Prof. V. V. Nalawade, Presentation on Centroid, Centre off Gravity and Moment of Inertia

1. Prepared By
Prof. V. V. Nalawade
CENTER OF
GRAVITY AND
MOMENT OF
INERTIA
Prof. V.V.Nalawade 1

3. Class
No.
Portion Covered Per hour
1 Definitions of center of gravity, centroid, radius of gyration. Difference between
centroid and center of gravity. Difference between centroidal axis and reference
axis.
2
Centroids of some common plane figures. Activity Based on to find centroid for
some common figures.
3
Numerical on centroid for different sections.
4
Definitions of moment of inertia, radius of gyration, Parallel axis theorem and its
application, Perpendicular axis theorem and its application.
5
Moment of Inertia of some common plane figures, Examples on M. I using
parallel axis theorem.
6 Numerical on Moment of Inertia ( I, L, T sections)
7 Numerical on Moment of Inertia ( Composite sections)

4. CENTER OF GRAVITY

5. CENTER OF GRAVITY
• Centre of gravity is a point where the whole
weight of the body is assumed to act. i.e., it is
a point where entire distribution of
gravitational force is supposed to be
concentrated
• It is generally denoted “G” for all three
dimensional rigid bodies.
• e.g. Sphere, table , vehicle, dam, human etc

6. • Centroid is a point where the whole area
of a plane lamina is assumed to act.
• It is a point where the entire length,
area & volume is supposed to be
concentrated.
• It is a geometrical centre of a figure.
• It is used for two dimensional figures.
• e.g. rectangle, circle, triangle,
semicircle
CENTROID

7. 7

8. A1X1+A2X2+A3X3+……+AnXn
A1+A2+A3+……+
An
A1Y1+A2Y2+A3Y3+…….+AnYn
A1+A2+A3+…….+
An
• Step 1.Divide the composite figure into regular standard figures.
• Step 2.calculate the area of each component also mark the c.g of respective
component.
• Step 3.draw the reference axis in x and y direction. Measure the distances of
centroids of each figure from given reference axis.
• Step 4.calculate the centroidal distance of composite figure by using following
formula.
Where, X1,X2,X3… are distances
of centroids of components from
reference axis drawn In y-
direction & Y1,Y2,Y3… are
distances of centrods of
components from reference axis
drawn in x-direction.
(note: For hollow areas use
negative sign)
Procedure To Locate Centroid Of
Composite Area
8

9. Problems on CG

11. 100 mm
Y- Axis
X-Axis
30 mm
120 mm
30 mm
G
G
y1
y2

24. Assignment 1

25. Assignment 2

28. MOMENT OF INERTIA
• The concept which gives a quantitative
estimate of the relative distribution of
area & mass of a body with respect to
some reference axis is termed as the
MOMENT OF INERTIA of the body.

29. RADIUS OF GYRATION
• The radius of gyration of a given area
about any axis is that distance from the
given axis at which the entire area
without changing the MI about the
given axis.
• It is denoted by K or r.

30. MI OF PLANE AREA ABOUT
THE CENTROIDAL AXES
a1
a2
a3
y1 y2 y3
x1
x2
x3
X Axis
Y Axis

31. • Let us consider an irregular area A to be
divided into small elementary areas a1,a2 &
a3….., an of known geometric shapes as shown
in fig.,
• Let x1,x2 & x3….., xn = respective distances of
the areas a1,a2 & a3….., an from Y-Y axis.
• Let y1,y2 & y3….., yn = respective distances of
the areas a1,a2 & a3….., an from X-X axis.
• Now , calculate the Moments of area of each
area
• First moment of area a1 about Y-Y axis = a1 .
X1
Continue……………

32. Continue……………
• Second moment of area a1 about Y-Y axis
= (a1 . x1) . x1 = a1 . x1
2
• Similarly the second moment of area a2 about
Y-Y axis = a2 . x2
2
• About a3 = a3 . x3
2
• Up to an = an . xn
2
Therefore sum of second moment of all
elementary areas about Y-Y axis
= a1 . x1
2 + a2 . x2
2 + a3 . x3
2
+…………………………+ an . xn
2 = Σax2
 Since the moment is taken @ Y-Y axis, Σax2
denotes the MI of an area @ Y-Y axis & is
denoted by I

33. Continue……………
 Since the moment is taken @ Y-Y axis, Σax2
denotes the MI of an area @ Y-Y axis & is
denoted by IYY.
 Hence IYY = Σax2
 Similarly the sum of second moment of all
elementary areas @ X-X axis is denoted by
Ixx.
 Hence Ixx = Σay2
 And unit of moment of inertia = unit of area .
(Unit of distance)2 = mm2 . mm2 = mm4
 Therefore unit of MI = mm4 / m4 / cm4… etc

34. Continue……………
From this Moment of Inertia (MI)
of a body about any axis is defined
as the sum of second moment of all
elementary areas about that axis.

35. Concept of radius of Gyration
K
X
y1 y2
y3
a1
a2
a3
AREA A
X X
Fig. i Fig. ii

36. Continue……………
 Let us consider an irregular area A to be
divided into infinite no. of small elementary
areas a1, a2, a3 ,………an of known geometric
shapes as shown in fig.
• Let y1,y2 & y3….., yn be the distances of the
centroid of the areas a1,a2 & a3….., an from X-
X axis.
• We know that, Ixx = Σay2
• = a1 . y1
2 + a2 . y2
2 + a3 . y3
2 +…………………………+ an .
yn
2 ………………………. eqn 1
• Now, let us assume that the entire area A is
concerned into a long narrow horizontal strip
as shown in fig. ii

37. Continue……………
• Now, let us assume that the entire area A is
concerned into a long narrow horizontal strip
as shown in fig. ii
• So that y1 = y2 = y3 =…………= yn =K
• & a1 + a2 + a3 +…………+ an = A
 therefore Ixx = K2A…………. eqn 2
• Where K is known as radius of gyration about
x-x axis.
• Now, from eqn 2, we have
 K2 = Ixx / A
 K = Root of Ixx / A
• From this, the radius of gyration of the area
A @ y-y axis can be calculated.

38. PARALLEL AXIS THEOREM
• It status that, “ The MI of a plane section
about any axis parallel to the centroidal axis is
equal to the MI of the section about the
centroidal axis plus the product of the area of
the section & the square of the distance
between the two axes.”
G
P
h
X
Q
X
AREA A

39. Continue……………
• Consider an irregular area A as shown in fig.
• Let G be its centroid,
• IG = MI of area about its centroidal axis
• IPQ = MI of the area about any axis PQ which
is parallel to the centroidal axis x-x
• A = Area of the section
• h = Distance between the two axes
• Then by parallel axis theorem,
• IPQ = IG + A.h2
• In this case, IG = Ixx
IPQ = IXX + A.h2
G
P
h
X
Q
X
AREA A

40. Example……………
• Consider rectangular section of width b and
depth d as shown in fig,
• Let h be the distance between x-x & P-Q axis.
• h’ = Distance between Y-Y & A-B axis.
• Now MI of a rectangle @ the axis PQ is given
by,
• IPQ = IG + Ah2
• = IXX + Ah2
• = bd3 / 12 + (b.d)h2
G
P
h
X
Q
X
B
A
h’
Y
Y
d
b

41. Continue……………
• And MI of a rectangle about the axis AB is
given by;
IAB= IG + A.h’2
 = IYY + A.h’2
 = db3 / 12 + (b.d) .h’2
• USE: This theorem is used to find the MI of
any plane figure about any axis located at
some distance away from the centroidal axis.

42. 42
Moment of Inertia of simple shape :-

43. PERPENDICULAR AXIS THEOREM
• It states that, “ If Ixx & Iyy are the
moments of inertia of a plane section about
the two mutually perpendicular axes meeting
at ‘O’, then the MI of Izz about the third
axis z-z perpendicular to the plane & passing
through the intersection of x-x & y-y is given
by,
• Izz = Ixx + Iyy
• This third axis zz is called as Polar axis,
therefore this theorem is also called as polar
axis theorem.
• It is denoted by Ip & is given by
• Izz = Ip = Ixx + Iyy

45. 2
Ixx Ig Ah
 
2
Iyy Ig Ah
 
/
Kxx Ixx A

/
Kyy Iyy A

/
Kzz Izz A

Im/
Kab m

Where , A= Area of plane body.
h= Perpendicular distance between reference axis and centroidal axis.
If mass moment of inertia is calculated about an axis ab then radius of
gyration about an axis ab is
Kab =
45
2.3 Parallel Axis Theorem:-
2.4 Radius of Gyration:-