Centroids, Centre of Mass & Centres of Gravity

1. Centroids, Centre of Mass& Centres of Gravity The points through which the resultant of distributed forces and the earth's gravitational attraction in a body passes

2. Centre of gravity of a body is the point through which the resultant of the earth's pull upon the body passes and at which the weight of the body can be considered to be acting • to determine the centre of gravity of a body, we need to determine the location of the resultant force representing the earth's effect on the body

3. Considered an arbitrary body of uniform thickness t and material density  • Every particle of the body will be subjected to the force of gravity • Sum of the effects of the earths pull on all the particles yields the weight of the body W, dividing the body in n parts so the total weight of the body W will be the sum of the weights of each of the parts,

4. Considered an arbitrary body of uniform thickness t and material density  • Every particle of the body will be subjected to the force of gravity • Sum of the effects of the earths pull on all the particles yields the weight of the body W, dividing the body in n parts so the total weight of the body W will be the sum of the weights of each of the parts,

5. Where Wi is the force on the elemental area i. Using the definition • Separate the arithmetic from the concept • Hence,

6. Applying Varignon's theorem, we know that the moment of W about any axis will be equal to the sum of the moments of the forces • about the same axis. Defining the co-ordinates of W as , we have that considering moments about the y-axis.

7. Observe that Ax or xdA is the product of area and a distance from a particular axis (analogous to the product of force and distance which is defined as the moment of force). Hence xdA can be called the moment of area. The equation suggests that the total area can be considered to be concentrated at x

8. Similarily, considering moments about the x-axis, we obtain that • the point defined by the co-ordinates is the centre of gravity of a homogeneous body. It is also known as the centroid of the area A of the body. The integral is known as the first moment of the area with respect to the y-axis and is denoted by Qy, i.e. Similarily, for moment of area about the x axis.

9. The last two expressions provide easy methods for computing the co-ordinates of the centroids of a two-dimensional body. For such areas, the sum of the first moments of areas about a particle is divided by the total area of the figure. Note that the centroid C of an area with an axis of symmetry must be along that axis of symmetry. Hence an area with axes of symmetry must have it's centroid located at the intersection of the two axes of symmetry. If x-axis is an axis of symmetry, for each area dA with co-ordinate y, there must be a corresponding area of co-ordinate -y. Hence, we have that which is the first moment of area about the axis of symmetry must be equal to zero, i.e.

10. Similarly, if the y-axis is an axis of symmetry, • If both x- and y- are axes of symmetry, then

11. Example 1 • Determine, using first principles , the centroid of the rectangle given below. Solution

12. Properties of First Moments of Areas, Lines, Masses and Volume • Observe that the integral may be interpreted as the first moment of area about the y axis, Qy. Similarily, is the first moment of area about the x-axis, Qx. Hence • Observe that the co-ordinates of the centriods of an area my be obtained by dividing the first moments of that area by the total area. First moments of areas are important in the computation of shear stresses under transverse loads.

13. Properties of First Moments of Areas, Lines, Masses and Volume • When an area or line or mass or volume possesses an axis of symmetry BB', its first moment with respect to BB' is zero and its cetroid is located on bb'. • If two axes of symmetry exist, the centroid must be loacted at the intersection of the two axes of symmetry.

14. Properties of First Moments of Areas, Lines, Masses and Volume • The centroid coincides with the centre of symmetry. The centre of symmetry is that point about which, for every elemental area dA of co-ordinates x and y, there exists a corresponding elemental area dA' of co-ordinates -x and -y.

15. Center of Mass • Center of mass of an object is the centroid of its mass, i.e. • Integration must be performed over the entire mass of the object. Note that • The weight of an object can be assumed to act at the centre of mass, i.e. the mass is concentrated at this point. Mass density,  of an object is the mass per unit volume, i.e. (For homogeneous objects)

16. Center of Mass (2) • Weight Density g of an object is the weight per unit volume, i.e. • The center of mass of a homogeneous body coincides with the centroid of its volume • The center of mass of a homogeneous plate of uniform thickness coincides with the centroid of its x-sectional area. • The ceter of mass of a homogeneous slender bar of uniform x-sectional area coincides approximately with the centroid of the axis of the bar.

17. Composite Area • A composite area can be considered to be an area that is made up of a number of distinct regular areas. Consider the areas shown below:

18. Composite Area • Each of these areas can be broken down into the constituent areas that make up the given area. • The centroid of a composite area is determined by finding the sum of the statical moments of it's constituent areas and dividing by the total area of the composite area. • In doing this, the correct sign for the co-ordinate of the centroid of each consituent area with respect to the co-ordinate axes must be used. Areas of holes are considered to be negative while all other areas must have a positive sign. • (See Page 175, Figure 5.8 a, for Centroids of Common Shapes of Areas).

19. Theorems of Pappus - Guldinus (1) • Named after the Greek geometer, Pappus and Swiss mathematician, Guldinus • These theorems establish the way a surface of revolution and a body revolution may be obtained by revolving a plane curve and a plane area, respectively • A surface of revolution is a surface that can be generated by rotating a plane curve about a fixed axis.

20. Theorems of Pappus - Guldinus (2) • A body of revolution is one that can be generated by rotating a plane area about a fixed axis e.g. A solid sphere from rotating a semi-circular area; solid cone from rotating a triangular area; solid torus by rotating a circular area. • Note that the Pappus - Guldinus theorems may be used to determine the centroids of a plane curve or a plane area with respect to the fixed axis of rotation if the surface area / volume as well as the length / area of generating plane curve and plane area, rexpectively are known.

21. Theorum 1 • The area of a surface of revolution is equal to the length of the generating plane curve times the distance travelled by the centroid of the plane curve while the surface is being generated, i.e.

22. Theorem 2 • The volume of a body of revolution is equal to the area of the generating plane area times the distance travelled by the centroid of the plane area while the body is being generated, i.e.

23. y h x b Example 2 Determine the centroid of the triangle given below using first principles. Solution

24. Example 3 Determine the centroid of the area shown by direct interpolation. y=mx b y=kx3 a Solution

25. y (1,1) y=x2 x x dx Example 4 Find x and y of the area between the lines y Solution

26. Example 5 Find x and y of the area between the lines y (300,27000300) y=x + x3 y x x dx Solution

27. Example 6 Find x of the area between the two lines. y y=x Y=x2-20 (5,5) y (-4,-4) Yel=-1/2(y)+x = x-1/2(x-x2+20) =1/2(x+x2-20) x dx Solution

28. Example 7 Find the centroid of the line y=2/3(x-1)3/2 dy dL y 5 x dx Solution

29. Example 8 Find the centroid by direct integration y y=kx3 a x a Solution

30. Example 9 Determine the centroid of the circular arc given below y Solution  x R

31. Example 10 A homogenious wire bent into the shape shown, find x and y. r 45o 45o Solution

32. dA=1/2(r[rd]) . r d 2/3(rsin)  (2/3)r Example 11 A) Solution

33. h a Example 12 Find the volume of the solid generated by rotating parabolic spandrel about a) x-axis b)about axis A1 A1 Solution y=kx2

34. r = 42mm r =52mm h=60mm r = 20 mm Example 13 Find volume and total surface area. Solution

35. 1m 2m VA=6N VB=10N Example 14 Find the x-coordinate of the centroid of the hole. Material has Mass per unit area of 2kg/m2 Solution

36. Example 15 Four students with weights shown and position shown, find the coordinates of the positions of students C&D if Mo=0 . . 660N 728N A(-3,3) C(Cx,Cy) . . B(-2.5,1.5) D(Dx,Dy) 488N 496N Solution

37. Example 16 Find the center of gravity of the homogenius wire. Solution 520 mm 200 mm 480 mm

38. tf tf bw Example 17 Determine the centriod of the area shown below. bw tf Solution tw h b

39. Example 18 Determine the centroid of the given area. Note: The cicular hole is treated as a negative area. y h r yc x xc bt bt bc Solution

40. 15 I II x 12 21 Example 19 Find the centriod of the object shown below. y Solution

41. y 20 mm r =15mm 30 mm 30 mm Example 20 Find the centroid of the given shape below Solution x

42. Example 21 y Find the centriod of the given area I II 60 30 24 x 20 Solution