Chapter 10 – Moment of Inertia (MoI)
Chapter 10 Objectives:
Students will be able to:
a) Define the moments of
inertia (MoI) for an area.
b) Determine the MoI for
an area by integration.
c) Determine the MoI for a
composite area.
d) Use the parallel-axis
theorem to find the MoI
about any axis.
The load that a beam can bear is related to the
moment of inertia (MOI) of its cross-section.
1
APPLICATIONS
Many structural members like beams and columns have cross
sectional shapes like an I, H, C, etc..
Why do they usually not have solid rectangular, square, or
circular cross sectional areas?
What primary property of these members influences design
decisions?
2
MOMENTS OF INERTIA FOR AREAS
Chapter 10 – Moments of Inertia
Moments of inertia are not actually used in Statics; however,
since the calculations for moments of inertia are quite similar
to those for centroids they are introduced at this point.
Moments of inertia are used in courses such as mechanics of
materials, dynamics, and fluid mechanics.
The moment of inertia of an object is a measure of its
resistance to change in rotation. Everyday experience tells
us that it is harder to start (or stop) a large wheel turning than a
small wheel. Mathematically, this is represented by the large
wheel having a larger moment of inertia.
3
MOMENTS OF INERTIA FOR AREAS
Moments of inertia are used in various engineering
calculations, including:
• Locating the resultant of hydrostatic pressure forces on
submerged bodies
• Calculating stresses in beams – they are at times related to
the moment of inertia of the cross-sectional area of the beam
(resistance to bending)
• Mass moments of inertia are used in studying the rotational
motion of objects
4
Moment of Inertia – Practical Application
If a driver has a large MoI, it is less likely to twist if a shot is
hit slightly off center. In 2006 the USGA approved the
implementation of a clubhead moment of inertia test and a
limit of 5900 gm-cm2 (about the face plane).
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Moment of Inertia – Practical Application
When ice skaters move their arms closer to their body, they
spin faster. This is due to conservation of angular momentum,
Iw = (moment of inertia)(angular velocity). When their arms
move closer to their body, I is reduced, so w increases.
There are several videos on YouTube illustrating this effect.
http://www.youtube.com/watch?v=5
ET-HdHn2XY
http://www.stevespanglerscience.co
m/experiment/ice-skating-spin
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Moment of Inertia – Demonstration
C
The block is easiest to
rotate about its centroid as
its MoI is minimum.
C
The block is harder to rotate
as the dowel is moved away
7
from the centroid.
DEFINITION OF MOMENTS OF INERTIA FOR AREAS (Section 10.1)
Consider a plate submerged in a liquid.
The pressure of a liquid at a distance y
below the surface is given by p =  y,
where  is the specific weight of the
liquid.
The force on the area dA at that point is dF = p dA.
The moment about the x-axis due to this force is y (dF).
The total moment is A y dF = A  y2 dA =  A( y2 dA).
This sort of integral term also appears in solid mechanics when
determining stresses and deflection.
The integral term, A( y2 dA), is referred to as the moment of
inertia of the area of the plate about an axis.
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DEFINITION OF MOMENTS OF INERTIA FOR AREAS
10cm
3cm
10cm
10cm
1cm
P
3cm
x
(A)
(B)
(C)
R
S
1cm
Consider three different possible cross sectional shapes and areas for the
beam RS. All have the same total area and, assuming they are made of same
material, they will have the same mass per unit length.
For the given vertical loading P on the beam, which shape will
develop less internal stress and deflection? Why?
The answer depends on the MoI of the beam about the x-axis. It turns
out that Section A has the highest MoI because most of the area is
farthest from the x axis. Hence, it has the least stress and deflection.
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DEFINITION OF MOMENTS OF INERTIA FOR AREAS
For the differential area dA, shown in the
figure:
d Ix = y2 dA ,
d Iy = x2 dA , and,
d JO = r2 dA , where JO is the polar
moment of inertia about the pole O or z axis.
The moments of inertia for the entire area are obtained by
integration.
Ix =
A y2 dA ; Iy =
JO = A r2 dA =
A x2 dA
A ( x2 + y2 ) dA =
Ix + Iy
The MoI is also referred to as the second moment of an area
and has units of length to the fourth power (m4 or in4).
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RADIUS OF GYRATION OF AN AREA (Section 10.3)
y
A
kx
For a given area A and its MoI, Ix , imagine
that the entire area is located at distance kx
from the x axis. Then,
Ix
2
I x  k x A or k x 
A
x kx = the radius of gyration of the area
about the x axis.
y
A
Similarly,
Iy  k y A
2
kY
or
ky 
Iy
A
ky = the radius of gyration of the area
about the y axis.
x
The radius of gyration has units of length and gives an indication
of the spread of the area from the axes.
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This characteristic is important when designing columns.
Methods for Determining the MoI of an Object
Moments of Inertia (MoI) can be found using three methods:
1. Composites – If an object can be divided up into relatively simple
shapes with known moments of inertia, then the moment of inertia of
the entire object is the sum of the moments of inertial of the parts.
2. Integration – If the area or volume of an object can be described by
mathematical equations, then the moment of inertia can be
determined through integration.
3. Solid modeling software – Software such as AutoCAD, Inventor, and
SolidWorks can be used to construct 3D models of objects. The
software can also determine the moment of inertia of the objects (as
well as volumes, radii of gyration, centroids, etc.)
12
Finding the MoI for an area by integration
For simplicity, the area element used has a
differential size in only one direction
(dx or dy). This results in a single integration
and is usually simpler than doing a double
integration with two differentials, i.e., dx·dy.
The step-by-step procedure is:
1. Choose the element dA: There are two choices: a vertical strip or a
horizontal strip. Some considerations about this choice are:
a) The element parallel to the axis about which the MoI is to be
determined usually results in an easier solution. For example,
we typically choose a horizontal strip for determining Ix and a
vertical strip for determining Iy.
13
Finding the MoI for an area by integration (continued)
b) If y is easily expressed in terms of x (e.g.,
y = x2 + 1), then choosing a vertical strip
with a differential element dx wide may
be advantageous.
2. Integrate to find the MoI. For example, given the element shown in
the figure above:
Iy =
 x2 dA =  x2 y dx
and
Ix =  d Ix =  (1 / 3) y3 dx (using the information for Ix for a
rectangle about its base from the inside back cover of the textbook).
Since in this case the differential element is dx, y needs to be
expressed in terms of x and the integral limit must also be in
terms of x. As you can see, choosing the element and integrating can
be challenging. It may require a trial and error approach plus
experience.
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Example: Finding the MoI using Integration
Solution - using a
horizontal strip to find Ix
 y2 
dA  (2 - x)dy   2 - dy
2 

2


y
2
2
I x   y dA   y  2 - dy
2 

0
2
2
Given: The shaded area shown in
the figure.
2 3 1 5
I x  y  y  2.13 m 4
3
10 0
Find: The MoI of the area about
the x- and y-axes.
Plan: Follow the steps given
earlier.
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EXAMPLE (continued)
- Now using a vertical strip to find Iy
2
I y   x 2 dA   x 2 ydx
0
2
Iy   x 2
 2x dx  2  x
2
0
2.5
dx
0
2
Alternate method for finding Ix:
2 2.5
Iy 
x
 4.57 m 4
3.5
0
In the above example, it would be difficult to determine Iy using a
horizontal strip. However, Ix in this example can be determined using a
3
vertical strip. So,
1 3
1
I x   y dx  
2x dx
3
3
 
(Since Ix = (1/3)bh3 for a rectangle with its edge on the x-axis
or dIx = (1/3)y3dx for a rectangular strip)
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Example: Find Ix, Iy, kx and ky for the triangle below using integration
Use a horizontal strip to find Ix and a vertical strip to find Iy.
Compare the results to those found in a table of common shapes.
y
h
b
x
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Finding Moment of Inertia (MoI) using Composite Shapes (Section 10.4)
Cross-sectional areas of structural
members are usually made of
simple shapes or combination of
simple shapes. To design these
types of members, we need to
find the moment of inertia (MoI).
Another example of a structural
member with a composite cross-area.
Tables of composite shapes in the text
provide us with moments of inertia’s for
common shapes about specific axes (note
that the values change if you use a
different axis of rotation!)
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Moments of Inertia of Common Shapes
19
Moments of Inertia of Common Shapes (continued)
20
Parallel-Axis Theorem
21
The tables on the previous
slides provide values of Ix
and Iy about specific axes
(usually the centroidal axes).
What if we want to use a use
a different axis (this is often
the case)? Ix and Iy change!
6”
6”
6”
x
x
2”
2”
Ix 
1 3 1
3
bh  62   4 in 4
12
12
I x  4 in
4
x
2”
I x  4 in 4
(We will see later
(We will see later
than I x  16 in 4 )
than I x  112 in 4 )
PARALLEL-AXIS THEOREM FOR AN AREA (Section 10.2)
The Parallel-Axis Theorem relates the
moment of inertia (MoI) of an area about
an axis passing through the area’s centroid
to the MoI of the area about a
corresponding parallel axis. This theorem
has many practical applications, especially
when working with composite areas.
Consider an area with centroid C. The x' and y' axes pass through C.
The MoI about the x-axis, which is parallel to, and distance dy from
the x ' axis, is found by using the parallel-axis theorem.
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PARALLEL-AXIS THEOREM (continued)
I x   y 2 dA   y'  d y  dA
2
A
zero
A
I x   y' 2 dA  2d y  y' dA  d y
A
A
 y' dA
but since y'  A
 dA
2
 dA
A

is the definition of a centroid
A
A
and since y' goes through t he centroid, then y'  0,
so
 y' dA  0 (or the middle integral on the second line must be 0).
A
So, I x   y' 2 dA  0  d y
A
2
2
dA

y'

 dA  d y A
2
A
A
Or I x  I x '  d y A (similar result for I y )
2
Ix  Ix '  d y A
2
Parallax-Axis
Theorem
Iy  Iy '  dx A
2
J0  JC  d 2A
where Ix‘ = IxC = centroidal value
(generally found in a table)
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Example: Parallel-Axis Theorem – Find Ix for each part below
24
6”
6”
6”
x
x
2”
2”
x
2”
MOMENT OF INERTIA FOR A COMPOSITE AREA
(Section 10.4)
Moments of inertia are additive.
A composite area is made by adding or
subtracting a series of “simple” shaped
areas like rectangles, triangles, and
circles.
For example, the area on the left can be
made from a rectangle minus a smaller
rectangle plus a triangle.
Ix(total) = Ix(large rectangle) + Ix(triangle) – Ix (small rectangle)
Iy(total) = Iy(large rectangle) + Iy(triangle) – Iy (small rectangle)
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Finding MoI of Common Shapes
Finding moments of inertia using composites:
When moments of inertia are found using composites, two types of problems
are typically considered:
1) Finding Ix and Iy around fixed axes (or specified axes).
2) Finding Ix and Iy around the centroidal axes for the entire object (so the
centroid must be found first).
y’
y
C
x’
x
Example: Find Ix and Iy
about the fixed x and y axes
shown.
Example: Find Ix and Iy
about the centroidal axes of
the object.
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STEPS FOR ANALYSIS – Finding MoI using Composite Shapes
1. Divide the given area into its simpler shaped parts.
2. Locate the centroid of each part and indicate the perpendicular
distance from each centroid to the desired reference axis.
3. If you wish to find the MoI about the centroidal axes of the entire
object, find the centroid of the entire object. If you wish to find the
MoI about fixed axes, proceed to the next step.
4. Determine the MoI of each simpler shaped part (generally about its
centroid) using the table provided in the text.
5. Use the Parallel-Axis Theorem to find the desired axes.
6. The MoI of the entire object is the sum of the MoI’s for each simpler
shape. Don’t forget that the MoI is negative for shapes to be
subtracted.
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Example: Determine the moments of inertia about the fixed x and y axes shown.
y
R3.00"
8.00"
R1.00"
6.00"
O
x
28
Example: Determine the moments of inertia about the centroidal x and y axes of
the entire object shown below.
y
30
12.5
R7
12.5
12.5
x
55
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MASS MOMENT OF INERTIA
The large flywheel in the picture is
connected to a large metal cutter. The
flywheel is used to provide a uniform motion
to the cutting blade while it is cutting
materials. Why is most of the mass of the
flywheel located near the flywheel’s
circumference?
The Mass Moment of Inertia (MMI) is
important when analyzing rotational motion
(in dynamics and other courses).
Finding the MMI is quite similar to finding the moment of inertia of an area.
The MMI can be found using integration techniques or using composites.
Moment of Inertia (MoI)
of a rectangular area
1 3
I x '  bh
12
2
I x  I x '  d y  A
(parallel - axis theorem)
Mass Moment of Inertia (MMI)
of a rectangular plate
1
I xx'  mb 2
12
2
I x  I G  d  m
(parallel - axis theorem)
Moments of Inertia of Common Shapes (continued)
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Moments of Inertia of Common Shapes (continued)
33
MASS MOMENT OF INERTIA
Consider a rigid body with a center of
mass at G. It is free to rotate about the z axis, which passes
through G. Now, if we apply a torque T about the z axis to
the body, the body begins to rotate with an angular
acceleration .
T and  are related by the equation T = I  . In this equation, I is the
mass moment of inertia (MMI) about the z axis.
The MMI of a body is a property that measures the resistance
of the body to angular acceleration. This is similar to the role of
mass in the equation F = m a. The MMI is often used when
analyzing rotational motion (done in dynamics).
DEFINITION OF THE MMI
Consider a rigid body and the arbitrary axis p
shown in the figure. The MMI about the p axis
is defined as I = m r2 dm, where r, the
“moment arm,” is the perpendicular distance
from the axis to the arbitrary element dm.
The MMI is always a positive quantity and
has a unit of kg ·m2 or slug · ft2.
RELATED CONCEPTS
Parallel-Axis Theorem
Just as with the MoI for an area, the
parallel-axis theorem can be used to find
the MMI about a parallel axis z that is a
distance d from the z’ axis through the
body’s center of mass G. The formula is
Iz = IG + (m) (d)2 (where m is the mass
of the body).
m
The radius of gyration is similarly defined as
k = (I / m)
Finally, the MMI can be obtained by integration or by the
method for composite bodies. The latter method is easier for
many practical shapes.
EXAMPLE 10-100
Given: The pendulum consists of a slender
rod with a mass 10 kg and sphere
with a mass of 15 kg.
Find:
The pendulum’s MMI about an axis
perpendicular to the screen and
passing through point O.
Plan:
Follow steps similar to finding the
MoI for a composite area.
Solution:
1. The wheel can be divided into a slender rod
(r) and sphere (s).
EXAMPLE (continued)
Gr
2. The center of mass for rod is at point Gr, 0.225 m
from Point O. The center of mass for sphere is at Gs,
0.55 m from point O.
O
3. The MMI data for a slender rod and sphere are given
on the inside back cover of the textbook. Using
those data and the parallel-axis theorem, calculate
the following.
Gs
IO = IG + (m) (d) 2
IOr = (1/12) (10)(0.45)2 +10 (0.225)2 = 0.675 kg·m2
IOs = (2/5) (15) (0.1)2 + 15 (0.55)2 = 4.598 kg·m2
4. Now add the three MMIs about point O.
IO = IOr + IOs = 5.27 kg·m2