CENTROID AND CENTRE OF GRAVITY SPP.pptx

CENTROID AND
CENTRE OF GRAVITY
Prof. Samirsinh P Parmar
Mail: samirddu@gmail.com
Asst. Professor, Department of Civil Engineering,
Faculty of Technology,
Dharmsinh Desai University, Nadiad-387001
Gujarat, INDIA
CL-101 ENGINEERING MECHANICS
B. Tech Semester-I
Lecture-02

Content of the presentation:
1. Introduction.
2. Centroid.
3. Methods for Centre of Gravity.
4. Centre of Gravity by Geometrical Considerations.
5. Centre of Gravity by Moments.
6. Axis of Reference.
7. Centre of Gravity of Plane Figures.
8. Centre of Gravity of Symmetrical Sections.
9. Centre of Gravity of Unsymmetrical Sections.
10. Centre of Gravity of Solid Bodies.
11. Centre of Gravity of Sections with Cut out Holes.
CL-101 Engg. Mechanics, DoCL, DDU- SPP 2

INTRODUCTION
• The Centre of gravity is a theoretical point in the body where the body’s total weight is thought to be
concentrated.
• It is important to know the centre of gravity because it predicts the behaviour of a moving body when
acted on by gravity.
• It is also useful in designing static structures such as buildings and bridges.
• In a uniform gravitational field, the centre of gravity is identical to the centre of mass. Yet, the two points
do not always coincide.
• For the Moon, the centre of mass is very close to its geometric centre. However, its centre of gravity is
slightly towards the Earth due to the stronger gravitational force on the Moon’s near side.
• In a symmetrically shaped object formed of homogenous material, the centre of gravity may match the
body’s geometric centre. However, an asymmetrical object composed of various materials with different
masses is likely to have its centre of gravity located at some distance away from its geometric centre.
In hollow bodies or irregularly shaped bodies, the centre of gravity lies at a point external to the
physical material.

INTRODUCTION → C.G.

Centre of Gravity

Centre of Gravity
Normal Load Over Load Imbalanced Over Load CG disturbed → Overturning

REAL LIFE APPLICATIONS
 It is for reasons of stability that the luggage compartment of a tour bus is
located at the bottom and not on the roof
 Extra passengers are similarly not allowed on the upper deck of a
crowded double-decker bus.

REAL LIFE APPLICATIONS
 Racing cars are built low and broad for stability
 Bunsen burners, table lamps and fans are designed with large, heavy
bases to make them stable.
 The legs of a baby’s highchair are set wide apart so that the chair is stable.

APPLICATIONS
• To design the structure for supporting a water tank, we will need to
know the weights of the tank and water as well as the locations
where the resultant forces representing these distributed loads
act.
• One concern about a sport utility vehicle (SUV) is that it might tip
over while taking a sharp turn.
• One of the important factors in determining its stability is the
SUV’s center of mass.
• Should it be higher or lower to make a SUV more stable?
• To design the ground support structure for the goal post, it is
critical to find total weight of the structure and the center of gravity
location.
• Integration must be used to determine total weight of the goal post
due to the curvature of the supporting member.

CENTROID
• The plane figures (like triangle, quadrilateral, circle etc.) have only areas, but no
mass.
• The center of area of such figures is known as centroid.
• The method of finding out the centroid of a figure is the same as that of finding
out the center of gravity of a body.
• In many books, the authors also write center of gravity for centroid and vice versa.

Centre of Gravity vs. Centroid:
Comparison Chart
Centre of Gravity Centroid
Centre of mass of a geometric object with
any density
Centre of mass of a geometric object of
uniform density
Point where weight of a body or system
may be considered to act
Geometrical centre
Denoted by g Denoted by c

METHODS FOR CENTRE OF GRAVITY
• The center of gravity (or centroid) may be found out by any one
of the following two methods:
1. By geometrical considerations
2. By moments
3. By graphical method
• As a matter of fact, the graphical method is a tedious and
cumbersome method for finding out the centre of gravity of
simple figures.

CENTRE OF GRAVITY BY GEOMETRICAL
CONSIDERATIONS
• The center of gravity of simple figures may be found
out from the geometry of the figure as given below.
1. The center of gravity of uniform rod is at its middle
point.
2. The center of gravity of a rectangle (or a
parallelogram) is at the point, where its diagonals
meet each other. It is also a middle point of the
length as well as the breadth of the rectangle as
shown in Fig. 1.
3. The center of gravity of a triangle is at the point,
where the three medians (a median is a line
connecting the vertex and middle point of the
opposite side) of the triangle meet as shown in
Fig.2.
Fig.1. Rectangle
Fig. 2. Triangle

CENTRE OF GRAVITY BY GEOMETRICAL CONSIDERATIONS
4. The center of gravity of a
trapezium with parallel
sides a and b is at a
distance of
𝒉
𝟑
×
𝒃+𝟐𝒂
𝒃+𝒂
measured form the side b
as shown in Fig.3.
5.The center of gravity of a semicircle
is at a distance of
𝟒𝒓
𝟑𝝅
from its base
measured along the vertical radius as
shown in Fig. 4.
Fig.3. Trapezium Fig.4. Semicircle

6. The center of gravity of a circular sector
making semi-vertical angle α is at a distance
of
𝟐
𝟑
×
𝑺𝒊𝒏𝜶
𝜶
from the center of the sector
measured along the central axis as shown
in Fig. 5.
Fig. 5. Circular sector Fig. 6. Hemisphere
7. The center of gravity of a hemisphere is at a
distance of
𝟑𝒓
𝟖
from its base, measured along
the vertical radius as shown in Fig.6.

8. The center of gravity of a cube is
at a distance of
𝒍
𝟐
from every face
(where, l is the length of each side).
9. The center of gravity of a
sphere is at a distance of d/2
from every point (where, d is
the diameter of the sphere).
10. The center of gravity of right circular solid
cone is at a distance of
𝒉
𝟒
from its base,
measured along the vertical axis as shown in
Fig. 7.
11. The center of gravity of a segment of sphere
of a height h is at a distance of
𝟑
𝟒
×
𝟐𝒓−𝒉 𝟐
𝟑𝒓−𝒉
from the center of the sphere measured along
the height. as shown in Fig.8.
Fig. 7. Right circular solid cone Fig. 8. Segment of a sphere

CENTRE OF GRAVITY BY MOMENTS
The centre of gravity of a body may also be found out by moments as discussed below:
Consider a body of mass M whose centre of gravity is required to be found out. Divide the body into
small masses, whose centres of gravity are known as shown in Fig. 6.9. Let m1, m2, m3....; etc. be
the masses of the particles and (x1, y1), (x2, y2), (x3, y3), ...... be the co-ordinates of the centres of
gravity from a fixed point O as shown in Fig. 6.9.
Let x and y be the co-ordinates of the centre of gravity of the body. From the principle of moments, we
know that
Fig. 6.9. Centre of gravity by moments

AXIS OF REFERENCE
• The center of gravity of a body is always calculated with reference to some assumed
axis known as axis of reference (or sometimes with reference to some point of
reference).
• The axis of reference, of plane figures, is generally taken as the lowest line of the figure
for calculating y and the left line of the figure for calculating x .

CENTRE OF GRAVITY OF LINEAR ELEMENTS

22
Example : Locate the centroid of the
uniform wire bent as shown in Fig
CL-101 Engg. Mechanics, DoCL, DDU- SPP

CENTRE OF GRAVITY OF PLANE FIGURES
• The plane geometrical figures (such as T-section, I-section, L-section etc.) have only
areas but no mass.
• The center of gravity of such figures is found out in the same way as that of solid bodies.
• The center of area of such figures is known as centroid, and coincides with the center of
gravity of the figure.
• It is a common practice to use center of gravity for centroid and vice versa.

CENTRE OF GRAVITY OF PLANE FIGURES
Let x and y be the co-ordinates of the center of gravity with respect to some axis of reference, then
Where, a1, a2, a3........ etc., are the areas into which the whole figure is divided x1, x2, x3 ..... etc.,
are the respective co-ordinates of the areas a1, a2, a3....... on X-X axis with respect to same axis of
reference.
y1, y2, y3....... etc., are the respective co-ordinates of the areas a1, a2, a3....... on Y-Y axis with
respect to same axis of the reference.
Note. While using the above formula, x1, x2, x3 ..... or y1, y2, y3 or x and y must be measured from the same axis of reference (or
point of reference) and on the same side of it.
However, if the figure is on both sides of the axis of reference, then the distances in one direction are taken as positive and those in
the opposite directions must be taken as negative.

CENTRE OF GRAVITY OF
SYMMETRICAL SECTIONS
• Sometimes, the given section,
whose centre of gravity is required
to be found out, is symmetrical
about X-X axis or Y-Y axis.
• In such cases, the procedure for
calculating the centre of gravity of
the body is very much simplified;
as we have only to calculate either
x or y .
• This is due to the reason that the
centre of gravity of the body will lie
on the axis of symmetry.
symmetrical
about X-X axis
symmetrical
about Y-Y axis.

Example 1. Find the center of gravity of a 100 mm × 150 mm × 30 mm T-section.
Solution. As the section is symmetrical about Y-Y axis, bisecting the web, therefore
its center of gravity will lie on this axis. (∴ 𝑋 = 50 mm)
Split up the section into two rectangles ABCH and DEFG as shown in Figure.
Answer (𝑋, 𝑌) = (50mm, 94.1mm)
𝑌 CL-101 Engg. Mechanics, DoCL, DDU- SPP 27

Example: 2. Find the center of gravity of a channel section 100 mm × 50 mm × 15 mm.
As the section is symmetrical about X-X axis, therefore its center of gravity
will lie on this axis.
Now split up the whole section into three rectangles ABFJ, EGKJ and
CDHK as shown in Figure.
We know that distance between the centr
the
section and left face of the section AC,

Example: 2. Find the center of gravity of a channel section 100 mm × 50 mm × 15 mm.
Cont……
Answer (𝑋, 𝑌) = (17.8 mm, 50 mm)

Example 6.3. 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 center of gravity of the section

CENTRE OF GRAVITY OF
UNSYMMETRICAL SECTIONS

Example: 3. Find the centroid of an unequal
angle section 100 mm × 80 mm × 20 mm.
Solution. As the section is not symmetrical about any axis, therefore
we have to find out the values of x and y for the angle section.
Split up the section into two rectangles as shown in Figure..
Let left face of the vertical section and bottom face of the horizontal
section be axes of reference.

Example: 3. Find the centroid of an unequal
angle section 100 mm × 80 mm × 20 mm.
Answer (𝑋, 𝑌) = (25 mm, 35 mm)
C.G
𝑋 = 25mm
𝑌 = 35mm

Example:4. A uniform lamina shown in Figure consists of
a rectangle, a circle and a triangle. Determine the center
of gravity of the lamina. All dimensions are in mm.
Answer (𝑿, 𝒀) = (71.1 mm, 32.2 mm)

Example:5. Locate the centroid of the plate area
Solution:
Composite Parts
Plate divided into 3 segments.
Area of small rectangle considered “negative”.

EXERCISE .1

CENTRE OF GRAVITY OF SOLID BODIES
The center of gravity of solid bodies (such as hemispheres, cylinders, right circular solid
cones etc.) is found out in the same way as that of plane figures.
The only difference, between the plane figures and solid bodies, is that in the case of solid
bodies, we calculate volumes instead of areas.
The volumes of few solid bodies are given below :

Volume formulas:

Example: A solid body formed by joining the base of
a right circular cone of height H to the equal base of a
right circular cylinder of height h. Calculate the
distance of the centre of mass of the solid from its
plane face, when H = 120 mm and h = 30 mm.
Solution: As the body is symmetrical about the vertical
axis, therefore its center of gravity will lie on this axis (i.e.
y-y axis) as shown in Figure. (∴ 𝑿 = r)
Let r be the radius of the cylinder base in cm. Now let
base
of the cylinder be the axis of reference.

Example: A solid body formed by joining the base of
a right circular cone of height H to the equal base of a
right circular cylinder of height h. Calculate the
distance of the centre of mass of the solid from its
plane face, when H = 120 mm and h = 30 mm.

Example: . A body consists of a right circular solid cone of height 40 mm and radius 30 mm placed on a solid
hemisphere of radius 30 mm of the same material. Find the position of center of gravity of the body.
Solution. As the body is symmetrical about Y-Y axis, therefore its centre of
gravity will lie on this axis as shown in Figure.
Let bottom of the hemisphere (D) be the point of reference.

Example: . A body consists of a right circular solid cone of height 40 mm and radius 30 mm placed on a solid
hemisphere of radius 30 mm of the same material. Find the position of center of gravity of the body.
We know that distance between center of gravity of
the body and bottom of hemisphere D,

Example : . A right circular cylinder of 12 cm diameter is joined with a hemisphere of the same
diameter face to face. Find the greatest height of the cylinder, so that centre of gravity of the
composite section coincides with the plane of joining the two sections. The density of the material of
hemisphere is twice that the material of cylinder.

EXERCISE - HW

CENTRE OF GRAVITY OF SECTIONS WITH CUT OUT HOLES
The center of gravity of such a section is found out by considering the main section,
first as a complete one, and then deducting the area of the cut out hole i.e., by taking
the area of the cut out hole as negative. Now substituting a2 (i.e., the area of the cut
out hole) as negative, in the general equation for the center of gravity, we get
Note. In case of circle the section will be symmetrical along the line joining the
centers of the bigger and the cut out circle.

Example : A square hole is punched out of circular lamina, the diagonal of the square
being the radius of the circle as shown in Figure. Find the center of gravity of the
remainder, if r is the radius of the circle.
Solution. As the section is symmetrical about X-X axis, therefore its
center of gravity will lie on this axis. Let A be the point of reference.

Example. A semicircle of 90 mm radius is cut out from a trapezium as shown in Figure. Find the
position of the center of gravity of the figure.
Solution. As the section is symmetrical about Y-Y axis, therefore its center of gravity will lie on this axis.
Now consider two portions of the figure viz., trapezium ABCD and semicircle EFH.
Let base of the trapezium AB be the axis of reference.

Example. A semicircular area is removed from a
trapezium as shown in Fig. (dimensions in mm) Determine
the centroid of the remaining area (shown hatched).
Solution. As the section in not symmetrical about
any axis, therefore we have to find out the values of
𝑋and 𝑌 for the area. Split up the area into three
parts as shown in Fig. Let left face and base of the
trapezium be the axes of reference.

Sample Problem
50
For the plane area shown, determine
the first moments with respect to the
x and y axes and the location of the
centroid.
SOLUTION:
• Divide the area into a triangle, rectangle,
and semicircle with a circular cutout.
• Compute the coordinates of the area
centroid by dividing the first moments by
the total area.
• Find the total area and first moments of
the triangle, rectangle, and semicircle.
Subtract the area and first moment of the
circular cutout.
• Calculate the first moments of each area
with respect to the axes.

Sample Problem
51
3
3
3
3
mm
10
7
.
757
mm
10
2
.
506






y
x
Q
Q
• Find the total area and first moments of the
triangle, rectangle, and semicircle. Subtract the
area and first moment of the circular cutout.

Sample Problem
52
2
3
3
3
mm
10
13.828
mm
10
7
.
757







A
A
x
X
mm
8
.
54

X
2
3
3
3
mm
10
13.828
mm
10
2
.
506







A
A
y
Y
mm
6
.
36

Y
• Compute the coordinates of the area
centroid by dividing the first moments by
the total area.

Example. A solid consists of a right circular cylinder and a
hemisphere with a cone cut out from the cylinder as
shown in Fig. Find the center of gravity of the body.
Answer (𝑋, 𝑌) = (93.75 mm, 60 mm)

Example. A frustum of a solid right circular cone has an
axial hole of 50 cm diameter as shown in Fig. Determine
the centre of gravity of the body.
Solution. As the body is
symmetrical about vertical axis,
therefore its center of gravity lie
on this axis. For the sake of
simplicity, let us assume a right
circular cone OCD, from which a
right circular cone OAB is cut off
as shown in Fig.

We know that distance between center of gravity of the
body and the base of the cone,

Example. A solid hemisphere of 20 mm radius supports a solid cone of the same base and 60 mm height as
shown in Fig. 6.30. Locate the center of gravity of the composite section. If the upper portion of the cone is
removed by a certain section, the center of gravity lowers down by 5 mm. Find the depth of the section plane (h)
below the apex.
Solution. As the body is symmetrical about Y-Y axis, therefore its centre of
gravity will lie on this axis. Let apex of the cone (O) be the axis of reference.
Centre of gravity of the composite section

We know that distance between center of gravity of the body and apex of the cone,
Depth of the section plane below the apex

EXERCISE

Determination of Centroids by Integration
5 - 61
 
 
ydx
y
dA
y
A
y
ydx
x
dA
x
A
x
el
el








2
 
 
 
 
dx
x
a
y
dA
y
A
y
dx
x
a
x
a
dA
x
A
x
el
el











2
























d
r
r
dA
y
A
y
d
r
r
dA
x
A
x
el
el
2
2
2
1
sin
3
2
2
1
cos
3
2












dA
y
dy
dx
y
dA
y
A
y
dA
x
dy
dx
x
dA
x
A
x
el
el
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.

Sample Problem
5 - 62
Determine by direct integration the
location of the centroid of a parabolic
spandrel.
SOLUTION:
• Determine the constant k.
• Evaluate the total area.
• Using either vertical or horizontal
strips, perform a single integration to
find the first moments.
• Evaluate the centroid coordinates.

Sample Problem
5 - 63
SOLUTION:
• Determine the constant k.























2
1
2
2
2
2
2
1
)
(
)
(
b
y
x
or
a
x
a
b
y
a
b
k
a
k
b
a
x
k
y
• Evaluate the total area.
3
3
)
(
0
2
2
3
2
0
2
2
ab
x
a
ax
x
a
b
dx
a
x
a
b
dx
y
dA
A
a
a


















Sample Problem
5 - 64
• Using vertical strips, perform a single integration
to find the first moments.
 
 
 
10
2
1
2
1
2
12
2
0
4
4
2
0
2
2
2
2
0
2
0
2
2
ab
dx
a
x
a
b
dx
a
x
a
b
dx
y
y
dA
y
Q
b
a
a
b
dx
a
x
a
b
x
dx
xy
dA
x
Q
a
a
el
x
a
a
el
y



























Sample Problem
5 - 65
• Or, using horizontal strips, perform a single
integration to find the first moments.
 
10
1
12
2
1
2
2
2
0
2
3
2
1
2
1
2
0
2
1
2
1
2
0
2
ab
dy
y
b
a
ay
dy
b
y
ya
yxdy
dA
y
Q
b
a
dy
y
b
a
a
dy
x
dy
x
x
dA
x
Q
b
el
x
b
b
el
y


















































Sample Problem
5 - 66
• Evaluate the centroid coordinates.
12
3
2
b
a
ab
x
Q
A
x y


4
a
x 
10
3
2
ab
ab
y
Q
A
y x


b
y
10
3


Centroids of common shapes of areas.

Centroids Useful equations
for Integration method

Centroids of common shapes of lines

CENTROID AND CENTRE OF GRAVITY SPP.pptx

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to CENTROID AND CENTRE OF GRAVITY SPP.pptx

Similar to CENTROID AND CENTRE OF GRAVITY SPP.pptx (20)

More from Samirsinh Parmar

More from Samirsinh Parmar (20)

Recently uploaded

Recently uploaded (20)

CENTROID AND CENTRE OF GRAVITY SPP.pptx