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Ishik University / Sulaimani

Civil Engineering Department

Engineering Mechanics I

Chapter -5-

Equilibrium of a Rigid Body 1

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CHAPTER OBJECTIVES

To develop the equations of equilibrium for a rigid

body.

To introduce the concept of the free-body diagram for

a rigid body.

To show how to solve rigid-body equilibrium

problems using the

equations of equilibrium.

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CHAPTER OUTLINE

Free-Body Diagrams

Equations of Equilibrium

Two and Three-Force Members

Constraints for a Rigid Body

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FBD is the best method to represent all the known and

unknown forces in a system.

FBD is a sketch of the outlined shape of the body, which

represents it being isolated from its surroundings.

Necessary to show all the forces and couple moments

that the surroundings exert on the body so that these

effects can be accounted for when equations of

equilibrium are applied.

Free-Body Diagrams

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Free-Body Diagrams

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Free-Body Diagrams

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Free-Body Diagrams

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Support Reactions

If the 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.

Consider the three ways a horizontal member, beam is

supported at the end.

- roller, cylinder

- pin

- fixed support

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Support Reactions

Roller or cylinder

Prevent the beam from translating in the vertical direction.

Roller can only exerts a force on the beam in the vertical

direction.

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Support Reactions

Pin

The pin passes through a hold in the beam and two leaves that

are fixed to the ground.

Prevents translation of the beam in any direction Φ.

The pin exerts a force F on the beam in this direction.

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Support Reactions

Fixed Support

This support prevents both translation and rotation of the beam.

A couple and moment must be developed on the beam at its

point of connection.

Force is usually represented in x and y components.

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Cable exerts a force on the bracket

Type 1 connections

Rocker support for this bridge girder

allows horizontal movements so that

the bridge is free to expand and

contract due to temperature

Type 5 connections

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Concrete Girder rest on the ledge

that is assumed to act as a smooth

contacting surface

Type 6 connections

Utility building is pin supported

at the top of the column

Type 8 connections

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Floor beams of this building are

welded together and thus form fixed

connections

Type 10 connections

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External and Internal Forces

A rigid body is a composition of particles, both external

and internal forces may act on it.

For FBD, internal forces act between particles which are

contained within the boundary of the FBD, are not

represented.

Particles outside this boundary exert external forces on the

system and must be shown on FBD.

FBD for a system of connected bodies may be used for

analysis.

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Weight and Center of Gravity

When a body is subjected to gravity, each particle has a

specified weight.

For entire body, consider gravitational forces as a system

of parallel forces acting on all particles within the

boundary.

The 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.

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Idealized Models

Consider a steel beam used to support the roof joists of a building.

For force analysis, reasonable to assume rigid body since small

deflections occur when beam is loaded.

Bolted connection at A will allow for slight rotation when load is

applied => use Pin.

Support at B offers no resistance to horizontal movement => use Roller.

Building code requirements used to specify the roof loading (calculations of the joist forces).

Large roof loading forces account for extreme loading cases and for dynamic or vibration effects.

Weight is neglected when it is small compared to the load the beam supports.

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Example 5.1

Draw the free-body diagram of the uniform beam. The beam has a

mass of 100kg.

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Solution

Free-Body Diagram

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Solution

Support at A is a fixed wall.

Three forces acting on the beam at A denoted as Ax, Ay, Az,

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.

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Draw the free-body diagram of member AB, which is supported

by a roller at A and a pin at B. Explain the significance of each

force on the diagram.

Example 5.2

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Solution

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Example 5.3

Draw the free-body diagram of the beam which supports the 80-

kg load and is supported by the pin at A and a cable which

wraps around the pulley at D. Explain the significance of each

force on the diagram.

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Solution

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Equations of Equilibrium

For equilibrium of a rigid body in 2D,

∑Fx = 0;

∑Fy = 0;

∑MO = 0

∑Fx and ∑Fy represent the algebraic sums of the x and y

components of all the forces acting on the body.

∑MO represents the algebraic sum of the couple moments and

moments of the force components about an axis perpendicular to

x-y plane and passing through arbitrary point O, which may lie

on or off the body.

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Alternative Sets of Equilibrium Equations

For coplanar equilibrium problems,

∑Fx = 0;

∑Fy = 0;

∑MO = 0 can be used

Two alternative sets of three independent equilibrium equations

may also be used.

∑Fa = 0;

∑MA = 0;

∑MB = 0

When applying these equations, it is required that a line passing

through points A and B is not perpendicular to the a axis.

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Alternative Sets of Equilibrium Equations

Consider FBD of an arbitrarily shaped body.

All the forces on FBD may be replaced by an

equivalent resultant force

FR = ∑F acting at point A and a

resultant moment MRA = ∑MA

If ∑MA = 0 is satisfied, MRA = 0

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Alternative Sets of Equilibrium Equations

A second set of alternative equations is

∑MA = 0;

∑MB = 0;

∑MC = 0

Points A, B and C do not lie on the same line.

Consider FBD, if ∑MA = 0, MRA = 0

∑MA = 0 is satisfied if line of action of FR passes through

point B.

∑MC = 0 where C does not lie on line AB.

FR = 0 and the body is in equilibrium.

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Example 5.4

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Solution

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Example 5.5

The cord supports a force of 500N and wraps over the

frictionless pulley. Determine the tension in the cord at C and

the horizontal and vertical components at pin A.

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Solution

Equations of Equilibrium

Tension remains constant as cord passes over the pulley (true for

any angle at which the cord is directed and for any radius of the pulley.

NT

mTmN

M A

500

0)2.0()2.0(500

;0

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NA

NNA

F

NAx

NA

F

y

y

y

x

x

933

030cos500500

;0

250

030sin500

;0

Solution

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Example 5.6

The link is pin-connected at a and rest a smooth support at B.

Compute the horizontal and vertical components of reactions

at pin A.

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Solution

FBD

Reaction NB is perpendicular to the link at B.

Horizontal and vertical components of reaction are represented

at A.

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Solution

Equations of Equilibrium;

NAx

NA

F

NN

mNmNmN

M

x

x

B

B

A

100

030sin200

;0

200

0)75.0()1(60.90

;0

NA

NNA

F

y

y

y

233

030cos20060

;0

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Example 5.7

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Solution

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Example 5.8

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Solution

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Example 5.9

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Solution

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Simplify some equilibrium problems by recognizing members

that are subjected top only 2 or 3 forces.

Two-Force Members

When a member is subject to no couple

moments and forces are applied at only two

points on a member, the member is called a

two-force member.

Two and Three-Force Members

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Two-Force Members

Hence, only the force magnitude must be determined or stated.

Other examples of the two-force members held in equilibrium

are shown in the figures to the right.

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Three-Force Members

If a member is subjected to only three forces, it

is necessary that the forces be either concurrent

or parallel for the member to be in equilibrium.

To show the concurrency requirement,

consider a body with any two of the three

forces acting on it, to have line of actions that

intersect at point O.

To satisfy moment equilibrium about O, the

third force must also pass through O, which

then makes the force concurrent.

If two of the three forces parallel, the point of

currency O, is considered at “infinity”.

Third force must parallel to the other two

forces to insect at this “point”.

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Example 5.10

Determine the normal reactions at A and B.

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Solution

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Example 5.11

Determine the tension in the cord and the horizontal and

vertical components of reaction at support A of the beam.

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Solution

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Determine the horizontal and vertical components of reaction at

C and the tension in the cable AB for the truss.

Example 5.12

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Solution

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Homework 5.1

Determine the horizontal and vertical components of reaction

at the pin A and the force in the cable BC. Neglect the

thickness of the members.

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Determine the horizontal and vertical components or reaction

at the pin A and the reaction on the beam fit C.

Homework 5.2

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Homework 5.3

Determine the normal reaction at the roller A and horizontal

and vertical components at pin B for equilibrium of the

member.

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