The document discusses the concepts of centroid and centre of gravity. It defines centroid as the point where the whole area of a plane figure can be assumed to be concentrated. The centre of gravity is the point where the entire weight of a body acts and can be balanced. The document then provides methods to determine the centroid of common geometric shapes such as rectangles, triangles, semicircles, and composite shapes using the principle of moments. It includes examples of calculating the x- and y-coordinates of the centroid for various geometric problems.

1. CENTROID
MODULE IV
SERIN ISSAC
ASSISTANT PROFESSOR
DEPARTMENT OF CIVIL ENGINEERING
NHCE

2. CENTRE OF GRAVITY
 It is the point where the whole weight of the body is assumed to be
concentrated. It is the point on which the body can be balanced.
 It is the point through which the weight of the body is assumed to act. This
point is usually denoted by ‘C.G.’ or ‘G’.
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
2

3. CENTROID
 Centroid is the point where the whole area of the plane figure is
assumed to be concentrated.
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
3

4. CENTROID
CENTRE OF GRAVITY
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
4

5. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
5

6.  It is easy to find the centroid of simple shapes.
 If the object has an axis of symmetry the centroid will always lie on
that axis.
 If the object has two axes of symmetry, the centroid will be at the
intersection of the two axes.
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
6

7. SIMPLE GEOMETRIC SHAPES
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 7

8. COMPOSITE GEOMETRIC SHAPES
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 8

9. COMPOSITE GEOMETRIC SHAPES
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 9

10.  CENTROID of a figure is always represented in a coordinate system as shown in figure
below. The calculation of centroid means the determination of 𝒙 and 𝒚.
y
x
𝒙
𝒚
𝑏
ℎ
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 10

11. Determination of Centroid by the
Method of Moments
 Let us consider a plane area A. The centre of gravity/ centroid of the area G
is located at a distance 𝒙 from the y-axis and at a distance 𝒚 from the x-axis
(the point through which the total weight W acts).
G𝒙
𝒚
A
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 11

12. Assume the area A is divided into infinite small areas 𝑎1, 𝑎2,
𝑎3, 𝑎4…etc and their corresponding centroids are 𝑔1, 𝑔2, 𝑔3,
𝑔4…etc.
Let (𝑥1, 𝑦1), (𝑥2, 𝑦2), (𝑥3, 𝑦3) , (𝑥4, 𝑦4)….etc be the coordinates
of the centroids w.r.t x axis and y axis.
𝑎1
𝑎2 𝑎3
𝑎4
𝑎6
𝑎5
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 12

13. Applying the principle of MOMENT of area,
Moment of Total area A about y axis = Area x centroidal distance
= A x 𝒙
Sum of moments of small areas about y axis
=𝑎1 𝑥1 + 𝑎2 𝑥2 + 𝑎3 𝑥3 + 𝑎4 𝑥4…. etc.
= 𝑎𝑥
Using Varignon’s theorem of moments,
A x 𝒙 = 𝑎𝑥
Therefore, 𝒙 =
𝒂𝒙
𝑨
Similarly, 𝒚 =
𝒂𝒚
𝑨
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
13

14. Axes of Reference
These are the axes with respect to
which the centroid of a given
figure is determined.
Centroidal Axis
The axis which passes through the
centroid of the given figure is known as
centroidal axis, such as
the axis X-X and the axis Y-Y shown in
Figure.
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
14

15. SYMMETRICAL AXES
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 15

16. SYMMETRICAL AXES
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 16

17. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 17

18. DERIVATION OF CENTROID OF
SOME IMPORTANT GEOMETRICAL
FIGURES
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 18

19. RECTANGLE
 Let us consider a rectangular lamina of area (b x d) as shown in
Figure.
 Now consider a horizontal elementary strip of area (b x dy),
which is at a distance y from the reference axis AB.
 Moment of area of elementary strip about AB
= (b x dy) . y
 Sum of moments of such elementary strips about AB is
given by,
𝟎
𝒅
(b . dy) . y
= b 𝟎
𝒅
ydy
= b .
𝒚 𝟐
𝟐 𝟎
𝒅
=
𝒃𝒅 𝟐
𝟐
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
19

20.  Moment of total area about AB = bd . 𝒚
 Apply the principle of moments about AB,
bd . 𝒚 =
𝒃𝒅 𝟐
𝟐
𝒚 =
𝒅
𝟐
By considering a vertical strip, similarly, we can prove that
𝒙 =
𝒃
𝟐
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
20

21. TRIANGLE
 Let us consider a triangular lamina of area ( 𝟏
𝟐 x b x d) as
shown in Figure.
 Now consider a horizontal elementary strip of area (𝒃 𝟏 x dy),
which is at a distance y from the reference axis AB.
 Using the property of similar triangles, we have
𝑏1
𝑏
=
𝑑 −𝑦
𝑑
𝑏1=
𝑑 −𝑦
𝑑
. b
 Area of the elementary strip =
𝒅 −𝒚
𝒅
. b . dy
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 21

22.  Moment of area of elementary strip about AB
= Area x y
=
𝒅 −𝒚
𝒅
. b . dy. y
Sum of moments of such elementary strips is given by
=
=
𝒃𝒅 𝟐
𝟔
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
22

23.  Whereas, 𝒙 =
𝒃
𝟐
triangle is symmetrical about y axis
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
23

24. y y
x x
b/3
d/3 d/3
2b/3
b
d
b
d
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
24

25. SEMI CIRCLE
 Let us consider a semi circular lamina of area (
𝝅𝑹 𝟐
𝟐
) as shown in Figure.
 Now consider a triangular elementary strip of area (
𝟏
𝟐
x R x Rdθ) at an
angle of θ from the x-axis.
 Its centre of gravity is
𝟐
𝟑
R from O.
 its projection on the x-axis =
𝟐
𝟑
R cosθ
 Moment of area of elementary strip about
the y-axis = (
𝟏
𝟐
x R x Rdθ) .
𝟐
𝟑
R cosθ
=
𝑹 𝟑 𝒄𝒐𝒔𝜽 d 𝜽
𝟑
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
25

26. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
26

27. QUATER CIRCLE
 Let us consider a quarter circular lamina of area (
𝝅𝑹 𝟐
𝟒
) as shown in Figure.
 Now consider a triangular elementary strip of area (
𝟏
𝟐
x R x Rdθ) at an
angle of θ from the x-axis.
 Its centre of gravity is
𝟐
𝟑
R from O.
 its projection on the x-axis =
𝟐
𝟑
R cosθ
 Moment of area of elementary strip about
the y-axis = (
𝟏
𝟐
x R x Rdθ) .
𝟐
𝟑
R cosθ
=
𝑹 𝟑 𝒄𝒐𝒔𝜽 d 𝜽
𝟑
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
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28. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
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29. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 29

30. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
30

31. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
31

32. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 32

33. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 33

34. 120mm
10mm
60mm
10mm
NUMERICAL 1
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 34

35. 120mm
10mm
60mm
10mm
x
y
1
2
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 35

36. COMPONENTS CENTROIDAL
x DISTANCE
(mm)
CENTROIDAL
y DISTANCE
(mm)
AREA (𝒎𝒎 𝟐
) ax ay
Rectangle 1 60 65 120 x 10 = 1200 72,000 78,000
Rectangle 2 60 30 60 x 10 = 600 36,000 18,000
𝑎 = 1800 𝑎𝑥 = 108000 𝑎𝑦 = 96,000
𝑥 =
𝑎𝑥
𝑎
=
108000
1800
= 60mm
𝑦 =
𝑎𝑦
𝑎
=
96000
1800
= 53.33mm
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 36

37. 80 mm
10mm
40mm
10mm
NUMERICAL 2
24mm
25mm
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 37

38. 80 mm
10mm
40mm
10mm
NUMERICAL 2
24mm
25mm
y
x
1
2
3
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 38

39. 𝑎 = 1800 𝑎𝑥 = 72,000 𝑎𝑦 = 80,000
𝑥 =
𝑎𝑥
𝑎
=
72,000
1800
= 40mm
𝑦 =
𝑎𝑦
𝑎
=
80,000
1800
= 44.44mm
COMPONENTS CENTROIDAL x
DISTANCE
(mm)
CENTROIDAL y
DISTANCE (mm)
AREA (𝒎𝒎 𝟐
) ax ay
RECTANGLE 1 40 5+40 +24 = 69 80 x 10 = 800 32,000 55,200
RECTANGLE 2 40 20 +24 = 44 40 x 10 = 400 16,000 17,600
RECTANGLE 3 40 12 25 x 24 = 600 24,000 7,200
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 39

40. 100 mm
20mm
100mm
20mm
NUMERICAL 3
20mm
150 mm
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 40

41. 60 mm
12mm
128mm
10mm
NUMERICAL 4
75 mm
10mm
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 41

42. NUMERICAL 5
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 42
80mm 80mm

43. NUMERICAL 5
y
x
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 43
80mm 80mm
1 2
2
3
x 80
1
3
x 80
𝟖𝟎 +
𝟒 𝐱 𝟖𝟎
𝟑𝛑
𝟒 𝐱 𝟖𝟎
𝟑𝛑

44. 𝑎 = 8226.54 𝑎𝑥 = 743430.23 𝑎𝑦 = 2,55,963.03
𝑥 =
𝑎𝑥
𝑎
=
743430.23
8226.54
= 90.37mm
𝑦 =
𝑎𝑦
𝑎
=
2,55,963.03
8226.54
= 31.11mm
COMPONENTS CENTROIDAL x
DISTANCE
(mm)
CENTROIDAL y
DISTANCE (mm)
AREA (𝒎𝒎 𝟐
) ax ay
TRIANGLE 1 2
3
x 80= 53.33
1
3
x 80 = 26.66
1
2
x 80 x 80 =
3200
1,70,656 85,312
QUARTER
CIRCLE 2
𝟖𝟎 +
𝟒 𝐱 𝟖𝟎
𝟑𝛑
=
113.95
𝟒 𝐱 𝟖𝟎
𝟑𝛑
= 33.95 𝜋x802
4
=
5026.54
5,72,774.23 170651.03
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 44

45. NUMERICAL 6
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 45
80mm 50mm150mm
150mm
150mm
150mm
50mm
50mm
50mm

46. NUMERICAL 6
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 46
80mm 50mm150mm
150mm
150mm
150mm
50mm
50mm
50mm
y
x
1
2
3
REQUIRED AREA = SQUARE 1 – RIGHT ANGLED TRIANGLE 2 – QUARTER CIRCLE 3

47. 𝑎 = 11,078.55
𝑎𝑥 = 1187535.5 𝑎𝑦 = 1187535.5
𝑥 =
𝑎𝑥
𝑎
= 107.19 mm
𝑦 =
𝑎𝑦
𝑎
= 107.19 mm
COMPONENTS CENTROIDAL x
DISTANCE (mm)
CENTROIDAL y
DISTANCE (mm)
AREA (𝒎𝒎 𝟐
) ax ay
SQUARE 1 100 100 200 x 200 = 40,000 40,00,000 40,00,000
RIGHT ANGLED
TRIANGLE 2
50 + (2/3 x 150)
= 150
50 + (2/3 x 150)
= 150
½ x 150 x150 =
-11,250
- 16,87,500 - 16,87,500
QUARTER
CIRCLE 3
4 x 150
3π
= 63.66
4 x 150
3π
= 63.66
𝜋x1502
4
= - 17671.45
- 11,24,964.5 - 11,24,964.5
SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE

48. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 48
NUMERICAL 7
1000 mm
800 mm 200 mm

49. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 49
4mm 6mm 3mm
6mm
3mm
NUMERICAL 8

50. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 50
COMPONENT CENTROIDAL x
DISTANCE (mm)
CENTROIDAL y
DISTANCE (mm)
AREA (m𝒎 𝟐
) ax ay
1 RECTANGLE 13
2
= 6.5
9
2
= 4.5 13 x 9 = 117 760.5 526.5
2 SEMI
CIRCLE
4
2
= 2 9 -
4 x 2
3𝜋
= 8.15 𝜋22
2
= - 6.28 -12.56 -51.182
3 TRIANGLE 13 −
3
3
= 12 3 +
6
3
= 5
1
2
x 3 x 6 = - 9 -108 -45
4 SQUARE 10 +
3
2
= 11.5
3
2
= 1.5 3 x 3 = - 9 -103.5 -13.5
5 QUARTER
CIRCLE
10 -
4 x 3
3𝜋
= 8.72
4 x 3
3𝜋
= 1.27 𝜋 x 32
4
= - 7.06 -61.56 -8.97
1
2
3
45
𝑎 = 85.66
𝑎𝑥 = 474.8768 𝑎𝑦 = 407.8518
𝑥 =
𝑎𝑥
𝑎
= 5.51 mm
𝑦 =
𝑎𝑦
𝑎
= 4.76 m

51. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 51
NUMERICAL 9
3 m
4 m
9 m
1.5 m 6 m
10.5 m
2 m1m
1m 1m

52. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 52
COMPONENT CENTROIDAL x
DISTANCE (m)
CENTROIDAL y
DISTANCE (m)
AREA (𝒎 𝟐
) ax ay
1 TRIANGLE 2 x 1.5
3
= 1
9
3
= 3
1
2
x 1.5 x 9 = 6.75 6.75 20.25
2 RECTANGLE 1.5 +
3
2
= 3
13
2
= 6.5 3 x 13 = 39 117 253.5
3 TRIANGLE 1.5 + 3 +
6
3
= 6.5
9
3
= 3
1
2
x 6 x 9 = 27 175.5 81
4 SEMICIRCLE 1.5 + 1 + 0.5 = 3 4 x 0.5
3𝜋
+ 2 +1 = 3.2 𝜋52
2
= - 0.3926 -1.1778 -1.26025
5 RECTANGLE 1.5 + 1 + 0.5 = 3 2
2
+ 1 = 2 1 x 2 = - 2 -6 -4
1
2
3
4
5
𝑎 = 70.3574
𝑎𝑥 = 292.0722
𝑎𝑦 = 349.4898
𝑥 =
𝑎𝑥
𝑎
= 4.15 m
𝑦 =
𝑎𝑦
𝑎
= 4.96 m

53. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 53
NUMERICAL 10
5mm
10mm
30mm
5mm 5mm
10mm 10mm 10mm
20 mm
25mm

54. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 54
y
x
1
2
3
4
5

55. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 55
COMPONENTS CENTROIDAL x
DISTANCE (mm)
CENTROIDAL y
DISTANCE (mm)
AREA (𝒎𝒎 𝟐
) ax ay
1 TRIANGLE 20 30
3
+ 25 = 35
1
2
x 40 x 30 = 600 12000 21000
2 RECTANGLE 20 25
2
= 12.5 30 x 25 = 750 15000 9375
3 RECTANGLE 20 20
2
= 10 10 x 20 = - 200 -4000 -2000
4 SEMICIRCLE 20 4 x 5
3𝜋
+ 20 = 22.12 𝜋52
2
= - 39.3 -786 -869.316
5 CIRCLE 20 10 + 25 = 35 𝜋x 52 = - 78.53 -1570.6 -2748.55
𝑎 = 1032.17
𝑎𝑥 = 20643.4
𝑎𝑦 = 24757.13
𝑥 =
𝑎𝑥
𝑎
= 20 mm
𝑦 =
𝑎𝑦
𝑎
= 23.98 mm

56. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 56
1
2
3
NUMERICAL 11

57. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 57
COMPONENTS CENTROIDAL x
DISTANCE (mm)
CENTROIDAL y
DISTANCE (mm)
AREA (𝒎𝒎 𝟐
) ax ay
1 SEMICIRCLE -
4 x 2.25
3𝜋
= -0.955 2.25 𝜋 𝑥 2.252
2
= 7.95 - 7.952 17.88
2 RECTANGLE 6
2
= 3
4.5
2
= 2.25 6 x 4.5= 27 81 60.75
3 RIGHT
ANGLED
TRIANGLE
6 +
3
3
= 7
4.5
3
= 1.5
3 x 4.5
2
= 6.75
47.25 10.125
𝑎 = 41.7
𝑎𝑥 = 120.298
𝑎𝑦 = 88.755
𝑥 =
𝑎𝑥
𝑎
= 2.88 mm
𝑦 =
𝑎𝑦
𝑎
= 2.12 mm

58. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 58
NUMERICAL 12

59. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE
COMPONENTS CENTROIDAL x
DISTANCE (mm)
CENTROIDAL y
DISTANCE (mm)
AREA (𝒎𝒎 𝟐
) ax ay
1 SEMICIRCLE 200 4 x 200
3𝜋
= 84.88 𝜋 𝑥 2002
2
= 62831.85 12,566,370 5,333,167.428
2 SEMICIRCLE 400 4 x 400
3𝜋
= -169.76 𝜋 𝑥 4002
2
= 251327.41 100,530,964.9 -42,665,341.51
3 SEMICIRCLE 400 + 200 = 600 4 x 200
3𝜋
= - 84.88 𝜋 𝑥 2002
2
=- 62831.85 -37,699,110 5,333,167.428
𝑎 =251327.41
𝑎𝑥 = 75,398,224.9
𝑎𝑦 = −31,999,006.65
𝑥 =
𝑎𝑥
𝑎
= 300 mm
𝑦 =
𝑎𝑦
𝑎
= - 127.32 mm
1
2
3
1 SEMICIRCLE + 2 SEMICIRCLE - 3 SEMICIRCLE
X
Y
-Y

60. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 60
NUMERICAL 13
X
Y

61. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 61
X = 46.11mm

62. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 62
NUMERICAL 14
DETERMINE THE CENTOID WITH RESPECT TO THE APEX

63. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 63
Y
X
-X

64. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 64