25. Centroid by integration
Chapter Objectives
 To discuss the concept of center of gravity, center of
mass and the centroid.
 To show how to determine the location of the center of
gravity and centroid for a system of discrete particles
and a body of arbitrary shape.
25. Centroid by integration
Center of Gravity, Center of Mass, and Centroid of a Body.
Center of Gravity of a System of Particles
z
z
W2
Wn
z
WR
W1
=
zn

G
z1

xn

y
z
x1


yn
y
_
z
x
y
y1
x
=
My
Mx

WR
y
_
x
_
y
x
x
The unique point G where the resultant weight of a n-particle system is reduced
to a single force (no moments) is called center of gravity.
25. Centroid by integration
Center of Gravity, Center of Mass, and Centroid of a Body. Center of Gravity
– Center of Mass
A rigid body can be considered as made up of an infinite number of
particles. Hence, using the same principles as in the previous slides,
we get the coordinates of G by simply replacing
,W
(discrete summation)
 x dW
 dW

(continuous summation)
~
x 
V
V
 , dW
 y dW
 dW
~
y 
V
V
z 
z dW
 ~
 dW
V
V
25. Centroid by integration
Considering that the weight acts in the z-direction,
then the equivalence between the particle and
reduced systems shows that
 Fz,
 My,
WR  W1  W2  ...  Wn  W
xW  ~
xW ~
x W  ...  ~
xW
R
1
1
2
2
n
n
~
x1W1  ~
x2W2  ...  ~
xnWn  ~
xW
x 

WR
W
The above formula can be generalized to
~
xW
x 
W
~
yW
y 
W
~
zW
z 
W
25. Centroid by integration
25. Centroid by integration
9.1 Center of Gravity, Center of Mass, and Centroid of a Body.
Center of Mass of a System of Particles
z
Wn

xn

x1
y
z
y1
WR
My
Mx
=
z1

yn
WR

zn
x
z
W1


z
W2
=
G
y
_
z
x
y
x
x
_
x
_
y
The center of gravity (G) is a point which locates the resultant weight of a
system of particles or body.
Similarly, the center of mass is a point which locates the resultant
mass of a system of particles or body. Generally, its location is the same
as that of G.
Since the acceleration due to gravity g is constant for every particle is
constant, then,
W = mg
Substituting into the center of gravity
formula and canceling g, yields to
~
xm
x 
m
~
ym
y 
m
~
zm
z 
m
25. Centroid by integration
y
25. Centroid by integration
25. Centroid by integration
Since the specific weight is defined as =W/V or
=dW/dV the center or gravity is expressed as
 x  dV
  dV
~
x 
V
 y  dV
  dV
~
y
V
V
z 
V
 ~z  dV
  dV
V
V
The center of mass is obtained when the density,
 =  g, is considered, therefore
x dm

x
 dm
~
V
V
y dm

y
 dm
~
V
V
~
z dm

z
 dm
V
V
25. Centroid by integration
Center of Gravity, Center of Mass, and Centroid of a Body. Centroid
The centroid C is a point which defines the geometric center of an object.
The centroid coincides with the center of
mass or the center of gravity only if the
material of the body is homogenous (density
or specific weight is constant throughout the
body).
25. Centroid by integration
x 


V
~
x dL
V
dL
;y 


V
~
y dL
V
dL
;z 


V
~
z dL
V
dL
25. Centroid by integration
x 


V
~
x dA
V
dA
;y 


V
~
y dA
V
dA
;z 


V
~
z dA
V
dA
25. Centroid by integration
x 


V
~
x dV
V
dV
;y 


V
~
y dV
V
dV
;z 


V
~
z dV
V
dV
25. Centroid by integration
Center of Gravity, Center of Mass, and Centroid of a Body.
Centroid Symmetry
If an object has an axis of symmetry, then the centroid of object
lies on that axis.
Not necessarily, the centroid is located on the object.
25. Centroid by integration
Center of Gravity, Center of Mass, and Centroid of a Body.
Steps for Determining Area Centroid
1.
Choose an appropriate differential element dA at a general point (x,y).
Rectangular of polar coordinates could be used. Hint: Generally, if y is easily
expressed in terms of x (e.g., y = x2 + 1), use a vertical rectangular
element. If the converse is true, then use a horizontal rectangular element.
2.
Express dA in terms of the differentiating element dx (or dy).
3.
Determine coordinates (x , y) of the centroid of the differential element in
terms of the general point (x,y).
Express all the variables and integral limits in the formula using either x or
y depending on whether the differential element is in terms of dx or dy,
respectively, and integrate.
4.
 
Note: Similar steps are used for determining CG, CM, etc. These steps will
become clearer by doing a few examples.
25. Centroid by integration
25. Centroid by integration
Example 1
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
25. Centroid by integration
Composite Bodies
y
 First moments of areas, like moments of
forces, can be positive or negative.
 For example: an area whose centroid is
located to the left of Y axis will have a
negative first moment with respect to that
axis.
y
 Also, the area of a hole should be
assigned a negative sign.
X
A XA
A1
-
+
-
A2
A3
+
+
+
-
+
-
25. Centroid by integration
Centroid of Common Areas
25. Centroid by integration
Centroid of Common Lines
25. Centroid by integration
Example 1
y
20 mm
30 mm
Locate the centroid of the plane
area shown.
36 mm
24 mm
x
25. Centroid by integration
Example 1
y
20 mm
30 mm
Locate the centroid of the plane
area shown.
36 mm
Several
points
should
be
emphasized when solving these
types of problems.
24 mm
x
1. Decide how to construct the given area from common shapes.
2. It is strongly recommended that you construct a table
containing areas or length and the respective coordinates of the
centroids.
3. When possible, use symmetry to help locate the centroid.
25. Centroid by integration
Example 1
y
20 + 10
Decide how to construct the
given area from common
shapes.
C1
C2
24 + 12
30
10
x
Dimensions in mm
25. Centroid by integration
Example 1
y
20 + 10
Construct a table containing areas
and respective coordinates of the
centroids.
C1
C2
Dimensions in mm
24 + 12
30
10
A, mm2
1 20 x 60 =1200
2 (1/2) x 30 x 36 =540

1740
x
x, mm
10
30
y, mm
30
36
xA, mm3
12,000
16,200
28,200
yA, mm3
36,000
19,440
55,440
25. Centroid by integration
Example 1
y
20 + 10
Then X A =  xA
X (1740) = 28,200
or X = 16.21 mm
C1
C2
and
24 + 12
30
10
Dimensions in mm
x
Y A =  yA
Y (1740) = 55,440
or Y = 31.9 mm
25. Centroid by integration
Example 2
For the plane area shown, determine (a) the first
moments with respect to the X and Y axes, (b) the
location of the centroid
Y
120 mm
60 mm
80 mm
60 mm
40 mm
X
25. Centroid by integration
Example 2
1. The area of the circle is negative, since it is to be
subtracted from the other areas.
2. The coordinate y of the centroid of the triangle is
negative for the axes shown
25. Centroid by integration
Example 2
Component
A, mm2
X, mm
Y, mm
XA, mm3
YA, mm3
(120)(80) =
9600
60
40
576000
384000
½ (120)(60) =
3600
40
- 20
144000
-72000
Semicircle
½ p(60)2 =
5655
60
105.46
339300
596400
Circle
- p(40)2 =
- 5027
60
80
-301600
- 402200
XA = 757700
YA = 506200
Rectangle
Triangle
A = 13828
25. Centroid by integration
Example 2
(a)First moments of the area:
Y
xA = 757700 mm3
yA = 506200 mm3
(b) Location of Centroid (C):
C
36.6 mm
X
XA = xA and YA = yA
X(13828 mm2) = 757700 mm3
X = 54.8 mm
54.8 mm
Y(13828 mm2) = 506200 mm3
Y = 36.6 mm
25. Centroid by integration
Example 3
Find the centroid of the given body
25. Centroid by integration
Example 3
To find the centroid,
1
x
AT
x A
1
y
AT
y A
i
i
i
i
A2
A1
Determine the area of the components
1
120 mm  60 mm   3600 mm2
2
A2  120 mm 100 mm   12000 mm2
A1 
25. Centroid by integration
Example 3
The centroid of each area
b 120 mm
x1  
 40 mm
3
3
h1
60 mm
y1  h1   60 mm 
 40 mm
3
3
b 120 mm
x2  
 60 mm
2
2
h2
100 mm
y2  h1   60 mm 
 110 mm
3
2
A2
A1
25. Centroid by integration
Example 3
The total area
AT  A1  A2  3600 mm2  12000 mm2
 15600 mm2
Compute the x centroid
x
1
AT
x A
i
i
A2
A1
1
2
2



40
mm
3600
mm

60
mm
12000
mm








2 

15600 mm
 55.38 mm
25. Centroid by integration
Example 3
Compute the y centroid
1
y
AT
yA
i
i
A2
A1
1
2
2


40
mm
3600
mm

110
mm
12000
mm








2 

15600 mm
 93.85 mm

25. Centroid by integration
Example 3
The problem can be done using a table to represent
the composite body.
Body
Triangle
Square
Area(mm2)
3600
12000
Sum
15600
centroid (x)
centroid (y)
x (mm)
40
60
3
3
x*Area
(mm
)
y*Area
(mm
)
y(mm)
40
144000
144000
110
720000
1320000
864000
1464000
55.38 mm
93.85 mm
25. Centroid by integration
Example 3
An alternative method of computing the
centroid is to subtract areas from a total area.
Assume that area is a large
square and subtract the small
triangular area.
A1
A2
25. Centroid by integration
Example 3
The problem can be done using a table to represent
the composite body.
Body
Square
Triangle
Area(mm2)
19200
-3600
Sum
15600
centroid (x)
centroid (y)
x (mm)
60
80
3
3
y(mm) x*Area (mm ) y*Area (mm )
80
1152000
1536000
20
-288000
-72000
864000
1464000
55.38 mm
93.85 mm
25. Centroid by integration
Example 4
Find the centroid of the given body
25. Centroid by integration
Example 4
Determine the area of the components
1
A1   90 mm  60 mm 
2
 2700 mm 2
A2  120 mm  90 mm 
 10800 mm
A3 
p
2
 40 mm 
A2
2
2
 2513.3 mm 2
A3
A1
25. Centroid by integration
Example 4
The centroid of each area
b 90 mm

 30 mm
3
3
h
60 mm
y1  h1  1  60 mm 
 40 mm
3
3
b 90 mm
x2  
 45 mm
2
2
h
120 mm
y2  h1  2  60 mm 
 120 mm
3
2
4  40 mm 
4r
x3  b 
 90 mm 
 73.02 mm
3p
3p
y3  60 mm  20 mm  40 mm  120 mm
x1 
A2
A3
A1
25. Centroid by integration
Example 4
Body
Triangle
Square
Hemisphere
Area(mm2)
2700
10800
-2513.27
Sum
10986.73
centroid (x)
centroid (y)
1
x
AT
x (mm)
30
45
73.02
y(mm)
40
120
120
x*Area (mm3)
81000
486000
-183528.00
y*Area (mm3)
108000
1296000
-301592.89
383472.00
1102407.11
34.90 mm
100.34 mm
383472.00 mm3
 xi Ai  10986.73 mm2
 34.90 mm
1
y
AT
1102407.11 mm3
 yi Ai  10986.73 mm2
 100.34 mm
25. Centroid by integration
Example 4
An Alternative Method would be to subtract areas
Body
Triangle
Square
Hemisphere
Area(mm2)
-2700
16200
-2513.27
Sum
10986.73
centroid (x)
centroid (y)
x (mm)
60
45
73.02
y(mm)
20
90
120
x*Area (mm3)
-162000
729000
-183528.00
y*Area (mm3)
-54000
1458000
-301592.89
383472.00
1102407.11
34.90 mm
100.34 mm
25. Centroid by integration
Example 5
Find the centroid of the body
25. Centroid by integration
Example 6
Find the centroid of the body
25. Centroid by integration