3.1 INTRODUCTION

3.2 FREE BODY

3.3 EQUILIBRIUM EQUATIONS FOR A RIGID BODY

**A: EQUILIBRIUM IN 2D**

3.4 EQUILIBRIUM EQUATIONS (2D)

3.5 FREE-BODY DIAGRAMS (2D)

3.6 SPECIAL SYSTEMS OF FORCES (2D)

3.7 CONSTRAINTS AND EQUILIBRIUM (2D)

3.8 SOLVING PROBLEMS (2D)

**B: EQUILIBRIUM IN 3D**

3.9 EQUILIBRIUM EQUATIONS (3D)

3.10 FREE-BODY DIAGRAMS (3D)

3.11 SPECIAL SYSTEMS OF FORCES (3D)

3.12 CONSTRAINTS AND EQUILIBRIUM (3D)

3.13 SOLVING PROBLEMS (3D)

**C: EQUILIBRIUM OF SPECIAL MEMBERS**

3.14 TWO-FORCE MEMBERS

3.15 THREE-FORCE MEMBERS

3.16 SUMMARY

**3.1 INTRODUCTION**

3.1.1 The Concept Of Equilibrium

The concept of equilibrium is introduced to describe a body which is stationary or which is moving with a constant velocity. A body under such a state is acted upon by balanced forces and balanced couples only. There is no unbalanced force or unbalanced couple acting on it. In statics, the concept of equilibrium is usually used in the analysis of a body which is stationary, or is said to be in the state of static equilibrium.

The concept of equilibrium is the most basic and most important concept in engineering analysis. The concept must be really understood by every student. The ability to understand mechanics and many other engineering disiciplines is dependent on mastering the concept of equilibrium.

3.1.2 Particles and Rigid Bodies

**Particles. ** A particle is a body whose size does not have any effect on the results of mechanical analyses on it and, therefore, its dimensions can be neglected. The size of a particle is very small compared to the size of the system being analysed.

**Rigid body.** A body is formed by a group of particles. The size of a body affects the results of any mechanical analysis on it. A body is said to be *rigid* when the relative positions of its particles are always fixed and do not change when the body is acted upon by any load (whether a force or a couple). Most bodies encountered in engineering work can be considered rigid from the mechanical analysis point of view becase the deformations that take place within these bodies under the action of loads can be neglected when compared to other effects produced by the loads. All bodies to be studied in this book are rigid, except for *springs*. Springs undergo deformations that cannot be neglected when acted upon by forces or moments. For the analyses ini this book, only the effects of the deformations of springs on a rigid body interacting with the springs are considered but the springs themselves will not be analysed as a body.

3.1.3 Effects Of The Action Of Forces on Particles and Rigid Bodies

In general, a force acting on a particle tends to cause the particle to translate. Also, a force on on a body not only tends to cause the body to translate (as in the case of the particle) but also tends to cause the body to rotate about any axis which does not intersect with or is not parallel to the line of action of the force. This tendency to rotate is called the “moment” and is measured quantitatively by the methods described in Chapter 2.

3.2 FREE BODY

Mechanical analysis of a structure, in general, is started by applying Newton’s law to the whole structure, or to part of the structure. To see what actually happens to any particular part of a structure, that part has to be isolated from the other parts of the body. The concept of isolating the part which is the target of analysis, is a very important concept in mechanical analysis. The part which is isolated is called a *free body*.

In this section, we will see how a targeted part is isolated and how the free body formed through the isolation process is displayed graphically.

3.2.1 Mechanical System

A mechanical system is defined as a body system that can be isolated from other bodies. The system can be formed by a single body, part of a body, or a group of connected bodies. The bodies forming the system can either be rigid or non-rigid. A mechanical system can be solid, fluid, or even a combination of solild and fluid.

3.2.2 Free-body Diagram (FBD)

The isolation of a mechanical system is achieved by cutting and isolating the system from its surroundings. The isolation enables us to see the interactions between the isolated part and the other parts. The part which has been cut (imaginarily), forms a free body. A diagram which portrays the free body, complete with the system of external forces acting on it due to its interaction with the parts which have been removed, is called the free-body diagram (FBD) of the isolated part.

The FBD of of a body system shows all loads acting on the external boundary of the isolated body. The loads include active forces applied on the free body by the parts removed from it. Note that the internal forces acting on a structure becomes external forces when exposed by the imaginery “cutting” process carried out to form the free body.

As an example, consider the structure of the arm of a lift truck, Figure 3.1. Assume that an analysis is to be carried-out on the whole structure of the arm when it is carrying a load as shown, where the weight of the component members of the arm can be nglected compared to the weight of the load. Assume also that all joints of the arm do not prevent rotation around the respective joints, i.e. every joint produces a reactive force only and does not produce any reactive moment. Because the direction and the sense of every reactive force are not known, the direction nad sense shall be assumed.

The arm can be isolated from the body of the lift truck at point A where it is pinned to the body of the lorry and at point C where it is acted upon by the active forceof the hydraulic piston rod. The isolated arm is shown in Figure 3.1(b). On the diagram, FA is the force produced by the interaction between point A on the arm and that on lorry body, FC is the interactive force between point C on the lorry arm and the hydraulic piston, *m***g** is the weight of the load, and *G* is the centre of gravity of the load. Figure 3.1(b) is the FBD for the entire arm for the conditions specified.

If what is to be analysed is the load container only, the FBD shown in Figure 3.1(c) is drawn. If member BC is what one wants to analyse, the FBD is shown in Figure 3.1(d).

Please note that, in the FBDs shown, the direction and sense of all the reactions are drawn arbitarily because they are assumed to be unknown. We will learn later how to determine the direction of some types of reactivce forces through observation.

3.2.3 Procedure To Draw A FBD

A FBD is drawn by following the following steps in order:

1. Show the body which has been isolated from its surroundings by drawing its outline.

2. Show, on the drawing, all loads acting on the body. The loads consist of the active forces couples (which cause the tendency to move) and the reactive forces and couples (caused by any constraint and tend to prevent motion).

3. Indicate the magnitude and sense of each load. Use letters to indicate unknown magnitudes. For unknown sense, choose the sense arbitarily but every pair of interacting loads must obey Newton’s third law.

4. Show all dimensions which are necessary to calculate moments.

It must be stressed here that only a correctly drawn FBD will produce the correct solution. The *most important* step taken to solve problems in mechanics is drawing the FBD.

——————————————————————————–

**EXAMPLE PROBLEM**

**EP3.1 **Figure MC3.1 shows the scoop of a tractor which contains a load of mass *m* and the link mechanism. The joint at A is a pin joint. Draw the FBD of the scoop which is isolated from all other members of the scoop mechanism.

Note that a pin joint can produce a reactive force only buat is not able to produce a reactive couple. Assume that the force on the hydraulic cylinder at B acts along the line of the member. For forces whose directions are unknown, they can be represented by two perpendicular components.

Neglect the weight of the scoop and the weight of the links.

Figure MC3.1

**Solution**

The weight of the load acts downwards at the centre of gravity* G*. The reactive force at B acts along the axis of the hydraulic cylinder. Its sense is unknown but can be assumed to act in any sense. The reaction at A is unknown both in terms of its direction and its sense. Hence it is represented by its two perpendicular components with the sense arbitarily set.

The FBD is as shown in Figure MC3.1(1)

——————– TUTORIAL PROBLEMS TP3.1-TP3.5 ——————–

3**.3 EQUILIBRIUM EQUATIONS FOR A RIGID BODY**

A rigid acted upon by any applied load will tend to translate dan rotate about a particular axis. The tendency to translate is due to the action ot the resultant force on the body and the tendency to rotate is due to the action of the resultant couple. Equilibrium will occur on the body if the resultant force as well as the resultant couple are both zero. Mathematically, equilibrium is determined by the conditions:

∑**F** = **0**

(3.1)

∑**M** = **0**

where ∑**F** is the resultant force and ∑**M** the resultant moment.(1)

The first expression of Equation 3.1 says that, for equilibrium, the sum of all forces acting on the body is zero. The second expression says that the sum of the moment about any axis must be zero. The two conditions manifested by Equation 3.1 are *necessry* and *sufficient* for equilibrium. It is said to be necessary because fulfilling the first condition only will result in inequilibrium in terms of rotation. Similarly, if only the second condition is fulfilled, inequilibrium occurs in terms of trlanslation. The two conditions are said to be sufficient because equilibrium takes place when both the conditions are met without the need for any additional condition.

For the purpose learning, the study on equilibrium in this book will be divided into two stages:

(1) the equilibrium of bodies acted upon by planar force systems, and

(2) the equilibrium of bodies acted upon by a general spatial force system.

Section 3.4-3.8 discusses equilibrium in two dimensions (planar force systems) whilst Section 3.9-3.11 discusses equilibrium in three dimensions (spatial force system).

*A: EQUILIBRIUM IN 2D*

**3.4 EQUILIBRIUM EQUATIONS (2D)**

3.4.1 General Equations

The resultant force vector for a planar force sysem acts on the pane of ation of the original force system. The resultant moment vector acts perpendicularto that plane. Hence the resultant force vector can be represented by its two perpendicular components acting on the plane. On the other hand, the resultan moment vector can be represented by a ircular arrow which indiccates it sense of action. For example, if the surface ofthis book is taken as the plane of action of the given force system, say plane x-y, the following relationships are obtained:

∑**F** = ∑**F**_{x} + ∑**F**_{y }

∑**M _{O}** = ∑|

**r**×

**F**| = ∑

*Fd*

where *d* is the perpendicular distance between any moment centre *O* and the line of action of **F**.

The second expression in this equation gives the magnitude of the resultant moment. The direction of the ∑**M _{O}** vector is perpendicular to this page; its sense of action is indicated by the circular arrow which acts in the clockwise or anti-clockwise direction, depending on the sense of

**r**and

**F**, Figure 3.2.

Figure 3.2

Using the relationships stated above, the equilibrium equation for the two-dimensional case can be written in the scalar form:

∑*F*_{x }= 0

∑*F*_{x} = 0 (3.2)

∑*M*_{O} = 0

The first two eqations in Equation (3.2) state that the sum of the force components in the x- and y- directions are both zero. The third expression states that the moment about any point *O* (i.e. about any axis which is perpendicular to the *x-y* plane and passing through *O*) is zero.

All three expressions of Equation (3.2) are necessary for equilibrium. In addition, if a body is supported in such a way that all forces acting on the body can be determined by using Equation (3.2) alone, then the equation is said to be sufficient. In this early stage of their study, the students are required to handle only problems that can be solved using Equation (3.2). Thus Equation (3.2) is necessary and sufficient for the equilibrium of all bodies to be analyse in this book.

Note that only **three** independent equations exist to describe equilibrium in a plane. Hence, only three unknown values can be solved. In the early stageof their study, students often make the mistake of by adding the number of equation through taking moments about additional points. It is sstressed here that such an effort is fruitless since it will produce equations that are not independent and duplicate one of the three basic equations given in Equation (3.2).

In general, the expressions in Equation (3.2) have to be solved simultaneously. However, this can be avoided by taking the point of intersection of two unknown forces as the moment centre *O*. By doing this, the third unknown can be solved directly. Besides, the process of getting the solution can be simplified by specifying the orientation of the x-y axis system in such a way as to produce the simplest resultant force system in the diretion of the axes.

3.4.2 Alternataive Equations

Equation (3.2) is the original equation most used to solve the problem involving the equilibrium of bodies. However, there are two more combinations of *three* other expressions that can be used to represent conditions manifested by Equation (3.2). These combinations can be used as alternatives to Equation (3.2).

The first alternative equation is:

*F _{x} = 0 or F_{y} = 0*

*M _{A} = 0 *(3.3)

*M _{B} = 0*

This equation involves one force expression and two moment expressions. To use this first alternative equation, point *A* and *B* must not lie on a line perpendicular to the direction of summation of forces. (For example, it cannot be used to solve the mechanism in Figure 3.3(a) if the x-direction is chosen to sum- up the forces).

The second alterntive equation involves tiga moment expressions i.e.:

∑*M _{A}* = 0

∑*M _{B}* = 0 (3.4)

∑*M _{C}* = 0

Equation (3.4) can be used on condition that points *A*, *B*, and *C* do not lie on the same straight line. It cannot, for example, be used to solve the mechanism in Figure 3.3(b).

It can be proved that Equation (3.3) and Equation (3.4) fulfill the same conditions manifested by Equation 3.2 *provided that* the conditions associated with their application are obeyed. Hence, any one of the alternative equations can be used as a replacemanta for Equation (3.2). As long as three, and not more than three, of the above equations area used, (i.e. two force equations and one moment equation, or one force equation and two moment equations, or tree moment equations), the combination is formed of independent equations.

Note that

3.5 FREE-BODY DIAGRAMS (2D)

3.6 SPECIAL SYSTEMS OF FORCES (2D)

3.7 CONSTRAINTS AND EQUILIBRIUM (2D)

3.8 SOLVING PROBLEMS (2D)

*B: EQUILIBRIUM IN 3D*

3.9 EQUILIBRIUM EQUATIONS (3D)

3.10 FREE-BODY DIAGRAMS (3D)

3.11 SPECIAL SYSTEMS OF FORCES (3D)

3.12 CONSTRAINTS AND EQUILIBRIUM (3D)

3.13 SOLVING PROBLEMS (3D)

*C: EQUILIBRIUM OF SPECIAL MEMBERS*

3.14 TWO-FORCE MEMBERS

3.15 THREE-FORCE MEMBERS

**3.16 SUMMARY**

This chapter introduces the concept of *equilibrium*. The conditions for equilibrium and the equations of equilibrium for particles and rigid bodies are given in the scalar and vector forms. The method of writing these equations using the free-body diagram (FBD) and the method of solving the equations are given.

The equilibrium problem is divided into two parts: first, equilibrium under the action of a planar force system and second, equilibrium under teh action of a spatial force system. The problems of planar force equilibrium are solved using scalar analysis whilst those involving spatial force systems are solved using vector analysis.

The last part of this chapter describes the equilibrium of structural members with special chracteristics from the point of view of the equilibrium analysis, namely* two-force members* and *three-force members*.

The most basic but very important matter discussed in this chapter is the concept of *free body*. The method of drawing the FBDs and using them in solving equilibrium problems are elaborated.

**Notes**

(1)