2. CO 3
Apply the conditions of equilibrium to various practical problems
involving different force
system.
CO 4
Choose appropriate theorems, principles or formulae to solve
problems of mechanics.
CO 5
Solve problems involving rigid bodies, applying the properties of
distributed areas and masses
Course Outcomes: After completion of the course the student will
be able to:
Module 3_Moment of Inertia 2
3. Module 3_Moment of Inertia 3
What is a Moment of Inertia?
It is a measure of an object’s resistance to changes to its rotation.
Also defined as the capacity of a cross-section to resist bending.
It must be specified with respect to a chosen axis of rotation.
It is usually quantified in m4 or kgm2
4. Area Moments of Inertia
• Moment of inertia also called second moment of area of plane figure with
respect to x and y axis in its plane is defined respectively by
• It is the integral of the second moment of an elemental area about
reference axis.
• Moment of Inertia of an area is purely a mathematical property of the area
5. Polar Moment of Inertia
• Moment of inertial of physical body is called mass moment of inertia
• Moment of inertia about an axis perpendicular to the plane of area.
Denoted by J or Izz
8. Module 3_Moment of Inertia 8
PARALLEL AXIS THEOREM
•It states that if the moment of inertia of a plane area about an axis in the plane of area through the C.G. of
the plane area be represented by IG, then the moment of the inertia of the given plane area about a parallel
axis AB in the plane of area at a distance h from the C.G. of the area is given by
IAB = IG + Ah2.
where IAB = Moment of inertia of the given area about AB
IG = Moment of inertia of the given area about C.G.
A = Area of the section
h = Distance between the C.G. of the section and the axis AB.
•It is a transfer theorem which is used to transfer moment of inertia from one axis to another axis
•These 2 axis should be parallel to each other and one of these axis should be centriodal axis
10. Module 3_Moment of Inertia 10
PERPENDICULAR AXIS THEOREM
Theorem of the perpendicular axis states that if IXX and IYY be the moment of inertia of a plane section
about two mutually perpendicular axis X-X and Y-Y in the plane of the section, then the moment of
inertia of the section IZZ about the axis Z-Z, perpendicular to the plane and passing through the
intersection of X-X and Y-Y is given by
IZZ = IXX + IYY.
The moment of inertia IZZ is also known as polar moment of inertia.
19. 1. Calculate the moment of inertia of the angle section having the dimensions as shown about X
and Y axis shown.
19
10 cm
10 cm
2 cm
2 cm
8 cm
10 cm
2 cm
1
2
2 cm
G2
G1
MOMENT OF INERTIA OF COMPOSITE AREAS
26. Module 3_Moment of Inertia 26
Fig. shows a T-section of dimensions 10 × 10 × 2 cm. Determine the moment of inertia of the section about the
horizontal and vertical axes, passing through the center of gravity of the section. Also find the polar moment of
inertia of the given T-section.
Ixx = 314.221
Iyy =172
Izz = Ixx + Iyy = 486.221
27. Module 3_Moment of Inertia 27
Find the moment of inertia of the section shown in Fig. about the centroidal axis X-X perpendicular to the web.
2166.667
29. Module 3_Moment of Inertia 29
Determine the moments of inertia of the section about horizontal and vertical axes passing through the
centroid of the section.
28.84 × 106
8.4 × 106
32. Module 3_Moment of Inertia 32
Find the moment of inertia of the lamina with a circular hole of 30 mm diameter about the axis AB as shown in
Fig. [Ans. 638.3 × 103 mm4]
39. Mass moment of inertia of a ring of Radius R
Let A be the cross-sectional area of the ring and 𝝆 be the mass density of the ring material
Consider an elemental length dl
Volume of this elemental length = A x dl
Mass of this elemental volume = density x volume = 𝝆 x A x dl
Second moment of this mass about ZZ axis = ⅆ𝒎 𝒙𝟐
= 𝝆 x A x dl x 𝑹𝟐
41
40. 𝑰𝒛𝒛 = 𝟎
𝟐𝝅𝑹
𝝆 x A x dl x 𝑹𝟐
= 𝑹𝟐𝝆 x A 𝒍 𝟎
𝟐𝝅𝑹
= 𝑹𝟐𝝆 x A x 𝟐𝝅𝑹
= 𝝆 x A x 𝟐𝝅𝑹x 𝑹𝟐
= m𝑹𝟐
𝑰𝒛𝒛= 𝑰𝑿𝑿 + 𝑰𝒀𝒀 = 2 𝑰𝑿𝑿
Therefore
𝑰𝑿𝑿 = 𝑰𝒀𝒀 =
m 𝑹𝟐
𝟐
42
41. Module 3_Moment of Inertia 43
Mass Moment of inertia of a Solid cylinder
Here we have to consider a few things:
•The solid cylinder has to be cut or split
into infinitesimally thin rings.
•Each ring consists of the thickness of dr
with length L.
•We have to sum up the moments of
infinitesimally these thin cylindrical
shells.
44. Theorem of Pappus’s and Guldinus
• Surface area and volume generated by rotating a curve and a plane
area about a non intersecting axis
• Another method for finding Centroid/Center of gravity.
• Surface of revolution
• The surface generated by rotating a plane curve about a fixed axis
• Body of revolution
• The body generated by rotating a plane area about a fixed axis
Module 3_Pappu's Theorem 46
45. Module 3_Pappu's Theorem 47
The first theorem of Pappus states that the surface area of a surface of
revolution generated by the revolution of a curve about an external axis is equal
to the product of the arc length of the generating curve and the distance
traveled by the curve's geometric centroid.
First theorem of Pappus
47. Module 3_Pappu's Theorem 49
Second theorem of Pappus
The second theorem of Pappus states that the volume of a solid of
revolution generated by the revolution of a lamina about an external axis is
equal to the product of the area A of the lamina and the distance traveled by
the lamina's geometric centroid.
50. • A small block of weight 1000 N as shown in Figure, is placed on a 30°
inclined plane with μ= 0.25. Determine the horizontal force to be
applied for impending motion down the plane
Module 3_Moment of Inertia 52
51. Module 3_Moment of Inertia 53
Find the centre of gravity of lamina from O.