UNIT-01 (Introduction to Mechanics)

1. Define force with its effect and characteristics?

Force: It is a push or a pull acting on an object which changes or tends to change the state of the object.
Effect of a force:
(i) External Effect (rigid body)
(a) To keep the body at rest (Reactions)
(b) To change the condition of motion (Acceleration, retardation, displacement, velocity, time)
(ii) Internal Effect (Deformable body)
(a) Induce the stress in the body (Stress)
(b) Produce the strain in the body (Strain)
Characteristics of a Force:
(i) Magnitude
(ii) Direction
(iii) Point of application 10N
(iv) Line of action

2. What do you understand by the moment of a force?

Moment: It is defined as the product of the force (F) and the moment arm (d).
The moment arm (d) is the perpendicular distance between the line of action of the force and the actual
point about which the force causes rotation.
Moment = Force x Moment arm
M = F x d (N-m)

M=FXd

3. What do you understand by the couple or coupling moment?

Couple: Two equal forces of opposite direction, with a distance between them will cause a moment,
called as couple.
The moment caused by a couple = Force x arm of the couple = F X d
4. What do you understand by principle of transmissibility of force?
Principle of transmissibility: It state that a force applied to a rigid body will have the same effect if it is
transmitted anywhere along its line of action within the body
Limitation: This principle is only applicable for rigid bodies

1 1 1

Proof: For example, consider a force ‘F’ acting at point ‘O’ on a rigid body as shown in fig (a).
There is another point ‘O1’ in the same line of action of the force ‘F’ suppose at this point O1 two equal
and opposite forces F1 and F2 of magnitude F are applied as shown in fig (b).
The force F and F2 being equal and opposite will cancel each other, leaving a force F1 at point O1 as
shown in fig(c).But force F1 is equal to force F. The original force F acting at point O has been
transferred to point O1, which is along the line of action of F without changing the effect of the force on
the rigid body. Hence any force acting at a point on a rigid body can be transmitted to act at any other
point along its line of action without changing its effect on the rigid body. This proves the principle of
transmissibility of forces.

5. What do you understand by Varignon’s theorem or principle of moments?

Varignon’s theorem: If a number of coplanar forces are acting on a body, then the algebraic sum of
their moments about a point in their plane is equal to the moment of their resultant about the same
point.“

Proof: Let F1, and F2, be the two forces represented by the lines AB and AD and diagonal AC
represent the resultant R.

Moment of force F1 about O = M1 = 2 x Area of △OAB…..(i)

Moment of force F2 about O = M2 = 2 x Area of △OAD……(ii)
Moment of resultant force R about O =MR = 2 x area of △OAC …..(iii)
Sum of moments of two forces (F1 & F2) about O
= 2 Area of △OAB + 2 x Area of △OAD
= 2 Area of △ADC + 2 x Area of △OAD
= 2 x area of △OAC……(iv)

So, from equation (iii) & (iv) the sum of the moment of two forces (F1 & F2) about O is equal to the
moment of the resultant force R about O.
6. What do you understand by force system? Also give its classification.
Force System: When two or more forces act on a body, it is known as force system.

7. What is the difference between coplanar and non coplanar force system?
Coplanar Force System: When the lines of action of a set of forces lie
in a single plane is called coplanar force system.

Non-Coplanar Force System: When the line of action of all the forces
do not lie in one plane, is called Non-coplanar force system
8. Define: (i) Coplanar collinear force system (ii) Coplanar concurrent force system (iii) Coplanar
non concurrent force system (iv) Coplanar parallel force system
(i) Coplanar Collinear Force system: When the line of action of
all the forces is in the same straight line and in the same plane
it is known as coplanar collinear force system.

(ii) Coplanar Concurrent Force system: When two or more forces

intersect at a single point on the same plane is called
coplanar concurrent force system

(iii) Coplanar Non Concurrent Force system: When two or

more forces do not intersect at a single point on the
same plane is called coplanar non concurrent force system

(iv)Coplanar parallel Force system : When two or more

forces acting on a body are parallel to each other on the
same plane, the force system is called coplanar parallel force system

9. What do you understand by principle of composition of forces?

Principle of composition of forces: The combined effect of any
force system (Resultant force) acting on a rigid body is the sum
of the effects of individual forces and the principle of finding
that resultant force is called as principle of composition.

10. What do you understand by principle of resolution of force?

Principle of resolution of force: A single force can be divided into
two components which are perpendicular to each other in
such a way that the effects remains same as that of
single force and this method is called as resolution of a force.
11. What do you understand by resultant? Also write the different methods of finding the Resultant of
coplanar force system.
Resultant: It is defined as a single force which replaces any coplanar force system in such a way that it
produces the same effect.
Methods of finding the resultant of a coplanar concurrent force system:
Graphical Method: (a) Parallelogram law of forces
(b)Triangle law of forces
(c) Polygon law of forces

Analytical method: (a) Parallelogram law of forces

(b) Method of resolution

12. What do you understand by parallelogram law of forces?

Parallelogram law: If two force vectors can be represented by the two adjacent sides of a parallelogram
drawn from a point, then their resultant sum vector is represented completely by the diagonal of the
parallelogram drawn from the same point.

Proof of Parallelogram law of forces:

Extend the vector P till D such that CD is perpendicular to OD. Since OB is parallel to AC, therefore the
angle AOB is equal to the angle CAD as they are corresponding angles, i.e., angle CAD = θ. Now, first,
we will derive the formula for the magnitude of the resultant vector R (side OC).
In right-angled triangle OCD, we have
OC2 = OD2 + DC2
⇒ OC2 = (OA + AD)2 + DC2 --- (1)
In the right triangle CAD, we have
cos θ = AD/AC and sin θ = DC/AC
⇒ AD = AC cos θ and DC = AC sin θ
⇒ AD = Q cos θ and DC = Q sin θ --- (2)
Substituting values from (2) in (1), we have
R2 = (P + Q cos θ)2 + (Q sin θ)2
⇒ R2 = P2 + Q2cos2θ + 2PQ cos θ + Q2sin2θ
⇒ R2 = P2 + 2PQ cos θ + Q2(cos2θ + sin2θ)
⇒ R2 = P2 + 2PQ cos θ + Q2 [cos2θ + sin2θ = 1]
⇒ R = √(P2 + 2PQ cos θ + Q2) → Magnitude of the resultant vector R
Next, we will determine the direction of the resultant vector. We have in right traingle ODC,
tan β = DC/OD
⇒ tan β = Q sin θ/(OA + AD) [From (2)]
⇒ tan β = Q sin θ/(P + Q cos θ) [From (2)]
⇒ β = tan-1[(Q sin θ)/(P + Q cos θ)] → Direction of the resultant vector R
13. Define: (i) Triangle law of forces (ii) Polygon law of forces
(i) Triangle law of forces: If two forces acting at a point are represented in magnitude and direction by
the two adjacent sides of a triangle taken in order, then the closing side of the triangle taken in the
reversed order represents the resultant of the forces in magnitude and direction

(ii) Polygon law of forces: If a number of forces acting simultaneously on a particle be represented in
magnitude and direction by the sides of a polygon taken in order, their resultant may be represented in
magnitude and direction by the closing side of the polygon taken in opposite order.

14. Explain the method of resolution of forces for finding the resultant of coplanar force system.
(1) Resolve all the forces along the horizontal x-axis and vertical y-axis.

(2) Calculate the algebraic sum of all the forces in +x-axis F X

and +y-axis F
y

(3) Magnitude of the resultant (R) is calculated by:R   F X

2
  FY 2 

(4) Direction of the resultant (α) with horizontal x-axis is calculated by:   tan 1
F Y

F X

(5) Resultant (R) may lie in any four quadrant depending on the sign of  FX and F y

15. What do you understand by static equilibrium? Also write the equations of equilibrium for
coplanar concurrent and non concurrent force system.
Static Equilibrium: When a body is at rest under the action of any force system then it is called as
static equilibrium.
(a) Equations of static equilibrium for a coplanar concurrent force system:
(i) Net horizontal component of force system = 0;

(ii) Net vertical component of force system = 0;

 (b) Equations of static equilibrium for a coplanar non-concurrent force system:
(i) Net horizontal component of force system = 0;

(ii) Net vertical component of force system = 0;

(iii) Net moment of the force system about any point in the same plane = 0;

16. What do you understand by free body diagram?

Free body diagram: A graphical representation of any object which is isolated from any system with all
the forces (External forces, body weight and reactions) is called as free body diagram.
Steps for FBD:
(i) Isolate the body to be analyze and draw its diagram
(ii) Identify all the forces acting in the body: External forces, body weight and support reactions
(iii) Incorporate all the forces in the diagram of the body

Example: Pulling a block in an inclined plane.

17. State and prove the Lami’s theorem.

Lami’s theorem: if three coplanar forces acting at a point be in static equilibrium, then each force is
proportional to the sine of the angle between the other two forces

Proof of Lami’s theorem: P, Q and R are the three forces acting at a point and α, β and γ are the
angles as shown in Fig. and under the action of these three forces the body is at equilibrium.
When we draw the sides of a polygon in an order by using P, Q, R then we get a complete triangle as
shown in the fig.

Applying sine rule in the triangle: P/sin(180−α)=Q/sin(180−β)=R/sin(180−ϒ)

P/(sinα)=Q/(sinβ)=R/(sinϒ)...... Hence Proved

18. What are the different support reactions? Explain with neat sketches.
1. Simple Support: When any object simply rest on the supporting surface then it is called a simple
support and the reactive force is normal to the surface known as normal reaction.

RY

2. Roller support: When any object rest on the supporting surface with rolling contact then it is called a
roller support and the reactive force is normal to the rolling surface known as normal reaction.

RY
3. Hinged or pinned support: When any object rest on the supporting surface with hinge or pin contact
then it is called a hinged or pinned support and the reactive force have horizontal and vertical
component.

RX
RY
4. Fixed support: When any object fixed with its supporting surface then it is called fixed support and
the reactive force have horizontal and vertical component as well as one additional reactive moment.

19. What are the different types of loading? Explain with neat sketches.
(i) Point or concentrated load: When the applied load is concentrated load at a point then it is called as
point or concentrated load.

(ii) Uniformly distributed load (UDL): When the applied load is distributed with uniform load
intensity then it is called as uniformly distributed load.
 (iii) Uniformly varying load (UVL): When the applied load is distributed with uniform varying load
intensity then it is called as uniformly varying load.

20. What do you understand by beam? Also give the classification of beams.
Beam: A beam is a structural element that primarily resists loads applied laterally to the beam's axis.

Beams are of two types: (i) Statically determinate beams (ii) Statically indeterminate beams
(i) Statically determinate beams: Support reactions can be determined by equations of static
equilibrium.
Example: Simply supported beam

(ii) Statically indeterminate beams: Support reactions cannot be determined by equations of static
equilibrium.
Example: Continuous beam

21. What do you understand by stress? Also define different types of stress?
Stress: - It is defined as the resistance developed by the body against applied load, and is defined as load
per unit area.
Types of stress: (1) Normal stress (2) Shear stress
(1)Normal Stress: -Stress that acts perpendicular to the area is called
normal stress.The normal stress is again subdivided into two parts.
(a)Tensile Stress: - The stress-induced in a body
when it is subjected to two equal and opposite pulls
is called tensile stress.
(b)Compressive Stress :- The stress which induced in a
body when it is subjected to two equal and opposite pushes
is called compressive stress.
(2)Shear Stress: - Shear stress induced in a body when it is subjected to two equal and opposite forces
that acts tangential to the area.
22. What do you understand by strain? Also define different types of strain?
Strain :- It is defined as the change in dimension to the original dimension of the specimen due to
external loading.
Types of strain:
(i) Longitudinal strain: The ratio of change in
length to its original length in the direction of
loading is known as longitudinal strain.
Longitudinal strain = ΔL/L

(ii) Lateral strain: The ratio of change in length

to its original length in the perpendicular or lateral direction
of the loading is called lateral strain.
Lateral strain = ΔD/D

23. What do you understand by Poisson’s ratio?

Poisson’s ratio (µ) : It is defined as the ratio of lateral strain to the longitudinal strain within elastic limit.

Poisson’s ratio (µ) = Lateral strain / Longitudinal strain

= (ΔD/D) / (ΔL/L)

24. Define Hooke’s law.

Hooke’s Law :- When a material is loaded, within its elastic limit, the stress is proportional to strain.

25. Define all elastic constants.

(a) Modulus of Elasticity (E): It is defined as the ratio of longitudinal stress and longitudinal strain
within the elastic limit.
Modulus of elasticity or Young’s modulus = longitudinal stress / longitudinal strain
(b) Modulus of Rigidity (G): It is defined as the ratio of shear stress and shear strain within the elastic
limit.
Modulus of rigidity or Shear modulus = shear stress / shear strain
(c) Bulk Modulus (K): when a body is subjected to three mutually perpendicular stresses of equal
intensity, the ratio of volumetric stress to corresponding volumetric strain is known as bulk modulus.
Bulk Modulus = volumetric stress / volumetric strain
26. Establish the relationship between Modulus of elasticity (Young’s modulus) and Modulus of
rigidity (Shear modulus).
Let us establish a relation among the elastic constants E, G and µ.
Consider a cube of material of side ‘a' subjected to the shear stress as shown in the figure and producing
the strained shape as shown in the figure below.
Assuming that the strains are small and the angle A C B may be taken as 450.
Therefore strain on the diagonal OA = Change in length / original length
The strain on diagonal OA = BC/OA = AC Cos45 / OA
εOA = a Tan ϒ. Cos45 / a √2
εOA = a.ϒ. Cos45 / a √2 = ϒ/2……(i)

Modulus of rigidity = G = Shear stress / Shear strain

G=τ/ϒ
Or ϒ = τ / G…………..(ii)
So from equation (i): εOA = τ / 2G……. .(iii)
Thus, for the direct state of stress system, which applied along the diagonals:

27. Establish the relationship between Modulus of elasticity (Young’s modulus) and Bulk modulus.
Consider a cube (side L) subjected to three equal stresses (σ) as shown in the figure below
Consider the deformation of the side of the cube under the action of three mutually perpendicular tensile
stress of equal intensity.
Side AB of the cube will suffer the following strains :-
(1)Tensile strain equal to σ /E due to stresses on the face AEHD
and BFGC .
(2)Compressive lateral strain equal to -µσ /E due to stresses on
the face AEFB and DHGC .
(3)Compressive lateral strain equal to -µσ /E due to stresses on face ABCD and
EFGH

Now, Net tensile strain in side AB will b

28. What do you understand by principle of superposition?
The principle of superposition states that when there are numbers of loads acting together on an elastic
material, the resultant deformation will be the algebric sum of individual deformations caused by each load
acting separately.

29. Explain the stress strain diagram of ductile material (Mild steel).
1.Proportional Limit Region: OA = The region from Point O to Point A is Proportional Limit
Region. In this region, Stress is directly proportional to Strain in this region.
2.Elastic Limit Region: AB = The region from Point A to Point B is Elastic Limit Region. In this
region, if we remove load in this region the material regain its original shape and size.
 3.Plastic Region: BC = After point B Plastic Region starts which means if we remove load after point
B the material don’t regain its original shape and size. The yielding of material starts at point C.
4.Upper and Lower Yield Point: CD = Point C shows the Upper Yield point and Point D shows the
Lower Yield point. Yielding of material starts at Upper Yield point C and ends at Lower Yield Point D.
5.Strain Hardening Region: DE = Point D to Point E is Strain Hardening Region. Strength of
material is increased with the application of load in this region therefore more stress is required for
deformation in this region.The point E is Ultimate Stress point. Stress at this point is known
as Ultimate Stress.
6.Breaking Point: EF = From point E to F Necking of the material begins. After point E the cross
sectional area of material starts decreasing at rapid rate. The specimen breaks at point F, known
as Breaking Point. The stress at this point is known as Nominal Breaking Stress.

30. Explain the stress strain diagram of brittle material (Cast iron).
Brittle materials such as cast iron don’t exhibit a
yielding phenomenon.
These materials skip the plasticity region and undergo a
fracture directly after the elasticity region.

31. Define factor of safety.

Factor of safety (FOS): It is defined as the ratio of maximum allowable stress to working or design
stress.

Factor of safety = Maximum allowable stress / Working or design stress

Factor of safety for ductile materials: The maximum allowable stress for ductile materials is the yield
stress so for ductile materials the factor of safety is given by:
Factor of safety for ductile material = Yield stress / Working stress

Factor of safety for brittle materials: The maximum allowable stress for brittle materials is the
ultimate stress so for brittle materials the factor of safety is given by:
Factor of safety for brittle material = Ultimate stress / Working stress

*******************************************************************************