Embed Size (px)
Citation preview
1
CHAPTER 9: Moments of Inertia! Moment of Inertia of Areas
! Second Moment, or Moment of Inertia, of anArea
! 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
2
9.1 Moment of Inertia: Definition
x
y
y
xdA=(dx)(dy)
∫=A
x dAyI 2)(
O
∫=A
y dAxI 2)(
3
x
y
x´= Centroidal axis
y´ = Centroidal axis
CG
dy
y´
dx x´dA
∫ +=A
yx dAdyI 2)'(
∫ ++=A
yy dAddyy ])())('(2)'[( 22
∫∫∫ ++=A
yA
yA
dAddAdydAy 22 )())('(2)'(
∫∫ ++=A
yA
yx dAddAydI 2'2
0, y´ = 0
AdII yxx20++=
AdII xyy20++=
9.2 Parallel-Axis Theorem of an Area
2AdJJ CO +=
O
4
y
x
kx
O
A
AJk
AI
kAIk O
Oy
yx
x ===
9.3 Radius of Gyration of an Area
2
The radius of gyration of an areaA with respect to the x axis isdefined as the distance kx, whereIx = kx A. With similar definitions forthe radii of gyration of A withrespect to the y axis and withrespect to O, we have
5
The rectangular moments of inertia Ixand Iy of an area are defined as
These computations are reduced to single integrations by choosing dA to be a thinstrip parallel to one of the coordinate axes. The result is
x
y
y
dx
x∫∫ == dAxIdAyI yx
22
dxyxdIdxydI yx23
31
==
9.4 Determination of the Moment of Inertia of anArea by Integration
6
x´
y´
b/2
h/2
� Moment of Inertia of a Rectangular Area.
x
y
b
hy
dy
∫=A
x dAyI 2
∫=h
bdyy0
2 )(
hby
0
3
3)(
=
ydy
∫==A
xx dAyII 2'
∫=h
dyby0
2 )2
(4
2/
0
3
3)
2(4
hyb
=
3
3bh=
12
3bh=
dA = bdydA = (b/2)dy
7
x´
y´
b/2
h/2
x
y b
h
∫=A
y dAxI 2
∫=b
hdxx0
2 )(
bhx
0
3
3)(
=
∫==A
yy dAxII 2'
∫=h
dxhx0
2 )2
(4
2/
0
3
3)
2(4
bxh
=
3
3hb=
12
3hb=
x dx
xdx
dA = hdxdA = (h/2)dx
8
x
y
b
h/2
h/212
3bhIx =
3
3bhIx =
2AdII xx +=
23
)2
)((12
hbhbh+=
412
33 bhbh+=
3
3bhIx =
9
dIx = y2 dA dA = l dy
Using similar triangles, we have
dyh
yhbdAh
yhblh
yhbl −
=−
=−
=
Integrating dIx from y = 0 to y = h, we obtain
∫∫
∫−=
−=
=hh
x
dyyhyhbdy
hyhby
dAyI
0
32
0
2
2
)(
12]
43[
3
0
43 bhyyhhb h =−=
� Moment of Inertia of a Triangular Area.
2AdII xx +=
36)
3)(
2(
12
32
3
2
bhhbhbhAdII xx
=−=
−=
y
x
h
b/2
h-y
b/2
dy
yl
10
Example 9.1
Determine the moment of inertia of the shaded area shown with respect toeach of the coordinate axes.
x
y
a
y = kx2
b
11
x
y
a
y = kx2
b
� Moment of Inertia Ix.
dydA = (a-x)dy
2kxy =2kab =
2abk =
2/12/1
22 y
baxorx
aby ==
Substituting x = a and y=b
∫=A
x dAyI 2
∫ −=b
dyxay0
2 )(
∫ −=b
dyybaay
0
2/12/1
2 )(
dyybadyya
bb
∫∫ −=0
2/52/1
0
2
bb
ybaay
0
2/72/1
0
3
)72(
3−=
)72(
32/7
2/1
3
bbaab
−=
72
3
33 abab−=
21
3ab=
12
x
y
a
y = kx2
b
� Moment of Inertia Iy.
22 x
aby = ∫=
Ay dAxI 2
∫=a
ydxx0
2
dx
dA = ydx
∫=a
dxxabx
0
22
2 )(
∫=a
dxxab
0
42
ax
ab
0
5
2 )5
)((=
)5
)((5
2
aab
=
5
3ba=
13
Example 9.2
Determine the moment of inertia of the shaded area shown with respect toeach of the coordinate axes.
x
y
y2 = x2
y1 = x
(a,b)
14
x
y
y2 = x2
y1 = x
(a,b)
dy
dA = (x2 - x1)dy
∫=A
x dAyI 2
∫ −=b
dyxxy0
2 )( 12
∫ −=b
dyyyy0
2/12 )(
∫∫ −=bb
dyydyy0
3
0
2/5 )()(
bb yy0
4
0
2/7
472
−=
� Moment of Inertia Ix.
472 4
2/7 bb −=
15
x
y
y2 = x2
y1 = x
(a,b)
� Moment of Inertia Iy.
∫=A
y dAxI 2
∫ −=a
dxyyx0
12 )( 2
dx
dA = (y1 - y2)dx ∫ −=a
dxxxx0
22 )(
∫∫ −=aa
dxxdxx0
4
0
3 )()(
aaxx
0
5
0
4
54−=
54
54 aa−=
16
The parallel-axis theorem is used very effectively to compute the moment ofinertia of a composite area with respect to a given axis.
d
c
ocentroid are related to the distance d between points C and O by the relationship
2AdJJ CO +=
9.5 Moment of Inertia of Composite Areas
A similar theorem can be used withthe polar moment of inertia. Thepolar moment of inertiaJO of an area about O and the polarmoment of inertia JC of the areaabout its
17
Example 9.3
Compute the moment of inertia of the composite area shown.
75 mm
75 mm
100 mm
25 mm
x
18
SOLUTION
75 mm
75 mm
100 mm
x
25 mm
= (dy)Cir
Ciryxctx AdIbhI )()3
( 2Re
3
+−=
Circt ])75)(25()25(41[])150)(100(
31[ 224
Re3 ×+−= ππ
= 101x106 mm4
19
Example 9.4
Determine the moments of inertia of the beam�s cross-sectional areashown about the x and y centroidal axes.
400
Dimension in mm
100
400
100 600
100
x
y
C
20
400
Dimension in mm
100
400
100 600
100
x
y
C
SOLUTION
AdyA
BdyD
D
CyxByxAyxx AdIAdIAdII )()()( 222 +++++=
])200)(300100()300)(100(121[
]0)100)(600(121[])200)(300100()300)(100(
121[
23
323
×++
++×+=
= 2.9x109 mm4
0
21
400
Dimension in mm
100
400
100 600
100
x
y
C
CxyBxyAxyy AdIAdIAdII )()()( 222 +++++=
C
BA
])250)(300100()100)(300(121[
]0)600)(100(121[])250)(300100()100)(300(
121[
23
323
×++
++×+=
= 5.6x109 mm4
AdxA
dxDD
0
22
Example 9.5 (Problem 9.31,33)
Determine the moments of inertia and the radius of gyration of theshaded area with respect to the x and y axes.
6 mm
O x
y
24 mm
24 mm
6 mm
12 mm12 mm
24 mm 24 mm
8 mm
23
6 mm
O x
y
24 mm
24 mm
6 mm
12 mm12 mm
24 mm 24 mm
8 mm
SOLUTION
A
dyA
B
Ix = 390x103 mm4
mmAIk x
x 9.21)]648()488()624[(
10390 3
=×+×+×
×==
CyxByxAyxx AdIAdIAdII )()()( 222 +++++=
Iy = 64.3x103 mm4
CBA ])48)(6(121[])8)(48(
121[])24)(6(
121[ 333 ++=
0 00
CxyBxyAxyy AdIAdIAdII )()()( 222 +++++=
mmAI
k yy 87.8
)]648()488()624[(103.64 3
=×+×+×
×==
0
C
B
A
])27)(648()6)(48(121[
]0)48)(8(121[
])27)(624()6)(24(121[
23
3
23
×++
++
×+=
C
dyC
24
Example 9.6 (Problem 9.32,34)
Determine the moments of inertia and the radius of gyration of theshaded area with respect to the x and y axes.
2 m
O1 m1 m1 m1 m
0.5 m0.5 m 2 m 2 m
0.5 m0.5 m
x
y
25
2 m
O1 m1 m1 m1 m
0.5 m0.5 m 2 m 2 m
0.5 m0.5 m
x
y
A
BdyB
CdyC
Ix = 46 m4
mAIk x
x 599.1)]14()24()65[(
46=
×−×−×==
142
242
652 )()()( ××× +−+−+= CyxByxAyxx AdIAdIAdII
0
C
BA
])5.1)(14()1)(4(121[
])2)(42()2)(4(121[]0)6)(5(
121[
23
233
×+−
×+−+=
Iy = 46.5 m4
CBA ])4)(1(121[])4)(2(
121[])5)(6(
121[ 333 −−=
0 00
CxyBxyAxyy AdIAdIAdII )()()( 222 +−+−+=
mAI
k yy 607.1
)]14()24()65[(5.46
=×−×−×
==
26
Example 9.7
Determine the moments of inertia and the radius of gyration of theshaded area with respect to the x and y axes and at the centroidal axes.
1 cm 1 cm
5 cm
1 cm
5 cm
x
y
27
1 cm 1 cm
5 cm
1 cm
5 cm
x
y
CG
Y
∑∑ = AyAY
cm
Y
5.2)15(3
)51)(5.0()]15)(5.3[(2
=×
×+×=
4
2
2
25.51)5.2)(15(145
cm
AdII yxx
=
−=
−=
� Moments of inertia about centroid
4
23
23
25.51
])2)(15()1)(5(121[(
])1)(15()5)(1(121[(2
cm
I
OR
x
=
×++
×+=
cmAIkk x
yx 848.115
25.51====
4
323
145
)1)(5(31])5.3)(15()5)(1(
121[(2
cm
Ix
=
+×+=
� Moments of inertia about x axis4
323
25.51
)5)(1(121])2)(15()1)(5(
121[(2
cm
II yy
=
+×+==
0.5
3.5
2
28
Example 9.8
The strength of a W360 x 57 rolled-steel beam is increased by attaching a229 mm x 19 mm plate to its upper flange as shown. Determine themoment of inertia and the radius of gyration of the composite section withrespect to an axis which is parallel to the plate and passes through thecentroid C of the section.
C
229 mm
19 mm
358 mm
172 mm
29
229 mm
172 mm
SOLUTION19 mm
358 mm xO
C x´Y
188.5 mm
The wide-flange shape of W360 x 57found by referring to Fig. 9.13A = 7230 mm2 42.160 mmI x =
AyAY Σ=Σ
Aplate = (229)(19) = 4351 mm2
)7230)(0()4351)(5.188()72304351( +=+Y
mmY 8.70=
� Centroid
� Moment of Inertia
flangewidexplatexx III −+= )()( '''
flangewidexplatex YAIAdI −+++= )()( 2'
2'
[ ]46
26
23
108.256)8.70)(7230(102.160
)8.705.188)(4351()19)(229(121
mm×=
+×+
−+=
� Radius of Gyration
)72304351(108.256 6
'2' +
×==
AIk x
x
mmkx 149' =
46' 10257 mmI x ×=
d
30
The polar moment of inertia ofan area A with respect to the poleO is defined as
The distance from O to the element of area dA is r. Observing that r 2 =x 2 + y 2 , weestablished the relation
y
xx
yr
A
dA
O ∫= dArJO2
yxO IIJ +=
9.6 Polar Moment of Inertia
31
Example 9.9
(a) Determine the centroidal polar moment of inertia of a circular area bydirect integration. (b) Using the result of part a, determine the moment ofinertia of a circular area with respect to a diameter.
y
x
r
O
32
udu
O
y
x
r
SOLUTIONa. Polar Moment of Inertia.
duudAdAudJO π22 ==
∫∫ ∫ ===rr
OO duuduuudJJ0
3
0
2 2)2( ππ
4
2rJO
π=
b. Moment of Inertia with Respect to a Diameter.
xyxO IIIJ 2=+=
xIr 22
4 =π
4
4rII xdiameter
π==
