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Chapter 9 Center of Gravity and Centroid
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CENTER OF GRAVITY AND CENTROID (Chapter 9)
Objective :
Students will:
a) Understand the concepts of center of
gravity, center of mass, and centroid.
b) Be able to determine the location of
these points for a system of particles
or a body.
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APPLICATIONS
To design the structure for supporting awater tank, we will need to know the
weights of the tank and water as well as the
locations where the resultant forcesrepresenting these distributed loads are
acting---center of gravity (CG).
.
One of the important factors indetermining SUVs stability is
its center of mass (CM).
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CONCEPT OF CG and CM
The center of gravity (G) is a point which
locates the resultant weight of a system of
particles or body.
From the definition of a resultant force, the sum of moments due to
individual particle weight about any point is the same as the moment
due to the resultant weight located at G. For the figure above, try taking
moments about A and B.
Also, note that the sum of moments due to the individual particles
weights about point G is equal to zero.
Similarly, the center of mass is a point which locates the resultant
mass of a system of particles or body. Generally, its location is the
same as that of G.
3m
4N
1m
A B1 N
3 N
G
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CG / CM FOR A SYSTEM OF PARTICLES
Consider a system of n particles as shown in
the figure. The net or the resultant weight is
given as WR = W.
Similarly, we can sum moments about the x and z-axes to find the
coordinates of G.
By replacing the W with a M in these equations, the coordinatesof the center of mass can be found.
Summing the moments about the y-axis, we get
x WR = x1W1 + x2W2 + .. + xnWn
where x1 represents x coordinate of W1, etc..
~~~
~
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CG / CM / CENTROID OF A BODY (Section 9.2)
A rigid body can be considered as madeup of an infinite number of particles.
Hence, using the same principles as in
the previous slide, we get thecoordinates of G by simply replacing the
discrete summation sign (
) by the
continuous summation sign (
) and W
by dW.
Similarly, the coordinates of the center of mass and the centroid
of volume, area, or length can be obtained by replacing W by m,
V, A, or L, respectively.
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CONCEPT OF CENTROID
The centroid C is a point which defines the
geometric center of an object.
The centroid coincides with the center
of mass or the center of gravity only if
the material of the body is homogenous
(density or specific weight is constant
throughout the body).
If an object has an axis of symmetry, then
the centroid of object lies on that axis.
In some cases, the centroid is not
located on the object.
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Centroids of volumes
If the density (mass per unit volume) of
the material that makes up a body isconstant, then
dVV,dmm ==
=
V
dVV
V
dVx
xdVxdMVx VCV
yzC === ,
V
dVy
ydVydMVyV
C
V
xzC
=== ,
V
dVz
zdVzdMVz VCV
xyC
=== ,
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Centroids of Areas
If the body consists of a homogeneous, thin
plate of uniform thickness t and surface areaA, then tdAdmtAm == ,
=A
dAAA
dAx
xdAxdMAx ACA
yzC
=== ,
A
dAy
ydAydMAy
A
CA
xzC
=== ,
A
dAz
zdAzdMAz ACA
xyC
=== ,
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Centroid of Lines
If the body consists of a homogeneous
curved wire with a small uniform cross-sectional area A and length L, then AdLdmLAm == ,
=
L
dLL
L
dLx
xdLxdMLx LCL
yzC
=== ,
L
dLy
ydLydMLyL
CL
xzC
=== ,
L
dLz
zdLzdMLz LCL
xyC
=== ,
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STEPS FOR DETERMING AREA CENTROID
1. Choose an appropriate differential element dA at a general point (x,y).
Hint: Generally, if y is easily expressed in terms of x
(e.g., y = x2 + 1), use a vertical rectangular element. If the converse
is true, then use a horizontal rectangular element.
2. Express dA in terms of the differentiating element dx (or dy).
Note: Similar steps are used for determining CG, CM, etc.. These
steps will become clearer by doing a few examples.
4. Express all the variables and integral limits in the formula using
either x or y depending on whether the differential element is interms of dx or dy, respectively, and integrate.
3. Determine coordinates (x , y ) of the centroid of the rectangularelement in terms of the general point (x,y).
~ ~
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EXAMPLE
Solution
1. Since y is given in terms of x, choose
dA as a vertical rectangular strip.
x,y
x , y~~2. dA = y dx = (9 x2) dx
3. x = x and y = y / 2~~
Given: The area as shown.
Find: The centroid location (x , y)
Plan: Follow the steps.
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EXAMPLE (continued)
4. x = ( A x dA ) / ( A dA )~
0
0
0 x ( 9 x2) d x [ 9 (x2)/2 (x4) / 4] 30
( 9 x2) d x [ 9 x (x3) / 3 ] 3
= ( 9 ( 9 ) / 2 81 / 4 ) / ( 9 ( 3 ) ( 27 / 3 ) )
= 1.13 ft
3= =
3
33.60 ftA y dA 0 ( 9 x
2) ( 9 x2) dx
A dA 0
( 9 x2) d x
3
=y = =~
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CONCEPT QUIZ
1. The steel plate with known weight and non-uniform thickness and density is supported
as shown. Of the three parameters (CG, CM,
and centroid), which one is needed for
determining the support reactions? Are allthree parameters located at the same point?
A) (center of gravity, no)
B) (center of gravity, yes)C) (centroid, yes)
D) (centroid, no)
2. When determining the centroid of the area above, which type ofdifferential area element requires the least computational work?
A) Vertical B) Horizontal
C) Polar D) Any one of the above.
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EXAMPLE
Given: The area as shown.
Find: The x of the centroid.
Plan: Follow the steps.
Solution
1. Choose dA as a horizontal rectangular
strip.(x1,,y) (x2,y)
2. dA = ( x2
x1) dy
= ((2 y) y2) dy
3. x = ( x1 + x2) / 2
= 0.5 (( 2 y) + y2 )
~
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EXAMPLE(continued)
4. x = ( A x dA ) / ( A dA )~
A dA = 0
( 2 y y2) dy
[ 2 y y2 / 2 y3/ 3] 1 = 1.167 m2
1
0
A x dA = 0
0.5 ( 2 y + y2 ) ( 2 y y2 ) dy
= 0.5 0
( 4 4 y + y2 y4 ) dy
= 0.5 [ 4 y 4 y2 / 2 + y3 / 3 y5/ 5 ] 1
= 1.067 m3
0
1
1
~
x = 1.067 / 1.167 = 0.914 m
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ATTENTION QUIZ
1. If a vertical rectangular strip is chosen as the
differential element, then all the variables,
including the integral limit, should be in
terms of _____ .
A) x B) y
C) z D) Any of the above.
2. If a vertical rectangular strip is chosen, then what are the values of
x and y?
A) (x , y) B) (x / 2 , y / 2)
C) (x , 0) D) (x , y / 2)
~ ~
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homework
9-7, 9-9, 9-13, 9-34
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COMPOSITE BODIES (Section 9.3)
Objective:
Students will be able to determine
the location of the center of
gravity, center of mass, or
centroid using the method of
composite bodies.
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CONCEPT OF A COMPOSITE BODY
Many industrial objects can be considered as composite bodies
made up of a series of connected simpler shaped parts or
holes, like a rectangle, triangle, and semicircle.
Knowing the location of the centroid, C, or center of gravity, G,
of the simpler shaped parts, we can easily determine the
location of the C or G for the more complex composite body.
a
be
d
ab
e
d
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This can be done by considering each part as a particle andfollowing the procedure as described in Section 9.1. This is a
simple, effective, and practical method of determining the
location of the centroid or center of gravity.
CONCEPT OF A COMPOSITE BODY (continued)
a
be
d
ab
e
d
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Centroids of common shapes of areas and volumes
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STEPS FOR ANALYSIS
1. Divide the body into pieces that are known shapes. Holes are
considered as pieces with negative weight or size.
2. Make a table with the first column for segment number, the second
column for weight, mass, or size (depending on the problem), the
next set of columns for the moment arms, and, finally, several
columns for recording results of simple intermediate calculations.
3. Fix the coordinate axes, determine the coordinates of the center of
gravity of centroid of each piece, and then fill-in the table.
4. Sum the columns to get x, y, and z. Use formulas like
x = (
xi Ai ) / (
Ai ) or x = (
xi Wi ) / (
Wi )
This approach will become clear by doing examples.
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EXAMPLE
Given: The part shown.
Find: The centroid of the part.
Plan: Follow the steps for analysis.
Solution:
1. This body can be divided into the following pieces
rectangle (a) + triangle (b) + quarter circular (c) semicircular
area (d)
a
bc
d
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EXAMPLE (continued)
39.8376.528.0
27
4.5
9- 2/3
54
31.5
90
1.5
1
4(3) / (3 )4(1) / (3 )
3
7
4(3) / (3 )0
18
4.5
9 / 4 / 2
Rectangle
Triangle
Q. CircleSemi-Circle
A y
( in3)
A x
( in3)
y
(in)
x
(in)
Area A
(in2)
Segment
Steps 2 & 3: Make up and fill
the table using parts a,
b, c, and d.
a
bc
d
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x = (
x A) / (
A ) = 76.5 in3/ 28.0 in2 = 2.73 in
y = ( y A) / ( A ) = 39.83 in3/ 28.0 in2 = 1.42 in
4. Now use the table data and these formulas to find the
coordinates of the centroid.
EXAMPLE (continued)
C
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EXAMPLE - 2
Given: Two blocks of differentmaterials are assembled as shown.
The specific weights of the
materials are
A = 150 lb / ft3 andB = 400 lb / ft3.
Find: The center of gravity of thisassembly.
Plan: Follow the steps for analysis
Solution
1. In this problem, the blocks A and B can be considered as two
segments.
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EXAMPLE 2 (continued)
Weight = w =
(Volume in ft3)
wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb
wB = 450 (6) (6) (2) / (12)3 = 18.75 lb
62.559.3831.2521.88
6.25
56.25
3.125
56.25
12.5
18.75
2
3
1
3
4
1
3.125
18.75
A
B
wz(lbin)
w y(lbin)
w x(lbin)
z (in)y (in)x (in)w (lb)Segment
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EXAMPLE 2(continued)
~x = ( x w) / ( w ) = 31.25/21.88 = 1.47 iny = ( y w) / ( w ) = 59.38/21.88 = 2.68 in
z = (
z w) / ( w ) = 62.5 /21.88 = 2.82 in~~
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homework
9-44, 9-55, 9-61, 9-83
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