EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS Todays Objectives: Students will be able to: a) Identify support reactions, and, b) Draw a free-body diagram. In-Class Activities: Check Homework Reading Quiz Applications Support Reactions

Free-Body Diagrams Concept Quiz Group Problem Solving Attention Quiz READING QUIZ 1. If a support prevents translation of a body, then the support exerts a ___________ on the body. A) Couple moment B) Force C) Both A and B. D) None of the above 2. Internal forces are _________ shown on the free body diagram of a whole body. A) Always

B) Often C) Rarely D) Never APPLICATIONS 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 ? How are the idealized model and the free body diagram used to do this? Which diagram above is the idealized model? APPLICATIONS

(continued) 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 ? Again, how can we make use of an idealized model and a free body diagram to answer this question? CONDITIONS FOR RIGID-BODY EQUILIBRIUM (Section 5.1) Forces on a particle 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). 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) Forces on a rigid body and MO = 0 (no rotation) THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS 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. Finally, we need to apply the equations of equilibrium to solve for any unknowns. FREE-BODY DIAGRAMS (Section 5.2) 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. 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. FREE-BODY DIAGRAMS (continued) Idealized model Free-body diagram 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 Ax, Ay, MA, etc.. Indicate any necessary dimensions. SUPPORT REACTIONS IN 2-D A few example sets of diagrams s are shown above. Other support reactions are given in your textbook (Table 5-1). 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. 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 freebody diagram of the foot pedal. CONCEPT QUIZ 1. The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C. In a FBD of only the beam, there are how many unknowns? A) 2 forces and 1 couple moment B) 3 forces and 1 couple moment C) 3 forces

D) 4 forces CONCEPT QUIZ 2. If the directions of the force and the couple moments are both reversed, what will happen to the beam? A) B) C) D) The beam will lift from A. The beam will lift at B. The beam will be restrained. The beam will break.

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

GROUP PROBLEM SOLVING (continued) FBD ATTENTION QUIZ 1. Internal forces are not shown on a free-body diagram because the internal forces are_____. (Choose the most appropriate answer.) A) Equal to zero B) Equal and opposite and they do not affect the calculations C) Negligibly small D) Not important

2. How many unknown support reactions are there in this problem? A) 2 forces and 2 couple moments B) 1 force and 2 couple moments C) 3 forces D) 3 forces and 1 couple moment End of the Lecture Let Learning Continue EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS Todays Objectives: Students will be able to: a) Apply equations of equilibrium

to solve for unknowns, and, b) Recognize two-force members. In-Class Activities: Check Homework, if any Reading Quiz Applications Equations of Equilibrium Two-Force Members Concept Quiz Group Problem

READING QUIZ 1. The three scalar equations FX = FY = MO = 0, are ____ equations of equilibrium in two dimensions. A) Incorrect B) The only correct C) The most commonly used D) Not sufficient 2. A rigid body is subjected to forces as shown. This body can be considered as a ______ member. A) Single-force force

B) Two- C) Three-force D) Six-force APPLICATIONS A 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 ? 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? EQUATIONS OF EQUILIBRIUM (Section 5.3) 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: Fx = 0 Fy = 0 MO = 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. TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4) 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). 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. 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). 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. 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 FX = 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. EXAMPLE 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: 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. EXAMPLE (Continued) FBD of the beam: AY AX A 4 kN

1.5 m 1.5 m C 45 B FC 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, + MA = FC sin 45 1.5 4 3 = 0 Fc = 11.31 kN or 11.3 kN + FX = AX + 11.31 cos 45 = 0; AX = 8.00 kN + FY = AY + 11.31 sin 45 4 = 0; AY = 4.00 kN Note that the negative signs means that the reactions have the opposite direction to that shown on FBD. CONCEPT QUIZ

1. For this beam, how many support reactions are there and is the problem statically determinate? A) (2, Yes) B) (2, No) C) (3, Yes) D) (3, No) F 2. The beam AB is loaded and supported as shown: a) how many support reactions are there on the beam, b) is A this problem statically determinate, and c) is the structure stable?

A) (4, Yes, No) B) (4, No, Yes) C) (5, Yes, No) D) (5, No, Yes) F F Fixed support

F F B GROUP PROBLEM SOLVING 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:

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. GROUP PROBLEM SOLVING (Continued) FAB 4m Cx 4 5 3

0.2 m 5m Cy 2000(9.81) N FBD of the beam First write a moment equation about Point C. Why point C? + MC = (3 / 5) FAB 4 + (4 / 5) FAB 0.2 2000(9.81) 5 = 0 FAB = 38320 N = 38.3 kN

GROUP PROBLEM SOLVING (Continued) FAB 4m Cx 4 5 3 0.2 m 5m

Cy 2000(9.81) N FBD of the beam FAB = 38320 N = 38.3 kN Now solve the FX and FY equations. + FX = Cx (4 / 5) 38320 = 0 + FY = Cy + (3 / 5) 38320 2000(9.81) = 0 Solving these two equations, we get Cx = 30656 N or 30.7 kN and Cy = 3372 N or 33.7 kN

ATTENTION QUIZ 1. Which equation of equilibrium allows you to determine FB right away? AX A) FX = 0 100 lb A B AY FB

B) FY = 0 C) MA = 0 D) Any one of the above. 2. A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable for all types of loadings? A) (3, Yes) B) (3, No) C) (4, Yes)

D) (4, No) End of the Lecture Let Learning Continue 3-D FREE-BODY DIAGRAMS, EQUILIBRIUM EQUATIONS, CONSTRAINTS AND STATICAL DETERMINACY Todays Objective: Students will be able to: In-Class Activities: a) Identify support reactions in 3-D Check Homework, if any

and draw a free body diagram, Reading Quiz and, b) apply the equations of equilibrium. Applications Support Reactions in 3D Equations of Equilibrium Concept Quiz Group Problem Solving Attention quiz READING QUIZ 1.

If a support prevents rotation of a body about an axis, then the support exerts a ________ on the body about that axis. A) Couple moment B) Force C) Both A and B. D) None of the above. 2. When doing a 3-D problem analysis, you have ________ scalar equations of equilibrium. A) 3

B) 4 C) 5 D) 6 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. APPLICATIONS (continued) 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. 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. 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 ? SUPPORT REACTIONS IN 3-D (Table 5-2) A few examples are shown above. Other support reactions are given in your text book (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. 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. 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 MO = 0 . These two vector equations can be written as six scalar equations of equilibrium (EofE). These are FX = FY

MX = MY = = FZ = 0 MZ = 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.

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? IMPROPER CONSTRAINTS Here, we have 6 unknowns but there is nothing restricting

rotation about the AB axis. In some cases, there may be as many unknown reactions as there are equations of equilibrium. M A 0 However, if the supports are not properly constrained, the body may become unstable for some loading cases. EXAMPLE

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: a) b) c) d) Establish the x, y and z axes. Draw a FBD of the rod. Write the forces using scalar equations.

Apply scalar equations of equilibrium to solve for the unknown forces. EXAMPLE (continued) FBD of the rod: Applying scalar equations of equilibrium in appropriate order, we get F X = AX = 0 ; AX = 0 F Z = AZ + FBC 80 = 0 ; M Y = 80 ( 1.5 ) + FBC ( 3.0 ) = 0 ;

EXAMPLE (continued) FBD of the rod: = 40 lb Now write scalar moment equations about what point? Point A! M X = ( MA) X + 40 (6) 80 (6) = 0 ; M Z = ( MA) Z = 0 ; (MA ) Z= 0 (MA ) X= 240 lb ft CONCEPT QUIZ 1. The rod AB is supported using two

cables at B and a ball-and-socket joint at A. How many unknown support reactions exist in this problem? A) 5 force and 1 moment reaction B) 5 force reactions C) 3 force and 3 moment reactions D) 4 force and 2 moment reactions 2. CONCEPT QUIZ (continued) If an additional couple moment in

the vertical direction is applied to rod AB at point C, then what will happen to the rod? A) The rod remains in equilibrium as the cables provide the necessary support reactions. B) The rod remains in equilibrium as the ball-and-socket joint will provide the necessary resistive reactions. C) The rod becomes unstable as the cables cannot support compressive forces. GROUP PROBLEM SOLVING

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: a) Draw a FBD of the rod. b) Apply scalar equations of equilibrium to solve for the unknowns. GROUP PROBLEM SOLVING (continued) A FBD of the rod:

Applying scalar equations of equilibrium in appropriate order, we get F Y = 450 cos 45 + CY = 0 ; CY = 318 N M Y = CZ (0.6) 300 = 0 ; CZ = 500 N M Z = BX ( 0.8 ) ( 318 ) ( 0.6 ) = 0 ; BX = 239 N

GROUP PROBLEM SOLVING (continued) A FBD of the rod: M X = BZ ( 0.8 ) 450 cos 45 (0.4) 450 sin 45 ( 0.8 + 0.4 ) + 318 ( 0.4 ) + 500 ( 0.8 + 0.4 ) = 0 ; BZ = 273 N F X = AX + 239 = 0 ; AX = 239 N F Z = AZ ( 273 ) + 500 450 sin 45 = 0 ;

AZ = 90.9 N ATTENTION QUIZ 1. A plate is supported by a ball-andsocket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem? A) 4 forces and 2 moments B) 6 forces C) 5 forces D) 4 forces and 1 moment ATTENTION QUIZ 2. What will be the easiest way to determine the force reaction BZ ?

A) Scalar equation FZ = 0 B) Vector equation MA = 0 C) Scalar equation MZ = 0 D) Scalar equation MY = 0 End of the Lecture Let Learning Continue