UNIT 1 (1).pdf

Course No Title of the Course Course Structure Pre-Requisite
FCME006 Basics of Mechanical Engineering L-T-P: 4-0-0 None
COURSE OUTCOMES (COs)
After completion of this course, the students are expected to be able to demonstrate the following
knowledge, skills and attitudes:
1. To know force, its nature and applications.
2. To know the basic principles of civil and mechanical structures.
3. To understand the fundamentals of thermodynamics and fluid mechanics.
4. To know the working principles of IC Engines.
5. To understand the importance of different engineering materials.
6. To understand the different manufacturing processes and machining operations.
7. To know the use of Automation in manufacturing.

COURSE CONTENT
Group A
Unit-I
Introduction to Engineering Mechanics: Rigid and Elastic bodies, Force and its type, Law of parallelogram of forces, Triangle law of forces, Polygon law
of forces, Lami’s theorem, Laws of motion, Moment, Couple, Varignon’s theorem, Conditions of equilibrium, Concept of free body diagram, Coulomb’s
friction, Plane trusses, Analysis of trusses, Numerical problems. (6 Hours)
Unit-II
Introduction to Strength of Materials: Simple stresses and strains, Direct, shear, and volumetric stresses and strains, Hooke’s law, Tension test, Elastic
constants, Poisson’s ratio, Factor of safety, Introduction to beam, Types of beams, Types of loads, Shear force and bending moment diagrams (SFD and BMD)
for Simple and Cantilever beams under various loading conditions, Numerical problems. (6 Hours)
Unit-III
Introduction to Manufacturing Engineering: Classification and use of engineering materials, Basic principles and applications of methods of
manufacturing such as casting, forming and joining; Working principles and applications of machining operations such as Turning, Thread cutting, Milling,
Shaping, Grinding, etc., Use of automation in manufacturing. (6 Hours)
Group B
Unit-IV
Introduction to Thermodynamics: Thermodynamic system, Cycle, Path, Thermodynamic properties, Extensive and intensive properties, Thermodynamic
equilibrium, Reversible and irreversible processes, isochoric, Isothermal, Isobaric, Isentropic and Polytropic processes, First law of thermodynamics applied
to a cycle and process, Kelvin-Planck and Clausius statements of Second law of thermodynamics, Carnot cycle, Entropy, Clausius inequality, Internal
combustion (IC) engines, IC engines terminology, Spark ignition (SI) and Compression ignition (CI) engines, Two and four stroke engines, Air standard
cycles such as Otto, Diesel, Dual and Brayton cycles, Numerical problems. (12 Hours)
Unit-V
Introduction to Fluid Mechanics: Properties of a fluid, Density, Specific volume, Specific weight, Specific gravity, Kinetic and Kinematic viscosity, Pascal’s
law and its applications, Laminar and turbulent flow, Use of continuity equation and Bernoulli’s equation, Numerical problems. (6 Hours)

MECHANICS
MECHANICS OF FLUIDS MECHANICS OF SOLIDS
MECHANICS OF
DEFORMABLE BODIES
MECHANICS OF RIGID
BODIES
STATICS DYNAMICS
KINEMATICS KINETICS

LAWS OF MECHANICS
Newton's three laws form a part of foundation of mechanics. These are:
First law: Every object in a state of uniform motion tends to remain in that state of motion unless an
external force is applied to it.
This law is also called “law of inertia”.
Second law: The change of motion is proportional to the natural force impressed and is made in a
direction of the straight line in which the force is impressed.
Mathematically, this law is stated as:
F = m a
where F is the applied force, m is the mass and a is the acceleration. Note, that force and acceleration are
vectors.
Third law : For every action, there is an equal and opposite reaction.
The first two laws of Newton are valid only in inertial frame of reference. An inertial frame of reference has a
constant velocity. That is, it is moving at a constant straight line.

SYSTEM OF FORCES
When more than one force acts on a body, they constitute a SYSTEM OF FORCES.
Various types of SYSTEM OF FORCES are possible by the relative orientation of forces within the system.
1. Concurrent System of forces
2. Non concurrent System of forces
Each such type of force system produces different effect on bodies.
CONCURRENT SYSTEM OF FORCES brings in TRANSLATIONAL MOTION
NON CONCURRENT SYSTEM OF FORCES brings in ROTATION in addition to TRANSLATION

EFFECT OF FORCE ON A DEFORMABLE BODY
F
F F
F
Scenario 1: Force causing tension in the rod
Scenario 2: Force causing compression in the rod
EFFECT OF FORCE ON A RIGID BODY
F
TABLE
F
TABLE
Scenario 1: Force PUSHING the block
Scenario 2: Force PULLING the block
Whether the force is PUSHING or PULLING, the effect of
the force is SAME which is MOTION or TRANSLATION
Line of action
Line of action
Forces can be moved to any point along the line of action
on the rigid body

SYSTEM OF FORCES
COPLANAR FORCES
(Lines of action lying on the same plane)
COLLINEAR
FORCES
(Lines of
action lying
on the same
line)
PARALLEL
FORCES
(Lines of
action
parallel
to each
other)
CONCURRENT
FORCES
(Lines of
action
intersecting at
a point)
NON
CONCURRENT
AND NON
PARALLEL
FORCES
(Lines of action
neither parallel
nor
intersecting at
a point)
NON COPLANAR FORCES
(Lines of action lying on different planes)
COLLINEAR
FORCES
(Not
possible)
PARALLEL
FORCES
CONCURRENT
FORCES
NON
CONCURRENT
AND NON
PARALLEL
FORCES

RESULTANT OF A FORCE SYSTEM
To study the effect of a system of forces acting on a body, it is customary to replace the system of forces by
its RESULTANT.
RESULTANT is defined as a SINGLE EQUIVALENT FORCE which produces the same effect on the body as that
of all the given forces..
RESULTANT is helpful in determining the MOTION of the body.
For instance,
If the RESULTANT OF A FORCE SYSTEM is ZERO, then the body will REMAIN at REST if it was already
at REST or MOVE with CONSTANT VELOCITY if it was already moving.
Hence it can be concluded that the BODY remains in EQUILIBRIUM and is grouped under STATICS
If the RESULTANT of a FORCE SYSTEM is NON ZERO, the body will have VARYING STATE OF MOTION
and is grouped as DYNAMICS

RESULTANT OF COPLANAR CONCURRENT SYSTEM OF FORCE
Various methods are available for the estimation of RESULTANT of COPLANAR CONCURRENT SYSTEM OF
FORCES
1. Graphical Methods
1. Parallelogram Law
2. Triangle Law
3. Polygon Law
2. Trigonometric Methods
1. Cosine Law
2. Sine Law
3. Analytical Method
1. Vector Approach

SOME PRACTICAL SITUATIONS OF COPLANAR CONCURRENT SYSTEM OF FORCES

PARALLELOGRAM LAW
The parallelogram law states that when two concurrent forces F1 and F2 acting on a body are represented
by two adjacent sides of a parallelogram, the diagonal passing through their point of concurrency
represents the RESULTANT force R in magnitude and direction.
𝜃
𝐹1
𝐹2
𝜃
𝐹2
𝐹1
𝑅
𝜶
𝑂 𝐴
𝐵 𝐶
Graphical Solution:
1. To obtain the resultant graphically, from the origin O, draw the two force vectors on a graph to a convenient scale and
in the direction specified i.e. OA and OB respectively.
2. Complete the parallelogram OACB with the two force vectors as adjacent sides.
3. Draw the diagonal passing through the origin.
4. The length of the diagonal OC gives the magnitude of the resultant and tis inclination with the reference axis OA gives
the direction of the RESULTANT.

LAW OF COSINE
The mathematical statement of the parallelogram law is called the LAW OF COSINE. The magnitude and direction of the
resultant can be determined by trigonometry
𝜃
𝐹1
𝐹2
𝜃
𝐹2
𝐹1
𝑅
𝜶
𝑂 𝐴
𝐵 𝐶
𝐷
𝜃
From Δ 𝑂𝐶𝐷,
𝑂𝐶2 = 𝑂𝐷2 + 𝐶𝐷2
We know that 𝑂𝐷 = 𝑂𝐴 + 𝐴𝐷
𝑂𝐷 = 𝐹1 + 𝐹2 cos θ and 𝐶𝐷 = 𝐹2sin θ
The magnitude of the resultant R is given by 𝑅 = 𝑂𝐶 = (𝐹1+ 𝐹2 cos θ)2 + (𝐹2 sin θ)2
𝑅 = 𝐹1
2
+ 𝐹2
2
+ 2𝐹1𝐹2 𝑐𝑜𝑠𝜃
The inclination of the
resultant R is given by
𝜶 = 𝑡𝑎𝑛−1 𝐶𝐷
𝑂𝐴+𝐴𝐷
𝜶 = 𝑡𝑎𝑛−1 𝐹2𝑠𝑖𝑛θ
𝐹1+𝐹2𝑐𝑜𝑠θ

COROLLARY TO LAW OF COSINE
Case 1: If θ = 90o - THE TWO FORCES ARE PERPENDICULAR TO EACH OTHER
𝜃
𝐹2
𝐹1
𝑅
𝜶
𝑂 𝐴
𝐵 𝐶
θ = 90o
𝐹2
𝐹1
𝑅
𝜶
𝑂 𝐴
𝐵 𝐶
𝑅 = 𝐹1
2
+ 𝐹2
2
𝜶 = 𝑡𝑎𝑛−1 𝐹2
𝐹1
Case 2: If θ = 0o - THE TWO FORCES ARE COLLINEAR AND ACT IN THE SAME DIRECTION TO EACH OTHER
𝐹1
𝐹2
𝑂
𝑅 = 𝐹1 + 𝐹2 𝜶 = 0o
Case 3:If θ = 180o- THE TWO FORCES ARE COLLINEAR BUT ACTING IN THE OPPOSITE DIRECTION WHERE 𝑭𝟏>𝑭𝟐
𝑂 𝐹1
𝐹2 𝑅 = 𝐹1 − 𝐹2 𝜶 = 0o
When there are more than 2 concurrent forces in a force system, the RESULTANT can be found out in steps. First consider
any 2 forces and find the RESULTANT. The RESULTANT thus obtained is added on to the next force and so on until all given
forces are ADDED on by the PARALLELOGRAM LAW. And thus the overall resultant can be obtained.

TRIANGLE LAW
The resultant can also be determined by the TRIANGLE LAW which states that If two forces 𝐹1𝑎𝑛𝑑 𝐹2 acting
simultaneously on a body can be represented by the tow sides of a triangle (in magnitude and direction) taken in order
then the third side (closing side) represents the RESULTANT in the OPPOSITE order.
GRAPHICAL SOLUTION
From the ORIGIN, draw one of the force vectors on a graph to a convenient scale and in the direction specified.
With the head of this vector as ORIGIN, draw the second vector to scale and in the specified direction.
Join the tail of the first vector with the head of the second vector and its length gives the magnitude of the RESULTANT
to scale and its inclination to the reference axis gives the direction.
NOTE: Due to COMMUTATIVE PROPERTY of vectors, the order in which we construct the forces do not affect the
resultant.
𝜃
𝐹1
𝐹2
𝜃
𝐹2
𝐹1
𝑅
𝜶
𝑂
(𝜃 − 𝛼)

SINE LAW
The mathematical statement of the TRIANGLE LAW is called SINE LAW.
For a triangle of sides and included angles as shown in the figure, sine law can be expressed as
𝐶
𝐵
𝐴
𝛼
𝛽
𝛾
𝐴
𝑠𝑖𝑛𝛼
=
𝐵
𝑠𝑖𝑛𝛽
=
𝐶
𝑠𝑖𝑛𝛾
𝜃 𝐹2
𝐹1
𝑅
𝜶
(𝜃 − 𝛼)
𝜃
𝐹1
𝐹2
EXAMPLE OF APPLICATION OF SINE LAW
𝐹1
𝑠𝑖𝑛(𝜃 − 𝛼)
=
𝐹2
𝑠𝑖𝑛𝛼
=
𝑅
𝑠𝑖𝑛(180 − 𝜃)
Where A, B and C are the sides of the triangle and the angle
opposite to these sides being α, β and γ respectively
For a triangle we know that
𝐴2 = 𝐵2 + 𝐶2 − 2𝐵𝐶 𝑐𝑜𝑠α
𝐵2 = 𝐴2 + 𝐶2 − 2𝐴𝐶 𝑐𝑜𝑠β
𝐶2 = 𝐴2 + 𝐵2 − 2𝐴𝐵 𝑐𝑜𝑠γ

POLYGON LAW
If a number of CONCURRENT FORCES acting simultaneously on a body are represented in MAGNITUDE and DIRECTION
by the sides of a POLYGON taken in order then the CLOSING SIDE of the polygon represent the RESULTANT in the
OPPOSITE ORDER.
This is called as the POLYGON LAW which is an extension of the TRIANGLE LAW.
POLYGON LAW is used to find the resultant of more than TWO CONCURRENT FORCES.
It should be noted that due to ASSOCIATIVE PROPERTY of vectors, the order in which we add the forces do not affect the
RESULTANT.
𝑂
𝐹1
𝐹2
𝐹3
𝐹5
𝐹4
𝐹1
𝐹2 𝐹3
𝐹4
𝐹5
𝑅

Two forces are applied at the point A of a hook support as shown in Figure. Determine the magnitude an direction of
the resultant force by the trigonometric method using (i) parallelogram law and (ii) triangle law
A
60 N
25 N
35o
20o

A
60 N
25 N
35o
20o
𝐹1= 60 N, 𝐹2 = 25 N
The total included angle between 𝐹1 and 𝐹2 will be θ = 20 + 35 = 55o
The resultant as per parallelogram law is given by
𝑅 = 𝐹1
2
+ 𝐹2
2
+ 2𝐹1𝐹2 𝑐𝑜𝑠𝜃
𝑅 = 60 2 + 25 2 + 2 60 25 cos 55
𝑅 = 77.11 N
The angle made by the resultant R with 𝐹1 is given by
𝜶 = 𝑡𝑎𝑛−1 𝐹2𝑠𝑖𝑛θ
𝐹1+𝐹2𝑐𝑜𝑠θ
𝜶 = 𝑡𝑎𝑛−1 25 sin 55
60+25 cos 55
= 15.4o
(I) USING PARALLELOGRAM LAW

A
60 N
25 N
35o
20o
𝐹1= 60 N, 𝐹2 = 25 N
Draw the force triangle with 𝐹1 and 𝐹2 taken in order.
It should be noted that due to associative property of the vectors, the order in
which we take the forces will not affect the result. The closing side represent the
RESULTANT taken in OPPOSITE order.
The included angles between the forces are shown.
While measuring the included angle between the two vectors, we must take the
vectors either pointing towards or pointing away from the point of concurrency.
The total included angle between 𝐹1 and 𝐹2 will be θ = 20 + 35 = 55o
We can see that the angle between them in the force triangle is 180 – 55 = 125.
The angle between the RESULTANT and the force 𝐹1 is taken as α.
(I) USING TRIANGLE LAW

A
60 N
25 N
35o
20o
𝐹1= 60 N, 𝐹2 = 25 N
(I) USING TRIANGLE LAW
𝐹2 = 25 N
𝐹1 = 60 N
125o
α
55o
55 − α
𝑅
𝑅2 = 𝐹1
2
+ 𝐹2
2
− 2𝐹1𝐹2 𝑐𝑜𝑠125
𝑅2 = 602 + 252 − 2 60 25 𝑐𝑜𝑠125
𝑅 = 77.11 N
By using sine law
𝑅
𝑠𝑖𝑛125
=
𝐹1
sin(55 − 𝛼)
=
𝐹2
𝑠𝑖𝑛𝛼
77.11
𝑠𝑖𝑛125
=
60
sin(55 − 𝛼)
=
25
𝑠𝑖𝑛𝛼
Solving we get 𝛼 = 15.4

When there are more than 3 CONCURRENT forces acting on a particle, then graphical and trigonometric methods
become tedious to work with.
In such case, we use the analytical method as it can be applied to any number of forces acting in a system.
RESOLUTION OF A FORCE INTO COMPONENTS
The parallelogram law states that forces acting on a body can be replaced by a single RESULTANT force.
The inverse of this law should be also be true.
Hence a single force acting on a body may be replaced by two or more forces, which produce the same effect on the
body.
These forces are called as COMPONENTS of the original force.
The COMPONENTS of a force is taken about the rectangular axes and hence termed as RECTANGULAR COMPONENTS
ANALYTICAL METHOD

Y
X
F
𝐹𝑥
𝐹𝑦
O
𝜃
𝐹𝑥 = 𝐹 𝑐𝑜𝑠𝜃 𝐹𝑦 = 𝐹 𝑠𝑖𝑛𝜃
𝐹 = 𝐹𝑥
2
+ 𝐹𝑦
2
θ= 𝑡𝑎𝑛−1 𝐹𝑦
𝐹𝑥
If θ is the angle of inclination of F with respect to the X axis
then the scalar components of F are
The magnitude and direction of F can be expressed as
Depending upon the angle (θ) made by the force vector F with respect to the positive X axis, the sign of the components
of force can vary

Y
F
O
𝜃
X
Y
F
O X
Y
F
O
𝜃
Y
F
O
𝜃
FIRST QUADRANT SECOND QUADRANT THIRD QUADRANT FOURTH QUADRANT
X
X
FORCES IN VARIOUS QUADRANT AND THE SIGN OF THE FORCE COMPONENTS
QUADRANT θ SIGN OF
x- component y - component
1 0 – 90o + ve + ve
2 90o – 180o - ve + ve
3 180o – 270o - ve - ve
4 270o – 360o + ve - ve
𝜃

However, if we measure θ with respect to the X – axis such that it is always an ACUTE ANGLE as shown in
figure and consider the above SIGN CONVENTION for the components depending upon the QUADRANT in
which the FORCE lies, then VECTOR APPROACH can be avoided.
Y
F
O
𝜃
X
Y
F
O
X
Y
F
O
𝜃
Y
F
O
𝜃
FIRST QUADRANT SECOND QUADRANT THIRD QUADRANT FOURTH QUADRANT
X X
𝜃

Note: When RESOLVING a FORCE into orthogonal components, the ORIENTATION of the COORDINATE AXES is
ARBITRARY. Hence as shown in the figure, it is CONVENIENT to choose the X-axis parallel to the inclined plane than the
ground plane.
F = 50 N F = 50 N

Determine the resultant of the concurrent coplanar system of forces as shown in figure
400 N
350 N
100 N 200 N
5
3
2
2
2
4
4
5

θ1= 𝑡𝑎𝑛−1 3
5
= 30.96o
θ2= 𝑡𝑎𝑛−1 2
2
= 45o
θ3= 𝑡𝑎𝑛−1 4
5
= 38.66o
θ4= 𝑡𝑎𝑛−1 2
4
= 26.57o
Force Inclination of the force
with respect to X axis
X component
of force
Y component
of force
400 N 30.96o 343.01 205.78
350 N 45o -247.49 247.79
100 N 38.66o -78.09 -62.47
200 N 26.57o 178.88 -89.46
Total 196.31 301.34
𝑅 = Σ𝐹𝑥
2
+ Σ𝐹𝑦
2
𝑅 = 196.312 + 301.342
𝑅 =359.64 N
θ= 𝑡𝑎𝑛−1 Σ𝐹𝑦
Σ𝐹𝑥
= 56.92o

SYSTEM OF FORCES AND RESULTANT – II
NON CONCURRENT SYSTEM OF FORCES
Whenever a BODY is subjected to NON CONCURRENT system of forces,
the BODY will have rotational motion in addition to translation motion.
Hence the body can no longer be idealized as a PARTICLE and has to be
treated as a RIGID BODY

MOMENT OF A FORCE
Consider a door which is hinged to the door frame.
Opening and Closing of the door happens when a FORCE is
applied at the door which is far away from the hinge.
The effect of this force to open/close the door is to bring in the
EFFECT of ROTATIONAL motion of the doors about its HINGE.
Similarly in the use of pipe wrench, a rotational effect is
brought in for loosening/tightening of the pipe fixture.
This rotational effect of the FORCE is measured by a physical
quantity called as MOMENT of a FORCE.
The MOMENT OF A FORCE about a point is defined as a
measure of the tendency of the forces to rotate a body about
that point.
A door which is hinged
to the door frame
Use of pipe
wrench

MOMENT OF A FORCE ABOUT THE ORIGIN
The magnitude of the moment of a force about a point is defined as the product of the force and the
perpendicular distance of the line of action of the force from that point
VARIGNON’S THEOREM (OR) PRINCIPLE OF MOMENTS
Varignon’s theorem is useful in estimation of moment of a force when its
moment arm cannot be easily determined.
Varignon’s theorem states that “the moment about a given point O of the
resultant of several concurrent forces is equal to the sum of the moment of
individual forces about the same point O
PROOF:
Consider n number of forces 𝐹1, 𝐹2, 𝐹3, … … … 𝐹𝑛 is CONCURRENT as shown
in Figure
The resultant force is 𝑅 is given a the vector sum of these individual forces
𝑅 = 𝐹1 + 𝐹2 + 𝐹3 + 𝐹4+. . . . . . . . . . 𝐹𝑛
𝐹1
𝐹2
𝐹3
𝐹4
𝐹𝑛
𝑅
𝑟
𝑋
𝑌
𝑂
𝐹1

The moment of the resultant 𝑅 about the origin is
𝑀𝑜 = 𝑟 𝑥 𝑅
𝑀𝑜 = 𝑟 𝑥 (𝐹1 + 𝐹2 + 𝐹3 + 𝐹4+. . . . . . . . . . 𝐹𝑛)
By the distributive property of vector multiplication,
𝑀𝑜 = 𝑟 𝑥 𝐹1 + 𝑟 𝑥 𝐹2 + 𝑟 𝑥 𝐹3 + 𝑟 𝑥 𝐹4 +. . . . . . . . . . (𝑟 𝑥 𝐹𝑛)
𝑀𝑜 = ෍
𝑖 −1
𝑛
𝑟 𝑥 𝐹𝑖
𝑴𝒐 = 𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒃𝒐𝒖𝒕 𝒑𝒐𝒊𝒏𝒕 𝑶
Hence it can be said that the MOMENT of the RESULTANT of SEVERAL CONCURRENT FORCES about a point is equal to the
SUM of the MOMENTS of INDIVIDUAL FORCES about the same point
Varignon’s theorem is useful when the moment arm of a force is CANNOT BE EASILY DETERMINED, which
arises usually in COPLANAR problems. In such cases, the force is resolved into orthogonal components and the
moment can be determined by applying Varignon’s Theorem.

When 2 forces F and – F having the same magnitude, PARALLEL LINES OF ACTION and in OPPOSITE SENSE
act on a body, then they are said to form a COUPLE.
The ONLY EFFECT of couple is to ROTATE the body about the axis PERPENDCIULAR to the plane of the
couple.
Eg:
1. While driving a vehicle, turning of steering wheel to turn the car, COUPLE is applied
2. Tightening a screw using a screw driver
F
- F
COUPLE

Consider 2 equal and opposite forces acting at points A and B whose
respective position vectors are rA and rB as shown in Figure.
The sum of the moments of these 2 forces about the origin O is given as
Mo = [ rA x F ] + [ rB x (- F) ]
= (rA – rB) x F
= rBA x F
The moment vector thus obtained is called MOMENT OF A COUPLE
The MOMENT OF A COUPLE is defined as the PRODUCT OF THE
MAGNITUDE OF ONE OF THE FORCES AND THE PERPENDICULAR
DISTANCE BETWEEN THEM.
Mo = Fd
MOMENT OF A COUPLE

FORCE – COUPLE SYSTEM
Consider a Force F acting on a body at point A, whose position vector is r as shown in
Figure 1.
According to principle of transmissibility, the force can be moved to any point along the
line of action as it produces the same effect on the body.
However if we want to move the force to a point NOT LYING on its LINE of ACTION, then
we must introduce a COUPLE such that it produces the same effect as the force.
- F
F
F
Mo = r x F
Line of ACTION
Without affecting the effect of the force on the rigid body, let us introduce 2 forces
F and – F at the origin as shown in Figure 2.
The forces F acting at point A and – F force acting at point O put together will form
a COUPLE.
Hence the moment of couple is given by Mo = r x F and is the same as that of the
MOMENT of F about O.
Hence any force F acting on a rigid body can be moved to an arbitrary point O
provided that a COUPLE is added whose MOMET is EQUAL to the MOMENT of F
about O.
This is known as FORCE – COUPLE SYSTEM
Figure 1 Figure 2
Figure 3

RESULTANT OF NON CONCURRENT FORCES
If there are n number of NON CURRENT FORCES acting on a rigid body, then all those
FORCES can be moved to a COMMON POINT by introducing FORCE-COUPLE for each
individual force.
Once the forces are CONCURRENT at a POINT, their RESULTANT can be obtained by
VECTOR ADDITION.
The RESULTANT of non concurrent forces is a single force couple system whose force is
equal to the summation of all individual forces and the couple is equal to the
summation of all individual couples.
Mathematically it is written as

100 N 600 N
750 N
500 N
A B
In the figure shown, reduce the given system of forces acting on the beam AB to an equivalent force – couple
system at location A
2 m 1 m 1 m 3 m

100 N 600 N
750 N
500 N
A B
Taking the summation of all forces acting on the beam along the X and Y direction
෍ 𝐹𝑥 = 0
෍ 𝐹𝑦 = 100 − 500 + 750 − 600
෍ 𝐹𝑦 = −250 𝑁
The –ve sign indicates that the RESULTANT points in the –ve Y direction
SIGN CONVENTIONS
+ve for upward acting force
-ve for downward acting force

100 N 600 N
750 N
500 N
A B
Equivalent force – couple system at location A
Taking the summation of the MOMENTS of all forces about point A
෍ 𝑀𝐴 = 100𝑋2 − 500𝑋3 + 750𝑋4 − (600𝑋7)
෍ 𝑀𝐴 = −2500 𝑁𝑚
෍ 𝐹𝑦 = −250 𝑁
The –ve sign indicates that the MOMENT is in CLOCKWISE direction
2 m 1 m 1 m 3 m
SIGN CONVENTIONS
+ve for Counter Clock wise moment
-ve for Clock wise moment

100 N 600 N
750 N
500 N
A B
2 m 1 m 1 m 3 m
250 N
A B
2500 Nm
GIVEN PROBLEM
EQUIVALENT FORCE – COUPLE SYSTEM AT A

CONDITIONS OF EQUILIBRIUM
If all the forces are CONCURRENT, then we can idealize the body as a particle and can replace the system of forces by a
RESULTANT FORCE at the point of CONCURRENCY.
The effect of a CONCURRENT FORCE SYSTEM will be to TRANLATE the particle in the direction of the RESULTANT.
Hence to keep the particle in EQUILIBRIUM, the RESULTANT FORCE acting on it must be ZERO which means that the force
polygon must close.
Mathematically the CONDITION FOR EQUIBRIUM of a PARTCILE is expressed as
𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑅 = ෍ 𝐹 = 0
THE VECTORIAL SUMMATION OF ALL FORCES ACTING ON THE FREE BODY IS A NULL VECTOR
The conditions of EQUILIBRIUM of a body are different for different type of force systems.
Hence check if the LINES OF ACTION of all the forces acting on the free body are CONCURRENT or NON CONCURRENT.
CONDITIONS OF EQUILIBRIUM FOR CONCURRENT SYSTEM OF FORCES

If the lines of action of all the forces acting on the free body is NOT CONCURRENT, then treat the body as RIGID BODY.
Forces acting in a rigid body can be replaced by a RESULTANT FORCE at a COMMON POINT and a MOMENT about the same
point.
The effect of such a force system is to bring in TRANSLATION as well as ROTATION.
To keep the rigid body in EQUILIBRIUM, the RESULTANT FORCE and the MOMENT must be ZERO.
Mathematically, the CONDITION OF EQUILIBRIUM OF A RIGID BODY can be expressed as
𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑅 = ෍ 𝐹 = 0
𝑀𝑜𝑚𝑒𝑛𝑡 𝑀 = ෍ 𝑟 𝑥 𝐹 = 0
1. THE VECTORAL SUMMATION OF INDIVIDUAL FORCES ABOUT A COMMON POINT IS A NULL VECTOR
2. THE VECTORAL SUMMATION OF MOMENTS OF THE FORCES ABOUT THE SAME POINT IS ALSO A NULL
VECTOR
CONDITIONS OF EQUILIBRIUM FOR NON CONCURRENT SYSTEM OF FORCES

CONDITIONS OF EQUILIBRIUM
FOR
COPLANAR CONCURRENT SYSTEM OF FORCES
For a COPLANAR CONCURRENT SYSTEM OF FORCE, expressing the RESULTANT in terms of RECTANGULAR
COMPONENTS of INDIVIDUAL FORCES, then the EQUILIBRIUM CONDITION is
𝑅 = σ 𝐹𝑥 + σ 𝐹𝑦 = 0
Hence the necessary condition for EQUILIBRIUM are
෍ 𝐹𝑥 = 0
෍ 𝐹𝑦 = 0

The theorem states that if three coplanar concurrent forces acting on a body keep it in EQUILIBRIUM, the each force is
proportional to the sine of the angle between the other
F1
F2
F3
𝛼
𝛽
𝛾
F1
F2
F3
𝛽
180 − 𝛾
𝛾
180 − 𝛽
180 − 𝛼
𝐹1
𝑠𝑖𝑛𝛼
=
𝐹2
𝑠𝑖𝑛𝛽
=
𝐹3
𝑠𝑖𝑛𝛾
𝐴
sin(180 − 𝛼)
=
𝐵
sin(180 − 𝛽)
=
𝐶
sin(180 − 𝛾)
𝐴
𝑠𝑖𝑛𝛼
=
𝐵
𝑠𝑖𝑛𝛽
=
𝐶
𝑠𝑖𝑛𝛾
As per sine law we already know that
We know that A = F1 B = F2 and C = F3
We know that
sin(180 − 𝛼) = sin 𝛼
sin(180 − 𝛽) = 𝑠𝑖𝑛𝛽
sin(180 − 𝛾) = 𝑠𝑖𝑛𝛾
Therefore we can write
LAMI’S THEOREM

Two persons lift a mass of 100 kg by cables passing over 2 pulleys as shown in the figure. Determine the
forces P and Q that must be applied by the 2 persons if the body is in equilibrium at that position as shown
in figure.

P
Q
O
T
T
100 kg
FBD of CABLE
FBD of MASS
30 45

T
100 kg
FBD of MASS
Consider FBD of mass and applying the condition of equilibrium
σ 𝐹𝑦 = 0
T – (100 x 9.81) = 0
T = 981 N
SIGN CONVENTIONS
+ve for upward acting force
-ve for downward acting force

P
Q
O
T
FBD of CABLE
30 45
P
Q
T
60
45
180-(60+45) = 75
Applying Lami’s Theorem
𝑇 = 981
𝑠𝑖𝑛75
=
𝑄
𝑠𝑖𝑛60
We already know T = 981
Therefore consider the following
𝑇 = 981
𝑠𝑖𝑛75
=
𝑃
𝑠𝑖𝑛45
Solve for P and we get
P = 718.14 N
Similarly consider the following
𝑇
𝑠𝑖𝑛75
=
𝑃
𝑠𝑖𝑛45
=
𝑄
𝑠𝑖𝑛60
Solve for Q and we get
Q = 879.54 N
Hence the force to be exerted by the person at location P is 718.14 N
and by the person at location Q is 879.54 N

A free body diagram is a tool used to solve engineering mechanics problems. As the name suggests, the
purpose of the diagram is to "free" the body from all other objects and surfaces around it so that it can be
studied in isolation.
We will also draw in any forces or moments acting on the body, including those forces and moments exerted
by the surrounding bodies and surfaces that we removed.
The diagram shows a ladder supporting a person and the
free body diagram of that ladder.
As you can see, the ladder is separated from all other
objects and all forces acting on the ladder are drawn in with
key dimensions and angles shown.
A ladder with a man standing on it is shown on the left. Assuming
friction only at the base, a free body diagram of the ladder is
shown on the right.
FREE BODY DIAGRAMS

To construct the diagram, the following procedure is adapted.
1. First draw the body being analyzed, separated from all other surrounding bodies and surfaces.
Pay close attention to the boundary, identifying what is part of the body, and what is part of the
surroundings.
2. Second, draw in all external forces and moments acting directly on the body.
Do not include any forces or moments that do not directly act on the body being analyzed.
Do not include any forces that are internal to the body being analyzed.
Some common types of forces seen in mechanics problems are:
(i) Gravitational force
(ii) Normal force or Reaction force
(iii) Joint reactions
(iv) Frictional force
(v) Tension in cable
CONSTRUCTING THE FREE BODY DIAGRAM

1. Gravitational Forces:
Unless otherwise noted, the mass of an
object will result in a gravitational weight
force applied to that body.
This weight is usually given in pounds in the
English system, and is modeled as 9.81 (g)
times the mass of the body in kilograms for
the metric system (resulting in a weight in
Newtons).
This force will always point down towards
the center of the earth and act on the center
of mass of the body.
Gravitational forces always act downward on the center of mass.
COM means center of mass in the figure.

2. Normal Forces (or Reaction Forces):
Every object in direct contact with the body will
exert a normal force on that body which
prevents the two objects from occupying the
same space at the same time.
Note that only objects in direct contact can
exert normal forces on the body. An object in
contact with another object or surface will
experience a normal force that is perpendicular
(hence normal) to the surfaces in contact.

3. Joint Reaction:
Joints or connections between bodies can also cause
reaction forces or moments, and we will have one force
or moment for each type of motion or rotation the
connection prevents.
The roller allows for rotation and movement along the
surface, but a normal force in the y direction prevents
motion vertically.
The pin joint allows for rotation, but normal forces in the
x and y directions prevent motion in all directions.
The fixed connection has a normal forces preventing
motion in all directions and a reaction moment
preventing rotation
(a) ROLLER JOINT (c) FIXED JOINT
(b) PIN JOINT

3. Friction Forces: Objects in direct contact
with the body can also exert friction forces
on the body, which will resist the two bodies
sliding against one another.
These forces will always be perpendicular to
the surfaces in contact.
Friction is the subject of an entire chapter,
but for simple scenarios we usually assume
rough or smooth surfaces.
(i) For smooth surfaces we assume that
there is no friction force.
(ii) For rough surfaces we assume that the
bodies will not slide relative to one
another no matter what. In this case the
friction force is always just large enough
to prevent this sliding.
NOTE: For a smooth surface we assume only a normal
force perpendicular to the surface. For a rough surface
we assume normal and friction forces are present.
(a) – Smooth surface in contact (b) – Rough surface in contact

4. Tension in Cables:
Cables, wires or ropes attached to the
body will exert a tension force on the
body in the direction of the cable.
These forces will always pull on the
body, as ropes, cables and other
flexible tethers cannot be used for
pushing.

SKETCH FREE BODY DIAGRAM
1000 N
10 N
50 N
(A) (B)

30 N
CABLE
PIN JOINT
SKETCH FREE BODY DIAGRAM
FREE BODY DIAGRAM OF ROLLER
FREE BODY DIAGRAM
OF LINK

FRICTION
Friction is a force that resists the movement of two contacting surfaces that slide relative to one another. 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.

The effects regarding friction can be summarized by referring
to the graph shown in Fig, which shows the variation of the
frictional force F versus the applied load P.
Here the frictional force is categorized in three different
ways:
• F is a static frictional force if equilibrium is maintained.
• F is a limiting static frictional force 𝐹𝑠when it reaches a
maximum value needed to maintain equilibrium.
• F is a kinetic frictional force 𝐹𝑘when sliding occurs at the
contacting surface.
Notice also from the graph that for very large values of P or
for high speeds, aerodynamic effects will cause 𝐹𝑘and
likewise μ𝑘to begin to decrease.
μ𝑘 are approximately 25 percent smaller than μ𝑠

As a result of experiments, the following rules apply to bodies subjected to dry friction.
• The frictional force acts tangent to the contacting surfaces in a direction opposed to the motion or tendency for
motion of one surface relative to another.
• The maximum static frictional force 𝐹𝑠that can be developed is independent of the area of contact, provided the
normal pressure is not very low nor great enough to severely deform or crush the contacting surfaces of the
bodies.
• The maximum static frictional force is generally greater than the kinetic frictional force for any two surfaces of
contact. However, if one of the bodies is moving with a very low velocity over the surface of another, 𝐹𝑘becomes
approximately equal to 𝐹𝑠, i.e., μ𝑠≃μ𝑘.
• When slipping at the surface of contact is about to occur, the maximum static frictional force is proportional to
the normal force, such that 𝐹𝑠= μ𝑠N.
• When slipping at the surface of contact is occurring, the kinetic frictional force is proportional to the normal
force, such that 𝐹𝑘= μ𝑘N.
CHARACTERISTICS OF DRY FRICTION

A truss is a structure composed of slender members joined together at their end points. The members are made of
wooden struts or metal bars.
The truss shown in Figure (a) is an example of a typical roof-supporting truss. In this figure, the roof load is transmitted to
the truss at the joints by means of a series of purlins.
Since this loading acts in the same plane as the truss, the analysis of the forces developed in the truss members will be
two-dimensional as shown in Figure (b)
SIMPLE TRUSSES

TRUSSES
m + 3 = 2j
m – No. of members
j – No. of joints
A
E
D
C
B
m = 7 and j = 5
m+3 = 2j
7 + 3 = 2 x 5
10 = 10
PIN/HINGE JOINT
ROLLER JOINT

A
B C D
E
F G H m = 11 and j = 8
m+3 = 2j
11 + 3 = 2 x 8
14 ≠ 16
Truss is DEFICIENT of
MEMBERS.
MEMBERS ARE LESS
WHILE JOINTS ARE
MORE.
Determine whether the truss given is a DEFICIENT TRUSS or NOT?
If deficient how will you make it to be PERFECT TRUSS?

A
B C D E
F G H m = 13 and j = 8
m+3 = 2j
13 + 3 = 2 x 8
16 = 16
Truss is PERFECT NOW
SUGGESTION TO MAKE DEFICIENT TRUSS INTO A PERFECT TRUSS
Add extra members BF and DH

A
B C D
E
F G H
m = 15 and j = 8
m+3 = 2j
15 + 3 = 2 x 8
18 ≠ 16
Truss is NOW OVER
RIGID (or) REDUNDANT
TRUSS
MORE MEMBERS AND
LESS JOINTS
OVER RIGID (or) REDUNDANT TRUSS
Add extra members BF and DH
Add extra member CF and CH

Truss analysis using the method of joints is greatly simplified if we can first identify those members which
support no loading.
These zero-force members are used to increase the stability of the truss during construction and to provide
added support if the loading is changed.
The zero-force members of a truss can generally be found by inspection of each of the joints.
ZERO FORCE MEMBERS
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.

1. ZERO FORCE MEMBERS IDENTIFICATION
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.
1. Members AB and AF are zero-force members
2. Members DC and DE are zero-force members

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.
Member DA is a zero-force member
Member CA is a zero-force member
Zero-force members support no load; however, they are necessary for stability, and are available when additional loadings are applied to the joints of the truss.
These members can usually be identified by inspection.
2. ZERO FORCE MEMBERS IDENTIFICATION

Using the method of joints, determine all the zero-force members of the Fink roof truss shown in Fig

SOLUTION
Look for joint geometries that have three members for which two are
collinear. We have
Joint G.
+↑ σ 𝐹𝑦 = 0
𝐹𝐺𝐶 = 0
Joint D.
𝐹𝐷𝐹 = 0

METHOD OF JOINTS
The following procedure provides a means for analyzing a truss using the method of joints.
• Draw the free-body diagram of a joint having at least one known force and at most two unknown forces.
(If this joint is at one of the supports, then it may be necessary first to calculate the external reactions at
the support).
• Use one of the two methods described above for establishing the sense of an unknown force.
• Orient the x and y axes such that the forces on the free-body diagram can be easily resolved into their x
and y components and then apply the two force equilibrium equations σ 𝐹𝑥 = 0 and σ 𝐹𝑦 = 0 . Solve for
the two unknown member forces and verify their correct sense.
• Using the calculated results, continue to analyze each of the other joints. Remember that a member in
compression “pushes” on the joint and a member in tension “pulls” on the joint. Also, be sure to choose a
joint having at most two unknowns and at least one known force.

Determine the reactions and the forces in each member of the truss shown in Figure.
A
E D
C
B
PIN/HINGE JOINT ROLLER JOINT
4 kN 2 kN
60o
60o 60o
60o
2 m 2 m

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
The reaction at the hinge support at
location A has 2 components acting
in the horizontal and vertical
directions. Since the applied loads
are only in VERTICAL direction, the
horizontal component of reaction
force at A becomes ZERO and there
will be only VERTICAL reaction RA
m = 7 and j = 5
m+3 = 2j
7 + 3 = 2 x 5
10 = 10
2 m 2 m

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
2 m 2 m
ESTIMATION OF SUPPORT REACTION
Taking moment about A
RC x 4 = (2 kN x 3) + (4 kN X 1)
RC = 2.5 kN
RA + RC = 4 kN + 2 kN
RA =6 kN – 2.5 kN
RA =3.5 kN

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
2 m 2 m
A B
MEMBER IS UNDER TENSION
MEMBER IS UNDER COMPRESSION
A
B

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
2 m 2 m
Consider JOINT A
Draw free body diagram of JOINT A
A
RA = 3.5 kN
60o
FAE
FAB
Equations of equilibrium are
Sum of all forces in X direction = 0
FAB - FAE cos 60 = 0
Sum of all forces in Y direction = 0
RA - FAE sin 60 = 0
Solving
FAE = 4.04 kN
FAB = 2.02 kN
If the magnitude of the forces are coming to be POSITIVE, then the
assumed direction of the forces are CORRECT.
MEMBER IS
UNDER TENSION
MEMBER IS
UNDER
COMPRESSION

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
2 m 2 m
Consider JOINT C
Draw free body diagram of JOINT C
C
RC = 2.5 kN
60o
FCD
FCB
FCD cos 60 - FCB = 0
RC - FCD sin 60 = 0
Solving
FCD = 2.88 kN
FCB = 1.44 kN

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
2 m 2 m
Consider JOINT B
Draw free body diagram of JOINT B
B
60o
FEB
FAB = 2.02 kN
FBC - FAB + FDB cos 60 + FEB cos 60 = 0
FDB sin 60 - FEB sin 60 = 0
Solving
FEB = 0.58 kN
FDB = 0.58 kN
60o
FDB
FBC = 1.44 kN

A
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
RA
RC
2 m 2 m
Consider JOINT D
Draw free body diagram of JOINT D
D
60o
2 kN
FED
FED - FBD cos 60 - FCD cos 60 = 0
Solving
FED = 1.73 kN
FCD = 2.88 kN
FBD = 0.58 kN

S.No Member Force
1 FAB 2.02 kN
2 FAE 4.04 kN
3 FCD 2.88 kN
4 FCB 1.44 kN
5 FEB 0.58 kN
6 FDB 0.58 kN
7 FED 1.73 kN
E D
C
B
4 kN 2 kN
60o
60o 60o
60o
2 m 2 m
A 2.02 kN
4.04 kN 2.88 kN
1.44 kN
0.58 kN 0.58 kN
1.73 kN
RA = 3.5 kN RC = 2.5 kN

Determine the force in each member of the truss shown in Fig

SOLUTION
Support Reactions. No joint can be analyzed until the support reactions are determined,
because each joint has at least three unknown forces acting on it. A free-body diagram of the
entire truss is given in Fig (b). Applying the equations of equilibrium, we have
The analysis can now start at either joint A or C. The choice is arbitrary since there are one
known and two unknown member forces acting on the pin at each of these joints.
Joint A. Fig. (c). As shown on the free-body diagram, FAB is assumed to be compressive and
FAD is tensile. Applying the equations of equilibrium, we have

Joint D. Fig. (d). Using the result for FAD and summing forces in the horizontal direction, Fig.(d),
we have
The negative sign indicates that FDB acts in the opposite sense to that shown in Fig.(d).* Hence,
FDB = 250 N (T) Ans
To determine FDC, we can either correct the sense of FDBon the free body diagram, and then apply
σ 𝐹𝑌 = 0 , or apply this equation and retain the negative sign for FDB, i.e.,
Joint C. Fig. (e).

When we need to find the force in only a few members of a
truss, we can analyze the truss using the method of sections.
It is based on the principle that if the truss is in equilibrium
then any segment of the truss is also in equilibrium.
The method of sections can be used to “cut” or section the
members of an entire truss.
If the section passes through the truss and the free-body
diagram of either of its two parts is drawn.
We can then apply the equations of equilibrium to that part to
determine the member forces at the “cut section.”
METHOD OF SECTIONS
Since only three independent equilibrium equations (σ 𝐹𝑥 = 0, σ 𝐹𝑦 = 0, σ 𝑀𝑜 = 0) can be
applied to the free-body diagram, we should try to select a section that, in general, passes
through not more than three members in which the forces are unknown.

PROCEDURE FOR ANALYSIS
The forces in the members of a truss may be determined by the method of sections using the following
procedure.
Free-Body Diagram.
• Make a decision on how to “cut” or section the truss where forces are to be determined.
• Before isolating the appropriate section, it may first necessary to determine the truss’s support reactions. If
this is done then the three equilibrium equations will be available to solve for member forces at the section.
• Draw the free-body diagram of that segment of the sectioned truss which has the least number of forces
acting on it.
Equations of Equilibrium.
• Moments should be summed about a point that lies at the intersection of the lines of action of two unknown
forces, so that the third unknown force can be determined directly from the moment equation.
• If two of the unknown forces are parallel, forces may be summed perpendicular to the direction of these
unknowns to determine directly the third unknown force.

Determine the force in members GE, GC, and BC of the truss shown in Fig.
SOLUTION
Section aa shown in Fig has been chosen since it cuts through the three members whose forces are to be determined. In
order to use the method of sections, however, it is first necessary to determine the external reactions at A or D. A free-
body diagram of the entire truss is shown in Fig. Applying the equations of equilibrium, we have

Free-Body Diagram. For the analysis the free-body diagram of the left portion of the
sectioned truss will be used, since it involves the least number of forces as shown in Fig

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