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CHAPTER 4

Equilibrium of Rigid Body

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　　 Chapter Objectives

i. To develop the equations of equilibrium for a rigid body.

ii. To introduce the concept of the free-body diagram for a rigid body.

iii. To solve rigid body equilibrium problems using the equations of equilibrium

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Outline Application Definition Free Body Diagram Free Body Diagram

exercises Reaction supports Two-force & 3-force

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APPLICATIONS

A 200 kg platform is suspended off an oil rig.

How do we determine the force reactions

at the joints and the forces in the cables?

How are the idealised model and

the free body diagram used to do this?

Which diagram above is the idealised model?

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APPLICATIONS (continued)

A steel beam is used to support roof joists.

How can we determine the support reactions at A & B?

Again, how can we make use of an idealized model

and a free body diagram to answer this question?

AB

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Definition

Static equilibrium for a rigid body:

A body (or any part of it) which is currently stationary will remain stationary if the resultant force and resultant moment are zero for all the forces and couples applied on it.

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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 and MO = 0

Forces on a rigid body

Forces on a particle

CONDITIONS FOR RIGID-BODY EQUILIBRIUM

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How can we make use of an idealized model and a free body diagram to answer this question?

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How are the idealized model and the free body diagram

used to do this?

Which diagram above is the idealized model and free body diagram?

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Idealized model FBD

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4.2 Free Body Diagram

In solving a problem concerning the equilibrium of a rigid body, it is essentialto consider all of the forces acting on the body.

It is equally important to exclude any forces which is not directly applied to thebody.

Omitting a force or adding an extraneous one would destroy the conditions of equilibrium.

Therefore, the first step in the solution of the problem should be to draw a free body diagram of the rigid body under consideration.

Free body diagram have already been used on many occasions in chapter 2.

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Sample Photo

A free body diagram of the tractor shown would include all of the external forces acting on the tractor:the weight of the tractorthe weight of the load in the bucketthe forces exerted by the ground on the tires

In chapter 6, we will discuss how to determine the internal forces in structures made of several connected pieces, such the forces in the members that support the bucket of the tractor

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Procedure for drawing an FBD

1. Draw outlined shape.- Isolate the body from its constraints and connections.2. Show all forces.- Identify all external forces and couple moments that

act on the body.- Place each force and couple at the point that it is

applied

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3. Identify each loading and give dimensions.- The forces and couple moments that are

known should be labelled with their magnitudes and directions.

Procedure for drawing an FBD

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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.

A few examples are shown above.

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Free Body Diagram

No equilibrium problem should be solved without first drawing the free-body

diagram.

Internal forces are never shown on the FBD

since they occur in equal but opposite collinear pairs and therefore cancel out.

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The weight of a body is an external force, and its effect is shown as a single resultant force acting through the body’s centre of gravity.

Couple moments can be placed anywhere on the FBD since they are free vectors. Forces can act at any point along their lines of action since they are sliding vectors.

Free Body Diagram (cont)

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4.3 REACTIONS AT SUPPORT AND CONNECTIONS FOR A TWO DIMENTIONAL STRUCTURE

1.Reactions Equivalent to a Force with known Line of Action Support and connections causing reactions of this type include rollers, rockers, frictionless surface, short links and cables, collars on frictionless rods, and frictionless pins in slots. Each of these supports and connections can prevent motion in one direction only. They are shown in figure 4.1 together with the reactions they produce. Each of these reactions involves one unknown, namely the magnitude of the reaction; this magnitude should be denoted by an appropriate letter.

The line of action of the reaction is known and should be indicated clearly in the free body diagram.

2.Reactions Equivalent to a Force of Unknown Direction and Magnitude Support and connections causing reactions of this type include frictionless pins in fitted holes, hinges, and rough surface. The can prevent translation of the free body in all directions, but they cannot prevent the body from rotating about the connection. Reactions of this group involve two unknowns and are usually represented by their x and y components. In the case of rough surface, the component normal to the surface must be directed away from the surface , and thus is directed toward the free body diagram.

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4.3 REACTIONS AT SUPPORT AND CONNECTIONS FOR A TWO DIMENTIONAL STRUCTURE

3. Reactions Equivalent to a Force and a Couple These reactions are caused by fixed supports, which oppose any motion of the free body and thus constrain it completely. Fixed supports actually produce forces over the entire surface of contact; these forces, however, form a system which can be reduced to a force and couple. Reactions of this group involve three unknowns, consisting usually of the two components of the force and the moment of the couple.

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Rollers Rocker

Support of Connection

Frictionlesssurface

Short cable

Reaction

Short link

Collar on frictionless rod Frictionless pin in slot

Rough surfaceFrictionless pin

or hinge

Fix support

1

Force and couple

Force of unknowndirection

Force with knownline of action

Force with knownline of action

Force with knownline of action

Number ofUnknowns

1

1

2

3

Figure 4.1

Reactions at support and connections

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Reactions at Supports and Connections for a Two-Dimensional Structure

• Reactions equivalent to a force with known line of action.

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Reactions at Supports and Connections for a Two-Dimensional Structure

• Reactions equivalent to a force of unknown direction and magnitude.

• Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude

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The abutment mounted rocker bearing shown is used to support the roadway of a bridge.

Picture shows the rocker expansion bearing of a plategirder bridge.The convex surface of the rocker allows the supportof the girder to move horizontally.

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4.4 EQUILIBRIUM OF RIGID BODY IN TWO DIMENTIONS

• For all forces and moments acting on a two-dimensional structure,

Ozyxz MMMMF 00

• Equations of equilibrium become

000 Ayx MFF

where A is any point in the plane of the structure.

• The 3 equations can be solved for no more than 3 unknowns.

• The 3 equations can not be augmented with additional equations, but they can be replaced 000 BAx MMF

P Q S

PyPx

QyQx

A

Ax

B

SySx

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Equations of Equilibrium

(a) Equilibrium of forces

R = ΣF = 0ΣFx = 0 ΣFy = 0

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Equations of Equilibrium

(b) Equilibrium of moment ΣMO = 0

ΣMA = 0 ΣMB = 0

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4.5 STATICALLY INDETERMINATE REACTION, PARTIAL CONSTRAINS

• More unknowns than equations

• Fewer unknowns than equations, partially constrained

• Equal number unknowns and equations but improperly constrained

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Sample problems 4.1:

A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The centre of gravity of the crane is located at G. Determine the components of the reactions at A and B.

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To determine force B,

calculate moments about point A and the summation of moment is equal to zero

+ ΣMA = 0

+B (1.5) – (9.81k)(2) – (23.5k)(6) = 0

B = + 107.1 kN

B = 107.1 kN →

Solution

2400 x 9.8 = 23.5kN

23.5kN

9.81kN

Ay

Ax

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To determine horizontal component of A,

we use horizontal equilibrium

+→ΣFX = 0;

AX + B = 0, B = 107.1

AX + 107.1kN = 0

AX = -107.1 kN

AX = 107.1 kN ←

Solution

2400 x 9.8 = 23.5kN

23.5kN

Ay

Ax

9.81kN

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To determine vertical component of A,

we use vertical equilibrium

+↑ΣFY = 0

AY – 9.81k – 23.5k = 0

AY = + 33.3kN

AY = 33.3 kN↑

Solution

23.5kN

Ay

Ax

9.81kN

2400 x 9.8 = 23.5kN

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To find force A, A2 = AX

2 + AY2

A =

A =

A = 112.2 kN

tan θ =

θ = 17.3

A = 112.2 kN 17.3

22YX AA

22 )3.33()1.107(

AX = 107.1

AY = 33.3A = 112.2

θ

1.107

3.33

Solution

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To check the answer, we calculate the moment about B

+ ΣMB = -(9.81kN)(2m)

-(23.5 kN)(6m)

+(107.1kN)(1.5m)

= 0

Check

23.5kN

107.1kN9.81kN

107.1kN

33.3kN

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Sample problems 4.2

Three loads are applied to a beam as shown. The beam is supported by a roller at A and by a pin at B. Neglecting the weight of the beam,

determine the reactions at A and B when P = 70kN.

27kNP 27kN

0.9m1.8m

0.6m 0.6m

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Solution

Free-Body Diagram

A free body diagram of the beam is drawn. The reaction at A is verticaland is denoted by A.

The reaction at B is represented by components Bx and By.

Each component is assumed to actin the direction shown

27kNP 27kN

1.8m

0.9m 0.6m 0.6m

27kN70kN 27kN

1.8m

0.9m 0.6m 0.6m

A By

Bx

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Solution

Equilibrium Equations

27kN70kN 27kN

1.8m

0.9m 0.6m 0.6m

A By

Bx

Bx = 0

We write the following three equilibrium equations and solve for the reactions indicated :

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27kN70kN 27kN

1.8m

0.9m 0.6m 0.6m

A By

Bx

Equilibrium Equations

; -(70N)(0.9m) + By(2.7m) – (27 kN)(3.3m) – (27kN)(3.9m) = 0

By = +95.33 kN By = 95.33 kN

Solution

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27kN70kN 27kN

1.8m

0.9m 0.6m 0.6m

A By

Bx

Equilibrium Equations

- A(2.7m) + (70N)(1.8m) – (27 kN)(0.6m) – (27kN)(1.2m) = 0

A = +28.67 kN A = 28.67 kN

= 0 ;

Solution

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27kN70kN 27kN

1.8m

0.9m 0.6m 0.6m

ABy

Bx

= +28.67kN= +95.33kN

0

Check

= +28.67kN - 70kN +95.33kN - 27kN - 27kN = 0

The result are checked by adding the vertical components of all of the external forces

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Sample problem 4.3

750mm625mm

625mm

600mm

A loading car is at rest on a track forming an angleof 25˚ with the vertical.

The gross weight of the car and its load is 25kN, andit is applied at a point 750mm from the track, halfway between the two axles.

The car is held by a cable attached 600mm from the track.

Determine the tension in the cable and the reaction at eachpair of wheels.

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600mm

625mm

625mm750mm

150mm

625mm

625mm

22.65 kN

10.5 kN

Solution

Free-Body Diagram

A free body diagram of the car is drawn. The reaction at each wheel is perpendicular to the track, and the tension force T is parallel to the track. For convenience, we choose the x axis parallelto the track and the y axis perpendicular to the track.

The 25 kN weight is then resolved into x and y components.

Wx = + (25 kN) cos 25˚ = +22.65 kN

Wy = - (25 kN) sin 25˚ = -10.5 kN

25˚

22.65 kN

10.5 kN

25 kN

25 kN

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600mm

625mm

625mm750mm

150mm

625mm

625mm

22.65 kN

10.5 kN

Equilibrium Equations

We take moments about A to eliminate T and R1 from the computation

- (10.5 kN)(625mm) – (22.65 kN)(150mm) + R2 (1250mm) = 0

R2 = + 8 kN R2 = + 8 kN

Now, taking moments about B to eliminate T and R2 from the computation, we write

(10.5 kN)(625mm) – (22.65 kN)(150mm) – R1 (1250mm) = 0

R1 = + 2.5 kN R1 = 2.50 kN

= 0 ;

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600mm

625mm

625mm750mm

150mm

625mm

625mm

22.65 kN

10.5 kN

Equilibrium Equations

The value of T is found by writing

+ 22.65 kN – T = 0 T = + 22.65

T = 22.65 kN

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150mm

625mm

625mm

22.65 kN

10.5 kN

150mm

625mm

625mm

22.65 kN

10.5 kN

2.50 kN

8 kN

22.65 kN

Check

The computations are verified by writing

+2.50 kN + 8 kN -10.5 kN = 0

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PROBLEMS

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Problem 4.1

The boom on a 4300kg truck is used to unload a pallet of singlesof mass 1600 kg. Determine the reaction at each of the two (a) rear wheels B (b) front wheel C

6m

4.3m 0.5m0.4m

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Solution

6m

4.3m 0.5m0.4m

A 15˚

free body diagram

A = 1600kg

truck = 4300kg

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6m

4.3m 0.5m0.4m

A 15˚

15.696 kN

42.183 kN

free body diagram

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6m

4.3m 0.5m0.4m

A 15˚

free body diagram

42.183 kN

15.696 kN

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33.616 kN24.266 kN

42.183 kN

15.696 kN

Check

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Problem 4.3

1.7 m

1.2 m1.8 m

2.7 m

0.75 m

Two crates each having a mass of 110 kg, are placed as shown in the bed of a 13.5 kNpick up truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B

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1.7 m

1.2 m1.8 m

2.7 m

0.75 m

Solution Free body diagram

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1.7 m

1.2 m1.8 m

2.7 m

0.75 m

Free body diagram

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1.7 m

1.2 m1.8 m

2.7 m

0.75 m

Free body diagram

Check

WDWC 2FA W 2FB

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Problem 4.9

Four boxes are placed on a uniform 14kg, wooden plank which rests on two sawhorses.Knowing that the masses of boxes B and D are 4.5kg and 45kg, respectively,

Determine the range of values of the mass of box A so that the plank remains in equilibriumwhen box C is removed.

56Free body diagram for MA (min)

4.5kg 45kg

WA

WG

WB WD

A

E=0 F

E G

DB

F

57Free body diagram for MA (max)

4.5kg 45kg

WA

WG

WB WD

A

E F=0

E G

DB

F

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• Consider a plate subjected to two forces F1 and F2

• For static equilibrium, the sum of moments about A must be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A.

• Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero.

• Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.

4.6 EQUILIBRIUM OF A TWO FORCE BODY

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EQUILIBRIUM OF A TWO FORCE BODY

• When a body is subjected to forces at two points on the body.

• No couple moment.

• These forces maintain the force equilibrium provided they are equal in magnitudes and opposite in direction.

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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).

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• Consider a rigid body subjected to forces acting at only 3 points.

• Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero.

• Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D.

• The lines of action of the three forces must be concurrent or parallel.

4.7 EQUILIBRIUM OF A THREE FORCE BODY

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• If a member is subjected to only 3 forces, then it is necessary that the forces be either concurrent of parallel for equilibrium condition.

• All 3 forces must have lines of action that intersect at one point.

EQUILIBRIUM OF A THREE FORCE BODY

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Sample problems 4.6

A man raises a 10kg joist, of length 4m, by pulling on a rope.

Find the tension T in the rope and the reaction at A.

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Free-body diagram

Solution

The joist is a three-force body, since it is actedupon by three forces;

it’s weight W, the force T exerted by the rope, and the reaction R of the ground at A.

We note that,

W = mg = (10kg)( 9.81 m/s ) = 98.1 N2

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Solution

α= 58.6˚

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T = 81.9 N R = 147.8 N

58.6˚

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Newton’s Law

Newton’s First Law

If the resultant force acting on a particle is zero, the particle will remain at restor will move with constant speed in a straight line.

Newton’s Second Law

If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force.

Newton’s Third Law

The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

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THE END