Chapter 5: Distributed Forces: Centroids
and Centers of Gravity
최해진
hjchoi@cau.ac.kr
School of Mechanical
Engineering
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
Centroids of Common Shapes of Areas
Centroids of Common Shapes of Lines
Center of Gravity of a 3D Body:
Centroid of a Volume
Composite Plates and Areas
Centroids of Common 3D Shapes
Sample Problem 5.1
Composite 3D Bodies
Determination of Centroids by
Integration
Sample Problem 5.12
Sample Problem 5.9
Sample Problem 5.4
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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.
• Determination of the area of a surface of revolution simplifies
the analysis of beams subjected to distributed loads
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Center of Gravity of a 2D Body
집중력은 없다 : 어떤 물체에 기계적으로 가한 모든 외력은 아무리 작아도 유한한
접촉면적에 분포되기 때문에 분포력으로 볼 수 있다.
※ “concentrated” forces does not exist in the exact sense
R : Resultant of
R
distributed forces
※ internal forces
C
C
C
Stress : internal distributed
forces in solids
C
◎ 물체의 중심(Center of gravity ) :
물체의 각 부분에 작용하는 중력의 합력인
작용점
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Center of Gravity of a 2D Body
• Center of gravity of a plate
å Fz
• Center of gravity of a wire
Û W = DW1 + DW2 + L + DWn
모멘트 원리 : 임의의 축에 대한 중력의 합력 W의 모멘트는
물체의 미소 요소로 생각한 각 질점에 작용한 중력 dW의 모멘트의 합과 같다.
åMy
Û x W = å xi DWi = x1 DW1 +x2 DW2 + L + xn DWn
Û x W = ò x dW
åMx
( x , y ) : 물체의
무게 중심의 좌표
Û yW = å yi DWi = y1 DW1 + y2 DW2 + L + yn DWn
Û yW = ò y dW
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5-5
Centroids and First Moment of Areas and Lines
• Centroid of a line
• Centroid of an area
3
3
specific weight(비중량) g = r g (lb/ft or N/m )
x W = ò x dW Û x (gAt ) = ò x (g t )dA
x A = ò x dA = Qy
( r g ) = (kg/m 3 )(m/sec2 ) = N/m 3
x W = ò x dW Þ x (g La ) = ò x (g a )dL
x L = ò x dL
Þ y축에 대한 면적 A의 1차 모멘트
yL = ò y dL
yA = ò y dA = Qx
Þ x축에 대한 면적 A의 1차 모멘트
\x =
1
xdA ,
ò
A
y=
1
ydA
ò
A
( x , y ) : 면적 A의 도심 C의 좌표
\x =
1
xdL ,
ò
L
y=
1
ydL
ò
L
( x , y ) : 선 L의 도심 C의 좌표
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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
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Centroids of Common Shapes of Areas
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5-9
Centroids of Common Shapes of Lines
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Composite Plates and Areas
• Composite plates
X åW = å x W
Y åW = å y W
• Composite area
X å A = å xA
Y å A = å yA
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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.
Q x = +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
å
X =
=
å A 13.828 ´ 103 mm 2
X = 54.8 mm
y A + 506.2 ´ 103 mm 3
å
Y =
=
å A 13.828 ´ 103 mm 2
Y = 36.6 mm
연습문제: 5.3, 5.13, 5.20
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Determination of Centroids by Integration
x A = ò xdA = òò x dxdy = ò xel dA
yA = ò ydA = òò y dxdy = ò yel dA
x A = ò xel dA
= ò x ( ydx )
yA = ò yel dA
y
= ò ( ydx )
2
• Double integration to find the first moment
may be avoided by defining dA as a thin
rectangle or strip.
x A = ò xel dA
a+x
[ (a - x )dx]
=ò
2
yA = ò yel dA
= ò y [(a - x )dx ]
x A = ò xel dA
=ò
2r
æ1
ö
cosq ç r 2 dq ÷
3
è2
ø
yA = ò yel dA
=ò
2r
æ1
ö
sin q ç r 2 dq ÷
3 School èof2Mechanical
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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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Sample Problem 5.4
SOLUTION:
• Determine the constant k.
y = k x2
b = k a2 Þ k =
y=
b
a2
x2
or
b
a2
x=
a
b1 2
y1 2
• Evaluate the total area.
A = ò dA
a
é b x3 ù
b 2
= ò y dx = ò 2 x dx = ê 2 ú
êë a 3 úû 0
0a
ab
=
3
a
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Sample Problem 5.4
• Using vertical strips, perform a single integration
to find the first moments.
a
æ b
ö
Q y = ò xel dA = ò xydx = ò xç 2 x 2 ÷dx
ø
0 èa
a
é b x4 ù
a 2b
=ê 2
ú =
4
êë a 4 úû 0
2
a
y
1æ b
ö
Q x = ò yel dA = ò ydx = ò ç 2 x 2 ÷ dx
2
ø
02èa
a
é b2 x5 ù
ab 2
=ê 4 ú =
êë 2a 5 úû 0 10
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Sample Problem 5.4
• Or, using horizontal strips, perform a single
integration to find the first moments.
b 2
a+x
a - x2
(a - x )dy = ò
Q y = ò xel dA = ò
dy
2
2
0
1 b æç 2 a 2
= ò a 2 0 çè
b
2
ö
a
b
y ÷dy =
÷
4
ø
a
æ
ö
Q x = ò yel dA = ò y (a - x )dy = ò yç a - 1 2 y1 2 ÷dy
è
ø
b
a 3 2ö
ab 2
æ
= ò ç ay - 1 2 y ÷dy =
10
ø
b
0è
b
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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
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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 = 2p yL
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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 = 2p y A
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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 acceleration.
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 r = 7.85 ´ 103 kg m 3
determine the mass and weight of the
rim.
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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 acceleration.
(
)(
)
3ö
æ
m = rV = 7.85 ´ 103 kg m 3 7.65 ´ 10 6 mm 3 ç10 -9 m 3 mm ÷
è
ø
2
W = mg = (60.0 kg ) 9.81 m s
(
연습문제: 5.35, 5.42, 5.59
)
m = 60.0 kg
W = 589 N
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Distributed Loads on Beams
L
W = ò wdx = ò dA = A
0
(OP )W = ò xdW
L
(OP ) A = ò 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.
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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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 summing
moments about the beam ends.
å M A = 0 : B y (6 m ) - (18 kN )(3.5 m ) = 0
B y = 10.5 kN
å M B = 0 : - Ay (6 m ) + (18 kN )(6 m - 3.5 m ) = 0
Ay = 7.5 kN
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Center of Gravity of a 3D Body: Centroid
of a Volume
• Center of gravity G
r
r
- W j = å (- DW j )
r
r
r
r
rG ´ (- W j ) = å [r ´ (- DW j )]
r
r
r
r
rGW ´ (- j ) = (å r DW ) ´ (- j )
W = ò dW
r
r
rGW = ò r dW
• Results are independent of body orientation,
x W = ò xdW
yW = ò ydW
z W = ò zdW
• For homogeneous bodies,
W = g V and dW = g dV
x V = ò xdV
yV = ò ydV
z V = ò zdV
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Centroids of Common 3D Shapes
연습문제: 5.71, 5.77,5.113
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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 åW = å xW
Y å W = å yW
Z åW = å zW
• For homogeneous bodies,
X åV = å xV
Y å V = å yV
Z åV = å zV
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Sample Problem 5.12
SOLUTION:
• Form the machine element from a
rectangular parallelepiped and a
quarter cylinder and then subtracting
two 0.03-m- diameter cylinders.
Locate the center of gravity of the
steel machine element. The diameter
of each hole is 1 cm.
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Sample Problem 5.12
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Sample Problem 5.12
X = å xV
Y = å yV
Z = å zV
åV = (5.024 ´ 10 ) (63.347 ´ 10 )
-7
-3
åV = (- 13.52 ´ 10 ) (63.347 ´ 10 )
-7
(
-7
V
=
25
.
22
´
10
å
-3
) (63.347 ´ 10 )
-3
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