Centroids
https://engineeringstatics.org/Chapter_07-12.html
Contents
Introduction
Theorems of Pappus-Guldinus
Center of Gravity of a 2D Body
Sample Problem 5.7
Centroids and First Moments of
Areas and Lines
Distributed Loads on Beams
Sample Problem 5.4
Center of Gravity of a 3D Body:
Centroid of a Volume
Centroids of Common Shapes of
Areas
Centroids of Common Shapes of
Lines
Composite Plates and Areas
Sample Problem 5.1
Determination of Centroids by
Integration
5-2
Sample Problem 5.9
Centroids of Common 3D
Shapes
Composite 3D Bodies
Sample Problem 5.12
Application
There are many examples in engineering analysis of distributed
loads. It is convenient in some cases to represent such loads as a
concentrated force located at the centroid of the distributed load.
5-3
Introduction
• The earth exerts a gravitational force on each of the particles
forming a body – consider how your weight is distributed
throughout your body. These forces can be replaced 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; it is the “center of area.” 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.
5-4
Center of Gravity of a 2D Body
• Center of gravity of a plate
 M y x W   x W
  x dW
My
y W   y W
  y dW
5-5
• Center of gravity of a wire
Centroids and First Moments of Areas
and Lines
• Centroid of an area
x W   x dW
x At    x t dA
x A   x dA  Q y
 first moment wit h respect to y
yA   y dA  Q x
 first moment wit h respect to x
5-6
• Centroid of a line
x W   x dW
x  La    x  a dL
x L   x dL
yL   y dL
Determination of Centroids by
Integration
• Double integration to find the first moment
x A   xdA   x dxdy   xel dA
yA   y dA   y dxdy   y el dA
x A   xel dA
x A   xel dA
yA   y el dA
ax
 a  x dx 
2
yA   y el dA
  x  ydx 
y
   ydx 
2
5-7
may be avoided by defining dA as a thin
rectangle or strip.

  y a  x dx 
x A   xel dA

2r
1

cos   r 2 d 
3
2

yA   y el dA

2r
1

sin   r 2 d 
3
2

Sample Problem 5.4
SOLUTION:
• Determine the constant k.
• Evaluate the total area.
• Using either vertical or horizontal
strips, perform a single integration to
find the first moments.
Determine by direct integration the
location of the centroid of a parabolic
spandrel.
First, estimate the location of the
centroid by inspection. Discuss with
a neighbor where it is located,
roughly, and justify your answer.
5-8
• Evaluate the centroid coordinates.
Sample Problem 5.4
SOLUTION:
• Determine the constant k.
y  k x2
b  k a2  k 
y
b
a2
x2
or
b
a2
x
a
b1 2
y1 2
• Evaluate the total area.
A   dA
a
 b x3 
b 2
  y dx  
x dx  
2
2 3 
 a
 0
0a
ab

3
a
5-9
Sample Problem 5.4
• Using vertical strips, perform a single integration
to find the first moments.
 b 2 
Qy   x el dA   xy dx   x 2 x dx

0 a
a
a
 b x 4  a 2 b
  2
 
4
a 4 0
2
a 1  b
y
2 
Qx   y el dA   y dx    2 x  dx
2

0 2 a
a
 b 2 x 5  ab 2
  4
 
2a 5 0 10
5 - 10

Sample Problem 5.4
• Or, using horizontal strips, perform a single
integration to find the first moments. Try
calculating Qy or Qx by this method, and confirm
that you get the same value as before.
b a2  x2
ax
Qy   x el dA  
a  x dy  
dy

2
2
0
1 b  2 a 2 
a 2b
  a 
y dy 
2 0 
b 
4

a 1 2 
Qx   y el dA   y a  x dy   ya  1 2 y dy
 b

b 
a 3 2 
ab 2
  ay  1 2 y dy 

10
b
0 
5 - 11

Sample Problem 5.4
• Evaluate the centroid coordinates.
xA  Q y
ab a 2 b
x

3
4
x
3
a
4
y
3
b
10
yA  Q x
ab ab 2
y

3
10
Is this “center of area” close to where
you estimated it would be?
5 - 12
Determination of Centroids by
Integration
Usually, the choice between using a vertical or horizontal strip is equally good,
but in some cases, one choice is much better than the other. For example, for
the area shown below, is a vertical or horizontal strip a better choice, and why?
Think about this and discuss your choice with a neighbor.
y
x
5 - 13
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).
5 - 14
• The centroid of the area coincides with the
center of symmetry.
Centroids of Common Shapes of Areas
5 - 15
Centroids of Common Shapes of Lines
5 - 16
Composite Plates and Areas
• Composite plates
X W   x W
Y W   y W
• Composite area
X  A   xA
Y  A   yA
5 - 17
Sample Problem 5.1
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.
For the plane area shown, determine
the first moments with respect to the
x and y axes and the location of the
centroid.
5 - 18
• 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.
Sample Problem 5.1
• Find the total area and first moments of the
triangle, rectangle, and semicircle. Subtract the
area and first moment of the circular cutout.
5 - 19
Q x  506 .2  10 3 mm 3
Q y  757 .7  10 3 mm 3
Sample Problem 5.1
• 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
5 - 20
How do you find the volume of
a cylinder?
Take the cross-sectional area, then
multiply it by height.
Consider a different crosssectional area
Find the volume from revolving
a cross-sectional area about an
axis:
Break the cross-sectional area into elements, and
consider the path of revolution that each element
takes:
The average path length is the distance
the center travels.
Find the Volume of a cone:
Area of triangle =
Location of centroid =
Volume of cone =
Find the Volume of a Rocket:
What is the best way to split up the
cross-sectional areas?
1
1
2
2
3
a.)
3
b.)
Find the Volumes of each revolved section:
1
2
3
Find the Volumes of each revolved section:
1
2
3
Theorems of Pappus-Guldinus
• Body of revolution is generated by rotating a plane
area about a fixed axis.
• Volume of a body of revolution is
equal to the generating area times
the distance traveled by the centroid
through the rotation.
V  2 y A
5 - 30
Theorems of Pappus-Guldinus
• Surface of revolution is generated by rotating a
plane curve about a fixed axis.
• Area of a surface of revolution is
equal to the length of the generating
curve times the distance traveled by
the centroid through the rotation.
A  2 yL
5 - 31
Sample Problem 5.7
SOLUTION:
• Apply the theorem of Pappus-Guldinus
to evaluate the volumes of revolution of
the pulley, which we will form as a
large rectangle with an inner
rectangular cutout.
The outside diameter of a pulley is 0.8
m, and the cross section of its rim is as
shown. Knowing that the pulley is
made of steel and that the density of
steel is   7.85  10 3 kg m 3
determine the mass and weight of the
rim.
5 - 32
• Multiply by density and acceleration
to get the mass and weight.
Sample Problem 5.7
SOLUTION:
• Apply the theorem of Pappus-Guldinus
to evaluate the volumes or revolution for
the rectangular rim section and the inner
cutout section.
• Multiply by density and acceleration to
get the mass and weight.


W  mg  60.0 kg 9.81 m s 

m  V  7.85 10 3 kg m3 7.65 10 6 mm 3 10 9 m3 /mm 3
2
5 - 33

m  60 .0 kg
W  589 N
Distributed Loads on Beams
5 - 34
Distributed Loads on Beams
5 - 35
Distributed Loads on Beams
L
W   wdx   dA  A
0
OP W   xdW
L
OP  A   xdA  x A
0
5 - 36
• A distributed load is represented by plotting the load
per unit length, w (N/m) . The total load is equal to
the area under the load curve.
• A distributed load can be replace by a concentrated
load with a magnitude equal to the area under the
load curve and a line of action passing through the
area centroid.
Sample Problem 5.9
SOLUTION:
• The magnitude of the concentrated load
is equal to the total load or the area under
the curve.
• The line of action of the concentrated
load passes through the centroid of the
area under the curve.
• Determine the support reactions by (a)
A beam supports a distributed load as
drawing the free body diagram for the
shown. Determine the equivalent
beam and (b) applying the conditions
concentrated load and the reactions at
of equilibrium.
the supports.
5 - 37
Sample Problem 5.9
SOLUTION:
• The magnitude of the concentrated load is equal to
the total load or the area under the curve.
F  18 .0 kN
• The line of action of the concentrated load passes
through the centroid of the area under the curve.
X 
5 - 38
63 kN  m
18 kN
X  3 .5 m
Sample Problem 5.9
• Determine the support reactions by applying the
equilibrium conditions. For example,
successively sum the moments at the two
supports:
 M A  0 : By 6 m 18 kN 3.5 m  0
B y  10.5 kN

 M B  0 :  Ay 6 m 18 kN 6 m 3.5 m  0
A y  7.5 kN

• And by summing forces in the x-direction:
 Fx  0 : Bx  0
5 - 39
Example
Homework
Forces on
Submerged Surfaces
42
43
Forces on
Submerged Surfaces
44
Forces on
Submerged Surfaces
45
46
8m
1.2 m
47
Example
Homework
Homework
0.45+0.48=0.93
Homework
0.45+0.48=0.93
Center of Gravity of a 3D Body:
Centroid of a Volume
• Center of gravity G


 W j    W j 




rG   W j    r   W j 




rGW   j    r W    j 
W   dW
5 - 52


rGW   r dW
• Results are independent of body orientation,
x W   xdW
yW   ydW
z W   zdW
• For homogeneous bodies,
W   V and dW   dV
x V   xdV
yV   ydV
z V   zdV
Centroids of Common 3D Shapes
5 - 53
Composite 3D Bodies
• Moment of the total weight concentrated at the
center of gravity G is equal to the sum of the
moments of the weights of the component parts.
X W   xW
Y  W   yW
Z W   zW
• For homogeneous bodies,
X V   xV
5 - 54
Y  V   yV
Z V   zV
Sample Problem 5.12
SOLUTION:
• Form the machine element from a
rectangular parallelepiped and a
quarter cylinder and then subtracting
two 1-in. diameter cylinders.
Locate the center of gravity of the
steel machine element. The diameter
of each hole is 1 in.
5 - 55
Sample Problem 5.12
5 - 56
Sample Problem 5.12
X   xV
V  3.08 in  5.286 in 
4
3
X  0.577 in.

Y   yV
V  5.047 in  5.286 in 
4
3
Y  0.577 in.

Z   zV
V  1.618 in  5.286 in 
4
3
Z  0.577 in.
5 - 57
Integration Book Examples
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er_07-12.html