1. MEC 0011 Statics Lecture 4 Prof. Sanghee Kim Fall_ 2012

2. 4.7 Simplification of a Force and Couple System An equivalent system is when the external effects are the same as those caused by the original force and couple moment system External effects of a system is the translating and rotating motion of the body Or refers to the reactive forces at the supports if the body is held fixed

3. -FF F F F at any point on stick

4. QUIZ 1.In statics, a couple is defined as __________ separated by a perpendicular distance. A) two forces in the same direction B) two forces of equal magnitude C) two forces of equal magnitude acting in the same direction D) two forces of equal magnitude acting in opposite directions 2.The moment of a couple is called a _____ vector. A) Free B) Spin C) Romantic D) Sliding

5. Equivalent resultant force acting at point O and a resultant couple moment is expressed as If force system lies in the x–y plane and couple moments are perpendicular to this plane,

6. Procedure for Analysis 1.Establish the coordinate axes with the origin located at point O and the axes having a selected orientation 2.Force Summation 3.Moment Summation

7. Example 4.16 A structural member is subjected to a couple moment M and forces F 1 and F 2. Replace this system with an equivalent resultant force and couple moment acting at its base, point O.

8. Solution Express the forces and couple moments as Cartesian vectors. MoMo

9. Force Summation.

10. Chapter Objectives 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

11. 5.1 Conditions for Rigid-Body Equilibrium 2. the resultant external force F i (gravitational, electrical, magnetic) 1. the resultant internal force and the resultant external force f i (interactions with adjacent particles) F i + f i = 0 ∑F i + ∑f i = 0 since internal forces between particles in the body occur in equal but opposite collinear pairs (Newton’s third law)

12. 5.1 Conditions for Rigid-Body Equilibrium The equilibrium of a body is expressed as Consider summing moments about some other point, such as point A, we require MRMR

13. 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.

14. Internal Forces External and internal forces can act on a rigid body For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented(canceled) Newton’s 3 rd law Particles outside this boundary exert external forces on the system

15. Weight and Center of Gravity Each particle has a specified weight System can be represented by a single resultant force, known as weight W of the body Location of the force application is known as the center of gravity

16. Procedure for Drawing a FBD 1. Draw Outlined Shape Imagine body to be isolated or cut free from its constraints Draw outline shape 2. Show All Forces and Couple Moments Identify all external forces and couple moments that act on the body 3. Identify Each Loading and Give Dimensions Indicate dimensions for calculation of forces Known forces and couple moments should be properly labeled with their magnitudes and directions

17. Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg. Free-Body Diagram

18. Support at A is a fixed wall Three forces acting on the beam at A denoted as A x, A y, A z, drawn in an arbitrary direction Unknown magnitudes of these vectors Assume sense of these vectors For uniform beam, Weight, W = 100(9.81) = 981N acting through beam ’ s center of gravity, 3m from A

19. 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

20. Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑F x = 0;∑F y = 0; ∑M O = 0 2 alternative sets of 3 independent equilibrium equations, ∑F a = 0; ∑M A = 0; ∑M B = 0

21. Procedure for Analysis Equations of Equilibrium Apply ∑M O = 0 about a point O Unknowns moments of are zero about O and a direct solution the third unknown can be obtained Orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components Negative result scalar is opposite to that was assumed on the FBD

22. Example 5.5 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.

23. Free Body Diagrams 600N represented by x and y components 200N force acts on the beam at B Solution

24. Equations of Equilibrium