Seventh Edition
5
CHAPTER
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
Distributed Forces:
Centroids and Centers
of Gravity
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Seventh
Edition
Vector Mechanics for Engineers: Statics
Contents
• Introduction
• Center of Gravity of a 2D Body
• Centroids and First Moments of
Areas and Lines
• 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
• Sample Problem 5.4
•
•
•
•
•
Theorems of Pappus-Guldinus
Sample Problem 5.7
Distributed Loads on Beams
Sample Problem 5.9
Center of Gravity of a 3D Body:
Centroid of a Volume
• Centroids of Common 3D Shapes
• Composite 3D Bodies
• Sample Problem 5.12
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5-2
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Vector Mechanics for Engineers: Statics
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.
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Vector Mechanics for Engineers: Statics
Center of Gravity of a 2D Body
• Center of gravity of a plate
M
y
xW 

M
y
yW 

• Center of gravity of a wire
 xW
 x dW
 yW
 y dW
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Vector Mechanics for Engineers: Statics
Centroids and First Moments of Areas and Lines
• Centroid of an area
• Centroid of a line
xW 
x W   x dW
 x dW
x  La    x  a dL
x  At    x  t dA
x A   x dA  Q y
 first moment wit
h respect to
y
h respect to
x
xL 
 x dL
yL 
 y dL
y A   y dA  Q x
 first moment wit
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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.
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Vector Mechanics for Engineers: Statics
Centroids of Common Shapes of Areas
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Vector Mechanics for Engineers: Statics
Centroids of Common Shapes of Lines
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Edition
Vector Mechanics for Engineers: Statics
Composite Plates and Areas
• Composite plates
X W   xW
Y W   yW
• Composite area
X  A   xA
Y  A   yA
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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.
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5 - 10
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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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
3
3
3
3
Q x   506 . 2  10 mm
Q y   757 . 7  10 mm
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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.
3
3
 x A   757 . 7  10 mm
X 
3
2
A
13.828  10 mm
X  54 . 8 mm
3
3
 y A   506 . 2  10 mm
Y 
3
2
A
13.828  10 mm
Y  36 . 6 mm
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Vector Mechanics for Engineers: Statics
Determination of Centroids by Integration
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
x A   x dA   x dx dy   x el dA
y A   y dA   y dx dy   y el dA
xA 
 x el dA
  x  ydx 
yA 

 y el dA

y
2
 ydx 
xA 

 x el dA

a x
2
  a  x dx 
yA 
 y el dA
  y  a  x dx 
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
xA 

yA 

 x el dA

1 2

cos   r d  
3
2

2r
 y el dA

1 2

sin   r d  
3
2

2r
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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.
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Vector Mechanics for Engineers: Statics
Sample Problem 5.4
SOLUTION:
• Determine the constant k.
y  kx
2
b  ka
2
 k 
b
a
y 
b
a
2
x
2
or
2
a
x 
b
1 2
y
1 2
• Evaluate the total area.
A
 dA
a
 b x3 
b 2
y dx  
x dx  
2
2 3 
 a
 0
0a
a



ab
3
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Vector Mechanics for Engineers: Statics
Sample Problem 5.4
• Using vertical strips, perform a single integration
to find the first moments.
a
Qy 

x el dA 

xy dx 
 b 2
x
  2 x  dx

0 a
a
2
 b x4 
a b
 

2 4 
4
 a
 0
Qx 
 y el dA  
a
2
1 b 2
y dx   
x  dx
2
2
2

0 a
y
a
2
 b2 x5 
ab
 

4 5 
10
 2 a
 0
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Vector Mechanics for Engineers: Statics
Sample Problem 5.4
• Or, using horizontal strips, perform a single
integration to find the first moments.
Q y   x el dA  
ax
2
b
 a  x dy  
0
a
2
x
2
dy
2
2
2
b

1  2 a
a b
 a 
y dy 

2 0
b 
4
a

1
Q x   y el dA   y  a  x dy   y  a 
y
1 2

b
b
a

3
   ay 
y
1 2

b
0
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2
ab

 dy 
10

2

 dy

2
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Vector Mechanics for Engineers: Statics
Sample Problem 5.4
• Evaluate the centroid coordinates.
xA  Q y
x
ab
2

3
a b
x 
4
3
a
4
yA  Q x
y
ab
3
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
ab
10
2
y 
3
b
10
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Vector Mechanics for Engineers: Statics
Distributed Loads on Beams
L
W   w dx   dA  A
0
OP W   x dW
L
OP  A   x dA  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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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.
A beam supports a distributed load as
shown. Determine the equivalent
concentrated load and the reactions at
the supports.
• Determine the support reactions by
summing moments about the beam
ends.
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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
X  3 .5 m
18 kN
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Vector Mechanics for Engineers: Statics
Sample Problem 5.9
• Determine the support reactions by summing
moments about the beam ends.
M
A
 0:
B y  6 m   18 kN
3 .5 m   0
B y  10 . 5 kN
 M B  0 :  A y  6 m   18 kN  6 m  3 .5 m   0
A y  7 . 5 kN
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Vector Mechanics for Engineers: Statics
Center of Gravity of a 3D Body: Centroid of a Volume
• Center of gravity G

W j 


rG   W j  


rG W   j  
W 
 dW

• Results are independent of body orientation,

  W j 


 r    W j 


  r  W    j 

rG W 

r
 dW
xW 
 xdW
yW 
 ydW
zW 
 zdW
• For homogeneous bodies,
W   V and dW   dV
xV 
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 xdV
yV 
 ydV
zV 
 zdV
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Vector Mechanics for Engineers: Statics
Centroids of Common 3D Shapes
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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
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Y
 V   yV
Z
 V   zV
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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.
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Seventh
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Vector Mechanics for Engineers: Statics
Sample Problem 5.12
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Vector Mechanics for Engineers: Statics
Sample Problem 5.12

X   x V  V  3 . 08 in
4
 5 .286 in 3 
X  0 . 577 in.

Y   y V  V   5.047 in
4
 5 .286 in 3 
Y  0 . 577 in.

Z   z V  V  1 .618 in
4
 5 .286 in 3 
Z  0 . 577 in.
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5 - 28