1. EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS Objectives: a) Identify support reactions, and, b) Draw a free-body diagram.

2. APPLICATIONS How are the idealized model and the free body diagram used to do this? Which diagram above is the idealized model? The truck ramp has a weight of 400 lb. The ramp is pinned to the body of the truck and held in the position by the cable. How can we determine the cable tension and support reactions ?

3. APPLICATIONS (continued) Again, how can we make use of an idealized model and a free body diagram to answer this question? Two smooth pipes, each having a mass of 300 kg, are supported by the tines of the tractor fork attachment. How can we determine all the reactive forces ?

4. CONDITIONS FOR RIGID-BODY EQUILIBRIUM (Section 5.1) In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces). Forces on a particle For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.  F = 0 (no translation) and  M O = 0 (no rotation) Forces on a rigid body

5. THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS Finally, we need to apply the equations of equilibrium to solve for any unknowns. For analyzing an actual physical system, first we need to create an idealized model (above right). Then we need to draw a free-body diagram (FBD) showing all the external (active and reactive) forces.

6. FREE-BODY DIAGRAMS (Section 5.2) 2.Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body. Idealized model Free-body diagram (FBD) 1.Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape.

7. FREE-BODY DIAGRAMS (continued) 3.Label loads and dimensions on the FBD: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like A x, A y, M A, etc.. Indicate any necessary dimensions. Idealized modelFree-body diagram

8. SUPPORT REACTIONS IN 2-D As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body in the opposite direction. A few example sets of diagrams s are shown above. Other support reactions are given in your textbook (Table 5-1).

9. EXAMPLE Given:The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at B is 20 lb. Draw:A an idealized model and free- body diagram of the foot pedal.

10. GROUP PROBLEM SOLVING Draw a FBD of the crane boom, which is supported by a pin at A and cable BC. The load of 1250 lb is suspended at B and the boom weighs 650 lb.

11. GROUP PROBLEM SOLVING (continued) FBD

12. GROUP PROBLEM SOLVING Draw a FBD of member ABC, which is supported by a smooth collar at A, roller at B, and link CD.

13. GROUP PROBLEM SOLVING (continued) FBD

14. EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS Objectives: a) Apply equations of equilibrium to solve for unknowns, and, b) Recognize two-force members.

15. APPLICATIONS The uniform truck ramp has a weight of 400 lb. The ramp is pinned at A and held in the position by the cable. How can we determine the forces acting at the pin A and the force in the cable ? A

16. APPLICATIONS (continued) A 850 lb of engine is supported by three chains, which are attached to the spreader bar of a hoist. You need to check to see if the breaking strength of any of the chains is going to be exceeded. How can you determine the force acting in each of the chains?

17. EQUATIONS OF EQUILIBRIUM (Section 5.3)  F x = 0  F y = 0  M O = 0 where point O is any arbitrary point. Please note that these equations are the ones most commonly used for solving 2-D equilibrium problems. There are two other sets of equilibrium equations that are rarely used. For your reference, they are described in the textbook. A body is subjected to a system of forces that lie in the x-y plane. When in equilibrium, the net force and net moment acting on the body are zero (as discussed earlier in Section 5.1). This 2-D condition can be represented by the three scalar equations:

18. TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4) If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B. The solution to some equilibrium problems can be simplified if we recognize members that are subjected to forces at only two points (e.g., at points A and B).

19. EXAMPLE OF TWO-FORCE MEMBERS In the cases above, members AB can be considered as two-force members, provided that their weight is neglected. This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and B are thus known (along the line joining points A and B).

20. STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS 1. If not given, establish a suitable x - y coordinate system. 2. Draw a free body diagram (FBD) of the object under analysis. 3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns.

21. IMPORTANT NOTES 1. If there are more unknowns than the number of independent equations, then we have a statically indeterminate situation. We cannot solve these problems using just statics. 2. The order in which we apply equations may affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force, then solving  F X = 0 first allows us to find the horizontal unknown quickly. 3. If the answer for an unknown comes out as negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.

22. EXAMPLE 1. Put the x and y axes in the horizontal and vertical directions, respectively. 2. Determine if there are any two-force members. 3. Draw a complete FBD of the boom. 4. Apply the E-of-E to solve for the unknowns. Given:The 4kN load at B of the beam is supported by pins at A and C. Find:The support reactions at A and C. Plan:

23. EXAMPLE (Continued) Note: Upon recognizing CD as a two-force member, the number of unknowns at C are reduced from two to one. Now, using E-o-f E, we get,  +  F X = A X + 11.31 cos 45  = 0; A X = – 8.00 kN  +  F Y = A Y + 11.31 sin 45  – 4 = 0; A Y = – 4.00 kN +  M A = F C sin 45   1.5 – 4  3 = 0 F c = 11.31 kN or 11.3 kN FBD of the beam: AXAX AYAY A 1.5 m C B 4 kN FCFC 45° 1.5 m Note that the negative signs means that the reactions have the opposite direction to that shown on FBD.

24. GROUP PROBLEM SOLVING a) Establish the x – y axes. b) Draw a complete FBD of the jib crane beam. c) Apply the E-of-E to solve for the unknowns. Given:The jib crane is supported by a pin at C and rod AB. The load has a mass of 2000 kg with its center of mass located at G. Assume x = 5 m. Find: Support reactions at B and C. Plan:

25. GROUP PROBLEM SOLVING (Continued) +  M C = (3 / 5) F AB  4 + (4 / 5) F AB  0.2 – 2000(9.81)  5 = 0 F AB = 38320 N = 38.3 kN FBD of the beam 2000(9.81) N F AB 5 m CxCx 5 4 3 CyCy 4 m 0.2 m First write a moment equation about Point C. Why point C?

26. GROUP PROBLEM SOLVING (Continued) Now solve the  F X and  F Y equations.  +  F X = C x – (4 / 5) 38320 = 0  +  F Y = – C y + (3 / 5) 38320 – 2000(9.81) = 0 Solving these two equations, we get C x = 30656 N or 30.7 kN and C y = 3372 N or 33.7 kN F AB = 38320 N = 38.3 kN FBD of the beam 2000(9.81) N F AB 5 m CxCx 5 4 3 CyCy 4 m 0.2 m

27. 3-D FREE-BODY DIAGRAMS, EQUILIBRIUM EQUATIONS, CONSTRAINTS AND STATICAL DETERMINACY Objective: a) Identify support reactions in 3-D and draw a free body diagram, and, b) apply the equations of equilibrium.

28. APPLICATIONS Ball-and-socket joints and journal bearings are often used in mechanical systems. To design the joints or bearings, the support reactions at these joints and the loads must be determined.

29. APPLICATIONS (continued) If A is moved to a lower position D, will the force in the rod change or remain the same? By making such a change without understanding if there is a change in forces, failure might occur. The tie rod from point A is used to support the overhang at the entrance of a building. It is pin connected to the wall at A and to the center of the overhang B.

30. APPLICATIONS (continued) The crane, which weighs 350 lb, is supporting a oil drum. How do you determine the largest oil drum weight that the crane can support without overturning ?

31. SUPPORT REACTIONS IN 3-D (Table 5-2) As a general rule, if a support prevents translation of a body in a given direction, then a reaction force acting in the opposite direction is developed on the body. Similarly, if rotation is prevented, a couple moment is exerted on the body by the support. A few examples are shown above. Other support reactions are given in your text book (Table 5-2).

32. IMPORTANT NOTE A single bearing or hinge can prevent rotation by providing a resistive couple moment. However, it is usually preferred to use two or more properly aligned bearings or hinges. Thus, in these cases, only force reactions are generated and there are no moment reactions created.

33. EQUATIONS OF EQUILIBRIUM (Section 5.6) As stated earlier, when a body is in equilibrium, the net force and the net moment equal zero, i.e.,  F = 0 and  M O = 0. These two vector equations can be written as six scalar equations of equilibrium (EofE). These are  F X =  F Y =  F Z = 0  M X =  M Y =  M Z = 0 The moment equations can be determined about any point. Usually, choosing the point where the maximum number of unknown forces are present simplifies the solution. Any forces occurring at the point where moments are taken do not appear in the moment equation since they pass through the point.

34. CONSTRAINTS AND STATICAL DETERMINACY (Section 5.7) Redundant Constraints: When a body has more supports than necessary to hold it in equilibrium, it becomes statically indeterminate. A problem that is statically indeterminate has more unknowns than equations of equilibrium. Are statically indeterminate structures used in practice? Why or why not?

35. IMPROPER CONSTRAINTS In some cases, there may be as many unknown reactions as there are equations of equilibrium. However, if the supports are not properly constrained, the body may become unstable for some loading cases. Here, we have 6 unknowns but there is nothing restricting rotation about the AB axis.

36. EXAMPLE a) Establish the x, y and z axes. b) Draw a FBD of the rod. c) Write the forces using scalar equations. d) Apply scalar equations of equilibrium to solve for the unknown forces. Given:The rod, supported by thrust bearing at A and cable BC, is subjected to an 80 lb force. Find:Reactions at the thrust bearing A and cable BC. Plan:

37. EXAMPLE (continued) FBD of the rod: Applying scalar equations of equilibrium in appropriate order, we get  F X = A X = 0 ; A X = 0  F Z = A Z + F BC – 80 = 0 ;  M Y = – 80 ( 1.5 ) + F BC ( 3.0 ) = 0 ; Solving these last two equations: F BC = 40 lb, A Z = 40 lb

38. EXAMPLE (continued) FBD of the rod: M X = ( M A ) X + 40 (6) – 80 (6) = 0 ; (M A ) X = 240 lb ft  M Z = ( M A ) Z = 0 ; (M A ) Z = 0 = 40 lb Now write scalar moment equations about what point?Point A!

39. GROUP PROBLEM SOLVING a) Draw a FBD of the rod. b) Apply scalar equations of equilibrium to solve for the unknowns. Given: A rod is supported by smooth journal bearings at A, B, and C. Assume the rod is properly aligned. Find: The reactions at all the supports for the loading shown. Plan:

40. GROUP PROBLEM SOLVING (continued) Applying scalar equations of equilibrium in appropriate order, we get  F Y = 450 cos 45  + C Y = 0 ; C Y = – 318 N  M Y = C Z (0.6) – 300 = 0 ; C Z = 500 N  M Z = – B X ( 0.8 ) – ( – 318 ) ( 0.6 ) = 0 ; B X = 239 N A FBD of the rod:

41. GROUP PROBLEM SOLVING (continued) ∑ M X = B Z ( 0.8 ) – 450 cos 45  (0.4) – 450 sin 45  ( 0.8 + 0.4 ) + 318 ( 0.4 ) + 500 ( 0.8 + 0.4 ) = 0 ; B Z = – 273 N  F X = A X + 239 = 0 ; A X = – 239 N  F Z = A Z – ( – 273 ) + 500 – 450 sin 45  = 0 ; A Z = 90.9 N A FBD of the rod: