  2. rigid body Consider a rigid body which is at rest or moving with x y z reference at constant velocity 5.1 Conditions for Rigid body equilibrium

  3. Free body diagram of ith particle of the body External force (外力) : gravitational, electrical, magnetic or contact force j Internal force (內力) i Forceequilibrium equation for particle i (Newton first law)

  4. Force equilibrium equation for the whole body (Newton’s 3rd law 作用力與反作用力) Moment of the forces action on the ith particle about pt. O

  5. Moment equilibrium equation for the body Equations of equilibrium for a rigid body are 力平衡 力矩平衡

  6. 5.2 Equilibrium in Two Dimensions (2) Support Reactions A .Type of support : see Table 5-1 1. Free-body Diagram (1) F.B.D A sketch of the outlined shape of the body represents it as being isolated or “free” from its surrounding , i.e ., a free body”. B . General rules for support reaction: If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction . Likewise, if rotation is prevented, a couple moment is exerted on the body.

  7. FAx M FAy Examples: (a) roller or cylinder support (b) pin support (c) Fixed support

  8. (3) External and Internal forces A. Internal force Not represented on the F.B.D. became their net effect on the body is zero. B. External force Must be shown on the F.B.D. (a) “Applied” loadings (b) Reaction forces 反作用力 (c) Body weights 重力 (4) Weight and the center of gravity The force resultant from the gravitational field is referred as the weight of the body, and the location of its point of application is the center of gravity G.

  9. w body 平行力 等效力系 (ch4) P=G(重心)

  10. y x Couple moment Here: algebraic sum of x components of all force on the body. algebraic sum of y components of all force on the body. algebraic sum of couple moments and moments of all the force components about an axis ⊥ xy plane and passing 0. 2. Equations of Equilibrium for 2D rigid body (1) Conditions of equilibrium

  11. (2) Alternative equilibrium equation (A) When the moment points A and B do not lie on a line that is “perpendicular” to the axis a.

  12. (B) Points A, B and C do not lie on the same line a C A B a

  13. y 600N 200N x A B Bx 2m 2m 3m By Ay 100N (3) Example 3 unknown Ax, Bx, By Equations of equilibrium 3 equations for 3 unknowns

  14. 5.3 Two-and Three-Force Members 1. Two-Force member A member subject to no couple moments and forces applied at only two points on the member. FA A A B B Equations of Equilibrium FB

  15. F2 F1 F3 2. Three-Force member A member subject to only three forces, which are either concurrent or parallel if the member is in equilibrium. (1)Concurrent (3力交於O點) (2)parallel (3力相交無限遠處) F2 F1 o F3

  16. Fz Fy 3 reaction forces Fx 5.4 Equilibrium in Three Dimensional Rigid Body 1. Free Body Diagrams (1)F.B.D Same as 2D equilibrium problems (2)Support Reactions A. Types of support:see Table 5-2 B. General rules for reaction Same as two-dimensional case Examples: (a) Ball and Socket joint • No translation along any direction • Rotate freely about any axis

  17. z (b) single journal bearing • Rotate freely about its longitudinal axis • Translate along its longitudinal direction Fz Mz y Mx Fx (c) single pin Only allow to rotate about a specific axis. two unkown forces and couple moments x z Mz Fz My y Fy Fx Three unkown forces and two couple moments x

  18. 2. Equations of Equilibrium A. Vector equations of equilibrium B. Scalar equations of equilibrium

  19. 2KN-m 500N 5.6 Constraints for a rigid body Ex: 1. Redundant constraints (1) Redundant constraints Redundant supports are more than necessarily to hold a body in equilibrium. Equation of motion=3 5 unknown reactions >3 equation of motion there are two support reactions which are redundant supports and more than necessarily.

  20. 500N A B Ax C y MA x 2KN-m Cy By Ay F.B.D of above example (2) Statically indeterminate 靜不定 There are more unknown loadings on the body than equations of equilibrium available for the solution. unknown loadings AX,AY,MA,BY,CY;5 Equations of equilibrium ΣFX=0,ΣFY=0,ΣMA =0;3 5>3 Statically indeterminate structure

  21. Equations of Equilibrium for above example are + ΣFX=0 AX=0 +ΣFY=0 500-AY-BY-CY=0 +ΣMA=0 MA-2-DYBY-DCCY=0 Need two more equations to solve the five unknown forces. (3)Solutions for statically indeterminate structure Additional equations are needed ,which are obtained from the deformation condition at the points of redundant support based on the mechanics of deformation, such as mechanics of materials.

  22. F.B.D FC C 100N 100N C 0.2m B o A B FB FA A 2.Improper Constraints (1) Reaction force = equations of equilibrium If this kind of improper constraint occurs then system is instable A. The lines of action of the reactive forces intersect points on a common axis (concurrent). o body will rotate about Z-axis or point O

  23. B C A 100N FB FC 100N B C FA A B. The reactive forces are all parallel F.B.D Body will translate along x direction.

  24. 100N 100N FA FB o Stable? (2) Reaction forces < equations of equilibrium If the body is partially constrained then it is in instable condition F.B.D Not in equilibrium

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