CHAPTER 9:
Moments of Inertia
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Moment of Inertia of Areas
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Second Moment, or Moment of Inertia, of an
Area
Parallel-Axis Theorem
Radius of Gyration of an Area
Determination of the Moment of Inertia of an
Area by Integration
Moments of Inertia of Composite Areas
Polar Moment of Inertia
1
9.1 Moment of Inertia: Definition
I x = ∫ ( y ) 2 dA
x
y
dA=(dx)(dy)
A
I y = ∫ ( x) 2 dA
A
y
x
O
2
9.2 Parallel-Axis Theorem of an Area
y´ = Centroidal axis
dx
y
x´
A
dA
y´
CG
x´= Centroidal axis
= ∫ [( y ' ) 2 + 2( y ' )(d y ) + (d y ) 2 ]dA
A
= ∫ ( y ' ) 2 dA + ∫ 2( y ' )(d y )dA + ∫ (d y ) 2 dA
dy
A
x
O
I x = ∫ ( y '+ d y ) 2 dA
A
0, y´ = 0
= I x + 2d y ∫ y ' dA + d y
A
2
A
∫ dA
A
2
Ix = Ix + 0 + dy A
2
I y = I y + 0 + dx A
J O = J C + Ad 2
3
9.3 Radius of Gyration of an Area
y
The radius of gyration of an area
A with respect to the x axis is
defined as the distance kx, where
Ix = kx A. With similar definitions for
2
the radii
of gyration of A with
respect to the y axis and with
respect to O, we have
A
kx
O
x
kx =
Ix
A
ky =
Iy
A
kO =
JO
A
4
9.4 Determination of the Moment of Inertia of an
Area by Integration
y
The rectangular moments of inertia Ix
and Iy of an area are defined as
y
x
I x = ∫ y 2 dA
dx
I y = ∫ x 2 dA
x
These computations are reduced to single integrations by choosing dA to be a thin
strip parallel to one of the coordinate axes. The result is
dI x =
1 3
y dx
3
dI y = x 2 y dx
5
• Moment of Inertia of a Rectangular Area.
y
y´
dA = (b/2)dy
dA = bdy
dy
h
y
dy
y
h/2
x´
x
b/2
b
I x = ∫ y 2 dA
A
h
= ∫ y (bdy )
2
0
h
I x = I x ' = ∫ y 2 dA
A
h
b
= 4 ∫ y 2 ( dy )
2
0
(by 3 )
=
3 0
b y3
= 4( )
2 3
bh3
=
3
bh 3
=
12
h/ 2
0
6
y
y´
b
x
dA = hdx
h
dx
dA = (h/2)dx
h/2
x´
x
x
b/2
dx
I y = ∫ x 2 dA
A
b
= ∫ x (hdx)
2
0
b
I y = I y ' = ∫ x 2 dA
A
h
h
= 4 ∫ x 2 ( dx)
2
0
(hx 3 )
=
3 0
h x3
= 4( )
2 3
hb 3
=
3
hb 3
=
12
b/2
0
7
y
bh 3
Ix =
12
h/2
h/2
b
bh 3
x Ix =
3
I x = I x + Ad 2
bh 3
h
=
+ (bh)( ) 2
12
2
bh 3 bh 3
=
+
12
4
bh 3
Ix =
3
8
• Moment of Inertia of a Triangular Area.
y
Integrating dIx from y = 0 to y = h, we obtain
I x = ∫ y 2 dA
h
2
dy
h
l
b y 3 y 4 h bh 3
= [h − ]0 =
4
12
h 3
y
x
b/2
dIx = y2 dA
I x = I x + Ad 2
b/2
I x = I x − Ad 2
dA = l dy
bh 3 bh h 2 bh 3
=
− ( )( ) =
12
2 3
36
Using similar triangles, we have
l h− y
=
b
h
l =b
h− y
h
h
h− y
b
=∫y b
dy = ∫ ( hy 2 − y 3 ) dy
h
h0
0
h-y
dA = b
h− y
dy
h
9
Example 9.1
Determine the moment of inertia of the shaded area shown with respect to
each of the coordinate axes.
y
y = kx2
b
x
a
10
• Moment of Inertia Ix.
I x = ∫ y 2 dA
y
A
y = kx2
b
= ∫ y 2 ( a − x) dy
0
dA = (a-x)dy
dy b
a
b
x
= ∫ y 2 (a −
0
a 1/ 2
y )dy
b1/ 2
b
a
= a ∫ y dy − 1/ 2
b
0
2
Substituting x = a and y=b
y = kx 2
b = ka 2
b
k= 2
a
y=
b 2
x
a2
or
b
∫y
5/ 2
dy
0
b
b
ay 3
a 2
=
− 1/ 2 ( y 7 / 2 )
3 0 b
7
0
ab 3
a 2
=
− 1/ 2 ( b 7 / 2 )
3 b
7
x=
a 1/ 2
y
b1/ 2
ab3 2ab 3
=
−
3
7
ab 3
=
21
11
• Moment of Inertia Iy.
b
y = 2 x2
a
y
y = kx2
A
a
= ∫ x 2 ydx
0
dA = ydx
a
I y = ∫ x 2 dA
a
b
dx
x
= ∫ x2 (
0
b 2
x ) dx
2
a
a
b
= 2 ∫ x 4 dx
a 0
a
b x5
= ( 2 )( )
a
5 0
b a5
= ( 2 )( )
a
5
a 3b
=
5
12
Example 9.2
Determine the moment of inertia of the shaded area shown with respect to
each of the coordinate axes.
y
(a,b)
y2 = x2
y1 = x
x
13
• Moment of Inertia Ix.
y
I x = ∫ y 2 dA
(a,b)
A
b
y2 = x2
0
b
dy
y1 = x
= ∫ y 2 ( x 2 − x1 )dy
= ∫ y 2 ( y1/ 2 − y )dy
dA = (x2 - x1)dy
0
x
b
= ∫(y
0
b
5/ 2
) dy − ∫ ( y 3 ) dy
0
b
=
4 b
2 7/2
y
y
−
7
4
0
0
2 7 / 2 b4
= b −
7
4
14
• Moment of Inertia Iy.
y
I y = ∫ x 2 dA
(a,b)
A
dx
a
y2 = x2
dA = (y1 - y2)dx
y1 = x
= ∫ x 2 ( y1 − y 2 )dx
0
a
= ∫ x 2 ( x − x 2 )dx
0
a
x
a
= ∫ ( x )dx − ∫ ( x 4 ) dx
3
0
0
a
x4
x5
=
−
4 0 5
a
0
a 4 a5
=
−
4
5
15
9.5 Moment of Inertia of Composite Areas
A similar theorem can be used with
the polar moment of inertia. The
polar moment of inertia
JO of an area about O and the polar
moment of inertia JC of the area
about its
c
d
o
centroid are related to the distance d between points C and O by the relationship
J O = J C + Ad 2
The parallel-axis theorem is used very effectively to compute the moment of
inertia of a composite area with respect to a given axis.
16
Example 9.3
Compute the moment of inertia of the composite area shown.
100 mm
25 mm
75 mm
75 mm
x
17
SOLUTION
100 mm
25 mm
75 mm
75 mm = (dy)Cir
x
bh 3
2
Ix = (
) Re ct − ( I x + Ad y ) Cir
3
1
1
= [ (100)(150) 3 ]Re ct − [ π ( 25) 4 + (π × 252 )(75) 2 ]Cir
3
4
= 101x106 mm4
18
Example 9.4
Determine the moments of inertia of the beam’s cross-sectional area
shown about the x and y centroidal axes.
100
y
400
x
100
C
600
400
100
Dimension in mm
19
SOLUTION
y
100
A
dyA
400
100
B
C dyD
x
400
D
600
100
Dimension in mm
0
2
2
2
I x = ( I x + Ad y ) A + ( I x + Ad y ) B + ( I x + Ad y ) C
1
1
= [ (100)(300) 3 + (100 × 300)(200) 2 ] + [ (600)(100) 3 + 0]
12
12
1
+ [ (100)(300) 3 + (100 × 300)(200) 2 ]
12
= 2.9x109 mm4
20
y
100
A
400
dxA
x
100
C
600
dxD
400
D
100
Dimension in mm
0
2
2
2
I y = ( I y + Ad x ) A + ( I y + Ad x ) B + ( I y + Ad x ) C
1
1
= [ (300)(100) 3 + (100 × 300)(250) 2 ] A + [ (100)(600) 3 + 0]B
12
12
1
+ [ (300)(100) 3 + (100 × 300)(250) 2 ]C
12
= 5.6x109 mm4
21
Example 9.5 (Problem 9.31,33)
Determine the moments of inertia and the radius of gyration of the
shaded area with respect to the x and y axes.
y
12 mm 12 mm
6 mm
8 mm
24 mm
x
O
24 mm
6 mm
24 mm 24 mm
22
0
SOLUTION
2
12 mm 12 mm
8 mm
dyA 24 mm
O
x
B
dyC 24 mm
C
2
1
= [ (24)(6)3 + ( 24 × 6)(27) 2 ] A
12
1
+ [ (8)(48) 3 + 0]B
12
1
+ [ ( 48)(6) 3 + (48 × 6)(27) 2 ]C
12
6 mm
A
2
I x = ( I x + Ad y ) A + ( I x + Ad y ) B + ( I x + Ad y ) C
y
6 mm
Ix = 390x103 mm4
24 mm 24 mm
Ix
390 × 103
=
= 21.9 mm
[(24 × 6) + (8 × 48) + (48 × 6)]
A
kx =
0
0
0
2
2
2
I y = ( I y + Ad x ) A + ( I y + Ad x ) B + ( I y + Ad x ) C
1
1
1
= [ (6)(24) 3 ] A + [ ( 48)(8) 3 ]B + [ (6)(48) 3 ]C
12
12
12
Iy = 64.3x103 mm4
ky =
Iy
64.3 × 103
=
= 8.87 mm
[(24 × 6) + (8 × 48) + ( 48 × 6)]
A
23
Example 9.6 (Problem 9.32,34)
Determine the moments of inertia and the radius of gyration of the
shaded area with respect to the x and y axes.
y
0.5 m
2m
2m
0.5 m
2m
O
1m
1m
x
1m
1m
0.5 m
0.5 m
24
y
0.5 m
2m
B
A O
C
2m
dyB
0
0.5 m
2m
1m
dyC
1m
1m
1m
0.5 m
2
0.5 m
2
2
I x = ( I x + Ad y ) A5×6 − ( I x + Ad y ) B 4×2 − ( I x + Ad y ) C 4×1
1
1
(5)(6) 3 + 0] A − [ (4)(2) 3 + ( 2 × 4)(2) 2 ]B
12
12
1
− [ ( 4)(1) 3 + (4 × 1)(1.5) 2 ]C
12
=[
x
Ix = 46 m4
kx =
Ix
46
=
= 1.599 m
A
[(5 × 6) − ( 4 × 2) − (4 × 1)]
0
0
0
2
2
2
I y = ( I y + Ad x ) A − ( I y + Ad x ) B − ( I y + Ad x ) C
=[
1
1
1
(6)(5) 3 ] A − [ ( 2)(4) 3 ]B − [ (1)(4) 3 ]C
12
12
12
Iy = 46.5 m4
ky =
Iy
A
=
46.5
= 1.607 m
[(5 × 6) − ( 4 × 2) − (4 × 1)]
25
Example 9.7
Determine the moments of inertia and the radius of gyration of the
shaded area with respect to the x and y axes and at the centroidal axes.
y
1 cm
1 cm
5 cm
1 cm
x
5 cm
26
y
1 cm
• Moments of inertia about centroid
1 cm
I x = I x − Ad y
2
= 145 − (15)(2.5) 2
2
= 51.25 cm 4
CG
5 cm
OR
3.5
1 cm
Y
0.5
x
5 cm
Y ∑ A = ∑ yA
2[(3.5)(5 ×1)] + (0.5)(1× 5)
3(5 ×1)
= 2.5 cm
Y =
• Moments of inertia about x axis
1
1
I x = 2[( (1)(5) 3 + (5 × 1)(3.5) 2 ] + (5)(1) 3
12
3
= 145 cm 4
1
I x = 2[( (1)(5) 3 + (5 ×1)(1) 2 ]
12
1
+ [( (5)(1) 3 + (5 ×1)(2) 2 ]
12
= 51.25 cm 4
1
1
I y = I y = 2[( (5)(1) 3 + (5 × 1)(2) 2 ] + (1)(5) 3
12
12
= 51.25 cm 4
kx = ky =
Ix
51.25
=
= 1.848 cm
15
A
27
Example 9.8
The strength of a W360 x 57 rolled-steel beam is increased by attaching a
229 mm x 19 mm 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.
229 mm
19 mm
C
358 mm
172 mm
28
• Moment of Inertia
SOLUTION
19 mm
I x ' = ( I x ' ) plate + ( I x ' ) wide − flange
229 mm
= ( I x ' + Ad 2 ) plate + ( I x ' + AY 2 ) wide − flange
d
C
358 mm
Y
O
1

=  ( 229)(19) 3 + ( 4351)(188.5 − 70.8) 2 
12

+ 160.2 ×106 + (7230)(70.8) 2
x´ 188.5 mm
x
[
]
= 256.8 ×106 mm 4
I x ' = 257 × 10 6 mm 4
172 mm
• Centroid
The wide-flange shape of W360 x 57
found by referring to Fig. 9.13
I x = 160.2 mm 4
A = 7230 mm2
Aplate = (229)(19) = 4351 mm2
Y ΣA = Σy A
• Radius of Gyration
I x'
256.8 × 10 6
k =
=
A (4351 + 7230)
2
x'
k x ' = 149 mm
Y (4351 + 7230) = (188.5)(4351) + (0)(7230)
Y = 70.8 mm
29
9.6 Polar Moment of Inertia
y
The polar moment of inertia of
an area A with respect to the pole
O is defined as
dA
r
O
y
x
x
J O = ∫ r 2 dA
A
The distance from O to the element of area dA is r. Observing that r 2 =x 2 + y 2 , we
established the relation
JO = I x + I y
30
Example 9.9
(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.
y
r
O
x
31
SOLUTION
a. Polar Moment of Inertia.
y
dA = 2πu du
dJ O = u 2 dA
r
du
u
O
r
x
r
J O = ∫ dJ O = ∫ u ( 2πu du ) = 2π ∫ u 3 du
2
0
JO =
π
2
0
r4
b. Moment of Inertia with Respect to a Diameter.
J O = I x + I y = 2I x
π
2
r 4 = 2I x
I diameter = I x =
π
4
r4
32