CE201 Statics Chap5

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STATICS: CE201

Chapter 5

Equilibrium of a Rigid Body

Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson

Dr M. Touahmia & Dr M. Boukendakdji

Civil Engineering Department, University of Hail

(Fall 2010)

5. Equilibrium of a Rigid Body________________________________________________________________________________________________________________________________________________

Main Goals:

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

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

rigid body.

3. To show how to solve rigid body equilibrium problems using

the equations of equilibrium.

Contents:

5.1 Conditions for Rigid-Body Equilibrium 5.2 Free-Body Diagrams

5.3 Equations of Equilibrium

5.4 Two- and Three-Force Members

5.5 Free-Body Diagrams


5.7 Constraints and Statical Determinacy

Chapter5 - Equilibrium of a Rigid Body 2

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5.1 Conditions for Rigid-Body Equilibrium

The force and couple system acting on a body can be

reduced to an equivalent resultant force and resultant

couple moment at an arbitrary point O.

If the resultant force and couple moment are both equal

to zero, then the body is said to be in equilibrium.

=


0FFR

0 OOR MM

5.1 Conditions for Rigid-Body Equilibrium

Chapter5 - Equilibrium of a Rigid Body

The two equations of equilibrium for a rigid body are:

where O is an arbitrary point

When applying the equations of equilibrium, we can assume

that the body will remain rigid and not deform under the

applied load.

Most engineering materials such as steel and concrete are

very rigid and so their deformation is usually very small.

0F

0 OM

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No equilibrium problem should be solved without first

drawing the free-body diagram, so as to account for all

the forces and couple moments that act on the body.

Support Reactions:

We consider the various types of reactions that occur at

supports and points of contact between bodies subjected

to coplanar force systems. As a general rule:

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.


Support Reactions

Let us consider three ways in which a horizontal

member, such as a beam, is supported at its end:

Roller: This support prevents the beam from translating

in the vertical direction, the roller will only exert a force

on the beam in this direction.



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Support Reactions:

Pin: The pin can prevent translation of the beam in any

direction.

For purposes of analysis, it is generally easier to

represent this resultant force F by its two rectangular

components Fx and Fy .



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Support Reactions:

Fixed Support: This support will prevent both translation

and rotation of the beam.

A force and couple moment must be developed on the

beam at its point of connection. The force is usually

represented by its two rectangular components Fx and Fy.



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Example 1

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

beam has a mass of 100 kg.


Solution 1

Free-body diagram of the beam:

Since the support at A is fixed, the wall exerts three reactions

on the beam, denoted as Ax, Ay, and MA.

The magnitudes of theses reactions are unknown, and their

sense has been assumed.

The weight of the beam, W = 100(9.81) N = 9.81 N, acts

through the beams center of gravity G, which is 3 m from A

since the beam is uniform.Chapter5 - Equilibrium of a Rigid Body 10

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The conditions for equilibrium in two dimensions can be

written in scalar form as:

represent, respectively the algebraic

sums of the x and y components of all the forces and

represents the algebraic sum of the couple

moments and moments of all the force components

about an axis perpendicular to the x y plane and

passing through an arbitrary point O.


0 xF 0 yF 0 OM

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xF yF

O

M


Alternative Sets of Equilibrium Equations:

Two alternative sets of three independent equilibrium

equations may be used.

The first set of equation is

When using these equations, it is required that a line

passing through points A and B is not parallel to the y

axis.


0 xF

0 AM

0 BM

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Alternative Sets of Equilibrium Equations:

A second alternative set of equilibrium equation is

Here, the only requirement is that points A, B and C do

not lie in the same line.


0 AM

0 CM

0 BM

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Example 3

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.


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Solution 3

Free-Body Diagram: Identify each of the forces shown

on the free-body diagram of the beam.

For simplicity, the 600-N force is represented by its x

and y components


Solution 3

Equations of Equilibrium: Summing forces in the xdirection yields.

A direct solution for Ay can be obtained by applying the

moment equation about point B:


0 xF

0 BM

NAy 319

0m7m2.0N45cos600m5N 45sin600m2N 100 yoo A

0N 45cos600 xo B N 424xB

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Solution 3

Summing forces in the y direction yields:

Note: We can check this result by summing moments aboutpoint A.


0 yF

NBy 405

0 AM

0200N 100N 45sin600N 319 yo B

0m7m7N200m5N100m2.0N45cos600m2N45sin600 yoo B

N 405yB

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Example 4

The member is pin-connected at A and rests against a

smooth support at B. Determine the horizontal and

vertical components of reaction at the pin A.


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Solution 4

Free-Body Diagram:


Solution 4

Equations of Equilibrium: Summing moments about A,

we obtain a direct solution for NB.


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Example 5

Determine the support reactions on the member. The

collar at A is fixed to the member and can slide

vertically along the vertical shaft.


Solution 5

Free-Body Diagram: The collar exerts a horizontal

forces Ax and a moment MA on the member. The

reaction NB of the roller on the member is vertical.


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Solution 5

Equations of Equilibrium:


Example 6


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Solution 6

Free-Body Diagram: The pin at A exerts two

components of reaction on the member, Ax and Ay.


Solution 6

Equation of Equilibrium:


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Problem

Determine the horizontal and vertical components of

reaction on the beam caused by the pin at A and the roller

at B. Neglect the weight of the beam.



The solutions to some equilibrium problems can be

simplified by recognizing members that are subjected to

only two or three forces.

Two-Force Members:

A two-force member has forces applied at only two

points on the member.


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Two-Force Members:

To satisfy force equilibrium, FA and FB must be equal in

magnitude, , but opposite in direction.


FFF BA

0F

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Two-Force Members:

Moment equilibrium requires that FA and FB share the same

line of action, which can only happen if they are directed

along the line joining points A and B.

or

Therefore, for any two-force member to be in equilibrium,

the two forces acting on the member must have the same

magnitude, act in opposite directions and have the same line

of action, directed along the line joining the two points where

these forces act.Chapter5 - Equilibrium of a Rigid Body

0 AM 0 BM

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Three-Force Members:

If a member is subjected to only three forces, it is called

a three-force member.

Moment equilibrium can be satisfied only if the three

forces form a concurrent or parallel force system.



Three-Force Members:

If the lines of action of F1 and F2 intersect at point O, then the

line of action of F3 must also pass through point O so that the

forces satisfy:

If the three forces are all parallel, the location of the point of

intersection O will approach infinity.


0 OM

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5.5 Free-Body Diagrams (Three Dimensions)

The first step in solving three-dimensional equilibrium

problems is to draw a free-body diagram.

Before we can do this, it is first necessary to discuss the

types of reactions that can occur at the supports.

Support Reactions:

As in the two-dimensional case:

A force is developed by a support that restricts thetranslation of its attached member.

A couple moment is developed when rotation of theattached member is prevented.



Support Reactions:

For example the ball-and-socket joint prevents any

translation of the connecting member; therefore, a force

must act on the member at the point of connection.

This force has three components having unknown

magnitudes: Fx, Fy, Fz

Then we can obtain the magnitude of force F,


222

zyx FFFF

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Support for Rigid Bodies Subjected to Three-Dimensional

Force Systems:




Force Systems:


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Force Systems:




Force Systems:


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Example 5

Consider the two rods and plate, along with their

associated free-body diagram. The x, y, z axes are

established on the diagram and the unknown reaction

components are indicated in the positive sense.


Free-body diagram

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Example 5


300N

300N

200N.m

200N.m

Free-body diagram

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Example 5


400N

400N

Free-body diagram

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5.6 Equations of Equilibrium (Three Dimensions)

The conditions for equilibrium of a rigid body

subjected to a three-dimensional force system

require that both the resultant force and resultant

couple moment acting on the body be equal to zero.


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5.6 Equations of Equilibrium (3 D)

Vector Equations of Equilibrium: The two conditions

for equilibrium of a rigid body may be expressed

mathematically in vector form as:

and

where:

is the vector sum of all the external forces

acting on the body

is the sum of the couple moments and the

moments of all the forces about any point O (located

either on or off the body).


0 OM0F

0F

0 OM

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5.6 Equations of Equilibrium (3-D)

Scalar Equations of Equilibrium: If all the external

forces and couple moments are expressed in Cartesian

vector form, we have:


0 kjiF zyx FFF

0 kjiM zyxO MMM

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5.6 Equations of Equilibrium (3 D)

Since the i, j, and k components are independent from one

another, the above equations are satisfied provided that:

and

These six scalar equilibrium equations may be used to solve

for at most six unknowns shown on the free-body diagram.


0 xF 0 yF 0 zF

0 xM 0 yM 0 zM

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To ensure the equilibrium of a rigid body, it is not only

necessary to satisfy the equations of equilibrium, but the

body must also be properly held or constrained by its

supports.

Redundant Constraints: When a body has redundant

supports, that is, more supports than are necessary to

hold it in equilibrium, it becomes statically

indeterminate.

Statically indeterminate: means that there will be more

unknown loadings on the body than equations of

equilibrium available for their solutions.


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Example: The beam is shown together with its free-

body diagram. The beam is statically indeterminate

because of additional (or redundant) supports

reactions.

There are five unknowns MA, Ax, Ay, By, and Cy, for

which only three equilibrium equations can be written:


0 xF 0 yF 0 OM

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Second example: The pipe is also statically indeterminatebecause of additional (or redundant) supports reactions.

The Pipe assembly has eight unknowns, for which only six

equilibrium equations can be written.


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The additional equations needed to solve statically

indeterminate problems are generally obtained from

the deformation conditions at the points of supports

This is done in courses dealing with mechanics of

Materials



Improper Constraints: Having the same number ofunknown reactive forces as available equations of

equilibrium does not always guarantee that a body will be

stable when subjected to a particular loading.

For example, the pin support at A and the roller support at B

for the beam are placed in such a way that the lines of

action the reactive forces are concurrent at point A.

Consequently, the applied loading P will cause the beam to

rotate slightly about A, and so the beam is improperly

constrained.


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In three dimensions, a body will be improperly

constrained if the lines of action of all the reactive

forces intersect a common axis.

For example, the reactive forces at the ball-and-socket

supports at A and B all intersect the axis passing through A

and B.

Note: Since the moments of these forces about A and B are

all zero, then the loading P will rotate the member about the

AB axis.



Another way in which improper constraining leads to

instability occurs when the reactive forces are all parallel.

Note: The summation of forces along the x axis will not be

equal zero.


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CE201 Statics Chap5