CHAPTER 9
Moments of Inertia
9.1 INTRODUCTION
• Previously considered distributed forces which were proportional to the
area or volume over which they act.
- The resultant was obtained by summing or integrating over the
areas or volumes.
- The moment of the resultant about any axis was determined by
computing the first moments of the areas or volumes about that
axis.
• Will now consider forces which are proportional to the area or volume
over which they act but also vary linearly with distance from a given axis.
- It will be shown that 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.
• Current chapter will present methods for computing the moments and
products of inertia for areas and masses.
9.2 MOMENTS OF INERTIA OF AREAS

• 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
9.3 Determination of the 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
dI y  x 2 dA  x 2 y dx
9.4 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
9.5 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 
kO2  k x2  k y2
Iy
A
JO
A
Example
For the rectangle shown in Fig 9.8, let us compute the radius
of gyration kx with respect to its base. Using formulas (9.5) and (9.2),
we write
1 3
bh
l
h2
2
3
x
kx  

A
bh
3
kx 
h
3
Sample Problem 9.1
Determine the moment of inertia of a triangle with respect to its base
SOLUTION:
• A differential strip parallel to the x axis is chosen for
dA.
dI x  y 2dA
dA  l dy
• For similar triangles,
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
bh3
I x
12
Sample Problem 9.2
a) Determine the centroidal polar moment of inertia of a circular area by direct
integration.
b) Using the result of part a, determine the moment of inertia of a circular area with
respect to a diameter.
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.
b) Using the result of part a,
determine the moment of inertia
of a circular area with respect to a
diameter.

2
r4
• From symmetry, Ix = Iy,
JO  I x  I y  2I x

2
r 4  2I x
I diameter  I x 

4
r4
Sample Problem 9.3
a) Determine the moment of inertia of the shaded area shown with
respect to each of the coordinates axes.( properties of this area were considered
Solution
Equation of the curve and the total area
1
A  ab
3
b 2
x
2
a
y
refer to page 225
text book SI 8E
Moment of Inertia IX
3
dI X
1 3
1 b 2
1 b3 6
 y dx   2 x  dx 
x dx
6
3
3 a
3
a

a
IX 
IX
 dI
ab 3

21
X


a
0
 1 b3 x 7 
1 b3 6
x dx  

6
6
3a
3
a
7

0
Moment of Inertia IY
b
 b 
dl y  x 2 dA  x 2  ydx   x 2  2 x 2 dx  2 x 4 dx
a
a

a b
 b x5  a
4
I y   dI y   2 x dx   2 
0 a
a 5  0
a 3b
Iy 
5
Radii of Gyration kx and ky
I x ab3 / 21 b 2
k  

A
ab / 3
7
kx 
1
b
7
ab3 / 5 3 2
k  
 a
A ab / 3 5
ky 
3
a
5
2
x
2
y
Iy
9.6 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
parallel axis theorem
• 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
1 bh3
 36
2
1 bh3
 12
 
2
1
1
 2 bh 3 h
9.7 Moment 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.
moment of inertia of common geometric shapes
9.7 Moment of Inertia of Composite Areas
Properties of Rolled Steel Shapes ( SI Units )
Sample Problem 9.4
229mm
19mm
352mm
171mm
The strength of a W360 x 44 rolled steel beam is increased by attaching a
229 x 19mm plate to its upper flange as shown.
Determine the moment of inertia and the radius of gyration of the
composite section with respect to an axis which is parallel to the plate and
passes through the centroid C of the section.
SOLUTION:
229mm
19mm
352mm
171mm
• 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
composite section centroidal axis.
• Calculate the radius of gyration from the
moment of inertia of the composite
section.
229mm
19mm
352mm
The origin O of the coordinates is placed at the centroid of the wide-flange shape,
and the distance Y to the centroid of the composite section is computed using the
methods of Chap. 5. The area of the wide-flange shape is found by referring to
Fig. 9.13A. The area and the y coordinates of the centroid of the plane are
A  (229 mm)(19 mm)  4351 mm 2
y
171mm
1
1
(352mm )  (19mm )  185.5mm
2
2
Section
2 2
A, inmm
3
y , in.
mm yA, in mm 3
Plate
6.75 4351 7185.5
.425 50.12 807111
0
Beam Section 11.20 5730 0 0 0
.95
.12
10081
807111
 A  17
 yA  50
185.5mm
Y  A
yA
Y (10081) 
Y  80.1
229mm
19mm
352mm
Moment of Inertia. The parallel-axis theorem is used to determine the moments
of inertia of the wide-flange shape and the plate with respect to the x’ axis. This axis
is a centroidal axis for the composite section but not for either of the elements considered
separately. The value of Ix for the wide-flange shape is obtained from Fig. ).9.13A
For the wide-flange shape,
2
I X '  I X  AY  122x10 6  (5730x(80.1) 2 (80.1) 2  1.59x108 mm 4
For the plate,
171mm
1
I X '  I X  Ad 2   (229)(19) 3  (4351)(185.5  80.1) 2  4.84x107 mm 4
 12 
For the composite area,
I X '  1.59x108  4.84x107  2.07x108 mm 4
Radius of Gyration. We have
k
2
X'
2.07 x108 mm 4


A
10081mm 2
IX'
k X '  143mm
Sample Problem 9.5
Determine the moment of inertia of the shaded area with respect to the x axis.
Solution
The given area can be obtained by subtracting a half circle from a rectangle.
The moments of Inertia of the rectangle and the half circle will be computed separately.
240mm
A
C
120mm
a
A’
x’
b
Moment of Inertia of Rectangle.
Referring to Fig. 9.12, we obtain
I x  13 bh3  13 240120  138.2  106 mm 4
Moment of Inertia of Half Circle.
Referring to Fig. 5.8, we determine the location of the centroid C
of the half circle with respect to diameter AA’.
The distance b from the centroid C to the axis is
Referring now to Fig. 9.12, we compute the moment of inertia of the half circle
with respect to diameter AA’; we also compute the area of the half circle
I AA  18 r 4  18  904  25.76  106 mm 4
A
1 2 1
πr  π(90mm ) 2  12.72x103 mm 2
2
2
Using the parallel axis theorem , we obtain the value of
I X'
I AA'  I X '  Aa 2
25.76 x 106 mm 4  I X'  (12.72x10 3 mm 2 )(38.2mm ) 2
I X'  7.20 x 10 6 mm 4
Again using the parallel axis theorem, we obtain the value of

IX

I x  I x  Ab 2  7.20 106 mm 4  12.72 103 mm 2 81.8mm 
 92.3 106 mm 4
2
• 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

I x  45.9  106 mm 4
138.2  106 mm 4

92.3  106 mm 4
PROBLEMS
X
X
9.8 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.
9.8 Product of Inertia
• Parallel axis theorem for products of inertia:
I xy   xy dA   ( x '  x)( y '  y ) dA
  x ' y ' dA  y  x ' dA  x
I xy  I xy  xyA
'
y
 dA  x y  dA
Sample Problem 9.6
Determine the product of inertia of the right triangle shown
(a) with respect to the x and y axes and
(b) with respect to centroidal axes parallel to the x and y axes.
Solution
a.
Product of Inertia Ixy
A vertical rectangular strip is chosen as the differential element
of area. Using the parallel-axis theorem, we write
dI xy  dI x ' y '  x el y el dA
Since the element is symmetrical with respect to the x’ and y’ axes, we note
that dIxy = 0. From the geometry of the triangle, we obtain
 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
2
 x 2 x3 x 4 
x3 
2 x x
h  

dx  h    2 
 2 b 2b 2 
0

 4 3b 8b  0
1 b 2h 2
I xy  24
• 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
THE END