KTU BE 100 Engineering Mechanics

This is the draft version of Teaching Guide for the first year course BE 100 Engineering Mechanics of APJ Abdul Kalam Kerala Technological University

1. APJ ABDUL KALAM
KERALA TECHNOLOGICAL
UNIVERSITY


BE 100 ENGINEERING MECHANICS

Teaching Guide

Draft Version (Released in May 2017)








Prepared By
JINSHAD U
Assistant Professor
Al Ameen Engineering College

2. 1

SYLLABUS
Module Contents Hrs
Sem.
Exam
Marks
Page
No
I
Statics: Fundamental concepts and laws of mechanics - Rigid body -
Principle of transmissibility of forces
2
15%
3
Coplanar force systems - Moment of a force - Principle of moments 2
Resultant of force and couple system 4
Equilibrium of rigid body - Free body diagram - Conditions of
equilibrium in two dimensions - Two force and three force members.
3
II
Types of supports - Problems involving point loads and uniformly
distributed loads only.
5
15% 35
Force systems in space - Degrees of freedom - Free body diagram -
Equations of equilibrium - Simple resultant and Equilibrium problems.
4
III
Properties of planar surfaces - Centroid and second moment of area
(Derivations not required) - Parallel and perpendicular axis theorem -
Centroid and Moment of Inertia of composite area.
3
15% 63 Polar Moment of Inertia - Radius of gyration - Mass moment of inertia
of cylinder and thin disc (No derivations required).
2
Product of inertia — Principal Moment of Inertia (conceptual level). 3
Theorems of Pappus and Guldinus. 1
IV
Friction - Characteristics of dry friction - Problems involving friction of
ladder, wedges and connected bodies.
6
15% 51
Definition of work and virtual work - Principle of virtual work for a
system of connection bodies - Problems on determinate beams only.
4
V
Dynamics: Rectangular and Cylindrical co-ordinate system 1
20% 79
Combined motion of rotation and translation - Concept of
instantaneous centre - Motion of connecting rod of piston and crank
of a reciprocating pump.
4
Rectilinear translation - Newton’s second law - D’Alembert’s Principle -
Application to connected bodies (Problems on motion of lift only).
4
VI
Mechanical vibrations - Free and forced vibration - Degree of freedom. 1
20% 92 Simple harmonic motion - Spring-mass model - Period - Stiffness -
Frequency - Simple numerical problems of single degree of freedom.
7

Question Papers KTU

101

3. 2

INSTRUCTIONS TO TEACHERS / STUDENTS
1. This Teaching Guide is prepared based on the course BE 100 ENGINEERING MECHANICS syllabus
laid down by APJ Kerala Technological University and released in 2016.
2. This draft version of the Teaching Guide prepared based on the above syllabus is not meant for
the study of Engineering Mechanics in general.
3. The Guide can be referred by both teachers and students for browsing the topics in general.
However, in no way, can it replace a standard textbook or reference of study.
4. This course kit is intended for private circulation only and is not meant for publication by any
means. The material contains information, formulae, diagrams, pictures, questions, etc from
various textbooks, references and online resources. The author does not claim any right to the
contents of the course kit.
5. The contents and questions in the textbook have been set to help students understand core
concepts in a simple lucid manner.
6. The entire course is divided into modules which have been further sub-divided into chapters
based on the topics of study. The modules have been presented in an order based on an easy
method of teaching. Modules and Chapter have been given appropriate names for references.
7. The contents within boxes are meant for extra information and understanding. The matter
within the boxes may also include topics that I felt need to be discussed for better
understanding.
8. The questions have been set on increasing standard meant for better understanding of
concepts. The answer kit for the questions in this course kit is in the process of completion.
9. Since there are infinite number of practical problems associated with mechanics, students are
advised grasp the concept rather than blindly know how to solve a particular type of problem.
10. A list of the University question papers has been attached at the end to help students
understand the Question Pattern.
11. I am highly indebted to my colleagues, students and critiques for motivating me to continue
updating this material for the betterment of this study of Mechanics
12. Finally, I advise students to critically analyse the concepts of each topic and question them if it
does not ring your bells since there lays the true process of learning.
JINSHAD U

4. 3

MODULE 1 – STATICS 1
CHAPTER 1 – RIGID BODY : FORCES AND FORCE SYSTEMS
A. MECHANICS
Mechanics can be defined as the branch of physics concerned with the state of rest or motion of
bodies that subjected to the action of forces.
Archimedes (287–212 BC), Galileo (1564–1642), Sir Issac Newton (1642–1727) and Einstein (1878–1955) have
contributed a lot to the development of mechanics. Contributions by Varignon, Euler, D. Alembert are also
substantial. The mechanics developed by these researchers 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 and experimental findings
developed around the state of rest and state of motion of the bodies and put forth them in the form of three
laws of motion as well as the law of gravitation. The mechanics based on these laws is called Classical mechanics
or Newtonian Mechanics.
Albert Einstein proved that Newtonian mechanics fails to explain the behaviour of high speed (speed of light)
bodies. He put forth the theory of Relativistic Mechanics.
Schrödinger (1887–1961) and Broglie (1892–1965) showed that Newtonian mechanics fails to explain the
behaviour of particles when atomic distances are concerned. They put forth the theory of Quantum Mechanics.

Classifications of Classical Mechanics

5. 4

B. RIGID BODY
A rigid body may be defined as a body in which the relative positions of any two particles do not
change under the action of forces means the distance between two points/particles remain same
before and after applying external forces.
OR
A body which does not deform under the influence of forces is known as a rigid body.
For a rigid body, relative positions of A’B’ and AB remains same before and after the application of
forces

Note :Physical bodies deform slightly under the action of loads. If the deformation is negligible compared to its
size, the body is termed as rigid. In Rigid Body mechanics, we assume bodies to be rigid bodies.
Rigid Body Mechanics
Rigid Body Mechanics can be divided into two branches.
1. Statics: It is the branch of mechanics that deals with the study of forces acting on a body in
equilibrium. Either the body at rest or in uniform motion is called statics
2. Dynamics: It is the branch of mechanics that deals with the study of forces on body in motion is
called dynamics. It is further divided into two branches.
Force
Force may be defined as any action that tends to change the state of rest or motion of a body to which it is
applied.
The three quantities required to completely define force are called its specification or characteristics.
1. Magnitude
2. Point of application
3. Direction of application/Line of action
Force is a vector quantity and its unit is Newton (N) in S.I. systems and dyne in C.G.S. system.
Types Of Quantities
1. Scalar : Scalar quantities are those with which only a magnitude is associated. Examples of scalar

6. 5

quantities are time, volume, density, speed, energy, and mass
2. Vector : Vector quantities, on the other hand, possess direction as well as magnitude, and must obey the
parallelogram law of addition. Examples of vector quantities are displacement, velocity, acceleration,
force, moment, and momentum.

Effects of Force on a body
Whenever a force acts on a body or particle, it may produce some external as well as internal effects or changes.
• A force may change the state or position of a body by inducing motion of the body. (External effect)
• A force may change the size or shape of an object when applied on it. It may deform the body thus
inducing internal effects on the body.
• A force may induce rotational motion into a body when applied at a point other than its center of
gravity.
Line of action of force
The direction of a force is the direction, along a straight-line through its point of application in which the force
tends to move a body when it is applied. This line is called line of action of force.
Graphical Representation of force
Graphically a force may be represented by the segment of a straight line.

C. LAWS OF MECHANICS
The following are the fundamental laws of mechanics:
1. Newton’s Laws of Motion
2. Newton’s Law of Gravitation
3. Law of transmissibility of forces
4. Parallelogram law of forces

Newton’s Laws of Motion
Law 1 :A particle remains at rest or continues to move with uniform velocity if there is no unbalanced
force acting on it.

7. 6


Law 2 :The second law states that the rate of change of momentum of a body, is directly proportional
to the force applied and this change in momentum takes place in the direction of the
applied force.
Newton's second law of motion explains how an object will change velocity if acted by a
force. For a body with constant mass, the second law can also be stated in terms of an
object's acceleration. The acceleration of a particle is proportional to the vector sum of
forces acting on it and occurs along a straight line in which the force acts.


Law 3 :To every action there is always an equal reaction: or the mutual interactions of two bodies are
always equal but directed in opposite direction

8. 7

Law of Gravitation
Two particles will be attracted towards each other along their connecting line with a force whose
magnitude is directly proportional to the product of the masses and inversely proportional to the
distance squared between the particles.

Triangular law of forces
“If the two forces acting on a body are represented in magnitude and direction as two sides of a
triangle in order then the third side or the closing side of the triangle would be the resultant in
opposite order.”

Polygon Law of Forces
If a number of concurrent forces acting on a rigid body are represented in magnitude and direction
by the sides of a polygon, taken in order, then their resultant is represented in magnitude and
direction by the closing side of polygon taken in reverse order.


Law of Transmissibility of Force
Principle of transmissibility states that a force may be applied at any point on a rigid body along its
given line of action without altering the effects of the force on which it acts.
For example, the force F acts on a rigid body at point A. According to the principle of transmissibility of
forces, this force has the same effect on the body as a force F applied at point B along its line of action.

9. 8


Examples

Fig : Rigid Bodies (valid)

Fig : Deformable Bodies (not valid)
Note :In engineering mechanics we deal with only rigid bodies. If deformation of the body is to be considered in
a problem, the law of transmissibility of forces will not hold good.
Parallelogram Law of Forces
If two forces acting at a point be represented in magnitude and direction by the two adjacent sides of
a parallelogram, then their resultant is represented in magnitude and direction by lite diagonal of
the parallelogram passing through that point

Analytical Proof

10. 9

Given OA = P force and OB = Q force. Construct parallelogram OBCA as shown and drop perpendicular
CD on extension of OA.
Thus, AD= Q cos θ,CD=Q sin θ
OCD is a right angled triangle,
:. OC
2
=OD
2
+CD
2

R
2
= (P + Q cos θ)
2
+ (Q sin θ)
2

= P
2
+ Q
2
cos
2
θ + 2PQ cos θ + Q
2
sin
2
θ
= p
2
+ Q
2
(cos
2
θ + sin
2
θ)+ 2PQ cos θ.
= P
2
+ Q
2
+ 2PQ cos θ [since cos
2
a+ sin
2
a= 1]

Direction of R :
Tanα = CD/OD =Q sin a/ P+Q cos θ

Note : Parallelogram Law is valid for all vectors and hence forces also
D. FORCE SYSTEMS
When several forces of different magnitude and direction act upon a body or particle, it is called
force system.


TYPES OF FORCE SYSTEMS
1. Collinear Force System : When a system of forces act along the same line, they are called
collinear forces.
Practical Example : Rope being pulled by 2 persons

11. 10

2. Coplanar and Non Coplanar Force Systems
a. Coplanar Force Systems: When the system of forces are in a plane, it is called coplanar
system of forces.

Practical Example : Forces in a truss or beam

b. Non-Coplanar Force System: When the system of forces are not in the same plane, it is
called non-coplanar system of forces.

Practical Examples : Body hanging from cables

Concurrent Non-Concurrent

12. 11

3. Concurrent and Non Concurrent Force System: A concurrent force system contains forces
whose lines-of action meet at same one point. They may either be
a. Coplanar Concurrent System

Practical Examples : Force acting on a gusset plate or hook.


b. Non-Coplanar Concurrent System

Practical Examples : Body hanging from cables

4. Parallel Force Systems: Forces whose line of action are parallel to each other are called parallel
force system. They are of 2 types

13. 12

a. Like Parallel Forces

Practical Examples : Force acting on the signals post or legs of table

Coplanar Like Parallel Non -Coplanar Like Parallel
b. Unlike Parallel Forces

Practical Example : Force applied on steering wheel or raft concrete foundation

Coplanar Unlike Parallel Non -Coplanar Unlike Parallel

14. 13

CHAPTER 2 – RESULTANT OF FORCE SYSTEMS, MOMENT & COUPLE
Resultant
Resultant is a single force that will replace a system of forces and produces the same effect on the
rigid body as that of the system of forces.
A. RESULTANT OF TWO CONCURRENT FORCES - Parallelogram Law of Forces
Sign Conventions
The following sign conventions shall be used throughout the book
1. Upward forces are considered as positive, whereas the downwards as negative.
2. Forces acting towards right are considered as positive, whereas those towards left as negative.
PROBLEMS
1. Find the magnitude of the two forces, such that if they act at right angles, their resultant is √10 N.
But if they Act at 60°, their resultant is √13 N.
2. The greatest and least resultants of two forces F1 and F2 are 17 N and 3 N respectively.
Determine the angles between them when their resultant is √149 N
3. A screw eye is subjected to two forces F1 and F2 as shown in figure. Determine the magnitude
and direction of the resultant force by parallelogram method

4. The two structural members, one of which is in tension and the other in compression, exert the
indicated forces on joint O. Determine the magnitude of the resultant R of the two forces and the
angle which R makes with the positive x-axis.

15. 14

Resolution Of Forces
The replacement of a single force by a several components which will be equivalent in action to the
given force is called resolution of a force.
Forces can be resolved in any 2 directions. However, it is convenient to resolve them into the two
orthogonal components (mutually perpendicular directions)
Resolution of Coplanar Forces in Rectangular Coordinates

B. RESULTANT OF CONCURRENT COPLANAR FORCE SYSTEMS
Procedure
1. Resolve all the forces into x and y components
2. Add the components of forces along the x and y axes with proper sense of direction.
3. Find the resultant and inclination of the forces
=

16. 15

Note : 1. If both FRx and FRy are positive, the resultant lies in the first quadrant
2. If both FRx and FRy are negative the resultant lies in the third quadrant
3. If FRxis positive and FRy is negative, the resultant lies in the fourth quadrant
4. If FRxis negative and FRy is positive, the resultant lies in the second quadrant

PROBLEMS
1. Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

2. If the magnitude of the resultant force is to be 9 kN directed along the positive x axis, determine
the magnitude of force T acting on the eyebolt and its angle.

3. Determine the resultant of the 3 forces acting on the bracket and its direction.

17. 16


4. The forces 20 N, 30 N, 40 N, 50 N and 60 N are acting at one of the angular points of a regular
hexagon, towards the other five angular points, taken in order. Find the magnitude and direction
of the resultant force.

5. If Φ = 30 and the resultant force acting on the gusset plate is directed along the positive x axis,
determine the magnitudes of F2 and the resultant force.

18. 17

Problems for Practice
1. Determine the resultant of the forces shown below

2. Determine the resultant of the forces acting on the ring shown in figure.

3. Find the resultant of the three concurrent forces as shown on figure.

4. Find the magnitude and direction of the resultant of the following forces.
i. 20 N inclined at 30° towards North of East.
ii. 25 N towards North.
iii. 30 N towards North West and
iv. 35 N inclined at 40° towards South of West.

19. 18

Moment
The tendency of a force to rotate the body in the direction of its application a force about a
point that is not on the line of action of the force is called Moment of force or simply moment.
Moment is also referred to as torque.

Scalar Formulation
Moment is a vector quantity whose direction is perpendicular to the plane of the body. The right-hand
rule is used to establish the sense of direction of moment. Throughout the text, clockwise moments
are taken as positive while anti-clockwise as negative.
Eg: Force applied perpendicular to the handle of wrench to rotate a pipe


Varignon’s Theorem or Principle of Moments
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.”

20. 19

PROBLEMS
1. Determine the moment produced by the force about point O.

2. Determine the resultant moment produced by the forces about point O.


3. Determine the moment of the force in figure about point O

Problems for Practice
1. Calculate the moment of the 250-N force on the handle of the monkey wrench about the center
of the bolt.

21. 20


2. Determine magnitude of moment about point A and C

Couple
The moment produced by two equal, opposite, and non-collinear forces is called a couple. The
perpendicular distance between the lines of action of the two and opposite parallel forces is known as
arm of the couple.


A couple can be represented by a vector with magnitude and direction equal to the moment of the
couple.

22. 21

Practical Examples : Force apllied to a handle of steering wheel


Differences between Moment and Couple
Moment Couple
1 Moment is the tendency of force to rotate a
body with the given point or axis
2 It is produced by forces not passing through
point of rotation axis
3 There is a resultant force acting on the body
in the direction of force and rotate the body.
4 To balance the force causing moment, equal
and opposite force is required.
5 For example,
To tight the nut by spanner
To open or close the door

1 Two equal and opposite forces whose lines of
action are different form a couple
2 It is produced by the two equal and opposite
parallel, non collinear forces.
3 Resultant force of couple is zero. Hence, body
does not move, but rotate only.
4 Couple cannot be balanced by a single force,
it can be balanced by a couple only.
5 For example,
To rotate the key in lock
To open or close the wheel valve of water line
To rotate the steering wheel of car.

C. RESULTANT OF COPLANAR NON-CONCURRENT FORCE SYSTEMS
A system of several forces and couple moments acting on a body can be reduced to equivalent single
resultant force acting at a point O and a resultant couple moment.

23. 22

X-Y Intercepts
Intercepts are the points/coordinates where the line of action meets the corresponding axes.

PROBLEMS
1. Determine and locate the resultant R of the forces acting on the beam.

2. Determine and locate the resultant R of the two forces and one couple acting on the beam.

3. Forces act on the sides of a regular hexagon of side 2 m as shown in figure. Determine the
resultant and its position from the centre of the hexagon

24. 23

4. Determine the resultant of the force

5. A dam is subjected to three forces as shown in figure. Determine the single equivalent force and
locate its point of intersection with the base AC, assuming all forces lie in the same plane.

6. Determine the resultant of the forces and couple which act on the plate shown. Also find the x
and y intercepts about point O.

Problems for Practice
1. Four parallel forces of magnitudes 100 N, 200 N, 50 N and 400 N are shown in Figure. Determine
the magnitude of the resultant and also the distance of the resultant from the A.

25. 24


2. ABCD is a square , each side being 20 cm and E is the middle point of AB. Forces of magnitude 7,
8, 12, 5, 9 and 6 kN act on lines of directions AB, EC, BC, BD, CA and DE respectively. Find the
magnitude and direction of resultant force.
3. A particle is acted upon by three forces equal to 50 N, 100 N and 130 N, along the three sides of
an equilateral triangle, taken in order. Find the magnitude and line of action of the resultant
force.

4. Four forces act on a rectangle as shown in figure. Find the magnitude and direction of the
resultant force. Also find the intersection of line of action of the resultant with x and y axes
assuming D as origin

5. Determine the resultant of a force system acting tangential to the circle of radius 1 m, as shown
in figure. Also find its direction and line of action.

26. 25


D. RESOLUTION OF FORCE INTO FORCE AND COUPLE SYSTEM
Suppose we have to sift he force from point A to B. The procedure to be followed
1. Apply 2 equal and opposite force at point A parallel to force B of the same magnitude
2. If the points are separated by a distance d, the opposite forces F and –F form a couple retaining
force F at point B in the same direction as A


PROBLEMS
1. Replace the force system acting on the beam by an equivalent force and couple at point B.

2. Reduce the following force system into
a) A single force

27. 26

b) Resultant force and couple acting at point A.
c) Resultant force and couple acting at point B.
d) Resultant force and couple acting at point C.

3. Replace the force system by a resultant force and couple moment at point O.

4. Two coplanar forces P and Q are shown in figure. Assume all squares of the same size.
i) If P = 4 kN, find the magnitude and direction of Q if their resultant passes through E
ii) If Q = 110 kN, find the magnitude and direction of P if their resultant passes through F

28. 27

Problems for Practice
1. A system of parallel forces are acting on a rigid bar as shown in Figure Reduce this system to:
a) a single force,
b) a single force and a couple at A
c) a single force and a couple at B.

2. A 2 m X 4 m plate is subjected to a system of three coplanar forces as shown in figure. Determine
the equivalent action at O which may replace the force system

3. Figure shows two vertical forces and couple of moment acting on a horizontal rod fixed end.
i) Determine the resultant of the system
ii) Determine an equivalent system through the fixed support

29. 28

CHAPTER 3 - EQUILBRIUM OF RIGID BODIES (COPLANAR CONCURRENT SYSTEM)
Types of Forces
Forces that act on a body can be divided into two general categories—
 Applied forces (action)
 Reactive forces (or, simply reactions)

Reaction
Reaction is the opposing force that a support offers whenever it is acted upon by external or inherent
forces.
Free Body Diagram
Free body diagram is a diagram in which a rigid body is isolated from the system and all active forces
applied to the body and reactive forces as a result of mechanical contact are represented.
Examples
Body resting a smooth surface

Suspended ball resting against a smooth wall

Weight suspended from a string

Body resting in groove


Steps for Drawing Free Body Diagram
1. A sketch of the body is drawn assuming that all supports (surfaces of contact, supporting
cables, etc.) have been removed.
2. All applied forces (including weight) and support reactions are drawn and labeled on the
sketch.
3. Apply the weight of the body to its center of gravity (if it is uniform, then apply it to the
centroid). If the sense of a reaction is unknown, it should be assumed
PROBLEMS
1. Draw FBD of the following crate at point As

30. 29


2. Draw the Free Body Diagram of the three spheres A, B and C shown in figure

A. EQUILBRIUM OF RIGID BODIES
A rigid body is said to be in equilibrium if the resultant of all external and reactive forces and
moments acting on it is zero.
Lami’s Theorem
If three coplanar concurrent forces acting on a body keep it in equilibrium, then each force is
proportional to the sine of the angle between the other two”

Note: Lami’s theorem is applicable only to 3 coplanar concurrent forces in equilibrium

Proof
By applying polygon law of forces, draw triangle OAB representing the system of forces shown with
external angles as indicated

31. 30


Applying Sine Law

General Equations of Equilibrium
1. The algebraic sum of all forces in a force system is zero.

2. The algebraic sum of all moments in a force system is zero.

Equations of Equilibrium For Coplanar Systems

32. 31

Solving Equilibrium Problems
1. Draw proper Free- Body Diagram
2. Resolve all the forces into x and y components
3. Apply Equilibrium conditions along the x and y directions
4. Solve the resultant algebraic equations
In case of moments, try to select the point you take moments around such that the line of action of at
least one unknown force passes through that point. This will eliminate one unknown from your
moment equation and will result in simpler equations to work with.
PROBLEMS
1. Three cables are joined at the junction ring C. Determine the tensions in cables AC and BC caused
by the weight of the 30-kg cylinder.

2. Determine the tension in the cables AB, BC and CD necessary to support the 10-kg and 15-kg
traffic lights at B and C, respectively. Also, find the angle θ.

3. Two smooth pipes, each having a mass of 300 kg, are supported by the forked tines of the
tractor. Find the reactions at the points of contact.

33. 32


4. Two smooth cylinders A and B each of weight 400 N and radius 20 cm are connected at their
centres by a link of length 80 cm and rest upon a horizontal plane as shown in Figure. The
cylinder C above has a weight of 800 N and radius 30 cm. Find the force in the string and pressure
produced in the floor at the points of contact.

5. Find the reactions at the points of contact if the weights of he balls A,B and C are 100 kN, 80 kN
and 120 kN respectively

6. A wheel wheghing 1000 N rest on a block on an inclined plane 30
0
. The height of the block is 10
cm and the diameter of the ball is 60 cm. Find the force that will move the with minimum itensity

34. 33


Problems for Practice
1. A light string whose extremity A is fixed, has weights W1 and W2 attached to it at B and C. It
passes round a small smooth peg at D carrying a weight of 300 N at the free end. Find
i. Tensions in the portion AB, BC and CD of the string
ii. Magnitudes of W1 and W2.

2. Two vertical masts AB and CD are guyed by the wires BF and DG, in the same vertical plane and
connected by a cable BD of length l = 20 m, from the middle point E of which is suspended a load
Q of 100 N as shown in figure. Find the tensile force S in each of the two guy wires BF and DG if
the sag d = 1 m.

3. A smooth circular cylinder of radius 1.5 meter is lying in a triangular groove, one side of which
makes 15° angle and the other 40° angle with the horizontal. Find the reactions at the surfaces of
contact, if there is no friction and the cylinder weights 100 N.

35. 34

4. Block P = 5 kg and Block Q of mass m kg are suspended through a chord, which is in equilibrium
as shown in figure. Determine the mass of block Q.

5. Three cylinders weighting 100 N each and of 80 mm diameter are placed in a channel of 180 mm
width as shown in figure. Determine the reactions at points of contact

6. A rope of 9 m long is connected at points A and B, two points in the same level, 8 m apart. A load
of 300 N is suspended from a point C on the rope, 3 m from A. What load connected to a point D,
on the rope, 2 m from B is necessary to keep portion CD parallel to AB

36. 35

MODULE 2 – STATICS 2
CHAPTER 4 - EQUILBRIUM OF RIGID BODIES (COPLANAR NON-CONCURRENT)
A. BEAMS
Beams are the best examples of a system of coplanar non-concurrent systems. We will restrict our
study only to finding reactions of such determinate beams
Determinate and Indeterminate structures
1. Determinate Structures : Structures in which the equilibrium equations are sufficient to find the
forces and stresses associated with the structure
r = 3n
2. Indeterminate Structures : Structures in which the equilibrium equations alone are not
sufficient to find the forces and stresses associated with the structure
r > 3n
where r - no. of reactions and n - no. of equilibrium equations
Type Of Supports
1. Simple Support : If one end of a beam simply rests on a support, the support is known as simple
support. The beam is free to slide and rotate at simple support and the reaction is
perpendicular to the surface. Eg : Scale resting on two tables
2. Roller Support : The end which is supported on a frictionless roller to permit contraction and
expansion due to temperature or lateral forces is known as roller support. The reaction in roller
support is normal to the surface.

3. Hinged/Pined Support : This support does not permit movement along any direction but is free
to rotate. The reaction components are taken along horizontal and vertical directions. Eg –
Hinges of doors and windows

4. Fixed Support : At fixed support, the beam is not free to slide or rotate along any direction. The
reaction components include horizontal, vertical and moment.

37. 36


Type of Support
Roller Support

Hinged/Pin Support

Fixed Support

Types of Loadings
1. Concentrated load or Point Load: If a load acts at a point, either horizontal, vertical or inclined,
it is called a concentrated load or point load. If the load acts at a very small area, it is taken as a
point load. It is expressed in N or Kn.

2. Uniformly Distributed Load :This type of load is spread over acertain length with the intensity of
load being constant. It is expressed in N/m or kN/m. The resultant load R=wL acts at a midpoint
of udl.

3. Uniformly Varying Load: This type of load is spread over a certain length with the intensity of
load varying linearly. The resultant load R=wL/2 acts at a distance of 1/3 from point of action of
w N/m.

4. Pure Moment : A pure moment force or couple can also act on beams. The unit of moment is
kNm.

38. 37

PROBLEMS
1. Determine the reactions at VA and VB

2. Determine the location of the two supports so that both the reactions are equal

3. Find the reactions at the supports

4. Find the reactions at the supports

5. Fid support reactions

39. 38

6. Find the reactions at the supports

7. A 5 m bar of negligible weight rests in a horizontal position on smooth planes as shown in figure.
Determine the load P and reactions at supports

Problems for Practice
1. A simply supported beam AB of span 4m is carrying point loads 5kN, 2kN and 3kN at 1 m, 2m and
3m respectively from the support A. Calculate the support reactions at A and B.
2. Find the reactions at the supports


8. Determine reactions at supports of the following beams

40. 39



3. Find the tension in the cable at A and reactions at C

B. EQUILBRIM OF GENERAL COPLANAR SYSEMS
Apart from beams there are a number of coplanar systems whose reactions is of interest. Here is
one question put forth for those who wish to solve it.
1. Calculate the force and moment reactions at the bolted base O of the overhead traffic-signal
assembly. Each traffic signal has a mass of 36 kg, while the masses of members OC and AC are
50 kg and 55 kg, respectively. The mass center of member AC is at G.

41. 40

CHAPTER 5–RESULTANT & EQUILBRIUM OF FORCE SYSTEMS IN SPACE (NON-
COPLANAR)
Representation of Force
There are two ways of representation of force. The method used depends on the type of problem
being solved and the easiest approach to finding a solution.
1. Scalar Notation
2. Vector Notation
Vector Notation Of Forces
1. Two Dimensional Force Systems (Coplanar Forces)
It is also possible to represent the x and y components of a force in terms of Cartesian unit
vectors i and j.

where the scalars Fx and Fy are the x and y scalar components of the vector F.
2. Three Dimensional Force Systems(Non-Coplanar Forces)

Unit Vector
Vectors having unit magnitude and represents only the direction of vectors is called a unit vector. It is
usually denoted by n.

42. 41

A vector V may be expressed mathematically by multiplying its magnitude V by a vector n whose
magnitude is one and whose direction coincides with that of V.

The unit vectors along the Rectangular Coordinate axis x, y and z are

PROBLEMS
1. A force vector F = 700i + 1500j is applied to a bolt. Determine the magnitude of force and the
angle it forms with the horizontal.
2. A force of 500 N forms angles 60
0
, 45
0
and 120
0
respectively with x, y and z axes. Write the vector
form of force.
Position Vector
A position vector r is defined as a fixed vector which locates a point in space relative to another
point.
a) Position Vector of P relative to origin

b) Position vector of B with respect to A

43. 42



A. RESULTANT OF NON COPLANAR FORCES -By Vector Notation
1. Resultant of Non Coplanar Concurrent Forces
In vector notation, the scalar components of the resultant vector can be obtained by adding
algebraically the sum of the corresponding scalar components of the force vectors.


where


PROBLEMS
1. Express following 900 N force in vector notation

2. A force vector is represented by a lie AB. The coordinates of point A are (2,4,3) and of point B are
(1,-5,2). If the magnitude of force is 10 N, determine
i. Specify the force vector

44. 43

ii. Components of force along x, y and z axis
iii. Angles with the x, y and z axis
3. Determine the magnitude and the coordinate direction angles of the resultant force acting on the
ring shown below.

4. Determine the magnitude of the resultant force at A.

Problems for Practice
1. A force of magnitude 50 kN is acting along the line joining A(2,0,6) and B(3,-2,0). Write the vector
form of the force.
2. Three forces 500 kN, 700 kN and 800 kN are acting along the three diagonals of adjacent faces of
a cube of size 3m as shown in figure. Determine the resultant

45. 44


B. MOMENT AND COUPLE – In Vector Notation
Moment
Moment is a vector quantity whose direction is perpendicular to the plane of the body. The right-hand
rule is used to establish the sense of direction of moment.
Vector Formulation

Varignon’s Theorem
Vector Formulation
Couple
Vector Formulation
Couple vectors are free vectors, i.e., the point of application is not significant.

46. 45

Note : Cross Product of vectors


PROBLEMS
1. Determine the magnitude of the moment of the force about the y axis.

2. Determine the moment of the forces about point O

Problems for Practice
1. A force of 360 N acts from point A to B as shown in figure. Find moment of the force C. (Points A
and C are in Y-Z and X-Z planes respectively)

47. 46


2. A tension T of magnitude 10 kN is applied on the cable attached to the top of a rigid mast and
secured to the ground at B as shown in figure. Determine the moment of tension force about z
axis passing through the base point O.

2. Resultant of Non Coplanar Non-Concurrent Forces
PROBLEMS
1. A table exerts the four forces shown on the floor surface. Reduce the force system to a force–
couple system at point O. Determine the resultant of the following force and its location

2. Replace the two forces acting on the post by a resultant force and couple moment at point O.

48. 47

Express the results in Cartesian vector form.

Problems for Practice
1. A rectangular concrete slab support loads at its four corners as shown in figure. Determine the
resultant of the forces and the point of application of the resultant.

2. A rectangular block is subjected to three forces as shown in figure. Reduce them into an
equivalent force couple system acting at A.

49. 48

C. EQUILIBRIUM OF NON COPLANAR FORCES – By Vector Notation

1. Equilibrium of Non-Coplanar Concurrent Forces
In vector notation, the equation of equilibrium can be summarized as

50. 49

PROBLEMS
1. A balloon is moored by three cables as shown. The lift on the balloon is 600 N. Draw the FBD of
the balloon and write a vector expression for the tension in each cord.

2. The square steel plate has a mass of 1800 kg withmass center at its center G. Calculate the
tension ineach of the three cables with which the plate islifted while remaining horizontal.

3. Detemine tension in the cables AB, AC and AD. All dimensions are in meters.

51. 50

Problems for Practice
1. A tripod supports a load of 2 kN as shown in figure. The ends A, B and C in the x-z plane. Find the
forces in the three legs of the tripod

2. Equilibrium of Non-Coplanar Non-Concurrent Forces
This following question is just to rise your curiosity regarding equilibrium of non-coplanar non
concurrent systems.
PROBLEMS
1. The vertical mast supports the 4-kN force and is constrained by the two fixed cables BC and BD and by
a ball-and-socket connection at A. Calculate the tension in BD.

52. 51

MODULE 4 – STATICS 3
CHAPTER 6 – FRICTION OF RIGID BODIES
A. FRICTION
In the preceding chapters we have usually assumed that the forces of action and reaction between
contacting surfaces act normal to the surfaces. This assumption is valid only between smooth surfaces.
No surface is perfectly smooth, a surface contains microscopic “bumps”. When two surfaces are
moving against each other, these bumps interlock against each other and oppose the motion. Friction
originates from this opposition.
Types of Friction
(a) Dry Friction : Dry friction occurs when the unlubricated surfaces of two solids are in contact
under a condition of sliding or a tendency to slide.
(b) Fluid Friction : Fluid friction occurs when adjacent layers in a fluid (liquid or gas) are moving
at different velocities.
(c) Internal Friction : Internal friction occurs in all solid materials which are subjected to cyclical
loading. For highly elastic materials the recovery from deformation occurs with very little loss
of energy due to internal friction.
Types of Dry Friction
1. Static Friction : It is the friction experienced by a body when it is at rest or on the verge of
motion. Eg – Body resting on an ramp with small inclination
2. Dynamic Friction : It is the friction experienced by a body when it is in motion. Eg – Resisting
force experienced by a rolling skater
Static Friction Dynamic Friction
1. Static friction is the frictional force acting between
two surfaces which are attempting to move, but
are not moving. The body remains in equilibrium
2. Static friction is proportional to the external
forces and increases linearly with the force
applied until it reaches a maximum value.
3. Static friction could have a value less or greater
than the value for kinetic friction. But it cannot
increase beyond a maximum value called the
limiting value of frictional force.
4. The coefficient of static friction is less than the
dynamic friction
5. Eg : Body resting on an ramp with small inclination
1 Kinetic friction is the frictional force acting
between two surfaces which are in motion
against each other

2. Kinetic friction remains constant regardless
of the force applied. It is independent of
mass and acceleration.
3. Kinetic friction has a value less than the
limiting value of static friction.



5. Eg : Resisting force experienced by a rolling
skater

53. 52


Laws of Coloumb’s Friction (Dry Friction)
1. When two bodies are in contact, the direction of force of friction on one of them is opposite to
the direction in which this body has a tendency to move tangent to the surface.
2. The frictional force is independent of the area of contact of the surfaces.
3. The force of friction is dependent upon the types of materials of the two bodies in contact.
4. The limiting frictional force bears a constant ratio with the normal reaction.
5. When one body is just on the verge of sliding over another body, the force of friction is
maximum and this maximum frictional force is called Limiting Static Frictional Force
6. Limiting static frictional force is greater than kinetic frictional force for any two surfaces of
contact.
7. The kinetic frictional force is independent of the relative velocities of the bodies in contact.
Limiting Force of Friction
The maximum value of frictional force at the surfaces of contact when the body is at the verge of
motion is called limiting static frictional force (Fs).
When this value is reached, the block is in unstable equilibrium since any further increase in P will
cause the block to move. It is experimentally found that friction directly varies as the applied force until
the movement produces in the body.

Where Fs = Limiting static frictional force
µs = coefficient of static friction
N = Normal reaction

54. 53


Coefficient of static friction
It is defined as the ratio of limiting force of friction (Fs) to the normal reaction (N) between the two
bodies. It is constant denoted by µs
µs = =
Angle of friction
It is defined as the angle made by the resultant of the normal reaction and frictional force (R) to the
normal reaction (N). It is denoted by φ

Cone of friction
It is defined as the right circular cone with vertex at the point of contact of the two surfaces, axis in
the direction of the normal reaction (N) and semi-vertical angle equal to the angle of fiction φ

Angle of repose
It is defined as the maximum inclination of a plane at which a body remains in equilibrium over the
inclined plane by assistance of friction alone.

55. 54

Proof

B. FRICTION OF CONNECTED BODIES
PROBLEMS
1. A pull of 20 N, inclined at 25
0
to the horizontal plane is required to just move a body placed on a
rough horizontal plane. But the push required to move the body is 25 N inclined at 25
0
. Find the
weight of the body and the coefficient of friction.
2. A horizontal force P=100 N is just sufficient to hold the crate from sliding down the plane, and a
horizontal force of is required to just push the crate up the plane. Determine the coefficient of
static friction between the plane and the crate, and find the mass of the crate.

3. Crates A and B weigh 500 N and 1000 respectively. They are connected together with a cable and
placed on the inclined plane. If the angle is gradually increased, determine when the crates begin

56. 55

to slide. The coefficient of friction between the inclined plane and the block A is 0.25 and that for
the B is 0.35.

4. Two identical blocks of weight W are supported by a rod inclined at 45° with the horizontal as
shown in figure. If both the blocks are in limiting equilibrium, find the coefficient of friction (μ),
assuming it to be the same at floor as well as at wall.

5. Two blocks A and B, connected by a horizontal rod and frictionless hinges are supported on two
rough planes as shown in Figure. The coefficients of friction are 0.3 between block A and the
horizontal surface, and 0.4 between block B and the inclined surface. If the block B weighs 100 N,
what is the smallest weight of block A, that will hold the system in equilibrium?

57. 56

Problems for Practice
1. An inclined plane as shown in Figure is used to unload slowly a body weighing 400 N from a truck
1.2 m high into the ground. The coefficient of friction between the underside of the body and the
plank is 0.3. State whether it is necessary to push the body down the plane or hold it back from
sliding down. What minimum force is required parallel to the plane for this purpose?

2. Two blocks A and B of weights 1 kN and 2 kN respectively are in equilibrium position as shown in
Figure. If the coefficient of friction between the two blocks as well as the block B and the floor is
0.3, find the force (P) required to move the block B.

3. Determine the least value of the force P to cause motion to impend rightwards. Assume the co-
efficient of friction under the blocks to be 0.2 and pulley to be frictionless.

4. What should he the value of the angle θ in Figure so that the motion of the 90 N block impends
down the plane? The co-efficient of friction for all the surfaces is 1/3.

58. 57


C. LADDER FRICTION
PROBLEMS
1. A uniform ladder of weight 100 kN rests against a smooth wall at B and the end A rests on the
rough horizontal plane for which the coefficient of static friction is 0.3. Determine the angle of
inclination of the ladder, the frictional force at A and the normal reactions at A and B

2. A uniform ladder 3 m long weighs 200 N. It is placed against a wall making an angle of 60° with
the floor as shown in figure. The coefficient of friction between the wall and the ladder is 0.25
and that between the floor and ladder is 0.35. The ladder, in addition to its own weight, has to
support a man of 1000 N at its top at B. Calculate:
a. The horizontal force P to be applied to ladder at the floor level to prevent slipping.
b. If the force P is not applied, what should be the minimum inclination of the ladder with the
horizontal, so that there is no slipping of it with the man at its top.

59. 58



3. A ladder of length 5m and weight 120 N is placed on a flat floor against a vertical wall. If the
coefficient of friction are 0.3 and 0.2 along the interface of floor and wall respectively, determine
the smallest angle θ with the horizontal so that the ladder can be placed without slipping.
4. Determine the distance s to which the 90-kg painter can climb without causing the 4-m ladder to
slip at its lower end A. The top of the 15-kg ladder and at the ground the coefficient of static
friction is 0.25.

5. A uniform ladder of 4 m length rests against a vertical wall with which it makes an angle of 45°
with the vertical. The coefficient of friction between the ladder and the wall is 0.4 and that
between ladder and the floor is 0.5. If a man, whose weight is one-half of that of the ladder
ascends it, how high will it be when the ladder slips?
Problems for Practice
1. A ladder shown in Figure is 4 m long and is supported by a horizontal floor and vertical wall. The
coefficient of friction at the wall is 0.25 and that at the floor is 0.5. The weight of the ladder is 30

60. 59

N and is considered to be concentrated at G. The ladder also supports a vertical load of 150 at C.
Compute the least value of (α) at which the ladder maybe placed without slipping to the left.

2. A ladder 5 meters long rests on a horizontal ground and leans against a smooth vertical wall at an
angle 70° with the horizontal. The weight of the ladder is 900 N and acts at its middle. The ladder
is at the point of sliding, when a man weighing 750 N stands on a rung 1.5 m from the bottom of
the ladder. Calculate the coefficient of friction between the ladder and the floor.
D. WEDGE FRICTION

PROBLEMS
1. A block weighing 1500 N, overlying a 10° wedge on a horizontal floor and leaning against a
vertical wall, is to be raised by applying a horizontal force to the wedge. Assuming the coefficient
of friction between all the surfaces in contact to be 0.3, determine the minimum horizontal force
required to raise the block.

61. 60


2. Determine the smallest horizontal force P required to pull out wedge A. The crate has a weight of
300 N and the coefficient of static friction at all contacting surfaces is 0.3. Neglect the weight of
the wedge.

3. A 15° wedge (A) has to be driven for tightening a body (B) loaded with 1000N weight as shown in
figure. If the angle of friction for all the surfaces is 14°, find graphically the force (P), which should
be applied to the wedge.

4. Two Blocks are resting on the wall as shown in figure. Find the range of value of force P applied
to the lower block for which the system remains in equilibrium. Coefficient of friction is 0.25 at
the floor, 0.3 at the wall and 0.2 between the blocks

62. 61

CHAPTER 7 – VIRTUAL WORK
Virtual work
If a body is in equilibrium under the action of a system of forces, the work done is zero since there are
no displacements. But if we assume that a body in equilibrium undergoes small imaginary
displacements (known as virtual displacements) consistent with the geometrical conditions,
imaginary work is said to be done by the system of forces. This imaginary work done is called virtual
work.
The body A shown below is in equilibrium under the action of forces. However it assumed that it
undergoes an imaginary displacement in the direction of force F.
Displacement in direction of force = Δs cos α
Virtual work done = F Δs cos α.

Principle of virtual work
If a system of forces acting on a body or a system of bodies is in equilibrium, the total virtual work
done by the forces acting on the body or system of bodies is zero for any virtual displacement
consistent with geometrical conditions.

For example, consider the free-body diagram of the particle (ball) that rests on the floor. If we
“imagine” the ball to be displaced downwards a virtual amount δy

Total virtual work done

63. 62

Since the system is in equilbrium


Since δy ≠ 0, then
A. APPLICATION OF THE PRINCIPLE OF VIRTUAL WORK ON BEAMS

PROBLEMS
1. Find support reactions of the following beams using virtual work

64. 63

MODULE 3 – PROPERTIES OF AREAS
CHAPTER 8 – CENTROID
Centroid
The point at which the total area of a plane figure or lamina is assumed to be concentrated is called
centroid
Centroid of a Line


Centroid of an Area


Characteristics of Centroid
 The centroid represents the geometric center of a body
 The centroid may be located at a point that does not lie on the line/area.
 The coordinates of centroid is calculated with reference to the chosen axis
 An area can have only one centroid for all positions of the figure.
 In case of symmetric figures, centroid is located along the axes of symmetry
Centroid Of Regular Figures
Shape Area xc yc
Rectangle

65. 64


Triangle



Semi-Circle




Quarter Circle




Centre of Gravity
Centre of gravity is the point about which the resultant of the whole weight of the body may be
considered to act. It is denoted by G


A. CENTROID OF COMPOSITE FIGURES

66. 65


Note : If the area has a hole or cut out portion, the first moment of inertia and area must be subtracted
to yield the centroid
PROBLEMS
1. Find the centroid of an unequal angle section 100 mm × 80 mm × 20 mm.

2. Locate the centroid of the following areas

3. An I-section has the following dimensions in mm units :Bottom flange = 300 × 100, Top flange =
150 × 50, Web = 300 × 50. Determine mathematically the position of centre of gravity of the
section.
4. Find centre of gravity of the following lamina.

67. 66


5. Determine the coordinates of the centroid of the given shaded area

Problems for Practice
1. Find centroid of the shaded area.

68. 67

B. PAPPUS GULDINUS THEOREMS
Pappus Guldinus Theorems are two theorems describing a simple way to calculate volumes (solids) and
surface areas (shells) of revolution.
First Theorem
The surface area A of a surface of revolution generated by rotating a plane curve about an axis
external to it and on the same plane is equal to the product of the arc length of the curve and the
distance y traveled by its geometric centroid.

Examples


Second Theorem
The second theorem states that the volume V of a solid of revolution generated by rotating a plane
area about an external axis is equal to the product of the area A and the distance y traveled by its
geometric centroid.

69. 68


Examples

PROBLEMS
1. Calculate the shaded area obtained by revolving the line ABC about the i) X axis and ii) Y axis

2. Obtain the expression for the area of a surface generated by rotation of a semicircular arc of
radius r about an axis passing through its end points.
3. Obtain the expression for the volume of a body generated by revolution of a triangle when its
base h is in touch with the axis of rotation and the length of the other side is r.

70. 69

CHAPTER 9 – MOMENT OF INERTIA
Moment of Inertia
The Second moment of area, also known as moment of inertia of plane area is a geometrical
property of an area which reflects how its points are distributed with regard to an arbitrary axis.
Moment of Inertia is given as the sum of the products of each area in the lamina with the square of its
distance from the axis of rotation.


It is denoted with I for an axis that lies in the plane or with a J for an axis perpendicular to the plane
(referred as polar moment of inertia).
Characteristics of Moment of Inertia
 MI is determined with reference to an axis and changes with change in axis position or
orientation
 Moment of Inertia about any axis is always a positive quantity
 MI is minimum when the axis of rotation passes through the centroid
 The SI units for area moments of inertia are expressed as m
4
or mm
4
.
Parallel Axis Theorem
Parallel Axis Theorem states that the moment of inertia of a plane lamina about any axis is equal to
the sum of moment of inertia about a parallel axis passing through the centroid and the product of
area and square of the distance between the two parallel axes.

Proof
In Figure, the x0-y0 axes pass through the centroid C of the area. If I’x and I’y are the moments of inertia
of the area about the centroidal axes x0-y0 and Ix and Iy the moments of inertia of the area about the
parallel axes x-y.

71. 70


By definition, the moment of inertia of the element dA about the x-axis is given by

Similarly

Perpendicular Axis Theorem
Perpendicular Axis Theorem states that the moment of inertia of a plane lamina about an axis
perpendicular to the lamina and passing through the centroid is equal to the sum of moment of
inertias of the lamina about two mutually perpendicular axes in the same plane and passing through
the centroid.
OR
Perpendicular Axis Theorem states that the polar moment of inertia is equal to the sum of moment of
inertias of the two centroidal axis in the plane of the lamina

Proof
In Figure, the x0-y0 axes pass through the centroid C of the area. If I’x and I’y are the moments of inertia
of the area about the centroidal axes

72. 71



Moment Of Inertia Of Regular Sections
Shape Ix Iy
Rectangle





Triangle



Circle

73. 72

Semi-Circle


Quarter Circle


A. MOMENT OF INERTIA OF COMPOSITE SECTIONS
Procedure
1. Divide the given area into its simpler shaped parts.
2. Locate the centroid of each part and indicate the perpendicular distance from each centroid to the
desired reference axis.
3. Determine the MI of each “simpler” shaped part about the desired reference axis using the parallel-axis
theorem
4. The MI of the entire area about the reference axis is determined by performing an algebraic summation
of the individual MIs obtained
PROBLEMS
1. Determine moment of inertia about the x and y axis.

2. Find the moment of inertia of a T-section with flange as 150 mm × 50 mm and web as 150 mm ×
50 mm about X-X and Y-Y axes through the centre of gravity of the section. Find the moment of
inertia of the section about the centroidal axis

74. 73


3. Determine the moments of inertia of the cast-iron beam section about horizontal and vertical
axes passing through the centroid of the section.

4. Find moments of inertia about the centroidal axes of the section shown in figure

5. A compound beam is made by welding two steel plates 160 mm × 12 mm one on each flange of
an ISLB 300 section. Find the moment of inertia the beam section about an axis passing through
its centre of gravity and parallel to X-X axis. Take moment of inertia of the ISLB 300 section about
X-X axis as 73.329 × 10
6
mm
4
.

75. 74


6. Find the value of x so that the centre of gravity of the uniform lamina shown in figure remains at
the centre of the rectangle ABCD

Problems for Practice
1. Find the MI of the lamina with a circular hole of 30 mm diameter about the axis AB

2. A rectangular hole is made in a triangular section as shown in figure. Determine the moment of
inertia of the section abut X-X axis passing through the centre of gravity of the section and
parallel to BC. Also find moment of inertia through BC

76. 75


3. Find moment of inertia with respect to centroidal axes

Polar Moment of Inertia
The moment of inertia of a plane lamina about an axis perpendicular to the lamina and passing
through the centroid is called polar moment of inertia

Radius of Gyration
The radius of gyration of an area about an axis is the distance to a long narrow strip whose area is
equal to the area of the lamina and whose moment of inertia remain the same as that of the original
area
Consider a lamina of area A whose moment of Inertia is Ix, Iy and Iz with respect to x, y and z axis.
Imagine the area is concentrated into a thin strip parallel to the axis and has the same moment of
inertia as that of the lamina

77. 76



Product of Inertia
The product of inertia is the sum of the product of each differential area and its moment arms about
the two perpendicular reference axis.

Characteristics of Product of Inertia
 The SI units for product of inertia are expressed as m
4
or mm
4
.
 Product of inertia may be positive, negative, or zero, depending on the location and orientation of
the coordinate axes.
 Product of inertia for an area will be zero if either x or y axis is an axis of symmetry for the area.

Rotation of Moments of Inertia
The moment of inertia and product of inertia of a lamina with respect to an inclined axis u and v
rotated an angle θ with respect to the reference axis x- y is given by

78. 77


Principal Axes and Principal moments of Inertia
The inclined set of axis u-v about which the product of inertia Ixy is zero is called the Principal Axes.
The angle made by the principal axes with respect to the reference axes is called principal angle θp.
The moments of inertia about the principal axes are called Principal Moment of Inertia and yield the
maximum and minimum values of moment of inertia of the lamina.


PROBLEMS
1. Determine the moments of inertia and the radius of gyration of the shaded area with respect to
the x and y axes.

2. For the given section, find
i. Polar Moment of Inertia about the centroidal Z-Z Axis

79. 78

ii. Radius of gyration about x, y and z axes
iii. Principle moments of Inertia

Mass moment of Inertia
Mass moment of inertia or rotational inertia of a body is a measure of the body’s resistance to
angular acceleration. Mass Moment of Inertia is the sum of the products of the mass of each particle
in the body with the square of its distance from the axis of rotation.
The mass moment of inertia about the axis O-O is given as


Mass Moment of Inertia of Circular Thin Disc

Mass Moment of Inertia of Cylinder

80. 79

MODULE 5 – DYNAMICS 1
CHAPTER 10 – KINEMATICS OF RIGID BODIES
DYNAMICS
Dynamics is that branch of mechanics which deals with the motion of bodies under the action of
forces. Dynamics has two distinct parts
1. Kinematics : Kinematics is the branch of dynamics which describes the motion of bodies
without reference to the forces which either cause the motion or are generated as a result of
the motion. Kinematics deals only with the geometric aspects of the motion. Some engineering
applications of kinematics include calculation of flight trajectories for aircraft, rockets, and
spacecraft
2. Kinetics : Kinetics is the study of the relationships between motion and the corresponding
forces which cause or accompany the motion.
Types Of Motion Of Rigid Bodies
1. Translation Motion : Any motion in which every line in the body remains parallel to its original
position at all times. In translation there is no rotation of any line in the body. When the paths
of motion for any two points on the body are parallel lines, the motion is called rectilinear
translation. If the paths of motion are along curved lines which are equidistant, the motion is
called curvilinear translation,
2. Rotation about a fixed axis : When a rigid body rotates about a fixed axis, all the particles of
the body, except those which lie on the axis of rotation, move along circular paths. All lines in
the body which are perpendicular to the axis of rotation (including those which do not pass
through the axis) rotate through the same angle in the same time.
3. General plane motion : When a body is subjected to general plane motion, it undergoes a
combination of translation and rotation

81. 80


A. RECTILINEAR MOTION
Basic Equations
1. Displacement: The displacement of the particle is defined as the change in its position.

2. Velocity: If the particle moves through a displacement during the time interval the average
velocity of the particle during this time interval is

Consequently, the instantaneous velocity is a vector defined as

82. 81

3. Acceleration: Provided the velocity of the particle is known at two points, the average
acceleration of the particle during the time interval is defined as

The instantaneous acceleration at time t is a vector that is found by

B. CURVILINEAR MOTION
Rectangular Coordinates
If all motion components such as displacement, velocity and acceleration are directly expressible in
terms of horizontal and vertical coordinates. Here the coordinates x and y are independently as
functions of time

83. 82

C. FIXED AXIS ROTATION

1. Angular Position: At the instant shown in the figure, the angular position of r is defined by the
angle θ, measured between a fixed reference line and r.
2. Angular Displacement: The change in the angular position, which can be measured as a
differential dθ, is called the angular displacement. This vector has a magnitude of measured in
degrees, radians, or revolutions per minute(rpm)
3. Angular Velocity: The time rate of change in the angular position is called the angular velocityω
(omega). Since occurs during an instant of time dt, then,

4. Angular Acceleration: The angular accelerationα measure the time rate of change of the
angular velocity.


Constant Angular Acceleration

Cylindrical/Polar Coordinates
Here the all components of motion such as displacement, velocity are expressed in terms of the e
radial distance to the origin, r, and the angle that the radial line makes with an arbitrary fixed line, such
as the x axis. Thus, the trajectory of a particle will be determined if we know r and θ as a function of t,
i.e. r(t), θ(t). The directions of increasing r and θ are defined by the orthogonal unit vectors e
r
and e
θ
.

84. 83


Linear Velocity/Acceleration of Rotating Body (Normal and Tangential Coordinates)


Relation Between Linear Motion And Angular Motion

85. 84

CHAPTER 11 – KINETICS OF RIGID BODIES - TRANSLATION
D-Alembert’s Principle
A rigid body in motion is acted upon be a system of forces, it can be transformed into an equivalent
static system with the addition of inertia force acting opposite to the direction of motion. This state
of equilibrium is called dynamic or kinetic equilibrium.


In other words, the body is in equilibrium under the action of the real force F and the fictitious force
called Inertia force acting in the opposite direction of applied force and resisting acceleration.
D'Alembert's principle is merely another way of writing Newton's second law, it has the advantage of
changing a problem in kinetics into a problem in statics.
Inertia Force
Inertia force is fictitious force acting in the opposite direction of applied force and resisting
acceleration. It is denoted by Fi is equal to the product of the mass m and acceleration a of the body.

A. MOTION OF CONNECTED BODIES (not clearly mentioned in syllabus)
PROBLEMS
1. Determine the tension in the strings and accelerations of two blocks of mass 150 kg and
50 kg connected by a string and a frictionless and weightless pulley as shown in Figure.

86. 85

2. Find the acceleration of a solid body A of mass 10 kg, when it is being pulled by another
body B of mass 5 kg along a smooth horizontal plane as shown in Figure. Also find the
tension in the string, assuming the string to be inextensible. Take g = 9.8 m/s
2
.

3. The system of bodies shown in Figure starts from rest. Determine the acceleration of body
B and the tension in the string supporting body A.

5. Two smooth inclined planes whose inclinations with the horizontal are 30
0
and 20
0
are
placed back to back. Two bodies of mass 10 kg and 6 kg are placed on them and are
connected by a light inextensible string passing over a smooth pulley as shown in Figure.
Find the tension in the string and acceleration of the masses. Take g = 9.8 m/s
2
.

6. Two blocks A and B of weight 150N and 100N are released from rest on a 30° inclined
plane, when they are 15m apart. The coefficient of friction between the upper block A
and the plane is 0.2 and that between the lower block B and the plane is 0.4. In what time
block A reach block B? After they touch and move as a single unit, what will be their
combined acceleration?

B. MOTION OF LIFT

87. 86

Lift Moving Upwards

Lift Moving Downwards

PROBLEMS
1. A lift carries a weight of 100Nandi is moving with a uniform acceleration of 2.45 m/s
2
. Determine
the tension in the cables supporting the lift, when
(i) lift is moving upwards, and
(ii) lift is moving downwards. Take g - 9.80 m/s
2
.
2. A life has an upward acceleration of 1.225 m/s
2
. What pressure will a man weighing 500 N exert
on the floor of the lift? What pressure would he exert if the lift had an acceleration of 1.225 m/s
2

downwards? What upward acceleration would cause his weight to exert a pressure of 600 N on
the floor? Take g = 9.8 m/s
2
.
3. An elevator weighs 2500 N and is moving vertically downwards with a constant acceleration.
Write the equation for the elevator cable tension. Starting from rest it travels a distance of 35
metres during an interval of 10 seconds. Find the cable tension during this time. Neglect all other
resistances to motion. What are the limits of cable tension?
4. A cage, carrying 10 men each weighing 500 N, starts moving downwards from rest in a mine
vertical shaft. The cage attains a speed of 12 m/s in 20 metres. Find the pressure exerted by each
man on the floor of the cage. Take g = 9.80 m/s
2
.
5. An elevator weighing 5000 N is ascending with an acceleration of 3 m/s
2
. During this ascent its
operator whose weight is 700 N is standing on the scales placed on the floor. What is the scale
reading? What will he the total tension in the cables of the elevator during this motion?

88. 87

CHAPTER 12 – KINETIC OF RIGID BODIES – GENERAL PLANE MOTION
Instantaneous Centre
A rigid body in plane motion undergoing both rotation and translation, at a given instant of time
appears as if undergoing pure rotation about a certain point in the plane of the body. This centre
point which is instantaneously at rest and has zero velocity is called the instantaneous centre.

Characteristics of I-Centre
1. The instantaneous centre is changing every instant and is not a fixed point
2. The velocity at any point in the body at that instant can be determined by assuming that the
body is undergoing pure rotation with angular velocity ω about the instantaneous centre
3. The velocity of the instantaneous centre is zero
Method of locating instantaneous centre
1. The velocity of a point A and the angular velocity ω are known
The Instantaneous Centre is located along the line drawn perpendicular to at A, such that the
distance from A to the IC is


2. The lines of action of two non-parallel velocities vA and vB are known
Draw perpendiculars to velocity at the points A and B whose point of intersection will yield IC at
the instant considered.

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3. The lines of action of two parallel velocities vA and vB are known
Here the location of the IC is determined by proportional triangles.


PROBLEMS
1. A cylinder of radius 1 m rolls without slipping along a horizontal plane. Its centre has a uniform
velocity of 20 m/s. Find the velocities of the points D and E on the circumference of the cylinder
as shown in Figure

2. A cylindrical roller is in contact at its top and bottom, with two conveyor belts AB and DE as
shown in Figure If the belts run at uniform speeds of 3 m/s and 2 m/s respectively, find the
angular velocity of the roller, when,
(i) the velocities are in the same direction and
(ii) the direction of velocities are opposite. The diameter of the roller is 40 cm.

90. 89


3. A compound wheel rolls without slipping as shown in Fig with a velocity of 2m/s. Find the
velocities of points A and B

4. Calculate the velocity at a point 2/3 rd radius from the top point of a rail car wheel of radius 250
mm, if the car moves without slipping on straight rails at 15 m/s.
A. MOTION OF A CONNECTING ROD AND PISTON OF A RECIPROCATING PUMP

Graphical Method for Linear Velocity of Piston
Consider the mechanism of a reciprocating pump in which BC be the crank, AB the connecting rod and
C the piston. Let crank radius be r, the rod length l, the crank angle θ
Steps
i. Find the angular velocity of crank about the dead centre C and linear velocity of crank at B


ii. Apply sine rule in triangle ABC and obtain angle Φ

91. 90


iii. Draw perpendiculars to velocity at the two ends of the connecting crank, one can locate the
instantaneous centre at O.
Angle AOB = 180 – 90 - θ
Angle OAB = 90 – Φ
iv. Find AO and BO by sine rule
v. Let ωo be the angular velocity of the connecting rod about the instantaneous centre


Analytical Method for Linear Velocity and Acceleration of Piston


PART VELOCITY ACCELERATION
Crank

Piston/Slider

Connecting Rod

92. 91

PROBLEMS
1. In a reciprocating pump, the lengths of connecting rod and crank are 1125 mm and 250 mm
respectively. The crank is rotating at 420 r.p.m. Find the velocity with which the piston will move,
when the crank has turned through an angle of 40° from the inner dead centre.
2. A reciprocating engine mechanism has its crank 10 cm long and the length of the connecting rod
is 40 cm. Find the velocity and acceleration of the piston by analytical method when the crank
rotates uniformly at angular speed of 10 rad/sec for the following positions a) at IDC b) at θ = 30
0

c) at θ = 90
0

93. 92

MODULE 6 - MECHANICAL VIBRATIONS
Degree of Freedom
The number of independent coordinates required to describe completely the motion/position of a
system at any time is called Degree of Freedom
A rigid body in space has six degrees of freedom, three co-ordinates (x, y and z) to define rectilinear
position and three parameters to define the angular positions. The constraints to the motion reduce
the degee of freedom of the system.
1. Single Degree of Freedom System

2. Multiple Degree of Freedom system

A. CLASSIFICATION OF VIBRATIONS
Classification Based on Exciting Force
1. Free Vibration : Vibrations in which after the initial displacement/velocity, no external forces
act on the body and the vibration is maintained by the internal elastic forces is called Natural
Vibration
Eg : vibration in string, motion of pendulum
2. Forced Vibration : When a body vibrates under the influence of a continuous periodic disturbing
internal or external force, then the body is said to be under forced vibration
Eg : vibration in rotating engines

94. 93

Classification Based on Direction of Motion
1. Longitudinal Vibration: When the particles of bar or disc move parallel to the axis of the shaft,
then the vibrations are called longitudinal vibrations.
2. Transverse Vibration: When the particles of the bar or disc move approximately perpendicular to
the axis of the shaft on either side in the transverse direction, then the vibrations are known as
transverse vibrations.
3. Torsional Vibration: When the particles of the bar or disc get alternately twisted and untwisted
on account of vibratory motion of suspended body, it is said to be undergoing torsional
vibrations.

Damped vibrations and resonance
1. Damped Vibrations :The vibrations of a body whose amplitude goes on reducing over every cycle
of vibrations are known as damped vibrations. This is due to the fact that a certain amount of
energy possessed by the vibrating body is always dissipated in overcoming frictional resistance to
the motion.
2. Resonance : When the frequency of external force is equal to the natural frequency of the
vibrations, resonance takes place, amplitude or deformation or displacement will reach to its
maximum at resonance and the system will fail due to breakdown. This state of disturbing force on
the vibrating body is known as the state of resonance.
Periodic and Oscillatory motion
1. Periodic Motion : A motion which repeats itself at regular intervals of time is called a periodic
motion. Eg – motion of pendulum, motion of planets around sun
2. Oscillatory Motion :If a body is moving back and forth repeatedly about a mean position, it is
said to possess oscillatory motion. Eg - motion of the pendulum, vibrations of the string
An oscillatory motion is always periodic. A periodic motion may or may not be oscillatory. For example,
the motion of planets around the Sun is always periodic but not oscillatory. The motion of the
pendulum of a clock is periodic as well as oscillatory.

95. 94

General Terms in Vibratory Motion
1. Time Period :The time interval after which the motion is repeated itself is called time period. It
is usually expressed in seconds.
2. Cycle :The motion completed during one time period is called cycle.
3. Frequency :The number of cycles executed in one second is called frequency. It is usually
expressed in hertz (Hz).
4. Amplitude : The maximum displacement from the mean position is called the amplitude

B. SIMPLE HARMONIC MOTION (SHM)
A body is said to move in Simple Harmonic motion if it satisfies the following conditions
 The motion should be oscillatory periodic motion with elastic restoring force acting on the
system.
 The acceleration of the system is directly proportional to its displacement from the mean
position at any instant of time
 The acceleration is always directed towards the mean position
Eg : motion of a pendulum, motion of body attached to elastic spring

Equations of SHM
Simple Harmonic Motion can be modelled as the projection of a particle rotating at a constant angular
velocity ω rotating about a point O with a radius r equal to the amplitude

Let P be the position of the particle at some instant after t sec from X. Therefore, angle turned by the
particle
Projection of point P about the y-axis is given as

Differentiating with respect to time t

96. 95


Further differentiating with respect to time t

Max Displacement, Velocity and Acceleration Equations


Similarly, the projection of the point P along the x-axis also satisfies the Simple Harmonic Motion.
Expressing the acceleration in terms of x.

Expressing it in the standard format

General Equations
1. Time Period

2. Frequency

C. UNDAMPED FREE VIBRATION
In this approach, the free undamped vibration of an object is modelled as mass-spring system.
Consider a compact mass m that slides over a frictionless horizontal surface. Suppose that the mass is
attached to one end of a light horizontal spring of stiffness k whose other end is anchored in an
immovable wall.

97. 96


The following forces act on the system
1. Restoring Force of the spring due to its stiffness
2. The weight of the body
3. The normal reaction

Since the system does not undergo vertical deflection, ΣFy = 0
Since the system vibrates along x axis



Rearranging the terms into a “standard form” gives

This equation is synonymous to the equation of motion of SHM. Hence, spring mass model can be
considered as a body executing SHM with a natural frequency given by

98. 97

General Equations
1. Natural Angular Frequency

2. Time Period

3. Frequency

COMBINATION OF SPRINGS

99. 98


PROBLEMS
1. A 80 N weight is hung on the end of a helical spring and is set vibrating vertically. The weight
makes 4 oscillations per second. Determine the stiffness of the spring
2. If a helical spring having a stiffness of 90 N/cm is available, what weight should be hung on it so
that it will oscillate with a periodic time of 1 sec.
3. A weight of 50 N suspended from a spring vibrates vertically with an amplitude of 8 cm and a
frequency of 1 oscillation per second. Find (a) the stiffness of the spring, (b) the maximum
tension induced in the spring and (c) the maximum velocity of the weight.
4. A body of mass 50 kg is suspended from two springs of stiffness 4 kN/m and 6 kN/m as shown in
Figures (a), (b) and (c). The body is pulled 50 mm down from its equilibrium position and then
released. Calculate the
i. frequency of oscillation
ii. maximum velocity and
iii. maximum acceleration

5. A particle moving with SHM has an amplitude of 4.5 m and period of oscillation is 3.5 second.
Find the time required by the particle to pass two points which are at a distance of 3.5 m and 1.5
m from the centre and on the same side of mean position.

100. 99

6. A helical spring under a weight of 20 N extends 0.30 mm. A weight of 700 N is supported on the
same spring. Determine the period and frequency of vibration of the weight and spring when
they are displaced vertically by a distance of 0.9 cm and released. Find the velocity of the weight
when the weight is 4 mm below its equilibrium position. Take the weight of spring as negligible.
7. A body, moving with simple harmonic motion, has an amplitude of 1 m and period of oscillation
is 2 seconds. Find the velocity and acceleration of the body at t = 0.4 second, when time is
measured from (i) the mean position and (ii) the extreme position.
8. The weight of an empty railway wagon is 24 kN. On loading it with goods weighing 32 kN, its
springs get compressed by 8 cm.
i. Calculate its natural period of vibrations when : (i) empty and (ii) loaded as above.
ii. It is set into natural vibrations with an amplitude of 10 cm when empty. Calculate the
velocity when its displacement is 4 cm from statical equilibrium position.
9. A particle is moving with simple harmonic motion and performs 8 complete oscillations per
minute. If the particle is 5 cm from the centre of the oscillation, determine the amplitude, the
velocity of the particle and maximum acceleration. Given that the velocity of the particle at a
distance of 7 cm from the centre of oscillation in 0.6 times the maximum velocity.
10. A body performing simple harmonic motion has a velocity = 12 m/s when the displacement is 50
mm and 3 m/s when the displacement is 100 mm, the displacement being measured from the
mid-point. Calculate the frequency and amplitude of the motion. What is the acceleration when
the displacement is 75 mm?
11. The piston of an IC engine moves with simple harmonic motion. The crank rotates at 420 length is
40 cm. Find the velocity and acceleration of the piston when is at a distance of 10 cm from the
mean position.