Centroid

SERIN ISSAC, ASSISTANT PROFESSOR, NEW HORIZON COLLEGE OF ENGINEERING

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

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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 and .   y x         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 from the y-axis and at a distance from the x-axis (the point through which the total weight W acts).   G     A SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 11

12. Assume the area A is divided into infinite small areas , , , … etc and their corresponding centroids are , , …etc. Let ( ), ), )…. etc be the coordinates of the centroids w.r.t x axis and y axis .               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 Sum of moments of small areas about y axis = + + …. e tc. = Using Varignon’s theorem of moments, A x = Therefore , = Similarly, =   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

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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, = b = b . =   SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 19

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

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

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

23. Whereas, = 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 ( ) as shown in Figure. Now consider a triangular elementary strip of area ( x R x Rd θ ) at an angle of θ from the x-axis. Its centre of gravity is R from O. its projection on the x-axis = R cos θ Moment of area of elementary strip about the y-axis = ( x R x Rd θ ) . R cos θ =   SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 25

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

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34. 120mm 10mm 6 0mm 10mm NUMERICAL 1 SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 34

35. 120mm 10mm 6 0mm 10mm x y 1 2 SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 35

36. COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) AREA ( 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 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) 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       = = 60mm   = = 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.       = = 4 0mm   = = 44.44mm   COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) AREA ( 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 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) 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 2 0mm 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 x 80   x 80      

44.       = = 90.37mm   = = 31.11mm   COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) AREA ( ax ay TRIANGLE 1 x 80 = 53.33 x 80 = 26.66 x 80 x 80 = 3200 1,70,656 85,312 QUARTER CIRCLE 2 = 113.95 = 33.95 = 5026.54 5,72,774.23 170651.03 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) ax ay TRIANGLE 1 1,70,656 85,312 QUARTER CIRCLE 2 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 50mm 150mm 150mm 150mm 150mm 50mm 50mm 50mm

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

47.       = = 107.19 mm   = = 107.19 mm   COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) AREA ( 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 = 63.66 = 63.66 = - 17671.45 - 11,24,964.5 - 11,24,964.5 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) 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 - 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 ( ax ay 1 RECTANGLE = 6.5 = 4.5 13 x 9 = 117 760.5 526.5 2 SEMI CIRCLE = 2 9 - = 8.15 = - 6.28 -12.56 -51.182 3 TRIANGLE 1 = 12 3 + = 5 x 3 x 6 = - 9 -108 -45 4 SQUARE 10 + = 11.5 = 1.5 3 x 3 = - 9 - 103.5 -13.5 5 QUARTER CIRCLE 10 - = 8.72 = 1.27 = - 7.06 - 61.56 -8.97 COMPONENT CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) ax ay 1 RECTANGLE 13 x 9 = 117 760.5 526.5 2 SEMI CIRCLE -12.56 -51.182 3 TRIANGLE -108 -45 4 SQUARE 3 x 3 = - 9 - 103.5 -13.5 5 QUARTER CIRCLE - 61.56 -8.97 1 2 3 4 5       = = 5.51 mm   = = 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 m 1m 1m 1m

52. SERIN ISSAC, DEPARTMENT OF CIVIL ENGINEERING, NHCE 52 COMPONENT CENTROIDAL x DISTANCE (m) CENTROIDAL y DISTANCE (m) AREA ( ax ay 1 TRIANGLE = 1 = 3 x 1.5 x 9 = 6.75 6.75 20.25 2 RECTANGLE 1.5 + = 3 = 6.5 3 x 13 = 39 117 253.5 3 TRIANGLE 1.5 + 3 + = 6.5 = 3 x 6 x 9 = 27 175.5 81 4 SEMICIRCLE 1.5 + 1 + 0.5 = 3 + 2 +1 = 3.2 = - 0.3926 -1.1778 -1.26025 5 RECTANGLE 1.5 + 1 + 0.5 = 3 + 1 = 2 = - 2 -6 -4 COMPONENT CENTROIDAL x DISTANCE (m) CENTROIDAL y DISTANCE (m) ax ay 1 TRIANGLE 6.75 20.25 2 RECTANGLE 3 x 13 = 39 117 253.5 3 TRIANGLE 175.5 81 4 SEMICIRCLE 1.5 + 1 + 0.5 = 3 -1.1778 -1.26025 5 RECTANGLE 1.5 + 1 + 0.5 = 3 -6 -4 1 2 3 4 5       = = 4.15 m   = = 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 ( ax ay 1 TRIANGLE 20 + 25 = 35 x 40 x 30 = 600 12000 21000 2 RECTANGLE 20 = 12.5 30 x 25 = 750 15000 9375 3 RECTANGLE 20 = 10 10 x 20 = - 200 -4000 -2000 4 SEMICIRCLE 20 + 20 = 22.12 = - 39.3 -786 -869.316 5 CIRCLE 20 10 + 25 = 35 = - 78.53 -1570.6 -2748.55 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) ax ay 1 TRIANGLE 20 12000 21000 2 RECTANGLE 20 30 x 25 = 750 15000 9375 3 RECTANGLE 20 10 x 20 = - 200 -4000 -2000 4 SEMICIRCLE 20 -786 -869.316 5 CIRCLE 20 10 + 25 = 35 -1570.6 -2748.55       = = 20 mm   = = 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 ( ax ay 1 SEMICIRCLE - = -0.955 = 7.95 - 7.952 17.88 2 RECTANGLE = 3 = 2.25 6 x 4.5 = 27 81 60.75 3 RIGHT ANGLED TRIANGLE 6 + = 7 = 1.5 = 6.75 47.25 10.125 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) ax ay 1 SEMICIRCLE - 7.952 17.88 2 RECTANGLE 6 x 4.5 = 27 81 60.75 3 RIGHT ANGLED TRIANGLE 47.25 10.125       = = 2.88 mm   = = 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 ( ax ay 1 SEMICIRCLE 200 = 84.88 = 62831.85 12,566,370 5,333,167.428 2 SEMICIRCLE = -169.76 = 251327.41 100,530,964.9 -42,665,341.51 3 SEMICIRCLE 400 + 200 = 600 = - 84.88 =- 62831.85 -37,699,110 5,333,167.428 COMPONENTS CENTROIDAL x DISTANCE (mm) CENTROIDAL y DISTANCE (mm) ax ay 1 SEMICIRCLE 200 12,566,370 5,333,167.428 2 SEMICIRCLE 100,530,964.9 -42,665,341.51 3 SEMICIRCLE 400 + 200 = 600 -37,699,110 5,333,167.428 251327.41       = = 300 mm   = = - 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