The document discusses the topic of centroid and moment of inertia. It provides definitions of centroid, explains how to find the centroid of one, two, and three dimensional objects. It also gives several examples of calculating the centroid of composite linear elements like wires and geometric shapes. The examples show how to break objects into elements, determine the centroid of each element, and calculate the overall centroid using a weighted average.

1. Department of
Mechanical
Engineering.
Prof. Malay Badodariya
+91 9429158833
Unit no.: 5
Subject Name: FMD
01ME0305
FundamentalOfMachineDesign (01ME0504)
Unit 5:Centroid and
Moment of inertia
(Part 1 -Centroid)

2. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Unit no.: 5
Subject Name: FMD
01ME0305
Prof. Malay Badodariya
+91 9429158833

3. a)
Definition
The centroid is the centre point of the object.
It is defined as a point about which the entire line, area or
volume is assumed to be concentrated.
The centroid of a body is the point where there is equal
volume on all sides.
The centroid of a solid body made from a single material is
the center of its mass.
If the mass of a body is distributed evenly, then
the centroid and center of mass are the same.
If you cut a shape out of a piece of card it will balance
perfectly on its centroid.

4. Centroid ofTriangle:
The point in which the three medians of the triangle
intersect is known as the centroid of a triangle.
It is also defined as the point of intersection of all the three
medians.
The median is a line that joins the midpoint of a side and the
opposite vertex of the triangle.

5. Centroid of Square:
 The point where the diagonals of the square intersect each
other is the centroid of the square.
 As we all know the square has all its sides equal, hence it is
easy to locate the centroid in it.
 See the below figure, where O is the centroid of the square.
O

6. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

7. b)
Axis of
reference
The CG or centroid of a body is always calculated with
reference to some assumed axis.This assumed axis is
known as axis of reference.

8. x x
y
y
x x
y
y
X
Y
X
Y
Y X

9. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

10. One Dimensional Centroid (Length)
Sr. No. Geometrical Shape Length 𝑿 𝒀
1 L Centre of Length (L/2)
2 2πr 𝑋 = r 𝑌 = r

11. Sr. No. Geometrical Shape Length 𝑿 𝒀
3 πr 𝑋 = r 𝑌 =
2𝑟
𝜋
4 πr
2
𝑋 =
2𝑟
𝜋
𝑌 =
2𝑟
𝜋

12. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

13. Example on Composite Linear Elements
Example 1: Find centroid of wire ABCD shown in figure.
Solution:
Given Data,
D
A B
C
20 cm
15cm
10 cm
From figure we can see that,
Element 1 = AB
Element 2 = BC
Element 3 = CD
Now draw axis of reference
A B
C
20 cm
15cm
10 cm
X-Axis
Y-Axis
D

14. Example on Composite Linear Elements
Example 1: Find centroid of wire ABCD shown in figure.
Solution:
Given Data,
Now Consider Element 1 = AB
Length of AB = 20 cm
𝑋1=
𝐿
2
=
20
2
= 10 𝑐𝑚
𝑌1 = 𝐿𝑖𝑛𝑒 𝐴𝐵 𝑙𝑖𝑒𝑠 𝑜𝑛 𝑋 𝑎𝑥𝑖𝑠,
So it’s Zero.
𝑌1 = 0
A B
C
20 cm
15cm
10 cm
X-Axis
Y-Axis
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
A B
20 cm
Y-Axis
D

15. Example on Composite Linear Elements
Example 1: Find centroid of wire ABCD shown in figure.
Solution:
Given Data,
Now Consider Element 2 = BC
Length of BC = 15 cm
𝑋2= 20 𝑐𝑚
𝑌2 =
𝐿
2
=
15
2
= 7.5 cm
A B
C
20 cm
15cm
10 cm
X-Axis
Y-Axis
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
B
Y-Axis
C
15cm
X2
Y2
D
X-Axis

16. Example on Composite Linear Elements
Example 1: Find centroid of wire ABCD shown in figure.
Solution:
Given Data,
Now Consider Element 3 = CD
Length of CD = 10 cm
𝑋3= 20 + 5 = 25 𝑐𝑚
𝑌3 = 15 cm
A B
C
20 cm
15cm
10 cm
X-Axis
Y-Axis
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Y-Axis
C
X3
Y3
D
10 cm D

17. Example on Composite Linear Elements
Example 1: Find centroid of wire ABCD shown in figure.
Solution:
Given Data,
A B
C
20 cm
15cm
X-Axis
Y-Axis
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
10 cm D
Element Length Xi Yi
AB (1) 20 cm X1=10 cm Y1=0 cm
BC (2) 15 cm X2=20 cm Y2=7.5 cm
CD (3) 10 cm X3=25 cm Y3=15 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝑙1 𝑋1+ 𝑙2 𝑋2+ 𝑙3 𝑋3
𝑙1+𝑙2+𝑙3
=
200+300+250
45
= 16.67 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝑙1 𝑌1+ 𝑙2 𝑌2+ 𝑙3 𝑌3
𝑙1+𝑙2+𝑙3
=
0+112.5+150
45
= 5.83 cm
16.67 cm
5.83
cm

18. Example on Composite Linear Elements
Example 2: Find centroid of a wire making equilateral triangle of 4 cm.
Solution:
Given Data,
A B
C
4 cm X-Axis
Y-Axis
From figure we can see that,
Element 1 = AB
Element 2 = BC
Element 3 = CD
A B
4 cm
Y-Axis
Now Consider Element 1 = AB
Length of AB = 4 cm
𝑋1=
𝐿
2
=
4
2
= 2 𝑐𝑚
𝑌1 = 𝐿𝑖𝑛𝑒 𝐴𝐵 𝑙𝑖𝑒𝑠 𝑜𝑛 𝑋 𝑎𝑥𝑖𝑠,
So it’s Zero.
𝑌1 = 0

19. Example on Composite Linear Elements
Example 2: Find centroid of a wire making equilateral triangle of 4 cm.
Solution:
Given Data,
A B
C
4 cm X-Axis
Y-Axis
A
4 cm
Y-Axis
Now Consider Element 3 = AC
Length of AC = 4 cm
Sin 60 =Y3/2 ,Y3 = 2 Sin 60 = 1.73 cm
Cos 60 = X3/2 , X3 = 2 Cos 60 = 1 cm
X3 = 1 & Y3 = 1.73 cm
C
Y3
X3
B
C
Y-Axis
Y2
Now Consider Element 2 = BC
Length of BC = 4 cm
X2 = 4 – X3 = 4 – 1 = 3 cm
Y2 =Y3 = 1.73 cm
X2 = 3 & Y2 = 1.73 cm
X2

20. Example on Composite Linear Elements
Example 2: Find centroid of a wire making equilateral triangle of 4 cm.
Solution:
Given Data,
A B
C
4 cm X-Axis
Y-Axis
Element Length Xi Yi
AB (1) 4 cm X1=2 cm Y1=0 cm
BC (2) 4 cm X2=3 cm Y2=1.73 cm
CA (3) 4 cm X3=1 cm Y3=1.73 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝑙1 𝑋1+ 𝑙2 𝑋2+ 𝑙3 𝑋3
𝑙1+𝑙2+𝑙3
=
8+12+4
12
= 2 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝑙1 𝑌1+ 𝑙2 𝑌2+ 𝑙3 𝑌3
𝑙1+𝑙2+𝑙3
=
0+6.92+6.92
12
= 1.15 cm2 cm
1.15 cm

21. Example on Composite Linear Elements
Example 3: Find centroid of wire as shown in figure.
Solution:
Given Data,
A
B
C
X-Axis
Y-Axis
D
100mm100mm
100 mm
100mm100mm
100 mm
From figure we can see that,
Element 1 = AC = Straight Line
Element 2 = BC = Arc
Element 3 = BD = Line
Now Consider Element 1 = AC
Length of AC = l1 = 200 mm
𝑋1 = 0
𝑌1 =
𝐿
2
=
200
2
= 100 mm

22. Example on Composite Linear Elements
Example 3: Find centroid of wire as shown in figure.
Solution:
Given Data,
B
C
X-Axis
Y-Axis
100mm100mm
100 mm
100mm100mm
Now Consider Element 2 = BC
Length of BC = π.r = (3.14)(50) = 157.08 mm
𝑋2 =
2.𝑟
𝜋
= 31.83 mm
𝑌2 = 100 + 50 = 150 mm
X2
Y2

23. Example on Composite Linear Elements
Example 3: Find centroid of wire as shown in figure.
Solution:
Given Data,
A
B
X-Axis
Y-Axis
100mm100mm
100 mm
100mm
100 mm
Now Consider Element 3 = BD
Length of BD = l3 = 1002 + 1002 = 141.42 mm
Here AB = AD = 100 mm
So, X3 =Y3 = L/2
X3 =Y3 = 50 mm
D

24. Example on Composite Linear Elements
Example 3: Find centroid of wire as shown in figure.
Element Length Xi Yi
AB (1) 200 mm X1=0 mm Y1=100 mm
BC (2) 157.08 mm X2=31.83 mm Y2=150 mm
CA (3) 141.42 mm X3=50 mm Y3=50 mm
A
B
C
X-Axis
Y-Axis
D
100mm100mm
100 mm
D
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝑙1 𝑋1+ 𝑙2 𝑋2+ 𝑙3 𝑋3
𝑙1+𝑙2+𝑙3
=
200 𝑋 0 + 157.08 𝑋 31.83 +(141.42 𝑋 50)
200+157.08+141.42
= 24.21 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝑙1 𝑌1+ 𝑙2 𝑌2+ 𝑙3 𝑌3
𝑙1+𝑙2+𝑙3
= 101.57 mm
101.57
mm
24.21
mm

25. Example on Composite Linear Elements
Example 4: Find centroid of wire as shown in figure.
Solution:
Given Data,
X-Axis
Y-Axis
100mm200mm
100mm200mm
A
B C
D
100
mm
100
mm
From figure we can see that,
Element 1 = AC = Straight Line
Element 2 = BC = Arc
Element 3 = CD = Line
Now draw axis of reference
Here figure is symmetric so CG of
figure pass from vertical axis so
value of 𝑋= 0 so only find length of
each element andY1,Y2 &Y3.

26. Example on Composite Linear Elements
Example 4: Find centroid of wire as shown in figure.
Solution:
Given Data,
X-Axis
Y-Axis
100mm200mm
A
B C
D100
mm
100
mm
Now Consider Element 1 = AB
Length of AB = l1 = 200 mm
𝑌1 =
𝐿
2
=
200
2
= 100 mm
Now Consider Element 3 = CD
Length of AB = l3 = 200 mm
𝑌3 =
𝐿
2
=
200
2
= 100 mm
Now Consider Element 2 = BC
Length of BC = l2 = π.r = 314.06 mm
𝑌2 = 200 +
2.𝑟
𝜋
= 263.66 mm

27. Example on Composite Linear Elements
Example 4: Find centroid of wire as shown in figure.
Solution:
Given Data,
X-Axis
Y-Axis
100mm200mm
A
B C
D100
mm
100
mm
Element Length Yi
AB (1) 200 mm Y1=100 mm
BC (2) 314.06 mm Y2=263.66 mm
CD (3) 200 mm Y3=100 mm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝑙1 𝑋1+ 𝑙2 𝑋2+ 𝑙3 𝑋3
𝑙1+𝑙2+𝑙3
= 0
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝑙1 𝑌1+ 𝑙2 𝑌2+ 𝑙3 𝑌3
𝑙1+𝑙2+𝑙3
= 171.99 mm
171.99 mm

28. Example on Composite Linear Elements
Example 5: Find centroid of wire as shown in figure.
Solution:
Given Data,
Element Length Xi Yi
AB (1) 400 mm X1=200 mm Y1= 0 mm
BC (2) 471.24 mm X2=550 mm Y2=95.49 mm
CD (3) 250 mm X3= 808.25 mm Y3=62.5 mm
C F
E CE = CD/2 = 125 mm
Angle C = 30°
Sin 30° =
𝐸𝐹
𝐶𝐸
,
EF = CE Sin 30 = 62.5 mm
CF = CE Cos 30 = 108.25 mm
X3 = 400 + 300 + 108.25 = 808.25 mm
Y3 = 62.5 mm

29. Example on Composite Linear Elements
Example 5: Find centroid of wire as shown in figure.
Solution:
Given Data,
Element Length Xi Yi
AB (1) 400 mm X1=200 mm Y1= 0 mm
BC (2) 471.24 mm X2=550 mm Y2=95.49 mm
CD (3) 250 mm X3= 808.25 mm Y3=62.5 mm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝑙1 𝑋1+ 𝑙2 𝑋2+ 𝑙3 𝑋3
𝑙1+𝑙2+𝑙3
𝑋 =
400 𝑋 200 + 471.24 𝑋 550 +(808.25 𝑋 250)
400+471.24+250
= 482.71 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝑙1 𝑌1+ 𝑙2 𝑌2+ 𝑙3 𝑌3
𝑙1+𝑙2+𝑙3
𝑌 =
400 𝑋 0 + 471.24 𝑋 95.49 +(250 𝑋 62.5)
400+471.24+250
= 54.07 mm

30. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

31. Two Dimensional Centroid (Area)
Sr. No. Geometrical Shape Area 𝑿 𝒀
1 A = B.D
𝐵
2
𝐷
2
2
A =
1
2
b.h 𝑋 =
1
3
b 𝑌 =
1
3
h

32. Sr. No. Geometrical Shape Area 𝑿 𝒀
3
A = (a+b)
ℎ
2
𝑏
2
ℎ
3
𝑏 + 2𝑎
𝑏 + 𝑎
4 A = π 𝑟2 𝑋 = r 𝑌 = r

33. Sr. No. Geometrical Shape Area 𝑿 𝒀
5 A =
𝜋 𝑟2
2
𝑋 = r 𝑌 =
4 𝑟
3 𝜋
6
A =
𝜋 𝑟2
4
𝑋 =
4 𝑟
3 𝜋
𝑌 =
4 𝑟
3 𝜋

34. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

35. Example on Composite Area (2D)
Example 1: Calculate centre of gravity of T-section having flange 20 X 2 cm and web 30 X 2 cm. Also show position of
CG on figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
Now draw axis of reference
20 cm
2 cm
30 cm
2 cm

36. Example on Composite Area (2D)
Example 1: Calculate centre of gravity of T-section having flange 20 X 2 cm and web 30 X 2 cm. Also show position of
CG on figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
Now Consider Element 1
Area of Element 1 = 2 x 20 = 40 𝐶𝑀2
𝑋1=
𝐿
2
=
20
2
= 10 cm
Y1 = 30 +
𝑊
2
= 30 +
2
2
= 30 + 1 = 31 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X1
Y1

37. Example on Composite Area (2D)
Example 1: Calculate centre of gravity of T-section having flange 20 X 2 cm and web 30 X 2 cm. Also show position of
CG on figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
Now Consider Element 2
Area of Element 2 = 2 x 30 = 60 𝑐𝑚2
𝑋2= 10 cm
Y2 =
𝑊
2
=
30
2
= 15 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X2 Y2

38. Example on Composite Area (2D)
Example 1: Calculate centre of gravity of T-section having flange 20 X 2 cm and web 30 X 2 cm. Also show position of
CG on figure.
Solution:
Given Data,
20 cm
2 cm
30 cm
2 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
𝑿
𝒀
Element Area (𝒄𝒎 𝟐 ) Xi Yi
1 40 X1=10 cm Y1=31 cm
2 60 X2=10 cm Y2=15 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2
𝐴1+𝐴2
=
400+600
100
= 10 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2
𝐴1+𝐴2
=
1240+900
100
= 21.40 cm
G

39. Example on Composite Area (2D)
Example 2: Find centroid of the section shown in figure.
Solution:
Given Data,
10 cm
2.5 cm
10 cm
Now draw axis of reference
2.5 cm
2.5
cm
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm

40. Example on Composite Area (2D)
Example 2: Find centroid of the section shown in figure.
Solution:
Given Data,
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Now Consider Part 1
Area of Element 1 = 2.5 x 10 = 25 𝑐𝑚2
𝑋1=
𝐿
2
=
10
2
= 5 cm
Y1 = 2.5 + 10 +
𝑊
2
= 2.5 + 10 +
2.5
2
= 13.75 cm
X1
Y1

41. Example on Composite Area (2D)
Example 2: Find centroid of the section shown in figure.
Solution:
Given Data,
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Now Consider Part 2
Area of Element 1 = 2.5 x 10 = 25 𝑐𝑚2
𝑋2=
𝐿
2
=
2.5
2
= 1.25 cm
Y2 = 2.5 +
𝑊
2
= 2.5 +
10
2
= 7.5 cm
X2
Y2

42. Example on Composite Area (2D)
Example 2: Find centroid of the section shown in figure.
Solution:
Given Data,
10 cm
2.5 cm
10 cm
2.5 cm
2.5
cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Now Consider part 3
Area of Element 3 = 2.5 x 10 = 25 𝑐𝑚2
𝑋3=
𝐿
2
=
10
2
= 5 cm
Y3 =
𝑊
2
=
2.5
2
= 1.25 cm
X3
Y3

43. Example on Composite Area (2D)
Example 2: Find centroid of the section shown in figure.
Solution:
Given Data,
10 cm
2.5 cm
10 cm
2.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Part Area (𝒄𝒎 𝟐 ) Xi Yi
1 25 X1=5 cm Y1=13.75 cm
2 25 X2=1.25 cm Y2=7.5 cm
3 25 X3=5 cm Y2=1.25 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2+𝐴3 𝑋3
𝐴1+𝐴2+𝐴3
=
125+31.25+125
75
= 3.75 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2+𝐴3 𝑌3
𝐴1+𝐴2+𝐴3
=
343.75+187.5+31.25
75
= 7.5 cm
𝑿
𝒀
G

44. Example on Composite Area (2D)
Example 3: Find centroid of the section shown in figure.
Solution:
Given Data,
12 cm
2 cm
2 cm
2 cm
8 cm
16 cm
Now draw axis of reference

45. Example on Composite Area (2D)
Example 3: Find centroid of the section shown in figure.
Solution:
Given Data,
12 cm
2 cm
2 cm
8 cm
16 cm
Now Consider part 1
Area of Element 1 = 8 x 2 = 16 𝑐𝑚2
𝑋1=
𝐿
2
=
8
2
= 4 cm
Y1 = 2 + 10 +
𝑊
2
= 2 + 10 +
2
2
= 13 cm
X1
Y1
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,

46. Example on Composite Area (2D)
Example 3: Find centroid of the section shown in figure.
Solution:
Given Data,
12 cm
2 cm
2 cm
8 cm
16 cm
Now Consider part 2
Area of Element 2 = 10 x 2 = 20 𝑐𝑚2
𝑋2=6 +
𝐿
2
= 6 +
2
2
= 7 cm
Y2 = 2 +
𝑊
2
= 2 +
10
2
= 7 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
2 cm
X2 Y2

47. Example on Composite Area (2D)
Example 3: Find centroid of the section shown in figure.
Solution:
Given Data,
12 cm
2 cm
2 cm
8 cm
16 cm
Now Consider part 3
Area of Element 3 = 16 x 2 = 32 𝑐𝑚2
𝑋3=6 +
𝐿
2
= 6 +
16
2
= 14 cm
Y3 =
𝑊
2
=
2
2
= 1 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X3
2 cm
Y3

48. Example on Composite Area (2D)
Example 3: Find centroid of the section shown in figure.
Solution:
Given Data,
2 cm
2 cm
2 cm
8 cm
16 cm
Part Area (𝒄𝒎 𝟐 ) Xi Yi
1 16 X1=4 cm Y1=13 cm
2 20 X2=7 cm Y2=7 cm
3 32 X3=14 cm Y2=1 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2+𝐴3 𝑋3
𝐴1+𝐴2+𝐴3
=
64+140+448
68
= 9.58 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2+𝐴3 𝑌3
𝐴1+𝐴2+𝐴3
=
208+140+32
68
= 5.58 cm
𝑿 𝒀
G

49. Example on Composite Area (2D)
Example 4: Find centroid of the section shown in figure.
Solution:
Given Data,
9cm
3 cm
Now draw axis of reference
6 cm
9cm
3 cm
6 cm
Now divide object in two element
1. Rectangle
2. Right angle triangle

50. Example on Composite Area (2D)
Example 4: Find centroid of the section shown in figure.
Solution:
Given Data,
9cm
3 cm
6 cm
Now Consider part 1
Area of Element 1 = 9 x 3 = 27 𝑐𝑚2
𝑋1=
𝐿
2
=
3
2
= 1.5 cm
Y1 =
𝑊
2
=
9
2
= 4.5 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X1 Y1

51. Example on Composite Area (2D)
Example 4: Find centroid of the section shown in figure.
Solution:
Given Data,
9cm
3 cm
6 cm
Now Consider part 2
Area of Element 2 =
1
2
𝑋 𝑏 𝑋 ℎ =
1
2
𝑋 3 𝑋 9 = 13.5 𝑐𝑚2
𝑋2= 3 +
1
3
𝑏 = 3 +
1
3
𝑋 3 = 4 cm
Y2 =
1
3
ℎ =
1
3
𝑋 9 = 3 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
X2
Y2

52. Example on Composite Area (2D)
Example 4: Find centroid of the section shown in figure.
Solution:
Given Data,
9cm
3 cm
6 cm
Note:
𝑋 = Distance between Centroid of element toY axis,
𝑌 = Distance between Centroid of element to X axis,
Element Area (𝒄𝒎 𝟐 ) Xi Yi
1 27 X1=1.5 cm Y1=4.5 cm
2 13.5 X2=4 cm Y2=3 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2
𝐴1+𝐴2
= 2.33 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2
𝐴1+𝐴2
= 4 cm
𝑿
𝒀
G

53. Example on Composite Area (2D)
Example 5: Find centroid of the section shown in figure.
Solution:
Given Data, Now draw axis of reference
Now divide object in four part
1. Rectangle
2. Half Circle
3. Right angleTriangle
4. Right angle triangle
6 cm
2 cm
2 cm
3cm4cm
6 cm
2 cm
2 cm
3cm4cm
1
2
3
4

54. Two Dimensional Centroid
Sr. No. Geometrical Shape Area 𝑿 𝒀
1 A = B.D
𝐵
2
𝐷
2
2
A =
1
2
b.h 𝑋 =
1
3
b 𝑌 =
1
3
h

55. Example on Composite Area (2D)
Example 5: Find centroid of the section shown in figure.
Solution:
Given Data,
6 cm
2 cm
2 cm
3cm4cm
1
2
3
4
Now Consider Part 1 (Rectangle)
Area of Element 1 = 7 x 2 = 14 𝑐𝑚2
𝑋1=
𝐿
2
=
2
2
= 1 cm
Y1 =
𝑊
2
=
7
2
= 3.5 cm
X1
Y1

56. Example on Composite Area (2D)
Example 5: Find centroid of the section shown in figure.
Solution:
Given Data,
6 cm
2 cm
2 cm
3cm4cm
1
2
3
4
Now Consider Part 2 (Half Circle)
Area of Element 2 = A2 =
𝜋 𝑟2
2
= 3.53 𝑐𝑚2
𝑋2=2 +
4.𝑟
3.π
= 2.64 cm
Y2 = 4 + r = 5.5 cm
X2
Y2

57. Example on Composite Area (2D)
Example 5: Find centroid of the section shown in figure.
Solution:
Given Data,
6 cm
2 cm
2 cm
3cm4cm
1
2
3
4
Now Consider Part 3 (Right AngleTriangle)
Area of Element 3 = A3 =
1
2
b.h = 6 𝑐𝑚2
𝑋3=2 +
1
3
(6) = 4 cm
Y3 = 2 +
1
3
(2) = 2.67 cm
X3
Y3

58. Example on Composite Area (2D)
Example 5: Find centroid of the section shown in figure.
Solution:
Given Data,
6 cm
2 cm
2 cm
3cm4cm
1
2
3
4
Now Consider Part 4 (Right AngleTriangle)
Area of Element 3 = A4 =
1
2
b.h = 6 𝑐𝑚2
𝑋4=2 +
1
3
(6) = 4 cm
Y4 =
2
3
(2) = 1.33 cm
X4
Y4

59. Example on Composite Area (2D)
Example 5: Find centroid of the section shown in figure.
Solution:
Given Data,
6 cm
2 cm
2 cm
3cm4cm
1
2
Element Area (𝒄𝒎 𝟐 ) Xi Yi
1 14 X1=1 cm Y1=3.5 cm
2 3.53 X2=2.64 cm Y2=5.5 cm
3 6 X3= 4 cm Y3=2.67 cm
4 6 X4= 4 cm Y4=1.33 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2+𝐴3 𝑋3+𝐴4 𝑋4
𝐴1+𝐴2+𝐴3+𝐴4
=
14+9.32+24+24
29.53
= 2.42 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2+𝐴3 𝑌3+𝐴4 𝑌4
𝐴1+𝐴2+𝐴3+𝐴4
=
49+19.42+16.02+7.98
29.53
= 3.13 cm
𝑿
𝒀

60. Example on Composite Area (2D)
Example 6: A circle of 100 mm diameter is cut out from circle of 500 mm diameter. The distance between centres of
circle is 150 mm on left side to centre of big circle. Find centroid of the given lamina.
Solution:
Given Data,
100
mm
500mm
Now Consider Part 1 (Big Circle)
Area of Part 1 = 𝐴 = 𝜋𝑟2
A1 = π 2502 = 196349.54 𝑚𝑚2
𝑋1= r = 250 mm
Y1 = r = 250 mm
X-Axis
Y-Axis
150
mm

61. Example on Composite Area (2D)
Example 6: A circle of 100 mm diameter is cut out from circle of 500 mm diameter. The distance between centres of
circle is 150 mm on left side to centre of big circle. Find centroid of the given lamina.
Solution:
Given Data,
100
mm
500mm
Now Consider Part 2 (Small Circle)
Area of Part 1 = 𝐴 = 𝜋𝑟2
A2 = π 502 = 7853.98 𝑚𝑚2
𝑋2= 250 – 150 = 100 mm
Y2 = 250 mm
X-Axis
Y-Axis
150
mm

62. Example on Composite Area (2D)
Example 6: A circle of 100 mm diameter is cut out from circle of 500 mm diameter. The distance between centres of
circle is 150 mm on left side to centre of big circle. Find centroid of the given lamina.
Solution:
Given Data,
100
mm
500mm
X-Axis
Y-Axis
150
mm
Element Area (𝒎𝒎 𝟐 ) Xi Yi
1 196349.54 X1= 250 mm Y1= 250 mm
2 7853.98 X2= 100 mm Y2= 250 mm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1− 𝐴2 𝑋2
𝐴1−𝐴2
= 256.25 mm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1− 𝐴2 𝑌2
𝐴1−𝐴2
= 250 mm
𝑿
𝒀

63. Example on Composite Area (2D)
Example 7: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 1 (Half Circle)
Area of Part 1 = A1 =
𝜋 𝑟2
2
= 628.32 𝑐𝑚2
𝑋1=20 cm
Y1 = 15 + 20 +
4.𝑟
3.𝜋
= 43.48 cm
2010 10
152020
10
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 20)
2. Rectangle (40 X 35)
3. Circle (Radius 10)
4. Rectangle (20 X 15)

64. Example on Composite Area (2D)
Example 7: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 2 (Rectangle)
Area of Part 2 = A2 = 40 X 35= 1400 𝑐𝑚2
𝑋2=20 cm
Y2 =
35
2
= 17.5 cm
2010 10
152020
10
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 20)
2. Rectangle (40 X 35)
3. Circle (Radius 10)
4. Rectangle (20 X 15)

65. Example on Composite Area (2D)
Example 7: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 3 (Circle)
Area of Part 3 = A3 = 𝜋𝑟2= 314.15 𝑐𝑚2
𝑋3=20 cm
Y3 = 35 cm
2010 10
152020
10
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 20)
2. Rectangle (40 X 35)
3. Circle (Radius 10)
4. Rectangle (20 X 15)

66. Example on Composite Area (2D)
Example 7: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 4 (Rectangle)
Area of Part 4 = 20 X 15 = 300 𝑐𝑚2
𝑋4=20 cm
Y4 =
15
2
= 7.5 cm
2010 10
152020
10
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 20)
2. Rectangle (40 X 35)
3. Circle (Radius 10)
4. Rectangle (20 X 15)

67. Example on Composite Area (2D)
Example 7: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
2010 10
152020
10
X-Axis
Y-Axis
Element Area (𝒄𝒎 𝟐
) Xi Yi
1 628.32 X1=20 cm Y1=43.48 cm
2 1400 X2=20 cm Y2=17.5 cm
3 314.15 X3= 20 cm Y3=35 cm
4 300 X4= 20 cm Y4=7.5 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2−𝐴3 𝑋3−𝐴4 𝑋4
𝐴1+𝐴2−𝐴3−𝐴4
= 20 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2−𝐴3 𝑌3−𝐴4 𝑌4
𝐴1+𝐴2−𝐴3−𝐴4
= 27.27 cm
𝑿
𝒀

68. Example on Composite Area (2D)
Example 8: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
X-Axis
Y-Axis
3 3 3
1.51.563
3
Now divide object in four part
1. Big Rectangle (3 X 12)
2. RightAngleTriangle
3. Semi Circle (Radius 1.5)
4. Small Rectangle (3 X 1.5)
24
1
Now Consider Part 1
Area of Part 1 = 3 X 12 = 36 𝑐𝑚2
𝑋1=
𝐿
2
=
3
2
= 1.5 cm
Y1 =
𝑊
2
=
12
2
= 6 cm
X1
Y1

69. Example on Composite Area (2D)
Example 8: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
X-Axis
Y-Axis
3 3 3
1.51.563
3
Now divide object in four part
1. Big Rectangle (3 X 12)
2. RightAngleTriangle
3. Semi Circle (Radius 1.5)
4. Small Rectangle (3 X 1.5)
24
1
Now Consider Part 2
Area of Part 2 =
1
2
𝑏 ℎ =
1
2
𝑋 6 𝑋 9= 27 𝑐𝑚2
𝑋2= 3 + (
1
3
𝑋 6) = 5 cm
Y2 =
1
3
ℎ = 3 cm
X2
Y2

70. Example on Composite Area (2D)
Example 8: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
X-Axis
Y-Axis
3 3 3
1.51.563
3
Now divide object in four part
1. Big Rectangle (3 X 12)
2. RightAngleTriangle
3. Semi Circle (Radius 1.5)
4. Small Rectangle (3 X 1.5)
24
1
Now Consider Part 3
Area of Part 3 = A3 =
𝜋 𝑟2
2
= 3.53 𝑐𝑚2
𝑋3= 3 + 1.5 = 4.5 cm
Y3 = 1.5 +
4.𝑟
3.𝜋
= 2.13 cm
X3
Y3

71. Example on Composite Area (2D)
Example 8: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
X-Axis
Y-Axis
3 3 3
1.51.563
3
Now divide object in four part
1. Big Rectangle (3 X 12)
2. RightAngleTriangle
3. Semi Circle (Radius 1.5)
4. Small Rectangle (3 X 1.5)
24
1
Now Consider Part 4
Area of Part 4 = A4 = 3 X 1.5 = 4.5 𝑐𝑚2
𝑋4= 3 +
𝐿
2
= 4.5 cm
Y4 =
𝑊
2
= 0.75 cm
X4
Y4

72. Example on Composite Area (2D)
Example 8: Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
X-Axis
Y-Axis
3 3 3
1.51.563
3
24
1
Element Area (𝒄𝒎 𝟐
) Xi Yi
1 36 X1=1.5 cm Y1= 6 cm
2 27 X2=5 cm Y2=3 cm
3 3.53 X3= 4.5 cm Y3= 2.13 cm
4 4.5 X4= 4.5 cm Y4= 0.75 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1+ 𝐴2 𝑋2−𝐴3 𝑋3−𝐴4 𝑋4
𝐴1+𝐴2−𝐴3−𝐴4
= 2.78 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1+ 𝐴2 𝑌2−𝐴3 𝑌3−𝐴4 𝑌4
𝐴1+𝐴2−𝐴3−𝐴4
= 5.20 cm
𝑿
𝒀

73. Example on Composite Area (2D)
Example 9 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 1 (Half Circle)
Area of Part 1 = A1 =
𝜋 𝑟2
2
= 157.08 𝑐𝑚2
𝑋1=10 cm
Y1 = 20 +
4.𝑟
3.𝜋
= 24.24 cm
101020
5
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 10)
2. Circular Hole (Radius 5)
3. Rectangle (20 X 20)
4. Triangle
X1
Y1

74. Example on Composite Area (2D)
Example 9 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 2 (Circular Hole)
Area of Part 2 = A2 = π 𝑟2= 78.53 𝑐𝑚2
𝑋2=10 cm
Y2 = 20 cm
101020
5
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 10)
2. Circular Hole (Radius 5)
3. Rectangle (20 X 20)
4. Triangle
X2
Y2

75. Example on Composite Area (2D)
Example 9 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 3 (Rectangle)
Area of Part 3 = A3 = 20 x 20= 400 𝑐𝑚2
𝑋3=10 cm
Y3 =10 cm
101020
5
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 10)
2. Circular Hole (Radius 5)
3. Rectangle (20 X 20)
4. Triangle
X3
Y3

76. Example on Composite Area (2D)
Example 9 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
Now Consider Part 4 (Rectangle)
Area of Part 4= A4 =
1
2
𝑋 𝑏 𝑋 ℎ= 100 𝑐𝑚2
𝑋4=
2
3
𝑏 =
2
3
𝑋 20 = 13.33 𝑐𝑚
Y4 =
1
3
ℎ =
1
3
𝑋 10 = −3.33 𝑐𝑚
101020
5
X-Axis
Y-Axis
Now divide object in four part
1. Half Circle (Radius 10)
2. Circular Hole (Radius 5)
3. Rectangle (20 X 20)
4. Triangle
X4
Y4

77. Example on Composite Area (2D)
Example 9 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
101020
5
X-Axis
Y-Axis
Element Area (𝒄𝒎 𝟐
) Xi Yi
1 157.08 X1=10 cm Y1= 24.24 cm
2 78.53 X2=10 cm Y2=20 cm
3 400 X3= 10 cm Y3= 10 cm
4 100 X4= 13.33 cm Y4= -3.33 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝐴1 𝑋1− 𝐴2 𝑋2+ 𝐴3 𝑋3+ 𝐴4 𝑋4
𝐴1 − 𝐴2+ 𝐴3+ 𝐴4
= 10.57 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝐴1 𝑌1− 𝐴2 𝑌2+ 𝐴3 𝑌3+ 𝐴4 𝑌4
𝐴1− 𝐴2+ 𝐴3+ 𝐴4
= 10.20 cm
𝑿
𝒀

78. Example on Composite Area (2D)
Example 10 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
100
100 110
75
100
Now divide object in four part
1. Rectangle (75 X 200)
2. Quarter Circle(Radius 100)
3. Triangle
X-Axis
Y-Axis
1
2 3
Now Consider Part 1 (Rectangle)
Area of Part 1 = A1 = 75 x 200 = 15000 𝑐𝑚2
𝑋1=37.5 cm
Y1 =100 cm
X1
Y1

79. Example on Composite Area (2D)
Example 10 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
100
100 110
75
100
Now divide object in four part
1. Rectangle (75 X 200)
2. Quarter Circle(Radius 100)
3. Triangle
X-Axis
Y-Axis
1
2 3
Now Consider Part 2
Area of Part 2 = A2 =
𝜋𝑟2
4
= 7853.98 𝑐𝑚2
𝑋2=
4.𝑟
3.𝜋
= 42.44 cm
Y2 =
4.𝑟
3.𝜋
= 42.44 cm
X2
Y2

80. Example on Composite Area (2D)
Example 10 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
100
100 110
75
100
Now divide object in four part
1. Rectangle (75 X 200)
2. Quarter Circle(Radius 100)
3. Triangle
X-Axis
Y-Axis
1
2 3
Now Consider Part 3
Area of Part 3 = A3 =
1
2
𝑋 𝑏 𝑋 ℎ = 13500 𝑐𝑚2
𝑋3=75 +
1
3
𝑏= 120 cm
Y3 =
1
3
h = 66.67 cm
X3
Y3

81. Example on Composite Area (2D)
Example 10 : Find the centroid of the plane area shown in fig. (All dimension in cm).
Solution:
Given Data,
100
100 110
75
100
X-Axis
Y-Axis
1
2 3
Element Area (𝒄𝒎 𝟐
) Xi Yi
1 15000 X1=37.5 cm Y1= 100 cm
2 7853.98 X2=42.44 cm Y2=42.44 cm
3 13500 X3= 120 cm Y3= 66.67 cm
𝑿
𝒀
𝑋 = Distance between Centroid of element toY axis,
𝑋 = 89.56 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 = 100.10 cm

82. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

83. Three Dimensional Centroid (Volume)
Sr. No. Geometrical Shape Volume 𝑿 𝒀
1 V = 𝜋. 𝑟2. ℎ
r ℎ
2
2 𝑉 =
𝜋.𝑟2.ℎ
3
r
ℎ
4

84. Sr. No. Geometrical Shape Volume 𝑿 𝒀
3 V =
2
3
𝜋. 𝑟3
r
3. 𝑟
8
4
V =
4
3
𝜋. 𝑟3
r r

85. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

86. Example on Composite Area (3D)
Example 1 : Find the C.G. of the system shown in fig. (All dimension in cm).
Solution:
Given Data,
6
5
8 cm X 8 cm
86
Cuboild
8 x 8 x 5
Cylinder
Cone
X-Axis
Y-Axis
Now divide object inThree part
1. Cuboild (8 x 8 x 5)
2. Cylinder ( Dia. 6 cm, Height 8 cm)
3. Cone (Base 6 cm, Height 6 cm)
Now Consider Part 1 (Cuboild)
Volume of Part 1 =V1 = L x b x h = 320 𝑐𝑚3
𝑋1= 4 cm
Y1 = 2.5 cm
X1
Y1

87. Example on Composite Area (2D)
Example 1 : Find the C.G. of the system shown in fig. (All dimension in cm).
Solution:
Given Data,
6
5
8 cm X 8 cm
86
Cuboild
8 x 8 x 5
Cylinder
Cone
X-Axis
Y-Axis
Now divide object inThree part
1. Cuboild (8 x 8 x 5)
2. Cylinder ( Dia. 6 cm, Height 8 cm)
3. Cone (Base 6 cm, Height 6 cm)
Now Consider Part 2 (Cylinder)
Volume of Part 2 =V2 = 𝜋. 𝑟2. ℎ = 226.19 𝑐𝑚3
𝑋2= 4 cm
Y2 = 5 + 4 = 9 cm
X2
Y2

88. Example on Composite Area (2D)
Example 1 : Find the C.G. of the system shown in fig. (All dimension in cm).
Solution:
Given Data,
6
5
8 cm X 8 cm
86
Cuboild
8 x 8 x 5
Cylinder
Cone
X-Axis
Y-Axis
Now divide object inThree part
1. Cuboild (8 x 8 x 5)
2. Cylinder ( Dia. 6 cm, Height 8 cm)
3. Cone (Base 6 cm, Height 6 cm)
Now Consider Part 3 (Cone)
Volume of Part 3 =V3 =
𝜋.𝑟2.ℎ
3
= 56.55 𝑐𝑚3
𝑋3= 4 cm
Y3 = 5 + 8 +
ℎ
4
= 14.5 cm
X3
Y3

89. Example on Composite Area (3D)
Example 1 : Find the C.G. of the system shown in fig. (All dimension in cm).
Solution:
Given Data,
6
5
8 cm X 8 cm
86
Cuboild
8 x 8 x 5
Cylinder
Cone
X-Axis
Y-Axis
Element Volume (𝒄𝒎 𝟑 ) Xi Yi
1 320 X1=4 cm Y1= 2.5 cm
2 226.19 X2=4 cm Y2= 9 cm
3 56.55 X3= 4 cm Y3= 14.5 cm
𝑋 = Distance between Centroid of element toY axis,
𝑋 =
𝑉1 𝑋1+ 𝑉2 𝑋2+𝑉3 𝑋3
𝑉1+𝑉2+𝑉3
= 4 cm
𝑌 = Distance between Centroid of element to X axis,
𝑌 =
𝑉1 𝑌1+ 𝑉2 𝑌2+𝑉3 𝑌3
𝑉1+𝑉2+𝑉3
= 6.06 cm
𝑿
𝒀

90. Topic will be cover:
1. Centroid:
a) Definition
b) Axis of Reference
c) One dimensional Centroid (Length)
d) Example (5 Example)
e) Two dimensional Centroid (Area)
f) Example (10 Example)
g) Three dimensional Centroid (Volume)
h) Example (1 Example)
i) Pappus Guldinus Theorems
Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 5
Subject Name: FMD
01ME0305

91. i)
Pappus
Guldinas
Theorem
These theorems are used for finding surface area and
volume of bodies.
Theorem 1:
 "The area of surface of revolution is equal to the length
generating curve times the distance travelled by the
centroid of the curve the surface is being generated.“
Solution: A = L.θ. 𝒀 Where,
L = Length of generating curve
θ =Angle of revolution
𝑌 = Distance of centroid of
curve from the axis about
which the curve is rotated

92. Proof:
 Let semcircular arc is rotated at 360° about its base.
 After one revolution spherical shell is obtained.
Where,
L = Length of generating curve
θ =Angle of revolution
𝑌 = Distance of centroid of
curve from the axis about
which the curve is rotated

93. Theorem 2:
 "The volume of a body of revolution is equal to the
generating area times the distance travelled by the
centroid of the area while the body' being generated."
Solution: V = A.θ. 𝒀

95. Department of Mechanical
Engineering.
Prof. Malay Badodariya
malay.badodariya@marwadie
ducation.edu.in