IZMIR INSTITUTE OF TECHNOLOGY
8.0
22.3.2016
Department of Architecture
AR231 Fall12/13
SECOND MOMENT OR MOMENT OF INERTIA
OF AN AREA
8.1
Introduction
8.2
Moment of Inertia of an Area
8.3
Moment of Inertia of an Area by Integration
8.4
Polar Moment of Inertia
8.5
Radius of Gyration
8.6
Parallel Axis Theorem
8.7
Moments of Inertia of Composite Areas
8.8
Product of Inertia
8.9
Principal Axes and Principal Moments of Inertia
Dr. Engin Aktaş
1
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.1. Introduction
• Forces which are proportional to the area or volume over which they act
but also vary linearly with distance from a given axis.
- the magnitude of the resultant depends on the first moment of the
force distribution with respect to the axis.
- The point of application of the resultant depends on the second
moment of the distribution with respect to the axis.
• Herein methods for computing the moments and products of inertia for
areas and masses will be presented
22.3.2016
Dr. Engin Aktaş
2
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.2. Moment of Inertia of an
Area

• Consider distributed forces F whose magnitudes are
proportional to the elemental areas A on which they
act and also vary linearly with the distance of A
from a given axis.
• Example: Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from the
neutral axis which passes through the section centroid.

F  kyA
R  k  y dA  0  y dA  Qx  first moment
2
M  k  y 2 dA
 y dA  second moment
• Example: Consider the net hydrostatic force on a
submerged circular gate.
F  pA  yA
R    y dA
M x    y 2 dA
22.3.2016
Dr. Engin Aktaş
3
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.3.Moment of Inertia of an Area by Integration
• Second moments or moments of inertia of
an area with respect to the x and y axes,
I x   y 2 dA
I y   x 2 dA
• Evaluation of the integrals is simplified by
choosing dA to be a thin strip parallel to
one of the coordinate axes.
• For a rectangular area,
h
I x   y dA   y 2bdy  13 bh3
2
0
• The formula for rectangular areas may also
be applied to strips parallel to the axes,
dI x  13 y 3dx
22.3.2016
Dr. Engin Aktaş
dI y  x 2 dA  x 2 y dx
4
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.4. Polar Moment of Inertia
• The polar moment of inertia is an important
parameter in problems involving torsion of
cylindrical shafts and rotations of slabs.
J 0   r 2 dA
• The polar moment of inertia is related to the
rectangular moments of inertia,


J 0   r 2 dA   x 2  y 2 dA   x 2 dA   y 2 dA
 I y  Ix
22.3.2016
Dr. Engin Aktaş
5
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.5. Radius of Gyration of an Area
• Consider area A with moment of inertia
Ix. Imagine that the area is
concentrated in a thin strip parallel to
the x axis with equivalent Ix.
I
I x  k x2 A
kx  x
A
kx = radius of gyration with respect
to the x axis
• Similarly,
Iy 
k y2 A
ky 
J O  kO2 A kO 
Iy
A
JO
A
kO2  k x2  k y2
22.3.2016
Dr. Engin Aktaş
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Examples
SOLUTION:
• A differential strip parallel to the x axis is chosen for
dA.
dI x  y 2dA
dA  l dy
• For similar triangles,
Determine the moment of
inertia of a triangle with respect
to its base.
l h y

b
h
l b
h y
h
dA  b
h y
dy
h
• Integrating dIx from y = 0 to y = h,


h y
bh 2
I x   y dA   y b
dy   hy  y 3 dy
h
h0
0
2
h
2
h
b  y3 y 4 
 h  
h 3
4
0
22.3.2016
Dr. Engin Aktaş
bh3
I x
12
7
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
SOLUTION:
• An annular differential area element is chosen,
dA  2 u du
dJ O  u 2dA
r
r
J O   dJ O   u 2 u du   2  u 3du
2
0
0
JO 
a) Determine the centroidal polar
moment of inertia of a circular
area by direct integration.
2
r4
• From symmetry, Ix = Iy,
JO  I x  I y  2I x
b) Using the result of part a,
determine the moment of inertia
of a circular area with respect to a
diameter.
22.3.2016


2
r 4  2I x
I diameter  I x 
Dr. Engin Aktaş

4
r4
8
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.6. Parallel Axis Theorem
• Consider moment of inertia I of an area A
with respect to the axis AA’
I   y 2 dA
• The axis BB’ passes through the area centroid
and is called a centroidal axis.
I   y 2 dA    y   d 2 dA
  y  2 dA  2d  y dA  d 2  dA
I  I  Ad 2
22.3.2016
Dr. Engin Aktaş
parallel axis theorem
9
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
• Moment of inertia IT of a circular area with
respect to a tangent to the circle,
 
I T  I  Ad 2  14  r 4   r 2 r 2
 54  r 4
• Moment of inertia of a triangle with respect to a
centroidal axis,
I AA  I BB  Ad 2
I BB  I AA  Ad
2
1 bh3
 12
 
2
1
1
 2 bh 3 h
1 bh3
 36
22.3.2016
Dr. Engin Aktaş
10
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.7. Moments of Inertia of Composite Areas
• The moment of inertia of a composite area A about a given axis is
obtained by adding the moments of inertia of the component areas
A1, A2, A3, ... , with respect to the same axis.
22.3.2016
Dr. Engin Aktaş
11
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example
225 mm
20 mm
358 mm
SOLUTION:
• Determine location of the centroid of
composite section with respect to a
coordinate system with origin at the
centroid of the beam section.
• Apply the parallel axis theorem to
determine moments of inertia of beam
section and plate with respect to
172 mm
composite section centroidal axis.
The strength of a W360x57 rolled steel
beam is increased by attaching a
• Calculate the radius of gyration from the
225x20 mm plate to its upper flange.
moment of inertia of the composite
section.
Determine the moment of inertia and
radius of gyration with respect to an
axis which is parallel to the plate and
passes through the centroid of the
section.
22.3.2016
Dr. Engin Aktaş
12
IZMIR INSTITUTE OF TECHNOLOGY
225 mm
20 mm
Department of Architecture
AR231 Fall12/13
SOLUTION:
• Determine location of the centroid of composite section
with respect to a coordinate system with origin at the
centroid of the beam section.
358 mm
172 mm
A  225 mm20 mm  4500 mm2
1
1
y  358 mm  20 mm  189 mm
2
2
Section
189 mm
3
mm3
y , in.
mm yA, in
Plate
6.75
7189
.425 50
.12 x 103
850.5
4500
7230
Beam Section 11.20
0
0
17.95
50.12 x 103
 A 11730
 yA  850.5
Y  A   yA
22.3.2016
2 2
mm
A, in
yA 850.5  10

Y 

 A 11730
Dr. Engin Aktaş
3
 72.51 mm.
13
IZMIR INSTITUTE OF TECHNOLOGY
225 mm
20 mm
Department of Architecture
AR231 Fall12/13
• Apply the parallel axis theorem to determine moments of
inertia of beam section and plate with respect to composite
section centroidal axis.
I x,beam section  I x  AY 2  160.2 106  723072.51
2
358 mm
 198.2 10 4
I x,plate  I x  Ad 2  121 22520  4500189  72.51
3
2
 61.2 106 mm 4
172 mm
I x  I x,beam section  I x,plate  192.6 106  61.2 106
I x  254 106 mm4
189 mm
• Calculate the radius of gyration from the moment of inertia
of the composite section.
k x 
22.3.2016
I x 253.8 106 mm 4

A
11730 mm 2
Dr. Engin Aktaş
k x  147.1 mm.
14
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example
SOLUTION:
• Compute the moments of inertia of the
bounding rectangle and half-circle with
respect to the x axis.
• The moment of inertia of the shaded area is
obtained by subtracting the moment of
inertia of the half-circle from the moment
of inertia of the rectangle.
Determine the moment of inertia
of the shaded area with respect to
the x axis.
22.3.2016
Dr. Engin Aktaş
15
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
SOLUTION:
• Compute the moments of inertia of the bounding
rectangle and half-circle with respect to the x axis.
Rectangle:
I x  13 bh3  13 240120  138.2  106 mm 4
Half-circle:
moment of inertia with respect to AA’,
I AA  18 r 4  18  904  25.76  106 mm 4
moment of inertia with respect to x’,
4r 4 90 
a

 38.2 mm
3
3
b  120 - a  81.8 mm
A  12 r  12  90 
2
2
 12.72  103 mm 2
22.3.2016


I x  I AA  Aa 2  25.76  106 12.72  103

 7.20  106 mm 4
moment of inertia with respect to x,


I x  I x  Ab 2  7.20  106  12.72  103 81.82
 92.3  106 mm 4
Dr. Engin Aktaş
16
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
• The moment of inertia of the shaded area is obtained by
subtracting the moment of inertia of the half-circle from
the moment of inertia of the rectangle.
Ix

138.2  106 mm 4

92.3  106 mm 4
I x  45.9  106 mm 4
22.3.2016
Dr. Engin Aktaş
17
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.8. Product of Inertia
• Product of Inertia:
I xy   xy dA
• When the x axis, the y axis, or both are an
axis of symmetry, the product of inertia is
zero.
• Parallel axis theorem for products of inertia:
I xy  I xy  xyA
22.3.2016
Dr. Engin Aktaş
18
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
8.9. Principal Axes and Principal Moments of Inertia
Given I x   y 2 dA I y   x 2 dA
I xy   xy dA
we wish to determine moments
and product of inertia with
respect to new axes x’ and y’.
Note: x  x cos  y sin 
y   y cos  x sin 
22.3.2016
• The change of axes yields
Ix  I y Ix  I y
I x 

cos 2  I xy sin 2
2
2
Ix  I y Ix  I y
I y 

cos 2  I xy sin 2
2
2
Ix  I y
I xy 
sin 2  I xy cos 2
2
• The equations for Ix’ and Ix’y’ are the
parametric equations for a circle,
I x  I ave 2  I x2y  R 2
Ix  I y
I ave 
2
 Ix  I y  2
  I xy
R  
 2 
• The equations for Iy’ and Ix’y’ lead to the
same circle.
Dr. Engin Aktaş
19
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
• At the points A and B, Ix’y’ = 0 and Ix’ is
a maximum and minimum, respectively.
I max, min  I ave  R
tan 2 m  
2 I xy
Ix  I y
• The equation for Qm defines two
angles, 90o apart which correspond to
the principal axes of the area about O.
I x  I ave 2  I x2y  R 2
I ave 
22.3.2016
Ix  I y
2
 Ix  I y  2
  I xy
R  
 2 
• Imax and Imin are the principal moments
of inertia of the area about O.
Dr. Engin Aktaş
20
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example
SOLUTION:
• Determine the product of inertia using
direct integration with the parallel axis
theorem on vertical differential area strips
• Apply the parallel axis theorem to
evaluate the product of inertia with respect
to the centroidal axes.
Determine the product of inertia of
the right triangle (a) with respect
to the x and y axes and
(b) with respect to centroidal axes
parallel to the x and y axes.
22.3.2016
Dr. Engin Aktaş
21
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
SOLUTION:
• Determine the product of inertia using direct integration
with the parallel axis theorem on vertical differential
area strips
 x
 x
y  h1   dA  y dx  h1  dx
 b
 b
 x
xel  x
yel  12 y  12 h1  
 b
Integrating dIx from x = 0 to x = b,
b
I xy   dI xy   xel yel dA   x
0
b

1
2
2
x
h 1   dx
 b
2
2
b
 x 2 x3 x 4 
x 
2 x x
h  

dx  h    2 
 2 b 2b 2 
0

 4 3b 8b  0
2
3
1 b 2h 2
I xy  24
22.3.2016
Dr. Engin Aktaş
22
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
• Apply the parallel axis theorem to evaluate the
product of inertia with respect to the centroidal axes.
x  13 b
y  13 h
With the results from part a,
I xy  I xy  x yA
 13 h12 bh
1 b2h 2  1 b
I xy  24
3
1 b 2h 2
I xy   72
22.3.2016
Dr. Engin Aktaş
23
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Example
SOLUTION:
• Compute the product of inertia with
respect to the xy axes by dividing the
section into three rectangles and applying
the parallel axis theorem to each.
• Determine the orientation of the
principal axes (Eq. 9.25) and the
principal moments of inertia (Eq. 9. 27).
For the section shown, the moments of
inertia with respect to the x and y axes
are Ix = 10.38 in4 and Iy = 6.97 in4.
Determine (a) the orientation of the
principal axes of the section about O,
and (b) the values of the principal
moments of inertia about O.
22.3.2016
Dr. Engin Aktaş
24
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
SOLUTION:
76 mm
• Compute the product of inertia with respect to the xy axes
by dividing the section into three rectangles.
13 mm
Apply the parallel axis theorem to each rectangle,

102 mm
I xy   I xy  x yA
13 mm
76 mm
31.5 mm
44.5 mm
Note that the product of inertia with respect to centroidal
axes parallel to the xy axes is zero for each rectangle.
Rectangle
I
II
III
Area, mm 2 x , mm.
988  31.5
988
0
988  31.5
44.5 mm
y , mm.
 44.5
0
 44.5
x yA, mm 4
 1384929
0
 1384929
 xyA  2769858
I xy   xyA  2770000mm4
31.5 mm
22.3.2016

Dr. Engin Aktaş
25
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
tan 2 m  
• Determine the orientation of the principal axes by
m = 127.5o
the principal moments of
inertia by
2 I xy
Ix  I y
I max, min  I ave  R
I x  I ave 2  I x2y  R 2
m =
I ave 
37.5o
I max, min 
Ix  Iy
2

 Ix  I y  2
  I xy
R  
 2 
2 m  74.9 and 254.9
4
 m  37.5 and  m  127.5
 Ix  I y 
  I xy2
 
 2 
2
2
 4.42 106  2.93 106 
4.42 10  2.93 10
   2.77 106

 
2
2


6

2  2.77 106
tan 2 m  

 3.72
6
6
Ix  Iy
4.42 10  2.93 10
I y  2.93 106 mm 4
I xy  2.77 10 mm
2
2 I xy
I x  4.42 106 mm 4
6
Ix  I y
6


2
I a  I max  6.54 mm 4
I b  I min  8.07 mm 4
22.3.2016
Dr. Engin Aktaş
26