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VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ninth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2010 The McGraw-Hill Companies, Inc. All rights reserved.
5Distributed Forces:
Centroids and Centers
of Gravity
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2010The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: StaticsNinth
Edition
Contents
5 - 2
Introduction
Center of Gravity of a 2D Body
Centroids and First Moments of
Areas and Lines
Centroids of Common Shapes ofAreas
Centroids of Common Shapes of
Lines
Composite Plates and AreasSample 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 BodiesSample Problem 5.12
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Vector Mechanics for Engineers: StaticsNinth
Edition
Introduction
5 - 3
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 gravityfor the body.
The centroid of an areais analogous to the center of
gravity of a body. The concept of thefirst moment of anareais used to locate the centroid.
Determination of the area of asurface of revolutionand
the volume of a body of revolutionare accomplished
with the Theorems of Pappus-Guldinus.
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Vector Mechanics for Engineers: StaticsNinth
Edition
Center of Gravity of a 2D Body
5 - 4
Center of gravity of a plate
dWy
WyWyM
dWx
WxWxM
y
y
Center of gravity of a wire
NE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Centroids and First Moments of Areas and Lines
5 - 5
x
QdAyAy
yQdAxAx
dAtxAtx
dWxWx
x
y
respect tohmoment witfirst
respect tohmoment witfirst
Centroid of an area
dLyLydLxLx
dLaxLax
dWxWx
Centroid of a line
NE
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Vector Mechanics for Engineers: StaticsNinth
Edition
First Moments of Areas and Lines
5 - 6
An area is symmetric with respect to an axis BB
if for every pointPthere exists a pointP
suchthat PPis perpendicular to BBand 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 dAat (x,y) there exists an
area dAof equal area at (-x,-y).
The centroid of the area coincides with the
center of symmetry.
NE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Centroids of Common Shapes of Areas
5 - 7 NE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Centroids of Common Shapes of Lines
5 - 8
f SNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Composite Plates and Areas
5 - 9
Composite plates
WyWY
WxWX
Composite area
AyAY
AxAX
V t M h i f E i St tiNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.1
5 - 10
For the plane area shown, determine
the first moments with respect to the
xandyaxes and the location of the
centroid.
SOLUTION:
Divide the area into a triangle, rectangle,
and semicircle with a circular cutout.
Compute the coordinates of the area
centroid by dividing the first moments by
the total area.
Find the total area and first moments ofthe triangle, rectangle, and semicircle.
Subtract the area and first moment of the
circular cutout.
Calculate the first moments of each area
with respect to the axes.
V t M h i f E i St tiNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.1
5 - 11
33
33
mm107.757
mm102.506
y
x
Q
Q Find the total area and first moments of the
triangle, rectangle, and semicircle. Subtract the
area and first moment of the circular cutout.
V t M h i f E i St tiNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.1
5 - 12
23
33
mm1013.828mm107.757
A
AxX
mm8.54X
23
33
mm1013.828
mm102.506
A
AyY
mm6.36Y
Compute the coordinates of the area
centroid by dividing the first moments bythe total area.
V t M h i f E i St tiNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Determination of Centroids by Integration
5 - 13
ydxy
dAyAy
ydxx
dAxAx
el
el
2
dxxay
dAyAy
dxxaxa
dAxAx
el
el
2
drr
dAyAy
drr
dAxAx
el
el
2
2
2
1sin
3
2
2
1cos
3
2
dAydydxydAyAy
dAxdydxxdAxAx
el
el Double integration to find the first moment
may be avoided by defining dAas a thin
rectangle or strip.
V t M h i f E i St tiNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.4
5 - 14
Determine by direct integration the
location of the centroid of a parabolic
spandrel.
SOLUTION: Determine the constant k.
Evaluate the total area.
Using either vertical or horizontal
strips, perform a single integration tofind the first moments.
Evaluate the centroid coordinates.
V t M h i f E i St tiNE
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.4
5 - 15
SOLUTION:
Determine the constant k.
2121
22
2
2
2
yb
axorxaby
a
bkakb
xky
Evaluate the total area.
3
30
3
20
2
2
ab
x
a
bdxx
a
bdxy
dAA
aa
V t M h i f E i St tiN
Ed
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.4
5 - 16
Using vertical strips, perform a single integration
to find the first moments.
1052
2
1
2
44
2
0
5
4
2
0
22
2
2
0
4
2
0
2
2
abx
a
b
dxxa
bdxy
ydAyQ
bax
a
b
dxxa
bxdxxydAxQ
a
a
elx
a
a
ely
V t M h i f E i St tiNi
Ed
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Vector Mechanics for Engineers: StaticsNinth
Edition
Sample Problem 5.4
5 - 17
Or, using horizontal strips, perform a single
integration to find the first moments.
10
42
1
22
2
0
23
21
21
21
2
0
22
0
22
abdyy
b
aay
dyyb
aaydyxaydAyQ
badyyb
aa
dyxa
dyxaxa
dAxQ
b
elx
b
b
ely
Vector Mechanics for Engineers StaticsNi
Ed
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Vector Mechanics for Engineers: Staticsinth
dition
Sample Problem 5.4
5 - 18
Evaluate the centroid coordinates.
43
2baabx
QAx y
ax4
3
103
2abab
y
QAy x
by10
3
Vector Mechanics for Engineers: StaticsNi
Ed
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Vector Mechanics for Engineers: Staticsinth
dition
Theorems of Pappus-Guldinus
5 - 19
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 generatingcurve times the distance traveled by
the centroid through the rotation.
LyA 2
Vector Mechanics for Engineers: StaticsNi
Ed
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Vector Mechanics for Engineers: Staticsnth
dition
Theorems of Pappus-Guldinus
5 - 20
Body of revolution is generated by rotating a planearea 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.
AyV 2
Vector Mechanics for Engineers: StaticsNi
Ed
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Vector Mechanics for Engineers: Staticsnth
dition
Sample Problem 5.7
5 - 21
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 ofsteel is
determine the mass and weight of the
rim.
33 mkg1085.7
SOLUTION:
Apply the theorem of Pappus-Guldinusto evaluate the volumes or revolution
for the rectangular rim section and the
inner cutout section.
Multiply by density and accelerationto get the mass and acceleration.
Vector Mechanics for Engineers: StaticsNin
Ed
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Vector Mechanics for Engineers: Staticsnth
dition
Sample Problem 5.7
5 - 22
SOLUTION:
Apply the theorem of Pappus-Guldinusto evaluate the volumes or revolution for
the rectangular rim section and the inner
cutout section.
3393633 mmm10mm1065.7mkg1085.7Vm kg0.60m
2
sm81.9kg0.60mgW N589
W
Multiply by density and acceleration to
get the mass and acceleration.
Vector Mechanics for Engineers: StaticsNin
Ed
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Vector Mechanics for Engineers: Staticsnth
dition
Distributed Loads on Beams
5 - 23
A distributed load is represented by plotting the loadper unit length, w(N/m) . The total load is equal to
the area under the load curve.
AdAdxwWL
0
AxdAxAOP
dWxWOPL
0
A distributed load can be replace by a concentratedload with a magnitude equal to the area under the
load curve and a line of action passing through the
area centroid.
Vector Mechanics for Engineers: StaticsNin
Ed
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Vector Mechanics for Engineers: Staticsnth
dition
Sample Problem 5.9
5 - 24
A beam supports a distributed load as
shown. Determine the equivalent
concentrated load and the reactions atthe supports.
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 thearea under the curve.
Determine the support reactions by
summing moments about the beam
ends.
Vector Mechanics for Engineers: StaticsNin
Ed
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Vector Mechanics for Engineers: Staticsnth
ition
Sample Problem 5.9
5 - 25
SOLUTION:
The magnitude of the concentrated load is equal to
the total load or the area under the curve.
kN0.18F
The line of action of the concentrated load passes
through the centroid of the area under the curve.
kN18
mkN63 X m5.3X
Vector Mechanics for Engineers: StaticsNin
Ed
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Vector Mechanics for Engineers: Staticsnth
ition
Sample Problem 5.9
5 - 26
Determine the support reactions by summing
moments about the beam ends.
0m.53kN18m6:0 yA BM
kN5.10yB
0m.53m6kN18m6:0 yB AM
kN5.7y
A
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Vector Mechanics for Engineers: Staticsnth
ition
Center of Gravity of a 3D Body: Centroid of a Volume
5 - 27
Center of gravity G
jWjW
jWrjWr
jWrjWr
G
G
dWrWrdWWG
Results are independent of body orientation,
zdWWzydWWyxdWWx
zdVVzydVVyxdVVx
dVdWVW and
For homogeneous bodies,
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Vector Mechanics for Engineers: Staticsnth
ition
Centroids of Common 3D Shapes
5 - 28
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Vector Mechanics for Engineers: Staticsnth
tion
Composite 3D Bodies
5 - 29
Moment of the total weight concentrated at the
center of gravity G is equal to the sum of themoments of the weights of the component parts.
WzWZWyWYWxWX
For homogeneous bodies,
VzVZVyVYVxVX
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Vector Mechanics for Engineers: Staticsthtion
Sample Problem 5.12
5 - 30
Locate the center of gravity of thesteel machine element. The diameter
of each hole is 1 in.
SOLUTION:
Form the machine element from arectangular parallelepiped and a
quarter cylinder and then subtracting
two 1-in. diameter cylinders.
Vector Mechanics for Engineers: StaticsNin
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Vector Mechanics for Engineers: Staticsthtion
Sample Problem 5.12
5 - 31
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Vector Mechanics for Engineers: Staticsthtion
Sample Problem 5.12
34 in.2865in08.3 VVxX
34 in.2865in5.047 VVyY
34 in.2865in.6181 VVzZ
in.577.0X
in.577.0Y
in5770
Z