1. CHAP 5 Equilibrium of a Rigid body

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

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
Force equilibrium 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
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”.
(2) Support Reactions A .Type of support : see Table 5-1
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. (a) roller or cylinder support
Examples:
(a) roller or cylinder support
(b) pin support
(c) Fixed support M
FAy
FAx

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. body
平行力
等效力系
(ch4) w
P=G(重心)

10. 2. Equations of Equilibrium for 2D rigid body
(1) Conditions of equilibrium
Couple moment y x
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.

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

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

13. (3) Example 600N 200N 2m 3m 100N y x Bx A B By Ay 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. 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) F1 F2 F3 F2 F1 o F3

16. 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 Fx Fy Fz
3 reaction forces

17. y x 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. A. Vector equations of equilibrium
B. Scalar equations of equilibrium

19. 5.6 Constraints for a rigid body
1. Redundant constraints
(1) Redundant constraints
Redundant supports are more than necessarily to hold a body in equilibrium.
Ex:
2KN-m
500N
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. (2) Statically indeterminate 靜不定
There are more unknown loadings on the body than equations of equilibrium available for the solution.
F.B.D of above example x y A B C
2KN-m Ay By Cy MA Ax
500N
unknown loadings AX,AY,MA,BY,CY;5
Equations of equilibrium ΣFX=0,ΣFY=0,ΣMA =0;3
5>3
Statically indeterminate structure

21. (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.
Equations of Equilibrium for above example are
+ ΣFX= AX=0
+ΣFY= AY-BY-CY=0
+ΣMA= MA-2-DYBY-DCCY=0
Need two more equations to solve the five unknown forces.

22. (1) Reaction force = equations of equilibrium
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). A o B C
100N
F.B.D FC FA FB A B C
0.2m
100N o
body will rotate about Z-axis or point O

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

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