Chapter 5: Equilibrium of a Rigid Body

Transcription

1 Chapter 5: Equilibrium of a Rigid Body Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body Solve rigid-body equilibrium problems using the equations of equilibrium

2 5.1 Conditions for Rigid-Body Equilibrium We will develop both the necessary and sufficient conditions for the equilibrium of the rigid body. The body is subjected to an external force and couple moment system that is the result of the effects of gravitational, electrical, magnetic, or contact forces caused by adjacent bodies.

3 5.1 Conditions for Rigid-Body Equilibrium The force and couple moment system can be reduced to an equivalent resultant force and resultant couple moment at any arbitrary point O. The equilibrium of a body is expressed as F R F 0 M R O M O 0

4 5.2 Free Body Diagrams Succesful application of the equations of equilibrium requires a complete specification of all the known and unknown external forces that act on the body. The best way to account for these forces is to draw a free-body diagram. This diagram is a sketch of the outlined shape of the body, which represents it as being isolated or free from its surroundings. On this sketch it is necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when the equations of equilibrium are applied.

5 5.2 Free Body Diagrams Support Reactions If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction. If rotation is prevented, a couple moment is exerted on the body. Roller Supports Since this support only prevents the beam from translating in the vertical direction, the roller can only exert a force on the beam in this direction.

6 5.2 Free Body Diagrams Hinge or Pin Supports The pin can prevent translation of the beam in any direction,, and so the pin must exert a force F on the beam in this direction. It is usually easier to represent this resultant force F by its two components F x and F y.

7 5.2 Free Body Diagrams Fixed Support This support will prevent both translation and rotation of the beam, and so to do this a force and a couple moment must be developed on the beam at its point of connection.

8 5.3 Equations of Equilibrium For equilibrium of a rigid body in 2D, F x = 0 F y = 0 M O = 0 F x and F y represent sums of x and y components of all the forces M O represents the sum of the couple moments and moments of the force components

9 5.3 Equations of Equilibrium Alternative Sets of Equilibrium Equations One such set is, F x = 0 M A = 0 M B = 0 It is required that a line passing through points A and B is not parallel to the y axis.

10 5.3 Equations of Equilibrium Alternative Sets of Equilibrium Equations A second alternatice set of equilibrium equations is, M A = 0 M B = 0 M C = 0 It is necessary that points A, B and C do not lie on the same line.

11 5.4 Two- and Three-Force Members Two-Force Members When forces are applied at only two points on a member, the member is called a two-force member Only force magnitude must be determined

12 5.4 Two- and Three-Force Members Two-Force Members For any two-force member to be in equilibrium, the two forces acting on the member must have the same magnitude, act in opposite directions, and have the same line of action, directed along the line joining the two points where these forces act.

13 5.4 Two- and Three-Force Members Three-Force Members If a member is subjected to only three forces, then it is necessary that the forces be either concurrent or parallel for the member to be in equilibrium.

14 EQUILIBRIUM in THREE DIMENSIONS 5.5 Free Body Diagrams Support Reactions As in the two-dimensional case: A force is developed by a support that restricts the translation of its attached member. A couple moment is developed when rotation of the attached member is prevented.

15 5.6 Equations of Equilibrium Vector Equations of Equilibrium For two conditions for equilibrium of a rigid body in vector form, F = 0 M O = 0 Scalar Equations of Equilibrium If all external forces and couple moments are expressed in Cartesian vector form F = F x i + F y j + F z k = 0 M O = M x i + M y j + M z k = 0

16 5.6 Equations of Equilibrium Since the i, j, and k components are independent from one another, the above equations are satisfied provided and F x = 0 F y = 0 F z = 0 M x = 0 M y = 0 M z = 0