Engineering Mechanics I (Statics) (Part 1/2)

Shieh-Kung
Huang
ENGINEERING
MECHANICS –
STATICS
Shieh-Kung Huang
黃謝恭
1

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Copyright © 2016 by Pearson Education, Inc. All rights reserved.
INTRODUCTION AND GENERAL PRINCIPLES
Chapter Objectives
10
Students will be able to
1.Explain mechanics / statics
2.Work with two types of units
3.Round the final answer appropriately
4.Apply problem-solving strategies
5.Resolve a 2-D or 3-D vector into components
6.Add 2-D or 2-D vectors using Cartesian vector notations
7.Determine an angle and the projection between two vectors
CHAPTER 1
1.1Mechanics
1.2Fundamental Concepts
1.3The International System of Units
1.4Numerical Calculations
1.5General Procedure for Analysis
1.6Scalars and Vectors
1.7Vector Operations
1.8Vector Addition of Forces
1.9Addition of a System of Coplanar Forces
1.10Cartesian Vectors
1.11Position Vectors
1.12Force Vector Directed Along a Line
1.13Dot Product
Chapter Outline

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1.1 MECHANICS
What is Mechanics?
11
Chapter 1 Introduction and General Principles
Study of what happens to a “thing” (the technical name is “BODY”) when FORCESare applied to it.
Either the body or the forces can be large or small.
Branches of Mechanics
Mechanics
Rigid Bodies
(Things that do not change shape)
Deformable Bodies
(Things that do change shape)
Fluids
Statics Dynamics IncompressibleCompressible

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1.2 FUNDAMENTAL CONCEPTS
Idealizations
12
Chapter 1 Introduction and General Principles
Three important idealizations used in statics
•Rigid Body
A rigid body can be considered as a combination of a large number of particles in which all the
particles remain at a fixed distance from one another, both before and after applying a load.
•Particle
A particle has a mass, but a size that can be neglected.
•Concentrated Force
A concentrated force represents the effect of a loading which is assumed to act at a point on a body.
Mechanics
Rigid Bodies
(Things that do not change shape)
Deformable Bodies
(Things that do change shape)
Fluids
Statics Dynamics IncompressibleCompressible

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1.2 FUNDAMENTAL CONCEPTS
Newton’s Three Laws of Motion
13
Chapter 1 Introduction and General Principles
Engineering mechanics is formulated on the basis of Newton’s three laws of motion
•First Law
A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this
state provided the particle is not subjected to an unbalancedforce.
•Second Law
A particle acted upon by an unbalancedforce experiences an acceleration a that has the same
direction as the force and a magnitude that is directly proportionalto the force.
•Third Law
The mutual forces of action and reaction between two particles are equal, opposite, and collinear.

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1.2 FUNDAMENTAL CONCEPTS
Basic Quantities
14
Chapter 1 Introduction and General Principles
Four fundamental physical quantities (or dimensions)
•Length
•Mass
•Time
•Force
Newton’s 2
nd
Law relates them:
We use this equation to develop systems of units.
Units are arbitrary names we give to the physical quantities.
Force, mass, time and acceleration are related by Newton’s 2
nd
law. Three of these are assigned units
(called base units) and the fourth unit is derived. Which one is derived varies by the system of units.
We will work with two unit systems in statics:
•International System (SI)
•U.S. Customary (FPS) mm==Fad

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1.2 FUNDAMENTAL CONCEPTS
Weight
15
Chapter 1 Introduction and General Principles
Weight of a body:
Newton’s Second Law:
By comparison with Newton’s Second Law, we can see that gis the acceleration due to gravity.m=Fa Wmg=

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1.3 THE INTERNATIONAL SYSTEM OF UNITS
Metric Prefixes
19
Chapter 1 Introduction and General Principles
−No plurals (e.g., m= 5 kg, not kgs)
−Separate units with a • (e.g., meter second = m • s)
−Most symbols are in lowercase
(some exceptions are M and G)
−Exponential powers apply to units, e.g., cm • cm = cm
2
−Compound prefixes should not be used
−Table 1-3 in the textbook shows some prefixes
For more information, please see https://en.wikipedia.org/wiki/Metric_prefix

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1.3 THE INTERNATIONAL SYSTEM OF UNITS
Metric Prefixes
20
Chapter 1 Introduction and General Principles

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1.3 THE INTERNATIONAL SYSTEM OF UNITS
Metric Prefixes
23
Chapter 1 Introduction and General Principles
Four fundamental physical
quantities (or dimensions)
•Length
•Mass
•Time
•Force

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1.3 THE INTERNATIONAL SYSTEM OF UNITS
Metric Prefixes
24
Chapter 1 Introduction and General Principles
Four fundamental physical
quantities (or dimensions)
•Length
•Mass
•Time
•Force

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1.4 NUMERICAL CALCULATIONS
25
Chapter 1 Introduction and General Principles
Some important aspects involved in all engineering calculations
•Dimensional Homogeneity
Must have dimensional “homogeneity.”
Dimensions have to be the same on both sides of the equal sign, (e.g., distance = speed x time).
•Significant Figures
Use an appropriate number of significant figures (3 for answer, at least 4 for intermediate calculations).
•Rounding Off Numbers
Be consistent when rounding off.
−greater than 5, round up (3528 →3530)
−smaller than 5, round down (0.03521 →0.0352)
−equal to 5, see your textbook for an explanation
−Calculations
When a sequence of calculations is performed, it is best to store the intermediate results in the calculator.
And, always be careful while calculation.

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1.5 GENERAL PROCEDURE FOR ANALYSIS
26
Chapter 1 Introduction and General Principles
1.Interpret:
Readcarefully and determine what is givenand
what is to be found/delivered.
Ask, if not clear. If necessary, make assumptions
and indicatethem.
2.Plan:
Think about major steps(or a road map) that you
will take to solve a given problem.
Think of alternative/creativesolutions and choose
the best one.
3.Execute:
Carry out your steps. Use appropriate diagrams
and equations.
Estimateyour answers.
Avoidsimple calculation mistakes.
Reflecton and then revise your work, if necessary.

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1.6 SCALARS AND VECTORS
30
Chapter 1 Introduction and General Principles
•Scalar
A quantity that can be fully characterized by its magnitude (a positive or negative number)
e.g. Mass, volume, and length
•Vector
A quantity that has magnitude and direction
e.g. Position, force and moment
Represent by a letter with an arrow over it, , or a bold letter, A,
Magnitude is designated as or
In this subject, vector is presented as ??????and its magnitude (positive quantity) as ??????
Scalars Vectors
Examples Mass, Volume, Length Force, Position, Velocity
CharacteristicsIt has a magnitude (Positive or negative)It has a magnitude and direction
Addition rule Simple arithmetic Parallelogram law
Special Notation None Bold font, a line, an arrow, or a “carrot”A A A

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1.7 VECTOR OPERATIONS
Vector Addition
31
Chapter 1 Introduction and General Principles
•Multiplication and Division of a Vector by a Scalar
Product of vector ??????and scalar ais
Magnitude of scaled vector
Law of multiplication applies e.g.
•Vector Addition
Additionof two vectors A and B gives a resultant vector R by the parallelogram law
Result R can be found by triangle construction
Communicative e.g.
Special case: Vectors A and B are collinear
(both have the same line of action)aA aA ()10aaa=AA =+=+RABBA

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1.7 VECTOR OPERATIONS
Vector Subtraction
32
Chapter 1 Introduction and General Principles
•Vector Subtraction
Special case of addition e.g.
Rules of Vector Addition Applies()'=−=+−RABAB

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1.8 VECTOR ADDITION OF FORCES
Finding a Resultant Force
33
Chapter 1 Introduction and General Principles
•Finding a Resultant Force
Parallelogram law is carried out to find the resultant force
•Resultant( )
12R
=+FFF

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1.8 VECTOR ADDITION OF FORCES
Finding the Components of a Force
34
Chapter 1 Introduction and General Principles
“Resolution” of a vector is breaking up a vector into components.
It is kind of like using the parallelogram law in reverse.

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1.8 VECTOR ADDITION OF FORCES
Addition of Several Forces
35
Chapter 1 Introduction and General Principles
Using the skills several times to get the resultant force or find the components of a force

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1.8 VECTOR ADDITION OF FORCES
36
Chapter 1 Introduction and General Principles

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1.9 ADDITION OF A SYSTEM OF COPLANAR FORCES
Scalar Notation
41
Chapter 1 Introduction and General Principles
•Scalar Notation
x and y axes are designated positive and negative
Components of forces expressed as algebraic scalars
and
andxy
=+FFF cos
x
FF = sin
y
FF = x
a
FF
c

=

 y
b
FF
c

=−



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1.9 ADDITION OF A SYSTEM OF COPLANAR FORCES
Cartesian Vector Notation
42
Chapter 1 Introduction and General Principles
•Cartesian Vector Notation
Cartesian unit vectors iand j are used to designate the x and y directions
Unit vectors iand j have dimensionless magnitude of unity ( = 1 )
Magnitude is always a positive quantity, represented by scalars F
xand F
y
The x and y axis are always perpendicular to each other.
Together, they can be directed at any inclination.xy
FF=+Fij

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1.9 ADDITION OF A SYSTEM OF COPLANAR FORCES
Cartesian Vector Notation
43
Chapter 1 Introduction and General Principles
•To determine resultant of several coplanar forces:
Resolve force into x and y components
Addition of the respective components using scalar algebra
Resultant force is found using the parallelogram law
Cartesian vector notation:
•Vector resultant is therefore
•If scalar notation are used11 1
2 2 2
33 3
xy
xy
xy
FF
FF
FF
=+
=−+
=−
Fij
Fij
Fij ()()123R Rx Ry
FF=++=+FFFFij 123
123
Rxxxx
Ryy yy
FFFF
FFFF
=−+
=+−

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1.9 ADDITION OF A SYSTEM OF COPLANAR FORCES
Cartesian Vector Notation
44
Chapter 1 Introduction and General Principles
•In all cases we have
•Magnitude of ??????
??????can be found by Pythagorean Theorem
and
You can always represent a 2-D vector with a magnitude and angle.Rx x
Ry y
FF
FF
=
=

 22
R RxRy
FFF=+ 1
tan
Ry
Rx
F
F

−
=

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1.10 CARTESIAN VECTORS
Cartesian Vector Representation
51
Chapter 1 Introduction and General Principles
•Right-Handed Coordinate System
A rectangular or Cartesian coordinate system is said to be right-handed provided:
•Thumb of right hand points in the direction of the positive z axis
•z-axis for the 2D problem would be perpendicular, directed out of the page.
•Rectangular Components of a Vector
A vector A may have one, two or three rectangular components along the x, y
and z axes, depending on orientation
By two successive application of the parallelogram law
Combing the equations, Acan be expressed as
•Unit Vector
Direction of A can be specified using a unit vector
Unit vector has a magnitude of 1 (dimensionless)
If A is a vector having a magnitude of ,
unit vector having the same direction as
A is expressed by . So that'
'
z
xy
=+
=+
AAA
AAA xyz
=++AAAA 0A A
A=uA A
A=Au

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1.10 CARTESIAN VECTORS
Cartesian Vector Representation
52
Chapter 1 Introduction and General Principles
•Cartesian Vector Representations
3 components of Aact in the positive i, jand kdirections
*Note the magnitude and direction of each components are separated, easing vector algebraic operations.xyz
AAA=++Aijk

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1.10 CARTESIAN VECTORS
Magnitude of a Cartesian Vector
53
Chapter 1 Introduction and General Principles
•Magnitude of a Cartesian Vector
From the colored triangle,
From the shaded triangle,
Combining the equations gives magnitude of A22
'
z
AAA==+A 22
'
xy
AAA=+ 222
xyz
AAAA==++A

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1.10 CARTESIAN VECTORS
Coordinate Direction Angles
54
Chapter 1 Introduction and General Principles
•Direction of a Cartesian Vector
Orientation of A is defined as the coordinate direction angles a, band g
measured between the tail of A and the positive x, y and z axes
0°≤ a, band g ≤ 180 °
The direction cosines of Ais
Angles a, band g can be determined by the inverse cosines
Given
then,
where
u
Acan also be expressed as
Since , we have
A as expressed in Cartesian vector form is222
x y z
AAAA=++ ()()()
yxz
A
AAA
AAAA
==++
A
uijk x y z
AAA=++Aijk coscoscos
yxz
AAA
AAA
abg=== coscoscos
A
abg=++uijk 1
A
=u coscoscos1abg++= coscoscos
A
x y z
A
AAA
AAA
abg
=
=++
=++
Au
ijk
ijk

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1.10 CARTESIAN VECTORS
Transverse and Azimuth Angles
55
Chapter 1 Introduction and General Principles
A can be specified using two angles, namely, a transverseazimuth angle and an azimuthzenith angle f
applying trigonometry first to the blue right triangle yields
Now applying trigonometry to the gray right triangle
A as expressed in Cartesian vector form issincossinsincos
A
x y z
A
AAA
AAA
fff
=
=++
=++
Au
ijk
ijk cos'sin
z
AAAAff== 'cossincos
'sinsinsin
x
y
AAA
AAA
f
f
==
==

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1.10 CARTESIAN VECTORS
Addition of Cartesian Vectors
56
Chapter 1 Introduction and General Principles
•Concurrent Force Systems
Force resultant is the vector sum of all the forces in the system
For example,( )( )( )
x x y y z z
ABABAB=+=+++++RABijk R x y z
FFF==++FFijk

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1.10 CARTESIAN VECTORS
57
Chapter 1 Introduction and General Principles
•Vector in Cartesian coordinate system
•Requirement for representing of a vectorcoscoscoscoscoscos
sincossinsincossincossinsincos
A
x y z
A
A
A
AAA
AAA
AAA
abgabg
ffffff
=
=++
=++ =++
=++ =++
Au
ijk
ijkuijk
ijkuijk and
,,and
,,,and
,,and
A
xy z
A
AAA
A
A
abg
f







u
A

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1.11 POSITION VECTORS
x, y, z Coordinates
62
Chapter 1 Introduction and General Principles
•x,y,zCoordinates
Right-handed coordinate system
Positive z axis points upwards, measuring the height of an object or the altitude of a point
Points are measured relative to the origin, O.

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1.11 POSITION VECTORS
Position Vector
63
Chapter 1 Introduction and General Principles
Position vector r is defined as a fixed vector which locates a point in space relative to another point.
E.g.
Vector addition gives
Solvingxyz=++rijk AB
+=rrr ( )( )( )
B A B A B A B A
xxyyzz=−=−+−+−rrrijk

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1.12 FORCE VECTOR DIRECTED ALONG A LINE
Position Vector
66
Chapter 1 Introduction and General Principles
In 3D problems, direction of Fis specified by 2 points, through which its line of action lies
Fcan be formulated as a Cartesian vector
Note that Fhas units of forces (N)
unlike r, with units of length (m)
Force Facting along the chain can be presented as a Cartesian vector by
•Establish x, y, z axes
•Form a position vector ralong length of chain
Unit vector, that defines the direction of both the chain and the force
We getFF
r

==


r
Fu r=ur F=Fu

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1.13 DOT PRODUCT
Laws of Operation
73
Chapter 1 Introduction and General Principles
•Dot product of vectors A and B is written as A·B (Read A dot B)
Define the magnitudes of A and B and the angle between their tails
where 0°≤ ≤180°
Referred to as scalar product of vectors as result is a scalar
•Laws of Operation
•Commutative law:
•Multiplication by a scalar:
•Distribution law:cosAB=AB =ABBA ()() ()()aaaa===ABABABAB ( )()()+=+ABCABAC

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1.13 DOT PRODUCT
Cartesian Vector Formulation
74
Chapter 1 Introduction and General Principles
•Cartesian Vector Formulation
Dot product of Cartesian unit vectors
Similarly
Dot product of 2 vectors A and B
•Applications
The angle formed between two vectors or intersecting lines.
The components of a vector parallel and perpendicular to a line.()()
()()
11cos01
11cos0
4

==
==
ii
ij 111
000
===
===
iijjkk
ijikjk ( )( )
() () ()
() () ()
() () ()
x y z x y z
xx xy xz
yx yy yz
zx zy zz
xx yy zz
AAABBB
ABABAB
ABABAB
ABABAB
ABABAB
=++++
=++
+++
+++
=++
ABijkijk
iiijik
jijjjk
kikjkk ()
1
cos
AB

− 
=


AB cos
aa
A=Au

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EQUILIBRIUM OF A PARTICLE
Chapter Objectives
80
CHAPTER 2
Students will be able to
1.Draw a free-body diagram (FBD)
2.Apply equations of equilibrium to solve a 2-D problem
3.Applying the three scalar equations
(based on one vector equation) of equilibrium
2.1Condition for the Equilibrium of a Particle
2.2The Free-Body Diagram
2.3Coplanar Force systems
2.4Three-Dimensional Force systems
Chapter Outline

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2.1 CONDITION FOR THE EQUILIBRIUM OF A PARTICLE
81
Chapter 2 Equilibrium of a Particle
•A particle is said to be in equilibriumif
−it remains at rest if originally at rest, or
−has a constant velocity if originally in motion.
•Newton’s first law of motion stated by the equation of equilibrium
where SFis the vector sum of all the forcesacting on the particle.
The equation is not only a necessary condition for equilibrium, it is also a sufficient condition.
•Newton’s second law of motion stated by the equation of equilibrium
When the force fulfill Newton's first law of motion,
therefore, the particle is moving in constant velocity or at rest.=F0 m=Fa m==a0a0

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2.2 THE FREE-BODY DIAGRAM
Springs
82
Chapter 2 Equilibrium of a Particle
•Free-Body Diagram (FBD)
Best representation of allthe unknown forces (SF) which act onthe body or particle
A sketch showing the body or particle “free” from the surroundings with all the forces acting on it
Consider three common connections in this subject –
−Spring
−Cables and Pulleys
−Smooth Contact
•Spring
Linear elastic spring: change in length is directly proportional to the force acting on it
Spring constantor stiffnessk: defines the elasticity of the spring
The magnitude of force, when spring is elongated or compressed, isFks=

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2.2 THE FREE-BODY DIAGRAM
Cables and Pulleys
83
Chapter 2 Equilibrium of a Particle
•Cables and Pulley
Cables (or cords) are assumed negligible weight and cannot stretch.
A cable can support onlya tension or “pulling” force, and therefore, the force always acts in the
direction of the cable.
The tension force must have a constantmagnitude for equilibrium, meaning that, for any angle , the
cable is subjected to a constant tension T.

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2.2 THE FREE-BODY DIAGRAM
Smooth Contact
84
Chapter 2 Equilibrium of a Particle
•Smooth Contact
If an object rests on a smooth surface(frictionless), then the surface will exert a force on the object
that is normal to the surface at the point of contact.
If the surface isn’t smooth, the friction force appears.

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2.2 THE FREE-BODY DIAGRAM
Procedure for Drawing a Free-Body Diagram
85
Chapter 2 Equilibrium of a Particle

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2.3 COPLANAR FORCE SYSTEMS
88
Chapter 2 Equilibrium of a Particle
If a particle or body is subjected to a system of
coplanar forces in the x-y plane, then each force can
be resolve into its iand jcomponents for equilibrium
Scalar equations of equilibrium require that the
algebraic sumof the x and y components to equal
to zero.00
xy
xy
FF
FF
=
+=
==



F0
ij0

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2.4 THREE-DIMENSIONAL FORCE SYSTEMS
95
Chapter 2 Equilibrium of a Particle
Similar to coplanar force systems, the necessary and
sufficient condition for particle equilibrium is
Each force can be resolving into i, j, and k
components
Three scalar equations representing algebraic sums
of the x, y, and z forces.=F0 x y z
FFF++=ijk0 000
x y z
FFF===

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SYSTEMS OF FORCES AND MOMENTS
Chapter Objectives
105
CHAPTER 3
Students will be able to
1.define and determine moment of a force in 2-D and 3-D cases
2.perform scalar analysis
3.perform vector analysis.
4.define and determine the moment of a couple
5.determine the effect of moving a force
6.find an equivalent force-couple system for a system of forces
and couples
7.determine an equivalent force for a distributed load
3.1Moment of a Force—Scalar Formulation
3.2Cross Product
3.3Moment of a Force—Vector Formulation
3.4Principle of Moments
3.5Moment of a Force about a Specified Axis
3.6Moment of a Couple
3.7Simplification of a Force and Couple System
3.8Further Simplification of a Force and Couple
System
3.9Reduction of a Simple Distributed Loading
Chapter Outline

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3.1 MOMENT OF A FORCE —SCALAR FORMULATION
Magnitude and Direction
106
Chapter 3 Systems of Forces and Moments
The momentof a force about a point or axis provides a measure of the tendency for rotationabout the
point or axis (sometimes called torque)
Torque is tendency of rotation caused by F
xor simple moment (M
O)
z
•Magnitude
The magnitude of the moment, M
O, is
where dis the moment armor perpendicular distancefrom
the axis at point Oto the line of actionof the force.
•Direction
The direction is determined using “right hand rule,” called
moment axis.O
MFd=

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3.1 MOMENT OF A FORCE —SCALAR FORMULATION
Resultant Moment
107
Chapter 3 Systems of Forces and Moments
For two-dimensional problems, where all the forces lie within the x–y plane, the resultant moment (M
O)
z
about point O(the z axis) can be determined by finding the algebraic sumof the moments caused by all
the forces in the system.
As a convention, we will generally consider positivemoments as counterclockwisesince they are directed
along the positive z axis (out of the page). Clockwisemoments will be negative.()
Oz
MFd= ()
11 22 33Oz
MFdFdFd=−+

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3.2 CROSS PRODUCT
Magnitude and Direction
111
Chapter 3 Systems of Forces and Moments
Cross product of two vectors Aand Byields C, which is written as
•Magnitude
The magnitudeof Cis defined as the product of the magnitudes of Aand Band the sine of the angle .
For angle θ, 0°≤ ≤ 180°
•Direction
Vector Chas a directionthat is perpendicular to the plane containing Aand Bsuch that Cis specified
by the right hand rule.
Expressing vector Cwhen magnitude and direction are known=CAB sinCAB = (sin)
C
AB==CABu

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3.2 CROSS PRODUCT
Laws of Operation
112
Chapter 3 Systems of Forces and Moments
•Laws of Operation
−Commutative law is not valid
Rather,
Cross product of two vectors Aand Byields a vector opposite in direction to C
−Multiplication by a Scalar
−Distributive Law
Proper orderof the cross product must be maintained
since they are not commutative( )() ()( )aaaa===ABABABAB ( )+=+ABDABAD ABBA =−BAC =−ABBA

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3.2 CROSS PRODUCT
Cartesian Vector Formulation
113
Chapter 3 Systems of Forces and Moments
Use on pair of Cartesian unit vectors, i, j, and k, to find the cross product.
A more compact determinant in the form as
The cross product of two general vectors Aand B, which are expressed in Cartesian vector form, issinCAB = x y z
x y z
ABAAA
BBB
=
ijk ( )( )( )
x y z
x y z
y z x y xz
y z x y xz
yz zy xz zx xy yx
ABAAA
BBB
AA AA AA
BB BB BB
ABABABABABAB
=
=−+
=−−−+−
ijk
ijk
i j k

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3.3 MOMENT OF A FORCE —VECTOR FORMULATION
Magnitude and Direction
114
Chapter 3 Systems of Forces and Moments
The moment of force Fabout point O, or actually about the moment axis passing through Oand
perpendicular to the plane containing Oand F, can be expressed using vector cross product.
•Magnitude
For magnitude of cross product,
Treat ras a sliding vector. Since the moment arm is .
•Direction
The direction and sense of M
Oare determined by right-hand rule
*Note:
−“curl” of the fingers indicates the sense of rotation
−Maintain proper orderof rand Fsince cross product is not commutativeO
=MrF sin(sin)
O
MrFFrFd=== sindr=

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3.3 MOMENT OF A FORCE —VECTOR FORMULATION
Principle of Transmissibility
115
Chapter 3 Systems of Forces and Moments
For force Fapplied at any point Aon the line of action of the force F, moment created about Ois .
Fhas the properties of a sliding vector at any point on the line, thusOA
=MrF 1 2 3O
===MrFrFrF

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3.3 MOMENT OF A FORCE —VECTOR FORMULATION
Cartesian Vector Formulation
116
Chapter 3 Systems of Forces and Moments
For the position vector and force expressed in Cartesian form, the moment is
where ris the position vector from the point Oto any point on the line of action of the forceF.
With the determinant is expended asO x y z
x y z
rrr
FFF
==
ijk
MrF ( )( )( )
O yzzy xzzx xyyx
rFrFrFrFrFrF=−−−+−M i j k

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3.3 MOMENT OF A FORCE —VECTOR FORMULATION
Resultant Moment of a System of Forces
117
Chapter 3 Systems of Forces and Moments
If a body is acted upon by a system of forces, the resultant moment of forces about point Ocan be
determined by vector addition of the moment of each force.()
O
=MrF

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3.4 PRINCIPLE OF MOMENTS
Varignon’sTheorem
122
Chapter 3 Systems of Forces and Moments
A concept often used in mechanics is the principle of moments, which is sometimes referred to as
Varignon’stheoremsince it was originally developed by the French mathematician Pierre Varignon
(1654–1722).
It states that the moment of a force about a point is equal to the sum of the moments of the components
of the force about the point.( )
1 2 1 2O
==+=+MrFrFFrFrF ( )
O x y x y x y
FyFx==+=+=+MrFrFFrFrF

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3.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS
Scalar Analysis
127
Chapter 3 Systems of Forces and Moments
For moment of a force about a point, the moment and its axis is always perpendicular to the plane.
A scalar or vector analysisis used to find the component of the moment along a specified axis that passes
through the point.
•Scalar Analysis
According to the right-hand rule, M
yis directed along the positive y axis.
In general, for any axis a, the moment is
Force will not contribute a moment if force line of action is parallel or passes through the axis. (cos)
yy
MFdFd == aa
MFd=

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3.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS
Vector Analysis
128
Chapter 3 Systems of Forces and Moments
•Vector Analysis
For magnitude of M
a,
where u
ais unit vector
In determinant form,
where u
arepresents the unit vector along the direction of the a axis
rrepresents the position vector from any pointOon the a axis to
any pointAon the line of action of the force
Frepresents the force vector
The above equation is also called the scalar triple product. cos
a o a o
MM == uM ()
( )( )( )
ax ay az
a a x y z
x y z
axyzzy ayxzzx azxyyx
uuu
M rrr
FFF
urFrFurFrFurFrF
==
=−−−+−
urF
For more information, please see https://en.wikipedia.org/wiki/Triple_product

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3.6 MOMENT OF A COUPLE
Scalar and Vector Formulation
134
Chapter 3 Systems of Forces and Moments
•Couple
A coupleis defined as two parallel forces that have the same magnitude, but
opposite directions, and are separated by a perpendicular distance d.
−two parallel forces
−same magnitude but opposite directions (resultant force is zero)
−separated by a perpendicular distance (tendency to rotate in specified direction)
The moment produced by a couple is called a couple moment. We can determine its value by finding the
sum of the moments of both couple forces about anyarbitrary point.
•Scalar Formulation
Magnitude of a couple moment
The directionare determined by right hand rule
In all cases, Macts perpendicular to the plane containing the forces
•Vector Formulation
It can also be expressed by the vector cross product
If moments are taken about point A, moment of –Fis zeroabout
this point, and therefore, ris crossed with the force Fto which
it is directedMFd= =MrF

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3.6 MOMENT OF A COUPLE
Equivalent Couples
135
Chapter 3 Systems of Forces and Moments
The two couples are equivalentif they produce the samemoment.
Forces of equal couples lie on the same plane or plane parallel to one another.

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3.6 MOMENT OF A COUPLE
Resultant Couple Moment
136
Chapter 3 Systems of Forces and Moments
Since couple moments are free vectors, we can join their tails at any arbitrary point and determine by
vector addition.
(A free vector can act at any pointsince Mdepends onlyupon the position vector rdirected betweenthe
forces and notthe position vectors r
Aand r
B, directed from the arbitrary point to the forces)
The resultant moment of two couples
If more than two couple moments act on the body, this concept can be generalized as12R
=+MMM R
=MrF

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3.6 MOMENT OF A COUPLE
144
Chapter 3 Systems of Forces and Moments
•Moment produced by a force on a point
Movement: Translation and Rotation
Magnitudeof moment: Will be different with different points
Summation of moment: Work if and only if there is the same point
•Moment produced by a force on a axis
Movement: Translation and Rotation
Magnitude of moment: Will be different with different axes
Summation of moment: Work if and only if there is the same point
•Moment produced by a couple (force) on a point
Movement: Only Rotation (it creates a pure moment)
Magnitude of moment: Will be the same with different points (slides 134)
Summation of moment: Work at any arbitrary point (slides 136)

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3.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
145
Chapter 3 Systems of Forces and Moments
An equivalent systemis when the external effectsare the same as those caused by the original force and
couple moment system.
In this context, the external effects of a system are the translating and rotating motionof the body, or refer
to the reactive forcesat the supports if the body is held fixed
Equivalent resultant force acting at point O and a resultant couple moment is expressed as
If force system lies in the x–y plane and couple moments are perpendicular to this plane,

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3.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
146
Chapter 3 Systems of Forces and Moments
An equivalent resultant force acting at point Oand a resultant couple moment are expressed as
If force system lies in the x–y plane and any couple moments are perpendicular to this plane, then the
above equations reduce to the following three scalar equations()
R
ROO
=
=+


FF
MMM ()
()
()
Rxx
Ryy
ROO
FF
FF
MMM
=
=
=+




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3.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
147
Chapter 3 Systems of Forces and Moments

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3.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
Concurrent Force System
154
Chapter 3 Systems of Forces and Moments
A concurrent force systemis one where lines of action of all the forces intersect at a common point O.
The force system produces no moment about this point.()
R
ROO
=
=+


FF
MMM

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3.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
Coplanar Force System
155
Chapter 3 Systems of Forces and Moments
In the case of a coplanar force system, the lines of action of all the forces lie in the same plane, and so
the resultant force of this system also lies in this plane.
The resultant moment and resultant force will be mutually perpendicular.
Furthermore, the resultant moment can be replaced by moving the resultant force a perpendicular or
moment arm distance daway from point Osuch that F
Rproduces the same moment about point O.()
R
RRO
MFd
=
=
FF

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3.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
Parallel Force System
156
Chapter 3 Systems of Forces and Moments
The parallel force systemconsists of forces that are all parallel to the z axis.
Thus, the resultant force F
Rat point Omust also be parallel to this axis.()
R
RRO
MFd
=
=
FF

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3.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
Reduction to a Wrench
157
Chapter 3 Systems of Forces and Moments
In general, a three-dimensional force and couple moment system will have an equivalent resultant force
F
Racting at point Oand a resultant couple moment (M
R)
Othat are not perpendicularto one another.

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3.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING
Loading Along a Single Axis
164
Chapter 3 Systems of Forces and Moments
A body may be subjected to a loading that is distributed over its surface, called distributed loadings.
The pressure exerted at each point on the surface indicates the intensity of the loading.
It is measured using pascals Pa (or N/m
2
) in SI units or lb/ft
2
in the U.S. Customary system.
•Loading Along a Single Axis
The most common type of distributed loading encountered in engineering practice can be represented
along a single axis.
It contains only one variable x, and for this reason, we can also represent it as a coplanar distributed
load.

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3.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING
Magnitude of Resultant Force
165
Chapter 3 Systems of Forces and Moments
•Magnitude of Resultant Force
The magnitude of dFis determined from differential area dAunder the loading curve.
For the entire length L,
Therefore, the magnitude of the resultant force F
Ris equal to the total area Aunder the loading diagram.()
R
LA
FwxdxdAA===

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3.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING
Location of Resultant Force
166
Chapter 3 Systems of Forces and Moments
•Location of Resultant Force
The location can be determined by equating the moments of the force .
Since dFproduces a moment of about O, then for the entire length L,
Solving for the location
This coordinate , locates the geometric center or centroidof the areaunder the distributed loading. In
other words, the resultant force has a line of action which passes through the centroid C(geometric center)
of the area under the loading diagram.()xdFxwxdx= ()
R
LL
xFxdFxwxdx−=−=− ()
()
LA
LA
xwxdxxdA
x
wxdxdA
==

 x ()
ROO
MM=

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EQUILIBRIUM OF A RIGID BODY
Chapter Objectives
173
CHAPTER 4
Students will be able to
1.define and determine moment of a force in 2-D and 3-D cases
2.perform scalar analysis
3.perform vector analysis.
4.define and determine the moment of a couple
5.determine the effect of moving a force
6.find an equivalent force-couple system for a system of forces
and couples
7.determine an equivalent force for a distributed load
3.1Moment of a Force—Scalar Formulation
3.2Cross Product
3.3Moment of a Force—Vector Formulation
3.4Principle of Moments
3.5Moment of a Force about a Specified Axis
3.6Moment of a Couple
3.7Simplification of a Force and Couple System
3.8Further Simplification of a Force and Couple
System
3.9Reduction of a Simple Distributed Loading
Chapter Outline

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4.1 CONDITIONS FOR RIGID -BODY EQUILIBRIUM
174
Chapter 4 Equilibrium of a Rigid Body
The necessary and sufficient conditions for the equilibriumof a body are expressed as
Consider summing moments about some other point, such as point A, we require()
R
ROO
==
==


FF0
MM0 ()0
A R R O
=+=MrFM

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4.2 FREE-BODY DIAGRAMS IN 2D
Support Reactions
175
Chapter 4 Equilibrium of a Rigid Body
The best way to account for these forces is to draw a free-body diagram.
A thorough understanding of how to draw a free-body diagram is of primary
importance for solving problems in mechanics.
•Support Reactions
If a support prevents the translationof a body in a given direction, then a
forceis developed on the body in the opposite direction.
If a support prevents the rotationof a body in a given direction, then a
couple moment is exerted on the body in the opposite direction.

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4.2 FREE-BODY DIAGRAMS IN 2D
Internal Forces
178
Chapter 4 Equilibrium of a Rigid Body
The internal forces that act between adjacent particles in a body always occur in collinear pairs such that
they have the same magnitude and act in opposite directions (Newton’s third law).
Since these forces cancel each other, they will not create an external effecton the body.
For free-body diagram (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.

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4.2 FREE-BODY DIAGRAMS IN 2D
Weight and the Center of Gravity
179
Chapter 4 Equilibrium of a Rigid Body
When a body is within a gravitational field, then each of its particles has a specified weight.
Such a system of forces can be reduced to a single resultant force acting through a specified point.
We refer to this force resultant as the weight Wof the body and to the location of its point of application as
the center of gravity.
When the body is uniformor made from the same material, the center of gravity will be located at the
body’s geometric center or centroid.

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4.2 FREE-BODY DIAGRAMS IN 2D
Idealized Models
180
Chapter 4 Equilibrium of a Rigid Body
For analyzing an actual physical system, first we need to create an idealized model(above right).
Then, we need to draw a free-body diagram (FBD) showing allthe external (active and reactive) forces.
Finally, we need to apply the equations of equilibriumto solve for any unknowns.

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4.3 EQUATIONS OF EQUILIBRIUM IN 2D
189
Chapter 4 Equilibrium of a Rigid Body
When the body is subjected to a system of forces, which all lie in the x–y plane,
−∑F
xand ∑F
yrepresent the algebraic sums of x and y components of all the forces
−∑M
Orepresents the algebraic sum of the couple moments and moments of all the force components
about the z axis
•Alternative Sets of Equilibrium Equations
The first one is
When using these equations it is required that a line passing through points Aand Bis not parallelto
the y axis.
The second one is
Here it is necessary that points A, B, and Cdo not lie on the same line.000
x x O
FFM=== 000
x A B
FMM=== 000
A B C
MMM===

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4.4 TWO-AND THREE-FORCE MEMBERS
Two-Force Members
197
Chapter 4 Equilibrium of a Rigid Body
As the name implies, a two-force memberhas forces applied at only two points on the member.
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.

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4.4 TWO-AND THREE-FORCE MEMBERS
Three-Force Members
198
Chapter 4 Equilibrium of a Rigid Body
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 concurrentor parallelforce system.

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4.5 FREE-BODY DIAGRAMS IN 3D
Support Reactions
201
Chapter 4 Equilibrium of a Rigid Body
Similar to Chapter 4.2, the best way to account for these forces is to draw a free-body diagram.
Again, a thorough understanding of how to draw a free-body diagram is of primary importance for solving
problems in mechanics.
•Support Reactions
A force is developed by a support that restricts the translation of its attached member.
A couple moment is developed when rotation of the attached member is prevented.
•Free-Body Diagrams
The general procedure for establishing the free-body diagram of a rigid body has been outlined in
Chapter 4.2.
Essentially, it requires first “isolating” the body by drawing its outlined shape.
This is followed by a careful labelingof allthe forces and couple moments with reference to an
established x, y, z coordinate system.
As a general rule, it is suggested to show the unknown components of reaction as acting on the free-
body diagram in the positive sense. In this way, if any negative values are obtained, they will
indicate that the components act in the negative coordinate directions.

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4.6 EQUATIONS OF EQUILIBRIUM IN 3D
Free-Body Diagrams
204
Chapter 4 Equilibrium of a Rigid Body
•Free-Body Diagrams
The general procedure for establishing the free-
body diagram of a rigid body has been
outlined in Chapter 4.2.
Essentially, it requires first “isolating” the body
by drawing its outlined shape.
This is followed by a careful labelingof allthe
forces and couple moments with reference to
an established x, y, z coordinate system.
As a general rule, it is suggested to show the
unknown components of reaction as acting on
the free-body diagram in the positive sense. In
this way, if any negative values are obtained,
they will indicate that the components act in
the negative coordinate directions.

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4.6 EQUATIONS OF EQUILIBRIUM IN 3D
Vector Equations of Equilibrium
205
Chapter 4 Equilibrium of a Rigid Body
the conditions for equilibrium of a rigid body subjected to a three-dimensional force system require that
both the resultant forceand resultant couple momentacting on the body be equal to zero.
•Vector Equations of Equilibrium
The two conditions for equilibrium of a rigid body may be expressed mathematically in vector form as
∑Fis the vector sum of all the external forces acting on the body
∑M
Ois the sum of the couple moments and the moments of all the forces about any point O located
either on or off the bodyO
==F0M0

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4.6 EQUATIONS OF EQUILIBRIUM IN 3D
Scalar Equations of Equilibrium
206
Chapter 4 Equilibrium of a Rigid Body
•Scalar Equations of Equilibrium
If all the external forces and couple moments are expressed in Cartesian vector form
Since the i, j, and kcomponents are independent from one another, the above equations are satisfied
These six scalar equilibrium equationsmay be used to solve for at most six unknowns shown on the
free-body diagram.
The moment equations can be determined about any point. Usually, choosing the point where the
maximum number of unknown forces are present simplifies the solution. Any forces occurring at the
point where moments are taken do not appear in the moment equation since they pass through the
point. x y z
O x y z
FFF
MMM
=++=
=++=


Fijk0
Mijk0 000
000
x y z
x y z
FFF
MMM
===
===



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4.7 CONSTRAINTS AND STATICALDETERMINACY
Redundant Constraints
207
Chapter 4 Equilibrium of a Rigid Body
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 (unknown reactions), that is, more supports than are necessary
to hold it in equilibrium, it becomes statically indeterminate.
A problem that is statically indeterminate has more unknowns than equations of equilibrium.

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4.7 CONSTRAINTS AND STATICALDETERMINACY
Improper Constraints
208
Chapter 4 Equilibrium of a Rigid Body
•Improper Constraints
Having the same number of unknown reactive forces as available equations of equilibrium does not
always guarantee that a body will be stable when subjected to a particular loading.
In three dimensions, a body will be improperly constrained if the lines of action of all the reactive
forces intersect a common axis.

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4.7 CONSTRAINTS AND STATICALDETERMINACY
Improper Constraints
209
Chapter 4 Equilibrium of a Rigid Body
•Improper Constraints
Another way in which improper constraining leads to instability occurs when the reactive forcesare all
parallel.
In some cases, a body may have fewer reactive forcesthan equations of equilibrium that must be
satisfied. The body then becomes only partially constrained.
To summarize these points, a body is considered improperly constrainedif all the reactive forces
intersect at a common point or pass through a common axis, or if all the reactive forces are parallel.
In engineering practice, these situations should be avoided at all times since they will cause an
unstable condition.

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ANALYSIS OF A STRUCTURE
Chapter Objectives
219
CHAPTER 5
Students will be able to
1.Define a simple truss
2.Determine the forces in members of a simple truss
3.Identify zero-force members
4.Determine the forces in truss members using the method of
sections
5.Draw the free-body diagram of a frame or machine and its
members
6.Determine the forces acting at the joints and supports of a
frame or machine
5.1Simple Trusses
5.2The Method of Joints
5.3Zero-Force Members
5.4The Method of Sections
5.5Space Trusses
5.6Frames and Machines
Chapter Outline

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5.1 SIMPLE TRUSSES
Overview of Trusses
220
Chapter 5 Analysis of a Structure
A trussis a structure composed of slender members joined together at their end points.
In particular, planartrusses lie in a single plane and are often used to support roofs and bridges.

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5.1 SIMPLE TRUSSES
Assumptions for Design
221
Chapter 5 Analysis of a Structure
•Assumptions for Design
−All loadings are applied at the joints
Frequently, the weight of the members is neglectedbecause the force supported by each member
is usually much larger than its weight.
−The members are joined together by smooth pins.
We can assume these connections act as pins provided the center lines of the joining members
are concurrent.
Because of these two assumptions, each truss member will act as a two-force member, and therefore the
force acting at each end of the member will be directed along the axis of the member.
If the force tends to elongatethe member, it is a tensile force; whereas if it tends to shortenthe member, it
is a compressive force.

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5.1 SIMPLE TRUSSES
Simple Truss
222
Chapter 5 Analysis of a Structure
If three members are pin connected at their ends, they form a triangular trussthat will be rigid.
Attaching two more members and connecting these members to a new joint Dforms a larger truss.
If a truss can be constructed by expanding the basic triangular truss in this way, it is called a simple truss.
For these trusses, the number of members (M) and the number of joints (J) are related by the equation.23MJ=−

Shieh-Kung
Huang
Copyright © 2016 by Pearson Education, Inc. All rights reserved.
5.2 THE METHOD OF JOINTS
223
Chapter 5 Analysis of a Structure
The method of jointsis based on the fact that if the entire truss is in equilibrium, then each of its joints is
also in equilibrium.
Since the members of a plane trussare straight two-force members lying in a single plane, each joint is
subjected to a force system that is coplanarand concurrent.00
xy
FF==

Shieh-Kung
Huang
Copyright © 2016 by Pearson Education, Inc. All rights reserved.
5.3 ZERO-FORCE MEMBERS
232
Chapter 5 Analysis of a Structure
These zero-force (no loading) members are used to increase the stability of the truss during construction
and to provide added support if the loading is changed.
From the observations, we can conclude that if only two non-collinear members form a truss joint and no
external load or support reaction is applied to the joint,
the two members must be zero-force members.
And, if three members form a truss joint for which two of the members are collinear, the third member is a
zero-force member provided no external
force or support reaction has a component
that acts along this member.

Shieh-Kung
Huang
Copyright © 2016 by Pearson Education, Inc. All rights reserved.
5.4 THE METHOD OF SECTIONS
235
Chapter 5 Analysis of a Structure
The method of sectionsis based on the principle that if the truss is in equilibrium then any segment of the
truss is also in equilibrium.
Since only threeindependent equilibrium equations ( ) can be applied to the
free-body diagram of any segment, then we should try to select a section that, in general, passes
through not more than threemembers in which the forces are unknown.0,0,0
x y O
FFM===

Shieh-Kung
Huang
Copyright © 2016 by Pearson Education, Inc. All rights reserved.
5.5 SPACE TRUSSES
Assumptions for Design
243
Chapter 5 Analysis of a Structure
A space trussconsists of members joined together at their ends to form a stable three-dimensional
structure.
The simplest form of a space truss is a tetrahedron,
formed by connecting six members together, and
a simple space trusscan be built from this basic
tetrahedral element.
•Assumptions for Design
−All loadings are applied at the joints
The weight of the members is neglected.
−The members are joined together by
ball-and-socket connections
The connections act as common points
provided the center lines of the joining
members are concurrent.

Shieh-Kung
Huang
Copyright © 2016 by Pearson Education, Inc. All rights reserved.
5.6 FRAMES AND MACHINES
Overview of Frames and Machines
246
Chapter 5 Analysis of a Structure
Frames and machines are two types of structures which are often composed of pin-connected multi-force
members.
Framesare generally stationary and used to support external loads, whereas machinescontain moving
parts and are designed to transmit and alter the effect of forces.

Engineering Mechanics I (Statics) (Part 1/2)

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