CENTROID AND CENTRE OF GRAVITY SPP.pptx

Centroid, Centre of gravity, center of mass,
CG of Line element,
CG of 2D elements,
Solved Numerical;
Practice problem CG;
CG Solved Problems;
Equations of CG for integration method.

1. CENTROID AND CENTRE OF GRAVITY Prof. Samirsinh P Parmar Mail: [email protected] 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

2. Content of the presentation: Introduction. Centroid. Methods for Centre of Gravity. Centre of Gravity by Geometrical Considerations. Centre of Gravity by Moments. Axis of Reference. Centre of Gravity of Plane Figures. Centre of Gravity of Symmetrical Sections. Centre of Gravity of Unsymmetrical Sections. Centre of Gravity of Solid Bodies. Centre of Gravity of Sections with Cut out Holes. CL-101 Engg. Mechanics, DoCL, DDU- SPP 2

3. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 3

4. INTRODUCTION → C.G. CL-101 Engg. Mechanics, DoCL, DDU- SPP 4

5. Centre of Gravity CL-101 Engg. Mechanics, DoCL, DDU- SPP 5

6. Centre of Gravity CL-101 Engg. Mechanics, DoCL, DDU- SPP 6

7. Centre of Gravity CL-101 Engg. Mechanics, DoCL, DDU- SPP 7

8. Centre of Gravity Normal Load Over Load Imbalanced Over Load CG disturbed → Overturning CL-101 Engg. Mechanics, DoCL, DDU- SPP 8

9. CL-101 Engg. Mechanics, DoCL, DDU- SPP 9 R EAL 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.

10. CL-101 Engg. Mechanics, DoCL, DDU- SPP 10 R EAL 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.

11. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 11

12. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 12

13. 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 CL-101 Engg. Mechanics, DoCL, DDU- SPP 13

14. METHODS FOR CENTRE OF GRAVITY The center of gravity (or centroid) may be found out by any one of the following two methods: By geometrical considerations By moments 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 14

15. 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. The center of gravity of uniform rod is at its middle point. 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. 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 CL-101 Engg. Mechanics, DoCL, DDU- SPP 15

16. CENTRE OF GRAVITY BY GEOMETRICAL CONSIDERATIONS The center of gravity of a trapezium with parallel sides a and b is at a distance of measured form the side b as shown in Fig.3.   5.The center of gravity of a semicircle is at a distance of from its base measured along the vertical radius as shown in Fig. 4.   Fig.3. Trapezium Fig.4. Semicircle CL-101 Engg. Mechanics, DoCL, DDU- SPP 16

17. 6. The center of gravity of a circular sector making semi-vertical angle α is at a distance of 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 from its base, measured along the vertical radius as shown in Fig.6.   CENTRE OF GRAVITY BY GEOMETRICAL CONSIDERATIONS CL-101 Engg. Mechanics, DoCL, DDU- SPP 17

18. 8. The center of gravity of a cube is at a distance of 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 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 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 GEOMETRICAL CONSIDERATIONS CL-101 Engg. Mechanics, DoCL, DDU- SPP 18

19. 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 CL-101 Engg. Mechanics, DoCL, DDU- SPP 19

20. 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 . CL-101 Engg. Mechanics, DoCL, DDU- SPP 20

21. CENTRE OF GRAVITY OF LINEAR ELEMENTS CL-101 Engg. Mechanics, DoCL, DDU- SPP 21

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

23. CL-101 Engg. Mechanics, DoCL, DDU- SPP 23

24. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 24

25. 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, a 1 , a 2 , a 3 ........ etc., are the areas into which the whole figure is divided x 1 , x 2 , x 3 ..... etc., are the respective co-ordinates of the areas a 1 , a 2 , a 3 ....... on X-X axis with respect to same axis of reference. y 1 , y 2 , y 3 ....... etc., are the respective co-ordinates of the areas a 1 , a 2 , a 3 ....... on Y-Y axis with respect to same axis of the reference. Note. While using the above formula, x 1 , x 2 , x 3 ..... or y 1 , y 2 , y 3 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 25

26. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 26

27. 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. ( = 50 mm ) Split up the section into two rectangles ABCH and DEFG as shown in Figure.   Answer ( , ) = (50mm, 94.1mm)     CL-101 Engg. Mechanics, DoCL, DDU- SPP 27

28. 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 centre of gravity of the section and left face of the section AC , CL-101 Engg. Mechanics, DoCL, DDU- SPP 28

29. Example: 2. Find the center of gravity of a channel section 100 mm × 50 mm × 15 mm . Cont … … Answer ( , ) = (17.8 mm, 50 mm)   CL-101 Engg. Mechanics, DoCL, DDU- SPP 29

30. 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 CL-101 Engg. Mechanics, DoCL, DDU- SPP 30

31. CENTRE OF GRAVITY OF UNSYMMETRICAL SECTIONS CL-101 Engg. Mechanics, DoCL, DDU- SPP 31

32. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 32

33. Example: 3 . Find the centroid of an unequal angle section 100 mm × 80 mm × 20 mm. Answer ( , ) = (25 mm, 35 mm)   C.G = 25mm   = 35mm   CL-101 Engg. Mechanics, DoCL, DDU- SPP 33

34. 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 ( , ) = (71.1 mm, 32.2 mm)   CL-101 Engg. Mechanics, DoCL, DDU- SPP 34

35. Example:5. Locate the centroid of the plate area Solution: Composite Parts Plate divided into 3 segments. Area of small rectangle considered “ negative ” . CL-101 Engg. Mechanics, DoCL, DDU- SPP 35

36. EXERCISE .1 CL-101 Engg. Mechanics, DoCL, DDU- SPP 36

37. 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 : CL-101 Engg. Mechanics, DoCL, DDU- SPP 37

38. Volume formulas: CL-101 Engg. Mechanics, DoCL, DDU- SPP 38

39. 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. (∴ = r ) Let r be the radius of the cylinder base in cm. Now let base of the cylinder be the axis of reference.   CL-101 Engg. Mechanics, DoCL, DDU- SPP 39

40. 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 . CL-101 Engg. Mechanics, DoCL, DDU- SPP 40

41. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 41

42. 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 , CL-101 Engg. Mechanics, DoCL, DDU- SPP 42

43. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 43

44. EXERCISE - HW CL-101 Engg. Mechanics, DoCL, DDU- SPP 44

45. 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 a 2 ( 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 45

46. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 46

47. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 47

48. 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 and 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.   CL-101 Engg. Mechanics, DoCL, DDU- SPP 48

49. CL-101 Engg. Mechanics, DoCL, DDU- SPP 49

50. 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.

51. Sample Problem 51 Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.

52. Sample Problem 52 Compute the coordinates of the area centroid by dividing the first moments by the total area.

53. 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 ( , ) = (93.75 mm, 60 mm)   CL-101 Engg. Mechanics, DoCL, DDU- SPP 53

54. 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. CL-101 Engg. Mechanics, DoCL, DDU- SPP 54

55. We know that distance between center of gravity of the body and the base of the cone, CL-101 Engg. Mechanics, DoCL, DDU- SPP 55

56. 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 CL-101 Engg. Mechanics, DoCL, DDU- SPP 56

57. We know that distance between center of gravity of the body and apex of the cone, Depth of the section plane below the apex CL-101 Engg. Mechanics, DoCL, DDU- SPP 57

58. CL-101 Engg. Mechanics, DoCL, DDU- SPP 58

59. EXERCISE CL-101 Engg. Mechanics, DoCL, DDU- SPP 59

60. EXERCISE CL-101 Engg. Mechanics, DoCL, DDU- SPP 60

61. Determination of Centroids by Integration 5 - 61 Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip.

62. 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.

63. Sample Problem 5 - 63 SOLUTION : Determine the constant k. Evaluate the total area.

64. Sample Problem 5 - 64 Using vertical strips, perform a single integration to find the first moments.

65. Sample Problem 5 - 65 Or, using horizontal strips, perform a single integration to find the first moments.

66. Sample Problem 5 - 66 Evaluate the centroid coordinates.

67. Centroids of common shapes of areas. CL-101 Engg. Mechanics, DoCL, DDU- SPP 67

68. CL-101 Engg. Mechanics, DoCL, DDU- SPP 68 Centroids Useful equations for Integration method

69. CL-101 Engg. Mechanics, DoCL, DDU- SPP 69

70. Centroids of common shapes of lines CL-101 Engg. Mechanics, DoCL, DDU- SPP 70

71. CL-101 Engg. Mechanics, DoCL, DDU- SPP 71