Announcements
Moments of Inertia - Chapter 10
Today’s 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.
In-Class Activities:
• Reading quiz
• Applications
• MoI: concept and definition
• MoI by integration
• Concept quiz
• Group problem solving
• Attention quiz
Engr222 Spring 2004 Chapter 10
1
Reading Quiz
1. The definition of the Moment of Inertia for an area involves an
integral of the form
A) x dA.
B)
x2 dA.
C) x2 dm.
D)
m dA.
2. Select the SI units for the Moment of Inertia for an area.
A) m4
B) m2
C) kg·m2
D) kg·m3
Applications
Many structural members like beams
and columns have cross sectional
shapes like 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?
How can we calculate this property?
Engr222 Spring 2004 Chapter 10
2
Applications - continued
Many structural members are
made of tubes rather than solid
squares or rounds. Why?
What parameters of the cross
sectional area influence the
designer’s selection? How can
we determine the value of these
parameters for a given area?
Moment of Inertia of an Area
Consider a plate submerged in a
liquid. The pressure of a liquid at
a distance z below the surface is
given by p = γ z, where γ 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 z (dF). The
total moment is A z dF = A γ z2 dA = γ A( z2 dA).
This sort of integral term also appears in solid mechanics when
determining stresses and deflection. This integral term is referred
to as the moment of inertia of the area of the plate about an axis.
Engr222 Spring 2004 Chapter 10
3
Moment of Inertia of an Area - continued
10cm
3cm
10cm
10cm
1cm
(A)
P
3cm
x
(B)
(C)
R
1cm
S
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.
Moment of Inertia - Definition
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 =
JO =
A
A
y2 dA ;
r2 dA =
Iy =
A
A
x2 dA
( 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).
Engr222 Spring 2004 Chapter 10
4
Radius of Gyration of an Area - Section 10.3
A
y
For a given area A and its MoI, Ix ,
imagine that the entire area is located at
distance kx from the x axis.
kx
x
Then, Ix = k2xA or kx = √ ( Ix / A). This
kx is called the radius of gyration of the
area about the x axis. Similarly;
kY = √ ( Iy / A ) and kO = √ ( JO / A )
The radius of gyration has units of length and gives an indication
of the spread of the area from the axes. This characteristic is
important when designing columns.
Moment of Inertia by Integration - Section 10.4
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, 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.
Engr222 Spring 2004 Chapter 10
5
Moment of Inertia 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 a
rectangle about its base from the inside back cover of the textbook
and from the parallel-axis theorem).
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.
Example
(x,y)
Given: The shaded area shown in the
figure.
Find: The MoI of the area about the
x- and y-axes.
Plan: Follow the steps given earlier.
Solution
Ix
=
y2 dA
dA = (4 – x) dy =
Ix
=
4
O
(4 – y2/4) dy
y2 (4 – y2/4) dy
4
= [ (4/3) y3 – (1/20) y5 ] 0 = 34.1 in4
Engr222 Spring 2004 Chapter 10
6
Example - continued
y
(x,y)
Iy
=
x2 dA =
=
x2 (2 √ x) dx
= 2
4
0
x2 y dx
x 2.5 dx
4
= [ (2/3.5) x 3.5 ] 0
= 73.1 in 4
In the above example, it will be difficult to determine Iy using
a horizontal strip. However, Ix in this example can be
determined using a vertical strip. So,
(1/3) y3 dx = (1/3) (2√x)3 dx .
Ix =
Concept Quiz
1. A pipe is subjected to a bending
moment as shown. Which property
of the pipe will result in lower stress
(assuming a constant cross-sectional
area)?
A) Smaller Ix
B) Smaller Iy
C) Larger Ix
D) Larger Iy
2. In the figure to the right, what is the
differential moment of inertia of the
element with respect to the y-axis (dIy)?
A) x2 y dx
C)
y2
x dy
Engr222 Spring 2004 Chapter 10
B) (1/12) x3 dy
D) (1/3) y dy
M
y
M
x
Pipe section
y
y=x3
x,y
x
7
Group Problem Solving
Given: The shaded area shown.
(x,y)
Find: Ix and Iy of the area.
Plan: Follow the steps described
earlier.
Solution
Ix
=
(1/3) y3 dx
=
(1/3) x dx = [x 2 / 6 ]0
8
0
8
= 10.7 in 4
Group Problem Solving - continued
(x,y)
IY =
x 2 dA
=
=
=
x 2 y dx
x 2 ( x (1/3)) dx
8
0
x (7/3) dx
= [(3/10) x
(10/3)
8
]0
= 307 in 4
Engr222 Spring 2004 Chapter 10
8
Attention Quiz
1. When determining the MoI of the
element in the figure, dIy equals
A) x 2 dy
B) x 2 dx
C) (1/3) y 3 dx D) x 2.5 dx
(x,y)
y2 = x
2. Similarly, dIx equals
A) (1/3) x 1.5 dx
B) y 2 dA
C) (1/12) x 3 dy
D) (1/3) x 3 dx
Textbook Problem 10-
Engr222 Spring 2004 Chapter 10
9
Announcements
• HW #4 and #5 due Friday
Moment of Inertia for Composite Areas
Today’s Objectives:
Students will be able to:
1. Apply the parallel-axis theorem.
2. Determine the moment of inertia
(MoI) for a composite area.
Engr222 Spring 2004 Chapter 10
In-Class Activities:
• Reading quiz
• Applications
• Parallel-axis theorem
• Method for composite
areas
• Concept quiz
• Group problem solving
• Attention quiz
10
Reading Quiz
1. The parallel-axis theorem for an area is applied between
A) an axis passing through its centroid and any corresponding
parallel axis.
B) any two parallel axis.
C) two horizontal axes only.
D) two vertical axes only.
2. The moment of inertia of a composite area equals the ____ of
the MoI of all of its parts.
A) vector sum
B) algebraic sum (addition or subtraction)
C) addition
D) product
Applications
Cross-sectional areas of structural
members are usually made of
simple shapes or combinations of
simple shapes.
Is there a simpler method for
determining the MoI of such
cross-sectional areas as compared
to the integration method? If yes,
can you describe the method?
Engr222 Spring 2004 Chapter 10
11
Applications - continued
This is another example of a
structural member with a
composite cross-area.
Design calculations typically
require use of the MoI for
these cross-sectional areas.
Can you describe a simple
method to calculate MoI?
Parallel-Axis Theorem for an Area - Section 10.2
This 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.
Engr222 Spring 2004 Chapter 10
12
Parallel-Axis Theorem for an Area - continued
IX =
=
A
A
y 2 dA =
A
y'2 dA + 2 dy
A
(y'+ dy)2 dA
y'dA + dy 2
A
dA
Using the definition of the centroid:
y = y'+ dy
y' =
( A y'dA) / ( A dA) . Now
since C is at the origin of the x'– y'axes,
y'= 0 , and hence
A
Thus IX = IX' +
y'dA = 0 .
A dy 2
Similarly, IY = IY'+ A dX 2
JO = JC
+
and
Ad2
Parallel-axis Theorem for a Rectangle
• See inside back cover of your textbook
Engr222 Spring 2004 Chapter 10
13
Moment of Inertia for a Composite Area
Section 10.5
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 triangle and circle.
The MoI of these “simpler” shaped areas about their centroidal
axes are found in most engineering handbooks as well as the
inside back cover of the textbook. Using these data and the
parallel-axis theorem, the MoI for a composite area can easily be
calculated.
Steps for Analysis
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. Determine the MoI of each “simpler” shaped part about the
desired reference axis using the parallel-axis theorem
IX = IX’ + A ( dy )2
4. The MoI of the entire area about the reference axis is
determined by performing an algebraic summation of the
individual MoIs obtained in Step 3. (Please note that MoI
of a hole is subtracted).
Engr222 Spring 2004 Chapter 10
14
Example
Given: The beam’s cross-sectional area.
[1] [2] [3]
Find:
The moment of inertia of the area
about the y-axis and the radius of
gyration ky.
Plan:
Follow the steps for analysis.
Solution
1. The cross-sectional area can be divided into three rectangles
( [1], [2], [3] ) as shown.
2. The centroids of these three rectangles are in their center.
The distances from these centers to the y-axis are 0 mm,
87.5 mm, and 87.5 mm, respectively.
Example - continued
3. From the inside back cover of the
book, the MoI of a rectangle about
its centroidal y axis is (1/12) hb3.
Iy[1] = (1/12) (25mm) (300mm)3
= 56.25 (106) mm4
[1] [2] [3]
Using the parallel-axis theorem,
IY[2] = IY[3] = IY’ + A (dX)2
= (1/12) (100) (25)3 + (25) (100) ( 87.5 )2
= 19.27 (106) mm 4
Engr222 Spring 2004 Chapter 10
15
Example - continued
4.
Iy =
Iy1 + Iy2 + Iy3
= 94.8 (106) mm 4
ky = √ ( Iy / A)
A = 300 (25) + 25 (100) + 25 (100) = 12,500 mm 2
ky = √ ( 94.79) (106) / (12500) = 87.1 mm
Concept Quiz
1. For the area A, we know the
centroid’s (C) location, area, distances
between the four parallel axes, and the
MoI about axis 1. We can determine
the MoI about axis 2 by applying the
parallel axis theorem ___ .
A
d3
d2
d1
C
•
Axis
4
3
2
1
A) directly between the axes 1 and 2.
B) between axes 1 and 3 and then
between the axes 3 and 2.
C) between axes 1 and 4 and then
axes 4 and 2.
D) None of the above.
Engr222 Spring 2004 Chapter 10
16
Concept Quiz - continued
2. For the same case,
consider the MoI about
each of the four axes.
About which axis will the
MoI be the smallest
number?
A) Axis 1
B) Axis 2
C) Axis 3
D) Axis 4
E) Can not tell
A
d3
d2
d1
C
•
Axis
4
3
2
1
Group Problem Solving
Given: The shaded area as shown in the
figure.
Find: The moment of inertia for the area
about the x-axis and the radius of
gyration kX.
Plan: Follow the steps for analysis.
(a) (b) (c)
Solution
1. The given area can be obtained by subtracting both the circle (b) and
triangle (c) from the rectangle (a).
2. Information about the centroids of the simple shapes can be
obtained from the inside back cover of the book. The perpendicular
distances of the centroids from the x-axis are: da = 5 in , db =
4 in, and dc = 8 in.
Engr222 Spring 2004 Chapter 10
17
Group Problem Solving - continued
3. IXa = (1/12) 6 (10)3 + 6 (10)(5)2
= 2000 in 4
IXb = (1/4) π (2)4 +
π (2)2 (4)2
= 213.6 in 4
IXc
(a) (b) (c)
IX =
IXa
= (1 /36) (3) (6)3 +
(½) (3) (6) (8)2 = 594 in 4
–
IXb
–
kX =
√ ( IX / A )
A =
10 ( 6 ) –
kX =
√ (1192 / 38.43) =
IXc
=
1190 in 4
π (2)2 – (½) (3) (6) =
38.43 in 2
5.57 in.
Attention Quiz
1. For the given area, the moment of inertia
about axis 1 is 200 cm4 . What is the MoI
about axis 3 (the centroidal axis)?
A) 90 cm 4
B) 110 cm 4
C) 60 cm 4
D) 40 cm 4
2. The moment of inertia of the rectangle about
the x-axis equals
A) 8 cm 4
B) 56 cm 4
C) 24 cm 4
D) 26 cm 4
Engr222 Spring 2004 Chapter 10
A=10 cm2
C
d2
3
2
C
••
d1
1
d1 = d2 = 2 cm
3cm
2cm
2cm
x
18
Textbook Problem 10-
Announcements
• WOP schedule next week. Class meets 10:45–11:30am.
Engr222 Spring 2004 Chapter 10
19
Mass Moment of Inertia - Section 10.9
Today’s Objectives:
Students will be able to :
a) Explain the concept of the mass
moment of inertia (MMI).
b) Determine the MMI of a
composite body.
In-Class Activities:
• Reading quiz
• Applications
• MMI: concept and definition
• Determining the MMI
• Concept quiz
• Group problem solving
• Attention quiz
Reading Quiz
1. The formula definition of the mass moment of inertia about
an axis is ___________ .
A)
r dm
B)
r2 dm
C)
m dr
D)
m2 dr
2. The parallel-axis theorem can be applied to determine
________ .
A) only the MoI
B) only the MMI
C) both the MoI and MMI
D) None of the above.
Note: MoI is the moment of inertia of an area and MMI is the
mass moment inertia of a body.
Engr222 Spring 2004 Chapter 10
20
Applications
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.
What property of the flywheel is
most important for this use?
How can we determine a value
for this property?
Why is most of the mass of the
flywheel located near the
flywheel’s circumference?
Applications - continued
If a torque M is applied to a
fan blade which is initially
at rest, its angular speed
begins to increase.
On which property (P) of
the fan blade does the
angular acceleration (α)
depend? How can we
determine a value for P?
What is the relationship
between M, P, and α?
Engr222 Spring 2004 Chapter 10
21
Concept of the Mass Moment of Inertia
T
G
·
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 α.
T and α are related by the equation T = I α . 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 = ma. The MMI is often used
when analyzing rotational motion (done in dynamics).
Definition of the Mass Moment of Inertia
p
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.
Engr222 Spring 2004 Chapter 10
22
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 p’
that is a distance d from the axis
through the body’s center of mass G.
The formula is Ip’ = IG + (m) (d)2
(where m is the mass of the body).
d
m
p’
G·
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
Given: The wheel consists of a thin ring
with a mass 10 kg and four spokes
(slender rods) with a mass of 2 kg
each.
q
r
p
Find:
The wheel’s MMI about an axis
perpendicular to the screen and
passing through point A.
Plan:
Follow steps similar to finding the
MoI for a composite area.
Solution:
1. The wheel can be divided into a thin ring (p) and two slender
rods (q and r). Will both rods be treated the same?
Engr222 Spring 2004 Chapter 10
23
Example - continued
2. The center of mass for each of the three
pieces is at point O, 0.5 m from Point A.
q
r
p
3. The MMI data for a thin ring and
slender rod are given on the inside
back cover of the textbook. Using
those data and the parallel-axis
theorem, calculate the following:
IA = IO + (m) (d) 2
IAp = 10 (0.5)2 + 10 (0.5)2 = 5.0 kg·m2
IAq = IAr = (1/12) (4) (1)2 + 4 (0.5)2 = 1.333 kg·m2
4. Now add the three MMIs about point A.
IA = IAp + IAq + IAr = 7.67 kg·m2
Concept Quiz
1. Consider a particle of mass 1 kg
located at point P, whose coordinates
are given in meters. Determine the MMI
of that particle about the z axis.
B) 16 kg·m2
A) 9 kg·m2
C) 25 kg·m2
D) 36 kg·m2
2. Consider a rectangular frame made of four
slender bars with four axes (zP, zQ, zR and zS)
perpendicular to the screen and passing
through the points P, Q, R, and S
respectively. About which of the four axes
will the MMI of the frame be the largest?
A) zP
B) zQ
C) zR
D) zS
E) Not possible to determine.
Engr222 Spring 2004 Chapter 10
z
·P(3,4,6)
y
x
P
•
S•
Q
•
•R
24
Group Problem Solving
R
P
Plan:
Given: The pendulum consists of a 24 lb
plate and a slender rod weighing
8 lb.
Find: The radius of gyration of
the pendulum about an axis
perpendicular to the screen and
passing through point O.
Determine the MMI of the pendulum using the method for
composite bodies. Then determine the radius of gyration
using the MMI and mass values (check units).
Solution
1. Separate the pendulum into a square plate (P) and a slender
rod (R).
Group Problem Solving - continued
R
P
2. The center of mass of the plate
and rod are 3.5 ft and 0.5 ft from
point O, respectively.
3. The MMI data on plates and slender rods are given on the
inside cover of the textbook. Using those data and the parallel-axis
theorem,
IP = (1/12) (24/32.2) (12 + 12) + (24/32.2) (3.5)2 = 9.254 slug·ft2
IR = (1/12) (8/32.2) (5)2 + (8/32.2) (0.5)2 = 0.5797 slug·ft2
4. IO = IP + IR = 9.254 + 0.5797 = 9.834 slug·ft2
5. Total mass (m) equals (24+8)/32.2 = 0.9938 slug
Radius of gyration k = √IO / m
Engr222 Spring 2004 Chapter 10
= 3.15 ft
25
Attention Quiz
1. A particle of mass 2 kg is located 1 m
down the y-axis. What are the MMI of
the particle about the x, y, and z axes,
respectively?
x
A) (2, 0, 2)
B) (0, 2, 2)
C) (0, 2, 2)
D) (2, 2, 0)
2. Consider a rectangular frame made of four
slender bars and four axes (zP, zQ, zR and zS)
perpendicular to the screen and passing
through points P, Q, R, and S, respectively.
About which of the four axes will the
MMI of the frame be the lowest?
A) zP
B) zQ
C) zR
D) zS
E) Not possible to determine.
z
1m
•
P
•
S•
y
Q
•
•R
Textbook Problem 10-
Engr222 Spring 2004 Chapter 10
26