Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body

7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body

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ME 1204: Engineering Statics

Dr. Faraz Junejo


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Objectives

Develop the equations of equilibrium for a rigid

body

Concept of the free-body diagram for a rigid

body

Solve rigid-body equilibrium problems using

the equations of equilibrium


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Outline

1. Concept of Rigid Body

2. Conditions for Rigid Equilibrium

3. Free-Body Diagrams

4. Equations of Equilibrium


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Rigid Body

A rigid body can be considered as a combination of a

large number of particles in which all particles remain at

a fixed distance from one another, both before and after

applying a load.

This model is important as material properties of any

body that is assumed to be rigid will not have to be

considered when studying the effects of forces acting on

the body.


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Rigid Body (contd.)

Staticsdeals primarily with the calculation of external

forces which act on rigid bodies in equilibrium.

Determination of the internal deformations belongs to the

study of the mechanics of deformable bodies, which

normally follows statics in the curriculum i.e. Mechanics

or Strength of materialscourse.


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CONDITIONS FOR RIGID-BODY EQUILIBRIUM

In Chapter 3 we only consideredforces acting on a particle

(concurrent forces). In this caserotation is not a concern, soequilibrium could be satisfied by:

Forces on a particle

We will now consider cases where forces arenot concurrent so we are also concerned thatthe rigid body does not rotate. In order for a

rigid body to be in equilibrium, the net forceas well as the net moment about anyarbitrary point O must be equal to zero.

Forces on a rigidbody

F= 0 (no translation)

and MO

= 0 (no rotation)

F= 0 (no translation)


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Conditions for Rigid-Body Equilibrium

The equilibrium of a body is expressed as

0

0

OOR

R

MM

FF


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Conditions for Rigid-Body Equilibrium

(contd.)

Consider summing moments

about some other point, such

as point A, we require

0ORRA MFrM

0

0

OOR

R

MM

FF


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Free Body Diagrams

Support Reactions

If a 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.


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Support Reactions (contd.)


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Summary: Support Reactions

A reactionis a force with known line of action, or a force

of unknown direction, or a moment.

The line of action of the force or direction of the moment

is directly related to the motion that is prevented.


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Recognizing support unknowns in FBDs

S t R ti i 2D (T bl 5 1) i thi t bl !


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Support Reactions in 2D (Table 5-1)

The 2D reactions shown below are the ones shown in Table 5-1 inthe text. As a general rule:

1) if a support prevents translation of a body in a givendirection, thena force is developed on the body in the oppositedirection.2) if rotation is prevented, a couple moment is exerted on thebody in the opposite direction

memorize this table!


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Free Body Diagrams (contd.)

Internal Forces

External and internal forces can act on a rigid body

In summary:

For FBD, internal forces act between adjacent particles which are

contained within the boundary of the FBD, are not represented

Particles outside this boundary exert external forces on the

system


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Internal Forces (contd.)


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Free Body Diagrams (contd.)

Weight and Center of Gravity

Each particle has a specified weight

System can be represented by a single resultant force,known as weight Wof the body

Location of the force application is known as the

center of gravity


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Procedure for Drawing a FBD

1. Draw Outlined Shape

Imagine body to be isolated or cut free from its

constraints

Draw outline shape

2. Show All Forces and Couple Moments

Identify all external forces and couple moments that

act on the body


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3. Identify Each Loading and Give Dimensions

Indicate dimensions for calculation of forces

Known forces and couple moments should be properly

labeled with their magnitudes and directions

Procedure for Drawing a FBD (contd.)


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Example: 1

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

beam has a mass of 100kg.


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Example: 1 (contd.)

Free-Body Diagram


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Free-Body Diagrams Description

Support at A is a fixed wall, the wall exerts three reactions

on the beam. These 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 beams center of gravity, 3m from A

Example: 1 (contd.)

EXAMPLE: 2


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EXAMPLE: 2

Smooth pin

Weightless link (see Table 5-1)

Given:The operator applies a verticalforce to the pedal so that thespring is stretched 1.5 in. and theforce in the short link atB is20 lb.

Draw:A an idealized model and free-body diagram of the foot pedal.


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Free-Body Diagrams Description

Example: 2 (contd.)


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Exercise: 1

Draw a FBD of member ABC, which is supported by a smoothcollar at A, rocker at B, and link CD.


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Equations of Equilibrium

For equilibrium of a rigid body in 2D,

Fx= 0; Fy= 0; MO= 0

Fxand Fyrepresent sums of x and y components of

all the forces

MOrepresents the sum of the couple moments and

moments of the force components


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Procedure for Analysis

Free-Body Diagram

Force or couple moment having an unknown magnitude

but known line of action can be assumed

Indicate the dimensions of the body necessary for

computing the moments of forces


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Apply MO= 0 about a point O, that l ies at the

in tersect ion of the lines of act ion of two unknown

forces.

In this way, moments of these unknowns are zero

about O and a direct solution forthe third unknown can

be obtained

Procedure for Analysis (contd.)


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When applying the force equilibrium equations, orient

the x and yaxes along the lines that will provide the

simplest resolution of the forces into theirxand ycomponents

Negative result implies scalar for a force or couple

moment magnitude is opposite to that was assumed on

the FBD

Procedure for Analysis (contd.)


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Example: 1

Determine the horizontal and vertical components of reaction

on the beam caused by the pin at B and the rocker at A.

Neglect the weight of the beam in the calculations.


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Free Body Diagram

600N represented by x and y components

200N force acts on the beam at B

Example: 1 (contd.)


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NBBNFxxx

424045cos600;0

NB

BNNNN

F

NA

mAmNmNmN

M

y

y

y

y

y

B

405

020010045sin600319

;0

319

0)7()2.0)(45cos600()5)(45sin600()2(100

;0

Example: 1 (contd.)


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Example: 2


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Example: 2 (contd.)


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Example: 2 (contd.)


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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.

Only force magnitude must be determined


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Two- Force Members (contd.)


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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 Aand B

are thus known (along the line joining points A and B).

Th F M b


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Three-Force Members When subjected to three forces, the forces are concurrent

or parallel


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Example: 1

The lever ABC is pin-supported at A and connected to a

short link BD. If the weight of the members are negligible,

determine the force of the pin on the lever at A.


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Example: 1 (contd.)

Free Body Diagrams BDis a two-force member

LeverABCis a three-force member


Solving,

kNF

kNFA

32.1

07.1

045sin3.60sin;0

040045cos3.60cos;0

3.604.0

7.0tan 1

FFF

NFFF

Ay

Ax


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A man raises a 10 kg joist, of

length 4 m, by pulling on a

rope.

Find the tension in the ropeand the reaction atA.

SOLUTION:

Create a free-body diagram of the joist.

Note that the joist is a 3 force body

acted upon by the rope, its weight, and

the reaction atA.

The three forces must be concurrent

for static equilibrium. Therefore, the

reaction Rmust pass through the

intersection of the lines of action of the

weight and rope forces. Determine the

direction of the reaction force R.

Utilize a force triangle to determine

the magnitude of the reaction force R.

Example: 2

Example: 2 (contd )


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Create a free-body diagram of the joist.

Determine the direction of the reaction

force R.

636.1

414.1

313.2tan

m2.313m515.0828.2

m515.020tanm414.1)2045cot(

m414.1

m828.245cosm445cos

2

1

AE

CE

BDBFCE

CDBD

AFAECD

ABAF

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Example: 2 (contd.)


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Determine the magnitude of the reactionforce R.

38.6sin

N1.98

110sin4.31sin

RT

N8.147

N9.81

R

T

Example: 2 (contd.)


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Exercises

Engineering Mechanics - Statics, R.C. Hibbeler, 12th

Edition

Q5.11Q5.24

Q5.32Q5.37

Q5.41Q5.45


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QUIZ

1. If a support prevents translation of a body, then thesupport 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 bodydiagram of a whole body.

A) AlwaysB) Often

C) Rarely

D) Never


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QUIZ (contd.)

3. 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


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QUIZ (contd.)

4. Internal forces are not shown on a free-body diagram

because the internal forces are_____.

A) Equal to zero

B) Equal and opposite and they do not affect thecalculations

C) Negligibly small

D) Not important


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QUIZ (contd.)

5. 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

6. A rigid body is subjected to forces.

This body can be considered

as a ______ member.A) Single-force B) Two-force

C) Three-force D) Six-force

Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body

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