Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
7/1 Introduction
First moments of areas, as presented in Chapter 6, dealt with the integrals
̅ and . In this chapter, we discuss the second moments of plane areas,
also known as moments of inertia, ̅ and . In this chapter will present
methods for computing the moment of inertia for areas.
7/2 Moment of Inertia of an Area by Integration
In Section. 6.2, the first moments of the area of a plane region about the x- and y-axes
were defined as
Q x = ∫ yel dA
Q y = ∫ x el dA
where A is the area of the region and x and y are the coordinates of the differential
area element dA, as shown in (Fig. 7.1).
Fig. (7-1)
The second moments of inertia of the area about the x- and y-axes, respectively, are
defined by (Fig. 7-2)
I x = ∫ y 2 dA
I y = ∫ x 2 dA
y
y
y
dA = ( a – x ) dy
dA = dx dy
dA = y d x
dx
x
dy
x
dy
y
y
x
y
x
x
dx
a
dIx = y2 dA dIy = x2 dA
dIx = y2 dA
dIy = x2 dA
(a)
(b)
(c)
Fig. (7-2)
70
x
Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
For example Moment of Inertia of rectangular area,
The formula for rectangular areas may also be applied to
strips parallel to the axes,
= = 2
h
I x = ∫ y dA = ∫ y 2 bdy = 1 bh3
0
3
7/3 Computing Ix and Iy Using the Same Elemental Strips:
dI x = 1 y 3dx
3
dI y = x 2 dA = x 2 y dx
7/4 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 2dA
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
71
Chapter Seven
A.Lecturer Saddam K. Kwais
7/5 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 x = k x2 A
kx =
Ix
A
kx = radius of gyration with respect to the x axis
Similarly,
I y = k 2y A
ky =
Iy
A
JO
2
J O = kO
A kO =
A
2
kO
= k x2 + k 2y
EXAMPLE 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 2 dA
dA = l dy
72
Moment of Inertia of Areas
Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
• 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
(
)
the
centroidal
bh 2
I x = ∫ y dA = ∫ y b
dy = ∫ hy − y 3 dy
h
h0
0
2
2 h− y
h
b  y3 y 4 
= h
−

h  3
4 
0
bh3
I x=
12
EXAMPLE 2. a) Determine
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 = 2π u du
dJ O = u 2 dA
r
r
J O = ∫ dJ O = ∫ u (2π u du ) = 2π ∫ u3du
2
0
JO =
0
π 4
r
2
• From symmetry, Ix = Iy,
JO = I x + I y = 2I x
⇒
π 4
π
r = 2 I x ⇒ I diameter = I x = r 4
2
4
73
Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
7/6 Parallel Axis Theorem:
• Consider moment of inertia I of an area A
with respect to the axis AA’
I = ∫ y 2dA
• 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,
( )
IT = I + Ad 2 = 1 π r 4 + π r 2 r 2
4
= 5π r4
4
• Moment of inertia of a triangle with respect to
a centroidal axis,
I AA′ = I BB ′ + Ad 2
( )2
I BB ′ = I AA′ − Ad 2 = 1 bh3 − 1 bh 1 h
12
2
3
= 1 bh3
36
7/7 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.
74
Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
Table (7-1) : Inertial properties of plane Areas : Part 1
75
Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
Table (7-2) : Inertial properties of plane Areas : Part 2
76
Chapter Seven
A.Lecturer Saddam K. Kwais
Moment of Inertia of Areas
EXAMPLE 3. Determine the moment of inertia
of the shaded area with respect to the x axis.
SOLUTION:
• Compute the moments of inertia of the bounding rectangle and half-circle with
respect to the x axis.
Rectangle:
I x = 1 bh3 = 1 (240 )(120 ) = 138.2 × 10 6 mm 4
3
3
Half-circle:
moment of inertia with respect to AA’,
I AA′ = 1 πr 4 = 1 π (90 )4 = 25.76 × 10 6 mm 4
8
8
moment of inertia with respect to x’,
(
)(
I x ′ = I AA′ − Aa 2 = 25.76 × 106 12.72 × 103
)
= 7.20 × 106 mm 4
4r (4 )(90 )
=
= 38.2 mm
3π
3π
b = 120 - a = 81.8 mm
a=
A = 1 πr 2 = 1 π (90 )2
2
2
= 12.72 × 103 mm 2
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
• 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
I x = 45 . 9 × 10 6 mm 4
77
−
92.3 × 106 mm 4