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Center of Gravity and Centroid - Chapter 9
Today’s 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
In-Class Activities:
or a body.
• Reading quiz
• Applications
• Center of gravity, etc.
• Concept quiz
• Group problem solving
• Attention quiz
Engr 222 Spring 2004 Chapter 9
1
Reading Quiz
1. The _________ is the point defining the geometric center
of an object.
A) Center of gravity
B) Center of mass
C) Centroid
D) None of the above
2. To study problems concerned with the motion of matter
under the influence of forces, i.e., dynamics, it is necessary
to locate a point called the ________.
A) Center of gravity
B) Center of mass
C) Centroid
D) None of the above
Applications
To design the structure for
supporting a water tank, we will
need to know the weights of the
tank and water as well as the
locations where the resultant
forces representing these
distributed loads are acting.
How can we determine these
weights and their locations?
Engr 222 Spring 2004 Chapter 9
2
Applications - continued
One concern about a sport utility
vehicle (SUV) is that it might tip
over while taking a sharp turn.
One of the important factors in
determining its stability is the
SUV’s center of mass.
Should it be higher or lower for
making a SUV more stable?
How do you determine its location?
Concept of CG and CM
4N
3m
A •
1N
1m
•
G
•
3N
B
The center of gravity (G) is a point which
locates the resultant weight of a system of
particles or a 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
particle’s 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.
Engr 222 Spring 2004 Chapter 9
3
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.
CG / CM for a System of Particles - Section 9.1
Consider a system of n particles as shown in
the figure. The net or the resultant weight is
given as WR = W.
Summing the moments about the y-axis, we get
~
~
~
x WR = x1W1 + x2W2 + ……….. + xnWn
~
where x1 represents x coordinate of W1, etc.
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 coordinates
of the center of mass can be found.
Engr 222 Spring 2004 Chapter 9
4
CG / CM / Centroid of a Body - Section 9.2
A rigid body can be considered as
composed of an infinite number of
particles. Hence, using the same
principles as in the previous slide, we
get the coordinates 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.
Steps for Determining the Centroidal Location
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).
3. Determine coordinates~ (x~, y) of the centroid of the rectangular
element in terms of the general point (x,y).
4. Express all the variables and integral limits in the formula
using either x or y, depending on whether the differential
element is in terms of dx or dy, respectively, and integrate.
Note: Similar steps are used for determining CG, CM, etc.
These steps will become clearer by doing a few examples.
Engr 222 Spring 2004 Chapter 9
5
Example
Given: The area as shown.
Find: The centroid location (x , y)
Plan: Follow the steps.
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
•
Example - continued
4. x = (
3
=
0
A
~
x dA ) / (
A
x ( 9 – x2) d x
dA )
=
3
x ( 9 – x2) d x
0
[ 9 (x2)/2 – (x4) / 4] 03
[ 9 x – (x3) / 3 ] 3
= ( 9 ( 9 ) / 2 – 81 / 4 ) / ( 9 ( 3 ) – ( 27 / 3 ) )
0
= 1.13 ft
y =
3
~
y dA
A
A
Engr 222 Spring 2004 Chapter 9
dA
=
½ 0 ( 9 – x2) ( 9 – x2) dx
3
0
( 9 – x2) d x
= 3.60 ft
6
Concept Quiz
1. The steel plate with known weight and nonuniform 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 all
three parameters located at the same point?
A)
B)
C)
D)
(center of gravity, no)
(center of gravity, yes)
(centroid, yes)
(centroid, no)
2. When determining the centroid of the area above, which type of
differential area element requires the least computational work?
A) Vertical
B) Horizontal
C) Polar
D) Any one of the above
Group Problem Solving
Given: The area as shown.
Find:
The x of the centroid.
Plan:
Follow the steps.
Solution
(x1,,y)
1. Choose dA as a horizontal rectangular
(x2,y)
strip.
2. dA = ( x2 – x1) dy
= ((2 – y) – y2) dy
~
3. x = ( x1 + x2) / 2
= 0.5 (( 2 – y) + y2 )
Engr 222 Spring 2004 Chapter 9
7
Group Problem Solving - continued
4.
A
x
~
= (
A
1
dA =
0
x dA ) / (
A
dA )
( 2 – y – y2) dy
[ 2 y – y2 / 2 – y3 / 3] 01 =
A
~x dA =
x
1
0
1.167 m2
0.5 ( 2 – y + y2 ) ( 2 – y – y2 ) dy
1
( 4 – 4 y + y2 – y4 ) dy
=
0.5
=
0.5 [ 4 y – 4 y2 / 2 + y3 / 3 – y5 / 5 ] 10
=
1.067 m3
=
0
1.067 / 1.167 = 0.914 m
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
~ y? ~
x and
A) (x , y)
B) (x / 2 , y / 2)
C) (x , 0)
D) (x , y / 2)
Engr 222 Spring 2004 Chapter 9
8
Textbook Problem 9-16
Locate the centroid of the shaded area shown below.
Textbook Problem 9-35
Locate the centroid of the solid.
Engr 222 Spring 2004 Chapter 9
9
Announcements
Composite Bodies - Section 9.3
Today’s 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.
In-Class Activities:
• Reading quiz
• Applications
• Method of composite bodies.
• Concept quiz
• Group problem solving
• Attention quiz
Engr 222 Spring 2004 Chapter 9
10
Reading Quiz
1. A composite body in this section refers to a body made of ____.
A)
B)
C)
D)
carbon fibers and an epoxy matrix
steel and concrete
a collection of “simple” shaped parts or holes
a collection of “complex” shaped parts or holes
2. The composite method for determining the location of the
center of gravity of a composite body requires _______.
A) integration
B) differentiation
C) simple arithmetic
D) All of the above.
Applications
The I - beam is commonly used in
building structures. When doing a
stress analysis on an I - beam, the
location of the centroid is very
important.
How can we easily determine the
location of the centroid for a given
beam shape?
Engr 222 Spring 2004 Chapter 9
11
Applications - continued
Cars, SUVs, bikes, etc., are
assembled using many
individual components.
When designing for stability
on the road, it is important to
know the location of the bikes’
center of gravity (CG).
If we know the weight and CG of individual components,
how can we determine the location of the CG of the
assembled unit?
Concept of a Composite Body
b
a
e
b
d
a
e
d
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.
Engr 222 Spring 2004 Chapter 9
12
Concept of a Composite Body - continued
b
a
e
b
a
e
d
d
This can be done by considering each part as a “particle” and
following 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.
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
∼
= ( Σ xi Ai ) / ( Σ Ai ) or x
∼
= ( Σ xi Wi ) / ( Σ Wi )
This approach will become clear by doing examples.
Engr 222 Spring 2004 Chapter 9
13
Example
Given: The part shown.
a
c
b
d
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)
Example - continued
a
Steps 2 & 3: Make up and fill
the table using parts a,
b, c, and d.
c
b
d
Segment
Area A
(in2)
x∼
(in)
∼y
(in)
A∼
x
( in3)
A∼
y
( in3)
Rectangle
Triangle
Q. Circle
Semi-Circle
18
4.5
9π/4
–π/2
3
7
– 4(3) / (3 π)
0
1.5
1
4(3) / (3 π)
4(1) / (3 π)
54
31.5
–9
0
27
4.5
9
- 2/3
76.5
39.83
Σ
Engr 222 Spring 2004 Chapter 9
28.0
14
Example - continued
·C
4. Now use the table data and these formulas to find the
coordinates of the centroid.
x = ( Σ x∼ A) / ( Σ A ) = 76.5 in3/ 28.0 in2 = 2.73 in
y = (Σ∼
y A) / (Σ A ) = 39.83 in3 / 28.0 in2 = 1.42 in
Concept Quiz
1.
3cm 1 cm
Based on the typical centroid information
available in handbooks, what are the
minimum number of segments you will
have to consider for determing the
centroid of the given area?
A) 1
B) 2
C) 3
Engr 222 Spring 2004 Chapter 9
B) 45 °
C) 60 °
3cm
D) 4
2. A storage box is tilted up to clean the rug
underneath the box. It is tilted up by pulling
the handle C, with edge A remaining on the
ground. What is the maximum angle of tilt
(measured between bottom AB and the
ground) possible before the box tips over?
A) 30°
1 cm
D) 90 °
C
G
B
30º
A
15
Group Problem Solving
Given: Two blocks of different
materials are assembled as shown.
The weight densities of the
materials are
ρA = 150 lb / ft3 and
ρB = 400 lb / ft3.
Find: The center of gravity of this
assembly.
Plan: Follow the steps for analysis
Solution
1. In this problem, the blocks A and B can be considered as two
segments.
Group Problem Solving - continued
Weight = w = ρ (Volume in ft3)
wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb
wB =
Segment w (lb) x∼ (in)
A
B
3.125
18.75
Σ
21.88
Engr 222 Spring 2004 Chapter 9
4
1
450 (6) (6) (2) / (12)3 = 18.75 lb
y∼ (in)
∼z (in)
w x∼
(lb·in)
∼
w ∼y
wz
(lb·in) (lb·in)
1
3
2
3
12.5
18.75
3.125
56.25
6.25
56.25
31.25
59.38
62.5
16
Group Problem Solving - continued
x = (Σ x~ w) / ( Σw ) = 31.25/21.88 = 1.47 in
y = (Σ ~y w) / ( Σw ) = 59.38/21.88 = 2.68 in
z = (Σ z~ w) / ( Σw ) = 62.5 /21.88 = 2.82 in
Attention Quiz
1.
2.
y
A rectangular area has semicircular and
triangular cuts as shown. For determining the
centroid, what is the minimum number of
pieces that you can use?
A) Two
B) Three
C) Four
D) Five
2cm
4cm
x
2cm 2cm
y 1m 1m
For determining the centroid of the area, two
square segments are considered; square ABCD A
D
and square DEFG. What are the coordinates
E
1m
~~
(x, y ) of the centroid of square DEFG?
G
F
1m
A) (1, 1) m
B) (1.25, 1.25) m
x
B
C
C) (0.5, 0.5 ) m
D) (1.5, 1.5) m
Engr 222 Spring 2004 Chapter 9
17
Textbook Problem 9-53
Determine the location, ybar, of the centroid of the beam’s
cross-sectional area. Neglect the size of the corner welds at A
and B for the calculation.
Engr 222 Spring 2004 Chapter 9
18