Tenth Edition
5
CHAPTER
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
David F. Mazurek
Lecture Notes:
John Chen
California Polytechnic State University
Distributed Forces:
Centroids and Centers
of Gravity
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Tenth
Edition
Vector Mechanics for Engineers: Statics
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 5.9
Centroids of Common 3D
Shapes
Composite 3D Bodies
Sample Problem 5.12
Sample Problem 5.1
Determination of Centroids by
Integration
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-2
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-3
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-4
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
y W   y W
  y dW
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-5
Tenth
Edition
Vector Mechanics for Engineers: Statics
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 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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-6
Tenth
Edition
Vector Mechanics for Engineers: Statics
Determination of Centroids by Integration
x A   xdA   x dxdy   xel dA
yA   ydA   y dxdy   yel dA
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
x A   xel dA
x A   xel dA
yA   yel dA
ax
 a  x dx
2
yA   yel dA
  x  ydx
y
   ydx
2

  y a  x dx
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
x A   xel dA

2r
1

cos  r 2 d 
3
2

yA   yel dA

2r
1

sin   r 2 d 
3
2

5-7
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
• Evaluate the centroid coordinates.
First, estimate the location of the
centroid by inspection. Discuss with
a neighbor where it is located,
roughly, and justify your answer.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-8
Tenth
Edition
Vector Mechanics for Engineers: Statics
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   2 x dx   2 
 a 3  0
0a
ab

3
a
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5-9
Tenth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 5.4
• Using vertical strips, perform a single integration
to find the first moments.
 b 2 
Qy   x el dA   xydx   x 2 x d x

0 a
a
a
 b x 4  a 2 b
  2
 
4
a 4 0
2
a 1  b

y
Qx   y el dA   y dx    2 x 2  dx
2

0 2 a
a
 b 2 x 5  ab2
  4
 
2a 5 0 10

© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 10
Tenth
Edition
Vector Mechanics for Engineers: Statics
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 d y
 b

b 
a 3 2 
ab2
  ay  1 2 y d y 

10
b
0 
© 2013The McGraw-Hill Companies,
Inc. All rights reserved.
5 - 11
Tenth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 5.4
• Evaluate the centroid coordinates.
xA  Q y
ab a 2b
x

3
4
3
x a
4
yA  Q x
ab ab 2
y

3
10
y
3
b
10
Is this “center of area” close to where
you estimated it would be?
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 12
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 13
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 14
Tenth
Edition
Vector Mechanics for Engineers: Statics
Centroids of Common Shapes of Areas
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 15
Tenth
Edition
Vector Mechanics for Engineers: Statics
Centroids of Common Shapes of Lines
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 16
Tenth
Edition
Vector Mechanics for Engineers: Statics
Composite Plates and Areas
• Composite plates
X W   x W
Y W   y W
• Composite area
X  A   xA
Y  A   yA
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 17
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 18
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
Qx  506.2  103 mm 3
Q y  757.7  103 mm 3
5 - 19
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 20
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 21
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 22
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
• Multiply by density and acceleration
to get the mass and weight.
5 - 23
Tenth
Edition
Vector Mechanics for Engineers: Statics
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 mm3 10 9 m3 /mm3
2
© 2013The McGraw-Hill Companies, Inc. All rights reserved.

m  60.0 kg
W  589 N
5 - 24
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 25
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 26
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
X  3.5 m
5 - 27
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 28
Tenth
Edition
Vector Mechanics for Engineers: Statics
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


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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
yV   ydV
z V   zdV
5 - 29
Tenth
Edition
Vector Mechanics for Engineers: Statics
Centroids of Common 3D Shapes
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 30
Tenth
Edition
Vector Mechanics for Engineers: Statics
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
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
Y  V   yV
Z V   zV
5 - 31
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 32
Tenth
Edition
Vector Mechanics for Engineers: Statics
Sample Problem 5.12
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 33
Tenth
Edition
Vector Mechanics for Engineers: Statics
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.
© 2013The McGraw-Hill Companies, Inc. All rights reserved.
5 - 34