MODULE I
Introduction to Engineering Mechanics
The state of rest and state of motion of the bodies under the action of different forces
hasengaged the attention of philosophers, mathematicians and scientists for many centuries.
Thebranch of physical science that deals with the state of rest or the state of motion is termed
as Mechanics. Starting from the analysis of rigid bodies under gravitational force and
simpleapplied forces the mechanics has grown to the analysis of robotics, aircrafts,
spacecrafts underdynamic forces, atmospheric forces, temperature forces etc.
Archimedes (287–212 BC), Galileo (1564–1642), Sir Issac Newton (1642–1727)
andEinstein (1878–1955) have contributed a lot to the development of mechanics.
Contributions by Varignon, Euler, D.Alembert are also substantial. The mechanics developed
by theseresearchers may be grouped as
(i) Classical mechanics/Newtonian mechanics
(ii) Relativistic mechanics
(iii) Quantum mechanics/Wave mechanics.
Sir Issac Newton, the principal architect of mechanics, consolidated the philosophy
andexperimental findings developed around the state of rest and state of motion of the bodies
andput forth them in the form of three laws of motion as well as the law of gravitation.
Themechanics based on these laws is called Classical mechanics or Newtonian mechanics.
Albert Einstein proved that Newtonian mechanics fails to explain the behaviour of
highspeed (speed of light) bodies. He put forth the theory of Relativistic Mechanics.
Schrödinger (1887–1961) and Broglie (1892–1965) showed that Newtonian
mechanicsfails to explain the behaviour of particles when atomic distances are concerned.
They put forththe theory of Quantum Mechanics.
Engineers are keen to use the laws of mechanics to actual field problems. Application
oflaws of mechanics to field problem is termed as Engineering Mechanics. For all the
problemsbetween atomic distances to high speed distances Classical/Newtonian mechanics
hasstood the test of time and hence that is the mechanics used by engineers.
CLASSIFICATION OF ENGINEERING MECHANICS
Depending upon the body to which the mechanics isapplied, the engineering
mechanicsis classified as
(a) Mechanics of Solids, and
(b) Mechanics of Fluids.
The solid mechanics is further classified as mechanics of rigid bodies and mechanics
ofdeformable bodies. The body which will not deform or the body in which deformation can
beneglected in the analysis are called as Rigid bodies. The mechanics of the rigid bodies
dealingwith the bodies at rest is termed as Statics and that dealing with bodies in motion is
calledDynamics. The dynamics dealing with the problems without referring to the forces
causingthe motion of the body is termed as Kinematics and if it deals with the forces causing
motionalso, is called Kinetics.
If the internal stresses developed in a body are to be studied, the deformation of the
body should be considered. This field of mechanics is called Mechanics of Deformable
Bodies/Strength of Materials/Solid Mechanics. This field may be further divided into
Theoryof Elasticity and Theory of Plasticity.
Liquid and gases deform continuously with application of very small shear forces.
Suchmaterials are called Fluids. The mechanics dealing with behaviour of such materials is
calledFluid Mechanics. Mechanics of ideal fluids, mechanics of viscous fluid and mechanics
ofincompressible fluids are further classification in this area. The classification of mechanics
issummarised below in flow chart.
BASIC TERMINOLOGIES IN MECHANICS
The following are the basic terms to study mechanics:
Mass
The quantity of the matter possessed by a body is called mass. The mass of a body
willnot change unless the body is damaged and part of it is physically separated. When a
body istaken out in a spacecraft, the mass will not change but its weight may change due to
change ingravitational force. Even the body may become weightless when gravitational force
vanishesbut the mass remain the same.
Time
Time is the measure of succession of events. The successive event selected is the
rotationof earth about its own axis and this is called a day. To have convenient units for
variousactivities, a day is divided into 24 hours, an hour into 60 minutes and a minute into
60seconds.Clocks are the instruments developed to measure time. To overcome difficulties
due to irregularitiesin the earth’s rotation, the unit of time is taken as second which is defined
as theduration of 9192631770 period of radiation of the cesium-133 atom.
Space
The geometric region in which study of body is involved is called space. A point in
thespace may be referred with respect to a predetermined point by a set of linear and
angularmeasurements. The reference point is called the origin and set of measurements as
‘coordinates’.If coordinates involve only in mutually perpendicular directions they are
known asCartesian coordinates. If the coordinates involve angle and distances, it is termed as
polarcoordinate system.
Length
It is a concept to measure linear distances. The diameter of a cylinder may be 300
mm,the height of a building may be 15 m. Actually metre is the unit of length. However
dependingupon the sizes involved micro, milli or kilo metre units are used for measurement.
A metre isdefined as length of the standard bar of platinum-iridium kept at the International
Bureau ofWeights and Measures. To overcome difficulties of accessibility and reproduction,
now meteris defined as 1690763.73 wavelength of krypton-86 atom.
Displacement
Displacement is defined as the distancemoved by a body/particle in the specified
direction.Referring to Fig. 1.1, if a body moves fromposition A to position B in the x-y plane
shown, itsdisplacement in x-direction is AB′ and its displacementin y-direction is B′B.
Velocity
The rate of change of displacement withrespect to time is defined as velocity.
Acceleration
Acceleration is the rate of change of velocitywith respect to time. Thus
a =dv/dt, where v is velocity
...(1.1)
Momentum
The product of mass and velocity is called momentum. Thus
Momentum = Mass × Velocity
...(1.2)
Continuum
A body consists of several matters. It is a well-known fact that each particle can
besubdivided into molecules, atoms and electrons. It is not possible to solve any
engineeringproblem by treating a body as a conglomeration of such discrete particles. The
body is assumedto consist of a continuous distribution of matter. In other words, the body is
treated ascontinuum.
Rigid Body
A body is said to be rigid, if the relative positions of any two particles in it do not
changeunder the action of the forces. In Fig. 1.2 (a) points A and B are the original position in
a body.After application of a system of forces F1, F2, F3, the body takes the position as
shown inFig. 1.2 (b). A′ andB′ are the new positions of A and B. If the body is treated as rigid,
therelative position of A′B′ andAB are the same i.e.,
A′B′ = AB.
(a)
(b)
Fig. 1.2
Particle
A particle may be defined as an object which has only mass and no size. Such a
bodycannot exist theoretically. However in dealing with problems involving distances
considerablylarger compared to the size of the body, the body may be treated as particle,
without sacrificingaccuracy. Examples of such situations are:
— A bomber aeroplane is a particle for a gunner operating from the ground.
— A ship in mid sea is a particle in the study of its relative motion from a control tower.
— In the study of movement of the earth in celestial sphere, earth is treated as a particle.
1.1.1 Sealer quantity
A quantity which can be described by its magnitude alone is called sealer quantity. Sealer
quantities have a unidirectional aspect. They may be added or subtracted arithmetically, Eg : mass, time, speed, area.
1.1.2 Vector quantity
A quantity which has got both magnitude and direction is called a vector quantity. Vectors
have two dimensional aspects. Vectors can be combined together using vector algebra.
Eg.force, velocity, moment .
1.2 IDEALIZATION OF MECHANICS
It will be required to represent an action using the known laws and also to be able to form
equations simple enough to be susceptible to mathematical computational techniques. For this
we must replace the actual physical action and participating bodies with hypothetical, highly
simplify substitutes. It is to be made sure, of course that the results of our substitutions have a
reasonable correlation with reality. The mathematical model to substitute the physical system
to be analyzed should be chosen with considerable amount of imagination, ingenuity and
insight into the physical behavior. For examples, for the study of the trajectory of a planet, it
is the mass of the planet and not its size that is significant. Hence we can consider planets as a
particle for such computations. A frictionless surface, a rigid body and so on are all such
simplifications that pervade mechanics.
LAWS OF MECHANICS
The following are the fundamental laws of mechanics:
Newton’s first law
Newton’s second law
Newton’s third law
Newton’s law of gravitation
Law of transmissibility of forces, and
Parallelogram law of forces.
Newton’s First Law
It states that every body continues in its state of rest or of uniform motion in a
straightline unless it is compelled by an external agency acting on it. This leads to the
definition offorce as the external agency which changes or tends to change the state of rest or
uniformlinear motion of the body.
Newton’s Second Law
It states that the rate of change of momentum of a body is directly proportional to
theimpressed force and it takes place in the direction of the force acting on it. Thus according
tothis law,
Force ∝ rate of change of momentum. But momentum = mass × velocity
As mass do not change,
Force ∝ mass × rate of change of velocity
i.e.,
Force ∝ mass × acceleration
F ∝m × a
...(1.3)
Newton’s Third Law
It states that for every action there is an equal and opposite reaction. Consider the
twobodies in contact with each other. Let one body applies a force F on another. According
to thislaw the second body develops a reactive force R which is equal in magnitude to force F
and actsin the line same as F but in the opposite direction. Figure1.3 shows the action of the
ball andthe reaction from the floor. In Fig. 1.4 the action of the ladder on the wall and the
floor and thereactions from the wall and floor are shown.
Fig 1.3
Fig 1.4
Newton’s Law of Gravitation
Everybody attracts the other body. The force of attraction between any two bodies
isdirectly proportional to their masses and inversely proportional to the square of the
distancebetween them. According to this law the force of attraction between the bodies of
mass m1 andmass m2 at a distance d as shown in Fig. 1.5 is
F=G
𝑚1 𝑚2
𝑑2
WhereG is the constant of proportionality and is known as constant of gravitation.
...(1.4)
Fig. 1.5
Law of Transmissibility of Force
According to this law the state of rest or motion of the rigid body is unaltered if a
forceacting on the body is replaced by another force of the same magnitude and direction but
actinganywhere on the body along the line of action of the replaced force.
Let F be the force acting on a rigid body at point A as shown in Fig. 1.6. According to
thelaw of transmissibility of force, this force has thesame effect on the state of body as the
force Fappliedat point B. For using law of transmissibility of forces, itshould be carefully
noted that it is applicable onlyif the body can be treated as rigid.
The law of transmissibility of forces can be proved using the law of superposition,
whichcan be stated as the action of a given system of forces on a rigid body is not changed by
addingor subtracting another system of forces in equilibrium.
Fig 1.6
Consider the rigid body shown in Fig. 1.7 (a). It is subjected to a force F at A. B
isanother point on the line of action of the force. From the law of superposition it is obvious
thatif two equal and opposite forces of magnitude F are applied at B along the line of action
ofgiven force F, [Ref. Fig. 1.7 (b)] the effect of given force on the body is not altered. Force
F at Aand opposite force F at B form a system of forces in equilibrium. If these two forces are
subtractedfrom the system, the resulting system is as shown in Fig. 1.7 (c). Looking at the
system offorces in Figs. 1.7 (a) and 1.7 (c), we can conclude the law of transmissibility of
forces is proved.
Fig 1.7
Parallelogram Law of Forces
The parallelogram law of forces enables us to determine the single force called
resultantwhich can replace the two forces acting at a point with the same effect as that of the
two forces.This law was formulated based on experimental results. Though Stevinces
employed it in1586, the credit of presenting it as a law goes to Varignon and Newton (1687).
This law statesthat if two forces acting simultaneously on a body at a point are presented in
magnitude anddirection by the two adjacent sides of a parallelogram, their resultant is
represented in magnitudeand direction by the diagonal of the parallelogram which passes
through the point ofintersection of the two sides representing the forces.
In Fig.1.8 the force F1 = 4 units and force F2 = 3 units are acting on a body at point
A.Then to get resultant of these forces parallelogram ABCD is constructed such that AB is
equalto 4 units to linear scale and AC is equal to 3 units. Then according to this law, the
diagonal ADrepresents the resultant in the direction and magnitude. Thus the resultant of the
forces F1and F2 on the body is equal to units corresponding to AD in the direction α to F1.
Fig. 1.8
Equations for the Magnitude and Direction of the Resultant.
Consider two forces P and Q acting at point, making an angle θ between them. Let these
forces be represented in magnitude and direction by the sides OA and OB of the
parallelogram OACB as in fig 2.1. Then according to the law, the diagonal OC of the
parallelogram represents the resultant of P and Q in magnitude and direction.
We can now determine the magnitude and direction of the resultant from the geometry of the
figure. Draw CD perpendicular to OA produced. Let R be the magnitude of the resultant of P
and Q and ф, its inclination with P
From the figure,
OC2 = OD2 + DC2
= (OA + AD)2 + DC2
= OA2 + 2 OA . AD + (AD2 + DC2)
= OA2 + 2 OA . AC Cos θ + AC2
R2 = P2 + Q2 + 2 PQ Cos θ
That is,
Or,
R
=
√𝑃2 + 𝑄 2 + 2𝑃𝑄𝑐𝑜𝑠𝜃
Also,
Tan ф
=
𝐶𝐷
𝑂𝐴+𝐴𝐷
=
𝑄𝑠𝑖𝑛𝜃
𝑃+𝑄𝑐𝑜𝑠𝜃
. . . . . . (1)
That is,
ф
𝑄 𝑠𝑖𝑛𝜃
= tan-1 [𝑃+𝑄𝑐𝑜𝑠𝜃 ]
. . . . . . (2)
And,
R sin ф
= Q sin θ
R cos ф
= P + Q cos θ
R
=
. . . . . . (3)
. . . . . . (4)
𝑃+𝑄𝑐𝑜𝑠𝜃
𝐶𝑜𝑠𝜙
Types of forces
Collinear : If several forces lie along the same line of action, they are said to be collinear.
Coplanar:When all forces acting on a body are in the same plane, the forces are coplanar.
Concurrent Forces
In a concurrent force system, all forces pass through a common point.. If three forces
are applied to a body, as shown in the figure, they must pass through a common point (O), or
else the condition, ∑Mo = 0, will not be satisfied and the body will rotate because of
unbalanced moment. Moreover, the magnitudes of the forces must be such that the force
equilibrium equations, ∑Fx = 0, ∑Fy= 0, are satisfied.
It is fairly easy to see the reasoning for the first condition. Consider the two forces, F1
and F2, intersecting at point O in the figure. The sum of moments of these two forces about
point O is obviously equal to zero because they both pass through O. If F3 does not pass
through O, on the other hand, it will have some nonzero moment about that point. Since this
nonzero moment will cause the body to rotate, the body will not be in equilibrium.
Therefore, not only do three non-parallel forces applied to a body have to be
concurrent for the body to be in an equilibrium state, but their magnitudes and directions
must be such that the force equilibrium conditions are satisfied (∑Fx = ∑Fy = 0). Notice that
there is no need for the moment equilibrium equation in this case since it is automatically
satisfied
A concurrent force system contains forces whose lines-of action meet at some one
point. Forces may be tensile (pulling)
Forces may be compressive (pushing)
Force exerted on a body has two effects:
The external effect, which is tendency to change the motion of the body or to develop
resisting forces in the body
The internal effect, which is the tendency to deform the body.
RESOLUTION
The process of reducing a force system to a simpler equivalent system is called a
reduction. The process of expanding a force or a force system into a less simple equivalent
system is called a resolution
If the force system acting on a body produces no external effect, the forces are said to
be in balance and the body experience no change in motion is said to be in equilibrium.
Resultant Forces
If two forces P and Q acting on a particle A may be replaced by a single force R,
which has the same effect on the particle. This force is called the resultant of the forces P and
Q and may be obtained by constructing a parallelogram, using P and Q as two sides of the
parallelogram. The diagonal that pass through A represents the resultant. This is known as
the parallelogram lawfor the addition of two forces. This law is based on experimental
evidence; it cannot be proved or derived mathematically.
For multiple forces action on a point, the forces can be broken into the components of x and
y.
Free-body diagrams
Free-body diagrams can be used as a convenient way to keep track of forces acting on
a system. Ideally, these diagrams are drawn with the angles and relative magnitudes of the
force vectors preserved so that graphical vector addition can be done to determine the
resultant.
As well as being added, forces can also be resolved into independent components at
right angles to each other. A horizontal force pointing northeast can therefore be split into
two forces, one pointing north, and one pointing east. Summing these component forces using
vector addition yields the original force. Resolving force vectors into components of a set of
basis vectors is often a more mathematically clean way to describe forces than using
magnitudes and directions.This is because, for orthogonal components, the components of the
vector sum are uniquely determined by the scalar addition of the components of the
individual vectors. Orthogonal components are independent of each other; forces acting at
ninety degrees to each other have no effect on each other. Choosing a set of orthogonal basis
vectors is often done by considering what set of basis vectors will make the mathematics
most convenient. Choosing a basis vector that is in the same direction as one of the forces is
desirable, since that force would then have only one non-zero component. Force vectors can
also be three-dimensional, with the third component at right-angles to the two other
components
.
Equilibria
Equilibrium occurs when the resultant force acting on an object is zero (that is, the
vector sum of all forces is zero). There are two kinds of equilibrium: static equilibrium and
dynamic equilibrium.
System in equilibrium
A system is in equilibrium when the sum of all forces is zero.For example consider a
system consisting of an object that is being lowered vertically by a string with tension, T, at a
constant velocity. The system has a constant velocity and is therefore in equilibrium because
the tension in the string (which is pulling up on the object) is equal to the force of gravity,
mg, which is pulling down on the object. (Assume up is positive and down is negative.)
Static equilibrium
The simplest case of static equilibrium occurs when two forces are equal in magnitude
but opposite in direction. For example, an object on a level surface is pulled (attracted)
downward toward the center of the Earth by the force of gravity. At the same time, surface
forces resist the downward force with equal upward force (called the normal force). The
situation is one of zero net force and no acceleration.
Pushing against an object on a frictional surface can result in a situation where the
object does not move because the applied force is opposed by static friction, generated
between the object and the table surface. For a situation with no movement, the static friction
force exactly balances the applied force resulting in no acceleration. The static friction
increases or decreases in response to the applied force up to an upper limit determined by the
characteristics of the contact between the surface and the object
A static equilibrium between two forces is the most usual way of measuring forces,
using simple devices such as weighing scales and spring balances. For example, an object
suspended on a vertical spring scale experiences the force of gravity acting on the object
balanced by a force applied by the "spring reaction force" which is equal to the object's
weight. Using such tools, some quantitative force laws were discovered: that the force of
gravity is proportional to volume for objects of constant density (widely exploited for
millennia to define standard weights); Archimedes' principle for buoyancy; Archimedes'
analysis of the lever; Boyle's law for gas pressure; and Hooke's law for springs. These were
all formulated and experimentally verified before Isaac Newton expounded his three laws of
motion.
Dynamical equilibrium
Dynamical equilibrium was first described by Galileo who noticed that certain
assumptions of Aristotelian physics were contradicted by observations and logic. Galileo
realized that simple velocity addition demands that the concept of an "absolute rest frame"
did not exist. Galileo concluded that motion in a constant velocity was completely equivalent
to rest. This was contrary to Aristotle's notion of a "natural state" of rest that objects with
mass naturally approached. Simple experiments showed that Galileo's understanding of the
equivalence of constant velocity and rest to be correct. For example, if a mariner dropped a
cannonball from the crow's nest of a ship moving at a constant velocity, Aristotelian physics
would have the cannonball fall straight down while the ship moved beneath it. Thus, in an
Aristotelian universe, the falling cannonball would land behind the foot of the mast of a
moving ship. However, when this experiment is actually conducted, the cannonball always
falls at the foot of the mast, as if the cannonball knows to travel with the ship despite being
separated from it. Since there is no forward horizontal force being applied on the cannonball
as it falls, the only conclusion left is that the cannonball continues to move with the same
velocity as the boat as it falls. Thus, no force is required to keep the cannonball moving at the
constant forward velocity.
Moreover, any object traveling at a constant velocity must be subject to zero net force
(resultant force). This is the definition of dynamical equilibrium: when all the forces on an
object balance but it still moves at a constant velocity.
A simple case of dynamical equilibrium occurs in constant velocity motion across a
surface with kinetic friction. In such a situation, a force is applied in the direction of motion
while the kinetic friction force exactly opposes the applied force. This results in a net zero
force, but since the object started with a non-zero velocity, it continues to move with a nonzero velocity. Aristotle misinterpreted this motion as being caused by the applied force.
However, when kinetic friction is taken into consideration it is clear that there is no net force
causing constant velocity motion.
MOMENT OF AFORCE
Moment of force (often just moment) is the tendency of a force to twist or rotate an
object; see the article torque for details. A moment is valued mathematically as the product of
the force and the moment arm. The moment arm is the perpendicular distance from the point
of rotation, to the line of action of the force. The moment may be thought of as a measure of
the tendency of the force to cause rotation about an imaginary axis through a point.
The moment of a force can be calculated about any point and not just the points in
which the line of action of the force is perpendicular. Image A shows the components, the
force F, and the moment arm x multiply when they are perpendicular to one another.
Mo=F*d
Varignon’s theorem of moments
French mathematician Varignon (1654-1722) gave the following theorem which is
also known as principle of moments. Varignon’s theorem states that the moment of a force
about any axis is equal to the sum of moments of its components about that axis.
Consider a force F acting at a point A. F1 and F2 are the components of F along any
two directions. The moment of F about an axis through an arbitrary point O is F × d, where d
is the arm of force F. d1 and d2 are the arm of forces of F1 and F2 respectively. Sum of
moments of the components F1 and F2 about O is F1d1 + F2d2. It is to be proved that F × d =
F1 × d1 + F2 × d2.
Join A and O. Draw a line through A and perpendicular to OA. Let AG, AJ and AE
be the rectangular components of F1, F2 and F respectively.
Referring Fig. 1.107
AE = AG + GE
= AG + BH
= AG + BD cos  2
= AG + AC cos  2
AD cos  = AB cos 1 + AC cos  2
F cos  = F1 cos 1  F2 cos 2 . (Theorem of resolved parts)
Multiplying by OA.
F × OA cos  = F1  OA cos 1  F2  OA cos 2
F × d = F1  d1  F2  d 2
Moment of F about O = moment of F1 about O + moment of F2 about O.
Since F1 and F2 are the components of the force F, and O is an arbitrary point,
moment of a given force about any point = sum of moments of its components about the
same point.
Example 1.31
Calculate the moment of the given force F = 10 kN about point O, as shown in fig.
1.108.
M0 = (10 sin 80) × (5 cos 20) - (10 cos 80) × (5 sin 20)
= 43.30 kN m.
Example 1.33
A uniform wheel 60 cm diameter weighing 1000 N rests against a rectangular
obstacle 15 cm height as shown in figure 1.112. Find the least force required, which when
acting through the centre of the wheel will just turn the wheel over the corner A of the block.
Also find the reaction of the block.
Solution
Let the inclination of the force P with AC be  . The inclination of line joining A
and C with vertical,  = 600 (refer example 1.21). Taking moments of P and W about A,
For
 M  0,(Psin )  AC  W  (ACsin )  0
P sin  = W sin 60 = 1000 sin 60
= 866 N.
P=
866
sin 
For P to be minimum, sin  should be maximum. For this  should be 900
P = 866 N
Resolving the forces along AC, RA - P cos  - W cos  = 0
RA = 866 cos 90 + 1000 cos 60
= 500 N
Couple
The forces whose lines of action are parallel to each other are called parallel forces.
Parallel forces may be like parallel forces or unlike parallel forces. Two parallel forces are
said to be like when they act in the same direction. Two parallel forces are said to unlike
when they act in opposite directions.
The resultant of two unlike parallel forces of same magnitude is zero. Hence these
forces cannot be replaced by another single force. Such two forces having the same
magnitude, parallel line of action, and opposite sense are said to form a couple. The plane in
which the forces act is called the plane of the couple. The distance between the line of action
of forces is called arm of couple.
Example 1.34
Resolve the force of 300 N shown in fig. 1.116, into two parallel components.
(i) at B and C and
Solution
(ii) at C and D
Let P and Q be the components
P + Q = 300
Case (i) When the components are at B and C
Moment of the given force about B is 300 × 2 = 600 Nm The magnitude of P and Q
should be such that the sum of their moments about B should be 600 Nm
P × 0 + Q × 3 = 600
Q = 200 N
P + Q = 300
P + 200 = 300
P =100 N
Case (ii) When the components are at C and D
P + Q = 300 N
Moment of the given force about D is 300 × 3 = 900 Nm The magnitude of P and Q
should be such that the sum of their moments about D should be 900 Nm.
P × 2 + Q × 0 = 900
P = 450 N
P + Q = 300
450 + Q = 300
Q = 300 - 450
= -150 N
Q is 150 N downward.
Example 1.35
Two unlike parallel forces are acting at distance of 450 mm from each other. The
forces are equivalent to a single force of 900 N, which acts at a distance of 200mm from the
greater of the two forces. Find the magnitude of the forces.
Solution
Let P > Q
P -Q = 900.................(i)
Moment of P about B must be equal to
moment of 900 N about B.
P × 450 = 900 × 650
P = 1300 N
From eqn (i)
1300 - Q = 900
Q = 1300 - 900
Q = 400 N.
Properties of force couple
The two forces constituting a couple are equal in magnitude and opposite in
direction. Therefore, the sum of forces of a couple is zero.
F  0
The moment of a couple about any point in the plane of couple is a constant and is
independent of the position of moment centre.
Moment of couple about O,
M = F (d+x) - F x
=Fd+Fx-Fx
=Fd
Since the moment of couple, F × d is independent of x, it is a constant.
Another property of couple is that the action of a couple on a rigid body will not be
changed if its arm is turned in the plane of couple through any angle about one of its ends, as
shown in fig. 1.122.
By successive rotation of the arm of a couple in its plane, first about one end and
then about the other, the couple can be put in any desired position in its plane. Thus we can
transpose a couple in its plane without changing its action on a body.
The action of a couple on a body does not change if both the magnitudes of the
forces and the arm of the couple are changed in such a day that the moment of the couple
remains unchanged. That is, without changing the action on a body, a given couple can be
replaced by another one with different forces with a different arm, provided the moments of
the two couples are equal.
The couple consisting of magnitude of forces F and arm of couple d can be replaced
by the couple of magnitude of force Q and arm of force d1, if F × d = Q × d1. Refer fig. 1.123.
Several couples in one plane can be replaced by a single couple acting in the same
plane such that the moment of this single couple is equal to the algebric sum of the moments
of the given couples. Refer fig. 1.124.
F × d + Q d1 = F × d + Q1 × d
= (F + Q1) d where Q1 is given by
Q1 × d = Q × d1
Q1 =
Qd1
d
A given couple can be resolved into several component couples by choosing the
component couples in such a manner that the algebric sum of their moments is equal to the
moment of the given couple.
Example 1.38
A rigid bar AB is acted upon by forces as shown in fig. 1.132. Reduce the force
system to (i) a single force (ii) force moment system at A, and (iii) force moment system at
D.
Solution
 Fv  8  6  8 12  6kN
R=
  FH     FV 
2
2
Case (i)
=
02  62  6kN
Let the resultant be at a distance x from A.
 MA  6  4  8  8 12 11  44kNm.
= 44 kNm. c.c.w.
Moment of resultant about A = 6 × x
6 x = 44
x=
44
 7.33m
6
Case (ii)
Sum of moments of forces about A is 44 kNm. The resultant force is 6 kN upwards.
Therefore, the system can be reduced to a force moment system at A as shown in fig.
1.134(a).
Case (iii)
The sum of moments of forces about D is 8 × 8 - 6 × 4 - 12 × 3 = 4 kNm, c.w. (Refer
fig. 3.38). Therefore, the given force system can be reduced to a single force of magnitude 6
kN along with a clockwise moment of 4 kNm at D as shown fig.3.40(b).
Example 1.39
Replace the given system of forces to (i) an equivalent force - couple system at A (ii)
an equivalent force - couple system at B. and (iii) a single force.
Solution.
 FV  500  200  300 100  500kN
= 500 N downward.
Since
 FH  0, resultant R =  FV  500N downward.
Case (i)
 MA  500  0  200  2  300  4 100  6
= - 200 kNm.
= 200 kNm. c.c.w.
The given system can be reduced to a single downward force of magnitude 500kN at
A along with a counter clockwise moment of 200 kNm as shown in fig. 1.136.
Case (ii)
 MB  500  6  200  4  300  2
= -3000 - 800 + 600
= -3200 kNm.
= 3200 kNm. c.c.w.
The given system of forces can be reduced to a single force of magnitude 500kN
along with a counter clockwise moment of 3200 kNm at B as shown in fig. 1.137.
Case (iii)
 MA  200kNm
= 200 kNm. c.c.w.
Since the sum of moments about A is c.c.w., the moment due to R should also be
counter clockwise. Therefore, R should be towards left of A.
Let it be at a distance x from A, as shown in fig. 1.138.
R × x = 200
x=
200
= 0.4m.
500
The given system of forces can be replaced by a single downward force of
magnitude 500 kN at a distance 0.4m towards left of A.
Equilibrium of a rigid body under coplanar forces
A rigid body is said to be in equilibrium when the external forces acting on it from
system of forces equivalent to zero. The necessary and sufficient conditions for the
equilibrium of a rigid body can be expressed analytically,
 Fx  0,  Fy  0 and  M  0 .
Example 1.40
A uniform beam AB of weight 100 kN and 6 m long had two bodies of weights 60
kN and 80 kN suspended from its two ends as shown in fig. 1.139. At what point the beam
should be supported that it may rest horizontally.
Solution.
The position of support should be such that the sum of moments of all the forces about that
point must be zero.
Let the required support be at a distance x towards left of B.
Taking moments of forces about the support,
80 × x - 100 (3 - x) - 60 (6 - x) = 0
80x = 300 - 100 x + 360 - 60 x
240 x = 660
x=
660
 2.75m .
240
The support should be 2.75m towards left of point B.
Example 1.41
A slender bar AB of weight 100 N and length 6 m rests on a small roller at D. The
end A rests on a smooth vertical wall as shown in fig. 1.140. Calculate the angle  that the
bar makes with the horizontal for the condition of equilibrium.
Solution.
For
 FH  0,
R A  R D sin   0 ..................(i)
For
 Fv  0,
R D cos   100  0
RD 
For
100
.....................(ii)
cos 
 M  0, taking moments about A,
R D  AD  W cos  AC  0
R D 
RD 
2
 100 cos  3  0
cos 
100 cos  3cos 
 150 cos 2 
2
100
 150 cos 2 
cos 
150cos2  cos   100
  tan 1
R AV
R AH
= tan 1
250
1250
= 11.310