Engineering Mechanics:
Statics in SI Units, 12e
9
Center of Gravity and Centroid
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Center of Gravity and Center of Mass for a System of
Particles 物體的重心與質心
2. Composite Bodies 組合體
3. Theorems of Pappus and Guldinus 巴伯與古丁定理
4. Resultants of a General Distributed Loading 一般分佈
負載的合力
5. Fluid Pressure 流體壓力
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Center of Gravity and Center of Mass for a System
of Particles
2. Composite Bodies
3. Theorems of Pappus and Guldinus
4. Resultants of a General Distributed Loading
5. Fluid Pressure
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass
for a System of Particles
Center of Gravity 重心
• Consider system of n particles fixed within a region of
space
• The weights of the particles can be replaced by a
single (equivalent) resultant weight having defined
point G of application
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9.1 Center of Gravity and Center of Mass
for a System of Particles
Center of Gravity
• Resultant weight = total weight of n particles
W   dW
•
Sum of moments of weights of all the particles about x,
y, z axes = moment of resultant weight about these
axes
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9.1 Center of Gravity and Center of Mass
for a System of Particles
Center of Gravity
x W   ~x dW
yW   ~ydW
z W   ~z dW
x
~x dW

 dW
~y dW

y
 dW
~z dW

z
 dW
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9.1 Center of Gravity and Center of Mass
for a System of Particles
Center Mass 質心
• Provided acceleration due to gravity g for every
particle is constant, then W = mg
• A rigid body is composed of an infinite number of
particles
x
~x dm

 dm
;y 
~ydm

 dm
;z 
~z dm

 dm
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9.1 Center of Gravity and Center of Mass
for a System of Particles
Centroid of a Volume 體積的形心
• Consider an object subdivided into volume elements
dV, for location of the centroid,
~x dV

x V
 dV
V
~ydV

;y V
 dV
V
~z dV

;z  V
 dV
V
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9.1 Center of Gravity and Center of Mass
for a System of Particles
Centroid of an Area 面積的形心
• For centroid for surface area of an object, such as
plate and shell, subdivide the area into differential
elements dA
x
~x dA

A
 dA
A
y
~y dA

A
 dA
A
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9.1 Center of Gravity and Center of Mass
for a System of Particles
Centroid of a Line 線的形心
• If the geometry of the object takes the form of a line,
the balance of moments of differential elements dL
about each of the coordinate system yields
~x dL
x

L
 dL
L
y
~y dL

L
 dL
L
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Example 9.1
Locate the centroid of the rod bent into the shape of a
parabolic arc.
x
~
 x dL
L
 dL
L
y
~y dL

L
 dL
L
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Example 9.1
For differential length of the element dL
dL 
dx 2  dy 2
2
 dx 
    1 dy
 dy 
Since x = y2 and then dx/dy = 2y
dL 
2 y 2  1 dy
The centroid is located at
~
x  x, ~
yy
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Example 9.1
Integrations
x
~
 x dL
L
 dL
L
1

0
1
0
x 4 y 2  1 dy
4 y 2  1 dy

1 2
y
0
1

0
4 y 2  1 dy
4 y 2  1 dy
0.6063
 0.410m
1.479
~
 ydL 1 y 4 y 2  1 dy
yL
 01
 dL  4 y 2  1 dy

L

0
0.8484
 0.574m
1.479
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Test
• Locate the centroid of the area (求面積的形心)
x
~
 x dA
A
 dA
A
y
~y dA

A
 dA
A
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Test
For differential element dA
dA  y dx
The centroid is located at
~
xx
~y  y
2
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Test
x
~
 x dA
A
 dA
A
1



xy dx
0
1
0
y dx
1


 x dx
0
1
x 3 dx
2
0
0.250
 0.75m
0.333
1 y
~
y
dA
A
0 ( 2 ) y dx
y
 1

 dA  y dx

0
A
x2 2
0 ( 2 ) x dx
1
2
x
 dx
1
0
0.100

 0.3m
0.333
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Test
• Locate the y centroid for the paraboloid of revolution
(求旋轉體的形心 y 座標)
~
 ydV
yV
 dV
V
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Test
For differential element dV
dV  (z ) dy
2
The centroid is located at
~
yy
Copyright © 2010 Pearson Education South Asia Pte Ltd
Test
~
 ydV
yV
 dV
V
100



0
y (z 2 ) dy
100
0
100



0
y ( (100 y )) dy
100
0
(z 2 ) dy
( (100 y )) dy
 66.7 mm
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Center of Gravity and Center of Mass for a System of
Particles
2. Composite Bodies
3. Theorems of Pappus and Guldinus
4. Resultants of a General Distributed Loading
5. Fluid Pressure
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.2 Composite Bodies 組合體
•
•
•
Consists of a series of connected “simpler” shaped
bodies, which may be rectangular, triangular or
semicircular
A body can be divided into its composite parts
Accounting for finite number of weights
~
xW
x
W
~
yW
y
W
~
zW
z
W
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9.2 Composite Bodies
Procedure for Analysis
1. Divide the body 切割
2. Determine the coordinates of centroid of each part
求出各零件的形心座標
3. Summations 累加
~
xW
x
W
~
yW
y
W
~
zW
z
W
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.10
Locate the centroid of the plate area.
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Solution
Plate divided into 3 segments.
Area of small rectangle considered “negative”.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Location of the centroid for each piece is determined and
indicated in the diagram.
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Solution
Summations
~
xA  4
x

 0.348mm
 A 11.5
~
y A 14
y

 1.22mm
 A 11.5
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Test
• Locate the centroid of the wire (求纜線的形心)
~
xL
x
L
 ~y L
y
L
 ~z L
z
L
y
2r

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Test
• 切割
y
2r

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Test
• 計算各個零件之形心
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Test
• 累加
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Chapter Outline
1. Center of Gravity and Center of Mass for a System of
Particles
2. Composite Bodies
3. Theorems of Pappus and Guldinus
4. Resultants of a General Distributed Loading
5. Fluid Pressure
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.3 Theorems of Pappus and Guldinus
•
The theorems of Pappus and Guldinus (巴伯與古丁定
理) are used to find the surfaces area and volume of
any object of revolution (旋轉物件) provided the
generating curves and areas do not cross the axis
they are rotated
•
•
巴伯與古丁定理是用來求任何旋轉物件的表面積與體積
旋轉的曲線及平面，不與旋轉軸相交
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9.3 Theorems of Pappus and Guldinus
Surface Area
• A surface area of revolution is generated by
revolving a plane curve about a non-intersecting
fixed axis in the plane of the curve
•
旋轉的曲線，不與旋轉軸相交
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9.3 Theorems of Pappus and Guldinus
•
Area of a surface of revolution = product of length of
the curve and distance traveled by the centroid in
generating the surface area
•
旋轉表面積(A)＝曲線長度(L)乘上曲線形心在旋轉過程
中移動的距離
A  r L
Note: θ以弳度表示
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Example 9.12
Show that the surface area of a sphere is A = 4πR2 and
its volume V = 4/3 πR3.
( 證明球體的表面積及體積公式為 A = 4πR2 , V = 4/3 πR3)
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.12 -- Solution
Surface Area -Generated by rotating semi-arc about the x axis
For centroid,
r  2R / 
For surface area,
A  r L
 2R 
2
A  2 
 R  4R
  
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9.3 Theorems of Pappus and Guldinus
Volume
• A volume of revolution is generated by revolving a
plane area bout a nonintersecting fixed axis in the
plane of area
•
旋轉的面積，不與旋轉軸相交
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9.3 Theorems of Pappus and Guldinus
•
Volume of a body of revolution = product of
generating area and distance traveled by the centroid
in generating the volume
•
旋轉體積(V)＝平面面積(A)乘上面積形心在旋轉過程中
移動的距離
V  r A
Note: θ以弳度表示
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.12
Show that the surface area of a sphere is A = 4πR2 and
its volume V = 4/3 πR3.
( 證明球體的表面積及體積公式為 A = 4πR2 , V = 4/3 πR3)
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.12 -- Solution
Volume -Generated by rotating semicircular area about the x axis
For centroid,
4R
r
3
For volume,
V   rA
 4R   1 2  4 3
V  2 
  R   R
 3   2
 3
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9.3 Theorems of Pappus and Guldinus
Composite shapes 組合形狀
• The total surface area or volume generated is the
addition of the surface area or volumes generated by
each of the composite parts
• 組合體的表面積或體積＝每一個零件旋轉產生的表面積
或積體之和
A  ~
rL
V  ~
rA
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Test
• Determine the surface area and volume of the full solid.
A  ~
rL
V  ~
rA
Copyright © 2010 Pearson Education South Asia Pte Ltd
Test -- Answer
(2)
(3)
(1)
(4)
A  ~
rL
A  2 ~
rL

 2 [(2.5)(2)  (3)( 12  12 )  (3.5)(3)  (3)(1)]  143 (m 2 )
(1)
(2)
(3)
(4)
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Test -- Answer
(1)
(2)
V  ~
rA
V  2 ~
rA

1
 2 {(3.1667)( 11)  (3)(2 1)}  47.6 (m3 )
2
(1)
(2)
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Center of Gravity and Center of Mass for a System of
Particles
2. Composite Bodies
3. Theorems of Pappus and Guldinus
4. Resultants of a General Distributed Loading
5. Fluid Pressure
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.4 Resultant of a General Distributed
Loading 一般分佈負載的合力
Pressure Distribution over a Surface
• The flat plate subjected to the loading function
p = p(x, y) [N/m2] Pa(Pascal)
• The force dF acting on the differential area dA m2 of
the plate, located at the differential point (x, y)
dF = p(x, y) [N/m2] dA [m2]
= p(x, y) dA [N]
= dV
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9.4 Resultant of a General Distributed
Loading
• Magnitude of resultant force (FR) = total volume under
the distributed loading diagram
• 合力的大小＝分佈負載圖下所涵蓋的所有體積
FR   p( x, y)dA   dV  V
A
V
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9.4 Resultant of a General Distributed
Loading
• Location of Resultant Force （合力的作用位置）is
xp ( x, y )dA  xdV

x

 p( x, y)dA  dV
yp ( x, y )dA  ydV

y

 p( x, y)dA  dV
A
V
A
V
A
V
A
V
合力的作用線通過分佈負載圖
下所涵蓋體積的形心
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Center of Gravity and Center of Mass for a System of
Particles
2. Composite Bodies
3. Theorems of Pappus and Guldinus
4. Resultants of a General Distributed Loading
5. Fluid Pressure
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.5 Fluid Pressure 流體壓力
• According to Pascal’s law, a fluid at rest creates a
pressure p at a point that is the same in all directions
• Magnitude of p depends on the specific weight γ or
mass density ρ of the fluid and the depth z of the point
from the fluid surface
p = γz = ρgz
– Valid for incompressible fluids
– Gas are compressible fluids and the above equation
cannot be used
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9.5 Fluid Pressure 流體壓力
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9.5 Fluid Pressure
Flat Plate of Constant Width 固定寬度的平版
• Consider flat rectangular plate of constant width
submerged in a liquid having a specific weight γ
• Plane of the plate makes an angle with the horizontal
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9.5 Fluid Pressure
Flat Plate of Constant Width
• As pressure varies linearly with depth, the distribution
of pressure over the plate’s surface is represented by a
trapezoidal volume having an intensity of p1= γz1 at
depth z1 and p2 = γz2 at depth z2
• Magnitude of the resultant force FR
= volume of this loading diagram
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Example 9.14
Determine the magnitude and location of the resultant
hydrostatic force acting on the submerged rectangular
plate AB. The plate has a width of 1.5m; ρw = 1000kg/m3.
(求平板AB受液體壓力的大小及作用位置)
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
The water pressures at depth A and B are
(AB點位置的壓力)
p A   w gz A  (1000kg / m3 )(9.81m / s 2 )(2m)  19.62kPa
pB   w gz B  (1000kg / m3 )(9.81m / s 2 )(5m)  49.05kPa
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Solution
The intensities of the load at A and B
(AB點位置的負載強度 (平版有固定寬度)，3D簡化為2D)
wA  bp A  (1.5m)(19.62kPa)  29.43kN / m
wB  bpB  (1.5m)(49.05kPa)  73.58kN / m
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Solution
The magnitude of the resultant force FR
FR  area of trapezoid
1
 (3)(29.4  73.6)  154.5 N
2
This force acts through the centroid
of the area (作用力通過面積形心)
1  2(29.43)  73.58 
h 
(3)  1.29m
3  29.43  73.58 
measured upwards from B
Copyright © 2010 Pearson Education South Asia Pte Ltd
Test
Determine the magnitude of the hydrostatic force acting on
gate AB, which has a width of 1.5m. Water has a density of
ρw = 1Mg/m3. (求平板AB受液體壓力的大小)
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
The water pressures at depth A and B are
(AB點位置的壓力)
p A   w gz A  (1000kg / m3 )(9.81m / s 2 )(0m)  0 Pa
pB   w gz B  (1000kg / m3 )(9.81m / s 2 )(2m)  19620 Pa
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Solution
The intensities of the load at A and B
(AB點位置的負載強度 (平版有固定寬度)，3D簡化為2D)
wA  bp A  (1.5m)(0 Pa )  0 N / m
wB  bpB  (1.5m)(19620 Pa )  29430 N / m
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
The magnitude of the resultant force FR
FR  area of trapezoid
1
 (29430)( 1.52  2 2 )
2
 36787.5 N
 36.8kN
Copyright © 2010 Pearson Education South Asia Pte Ltd
Test
• Determine the magnitude of the hydrostatic force acting
on gate AB, which has a width of 0.5m. The specific
weight of water is γ = 10KN/m3. (求平板AB受液體壓力的
大小)
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END OF CHAPTER 9
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