# Document 362166

```Chapter Seven
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
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
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
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
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
Moment of Inertia of Areas
Table (7-1) : Inertial properties of plane Areas : Part 1
75
Chapter Seven
Moment of Inertia of Areas
Table (7-2) : Inertial properties of plane Areas : Part 2
76
Chapter Seven
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
```