1. CENTROIDS AND CENTERS OF GRAVITY
CHAPTER 5
CENTROIDS AND CENTERS OF GRAVITY

2. CENTROIDS AND CENTERS OF GRAVITY
CHAPTER 5
CENTROIDS AND CENTERS OF GRAVITY
5.1 Introduction
5.2 Centers of Gravity of a Two Dimensional Body
5.3 Centroids of Areas and Lines
5.4 First Moments of Area and Lines
5.5 Composite Plates and wires

3. Introduction
The earth exerts a gravitational force on each of the particles forming a body. These forces can be replace by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body.
The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid.
Determination of the area of a surface of revolution and the volume of a body of revolution are accomplished with the Theorems of Pappus-Guldinus.

4. 5.2 Center of Gravity of a Two Dimensional Body
Center of gravity of a plate
Center of gravity of a wire

5. 5.3 Centroids of Area and Lines
Centroid of an area
Centroid of a line

6. 5.4 First Moments of Areas and Lines
An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’.
The first moment of an area with respect to a line of symmetry is zero.
If an area possesses two lines of symmetry, its centroid lies at their intersection.
If an area possesses a line of symmetry, its centroid lies on that axis
An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y).
The centroid of the area coincides with the center of symmetry.

7. Shape x y Area h 3 bh 2 4r 3π 4r 3π πr 4 4r 3π πr 2 4a 3π 4b 3π πab 4
Triangular area 4r 3π 4r 3π πr 4 2
Quarter-circular
area 4r 3π πr 2
Semicircular
area 4a 3π 4b 3π
πab 4
Quarter-elliptical
area 4b 3π
πab 2
Semi-elliptical
area
Figure A : Centroids of common shapes of areas

8. Shape x y Area 3a 3h ah ah n+1 n+1 n+2 a n+1 4n+2 h 3a 8 3h 5 2ah 3 3h
Semi-parabolic
area 3a 8 3h 5
2ah 3
parabolic
area 3h 5
4ah 3
Parabolic
spandrel 3a 4 3h 10 ah 3 ah
n+1
n+1
n+2 a
n+1
4n+2 h
General
spandrel
Figure A : Centroids of common shapes of areas

9. Shape x y Area 2r sin α αr 3α Circular sector
Figure A : Centroids of common shapes of areas

10. Shape x y Area 2r π 2r π πr 2 2r π πr r sin α α 2αr Quarter-circular
arc 2r π 2r π πr 2
Semicircular
arc 2r π πr
Arc of circle
r sin α α
2αr
Figure B : Centroids of common shapes of lines

11. 5.5 Composite Plates and Wires
In many instances, a flat plate can be divided in rectangles, triangles,or the other common shapes
shown in figure A. The abscissa X of its center of gravity G can be determine from the abscissas
X1,X2,…..Xn of the centers of gravity of the various parts by expressing that the moment of the weight
of the whole plate about the y axis is equal to the sum of the moments of the weights of the various parts
about the same axis (Figure C). The ordinate Y of the center of gravity of the plate is found in a similar
way by equating moments about the x axis. We write
: X ( W1 + W2 + …..+ Wn ) = X1W1 + X2W2 + …..+ XnWn My
: Y ( W1 + W2 + …..+ Wn ) = y1W1 + y2W2 + …..+ ynWn Mx

12. Composite plates Composite area
These equations can be solved for the coordinates X and Y of the center
of gravity of the plate
Composite plates
Composite area
Figure C : Center of gravity of a composite plate

13. Qy Qx : X ( A1 + A2 + …..+ An ) = X1A1 + X2A2 + …..+ XnAn
If the plate is homogeneous and of uniform thickness, the center of gravity concides
with the centroid C of its area.The abscissa X of the centroid of the area can be determined
by noting that the first moment Qy of the composite area with respect to th ey axis can be
expressed both as the product of X and the total area and as the sum of the first moments
of the elementary areas with with respect to the y axis ( figure C ). The ordinate Y of the
centroid is found in a similar way by considering the first moment Qx of the composite
area. We have
: X ( A1 + A2 + …..+ An ) = X1A1 + X2A2 + …..+ XnAn Qy
: Y ( A1 + A2 + …..+ An ) = y1A1 + y2A2 + …..+ ynAn Qx
or for short

14. Sample Problem 5.1
For the plane area shown, determine
the first moments with respect to the x and y axes
(b) the location of the centroid

15. Calculate the first moments of each area with respect to the axes.
solution
SOLUTION:
Divide the area into a triangle, rectangle, and semicircle with a circular cutout.
Calculate the first moments of each area with respect to the axes.
Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.
Compute the coordinates of the area centroid by dividing the first moments by the total area.

16. rectangle
triangle
semicircle
circle

17. a) First Moments of Area
b) Location of centroids
Substituting the values given in the table into the equations
defining the centroid of a composite area, we obtain

18. Location of the centroids
Compute the coordinates of the area centroid by dividing the first moments by the total area.

19. Sample problem 5.2
650 mm
600 mm
250 mm
The figure shown is made from a piece of thin, homogeneous wire.
determine the location of its center of gravity.

20. Solution Since the figure is formed of homogeneous wire,
its center of gravity coincides with the centroid of the
corresponding line.
Therefore, that centroid will be determined.
Choosing the coordinates axes shown, with origin at A,
we determine the coordinates of the centroid of each line
segment and compute the first moments with respect to the
coordinate axes
600 mm
300 mm
125 mm
250 mm

21. Sample Problem 5.3
A uniform semicircular rod of weight W and radius r is attached to a pin at A
and rest against a frictionless at B.
Determine the reactions at A and B.

22. Solution Ay Ax Free Body diagram
A free body diagram of the rod is drawn.
The forces acting on the rod are its weight W, which is applied
at the center of gravity G.
(whose position is obtained fro Fig B).
A reaction at A, represented by its components Ax and Ay
and a horizontal reaction at B.
Ay = W
Ax = W π

23. Ax Ay
Ay = W
Ax = W π
Solution

24. PROBLEMS

25. Problem 5.1 Locate the centroid of the plane area shown 300 mm 150 mm

26. Solution 1 2 C1 C2 300 mm 150 mm 200mm 400 mm (150mm + 75mm) = 225 mm

27. Problem 5.2
Locate the centroid of the plane area shown

28. ６0 mm mm
Solution 2 1 C2
75mm C1
37.5mm
h = 3
25mm
40 mm

29. Problem 5.3 Locate the centroid of the plane area shown 250mm 400mm

30. 250mm 400mm 600mm Solution C1 C2 1 2 250 + 1 ( 4000 )mm 3 2 ( 250 )mm

31. Problem 5.4
Locate the centroid of the plane area shown

32. Solution Y
1.5mm
11mm
22mm C1 1 X
2mm
8mm Y X X Y
2mm
6mm 2 C2

33. Problem 5.5
Locate the centroid of the plane area shown

34. 100mm
60mm 1 C1 2 C2
120mm Y X
120 – 4 x 60 = 94.5 mm 3π
5655.6
18 344
18344
100

35. THE END