Introduction to Mechanics

UNIT-1
 SYLLABUS

 Force moment and couple, principle of transmissibility,

Varignon's theorem. Resultant of force system- concurrent
and non-concurrent coplanar forces, Types of supports
(Hinge, Roller) and loads (Point, UDL, UVL), free body
diagram, equilibrium equations and Support Reactions.

 Normal and shear Stress, strain, Hooke's law, Poisson’s

ratio, elastic constants and their relationship, stress-strain
diagram for ductile and brittle materials, factor of safety.
Lecture No. 1
 Engineering Mechanics
 The branch of science which deals with the forces and their effects on the
bodies on which they act is called mechanics.

 Engineering Mechanics can be classified as-

 Classification of Engg. Mechanics
1. Statics: The branch of applied mechanics which deals with
the forces and their effects while acting upon bodies which
are at rest is called statics.

2. Dynamics: The branch of applied mechanics which deals

with the forces and their effects while acting upon bodies
which are in motion is called dynamics.

It is further divided into two types:

Kinematics: The branch of dynamics which deals with
motion of bodies without considering the forces which
cause motion is called kinematics.
Kinetics: The branch of dynamics which deals with the
relationship between motion of bodies and forces
causing motion is called kinetics.
 FORCE
 Force may be defined as a push or pull which either changes
or tend to change the state of rest or of uniform motion of a
body.
 Force is a vector quantity.
 Unit- Newton (N)
 Characteristics of a force:
The followings are the characteristics of a force:
1. Magnitude: The quantity of a force is called its magnitude such as
50 N, 80 N, 25 kg etc.
2. Direction: The direction of a force is the direction of the line along
which it acts.
3. Nature of the force or sense: Nature of the force means whether
the force is a push or a pull at the point of application.
4. Point of application
 Force System
 A force system is a collection of several forces acting
simultaneously on a body in one or more plane.
 Classification of Forces System:
 Force System
 Coplanar forces: The forces, whose lines of action lie on the
same plane.
 Non-coplanar Forces: The forces, whose lines of action lie on
the different plane.
 Collinear forces: The forces acting along the same line of
action.
 Non-collinear forces: The forces acting along the different
lines of action.
 Parallel Forces: The forces line of action are parallel to each
other is called parallel forces.
 Force System
 Concurrent forces: If the forces applied to a body are
such that their lines of action meet at a single point,
then they are called concurrent forces.

 Non-Concurrent forces: When the forces acting on a

body do not intersect at a common point, the system
of forces is said to be non- concurrent forces.
Force System
Force System
 Moment of Force
 The tendency of forces is not only to move the body but also to
rotate the body. This rotational tendency of a force is called
moment.
 The force multiplied by the perpendicular distance from the
point to the line of action of the force is called moment about
that point.
 Unit of moment N-m.
 Moment of Couple
 A couple is pair of two equal and opposite forces acting on a
body in a such a way that the lines of action of the two forces
are not in the same straight line.

 The moment of a couple is known as torque which is equal to

one of the forces forming the couple multiplied by arm of the
couple.
 Moment of couple
The following are the examples of couples in every day life.
1. Opening or closing a water tap. The two forces constitute a
couple.
2. Turning the cap of a pen.
3. Unscrewing the cap of an ink bottle.
4. Twisting a screw driver.
5. Steering a motor-car.
 Principal of transmissibility of forces

It states that the point of application of a force may be

changed to any other point along the line of action of the
force without any change in the external effects produced
by the force.
 Law of superposition of forces

"The principle of superposition of forces states that the

resultant force has the same effect as the sum of the
individual forces acting on an object."
Lecture No. 2
 Varignon’s Theorem
The Principle of Moments, also known as Varignon's
Theorem, states that the moment of any force is equal to
the algebraic sum of the moments of the components of
that force.
OR

If many coplanar forces are acting on a body, then algebraic

sum of moments of all the forces about a point in the plane
of the forces is equal to the moment of their resultant about
the same point.
Varignon’s Theorem Proof
Varignon’s Theorem Proof
Varignon’s Theorem Proof
Varignon’s Theorem Problem
 Resultant of Forces
 A resultant force is a single force which can replace two or
more forces and produce the same effect on the body as the
forces.
 LAWS OF FORCES
The method of determination of the resultant of some forces
acting simultaneously on a particle called composition of
forces.
The various laws used for the composition of forces are given
as under :
1. Parallelogram law of forces,
2. Triangle law of forces,
3. Polygon law of forces.
 Parallelogram law of forces
 If two forces, acting simultaneously on a particle, be represented in
magnitude and direction by the two adjacent sides of a parallelogram
then their resultant may be represented in magnitude and direction by
the diagonal of the parallelogram which passes through their point of
intersection.”
 Let two forces P and Q acting simultaneously on a particle be
represented in magnitude and direction by the adjacent sides oa and ob
of a parallelogram oacb drawn from a point o, their resultant R will be
represented in magnitude and direction by the diagonal oc of the
parallelogram.
Parallelogram law of forces
Parallelogram law of forces
 Triangle law of forces
 “If two forces acting simultaneously on a body are
represented in magnitude and direction by the two sides of
triangle taken in order then their resultant may be
represented in magnitude and direction by the third side
taken in opposite order.”
 Let P and Q be the two coplanar concurrent forces. The
resultant force R in this case can be obtained with the help of
the triangle law of forces both graphically and analytically as
given below :
 Polygon law of forces
 “If a number of coplanar concurrent forces, acting simultaneously on a
body are represented in magnitude and direction by the sides of a
polygon taken in order, then their resultant may be represented in
magnitude and direction by the closing side of a polygon, taken in the
opposite order”.
 If the forces P1, P2, P3, and P4 acting simultaneously on a particle be
represented in magnitude and direction by the sides oa, ab, bc and cd of a
polygon respectively, their resultant is represented by the closing side do
in the opposite direction as shown in Fig.
 RESULTANT OF SEVERAL COPLANAR
CONCURRENT FORCES
To determine the resultant of a number of coplanar concurrent
forces any of the following two methods may be used :
1. Graphical method (Polygon law of forces),
2. Analytical method (Principle of resolved parts).

Resultant by analytical method

Resultant by analytical method
 LAMI’S THEOREM
It states that “If three coplanar forces acting on a point in a
body keep it in equilibrium, then each force is proportional to
the sine of the angle between the other two forces.”
Fig. shows three forces P, Q and R
acting at a point O. Let the angle
between P and Q be 𝛼, between Q
and R be 𝛽 and between R and P
be 𝛾 . If these forces are in
equilibrium then according to
Lam’s theorem-
Numerical
Lecture No. 3
 Types of Supports
When a number of forces are acting on a beam then
support of the beam will provide the reactions called
Support Reactions.

Following are types of supports:

1. Roller support

2. Hinge(Pin) Support

3. Fixed (Built-in) Support

 1. Roller support
 Axial motion is permitted
 Rotation is permitted
 Only vertical motion is restricted
 No of reaction = 1, RV
Roller support real life example
 2. Pin(Hinged) support
 Vertical motion is restricted
 Axial motion is restricted
 Rotation is permitted
 No of reaction = 2, RAV and RAH
Pin support real life example
 3. Fixed support
 Vertical motion is restricted.
 Axial motion is restricted.
 Rotation is restricted.
 No of reaction = 3,
 RAV, RAH and Mz
Fixed support real life example
 Type of Loading on the beam
Following are the types of load which are subjected to a
beam
1. Concentrated load(point load)
2. Uniformly Distributed load(UDL)
3. Uniformly Varying load(UVL)
4. Moment (M)
 1. Concentrated load(point load)

 This load acts at a point.

It is represented by an arrow as shown in Fig.

2. Uniformly Distributed load(UDL)
 Load acts over a certain length
 Intensity of load is uniform
 It is represented as shown in Fig.
 Total load = Area of plane fig (rectangle)
 Total load acts at middle of the loaded length.

 Given load may be replaced by a 20×4 = 80 kN,

concentrated load acting at a distance 2 m from the left
support.
 3. Uniformly Varying load(UVL)
 The load varies Uniformly from C to D.
Total load = Area of plane fig (triangle)
Centroid of the triangle represents the center of gravity of the
load. (1/3rd from D or 2/3rd from C)

E.g.
Its intensity is zero at C and is 20 kN/m (maximum) at D.
Total load is 1 /2 × 3 × 20 = 30 kN
This load is equivalent to 30 kN acting at 3 m from A.
 4. Moment
A beam may be subjected to external moment at certain
points.

In Fig. the beam is subjected to clockwise moment of

30 kN-m at a distance of 2 m from the left support.
 Equilibrium Equations
 The body is said to be in equilibrium if the resultant of all
forces acting on it is zero.
 There are two major types of static equilibrium, namely,
translational equilibrium and rotational equilibrium.

Concurrent force system

ΣFx=0
ΣFy=0

Parallel Force System

ΣF=0
ΣM=0

Non-Concurrent Non-Parallel Force System

ΣFx=0
ΣFy=0
ΣM=0
Example - 1
The beam AB of span 12 m shown in Fig. is hinged at A and is on rollers at B.
Determine the reactions at A and B for the loading shown in the Fig.
Solution:
Example - 2
Find the reactions at supports A and B of the loaded beam shown in Fig.
Example - 3
The cantilever shown in Fig. is fixed at A and is free at B. Determine the reactions
when it is loaded as shown in the Figure.
Solution
Example - 4
Compute the reaction developed at support in the cantilever beam shown in Fig.
Example - 5
Determine the reactions at supports A and B of the overhanging beam shown in
Fig
Solution:
As supports A and B are simple supports and loading is only in vertical direction, the reactions RA
and RB are in vertical directions only
Example - 6
Find the reactions at supports A and B of the beam as shown in Fig.
Lecture No. 4
Example - 7
Determine the reactions at A and B of the overhanging beam shown in Fig.
Solution
Example - 8
A beam AB 20 m long supported on two intermediate supports 12 m apart, carries
a uniformly distributed load of 6 kN/m and two concentrated loads of 30 kN at
left end A and 50 kN at the right end B as shown in Fig. How far away should the
first support C be located from the end A so that the reactions at both the
supports are equal ?
Example – 9
A cantilever beam is loaded as shown. Determine all reactions at support A.
SOLUTION
 Free Body Diagram(FBD)
A free body diagram consists of a diagrammatic
representation of a single body or a subsystem of bodies
isolated from its surroundings showing all the forces acting
on it.
FBD Standard cases
Lecture No. 5
 Stress(σ)
 Stress is defined as the internal resisting force
developed at a point offered by a body against
deformation when it is subjected to external load.

𝑹 𝑷
σ= =
𝑨 𝑨

 Unit- N/m2, N/mm2

1 Pascal = N/m2 and 1MPa = N/mm2
 Stress Vs Pressure
 Mag. Of IRF developed  Mag. Of External
at a point. Applied Force at a point.
 Stress acts normal or  Always acts normal to
parallel to surface. the surface.
 Mag. Of stress at a point  Mag. Of Pressure at a
in different direction is point in all the direction
different. remains same.
 Vector quantity.  Scalar quantity.
 It can not be measured.  It can be measured.
 Due to Stress, pressure  Due to pressure, Stress
will not be developed. will be developed.
 Types of Stress

Stress is of following types:

Stress

Shear Stress
 Normal Stress
 Stress is said to be Normal stress when the direction of

the deforming force is perpendicular to the cross-

sectional area of the body.
 Tensile Stress(𝝈𝒕 )
 When a structural member is subjected to two equal and

opposite tensile forces, the stress produced is called

tensile stress.

 The tensile stress at any cross-section X-X is given

𝑷
as𝝈𝒕 =
𝑨
 Compressive Stress(𝝈𝒄 )
 When a structural member is subjected to two equal and

opposite compressive forces, the stress produced is

called compressive stress.

 The compressive stress at any cross-section X-X is

𝑷
given as 𝝈𝒄 =
𝑨
 Volumetric Stress(𝝈v)
 When the deforming force or applied force acts from
all dimension resulting in the change of volume of the
object then such stress in called volumetric stress or
Bulk stress.
 In short, When the volume of body changes due to
the deforming force it is termed as Volume stress.
 Shear Stress()
 Stress produced by a force tangential to the surface of a

body is known as shear stress.

 It is represented by .
 Consider a rectangular block ABCD fixed at the bottom
plane and subjected to tangential force P at the upper
plane.
 Strain(𝜺)
 When a body is subjected to tensile or compressive
load, its dimension will increase or decrease along the
line of action of load applied.

 It can be defined as the ratio of change in dimension to

the original dimension.
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏
 𝑺𝒕𝒓𝒂𝒊𝒏 =
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏

 Strain is dimensionless quantity.

 Types of Strain

Strain

Normal Strain Shear Strain

Lateral
Strain
Longitudinal Volumetric
Strain Strain
 Types of Strain
 Strain in the direction of applied load is called
longitudinal or primary or linear strain.

 Strain in the perpendicular direction of applied load is

called lateral strain.

 Longitudinal and lateral strain are always opposite in

nature.
 Volumetric Strain(𝜺v)
 When a body is immersed in a
fluid to a large depth, the body is
subjected to equal external
pressure at all points on the
body.
 Due to this external pressure,
stress is produced within the
body which is called hydrostatic
stress.
 This external pressure causes
change in volume of the body.
 This change in volume per unit
volume is called volumetric
strain, 𝜺v.
 Shear Strain(φ)
 Strain produced by a force tangential to the surface of

a body is known as Shear strain.

 Under the action of tangential force, the block ABCD
gets distorted and takes the shape ABC’D’ by
deforming through an angle φ.

𝑪𝑪′
 tan φ = φ =
𝑩𝑪

 The angular deformation φ in radians represents the

shear strain.
 Hooke’s Law
 Hooke’s law states that when a material is loaded within
proportional limit, stress is directly proportional to
strain.
Mathematically,
σ α𝜺
σ=E𝜺
Where, E = constant of proportionality and called as
Young’s Modulus or Modulus of Elasticity
σ 𝒔𝒕𝒓𝒆𝒔𝒔
Now, E= ; 𝐄= 𝒔𝒕𝒓𝒂𝒊𝒏
𝜺
𝑷 ∆𝑳
 We know that, Stress (σ) = and Strain (ε) =
𝑨 𝑳
𝑃𝐿
∆𝐿 =
𝐴𝐸
 𝒔𝒕𝒓𝒆𝒔𝒔
 Young Modulus, 𝐄= 𝒔𝒕𝒓𝒂𝒊𝒏

 Unit of Young’s modulus is same as unit of stress

because strain is dimensionless quantity.

 N/m2, Pa, Kpa, Mpa, Gpa

 E is a property of the material

 Poisson’s Ratio(μ)
 It is defined as the ratio of lateral strain to
longitudinal strain.

𝑳𝒂𝒕𝒆𝒓𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏
 Poisson’s Ratio(μ) =
𝑳𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏

 Poisson’s ratio is dimensionless.

 The value of μ lies between 0.25 to 0.33 for most of the

engineering materials.
Lecture No. 6
 Elastic Constants
 Different types of stresses and their corresponding
strains within elastic limit are related which are referred
to as elastic constants.

 For homogenous and isotropic material no. of elastic

constants are four.(μ, E, G, K).
1) Poisson’s Ratio(μ)
2) Modulus Of Elasticity or Young’s modulus(E)
3) Modulus Of Rigidity or Shear Modulus(G)
4) Bulk Modulus(K)
 Elastic constants are used to-
i. determine strain theoretically
ii. obtain relationship between stress and strain
Modulus Of Elasticity or Young’s
Modulus(E)
 It is defined as the ratio of normal stress and normal

strain, when material is loaded within elastic limit.

σ 𝒔𝒕𝒓𝒆𝒔𝒔
 E= =
𝜺 𝒔𝒕𝒓𝒂𝒊𝒏

 Unit of Young’s modulus is same as unit of stress

because strain is dimensionless quantity.

 N/m2, Pa, Kpa, Mpa, Gpa

 E is a property of the material.

 Modulus of Rigidity or Shear
Modulus(G)
 It is defined as the ratio of shear
stress and shear strain, when
material is loaded within elastic limit.

𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒆𝒔𝒔 
G= =
𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒂𝒊𝒏 φ

 Unit of shear modulus is same as

unit of stress because strain is
dimensionless quantity.
 N/m2, Pa, KPa, MPa, GPa
 Bulk Modulus(K)
 It is defined as the ratio of volumetric stress and
volumetric strain, when material is loaded within elastic
limit.

𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈v
G= =
𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝑺𝒕𝒓𝒂𝒊𝒏 𝜺v

 Unit of Bulk modulus is same as unit of stress because

strain is dimensionless quantity.

 N/m2, Pa, KPa, MPa, GPa

 Relationship between E, G and µ

Consider a cubic element ABCD

When the block is subjected to tangential force it distorts to a new
shape ABC’D’.

Longitudinal stain in diagonal

AC
= =
 Relationship between E, G and µ
Extension CC’ is very small , AC’B is assumed to be equal to
ACB =45°.
EC’ = CC’ COS 45
= CC’/ √2

From triangle BCC’


2

Where  represents the shear strain.

 Relationship between E, G and µ
In terms of shear stress  and modulus of rigidity G
Shear strain() = /G
Longitudinal stain for diagonal AC = /2G

The strain in diagonal AC is also given by =

Strain due to tensile stress in AC – strain due to compressive stress in
BD
Note: In case of pure shear stress (σ = )

/2G
 Relationship between E, K and µ
 Consider a cubical element subjected to volumetric stress  which

acts simultaneously along mutually perpendicular x , y and z

direction.

 𝜺x = strain in x-direction due to𝝈𝑥 - strain in x direction due to𝝈𝑦 - Strain in x-

direction due to 𝝈𝑧

 Strain in x direction-
Relationship between E, K and µ
Lecture No. 7
 Stress-Strain diagram
A. For Ductile material(e.g. Mild steel)
 Proportional limit(A):
 Upto this limit, stress is a linear function of strain
and material obeys Hook’s law.
 0-A is a straight line of the curve and its slope
represents the value of modulus of elasticity.

 Elastic limit(B):
 It represents maximum stress upto which
material is still able to regain its original shape
and size after removal of load i.e. upto this point
deformation is recoverable.
 Upper yield point(C) and Lower yield point(D):
 Beyond elastic limit, the material shows
considerable strain even though there is no
increase in load or stress.
 Deformation is not fully recoverable i.e. the
behaviour of material is inelastic.
 This phenomenon from C to D is called yielding.
 Ultimate stress point(E):
 After yielding has taken place, the material
becomes hardened and increase in load is required
to take the material to its maximum stress at point
E.
 Point E represents the maximum stress of this
curve and this point is known as ultimate stress
point.
 Breaking point(F):

 In the portion EF, there is falling off the load(stress)

from the maximum until fracture takes place at F.

 The point F is known as fracture or breaking point

and corresponding stress is called the breaking
stress.
 Stress-Strain diagram
B. For Brittle material (e.g. Cast Iron)
 For brittle materials, like cast Iron, no appreciable
deformation is obtained and the failure occurs without
yielding.

 Brittle material fails suddenly with out any indication.

 Ductile materials give indication before failure.
 Factor of safety(FoS)
 FoS is defined as Ultimate stress to working stress.
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
FoS =
𝑊𝑜𝑟𝑘𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
 A factor of safety increases the safety of people and
reduces the risk of failure of a product.
 If a structure fails there is a risk of injury and death as
well as a company's financial loss.
 How much FoS is required depends on the materials and
its applications.
 As the FoS increases, the cost of the product also
increases therefore cost is also a considerable
parameter.
 For safe design FoS must be greater than unity.
 Factor of safety(FoS)
 Ultimate Stress: It is the maximum stress that a
material can withstand while being stretched or pulled
before breaking.

 Working Stress: working stress is known as the

maximum allowable stress that a material or object will
be subjected to when it is in service. This stress is
always lower than the Yield stress and the ultimate
tensile stress.
Lecture No. 8
 Question 1.1
A steel bar of 1.5 m long, 50 mm wide and 20 mm thick
is subjected to an axial tensile load of 120 kN. If the
extension in the length of the bar is 0.9 mm, make
calculation for intensity of stress, strain and modulus
of elasticity of the bar material.

Ans: σ = 120 N/mm2, ε = 0.0006, E = 200 GN/m2

 Question 1.2
A hollow right circular cylinder is made of cast iron and
has a outside diameter of 75 mm and an inside
diameter of 60 mm. The cylinder measures 600 mm in
length and is subjected to axial compressive load of
50kN.Determine normal stress and shortening in length
of the cylinder under this load. Take the modulus of
elasticity of cast iron to be 100 GPa.

Ans: σ = 31.45 N/mm2, ɗl = 0.1887 mm

 Question 1.3
A tensile load of 56 kN was applied to a bar of 30 mm
diameter with 300 mm gauge length. Measurements
showed 0.12 mm increase in length and the
corresponding 0.0036 mm contraction in diameter.
Make calculations for the Poisson’s ratio and the
values of three moduli(elastic constants).

Ans: μ = 0.3, E = 1.9815x105 N/mm2 , G = 0.762x105

N/mm2,K = 1.65x105 N/mm2
Thank You