Chapter 5
Distributed Forces:
Centroids and Centers of Gravity
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.
5-2
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.
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Center of Mass
The center of mass of a system is the point where
the system can be balanced in a uniform
gravitational field.
9.2 The Center of Mass: A System of Particles
Consider a situation in which n particles are strung out along
the x axis. Let the mass of the particles are m1, m2, ….mn, and
let them be located at x1, x2, …xn respectively. Then if the
total mass is M = m1+ m2 + . . . + mn, then the location of the
center of mass, xcom, is
Center of Mass
For two objects:
The center of mass is closer to the more
massive object.
Center of Mass
The center of mass need not be within the object:
Moment
The moment of a mass is a measure of its tendency to rotate about a point.
Clearly, the greater the mass (and the greater the distance from the point), the
greater will be the tendency to rotate.
The moment is defined as:
Moment = mass × distance from a point
Example 1
In this case, there will be a total moment about O of:
(Clockwise is regarded as positive in this work.)
M=2×1−10×3=−28 kgm
Center of Mass
We now aim to find the center of mass of the system and this will lead to a more
general result.
We have 3 masses of 10 kg, 5 kg and 7 kg at 2 m, 2 m and 1 m distance from O as
shown.
We wish to replace these masses with one single mass to give an equivalent
moment. Where should we place this single mass?
Total moment =10×2+5×4+7×5=75 kg.m
If we put the masses together, we have: 10+5+7=22 kg
For an equivalent moment, we need:
22×d =75 where d is the distance from the center of mass to the point of rotation.
i.e. d =75 22 ≈3.4 m
So our equivalent system (with one mass of 22 kg ) would have:
Center of Gravity of a 2D Body
• Center of gravity of a plate
• Center of gravity of a wire
 M y x W   xW
  x dW
 M y yW   yW
  y dW
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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
• 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 wit h respect to x
5 - 11
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.
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Centroids of Common Shapes of Areas
5 - 13
Centroids of Common Shapes of Lines
5 - 14
Composite Plates and Areas
• Composite plates
X W   x W
Y W   y W
• Composite area
X A   xA
Y  A   yA
5 - 15
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.
• 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.
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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.
Qx  506.2  103 mm 3
Q y  757.7  103 mm 3
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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 - 18
Theorems of Pappus-Guldinus:
The storage tanks shown are all bodies of revolution. Thus, their surface area and
volumes can be determined using Theorems of Pappus-Goldinus.
2 - 19
Theorems of Pappus-Guldinus----Surface Area
Calculation
• 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 - 20
Pappus's Centroid Theorem—Surface Area Calculation
The first theorem of Pappus states that the surface area
of a
surface of revolution generated by the revolution of a curve about
an external axis is equal to the product of the arc length of the
generating curve and the distance traveled by the curve's
geometric centroid
,
The following table summarizes the surface areas calculated
using Pappus's centroid theorem for various surfaces of
revolution.
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Pappus's Centroid Theorem—Surface Area Calculations
solid
generating curve
cone
inclined line segment
cylinder parallel line segment
sphere
semicircle
5 - 22
Theorems of Pappus-Guldinus---Volume Calculation
• 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
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Pappus's Centroid Theorem—Volume Calculations
Volume calculations:
Similarly, the second theorem of Pappus states that the volume of a solid of revolution generated by the revolution
of a lamina about an external axis is equal to the product of the area of the lamina and the distance traveled by
the lamina's geometric centroid ,
The following table summarizes the surface areas and volumes calculated using Pappus's centroid theorem for
various solids and surfaces of revolution.
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Pappus's Centroid Theorem---Volume Calculation
solid
generating lamina
cone
right triangle
cylinder rectangle
sphere
semicircle
5 - 25
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 103 kg m3
determine the mass and weight of the
rim.
• Multiply by density and acceleration
to get the mass and weight.
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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103 kg m3 7.65106 mm3 109 m3 /mm3
2

m  60.0 kg
W  589 N
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Distributed Loads on Beams
L
W   wdx   dA  A
0
OP W   xdW
L
OP  A   xdA  x A
0
• 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.
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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.
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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
63 kN  m
18 kN
X  3.5 m
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Sample Problem 5.9
• Determine the support reactions by applying the
equilibrium conditions. For example,
successively sum the moments at the two
supports:
 MA  0 : By 6 m 18 kN3.5 m 0
B y  10.5 kN

 MB  0 :  Ay 6 m 18 kN6 m 3.5 m 0
Ay  7.5 kN

• And by summing forces in the x-direction:
 Fx  0 : Bx  0
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