Engineering Mechanics I (Statics)

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

Shieh-Kung
Huang
Copyright © 2016 by Pearson Education, Inc. All rights reserved.
ENGINEERING MECHANICS – STATICS
Position in Civil Engineering
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Engineering
Mechanics –
Statics
Engineering
Mechanics –
Dynamics
Structural
Theory
I
Structural
Theory
II
Mechanics of
Materials
Engineering
Materials
Reinforced
Concrete
Advanced
Mechanics of
Material
Structural
Theory
III
Dynamics of
Structures
Design of
Reinforced
Concrete
Design of Steel
Structures
I
Design of Steel
Structures
II
Prestressed
Concrete

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Basic Information about This Course
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• Instructor
Shieh-Kung Huang (黃謝恭)
Assistant Professor, Civil Engineering, National Chung Hsing University
Office: 508 Concrete Technology Building (混凝土科技研究中心)
E-mail: skhuang@nchu.edu.tw
Tel: (04) 2287-2221 ext. 508
• Course Hour
Monday 15:10 – 18:00
• Classroom
103 Civil & Environmental Engineering Building (土木環工大樓)
• Office Hour
Wednesday 2:00 – 4:00, or by appointment
• Teaching Assistant
TBD (To be determined)
• Prerequisites
Engineering Materials

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Course Objectives and Text Book
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• Hibbeler, Russell C. Engineering mechanics: Statics. Pearson Prentice Hall, 2016.
• Beer, Ferdinand, E. Johnston, and David Mazurek. Vector Mechanics for Engineers: Statics. McGraw-
Hill, 2016.

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INTRODUCTION AND GENERAL PRINCIPLES
Chapter Objectives
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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
1.1 Mechanics
1.2 Fundamental Concepts
1.3 The International System of Units
1.4 Numerical Calculations
1.5 General Procedure for Analysis
1.6 Scalars and Vectors
1.7 Vector Operations
1.8 Vector Addition of Forces
1.9 Addition of a System of Coplanar Forces
1.10 Cartesian Vectors
1.11 Position Vectors
1.12 Force Vector Directed Along a Line
1.13 Dot Product
Chapter Outline

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1.1 MECHANICS
What is Mechanics?
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Chapter 1 Introduction and General Principles
Study of what happens to a “thing” (the technical name is “BODY”) when FORCES are 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 Incompressible Compressible

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1.2 FUNDAMENTAL CONCEPTS
Idealizations
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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 Incompressible Compressible

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Newton’s Three Laws of Motion
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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 unbalanced force.
• Second Law
A particle acted upon by an unbalanced force experiences an acceleration a that has the same
direction as the force and a magnitude that is directly proportional to the force.
• Third Law
The mutual forces of action and reaction between two particles are equal, opposite, and collinear.

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Basic Quantities
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Four fundamental physical quantities (or dimensions)
• Length
• Mass
• Time
• Force
Newton’s 2nd 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 2nd 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)
m m
= =
F a d

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Weight
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Weight of a body:
Newton’s Second Law:
By comparison with Newton’s Second Law, we can see that g is the acceleration due to gravity.
m
=
F a
W mg
=

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What is the speed limit on the US freeway compared to Taiwan’s one?
(1 mile = 1.609 km)
80 mile 1.609 km
80 mile h
1h 1mile
128.72 km
128.72 km h
1h
65 mile 1.609 km
65 mile h
1h 1mile
104.585 km
104.585 km h
1h
 
=  
 
= =
 
=  
 
= =

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1.3 THE INTERNATIONAL SYSTEM OF UNITS
Metric Prefixes
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− 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 = cm2
− 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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Metric Prefixes
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Metric Prefixes
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Four fundamental physical
quantities (or dimensions)
• Length
• Mass
• Time
• Force

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Metric Prefixes
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Four fundamental physical
quantities (or dimensions)
• Length
• Mass
• Time
• Force

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1.4 NUMERICAL CALCULATIONS
26
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).
• 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.
• 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

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1.5 GENERAL PROCEDURE FOR ANALYSIS
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1. Interpret:
Read carefully and determine what is given and
what is to be found/delivered.
Ask, if not clear. If necessary, make assumptions
and indicate them.
2. Plan:
Think about major steps (or a road map) that you
will take to solve a given problem.
Think of alternative/creative solutions and choose
the best one.
3. Execute:
Carry out your steps. Use appropriate diagrams
and equations.
Estimate your answers.
Avoid simple calculation mistakes.
Reflect on and then revise your work, if necessary.

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1.6 SCALARS AND VECTORS
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• 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 A and its magnitude (positive quantity) as A
Scalars Vectors
Examples Mass, Volume, Length Force, Position, Velocity
Characteristics It 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
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• Multiplication and Division of a Vector by a Scalar
Product of vector A and scalar a is aA
Magnitude of scaled vector
Law of multiplication applies e.g.
• Vector Addition
Addition of 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
( )
1 0
a a a
= 
A A
= + = +
R A B B A

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1.7 VECTOR OPERATIONS
Vector Subtraction
33
• Vector Subtraction
Special case of addition e.g.
Rules of Vector Addition Applies
( )
' = − = + −
R A B A B

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1.8 VECTOR ADDITION OF FORCES
Finding a Resultant Force
34
• Finding a Resultant Force
Parallelogram law is carried out to find the resultant force
• Resultant
( )
1 2
R = +
F F F

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Finding the Components of a Force
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“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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Addition of Several Forces
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Using the skills several times to get the resultant force or find the components of a force

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1.9 ADDITION OF A SYSTEM OF COPLANAR FORCES
Scalar Notation
42
• Scalar Notation
x and y axes are designated positive and negative
Components of forces expressed as algebraic scalars
and
and
x y
= +
F F F
cos
x
F F 
= sin
y
F F 
=
x
a
F F
c
 
=  
 
y
b
F F
c
 
= −  
 

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Cartesian Vector Notation
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• Cartesian Vector Notation
Cartesian unit vectors i and j are used to designate the x and y directions
Unit vectors i and j have dimensionless magnitude of unity ( = 1 )
Magnitude is always a positive quantity, represented by scalars Fx and Fy
The x and y axis are always perpendicular to each other.
Together, they can be directed at any inclination.
x y
F F
= +
F i j

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• 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 used
1 1 1
2 2 2
3 3 3
x y
x y
x y
F F
F F
F F
= +
= − +
= −
F i j
F i j
F i j
( ) ( )
1 2 3
R Rx Ry
F F
= + + = +
F F F F i j
1 2 3
1 2 3
Rx x x x
Ry y y y
F F F F
F F F F
= − +
= + −

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• In all cases we have
• Magnitude of 𝐹𝑅 can be found by Pythagorean Theorem (畢氏定理)
and
You can always represent a 2-D vector with a magnitude and angle.
Rx x
Ry y
F F
F F
=
=


2 2
R Rx Ry
F F F
= + 1
tan Ry
Rx
F
F
 −
=

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1.10 CARTESIAN VECTORS
Cartesian Vector Representation
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• 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, A can 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
x y
= +
= +
A A A
A A A
x y z
= + +
A A A A
0
A 
A A
=
u A A
A
=
A u

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Cartesian Vector Representation
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• Cartesian Vector Representations
3 components of A act in the positive i, j and k directions
*Note the magnitude and direction of each components are separated, easing vector algebraic operations.
x y z
A A A
= + +
A i j k

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Magnitude of a Cartesian Vector
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• Magnitude of a Cartesian Vector
From the colored triangle,
From the shaded triangle,
Combining the equations gives magnitude of A
2 2
' z
A A A
= = +
A
2 2
' x y
A A A
= +
2 2 2
x y z
A A A A
= = + +
A

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Coordinate Direction Angles (Direction Cosine)
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• Direction of a Cartesian Vector
Orientation of A is defined as the coordinate direction angles a, b and g
measured between the tail of A and the positive x, y and z axes
0° ≤ a, b and g ≤ 180 °
The direction cosines of A is
Angles a, b and g can be determined by the inverse cosines
Given
then,
where
uA can also be expressed as
Since , we have
A as expressed in Cartesian vector form is
2 2 2
x y z
A A A A
= + +
( ) ( ) ( )
y
x z
A
A
A A
A A A A
= = + +
A
u i j k
x y z
A A A
= + +
A i j k
cos cos cos
y
x z
A
A A
A A A
a b g
= = =
cos cos cos
A
a b g
= + +
u i j k
1
A
=
u cos cos cos 1
a b g
+ + =
cos cos cos
A
x y z
A
A A A
A A A
a b g
=
= + +
= + +
A u
i j k
i j k

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Azimuth and Zenith Angles
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A can be specified using two angles, namely, a transverse azimuth angle  and an azimuth zenith 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 is
sin cos sin sin cos
A
x y z
A
A A A
A A A
f  f  f
=
= + +
= + +
A u
i j k
i j k
cos ' sin
z
A A A A
f f
= =
'cos sin cos
'sin sin sin
x
y
A A A
A A A
 f 
 f 
= =
= =

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Addition of Cartesian Vectors
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• Concurrent Force Systems
Force resultant is the vector sum of all the forces in the system
For example,
( ) ( ) ( )
x x y y z z
A B A B A B
= + = + + + + +
R A B i j k
R x y z
F F F
= = + +
   
F F i j k

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• Vector in Cartesian coordinate system
• Requirement for representing of a vector
cos cos cos cos cos cos
sin cos sin sin cos sin cos sin sin cos
A
x y z
A
A
A
A A A
A A A
A A A
a b g a b g
f  f  f f  f  f
=
= + +
= + +  = + +
= + +  = + +
A u
i j k
i j k u i j k
i j k u i j k
and
, , and
, , , and
, , and
A
x y z
A
A A A
A
A
a b g
f 



 



u
A

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1.11 POSITION VECTORS
x, y, z Coordinates
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• x,y,z Coordinates
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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Position Vector
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Position vector r is defined as a fixed vector which locates a point in space relative to another point.
E.g.
Vector addition gives
Solving
x y z
= + +
r i j k
A B
+ =
r r r
( ) ( ) ( )
B A B A B A B A
x x y y z z
= − = − + − + −
r r r i j k

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1.12 FORCE VECTOR DIRECTED ALONG A LINE
Position Vector
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In 3D problems, direction of F is specified by 2 points, through which its line of action lies
F can be formulated as a Cartesian vector
Note that F has units of forces (N)
unlike r, with units of length (m)
Force F acting along the chain can be presented as a Cartesian vector by
• Establish x, y, z axes
• Form a position vector r along length of chain
Unit vector, that defines the direction of both the chain and the force
We get
F F
r
 
= =  
 
r
F u
r
=
u r
F
=
F u

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1.13 DOT PRODUCT
Laws of Operation
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• 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:
cos
AB 
 =
A B
 = 
A B B A
( ) ( ) ( ) ( )
a a a a
 =  =  = 
A B A B A B A B
( ) ( ) ( )
 + =  + 
A B C A B A C

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1.13 DOT PRODUCT
Cartesian Vector Formulation
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• 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.
( )( )
( )( )
1 1 cos0 1
1 1 cos 0
2

 = =
 = =
i i
i j
1 1 1
0 0 0
 =  =  =
 =  =  =
i i j j k k
i j i k j k
( )( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
x y z x y z
x x x y x z
y x y y y z
z x z y z z
x x y y z z
A A A B B B
A B A B A B
A B A B A B
A B A B A B
A B A B A B
 = + + + +
=  +  + 
+  +  + 
+  +  + 
= + +
A B i j k i j k
i i i j i k
j i j j j k
k i k j k k
( )
1
cos
AB
 − 
 
=  
 
A B
cos
a a
A 
=
A u

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1.13 DOT PRODUCT
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1.13 DOT PRODUCT
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1.13 DOT PRODUCT
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1.13 DOT PRODUCT
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1.13 DOT PRODUCT
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EQUILIBRIUM OF A PARTICLE
Chapter Objectives
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CHAPTER 2
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
2.1 Condition for the Equilibrium of a Particle
2.2 The Free-Body Diagram
2.3 Coplanar Force systems
2.4 Three-Dimensional Force systems
Chapter Outline

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2.1 CONDITION FOR THE EQUILIBRIUM OF A PARTICLE
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Chapter 2 Equilibrium of a Particle
• A particle is said to be in equilibrium if
− 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 SF is the vector sum of all the forces acting 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.
=
F 0
m
=
F a
m =  =
a 0 a 0

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2.2 THE FREE-BODY DIAGRAM
Springs
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• Free-Body Diagram (FBD)
Best representation of all the unknown forces (SF) which act on the 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 constant or stiffness k: defines the elasticity of the spring
The magnitude of force, when spring is elongated or compressed, is
F ks
=

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Cables and Pulleys
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• Cables and Pulley
Cables (or cords) are assumed negligible weight and cannot stretch.
A cable can support only a tension or “pulling” force, and therefore, the force always acts in the
direction of the cable.
The tension force must have a constant magnitude for equilibrium, meaning that, for any angle , the
cable is subjected to a constant tension T.

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Smooth Contact
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• 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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Procedure for Drawing a Free-Body Diagram
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2.3 COPLANAR FORCE SYSTEMS
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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 i and j components for equilibrium
Scalar equations of equilibrium require that the
algebraic sum of the x and y components to equal
to zero.
0 0
x y
x y
F F
F F
=
+ =
= =

 
 
F 0
i j 0

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2.4 THREE-DIMENSIONAL FORCE SYSTEMS
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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.
=
F 0
x y z
F F F
+ + =
  
i j k 0
0 0 0
x y z
F F F
= = =
  

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SYSTEMS OF FORCES AND MOMENTS
Chapter Objectives
108
CHAPTER 3
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
3.1 Moment of a Force—Scalar Formulation
3.2 Cross Product
3.3 Moment of a Force—Vector Formulation
3.4 Principle of Moments
3.5 Moment of a Force about a Specified Axis
3.6 Moment of a Couple
3.7 Simplification of a Force and Couple System
3.8 Further Simplification of a Force and Couple
System
3.9 Reduction of a Simple Distributed Loading
Chapter Outline

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3.1 MOMENT OF A FORCE—SCALAR FORMULATION
Magnitude and Direction
109
Chapter 3 Systems of Forces and Moments
The moment of a force about a point or axis provides a measure of the tendency for rotation about the
point or axis (sometimes called torque)
Torque is tendency of rotation caused by Fx or simple moment (MO)z
• Magnitude
The magnitude of the moment, MO, is
where d is the moment arm or perpendicular distance from
the axis at point O to the line of action of the force.
• Direction
The direction is determined using “right hand rule,” called
moment axis.
O
M Fd
=

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Resultant Moment
110
For two-dimensional problems, where all the forces lie within the x–y plane, the resultant moment (MO)z
about point O (the z axis) can be determined by finding the algebraic sum of the moments caused by all
the forces in the system.
As a convention, we will generally consider positive moments as counterclockwise since they are directed
along the positive z axis (out of the page). Clockwise moments will be negative.
( )
O z
M Fd
= 
( ) 1 1 2 2 3 3
O z
M Fd F d F d
= − +

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3.2 CROSS PRODUCT
114
Cross product of two vectors A and B yields C, which is written as
• Magnitude
The magnitude of C is defined as the product of the magnitudes of A and B and the sine of the angle .
For angle θ, 0° ≤  ≤ 180°
• Direction
Vector C has a direction that is perpendicular to the plane containing A and B such that C is specified
by the right hand rule.
Expressing vector C when magnitude and direction are known
= 
C A B
sin
C AB 
=
( sin ) C
AB 
=  =
C A B u

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3.2 CROSS PRODUCT
Laws of Operation
115
• Laws of Operation
− Commutative law is not valid
Rather,
Cross product of two vectors A and B yields a vector opposite in direction to C
− Multiplication by a Scalar
− Distributive Law
Proper order of the cross product must be maintained
since they are not commutative
( ) ( ) ( ) ( )
a a a a
 =  =  = 
A B A B A B A B
( )
 + =  + 
A B D A B A D
  
A B B A
 = −
B A C
 = − 
A B B A

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3.2 CROSS PRODUCT
116
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 A and B (in Cartesian vector form) is
sin
C AB 
=
x y z
x y z
A B A A A
B B B
 =
i j k
( ) ( ) ( )
x y z
x y z
y z x y
x z
y z x y
x z
y z z y x z z x x y y x
A B A A A
B B B
A A A A
A A
B B B B
B B
A B A B A B A B A B A B
 =
= − +
= − − − + −
i j k
i j k
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3.3 MOMENT OF A FORCE—VECTOR FORMULATION
117
The moment of force F about point O, or actually about the moment axis passing through O and
perpendicular to the plane containing O and F, can be expressed using vector cross product.
• Magnitude
For magnitude of cross product,
Treat r as a sliding vector. Since the moment arm is .
• Direction
The direction and sense of MO are determined by right-hand rule
*Note:
− “curl” of the fingers indicates the sense of rotation
− Maintain proper order of r and F since cross product is not commutative
O
= 
M r F
sin ( sin )
O
M rF F r Fd
 
= = =
sin
d r 
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Principle of Transmissibility
118
For force F applied at any point A on the line of action of the force F, moment created about O is .
F has the properties of a sliding vector at any point on the line, thus
O A
= 
M r F
1 2 3
O
=  =  = 
M r F r F r F

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For the position vector and force expressed in Cartesian form, the moment is
where r is the position vector from the point O to any point on the line of action of the force F.
With the determinant is expended as
O x y z
x y z
r r r
F F F
=  =
i j k
M r F
( ) ( ) ( )
O y z z y x z z x x y y x
r F r F r F r F r F r F
= − − − + −
M i j k

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Resultant Moment of a System of Forces
120
If a body is acted upon by a system of forces, the resultant moment of forces about point O can be
determined by vector addition of the moment of each force.
( )
O
= 

M r F

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x
or

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3.4 PRINCIPLE OF MOMENTS
Varignon’s Theorem
125
A concept often used in mechanics is the principle of moments, which is sometimes referred to as
Varignon’s theorem since 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 2
O
=  =  + =  + 
M r F r F F r F r F
( )
O x y x y x y
F y F x
=  =  + =  +  = +
M r F r F F r F r F

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3.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS
Scalar Analysis
130
For moment of a force about a point, the moment and its axis is always perpendicular to the plane.
A scalar or vector analysis is 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, My is 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 )
y y
M Fd F d 
= =
a a
M Fd
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Vector Analysis
131
• Vector Analysis
For magnitude of Ma,
where ua is unit vector
In determinant form,
where ua represents the unit vector along the direction of the a axis
r represents the position vector from any point O on the a axis to
any point A on the line of action of the force
F represents the force vector
The above equation is also called the scalar triple product.
cos
a o a o
M M 
= = 
u M
( )
( ) ( ) ( )
ax ay az
a a x y z
x y z
ax y z z y ay x z z x az x y y x
u u u
M r r r
F F F
u r F r F u r F r F u r F r F
=   =
= − − − + −
u r F
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
137
• Couple
A couple is 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 any arbitrary point.
• Scalar Formulation
Magnitude of a couple moment
The direction are determined by right hand rule
In all cases, M acts 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 –F is zero about
this point, and therefore, r is crossed with the force F to which
it is directed
M Fd
=
= 
M r F

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Equivalent Couples
138
The two couples are equivalent if they produce the same moment.
Forces of equal couples lie on the same plane or plane parallel to one another.

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Resultant Couple Moment
139
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 point since M depends only upon the position vector r directed between the
forces and not the position vectors rA and rB, 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 as
1 2
R
= +
M M M
R
= 

M r F

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• Moment produced by a force on a point (Chapter 3.3)
Movement: Translation and Rotation
Magnitude of 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 (Chapter 3.5)
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 (Chapter 3.6)
Movement: Only Rotation (it creates a pure moment)
Magnitude of moment: Will be the same with different points (slides 137)
Summation of moment: Work at any arbitrary point (slides 139)

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3.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
148
An equivalent system is when the external effects are 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 motion of the body, or refer
to the reactive forces at 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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An equivalent resultant force acting at point O and 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
R O
O
=
= +

 
F F
M M M
( )
( )
( )
R x
x
R y
y
R O
O
F F
F F
M M M
=
=
= +


 

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3.8FURTHERSIMPLIFICATIONOFAFORCEANDCOUPLESYSTEM
Concurrent Force System
157
A concurrent force system is 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
R O
O
=
= +

 
F F
M M M

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Coplanar Force System
158
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 d away from point O such that FR produces the same moment about point O.
( )
R
R R
O
M F d
=
=

F F

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Parallel Force System
159
The parallel force system consists of forces that are all parallel to the z axis.
Thus, the resultant force FR at point O must also be parallel to this axis.
( )
R
R R
O
M F d
=
=

F F

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Reduction to a Wrench
160
In general, a three-dimensional force and couple moment system will have an equivalent resultant force
FR acting at point O and a resultant couple moment (MR)O that are not perpendicular to one another.

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kN

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3.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING
Loading Along a Single Axis
167
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/m2) in SI units or lb/ft2 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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Magnitude of Resultant Force
168
• Magnitude of Resultant Force
The magnitude of dF is determined from differential area dA under the loading curve.
For the entire length L,
Therefore, the magnitude of the resultant force FR is equal to the total area A under the loading diagram.
( )
R L A
F w x dx dA A
= = =
 

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Location of Resultant Force
169
• Location of Resultant Force
The location can be determined by equating the moments of the force .
Since dF produces a moment of about O, then for the entire length L,
Solving for the location
This coordinate , locates the geometric center or centroid of the area under 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.
( )
xdF xw x dx
=
( )
R L L
xF xdF xw x dx
− = − = −
 
( )
( )
L A
L A
xw x dx xdA
x
w x dx dA
= =
 
 
x
( )
R O
O
M M
= 

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EQUILIBRIUM OF A RIGID BODY
Chapter Objectives
176
CHAPTER 4
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
3.1 Moment of a Force—Scalar Formulation
3.2 Cross Product
3.3 Moment of a Force—Vector Formulation
3.4 Principle of Moments
3.5 Moment of a Force about a Specified Axis
3.6 Moment of a Couple
3.7 Simplification of a Force and Couple System
3.8 Further Simplification of a Force and Couple
System
3.9 Reduction of a Simple Distributed Loading
Chapter Outline

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4.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM
177
Chapter 4 Equilibrium of a Rigid Body
The necessary and sufficient conditions for the equilibrium of a body are expressed as
Consider summing moments about some other point, such as point A, we require
( )
R
R O
O
= =
= =


F F 0
M M 0
( ) 0
A R R O
=  + =
M r F M

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4.2 FREE-BODY DIAGRAMS IN 2D
Support Reactions
178
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 translation of a body in a given direction, then a
force is developed on the body in the opposite direction.
If a support prevents the rotation of a body in a given direction, then a
couple moment is exerted on the body in the opposite direction.

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Support Reactions
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Support Reactions
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Internal Forces
181
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 effect on 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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Weight and the Center of Gravity
182
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 W of the body and to the location of its point of application as
the center of gravity.
When the body is uniform or made from the same material, the center of gravity will be located at the
body’s geometric center or centroid.

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Idealized Models
183
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 all the external (active and reactive) forces.
Finally, we need to apply the equations of equilibrium to solve for any unknowns.

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Idealized Models
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4.3 EQUATIONS OF EQUILIBRIUM IN 2D
192
When the body is subjected to a system of forces, which all lie in the x–y plane,
− ∑Fx and ∑Fy represent the algebraic sums of x and y components of all the forces
− ∑MO represents 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 A and B is not parallel to
the y axis.
The second one is
Here it is necessary that points A, B, and C do not lie on the same line.
0 0 0
x y O
F F M
= = =
  
0 0 0
x A B
F M M
= = =
  
0 0 0
A B C
M M M
= = =
  

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4.4 TWO- AND THREE-FORCE MEMBERS
Two-Force Members
200
As the name implies, a two-force member has 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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Three-Force Members
201
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.

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Support Reactions
204
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 2.1 and Chapter 4.2.
Essentially, it requires first “isolating” the body by drawing its outlined shape.
This is followed by a careful labeling of all the 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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Support Reactions
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Free-Body Diagrams
207
• 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 labeling of all the
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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Vector Equations of Equilibrium
208
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.
• Vector Equations of Equilibrium
The two conditions for equilibrium of a rigid body may be expressed mathematically in vector form as
∑F is the vector sum of all the external forces acting on the body
∑MO 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
O
= =
 
F 0 M 0

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Scalar Equations of Equilibrium
209
• Scalar Equations of Equilibrium
If all the external forces and couple moments are expressed in Cartesian vector form
Since the i, j, and k components are independent from one another, the above equations are satisfied
These six scalar equilibrium equations may be used to solve for at most six unknowns shown on the
free-body diagram.
The moment equations can be determined about any point or line. Usually, choosing the point where the
maximum number of unknown forces are present simplifies the solution. Any forces occurring at the
point or the line where moments are taken do not appear in the moment equation since they pass
through it.
x y z
O x y z
F F F
M M M
= + + =
= + + =
   
   
F i j k 0
M i j k 0
0 0 0
0 0 0
x y z
x y z
F F F
M M M
= = =
= = =
  
  

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4.7 CONSTRAINTS AND STATICAL DETERMINACY
Redundant Constraints
210
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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Improper Constraints
211
• 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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Improper Constraints
212
• Improper Constraints
Another way in which improper constraining leads to instability occurs when the reactive forces are all
parallel.
In some cases, a body may have fewer reactive forces than equations of equilibrium that must be
satisfied. The body then becomes only partially constrained.
To summarize these points, a body is considered improperly constrained if 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
222
CHAPTER 5
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
5.1 Simple Trusses
5.2 The Method of Joints
5.3 Zero-Force Members
5.4 The Method of Sections
5.5 Space Trusses
5.6 Frames and Machines
Chapter Outline

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5.1 SIMPLE TRUSSES
Overview of Trusses
223
Chapter 5 Analysis of a Structure
A truss is a structure composed of slender members joined together at their end points.
In particular, planar trusses 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
224
• Assumptions for Design
− All loadings are applied at the joints
Frequently, the weight of the members is neglected because 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 elongate the member, it is a tensile force; whereas if it tends to shorten the member, it
is a compressive force.

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5.1 SIMPLE TRUSSES
Simple Truss
225
If three members are pin connected at their ends, they form a triangular truss that will be rigid.
Attaching two more members and connecting these members to a new joint D forms 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.
2 3
M J
= −

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5.2 THE METHOD OF JOINTS
226
The method of joints is 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 truss are straight two-force members lying in a single plane, each joint is
subjected to a force system that is coplanar and concurrent.
0 0
x y
F F
= =
 

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5.3 ZERO-FORCE MEMBERS
235
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.

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5.4 THE METHOD OF SECTIONS
238
The method of sections is based on the principle that if the truss is in equilibrium then any segment of the
truss is also in equilibrium.
Since only three independent 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 three members in which the forces are unknown.
0, 0, 0
x y O
F F M
= = =
  

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5.5 SPACE TRUSSES
Assumptions for Design
246
A space truss consists 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 truss can 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.

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5.5 SPACE TRUSSES
247

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5.5 SPACE TRUSSES
248

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5.6 FRAMES AND MACHINES
Overview of Frames and Machines
249
Frames and machines are two types of structures which are often composed of pin-connected multi-force
members.
Frames are generally stationary and used to support external loads, whereas machines contain moving
parts and are designed to transmit and alter the effect of forces.

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5.6 FRAMES AND MACHINES
Free-Body Diagrams
250

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INTERNAL FORCES AND MOMENTS
Chapter Objectives
251
CHAPTER 6
Use the method of sections for determining internal forces
in 2-D load cases
6.1 Internal Loadings Developed in Structural
Members
6.2 Shear and Moment equations and Diagrams
6.3 Relations between Distributed Load, Shear,
and Moment
6.4 Cables
Chapter Outline

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6.1INTERNALLOADINGSDEVELOPEDINSTRUCTURALMEMBERS
Overview of Internal Loadings
252
Chapter 6 Internal Forces and Moments
Internal loadings can be determined by using the method of sections.
− The force component NB that acts perpendicular to the cross section is termed the normal force.
− The force component VB that is tangent to the cross section is called the shear force.
− The couple moment MB is referred to as the bending moment.

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Sign Convention
253

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6.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS
262
Beams are structural members designed to support loadings applied perpendicular to their axes.
A simply supported beam is pinned at one end and roller supported at the other, whereas a cantilevered
beam is fixed at one end and free at the other.
If the resulting functions of x are plotted, the graphs are termed the shear diagram and bending-moment
diagram, respectively.

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6.3RELATIONSBETWEENDISTRIBUTEDLOAD,SHEAR,ANDMOMENT
Distributed Load
267
If a beam is subjected to several concentrated forces, couple moments, and distributed loads, the method
of constructing the shear and bending-moment may become quite tedious.
Here, a simpler method for constructing these diagrams is discussed—a method based on differential
relations that exist between the load, shear, and bending moment.
• Distributed Load
The distributed load will be considered positive when the loading acts upward as shown.
The internal shear force and bending moment shown on the free-body diagram are assumed to act in
the positive sense according to the established sign convention in Chapter 6.1.

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Relation between the Distributed Load and Shear
268
The relationship between the distributed load and shear is differentiation and integration.

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Relation between the Shear and Moment
269
The relationship between the shear and moment is also differentiation and integration.

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6.4 CABLES
Cable Subjected to Concentrated Loads
276
Flexible cables and chains combine strength with lightness and often are used in structures for support
and to transmit loads (tension) from one member to another.
• Assumption for Cables
The weight of the cables is neglected, and therefore, it subjects to a constant tensile force.
Generally, we will make the assumption that the cable is perfectly flexible (no shear and moment) and
inextensible (constant length).
• Cable Subjected to Concentrated Loads

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6.4 CABLES
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6.4 CABLES
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6.4 CABLES
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6.4 CABLES
Cable Subjected to a Distributed Load
280
Consider weightless cable subjected to a loading function measured in the x direction.
( )
w w x
=
2
0
2
( )
H
d y w x
dx F
= 

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6.4 CABLES
Cable Subjected to Its own Weight
281
When the weight of a cable becomes important in the force analysis, the loading function along the cable
will be a function of the arc length s rather than the projected length x.
2
1
dy ds
dx dx
 
= − 
 
 

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FRICTION
Chapter Objectives
282
CHAPTER 7
1. Understand the characteristics of dry friction
2. Draw a FBD including friction
3. Solve problems involving friction
4. Determine the forces on a wedge
5. Determine tension in a belt
7.1 Characteristics of Dry Friction
7.2 Problems Involving Dry Friction
7.3 Wedges
7.4 Frictional Forces on Screws
7.5 Frictional Forces on Flat Belts
7.6 Frictional Forces on Collar Bearings,
7.7 Frictional Forces on Journal Bearings
7.8 Rolling Resistance
Chapter Outline

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7.1 CHARACTERISTICS OF DRY FRICTION
Theory of Dry Friction
283
Chapter 7 Friction
Friction is a force that resists the movement of two contacting surfaces that slide relative to one another.
Two types of friction, fluid and dry friction, are common; however, in this chapter, we will study the
effects of dry friction, which is sometimes called Coulomb friction since its characteristics were studied
extensively by the French physicist Charles-Augustin de Coulomb in 1781.
This force always acts tangent to the surface at the points of contact and is directed so as to oppose the
possible or existing motion between the surfaces.
• Theory of Dry Friction
As shown on the free-body diagram of the block, the floor exerts an uneven distribution of both normal
force DNn and frictional force DFn along the contacting surface. As the result,
The Angle fs that Rs makes with N is called the angle of static friction.
s
x h
= = =
W N P F W P
1 1 1
tan tan tan
s s
s s
F N
N N

f 
− − −
   
= = =
   
   

Engineering Mechanics I (Statics)

Engineering Mechanics I (Statics)

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