# 9.3

```Centroid/CG/CM...so far
• Apply
to differential elements of mass (dm)
• dm can either be (assuming constant r)
– dA (for 2D problem)
– dl (for wire problem)
– dV (for volume problem)
– OR r(x,y,z) times any of above
(for inhomogeneous density)
Example 2D Problem (dm = dA)
Given: The area as shown.
Find:
The x of the centroid.
Plan:
Solution
(x1,,y)
(x2,y)
1. Choose dA as a horizontal
rectangular strip.
2. dA = ( x2 – x1) dy
= ((2 – y) – y2) dy
~
3. x
= ( x1 + x2) / 2
= 0.5 (( 2 – y) + y2 )
Example 2D Problem (continued)
4.
x
~
( A x dA ) / ( A dA )
=
A dA =
0
1
( 2 – y – y2) dy
[ 2 y – y2 / 2 – y3 / 3] 01
=
1.167 m2
1
~
A x dA = 0 0.5 ( 2 – y + y2 ) ( 2 – y – y2 ) dy
x =
1
=
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
1.067 / 1.167 = 0.914 m
Example 1D Problem (dm = dl)
Given: The wire segment shown.
Find:
The x of the centroid.
Solution
1. Choose dl as a tiny piece at (x,y)
Mass dm = length of dl
~
x=x=y2/4
2. dl=√ dx2+dy2 = √1 + dx2/dy2 dy
since y2=4x, dx = y/2 dy
dl = √1 + (y/2)2 dy = (&frac12;) √4 +y2 dy
4
~
3. L = ʃ dl = ʃ0 (&frac12;) √4 +y2 dy = 5.92
4
~
4. x = ʃ x dl /L = ʃ (y2/4) (&frac12;) √4 +y2 dy/L
= 1.64
0
HELP WITH INTEGRATION
• 1D problems especially can be difficult
• Use integral tables (link online)
• or www.WolframAlpha.com
(matlab, freemat also work, little more work)
Example 3D Problem (dm = dV)
Given: The volume shown.
Find:
The y of the centroid.
Solution
1. Choose dV as a vertical disk slice
2. dV = π z2 dy
(x1,,y)
(x2,y)
= π (y2 – 9 ) dy
5
3. V = ʃdV= π ʃ3 (y2 – 9 ) dy
= 46.08
5
~
4. y = ʃ y dV/ V = π ʃ3 y (y2 – 9 ) dy / V
= 4.36
Today’s Objective:
COMPOSITE BODIES
Students will be able to determine:
a)
The location of the center of
gravity.
In-Class Activities:
b) The location of the center of mass.
• Check homework, if any
c) The location of the centroid using
the method of composite bodies.
• Applications
• Method of Composite Bodies.
• Concept Quiz
• Group Problem Solving
• Attention Quiz
1. A composite body in this section refers to a body made of ____.
A) carbon fibers and an epoxy matrix
B) steel and concrete
C) a collection of “simple” shaped parts or holes
D) 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?
APPLICATIONS
(continued)
Cars, trucks, 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
a
e
b
a
b
e
d
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.
CONCEPT OF A COMPOSITE BODY
(continued)
a
e
b
a
b
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 = (  xi Ai ) / (  Ai ) or x = (  xi Wi ) / (  Wi )
This approach will become clear by doing examples!
EXAMPLE
Given: The part shown.
Find:
The centroid of
the part.
a
c
b
for analysis.
d
Solution:
1. This body can be divided into the following pieces:
rectangle (a) + triangle (b) + quarter circular (c) –
semicircular area (d)
EXAMPLE (continued)
Steps 2 &amp; 3: Make up and fill the
table using parts a, b,
c, and d.
a
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
9/4
–/2
3
7
– 4(3) / (3 )
0
1.5
1
4(3) / (3 )
4(1) / (3 )
54
31.5
–9
0
27
4.5
9
- 2/3
76.5
39.83

28.0
EXAMPLE
(continued)
&middot;C
4. Now use the table data and these formulas to find the coordinates
of the centroid.
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
CONCEPT QUIZ
1.
3cm
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?
1 cm
1 cm
3cm
1, 2, 3, or 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&deg;
B) 45 &deg;
C) 60 &deg;
D) 90 &deg;
C
G
B
30&ordm;
A
GROUP PROBLEM SOLVING
Given: Two blocks of different
materials are assembled as
shown.
The densities of the materials are:
rA = 150 lb / ft3 and
rB = 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 = r (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
16.67

19.79
4
1
400 (6) (6) (2) / (12)3 = 16.67 lb
y (in)
z (in)
1
3
2
3

w x
w y
wz
(lb&middot;in) (lb&middot;in) (lb&middot;in)
12.5
16.67
3.125
50.00
6.25
50.00
29.17
53.12
56.25
GROUP PROBLEM SOLVING (continued)
x = ( x~ w) / ( w ) = 29.17/19.79 = 1.47 in
y = ( y~ w) / ( w ) = 53.12/ 19.79 = 2.68 in
z = ( z~ w) / ( w ) = 56.25 / 19.79 = 2.84 in
ATTENTION QUIZ
y
1. 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?
2.
A) Two
B) Three
C) Four
D) Five
2cm
4cm
x
2cm 2cm
For determining the centroid of the area, two
y 1m 1m
square segments are considered; square ABCD
D
and square DEFG. What are the coordinates A
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
```