CHAPTER 5
CENTROIDS AND CENTERS OF GRAVITY
1
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
2
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.
3
5.2 Center of Gravity of a Two Dimensional Body
• Center of gravity of a plate
• Center of gravity of a wire
 M y x W   x W
  x dW
M y
y W   y W
  y dW
4
5.3 Centroids of Area and Lines
• Centroid of an area
x W   x dW
x At    x t dA
x A   x dA  Q y
 first moment with respect to y
• Centroid of a line
x W   x dW
x  La    x  a dL
x L   x dL
yL   y dL
yA   y dA  Qx
 first moment with respect to x
5
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 a line of symmetry, its
centroid lies on that axis
• If an area possesses two lines of symmetry, its
centroid lies at their intersection.
• 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.
6
Shape
x
y
Area
h
3
bh
2
4r
3π
4r
3π
πr
4
2
Semicircular
area
πr
2
2
0
4r
3π
Quarter-elliptical
area
4a
3π
4b
3π
πab
4
Semi-elliptical
area
0
4b
3π
πab
2
Triangular area
Quarter-circular
area
Figure A : Centroids of common shapes of areas
7
Shape
x
y
Area
Semi-parabolic
area
3a
8
3h
5
2ah
3
3h
5
4ah
3
3h
10
ah
3
parabolic
area
Parabolic
spandrel
General
spandrel
Figure A : Centroids of common shapes of areas
3a
4
n+1
a
n+2
n+1
h
4n+2
ah
n+1
8
Shape
x
y
Circular sector
2r sin α
3α
0
Area
αr
2
Figure A : Centroids of common shapes of areas
9
Shape
x
y
Area
Quarter-circular
arc
2r
π
2r
π
πr
2
Semicircular
arc
0
2r
π
πr
0
2αr
Arc of circle
r sin α
α
Figure B : Centroids of common shapes of lines
10
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
My : X ( W1 + W2 + …..+ Wn ) = X1W1 + X2W2 + …..+ XnWn
Mx : Y ( W1 + W2 + …..+ Wn ) = y1W1
+ y2W2 + …..+ ynWn
11
These equations can be solved for the coordinates X and Y of the center
of gravity of the plate
• Composite plates
X W   x W
Y W   y W
• Composite area
X  A   xA
Y  A   yA
Figure C : Center of gravity of a composite plate
12
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
Qy : X ( A1 + A2 + …..+ An ) = X1A1 + X2A2 + …..+ XnAn
Qx : Y ( A1 + A2 + …..+ An ) = y1A1
+ y2A2 + …..+ ynAn
or for short
13
Sample Problem 5.1
For the plane area shown, determine
(a) the first moments with respect to the x and y axes
(b) the location of the centroid
14
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.
15
rectangle
triangle
semicircle
circle
16
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
17
Location of the centroids
• Compute the coordinates of the area
centroid by dividing the first moments by
the total area.
x A  757.7 103 mm 3

X

 A 13.828103 mm 2
X  54.8 mm
y A  506.2 103 mm 3

Y 

 A 13.828103 mm 2
Y  36.6 mm
18
Sample problem 5.2
650 mm
250 mm
600 mm
The figure shown is made from a piece of thin, homogeneous wire.
determine the location of its center of gravity.
19
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
300 mm
250 mm
125 mm
600 mm
20
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.
21
Ay
Solution
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
22
Ay
Solution
Ax
Ay = W
Ax = W
π
23
PROBLEMS
24
Problem 5.1
300 mm
150 mm
200mm
400 mm
Locate the centroid of the plane area shown
25
Solution
1
C1
100 mm
(150mm + 75mm)
= 225 mm
150 mm
2
C2
150 mm
200 mm
200mm
300 mm
400 mm
26
Problem 5.2
Locate the centroid of the plane area shown
27
６0 mm + 52.5mm
Solution
2
1
C2
75mm
C1
37.5mm
h =
3
25mm
40 mm
28
Problem 5.3
600mm
250mm
400mm
Locate the centroid of the plane area shown
29
250 + 1 ( 4000 )mm
3
Solution
2 ( 250 )mm
3
1
C1
2
C2
600mm
250mm
400mm
30
Problem 5.4
Locate the centroid of the plane area shown
31
Solution
Y
1.5mm
C1
Y
22mm
Y
C2
11mm
8mm
2
1
X
X
2mm
6mm
X
2mm
32
Problem 5.5
Locate the centroid of the plane area shown
33
Y
Y
120 – 4 x 60 = 94.5 mm
3π
1
2
C2
C1
100mm
120mm
X
60mm
X
905 546
100
5655.6
2 400 000
534 454
18344
678 672
1 721 328
18 344
905 546
1 721 328
18344
34
THE END
35