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STATICS: CE201
Chapter 10
Moments of Inertia
Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson
Dr M. Touahmia & Dr M. Boukendakdji
Civil Engineering Department, University of Hail
(2012/2013)
10.
Moments of Inertia
________________________________________________________________________________________________________________________________________________
Chapter Objective:
1.
2.
Define the moments of inertia (MoI) for an area.
Determine the MoI for an area by integration.
Contents:
10.1
10.2
10.3
10.4
Definition of Moment of Inertia for Areas
Parallel-Axis Theorem for an Area
Radius of Gyration of an Area
Moments of Inertia for Composite Areas
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10.1 Definition of Moments of Inertia for Areas

Many structural members like
beams and columns have cross
sectional shapes like I, L, C, etc.
Some others are made of tubes
rather than solid squares or rounds.

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?
10.1 Definition of Moments of Inertia for Areas

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 F on the beam, which shape will
develop less internal stress and deflection? Why?

The answer depends on a property called Moment of Inertia (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: (σ = M.y/I); as I (MoI)
increases, σ (stress) decreases and deflection decreases also.
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10.1 Definition of Moments of Inertia for Areas

The Area Moment of Inertia of a beams cross-sectional area
measures the beams ability to resist bending.

The MoI is a geometrical property of a beam and depends
on a reference axis. It is frequently used in formulas related
to the strength and stability of structural members.

The larger the MoI the less the beam will bend.

The smallest MoI about any axis passes through the
centroid.
10.1 Definition of Moments of Inertia for Areas


The area moment of Inertia represents the second moment
of the area about an axis.
The moments of inertia of a differential
area dA about the x and y are:
dI x  y 2 dA
dI y  x 2 dA

The moment of inertia of dA about the
“pole” O or z axis is then: dJ O  r 2 dA

For the entire area A, the moments of inertia are determined
by integration:
I x   y 2 dA
A

I y   x 2 dA
A
2
The polar moment of inertia is: J O  Ar dA  I x  I y
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The step-by-step procedure for analysis 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.
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.
Area Moment of Inertia of Common Shapes:
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10.2 Parallel-Axis Theorem for an Area

The parallel-axis theorem can be used to find the moment of
inertia of an area about any axis that is parallel to an axis passing
through the Centroid and about which the moment of inertia is
known ( I x  I J ).
y

C
The MoI for an area about any axis parallel to the Centroid axis is
equal to its MoI about an axis passing through the area’s Centroid
plus the product of the area and the square of the perpendicular
distance between the axes.
I x  I x  Ad y2
I y  I y  Ad x2
J O  J C  Ad 2
10.3 Radius of Gyration of an area

For a given area A and its MoI, Ix , imagine that the entire
area is located at distance kx from the x axis:
I x  k x2 A

kx 
Ix
A
This kx is called the radius of gyration of the area about the
x axis. Similarly:
y
A
ky 
Iy
A
kx
kO 
JO
A
x
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Example 1
Determine the moment of inertia for the rectangle area
shown in the figure with respect to:
(a)
The Centroid x’ axis.
(b)
The axis xb passing through
the base of the rectangle.
(c)
The pole or z’ axis perpendicular
to the x’- y’ plan and passing
through the Centroid C.
Solution 1
(a) MoI about x’ axis: Integration from y’=- h/2 to y’= h/2

I x   y  2 dA 
A
h2
2
 y  bdy   b
h 2
h2
 y
2
dy  
h 2
1 3
bh
12

(b) MoI about xb:
I xb
1
1
h
 I x  Ad  bh 3  bh   bh 3
12
3
 2

(c) Polar MoI about point C:
2
2
y
I y 
1
hb 3
12
J C  I x  I y  

1
bh h 2  b 2
12

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Example 2
Determine the moment of inertia for the shaded area shown
in the figure about the x axis.
Solution 2

A differential element dA parallel to the x axis is chosen for
integration:
dA  100  x dy

Integration with respect to y, from
y = 0 to y = 200 mm, yields:
I x   y 2 dA 
A
2
 y 100  x dy 
0
200mm

200mm

0
200m

0

y2 
dy
y 2 100 
400 


y 
100y 2 
dy  107.10 6 mm 4
400


4
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10.4 Moments of Inertia for Composite Areas


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 figure 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.
Using these data and the parallel-axis
theorem, the MoI for a composite area can
easily be calculated.
The MoI of the composite area is equal to the
algebraic sum of the moments of inertia of
each of its parts.
Example 3
Determine the moment of inertia of the area shown in the
figure about the x axis.
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Solution 3

Composite Parts:
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.
=
Solution 3

Parallel-Axis Theorem:
1
(a) Circle: I  I  Ad   25   25 75  11.410  mm
4
4
2
x
x
2
2
6
4
y
(b) Rectangle: I x  I x  Ad y2  1 1001503  100150752  112.510 6  mm 4
12

Algebraic Summation: The MoI for the entire area is the
algebraic summation of the individual MoI:
I  112.510   11.410   101.110  mm
6
6
6
4
x
9