CIVL 2131
- Statics
Moment
of Inertia
Composite Areas
A math professor in an unheated room
is cold and calculating.
Radius of Gyration
¢  This
actually sounds like some sort of rule
for separation on a dance floor.
¢  It actually is just a property of a shape and
is used in the analysis of how some
shapes act in different conditions.
2
Moment of Inertia - Composite Area
Monday, November 26, 2012
1
Radius of Gyration
¢  The
radius of gyration, k, is the square root
of the ratio of the moment of inertia to the
area
kx =
ky =
kO =
3
Ix
A
Iy
A
JO
=
A
Ix + Iy
A
Moment of Inertia - Composite Area
Monday, November 26, 2012
Parallel Axis Theorem
¢  If
you know the moment of inertia about a
centroidal axis of a figure, you can
calculate the moment of inertia about any
parallel axis to the centroidal axis using a
simple formula
I y = I y + Ax
2
I x = I x + Ay 2
4
Moment of Inertia - Composite Area
Monday, November 26, 2012
2
Parallel Axis Theorem
¢  Since
we usually use the bar over the
centroidal axis, the moment of inertia
about a centroidal axis also uses the bar
over the axis designation
I y = I y + Ax 2
I x = I x + Ay 2
5
Moment of Inertia - Composite Area
Monday, November 26, 2012
Parallel Axis Theorem
¢  If
you look carefully at the expression, you
should notice that the moment of inertia
about a centroidal axis will always be the
minimum moment of inertia about any axis
that is parallel to the centroidal axis.
I y = I y + Ax
2
I x = I x + Ay 2
6
Moment of Inertia - Composite Area
Monday, November 26, 2012
3
Parallel Axis Theorem
¢  In
a manner similar to that which we used
to calculate the centroid of a figure by
breaking it up into component areas, we
can calculate the moment of inertia of a
composite area
I y = I y + Ax 2
I x = I x + Ay 2
7
Moment of Inertia - Composite Area
Monday, November 26, 2012
Parallel Axis Theorem
¢  Inside
the back cover of the book, in the
same figure that we used for the centroid
calculations we can find calculations for
moments of inertia
I y = I y + Ax
2
I x = I x + Ay 2
8
Moment of Inertia - Composite Area
Monday, November 26, 2012
4
Parallel Axis Theorem
I y = I y + Ax 2
I x = I x + Ay 2
¢  HERE
IS A CRITICAL MOMENT OF
CAUTION
¢  REMEMBER HOW THE PARALLEL AXIS
IS WRITTEN
¢  IF THE AXIS SHOWN IN THE TABLE IS
NOT THROUGH THE CENTROID, THEN
THE FORMULA DOES NOT GIVE YOU
THE MOMENT OF INERTIA THROUGH
THE CENTROIDAL AXIS
9
Moment of Inertia - Composite Area
Parallel Axis Theorem
Monday, November 26, 2012
I y = I y + Ax 2
I x = I x + Ay 2
¢  By
example
¢  The Iy given for the Semicircular area in
the table is about the centroidal axis
¢  The Ix given for the same Semicircular
area in the table is not about the centroidal
axis
10
Moment of Inertia - Composite Area
Monday, November 26, 2012
5
Using The Table
¢  We
want to locate the moment of inertia in
the position shown of a semicircular area
as shown about the x and y axis, Ix and Iy
y
10"
x
11
Moment of Inertia - Composite Area
Monday, November 26, 2012
Using the Table
¢  First,
we can look at the table and find the
Ix and Iy about the axis as shown
y
10"
x
12
Moment of Inertia - Composite Area
Monday, November 26, 2012
6
Using the Table
¢  In
this problem, the y axis is 8” from the y
centroidal axis and x axis is 6” below the
base of the semicircle, this would be
usually evident from the problem
description
5"
y
8"
10"
6"
x
13
Moment of Inertia - Composite Area
Monday, November 26, 2012
Using the Table
¢  Calculating
the Iy you should notice that
the y axis in the table is the centroid axis
so we won’t have to move it yet
1
Iy = π r4
8
1
4
I y = π ( 5in )
8
I y = 245.44in 4
14
y
Moment of Inertia - Composite Area
5"
10"
x
Monday, November 26, 2012
7
Using the Table
¢  Next
A=
we can calculate the area
π ( 5in )
2
y
2
A = 39.27in 2
15
5"
10"
x
Moment of Inertia - Composite Area
Monday, November 26, 2012
Using the Table
¢  If
we know that distance between the y
axis and the ybar axis, we can calculate
the moment of inertia using the parallel
axis theorem
5"
y
2
8 in
y
x
I y = I + Ad
16
6in
I x = I x + Ad y 2
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
8
Using the Table
¢  I
changed the notation for the distances
moved to avoid confusion with the
distance from the origin
I y = I y + Ad x 2
8 in
10"
6in
I x = I x + Ad y 2
17
5"
y
Moment of Inertia - Composite Area
2.12"
x
Monday, November 26, 2012
Using the Table
¢  The
axis we are considering may not
always be a the origin.
I y = I y + Ad x 2
18
8 in
6in
I x = I x + Ad y 2
5"
y
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
9
Using the Table
¢  If
the y axis is 8 inches to the left of the
centroidal axis, then the moment of inertia
about the y axis would be
I y = I y + Ad x2
I y = 245.44in 4 + ( 39.27in 2 ) (8in )
2
8 in
10"
6in
I y = 2758.72in 4
19
5"
y
Moment of Inertia - Composite Area
2.12"
x
Monday, November 26, 2012
Using the Table
¢  The
moment of inertia about the x axis is a
slightly different case since the formula
presented in the table is the moment of
inertia about the base of the semicircle,
not the centroid
5"
y
8 in
6in
20
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
10
Using the Table
¢  To
move it to the moment of inertia about
the x-axis, we have to make two steps
I x = I base − A ( d base to centroid )
21
2
5"
y
8 in
10"
6in
I x = I x + A ( d centroid to x-axis )
2
Moment of Inertia - Composite Area
2.12"
x
Monday, November 26, 2012
Using the Table
¢  We
can combine the two steps
I x = I base − A ( d base to centroid )
I x = I x + A ( d centroid to x-axis )
2
2
I x = I base − A ( d base to centroid ) + A ( d centroid to x-axis )
2
2
5"
y
8 in
6in
22
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
11
Using the Table
¢  Don’t
try and cut corners here
¢  You have to move to the centroid first
I x = I base − A ( d base to centroid )
I x = I x + A ( d centroid to x-axis )
2
2
I x = I base − A ( d base to centroid ) + A ( d centroid to x-axis )
2
2
5"
y
8 in
6in
23
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
Using the Table
¢  In
this problem, we have to locate the y
centroid of the figure with respect to the
base
¢  We can use the table to determine this
4r 4 ( 5in )
y=
=
3π
3π
y = 2.12in
This ybar is with respect the base of
the object, not the x-axis.
5"
y
8 in
6in
24
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
12
Using the Table
¢  Now
the Ix in the table is given about the
bottom of the semicircle, not the centroidal
axis
¢  That is where the x axis is shown in the
table
5"
y
8 in
6in
10"
25
Moment of Inertia - Composite Area
2.12"
x
Monday, November 26, 2012
Using the Table
¢  So
you can use the formula to calculate
the Ix (Ibase) about the bottom of the
semicircle
26
5"
y
Moment of Inertia - Composite Area
8 in
6in
1
I base = π r 4
8
1
4
I base = π ( 5in )
8
I base = 245.44in 4
10"
2.12"
x
Monday, November 26, 2012
13
Using the Table
¢  Now
we can calculate the moment of
inertia about the x centroidal axis
2
I base = I x + Ad base
to centroid
2
I x = I base − Ad base
to centroid
I x = 245.44in 4 − ( 39.27in 2 ) ( 2.12in )
2
I x = 68.60in 4
5"
y
8 in
6in
10"
27
Moment of Inertia - Composite Area
2.12"
x
Monday, November 26, 2012
Using the Table
¢  And
we can move that moment of inertia
the the x-axis
2
I x = I x + Ad centroid
to x-axis
I x = 68.60in 4 + ( 39.27in 2 ) ( 6in + 2.12in )
I x = 2657.84in 4
5"
y
8 in
6in
28
Moment of Inertia - Composite Area
2
10"
2.12"
x
Monday, November 26, 2012
14
Using the Table
¢  The
polar moment of inertia about the
origin would be
JO = I x + I y
J O = 2657.84in 4 + 2758.72in 4
J O = 5416.56in 4
5"
y
8 in
6in
29
Moment of Inertia - Composite Area
10"
2.12"
x
Monday, November 26, 2012
Another Example
¢  We
can use the parallel axis theorem to
find the moment of inertia of a composite
figure
30
Moment of Inertia - Composite Area
Monday, November 26, 2012
15
Another Example
y
6"
3"
6"
x
6"
31
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
¢  We
can divide up the area into smaller
areas with shapes from the table
y
6"
6"
I
3"
II
x
III
6"
32
Moment of Inertia - Composite Area
Monday, November 26, 2012
16
Another Example
Since the parallel axis theorem will require the area for
each section, that is a reasonable place to start
ID
I
II
III
Area
(in2)
36
9
27
y
6"
6"
I
3"
II
x
III
6"
33
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
We can locate the centroid of each area with respect
the y axis.
ID
Area
xbari
(in2)
(in)
I
36
3
II
9
7
III
27
6
y
6"
6"
I
3"
II
x
III
6"
34
Moment of Inertia - Composite Area
Monday, November 26, 2012
17
Another Example
From the table in the back of the book we find that the
moment of inertia of a rectangle about its y-centroid
axis is
3
y
6"
3"
y
1
I = bh
12
ID
Area
xbari
(in2)
(in)
I
36
3
II
9
7
III
27
35
6"
I
II
x
III
6"
6
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
In this example, for Area I, b=6” and h=6”
1
3
6
in
6
in
( )( )
12
I y = 108in 4
Iy =
ID
36
Area
xbari
(in2)
(in)
I
36
3
II
9
7
III
27
y
6"
6"
I
3"
II
x
III
6"
6
Moment of Inertia - Composite Area
Monday, November 26, 2012
18
Another Example
For the first triangle, the moment of inertia calculation
isn’t as obvious
y
6"
6"
I
3"
II
x
III
6"
37
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
The way it is presented in the text, we can only find
the Ix about the centroid
h
y
6"
x 6"
I
3"
II
x
III
b
38
6"
Moment of Inertia - Composite Area
Monday, November 26, 2012
19
Another Example
The change may not seem obvious but it is
just in how we orient our axis. Remember an
axis is our decision.
h
x
y
6"
6"
I
3"
b
x
II
x
III
b
6"
h
39
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
So the moment of inertia of the II triangle can
be calculated using the formula with the
correct orientation.
1 3
bh
36
1
3
I y = ( 6in )( 3in )
36
I y = 4.5in 4
Iy =
40
y
6"
6"
I
Moment of Inertia - Composite Area
3"
II
x
III
6"
Monday, November 26, 2012
20
Another Example
The same is true for the III triangle
1 3
bh
36
1
3
I y = ( 6in )( 9in )
36
I y = 121.5in 4
Iy =
41
y
6"
6"
I
3"
II
x
III
6"
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
Now we can enter the Iybar for each sub-area into the
table
SubArea
42
Area
xbari
Iybar
(in2)
(in)
(in4)
I
36
3
108
II
9
7
4.5
III
27
6
121.5
y
6"
6"
I
Moment of Inertia - Composite Area
3"
II
x
III
6"
Monday, November 26, 2012
21
Another Example
We can then sum the Iy and the A(dx)2 to get
the moment of inertia for eachy sub-area
6"
3"
I
6"
II
x
III
SubArea
Sub-Area
43
I
II
III
6"
Area
xbari
Iybar
A(dx)2
Iybar +
A(dx)2
(in2)
(in)
(in4)
(in4)
(in4)
I
36
3
108
324
432
II
9
7
4.5
441
445.5
III
27
6
121.5
972
1093.5
Area
xbari
(in2)
36
9
27
(in)
3
7
6
A(dx )2 Iy bar + A(dx )2
Iy bar
Moment of Inertia - Composite Area
(in4)
108
4.5
121.5
Monday, November 26, 2012
(in4)
324
441
972
(in4)
432
445.5
1093.5
1971
Another Example
And if we sum that last column, we have the
Iy for the composite figure
y
6"
6"
I
3"
II
x
III
SubArea
Area
xbari
Iybar
A(dx)2
Iybar +
A(dx)2
(in2)
(in)
(in4)
(in4)
(in4)
36
3
108
324
432
II
9
7
4.5
441
445.5
III
27
6
121.5
972
1093.5
I
6"
1971
44
Moment of Inertia - Composite Area
Monday, November 26, 2012
22
Another Example
We perform the same type analysis for the Ix
ID
y
6"
6"
I
II
Area
(in2)
36
9
27
I
II
III
45
3"
x
III
6"
Moment of Inertia - Composite Area
Monday, November 26, 2012
Another Example
Locating the y-centroids from the x-axis
Sub-Area
I
II
III
46
Area
(in2)
36
9
27
ybari
(in)
3
2
-2
Moment of Inertia - Composite Area
y
6"
6"
I
3"
II
x
III
6"
Monday, November 26, 2012
23
Another Example
Determining the Ix for each sub-area
Sub-Area
Area
ybari
Ixbar
(in2)
(in)
(in4)
I
36
3
108
II
9
2
18
III
27
-2
54
47
Moment of Inertia - Composite Area
y
6"
6"
I
3"
II
x
III
6"
Monday, November 26, 2012
Another Example
Making the A(dy)2 multiplications
SubArea
48
y
6"
6"
I
3"
II
Area
ybari
Ixbar
A(dy)2
(in2)
(in)
(in4)
(in4)
I
36
3
108
324
II
9
2
18
36
III
27
-2
54
108
Moment of Inertia - Composite Area
x
III
6"
Monday, November 26, 2012
24
Another Example
Summing and calculating Ix
SubArea
Area
)2
A(dy
Ixbar +
A(dy)2
ybari
Ixbar
(in2)
(in)
(in4)
(in4)
(in4)
I
36
3
108
324
432
II
9
2
18
36
54
III
27
-2
54
108
162
y
6"
6"
I
3"
II
x
III
6"
648
49
Moment of Inertia - Composite Area
Monday, November 26, 2012
Homework
¢  Problem
10-27
¢  Problem 10-29
¢  Problem 10-47
50
Moment of Inertia - Composite Area
Monday, November 26, 2012
25