WACHEMO UNIVERSITY
DURAME CAMPUS
Department of Construction Technology and Management
Course Title: Engineering Mechanics I (Statics)
Course Code: CEng2051
Pre-requisites: General Physics (Phys 1011)
Course Objective and Learning Outcomes
Up on successful completion of the course, students will be able to:
 Introduce the dot product in order to determine the angle between two vectors
 Understand and be able to apply Newton’s laws of motion
 Distinguish between concurrent, coplanar and space force systems
 Compute the resultant of coplanar and space force systems
 Present methods for determining the resultants of non-concurrent force systems.
 Indicate how to reduce a simple distributed loading to a resultant force having
a specified location.
 Draw free body diagrams, analyze reactions and pin forces induced in coplanar
and space systems using equilibrium equations and free body diagrams
 Determine the centroid and center of mass of plane areas & volumes
 Determine friction forces and their influence up on equilibrium of systems
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 Apply sound analytical techniques and logical procedure in the solution of
engineering problems
Competences to be acquired/course level competences
Student gets basic knowledge to
 Ability to define and apply the concepts of equilibrium;
 Demonstrate familiarity with structural analysis of trusses, frames and beams and
application of mechanics to engineering problems.
 Show how to determine the forces in the members of a truss using the method of joints and
the method of sections.
 Analyze the forces acting on the members of frames and machines composed of pinconnected members.
Course Description
This course presents the fundamental physical concepts, laws and Statics of particles: Resultants
of coplanar and none-coplanar force systems, Equitation of equilibrium for coplanar and nonecoplanar force systems. Statics of rigid bodies: Equilibrium of simple structures: trusses beams,
frames and machines. Analysis of structures (truss, Frames and machines). Centroid & center of
gravity, moment of inertial and Static friction.
Course outline
Chapter 1: Scalars and Vectors
Chapter 2: Force Systems
1.1 Introduction
2.1 Introduction
1.2 Scalars and Vectors
I. Two Dimensional Force Systems
1.3 Operation with Vectors
2.2 Rectangular Resolution of Forces
 Vector Addition or Composition
2.3 Moment and Couple
 Vector Multiplication: Dot & Cross
2.4 Resultants of general coplanar force systems
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II. Three Dimensional Force Systems
2.5 Rectangular Components
2.6 Moment and Couple
2.7 Resultants
Chapter 3: Equilibrium
Chapter 4: Analysis of simple Structures
3.1 Introduction
4.1 Introduction
I. Equilibrium in Two Dimensions
4.2 Plane Trusses
3.2 System Isolation
4.2.1 Method of Joints
3.3 Equilibrium Conditions
4.2.2 Method of Sections
II. Equilibrium in Three Dimensions
4.3 Frames and Simple Machines
3.4 System Isolation
3.5 Equilibrium Conditions
Chapter 5: Centroids
5.1 Introduction,
5.2 Center of gravity
5.3 Centroids of lines, Areas & Volumes
5.4 Centroids of composite bodies
5.5 Determination of centroid by integrations
Chapter 6: Area Moments of Inertia
Chapter 7: Friction
6.1 Introduction to area moments of inertia
7.1 Introduction
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6.2 Moment of inertia of plane areas and curves
7.2 Types of Friction
6.3 Moments of inertia of Composite areas
7.3 Characteristics of dry friction
6.4 Products of Inertia and Rotation of Axes
7.4 Application of Friction in Machines
Literatures
1. Meriam, J.L. and Kraige, L.G., Engineering mechanics, 7th edition
2. Meriam, J.L. and Kraige, L.G., Engineering mechanics, 6th edition
3. Engineering Mechanics: Statics & Dynamics by Anthony M. Bedford, Wallace Fowler, Prentice
Hall; 5th edition (July 2007)
4. Engineering Mechanics: Statics by Russell C. Hibbeler, Prentice Hall; 12th edition
(January 7, 2009)
5. Schaum's Outline of Engineering Mechanics by E. W. Nelson, Charles L. Best, William G.
McLean, McGraw-Hill; 5th edition (May 1997)
6. Engineering Mechanics- Statics and Dynamics by Anthony M Bedford, Wallace Fowler,
Prentice Hall; 4th edition (August 2004)
7. Andrew Pytel, Jaan Kiusalaas, Engineering Mechanics: Statics (SI Edition), 3rd Edition,
CEngage Learning, 2010
8. Francesco Costanzo and Michael E. Plesha, Engineering mechanics: Statics; 1st Edition,
McGraw-Hill Higher Education, 2010
9. J. L. Meriam and L.G. Kraige., Engineering mechanics: Statics (SI version), 8th
Edition, John Wiley & Sons Ltd., 2016
10. R. C. Hibbeler, Engineering Mechanics: Statics, 14th Edition, Prentice Hall; 2016
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Chapter 1: Scalars and Vectors
1.1 General Introduction
1.1.1 Introduction to Mechanics
Mechanics is the branch of physical science which describes and predicts the conditions of rest or
motion of bodies under the action of forces.
 It is the foundation of most engineering sciences and is an indispensable prerequisite to their
study. The purpose of mechanics is to explain and predict physical phenomena and thus to lay
the foundations for engineering applications.
Mechanics is divided into three parts:
i.
Mechanics of rigid bodies
ii.
Mechanics of deformable bodies and
iii.
Mechanics of fluids
In this course, we will study the mechanics of rigid bodies. Here bodies are assumed to be perfectly
rigid. The mechanics of rigid bodies is subdivided into:
a) Statics: - dealing with bodies at rest, concerns with equilibrium of rigid bodies under the action
of forces.
b) Dynamics: - dealing with bodies in motion.
*The basic principles of mechanics are relatively few in number, but they have exceedingly wide
application, and the methods employed in mechanics carry over into many fields of engineering
endeavor.
1.1.2 Basic Concepts in Mechanics
Space: - Geometric region occupied by bodies whose positions are described by linear and angular
measurements relative to a co-ordinate system.
Time: - the measure of succession of events, and it is basic in dynamics.
Mass: - it is the measure of inertia of a body, resistance to a change in velocity. It is the quantity of
matter in a body.
Force: - the action of one body on another body, it tends to move a body in the direction of its action.
Particle: - it is a body of negligible dimension; it can be analyzed as a point mass.
Rigid Body: - a body that is not deformed when acted on by external forces.
Free Body Diagram (FBD):-a diagrammatic representation of a portion of body/bodies completely
isolated from other bodies showing the forces applied by all other bodies that are assumed to be
removed on the one being considered.
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1.2. Scalars and Vectors
Mechanics deals with two kinds of quantities, scalars and vectors.
Scalars:-those with which the magnitude alone is associated.
E.g. Time, volume, mass, density, speed and energy
Vectors: - possess direction as well as magnitude and must obey the parallelogram law of addition.
E.g. Velocity, displacement, acceleration, force, moment and momentum
Physical quantities represented by vectors fall into one of the three classifications below;
i.
Free vector: - one whose action is not confined to or associated with a unique line in space.
E.g. If a body moves without rotation, we may represent the displacement of such a body by a
free vector.
ii.
Sliding vector: - one for which a unique line in space must be maintained along which the
quantity acts.
E.g. When we deal with the external action of a force on a rigid body, the force may be applied
at any point along its line of action without changing its effect on the body as a whole (the socalled principle of transmissibility) and hence may be considered a sliding vector.
iii.
Fixed vector: - one for which a unique point of application is specified, and therefore the vector
occupies a particular position in space.
E.g. The action of a force on a deformable or non-rigid body must be specified by a fixed vector
at the point of application of the force. In this problem the forces and deformations internal to the
body will be dependent on the point of application of the force, as well as its magnitude and line
of action.
V
-A vector quantity V
is represented by a line segment, having the
direction of the vector and having an arrow  head to indicate the sense.
-The length of the directed line segment represents to some convenient
scale the magnitude | V | of the vector.
-V
(V)
+ - (V) = 0 (equal and opposite vectors).
The direction of the vector V may be measured by an angle  from some known reference direction as
indicated above.
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1.2.1Vector Addition and Subtraction
Vector Addition
V2
V2
V1
V1
V = V1 + V2
V=V1+V2
V2
V1
Parallelogram Law
Triangle Law
Vector Subtraction
V1
V1
V
-V2
V=V1 - V2
V
-V2
V = V1 – V2
Parallelogram Law
Triangle Law
Vector Multiplication: Dot & Cross
Scalar/dot product of vectors
Definition (Scalar product of vectors)
Let
and
be any two vectors in
then scalar product of
A and B is given by
Note: The scalar product is also called a dot product or inner product.
Properties of dot product
The scalar product satisfies many of the laws that hold for real numbers.
Theorem 1.2: Let A, B, and C be vectors in
, then
1. A.B  B. A
2. t ( A.B)  (tA).B  A.(tB)
3. ( A  B).C  A.C  B.C
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Cross product of vectors
The second type of product of two vectors is the cross product. Unlike the dot product, the cross product
of two vectors is a vector.
Definition (The cross product)
The cross product
of two vectors
and
is given by
or, equivalently ,
Note:
 The cross product of two vectors is a vector and for this sometimes it is called vector product.
 The notion of vector product is defined only in the 3-spaces.
 This vector is perpendicular to both A and B and the plane containing A and B.
The following Properties are the basic properties of cross product
Theorem: if A, B, and C be vectors in the same space, then
Lagrange’s identity.
is orthogonal
Theorem: Let A and B be two non-zero vectors and
be the angle between them, then
From the six properties of cross product
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2
2
2
2
AxB  A B   A.B 


A
2
B
2
 A
2
CoTM Department
B

 A
2
B 1  cos2 
 A
2
B sin 2 
2
2
cos 2  (  is the angle between A and B)

2
AxB  A B sin  (For 0    , sin is non- negative)
Components of a Vector
Any two or more vectors whose sum equals a certain vector V are said to be the components of that
vector.
y
y’
E.g. i/
V
Vy’
ii/
V
Vy

x’
x
Vx’
Vx
Vx’ - the component of V in the direction of x’
Vy’- the component of V in the direction of y’
Vx and Vy are x and y components of V respectively.
Vx and Vy are called rectangular components (vector components that are mutually
perpendicular).
 is the direction of the vector V with respect to the x-axis.
 = tan-¹ | Vy | = tan-¹ Vy / Vx
Vx = Vcos
&
Vy = Vsin
| Vx |
Unit Vector
A unit vector in the direction of a vector V is defined as a vector whose magnitude is one and whose
direction coincides with that of V.
E.g. unit vectors i, j, and k in the x, y, and z directions respectively, with magnitudes of unity.
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V
y
V = Vxi + Vyj + Vzk
y
Vzk
z
Vx = Vcosx, Vy = Vcosy, Vz = Vcosz
Vyj
x
cosx, cosy and cosz are called the direction cosines.
Vxi
x
-The magnitude of the vector V is;
V=
Vx2 + Vy2 + Vz2
(cosx)2 + (cosy)2 + (cosz)2 = 1
-The unit vector in the direction of the vector V is;
n
= V/|V
If n is a unit vector in the  direction, the moment M of F about any axis  through O is expressed by;
M = Mo. n = (r X F. n) --- which is the scalar magnitude.
Or
M = (Mo. n) n = (r X F. n) n ----vector expression for the moment of F about an axis  through
O.
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CHAPTER 2. FORCE SYSTEMS
2.1. Introduction
The foundation for a basic understanding of not only statics but also of the entire subject of mechanics is
laid in this chapter.
A force has been defined as the action of one body on another. It is a vector quantity and its SI unit is
Newton (N).
External
Applied forces
Internal
Reaction forces
Action of a force
Stresses and strains
The relation between internal forces and internal strains involves the material properties of the body and is
studied in strength of materials, elasticity and plasticity.
Since the course deals essentially with the mechanics of rigid bodies, we will treat almost all forces as
sliding vectors for the rigid body on which they act.
When only the resultant external effects of a force are to be investigated, the force may be treated as a
sliding vector, and it is necessary and sufficient to specify the magnitude, direction and line of action of the
force.
Principle of transmissibility
It states that a force may be applied at any point on its given line of action without altering the resultant
effects of the force external to the rigid body on which it acts.
Contact forces
Forces
Distributed
Forces
Body forces
Concentrated
Contact forces:-generated through direct physical contact between two bodies.
Body forces:- those applied by remote action, such as gravitational and magnetic forces.
A force may be measured either by comparison with other known forces, using a mechanical balance, or by
the calibrated movement of an elastic element.
The action of a force is always accompanied by an equal and opposite reaction.
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F2
F2
R
- principle of transmissibility
R = F1 + F2
F1
-The
resultant
R
replaces
F1
and
F2
without
altering
the
external
effects
on
the body upon which they act.
In addition to the need for combining forces to obtain their resultant, we often have occasion to replace a
force by its vector components which act in specified directions.
F2
R is resolved into F1 and F2.
R
R = F1 + F2
F1
F1 and F2 are vector components of R.
The relationship between a force and its vector components along given axes must not be confused with the
relationship between a force and its orthogonal projections onto the same axes.
Only when the axes are perpendicular are the components and projections of the vector equal.
.Fa and Fb are orthogonal projections.
Fa
.F1
and
F2
are
vector
components
in
the
a
and
directions
F2
R
b
F1
Fb
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2.2. Coplanar (Two-Dimensional) Force Systems
2.2.1. Resolution of a Force
The most common two-dimensional resolution of a force vector is into rectangular components.
y
F = Fx + Fy= Fxi + Fyj
Fy
F
θ
Fx
x
The scalar components Fx and Fy may in general be positive or negative, depending on the quadrant into
which F points.
Fx = Fcosθ
,
Fy = Fcosθ ,
(Fx2 + Fy2)
F=
θ = tan-¹ (Fy / Fx)
It is essential that we be able to determine the correct components of a force no matter how the axes are
oriented or how the angles are measured.
F
β
Eg.
y
F
α
x
y
β
F
y
β
x
x
Fx = Fsin β
Fx = - Fcosβ
Fx = Fcos(β – α)
Fy = Fcos β
Fy = - Fsinβ
Fy = Fsin(β – α)
It is often convenient to utilize rectangular components in finding the sum or resultant force R.
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y
F2y
F2
F1
F1y
R
Ry
F1x
F2x
x
Rx
R = F1 + F2 = (F1xi + F1yj) + (F2xi + F2yj) or
R = Rxi + Ryj = (F1x + F2x)i + (F1y + F2y)j
Rx = F1x + F2x = Σ Fx
Ry = F1y + F2y = ΣFy
R=
(Rx2 + Ry2)
Sine and Cosine Laws
Cosine law
R2 = F12 + F22 – 2F1.F2cosα
R
γ
F1
θ
β
= F12 + F22 + 2F1.F2cosθ
α
θ
F2
Sine law
R/sinα = F2/sinγ = F1/sinβ
2.2.2. Moment, Couple and Force-Couple Systems
1 Moment
A force may tend to rotate a body about an axis. This rotational tendency is known as the moment M of the
force. It is also referred to as torque.
Moment is a vector quantity having direction perpendicular to the plane of the body and its sense depends
on the direction in which the force tends to rotate the body. The right-hand rule is used to identify the sense.
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The magnitude of the moment or tendency of the force to rotate the body about an axis normal to the plane
of the body is proportional to the magnitude of the force and the moment arm, which is the perpendicular
distance from the axis to the line of action of the force.
E.g.
o M
F
A
r
α
d
o
The magnitude of the moment is: M = F.d Its SI unit is Nm.
Sign convention consistency within a given problem is essential.
In some two-dimensional and many three-dimensional problems to follow, it is convenient to use a vector
approach for moment calculations. The moment of F about point A of the above fig. may be represented by
the cross-product expression:
M=rXF
,
rXF=-FXr
Where, r = a position vector which runs from the moment reference point A to any point on the line of
action of F.
The magnitude of this expression is given by;
M = Frsinα = Fd.
Note that the moment arm d = rsinα does not depend on the particular point on the line of action of F to
which the vector r is directed. The direction and sense of M are correctly established by applying the righthand rule to the sequence r X F. The moment is perpendicular to the plane containing the vectors r and F.
In evaluating the moment of a force about a given point, the choice between using the vector cross product
or using the scalar expression will depend largely on how the geometry of the problem is specified. If the
perpendicular distance between the line of action of the force and the moment center is given or is easily
determined, then the scalar approach is generally simpler. If, however, F and r are not perpendicular and are
easily expressible in vector addition, then the cross-product expression is often preferred.
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Varignon’s Theorem
It states that the moment of a force about any point is equal to the sum of the moments of the components of
the force about the same point.
P
R
Q
r
o
r
Proof:
Mo = r X R , R = P + Q
r X R = r X (P + Q)
Using the distributive law of cross product,
Mo = r X R = (r X P) + (r X Q)
2. Couple
Couple is the moment produced by two equal and opposite and non collinear forces.
-F
- These two forces cannot be combined into a single force, since
F
their sum in every direction is zero.
a
- The effect of the two forces is entirely to produce a tendency.
do
of rotation.
The magnitude of the couple M about an axis passing through any point such as o is;
M = F(a + d) – Fd = Fd (the magnitude contains no reference to the dimension a).The direction of the
couple is ccw.
The moment of a couple has the same value for all moment centers.
The moment of a couple may be represented by vector algebra using the cross product notation.
-F
M = rA X F + rB X (-F)
rB
r
F
= (rA - rB) X F
o
rA
= r X F (the expression contains no reference to the moment center o
to arbitrary points A and B on the line of action of F and –F, respectively.
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The direction of a couple is normal to the plane of the couple and the sense is established by the right-hand
rule. A couple is unchanged as long as the magnitude and direction of the vector remain constant.
Force-Couple Systems
The effect of force acting on a body has been described in terms of the tendency to push or pull the body in
the direction of the force and to rotate the body about any axis which does not intersect the line of the force.
The representation of this dual effect is often facilitated by replacing the given force by an equal parallel
force and a couple to compensate for the change in the moment of the force.
It is possible to replace a force at a point by the same force acting at a different point and a couple without
altering the external effects of the original force on the body. The combination of the force and couple is
referred to as a force-couple system.
E.g.
F
d
=
F
M = F.d
A given couple and a force which lies in the plane of the couple (normal to the couple vector) may be
combined to produce a single force by reversing the foregoing procedure.
2.2.3. Resultants
Most problems in mechanics deal with a system of forces, and it is usually necessary to reduce the system to
its simplest form in describing its action.
The resultant of a system of forces is the simplest force combination that can replace the original forces
without altering the external effect of the system on the rigid body to which the forces are applied.
For any system of coplanar forces F1, F2, F3, etc. in the x-y plane, the resultant R of the forces is given by:
R = F1 + F2 + F3 + ….. = ∑ Fi
= (F1xi + F1yj) + (F2xi + F2yj) + (F3xi + F3yj) + ……
= ∑ Fxi + ∑ Fyj
= Rxi + Ryj
Rx = ∑ Fx ; Ry = ∑ Fy , R =
((∑Fx)2 + (∑Fy)2)
θ = tan-1(Ry/Rx) = tan-1(∑Fy/ ∑Fx)
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It is possible to reduce a given system of forces to a force-couple system at an arbitrary chosen point.
F1
F2
Mo = ∑Fi*di
F1
M1
3
M3
3
F3
M2
3
F2
R = ∑F
F3
Mo = M1 + M2 + M3
M1 = F1*d1
M2 = F2*d2
M3 = F3*d3
R
d
`
d = Mo/R
R = ∑Fi
Mo = ∑Mi = ∑Fi*di
Mo = R*d
The moment of the resultant force about any point equals the sum of the moments of the original forces of
the system about the same point.
If the resultant force R for a given force system is zero, the resultant of the system need not be zero as it
may be a couple.
Eg.
F1
F1 + F2 = - F3
F2
R = 0,
M = F3*d
d
F3
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2.3. Non-coplanar (Three Dimensional) Force Systems
2.3.1. Resolution of a Force
z
Fzk
F
y
θz
θy
Fyj
θx
Fxi
x
Fxi, Fyj and Fzk are rectangular components of F.
F = Fxi + Fyj + Fzk,
Fx = Fcosθx,
Fy = Fcosθy,
Fz = Fcosθz
F = Fcosθxi + Fcosθyj + Fcosθzk = F (cosθxi + cosθyj + cosθzk)
n = unit vector in the direction of F
F = n F, F =
(Fx2 + Fy2 + Fz2)
In solving three-dimensional problems, one must usually find the x, y and z scalar components of a given or
unknown force. In most cases, the direction of a force is described;
i/ by two points on the line of action of the force, or
ii/ by two angles which orient the line of action.
i/ If the coordinates of points A(x1,y1,z1) and B(x2,y2,z2) on the line of action of the force are known and the
direction of the force is from A to B, the force may be written as;
F = FnAB = F.AB/| AB | = F. (x2 – x1)i + (y2 – y1)j + (z2 – z1)k
((x2-x1)2 + (y2-y1)2 + (z2-z1)2)
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z
ii/
Fz
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F
Φ
Fy
Fx
θ
Fxy
y
x
F = Fxi + Fyj + Fzk
Fxy = F.cosΦ, Fz = F.sinΦ, Fx = Fxy.cosθ = F.cosΦ.cosθ, Fy = Fxy.sinθ = F.cosΦ.sinθ
Rectangular components of a force F may be written with the aid of dot or scalar product operation.
If the unit vector n = αi +βj +γk and F = F(li + mj + nk), the projection of F in the n direction is given by:
i/ As a scalar
Fn = F. n = F(li + mj + nk) . (αi + βj + γk) = F(lα + mβ + nγ)
ii/ As a vector
Fn = (F.n) n
If θ is the angle between F and n,
F. n = Fncosθ, θ = cos-1((F . n)/ | F |)
It should be observed that the dot product relationship applies to non intersecting vectors as well as to
intersecting vectors.
2.3.2. Moment, Couple and Force-couple Systems
In three dimensions, the determination of the perpendicular distance between a point or line and the line of
action of the force can be a tedious computation. The use of a vector approach using cross-product
multiplication becomes advantageous.
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i/ Moment
z
Fz
F
Fy
r
Fx
O
y
rz
rx
ry
x
The moment Mo of F about an axis through O is given by;
Mo = r X F = (rxi + ryj + rzk) X (Fxi + Fyj + Fzk) =
i
j
k
rx
ry
rz
Fx
Fy
Fz
= (ry.Fz – rz.Fy)i + (rz.Fx – rx.Fz)j + (rx.Fy – ry.Fx)k
= Mxi + Myj + Mzk
If n is a unit vector in the  direction, the moment M of F about any axis  through O is expressed by;
M = Mo .n = (r X F. n) --- which is the scalar magnitude.
Or
M = (Mo. n) n = (r X F. n) n ------ vector expression for the moment of F about an axis 
through O.
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Varignon’s Theorem:- the sum of the moments of a system of concurrent forces about a given point equals
the moment of their sum about the same point.
F3
-The sum of the moments of the forces about O is;
F2
Mo = r X F1 + r X F2 + r X F3 + ------= r X (F1 + F2 + F3 + ----)
= r X  Fi
F1
=  (r X Fi)
r
=rXR
ii. Couple
M
- F
d
F
A
B r
O
rB
rA
-If the vector r joins any point B on the line of action of –F to any point A on the line of action of F. The
combined moment (couple) of the two forces about O is;
M = rA X F + rB X (-F) = (rA – rB) X F = r X F
The moment of the couple, M = r X F. It is the same about all points.
-
The moment of a couple is a free vector, whereas the moment of a force about a point (which is also
the moment about a defined axis through the point) is a sliding vector whose direction is along the
axis through the point.
-
A couple tends to produce a pure rotation of the body about an axis normal to the plane of the forces
which constitute the couple.
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Force-Couple System
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F
M=rXF
F
=
B
r
B
The force F at point A is replaced by an equal force F at point B and the couple
M = r X F.

Couple vectors obey all of the rules, which govern vector quantities.
2.3.3. Resultants
-Any system of forces may be replaced by its resultant force R and the resultant couple
M. For the system of forces F1, F2, F3, ----- acting on a rigid body, the resultant force R and the resultant
couple M is given by;
R = F1 + F2 + F3 + -------- =  Fi
M = M1 + M2 + M3 + ----- =  Mi
M1
Eg.
M2
F1
F1
=
O
F2
F2
M = M1 + M2
=
M1 = r1 X F1
,
M2 = r2 X F2
O
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R = F1 + F2
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. r1, r2 = vectors from O to any point on the line of action of F1 and F2 respectively.
In three dimensions, the magnitudes of the resultants and their components are;
Rx =  Fx, Ry =  Fy, Rz =  Fz
R=
((Fx)2 + (Fy)2 + (Fz)2)
Mx =  (r X F)x , My =  (r X F)y , Mz =  (r X F)z , M =
(Mx2 + My2 + Mz2)
.The magnitude and direction of M depends on the particular point selected. The magnitude and direction of
R, however, are the same no matter which point is selected.
Wrench Resultant
When the resultant couple vector M is parallel to the resultant force R, the resultant is said to be a wrench.
R
R
M
M
Positive Wrench
Negative Wrench
Eg. Screw driver application
-Any general force system may be represented by a wrench applied along a unique line of action.
Eg. Take the resultant force R and resultant couple M acting at some point O of the general force system.
M = M1 + M2
M
R
R
R
M2
O
M1
d
R
O
-R
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CHAPTER 3. EQUILIBRIUM
3.1. Introduction
When the resultant of all the forces acting on the rigid body and the resultant couple are both equal to zero, the
external forces form a system equivalent to zero and the body is said to be in equilibrium.
-
The necessary and sufficient conditions for the equilibrium of a body may be obtained by setting the
resultant force R and the resultant moment M equal to zero, and we have the equilibrium equations;
R =  Fi = 0
and
M =  Mi = 0
3.2. Equilibrium in Two-Dimensions
Mechanical System Isolation (Free- body Diagram)
-
A free-body diagram is a diagrammatic representation of the isolated body or combination of bodies
treated as a single body, showing all forces applied to it by mechanical contact with other bodies that are
imagined to be removed.
-
The free-body diagram is the most important single step in the solution of problems in mechanics.
-
Equilibrium equations should be written after a free-body diagram has been carefully drawn.
Common types of force application on mechanical system for analysis in two dimensions
Type of contact and force origin
Action on body to be isolated
1/ Flexible cable, belt chain or rope


T
2/ Smooth Surface
N
3/ Rough Surface
F
R
N
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4/ Roller Support
. Roller, rocker or ball support transmits a force normal to the supporting surface.
5/ Freely Sliding Guide
N
6/ Pin Connection
-Pin free to turn
Rx
Ry
Rx
-Pin not free to turn
M
Ry
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7/ Built-in or fixed Support
or
M
F
Weld
V
8/ Gravitational Attraction
Center of gravity, G
m
W = mg
9/ spring action
F
F
Construction of Free-body Diagrams
Procedures
1/ Choose the body or combinations of bodies to be isolated. The body chosen will usually involve one
or more of the desired unknown quantities.
2/ the body or combination chosen is next isolated by a diagram that represents its complete external
boundary.
3/ all forces (known or unknown) that act on the isolated body are next represented in their proper
positions on the diagram of the isolated body. Weights where appreciable, must be included.
4/ the choice of coordinate axes should be indicated directly on the diagram. Pertinent dimensions may
also be represented for convenience.
-
It is only through complete isolation and a systematic representation of all external forces that a reliable
accounting of the effects of all applied and reactive forces can be made.
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Examples
Mechanical System
Free-Body Diagram of Isolated Body
1/ Plane truss (weight of truss assumed negligible compared with P)
y
P
P
A
B
Ax
x
By
Ay
2/ Cantilever beam
F3
F2
F1
F3
F2
F1
y
F
Mass m
M
x
V
3/
W=mg
M
M
y
A
Smooth surface at A
N
P
B
W=mg
P
x
Bx
By
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4/ Rigid system of interconnected bodies analyzed as a single unit
FBD
P
P
y
m
mg
A
B
Ay
Bx
By
-Weight of mechanism neglected.
Equilibrium Conditions
-We defined equilibrium as the condition in which the resultant of all forces acting on a body is zero. The
necessary and sufficient conditions for complete equilibrium in two dimensions are;
 Fx = 0,  Fy = 0 and  Mo = 0
- Mo = 0 is the zero sum of the moments of all forces about any point o on or off the body.
y
Categories of Equilibrium in Two-Dimensions
Force System
FBD
x
Independent Equations
x
F3
1/ Collinear
 Fx = 0
F2
F1
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2/ Concurrent at a point
F1
F3
 Fx = 0
 Fy = 0
F2
F4
3/ Parallel
F1
 Fx = 0
F2
F3
F4
 Mz = 0
4/ General
F1
F2
 Fx = 0
 Fy = 0
F3
 Mz = 0
M
F4
3.3. Equilibrium in Three-Dimensions
Equilibrium Conditions
-The necessary and sufficient conditions for complete equilibrium in three dimensions are;
F=0 or
and M=0 or
Fx=0, Fy =0 and Fz=0
Mx=0, My=0 and Mz=0
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Notes: -In applying the vector form of the above equations, we first express each of the forces in terms of the
coordinate unit vectors i, j and k.
- For the first equation, F=0, the vector sum will be zero only if the coefficients of i, j and k in the
expression are, respectively, zero. These three sums when each is set equal to zero yield precisely the three
scalar equations of equilibrium,
Fx = 0, Fy = 0 and Fz = 0
For the second equation, M = 0, where the moment sum may be taken about any convenient point o, we
express the moment of each force as the cross product r X F, where r is the position vector from o to any point
on the line of action of the force F.
Thus, M = (rXF) = 0. The coefficients of i, j and k in the resulting moment equation when set equal to zero,
respectively, produce the three scalar moment equations Mx=0, My=0 and Mz=0.
Modeling the action of forces in Three-Dimensional analysis
Type of contact and force origin
Action on body to be isolated
1/ Member in contact with smooth surface, or ball supported member
z
z
z
x
y
y
y
x
x
2/ Member in contact with rough surface
z
z
F
y
x
x
y
N
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3/ Roller or wheel support with lateral constraint
z
z
y
x
P
x
y
N
4/ Ball-and-socket joint
z
z
y
x
x
y
Rx
Rz
Ry
5/ Fixed connection (embedded or welded)
z
z
Rx
x
y
Ry
Rz
x
Mx
Mz
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My
y
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6/ Thrust-bearing support
z
z
Ry
y
Rz
x
Mx
Rx
y
x
Mz
Categories of Equilibrium in Three-Dimensions
Force System
FBD
Independent equations
F1
F2
y
1/ Concurrent at a point
z
x
 Fx = 0
 Fy = 0
 Fz = 0
F3
F5
F4
y
x
 Fx = 0
F2
 Fy = 0
2/ Concurrent with a line
F1
F3
z
 Fz = 0
 My = 0
F4
 Mz = 0
F5
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y
3/ Parallel
 Fx = 0
x
F1
F2
F5
z
 My = 0
F3
 Mz = 0
F4
y
4/ General
F2
 Fx = 0
x
 Fy = 0
F1
 Fz = 0
M
z
 Mx = 0
F4
F3
 My = 0
 Mz = 0
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CHAPTER 4
4. ANALYSIS OF SIMPLE STRUCTURES
4.1 Introduction
An engineering structure is any connected system of members built to support or transfer forces
and to safely withstand the loads applied to it. In this chapter we shall analyze the internal forces
acting in several types of structures, namely, trusses, frames and simple machines.
Constraints and Statical Determinacy
Equilibrium equations, once satisfied, are both necessary and sufficient conditions to establish the
equilibrium of a body. However they don’t necessarily provide all the information that is required
to determine all the unknown forces that may act on a body in equilibrium.
If the number of unknown forces is more than the number of independent equilibrium equations,
the equilibrium equations alone are not enough to determine the unknown forces, possibly reaction
forces at the constraints.
The adequacy of the constraints to prevent possible movement of the body depends on the number,
arrangement and characteristics of the constraints.
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Problem Solution
It is found important to develop a logical and systematic approach in the solution of problems of
mechanics, which includes the following steps:
 Identify clearly the quantities that are known and unknown.
 Make an unambiguous choice of the body/group of bodies/ to be isolated and draw its
complete FBD, labeling all external known and unknown forces and couples which act on
it.
 Designate a convenient set of axes and choose moment centers with a view to simplifying
the calculations.
 Identify and state the applicable force and moment principles or equations which govern
the equilibrium condition of problem.
 Match the number of independent equations with the number of unknowns in each
problem.
 Carry out the solution and check the results.
4.2 Plane Trusses
A truss is a framework composed of members joined at their ends to form a rigid structure. When
the members of the truss lie essentially in a single plane, the truss is known as a plane truss.
Examples of commonly used trusses that can be analyzed as plane as plane trusses are: -
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The basic element of a plane truss is the triangle. Structures that are built from a basic triangle in
the manner described are known as simple trusses. When more members are present than are
needed to prevent collapse, the truss is statically indeterminate. A statically indeterminate truss
cannot be analyzed by the equations of equilibrium alone. Additional members or supports that
are not necessary for maintaining the equilibrium position are called redundant.
 Three bars joined by pins at their ends constitute a rigid frame.
 Four or more bars pin-jointed to form a polygon of as many sides
constitute a non rigid frame.
 We can make the non rigid frame stable or rigid by adding diagonal
bars.
The term rigid is used in the sense of non-collapsible and also in the sense that deformation of the
members due to induced internal strains is negligible.
All members in a simple truss are assumed to be two-force members. The members may be in
tension (T) or in compression ( C ).
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The weight of truss members is assumed small compared with the force it supports. If it is not, or
if the small effect of the weight is to be accounted for, the weight W of the member may be replaced
by two forces, each W/2 if the member is uniform, with one force acting at each end of the member.
These forces, in effect, are treated as loads externally applied to the pin connections. Accounting
for the weight of a member in this way gives the correct result for the average tension or
compression along the member but will not account for the effect of bending of the member.
 When welded or riveted connections are used to join structural members, the assumption
of a pin-jointed connection is usually satisfactory if the centerlines of the members are
concurrent at the joint.
 We also assume in the analysis of simple trusses that all external forces are applied at the
pin connections.
This condition is satisfied in most trusses. In bridge trusses the deck is usually laid on cross beams
that are supported at the joints.
Force analysis of plane trusses
Two methods for the force analysis of simple trusses will be given. The external reactions are
usually determined by computation from the equilibrium equations applied to the truss as a whole
before the force analysis of the remainder of the truss is begun.
4.2.1 Method of joints
This method for finding the forces in the members of a simple truss consists of satisfying the
conditions of equilibrium for the forces acting on the connecting pin of each joint.
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 The method deals with the equilibrium of concurrent forces, and only two independent
equilibrium equations are involved. ( Σ Fx = 0 and Σ Fy = 0 for each joint)
 We begin the analysis with any joint where at least one known load exists and where not
more than two unknown forces are present. Taking free body diagram of a joint, tension
will always be indicated by an arrow away from the pin, and compression will always be
indicated by an arrow toward the pin.
In some instances it is not possible to initially assign the correct direction of one or both of the
unknown forces acting on a given pin. In this event we may make an arbitrary assignment. A
negative value from the computation indicates that the assumed direction is incorrect.
4.2.2 Method of sections
On the analysis of plane trusses by the method of joints, we took advantage of only two of the
three equilibrium equations, since the procedures involve concurrent forces at each joint.
We may take advantage of the third or moment equation of equilibrium by selecting an entire
section of the truss for the free body in equilibrium under the action of a non-concurrent system of
forces. This method of sections has the basic advantage that the force in almost any desired
member may be found directly from an analysis of a section, which has cut that member. Thus it
is not necessary to proceed with the calculation from joint to joint until the member in question
has been reached.
In choosing a section of the truss, we note that, in general, not more than three members whose
forces are unknown may be cut, since these are only three available equilibrium relations which
are independent.
It is essential to understand that in the method of sections an entire portion of the truss is considered
a single body in equilibrium. Thus, the forces in members internal to the section are not involved
in the analysis of the section as a whole.
To classify the free body and the forces acting externally on it, the section is preferably passed
through the members and not the joints.
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In some cases, the methods of sections and joints can be combined for an efficient solution. The
moment equations are used to great advantage in the method of sections. One should choose a
moment center, either on or off the section, through which as many unknown forces as possible
pass.
It is not always possible to assign the proper sense of an unknown force when the free-body
diagram of a section is initially drawn. With an arbitrary assignment made, a positive answer will
verify the assumed sense and a negative result will indicate that the force is in the sense opposite
to that assumed.
4.3 Frames and Simple Machines
A structure is called a frame or machine if at least one of its individual members is a multiforce
member. A multiforce member is defined as one with three or more forces acting on it or one with
two or more forces and one or more couples acting on it.
Frames are structures which are designed to support applied loads and are usually fixed in position.
Machines are structures which contain moving parts and are designed to transmit forces or couples
from input values to output values.
In this article attention is focused on the equilibrium of interconnected rigid bodies which contain
multi force members. The forces acting on each member of a connected system are found by
isolating the member with a free-body diagram and applying the established equations of
equilibrium.
The principle of action and reaction must be carefully observed when we represent the forces of
interaction on the separate free-body diagrams.
If the frame or machine constitutes a rigid unit by itself when removed from its supports, the
analysis is best begun by establishing all the forces external to the structure considered as a single
rigid body. We then dismember the structure and consider the equilibrium of each part separately.
The equilibrium equations for the several parts will be related through the terms involving the
forces of interaction.
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If the structure is not a rigid unit by itself but depends on its external supports for rigidity, as in
the figure below, then the calculation of the external support reactions cannot be completed until
the structure is dismembered and the individual parts are analyzed.
In most cases we find that the analysis of frames and machines is facilitated by representing the
forces in terms of their rectangular components.
It is not always possible to assign every force or its components in the proper sense when drawing
the free body diagrams and it becomes necessary for us to make an arbitrary assignment.
-In any event it is absolutely necessary that a force be consistently represented on the diagrams for
interacting bodies, which involve the force in question. For example, for two bodies connected by
the pin in the figure below the force components must be consistently represented in opposite
directions on the separate free-body diagrams.
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If we choose to use vector notation in labeling the forces, then we must be careful to use a plus
sign for an action and a minus sign for the corresponding reaction.
-Situations occasionally arise where it is necessary to solve two or more equations simultaneously
in order to separate the unknowns. In most instances, however, we may avoid simultaneous
solutions by careful choice of the member or group of members for the free-body diagram and by
a careful choice of moment axes which will eliminate undesired terms from the equations.
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CHAPTER 5
5. CENTROIDS
5.1Introduction
Actually, “concentrated” forces do not exist in the exact sense, since every external force applied
mechanically to a body is distributed over a finite contact area however small.
When forces are applied over a region whose dimensions are not negligible compared with other
pertinent dimensions, then we must account for the actual manner in which the force is distributed
by summing up the effects of the distributed force over the entire region. We carry out this process
by using the procedures of mathematical integration. For this purpose we need to know the
intensity of the force at any location. There are three categories into which such problems fall;
1) Line Distribution:- when a force is distributed along a line. The loading is expressed as
force per unit length of line(N/m).
2) Area Distribution:- when a force is distributed over an area. The loading is expressed as
force per unit area (N/m2 ).
3) Volume Distribution :- when a force is distributed over the volume of a body (body force).
(N/m3).
The body force due to the earth’s gravitational attraction (weight) is by far the most commonly
encountered distributed force. The following sections of the chapter deal with the determination
of the point in a body through which the resultant gravitational forces acts and the associated
geometrical properties of lines, areas and volumes.
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5.2. Centroids of lines, Areas, and Volumes of Figures and Bodies
Center of Mass or Center of Gravity
Applying the principle of moments, the moment of the resultant gravitational force W about any
axis equals the sum of the moments about the same axis of the gravitational forces dw acting on
all particles treated as infinitesimal elements of the body.
W = ∫ dw
Applying moment principle about the y-axis, the moment about the y-axis of the elemental weight,
dw is x .dw.
The sum of these moments for all elements of the body is ∫ xdw .
𝑥̅ .W = ∫ xdw 𝑥̅ = ∫ Wxdw
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If the density of the body is not constant, dm = ρdv
The point (𝑥̅ , 𝑦̅, 𝑧̅) is known as the center of mass, and coincides with the center of gravity as long
as the gravity field is treated as uniform and parallel.
-The center of mass has a special significance in calculating the dynamic response of a body to
unbalanced forces.
In most problems the calculation of the position of the center of mass may be simplified by an
intelligent choice of reference axes. In general the axes should be placed so as to simplify the
equations of the boundaries as much as possible. Thus polar coordinates will be useful for bodies
having circular boundaries.
Another important clue may be taken from considerations of symmetry. Whenever there exists a
line or plane of symmetry in a homogenous body, a coordinate axis or plane should be chosen to
coincide with this line or plane.
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The center of mass will always lie on such a line or plane, since the moments due to symmetrically
located elements will always cancel, and the body can be considered composed of pairs of these
elements.
-The location of the center of mass is always facilitated by the observation of symmetry when it
exists.
5.3 Centroids of Lines, Areas and Volumes
When speaking of an actual physical body, we use the term center of mass. If the density is uniform
throughout the body, the positions of the centroid and the center of mass are identical, whereas if
the density varies, these two points will, in general, not coincide.
The term centroid is used when the calculation concerns a geometrical shape only.
The calculation of centroids fall within three distinct categories, depending on whether the shape
of the body involved can be modeled as a line, an area, or a volume.
1) Centroids of Lines
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It should be noted that, in general, the centroid C will not lie on the line. If the rod lies in a single
plane, such as the x-y plane, only two coordinates will require calculation.
2) Centroids of Areas
The centroid C for the curved surface will in general not lie on the surface. If the area is a flat
surface, say, the x-y plane, only the coordinates of C in that plane will be unknown.
3) Centroids of Volumes
For a general body of volume V and constant density, the coordinates of the center of mass also
become the coordinates of the centroid C of the body.
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Choice of Element for Integration
-With mass centers and centroids the concept of the moment principle is simple enough; the
difficulties reside primarily with the choice of the differential element and with setting up the
integrals. In particular there are five guidelines to be specially observed.
i.
Order of element ;- whenever possible, a first order differential element should be selected
in preference to a higher order element so that only one integration will be required to cover
the entire figure.
ii.
Continuity ;- whenever possible, we choose an element which can be integrated in one
continuous operation to cover the figure.
The horizontal strip in fig (a), would be preferable to the vertical strip in fig (b), which, if
used, would require two separate integrals because of the discontinuity in the expression
for the height of the strip at x = x1.
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iii.
Discarding higher-order terms
Higher-order terms may always be dropped compared with the lower-order terms.
iv.
Choice of Coordinates ;-choose the coordinate system which best matches the boundaries
of the figure.
The boundaries of the area in (a) are most easily described in rectangular coordinates, whereas the
boundaries of the circular sector in (b) are best suited to polar coordinates.
v.
Centroidal Coordinate of Element:- When a 1st or 2nd order differential element is
adopted, it is essential to use the coordinate of the centroid of the element for the moment
arm in setting up the moment of the differential element.
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It is essential to recognize that the subscript C serves as a reminder that the moment arms appearing
in the numerators of the integral expressions for moments are always the coordinates of the
centroids of the particular elements chosen.
5.3. Composite bodies and Figures
-Consider a body whose parts have masses m1, m2, m3 with the respective mass center coordinates
x1, x2, x3 in the x-direction;
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The moment principle gives;
Analogous relations hold for composite lines, areas and volumes, where the m’s are replaced by
L’s, A’s and V’s respectively.
It should be pointed out that if a hole or cavity is considered in one of the component parts of a
composite body or figure, the corresponding mass represented by the cavity or hole is treated as a
negative quantity.
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