CHAPTER 93
PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
By observation
Now
h
x
b
dI y = x 2 dA = x 2 [( h - y )dx ]
y=
xˆ
Ê
= hx 2 Á1 - ˜ dx
Ë b¯
Then
b
xˆ
Ê
I y = Ú dI y = Ú hx 2 Á1 - ˜ dx
0
Ë b¯
b
È1
x4 ˘
= h Í x3 - ˙ 4b ˚ 0
Î3
or
Iy =
1 3
bhb
12
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3
PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
At x = a,
Then
Now
y = a:
a=
k
a
or k = a2
a2
x
dIy = x 2 dA = x 2 ( y dx )
y=
Ê a2 ˆ
= x 2 Á dx˜ = a2 x dx
Ë x ¯
Then
Iy = Ú dI y = Ú
2a 2
a x dx
a
2a
a2
È1 ˘
= a2 Í x 2 ˙ = [(2a)2 - ( a)2 ]
2
Î2 ˚ a
or
Iy =
3 4
a b
2
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4
PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
y = kx 2
For x = a :
b = ka2
k=
Thus:
y=
b
a2
b
2
x2
a
dA = (b - y )dx
b ˆ
Ê
dIy = x 2 dA = x 2 (b - y )dx = x 2 Á b - 2 x 2 ˜ dx
¯
Ë
a
aÊ
1
1
b ˆ
Iy = Ú dI y = Ú Á bx 2 - 2 x 4 ˜ dx = a3b - a3b 0Ë
¯
3
5
a
Iy =
2 3
abb
15
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5
PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
Ê x x2 ˆ
y = 4h Á - 2 ˜
Ëa a ¯
dA = y dx
Ê x x2 ˆ
dI y = x 2 dA = 4hx 2 Á - 2 ˜ dx
Ëa a ¯
a Ê x3
x4 ˆ
I y = 4hÚ Á - 2 ˜ dx
0Ë a
a ¯
a
Ê a3 a3 ˆ
È x4
x5 ˘
I y = 4h Í - 2 ˙ = 4h Á - ˜ 5¯
Ë 4
Î 4 a 5a ˚ 0
1
I y = ha3 b
5
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6
PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
By observation
or
Now
h
x
b
b
x= y
h
dI x = y 2 dA = y 2 ( x dy )
y=
ˆ
Êb
= y 2 Á y dy˜
¯
Ëh
=
Then
b 3
y dy
h
I x = Ú dI x = Ú
hb
0
h
y 3 dy
h
b È1 ˘
= Í y4 ˙ h Î 4 ˚0
or
Ix =
1 3
bh b
4
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7
PROBLEM 9.6
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
At x = a,
Then
y = a:
a=
k
a
y=
a2
x
or k = a2
3
Now
1
1 Ê a2 ˆ
dI x = y 3 dx = Á ˜ dx
3
3Ë x ¯
=
Then
1 a6
dx
3 x3
I x = Ú dI x = Ú
a
2a
a6
1 È 1 1˘
dx = a6 Í3
3x
3 Î 2 x 2 ˙˚ a
a1
1 È 1
1 ˘
= - a6 Í
˙
2
6 Î ( 2a)
( a)2 ˚
or
1
I x = a4 b
8
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8
PROBLEM 9.7
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
See figure of solution of Problem 9.3.
b
x
a2
1
1
1
1 b3 6
dI x = b3 dx - y 3 dx = b3 dx x dx
3
3
3
3 a6
y=
aÊ 1
1 b3 6 ˆ
I x = Ú dI x = Ú Á b3 x dx
0 Ë3
3 a6 ˜¯
1
1 b3 a 7 Ê 1 1 ˆ 3
= b3 a = Á - ˜ ab
3
3 a6 7 Ë 3 21¯
6
1ˆ
Ê 7
= Á - ˜ ab3 = ab3
Ë 21 21¯
21
Ix =
2 3
ab b
7
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9
PROBLEM 9.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
See figure of solution of Problem 9.4.
y = h1 + ( h2 - h1 )
dI x =
x
a
1 3
y dx
3
I x = Ú dI y =
1
=
12
3
x˘
1 aÈ
h1 + ( h2 - h1 ) ˙ dx
Ú
Í
0
a˚
3 Î
a
4
x˘ Ê a ˆ
È
h
(
h
h
)
+
2
1
ÍÎ 1
a ˙˚ ÁË h2 - h1 ˜¯
0
(
)
2
2
a
a h2 + h1 ( h2 + h1 )( h2 - h1 )
4
4
=
h2 - h1 = ◊
h2 - h1
12( h2 - h1 )
12
(
)
Ix =
(
)
a 2
h1 + h22 ( h1 + h2 ) b
12
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10
PROBLEM 9.9
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
At x = 0, y = b :
b = c(1 - 0)
or
c=b
At x = a, y = 0 :
0 = b(1 - ka1/ 2 )
or k =
Then
Ê
x1/ 2 ˆ
y = b Á1 - 1/ 2 ˜
Ë a ¯
1
a
1/ 2
3
Now
x1/ 2 ˆ ˘
1
1È Ê
dI x = y 3 dx = Íb Á1 - 1/ 2 ˜ ˙ dx
3
3 ÍÎ Ë a ¯ ˙˚
or
1 Ê
x1/ 2
x x 3/ 2 ˆ
dI x = b3 Á1 - 3 1/ 2 + 3 - 3/ 2 ˜ dx
3 Ë
a a ¯
a
Then
I x = Ú dI x = 2Ú
a1 3Ê
x1/ 2
b Á1 - 3 1/ 2
0 3
a
Ë
x x 3/ 2 ˆ
+ 3 - 3/ 2 ˜ dx
a a ¯
a
2 È
x 3/ 2 3 x 2 2 x 5 / 2 ˘
= b3 Í x - 2 1/ 2 +
˙ 3 Î
2 a 5 a3 / 2 ˚ 0
a
or
Ix =
1 3
ab b
15
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11
PROBLEM 9.10
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
At x = a, y = b :
Then
Now
Then
b = ke a /a
or k =
b
e
b x /a
e = be x /a -1
e
1 3
1
dI x = y dx = (be x /a -1 )3 dx
3
3
1 3 3( x /a -1)
= be
dx
3
a
a 1 3 3( x /a -1)
b3 È a 3( x /a -1) ˘
I x = Ú dI x = Ú b e
dx = Í e
˙˚
0 3
3 Î3
0
1 3
= ab (1 - e -3 ) or
9
y=
I x = 0.1056ab3 b
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12
PROBLEM 9.11
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
At x = 3a,
or
Then
Now
y = b:
b = k (3a - a)3
k=
b
8a3
b
y = 3 ( x - a )3
8a
1 3
dI x = y dx
3
3
1È b
˘
= Í 3 ( x - a)3 ˙ dx
3 Î 8a
˚
b3
( x - a)9 dx
1536a9
3a
3a
b3
b3 È 1
9
10 ˘
I x = Ú dI x = Ú
( x - a) dx =
( x - a) ˙
a 1536a9
1536a9 ÍÎ10
˚a
=
Then
=
b3
[(3a - a)10 - 0]
15, 360a9
or
Ix =
1 3
ab b
15
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13
PROBLEM 9.12
Determine by direct integration the moment of inertia of the shaded
area with respect to the y axis.
SOLUTION
At x = 0,
y = b:
b = c(1 - 0)
or c = b
x = a,
y = 0:
0 = c(1 - ka1/ 2 )
or k =
1
a
1/ 2
Ê
x1/ 2 ˆ
y = b Á1 - 1/ 2 ˜ b
Ë a ¯
Then
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14
PROBLEM 9.13
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
At x = a,
b = ke a /a
y = b:
b
e
b
y = e x /a = be x /a -1
e
k=
or
Then
dI y = x 2 dA = x 2 ( y dx )
Now
= x 2 (be x /a -1dx )
a
I y = Ú dI y = Ú bx 2 e x /a -1dx
Then
0
Now use integration by parts with
dv = e x /a -1dx
u = x2
v = ae x /a -1
du = 2 x dx
Then
a 2 x /a -1
Ú0 x e
a
a
0
0
dx = ÈÎ x 2 ae x /a -1 ˘˚ - Ú ( ae x /a -1 )2 x dx
a
= a3 - 2a Ú xe x /a -1dx
0
Using integration by parts with
Then
u=x
dv = e x /a -1dx
du = dx
v = ae x /a -1
{
}
a
I y = b a3 - 2a ÈÍ( xae x /a -1 ) |0a - Ú ( ae x /a -1 )dx ˘˙
0
Î
˚
{
= b {a
}
)˘˚} = b a3 - 2a ÈÎa2 - ( a2 e x /a -1 ) |0a ˘˚
3
- 2a ÈÎa2 - ( a2 - a2 e -1
or
I y = 0.264a3b b
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15
PROBLEM 9.14
Determine by direct integration the moment of inertia of the shaded
area with respect to the y axis.
SOLUTION
At x = 3a,
y = b:
b = k (3a - a)3
or
k=
Then
y=
Now
Then
b
8a3
b
8a3
( x - a )3
È b
˘
dI y = x 2 dA = x 2 ( y dx ) = x 2 Í 3 ( x - a)3 dx ˙
Î 8a
˚
b 2 3
= 3 x ( x - 3x 2 a + 3xa2 - a3 )dx
8a
3a b
I y = Ú dI y = Ú
( x 5 - 3x 4 a + 3x 3 a2 - a3 x 2 )dx
a 8a 3
3a
=
b È1 6 3 5 3 2 4 1 3 3 ˘
x - ax + a x - a x ˙
3
5
4
8a3 ÍÎ 6
˚a
=
b ÏÈ 1
3
3
1
˘
(3a)6 - a(3a)5 + a2 (3a) 4 - a3 (3a)3 ˙
3 ÌÍ
5
4
3
8a Ó Î 6
˚
3
3
1
È1
˘¸
- Í ( a)6 - a( a)5 + a2 ( a) 4 - a3 ( a)3 ˙ ˝
5
4
3
Î6
˚˛
or I y = 3.43a3b b
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16
PROBLEM 9.15
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
At x = a,
y1 = y2 = b :
y2 : b = ka3
Then
or
and
or
Now
b
a
b
or k = 3
a
y1 : b = ma or m =
b
x
a
a
x1 = y
b
b
y2 = 3 x 3
a
a
x2 = 1/ 3 y1/ 3
b
dA = ( x2 - x1 )dy
y1 =
a ˆ
Ê a
= Á 1/ 3 y1/ 3 - y˜ dy
Ëb
b ¯
Then
b Ê y1/ 3
1
A = Ú dA = 2Ú a Á 1/ 3 0 Ëb
b
ˆ
y˜ dy
¯
b
1
1 2˘
È3 1
= 2a Í 1/ 3 y 4 / 3 y ˙ = ab
2b ˚ 0 2
Î4 b
Now
a
ÈÊ a
dI x = y 2 dA = y 2 ÍÁ 1/ 3 y1/ 3 b
ÎË b
ˆ ˘
y˜ dy ˙
¯ ˚
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17
PROBLEM 9.15 (Continued)
Then
b Ê 1
1 ˆ
I x = 2Ú a Á 1/ 3 y 7 / 3 - y 3 ˜ dy
0 Ëb
b ¯
b
È 3 1 10 / 3 1 4 ˘
= 2a Í
y
y 4b ˙˚0
Î10 b1/ 3
and
k x2 =
Ix
=
A
3
1
10 ab
1
2 ab
1
= b2 5
or
or
Ix =
1 3
ab b
10
kx =
b
b
5
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18
PROBLEM 9.16
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
y1 = k1 x 2
For
Thus,
y2 = k2 x1/ 2
x = 0 and
b = k1a2
b
k1 = 2
a
y1 = y2 = b
b = k2 a1/ 2
b
k2 = 1/ 2
a
b 2
b
x
y2 = 1/ 2 x1/ 2
a2
a
dA = ( y2 - y1 )dx
y1 =
aÈ b
b ˘
A = Ú Í 1/ 2 x1/ 2 - 2 x 2 ˙ dx
0 Îa
a
˚
2 ba3/ 2 ba3
3 a1/ 2 3a2
1
A = ab
3
A=
1 3
1
y2 dx - y13 dx
3
3
3
1 b
1 b3 6
x 3/ 2 dx x dx
=
3/ 2
3a
3 a6
dI x =
b3 a 3 / 2
b3 a 6
x
dx
Ú
Ú x dx
3a3/ 2 0
3a6 0
b3 a5 / 2 b3 a 7 Ê 2 1 ˆ 3
= Á - ˜ ab = 3/ 2 5 - 6
3a
3a 7 Ë 15 21¯
2
I x = Ú dI x =
k x2
I
= x =
A
(
()
3
3
35 ab
ab
b
)
Ix =
3 3
ab b
35
kx = b
9
b
35
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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19
PROBLEM 9.17
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
At x = a,
Then
Now
Then
y1 = y2 = b :
b
a3
b
y2 : b = ma or m =
a
b 3
y1 = 3 x
a
b
y2 = x
a
y1 : b = ka3
or k =
b ˆ
Êb
dA = ( y2 - y1 )dx = Á x - 3 x 3 ˜ dx
¯
Ëa
a
A = Ú dA = 2Ú
a
0
bÊ
1 3ˆ
ÁË x - 2 x ˜¯ dx
a
a
a
b È1
1
1
˘
= 2 Í x 2 - 2 x 4 ˙ = ab
a Î2
4a
˚0 2
Now
Then
b ˆ ˘
ÈÊ b
dI y = x 2 dA = x 2 ÍÁ x - 3 x 3 ˜ dx ˙
¯ ˚
a
ÎË a
I y = Ú dI y = 2Ú
a
0
b 2Ê
1 ˆ
x Á x - 2 x 3 ˜ dx
¯
Ë
a
a
a
b È1
1 1 6˘
= 2 Í x4 x a Î4
6 a2 ˙˚ 0
and
k y2 =
Iy
A
=
1 3
6a b
1
2 ab
1
= a2 3
or
1
I y = a3 b b
6
or k y =
a
b
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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20
PROBLEM 9.18
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 9.16.
1
A = ab
3
dI y = x 2 dA = x 2 ( y2 - y1 )dx
a
b ˆ
b
Ê b
I y = Ú x 2 Á 1/ 2 x1/ 2 - 2 x 2 ˜ dx = 1/ 2
0
¯
Ëa
a
a
Iy =
k y2
=
b b7/2 b a5 Ê 2 1 ˆ 3
◊
- 2◊
= Á - ˜ a b
a1/ 2 72
a 5 Ë 7 5¯
()
Iy
A
(
=
3 3
35 a b
ab
3
)
a 5/ 2
Ú0 x
dx -
b
a2
a 4
Ú0 x
dx
Iy =
3 3
abb
35
ky = a
9
b
35
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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21
PROBLEM 9.19
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
y1 :
At x = 2a,
y = 0:
0 = c sin k (2a)
2ak = p
y2 :
At x = a,
y = h:
At x = a,
y = 2h :
p
( a)
2a
2h = ma - b
At x = 2a,
y = 0:
0 = m(2a) + b
Solving yields
Then
or k =
p
2a
h = c sin
or
c=h
2h
, b = 4h
a
2h
p
y1 = h sin x
y2 = - x + 4 h
2a
a
m=-
=
Now
Then
2h
( - x + 2 a)
a
p ˘
È 2h
dA = ( y2 - y1 )dx = Í ( - x + 2a) - h sin x ˙ dx
2a ˚
Îa
A = Ú dA = Ú
2a
a
p ˘
È2
h Í ( - x + 2a) - sin x ˙ dx
2a ˚
Îa
2a
p ˘
2a
È 1
= h Í- ( - x + 2a)2 + cos x ˙
p
2a ˚ a
Î a
2ˆ
ÈÊ 2a ˆ 1
˘
Ê
= h ÍÁ - ˜ + ( - a + 2a)2 ˙ = ah Á1 - ˜
Ë p¯
ÎË p ¯ a
˚
= 0.36338ah
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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22
PROBLEM 9.19 (Continued)
Find:
We have
I x and k x
3
3
1 ˆ
1 ÏÔ È 2h
p ˆ ¸Ô
˘ Ê
Ê1
2
=
+
dI x = Á y2 - y1 ˜ dx
( x a) ˙ Á h sin x˜ ˝ dx
Ì
Ë3
3 ¯
3 ÔÓ ÍÎ a
2a ¯ Ô˛
˚ Ë
=
Then
Now
Then
p ˘
h3 È 8
( - x + 2a)3 - sin3
x
3 ÍÎ a3
2a ˙˚
p ˘
h3 È 8
( - x + 2a)3 - sin3
x dx
3
a 3 Í
2a ˙˚
Îa
sin3 q = sin q (1 - cos2 q ) = sin q - sin q cos2 q
I x = Ú dI x = Ú
2a
Ix =
h3
3
Úa
=
h3
3
2a
p
2a
È 2
˘
4
3 p
ÍÎ- a3 ( - x + 2a) + p cos 2a x - 3p cos 2a x ˙˚
a
8
p
p
3 Ê
2 p ˆ˘
Í 3 ( - x + 2a) - ÁË sin 2a x - sin 2a x cos 2a x˜¯ ˙ dx
Îa
˚
2a È
2a
h3
3
ÈÊ 2a 2a ˆ 2
4˘
ÍÁË - p + 3p ˜¯ + 3 ( - a + 2a) ˙
a
Î
˚
2 3Ê
2ˆ
= ah Á1 - ˜
Ë 3p ¯
3
=
I x = 0.52520ah3 and
k x2 =
I x 0.52520 ah3
=
A
0.36338ah
or
I x = 0.525ah3 b
or
k x = 1.202h b
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23
PROBLEM 9.20
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
y1 :
At x = 2a,
y = 0:
0 = c sin k (2a)
2ak = p
y2 :
At x = a,
y = h:
At x = a,
y = 2h :
p
( a)
2a
2h = ma - b
At x = 2a,
y = 0:
0 = m(2a) + b
Solving yields
Then
Now
Then
or k =
p
2a
h = c sin
or
c=h
2h
, b = 4h
a
2h
p
y1 = h sin x
y2 = - x + 4 h
2a
a
2h
= ( - x + 2a)
a
p ˘
È 2h
dA = ( y2 - y1 )dx = Í ( - x + 2a) - h sin x ˙ dx
2a ˚
Îa
m=-
A = Ú dA = Ú
2a
a
p ˘
È2
h Í ( - x + 2a) - sin x ˙ dx
a
a ˚
2
Î
2a
p ˘
2a
È 1
= h Í- ( - x + 2a)2 + cos x ˙
p
2a ˚ a
Î a
2ˆ
ÈÊ 2a ˆ 1
˘
Ê
= h ÍÁ - ˜ + ( - a + 2a)2 ˙ = ah Á1 - ˜
Ë
¯
Ë
p¯
a
Î p
˚
= 0.36338ah
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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24
PROBLEM 9.20 (Continued)
Find: I y and k y
p ˘ ¸
Ï È 2h
dI y = x 2 dA = x 2 Ì Í ( - x + 2a) - h sin x ˙ dx ˝
2a ˚ ˛
ÓÎ a
p ˘
È2
= h Í ( - x 3 + 2ax 2 ) - x 2 sin x ˙ dx
2a ˚
Îa
We have
I y = Ú dI y = Ú
Then
2a
a
p ˘
È2
h Í ( - x 3 + 2ax 2 ) - x 2 sin x ˙ dx
2a ˚
Îa
Now using integration by parts with
u = x2
du = 2 x dx
Then
Úx
2
sin
Úx
2
sin
Now let
Then
p
x dx
2a
p
2a
v = - cos x
p
2a
dv = sin
p
p ˆ
p ˆ
Ê 2a
Ê 2a
x dx = x 2 Á - cos x˜ - Ú Á - cos x˜ (2 x dx )
Ë
¯
Ë
2a
p
2a
p
2a ¯
p
u=x
dv = cos x dx
2a
p
2a
du = dx
v=
sin x
p
2a
p
2a
p
4 a È Ê 2a
p ˆ
p ˆ ˘
Ê 2a
x dx = - x 2 cos x +
x Á sin x ˜ - Ú Á sin x˜ dx ˙
Í
Ëp
2a
p
2a
p Î Ëp
2a ¯
2a ¯ ˚
2
È2 Ê 1
ˆ
I y = h Í Á - x 4 + ax 3 ˜
¯
Ë
4
3
a
Î
2a
Ê 2a
8a2
16a3
p
p
p ˆ˘
- Á - x 2 cos x + 2 x sin x + 3 cos x˜ ˙
2a
2a
2a ¯ ˙˚
p
p
Ë p
a
2
16a3 Ô¸
ÔÏ 2 È 1
˘ 2a
= h Ì Í - ( 2 a ) 4 + a( 2 a ) 3 ˙ - ( 2 a ) 2 + 3 ˝
3
p ˛Ô
˚ p
ÔÓ a Î 4
2
¸Ô
ÏÔ 2 È 1
2
˘ 8a
- h Ì Í- ( a)4 + a( a)3 ˙ - 2 ( a) ˝
3
˚ p
˛Ô
ÓÔ a Î 4
= 0.61345a3 h
and
k y2 =
Iy
A
=
0.61345a3 h
0.36338ah
or
I y = 0.613a3 h b
or
k y = 1.299a b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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25
PROBLEM 9.21
Determine the polar moment of inertia and the polar radius of gyration of
the shaded area shown with respect to Point P.
SOLUTION
We have
Then
dI x = y 2 dA = y 2 [( x2 - x1 )dy ]
a
2a
I x = 2 ÈÍ Ú y 2 (2a - a)dy + Ú y 2 (2a - 0)dy ˘˙
a
˚
Î 2
a
2
a
ÏÔ È 1 ˘
È 1 ˘ ¸Ô
= 2 Ì a Í y 3 ˙ + 2a Í y 3 ˙ ˝
Î 3 ˚ a ˛Ô
ÓÔ Î 3 ˚0
Ï È1
¸
˘ 2
= 2 Ìa Í ( a)3 ˙ + a ÈÎ(2a)3 - ( a)3 ˘˚ ˝
˚ 3
Ó Î3
˛
= 10 a 4
Also
Then
dI y = x 2 dA = x 2 [( y2 - y1 )dx ]
a
2a
I y = 2 ÈÍ Ú x 2 (2a - a)dx + Ú x 2 (2a - 0)dx ˘˙
a
˚
Î 0
= 10a 4
Now
JP = I x + I y = 10 a 4 + 10 a 4
and
k P2 =
JP
20 a 4
=
A ( 4 a)(2a) - (2a)( a)
10 2
= a 3
or
J P = 20a 4 b
or
k P = 1.826a b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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26
PROBLEM 9.22
Determine the polar moment of inertia and the polar radius of gyration of the
shaded area shown with respect to Point P.
SOLUTION
a ay 1
+
= (a + y)
2 2a 2
1
dA = x dy = ( a + y )dy
2
x=
a
a1
y2
1
1
A = Ú dA = Ú ( a + y )dy = ay +
0
0 2
2
2
2
a
a2 ˆ 3
1Ê
= Á a2 + ˜ = a2 2Ë
2¯ 4
0
A=
3 2
a v
2
1
1
1
I x = Ú y 2 dA = Ú y 2 ( a + y )dy = Ú ( ay 2 + y 3 )dy
0
2
2
2 0
a
1 y3 y 4
= a +
2 3
4
a
a
0
1 7 4
1 Ê 1 1ˆ
a
= Á + ˜ a4 =
2 12
2 Ë 3 4¯
Ix =
7 4
a v
12
Iy =
5 4
a v
16
JO =
43 4
a b
48
3
a1
1
1 aÊ 1
ˆ
I y = Ú x 3 dy = Ú Á ( a + y )˜ dy
0 3
¯
2
3 0 Ë2
1
1 a
1 1
Iy =
( a + y )3 dy =
( a + y )4
Ú
0
2
24
24 4
a
=
0
1
15
È(2a)4 - a 4 ˘ = a 4
Î
˚
96
96
1
5
I y = a4 2
32
From Eq. (9.4):
JO = I x + I y =
J O = kO2 A
7 4 5 4 Ê 28 + 15 ˆ 4
a + a =Á
a Ë 48 ˜¯
12
16
kO2 =
JO
=
A
43 4
48 a
3 2
2a
=
43 2
a
72
kO = a
43
72
kO = 0.773a b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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27
PROBLEM 9.23
(a) Determine by direct integration the polar moment of inertia of the annular
area shown with respect to Point O. (b) Using the result of Part a, determine the
moment of inertia of the given area with respect to the x axis.
SOLUTION
dA = 2p u du
(a)
dJ O = u 2 dA = u 2 (2p u du)
= 2p u3 du
R2
J O = Ú dJ O = 2p Ú u3 du
R1
= 2p
(b)
From Eq. (9.4):
(Note by symmetry.)
1 4
u
4
R2
JO =
p 4
È R2 - R14 ˘ b
˚
2Î
Ix =
p 4
È R2 - R14 ˘ b
˚
4Î
R1
Ix = I y
J O = I x + I y = 2I x
Ix =
p
1
J O = ÈÎ R24 - R14 ˘˚ 2
4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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28
PROBLEM 9.24
(a) Show that the polar radius of gyration kO of the annular area shown is
approximately equal to the mean radius Rm = ( R1 + R2 ) 2 for small values of the
thickness t = R2 - R1 . (b) Determine the percentage error introduced by using Rm
in place of kO for the following values of t/Rm: 1, 12 , and 101 .
SOLUTION
(a)
From Problem 9.23:
p 4
È R2 - R14 ˘
˚
2Î
4
4
p
J O 2 ÈÎ R2 - R1 ˘˚
2
kO =
=
A p R22 - R12
1
kO2 = ÈÎ R22 + R12 ˘˚
2
JO =
(
)
Thickness: t = R2 - R1
Mean radius:
Thus,
1
Rm = ( R1 + R2 )
2
1
1
R1 = Rm - t and R2 = Rm + t
2
2
2
2
1 ÈÊ
1 ˆ
1 ˆ ˘
1
Ê
kO2 = ÍÁ Rm + t ˜ + Á Rm - t ˜ ˙ = Rm2 + t 2
Ë
2 ÍÎË
2 ¯
2 ¯ ˙˚
4
For t small compared to Rm : kO2 ª Rm2 (b)
Percentage error =
kO ª Rm b
(exact value) - (approximation value)
100
(exact value)
P.E. = -
Rm2 + 14 t 2 - Rm
Rm2 + 14 t 2
(100) = -
( ) - 1100
1+ ( )
1 + 14
t
Rm
2
1 t
4 Rm
2
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29
PROBLEM 9.24 (Continued)
For
t
= 1:
Rm
1
t
For
= :
Rm 2
1
t
For
= :
Rm 10
P.E. =
P.E. = -
P.E. =
1 + 14 - 1
(100) = -10.56% b
( 12 )2 - 1 (100) = -2.99%
2
1 + 14 ( 12 )
b
(101 )2 - 1 (100) = -0.125%
2
1 + 14 ( 101 )
b
1 + 14
1 + 14
1 + 14
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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30
PROBLEM 9.25
Detetmine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point P.
SOLUTION
y1 :
At x = 2a,
y = 2a :
2a = k1 (2a)2
y2 :
At x = 0,
At x = 2a,
y = a:
or k1 =
1
2a
a=c
y = 2a :
2 a = a + k2 ( 2 a ) 2
1 2
x
2a
or
y2 = a +
k2 =
1
4a
1 2
x
4a
Then
y1 =
Now
1
( 4a2 + x 2 )
4a
1 2˘
È1
dA = ( y2 - y1 )dx = Í ( 4 a2 + x 2 ) x dx
a
a ˙˚
4
2
Î
1
=
( 4 a2 - x 2 )dx
4a
=
Then
Now
A = Ú dA = 2Ú
2a
0
2a
1
1 È 2
1 ˘
8
( 4 a2 - x 2 )dx =
4 a x - x 3 ˙ = a2
Í
4a
2a Î
3 ˚0
3
3
3
1 ˆ
1 ÔÏ È 1
˘ È 1 ˘ ¸Ô
Ê1
dI x = Á y23 - y13 ˜ dx = Ì Í ( 4a2 + x 2 ) ˙ - Í x 2 ˙ ˝ dx
Ë3
3 ¯
3 ÔÓ Î 4a
˚ Î 2a ˚ Ô˛
1È 1
1
˘
= Í
(64a6 + 48a 4 x 2 + 12a2 x 4 + x 6 ) - 3 x 6 ˙ dx
3
3 Î 64a
8a
˚
1
=
(64a6 + 48a 4 x 2 + 12a2 x 4 - 7 x 6 )dx
3
192a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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31
PROBLEM 9.25 (Continued)
Then
I x = Ú dI x = 2Ú
2a
1
192a3
0
(64a6 + 48a 4 x 2 + 12a2 x 4 - 7 x 6 )dx
2a
È
6
4 3 12 2 5
7˘
ÍÎ64a x + 16a x + 5 a x - x ˙˚
0
12
1 È
˘
=
64 a6 (2a) + 16a 4 (22a)3 + a2 (2a)5 - (2a)7 ˙
5
96a3 ÍÎ
˚
1
12
ˆ 32
Ê
= a 4 Á128 + 128 + ¥ 32 - 128˜ = a 4
¯ 15
96 Ë
5
=
Also
Then
1
96a3
È1
˘
dI y = x 2 dA = x 2 Í ( 4 a2 - x 2 )dx ˙
Î 4a
˚
2a
1 2
1 È4 2 3 1 5˘
I y = Ú dI y = 2Ú
x ( 4a2 - x 2 )dx =
a x - x ˙
0 4a
2a ÍÎ 3
5 ˚0
1
1 È4 2
˘ 32 Ê 1 1 ˆ 32
a ( 2 a )3 - ( 2 a )5 ˙ = a 4 Á - ˜ = a 4
=
Ë 3 5 ¯ 15
5
2a ÍÎ 3
˚ 2
2a
Now
JP = I x + I y =
and
k P2
J
= P =
A
32 4 32 4
a + a 15
15
64 4
15 a
8 2
3a
or J P =
8
= a2 5
64 4
a b
15
or k P = 1.265a b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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32
PROBLEM 9.26
Determine the polar moment of inertia and the polar radius of gyration of the
shaded area shown with respect to Point P.
SOLUTION
The equation of the circle is
x2 + y2 = r2
So that
Now
x = r2 - y2
dA = x dy = r 2 - y 2 dy
r
Then
A = Ú dA = 2Ú
Let
y = r sin q ; dy = r cos q dq
Then
A = 2Ú
= 2Ú
p /2
-p / 6
r 2 - y 2 dy
-r /2
r 2 - ( r sin q )2 r cos dq
p /2
Èq sin 2q ˘
r 2 cos2 q dq = 2r 2 Í +
-p / 6
4 ˙˚ -p / 6
Î2
p /2
È p Ê - p sin - p3 ˆ ˘
Êp
3ˆ
= 2r 2 Í 2 - Á 6 +
˙ = 2r 2 Á +
˜
˜
4 ¯ ˙˚
Ë3 8 ¯
ÍÎ 2 Ë 2
= 2.5274 r 2
Now
dI x = y 2 dA = y 2
Then
I x = Ú dI x = 2Ú
Let
Then
r 2 - y 2 dy
r
-r /2
)
y 2 r 2 - y 2 dy
y = r sin q ; dy = r cos q dq
I x = 2Ú
= 2Ú
Now
(
p /2
-p / 6
p /2
-p / 6
(r sin q )2 r 2 - ( r sin q )2 r cos q dq
r 2 sin 2 q ( r cos q )r cos q dq
sin 2q = 2 sin q cos q fi sin 2 q cos2 q =
1
sin 2q
4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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33
PROBLEM 9.26 (Continued)
Then
Also
Then
Let
Then
Now
Then
I x = 2Ú
dI y =
p /2
r 4 Èq sin 4q ˘
ˆ
Ê1
r 4 Á sin 2 2q ˜ dq = Í -p / 6
¯
Ë4
2 Î2
8 ˙˚ -p / 6
p /2
1 3
1
x dy =
3
3
I y = Ú dI y = 2Ú
(
r2 - y2
È p2 Ê p6 sin - 23p ˆ ˘
˙
Í -Á 8 ˜¯ ˙˚
ÍÎ 2 Ë 2
=
r4
2
=
r4 Ê p
3ˆ
Á
2 Ë 3 16 ˜¯
) dy
3
1 2
( r - y 2 )3/ 2 dy
-r /2 3
r
y = r sin q ; dy = r cos q dq
2 p /2 2
[r - ( r sin q )2 ]3/ 2 r cos q dq
3 Ú -p / 6
2 p /2 3
Iy = Ú
( r cos3 q )r cos q dq
3 -p / 6
1
cos 4 q = cos2 q (1 - sin 2 q ) = cos2 q - sin 2 2q
4
Iy =
Iy =
2 p /2 4 Ê 2
1
ˆ
r Á cos q - sin 2 2q ˜ dq
Ú
6
p
/
¯
Ë
3
4
p /2
=
2 4 ÈÊ q sin 2q ˆ 1 Ê q sin 4q ˆ ˘
r Á +
˜- Á ˜
3 ÍÎË 2
4 ¯ 4Ë2
8 ¯ ˙˚ -p / 6
2 4 ÏÔ È p2 1 Ê p2 ˆ ˘ È - p6 sin - p3 1 Ê - p6 sin - 23p ˆ ˘ ¸Ô
+
- Á
r ÌÍ ˙˝
˙-Í
3 Ô ÍÎ 2 4 ÁË 2 ˜¯ ˙˚ ÍÎ 2
4
4Ë 2
8 ˜¯ ˙˚ Ô
Ó
˛
2 Èp p p 1 Ê 3 ˆ p
1 Ê 3ˆ˘
= r4 Í - + + Á ˜ + Á ˜˙
3 ÍÎ 4 16 12 4 Ë 2 ¯ 48 32 Ë 2 ¯ ˙˚
2 Ê p 9 3ˆ
= r4 Á +
3 Ë 4 64 ˜¯
=
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34
PROBLEM 9.26 (Continued)
Now
JP = Ix + I y =
3ˆ 2 4 Ê p 9 3ˆ
r4 Ê p
+ r
+
Á
2 Ë 3 16 ˜¯ 3 ÁË 4 64 ˜¯
Êp
3ˆ
4
= r4 Á +
˜ = 1.15545r Ë 3 16 ¯
and
k P2 =
J P 1.15545r 4
=
A
2.5274r 2
or
J P = 1.155r 4 b
or
k P = 0.676r b
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35
PROBLEM 9.27
Determine the polar moment of inertia and the polar radius of gyration of the
shaded area shown with respect to Point O.
SOLUTION
At q = p , R = 2a :
2a = a + k (p )
a
p
or
k=
Then
R=a+
a
Ê qˆ
q = a Á1 + ˜
Ë p¯
p
dA = ( dr )( r dq )
= rdr dq
Now
A = Ú dA = Ú
Then
A= Ú
p
0
p
0
a (1+q /p )
Ú0
rdr dq = Ú
p
0
a (1+q /p )
È1 2 ˘
ÍÎ 2 r ˙˚
0
2
3
1 2Ê qˆ
1 Èp Ê q ˆ ˘
a Á1 + ˜ dq = a2 Í Á1 + ˜ ˙
2 ÍÎ 3 Ë p ¯ ˙˚
2 Ë p¯
dq
p
0
ÈÊ p ˆ
˘ 7
1
= p a2 ÍÁ1 + ˜ - (1)3 ˙ = p a2
6
ÍÎË p ¯
˙˚ 6
3
Now
Then
dJ O = r 2 dA = r 2 ( rdr dq )
J O = Ú dJ O = Ú
=Ú
p
0
p
a (1+q /p ) 3
0 Ú0
r dr dq
a(1+q /p )
È1 4 ˘
ÍÎ 4 r ˙˚
0
dq = Ú
p
0
4
1 4Ê qˆ
a Á1 + ˜ dq
4 Ë p¯
p
5
ÈÊ p ˆ 5
˘
1
1 Èp Ê q ˆ ˘
= a 4 Í Á1 + ˜ ˙ = p a 4 ÍÁ1 + ˜ - (1)5 ˙ 20
4 ÍÎ 5 Ë p ¯ ˙˚
ÍÎË p ¯
˙˚
0
and
kO2 =
JO
=
A
4
31
20 p a
2
7
6 pa
=
93 2
a
70
or
JO =
31 4
pa b
20
or kO = 1.153a b
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36
PROBLEM 9.28
Determine the polar moment of inertia and the polar radius of gyration of the
isosceles triangle shown with respect to Point O.
SOLUTION
By observation:
y=
or
x=
h
b
2
x
b
y
2h
Now
Ê b ˆ
dA = xdy = Á
y dy
Ë 2h ˜¯
and
dI x = y 2 dA =
Then
b 3
y dy
2h
I x = Ú dI x = 2Ú
h
0
b 3
y dy
2h
b y4
=
h 4
From above:
Now
and
y=
h
=
0
1 3
bh
4
2h
x
b
2h ˆ
Ê
dA = ( h - y )dx = Á h x˜ dx
Ë
b ¯
h
= (b - 2 x )dx
b
h
dI y = x 2 dA = x 2 (b - 2 x )dx
b
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37
PROBLEM 9.28 (Continued)
I y = Ú dI y = 2Ú
Then
b/2 h
0
b
x 2 (b - 2 x )dx
b/ 2
1 ˘
h È1
= 2 Í bx 3 - x 4 ˙
2 ˚0
b Î3
Now
and
3
4
1 Ê bˆ ˘ 1
h Èb Ê bˆ
= 2 Í Á ˜ - Á ˜ ˙ = b3 h
2 Ë 2 ¯ ˙˚ 48
b ÍÎ 3 Ë 2 ¯
1
1
J O = I x + I y = bh3 + b3 h 4
48
kO2 =
JO
=
A
2
2
bh
48 (12h + b )
1
2 bh
=
1
(12h2 + b2 )
24
or
or
JO =
bh
(12h2 + b2 ) b
48
kO =
12h2 + b2
b
24
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38
PROBLEM 9.29*
Using the polar moment of inertia of the isosceles triangle of Problem 9.28,
show that the centroidal polar moment of inertia of a circular area of radius r is
p r 4 /2. (Hint: As a circular area is divided into an increasing number of equal
circular sectors, what is the approximate shape of each circular sector?)
PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of
gyration of the isosceles triangle shown with respect to Point O.
SOLUTION
First the circular area is divided into an increasing number of identical circular sectors. The sectors can be
approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of the
isosceles triangles are as shown.
For an isosceles triangle (see Problem 9.28):
JO =
b = r Dq
Then with
( D J O )sector ;
=
Now
bh
(12h2 + b2 )
48
dJ O sector
dq
and h = r
1
( r Dq )( r ) ÈÎ12r 2 + ( r Dq )2 ˘˚
48
1 4
r Dq ÈÎ(12 + Dq 2 )˘˚
48
Ê D J O sector ˆ
Ï1
¸
= lim Á
= lim Ì r 4 ÈÎ12 + ( Dq )2 ˘˚ ˝
˜
Dq Æ 0 Ë
D
q
Æ
0
Dq ¯
Ó 48
˛
=
Then
( J O )circle = Ú dJ O sector = Ú
2p
0
1 4
r
4
1 4
1
2p
r dq = r 4 [q ]0
4
4
or
( J O )circle =
p 4
r b
2
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39
PROBLEM 9.30*
Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 / 2p. (Hint: Compare
the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the
same centroid.)
SOLUTION
From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is
( J C )cir =
p 4
r
2
The area of the circle is
Acir = p r 2
A2
2p
Two methods of solution will be presented. However, both methods depend upon the observation that as a
given element of area dA is moved closer to some Point C, The value of J C will be decreased ( J C = Ú r 2 dA; as
r decreases, so must J C ).
[ J C ( A)]cir =
So that
Solution 1
Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its
area is equal to A. To minimize the value of ( J C ) A, the area would have to be distributed as closely as possible
about C. This is accomplished by winding the strip into a tightly wound roll with C as its center; any voids in
the roll would place the corresponding area farther from C than is necessary, thus increasing the value of ( J C ) A.
(The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll is circular,
with the centroid of its area at C, it follows that
( JC ) A ≥
A2
Q.E.D. b
2p
where the equality applies when the original area is circular.
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40
PROBLEM 9.30* (Continued)
Solution 2
Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid) at
Point C. Without loss of generality, assume that
A1 = A2
A3 = A4
It then follows that
( J C ) A = ( J C )cir + [ J C ( A1 ) - J C ( A2 ) + J C ( A3 ) - J C ( A4 )]
Now observe that
J C ( A1 ) - J C ( A2 ) 0
J C ( A3 ) - J C ( A4 ) 0
since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase.
( J C ) A ( J C )cir or ( J C ) A A2
Q.E.D. b
2p
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41
PROBLEM 9.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 + A2 + A3
= [(24)(6) + (8)( 48) + ( 48)(6)] mm2
= (144 + 384 + 288) mm2
= 816 mm2
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
1
(24 mm)(6 mm)3 + (144 mm2 )(27 mm)2
12
= ( 432 + 104,976) mm 4
( I x )1 =
= 105, 408 mm 4
1
(8 mm)(48 mm)3 = 73,728 mm 4
12
1
( I x )3 = ( 48 mm)(6 mm)3 + (288 mm2 )(27 mm)2
12
= (864 + 209, 952) mm 4 = 210, 816 mm 4
( I x )2 =
Then
I x = (105, 408 + 73, 728 + 210, 816) mm 4
= 389, 952 mm 4 and
k x2 =
I x 389, 952 mm 4
=
A
816 mm2
or
I x = 390 ¥ 103 mm 4 b
or
k x = 21.9 mm b
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42
PROBLEM 9.32
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 - A2 - A3
= [(120)(144) - ( 48)(96) - (96)(24)] mm2
= (17280 - 4608 - 2304) mm2
= 10368 mm2
Now
where
I x = ( I x )1 - ( I x )2 - ( I x )3
( I x )1 =
( I x )2 =
1
(120 mm)(144 mm)3 = 29.8594 ¥ 106 mm 4
12
1
(96 mm)(48 mm)3 + ( 4608 mm)2 ( 48 mm)2
12
= 11.501568 ¥ 106 mm 4
1
2
(96 mm)(24 mm)3 + (2304 mm)2 (36 mm )
12
= 3.096576 ¥ 106 mm 4
( I x )3 =
Then
I x = (29.8594 - 11.501568 - 3.096576) ¥ 106 mm 4
=15.261256 ´ 106 mm4
and
k x2 =
I x 15.261256 ¥ 106 mm 4
=
A
10368 mm2
or
I x = 15.26 ¥ 106 mm 4 b
or kx = 38.4 mm b
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43
PROBLEM 9.33
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 + A2 + A3
= [(24 ¥ 6) + (8)( 48) + ( 48)(6)] mm2
= (144 + 384 + 288) mm2
= 816 mm2
Now
I y = ( I y )1 + ( I y )2 + ( I y )3
where
1
(6 mm)(24 mm)3 = 6912 mm 4
12
1
( I y )2 = ( 48 mm)(8 mm)3 = 2048 mm 4
12
1
( I y )3 = (6 mm)(48 mm)3 = 55, 296 mm 4
12
( I y )1 =
Then
I y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4
and
k y2 =
Iy
A
=
64, 256 mm 4
816 mm2
or
I y = 64.3 ¥ 103 mm 4 b
or
k y = 8.87 mm b
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44
PROBLEM 9.34
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 - A2 - A3
= [(120)(144) - ( 48)(96) - (96)(24)] mm2
= (17280 - 4608 - 2304) mm2
= 10368 mm2
Now
where
I y = ( I y )1 - ( I y )2 - ( I y )3
1
(144 mm)(120 mm)3 = 20.736 ¥ 106 mm 4
12
1
( I y )2 = (48 mm)(96 mm)3 = 3.538944 ¥ 106 mm 4
12
1
( I y )3 = (24 mm)(96 mm)3 = 1.769472 ¥ 106 mm 4
12
( I y )1 =
Then
and
I y = (20.736 - 3.53844 - 1.769472) ¥ 106 mm 4
= 15.428088 ¥ 106
k y2 =
Iy
A
=
or
15.428088 ¥ 106 mm 4
10368 mm2
I y = 15.43 ¥ 106 mm 4 b
or k y = 38.6 mm b
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45
PROBLEM 9.35
Determine the moments of inertia of the shaded area shown with respect to the x and
y axes when a = 20 mm.
SOLUTION
We have
where
I x = ( I x )1 + 2( I x )2
1
( 40 mm)(40 mm)3
12
= 213.33 ¥ 103 mm 4
( I x )1 =
2
Èp
p
ˆ ˘
Ê 4 ¥ 20
mm˜ ˙
( I x )2 = Í (20 mm)4 - (20 mm)2 Á
¯ ˙
Ë 3p
2
ÍÎ 8
˚
2
p
ÈÊ 4 ¥ 20
˘
ˆ
+ (20 mm)2 ÍÁ
+ 20˜ mm ˙
¯
Ë
2
3
p
Î
˚
= 527.49 ¥ 103 mm 4
Then
I x = [213.33 + 2(527.49)] ¥ 103 mm 4
Also
where
Then
or
I x = 1.268 ¥ 106 mm 4 b
I y = ( I y )1 + 2( I y )2
1
( 40 mm)(40 mm)3 = 213.33 ¥ 103 mm 4
12
p
( I y )2 = (20 mm)4 = 62.83 ¥ 103 mm 4
8
I y = [213.33 + 2(62.83)] ¥ 103 mm 4
( I y )1 =
or
I y = 339 ¥ 103 mm 4 b
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46
PROBLEM 9.36
Determine the moments of inertia of the shaded area shown with respect
to the x and y axes when a = 20 mm.
SOLUTION
Area = Square - 2(Semicircles)
Given:
For a = 20 mm, we have
Ix = I y =
Square:
Semicircle
:
1
(60) 4 = 1080 ¥ 103 mm 4
12
p
(20) 4 = 62.83 ¥ 103 mm 4
8
p
Êpˆ
(20) 4 = I y ¢ + Á ˜ (20)2 (8.488)2
= I y ¢ + Ad 2 ;
Ë 2¯
8
3
3
= 62.83 ¥ 10 - 45.27 ¥ 10
Ix =
I AA¢
I y¢
I y¢ = 17.56 ¥ 103 mm 4
I y = I y ¢ + A(30 - 8.488)2 = 17.56 ¥ 103 +
p
(20)2 (21.512)2
2
I y = 308.3 ¥ 103 mm 4
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47
PROBLEM 9.36 (Continued)
Semicircle
: Same as semicircle
.
Entire Area:
I x = 1080 ¥ 103 - 2(62.83 ¥ 103 )
= 954.3 ¥ 103 mm 4
I x = 954 ¥ 103 mm 4 b
I y = 463 ¥ 103 mm 4 b
I y = 1080 ¥ 103 - 2(308.3 ¥ 103 )
= 463.3 ¥ 103 mm 4
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48
PROBLEM 9.37
For the 4000-mm2 shaded area shown, determine the distance d2
and the moment of inertia with respect to the centroidal axis
parallel to AA¢ knowing that the moments of inertia with respect to
AA¢ and BB¢ are 12 ¥ 106 mm 4 and 23.9 ¥ 106 mm4, respectively,
and that d1 = 25 mm.
SOLUTION
We have
I AA¢ = I + Ad12 and
I BB¢ = I + Ad22
Subtracting
or
Using Eq. (1):
(
I AA¢ - I BB ¢ = A d12 - d22
(1)
)
I AA¢ - I BB ¢
A
(12 - 23.9)106 mm 4
= (25 mm)2 4000 mm2
= 3600 mm2
d22 = d12 -
I = 12 ¥ 106 mm 4 - ( 4000 mm2 )(25 mm)2 or d2 = 60.0 mm
or
I = 9.50 ¥ 106 mm 4 b
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49
PROBLEM 9.38
Determine for the shaded region the area and the moment of inertia
with respect to the centroidal axis parallel to BB¢, knowing that
d1 = 25 mm and d2 = 15 mm and that the moments of inertia with
respect to AA¢ and BB¢ are 7.84 ¥ 106 mm 4 and 5.20 ¥ 106 mm4,
respectively.
SOLUTION
We have
I AA¢ = I + Ad12 and
I BB¢ = I + Ad22
Subtracting
or
Using Eq. (1):
(
I AA¢ - I BB ¢ = A d12 - d22
A=
(1)
)
I AA¢ - I BB ¢
d12 - d22
=
(7.84 - 5.20)106 mm 4
(25 mm)2 - (15 mm)2
or
A = 6600 mm2 b
I = 7.84 ¥ 106 mm 4 - (6600 mm2 )(25 mm)2
= 3.715 ¥ 106 mm 4
or I = 3.72 ¥ 106 mm 4 b
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50
PROBLEM 9.39
The shaded area is equal to 3.125 ¥ 104 mm2. Determine its centroidal
moments of inertia I x and I y , knowing that I y = 2 I x and that the polar moment
of inertia of the area about Point A is JA = 878.9 ¥ 106 mm 4 .
SOLUTION
Given:
A = 3.125 ¥ 10 4 mm2
I y = 2 I x , J A = 878 ¥ 106 mm 4
J A = J C + A(150 mm)2
878.9 ¥ 106 mm 4 = J C + (3.125 ¥ 10 4 mm2 )(150 mm)2
J C = 175.775 ¥ 106 mm 4
JC = I x + I y
with I y = 2 I x
175.775 ¥ 106 mm 4 = I x + 2 I x I x = 58.6 ¥ 106 mm 4 b
I y = 2 I x = 117.2 ¥ 106 mm 4 b
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51
PROBLEM 9.40
The polar moments of inertia of the shaded area with respect to Points
A, B, and D are, respectively, JA = 1125 ¥106 mm 4, JB = 2625 ¥ 106 mm 4, and
JD = 1781.25 ¥ 106 mm4. Determine the shaded area, its centroidal moment
of inertia JC , and the distance d from C to D.
SOLUTION
See figure at solution of Problem 9.39.
J A = 1125 ¥ 106 mm 4 , J B = 2625 ¥ 106 mm 4 , J D = 1781.25 ¥ 106 mm 4
Given:
Eq. (1) subtracted by Eq. (2):
J B = J C + A(CB)2 ; 2625 ¥ 106 mm 4 = J C + A(1502 + d 2 ) (1)
J D = J C + A(CD )2 ; 1781.25 ¥ 106 mm 4 = J C + Ad 2 (2)
J B - J D = 843.75 ¥ 106 mm 4 = A(150)2 A = 37500 mm2 b
J A = J C + A( AC )2 ; 1125 ¥ 106 mm 4 = J C + (37500 mm2 )(150 mm)2
J C = 281.25 ¥ 1066mm 44
J C = 281.25 ¥ 10 mm
Eq. (2):
1781.25 ¥ 1066 mm 44 = 281.25 ¥ 1066mm 44 + (37500 mm22)d 22
1781.25 ¥ 10 mm = 281.25 ¥ 10 mm + (37500 mm )d J C = 281 ¥ 106 mm 4 b
d = 200 mm b
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52
PROBLEM 9.41
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
Dimensions in mm
First locate centroid C of the area.
Symmetry implies Y = 30 mm.
A, mm2
Then
xA, mm3
1
108 ¥ 60 = 6480
54
349,920
2
1
- ¥ 72 ¥ 36 = -1296
2
46
–59,616
S
5184
290,304
X SA = S xA : X (5184 mm2 ) = 290, 304 mm3
or
X = 56.0 mm
Now
I x = ( I x )1 - ( I x )2
where
x, mm
1
(108 mm)(60 mm)3 = 1.944 ¥ 106 mm 4
12
È1
˘
ˆ
Ê1
( I x )2 = 2 Í (72 mm)(18 mm)3 + Á ¥ 72 mm ¥ 18 mm˜ (6 mm)2 ˙
¯
Ë
36
2
Î
˚
( I x )1 =
= 2(11, 664 + 23, 328) mm 4 = 69.984 ¥ 103 mm 4
[( I x )2 is obtained by dividing A2 into
Then
]
I x = (1.944 - 0.069984) ¥ 106 mm 4
or
I x = 1.874 ¥ 106 mm 4 b
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53
PROBLEM 9.41 (Continued)
Also
where
Then
I y = ( I y )1 - ( I y )2
1
(60 mm)(108 mm)3 + (6480 mm2 )[(56.54) mm]2
12
= (6, 298, 560 + 25, 920) mm 4 = 6.324 ¥ 106 mm 4
1
( I y )2 = (36 mm)(72 mm)3 + (1296 mm2 )[(56 - 46) mm]2
36
= (373, 248 + 129, 600) mm 4 = 0.502 ¥ 106 mm 4
( I y )1 =
I y = (6.324 - 0.502)106 mm 4
or
I y = 5.82 ¥ 106 mm 4 b
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54
PROBLEM 9.42
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate C of the area:
Symmetry implies X = 12 mm.
Then
A, mm2
y, mm
yA, mm3
1
12 ¥ 22 = 264
11
2904
2
1
(24)(18) = 216
2
28
6048
S
480
8952
Y SA = S yA: Y ( 480 mm2 ) = 8952 mm3
Y = 18.65 mm
Now
where
Then
I x = ( I x )1 + ( I x )2
1
(12 mm)(22 mm)3 + (264 mm2 )[(18.65 - 11) mm]2
12
= 26, 098 mm 4
1
( I x )2 = (24 mm)(18 mm)3 + (216 mm2 )[(28 - 18.65) mm]2
36
= 22, 771 mm 4
( I x )1 =
I x = (26.098 + 22.771) ¥ 103 mm 4
or
I x = 48.9 ¥ 103 mm 4 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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55
PROBLEM 9.42 (Continued)
Also
where
I y = ( I y )1 + ( I y )2
1
(22 mm)(12 mm)3 = 3168 mm 4
12
È1
˘
ˆ
Ê1
( I y )2 = 2 Í (18 mm)(12 mm)3 + Á ¥ 18 mm ¥ 12 mm˜ ( 4 mm)2 ˙
¯
Ë2
Î 36
˚
( I y )1 =
= 5184 mm 4
[( I y )2 is obtained by dividing A2 into
Then
]
I y = (3168 + 5184) mm 4 or
I y = 8.35 ¥ 103 mm 4 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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56
PROBLEM 9.43
Determine the moments of inertia I x and I y of the area shown with respect to
centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid C of the area.
A, mm2
Then
x, mm
y, mm
100 ¥ 160 = 16 ¥ 103
50
80
800 ¥ 103
1280 ¥ 103
2
– 100 ¥ 40 = – 4 ¥ 103
38
86
–152 ¥ 103
–344 ¥ 103
S
12 ¥ 103
648 ¥ 103
936 ¥ 103
X SA = S xA: X (12 ¥ 103 mm2 ) = 648 ¥ 103 mm3
X = 54.0 mm
Y SA = S yA: Y (12 ¥ 103 mm2 ) = 936 ¥ 103 mm3
Y = 78.0 mm
or
Now
where
yA, mm3
1
or
and
xA, mm3
I x = ( I x )1 - ( I x )2
1
(100 mm)(160 mm)3 + (16 ¥ 103 mm2 )[(80 - 78) mm]2
12
= (34.133 + 0.064) ¥ 106 mm 4
( I x )1 =
= 34.19733 ¥ 106 mm 4
1
( I x )2 = ( 40 mm)(100 mm)3 + ( 4 ¥ 103 mm2 )[(86 - 78) mm]2
12
= (3.3333 + 0.256) ¥ 106 = (3.58933 ¥ 10 6 ) mm 4
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57
PROBLEM 9.43 (Continued)
Then
Also
where
Then
I x = (34.19733 - 3.58933) ¥ 10 6 mm 4
= 30.608 ¥ 106 mm 4
or I x = 30.6 ¥ 106 mm 4 b
I y = ( I y )1 - ( I y )2
1
(160 mm)(100 mm)3 + (16 ¥ 103 mm2 )[(54 - 50) mm]2
12
= (13.33333 + 0.256) ¥ 106 mm 4 = 13.58933 mm 4
1
( I y )2 = (100 mm)( 40 mm)3 + ( 4 ¥ 103 mm2 )[(54 - 38) mm]2
12
= (0.53333 + 1.024) ¥ 106 mm 4 = 1.55733 ¥ 106 mm 4
( I y )1 =
I y = (13.58933 - 1.55733) ¥ 106 mm 4
= 12.032 ¥ 106 mm 4
or I y = 12.03 ¥ 106 mm 4 b
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58
PROBLEM 9.44
Determine the moments of inertia I x and I y of the area shown with respect to
centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid C of the area.
Then
y, mm
x, mm
A, mm2
yA, mm3
1
72 ¥ 10 = 720
36
5
25.92 ¥ 103
3.6 ¥ 103
2
10 ¥ 76 = 760
5 mm
48 mm
3.8 ¥ 103
36.48 ¥ 103
3
26 ¥ 20 = 520
13 mm
96 mm
676 ¥ 103
49.92 ¥ 103
S
2000
36.48 ¥ 103
90 ¥ 103
X SA = S xA: X (2000 mm2 ) = 36.48 ¥ 103 mm3
X = 18.24 mm
or
and
xA, mm3
Y SA = S yA: Y (2000 mm2 ) = 90 ¥ 103 mm3
Y = 45 mm
or
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59
PROBLEM 9.44 (Continued)
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
1
(72 mm)(10 mm)3 + (720 mm2 )[( 45 - 5) mm]2
12
= (6 + 1152) ¥ 103 mm 4 = 1158 ¥ 103 mm 4
1
( I x )2 = (10 mm)(76 mm)3 + (760 mm2 )[( 48 - 45) mm]2
12
= (365.8133 + 6.84) ¥ 103 mm 4 = (372.6533) ¥ 103 mm 4
( I x )1 =
1
(26 mm)(20 mm)3 + (520 mm2 )[(96 - 45) mm)]2
12
= (17.3333) + 1352.52) ¥ 103 mm 4 = 1369.8533 ¥ 103 mm 4
( I x )3 =
Then
I x = (1158 + 372.6533 + 1369.8533) ¥ 103 mm 4 = 2900.5066 ¥ 103 mm 4
Also
where
Then
or
I x = 2.900 ¥ 106 mm 4 b
I y = ( I y )1 + ( I y )2 + ( I y )3
1
(10 mm)(72 mm)3 + (720 mm2 )[(36 - 18.24) mm]2
12
= (311.04 + 227.1) ¥ 103 mm 4 = 538.14 ¥ 103 mm 4
1
( I y )2 = (76 mm)(10 mm)3 + (760 mm2 )[(18.24 - 5) mm]2
12
= (6.3333 + 133.2262) ¥ 103 mm 4 = 139.5595 ¥ 103 mm 4
1
( I y )3 = (20 mm)(26 mm)3 + (520 mm2 )[(18.24 - 13) mm]2
12
= (29.2933 + 14.2780) ¥ 103 mm 4 = 43.5713 ¥ 103 mm 4
( I y )1 =
I y = (538.14 + 139.5595 + 43.5713) ¥ 103 mm 4 = 721.2708 ¥ 103 mm 4
or
I y = 0.721 ¥ 106 mm 4 b
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60
PROBLEM 9.45
Determine the polar moment of inertia of the area shown with respect to (a) Point O,
(b) the centroid of the area.
SOLUTION
Dimensions in mm Symmetry: X = Y
Determination of centroid, C, of entire section.
Section
Area, mm2
yA, mm3
1
p
(100)2 = 7.854 ¥ 103
4
42.44
333.3 ¥ 103
2
(50)(100) = 5 ¥ 103
50
250 ¥ 103
3
(100)(50) = 5 ¥ 103
–25
-125 ¥ 103
S
17.854 ¥ 103
y, mm
458.3 ¥ 103
Y SA = S yA: Y (17.854 ¥ 103 mm2 ) = 458.3 ¥ 103 mm3
Y = 25.67 mm
Distance O to C:
(a)
X = Y = 25.67 mm
OC = 2Y = 2 (25.67) = 36.30 mm
p
(100)4 = 39.27 ¥ 106 mm 4
8
Section 1:
JO =
Section 2:
JO = J + A(OD )2 =
ÈÊ 50 ˆ 2 Ê 100 ˆ 2 ˘
1
(50)(100)[502 + 1002 ] + (50)(100) ÍÁ ˜ + Á
˙
Ë 2 ˜¯ ˙
12
ÍÎË 2 ¯
˚
JO = 5.208 ¥ 106 + 15.625 ¥ 106 = 20.83 ¥ 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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61
PROBLEM 9.45 (Continued)
Section 3: Same as Section 2;
J O = 20.83 ¥ 106 mm 4
Entire section:
J O = 39.27 ¥ 106 + 2(20.83 ¥ 106 )
= 80.94 ¥ 106
(b)
Recall that,
OC = 36.30 mm and
JO = 80.9 ¥ 106 mm 4 b
A = 17.854 ¥ 103 mm2
J O = J C + A(OC )2
80.94 ¥ 106 mm 4 = J C + (17.854 ¥ 103 mm2 )(36.30 mm)2
J C = 57.41 ¥ 106 mm 4 J C = 57.4 ¥ 106 mm 4 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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62
PROBLEM 9.46
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
Note that symmetry implies Y = 0.
Dimensions in mm
A, mm2
1
p
(84)( 42) = 5541.77
2
2
p
( 42)2 = 2770.88
2
3
4
-
-
p
(54)(27) = -2290.22
2
-
XA, mm3
112
= -35.6507
p
-197, 568
-
p
(27)2 = -1145.11
2
S
X , mm
56
= 17.8254
p
49,392
72
= -22.9183
p
52,488
36
= 11.4592
p
–13,122
4877.32
–108,810
X SA = S xA: X ( 4877.32 mm2 ) = -108, 810 mm3
Then
X = -22.3094
or
J O = ( J O )1 + ( J O )2 - ( J O )3 - ( J O ) 4
(a)
where
p
(84 mm)( 42 mm)[(84 mm)2 + ( 42 mm)2 ]
8
= 12.21960 ¥ 106 mm 4
( J O )1 =
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63
PROBLEM 9.46 (Continued)
p
( 42 mm)4
4
= 2.44392 ¥ 106 mm 4
( J O )2 =
p
(54 mm)(27 mm)[(54 mm)2 + (27 mm)2 ]
8
= 2.08696 ¥ 106 mm 4
p
( J O ) 4 = (27 mm) 4
4
= 0.41739 ¥ 106 mm 4
( J O )3 =
Then
J O = (12.21960 + 2.44392 - 2.08696 - 0.41739) ¥ 106 mm 4
= 12.15917 ¥ 106 mm 4
or
J O = 12.16 ¥ 106 mm 4 b
J O = J C + AX 2
(b)
or
J C = 12.15917 ¥ 106 mm 4 - ( 4877.32 mm2 )( -22.3094 mm)2
or
J C = 9.73 ¥ 106 mm 4 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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64
PROBLEM 9.47
Determine the polar moment of inertia of the area shown with respect to (a) Point O,
(b) the centroid of the area.
SOLUTION
Section
Area, mm2
x, mm
xA, mm3
1
p
(90)2 = 6.3617 ¥ 103
4
38.1972
243.000 ¥ 103
p
(60)2 = -2.8274 ¥ 103
4
25.4648
–72.0000 ¥ 103
2
S
Then
Then
-
3.5343 ¥ 103
171 ¥ 103
XA = S xA: X (3.5343 ¥ 103 mm2 ) = 171 ¥ 103 mm3
X = 48.3830 mm
p
p
J O = (90 mm)4 - (60 mm)4 = 20.6756 ¥ 106 mm 4 8
8
J O = 20.7 ¥ 106 mm 4 b
OC = 2 X = 2 ( 48.383 mm) = 68.4239 mm
J O = J C + A(OC )2 :
20.6756 ¥ 106 mm 4 = J C + (3.5343 mm2 )(68.4239 mm)2
J C = 4.1286 ¥ 106 mm 4
J C = 4.13 ¥ 106 mm 4 b
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65
PROBLEM 9.48
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
A, mm2
y, mm
yA, mm3
1
p
(120)2 = 22.61947 ¥ 103
2
160
= 50.9296
p
1152 ¥ 103
2
1
- (240)(90) = -10.8 ¥ 103
2
30
–324 ¥ 103
S
11.81947 ¥ 103
828 ¥ 103
Y SA = S yA: Y (11.81947 ¥ 103 mm2 ) = 828 ¥ 103 mm3
Then
Y = 70.0539 mm
or
J O = ( J O )1 - ( J O )2
(a)
where
p
(120 mm)4 = 162.86016 ¥ 106 mm 4
4
( J O )2 = ( I x ¢ )2 + ( I y ¢ )2
( J O )1 =
1
(240 mm)(90 mm)3 = 14.58 ¥ 106 mm 4
12
Now
( I x¢ )2 =
and
È1
˘
( I y¢ )2 = 2 Í (90 mm)(120 mm)3 ˙ = 25.92 ¥ 106 mm 4
Î12
˚
[Note: ( I y¢ )2 is obtained using
]
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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66
PROBLEM 9.48 (Continued)
Then
( J O )2 = (14.58 + 25.92) ¥ 10 6 mm 4
= 40.5 ¥ 106 mm 4
Finally
J O = (162.86016 - 40.5) ¥ 106 mm 4 = 122.36016 ¥ 106 mm 4
or
J O = 122.4 ¥ 106 mm 4 b
J O = J C + AY 2
(b)
or
J C = 122.36016 ¥ 106 mm 4 - (11.81947 ¥ 103 mm2 )(70.0539 mm)2
= 64.3555 ¥ 106 mm 4
or
J C = 64.4 ¥ 106 mm 4 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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67
PROBLEM 9.49
Two 20-mm steel plates are welded to a rolled S section as shown. Determine the
moments of inertia and the radii of gyration of the combined section with respect
to the centroidal x and y axes.
SOLUTION
S section:
A = 6010 mm2
I x = 90.3 ¥ 106 mm 4
I y = 3.88 ¥ 106 mm 4
Note:
Atotal = AS + 2 Aplate
= 6010 mm2 + 2(160 mm)(20 mm)
= 12, 410 mm2
Now
where
I x = ( I x )S + 2( I x ) plate
( I x ) plate = I xplate + Ad 2
1
(160 mm)(20 mm)3 + (3200 mm2 )[(152.5 + 100) mm]2
12
= 84.6067 ¥ 106 mm 4
=
Then
I x = (90.3 + 2 ¥ 84.6067) ¥ 106 mm 4
= 259.5134 ¥ 106 mm 4 and
k x2 =
Ix
259.5134 ¥ 106 mm 4
=
Atotal
12410 mm2
or
I x = 260 ¥ 106 mm 4 b
or
k x = 144.6 mm b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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68
PROBLEM 9.49 (Continued)
Also
I y = ( I y )S + 2( I y ) plate
È1
˘
= 3.88 ¥ 106 mm 4 + 2 Í (20 mm)(160 mm)3 ˙
Î12
˚
= 17.5333 ¥ 106 mm 4
and
k y2 =
Iy
Atotal
=
17.5333 ¥ 106 mm 4
12, 410 mm2
or
or
I y = 17.53 ¥ 106 mm 4 b
k y = 37.6 mm b
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or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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69
PROBLEM 9.50
Two channels are welded to a rolled W section as shown. Determine the
moments of inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
W section:
A = 5880 mm2
I x = 45.8 ¥ 106 mm 4
I y = 15.4 ¥ 106 mm 4
Channel:
A = 2170 mm2
I x = 13.5 ¥ 106 mm 4
I y = 0.545 ¥ 106 mm 4
Atotal = AW + 2Achan
= 5880 + 2(2170) = 10220 mm2
Now
where
I x = ( I x )W + 2( I x )chan
( I x )chan = I xchan + Ad 2
= 13.5 ¥ 106 mm 4 + (2170 mm2 )(114.5 mm)2 = 41.94922 ¥ 106 mm 4
Then
I x = [( 45.8 + 2 ¥ ( 41.9492)] ¥ 106 mm 4 = 129.6984 ¥ 106 mm 4
and
k x2 =
Also
I y = ( I y ) W + 2( I y )chan
Ix
129.6984 ¥ 106 mm 4
=
Atotal
10220 mm2
= [15.4 + 2(0.545)] ¥ 106 mm 4 = 16.49 ¥ 106 mm 4 and
k y2 =
Iy
Atotal
=
16.49 ¥ 106 mm 4
10220 mm2
I x = 129.7 ¥ 106 mm 4 b
k x = 112.7 mm b
I y = 16.49 ¥ 106 mm 4 b
k y = 40.2 mm b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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70
PROBLEM 9.51
To form a reinforced box section, two rolled W sections and two plates
are welded together. Determine the moments of inertia and the radii of
gyration of the combined section with respect to the centroidal axes shown.
SOLUTION
(Note: x and y interchanged as w.r.t. Fig. (19.3A)
W section:
A = 5880 mm2
I x = 15.4 ¥ 106 mm 4
I y = 45.8 ¥ 106 mm 4
Note:
Atotal = 2 AW + 2 Aplate
= 2(5880 mm2 ) + 2(203 mm)(6 mm)
= 14,196 mm2
Now
where
I x = 2( I x )W + 2( I x ) plate
( I x )W
ˆ
Ê 203
= I x + Ad = 15.4 ¥ 10 mm + (5880 mm ) Á
mm˜
¯
Ë 2
2
6
4
2
2
= 75.9772 ¥ 106 mm 4
( I x ) plate = I xplate + Ad 2
1
(203 mm)(6 mm)3 + (1218 mm2 )[(203 + 3) mm]2
12
= 51.6907 ¥ 106 mm 4
=
Then
I x = [2(75.9772) + 2(51.6907)] ¥ 106 mm 4
= 255.336 ¥ 106 mm 4
and
or
k x2 =
I x = 255 ¥ 106 mm 4 
Ix
255.336 ¥ 106 mm 4
=
Atotal
14,196 mm2
or k x = 134.1 mm 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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71
PROBLEM 9.51 (Continued)
also
where
I y = 2( I y ) W + 2( I y ) plate
( I y )W = I y
( I y ) plate =
Then
1
(6 mm)(203 mm)3 = 4.1827 ¥ 106 mm 4
12
I y = [2( 45.8) + 2( 4.1827)] ¥ 106 mm 4
= 99.9654 ¥ 106 mm 4
or I y = 100.0 ¥ 106 mm 4 
and
k y2 =
Iy
Atotal
=
99.9654 ¥ 106 mm 4
14,196 mm2
or
k y = 83.9 mm 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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72
PROBLEM 9.52
Two channels are welded to a d ¥ 300 mm steel plate as shown. Determine the
width d for which the ratio I x /I y of the centroidal moments of inertia of the
section is 16.
SOLUTION
A = 2890 mm2
Channel:
I x = 28.0 ¥ 106 mm 4
I y = 0.945 ¥ 106 mm 4
I x = 2( I x )C + ( I x ) plate
Now
1
d (300 mm)3
12
= (56 ¥ 106 mm 4 + 2.25 ¥ 106 d )
= 2(28.0 ¥ 106 mm 4 ) +
I y = 2( I y )C + ( I y ) plate
and
( I y )C = I y + Ad 2
where
ˆ
Êd
= 0.945 ¥ 106 mm 4 + (2890 mm2 ) Á + 16.1 mm˜
¯
Ë2
2
Ê d2
ˆ
= (0.945 ¥ 106 + (2890) Á
+ 16.1 d + 16.12 ˜
Ë 4
¯
( I y ) platee
= (722.5 d 2 + 46529 d + 1.69412 ¥ 106 )
1
= (300 mm)d 3 = 25 d 3
12
I y = 25 d 3 + 2(722.5 d 2 + 46529 d + 1.69412 ¥ 106 )
= (25 d 3 + 1445d 2 + 93058 d + 3388240
I x = 16 I y fi 56 ¥ 106 + 2.25 ¥ 106 d = 16(25 d 3 + 1445 d 2 + 93058 d + 3.38824 ¥ 106 )
400 d 3 + 23120 d 2 - 761.072 ¥ 103 d - 1.78816 ¥ 106 = 0
Solving yields d = 25.0999 mm (the other roots are negative)
d = 25.10 mm 
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73
PROBLEM 9.53
Two L76 ¥ 76 ¥ 6.4-mm angles are welded to a C250 ¥ 22.8 channel.
Determine the moments of inertia of the combined section with respect to
centroidal axes respectively parallel and perpendicular to the web of the
channel.
SOLUTION
A = 929 mm2
Angle:
I x = I y = 0.512 ¥ 106 mm 4
A = 2890 mm2
Channel:
I x = 0.945 ¥ 106 mm 4
I y = 28.0 ¥ 106 mm 4
First locate centroid C of the section
A, mm2
y, mm
yA, mm3
2(929) = 1858
21.2
39,389.6
Channel
2890
-16.1
-46,529
S
4748
Angle
Then
-7139.4
Y S A = S y A: Y ( 4748 mm2 ) = -7139.4 mm3
Y = -1.50366 mm
or
Now
where
I x = 2( I x ) L + ( I x )C
( I x ) L = I x + Ad 2 = 0.512 ¥ 106 mm 4 + (929 mm2 )[(21.2 + 1.50366) mm]2
= 0.990859 ¥ 106 mm 4
( I x )C = I x + Ad 2 = 0.949 ¥ 106 mm 4 + (2890 mm2 )[(16.1 - 1.50366) mm]2
= 1.56472 ¥ 106 mm 4
Then
I x = [2(0.990859) + 1.56472] ¥ 106 mm 4
or
I x = 3.55 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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74
PROBLEM 9.53 (Continued)
Also
where
I y = 2( I y ) L + ( I y )C
( I y ) L = I y + Ad 2 = 0.512 ¥ 106 mm 4 + (929 mm2 )[(127 - 21.2) mm]2
= 10.9109 ¥ 106 mm 4
( I y )C = I y
Then
I y = [2(10.9109) + 28.0] ¥ 106 mm 4
or
I y = 49.8 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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75
PROBLEM 9.54
Two L102 ¥ 102 ¥ 12.7-mm angles are welded to a steel plate as shown.
Determine the moments of inertia of the combined section with respect to
­centroidal axes respectively parallel and perpendicular to the plate.
SOLUTION
For 102 ¥ 102 ¥ 12.7 mm
A = 2420 mm2 , I x = I y = 2.30 ¥ 106 mm 4
Section
Plate
Two angles
S
Area, mm2
y, mm
yA, mm3
(12)(240) = 2880
120
345.6 ¥ 103
2(2420) = 4840
30
145.2 ¥ 103
490.8 ¥ 103
7720
YA = S y A
Y (7720 mm2 ) = 490.8 ¥ 103 mm3
Y = 63.5751 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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76
PROBLEM 9.54 (Continued)
Entire section:
È1
˘
I x = S( I x ¢ + Ad 2 ) = Í (12)(240)3 + (12)(240)(56.4249)2 ˙ + 2[2.30 ¥ 106 + (2420)(33.5751)2 ]
Î12
˚
= 13.824 ¥ 106 + 9.1693 ¥ 106 + 4.6 ¥ 106 + 2.7280 ¥ 106
I x = 30.3 ¥ 106 mm 4 
= 30.3213 ¥ 106 mm 4
1
(240)(12)3 + 2[2.30 ¥ 106 + (2420)(36)2 ]
12
= (34.56 ¥ 103 + 4.6 ¥ 106 + 6.27264 ¥ 106 )
Iy =
= 10.9072 ¥ 106 mm 4
I y = 10.91 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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77
PROBLEM 9.55
The strength of the rolled W section shown is increased by welding a channel to
its upper flange. Determine the moments of inertia of the combined section with
respect to its centroidal x and y axes.
SOLUTION
A = 14, 400 mm2
W Section:
I x = 554 ¥ 106 mm 4
I y = 63.3 ¥ 106 mm 4
A = 2890 mm2
Channel:
I x = 0.945 ¥ 106 mm 4
I y = 28.0 ¥ 106 mm 4
First locate centroid C of the section.
Then
A, mm2
y, mm
yA, mm3
W Section
14400
-231
-3326400
Channel
2890
49.9
144211
S
17290
-3182189
Y S A = S y A: Y (17, 290 mm 2 ) = -3,182,189 mm3
Y = -184.047 mm
or
Now
where
I x = ( I x ) W + ( I x )C
( I x )W = I x + Ad 2
= 554 ¥ 106 mm 4 + (14, 400 mm2 )(231 - 184.047)2 mm2
= 585.755 ¥ 106 mm 4
( I x )C = I x - Ad 2
= 0.945 ¥ 106 mm 4 + (2890 mm2 )( 49.9 + 184.047)2 mm2
= 159.12 ¥ 106 mm 4
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78
PROBLEM 9.55 (Continued)
Then
I x = (585.75 + 159.12) ¥ 106 mm 4
also
or
I x = 745 ¥ 106 mm 4 
or
I y = 91.3 ¥ 106 mm 4 
I y = ( I y ) W + ( I y )C
= (63.3 + 28.0) ¥ 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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79
PROBLEM 9.56
Two L127 ¥ 76 ¥ 12.7-mm angles are welded to a 12 mm steel
plate. Determine the distance b and the centroidal moments of
inertia I x and I y of the combined section, knowing that I y = 4 I x .
SOLUTION
A = 93.75 mm2
Angle:
I x = 235.75 mm 4
I y = 63.75 mm 4
First locate centroid C of the section.
Area, mm2
Angle
2(93.75) = 187.50 43.50 / 326.25
(10)(12.5) = 125
Plate
S
Then
y, mm
- 6.25
where
29.50
Y S A = S yA: Y (312 mm2 ) = 29.50 mm3
Y = 23.6 mm
I x = 2( I x )angle + ( I x ) plate
( I x )angle = I x + Ad 2 = 235.75 mm 4 + (93.75 mm2 )[( 43.50 - 23.6) mm]2
( I x ) plate
Then
- 31.25
312.50
or
Now
yA, mm3
13.05
= 295..15 mm 4
1
= I x + Ad 2 = (250 mm)(12.5 mm)3 + (125 mm2 )[(6.25 + 23.6) mm]2
12
= 180.8 mm 4
I x = [2(295.15) + 180.8] mm 4 = 771.1125 mm 4
or
I x = 771.1 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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80
PROBLEM 9.56 (Continued)
We have
I y = 4 I x = 4(771.1125 mm 4 ) = 3084.45 mm 4
Now
where
or
I y = 2( I y )angle + ( I y ) plate
( I y )angle = I y + Ad 2 = 63.75 mm 4 + (93.75 mm2 )[(b - 18.65) mm]2
( I y ) plate =
Then
I y = 3084.5 mm 4 
1
(12.5 mm)(250 mm)3 = 1041.6675 mm 4
12
3084.45 mm 4 = 2[63.75 + 93.75(b - 18.65)2 ] mm 4 + 1041.6675 mm 4
or b = 98.5 mm 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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81
PROBLEM 9.57
The panel shown forms the end of a trough that is filled with water to
the line AA¢. Referring to Section 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel
(the center of pressure).
SOLUTION
From Section 9.2:
R = g Ú y dA, M AA¢ = g Ú y 2 dA
Let yP = distance of center of pressure from AA¢. We must have
2
M AA¢ g Ú y dA I AA¢
RyP = M AA¢ : yP =
=
=
R
yA
g Ú y dA
(1)
For semicircular panel:
I AA¢ =
p 4
r
8
y=
4r
3p
A=
p 4
r
I
8
yP = AA¢ =
yA Ê 4r ˆ p 2
ÁË ˜¯ r
3p 2
p 2
r
2
yP =
3p
r 
16
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82
PROBLEM 9.58
The panel shown forms the end of a trough that is filled with water to the line
AA¢. Referring to Section 9.2, determine the depth of the point of application
of the resultant of the hydrostatic forces acting on the panel (the center of
pressure).
SOLUTION
See solution of Problem 9.57 for derivation of Eq. (1):
yP =
I AA¢
yA
(1)
For quarter ellipse panel:
I AA¢ =
p 3
ab
16
y=
4b
3p
p 3
ab
I AA¢
16
yP =
=
yA Ê 4b ˆ Ê p ˆ
ab
ÁË ˜¯ ÁË
˜
3p 4 ¯
A=
p
ab
4
yP =
3p
b 
16
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83
PROBLEM 9.59
The panel shown forms the end of a trough that is filled with water to
the line AA¢. Referring to Section 9.2, determine the depth of the point
of application of the resultant of the hydrostatic forces acting on the
panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text:
yP =
Now
I AA¢
yA
YA = S yA
4 Ê1
h
ˆ
= (2b ¥ h) + h Á ¥ 2b ¥ h˜
¯
2
3 Ë2
7
= bh2
3
and
I AA¢ = ( I AA¢ )1 + ( I AA¢ )2
where
1
2
( I AA¢ )1 = (2b)( h)3 = bh3
3
3
1
ˆÊ4 ˆ
Ê1
( I AA¢ )2 = I x + Ad = (2b)( h)3 + Á ¥ 2b ¥ h˜ Á h˜
¯Ë3 ¯
Ë
36
2
11
= bh3
6
2
2
Then
Finally,
2
11
5
I AA¢ = bh3 + bh3 = bh3
3
6
2
5 3
bh
yP = 2
7 2
bh
3
or
yP =
15
h 
14
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84
PROBLEM 9.60*
The panel shown forms the end of a trough that is filled with water to
the line AA¢. Referring to Section 9.2, determine the depth of the point
of application of the resultant of the hydrostatic forces acting on the
panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text:
For a parabola:
Now
By observation
yP =
I AA¢
yA
y=
2
h
5
A=
4
ah
3
1
dI AA¢ = ( h - y )3 dx
3
y=
h 2
x
a2
3
So that
1Ê
1 h3 2
h ˆ
dI AA¢ = Á h - 2 x 2 ˜ dx =
( a - x 2 )3 dx
¯
3Ë
3 a6
a
=
Then
Finally,
I AA¢ = 2Ú
1 h 6
( a - 3a 4 x 2 + 3a2 x 4 - x 6 )dx
6
3a
h3 6
( a - 3a 4 x 2 + 3a2 x 4 - x 6 )dx
3 a6
a1
0
a
=
2 h3
3 a6
=
32 3
ah
105
3 2 5 1 7˘
È 6
4 3
ÍÎa x - a x + 5 a x - 7 x ˙˚
0
32 3
ah
y p = 105
2
4
h ¥ ah
5
3
or y p =
4
h
7
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85
PROBLEM 9.61
The cover for a 0.5 m-diameter access hole in a water storage tank
is attached to the tank with four equally spaced bolts as shown.
Determine the additional force on each bolt due to the water
pressure when the center of the cover is located 1.4 m below the
water surface.
SOLUTION
From Section 9.2:
R = g yA yP =
I AA¢
yA
where R is the resultant of the hydrostatic forces acting on the cover
and yP is the depth to the point of application of R.
Recalling that g = r g, we have
R = (103 kg/m3 ¥ 9.81 m/s2 )(1.4 m)[p (0.25 m)2 ]
= 2696.67 N
p
(0.25 m) 4 + [p (0.25 m)2 ](1.4 m)2
4
= 0.387913 m 4
I AA¢ = I x + Ay 2 =
Also
yP =
Then
0.387913 m 4
(1.4 m)[p (0.25 m)2 ]
= 1.41116 m
Now note that symmetry implies
FA = FB
FC = FD
Next consider the free-body of the cover.
We have
+
SM CD = 0 : [2(0.32 m) sin 45∞](2 FA )
- [0.32 sin 45∞ - (1.41116 - 1.4)] m
¥ (2696.67 N ) = 0
or
Then
+
FA = 640.92 N
SFz = 0 : 2(640.92 N ) + 2 FC - 2696.67 N = 0
FC = 707.42 N
or
FA = FB = 641 N 
and
FC = FD = 707 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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86
PROBLEM 9.62
A vertical trapezoidal gate that is used as an automatic valve is held shut
by two springs attached to hinges located along edge AB. Knowing that
each spring exerts a couple of magnitude 1470 N · m, determine the
depth d of water for which the gate will open.
SOLUTION
From Section 9.2:
R = g yA
yP =
I SS ¢
yA
where R is the resultant of the hydrostatic forces acting on the gate
and yP is the depth to the point of application of R. Now
ˆ
ˆ
Ê1
Ê1
yA = S yA = [( h + 0.17) m] Á ¥ 1.2 m ¥ 0.51 m˜ + [( h + 0.34) m] Á ¥ 0.84 ¥ 0.51 m˜
¯
¯
Ë2
Ë2
= (0.5202h + 0.124848) m3
Recalling that g = r g, we have
R = (103 kg/m3 ¥ 9.81 m/s2 )(0.5202h + 0.124848) m3
= 5103.162( h + 0.24) N
Also
where
I SS ¢ = ( I SS ¢ )1 + ( I SS ¢ )2
( I SS ¢ )1 = I x + Ad 2
=
1
ˆ
Ê1
(1.2 m)(0.51 m)3 + Á ¥ 1.2 m ¥ 0.51 m˜ [( h + 0.17) m]2
¯
Ë2
36
= [0.0044217 + 0.306( h + 0.17)2 ] m 4
= (0.306h2 + 0.10404h + 0.0132651) m 4
( I SS ¢ )2 = I X 2 + Ad 2
=
1
ˆ
Ê1
(0.84 m)(0.51 m)3 + Á ¥ 0.84 m ¥ 0.51 m˜ [( h + 0.34) m]2
¯
Ë
36
2
= [0.0030952 + 0.2142( h + 0.34)2 ] m 4
= (0.2142h2 + 0.145656h + 0.0278567) m 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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87
PROBLEM 9.62 (Continued)
Then
I SS ¢ = ( I SS ¢ )1 + ( I SS ¢ )2
= (0.5202h2 + 0.249696h + 0.0411218) m 4
and
yP =
=
(0.5202h2 + 0.244696h + 0.0411218) m 4
(0.5202h + 0.124848) m3
h2 + 0.48h + 0.07905
m
h + 0.24
For the gate to open, require that
SM AB : M open = ( yP - h) R
Substituting
or
Ê h2 + 0.48h + 0.07905
ˆ
2940 N ◊ m = Á
- h˜ m ¥ 5103.162( h + 0.24) N
h + 0.24
Ë
¯
5103.162(0.24h + 0.07905) = 2940
or
h = 2.0711 m
Then
d = (2.0711 + 0.79) m
or
d = 2.86 m 
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88
PROBLEM 9.63*
Determine the x coordinate of the centroid of the volume shown.
(Hint: The height y of the volume is proportional to the x coordinate;
consider an analogy between this height and the water pressure on a
submerged surface.)
SOLUTION
y=
First note that
Now
h
x
a
x Ú dV = Ú xEL dV
Êh ˆ
xEL = x dV = y dA = Á x˜ dA
Ëa ¯
where
Ú x ( ah x dA) = Ú x dA = ( I z ) A
Ú ah x dA Ú x dA ( xA) A
2
x=
Then
where ( I z ) A and ( x A ) A pertain to area.
OABC: ( I z ) A is the moment of inertia of the area with respect to the z axis, x A is the x coordinate of the
centroid of the area, and A is the area of OABC. Then
( I z ) A = ( I z ) A1 + ( I z ) A2
1
1
= (b)( a)3 + (b)( a)3
3
12
5 3
= ab
12
and
( xA) A = S xA
ÈÊ a ˆ
˘ ÈÊ a ˆ Ê 1
ˆ˘
= ÍÁ ˜ ( a ¥ b) ˙ + ÍÁ ˜ Á ¥ a ¥ b˜ ˙
¯˚
Ë
¯
Ë
¯
Ë
Î 2
˚ Î 3 2
2
= a2 b
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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89
PROBLEM 9.63* (Continued)
Finally,
x=
5 3
12 a b
2 2
3a b
or
5
x= a 
8
Analogy with hydrostatic pressure on a submerged plate:
Recalling that P = g y, it follows that the following analogies can be established.
Height y ~ P
dV = y dA ~ p dA = dF
xdV = x( y dA) ~ y dF = dM
Recalling that
It can then be concluded that
yP =
Mx
R
Ê
Ú dM ˆ˜
Á=
ÁË Ú dF ˜¯
x ~ yP
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90
PROBLEM 9.64*
Determine the x coordinate of the centroid of the volume shown;
this volume was obtained by intersecting an elliptic cylinder with
an oblique plane. (Hint: The height y of the volume is proportional
to the x coordinate; consider an analogy between this height and the
water pressure on a submerged surface.)
SOLUTION
Following the “Hint,” it can be shown that (see solution to Problem 9.63)
x=
(I z )A
x
( xA) A
where ( I z ) A and ( xA) A are the moment of inertia and the first moment of the area, respectively, of the elliptical
area of the base of the volume with respect to the z axis. Then
( I z ) A = I z + Ad 2
p
= (39 mm)(64 mm)3 + [p (64 mm)(39 mm)](64 mm)2
4
= 12.779520p ¥ 106 mm 4
( xA) A = (64 mm)[p (64 mm)(39 mm)]
= 0.159744p ¥ 106 mm3
Finally
x=
12.779520p ¥ 106 mm 4
0.159744p ¥ 106 mm 4
or
x = 80.0 mm 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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91
PROBLEM 9.65*
Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a
force P at the centroid C of the area and two couples. The force P is perpendicular to the area and is of
magnitude P = g gAy sin q , where g is the density of the liquid, and the couples are M x ¢ = (g gI x ¢ sin q )i and
M y ¢ = (g gI x ¢y ¢ sin q ) j, where I x ¢y ¢ = Ú x ¢y ¢ dA (see Section 9.8). Note that the couples are independent of the
depth at which the area is submerged.
SOLUTION
Let g be the specific weight = rg
The pressure p at an arbitrary depth ( y sin q ) is
p = g ( y sin q )
so that the hydrostatic force dF exerted on an infinitesimal area dA is
dF = ( g y sin q )dA
Equivalence of the force P and the system of infinitesimal forces dF requires
SF : P = Ú dF = g y sin q dA = g sin q Ú y dA
or P = g Ay sinq 
Equivalence of the force and couple ( P, M x ¢ + M y ¢ ) and the system of infinitesimal hydrostatic forces requires
SM x : - yP - M x ¢ = Ú ( - y dF )
Now
- Ú y dF = - Ú y(g y sin q )dA = -g sin q Ú y 2 dA
= -( g sin q ) I x
Then
or
- yP - M x ¢ = -( g sin q ) I x
M x ¢ = ( g sin q ) I x - y ( g Ay sin q )
= g sin q ( I x - Ay 2 )
or M x ¢ = g I x ¢ sinq 
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92
PROBLEM 9.65 (Continued)
SM y : xP + M y ¢ = Ú x dF
Now
Ú x dF = Ú x( g y sin q )dA = g sin q Ú xy dA
= ( g sin q ) I xy Then
or
( Equation 9.12)
xP + M y ¢ = ( g sin q ) I xy
M y ¢ = ( g sin q ) I xy - x ( g Ay sin q )
= g sin q ( I xy - Ax y )
or M y ¢ = g I x ¢y ¢ sinq 
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93
PROBLEM 9.66*
Show that the resultant of the hydrostatic forces acting on a
submerged plane area A is a force P perpendicular to the area and
of magnitude P = g gAy sin q = pA, where g is the density of the
liquid and p is the pressure at the centroid C of the area. Show
that P is applied at a point CP, called the center of pressure, whose
coordinates are xP = I xy /Ay and yP = I x /Ay , where I xy = Ú xy dA
(see Section. 9.8). Show also that the difference of ordinates yP - y
is equal to k x¢2 / y and thus depends upon the depth at which the area
is submerged.
SOLUTION
Let g be the specific weight = rg
The pressure P at an arbitrary depth ( y sin q ) is
P = g ( y sin q )
so that the hydrostatic force dP exerted on an infinitesimal area dA is
dP = ( g y sin q )dA
The magnitude P of the resultant force acting on the plane area is then
P = Ú dP = Ú g y sin q dA = g sin q Ú y dA
= g sin q ( yA)
Now
P = pA 
p = g y sinq Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then
requires
SM x : - yP P = - Ú y dP
Now
Ú y dP = Ú y( g y sin q )dA = g sin q Ú y
2
dA
= ( g sin q ) I x
Then
or
yP P = ( g sin q ) I x
yP =
(g sin q ) I x
g sin q ( yA)
or yP =
Ix

Ay
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94
PROBLEM 9.66* (Continued)
SM y : xP P = Ú x dP
Now
Ú x dP = Ú x( g y sin q )dA = g sin q Ú xy dA
= ( g sin q ) I xy Then
or
(Equation 9.12)
xP P = ( g sin q ) I xy
xP =
( g sin q ) I xy
g sin q ( yA)
or xP =
Now
I x = I x ¢ + Ay 2
From above
I x = ( Ay ) yP
By definition
I x¢ = k x¢ A
Substituting
( Ay ) yP = k x2¢ A + Ay 2
I xy
Ay

2
yP - y =
Rearranging yields
k x2¢

y
Although k x¢ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on
the depth.
( yP - y ) = f (depth)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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95
PROBLEM 9.67
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
y = a 1-
First note
=
We have
where
1
4a2 - x 2
2
dI xy = dI x ¢y ¢ + xEL yEL dA
dI x ¢y ¢ = 0 (symmetry) xEL = x
yEL =
1
1
y=
4a2 - x 2
2
4
dA = ydx =
Then
x2
4a2
1
4a2 - x 2 dx
2
I xy = Ú dI xy = Ú
2a
0
ˆ
ˆÊ1
Ê1
xÁ
4a2 - x 2 ˜ Á
4a2 - x 2 ˜ dx
¯
¯Ë2
Ë4
2a
=
1 2a
1È
1 ˘
( 4a2 x - x 3 )dx = Í2a2 x 2 - x 4 ˙
Ú
4 ˚0
8 0
8Î
=
a4
8
1 4˘
È
2
ÍÎ2(2) - 4 (2) ˙˚
or I xy =
1 4
a 
2
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96
PROBLEM 9.68
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
At
or
Then
We have
where
x = a, y = b: a = kb2
a
b2
a
x = 2 y2
b
k=
dI xy = dI x ¢y ¢ + xEL yEL dA
dI x ¢y ¢ = 0 (symmetry) xEL =
yEL = y dA = x dy =
Then
1
a
x = 2 y2
2
2b
a 2
y dy
b2
bÊ a
ˆ
Ê a
ˆ
I xy = Ú dI xy = Ú Á 2 y 2 ˜ ( y ) Á 2 y 2 dy˜
0 Ë 2b
¯
Ëb
¯
=
a2
2b 4
b
Ú0
y 5 dy =
a2
2b 4
b
È1 6 ˘
ÍÎ 6 y ˙˚
0
or I xy =
1 2 2
a b 
12
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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97
PROBLEM 9.69
Determine by direct integration the product of inertia of the given area with respect to
the x and y axes.
SOLUTION
First note that
h
y=- x
b
Now
where
dI xy = dI x ¢y ¢ + xEL yEL dA
dI x ¢y ¢ = 0 (symmetry)
xEL = x
yEL =
1
1h
y=x
2
2b
h
dA = y dx = - x dx
b
Then
I xy = Ú dI xy = Ú
=
1 h2
2 b2
0
Ú -b
0
-b
ˆ
Ê 1 h ˆÊ h
xÁx - x dx˜
¯
Ë 2 b ˜¯ ÁË b
0
x 3 dx =
1 h2 È 1 4 ˘
x
2 b2 ÍÎ 4 ˙˚ - b
1
or I xy = - b2 h2 
8
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98
PROBLEM 9.70
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
y = h1 + ( h2 - h1 )
x
a
dI xy = dI x ¢y ¢ + xEL yEL dA
dI x ¢y ¢ = 0 (by symmetry)
xEL = x
yEL =
1
y
2
dA = y dx
a Ê1 ˆ
1 a
I xy = Ú dI xy - Ú x Á y˜ y dx = Ú xy 2 dx
0 Ë2 ¯
2 0
2
=
x˘
1 a È
x h1 + ( h2 - h1 ) ˙ dx
a˚
2 Ú 0 ÍÎ
=
3
x2
1 aÈ 2
2 x ˘
+
+
h
x
h
(
h
h
)
(
h
h
)
2
Í
˙ dx
1
1
2
1
2
1
a
2 Ú0 Î
a2 ˚
4
1 È 2 a2
a3
2 a ˘
2
h
+
h
(
h
h
)
+
(
h
h
)
Í1
˙
1 2
1
2
1
2Î 2
3a
4a2 ˚
˘
1 È a2 2
2
1
1
1
= Íh12
+ h1h2 a2 - h12 a2 + h22 a2 - h2 h1a + h12 a2 ˙
2Î 2 3
3
4
2
4
˚
=
=
a2 È 2 Ê 1 2 1 ˆ
Ê 1ˆ ˘
Ê 2 1ˆ
h1 Á - + ˜ + h1h2 Á - ˜ + h22 Á ˜ ˙
Í
Ë 4¯ ˚
¯
Ë
¯
Ë
3 2
2 Î
2 3 4
=
a2
2
È 2Ê 1 ˆ
Ê 1ˆ
2 Ê 1ˆ ˘
Íh1 ÁË 12 ˜¯ + h1h2 ÁË 6 ˜¯ + h2 ÁË 4 ˜¯ ˙
Î
˚
I xy =
(
)
a2 2
h1 + 2h1h2 + 3h22 
24
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99
PROBLEM 9.70 (Continued)
The following table is provided for the convenience of the instructor, as many problems in this and the next
lesson are related.
Type of Problem
Compute
I x and I y
Figure 9.12
Compute
I xy
9.67
9.72
9.73
9.74
9.75
9.78
9.79
9.80
9.81
9.83
9.82
9.84
9.85
9.86
9.87
9.89
9.88
9.90
9.91
9.92
9.93
9.95
9.94
9.96
9.97
9.98
9.100
9.101
9.103
9.106
I x ¢ , I y ¢ , I x ¢y ¢
by equations
Principal axes by
equations
Figure 9.13B
Figure 9.13A
I x ¢ , I y ¢ , I x ¢y ¢
by
Mohr’s circle
Principal axes by
Mohr’s circle
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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100
PROBLEM 9.71
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
Dimensions in mm
I xy = ( I xy )1 + ( I xy )2 + ( I xy )3
We have
Now symmetry implies
( I xy )2 = 0
and for the other rectangles
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
where
I xy = ( x yA)1 + ( x yA)3
Thus
A, mm2
x, mm
y, mm
x yA, mm 4
1
10 ¥ 80 = 800
-55
20
-880,000
3
10 ¥ 80 = 800
55
-20
-880,000
-1, 760,000
S
I xy = -1.760 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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101
PROBLEM 9.72
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
Dimensions in mm
I xy = ( I xy )1 - ( I xy )2 - ( I xy )3
We have
( I xy )1 = 0
Now symmetry implies
I xy = I x ¢y ¢ + x yA
and for each triangle
1 2 2
where, using the results of Sample Problem 9.6, I x ¢y ¢ = - 72
b h for both triangles. Note that the sign of I x ¢y ¢
is unchanged because the angles of rotation are 0∞ and 180∞ for triangles 2 and 3, respectively. Now
A, mm2
x, mm
y, mm
x yA, mm 4
2
1
(120)(60) = 3600
2
-80
-40
11.520 ¥ 106
3
1
(120)(60) = 3600
2
80
40
11.520 ¥ 106
S
Then
23.040 ¥ 106
Ï È 1
¸
˘
I xy = - Ì2 Í- (120 mm)2 (60 mm)2 ˙ + 23.040 ¥ 106 mm 4 ˝
72
˚
Ó Î
˛
or
I xy = -21.6 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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102
PROBLEM 9.73
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy )2
For each semicircle
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
and
I xy = S x yA
Thus
A, mm2
x, mm
1
p
(150)2 = 11250 p
2
-75
2
p
(150)2 = 11250 p
2
75
S
y, mm
xyA, mm4
200
p
168.75 ¥ 106
-
200
p
168.75 ¥ 106
337.5 ¥ 106
I xy = 338 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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103
PROBLEM 9.74
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy )2
We have
For each rectangle
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
and
I xy = S x yA
Thus
A, mm2
x, mm
y, mm
x yA, mm 4
1
7.6 ¥ 6.4 = 486.4
-13.7
9.2
-58620.928
2
6.4 ¥ 44.6 = 285.44
21.7
-16.3
-100962.982
S
-159583.91
I xy = 159.6 ¥ 103 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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104
PROBLEM 9.75
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy )2 + ( I xy )3
We have
Now symmetry implies
( I xy )1 = 0
and for the other rectangles
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
where
I xy = ( x yA)2 + ( x yA)3
Thus
A, mm2
X , mm
y, mm
x yA, mm 4
2
8 ¥ 32 = 256
– 46
-20
235,520
3
8 ¥ 32 = 256
46
20
235,520
S
471,040
I xy = 471 ¥ 103 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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105
PROBLEM 9.76
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy )2 + ( I xy )3
We have
Now, symmetry implies
( I xy )1 = 0
and for each triangle
I xy = I x ¢y ¢ + x yA
where, using the results of Sample Problem 9.6, I x ¢y ¢ = - 72 b2 h2 for both triangles. Note that the sign of I x ¢y ¢
1
is unchanged because the angles of rotation are 0∞ and 180∞ for triangles 2 and 3, respectively.
Now
A, mm2
x, mm
y, mm
x yA, mm 4
2
1
(228)(380) = 43.32 ¥ 103
2
-228
533
3
1.75481 ¥ 109
3
1
(228)(381) = ( 43.434 ¥ 103 )
2
228
-178
1.76273 ¥ 109
S
Then
3.51754 ¥ 109
¸
Ï 1
¸ Ï1
I xy = Ì- ¥ (228)2 ¥ (380)2 ˝ - Ì (228)2 (381)2 ˝ - 3.51754 ¥ 109
72
72
˛
Ó
˛ Ó
= - 3.7266 ¥ 109
or
I xy = -3.73 ¥ 109 mm 4 
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106
PROBLEM 9.77
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy )2 + ( I xy )3
For each rectangle
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
and
I xy = S x yA
Thus
A, mm2
x, mm
y, mm
x yA, mm 4
1
91 ¥ 13 = 1183
22.5
-50.5
-1.3442 ¥ 106
2
13 ¥ 96 = 1248
-16.5
4
-82.368 ¥ 103
3
33 ¥ 25 = 825
-6.5
64.5
-345.88125 ¥ 103
-1.7724 ¥ 106
S
I xy = -1.772 ¥ 106 mm 4 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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107
PROBLEM 9.78
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy )2
For each rectangle
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
and
I xy = S x yA
Thus
A, mm2
1
76 ¥ 12.7 = 965.2
2
12.7 ¥ (127 - 12.7) = 1451.61
x, mm
–19.1
12.55
S
y, mm
x yA, mm 4
–37.85
697,777
25.65
467,284
1,165,061
I xy = 1.165 ¥ 106 mm 4 
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108
PROBLEM 9.79
Determine for the quarter ellipse of Problem 9.67 the moments of inertia
and the product of inertia with respect to new axes obtained by rotating the
x and y axes about O (a) through 45∞ counterclockwise, (b) through 30∞
clockwise.
SOLUTION
From Figure 9.12:
From Problem 9.67:
First note
p
(2a)( a)3
16
p
= a4
8
p
I y = ( 2 a )3 ( a )
16
p
= a4
2
Ix =
I xy =
1 4
a
2
1 Ê p 4 p 4ˆ 5
4
ÁË a + a ˜¯ = p a
2 8
2
16
p
p
3
1
1Ê
ˆ
( I x - I y ) = Á a4 - a4 ˜ = - p a4
16
2
2Ë 8
2 ¯
1
(I x + I y ) =
2
Now use Equations (9.18), (9.19), and (9.20).
Equation (9.18):
1
1
I x ¢ = ( I x + I y ) + ( I x - I y ) cos 2q - I xy sin 2q
2
2
=
Equation (9.19):
1
1
I y ¢ = ( I x + I y ) - ( I x - I y ) cos 2q + I xy sin 2q
2
2
=
Equation (9.20):
5
3
1
p a 4 - p a 4 cos 2q - a 4 sin 2q
16
16
2
5
3
1
p a 4 + p a 4 cos 2q + a 4 sin 2q
16
16
2
1
I x ¢y ¢ = ( I x - I y )sin 2q + I xy cos 2q
2
3
1
= - p a 4 sin 2q + a 4 cos 2q
16
2
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109
PROBLEM 9.79 (Continued)
(a)
q = +45∞ :
I x¢ =
I y¢ =
5
3
1
p a 4 - p a 4 cos 90∞ - a 4 sin 90∞
16
16
2
q = - 30∞ :
I x¢ =
I y¢ =
I x¢ = 0.482a 4 
or
I y¢ = 1.482a 4 
or
I x ¢y ¢ = -0.589a 4 
or
I x¢ = 1.120 a 4 
or
I y¢ = 0.843a 4 
or
I x ¢y ¢ = 0.760a 4 
5
3
1
p + p a 4 cos 90∞ + a 4
16
16
2
I x ¢y ¢ = -
(b)
or
3
1
p a 4 sin 90∞ + a 4 cos 90∞
16
2
5
3
1
p a 4 - p a 4 cos( -60∞) - a 4 sin( -60∞)
16
16
2
5
3
1
p a 4 + p a 4 cos( -60∞) + a 4 sin( -60∞)
16
16
2
I x ¢y ¢ = -
3
1
p a 4 sin( -60∞) + a 4 cos( -60∞)
16
2
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110
PROBLEM 9.80
Determine the moments of inertia and the product of inertia of the
area of Problem 9.72 with respect to new centroidal axes obtained by
rotating the x and y axes 30∞ counterclockwise.
SOLUTION
From Problem 9.72:
I xy = -21.6 ¥ 106 mm 4
Now
I x = ( I x )1 - ( I x )2 - ( I x )3
where
1
(240 mm)(160 mm)3
12
= 81.920 ¥ 106 mm 4
( I x )1 =
( I x ) 2 = ( I x )3 =
1
(120 mm)(60 mm)3
36
È1
˘
+ Í (120 mm)(60 mm) ˙ ( 40 mm)2
Î2
˚
= 6.480 ¥ 106 mm 4
Then
I x = [81.920 - 2(6.480)] ¥ 106 mm 4 = 68.96 ¥ 106 mm 4
Also
I y = ( I y )1 - ( I y )2 - ( I y )3
where
1
(160 mm)(240 mm)3 = 184.320 ¥ 106 mm 4
12
1
m)(120 mm)3
( I y )2 = ( I y )3 = (60 mm
36
È1
˘
+ Í (120 mm)(60 mm)˙ (80 mm)2
Î2
˚
( I y )1 =
= 25.920 ¥ 106 mm 4
Then
I y = [184.320 - 2(25.920)] ¥ 106 mm 4 = 132.48 ¥ 106 mm 4
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111
PROBLEM 9.80 (Continued)
Now
1
(I x + I y ) =
2
1
(I x - I y ) =
2
1
(68.96 + 132.48) ¥ 106 = 100.72 ¥ 106 mm 4
2
1
(68.96 - 132.48) ¥ 106 = -31.76 ¥ 106 mm 4
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
I x = ( I x + I y ) + ( I x - I y ) cos 2q - I xy sin 2q
2
2
= [100.72 + ( -31.76) cos 60∞ - ( -21.6)sin 60∞] ¥ 106 mm 4
or
Eq. (9.19):
1
1
I y ¢ = ( I x + I y ) - ( I x - I y ) cos 2q + I xy sin 2q
2
2
= [100.72 - ( -31.76) cos 60∞ + ( -21.6)sin 60∞] ¥ 106 mm 4
or
Eq. (9.20):
I x¢ = 103.5 ¥ 106 mm 4 
I y¢ = 97.9 ¥ 106 mm 4 
1
I x ¢y ¢ = ( I x - I y )sin 2q + I xy sin 2q
2
= [-31.76 sin 60∞ + ( -21.6) cos 60∞] ¥ 106 mm 4
or
I x ¢y ¢ = -38.3 ¥ 106 mm 4 
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112
PROBLEM 9.81
Determine the moments of inertia and the product of inertia of the
area of Problem 9.73 with respect to new centroidal axes obtained by
rotating the x and y axes 60∞ counterclockwise.
SOLUTION
From Problem 9.73:
I xy = 337.5 ¥ 106 mm 4
Now
where
I x = ( I x )1 + ( I x )2
p
(150 mm)4
8
= (63.28125 ¥ 106 )p mm 4
( I x )1 = ( I x )2 =
Then
I x = 2 I xj = (126.5625 ¥ 106 )p mm 4
Also
I y = ( I y )1 + ( I y )2
where
Then
Now
( I y )1 = ( I y )2 =
p
Èp
˘
(150 mm)4 + Í (150 mm)2 ˙ (75 mm)2 = (126.5625 ¥ 106 )p mm 4
8
2
Î
˚
I y = 2(126.5625 ¥ 106 )p mm 4 = (253.125 ¥ 106 )p mm 4
p
1
( I x + I y ) = (126.5625 ¥ 106 + 253.125 ¥ 106 ) = 189.84375 p ¥ 106 mm 4
2
2
p
1
( I x - I y ) = (126.5625 ¥ 106 - 253.125 ¥ 106 ) - 63.28125 ¥ 106 p mm 4
2
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
I x ¢ = ( I x + I y ) + ( I x - I y ) cos 2q - I xy sin 2q
2
2
= [{189.843751 + ( -63.28125) cos120∞}p - 337.5(sin 120∞)] ¥ 106 mm 4
= 403.53 ¥ 106 mm 4
or
I x¢ = 404 ¥ 106 mm 4 
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113
PROBLEM 9.81 (Continued)
Eq. (9.19):
Eq. (9.20):
1
1
I y ¢ = ( I x + I y ) - ( I x - I y ) cos 2q + I xy sin 2q
2
2
= [{189.84375 - ( -63.28125) cos120∞}p + 337.5 sin 120∞] ¥ 106 mm 4 = 789.29 ¥ 106 mm 4
or
I y¢ = 789 ¥ 106 mm 4 
or
I x ¢y ¢ = -341 ¥ 106 mm 4 
1
I x ¢y ¢ = ( I x - I y )sin 2q + I xy cos 2q
2
= [( -63.28125p )sin 120∞ + 337.5 cos120∞] ¥ 106 mm 4
= -340.919 ¥ 106 mm 4
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114
PROBLEM 9.82
Determine the moments of inertia and the product of inertia of the
area of Problem 9.75 with respect to new centroidal axes obtained by
rotating the x and y axes 45° clockwise.
SOLUTION
From Problem 9.75:
I xy = 471, 040 mm 4
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
1
(100 mm)(8 mm)3
12
= 4266.67 mm 4
( I x )1 =
1
(8 mm)(32 mm)3 + [(8 mm)(32 mm)](20 mm)2
12
= 124, 245.33
( I x ) 2 = ( I x )3 =
Then
I x = [4266.67 + 2(124, 245.33)] mm 4 = 252,757 mm 4
Also
I y = ( I y )1 + ( I y )2 + ( I y )3
where
Then
Now
1
(8 mm)(100 mm)3 = 666,666.7 mm 4
12
1
( I y )2 = ( I y )3 = (32 mm)(8 mm)3 + [(8 mm)(32 mm)]( 46 mm)2
12
= 543,061.3 mm 4
( I y )1 =
I y = [666,666.7 + 2(543,061.3)] mm 4 = 1,752,789 mm 4
1
1
( I x + I y ) = (252,757 + 1,752,789) mm 4 = 1,002,773 mm 4
2
2
1
1
( I x - I y ) = (252,757 - 1,752,789) mm 4 = -750,016 mm 4
2
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
I x ¢ = ( I x + I y ) + ( I x - I y ) cos 2q - I xy sin 2q
2
2
= [1,002,773 + ( -750, 016) cos( -90∞) - 471,040 sin( -90∞)]
or
I x¢ = 1.474 ¥ 106 mm 4 
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115
PROBLEM 9.82 (Continued)
Eq. (9.19):
Eq. (9.20):
1
1
I y ¢ = ( I x + I y ) - ( I x - I y ) cos 2q + I xy sin 2q
2
2
= [1, 002,773 - ( -750, 016) cos( -90∞) + 471,040 sin( -90∞)]
or
I y¢ = 0.532 ¥ 106 mm 4 
or
I x ¢y ¢ = 0.750 ¥ 106 mm 4 
1
I x ¢y ¢ = ( I x - I y )sin 2q + I xy cos 2q
2
= [( -750,016)sin( -90∞) + 471,040 cos( -90∞)]
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116
PROBLEM 9.83
Determine the moments of inertia and the product of inertia of the
L76 ¥ 51 ¥ 6.4 mm angle cross section of Problem 9.74 with respect
to new centroidal axes obtained by rotating the x and y axes 30°
clockwise.
SOLUTION
From Figure 9.13:
I x = 0.162 ¥ 106 mm 4
From Problem 9.74:
I y = 0.454 ¥ 106 mm 4
I xy = -0.159584 ¥ 106 mm 4
Now
1
(I x + I y ) =
2
1
(I x - I y ) =
2
1
(0.162 + 0.454) ¥ 106 mm 4 = 0.308 ¥ 106 mm 4
2
1
(0.162 - 0.454) ¥ 106 mm 4 = -0.146 ¥ 106 mm 4
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
I x ¢ = ( I x + I y ) + ( I x - I y ) cos 2q - I xy sin 2q
2
2
= [0.308 + ( -0.146) cos( -60∞) - ( -0.159584)sin( -60∞) ¥ 106 mm 4
or
Eq. (9.19):
1
1
I y ¢ = ( I x + I y ) - ( I x - I y ) cos 2q + I xy sin 2q
2
2
= [0.308 - ( -0.146) cos( -60∞) + ( -0.159584)sin( -60∞)] ¥ 106 mm 4
or
Eq. (9.20):
I x¢ = 96.8 ¥ 103 mm 4 
I y¢ = 519 ¥ 103 mm 4 
1
I x ¢y ¢ = ( I x - I y )sin 2q + I xy cos 2q
2
= [( -0.146)sin( -60∞) + ( -0.159584) cos( -60∞)] ¥ 106 mm 4
or I x ¢y ¢ = 46.6 ¥ 103 mm 4 
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117
PROBLEM 9.84
Determine the moments of inertia and the product of inertia of the
L127 ¥ 76 ¥ 12.7 mm angle cross section of Problem 9.78 with respect
to new centroidal axes obtained by rotating the x and y axes 45∞
counterclockwise.
SOLUTION
From Figure 9.13:
I x = 3.93 ¥ 106 mm 4
I y = 1.06 ¥ 106 mm 4
From Problem 9.78:
I xy = 1.165061 ¥ 106 mm 4
Now
1
1
( I x + I y ) = (3.93 + 1.06) ¥ 106 mm 4
2
2
= 2.495 ¥ 106 mm 4
1
1
( I x - I y ) = (3.93 - 1.06) ¥ 106 mm 4 = 1.435 ¥ 106 mm 4
2
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
I x ¢ = ( I x + I y ) + ( I x - I y ) cos 2q - I xy sin 2q
2
2
= [2.495 + 1.435 cos 90∞ - 1.165061sin 90∞] ¥ 106 mm 4
or
Eq. (9.19):
1
1
I y ¢ = ( I x + I y ) - ( I x - I y ) cos 2q + I xy sin 2q
2
2
= [2.495 - 1.435 cos 90∞ + 1.165061sin 90∞] ¥ 106 mm 4
or
Eq. (9.20):
I x¢ = 1.330 ¥ 106 mm 4 
I y¢ = 3.66 ¥ 106 mm 4 
1
I x ¢y ¢ = ( I x - I y )sin 2q + I xy cos 2q
2
= [(1.435 sin 90∞ + 1.165061cos 90∞] ¥ 106 mm 4
or
I x ¢y ¢ = 1.435 ¥ 106 mm 4 
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118
PROBLEM 9.85
For the quarter ellipse of Problem 9.67, determine the orientation of the
principal axes at the origin and the corresponding values of the moments
of inertia.
SOLUTION
From Problem 9.79:
Ix =
p 4
a
8
Problem 9.67:
I xy =
1 4
a
2
Now Eq. (9.25):
tan2q m = -
Iy =
2 I xy
Ix - I y
=-p
8
=
Then
p 4
a
2
2
(a)
1
2
4
a 4 - p2 a 4
8
= 0.84883
3p
2q m = 40.326∞ and 220.326∞
or
Also Eq. (9.27):
I max, min =
Ix + I y
2
q m = 20.2∞ and 110.2∞ 
2
Ê Ix - I y ˆ
2
+ I xy
± Á
˜
Ë 2 ¯
1Êp
p ˆ
= Á a4 + a4 ˜
2Ë 8
2 ¯
2
p ˘ Ê1 ˆ
È1 Ê p
± Í Á a4 - a4 ˙ + Á a4 ˜
Ë
2 ˚ Ë2 ¯
Î2 8
2
= (0.981,748 ± 0.772,644)a 4
or
I max = 1.754a 4 
and
I min = 0.209a 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
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119
PROBLEM 9.86
For the area indicated, determine the orientation of the principal axes
at the origin and the corresponding values of the moments of inertia.
Area of Problem 9.72.
SOLUTION
From Problem 9.80:
I x = 68.96 ¥ 106 mm 4
I y = 132.48 ¥ 106 mm 4
Problem 9.72:
Now Eq. (9.25):
I xy = -21.6 ¥ 106 mm 4
tan2q m = -
2 I xy
Ix - I y
=-
2( -21.6 ¥ 106 )
(68.96 - 132.48) ¥ 106
= -0.68010
Then
2q m = -34.220∞ and 145.780∞
or
Ix + I y
q m = -17.11∞ and 72.9∞ 
2
Ê Ix - I y ˆ
2
± Á
˜ + I xy
Ë 2 ¯
Also Eq. (9.27):
I max, min =
Then
2
˘
È 68.96 + 132.48
Ê 68.96 - 132.48 ˆ
+ ( -21.6)2 ˙ ¥ 106
I max, min = Í
± Á
˜
¯
Ë
2
2
Í
˙
˚
Î
2
= (100.72 ± 38.409) ¥ 106 mm 4
or
and
I max = 139.1 ¥ 106 mm 4 
I min = 62.3 ¥ 106 mm 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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120
PROBLEM 9.87
For the area indicated, determine the orientation of the principal axes
at the origin and the corresponding values of the moments of inertia.
Area of Problem 9.73.
SOLUTION
I x = (126.5625 ¥ 106 )p mm 4
From Problem 9.81:
I y = (253.125 ¥ 106 )p mm 4
I xy = 337.5 ¥ 106
Problem 9.73:
tan2q m = -
Now Eq. (9.25):
2 I xy
Ix - I y
=-
2(337.5 ¥ 106 )
(126.5625 - 253.125) ¥ 106 p
= 1.69765
2q m = 59.500∞ and 239.500∞
Then
or
I max, min =
Also Eq. (9.27):
Then
I max, min =
Ix + I y
2
q m = 29.7∞ and 119.7∞ 
2
Ê Ix - I y ˆ
2
± Á
˜ + I xy
Ë 2 ¯
ÏÔ (126.5625 - 253.125) ¥ 106 p ¸Ô
(126.5625 + 253.125) ¥ 106 p
6 2
± Ì
˝ + (337.5 ¥ 10 )
2
2
˛Ô
ÓÔ
= (596.412 ¥ 106 ± 391.700 ¥ 106 ) mm 4
or
and
I max = 988 ¥ 106 mm 4 
I min = 205 ¥ 106 mm 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
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121
PROBLEM 9.88
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments
of inertia.
Area of Problem 9.75.
SOLUTION
From Problem 9.82:
I x = 252,757 mm 4
I y = 1,752,789 mm 4
Problem 9.75:
Now Eq. (9.25):
I xy = 471,040 mm 4
tan 2q m = -
2 I xy
Ix - I y
2( 471,040)
252,757 - 1,752,789
= 0.62804
=-
Then
2q m = 32.130∞ and 212.130∞
or
Also Eq. (9.27):
I max, min =
Then
I max, min =
Ix - I y
2
q m = 16.07∞ and 106.1∞ 
2
Ê Ix - I y ˆ
2
± Á
˜ + I xy
Ë 2 ¯
2
252,757 + 1,752,789
Ê 252,757 - 1,752,789 ˆ
2
± Á
˜¯ + 471040
Ë
2
2
= (1,002,773 ± 885,665) mm 4
or
I max = 1.888 ¥ 106 mm 4 
and
I min = 0.1171 ¥ 106 mm 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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122
PROBLEM 9.89
For the angle cross section indicated, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
The L76 ¥ 51 ¥ 6.4 mm angle cross section of Problem 9.74.
SOLUTION
From Problem 9.83:
Problem 9.74:
Now Eq. (9.25):
I x = 0.162 ¥ 106 mm 4
I y = 0.454 ¥ 106 mm 4
I xy = -0.159584 ¥ 106 mm 4
tan 2q m = -
2 I xy
Ix - I y
=-
2( - 0.159584 ¥ 106 )
(0.162 - 0.454) ¥ 106
= -1.09304
Then
2q m = -47.545∞ and 132.455∞
or
Also Eq. (9.27):
I max, min =
Ix + I y
2
q m = -23.8∞ and 66.2∞ 
2
Ê Ix - I y ˆ
2
± Á
˜ + I xy
2
Ë
¯
2
Then
I max, min
6
(0.162 + 0.454) ¥ 106
ÔÏ (0.162 - 0.454) ¥ 10 Ô¸
6 2
=
± Ì
˝ + ( -0.159584 ¥ 10 )
2
2
Ô˛
ÔÓ
= (0.308 ¥ 106 ± 0.21629) ¥ 106 mm 4
or
and
I max = 0.524 ¥ 106 mm 4 
I min = 0.0917 ¥ 106 mm 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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123
PROBLEM 9.90
For the angle cross section indicated, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
The L127 ¥ 76 ¥ 12.7-mm angle cross section of Problem 9.78.
SOLUTION
From Problem 9.84:
I x = 3.93 ¥ 106 mm 4
I y = 1.06 ¥ 106 mm 4
Problem 9.78:
Now Eq. (9.25):
I xy = 1.165061 ¥ 106 mm 4
tan 2q m = -
2 I xy
Ix - I y
=-
2(1.165061 ¥ 106 )
(3.93 - 1.06) ¥ 106
= -0.81189
Then
2q m = -39.073∞ and 140.927∞
or
Ix + I y
q m = -19.54∞ and 70.5∞ 
2
Ê Ix - I y ˆ
+ I xy2
± Á
˜
Ë 2 ¯
Also Eq. (9.27):
I max, min =
Then
2
˘
È 3.93 + 1.06
Ê 3.93 - 1.06 ˆ
2˙
1
165061
¥ 106 mm 4
+
I max, min = Í
± Á
.
˜¯
Ë
2
2
˙
Í
˚
Î
2
= (2.495 ± 1.84840) ¥ 106 mm 4
or
I max = 4.34 ¥ 106 mm 4 
and
I min = 0.647 ¥ 106 mm 4 
By inspection, the a axis corresponds to I max and the b axis corresponds to I min .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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124
PROBLEM 9.91
Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67
the moments of inertia and the product of inertia with respect to new
axes obtained by rotating the x and y axes about O (a) through 45°
counterclockwise, (b) through 30∞ clockwise.
SOLUTION
From Problem 9.79:
Problem 9.67:
Ix =
p 4
a
8
Iy =
p 4
a
2
I xy =
1 4
a
2
The Mohr’s circle is defined by the diameter XY, where
1 ˆ
Êp
X Á a4, a4 ˜
Ë8
2 ¯
Now
1
I ave = ( I x + I y ) =
2
1 ˆ
Êp
and Y Á a 4 , - a 4 ˜
Ë2
2 ¯
1 Ê p 4 p 4ˆ 5
4
4
Á a + a ˜¯ = p a = 0.98175a
2Ë 8
2
16
2
and
2
2
Ê Ix - I y ˆ
È1 Ê p 4 p 4 ˆ ˘ Ê 1 4 ˆ
2
+
+
=
R= Á
I
a
a
a
˜
Á
˜
Á
xy
Í2 Ë 8
2 ¯ ˙˚ Ë 2 ¯
Ë 2 ˜¯
Î
= 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
=-p
8
2
(a)
1
2
4
a 4 - p2 a 4
= 0.84883
or
2q m = 40.326∞
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125
PROBLEM 9.91 (Continued)
a = 90∞ - 40.326∞
= 49.674∞
Then
b = 180∞ - ( 40.326∞ + 60∞)
= 79.674∞
(a)
q = +45∞ :
I x¢ = I ave - R cos a = 0.98175a 4 - 0.77264 a 4 cos 49.674∞
or I x¢ = 0.482a 4 
I y¢ = I ave + R cos a = 0.98175a 4 + 0.77264 a 4 cos 49.674∞
or I y¢ = 1.482a 4 
I x ¢y ¢ = - R sin a = -0.77264a 4 sin 49.674∞
(b)
or
q = -30∞ :
I x ¢y ¢ = -0.589a 4 
I x¢ = I ave + R cos b = 0.98175a 4 + 0.77264a 4 cos 79.674∞
or I x¢ = 1.120 a 4 
I y¢ = I ave - R cos b = 0.98175a 4 - 0.77264a 4 cos 79.674∞
or I y¢ = 0.843a 4 
I x ¢y ¢ = R sin b = 0.77264a 4 sin 79.674∞
or I x ¢y ¢ = 0.760a 4 
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126
PROBLEM 9.92
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the area of Problem 9.72 with respect to new centroidal
axes obtained by rotating the x and y axes 30∞ counterclockwise.
SOLUTION
From Problem 9.80:
I x = 68.96 ¥ 106 mm 4
I y = 132.48 ¥ 106 mm 4
Problem 9.72:
I xy = -21.6 ¥ 106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (68.96 ¥ 106 , -21.6 ¥ 106 ) and Y (132.48 ¥ 106,
21.6 ¥ 106 ).
Now
and
1
1
I ave = ( I x - I y ) = (68.96 + 132.48) ¥ 106 = 100.72 ¥ 106 mm 4
2
2
2
2
¸Ô
ÏÔ 1
È1
˘
È
˘
2
R = Í ( I x - I y ) ˙ + I xy
= Ì Í (68.96 - 132.48) ˙ + ( -21.6)2 ˝ ¥ 106 mm 4
Î2
˚
˚
ÔÓ Î 2
Ô˛
= 38.409 ¥ 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2( -21.6 ¥ 106 )
(68.96 - 132.48) ¥ 106
= -0.68010
=-
or
Then
2q m = -34.220∞
a = 180∞ - (34.220∞ + 60∞)
= 85.780∞
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127
PROBLEM 9.92 (Continued)
Then
I x¢ = I ave + R cos a = (100.72 + 38.409 cos 85.780∞) ¥ 106
or I x¢ = 103.5 ¥ 106 mm 4 
I y¢ = I ave - R cos a = (100.72 - 38.409 cos 85.780∞) ¥ 106
or I y¢ = 97.9 ¥ 106 mm 4 
I x ¢y ¢ = - R sin a = -(38.409 ¥ 106 )sin 85.780∞
or
I x ¢y ¢ = -38.3 ¥ 106 mm 4 
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128
PROBLEM 9.93
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the area of Problem 9.73 with respect to new centroidal
axes obtained by rotating the x and y axes 60∞ counterclockwise.
SOLUTION
From Problem 9.81:
I x = (126.5625 ¥ 106 )p mm 4
I y = (253.125 ¥ 106 )p mm 4
Problem 9.73:
I xy = 337.5 ¥ 106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (126.5625 ¥ 106 p , 337.5 ¥ 106 ) and Y (253.125 ¥ 106 p ,
- 337.5 ¥ 106 ).
Now
and
1
I ave = ( I x + I y ) = (189.84375 ¥ 106 )p mm 4
2
= 596.412 ¥ 106 mm 4
È1
R = Í ( I x - I y )2 + I xy2
Î2
= 391.700 ¥ 106 mm 4
(from solution of Problem 9.87)
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
=-
-2(337.5 ¥ 106 )
(126.5625 - 253.125) ¥ 106 ¥ p
= 1.69765
or
Then
2q m = 59.500∞
a = 120∞ - 59.500∞
= 60.500∞
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129
PROBLEM 9.93 (Continued)
Then
I x¢ = I ave - R cos a = 596.412 ¥ 106 - 391.700 ¥ 106 cos 60.500∞ = 403.53 ¥ 106
or I x¢ = 404 ¥ 106 mm 4 
I y = I ave + R cos a = 596.412 ¥ 106 + 391.7 ¥ 106 cos(60.5) = 789.294 ¥ 106 mm 4
or I y¢ = 789 ¥ 106 mm 4 
I x ¢y ¢ = - R sin a = -391.700 ¥ 106 sin(60.5∞)
or
I x ¢y ¢ = -341 ¥ 106 mm 4 
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130
PROBLEM 9.94
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the area of Problem 9.75 with respect to new
centroidal axes obtained by rotating the x and y axes 45∞ clockwise.
SOLUTION
From Problem 9.82:
I x = 252,757 mm 4
I y = 1,752,789 mm 4
Problem 9.75:
I xy = 471,040 mm 4
The Mohr’s circle is defined by the diameter XY, where X (252,757; 471,040) and Y (1,752,789; -471, 040).
Now
1
1
I ave = ( I x + I y ) = (252,757 + 1,752,789)
2
2
= 1,002,773 mm 4
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
È1
˘
= Í (252,757 - 1,752,789)2 ˙ + 471,0402
2
Î
˚
= 885,665 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = =-
2 I xy
Ix - I y
2( 471,040)
252,757 - 1,752,789
= 0.62804
or
Then
2q m = 32.130∞
a = 180∞ - (32.130 + 90∞)
= 57.870∞
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131
PROBLEM 9.94 (Continued)
Then
I x¢ = I ave + R cos a = 1, 002, 773 + 885, 665 cos 57.870∞
or I x¢ = 1.474 ¥ 106 mm 4 
I y¢ = I ave - R cos a = 1, 002, 773 - 885, 665 cos 57.870∞
or I y¢ = 0.532 ¥ 106 mm 4 
I x ¢y ¢ = R sin a = 885, 665 sin 57.870∞
or
I x ¢y ¢ = 0.750 ¥ 106 mm 4 
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132
PROBLEM 9.95
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the L76 ¥ 51 ¥ 6.4 mm angle cross section of Problem 9.74
with respect to new centroidal axes obtained by rotating the x and y
axes 30° clockwise.
SOLUTION
From Problem 9.83:
I x = 0.162 ¥ 106 mm 4
I y = 0.454 ¥ 106 mm 4
Problem 9.74:
I xy = -0.159584 ¥ 106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (0.162 ´ 106,- 0.159584 ´ 106 ) and Y (0.454 ´ 106,
0.159584 ´ 106 ).
Now
1
I ave = ( I x + I y )
2
= 0.308 ¥ 106 mm 4
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
= 0.21629 ¥ 106 mm 4
(from Problem 9.89)
The Mohr’s circle is then drawn as shown.
tan 2q m = =-
2 I xy
Ix - I y
2( -0.159584 ¥ 106 )
(0.162 - 0.454) ¥ 106
= -1.09304
or
Then
2q m = - 47.545∞
a = 60∞ - 47.545∞ = 12.455∞
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133
PROBLEM 9.95 (Continued)
Then
I x¢ = I ave - R cos a = 0.308 ¥ 106 - 0.21629 ¥ 106 cos12.455∞ = 0.0968 ¥ 106 mm 4
or I x¢ = 0.0968 ¥ 106 mm 4 
I y¢ = I ave + R cos a = 0.308 ¥ 106 + 0.21629 ¥ 106 cos 12.455∞ = 0.5192 ¥ 106 mm 4
or I y = 0.519 ¥ 106 mm 4 
I x ¢y ¢ = R sin a = 0.21629 ¥ 106 sin(12.455∞)
or
I x ¢y ¢ = 0.0466 ¥ 106 mm 4 
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134
PROBLEM 9.96
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the L127 ¥ 76 ¥ 12.7-mm angle cross section of
Problem 9.78 with respect to new centroidal axes obtained by rotating
the x and y axes 45∞ counterclockwise.
SOLUTION
I x = 3.93 ¥ 106 mm 4
From Problem 9.84:
I y = 1.06 ¥ 106 mm 4
I xy = 1.165061 ¥ 106 mm 4
Problem 9.78:
The Mohr’s circle is defined by the diameter XY, where X (3.93 ¥ 106 , 1.165061 ¥ 106 ),
Y (1.06 ¥ 106 , -1.165061 ¥ 106 ).
Now
1
1
I ave = ( I x + I y ) = (3.93 + 1.06) ¥ 106 = 2.495 ¥ 106 mm 4
2
2
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
2
¸Ô
ÏÔ 1
È
˘
= Ì Í (3.93 - 1.06)˙ + 1.1650612 ˝ ¥ 106 mm 4
Î2
˚
Ô˛
ÓÔ
= 1.84840 ¥ 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2(1.165061 ¥ 106 )
(3.93 - 1.06) ¥ 106
= -0.81189
=-
or
2q m = -39.073∞
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135
PROBLEM 9.96 (Continued)
Then
Then
a = 180∞ - (39.073∞ + 90∞)
= 50.927∞
I x¢ = I ave - R cos a = (2.495 - 1.84840 cos 50.927∞) ¥ 106
or
I x¢ = 1.330 ¥ 106 mm 4 
I y¢ = I ave + R cos a = (2.495 + 1.84840 cos 50.927∞) ¥ 106
or
I y¢ = 3.66 ¥ 106 mm 4 
I x ¢y ¢ = R sin a = (1.84840 ¥ 106 )sin 50.927∞
or I x ¢y ¢ = 1.435 ¥ 106 mm 4 
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136
PROBLEM 9.97
For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine the
orientation the principal axes at the origin and the corresponding values of
the moments of inertia.
SOLUTION
From Problem 9.79:
Ix =
p 4
a
8
Iy =
p 4
a
2
1 4
a
2
The Mohr’s circle is defined by the diameter XY, where
Problem 9.67:
I xy =
1 ˆ
Êp
X Á a4, a4 ˜
Ë8
2 ¯
Now
1 ˆ
Êp
and Y Á a 4 , - a 4 ˜
Ë2
2 ¯
1
1Êp
p ˆ
I ave = ( I x + I y ) = Á a 4 + a 4 ˜ = 0.98175a 4
2
2Ë 8
2 ¯
and
2
È1
˘
2
R = Í ( I x - I y ) ˙ + I xy
Î2
˚
2
p ˆ˘ Ê 1 ˆ
È1 Ê p
= Í Á a4 - a4 ˜ ˙ + Á a4 ˜
2 ¯˚ Ë 2 ¯
Î2 Ë 8
2
= 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
=-p
8
2
(a)
4
1
2
a 4 - p2 a 4
= 0.84883
or
and
2q m = 40.326∞
q m = 20.2∞
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137
PROBLEM 9.97 (Continued)
The Principal axes are obtained by rotating the xy axes through
20.2∞ counterclockwise 
about O.
Now
I max, min = I ave ± R = 0.98175a 4 ± 0.77264 a 4
or
I max = 1.754 a 4 
and
I min = 0.209 a 4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
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138
PROBLEM 9.98
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.72.
SOLUTION
I x = 68.96 ¥ 106 mm 4
From Problem 9.80:
I y = 132.48 ¥ 106 mm 4
I xy = -21.6 ¥ 106 mm 4
Problem 9.72:
The Mohr’s circle is defined by the diameter XY, where X (68.96 ¥ 106 , - 21.6 ¥ 106 )
and Y (132.48 ¥ 106 , 21.6 ¥ 106 )
Now
1
I ave = ( I x + I y )
2
1
= (68.96 + 132.48) ¥ 106
2
= 100.72 ¥ 106 mm 4
2
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
and
2
¸Ô
ÏÔ 1
È
˘
= Ì Í (68.96 - 132.48) ˙ + ( -21.6)2 ˝ ¥ 106
˚
ÔÓ Î 2
Ô˛
= 38.409 ¥ 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2( -21.6 ¥ 106 )
(68.96 - 132.48) ¥ 106
= -0.68010
=-
or
and
2q m = -34.220∞
q m = -17.11∞
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139
PROBLEM 9.98 (Continued)
The principal axes are obtained by rotating the xy axes through
17.11° clockwise 
about C.
Now
I max, min = I ave ± R = (100.72 + 38.409) ¥ 106
or
I max = 139.1 ¥ 106 mm 4 
and
I min = 62.3 ¥ 106 mm 4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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140
PROBLEM 9.99
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.76.
SOLUTION
From Problem 9.76:
I xy = -3.7266 ¥ 109 mm 4
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )1 =
1
(608 mm)(102 mm)3 = 53.767872 ¥ 106 mm 4
12
( I x )2 =
1
È1
˘ Ê 533
ˆ
(228 mm)(380 mm)3 + Í (228 mm)(380 mm)˙ Á
mm˜
¯
36
Î2
˚Ë 3
2
= 1.71494 ¥ 109 mm 4
( I x )3 =
1
2
È1
˘
(228 mm)(381mm)3 + Í (228 mm)(381mm) ˙ (178 mm )
36
Î2
˚
= 1.72644 ¥ 109 mm 4
Then
I x = [0.053768 + 1.71494 + 1.72644) ¥ 109 mm 4
= 3.495148 ¥ 109 mm 4
Also
where
Then
I y = ( I y )1 + ( I y )2 + ( I y )3
( I y )1 =
1
(102 mm)(608 mm)3 = 1.91042 ¥ 109 mm 4
12
( I y )2 =
È1
˘
1
(380 mm)(228 mm)3 + Í (228 mm)(380 mm) ˙ (228 mm)2 = 2.37706 ¥ 109 mm 4
36
2
Î
˚
( I y )3 =
È1
˘
1
(381 mm)(228 mm)3 + Í (228 mm)(381mm)˙ (228 mm)2 = 2.38331 ¥ 109 mm 4
36
Î2
˚
I y = (1.91042 + 2.37706 + 2.38331) ¥ 109 mm4 = 6.67079 ¥ 109 mm 4
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141
PROBLEM 9.99 (Continued)
The Mohr’s circle is defined by the diameter XY, where
Y (6.67079 ¥ 109 , 3.7266 ¥ 109 )
Now
X (3.495148 ¥ 109 , - 3.7266 ¥ 109 ) and
1
1
I ave = ( I x + I y ) = (3.495148 + 6.67079) ¥ 109 = 5.082969 ¥ 109
2
2
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
2
È1
˘
= Í (3.495148 - 6.67079) ¥ 109 ˙ + ( -3.7266 ¥ 109 )2
Î2
˚
= 4.05077 ¥ 109 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
=-
2( -3.7266 ¥ 109 )
(3.495148 - 6.67079) ¥ 109
= -2.34699
or
and
2q m = -66.922∞
q m = -33.5∞
The principal axes are obtained by rotating the xy axes through
33.5∞ clockwise 
about C.
Now
I max, min = I ave ± R = (5.082969 ± 4.05077) ¥ 109 mm 4
or
I max = 9.13 ¥ 109 mm 4 
and
I min = 1.032 ¥ 109 mm 4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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142
PROBLEM 9.100
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.73.
SOLUTION
From Problem 9.81:
I x = (126.5625 ¥ 106 )p mm 4
Problem 9.73:
I xy = 337.5 ¥ 106 mm 4
I y = (253.125 ¥ 106 )p mm 4
The Mohr’s circle is defined by the diameter XY, where X (126.5625 ¥ 106 p , 337.5 ¥ 106 ) and Y (253.125 ¥ 106 p ,
-337.5 ¥ 106 ).
1
Now
I ave = ( I x + I y ) = 596.412 ¥ 106 mm 4
2
and
È1
˘ 2
R = Í ( I x - I y )2 ˙ + I xy
Î2
˚
= 391.700 ¥ 106 mm 4
(from Solution of Problem 9.87)
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
=-
2(337.5 ¥ 106 )
(126.5625 - 253.125) ¥ 106 p
= 1.69765
or
and
2q m = 59.4998∞
q m = 29.7∞
The principal axes are obtained by rotating the xy axes through
29.7° counterclockwise 
about C.
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143
PROBLEM 9.100 (Continued)
Now
I max, min = I ave ± R = 596.412 ¥ 106 ± 391.700 ¥ 106
or
I max = 988 ¥ 106 mm 4 
and I min = 205 ¥ 106 mm 4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
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144
PROBLEM 9.101
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.74.
SOLUTION
From Problem 9.83:
I x = 0.162 ¥ 106 mm 4
I y = 0.454 ¥ 106 mm 4
Problem 9.74:
I xy = -0.159584 ¥ 106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (0.162 ¥ 106 , - 0.159584 ¥ 106 ) and Y (0.454 ¥ 106 ,
0.159584 ¥ 106 ).
Now
1
I ave = ( I x + I y ) = 0.308 ¥ 106 mm 4
2
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
= 0.21629 ¥ 106 mm 4
(from Solution of Problem 9.89)
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2( -0.159584 ¥ 106 )
(0.162 - 0.454) ¥ 106
= -1.09304
=-
Then
and
2q m = -47.545
q m = -23.8∞
The principal axes are obtained by rotating the xy axes through
23.8° clockwise b
about C.
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145
PROBLEM 9.101 (Continued)
Now
I max, min = I ave ± R = 0.308 ¥ 106 ± 0.21629 ¥ 106
or
and
I max = 0.524 ¥ 106 mm 4 b
I min = 0.0917 ¥ 106 mm 4 b
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
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146
PROBLEM 9.102
Using Mohr’s circle, determine for the area indicated the orientation of the
principal centroidal axes and the corresponding values of the moments
of inertia.
Area of Problem 9.77
(The moments of inertia I x and I y of the area of Problem 9.102 were
determined in Problem 9.44).
SOLUTION
From Problem 9.44:
I x = 2.90051 ¥ 106 mm 4
I y = 0.721271 ¥ 106 mm 4
Problem 9.77:
I xy = -0.680632 ¥ 106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (2.90051 ¥ 106 , - 0.680632 ¥ 106 ) and Y (0.72127 ¥ 106 ,
0.680632 ¥ 106 ).
Now
1
I ave = ( I x + I y ) = 1.81089 ¥ 106 mm 4
2
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
2
Î
˚
2
È1
˘
= Í (2.90051 - 0.721271) ¥ 106 ˙ + ( -0.680632 ¥ 106 )2
2
Î
˚
= 1.28473 ¥ 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2( -0.680632 ¥ 106 )
(2.90051 - 0.721271) ¥ 106
= 0.6246
2q m = 31.99∞
=-
or
and
q m = 16.00∞
The principal axes are obtained by rotating the xy axes through
16.00° counterclockwise b
about C.
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147
PROBLEM 9.102 (Continued)
Now
I max, min = I ave ± R = (1.81089 ± 1.28473) ¥ 106
I max = 3.10 ¥ 106 mm 4 b
or
I min = 0.526 ¥ 106 mm 4 b
and
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
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148
PROBLEM 9.103
The moments and product of inertia of an L102 ¥ 102 ¥ 12.7-mm angle cross section with respect to two
rectangular axes x and y through C are, respectively, I x = 2.30 ¥ 106 mm4, I y = 2.30 ¥ 106 mm4, and I xy 0,
with the minimum value of the moment of inertia of the area with respect to any axis through C being
I min = 1.15 ¥ 106 mm4. Using Mohr’s circle, determine (a) the product of inertia I xy of the area, (b) the
orientation of the principal axes, (c) the value of I max .
SOLUTION
(Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when
the 104-mm leg of the angle is parallel to the x axis. Further, for I xy 0, the angle must be oriented as shown.)
Now
1
1
I ave = ( I x + I y ) = (2.30 ¥ 106 ) ¥ 2 = 2.30 ¥ 106 mm 4
2
2
and
I min = I ave - R or
R = 2.30 ¥ 106 - 1.15 ¥ 106
= 1.15 ¥ 106 mm 4
Using I ave and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, X (2.30 ¥ 106 , I xy )
and Y (2.30 ¥ 106 , | I xy |).
2
(a)
We have
È1
˘
R2 = Í ( I x - I y ) ˙ + I xy2 ; Since I x = I y
2
Î
˚
R = I xy
or
I xy = 1.15 ¥ 106 mm 4
Solving for I xy and taking the negative root (since I xy 0) yields
I xy = -1.15 ¥ 106 mm 4
I xy = -1.150 ¥ 106 mm 4 b
(b)
We have from formula or otherwise clearly
2q m = - 90∞
or
q m = - 45∞
The principal axes are obtained by rotating the xy axes through
45° clockwise b
(c)
about C.
We have
I max = I ave + R = 2.30 ¥ 106 + 1.15 ¥ 106
= 3.45 ¥ 106 mm 4
or
I max = 3.45 ¥ 106 mm 4 b
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149
PROBLEM 9.104
Using Mohr’s circle, determine for the cross section of the rolled-steel
angle shown the orientation of the principal centroidal axes and the
corresponding values of the moments of inertia. (Properties of the
cross sections are given in Figure 9.13.)
SOLUTION
I x = 0.162 ¥ 106 mm 4
From Figure 9.13B:
I y = 0.454 ¥ 106 mm 4
We have
I xy = ( I xy )1 + ( I xy )2
For each rectangle
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0
and
I xy = S x yA
(symmetry)
A, mm2
x, mm
y, mm
x yA, mm 4
1
76 ¥ 6.4 = 486.4
-13.1
9.2
-58620.93
2
6.4 ¥ (51 - 6.4) = 285.44
21.7
–16.3
–100962.98
Â
–159583.91
I xy = -159584 mm 4
The Mohr’s circle is defined by the diameter XY where X (0.162 ¥ 106 , - 0.159584 ¥ 106 )
and Y (0.454 ¥ 106 , 0.159584 ¥ 106 )
Now
1
1
I ave = ( I x + I y ) = (0.162 + 0.454) ¥ 106
2
2
= 0.3080 ¥ 106 mm 4
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150
PROBLEM 9.104 (Continued)
and
2
2
ÏÔ 1
¸Ô
È1
˘
È
˘
2
R = Í ( I x - I y ) ˙ + I xy
= Ì Í (0.162 - 0.454)˙ + ( -0.159584)2 ˝ ¥ 106
Î2
˚
˚
ÔÓ Î 2
Ô˛
= 0.21629 ¥ 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2( -0.159584 ¥ 106 )
(0.162 - 0.454) ¥ 106
= -1.09304
=-
or
and
2q m = - 47.545
q m = -23.8∞
The principal axes are obtained by rotating the xy axes through
23.8° clockwise b
Now
About C.
I max, min = I ave ± R = (0.3080 ± 0.21629) ¥ 106
or I max = 0.524 ¥ 106 mm 4 b
and
I min = 0.0917 ¥ 106 mm 4 b
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
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151
PROBLEM 9.105
Using Mohr’s circle, determine for the cross section of the rolled-steel
angle shown the orientation of the principal centroidal axes and the
corresponding values of the moments of inertia. (Properties of the
cross sections are given in Figure 9.13.)
SOLUTION
From Figure 9.13B:
I x = 3.93 ¥ 106 mm 4
I y = 1.06 ¥ 106 mm 4
Problem 9.7B:
I xy = -1.165061 ¥ 106 mm 4
(Note that the figure of Problem 9.105 is obtained by replacing x with –x in the figure of Problem 9.78; thus
the change in sign of I xy .)
The Mohr’s circle is defined by the diameter XY, where X (3.93 ¥ 106 , - 1.165061 ¥ 106 ) Y (1.06 ¥ 106 , and
1.165061 ¥ 106 ).
Now
1
I ave = ( I x + I y )
2
1
= (3.93 + 1.06) ¥ 106
2
= 2.495 ¥ 106 mm 4
2
and
È1
˘
R = Í ( I x - I y ) ˙ + I xy2
Î2
˚
2
¸Ô
ÏÔ 1
È
˘
= Ì Í (3.93 - 1.06)˙ + ( -1.165061)2 ˝ ¥ 106
˚
ÔÓ Î 2
Ô˛
6
4
= 1.84840 ¥ 10 mm
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152
PROBLEM 9.105 (Continued)
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
2( -1.165061 ¥ 106 )
(3.93 - 1.06) ¥ 106
= 0.81189
=-
or
and
2q m = 39.073∞
q m = 19.54∞
The principal axes are obtained by rotating the xy axes through
19.54° counterclockwise b
Now
about C.
I max, min = I ave ± R = (2.495 ± 1.84840) ¥ 106
or
I max = 4.34 ¥ 106 mm 4 b
and
I min = 0.647 ¥ 106 mm 4 b
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
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153
PROBLEM 9.106*
For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are
I x = 12 × 106 mm4 and I y = 3 × 106 mm4, respectively. Knowing that after rotating the x and y axes about the
centroid 30∞ counterclockwise, the moment of inertia relative to the rotated x axis is 14.5 × 106 mm4, use Mohr’s
circle to determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.
SOLUTION
1
1
I ave = ( I x + I y ) = (12 + 3) ¥ 106 = 7.5 ¥ 106 mm 4
2
2
We have
Now observe that I x I ave , I x ¢ I x , and 2q = +60∞. This is possible only if I xy < 0. Therefore, assume
I xy 0 and (for convenience) I x ¢y ¢ 0. Mohr’s circle is then drawn as shown.
2q m + a = 60∞
We have
Now using DABD:
R=
I x - I ave 12 ¥ 106 - 7.5 ¥ 106
=
cos 2q m
cos 2q m
=
Using DAEF:
R=
or
Expanding:
or
or
( mm 4 )
I x ¢ - I ave 14.5 ¥ 106 - 7.5 ¥ 106
=
cos a
cos a
=
Then
4.5 ¥ 106
cos 2q m
7 ¥ 106
cos a
4.5 ¥ 106 7 ¥ 106
=
cos 2q m
cos a
( mm 4 )
a = 60∞ - 2q m
9 cos (60∞ - 2q m ) = 14 cos 2q m
9(cos 60∞ cos 2q m + sin 60∞ sin 2q m ) = 14 cos 2q m
tan 2q m =
14 - 9 cos 60∞
= 1.21885
9 sin 60∞
2q m = 50.633∞ and q m = 25.3∞
(Note: 2q m 60∞ implies assumption I x ¢y ¢ 0 is correct.)
Finally,
R=
4.5 ¥ 106
= 7.0946 ¥ 106 mm 4
cos 50.633∞
(a)From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given centroidal
x and y axes through q m about the centroid C or
25.3° counterclockwise b
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154
PROBLEM 9.106* (Continued)
(b)
We have
I max, min = I ave ± R = (7.5 ± 7.0946) ¥ 106
or
I max = 14.59 ¥ 106 mm 4 b
and
I min = 0.405 ¥ 106 mm 4 b
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
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155
PROBLEM 9.107
It is known that for a given area I y = 48 × 106 mm4 and I xy = –20 × 106 mm4, where the x and y axes are
rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating
the x axis 67.5∞ counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia I x of the area,
(b) the principal centroidal moments of inertia.
SOLUTION
First assume I x I y and then draw the Mohr’s circle as shown. (Note: Assuming I x I y is not consistent with the
requirement that the axis corresponding to ( I xy ) max is obtained after rotating the x axis through 67.5° CCW.)
From the Mohr’s circle we have
2q m = 2(67.5∞) - 90∞ = 45∞
(a)
From the Mohr’s circle we have
Ix = I y + 2
(b)
We have
and
Now
| I xy |
tan 2q m
= 48 ¥ 106 + 2
20 ¥ 106
tan 45∞
or
I x = 88.0 ¥ 106 mm 4 b
or
I max = 96.3 ¥ 106 mm 4 b
and
I min = 39.7 ¥ 106 mm 4 b
1
1
I ave = ( I x + I y ) = (88.0 + 48) ¥ 106
2
2
6
= 68.0 ¥ 10 mm 4
R=
| I xy |
sin 2q m
=
20 ¥ 106
= 28.284 ¥ 106 mm 4
sin 45∞
I max, min = I ave ± R = (68.0 ± 28.284) ¥ 106
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156
PROBLEM 9.108
Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with
respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of
rectangular axes through the centroid is zero.
SOLUTION
Consider the regular pentagon shown, with centroidal axes x and y.
Because the y axis is an axis of symmetry, it follows that I xy = 0. Since I xy = 0, the x and y axes must be
principal axes. Assuming I x = I max and I y = I min , the Mohr’s circle is then drawn as shown.
Now rotate the coordinate axes through an angle a as shown; the resulting moments of inertia, I x¢ and I y¢ ,
and product of inertia, I x ¢y ¢ , are indicated on the Mohr’s circle. However, the x ¢ axis is an axis of symmetry,
which implies I x ¢y ¢ = 0. For this to be possible on the Mohr’s circle, the radius R must be equal to zero (thus,
the circle degenerates into a point). With R = 0, it immediately follows that
(a)
I x = I y = I x ¢ = I y ¢ = I ave (for all moments of inertia with respect to an axis through C ) b
(b)
I xy = I x ¢y ¢ = 0 (for all products of inertia with respect to all pairs of rectangular axes with origin at C ) b
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157
PROBLEM 9.109
Using Mohr’s circle, prove that the expression I x ¢ I y ¢ - I x2¢y ¢ is independent of the orientation of the x¢ and y¢
axes, where Ix¢, Iy¢, and Ix¢y¢ represent the moments and product of inertia, respectively, of a given area with
respect to a pair of rectangular axes x¢ and y¢ through a given Point O. Also show that the given expression is
equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr’s circle.
SOLUTION
First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R
are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the moments of
inertia, I x¢ and I y¢ , and the product of inertia, I x ¢y ¢ , having been computed for an arbitrary orientation of the
x ¢y ¢ axes.
From the Mohr’s circle
I x ¢ = I ave + R cos 2q
I y ¢ = I ave - R cos 2q
I x ¢y ¢ = R sin 2q
Then, forming the expression
I x ¢ I y ¢ - I x2¢y ¢
I x ¢ I y ¢ - I x2¢y ¢ = ( I ave + R cos 2q )( I ave - R cos 2q ) - ( R sin 2q )2
(
)
2
= I ave
- R2 cos2 2q - ( R2 sin 2 2q )
2
= I ave
- R2
which is a constant
I x ¢ I y ¢ - I x2¢y ¢ is independent of the orientation of the coordinate axes Q.E.D. b
Shown is a Mohr’s circle, with line OA, of length L, the required tangent.
Noting that
OAC is a right angle, it follows that
2
L2 = I ave
- R2
or
L2 = I x ¢ I y ¢ - I x2¢y ¢ Q.E.D. b
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158
PROBLEM 9.110
Using the invariance property established in the preceding problem, express the product
of inertia Ixy of an area A with respect to a pair of rectangular axes through O in terms
of the moments of inertia Ix and Iy of A and the principal moments of inertia Imin and
Imax of A about O. Use the formula obtained to calculate the product of inertia Ixy of the
L76 × 51 × 6.4 mm angle cross section shown in Figure 9.13A, knowing that its
maximum moment of inertia is 0.6 × 106 mm4.
SOLUTION
Consider the following two sets of moments and products of inertia, which correspond to two different
orientations of the coordinate axes whole origin is at Point O.
Case 1:
I x ¢ = I x , I y ¢ = I y , I x ¢y ¢ = I xy
Case 2:
I x ¢ = I max , I y ¢ = I min , I x ¢y ¢ = 0
The invariance property then requires
2
I x I y - I xy
= I max I min
From Figure 9.13A:
or
I xy = ± I x I y - I max I min b
I x = 0.454 ¥ 106 mm 4
I y = 0.162 ¥ 106 mm 4
Using Eq. (9.21):
Substituting
or
Then
I x + I y = I max + I min
(0.454 + 0.162) ¥ 106 = 0.6 ¥ 106 + I min
I min = 0.016 ¥ 106 mm 4
I xy = ± (0.454 ¥ 106 )(0.162 ¥ 106 ) - (0.6 ¥ 106 )(0.016 ¥ 106 )
= ±0.25288 ¥ 106 mm 4
The two roots correspond to the following two orientations of the cross section.
I xy = -0.253 ¥ 106 mm 4 b
For
I xy = 0.253 ¥ 106 mm 4 b
and for
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159
PROBLEM 9.111
Determine the mass moment of inertia of a ring of mass m, cut from a thin
uniform plate, with respect to (a) the axis AA¢, (b) the centroidal axis CC¢ that
is perpendicular to the plane of the ring.
SOLUTION
Mass = m = rV = rtA
m
I mass = rt I area = I area
A
We first determine
(
A = p r22 - p r12 = p r22 - r12
I AA¢ , area =
(a)
I AA¢ , mass =
(
p 4 p 4 p 4
r2 - r1 = r2 - r14
4
4
4
(
)
)
(
1
m
m
p 4
r2 - r14 = m r22 + r12
I AA¢ , area =
2
2
4
A
p r2 - r1 4
(
)
(b)
)
By symmetry:
I BB ¢ = IAA¢
Eq. (9.38):
I CC ¢ = IAA¢ + I BB ¢ = 2 IAA¢ )
(
)
(
)
I AA¢ =
1
m r12 + r22 b
4
I CC ¢ =
1
m r12 + r22 b
2
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160
PROBLEM 9.112
A thin semicircular plate has a radius a and a mass m. Determine the mass
moment of inertia of the plate with respect to (a) the centroidal axis B¢, (b) the
centroidal axis CC¢ that is perpendicular to the plate.
SOLUTION
mass = m = rtA
I mass = rt I area =
m
I area
A
1
A = p a2
2
Area:
1Êp ˆ 1
I AA¢ , area = I DD ¢ , area = Á a 4 ˜ = p a 4
2Ë 4 ¯ 8
I AA¢ , mass = I DD ¢ , mass =
I BB ¢ = I DD ¢
(a)
m
m Ê1
ˆ 1
I AA¢ , area = 1 2 Á p a 4 ˜ = ma2
¯ 4
Ë
A
8
2 pa
1
Ê 4a ˆ
- m( AC ) = ma2 - m Á ˜
Ë 3p ¯
4
2
= (0.25 - 0.1801)ma2 (b)
Eq. (9.38):
I CC ¢ = I AA¢ + I BB ¢ =
2
I BB¢ = 0.0699ma2 b
1
ma2 + 0.0699ma2
4
I CC ¢ = 0.320ma2 b
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161
PROBLEM 9.113
The quarter ring shown has a mass m and was cut from a thin, uniform
plate. Knowing that r1 = 43 r2, determine the mass moment of inertia of the
quarter ring with respect to (a) the axis AA¢, (b) the centroidal axis CC¢ that
is perpendicular to the plane of the quarter ring.
SOLUTION
First note
mass = m = rV = rtA
p
= rt r22 - r12
4
I mass = rt I area
(
Also
=
(a)
(
m
)
)
p 2
r2 - r12
4
p 4
=
r2 - r14
16
(
Using Figure 9.12,
I AA¢ , area
Then
I AA¢ , mass =
I area
)
(
(
)
)
mÈ
Ê3 ˆ
= Ír22 + Á r2 ˜
Ë4 ¯
4 ÍÎ
(b)
(
m
p 4
¥
r2 - r14
p 2
16
2
r2 - r1
4
m 2
=
r2 + r12
4
2˘
˙
˙˚
)
or
I AA¢ =
25 2
mr2 b
64
Symmetry implies
I BB ¢ , mass = I AA¢ , mass
Then
I DD ¢ = I AA¢ + I BB ¢
ˆ
Ê 25
= 2 Á mr22 ˜
¯
Ë 64
=
25 2
mr2
32
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162
PROBLEM 9.113 (Continued)
Now locate centroid C.
X SA = Sx A
or
p ˆ 4r Ê p ˆ 4r Ê p ˆ
Êp
X Á r22 - r12 ˜ = 2 Á r22 ˜ - 1 Á r12 ˜
Ë4
4 ¯ 3p Ë 4 ¯ 3p Ë 4 ¯
4 r23 - r13
3p r22 - r12
or
X=
Now
r=X 2
3
Ê3 ˆ
r3 - r
4 2 2 ÁË 4 2 ˜¯
=
3p 2 Ê 3 ˆ 2
r2 - Á r2 ˜
Ë4 ¯
=
Finally,
37 2
r2
21p
I DD ¢ = I CC ¢ + mr 2
2
or
Ê 37 2 ˆ
25 2
mr2 = I CC ¢ + m Á
r2 ˜ 32
Ë 21p ¯
or
I CC ¢ = 0.1522mr22 b
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163
PROBLEM 9.114
The parabolic spandrel shown was cut from a thin, uniform plate.
Denoting the mass of the spandrel by m, determine its mass moment
of inertia with respect to (a) the axis BB ¢, (b) the axis DD¢ that is
perpendicular to the spandrel. (Hint: See Sample Problem 9.3.)
SOLUTION
First note
Also
mass = m = rV = rtA
I mass
Êp ˆ
= rt Á ab˜
Ë2 ¯
= rt I area
=
(a)
2m
I area
p ab
We have
I x , area = I BB ¢ , area + Ay 2
or
(a)
I BB ¢ , area =
p 3 Ê p ˆ Ê 4b ˆ
ab - Á ab˜ Á ˜
Ë 2 ¯ Ë 3p ¯
8
From Sample Problem 9.3:
Then
I BB ¢ , area =
1 3
ab
21
I BB ¢ , mass =
3m 1 3
¥ ab
ab 21
(b)
2
or
I BB¢ =
1 2
mb b
7
From Sample Problem 9.3:
1
I AA¢ , area = a3b
5
Now
I AA¢ , area = I11¢ , area
Ê3 ˆ
+ A Á a˜
Ë4 ¯
2
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164
PROBLEM 9.114 (Continued)
and
Then
and
Finally,
I 22¢ , area = I11¢ , area
Ê1 ˆ
+ A Á a˜
Ë4 ¯
2
ÈÊ 1 ˆ 2 Ê 3 ˆ 2 ˘
I 22¢ , area = I AA¢ , area + A ÍÁ a˜ - Á a˜ ˙
Ë4 ¯ ˙
ÍÎË 4 ¯
˚
1 3
1 Ê 1 2 9 2ˆ
= a b + ab Á a - a ˜
3 Ë 16
16 ¯
5
1 3
=
ab
30
3m 1 3
¥ ab
ab 30
1
= ma2
10
I 22¢ , mass =
I DD ¢ , mass = I BB ¢ , mass + I 22¢ , mass
=
1 2 1
mb + ma2 7
10
or
I DD¢ =
1
m(7a2 + 10b2 ) b
70
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165
PROBLEM 9.115
A thin plate of mass m was cut in the shape of a parallelogram as shown.
Determine the mass moment of inertia of the plate with respect to (a) the
x axis, (b) the axis B¢, which is perpendicular to the plate.
SOLUTION
mass = m = rt A
I mass = rt I area =
m
I area
A
(a)Consider parallelogram as made of horizontal strips and slide strips to form a square since distance from
each strip to x axis is unchanged.
1
I x, area = a 4
3
m
m Ê1 ˆ
I x , mass = I x , area = 2 Á a 4 ˜ A
a Ë3 ¯
(b)
1
I x = ma2 b
3
For centroidal axis y¢:
È1 ˘ 1
I y ¢ , area = 2 Í a 4 ˙ = a 4
Î12 ˚ 6
m
m Ê1 ˆ 1
I y ¢ , area = 2 Á a 4 ˜ = ma2
A
a Ë6 ¯ 6
7
1
= I y ¢ + ma2 = ma2 + ma2 = ma2
6
6
I y ¢ , mass =
I y ¢¢
For axis BB¢ ^ to plate, Eq. (9.38):
1
7
I BB ¢ = I x + I y ¢¢ = ma2 + ma2 3
6
I BB¢ =
3 2
ma b
2
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166
PROBLEM 9.116
A thin plate of mass m was cut in the shape of a parallelogram as shown.
Determine the mass moment of inertia of the plate with respect to (a) the
y axis, (b) the axis AA¢, which is perpendicular to the plate.
SOLUTION
See sketch of the solution of Problem 9.115.
(a)
From Part b of solution of Problem 9.115:
1
I y¢ = ma2
6
1
I y = I y ¢ + ma2 = ma2 + ma2 6
(b)
Iy =
7 2
ma b
6
I AA¢ =
1 2
ma b
2
From solution of Problem 9.115:
1
I y ¢ = ma2
6
Eq. (9.38):
and
1
I x = ma2
3
I AA¢ = I y ¢ + I x
1
1
= ma2 + ma2 6
3
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167
PROBLEM 9.117
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the x axis,
(b) the y axis.
SOLUTION
First note
mass = m = rV = rtA
1
È
˘
= rt Í(2a)( a) + (2a)( a)˙ = 3rta2
2
Î
˚
I mass = rtI area =
Also
(a)
Now
m
3a2
I area
I x , area = ( I x )1, area + ( I x )2, area
1
1
= (2a)( a)3 + (2a)( a)3
12
3
Then
(b)
I x, mass
5
= a4
6
m 5
= 2 ¥ a4 6
3a
or
I x, mass =
5
ma2 b
18
We have
I z , area = ( I z )1, area + ( I z )2, area
1
1
ˆ
È1
Ê
˘ È1
= Í ( a)(2a)3 ˙ + Í ( a)(2a)3 + (2a)( a) Á 2a + ¥ 2a˜
¯
Ë
2
3
Î3
˚ ÍÎ 36
m
¥ 10a 4 =
I x, mass =
Finally,
I y , mass = I x , mass + I z , mass
3a
˙ = 10 a
˙˚
4
10 2
ma
3
Then
2
2˘
=
5
10
ma2 + ma2
18
3
=
65 2
ma 18
or I y, mass = 3.61ma2 b
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168
PROBLEM 9.118
A thin plate of mass m has the trapezoidal shape shown. Determine the
mass moment of inertia of the plate with respect to (a) the centroidal
axis CC¢ that is perpendicular to the plate, (b) the axis AA¢ that is
parallel to the x axis and is located at a distance 1.5a from the plate.
SOLUTION
First locate the centroid C.
1
ˆ
Ê
X S A = S xA: X (2a2 + a2 ) = a(2a2 ) + Á 2a + ¥ 2a˜ ( a2 )
¯
Ë
3
X=
or
14
a
9
Ê1 ˆ
Ê1 ˆ
Z S A = S zA: Z (2a2 + a2 ) = Á a˜ (2a2 ) + Á a˜ ( a2 )
Ë2 ¯
Ë3 ¯
Z=
or
(a)
We have
4
a
9
I y , mass = I CC ¢ , mass + m( X 2 + Z 2 )
From the solution to Problem 9.117:
I y, mass =
Then
I cc¢ , mass
65 2
ma
18
ÈÊ 14 ˆ 2 Ê 4 ˆ 2 ˘
65 2
= ma - m ÍÁ a˜ + Á a˜ ˙
Ë9 ¯ ˙
18
ÍÎË 9 ¯
˚
or
(b)
We have
I cc¢ = 0.994ma2 b
I x , mass = I BB ¢ , mass + m( Z )2
and
I AA¢ , mass = I BB ¢ , mass + m(1.5a)2
Then
2
È
Ê4 ˆ ˘
I AA¢ , mass = I x , mass + m Í(1.5a)2 - Á a˜ ˙
Ë9 ¯ ˙
ÍÎ
˚
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169
PROBLEM 9.118 (Continued)
From the solution to Problem 9.117:
I x, mass =
Then
I AA¢ , mass =
5
ma2
18
2
È
5
Ê4 ˆ ˘
ma2 + m Í(1.5a)2 - Á a˜ ˙
Ë9 ¯ ˙
18
ÍÎ
˚
or
I AA¢ = 2.33ma2 b
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170
PROBLEM 9.119
The area shown is revolved about the x axis to form a homogeneous
solid of revolution of mass m. Using direct integration, express the
mass moment of inertia of the solid with respect to the x axis in
terms of m and h.
SOLUTION
We have
y=
2h - h
x+h
a
so that
r=
h
( x + a)
a
For the element shown:
dm = rp r 2 dx dI x =
1 2
r dm
2
2
Èh
˘
= rp Í ( x + a) ˙ dx
Îa
˚
h2
h2
1
2
x
+
a
dx
=
rp
(
)
[( x + a)3 ]0a
0
3
a2
a2
h2
7
1
= rp 2 (8a3 - a3 ) = rp ah2
3
3
a
a
Then
m = Ú dm = Ú rp
Now
a Èh
1
1
˘
I x = Ú dI x = Ú r 2 ( rp r 2 dx ) = rp Ú Í ( x + a)˙ dx
0 Îa
2
2
˚
4
h4
1
1
1 h4
5 a
rp ¥
[
(
x
+
a
)
]
=
rp
(32a5 - a5 )
0
4
10
2
5 a4
a
31
= rp ah4
10
=
From above:
Then
rp ah2 =
Ix =
3
m
7
31 Ê 3 ˆ 2 93 2
mh
Á m˜ h =
10 Ë 7 ¯
70
or
I x = 1.329 mh2 b
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171
PROBLEM 9.120
Determine by direct integration the mass moment of inertia with respect
to the y axis of the right circular cylinder shown, assuming that it has a
uniform density and a mass m.
SOLUTION
For element shown:
dm = r dV = rp a2 dx
dI y = dI y + x 2 dm
=
1 2
a dm + x 2 dm
4
ˆ
Ê1
= Á a2 + x 2 ˜ rp a2 dx
¯
Ë4
I y = Ú dI y = Ú
+ L/2
- L/2
ˆ
Ê1
rp a2 Á a2 + x 2 ˜ dx
¯
Ë4
x3
1 2
= rp a
a x+
4
3
+ L/2
2
Iy =
But for whole cylinder:
- L/2
È1 L 1 Ê Lˆ3 ˘
= 2rp a2 Í a2 + Á ˜ ˙
ÍÎ 4 2 3 Ë 2 ¯ ˙˚
1
rp a2 L(3a2 + L2 ) 
12
m = rV = rp a2 L
Iy =
Thus,
1
m(3a2 + L2 ) b
12
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172
PROBLEM 9.121
The area shown is revolved about the x axis to form a homogeneous
solid of revolution of mass m. Determine by direct integration
the mass moment of inertia of the solid with respect to (a) the
x axis, (b) the y axis. Express your answers in terms of m and the
dimensions of the solid.
SOLUTION
We have at
( a, h): h =
k
a
k = ah
or
For the element shown:
r=4
dm = rp r 2 dx
2
Ê ah ˆ
= rp Á ˜ dx
Ë x¯
m = Ú dm = Ú
Then
3a
a
2
Ê ah ˆ
rp Á ˜ dx
Ë x¯
3a
È 1˘
= rp a2 h2 Í- ˙
Î x ˚a
È 1 Ê 1ˆ ˘ 2
= rp a2 h2 Í- - Á - ˜ ˙ = rp ah2
Î 3a Ë a ¯ ˚ 3
(a)
For the element:
Then
dI x =
1 2
1
r dm = rp r 4 dx
2
2
I x = Ú dI x = Ú
4
a
3a
1
È 1 1˘
Ê ah ˆ
rp Á ˜ dx = rp a 4 h4 Í- 3 ˙
¯
Ë
x
2
2
Î 3 x ˚a
3a 1
ÈÊ 1 ˆ 3 Ê 1 ˆ 3 ˘ 1 26
1
= - rp a 4 h4 ÍÁ ˜ - Á ˜ ˙ = ¥ rp ah4
6
ÍÎË 3a ¯ Ë a ¯ ˙˚ 6 27
=
13
13 2
1 2
¥ rp ah2 ¥ h2 =
mh
6 3
9
54
or
I x = 0.241 mh2 b
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173
PROBLEM 9.121 (Continued)
(b)
For the element:
dI y = dI y + x 2 dm
=
1 2
r dm + x 2 dm
4
˘ Ê ah ˆ
3a È 1 Ê ah ˆ
I y = Ú dI y = Ú Í Á ˜ + x 2 ˙ rp Á ˜ dx
a
ÍÎ 4 Ë x ¯
˙˚ Ë x ¯
2
Then
=
3a Ê 1
rp a2 h2
a Á
Ë4
Ú
2
3a
2 2
˘
a 2 h2 ˆ
2 2È 1 a h
1
rp
dx
a
h
+
=
+ x˙
Í
˜
4
3
x
¯
Î 12 x
˚a
2 2
˘ Ô¸
˘ È 1 a 2 h2
Ê 3 ˆ ÔÏ È 1 a h
- Í+ a˙ ˝
3
a
+
= Á m˜ a Ì Í˙
3
3
Ë 2 ¯ Ô Î 12 (3a)
˚ ˛Ô
˚ Î 12 ( a)
Ó
=
Ê 1 26 h2
ˆ
3
ˆ
Ê 13 2
ma Á ¥
+ 2 a˜ = m Á
h + 3a2 ˜
¯
Ë 108
2
Ë 12 27 a
¯
or
I y = m(3a2 + 0.1204h2 ) b
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174
PROBLEM 9.122
Determine by direct integration the mass moment of inertia with respect
to the x axis of the tetrahedron shown, assuming that it has a uniform
density and a mass m.
SOLUTION
We have
x=
a
Ê
y + a = a Á1 +
Ë
h
yˆ
˜
h¯
and
z=
b
Ê
y + b = b Á1 +
Ë
h
yˆ
˜
h¯
For the element shown:
Then
2
yˆ
˜ dy
h¯
Ê
ˆ 1
Ê1
dm = r Á xz dy˜ = r ab Á1 +
Ë
¯ 2
Ë2
2
yˆ
1
Ê
r ab Á1 + ˜ dy
-h 2
Ë
h¯
m = Ú dm = Ú
0
h ÈÊ
1
= r ab ¥ ÍÁ1 +
2
3 ÍÎË
yˆ
˜
h¯
3 ˘0
˙
˙˚ - h
1
r abh ÈÎ(1)3 - (1 - 1)3 ˘˚
6
1
= r abh
6
=
Now, for the element:
Then
I AA¢ , area =
1 3 1 3Ê
xz = ab Á1 +
Ë
36
36
yˆ
˜
h¯
4
4
È1
yˆ ˘
Ê
dI AA¢ , mass = rtI AA¢ , area = r( dy ) Í ab3 Á1 + ˜ ˙
Ë
h ¯ ˚˙
ÍÎ 3
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175
PROBLEM 9.122 (Continued)
Now
2
È
Ê1 ˆ ˘
dI x = dI AA¢, mass + Í y 2 + Á z ˜ ˙ dm
Ë3 ¯ ˙
ÍÎ
˚
=
1
Ê
r ab3 Á1 +
Ë
36
4
yˆ
˜ dy
h¯
ÏÔ
È1 Ê
+ Ì y 2 + Í b Á1 +
Î3 Ë
ÓÔ
=
Now
Then
and
m=
1
Ê
r ab3 Á1 +
Ë
12
yˆ˘
˜
h ¯ ˙˚
2¸
Ô È1
Ê
˝ Í r ab ÁË1 +
2
˛Ô ÍÎ
2
˘
yˆ
˜¯ dy ˙
h
˙˚
4
Ê 2
1
yˆ
y3 y 4 ˆ
+
+
2
+
dy
r
ab
y
dy
˜
Á
2
h¯
h h2 ˜¯
Ë
1
r abh
6
È 1 b2 Ê
dI x = Í m Á1 +
ÍÎ 2 h Ë
4
3m Ê 2
yˆ
y3 y 4 ˆ ˘
y + 2 + 2 ˜ ˙ dy
˜¯ +
Á
h
h Ë
h h ¯ ˙˚
4
Ê 2
m È 2Ê
yˆ
y3 y 4 ˆ ˘
Íb Á1 + ˜ + 6 Á y + 2 + 2 ˜ ˙ dy
- h 2h
h¯
h h ¯ ˙˚
Ë
ÍÎ Ë
I x = Ú dI x = Ú
0
m È 2 hÊ
=
Íb ¥ Á 1 +
2h ÍÎ
5Ë
=
0
5
Ê 1 3 1 y4
yˆ
y5 ˆ ˘
+ 2 ˜˙
˜¯ + 6 Á y +
2 h 5h ¯ ˙˚
h
Ë3
-h
1
1
m Ï1 2
È1
5
3
4
5 ˘¸
Ì b h(1) - 6 Í ( - h) + ( - h) + 2 ( - h) ˙ ˝
2h Ó 5
2h
5h
Î3
˚˛
or
Ix =
1
m(b2 + h2 ) b
10
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176
PROBLEM 9.123
Determine by direct integration the mass moment of inertia with respect
to the y axis of the tetrahedron shown, assuming that it has a uniform
density and a mass m.
SOLUTION
We have
x=
a
yˆ
Ê
y + a = a Á1 + ˜
Ë
h
h¯
and
z=
b
yˆ
Ê
y + b = b Á1 + ˜
Ë
h
h¯
For the element shown:
Then
Ê
ˆ 1
Ê1
dm = r Á xz dy˜ = r ab Á1 +
Ë
¯ 2
Ë2
1
Ê
r ab Á1 +
-h 2
Ë
m = Ú dm = Ú
0
2
yˆ
˜ dy
h¯
2
yˆ
˜ dy
h¯
h ÈÊ
1
= r ab ¥ ÍÁ1 +
2
3 ÍÎË
0
3
yˆ ˘
˜ ˙
h ¯ ˙˚
-h
1
r abh ÈÎ(1)3 - (1 - 1)3 ˘˚
6
1
= r abh
6
=
Also
Then, using
we have
I BB ¢ , area =
1 3
xz
12
I DD ¢ , area =
1 3
zx
12
I mass = rtI area
ˆ
Ê1
dI DD ¢ , mass = r( dy ) Á zx 3 ˜
¯
Ë 12
ˆ
Ê1
dI BB ¢ , mass = r( dy ) Á xz 3 ˜
¯
Ë 12
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177
PROBLEM 9.123 (Continued)
Now
dI y = dI BB ¢ , mass + dI DD ¢ , mass
1
r xz ( x 2 + z 2 )dy
12
2
2
yˆ È
yˆ ˘
1
Ê
Ê
= r ab Á1 + ˜ Í( a2 + b2 ) Á1 + ˜ ˙ dy
Ë
Ë
h ¯ ÍÎ
h ¯ ˚˙
12
=
4
We have
m=
1
m
yˆ
Ê
r abh fi dI y = ( a2 + b2 ) Á1 + ˜ dy
Ë
6
2h
h¯
Then
m 2
Ê
( a + b2 ) Á1 +
- h 2h
Ë
I y = Ú dI y = Ú
0
4
yˆ
˜ dy
h¯
0
5
m 2
h ÈÊ
yˆ ˘
=
a + b2 ¥ ÍÁ1 + ˜ ˙
2h
5 ÎÍË
h ¯ ˙˚
-h
m 2
=
a + b2 ÈÎ(1)5 - (1 - 1)5 ˘˚
10
(
)
(
)
or
Iy =
1
m( a2 + b2 ) b
10
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178
PROBLEM 9.124*
Determine by direct integration the mass moment of inertia with
respect to the z axis of the semiellipsoid shown, assuming that it
has a uniform density and a mass m.
SOLUTION
First note that when
For the element shown:
Then
Ê
x2 ˆ
z = 0 : y = b Á1 - 2 ˜
Ë a ¯
1/ 2
Ê
x2 ˆ
y = 0 : z = c Á1 - 2 ˜
Ë a ¯
1/ 2
Ê
x2 ˆ
dm = r(p yzdx ) = prbc Á1 - 2 ˜ dx
Ë a ¯
Ê
a
x2 ˆ
m = Ú dm = Ú prbc Á1 - 2 ˜ dx
0
Ë a ¯
a
1
2
È
˘
= prbc Í x - 2 x 3 ˙ = pr abc
3a
Î
˚0 3
For the element:
Then
Now
I AA¢ , area =
p 3
zy
4
ˆ
Êp
dI AA¢ , mass = rtI AA¢ , area = r( dx ) Á zy 3 ˜
¯
Ë4
dI z = dI AA¢, mass + x 2 dm
2
È
Ê
Ê
p
x2 ˆ
x2 ˆ ˘
= rb3c Á1 - 2 ˜ dx + x 2 Íprbc Á1 - 2 ˜ dx ˙
4
Ë a ¯
Ë a ¯ ˙˚
ÍÎ
=
3m È b2 Ê
x2 x4 ˆ Ê 2 x4 ˆ ˘
Í Á1 - 2 2 + 4 ˜ + Á x - 2 ˜ ˙ dx
2a ÍÎ 4 Ë
a
a ¯ Ë
a ¯ ˙˚
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179
PROBLEM 9.124* (Continued)
Finally,
I z = Ú dI z
=
3m a È b2 Ê
x2 x4 ˆ Ê 2 x4 ˆ ˘
1
2
+ ˜ + Á x - 2 ˜ ˙ dx
Í
2a Ú0 ÍÎ 4 ÁË
a2 a 4 ¯ Ë
a ¯ ˙˚
a
3m È b2 Ê
2 x3 1 x5 ˆ Ê 1 3 1 x5 ˆ ˘
=
+
+ x Í Áx˙
2a ÍÎ 4 Ë
3 a2 5 a 4 ˜¯ ÁË 3
5 a2 ˜¯ ˙˚
0
=
3 È b2 Ê 2 1 ˆ
Ê 1 1ˆ ˘
m Í Á1 - + ˜ + a2 Á - ˜ ˙
Ë 3 5¯ ˚
2 Î 4 Ë 3 5¯
or
(
)
1
I z = m a2 + b2 b
5
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180
PROBLEM 9.125*
A thin steel wire is bent into the shape shown. Denoting the mass per
unit length of the wire by m¢, determine by direct integration the mass
moment of inertia of the wire with respect to each of the coordinate axes.
SOLUTION
First note
dy
= - x -1/ 3 ( a2 / 3 - x 2 / 3 )1/ 2
dx
2
Then
Ê dy ˆ
1 + Á ˜ = 1 + x -2 / 3 ( a2 / 3 - x 2 / 3 )
Ë dx ¯
2/3
Ê aˆ
=Á ˜
Ë x¯
2
For the element shown:
Ê dy ˆ
dm = m¢ dL = m¢ 1 + Á ˜ dx
Ë dx ¯
Ê aˆ
= m¢ Á ˜
Ë x¯
Then
Now
1/ 3
dx
a
a1/ 3
3
3
dx = m¢ a1/ 3 ÈÎ x 2 / 3 ˘˚ = m¢ a
1/ 3
0
0
2
2
x
1
/
3
Ê a
ˆ
a
I x = Ú y 2 dm = Ú ( a2 / 3 - x 2 / 3 )3 Á m¢ 1/ 3 dx˜
0
Ë x
¯
a
m = Ú dm = Ú m¢
ˆ
a Ê a2
= m¢ a1/ 3 Ú Á 1/ 3 - 3a 4 / 3 x1/ 3 + 3a2 / 3 x - x 5 / 3 ˜ dy
0Ëx
¯
1/ 3
= m¢a
a
È 3 2 2 / 3 9 4 / 3 4 / 3 3 2 / 3 2 3 8/ 3 ˘
ÍÎ 2 a x - 4 a x + 2 a x - 8 x ˙˚
0
Ê 3 9 3 3ˆ 3
= m ¢ a3 Á - + - ˜ = m ¢ a3
Ë 2 4 2 8¯ 8
or
Symmetry implies
1 2
ma b
4
1
I y = ma2 b
4
Ix =
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181
PROBLEM 9.125* (Continued)
Alternative solution:
Ê a1/ 3 ˆ
a
a
I y = Ú x 2 dm = Ú x 2 Á m¢ 1/ 3 dx˜ = m¢ a1/ 3 Ú x 5 / 3 dx
0
0
Ë x
¯
a
3
3
= m¢ a1/ 3 ¥ ÈÎ x8 / 3 ˘˚ = m¢ a3
0
8
8
1 2
= ma
4
Also
(
)
I z = Ú x 2 + y 2 dm = I y + I x
or
Iz =
1 2
ma b
2
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182
PROBLEM 9.126
A thin triangular plate of mass m is welded along its base AB to a block as
shown. Knowing that the plate forms an angle q with the y axis, determine
by direct integration the mass moment of inertia of the plate with respect
to (a) the x axis, (b) the y axis, (c) the z axis.
SOLUTION
For line BC:
z =-
=
h
x+h
a
2
h
(a - 2 x)
a
Ê1 ˆ
m = rV = rt Á ah˜
Ë2 ¯
Also
1
rtah
2
=
2
(a)
1 2
Êz ˆ
z dm¢ + Á ˜ dm¢
Ë 2¯
12
1
= z 2 dm¢
3
We have
dI x =
where
dm¢ = rtz dx
Then
I x = Ú dI x = 2Ú
a/2 1
0
3
z 2 ( rtz dx )
3
=
a/2 È h
2
˘
rt Ú Í ( a - 2 x )˙ dx
0 Îa
3
˚
=
a/2
2 h3 1 Ê 1 ˆ
rt 3 ¥ Á - ˜ ÈÎ( a - 2 x ) 4 ˘˚
0
¯
Ë
3 a
4 2
1
h3
rt 3 ÈÎ( a - a)4 - ( a)4 ˘˚
12 a
1
= rtah3
12
=-
or
1
I x = mh2 b
6
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183
PROBLEM 9.126 (Continued)
Now
Iz = Ú x 2 dm
and
Iz = Ú x 2 dm¢ = 2 Ú
a/2 2
x ( rtz dx )
0
a/2 2
x
0
= 2 rt Ú
Èh
˘
ÍÎ a ( a - 2 x ) ˙˚ dx
a/2
h Èa
1 ˘
= 2 rt Í x 3 - x 4 ˙
a Î3
4 ˚0
3
4
h Èa Ê aˆ
1 Ê aˆ ˘
Í Á ˜ - Á ˜ ˙
a ÍÎ 3 Ë 2 ¯
4 Ë 2 ¯ ˙˚
1
1
=
rta3 h =
ma2
48
24
= 2 rt
(b)
We have
I y = Ú ry2 dm = Ú ÈÎ x 2 + (z sin q )2 ˘˚ dm
= Ú x 2 dm + sin 2 q Ú z 2 dm
Now
I x = Ú z 2 dm fi I y = Iz + I x sin 2 q
=
1
1
ma2 + mh2 sin 2 q
24
6
(c)
We have
or
Iy =
m 2
( a + 4 h2 sin 2 q ) b
24
or
Iz =
m 2
( a + 4h2 cos2 q ) b
24
I z = Ú rz2 dm = Ú ( x 2 + y 2 )dm
= Ú ÈÎ x 2 + (z cos q )2 ˘˚ dm
= Ú x 2 dm + cos2 q Ú z 2 dm
= Iz + I x cos2 q
=
1
1
ma2 + mh2 cos2 q
24
6
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184
PROBLEM 9.127
Shown is the cross section of a molded flat-belt pulley.
Determine its mass moment of inertia and its radius of gyration
with respect to the axis AA¢. (The density of brass is 8650 kg/ m3
and the density of the fiber-reinforced polycarbonate used is
1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that
m = rV = rt ¥
and, using Figure 9.28, that
(
p 2
d2 - d12
4
2
I AA¢ =
)
1 Ê d2 ˆ
1 Êd ˆ
m2 Á ˜ - m1 Á 1 ˜
2 Ë 2¯
2 Ë 2¯
2
1 ÈÊ
p ˆ
p ˆ ˘
Ê
= ÍÁ rt ¥ d22 ˜ d22 - Á rt ¥ d12 ˜ d12 ˙
Ë
8 ÎË
4 ¯
4 ¯ ˚
1Êp ˆ
= Á rt ˜ d24 - d14
8Ë 4 ¯
(
)
(
)(
1Êp ˆ
= Á rt ˜ d22 - d12 d22 + d12
8Ë 4 ¯
1
= m d12 + d22
8
(
)
)
Now treat the pulley as four concentric rings and, working from the brass outward, we have
p
8650 kg/m3 ¥ (0.0175 m) ¥ (0.0112 - 0.0052 ) m2
m=
4
+ 1250 kg/m3 ÈÎ(0.0175 m) ¥ (0.0172 - 0.0112 ) m2
{
+ ( 0.002 m) ¥ ( 0.0222 - 0.0172 ) m2
}
+ ( 0.0095 m) ¥ (0.0282 - 0.0222 ) m2 ˘˚
= (11.4134 + 2.8863 + 0.38288 + 2.7980) ¥ 10 -3 kg
= 17.4806 ¥ 10 -3 kg
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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185
PROBLEM 9.127 (Continued)
and
1
I AA¢ = ÈÎ(11.4134)(0.0052 + 0.0112 ) + (2.8863)(0.0112 + 0.0172 )
8
+ (0.38288)(0.0172 + 0.0222 )
+(2.7980)(0.0222 + 0.0282 ) ˘˚ ¥ 10 -3 kg ◊ m2
= (208.29 + 147.92 + 37.00 + 443.48) ¥ 10 -9 kg ◊ m2
= 836.69 ¥ 10 -9 kg ◊ m2
2
-9
or I AA¢ = 837 ¥ 10 kg ◊ m b
Now
2
k AA
¢ =
I AA¢ 836.69 ¥ 10 -9 kg ◊ m2
=
= 47.864 ¥ 10 -6 m2
m
17.4806 ¥ 10 -3 kg
or
k AA¢ = 6.92 mm b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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186
PROBLEM 9.128
Shown is the cross section of an idler roller. Determine its mass
moment of inertia and its radius of gyration with respect to the axis
AA¢. (The specific weight of bronze is 8580 kg/m3; of aluminum,
2770 kg/m3; and of neoprene, 1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that
m = rV = rt ¥
(
)
(
p 2
p
d2 - d12 = rt d22 - d12
4
4
)
and, using Figure 9.28, that
2
I AA¢ =
1 Ê d2 ˆ
1 Êd ˆ
m2 Á ˜ - m1 Á 1 ˜
2 Ë 2¯
2 Ë 2¯
2
1 ÈÊ
p ˆ
p ˆ ˘
Ê
= ÍÁ rt ¥ d22 ˜ d22 - Á rt ¥ d12 ˜ d12 ˙
Ë
8 ÎË
4 ¯
4 ¯ ˚
1Êp ˆ
= Á rt ˜ d24 - d14
8Ë 4 ¯
(
)
(
)(
1Êp ˆ
= Á rt ˜ d22 - d12 d22 + d12
8Ë 4 ¯
(
)
)
1
= m d12 + d22
8
Now treat the roller as three concentric rings and, working from the
bronze outward, have
Have
m=
{
p
¥ (8580 kg/m3 ) (0.0195 m ) ÈÎ(0.009 m)2 - (0.006 m)2 ˘˚
4
+ 2770 kg/m3 (0.0165 m ) ÈÎ(0.012 m)2 - (0.009 m)2 ˘˚
(
)
+ (1250 kg/m ) (0.0165 m ) ÎÈ(0.027 m)
3
2
}
- (0.012 m)2 ˘˚
p
[7.52895 + 2.87942 + 12.06563] ¥ 10-3 kg
4
= 5.9132 ¥ 10 -3 kg + 2.26149 ¥ 10 -3 kg + 9.47632 ¥ 10 -3 kg
=
= 17.6510 ¥ 10 -3 kg
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187
PROBLEM 9.128 (Continued)
And
1
I AA¢ = {(5.9132 ¥ 10 -3 kg) ÈÎ(0.006)2 + (0.009)2 ˘˚ m2
8
+ (2.26149 ¥ 10 -3 kg) ÈÎ(0.009)2 + (0.012)2 ˘˚ m 2
+ (9.47632 ¥ 10 -3 kg) ÈÎ(0.012)2 + (0.0027)2 ˘˚ m2 }
1
= (691.844 + 508.835 + 8272.827)10 -9 kg ◊ m2
8
= 1.18419 ¥ 10 -6 kg ◊ m2
Now
or
2
k AA
¢ =
I AA¢ = 1.184 ¥ 10 -6 kg ◊ m2 b
I AA¢ 1.18419 ¥ 10 -6 kg ◊ m2
=
m
17.6510 ¥ 10 -3 kg
= 67.08902 ¥ 10 -6 m 2
k AA¢ = 8.19079 ¥ 10 -3 m
or
k AA¢ = 8.19 mm b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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188
PROBLEM 9.129
Knowing that the thin cylindrical shell shown is of mass m, thickness t, and
height h, determine the mass moment of inertia of the shell with respect
to the x axis. (Hint: Consider the shell as formed by removing a cylinder
of radius a and height h from a cylinder of radius a + t and height h; then
neglect terms containing t2 and t3 and keep those terms containing t.)
SOLUTION
From Figure 9.28 for a solid cylinder (change orientation of axes):
I x¢ =
1
m(3a2 + h2 )
12
Ê hˆ
I x = I x¢ + m Á ˜
Ë 2¯
2
1
h2
m(3a2 + h2 ) + m
12
4
2
Ê 3
h ˆ
= m Á a2 + ˜
3¯
Ë 12
=
Shell = Cylinder
Ix =
1
m(3a2 + 4h2 ) 
12
– Cylinder
1
1
m1[3( a + t )2 + 4h2 ] - m2 (3a2 + 4h2 )
12
12
1
1
= [ rp ( a + t )2 h][3( a + t )2 + 4 h2 ] - ( rp a2 h)(3a2 + 4h2 )
12
12
Ix =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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189
PROBLEM 9.129 (Continued)
Expand ( a + t ) factors and neglecting terms in t 2, t 3, and t 4 :
(
)(
)
rp h È 2
a + 2at + t 2 3a2 + 6at + 3t 2 + 4h2 ˘ - 3a 4 - 4a2 h2
˚
12 Î
rp h
3a 4 + 6a3t + 3a2t 2 + 4 a2 h2 + 6a3t + 8ath2 - 3a 4 - 4 a2 h2
=
12
rp h
12a3t + 8ath2
=
12
rp hat
=
12a2 + 8h2
12
Ix =
(
(
)
)
(
Mass of shell:
)
m = r(2p ath) = 2rp hat
(
2rp hat
12a2 + 8h2
24
4m 2
=
3a + 2h2
24
Ix =
(
)
)
or
Ix =
(
)
m 2
3a + 2h2 b
6
Checks:
1
¸
a = 0; I x = mh2 - slender rod Ô
Ô
3
˝ OK
1 2
h = 0; I x = ma - thin ring Ô
Ô˛
2
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190
PROBLEM 9.130
The machine part shown is formed by machining a conical
surface into a circular cylinder. For b = 12 h, determine the mass
moment of inertia and the radius of gyration of the machine part
with respect to the y axis.
SOLUTION
Mass:
h 1
1
mcone = rp a2 = rp a2 h
3
2 6
mcyl = rp a2 h
1
mcyl a2
2
1
= rp a 4 h
2
3
mcone a2
10
1
rp a 4 h
=
20
I y : I cyl =
I cone =
For entire machine part:
1
5
m = mcyl - mcone = rp a2 h - rp a2 h = rp a2 h
6
6
1
9
1
rp a 4 h =
rp a 4 h
I y = I cyl - I cone = rp a 4 h 2
200
20
or
ˆ Ê 6ˆ Ê 9 ˆ
Ê5
I y = Á rp a2 h˜ Á ˜ Á ˜ a2 ¯ Ë 5 ¯ Ë 20 ¯
Ë6
Then
k y2 =
I 27 2
=
a m 50
Iy =
27 2
ma b
50
k y = 0.735a b
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191
PROBLEM 9.131
After a period of use, one of the blades of a shredder has been worn to
the shape shown and is of mass 0.18 kg. Knowing that the mass moments
of inertia of the blade with respect to the AA¢ and B¢ axes are 0.320 g ◊ m2
and 0.680 g ◊ m2 , respectively, determine (a) the location of the centroidal
axis GG¢, (b) the radius of gyration with respect to axis GG¢.
SOLUTION
(a)
d B = (0.08 - d A ) m
We have
and, using the parallel axis
I AA¢ = I GG ¢ + md A2
I BB ¢ = I GG ¢ + md B2
Then
(
I BB ¢ - I AA¢ = m d B2 - d A2
)
= m ÈÎ(0.08 - d A )2 - d A2 ˘˚
= m(0.0064 - 0.16d A )
Substituting:
(0.68 - 0.32) ¥ 10 -3 kg ◊ m2 = 0.18 kg(0.0064 - 0.16d A ) m2
or
(b)
We have
I AA¢ = I GG ¢ + md A2
or
I GG¢ = 0.32 ¥ 10 -3 kg ◊ m2 - 0.18 kg ◊ (0.0275 m)2
d A = 27.5 mm b
= 0.183875 ¥ 10 -3 kg ◊ m 2
Then
2
kGG
¢ =
I GG ¢ 0.183875 ¥ 10 -3 kg ◊ m2
=
= 1.02153 ¥ 10 -3 m2
m
0.18 kg
or kGG¢ = 32.0 mm b
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192
PROBLEM 9.132
Determine the mass moment of inertia of the 400-g machine component shown with respect to
the axis AA¢.
SOLUTION
First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger cone
as indicated.
h h + 60
=
30
10
Now
or h = 30 mm
The weight of the body is given by
m = ( m1 + m2 - m3 ) = r(V1 + V2 - V3 )
p
p
˘
È
0.4 kg = r Íp (0.02)2 (0.06) + (0.005)2 (0.03) - (0.015)2 (0.09)˙ m3
3
3
Î
˚
or
= r(7.54 + 0.079 - 2.12) ¥ 10 -5 m3
or
r = 7274 kg/m3
Then
m1 = (7274)(7.54 ¥ 10 -5 ) = 0.5485 kg
m2 = (7274)(7.9 ¥ 10 -7 ) = 0.00575 kg
m3 = (7274)(2.12 ¥ 10 -5 ) = 0.1542 kg
Finally, using Figure 9.28 have,
I AA¢ = ( I AA¢ )1 + ( I AA¢ )2 - ( I AA¢ )3
1
3
3
m1a12 + m2 a22 - m3 a32
2
10
10
3
3
È1
˘
= Í (0.5485)(0.02)2 + (0.00575)( 0.005)2 - ( 0.1542)( 0.015)2 ˙
10
10
Î2
˚
=
= (10.97 + 0.00431 - 1.042) ¥ 10 -5 kg ◊ m2
or
I AA¢ = 9.93 ¥ 10 -5 kg ◊ m2 b
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193
PROBLEM 9.133
A square hole is centered in and extends through the aluminum machine
component shown. Determine (a) the value of a for which the mass
moment of inertia of the component with respect to the axis AA¢, which
bisects the top surface of the hole, is maximum, (b) the corresponding
values of the mass moment of inertia and the radius of gyration with
respect to the axis AA¢. (The density of aluminum is 2770 kg/ m3.)
SOLUTION
First note
m1 = rV1 = rb2 L
And
m2 = rV2 = r a2 L
(a)
Using Figure 9.28 and the parallel-axis theorem, have
I AA¢ = ( I AA¢ )1 - ( I AA¢ )2
2
È1
Ê aˆ ˘
= Í m1 (b2 + b2 ) + m1 Á ˜ ˙
Ë 2¯ ˙
ÍÎ12
˚
2
È1
Ê aˆ ˘
- Í m2 ( a2 + a2 ) + m2 Á ˜ ˙
Ë 2¯ ˙
ÍÎ12
˚
1 ˆ
Ê1
Ê 5 ˆ
= ( rb2 L) Á b2 + a2 ˜ - ( r a2 L) Á a2 ˜
¯
Ë6
Ë 12 ¯
4
rL
=
(2b 4 + 3b2 a2 - 5a 4 )
12
dI AA¢ r L
=
(6b2 a - 20a3 ) = 0
da
12
Then
or
a=0
Also
d 2 I AA¢
da
a=b
and
2
=
3
10
1
rL
(6b2 - 60a2 ) = r L(b2 - 10a2 )
12
2
Now, for
a = 0,
d 2 I AA¢
da2
0
and for a = b
\ ( I AA¢ ) max occurs when
or
a = 105
a=b
3
,
10
d 2 I AA¢
da2
0
3
10
3
= 57.511 mm
10
a = 57.5 mm b
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194
PROBLEM 9.133 (Continued)
(b)
From part (a)
( I AA¢ ) mass
2
4
Ê
rL È 4
3ˆ
3ˆ ˘
2Ê
Í
2b + 3b Á b
=
˜ - 5 Á b 10 ˜ ˙˙
12 Í
Ë 10 ¯
Ë
¯
˚
Î
49
49
=
rLb 4 =
(2770 kg/m3 )(0.375 m)(0.105 m) 4
240
240
= 25.778 ¥ 10 -3 kg ◊ m2
or (I AA¢ ) mass = 25.8 ¥ 10 -3 kg ◊ m2 b
and
where
2
k AA
¢ =
( I AA¢ ) mass
m
2
È
Ê
3ˆ ˘ 7
2
2
m = m1 - m2 = r L(b - a ) = r L Íb - Á b
˜ ˙ = r Lb
Í
Ë 10 ¯ ˙ 10
Î
˚
2
2
Then
2
k AA¢
49
r Lb 4
7 2 7
240
=
=
b = (105 mm)2 = 3215.625 mm2
7
24
24
r Lb2
10
k AA¢ = 56.706 mm
or k AA¢ = 56.7 mm b
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195
PROBLEM 9.134
The cups and the arms of an anemometer are fabricated from a
material of density r. Knowing that the mass moment of inertia
of a thin, hemispherical shell of mass m and thickness t with
respect to its centroidal axis GG¢ is 5ma2 /12, determine (a) the
mass moment of inertia of the anemometer with respect to the
axis AA¢, (b) the ratio of a to l for which the centroidal moment of
inertia of the cups is equal to 1 percent of the moment of inertia
of the cups with respect to the axis AA¢.
SOLUTION
(a)
First note
and
marm = rVarm = r ¥
p 2
d l
4
dmcup = r dVcup
= r[(2p a cos q )(t )( adq )]
Then
mcup = Ú dmcup = Ú
p /2
0
2pr a2t cos q dq
= 2pr a2t [sin q ]p0 / 2
= 2pr a2t
Now
( I AA¢ )anem = ( I AA¢ )cups + ( I AA¢ )arms
Using the parallel-axis theorem and assuming the arms are slender rods, we have
2 ˘
È
˘
( I AA¢ )anem = 3 ÈÎ( I GG ¢ )cup + mcup d AG
˚ + 3 Î I arm + marm d AGarm ˚
2
2
ÏÔ 5
È
È1
Ê a ˆ ˘ ¸Ô
Ê lˆ ˘
= 3 Ì mcup a2 + mcup Í(l + a)2 + Á ˜ ˙ ˝ + 3 Í marm l 2 + marm Á ˜ ˙
Ë 2 ¯ ˙Ô
Ë 2¯ ˙
ÍÎ
ÍÎ 2
ÔÓ12
˚˛
˚
ˆ
Ê5
= 3mcup Á a2 + 2la + l 2 ˜ + marm l 2
¯
Ë3
ˆ
ˆ Êp
Ê5
= 3(2pr
r a2t ) Á a2 + 2la + l 2 ˜ + Á r d 2 l ˜ (l 2 )
¯
¯ Ë4
Ë3
or
È
Ê 5 a2
a ˆ d 2l ˘
( I AA¢ )anem = prl 2 Í6a2t Á 2 + 2 + 1˜ +
˙b
l ¯
4 ˙˚
Ë3 l
ÍÎ
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196
PROBLEM 9.134 (Continued)
(b)
We have
or
( I GG ¢ )cup
( I AA¢ )cup
= 0.01
5
ˆ
Ê5
mcup a2 = 0.01mcup Á a2 + 2la + l 2 ˜ (from Part a)
¯
Ë3
12
a
Now let z = .
l
Then
or
ˆ
Ê5
5z 2 = 0.12 Á z 2 + 2z + 1˜
¯
Ë3
40z 2 - 2z - 1 = 0
2 ± ( -2)2 - 4( 40)( -1)
2( 40)
Then
z=
or
z = 0.1851 and z = - 0.1351
a
= 0.1851 b
l
To the instructor:
The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at
this time for use in the solutions of Problems 9.135 through 9.140.
Thin rectangular plate
( I x ) m = ( I x ¢ )m + md 2
=
ÈÊ b ˆ 2 Ê h ˆ 2 ˘
1
m(b2 + h2 ) + m ÍÁ ˜ + Á ˜ ˙
Ë 2¯ ˙
12
ÍÎË 2 ¯
˚
1
= m(b2 + h2 )
3
( I y ) m = ( I y ¢ )m + md 2
2
=
1
1
Ê bˆ
mb2 + m Á ˜ = mb2
Ë 2¯
12
3
I z = ( I z ¢ ) m + md 2
2
1
1
Ê hˆ
= mh2 + m Á ˜ = mh2
¯
Ë
12
2
3
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197
PROBLEM 9.134 (Continued)
Thin triangular plate
We have
ˆ
Ê1
m = rV = r Á bht ˜
¯
Ë2
1 3
bh
36
and
I z, area =
Then
I z , mass = rtI z , area
= rt ¥
1 3
bh
36
=
1
mh2
18
Similarly,
I y, mass =
1
mb2
18
Now
I x , mass = I y , mass + I z , mass =
1
m(b2 + h2 )
18
Thin semicircular plate
We have
ˆ
Êp
m = rV = r Á a2t ˜
¯
Ë2
p 4
a
8
and
I y , area = I z , area =
Then
I y , mass = I z , mass = rtI y , area
= rt ¥
=
p 4
a
8
1 2
ma
4
1 2
ma
2
Now
I x , mass = I y , mass + I z , mass =
Also
I x , mass = I x ¢ , mass + my 2
or
Ê 1 16 ˆ
I x ¢ , mass = m Á - 2 ˜ a2
Ë 2 9p ¯
and
I z , mass = I z ¢ , mass + my 2
or
Ê 1 16 ˆ
I z ¢, mass = m Á - 2 ˜ a2
Ë 4 9p ¯
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198
PROBLEM 9.134 (Continued)
y=z=
4a
3p
Thin Quarter-Circular Plate
We have
ˆ
Êp
m = rV = r Á a2t ˜
¯
Ë4
p 4
a
16
and
I y , area = I z , area =
Then
I y , mass = I z , mass = rtI y , area
= rt ¥
=
p 4
a
16
1 2
ma
4
1 2
ma
2
Now
I x , mass = I y , mass + I z , mass =
Also
I x , mass = I x ¢ , mass + m( y 2 + z 2 )
or
Ê 1 32 ˆ
I x¢ , mass = m Á - 2 ˜ a2
Ë 2 9p ¯
and
I y , mass = I y ¢ , mass + mz 2
or
Ê 1 16 ˆ
I y¢ , mass = m Á - 2 ˜ a2
Ë 4 9p ¯
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199
PROBLEM 9.135
A 2 mm thick piece of sheet steel is cut and bent into the
machine component shown. Knowing that the density of steel
is7850 kg/m3, determine the mass moment of inertia of the
component with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m = rSTV = rST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.3 m)2
= 1.413 kg
m2 = m3 = (7850 kg/m3 )(0.002 m) ¥ (0.15 ¥ 0.12) m2
= 0.2826 kg
Using Figure 9.28 and the parallel-axis theorem, we have
I x = ( I x )1 + 2( I x )2
È1
˘
= Í (1.413 kg)(0.3 m)2 ˙
12
Î
˚
1
È
˘
+ 2 Í (0.2826 kg)(0.12 m)2 + (0.2828 kg)(0.152 + 0.062 ) m2 ˙
Î12
˚
= [(0.0105975) + 2(0.0003391 + 0.0073759)] kg ◊ m2
= [(0.0105975) + 2(0.0077150)] kg ◊ m2
or
I x = 26.0 ¥ 10 -3 kg ◊ m2 b
I y = ( I y )1 + 2( I y )2
È1
˘
= Í (1.413 kg)(0.32 + 0.32 ) m2 ˙
Î12
˚
È1
˘
+ 2 Í (0.2826 kg)(0.15 m)2 + (0.2826 kg)(0.0752 + 0.152 ) m2 ˙
Î12
˚
= [(0.0211950) + 2(0.0005299 + 0.0079481)] kg ◊ m2
= [(0.0211950) + 2(0.0084780)] kg ◊ m2
or
I y = 38.2 ¥ 10 -3 kg ◊ m2 b
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200
PROBLEM 9.135 (Continued)
I z = ( I z )1 + 2( I z )2
È1
˘
= Í (1.413 kg)(0.3 m)2 ˙
12
Î
˚
1
È
˘
+ 2 Í (0.2826 kg)(0.152 + 0.122 ) m2 + (0.2826 kg)(0.0752 + 0.062 ) m2 ˙
Î12
˚
= [(0.0105975) + 2(0.0008690 + 0.0026070)] kg ◊ m 2
= [(0.0105975) + 2(0.0034760)] kg ◊ m2
or I z = 17.55 ¥ 10 -3 kg ◊ m2 b
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201
PROBLEM 9.136
A 2 mm thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/ m3,
determine the mass moment of inertia of the component with respect
to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m = rSTV = rST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.35 ¥ 0.39) m2
= 2.14305 kg
ˆ
Êp
m2 = (7850 kg/m3 )(0.002 m) Á ¥ 0.1952 ˜ m 2 = 0.93775 kg
¯
Ë2
ˆ
Ê1
m3 = (7850 kg/m3 )(0.002 m) Á ¥ 0.39 ¥ 0.15˜ m2 = 0.45923 kg
¯
Ë2
Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have
I x = ( I x )1 + ( I x )2 + ( I x )3
2
È1
Ê 0.39 ˆ ˘
= Í (2.14305 kg)(0.39 m)2 + (2.14305 kg) Á
m˜ ˙
¯ ˙
Ë 2
ÍÎ12
˚
ÏÔÊ 1 16 ˆ
ÈÊ 4 ¥ 0.195 ˆ 2
˘ ¸Ô
+ (0.195)2 ˙ m2 ˝
+ ÌÁ - 2 ˜ (0.93775 kg)(0.195 m)2 + (0.93775 kg) ÍÁ
˜
ÍÎË 3p ¯
˙˚ Ô˛
ÔÓË 2 9p ¯
ÏÔ 1
ÈÊ 0.39 ˆ 2 Ê 0.15 ˆ 2 ˘ 2 ¸Ô
+ Ì (0.45923 kg)[(0.39)2 + (0.15)2 ] m2 + (0.45923 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
18
3 ¯ ˙˚ Ô
ÍÎË 3 ¯
ÓÔ
˛
= [(0.027163 + 0.081489) + (0.011406 + 0.042081) + (0.004455 + 0.0089009)] kg ◊ m2
= (0.108652 + 0.053487 + 0.013364) kg ◊ m2
= 0.175503 kg ◊ m2
or
I x = 175.5 ¥ 10 -3 kg ◊ m2 b
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202
PROBLEM 9.136 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3
ÏÔ 1
ÈÊ 0.35 ˆ 2 Ê 0.39 ˆ 2 ˘ 2 ¸Ô
= Ì (2.14305 kg)[(0.35)2 + (0.39)2 ] m2 + (2.14305 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
2 ¯ ˙˚ Ô
ÍÎË 2 ¯
ÔÓ12
˛
È1
˘
+ Í (0.93775 kg)(0.195 m)2 + (0.93775 kg)(0.195 m)2 ˙
Î4
˚
2
ÏÔ 1
È
Ê 0.39 ˆ ˘ 2 ¸Ô
+ Ì (0.45923 kg)(0.39 m)2 + (0.45923 kg) Í(0.35)2 + Á
˙m
Ë 3 ˜¯ ˙ ˝Ô
ÍÎ
ÔÓ18
˚ ˛
= [(0.049040 + 0.147120) + (0.008914 + 0.035658)
+ (0.003880 + 0.064017)] kg ◊ m 2
= (0.196160 + 0.044572 + 0.067897) kg ◊ m2
= 0.308629 kg ◊ m2
or
I y = 309 ¥ 10 -3 kg ◊ m2 b
I z = ( I z )1 + ( I z )2 + ( I z )3
2
È1
Ê 0.35 ˆ ˘
= Í (2.14305 kg)(0.35 m)2 + (2.14305 kg) Á
m˜ ˙
¯ ˙
Ë 2
ÍÎ12
˚
È1
˘
+ Í (0.93775 kg)(0.195 m)2 ˙
4
Î
˚
2
ÏÔ 1
È
Ê 0.15 ˆ ˘ 2 ¸Ô
+ Ì (0.45923 kg)(0.15 m)2 + (0.45923 kg) Í(0.35)2 + Á
˙m
Ë 3 ˜¯ ˙ ˝Ô
18
ÍÎ
˚ ˛
ÓÔ
= [(0.021877 + 0.065631) + 0.008914) + (0.000574 + 0.057404)] kg ◊ m2
= (0.087508 + 0.008914 + 0.057978) kg ◊ m2
= 0.154400 kg ◊ m2
or
I z = 154.4 ¥ 10 -3 kg ◊ m2 b
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203
PROBLEM 9.137
The cover for an electronic device is formed from sheet aluminum that
is 1.25 mm thick. Determine the mass moment of inertia of the cover
with respect to each of the coordinate axes. (The density of aluminum
is 2770 kg/m3 .)
SOLUTION
Have
Now
m = rV
m1 = (2770 kg / m3 )(0.00125 m)(0.060 m)(0.075 m)
= 0.015581 kg
m2 = (2770 kg / m3 )(0.00125 m)(0.075 m)(0.155 m)
= 0.040252 kg
m3 = (2770 kg / m3 )(0.00125 m)(0.060 m)(0.155 m)
= 0.032201 kg
Using Figure 9.28 and the parallel-axis theorem, have
( I x ) = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4
and ( I x )3 = ( I x )4
È
È
Ê 1 1ˆ ˘
Ê 1 1ˆ ˘
= (0.015581 kg) Í(0.06)2 Á + ˜ ˙ m2 + (0.040252 kg) Í(0.155)2 Á + ˜ ˙ m2
Ë 12 4 ¯ ˚
Ë 12 4 ¯ ˚
Î
Î
Ï
Ê 1 1ˆ ¸
+ 2(0.032201 kg) Ì ÈÎ(0.06)2 + (0.155)2 ˘˚ Á + ˜ ˝ m2
Ë 12 4 ¯ ˛
Ó
= (1.86972 ¥ 10 -5 + 3.2235 ¥ 10 -4 + 5.9304 ¥ 10 -4 ) kg ◊ m2
= 934.087 ¥ 10 -6 kg ◊ m2
or
I x = 934 ¥ 10 -6 kg ◊ m2 b
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204
PROBLEM 9.137 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4
Ï
È1
Ê 1ˆ ¸
˘
= (0.015581 kg) Í (0.075)2 ˙ m2 + (0.040252 kg) Ì ÈÎ(0.075)2 + (0.155)2 ˘˚ Á ˜ ˝ m2
Ë 3¯ ˛
Î3
˚
Ó
1
È
Ê 1ˆ ˘
È1
˘
+ (0.032201 kg) Í(0.155)2 Á ˜ ˙ m2 + (0.032201 kg) Í (0.155)2 + (0.075)2 + (0.155)2 ˙ m2
¯
Ë
3
2
4
Î
˚
Î
˚
= (2.9214 ¥ 10 -5 + 3.9782 ¥ 10 -4 + 2.5788 ¥ 10 -4 + 4.3901 ¥ 10 -4 ) kg ◊ m2 = 1123.924 ¥ 10 -6 kg ◊ m2
or
I y = 1124 ¥ 10 -6 kg ◊ m2 b
I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4
Ï
È
Ê 1ˆ ¸
Ê 1ˆ ˘
= (0.015581 kg) Ì ÈÎ(0.075)2 + (0.06)2 ˘˚ Á ˜ ˝ m2 + (0.040252 kg) Í(0.075)2 Á ˜ ˙ m2
Ë 3¯ ˛
Ë 3¯ ˚
Î
Ó
È
ÈÊ 1 ˆ
˘
Ê 1ˆ ˘
+ (0.032201 kg) Í(0.06)2 Á ˜ ˙ m2 + (0.032201 kg) ÍÁ ˜ (0.06)2 + (0.075)2 ˙ m2
Ë 3¯ ˚
Î
ÎË 3 ¯
˚
= ( 4.7912 ¥ 10 -5 + 7.5473 ¥ 10 -5 + 3.8641 ¥ 10 -5 + 2.1977 ¥ 10 -4 ) kg ◊ m2
= 381.796 ¥ 10 -6 kg ◊ m2
or
I z = 382 ¥ 10 -6 kg ◊ m2 b
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205
PROBLEM 9.138
A framing anchor is formed of 2-mm-thick galvanized steel. Determine the mass
moment of inertia of the anchor with respect to each of the coordinate axes. (The
density of galvanized steel is 7530 kg/m3 .)
SOLUTION
First compute the mass of each component.
Have
m = rV = r At
Then
m1 = (7530 kg/m3 )(0.002 m)(0.045 m)(0.070 m)
= 0.047439 kg
m2 = (7530 kg/m3 )(0.002 m)(0.045 m)(0.045 m)
= 0.013554 kg
1
m3 = (7530 kg/m3 )(0.002 m) ¥ (0.04 m)(0.095 m)
2
= 0.028614 kg
Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, have
I x = ( I x )1 + ( I x )2 + ( I x )3
È1
Ï1
¸
˘
= (0.047439 kg) Í (0.07)2 ˙ m2 + (0.0135554 kg) Ì (0.020)2 + [(0.07)2 + (0.01)2 ]˝ m2
3
12
Î
˚
Ó
˛
1
Ï1
¸
+ (0.028614 kg) Ì [(0.095)2 + (0.04)2 ] + [(2 ¥ 0.095)2 + (0.040)2 ]˝ m2
9
Ó18
˛
= [(0.077484 + 0.068222 + 0.136751) ¥ 10 -3 ]kg ◊ m2 = 0282457 ¥ 10 -3 kg ◊ m2
or
I x = 0.2825 ¥ 10 -3 kg ◊ m2 b
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206
PROBLEM 9.138 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3
Ï
È1
Ê 1ˆ ¸
˘
= (0.047439 kg) Í (0.045)2 ˙ m 2 + (0.0135554 kg) Ì[(0.045)2 + (0.02)2 ] Á ˜ ˝ m2
Ë 3¯ ˛
Î3
˚
Ó
1
È1
˘
+ (0.028614 kg) Í (0.04)2 (0.045)2 + (0.04)2 ˙ m2
9
Î18
˚
= [(0.03202 + 0.010956 + 0.065574) ¥ 10 -3 ]kg ◊ m2
= 0.10855 ¥ 103 kg ◊ m2
I y = 0.1086 ¥ 103 kg ◊ m2 b
or
I z = ( I z )1 + ( I z )2 + ( I z )3
Ï
Ê 1ˆ ¸
Ê 1ˆ
= (0.047439 kg) Ì[(0.045)2 + (0.070)2 ] Á ˜ ˝ m2 + (0.013554 kg)[( 0.045)2 Á ˜
Ë
¯
Ë 3¯
3 ˛
Ó
+ (0.070)2 ]m2
2
È1
ˆ ˘
Ê2
+ (0.0228614 kg) Í (0.095)2 + (0.045)2 + Á 0.095˜ ˙ m2
¯ ˙
Ë3
ÍÎ18
˚
= [(0.0109505 + 0.075564 + 0.187064) ¥ 10 -3 ]kg ◊ m3
= 0.37213 ¥ 10 -3 kg ◊ m2
or
I z = 0.372 ¥ 10 -3 kg ◊ m2 b
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207
PROBLEM 9.139
A subassembly for a model airplane is fabricated from three pieces of 1.5 mm
plywood. Neglecting the mass of the adhesive used to assemble the three
pieces, determine the mass moment of inertia of the subassembly with respect
to each of the coordinate axes. (The density of the plywood is 780 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = rV = rtA
Then
ˆ
Ê1
m1 = m2 = (780 kg/m3 )(0.0015 m) Á ¥ 0.3 ¥ 0.12˜ m2
¯
Ë2
= 21.0600 ¥ 10 -3 kg
ˆ
Êp
m3 = (780 kg/m3 )(0.0015 m) Á ¥ 0.122 ˜ m2 = 13.2324 ¥ 10 -3 kg
¯
Ë4
Using the equations derived above and the parallel-axis theorem, we have
( I x )1 = ( I x )2
I x = ( I x )1 + ( I x )2 + ( I x )3
2
È1
Ê 0.12 ˆ ˘
m˜ ˙
= 2 Í (21.0600 ¥ 10 -3 kg)(0.12 m)2 + (21.0600 ¥ 10 -3 kg) Á
¯ ˙
Ë 3
ÍÎ18
˚
È1
˘
+ Í (13.2324 ¥ 10 -3 kg)(0.12 m)2 ˙
Î2
˚
= [2(16.8480 + 33.6960) + (95.2733)] ¥ 10 -6 kg ◊ m2
= [2(50.5440) + (95.2733)] ¥ 10 -6 kg ◊ m2
or
I x = 196.4 ¥ 10 -6 kg ◊ m2 b
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208
PROBLEM 9.139 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3
2
È1
Ê 0.3 ˆ ˘
= Í (21.0600 ¥ 10 -3 kg)(0.3 m)2 + (21.0600 kg) Á
m˜ ˙
Ë 3 ¯ ˙
ÍÎ18
˚
Ï1
+ Ì (21.0600 ¥ 10 -3 kg)[(0.3)2 + (0.12)2 ] m2
Ó18
ÈÊ 0.3 ˆ 2 Ê 0.12 ˆ 2 ˘ 2 ¸Ô
+ (21.0600 ¥ 10 kg) ÍÁ ˜ + Á
˙m ˝
Ë 3 ˜¯ ˙
ÍÎË 3 ¯
˚
˛Ô
-3
È1
˘
+ Í (13.2324 ¥ 10 -3 kg)(0.12 m)2 ˙
4
Î
˚
= [(105.300 + 210.600) + (122.148 + 244.296)
+ ( 47.637)] ¥ 10 -6 kg ◊ m2
= (315.900 + 366.444 + 47.637) ¥ 10 -6 kg ◊ m2
or
I y = 730 ¥ 10 -6 kg ◊ m2 b
Symmetry implies I y = I z
I z = 730 ¥ 10 -6 kg ◊ m2 b
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209
PROBLEM 9.140*
A farmer constructs a trough by welding a rectangular piece of
2 mm-thick sheet steel to half of a steel drum. Knowing that the
density of steel is 7850 kg/m3 and that the thickness of the walls of
the drum is 1.8 mm, determine the mass moment of inertia of the
trough with respect to each of the coordinate axes. Neglect the mass
of the welds.
SOLUTION
First compute the mass of each component. We have
m = rSTV = rST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.84 ¥ 0.21) m2
= 2.76948 kg
m2 = (7850 kg/m3 )(0.0018 m)(p ¥ 0.285 ¥ 0.84) m2
= 10.62713 kg
ˆ
Êp
m3 = m4 = (7850 kg/m3 )(0.0018 m) Á ¥ 0.2852 ˜ m2
¯
Ë2
= 1.80282 kg
Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have
( I x )3 = ( I x ) 4
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4
2
È1
˘
0.21ˆ
Ê
m2 ˙
= Í (2.76948 kg)(0.21 m)2 + (2.76948 kg) Á 0.285 ˜
Ë
2 ¯
ÍÎ12
˙˚
È1
˘
+ [(10.62713 kg)(0.285 m)2 ] + 2 Í (1.80282 kg)(0.285 m)2 ˙
Î2
˚
= [(0.01018 + 0.08973) + (0.86319) + 2(0.07322)] kg ◊ m2
= [(0.09991 + 0.86319 + 2(0.07322)] kg ◊ m2
or
I x = 1.110 kg ◊ m2 b
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210
PROBLEM 9.140* (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4
Ï1
= Ì (2.76948 kg)[(0.84)2 + (0.21)2 ] m2
Ó12
2
ÈÊ 0.84 ˆ 2 Ê
0.21ˆ ˘ 2 ¸Ô
+ Á 0.285 + (2.76948 kg) ÍÁ
˙m ˝
˜
˜
Ë
2 ¯ ˙˚
ÍÎË 2 ¯
Ô˛
2
ÏÔ 1
Ê 0.84 ˆ ¸Ô
+ Ì (10.62713 kg)[(0.84)2 + 6(0.285)2 ] m2 + (10.62713 kg) Á
m˜ ˝
¯ Ô
Ë 2
ÓÔ12
˛
È1
˘
+ Í (1.80282 kg)(0.285 m)2 ˙
Î4
˚
È1
˘
+ Í (1.80282 kg)(0.285 m)2 + (1.80282 kg)(0.84 m)2 ˙
4
Î
˚
= [(0.17302 + 0.57827) + (1.05647 + 1.87463)
+ (0.03661) + (0.03661 + 1..27207)] kg ◊ m2
= (0.75129 + 2.93110 + 0.03661 + 1.30868) kg ◊ m2
or
I y = 5.03 kg ◊ m2 b
I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4
2
È1
Ê 0.84 ˆ ˘
= Í (2.76948 kg)(0.84 m)2 + (2.76948 kg) Á
m˜ ˙
¯ ˙
Ë 2
ÍÎ12
˚
2
ÔÏ 1
Ê 0.84 ˆ Ô¸
+ Ì (10.62713 kg)[(0.84)2 + 6(0.285)2 ] m2 + (10.62713 kg) Á
m˜ ˝
¯ Ô
Ë 2
ÓÔ12
˛
È1
˘
+ Í (1.80282 kg)(0.285 m)2 ˙
Î4
˚
È1
˘
+ Í (1.80282 kg)(0.285 m)2 + (1.80282 kg)(0.84 m)2 ˙
4
Î
˚
= [(0.16285 + 0.48854) + (1.05647 + 1.87463)
+ (0.03661) + (0.03661 + 1..27207)] kg ◊ m2
= (0.65139 + 2.93110 + 0.03661 + 1.30868) kg ◊ m2
or
I z = 4.93 kg ◊ m2 b
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211
PROBLEM 9.141
The machine element shown is fabricated from steel. Determine
the mass moment of inertia of the assembly with respect to
(a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is
7850 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = rSTV
Then
m1 = (7850 kg/m3 )[(p (0.08 m)2 (0.04 m)]
= 6.31334 kg
m2 = (7850 kg/m3 )[p (0.02 m)2 (0.06 m)] = 0.59188 kg
m3 = (7850 kg/m3 )[p (0.02 m)2 (0.04 m)] = 0.39458 kg
Using Figure 9.28 and the parallel-axis theorem, we have
(a)
I x = ( I x )1 + ( I x )2 - ( I x )3
Ï1
¸
= Ì (6.31334 kg)[3(0.08)2 + (0.04)2 ] m2 + (6.31334 kg)(0.02 m)2 ˝
Ó12
˛
Ï1
¸
+ Ì (0.59188 kg)[3(0.02)2 + (0.06)2 ] m2 + (0.59188 kg)(0.03 m)2 ˝
12
Ó
˛
Ï1
¸
- Ì (0.39458 kg)[3(0.02)2 + (0.04)2 ] m2 + (0.39458 kg)(0.02 m)2 ˝
Ó12
˛
= [(10.94312 + 2.52534) + (0.23675 + 0.53269)
- (0.09207 + 0.15783)] ¥ 10 -3 kg ◊ m2
= (13.46846 + 0.76944 - 0.24990) ¥ 10 -3 kg ◊ m2
= 13.98800 ¥ 10 -3 kg ◊ m2
or
I x = 13.99 ¥ 10 -3 kg ◊ m2 b
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212
PROBLEM 9.141 (Continued)
(b)
I y = ( I y )1 + ( I y )2 - ( I y )3
È1
˘
= Í (6.31334 kg)(0.08 m)2 ˙
Î2
˚
È1
˘
+ Í (0.59188 kg)(0.02 m)2 + (0.59188 kg)(0.04 m)2 ˙
2
Î
˚
È1
˘
- Í (0.39458 kg)(0.02 m2 ) + (0.39458 kg)(0.04 m)2 ˙
Î2
˚
= [(20.20269) + (0.11838 + 0.94701)
- (0.07892 + 0.63133)] ¥ 10 -3 kg ◊ m2
= (20.20269 + 1.06539 - 0.71025) ¥ 10 -3 kg ◊ m2
= 20.55783 ¥ 10 -3 kg ◊ m2
(c)
or
I y = 20.6 ¥ 10 -3 kg ◊ m2 b
I z = ( I z )1 + ( I z )2 - ( I z )3
Ï1
¸
= Ì (6.31334 kg)[3(0.08)2 + (0.04)2 ] m2 + (6.31334 kg)(0.02 m)2 ˝
Ó12
˛
Ï1
¸
+ Ì (0.59188 kg)[3(0.02)2 + (0.06)2 ] m2 + (0.59188 kg)[(0.04)2 + (0.03)2 ] m2 ˝
Ó12
˛
Ï1
¸
- Ì (0.39458 kg)[3(0.02)2 + (0.04)2 ] m2 + (0.03958 kg)[(0.04)2 + (0.02)2 ] m2 ˝
12
Ó
˛
= [(10.94312 + 2.52534) + (0.23675 + 1.47970)
- (0.09207 + 0.78916)] ¥ 10 -3 kg ◊ m2
= (13.46846 + 1.71645 - 0.88123) ¥ 10 -3 kg ◊ m2
= 14.30368 ¥ 10 -3 kg ◊ m2
To the Instructor:
or
I z = 14.30 ¥ 10 -3 kg ◊ m2 b
The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions
of Problems 9.142 through 9.145.
From Figure 9.28:
Cylinder
( I x ) cyl =
1
mcyl a2
2
( I y )cyl = ( I z )cyl =
1
mcyl (3a2 + L2 )
12
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213
PROBLEM 9.141 (Continued)
Symmetry and the definition of the mass moment of inertia ( I = Ú r 2 dm) imply
1
( I )semicylinder = ( I )cylinder
2
1Ê1
ˆ
( I x )sc = Á mcyl a2 ˜
¯
2Ë2
and
However,
( I y )sc = ( I z )sc =
msc =
1È1
˘
mcyl (3a2 + L2 ) ˙
Í
2 Î12
˚
1
mcyl
2
1
msc a2
2
Thus,
( I x )sc =
and
( I y )sc = ( I z )sc =
1
msc (3a2 + L2 )
12
Also, using the parallel axis theorem find
Ê 1 16 ˆ
I x¢ = msc Á - 2 ˜ a2
Ë 2 9p ¯
1 ˘
ÈÊ 1 16 ˆ
I z¢ = msc ÍÁ - 2 ˜ a2 + L2 ˙
12 ˚
ÎË 4 9p ¯
where x ¢ and z ¢ are centroidal axes.
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214
PROBLEM 9.142
Determine the mass moment of inertia of the steel machine element
shown with respect to the y axis. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = rSTV
Then
m1 = 7850 kg / m3 ¥ (0.18 ¥ 0.074 ¥ 0.054) m3
= 5.6463 kg
Èp
˘
m2 = 7850 kg / m3 Í ¥ (0.027)2 ¥ (0.028)˙ m3
Î2
˚
= 0.25170 kg
m3 = 7850 kg / m3 ¥ ÈÎp ¥ (0.012)2 ¥ (0.024)˘˚ m3
= 0.085230 kg
m4 = 7850 kg / m3 ¥ (0.018 ¥ 0.18 ¥ 0.012) m3
= 0.30521 kg
The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations
derived above (component 2).
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215
PROBLEM 9.142 (Continued)
Find: I y
We have
I y = ( I y )1 + ( I y )2 + ( I y )3 - ( I y ) 4
Ï1
= Ì (5.6463 kg) ÈÎ(0.054)2 + (0.18)2 ˘˚ m2
Ó12
ÈÊ 0.054 ˆ 2 Ê 0.18 ˆ 2 ˘ 2 ¸Ô
+ (5.6463 kg) ÍÁ
˜ ˙ m ˝
˜ + ÁË
2 ¯ ˙˚
ÍÎË 2 ¯
Ô˛
Ï1
+ Ì (0.25170 kg)[3(0.027)2 + (0.028)2 ] m2
Ó12
2
È
0.028 ˆ ˘ 2 ¸Ô
Ê
+ (0.25170 kg) Í(0.027)2 + Á 0.18 ˜ ˙ m ˝
Ë
2 ¯ ˙˚
ÍÎ
Ô˛
Ï1
+ Ì (0.08523 kg)[3(0.012)2 + (0.024)2 ]m2
Ó12
2
¸Ô
0.024 ˆ
Ê
+ (0.08523 kg)[(0.027)2 + Á 0.18 +
m2 ˝
˜
Ë
2 ¯
˛Ô
Ï1
2
2
2
- Ì (0.30521 kg)[(0.018) + (0.18) ] m
Ó12
2
È
Ê 0.180 ˆ ˘ 2 ¸Ô
+ (0.30521 kg) Í(0.027)2 + Á
˙ m ˝
Ë 2 ˜¯ ˙
ÍÎ
Ô˛
˚
= [(16617.1 + 49851.2) + (62.317 + 7119.3)
+ (7.1593 + 3204.1) - (832.311 + 2694.7)] ¥ 10 -6 kg ◊ m2
= 73.334 ¥ 10 -3 kg ◊ m2
or
I y = 73.3 ¥ 10 -3 kg ◊ m2 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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216
PROBLEM 9.143
Determine the mass moment of inertia of the steel machine
element shown with respect to the z axis. (The density of steel
is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = rSTV
Then
m1 = 7850 kg/m3 ¥ (0.18 ¥ 0.074 ¥ 0.054) m3
= 5.6463 kg
Èp
˘
m2 = 7850 kg/m3 Í ¥ (0.027)2 ¥ (0.028)˙ m3
Î2
˚
= 0.25170 kg
m3 = 7850 kg/m3 ¥ ÈÎp ¥ (0.012)2 ¥ (0.024)˘˚ m3
= 0.085230 kg
m4 = 7850 kg/m3 ¥ (0.018 ¥ 0.18 ¥ 0.012) m3
= 0.30521 kg
The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations
derived above (component 2).
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217
PROBLEM 9.143 (Continued)
Find: I z
We have
I z = ( I z )1 + ( I z )2 + ( I z )3 - ( I z )4
Ï1
= Ì (5.6463 kg) ÎÈ(0.054)2 + (0.074)2 ˘˚ m2
Ó12
ÈÊ 0.054 ˆ 2 Ê 0.074 ˆ 2 ˘ 2 ¸Ô
+ (5.6463 kg) ÍÁ
˜ ˙ m ˝
˜ + ÁË
2 ¯ ˙˚
ÍÎË 2 ¯
Ô˛
Ï
Ê 1 16 ˆ
+ Ì(0.2517 kg) Á - 2 ˜ (0.027)2 m2
Ë 2 9p ¯
Ó
2
È
4 ¥ 0.027 ˆ ˘ 2 ¸Ô
Ê
+(0.2517 kg) Í(0.027)2 + Á 0.074 +
˜¯ ˙ m ˝
Ë
3p
ÍÎ
˙˚
Ô˛
Ï1
+ Ì (0.08523 kg)(0.012 m)2
Ó2
+(0.08523 kg)[(0.027)2 + (0.074)2 ]m2
}
Ï1
- Ì (0.30521 kg)[(0.018)2 + (0.012)2 ] m2
Ó2
2
È
Ê 0.012 ˆ ˘ 2 ¸Ô
+(0.30521 kg) Í(0.027)2 + Á
˙ m ˝
Ë 2 ˜¯ ˙
ÍÎ
Ô˛
˚
I z = [(3948.6 + 11845.9) + (58.693 + 2021.7)
+ (6.1366 + 528.85) - (11.9003 + 233.49)] ¥ 10 -6 kg ◊ m2
= 18.1645 ¥ 10 -3 kg ◊ m2
or
I z = 18.16 ¥ 10 -3 kg ◊ m2 b
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218
PROBLEM 9.144
Determine the mass moment of inertia and the radius of gyration of the
steel machine element shown with respect to the x axis. (The density
of steel is 7850 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = rSTV
Then
m1 = (7850 kg/m3 )(0.09 ¥ 0.03 ¥ 0.15) m3
= 3.17925 kg
Èp
˘
m2 = (7850 kg/m3 ) Í (0.045)2 ¥ 0.04 ˙ m3
2
Î
˚
= 0.998791 kg
m3 = (7850 kg/m3 )[p (0.038)2 ¥ 0.03] m3 = 1.06834 kg
Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to Problem 9.142)
for a semicylinder, we have
I x = ( I x )1 + ( I x )2 - ( I x )3
È1
˘
= Í (3.17925 kg)(0.032 + 0.152 ) m2 + (3.17925 kg)(0.075 m)2 ˙
Î12
˚
Ï
1
ÈÊ 1 16 ˆ
˘
+ Ì(0.998791 kg) ÍÁ - 2 ˜ (0.045 m)2 + (0.04 m)2 ˙
Ë
¯
12
Î 4 9p
˚
Ó
2
¸Ô
È
ˆ ˘
Ê 4 ¥ 0.045
+ (0.998791 kg) Í(0.13)2 + Á
+ 0.015˜ ˙ m2 ˝
¯ ˙
Ë 3p
ÍÎ
Ô˛
˚
Ï1
¸
- Ì (1.06834 kg)[3(0.038 m)2 + (0.03 m)2 ] + (1.06834 kg)(0.06 m)2 ˝
Ó12
˛
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219
PROBLEM 9.144 (Continued)
= [(0.0061995 + 0.0178833) + (0.0002745 + 0.0180409)
- (0.0004658 + 0..0038460)] kg ◊ m2
= (0.0240828 + 0.0183154 - 0.0043118) kg ◊ m2
= 0.0380864 kg ◊ m2
Now
or
I x = 38.1 ¥ 10 -3 kg ◊ m2 b
m = m1 + m2 - m3 = (3.17925 + 0.998791 - 1.06834) kg
= 3.10970 kg
and
k x2 =
I x 0.0380864 kg ◊ m2
=
m
3.10970 kg
or k x = 110.7 mm b
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220
PROBLEM 9.145
Determine the mass moment of inertia of the steel fixture
shown with respect to (a) the x axis, (b) the y axis, (c) the z axis.
(The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. we have
m = rSTV
m1 = 7850 kg/m3 ¥ (0.08 ¥ 0.05 ¥ 0.160) m3
= 5.02400 kg
Then
m2 = 7850 kg/m3 ¥ (0.08 ¥ 0.038 ¥ 0.07) m3 = 1.67048 kg
ˆ
Êp
m3 = 7850 kg/m3 ¥ Á ¥ 0.0242 ¥ 0.04˜ m3 = 0.28410 kg
¯
Ë2
Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
(a)
I x = ( I x )1 - ( I x )2 - ( I x )3
ÏÔ 1
ÈÊ 0.05 ˆ 2 Ê 0.16 ˆ 2 ˘ 2 ¸Ô
= Ì (5.02400 kg)[(0.05)2 + (0.16)2 ] m2 + (5.02400 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
2 ¯ ˙˚
12
ÍÎË 2 ¯
ÓÔ
˛Ô
2
2
ÏÔ 1
ÈÊ
0.038 ˆ
0.07 ˆ ˘ 2 ¸Ô
Ê
+
+
- Ì (1.67048 kg)[(0.038)2 + (0.07)2 ] m2 + (1.67048 kg) ÍÁ 0.05 .
0
05
˜ ˙m ˝
ÁË
˜
2 ¯
2 ¯ ˙˚
ÍÎË
ÔÓ12
Ô˛
Ï
1
ÈÊ 1 16 ˆ
˘
- Ì(0.28410 kg) ÍÁ - 2 ˜ (0.024)2 + (0.04)2 ˙ m2
Ë
¯
4
12
9p
Î
˚
Ó
2
2
ÈÊ
4 ¥ 0.024 ˆ
0.04 ˆ ˘ 2 ¸Ô
Ê
0
.
16
+
+ (0.28410 kg) ÍÁ 0.05 ˜ ˙m ˝
ÁË
˜¯
3p
2 ¯ ˙˚
ÍÎË
Ô˛
= [(11.7645 + 35.2936) - (0.8831 + 13.6745) - (0.0493 + 6.0187)] ¥ 10 -3 kg ◊ m2
= ( 47.0581 - 14.5576 - 6.0680) ¥ 10 -3 kg ◊ m2
= 26.4325 ¥ 10 -3 kg ◊ m2 or
I x = 26.4 ¥ 10 -3 kg ◊ m2 b
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221
PROBLEM 9.145 (Continued)
(b)
I y = ( I y )1 - ( I y )2 - ( I y )3
ÏÔ 1
ÈÊ 0.08 ˆ 2 Ê 0.16 ˆ 2 ˘ 2 ¸Ô
= Ì (5.02400 kg)[(0.08)2 + (0.16)2 ] m2 + (5.02400 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
2 ¯ ˙˚
12
ÍÎË 2 ¯
ÓÔ
˛Ô
2
ÏÔ 1
ÈÊ 0.08 ˆ 2 Ê
0.07 ˆ ˘ 2 ¸Ô
- Ì (1.67048 kg)[(0.08)2 + (0.07)2 ] m2 + (1.67048 kg) ÍÁ
0
.
05
+
+
˜ ˙m ˝
ÁË
˜
2 ¯ ˙˚
ÍÎË 2 ¯
ÔÓ12
Ô˛
2
ÏÔ 1
ÈÊ 0.08 ˆ 2 Ê
0.04 ˆ ˘ 2 ¸Ô
+
- Ì (0.28410 kg)[3(0.024)2 + (0.04)2 ] m2 + (0.28410 kg) ÍÁ
0
.
16
˜ ˙m ˝
ÁË
˜
12
2 ¯ ˙˚
ÍÎË 2 ¯
ÓÔ
˛Ô
= [(13.3973 + 40.1920) - (1.5730 + 14.7420) - (0.0788 + 6.0229)] ¥ 10 -3 kg ◊ m2
= (53.5893 - 16.3150 - 6.1017) ¥ 10 -3 kg ◊ m2
= 31.1726 ¥ 10 -3 kg ◊ m2 (c)
or
I y = 31.2 ¥ 10 -3 kg ◊ m2 b
I z = ( I z )1 - ( I z )2 - ( I z )3
ÏÔ 1
ÈÊ 0.08 ˆ 2 Ê 0.05 ˆ 2 ˘ 2 ¸Ô
= Ì (5.02400 kg)[(0.08)2 + (0.05)2 ] m2 + (5.02400 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
2 ¯ ˙˚
12
ÍÎË 2 ¯
ÓÔ
˛Ô
2
ÏÔ 1
ÈÊ 0.08 ˆ 2 Ê
0.038 ˆ ˘ 2 ¸Ô
+
- Ì (1.67048 kg)[(0.08)2 + (0.038)2 ] m2 + (1.67048 kg) ÍÁ
0
.
05
˜ ˙m ˝
ÁË
˜
2 ¯ ˙˚
ÍÎË 2 ¯
ÔÓ12
Ô˛
2
ÏÔ
ÈÊ 0.08 ˆ 2 Ê
4 ¥ 0.024 ˆ ˘ 2 ¸Ô
Ê 1 16 ˆ
+
- Ì(0.28410 kg) Á - 2 ˜ (0.024 m)2 + (0.28410 kg) ÍÁ
0
.
05
˜¯ ˙ m ˝
ÁË
˜
Ë 2 9p ¯
3p
ÍÎË 2 ¯
˙˚
ÔÓ
Ô˛
= [(3.7261 + 11.1784) - (1.0919 + 4.2781) - (0.0523 + 0.9049)] ¥ 10 -3 kg ◊ m2
= (14.9045 - 5.3700 - 0.9572) ¥ 10 -3 kg ◊ m2
= 8.5773 ¥ 10 -3 kg ◊ m2 or
I z = 8.58 ¥ 10 -3 kg ◊ m2 b
To the instructor:
The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use
in the solutions of Problems 9.146 through 9.148.
Slender Rod
Ix = 0
I y¢ = I z¢ =
1
mL2 (Figure 9.28)
12
1
I y = I z = mL2 (Sample Problem 9.9)
3
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222
PROBLEM 9.145 (Continued)
Circle
We have
I y = Ú r 2 dm = ma2
Now
I y = Ix + Iz
And symmetry implies
Ix = Iz
Ix = Iz =
1 2
ma
2
Semicircle
Following the above arguments for a circle, we have
Ix = Iz =
1 2
ma
2
I y = ma2
Using the parallel-axis theorem
I z = I z ¢ + mx 2
or
x=
2a
p
Ê1 4 ˆ
I z¢ = m Á - 2 ˜ a2
Ë2 p ¯
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223
PROBLEM 9.146
Aluminum wire with a weight per unit length of 0.05 kg/m is used to
form the circle and the straight members of the figure shown. Determine
the mass moment of inertia of the assembly with respect to each of the
coordinate axes.
SOLUTION
First compute the mass of each component.
m = rL
Have
Then
m1 = (0.05 kg/m)[2p (0.4 m)]
= 0.125664 kg
m2 = m3 = m4 = m5
= (0.05 kg/m)(0.2 m) = 0.01kg
Using the equations given above and the parallel-axis theorem, have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 + ( I x )5
ÈÊ 1 ˆ
ÈÊ 1 ˆ
˘
˘
= (0.125664 kg) ÍÁ ˜ (0.4 m))2 ˙ + (0.01 kg) ÍÁ ˜ (0.2 m)2 ˙
ÎË 2 ¯
˚
ÎË 3 ¯
˚
2
+ (0.01 kg) ÈÎ0 + (0.2 m) ˘˚
ÈÊ 1 ˆ
˘
+ (0.01 kg) ÍÁ ˜ (0.2 m)2 + (0.1 m)2 + (0.4 m)2 ˙
¯
Ë
Î 12
˚
ÈÊ 1 ˆ
˘
+ (0.01 kg) ÍÁ ˜ (0.2 m)2 + (0.2 m)2 + (0.4 m - 0.1m)2 ˙
¯
Ë
Î 12
˚
= [(10.053 + 0.13333 + 0.4 + 1.7333 + 1.3333) ¥ 10 -3 ] kg ◊ m2
= 13.6529 ¥ 10 -3 kg ◊ m2
or
I x = 13.65 ¥ 10 -3 kg ◊ m2 b
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224
PROBLEM 9.146 (Continued)
Have
where
Then
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 + ( I y )5
( I y )2 = ( I y ) 4
and ( I y )3 = ( I y )5
I y = (0.125664 kg)[(0.4 m)2 ] + 2(0.01 kg)[0 + (0.4 m)2 ]
È1
˘
+ 2 (0.01 kg) Í (0.2 m)2 + (0.3 m)2 ˙
Î12
˚
= [20.106 + 2(1.6) + 2(0.9333)] ¥ 10 -3 kg ◊ m2
= 25.1726 ¥ 10 -3 kg ◊ m2
or I y = 25.2 ¥ 10 -3 kg ◊ m2 b
By symmetry
Iz = Ix
or
I z = 13.65 ¥ 10 -3 kg ◊ m2 b
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225
PROBLEM 9.147
The figure shown is formed of 3-mm steel wire. Knowing that the density of the steel
is 7850 kg/m3 determine the mass moment of inertia of the wire with respect to each
of the coordinate axes.
SOLUTION
m = rV = r AL
Have
Then
m1 = m2 = (7850 kg/m3 ) ÈÎp (0.0015 m)2 ˘˚ ¥ (p ¥ 0.45 m)
m2 = m1 = 0.078445 kg
m3 = m4 = (7850 kg/m3 ) ÎÈp (0.0015 m)2 ˘˚ ¥ (0.45 m)
= 0.024970 kg
Using the equations given above and the parallel-axis theorem, have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4
where
( I x )3 = ( I x ) 4
Then
È1
˘
È1
˘
I x = (0.078445 kg) Í (0.45 m)2 ˙ + (0.078445 kg) Í (0.45 m)2 + (0.45 m)2 ˙
Î2
˚
Î2
˚
È1
˘
+ 2(0.02497) Í (0.45 m)2 + (0.225 m)2 + (0.45 m)2 ˙
Î12
˚
= [794256 + 23.8277 + 2(6.7419)] ¥ 10 -3 kg ◊ m2
= 45.254 ¥ 10 -3 kg ◊ m2
Have
where
or
I x = 45.3 ¥ 10 -3 kg ◊ m2 b
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4
( I y )1 = ( I y )2
and ( I y )3 = ( I y ) 4
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226
PROBLEM 9.147 (Continued)
Then
I y = 2(0.078445 kg) ÈÎ(0.45 m)2 ˘˚ + 2(0.02497 kg)[0 + (0.45 m)2 ]
= 2(15.8851 ¥ 10 -3 kg ◊ m2 ) + 2(5.0564 ¥ 10 -3 kg ◊ m2 )
= 41.883 ¥ 10 -3 kg ◊ m2
or
I y = 41.9 ¥ 10 -3 kg ◊ m2 b
Have
I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4
where
( I z )3 = ( I z ) 4
Then
È1
˘
I z = (0.078445 kg) Í (0.45 m)2 ˙
Î2
˚
2
˘
ÈÊ 1 4 ˆ
Ê 2 ¥ 0.45 m ˆ
+ (0.45 m)2 ˙
+(0.078445 kg) ÍÁ - 2 ˜ (0.45 m)2 + Á
˜
Ë
¯
p
˙˚
ÍÎË 2 p ¯
È1
˘
+2(0.02497 kg) + Í (0.45 m)2 ˙
Î3
˚
= [7.94256 + 23.8277 + 2(1.6855)] ¥ 10 -3 kg ◊ m2
= 35.141 ¥ 10 -3 kg ◊ m2
or
I z = 35.1 ¥ 10 -3 kg ◊ m2 b
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227
PROBLEM 9.148
A homogeneous wire with a mass per unit length of 0.056 kg/m is
used to form the figure shown. Determine the mass moment of inertia
of the wire with respect to each of the coordinate axes.
SOLUTION
First compute the mass m of each component. We have
m = ( m /L ) L
= 0.056 kg/m ¥ 1.2 m
= 0.0672 kg
Using the equations given above and the parallel-axis theorem, we have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 + ( I x )5 + ( I x )6
Now
Then
( I x )1 = ( I x )3 = ( I x )4 = ( I x )6
and ( I x )2 = ( I x )5
È1
˘
I x = 4 Í (0.0672 kg)(1.2 m)2 ˙ + 2[0 + (0.0672 kg)(1.2 m)2 ]
Î3
˚
= [4(0.03226) + 2(0.09677)] kg ◊ m 2
= 0.32258 kg ◊ m2 or
I x = 0.323 kg ◊ m2 b
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 + ( I y )5 + ( I y )6
Now
Then
( I y )1 = 0, ( I y )2 = ( I y )6 , and ( I y ) 4 = ( I y )5
È1
˘
I y = 2 Í (0.0672 kg)(1.2 m)2 ˙ + [0 + (0.0672 kg)(1.2 m)2 ]
Î3
˚
È1
˘
+ 2 Í (0.0672 kg)(1.2 m)2 + (0.0672 kg)(1.22 + 0.62 ) m2 ˙
Î12
˚
= [2(0.03226) + (0.09677) + 2(0.00806 + 0.12096)] kg ◊ m2
= [2(0.03226) + (0.09677) + 2(0.12902)] kg ◊ m2
= 0.41933 kg ◊ m2 Symmetry implies
or
I y = 0.419 kg ◊ m2 b
I z = 0.419 kg ◊ m2 b
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228
PROBLEM 9.149
Determine the mass products of inertia Ixy, Iyz, and Izx of the
steel fixture shown. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = rV
m1 = 7850 kg/m3 ¥ (0.08 ¥ 0.05 ¥ 0.16) m3 = 5.02400 kg
Then
m2 = 7850 kg/m3 ¥ (00.08 ¥ 0.038 ¥ 0.07) m3 = 1.67048 kg
ˆ
Êp
m3 = 7850 kg/m3 ¥ Á ¥ 0.0242 ¥ 0.04˜ m3 = 0.28410 kg
¯
Ë2
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of each component are zero because of
symmetry. Now
0.038 ˆ
Ê
y2 = Á 0.05 ˜ m = 0.031 m
Ë
2 ¯
4 ¥ 0.024 ˆ
Ê
y3 = Á 0.05 ˜¯ m = 0.039814 m
Ë
3p
and then
m, kg
y, m
0.025
z, m
0.08
mx y, kg ◊ m2
5.0240 ¥ 10 -3
my z , kg ◊ m2
10.0480 ¥ 10 -3
mz x, kg ◊ m2
16.0768 ¥ 10 -3
5.6796 ¥ 10 -3
1.5910 ¥ 10 -3
1
5.02400
x, m
0.04
2
1.67048
0.04
0.031
0.085
2.0714 ¥ 10 -3
4.4017 ¥ 10 -3
3
0.28410
0.04
0.039814
0.14
0.4524 ¥ 10 -3
1.5836 ¥ 10 -3
Finally,
I xy = ( I xy )1 - ( I xy )2 - ( I xy )3 = [(0 + 5.0240) - (0 + 2.0714) - (0 + 0.4524)] ¥ 10 -3
= 2.5002 ¥ 10 -3 kg ◊ m2 or
I xy = 2.50 ¥ 10 -3 kg ◊ m2 b
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229
PROBLEM 9.149 (Continued)
I yz = ( I yz )1 - ( I yz )2 - ( I yz )3 = [(0 + 10.0480) - (0 + 4.4017) - (0 + 1.5836))] ¥ 10 -3
= 4.0627 ¥ 10 -3 kg ◊ m2
or
I yz = 4.06 ¥ 10 -3 kg ◊ m2 b
I zx = ( I zx )1 - ( I zx )2 - ( I zx )3 = [(0 + 16.0768) - (0 + 5.6796) - (0 + 1.5910))] ¥ 10 -3
= 8.8062 ¥ 10 -3 kg ◊ m2
or
I zx = 8.81 ¥ 10 -3 kg ◊ m2 b
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230
PROBLEM 9.150
Determine the mass products of inertia Ixy, Iyz, and Izx of
the steel machine element shown. (The density of steel is
7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = rSTV
m1 = 7850 kg/m3 ¥ (0.1 ¥ 0.018 ¥ 0.09) m3 = 1.27170 kg
Then
m2 = 7850 kg/m3 ¥ (00.016 ¥ 0.06 ¥ 0.05) m3 = 0.37680 kg
m3 = 7850 kg/m3 ¥ (p ¥ 0.0122 ¥ 0.01) m3 = 0.03551 kg
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of each component are zero because of
symmetry. Now
my z , kg ◊ m2
mz x , kg ◊ m 2
0.045
mx y, kg ◊ m2
0.57227 ¥ 10 -3
0.51504 ¥ 10 -3
2.86133 ¥ 10 -3
0.048
0.045
1.66395 ¥ 10 -3
0.81389 ¥ 10 -3
1.55995 ¥ 10 -3
0.054
0.045
0.20134 ¥ 10 -3
0.08629 ¥ 10 -3
0.16778 ¥ 10 -3
2.43756 ¥ 10 -3
1.41522 ¥ 10 -3
4.58906 ¥ 10 -3
m, kg
x, m
y, m
z, m
1
1.27170
0.05
0.009
2
0.37680
0.092
3
0.03551
0.105
Â
Then
0
I xy = S( Ix ¢y ¢ + mx y )
or I xy = 2.44 ¥ 10 -3 kg ◊ m2 b
0
I yz = S ( I y ¢z ¢ + my z ) or
I zx
0
= S( I z ¢x ¢ + mzx )
I yz = 1.415 ¥ 10 -3 kg ◊ m2 b
or I zx = 4.59 ¥ 10 -3 kg ◊ m2 b
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231
PROBLEM 9.151
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast
aluminum machine component shown. (The specific weight of
aluminum is 2770 kg/m3)
SOLUTION
First compute the mass of each component. We have
m = rALV Then
kg
¥ (0.156 ¥ 0.016 ¥ 0.032) m3 m3
= 221.245 ¥ 10 -3 kg
m1 = 2770
kg
¥ (0.048 ¥ 0.016 ¥ 0.04) m3
m3
= 85.0944 ¥ 10 -3 kg
m2 = 2770
kg È p
˘
¥
¥ (0.016)2 ¥ 0.016˙ m3 = 17.8221 ¥ 10 -3 kg m3 ÍÎ 2
˚
kg
m4 = 2770 3 ¥ ÈÎp ¥ (0.011)2 ¥ 0.012˘˚ m3 = 12.6356 ¥ 10 -3 kg
m
m3 = 2770
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of each component are zero because
of symmetry. Now
4 ¥ 16 ˆ
Ê
x3 = - Á156 +
˜ mm = -162.791 mm
Ë
3p ¯
= -0.162791 m
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232
PROBLEM 9.151 (Continued)
and then
m, kg
x, m
y, m
z, m
mx y, kg ◊ m2
my z , kg ◊ m2
1
221.245 ¥ 10 -3
-0.078
0.008
-0.016
- 138.057 ¥ 10 -6
- 28.3194 ¥ 10 -6
276.114 ¥ 10 -6
2
85.0944 ¥ 10 -3
-0.024
0.008
-0.052
- 16.3381 ¥ 10 -6
-35.3993 ¥ 10 -6
106.198 ¥ 10 -6
3
17.8221 ¥ 10 -3 -0.1628
0.008
-0.016
- 23.2115 ¥ 10 -6
- 2.28123 ¥ 10 -6
46.423 ¥ 10 -6
4
12.6356 ¥ 10 -3
0.022
-0.016
- 43.3654 ¥ 10 -6
- 4.44773 ¥ 10 -6
31.5385 ¥ 10 -6
- 220.972 ¥ 10 -6
- 70.448 ¥ 10 -6
460.274 ¥ 10 -6
-0.156
Â
mz x, kg ◊ m2
0
Then
I xy = S( I x ¢y ¢ + mx y )
0
or
I xy = -221 ¥ 10 -6 kg ◊ m2 
I yz = S( I y ¢z ¢ + m y z )
or
I yz = -70.4 ¥ 10 -6 kg ◊ m2 
or
I zx = 460 ¥ 10 -6 kg ◊ m2 
0
I zx = S( I z ¢x ¢ + mz x )
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233
PROBLEM 9.152
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast
aluminum machine component shown. (The density of aluminum is
2770 kg/m3.)
SOLUTION
Have
m = rV
Then
m1 = (2770 kg/m3 )(0.118 ¥ 0.036 ¥ 0.044) m3 = 0.51775 kg Èp
˘
m2 = (2770 kg/m3 ) Í (0.022)2 ¥ 0.036˙ m3 = 0.075814 kg
Î2
˚
m3 = (2770 kg/m3 )(0.028 ¥ 0.022 ¥ 0.024) m3 = 0.04095 kg
Now observe that I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ are zero because of symmetry
Now
4 ¥ 0.023 ˆ
Ê
x2 = - Á 0.118 +
˜ m = -0.12734 m
Ë
3p ¯
0.062 ˆ
Ê
y3 = Á 0.036 ˜ m = 0.025 m
Ë
2 ¯
m, kg
x, m
y, m
z, m
mx y, kg ◊ m2
my z , kg ◊ m2
mz x, kg ◊ m2
1
0.51775
-0.059
0.018
0.022
- 0.54985 ¥ 10 -3
0.20503 ¥ 10 -3
- 0.67204 ¥ 10 -3
2
0.075814
-0.12734
0.018
0.022
- 0.17377 ¥ 10 -3
0.03002 ¥ 10 -3
- 0.21239 ¥ 10 -3
3
0.04095 kg
-0.041
0.025
0.026
- 0.014198 ¥ 10 -3
0.02662 ¥ 10 -3
- 0.04365 ¥ 10 -3
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234
And
Finally,
PROBLEM 9.152 (Continued)
0
I xy = S ( I x ¢y ¢ + mx y )
0
I yz = S ( I y ¢z ¢ + my z )
0
I zx = S ( I z ¢x ¢ + mx z )
I xy = ( I xy )1 + ( I xy )2 - ( I xy )3 = -0.70942 ¥ 10 -3 kg ◊ m2 or
I xy = - 0.709 ¥ 10 -3 kg ◊ m2 
I yz = ( I yz )1 + ( I yz )2 - ( I yz )3 = 0.20843 ¥ 10 -3 kg ◊ m2
or I yz = 0.208 ¥ 10 -3 kg ◊ m2 
I zx = ( I zx )1 + ( I zx )2 - ( I zx )3 = -0.84078 ¥ 10 -3 kg ◊ m2
or I zx = -0.841 ¥ 10 -3 kg ◊ m2 
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235
PROBLEM 9.153
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of
steel is 7850 kg/m3, determine the mass products of inertia
Ixy, Iyz, and Izx of the component.
SOLUTION
m1 = rV = 7850 kg/m3 (0.2 m)(0.3 m)(0.002 m) = 0.942 kg
m2 = rV = 7850 kg/m3 (0.4 m)(0.3 m)(0.002 m) = 1.884 kg
For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes.
Â
m, kg
x, m
y, m
z, m
mx y
my z
0.942
0.15
0.1
0.4
+ 14.13 ¥ 10 -3
+ 37.68 ¥ 10 -3
+ 56.52 ¥ 10 -3
1.884
0.15
0
0.2
0
0
+ 56.52 ¥ 10 -3
+ 14.13 ¥ 10 -3
+ 37.68 ¥ 10 -3
mz x
+ 113.02 ¥ 10 -3
I xy = + 14.13 ¥ 10 -3 kg ◊ m2 I xy = + 14.13 g ◊ m2 
I yz = + 37.68 ¥ 10 -3 kg ◊ m2 I yz = + 37.7 g ◊ m2 
I zx = + 113.02 ¥ 10 -3 kg ◊ m2 I zx = + 113.0 g ◊ m2 
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236
PROBLEM 9.154
A section of sheet steel 2-mm thick is cut and bent into the
machine component shown. Knowing that the density of steel is
7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and
Izx of the component.
SOLUTION
First compute the mass of each component.
We have
m = rSTV = rST tA Then
m1 = (7850 kg/m3 )[(0.002)(0.4)(0.45)]m3
= 2.8260 kg
Now observe that
È
ˆ˘
Ê1
m2 = 7850 kg/m3 Í(0.002) Á ¥ 0.45 ¥ 0.18˜ ˙ m3
¯˚
Ë
2
Î
= 0.63585 kg
( I x ¢y ¢ )1 = ( I y ¢z ¢ )1 = ( I z ¢x ¢ )1 = 0
( I y ¢z ¢ )2 = ( I z ¢x ¢ )2 = 0
From Sample Problem 9.6:
Then
Also
( I x ¢y ¢ )2,area = -
1 2 2
b2 h2
72
1
ˆ
Ê 1
( I x ¢y ¢ )2 = rST t ( I x ¢y ¢ )2,area = rST t Á - b22 h22 ˜ = - m2 b2 h2 ¯
Ë 72
36
x1 = y1 = z2 = 0
0.45 ˆ
Ê
x2 = Á - 0.225 +
˜ m = - 0.075 m
Ë
3 ¯
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237
PROBLEM 9.154 (Continued)
Finally,
È 1
I xy = S( I xy + mx y ) = (0 + 0) + Í- (0.63585 kg)(0.45 m)(0.18 m)
Î 36
Ê 0.18 m ˆ ˘
+ (0.63585 kg)( - 0.075 m) Á
Ë 3 ˜¯ ˙˚
= ( - 1.43066 ¥ 10 -3 - 2.8613 ¥ 10 -33 ) kg ◊ m2
and
or
I xy = - 4.29 ¥ 10 -3 kg ◊ m2 
I yz = S( I y ¢z ¢ + m y z ) = (0 + 0) + (0 + 0) = 0
or
I yz = 0 
or
I zx = 0 
I zx = S( I z ¢x ¢ + m z x ) = (0 + 0) + (0 + 0) = 0
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238
PROBLEM 9.155
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of steel is
7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and
Izx of the component.
SOLUTION
First compute the mass of each component. We have
m = rSTV = rST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.35 ¥ 0.39) m2
= 2.14305 kg
ˆ
Êp
m2 = (7850 kg/m3 )(0.002 m) Á ¥ 0.1952 ˜ m2
¯
Ë2
= 0.93775 kg
ˆ
Ê1
m3 = (7850 kg/m3 )(0.002 m) Á ¥ 0.39 ¥ 0.15˜ m2
¯
Ë2
= 0.45923 kg
Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero and
( I x ¢y ¢ )3 = ( I z ¢x ¢ )3 = 0 Also
( I y ¢z ¢ )3, mass = rST t ( I y ¢z ¢ )3, area
Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to
a 90° rotation, we have
1
( I y ¢z ¢ )3, area = b32 h32
72
ˆ 1
Ê 1
Then
( I y ¢z ¢ )3 = rST t Á b32 h32 ˜ = m3b3 h3
¯ 36
Ë 72
Also
y1 = x2 = 0
4 ¥ 0.195
m = 0.082761 m
y2 =
3p
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239
PROBLEM 9.155 (Continued)
Finally,
I xy = S( I x ¢y ¢ + mx y )
È
Ê 0.15 ˆ ˘
= (0 + 0) + (0 + 0) + Í0 + (0.45923 kg)(0.35 m) Á m˜ ˙
Ë
¯˚
3
Î
= - 8.0365 ¥ 10 -3 kg ◊ m2 or I xy = - 8.04 ¥ 10 -3 kg ◊ m2 
I yz = S( I y ¢z ¢ + my z )
= (0 + 0) + [0 + (0.93775 kg)(0.082761 m)(0.195 m)]
È1
Ê 0.15 ˆ Ê 0.39 ˆ ˘
m˜ ˙
m˜ Á
+ Í (0.45923 kg)(0.39 m)(0.15 m) + (0.45923 kg) Á ¯Ë 3
¯˚
Ë
3
Î 36
= [(15.1338) + (0.7462 - 2.9850)] ¥ 10 -3 kg ◊ m2
= (15.11338 - 2.2388) ¥ 10 -3 kg ◊ m2
= 12.8950 ¥ 10 -3 kg ◊ m2 or I yz = 12.90 ¥ 10 -3 kg ◊ m2 
I zx = S( I z ¢x ¢ + mz x )
= [0 + (2.14305 kg)(0.175 m)(0.195 m)] + (0 + 0)
È
˘
Ê 0.39 ˆ
m˜ (0.35 m)˙
+ Í0 + (0.45923 kg) Á
¯
Ë
3
Î
˚
= (73.1316 + 20.8950) ¥ 10 -33 kg ◊ m2
= 94.0266 ¥ 10 -3 kg ◊ m2 or I zx = 94.0 ¥ 10 -3 kg ◊ m2 
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240
PROBLEM 9.156
A section of sheet steel 2 mm thick is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass products of inertia Ixy, Iyz, and Izx of the component.
SOLUTION
First compute the mass of each component. We have
m = rSTV = rST tA Then
ˆ
Ê1
m1 = m3 = (7850 kg/m3 )(0.002 m) Á ¥ 0.225 ¥ 0.135˜ m2 = 0.23844 kg
¯
Ë2
ˆ
Êp
m2 = (7850 kg/m3 )(0.002 m) Á ¥ 0.2252˜ m2 = 0.62424 kg
¯
Ë 4
ˆ
Êp
m4 = (7850 kg/m3 )(0.002 m) Á ¥ 0.1352˜ m2 = 0.22473 kg
¯
Ë 4
Now observe that the following centroidal products of inertia are zero because of symmetry.
Also
Now
( I x ¢y ¢ )1 = ( I y ¢z ¢ )1 = 0
( I x ¢y ¢ )2 = ( I y ¢z ¢ )2 = 0
( I x ¢y ¢ )3 = ( I z ¢x ¢ )3 = 0
( I x ¢y ¢ )4 = ( I z ¢x ¢ ) = 0
y1 = y2 = 0
x3 = x4 = 0
I xy = S( I x ¢y ¢ + mx y )
I xy = 0 
so that
Using the results of Sample Problem 9.6, we have
1 2 2
I uv, area =
b h 24
Now
I uv , mass = rST t I uv , area
ˆ
Ê 1
= rST t Á b2 h2 ˜
¯
Ë 24
=
Thus,
( I zx )1 =
1
mbh
12
1
m1b1h1 12
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241
PROBLEM 9.156 (Continued)
( I yz )3 = -
while
1
m3b3 h3
12
because of a 90° rotation of the coordinate axes.
To determine I uv for a quarter circle, we have
0
dI uv = d I u ¢v ¢ + uEL vEL dm where
1
1 2
v=
a - u2
2
2
uEL = u vEL =
dm = rST t dA = rST tv du = rST t a2 - u 2 du
Then
)
(
a
ˆ
Ê1 2
I uv = Ú dI uv = Ú (u ) Á
a - u 2 ˜ rST t a2 - u 2 du 0
¯
Ë2
=
a
1
rST t Ú u( a2 - u 2 ) du
0
2
=
1
1 ˘
1
1
È1
ma 4
rST t Í a2 u 2 - u 4 ˙ = rST t a 4 =
2
4 ˚0 8
2p
Î2
a
Thus
( I zx )2 = -
1
m2 a22 2p
because of a 90° rotation of the coordinate axes. Also
( I yz ) 4 =
Finally,
1
m4 a42 2p
0
0
I yz = S( I yz ) = [( I y ¢z ¢ ) + m1 y1 z1 ] + [( I y ¢z ¢ ) + m2 y2 z2 ] + ( I yz )3 + ( I yz )4
È 1
˘ È1
˘
= Í- (0.23844 kg)(0.225 m)(0.135 m) ˙ + Í (0.22473 kg)(0.135 m)2 ˙
Î 12
˚ Î 2p
˚
= ( - 0.60355 + 0.65185) ¥ 10 -3 kg ◊ m2
0
or I yz = 48.3 ¥ 10 -6 kg ◊ m2 
0
I zx = S( I zx ) = ( I zx )1 + ( I zx )2 + [( I z ¢x ¢ )3 + m3 z3 x3 ] + [( I z ¢x ¢ ) 4 + m4 z4 x4 ]
È1
˘ È 1
˘
= Í (0.23844 kg)(0.225 m)(0.135 m)˙ + Í(0.62424 kg)(0.225 m)2 ˙
Î12
˚ Î 2p
˚
= (0.60355 - 5.02964) ¥ 10 -3 kg ◊ m2
or
I zx = - 4.43 ¥ 10 -3 kg ◊ m2 
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242
PROBLEM 9.157
Brass wire with a weight per unit length w is used to form the figure shown.
Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m=
W 1
= wL g g
W
w
(2p ¥ a) = 2p a g
g
w
w
m2 = ( a) = a
g
g
w
w
m3 = (2a) = 2 a
g
g
m1 =
Then
m4 =
w
wÊ
3 ˆ
ÁË 2p ¥ a˜¯ = 3p a
g
g
2
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of each component are zero because
of symmetry.
1
2
2p
m
x
y
z
w
a
g
2a
a
-a
2a
1
a
2
0
w 3
a
g
0
2a
0
a
0
0
2a
3
- a
2
2a
w
a
g
3
2
4
3p
w
a
g
w
a
g
mx y
4p
w 3
a
g
- 9p
w 3
a
g
w
(1 - 5p )a3
g
Â
my z
mz x
w 3
a
g
-2p
- 9p
w 3
a
g
-11p
w 3
a
g
- 4p
w 3
a
g
0
4
w 3
a
g
12p
4
w 3
a
g
w
(1 + 2p )a3
g
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243
PROBLEM 9.157 (Continued)
0
Then
I xy = S( I x ¢y ¢ + mx y )
0
or
I xy =
I yz = S( I y ¢z ¢ + my z )
0
or
I yz
I zx = S( I z ¢x ¢ + mz x )
or
I zx = 4
w 3
a (1 - 5p ) 
g
w
= -11p a3 
g
w 3
a (1 + 2p ) 
g
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244
PROBLEM 9.158
Brass wire with a weight per unit length w is used to form the figure shown.
Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m=
W 1
= wL g g
WÊ
3 ˆ 3 w
ÁË p ¥ a˜¯ = p a g
2
2 g
w
w
m2 = (3a) = 3 a
g
g
w
w
m3 = (p ¥ a) = p a
g
g
m1 =
Then
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of each component are zero because
of symmetry.
m
1
3 w
p a
2 g
x
-
2Ê3 ˆ
Á a˜
p Ë2 ¯
y
z
2a
1
a
2
2
3
w
a
g
0
1
a
2
2a
3
p
w
a
g
2
( a)
p
-a
a
mx y
-9
w 3
a
g
0
-2
w 3
a
g
-11
Â
w 3
a
g
my z
3 w 3
p a
2 g
3
mz x
-
9w 3
a
4g
w 3
a
g
-p
w 3
a
g
wÊp
ˆ 3
Á + 3˜¯ a
gË2
0
2
-
w 3
a
g
1w 3
a
4g
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245
PROBLEM 9.158 (Continued)
0
Then
I xy = S( I x ¢y ¢ + mx y )
I xy = -11
or
0
I yz = S( I y ¢z ¢ + my z )
0
or
I zx = S( I z ¢x ¢ + mz x )
or
I yz =
w 3
a 
g
1w 3
a (p + b) 
2g
I zx = -
1w 3
a 
4g
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246
PROBLEM 9.159
The figure shown is formed of 1.5-mm-diameter aluminum wire.
Knowing that the density of aluminum is 2800 kg/m3, determine
the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m = rALV = rAL AL
Then
Èp
˘
m1 = m4 = (2800 kg/m3 ) Í (0.0015 m)2 ˙ (0.25 m)
4
Î
˚
= 1.23700 ¥ 10 -3 kg
Èp
˘
m2 = m5 = (2800 kg/m3 ) Í (0.0015 m)2 ˙ (0.18 m)
Î4
˚
= 0.89064 ¥ 10 -3 kg
Èp
˘
m3 = m6 = (2800 kg/m3 ) Í (0.0015 m)2 ˙ (0.3 m)
Î4
˚
= 1.48440 ¥ 10 -3 kg
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247
PROBLEM 9.159 (Continued)
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of each component are zero because
of symmetry.
m, kg
x, m
y, m
z, m
mx y, kg ◊ m2
my z , kg ◊ m2
mz x, kg ◊ m2
1
1.23700 ¥ 10 -3
0.18
0.125
0
27.8325 ¥ 10 -6
0
0
2
0.89064 ¥ 10
-3
-6
0
0
3
1.48440 ¥ 10 -3
4
-3
0
0.125
5
0.89064 ¥ 10
-3
0.09
6
1.48440 ¥ 10 -3
0.18
1.23700 ¥ 10
20.0394 ¥ 10
0.09
0.25
0
0
0.25
0.15
0
55.6650 ¥ 10 -6
0
0.3
0
-6
0
0
0.3
0
0
24.0473 ¥ 10 -6
0
0.15
0
0
40.0788 ¥ 10 -6
47.8719 ¥ 10 -6
102.0525 ¥ 10 -6
64.1261 ¥ 10 -6
Â
Then
46.3875 ¥ 10
0
I xy = S( I x ¢y ¢ + mx y )
or
I xy = 47.9 ¥ 10 -6 kg ◊ m2 
0
= S( I y ¢z ¢ + my z )
or
I yz = 102.1 ¥ 10 -6 kg ◊ m2 
0
I zx = S( I z ¢x ¢ + mz x )
or
I zx = 64.1 ¥ 10 -6 kg ◊ m2 
I yz
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248
PROBLEM 9.160
Thin aluminum wire of uniform diameter is used to form the figure shown. Denoting
by m¢ the mass per unit length of the wire, determine the mass products of inertia
Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
Ê mˆ
m = Á ˜ L = m¢ L Ë L¯
Then
Êp ˆ p
m1 = m5 = m¢ Á R1 ˜ = m¢ R1 Ë2 ¯ 2
m2 = m4 = m¢( R2 - R1 )
Êp ˆ p
m3 = m¢ Á R2 ˜ = m¢ R2
Ë2 ¯ 2
Now observe that because of symmetry the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of components
2 and 4 are zero and
( I x ¢y ¢ )1 = ( I z ¢x ¢ )1 = 0
( I x ¢y ¢ )3 = ( I y ¢z ¢ )3 = 0
( I y ¢z ¢ )5 = ( I z ¢x ¢ )5 = 0
Also
x1 = x2 = 0
y2 = y3 = y4 = 0
z4 = z5 = 0 Using the parallel-axis theorem [Equations (9.47)], it follows that I xy = I yz = I zx for components 2 and 4.
To determine I uv for one quarter of a circular arc, we have dI uv = uvdm
where
and
u = a cos q
v = a sin q
dm = r dV = r [ A( adq )]
where A is the cross-sectional area of the wire. Now
so that
and
Êp ˆ
Êp ˆ
m = m ¢ Á a˜ = r A Á a˜
Ë2 ¯
Ë2 ¯
dm = m¢ adq
dI uv = ( a cos q )( a sin q )( m¢ adq )
= m¢ a3 sin q cos q dq
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249
PROBLEM 9.160 (Continued)
I uv = Ú dI uv = Ú
Then
p /2
0
m¢ a3 sin q cos qdq
p /2
È1
˘
= m¢ a3 Í sin 2 q ˙
2
Î
˚0
1
( I yz )1 = m¢ R13
2
Thus,
1
( I zx )3 = - m¢ R23
2
and
=
1
m ¢ a3
2
1
( I xy )5 = - m¢ R13
2
because of 90∞ rotations of the coordinate axes. Finally,
0
0
I xy = S( I xy ) = [( I x ¢y ¢ )1 + m1 x1 y1 ] + [( I x ¢y ¢ )3 + m3 x3 y3 ] + ( I xy )5
0
0
I yz = S( I yz ) = ( I yz )1 + [( I y ¢z ¢ )3 + m3 y3 z3 ] + [( I y ¢z ¢ )5 + m5 y5 z5 ]
I zx
0
0
= S( I zx ) = [( I z ¢x ¢ )1 + m1 z1 x1 ] + ( I zx )3 + [( I z ¢x ¢ )5 + m5 z5 x5 ]
or
or
or
1
I xy = - m¢ R13 
2
I yz =
1
m¢ R13 
2
1
I zx = - m¢ R23 
2
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250
PROBLEM 9.161
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of inertia.
SOLUTION
We have
and
Consider
I xy = Ú xydm I yz = Ú yzdm I zx = Ú zxdm x = x¢ + x
y = y¢ + y
(9.45)
z = z¢ + z (9.31)
I xy = Ú xydm
Substituting for x and for y
I xy = Ú ( x ¢ + x )( y ¢ + y ) dm
= Ú x ¢y ¢dm + y Ú x ¢dm + x Ú y ¢dm + x y Ú dm
By definition
and
I x ¢y ¢ = Ú x ¢y ¢dm
Ú x ¢dm = mx ¢
Ú y ¢dm = my ¢
However, the origin of the primed coordinate system coincides with the mass center G, so that
x¢ = y¢ = 0
I xy = I x ¢y ¢ + mx y Q.E.D. 
The expressions for I yz and I zx are obtained in a similar manner.
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251
PROBLEM 9.162
For the homogeneous tetrahedron of mass m shown, (a) determine by direct
integration the mass product of inertia Izx, (b) deduce Iyz and Ixy from the
result obtained in Part a.
SOLUTION
(a)
First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown.
a
zˆ
Ê
Now
x = - z + a = a Á1 - ˜
Ë
c
c¯
and
b
zˆ
Ê
y = - z + b = b Á1 - ˜
Ë c¯
c
The mass dm of the slab is
2
zˆ
Ê
ˆ 1
Ê1
dm = rdV = r Á xydz ˜ = r ab Á1 - ˜ dz Ë c¯
¯ 2
Ë2
Then
m = Ú dm = Ú
2
zˆ
Ê
r ab Á1 - ˜ dz
Ë
c¯
2
c1
0
c
3
ÈÊ c ˆ Ê
zˆ ˘
1
1
= r ab ÍÁ - ˜ Á1 - ˜ ˙ = r abc
2
6
ÍÎË 3 ¯ Ë c ¯ ˙˚
0
Now
where
and
Then
dI zx = dI z ¢x ¢ + zEL xEL dm
dI z ¢x ¢ = 0 (symmetry)
zEL = z
xEL =
1
1 Ê
zˆ
x = a Á1 - ˜
3
3 Ë c¯
2
˘
c È1 Ê
z ˆ ˘ È1
zˆ
Ê
I zx = Ú dI zx = Ú z Í a Á1 - ˜ ˙ Í r ab Á1 - ˜ dz ˙
0
Ë c¯
Î 3 Ë c ¯ ˚ ÍÎ 2
˚˙
=
z2
z3 z 4 ˆ
1 2 cÊ
ra b Ú Á z - 3 + 3 2 - 3 ˜ dz
0Ë
c
6
c
c ¯
c
m È1
z3 3 z 4 1 z5 ˘
= a Í z2 - +
˙
c Î2
c 4 c 2 5 c3 ˚ 0
or
I zx =
1
mac 
20
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252
PROBLEM 9.162 (Continued)
(b)Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular permutation
of ( x, y, z ) and ( a, b, c) .
Thus,
I xy =
1
mab 
20
I yz =
1
mbc 
20
Alternative solution for Part a:
First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown.
Now
x=-
a
Ê
y + a = a Á1 Ë
b
yˆ
˜
b¯
and
z=-
c
Ê
y + c = c Á1 Ë
b
yˆ
˜
b¯
The mass dm of the slab is
Now
where
and dI zx, area =
Then
Ê
ˆ 1
Ê1
dm = r dV = r Á xzdy ˜ = r ac Á1 Ë
¯ 2
Ë2
dI zx = rtdI zx , area
t = dy
1 2 2
x z from the results of Sample Problem 9.6.
24
2
2
ÏÔ 1 È Ê
yˆ ˘ È Ê
y ˆ ˘ ¸Ô
dIzx = r( dy ) Ì Ía Á1 - ˜ ˙ Íc Á1 - ˜ ˙ ˝
Ë
b¯ ˚Î Ë
b¯˚ Ô
ÔÓ 24 Î
˛
1
Ê
=
ra2 c 2 Á1 Ë
24
Finally,
2
yˆ
˜ dy b¯
I zx = Ú dI zx = Ú
b
0
4
1m Ê
yˆ
ac Á1 ˜¯ dy =
4b Ë
b
4
yˆ
˜ dy
b¯
4
yˆ
1m Ê
ac Á1 - ˜ dy
b¯
4b Ë
1 m ÈÊ b ˆ Ê
=
ac ÍÁ - ˜ Á1 4 b ÎË 5 ¯ Ë
b
5
yˆ ˘
˜¯ ˙ b ˚0
or
I zx =
1
mac 
20
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253
PROBLEM 9.162 (Continued)
Alternative solution for Part a:
The equation of the included face of the tetrahedron is
x y z
+ + = 1
a b c
x zˆ
Ê
y = b Á1 - - ˜
Ë a c¯
so that
For an infinitesimal element of sides dx, dy, and dz :
dm = r dV = r dydxdz zˆ
Ê
x = a Á1 - ˜
Ë c¯
From Part a
Now
I zx = Ú zxdm = Ú
= rÚ
0 Ú0
c
0
a (1- z /c )
0 Ú0
a (1- z /c )
c
= rb Ú
c
b (1- x /a - z /c )
Ú0
(
zx( r dydxdz )
)
zx ÈÎb 1 - ax - cz ˘˚ dxdz
a (1- z /c )
È1
1 x3 1 z 2 ˘
z Í x2 x ˙
3 a 2 c ˚0
Î2
dz
2
3
2
c È1
1 Ê
1 z 2Ê
zˆ ˘
zˆ
zˆ
Ê
= r b Ú z Í a 2 Á 1 - ˜ - a3 Á 1 - ˜ a Á1 - ˜ ˙ dz
0
2 c Ë c ¯ ˙˚
3a Ë c ¯
ÍÎ 2 Ë c ¯
= rb Ú
=
0
3
zˆ
Ê
a2 z Á1 - ˜ dz
Ë c¯
6
c1
z2
z3 z 4 ˆ
1 2 cÊ
r a b Ú Á z - 3 + 3 2 - 3 ˜ dz
0Ë
c
6
c
c ¯
c
m È1
z3 3 z 4 1 z5 ˘
= a Í z2 - +
˙
c Î2
c 4 c 2 5 c3 ˚ 0
or
I zx =
1
mac 
20
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254
PROBLEM 9.163
The homogeneous circular cylinder shown has a mass m. Determine the mass moment
of inertia of the cylinder with respect to the line joining the origin O and Point A that is
located on the perimeter of the top surface of the cylinder.
SOLUTION
Iy =
From Figure 9.28:
1 2
ma
2
and using the parallel-axis theorem
2
Ix = Iz =
Symmetry implies
1
1
Ê hˆ
m(3a2 + h2 ) + m Á ˜ = m(3a2 + 4 h2 ) Ë 2¯
12
12
I xy = I yz = I zx = 0
For convenience, let Point A lie in the yz plane. Then
lOA =
1
2
h + a2
( hj + ak )
With the mass products of inertia equal to zero, Equation (9.46) reduces to
0
I OA = I x l x2 + I y l y2 + I z l z2 2
2
Ê
ˆ
Ê
ˆ
1
1
h
a
+ m(3a2 + 4h2 ) Á
= ma2 Á
˜ ˜
2
12
Ë h2 + a 2 ¯
Ë h2 + a 2 ¯
or
1
10 h2 + 3a2

ma2
12
h2 + a 2
is given by
I OA =
Note: For Point A located at an arbitrary point on the perimeter of the top surface,lOA
1
lOA =
( a cos f i + hj + a sin f k ) 2
h + a2
which results in the same expression for I OA .
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255
PROBLEM 9.164
The homogeneous circular cone shown has a mass m. Determine the
mass moment of inertia of the cone with respect to the line joining
the origin O and Point A.
SOLUTION
2
First note that
9
Ê3 ˆ
dOA = Á a˜ + ( -3a)2 + (3a)2 = a
Ë2 ¯
2
Then
lOA =
1 Ê3
9 Á
Ë2
2a
ˆ 1
ai - 3aj + 3ak ˜ = (i - 2 j + 2k )
¯ 3
For a rectangular coordinate system with origin at Point A and axes aligned with the given x, y, z axes,
we have (using Figure 9.28)
3
3 È1
˘
I x = I z = m Í a2 + (3a)2 ˙
I y = ma2
5 Î4
10
˚
=
Also, symmetry implies
111 2
ma
20
I xy = I yz = I zx = 0
With the mass products of inertia equal to zero, Equation (9.46) reduces to
I OA = I x l x2 + I y l y2 + I z l z2 2
2
2
=
111 2 Ê 1 ˆ
3
111 2 Ê 2 ˆ
Ê 2ˆ
ma Á ˜ + ma2 Á - ˜ +
ma Á ˜ Ë 3¯
Ë 3¯
Ë 3¯
20
10
20
=
193 2
ma 60
or
I OA = 3.22ma2 
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256
PROBLEM 9.165
Shown is the machine element of Problem 9.141. Determine its
mass moment of inertia with respect to the line joining the origin O
and Point A.
SOLUTION
First compute the mass of each component.
We have
Then
m = rSTV m1 = (7850 kg/m3 )[p (0.08 m)2 (0.04 m)] = 6.31334 kg
m2 = (7850 kg/m3 )[p (0.02 m)2 (0.06 m)] = 0.59188 kg m3 = (7850 kg/m3 )[p (0.02 m)2 (0.04 m)] = 0.39458 kg
Symmetry implies
and
Now
I yz = I zx = 0
( I xy )1 = 0
( I x ¢y ¢ )2 = ( I x ¢y ¢ )3 = 0
I xy = S( I x ¢y ¢ + mx y ) = m2 x2 y2 - m3 x3 y3
= [0.59188 kg (0.04 m)(0.03 m)] - [0.39458 kg (- 0.04 m)(- 0.02 m)]
= (0.71026 - 0.31566) ¥ 10 -3 kg ◊ m2
= 0.39460 ¥ 10 -3 kg ◊ m2
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257
PROBLEM 9.165 (Continued)
From the solution to Problem 9.141, we have
I x = 13.98800 ¥ 10 -3 kg ◊ m2
I y = 20.55783 ¥ 10 -3 kg ◊ m2 By observation
I z = 14.30368 ¥ 10 -3 kg ◊ m2
1
lOA =
(2i + 3j)
13
Substituting into Eq. (9.46)
I OA =
I x l x2
+
I y l y2
0
+
I z l z2
0
0
- 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x 2
2
È
Ê 3 ˆ
Ê 2 ˆ
(
20
.
55783
)
+
= Í(13.98800) Á
˜
ÁË
Ë 13 ˜¯
13 ¯
ÍÎ
Ê 2 ˆ Ê 2 ˆ˘
- 2(0.39460) Á
¥ 10 -3 kg ◊ m2
Ë 13 ˜¯ ÁË 13 ˜¯ ˙˚
= ( 4.30400 + 14.23234 - 0.36425) ¥ 10 -3 kg ◊ m2
or
I OA = 18.17 ¥ 10 -3 kg ◊ m2 
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258
PROBLEM 9.166
Determine the mass moment of inertia of the steel fixture of
Problems 9.145 and 9.149 with respect to the axis through the
origin that forms equal angles with the x, y, and z axes.
SOLUTION
From the solutions to Problems 9.145 and 9.149, we have
I x = 26.4325 ¥ 10 -3 kg ◊ m2
Problem 9.145:
I y = 31.1726 ¥ 10 -3 kg ◊ m2
I z = 8.5773 ¥ 10 -3 kg ◊ m2
Problem 9.149:
I xy = 2.5002 ¥ 10 -3 kg ◊ m2
I yz = 4.0627 ¥ 10 -3 kg ◊ m2
I zx = 8.8062 ¥ 10 -3 kg ◊ m2
From the problem statement it follows that
lx = l y = lz
Now
or
l x2 + l 2y + l z2 = 1 fi 3l x2 = 1
1
lx = l y = lz =
3
Substituting into Eq. (9.46)
Noting that
We have
I OL = I x l x2 + I y l y2 + I z l z2 - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
1
l x2 = l y2 = l z2 = l x l y = l y l z = l z l x =
3
1
I OL = [26.4325 + 31.1726 + 8.5773
3
- 2(2.5002 + 4.0627 + 8.8062)] ¥ 10 -3 kg ◊ m2
or
I OL = 11.81 ¥ 10 -3 kg ◊ m2 
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259
PROBLEM 9.167
The thin bent plate shown is of uniform density and weight W. Determine
its mass moment of inertia with respect to the line joining the origin O and
Point A.
SOLUTION
First note that
m1 = m2 =
1W
2 g
1
(i + j + k ) 3
Using Figure 9.28 and the parallel-axis theorem, we have
and that
lOA =
I x = ( I x )1 + ( I x )2
È 1 Ê1W ˆ 2 1W
=Í Á
˜a +2 g
ÍÎ12 Ë 2 g ¯
2
Ê aˆ ˘
ÁË ˜¯ ˙
2 ˙˚
Ï 1 Ê1W ˆ 2
1W
( a + a2 ) +
+Ì Á
˜
2 g
ÓÔ12 Ë 2 g ¯
=
1W
2 g
ÈÊ a ˆ 2 Ê a ˆ 2 ˘ ¸Ô
ÍÁ ˜ + Á ˜ ˙ ˝
Ë 2 ¯ ˙Ô
ÍÎË 2 ¯
˚˛
ÈÊ 1 1 ˆ 2 Ê 1 1 ˆ 2 ˘ 1 W 2
ÍÁË 12 + 4 ˜¯ a + ÁË 6 + 2 ˜¯ a ˙ = 2 g a
Î
˚
I y = ( I y )1 + ( I y )2
ÏÔ 1 Ê 1 W ˆ
1W
2
2
=Ì Á
˜¯ ( a + a ) + 2 g
Ë
12
2
g
ÔÓ
ÈÊ a ˆ 2 Ê a ˆ 2 ˘ ¸Ô
ÍÁ ˜ + Á ˜ ˙ ˝
Ë 2 ¯ ˙Ô
ÍÎË 2 ¯
˚˛
Ï 1 Ê 1 W ˆ 2 1 W È 2 Ê a ˆ 2 ˘ ¸Ô
+Ì Á
˜ a + 2 g Í( a) + ÁË 2 ˜¯ ˙ ˝
ÔÓ12 Ë 2 g ¯
ÍÎ
˙˚ ˛Ô
=
1 W ÈÊ 1 1 ˆ 2 Ê 1 5 ˆ 2 ˘ W 2
Á + ˜ a + ÁË + ˜¯ a ˙ = a
2 g ÍÎË 6 2 ¯
12 4
˚ g
I z = ( I z )1 + ( I z )2
È 1 Ê 1 W ˆ 2 1 W Ê aˆ 2 ˘
=Í Á
˜ a + 2 g ÁË 2 ˜¯ ˙
ÍÎ12 Ë 2 g ¯
˚˙
Ï 1 Ê 1 W ˆ 2 1 W È 2 Ê a ˆ 2 ˘ ¸Ô
+Ì Á
˜ a + 2 g Í( a) + ÁË 2 ˜¯ ˙ ˝
ÔÓ12 Ë 2 g ¯
˙˚ ˛Ô
ÎÍ
=
1W
2 g
ÈÊ 1 1 ˆ 2 Ê 1 5 ˆ 2 ˘ 5 W 2
ÍÁË 12 + 4 ˜¯ a + ÁË 12 + 4 ˜¯ a ˙ = 6 g a Î
˚
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260
PROBLEM 9.167 (Continued)
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of both components are zero because
of symmetry. Also, y1 = 0
0
1W
Ê aˆ 1 W 2
Then
I xy = S( I x ¢y ¢ + mx y ) = m2 x2 y2 =
( a) Á ˜ =
a Ë 2¯ 4 g
2 g
I yz
I zx
0
1 W Ê aˆ Ê aˆ 1 W 2
= S( I y ¢z ¢ + my z ) = m2 y2 z2 =
a Á ˜Á ˜ =
2 g Ë 2¯ Ë 2¯ 8 g
0
= S( I z ¢x ¢ + mz x ) = m1z1 x1 + m2 z2 x2
=
1W
2 g
Ê aˆ Ê aˆ 1 W
ÁË ˜¯ ÁË ˜¯ +
2 2
2 g
3W 2
Ê aˆ
a
ÁË ˜¯ ( a ) =
2
8 g
Substituting into Equation (9.46)
I OA = I x l x2 + I y l y2 + I z l z2 - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
Noting that
l x2 = l y2 = l z2 = l x l y = l y l z = l z l x =
1
3
We have
Ê 1 W 2 1 W 2 3W 2ˆ˘
1 È1 W 2 W 2 5 W 2
I OA = Í
a + a +
a - 2Á
a +
a +
a ˙
3 Î2 g
6 g
8 g
8 g ˜¯ ˚
Ë4 g
g
1 È14
Ê 3ˆ ˘ W
= Í - 2 Á ˜ ˙ a2
Ë 4¯ ˚ g
3Î 6
or
I OA =
5W 2
a 
18 g
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261
PROBLEM 9.168
A piece of sheet steel of thickness t and density r is cut and bent into the
machine component shown. Determine the mass moment of inertia of
the component with respect to the line joining the origin O and Point A.
SOLUTION
Let g be the specific weight = rg
First note that
lOA =
1
(i + 2 j + k )
6
Next compute the mass of each component. We have
m = rV =
Then
g
(tA)
g
m1 =
g
gt
t ( 2a ¥ 2a) = 4 a 2
g
g
m2 =
g Êp
ˆ p gt 2
t Á ¥ a2 ˜ =
a
¯ 2 g
g Ë2
Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem 9.134)
for a semicircular plate for component 2, we have
I x = ( I x )1 + ( I x )2
Ï 1 Ê gt ˆ
¸
gt
= Ì Á 4 a2 ˜ [(2a)2 + (2a)2 ] + 4 a2 ( a2 + a2 ) ˝
g
ÔÓ12 Ë g ¯
Ô˛
Ï1 Ê p g t 2ˆ 2 p g t 2
¸
a ˜a +
a [(2a)2 + ( a)2 ]˝
+Ì Á
2 g
ÔÓ 4 Ë 2 g ¯
Ô˛
p gt 2Ê1 ˆ 2
gt Ê2
ˆ
= 4 a2 Á + 2˜ a2 +
a Á + 5˜ a
g Ë3 ¯
2 g Ë4 ¯
gt
= 18.91335 a 4
g
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262
PROBLEM 9.168 (Continued)
I y = ( I y )1 + ( I y )2
È 1 Ê gt ˆ
˘
gt
= Í Á 4 a 2 ˜ ( 2a)2 + 4 a 2 ( a 2 ) ˙
g
Î12 Ë g ¯
˚
2
ÏÔ p g t Ê 1 16 ˆ
˘ ¸Ô
p g t 2 ÈÊ 4 a ˆ
+Ì
a2 Á - 2 ˜ a2 +
a ÍÁ ˜ + ( a)2 ˙ ˝
2 g
ÍÎË 3p ¯
˙˚ Ô˛
ÔÓ 2 g Ë 2 9p ¯
16
gt Ê1 ˆ
p g t 2 Ê 1 16
ˆ
= 4 a2 Á + 1˜ a2 +
a Á - 2 + 2 + 1˜ a2
¯
2 g Ë 2 9p
g Ë3 ¯
9p
gt
= 7.668953 a 4
g
I z = ( I z )1 + ( I z )2
È 1 Ê gt ˆ
˘
gt
= Í Á 4 a 2 ˜ ( 2a)2 + 4 a 2 ( a 2 ) ˙
g
Î12 Ë g ¯
˚
2
ÏÔ p g t Ê 1 16 ˆ
˘ Ô¸
p g t 2 ÈÊ 4 a ˆ
+Ì
a2 Á - 2 ˜ a2 +
a ÍÁ ˜ + (2a)2 ˙ ˝
2 g
ÍÎË 3p ¯
˙˚ Ô˛
ÔÓ 2 g Ë 4 9p ¯
16
gt Ê1 ˆ
p g t 2 Ê 1 16
ˆ
= 4 a2 Á + 1˜ a2 +
a Á - 2 + 2 + 4˜ a 2
¯
2 g Ë 4 9p
g Ë3 ¯
9p
gt
= 12.00922 a 4
g
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , of both components are zero because
of symmetry. Also x1 = 0.
0
p g t 2 Ê 4a ˆ
Then
I xy = S( I x ¢y ¢ + mx y ) = m2 x2 y2 =
a Á ˜ ( 2a) 2 g Ë 3p ¯
gt
= 1.33333 a 4
g
0
I yz = S( I y ¢z ¢ + my z ) = m1 y1 z1 + m2 y2 z2 gt 2
p gt 2
a ( a)( a) +
a (2a)( a)
2 g
g
gt
= 7.14159 a 4
0 g
p g t 2 Ê 4a ˆ
I zx = S( I z ¢x ¢ + mz x ) = m2 z2 x2 =
a ( a) Á ˜
Ë 3p ¯
2 g
gt
= 0.66667 a 4
g
=4
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263
PROBLEM 9.168 (Continued)
Substituting into Eq. (9.46)
I OA = I x l x2 + I y l y2 + I z l z2 - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
2
gt Ê 2 ˆ
gt Ê 1 ˆ
+ 7.68953 a 4 Á
= 18.91335 a 4 Á
˜
Ë 6 ˜¯
¯
Ë
g
g
6
2
2
Ê
gt Ê 1 ˆ
gt ˆÊ 1 ˆÊ 2 ˆ
- 2 Á1.33333 a 4 ˜ Á
+ 12.00922 a 4 Á
˜
Ë 6¯
Ë
g
g ¯ Ë 6 ˜¯ ÁË 6 ˜¯
Ê
Ê
gt ˆÊ 2 ˆÊ 2 ˆ
gt ˆÊ 1 ˆÊ 1 ˆ
- 2 Á 0.66667 a 4 ˜ Á
- 2 Á 7.14159 a 4 ˜ Á
˜
Á
˜
Ë
Ë
g ¯ Ë 6¯ Ë 6¯
g ¯ Ë 6 ˜¯ ÁË 6 ˜¯
gt
= (3.15223 + 5.12635 + 2.00154 - 0.88889 - 4.76106 - 0.22222) a 4 g
or
I OA = 4.41 r t a 4 
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264
PROBLEM 9.169
Determine the mass moment of inertia of the machine component
of Problems 9.136 and 9.155 with respect to the axis through the
origin characterized by the unit vector l = ( - 4i + 8 j + k ) / 9.
SOLUTION
From the solutions to Problems 9.136 and 9.155, we have
Problem 9.136:
I x = 175.503 ¥ 10 -3 kg ◊ m2
I y = 308.629 ¥ 10 -3 kg ◊ m2
I z = 154.400 ¥ 10 -3 kg ◊ m2
Problem 9.155:
I xy = -8.0365 ¥ 10 -3 kg ◊ m2
I yz = 12.8950 ¥ 10 -3 kg ◊ m2
I zx = 94.0266 ¥ 10 -3 kg ◊ m2
Substituting into Eq. (9.46)
I OL = I x l x2 + I y l y2 + I z l z2 - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
2
2
2
È
Ê 1ˆ
Ê 8ˆ
Ê 4ˆ
= Í175.503 Á - ˜ + 308.629 Á ˜ + 154.400 Á ˜
Ë 9¯
Ë 9¯
Ë 9¯
ÍÎ
Ê 8ˆ Ê 1ˆ
Ê 4ˆ Ê 8ˆ
- 2( -8.0365) Á - ˜ Á ˜ - 2(12.8950) Á ˜ Á ˜
Ë 9¯ Ë 9¯
Ë 9¯ Ë 9¯
Ê 1ˆ Ê 4ˆ ˘
- 2(944.0266) Á ˜ Á - ˜ ˙ ¥ 10 -3 kg ◊ m2
Ë 9¯ Ë 9 ¯ ˚
= (34.6673 + 243.855 + 1.906 - 6.350
- 2.547 + 9.287) ¥ 10 -3 kg ◊ m2
or
I OL = 281 ¥ 10 -3 kg ◊ m2 
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265
PROBLEM 9.170
For the wire figure of Problem 9.148, determine the mass moment
of inertia of the figure with respect to the axis through the origin
characterized by the unit vector l = ( -3i - 6 j + 2k ) / 7.
SOLUTION
First compute the mass of each component. We have
Ê mˆ
m = Á ˜ L = 0.056 kg/m ¥ 1.2 m
Ë L¯
= 0.0672 kg
Now observe that the centroidal products of inertia, I x ¢y ¢ , I y ¢z ¢ , and I z ¢x ¢ , for each
component are zero because of symmetry.
Also
Then
x1 = x6 = 0
I xy
y4 = y5 = y6 = 0
z1 = z2 = z3 = 0
0
= S( I x ¢y ¢ + mx y ) = m2 x2 y2 + m3 x3 y3
= (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m)
= 0.096768 kg ◊ m2
0
I yz = S( I y ¢z ¢ + my z ) = 0
0
I zx = S( I z ¢x ¢ + mz x ) = m4 z4 x4 + m5 z5 x5
= (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m)
= 0.096768 kg ◊ m2
From the solution to Problem 9.148, we have
I x = 0.32258 kg ◊ m2
I y = I z = 0.41933 kg ◊ m2
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266
PROBLEM 9.170 (Continued)
Substituting into Eq. (9.46)
0
I OL = I x l x2 + I y l y2 + I z l z2 - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
2
2
2
È
Ê 2ˆ
Ê 6ˆ
Ê 3ˆ
.
.
+
+
= Í0.32258 Á - ˜
0 41933 Á ˜
0 41933 Á ˜
Ë 7¯
Ë 7¯
Ë 7¯
ÍÎ
Ê 2ˆ Ê 3ˆ ˘
Ê 3ˆ Ê 6ˆ
- 2(0.096768) Á - ˜ Á - ˜ - 2(0.096768) Á ˜ Á - ˜ ˙ kg ◊ m 2
Ë 7¯ Ë 7¯ ˚
Ë 7¯ Ë 7¯
I OL = (0.059249 + 0.30808 + 0.034231 - 0.071095 + 0.023698) kg ◊ m2
or
I OL = 0.354 kg ◊ m2 
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267
PROBLEM 9.171
For the wire figure of the problem indicated, determine the mass moment of inertia of
the figure with respect to the axis through the origin characterized by the unit vector
l = (- 3i - 6j + 2k )/7.
SOLUTION
Have
m = rV = r AL
Then
m1 = m2 = (7850 kg/m3 ) ÎÈp (0.0015 m)2 ˘˚ ¥ (p ¥ 0.45 m)
m2 = m1 = 0.078445 kg
m3 = m4 = (7850 kg/m3 ) ÎÈp (0.0015 m)2 ˘˚ ¥ (0.45 m)
= 0.024970 kg
Now observe that the centroidal products of inertia, I x ¢y ¢ = I y ¢z ¢ = I z ¢x ¢ = 0 for
each component.
Also
Then
x3 = x4 = 0,
I xy
y1 = 0,
z1 = z2 = 0
0
= S( I x ¢y ¢ + mx y ) = m2 x2 y2
Ê 2 ¥ 0.45 m ˆ
= (0.078445 kg) Á ˜¯ (0.45 m)
Ë
p
= -10.1128 ¥ 10 -3 kg ◊ m2
0
I yz = S( I y ¢z ¢ + my z ) = m3 y3 z3 + m4 y4 z4
where
m3 = m4 ,
y3 = y4 ,
z4 = - z3 ,
so that
I yz = 0
I zx = S( I z ¢x ¢ + mz x ) = m1 z1 x1 + m2 z2 x2 = 0
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268
PROBLEM 9.171 (Continued)
From the solution to Problem 9.147
I x = 45.254 ¥ 10 -3 kg ◊ m2
I y = 41.883 ¥ 10 -3 kg ◊ m2
I z = 35.141 ¥ 10 -3 kg ◊ m 2
Now
1
lOL = ( -3i - 6 j + 2k )
7
Have
0
0
I OL = I x l x2 + I y l y2 + I z l z2 - 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
[Eq. (9.46))]
2
2
2
È
Ê 2ˆ
Ê 6ˆ
Ê 3ˆ
= Í45.254 Á - ˜ + 41.883 Á - ˜ + 35.141Á ˜
Ë 7¯
Ë 7¯
Ë 7¯
ÍÎ
Ê 3ˆ Ê 6ˆ ˘
- 2( -10.1128) Á - ˜ Á - ˜ ˙ ¥ 10 -3 kg ◊ m2
Ë 7¯ Ë 7¯ ˚
= (8.3120 + 30.7712 + 2.8687 + 7.4298) ¥ 10 -3 kg ◊ m2
= 49.3817 ¥ 10 -3 kg ◊ m2
or
I OL = 49.4 ¥ 10 -3 kg ◊ m2 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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269
PROBLEM 9.172
For the wire figure of Problem indicated, determine the mass moment
of inertia of the figure with respect to the axis through the origin
characterized by the unit vector l = ( - 3i - 6j + 2k )/7.
SOLUTION
First compute the mass of each component.
Have
Then
m = rL
m1 = (0.05 kg/m)[2p (0.4 m)]
= 0.125664 kg
m2 = m3 = m4 = m5
= (0.05 kg/m)(0.2 m) = 0.01kg
Now observe that
Also,
Then
I x ¢y ¢ = I y ¢z ¢ = I z ¢x ¢ = 0 for each component.
x1 = x4 = x5 = 0,
y1 = 0,
z1 = z2 = z3 = 0
0
I xy = S( I x ¢y ¢ + mx y ) = m2 x2 y2 + m3 x3 y3
= (0.01 kg)[(0.40 m)(0.1m) + (0.3m)(0.2 m)]
= 1.000 ¥ 10 -3 kg ◊ m2
I yz = 1.000 ¥ 10 -3 kg ◊ m2
By symmetry
I yz = I xy
Now
I zx = S( I z ¢x ¢ + mz x ) = 0
From the solution to Problem 9.146
I x = I z = 13.6529 ¥ 10 -3 kg ◊ m2
I y = 25.1726 ¥ 10 -3 kg ◊ m2
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270
PROBLEM 9.172 (Continued)
1
lOL = ( -3i - 6 j + 2k )
7
Now
Have
I OL =
I x l x2
+
I y l y2
+
I z l z2
0
- 2 I xy l x l y - 2 I yz l y l z - 2 I zx l z l x
[Eq.(9.46)]]
ÏÔ
ÈÊ 3 ˆ 2 Ê 2 ˆ 2 ˘ ¸Ô
= Ì(13.6529) ÍÁ - ˜ + Á ˜ ˙ ˝ ¥ 10 -3 kg ◊ m2
Ë 7¯ ˙Ô
ÍÎË 7 ¯
˚˛
ÓÔ
2
ÔÏ
Ê -6 ˆ
+ Ì{(25.1726) Á ˜ ¥ 10 -3 kg ◊ m2
Ë 7¯
ÓÔ
ÈÊ 3 ˆ Ê 6 ˆ Ê 6 ˆ Ê 2 ˆ ˘ ¸
-2(1.000) ÍÁ - ˜ Á - ˜ + Á - ˜ Á ˜ ˙ ˝ ¥ 10 -3 kg ◊ m2
ÎË 7 ¯ Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ ˚ ˛
= (3.6222 + 18.44942 - 0.2449) ¥ 10 -3 kg ◊ m2
= 21.8715 ¥ 10 -3 kg ◊ m2
or
I OL = 21.9 ¥ 10 -3 kg ◊ m2 
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271
PROBLEM 9.173
For the rectangular prism shown, determine the values of the ratios b/a
and c/a so that the ellipsoid of inertia of the prism is a sphere when
computed (a) at Point A, (b) at Point B.
SOLUTION
(a)
Using Figure 9.28 and the parallel-axis theorem, we have
at Point A
1
I x ¢ = m(b2 + c 2 )
12
I y¢
1
Ê aˆ
= m( a2 + c 2 ) + m Á ˜
Ë 2¯
12
1
= m( 4 a2 + c 2 )
12
2
2
I z¢ =
1
1
Ê aˆ
m( a2 + b2 ) + m Á ˜ = m( 4a2 + b2 )
Ë 2¯
12
12
Now observe that symmetry implies
I x ¢y ¢ = I y ¢z ¢ = I z ¢x ¢ = 0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
1
1
1
m(b2 + c 2 ) x 2 + m( 4a2 + c 2 ) y 2 + m( 4a2 + b2 ) z 2 = 1
12
12
12
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore
I x ¢ x 2 + I y ¢ y 2 + I z ¢ z 2 = 1:
1
1
m(b2 + c 2 ) = m( 4a2 + c 2 )
12
12
1
= m( 4a2 + b2 )
12
Then
b2 + c 2 = 4 a2 + c 2 or
and
b2 + c 2 = 4 a2 + b2 or
b
=2 
a
c
=2 
a
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272
PROBLEM 9.173 (Continued)
(b)
Using Figure 9.28 and the parallel-axis theorem, we have at Point B
2
I x ¢¢ =
1
1
Ê cˆ
m(b2 + c 2 ) + m Á ˜ = m(b2 + 4c 2 )
Ë 2¯
12
12
2
1
1
Ê cˆ
m( a2 + c 2 ) + m Á ˜ = m( a2 + 4c 2 )
¯
Ë
12
2
12
1
= m( a2 + b2 )
12
I y ¢¢ =
I z ¢¢
Now observe that symmetry implies
I x ¢¢y ¢¢ = I y ¢¢z ¢¢ = I z ¢¢x ¢¢ = 0
From Part a it then immediately follows that
1
1
1
m(b2 + 4c 2 ) = m( a2 + 4c 2 ) = m( a2 + b2 )
12
12
12
Then
b 2 + 4c 2 = a 2 + 4c 2
or
and
b 2 + 4c 2 = a 2 + b 2
or
b
=1 
a
c 1
= 
a 2
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273
PROBLEM 9.174
For the right circular cone of Sample Problem 9.11, determine the value
of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere
when computed (a) at the apex of the cone, (b) at the center of the base
of the cone.
SOLUTION
(a)
From Sample Problem 9.11, we have at the apex A
3
ma2
10
3 Ê1
ˆ
I y = I z = m Á a 2 + h2 ˜
¯
5 Ë4
Ix =
Now observe that symmetry implies
I xy = I yz = I zx = 0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x x 2 + I y y 2 + I z z 2 = 1:
3
3 Ê1
3 Ê1
ˆ
ˆ
ma2 x 2 + m Á a2 + h2 ˜ y 2 + m Á a2 + h2 ˜ z 2 = 1
¯
Ë
¯
Ë
10
5
4
5
4
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore,
(b)
3
3 Ê1
ˆ
ma2 = m Á a2 + h2 ˜
¯
10
5 Ë4
From Sample Problem 9.11, we have
and at the centroid C
I x¢ =
3
ma2
10
I y¢¢ =
3 Ê 2 1 2ˆ
mÁ a + h ˜
20 Ë
4 ¯
or
a
=2 
h
or
a
2

=
h
3
2
Then
I y¢ = I z¢ =
3 Ê 2 1 2ˆ
1
Ê hˆ
mÁ a + h ˜ + mÁ ˜ =
m(3a2 + 2h2 )
¯
Ë
¯
Ë
20
4
4
20
Now observe that symmetry implies
I x ¢y ¢ = I y ¢z ¢ = I z ¢x ¢ = 0
From Part a it then immediately follows that
3
1
ma2 =
m(3a2 + 2h2 ) 10
20
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274
PROBLEM 9.175
For the homogeneous circular cylinder shown, of radius a and length L,
determine the value of the ratio a/L for which the ellipsoid of inertia of the
cylinder is a sphere when computed (a) at the centroid of the cylinder, (b) at
Point A.
SOLUTION
(a)
From Figure 9.28:
1 2
ma
2
1
I y = I z = m(3a2 + L2 )
12
Ix =
Now observe that symmetry implies
I xy = I yz = I zx = 0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
1 2 2 1
1
ma x + m(3a2 + L2 ) y 2 + m(3a2 + L2 ) = 1
2
12
12
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore
I x x 2 + I y y 2 + I z z 2 = 1:
(b)
1 2 1
ma = m(3a2 + L2 )
2
12
Using Figure 9.28 and the parallel-axis theorem, we have
I x¢ =
or
a
1

=
L
3
or
7
a
=

L
12
1 2
ma
2
2
I y¢ = I z¢ =
1
7 2ˆ
Ê1
Ê Lˆ
m(3a2 + L2 ) + m Á ˜ = m Á a2 +
L˜
Ë4
Ë 4¯
12
48 ¯
Now observe that symmetry implies
I x ¢y ¢ = I y ¢z ¢ = I z ¢x ¢ = 0
From Part a it then immediately follows that
1 2
7 2ˆ
Ê1
ma = m Á a2 +
L ˜
Ë
2
4
48 ¯
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275
PROBLEM 9.176
Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of the
body with respect to any one of the three axes cannot be larger than the sum of the mass moments of inertia of
the body with respect to the other two axes. That is, prove that the inequality I x I y + I z and the two similar
inequalities are satisfied. Further, prove that I y 12 I x if the body is a homogeneous solid of revolution, where x
is the axis of revolution and y is a transverse axis.
SOLUTION
(i)
I y + Iz Ix
To prove
I y = Ú ( z 2 + x 2 )dm
By definition
I z = Ú ( x 2 + y 2 )dm
I y + I z = Ú ( z 2 + x 2 )dm + Ú ( x 2 + y 2 )dm
Then
= Ú ( y 2 + z 2 )dm + 2Ú x 2 dm
Now
Ú(y
2
+ z 2 )dm = I x
and
Úx
2
dm 0
I y + Iz Ix
Q.E.D.
1
I y Ix
2
Q.E.D.
The proofs of the other two inequalities follow similar steps.
(ii)
If the x axis is the axis of revolution, then
I y = Iz
and from Part (i)
or
I y + Iz Ix
2I y I x
or
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276
PROBLEM 9.177
Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a sphere,
and use this property to determine the moment of inertia of the cube with respect to one of its diagonals.
(b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution, and determine
the principal moments of inertia of the cube at that point.
SOLUTION
(a)
At the center of the cube have (using Figure 9.28)
1
1
I x = I y = I z = m( a2 + a2 ) = ma2
12
6
Now observe that symmetry implies
I xy = I yz = I zx = 0
Using Equation (9.48), the equation of the ellipsoid of inertia is
Ê 1 2ˆ 2 Ê 1 2ˆ 2 Ê 1 2ˆ 2
ÁË ma ˜¯ x + ÁË ma ˜¯ y + ÁË ma ˜¯ z = 1
6
6
6
or
x2 + y2 + z2 =
which is the equation of a sphere.
6
( = R2 ) 
ma2
Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through the
center O of the cube must always
Ê
1 ˆ
1
be the same Á R =
I OL = ma2 
˜ .
6
I OL ¯
Ë
(b)The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes through
corner A. For a rectangular coordinate system at A and with one of the coordinate axes aligned with the
diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then rotated 120∞ about the
diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid will also remain unchanged
after it is rotated. As noted at the end of Section 9.17, this is possible only if the ellipsoid is an ellipsoid
of revolution, where the diagonal is both the axis of revolution and a principal axis.
1
I x ¢ = I OL = ma2 
6
In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal
and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallel-axis
theorem between the center of the cube and corner A for any perpendicular axis
It then follows that
I y¢
Ê 3 ˆ
1
= I z ¢ = ma2 + m Á
a˜
6
Ë 2 ¯
2
or
I y¢ = I z¢ =
11 2
ma 
12
(Note: Part b can also be solved using the method of Section 9.18.)
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277
PROBLEM 9.177 (Continued)
First note that at corner A
Ix = I y = Iz =
I xy = I yz = I zx
2 2
ma
3
1
= ma2
4
Substituting into Equation (9.56) yields
k 3 - 2ma2 k 2 +
55 2 6
121 3 9
m a km a =0
48
864
For which the roots are
1
k1 = ma2
6
11
k2 = k3 = ma2
12
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278
PROBLEM 9.178
Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin
at O, prove that the sum Ix + Iy + Iz of the mass moments of inertia of the body cannot be smaller than the similar
sum computed for a sphere of the same mass and the same material centered at O. Further, using the result of
Problem 9.176, prove that if the body is a solid of revolution, where x is the axis of revolution, its mass moment
of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10, where a is the radius of the sphere of the
same mass and the same material.
SOLUTION
(i)
Using Equation (9.30), we have
I x + I y + I z = Ú ( y 2 + z 2 )dm + Ú ( z 2 + x 2 ) dm + Ú ( x 2 + y 2 )dm
= 2Ú ( x 2 + y 2 + z 2 )dm
= 2Ú r 2 dm
where r is the distance from the origin O to the element of mass dm. Now assume that the given body can
be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m and radius
a which is centered at O; it then follows that
m1 = m2 ( mbody = msphere = m)
Then
( I x + I y + I z ) body = ( I x + I y + I z )sphere + ( I x + I y + I z )V1
- ( I x + I y + I z )V2
or
( I x + I y + I z ) body = ( I x + I y + I z )sphere + 2Ú r 2 dm - 2 Ú r 2 dm
m1
m2
Now, m1 = m2 and r1 r2 for all elements of mass dm in volumes 1 and 2.
Úm r
1
2
dm - Ú r 2 dm 0
m2
( I x + I y + I z ) body ≥ ( I x + I y + I z )sphere
so that
Q.E.D.
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279
PROBLEM 9.178 (Continued)
(ii)
First note from Figure 9.28 that for a sphere
Ix = I y = Iz =
2 2
ma
5
6
( I x + I y + I z )sphere = ma2
5
For a solid of revolution, where the x axis is the axis of revolution, we have
Thus
I y = Iz
Then, using the results of Part (i)
From Problem 9.178 we have
or
6
( I x + 2 I y ) body ma2
5
1
I y Ix
2
(2 I y - I x ) body 0
Adding the last two inequalities yields
6
( 4 I y ) body ma2
5
or
( I y ) body 3
ma2
10
Q.E.D.
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280
PROBLEM 9.179*
The homogeneous circular cylinder shown has a mass m, and the
diameter OB of its top surface forms 45∞ angles with the x and
z axes. (a) Determine the principal mass moments of inertia of the
cylinder at the origin O. (b) Compute the angles that the principal
axes of inertia at O form with the coordinate axes. (c) Sketch the
cylinder, and show the orientation of the principal axes of inertia
relative to the x, y, and z axes.
SOLUTION
(a)
First compute the moments of inertia using Figure 9.28 and the
parallel-axis theorem.
ÈÊ a ˆ 2 Ê a ˆ 2 ˘ 13 2
1
2
2
I x = I z = m(3a + a ) + m ÍÁ
˜ + ÁË 2 ˜¯ ˙ = 12 ma
12
ÍÎË 2 ¯
˙˚
Iy =
1 2
3
ma + m( a)2 = ma2
2
2
Next observe that the centroidal products of inertia are zero because of symmetry. Then
0
1
Ê a ˆ Ê aˆ
- ˜ =I xy = I x ¢y ¢ + mx y = m Á
ma2
Á
˜
Ë 2 ¯ Ë 2¯
2 2
0
1
Ê a ˆ Ê aˆ
=I yz = I y ¢z ¢ + my z = m Á ma2
˜
Á
˜
¯
Ë
Ë
2¯ 2
2 2
0
Ê a ˆÊ a ˆ 1 2
= ma
I zx = I z ¢x ¢ + mz x = m Á
Ë 2 ˜¯ ÁË 2 ˜¯ 2
Substituting into Equation (9.56)
Ê 13 3 13 ˆ
K 3 - Á + + ˜ ma2 K 2
Ë 12 2 12 ¯
2
2
1 ˆ
1 ˆ
ÈÊ 13 3 ˆ Ê 3 13 ˆ Ê 13 13 ˆ Ê
Ê
Ê 1ˆ
-Á ˜
+ ÍÁ ¥ ˜ + Á ¥ ˜ + Á ¥ ˜ - Á - Á˜
˜
Ë 2¯
Ë 2 2¯
ÎË 12 2 ¯ Ë 2 12 ¯ Ë 12 12 ¯ Ë 2 2 ¯
2
1 ˆ Ê 3ˆ Ê 1 ˆ
ÈÊ 13 3 13 ˆ Ê 13 ˆ Ê
- ÍÁ ¥ ¥ ˜ - Á ˜ Á ˜ -Á ˜Á ˜
¯
Ë
¯
Ë
Ë
12
2
12
12
2 2 ¯ Ë 2¯ Ë 2¯
Î
2
2˘
˙ ( ma2 )2 K
˙˚
2
1 ˆ
1 ˆÊ
1 ˆ Ê 1ˆ ˘
Ê
Ê 13 ˆ Ê
- 2Á - Á ˜ Á( ma2 )3 = 0
Á
˜
˜
Ë 12 ¯ Ë 2 2 ¯
Ë 2 2 ¯ Ë 2 2 ˜¯ ÁË 2 ˜¯ ˙˚
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281
PROBLEM 9.179* (Continued)
Simplifying and letting K = ma2z yields
z3 -
11 2 565
95
z +
z=0
3
144
96
Solving yields
z1 = 0.363383
z2 =
19
12
z 3 = 1.71995
The principal moments of inertia are then
K1 = 0.363ma2 
K 2 = 1.583ma2 
K3 = 1.720ma2 
(b)To determine the direction cosines l x , l y , l z of each principal axis, we use two of the equations of
Equations (9.54) and (9.57).
Thus,
( I x - K )l x - I xy l y - I zx l z = 0 (9.54a)
- I zx l x - I yz l y + ( I z - K )l z = 0 (9.54c)
l x2 + l 2y + l z2 = 1 (9.57)
(Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of l y
during the solution process.)
Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)
1
ˆ
Ê
ˆ
Ê1
ˆ
Ê 13 2
ma2 ˜ l y - Á ma2 ˜ l z = 0
ÁË ma - K ˜¯ l x - ÁË ¯
Ë
¯
12
2
2 2
1
ˆ
Ê
ˆ
Ê 13
ˆ
Ê1
- Á ma2 ˜ l x - Á ma2 ˜ l y + Á ma2 - K ˜ l z = 0
¯
Ë 12
¯
Ë2
¯
Ë 2 2
or
and
1
1
ˆ
Ê 13
l y - lz = 0
ÁË - z ˜¯ l x +
12
2
2 2
(i)
1
1
ˆ
Ê 13
l y + Á - z ˜ lz = 0 - lx +
¯
Ë 12
2
2 2
(ii)
Observe that these equations will be identical, so that one will need to be replaced, if
13
1
-z = 12
2
or
z=
19
12
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282
PROBLEM 9.179* (Continued)
Thus, a third independent equation will be needed when the direction cosines associated with K 2 are
determined. Then for K1 and K3
È13
È 1 Ê 13
ˆ˘
Ê 1ˆ ˘
Í12 - z - ÁË - 2 ˜¯ ˙ l x + Í- 2 - ÁË 12 - z ˜¯ ˙ l z = 0
Î
˚
Î
˚
Eq. (i) through Eq. (ii):
lz = lx
or
Substituting into Eq. (i):
1
1
ˆ
Ê 13
l y - lx = 0
ÁË - z ˜¯ l x +
12
2
2 2
7ˆ
Ê
l y = 2 2 Áz - ˜ lx
Ë
12 ¯
or
Substituting into Equation (9.57):
2
7ˆ ˘
È
Ê
l x2 + Í2 2 Á z - ˜ l x ˙ + ( l x )2 = 1
Ë
12 ¯ ˚
Î
or
2
È
7ˆ ˘ 2
Ê
Í2 + 8 Á z - ˜ ˙ l x = 1 Ë
12 ¯ ˙˚
ÍÎ
(iii)
K1 : Substituting the value of z1 into Eq. (iii):
2
È
7ˆ ˘
2
Ê
Í2 + 8 Á 0.363383 - ˜ ˙ ( l x )1 = 1
Ë
12 ¯ ˙˚
ÍÎ
or
and then
( l x )1 = ( l z )1 = 0.647249
7ˆ
Ê
( l y )1 = 2 2 Á 0.363383 - ˜ (0.647249)
Ë
12 ¯
= -0.402662
(q x )1 = (q z )1 = 49.7∞
(q y )1 = 113.7∞ 
K 3 : Substituting the value of z 3 into Eq. (iii):
2
È
7ˆ ˘
Ê
2
Í2 + 8 Á1.71995 - ˜ ˙ ( l x )3 = 1
¯
Ë
12
ÍÎ
˙˚
or
and then
( l x )3 = ( l z )3 = 0.284726
7ˆ
Ê
( l y )3 = 2 2 Á1.71995 - ˜ (0.284726)
Ë
12 ¯
= 0.915348
(q x )3 = (q z )3 = 73.5∞
(q y )3 = 23.7∞ 
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283
PROBLEM 9.179* (Continued)
K2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57).
Now
- I xy l x + ( I y - K )l y - I yz l z = 0 (9.54b)
Substituting for the moments and products of inertia.
1
1
ˆ
Ê
ˆ
Ê
ˆ
Ê3
-Áma2 ˜ l x + Á ma2 - K ˜ l y - Á ma2 ˜ l z = 0
¯
Ë2
¯
Ë 2 2
¯
Ë 2 2
1
ˆ
Ê3
lx + Á - x˜ l y +
lz = 0 ¯
Ë
2
2 2
2 2
1
or
(iv)
Substituting the value of x2 into Eqs. (i) and (iv):
1
1
Ê 13 19 ˆ
( l y )2 - ( l z )2 = 0
ÁË - ˜¯ ( l x )2 +
12 12
2
2 2
1
1
Ê 3 19 ˆ
( l x )2 + Á - ˜ ( l y )2 +
( l z )2 = 0
Ë
¯
2
12
2 2
2 2
or
and
-( l x ) 2 +
( l x )2 -
Adding yields
1
( l y )2 - ( l z )2 = 0
2
2
( l y )2 + ( l z )2 = 0
6
( l y )2 = 0
( l y ) 2 = -( l x ) 2
and then
Substituting into Equation (9.57)
0
( l x )22
or
( l x )2 =
+ ( l y )22
1
2
+ ( - l x )22 = 1
1
2
(q x )2 = 45.0∞ (q y )2 = 90.0∞ (q z )2 = 135.0∞ 
and ( l z )2 = -
(c)Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B. Principal axis
2 lies in the xz plane.
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284
PROBLEM 9.180
For the component described in Problem 9.165, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes
of inertia at the origin. Sketch the body and show the orientation of
the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.141 and 9.165 we have
I x = 13.98800 ¥ 10 -3 kg ◊ m2
Problem 9.141:
I y = 20.55783 ¥ 10 -3 kg ◊ m2
I z = 14.30368 ¥ 10 -3 kg ◊ m2
I yz = I zx = 0
Problem 9.165:
I xy = 0.39460 ¥ 10 -3 kg ◊ m2
Eq. (9.55) then becomes
Ix - K
- I xy
0
Thus
Substituting:
- I xy
Iy - K
0
0
2
0 = 0 or ( I x - K )( I y - K )( I z - K ) - ( I z - K ) I xy
=0
Iz - K
Iz - K = 0
2
I x I y - ( I x + I y ) K + K 2 - I xy
=0
K1 = 14.30368 ¥ 10 -3 kg ◊ m2
or
K1 = 14.30 ¥ 10 -3 kg ◊ m2 
and (13.98800 ¥ 10 -3 )(20.55783 ¥ 10 -3 ) - (13.98800 + 20.55783)(10 -3 ) K + K 2 - (0.39460 ¥ 10 -3 )2 = 0
K 2 - (34.54583 ¥ 10 -3 ) K + 287.4072 ¥ 10 -6 = 0
or
Solving yields
and
K 2 = 13.96438 ¥ 10 -3 kg ◊ m2
K3 = 20.58145 ¥ 10 -3 kg ◊ m2
or
K 2 = 13.96 ¥ 10 -3 kg ◊ m2 
or
K3 = 20.6 ¥ 10 -3 kg ◊ m2 
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285
PROBLEM 9.180 (Continued)
(b)To determine the direction cosines l x , l y , l z of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K1:
Begin with Eqs. (9.54a) and (9.54b) with I yz = I zx = 0.
( I x - K1 )( l x )1 - I xy ( l y )1 = 0
- I xy ( l x )1 + ( I y - K1 )( l y )1 = 0
Substituting:
[(13.98800 - 14.30368) ¥ 10 -3 ]( l x )1 - (0.39460 ¥ 10 -3 )( l y )1 = 0
-(0.39460 ¥ 10 -3 )( l x )1 + [(20.55783 - 14.30368) ¥ 10 -3 ]( l y )1 = 0
Adding yields
0
( l x )1 = ( l y )1 = 0
0
( l x )12 + ( l y )12 + ( l z )12 = 1
Then using Eq. (9.57)
( l z )1 = 1
or
(q x )1 = 90.0∞ (q y )1 = 90.0∞ (q z )1 = 0∞ 
K2: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0.
- I xy ( l x )2 + ( I y - K 2 )( l y )2 = 0 (i)
( I z - K 2 )( l z )2 = 0
I z π K 2 fi ( l z )2 = 0
Now
Substituting into Eq. (i):
-(0.39460 ¥ 10 -3 )( l x ) z + [(20.55783 - 13.96438) ¥ 10 -3 ]( l y )2 = 0
or
( l y )2 = 0.059847( l x )2
Using Eq. (9.57):
( l x )22
or
( l x )2 = 0.998214
and
( l y )2 = 0.059740
2
+ [0.05984]( l x )2 ]
(q x )2 = 3.4∞
0
=1
+ ( l z )22
(q y )2 = 86.6∞
(q z )2 = 90.0∞ 
K3: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0.
- I xy ( l x )3 + ( I y - K3 )( l y )3 = 0 (ii)
( I z - K3 )( l z )3 = 0
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286
PROBLEM 9.180 (Continued)
I z π K3 fi ( l z )3 = 0
Now
Substituting into Eq. (ii):
-(0.39460 ¥ 10 -3 )( l x )3 + [(20.55783 - 20.58145) ¥ 10 -3 ]( l y )3 = 0
or
Using Eq. (9.57):
( l y )3 = -16.70618( l x )3
0
( l x )32 + [-16.70618( l x )3 ]2 + ( l z )32 = 1
or
( l x )3 = -0.059751 (axes right-handed set fi "_" )
and
( l y )3 = 0.998211
(q x )3 = 93.4∞
(q y )3 = 3.43∞
(q z )3 = 90.0∞ 
Note: Since the principal axes are orthogonal and (q z )2 = (q z )3 = 90∞, it follows that
|( l x )2 | = |(l y )3 |
|( l y )2 | = | ( l z )3 |
The differences in the above values are due to round-off errors.
(c)
Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.
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287
PROBLEM 9.181*
For the component described in Problems 9.145 and 9.149,
determine (a) the principal mass moments of inertia at the origin,
(b) the principal axes of inertia at the origin. Sketch the body and
show the orientation of the principal axes of inertia relative to
the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
I x = 26.4325 ¥ 10 -3 kg ◊ m2
I xy = 2.5002 ¥ 10 -3 kg ◊ m2
Problem 9.149:
I y = 31.1726 ¥ 10 -3 kg ◊ m2
I yz = 4.0627 ¥ 10 -3 kg ◊ m2
I z = 8.5773 ¥ 10 -3 kg ◊ m2
I zx = 8.8062 ¥ 10 -3 kg ◊ m2
Substituting into Eq. (9.56):
K 3 - [(26.4325 + 31.1726 + 8.5773)(10 -3 )]K 2
+ [(26.4325)(31.1726) + (31.1726)(8.5773) + (8.5773)(26.4325)
- (2.5002)2 - ( 4.0627)2 - (8.8062)2 ](10 -6 ) K
- [(26.4325)(31.1726)(8.5773) - (26.4325)( 4.0627)2
- (31.1726)(8.8062)2 - (8.5773)(2.5002)2
- 2(2.5002)( 4.0627)(8.8062)](10 -9 ) = 0
or
K 3 - (66.1824 ¥ 10 -3 ) K 2 + (1217.76 ¥ 10 -6 ) K - (3981.23 ¥ 10 -9 ) = 0
Solving yields
K1 = 4.1443 ¥ 10 -3 kg ◊ m2
or
K1 = 4.14 ¥ 10 -3 kg ◊ m2 
K 2 = 29.7840 ¥ 10 -3 kg ◊ m2 or
K 2 = 29.8 ¥ 10 -3 kg ◊ m2 
K3 = 32.2541 ¥ 10 -3 kg ◊ m2 or
K3 = 32.3 ¥ 10 -3 kg ◊ m2 
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288
PROBLEM 9.181* (Continued)
(b)To determine the direction cosines l x , l y , l z of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K 1:
Begin with Eqs. (9.54a) and (9.54b).
( I x - K1 )( l x )1 - I xy ( l y )1 - I zx ( l z )1 = 0
I xy ( l x )1 + ( I y - K1 )( l y )1 - I yz ( l z )1 = 0
Substituting:
[(26.4325 - 4.1443)(10 -3 )]( l x )1 - (2.5002 ¥ 10 -3 )( l y )1 - (8.8062 ¥ 10 -3 )( l z )1 = 0
-(2.5002 ¥ 10 -3 )( l x )1 + [(31.1726 - 4.1443)(10 -3 )]( l y )1 - ( 4.0627 ¥ 10 -3 )( l z )1 = 0
Simplifying
8.9146( l x )1 - ( l y )1 - 3.5222( l z )1 = 0
-0.0925( l x )1 + ( l y )1 - 0.1503( l z )1 = 0
Adding and solving for ( l z )1 :
( l z )1 = 2.4022( l x )1
( l y )1 = [8.9146 - 3.5222(2.4022)]( l x )1
and then
= 0.45357( l x )1
Now substitute into Eq. (9.57):
( l x )12 + [0.45357( l x )1 ]2 + [2.4022( l x )1 ]2 = 1
( l x )1 = 0.37861
or
( l z )1 = 0.90950
( l y )1 = 0.17173
and
(q x )1 = 67.8∞ (q y )1 = 80.1∞ (q z )1 = 24.6∞ 
K 2:
Begin with Eqs. (9.54a) and (9.54b).
( I x - K 2 )( l x )2 - I xy ( l y )2 - I zx ( l z )2 = 0
- I xy ( l x )2 + ( I y - K 2 )( l y )2 - I yz ( l z )2 = 0
Substituting:
[(26.4325 - 29.7840)(10 -3 )]( l x )2 - (2.5002 ¥ 10 -3 )( l y )2 - (8.8062 ¥ 10 -3 )( l z )2 = 0
-(2.5002 ¥ 10 -3 )( l x )2 + [(31.1726 - 29.7840)(10 -3 )]( l y )2 - ( 4.0627 ¥ 10 -3 )(l z )2 = 0
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289
PROBLEM 9.181* (Continued)
Simplifying
-1.3405( l x )2 - ( l y )2 - 3.5222( l z )2 = 0
-1.8005( l x )2 + ( l y )2 - 2.9258( l z )2 = 0
Adding and solving for ( l z )2 :
( l z )2 = -0.48713( l x )2
( l y )2 = [-1.3405 - 3.5222( -0.48713)]( l x )2
and then
= 0.37527( l x )2
Now substitute into Eq. (9.57):
( l x )22 + [0.37527( l x )2 ]2 + [-0.48713( l x )2 ]2 = 1
( l x )2 = 0.85184
or
( l y )2 = 0.31967
and
( l z )2 = -0.41496
(q x )1 = 31.6∞ (q y )2 = 71.4∞ (q z )2 = 114.5∞ 
K 3:
Begin with Eqs. (9.54a) and (9.54b).
( I x - K3 )( l x )3 - I xy ( l y )3 - I zx ( l z )3 = 0
- I xy ( l x )3 + ( I y - K3 )( l y )3 - I yz ( l z )3 = 0
Substituting:
[(26.4325 - 32.2541)(10 -3 )]( l x )3 - (2.5002 ¥ 10 -3 )( l y )3 - (8.8062 ¥ 10 -3 )( l z )3 = 0
-(2.5002 ¥ 10 -3 )( l x )3 + [(31.1726 - 32.2541)(10 -3 )]( l y )3 - ( 4.0627 ¥ 10 -3 )( l z )3 = 0
Simplifying
-2.3285( l x )3 - ( l y )3 - 3.5222( l z )3 = 0
2.3118( l x )3 + ( l y )3 + 3.7565( l z )3 = 0
Adding and solving for ( l z )3 :
( l z )3 = 0.071276( l x )3
and then
( l y )3 = [-2.3285 - 3.5222(0.071276)]( l x )3
= -2.5795( l x )3
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290
PROBLEM 9.181* (Continued)
Now substitute into Eq. (9.57):
( l x )32 + [-2.5795( l x )3 ]2 + [0.071276( l x )3 ]2 = 1 ( l x )3 = 0.36134
or
and
(i)
( l y )3 = 0.93208
( l z )3 = 0.025755
(q x )3 = 68.8∞ (q y )3 = 158.8∞ (q z )3 = 88.5∞ 
(c)Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,
( l x )3 = -0.36134.
(q x )3 = 111.2∞ (q y )3 = 21.2∞ (q z )3 = 91.5∞ 
Then
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291
PROBLEM 9.182*
For the component described in Problem 9.167, determine (a) the principal
mass moments of inertia at the origin, (b) the principal axes of inertia at the
origin. Sketch the body and show the orientation of the principal axes of
inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solution of Problem 9.167, we have
1W 2
a
2 g
W
I y = a2
g
5W 2
Iz =
a
6 g
Ix =
1W 2
a
4 g
1W 2
=
a
8 g
3W 2
=
a
8 g
I xy =
I yz
I zx
Substituting into Eq. (9.56):
ÈÊ 1
5ˆ Ê W ˆ ˘
K 3 - ÍÁ + 1 + ˜ Á a2 ˜ ˙ K 2
6¯ Ë g ¯ ˚
ÎË 2
2
2
2
2
ÈÊ 1 ˆ
Ê 3ˆ ˘ Ê W 2 ˆ
Ê 1ˆ
Ê 5ˆ Ê 5ˆ Ê 1ˆ Ê 1 ˆ
(
)
(
)
+
+ ÍÁ ˜ 1 1 Á ˜ + Á ˜ Á ˜ Á ˜
ÁË ˜¯ ˙ Á a ˜ K
ÁË ˜¯
Ë 6¯ Ë 6¯ Ë 2¯ Ë 4¯
8
8 ˙˚ Ë g ¯
ÍÎË 2 ¯
3
ÈÊ 1 ˆ Ê 5 ˆ Ê 1 ˆ Ê 1 ˆ
Ê 3ˆ
Ê 1 ˆ Ê 1 ˆ Ê 3ˆ ˘ Ê W ˆ
Ê 5ˆ Ê 1 ˆ
- ÍÁ ˜ (1) Á ˜ - Á ˜ Á ˜ - (1) Á ˜ - Á ˜ Á ˜ - 2 Á ˜ Á ˜ Á ˜ ˙ Á a2 ˜ = 0
Ë 8¯
Ë 4 ¯ Ë 8¯ Ë 8¯ ˙ Ë g ¯
Ë 6¯ Ë 4¯
ÍÎË 2 ¯ Ë 6 ¯ Ë 2 ¯ Ë 8 ¯
˚
2
Simplifying and letting K =
2
2
W 2
a K yields
g
K 3 - 2.33333K 2 + 1.53125 K - 0.192708 = 0
Solving yields
K1 = 0.163917 K 2 = 1.05402 K3 = 1.11539
K1 = 0.1639
W 2
a 
g
K 2 = 1.054
W 2
a 
g
K3 = 1.115
W 2
a 
g
The principal moments of inertia are then
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292
PROBLEM 9.182* (Continued)
(b)To determine the direction cosines l x , l y , l z of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K 1:
Begin with Eqs. (9.54a) and (9.54b).
( I x - K1 )( l x )1 - I xy ( l y )1 - I zx ( l z )1 = 0
- I xy ( l x )1 + ( I y - K 2 )( l y )1 - I yz ( l z )1 = 0
Substituting
ÈÊ 1
Ê 3W 2ˆ
Ê 1 W 2ˆ
ˆ ÊW 2ˆ ˘
a ˜ ( l y )1 - Á
a ( l z )1 = 0
ÍÁË - 0.163917˜¯ Á a ˜ ˙ ( l x )1 - Á
Ë g ¯˚
Ë 8 g ˜¯
Ë4 g ¯
Î 2
È
ÊW ˆ ˘
Ê 1 W 2ˆ
Ê 1 W 2ˆ
-Á
a ˜ ( l x )1 + Í(1 - 0.163917) Á a2 ˜ ˙ ( l y )1 - Á
a ( l z )1 = 0
Ë g ¯˚
Ë4 g ¯
Ë 8 g ˜¯
Î
Simplifying
1.34433( l x )1 - ( l y )1 - 1.5( l z )1 = 0
-0.299013( l x )1 + ( l y )1 - 0.149507( l z )1 = 0
Adding and solving for ( l z )1 :
( l z )1 = 0.633715( l x )1
( l y )1 = [1.34433 - 1.5(0.633715)]( l x )1
and then
= 0.393758( l x )1
Now substitute into Eq. (9.57):
( l x )12 + [0.393758( l x )1 ]2 + (0.633715( l x )1 ]2 = 1
( l x )1 = 0.801504
or
and
( l y )1 = 0.315599
( l z )1 = 0.507925
(q x )1 = 36.7∞ (q y )1 = 71.6∞ (q z )1 = 59.5∞ 
K 2:
Begin with Eqs. (9.54a) and (9.54b):
( I x - k2 )( l x )2 - I xy ( l y )2 - I zx ( l z )2 = 0
- I xy ( l x )2 + ( I y - k2 )( l y )2 - I yz ( l z )2 = 0
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293
PROBLEM 9.182* (Continued)
Substituting
ÈÊ 1
Ê 1 W 2ˆ
Ê 3W 2ˆ
ˆ ÊW 2ˆ˘
a ˜ ( l y )2 - Á
a ( l z )2 = 0
ÍÁË - 1.05402˜¯ Á a ˜ ˙ ( l x )2 - Á
Ë4 g ¯
Ë g ¯˚
Ë 8 g ˜¯
Î 2
È
ÊW ˆ˘
Ê 1 W 2ˆ
Ê 1 W 2ˆ
-Á
a ( l z )2 = 0
a ˜ ( l x )2 + Í(1 - 1.05402) Á a2 ˜ ˙ ( l y )2 - Á
Ë g ¯˚
Ë4 g ¯
Ë 8 g ˜¯
Î
Simplifying
-2.21608( l x )2 - ( l y )2 - 1.5( l z )2 = 0
4.62792( l x )2 + ( l y )2 + 2.31396( l z )2 = 0
Adding and solving for ( l z )2
( l z )2 = -2.96309( l x )2
( l y )2 = [-2.21608 - 1.5( -2.96309)]( l x )2
and then
= 2.22856(l x )2
Now substitute into Eq. (9.57):
( l x )22 + [2.22856( l x )2 ]2 + [-2.96309( l x )2 ]2 = 1
( l x )2 = 0.260410
or
( l y )2 = 0.580339
and
( l z )2 = -0.771618
(q x )2 = 74.9∞ (q y )2 = 54.5∞ (q z )2 = 140.5∞ 
K 3:
Begin with Eqs. (9.54a) and (9.54b):
( I x - K3 )( l x )3 - I xy ( l y )3 - I zx ( l z )3 = 0
- I xy ( l x )3 + ( I y - K3 )( l y )3 - I yz ( l z )3 = 0
Substituting
ÈÊ 1
Ê 1 W 2ˆ
Ê 3W 2ˆ
ˆ ÊW 2ˆ ˘
a ˜ ( l y )3 - Á
a ( l z )3 = 0
ÍÁË - 1.11539˜¯ Á a ˜ ˙ ( l x )3 - Á
Ë4 g ¯
Ë g ¯˚
Ë 8 g ˜¯
Î 2
È
ÊW ˆ ˘
Ê 1 W 2ˆ
Ê 1 W 2ˆ
-Á
a ( l z )3 = 0
a ˜ ( l x )3 + Í(1 - 1.11539) Á a2 ˜ ˙ ( l y )3 - Á
¯
Ë
Ë4 g ¯
Ë 8 g ˜¯
g
Î
˚
Simplifying
-2.46156( l x )3 - ( l y )3 - 1.5( l z )3 = 0
2.16657( l x )3 + ( l y )3 + 1.08328( l z )3 = 0
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294
PROBLEM 9.182* (Continued)
Adding and solving for ( l z )3
( l z )3 = -0.707885( l x )3
and then
( l y )3 = [-2.46156 - 1.5( -0.707885)]( l x )3
= -1.39973( l x )3
Now substitute into Eq. (9.57):
( l x )32 + [-1.39973( l x )3 ]2 + [-0.707885( l x )3 ]2 = 1 or
( l x )3 = 0.537577
and
( l y )3 = -0.752463
(i)
( l z )3 = -0.380543
(q x )3 = 57.5∞ (q y )3 = 138.8∞ (q z )3 = 112.4∞ 
(c)Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,
( l x )3 = -0.537577
(q x )3 = 122.5∞ (q y )3 = 41.2∞ (q z )3 = 67.6∞ 
Then
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295
PROBLEM 9.183*
For the component described in Problem 9.168, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes
of inertia at the origin. Sketch the body and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solution to Problem 9.168, we have
gt 4
a
g
gt
I y = 7.68953 a 4
g
gt
I z = 12.00922 a 4
g
I x = 18.91335
gt 4
a
g
gt
= 7.14159 a 4
g
gt
= 0.66667 a 4
g
I xy = 1.33333
I yz
I zx
Substituting into Eq. (9.56):
È
Êg t ˆ˘
K 3 - Í(18.91335 + 7.68953 + 12.00922) Á a 4 ˜ ˙ K 2
Ë g ¯˚
Î
+ [(18.91335)(7.68953) + (7.68953)(12.00922) + (12.00922)(18.91335)
2
Êgt ˆ
- (1.33333)2 - (7.14159)2 + (0.66667)2 ] Á a 4 ˜ K
Ë g ¯
- [(18.91335)(7.68953)(12.00922) - (18.91335)(7.14159)2
- (7.68953)(0.66667)2 - (12.00922)(1.33333)2
3
Êgt ˆ
- 2(1.33333)(7.14159)(0.66667)] Á a 4 ˜ = 0
Ë g ¯
Simplifying and letting
K=
gt 4
a K
g
yields
K 3 - 38.61210 K 2 + 411.69009 K - 744.47027 = 0
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296
PROBLEM 9.183* (Continued)
Solving yields
K1 = 2.25890 K 2 = 17.27274 K3 = 19.08046
K1 = 2.26
gt 4
a 
g
K 2 = 17.27
gt 4
a 
g
K3 = 19.08
gt 4
a 
g
The principal moments of inertia are then
(b)To determine the direction cosines l x , l y , l z of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K1 :
Begin with Eqs. (9.54a) and (9.54b):
( I x - K1 )( l x )1 - I xy ( l y )1 - I zx ( l z )1 = 0
- I xy ( l x )1 + ( I y - K 2 )( l y )1 - I yz ( l z )1 = 0
Substituting
È
Ê g t 4ˆ˘
Ê
g t 4ˆ
g t 4ˆ
Ê
Í(18.91335 - 2.25890) Á a ˜ ˙ ( l x )1 - ÁË1.33333 a ¯˜ ( l y )1 - Á 0.66667 a ˜ ( l z )1 = 0
¯
g
a
Ë
Ë
g ¯
Î
˚
È
Êg t ˆ˘
Ê
Ê
gt ˆ
gt ˆ
- Á1.33333 a 4 ˜ ( l x )1 + Í(7.68953 - 2.25890) Á a 4 ˜ ˙ ( l y )1 - Á 7.14159 a 4 ˜ ( l z )1 = 0
Ë g ¯˚
Ë
Ë
g ¯
g ¯
Î
Simplifying
12.49087( l x )1 - ( l y )1 - 0.5( l z )1 = 0
-0.24552( l x )1 + ( l y )1 - 1.31506( l z )1 = 0
Adding and solving for ( l z )1
( l z )1 = 6.74653( l x )1
and then
( l y )1 = [12.49087 - (0.5)(6.74653)]( l x )1
= 9.11761( l x )1
Now substitute into Eq. (9.57):
( l x )12 + [9.11761( l x )1 ]2 + [6.74653( l x )1 ]2 = 1
or
( l x )1 = 0.087825
and
( l y )1 = 0.80075
( l z )1 = 0.59251
(q x )1 = 85.0∞ (q y )1 = 36.8∞ (q z )1 = 53.7∞ 
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297
PROBLEM 9.183* (Continued)
K2 :
Begin with Eqs. (9.54a) and (9.54b):
( I x - K 2 )( l x )2 - I xy ( l y )2 - I zx ( l z )2 = 0
- I xy ( l x )2 + ( I y - K 2 )( l y )2 - I yz ( l z )2 = 0
Substituting
Ê
Êgt ˆ
Ê
gt ˆ
gt ˆ
[(18.91335 - 17.27274) Á a 4 ˜ ( l x )2 - Á1.33333 a 4 ˜ ( l y )2 - Á 0.66667 a 4 ˜ ( l z )2 = 0
g ¯
g ¯
Ë
Ë g ¯
Ë
È
Ê
Êg t ˆ˘
Ê
gt ˆ
gt ˆ
- Á1.33333 a 4 ˜ ( l x )2 + Í(7.68953 - 17.27274) Á a 4 ˜ ˙ ( l y )2 - Á 7.14159 a 4 ˜ ( l z )2 = 0
g ¯
Ë
g ¯
Ë g ¯˚
Ë
Î
Simplifying
1.23046( l x )2 - ( l y )2 - 0.5( l z )2 = 0
0.13913( l x )2 + ( l y )2 + 0.74522( l z )2 = 0
Adding and solving for ( l z )2
( l z )2 = -5.58515( l x )2
and then
( l y )2 = [1.23046 - (0.5)( -5.58515)]( l x )2
= 4.02304 (l x )2
Now substitute into Eq. (9.57):
( l x )22 + [4.02304( l x )2 ]2 + [-5.58515( l x )2 ]2 = 1
or
( l x )2 = 0.14377
and
( l y )2 = 0.57839
( l z )2 = -0.80298
(q x )2 = 81.7∞ (q y )2 = 54.7∞ (q z )2 = 143.4∞ 
K3 :
Begin with Eqs. (9.54a) and (9.54b):
( I x - K3 )( l x )3 - I xy ( l y )3 - I zx ( l z )3 = 0
- I xy ( l x )3 + ( I y - K3 )( l y )3 - I yz ( l y )3 = 0
Substituting
Ê
Êgt ˆ
Ê
gt ˆ
gt ˆ
[(18.91335 - 19.08046) Á a 4 ˜ ( l x )3 - Á1.33333 a 4 ˜ ( l y )3 - Á 0.66667 a 4 ˜ ( l z )3 = 0
g ¯
Ë
g ¯
Ë g ¯
Ë
È
Êg t ˆ˘
Ê
Ê
gt ˆ
gt ˆ
- Á1.33333 a 4 ˜ ( l x )3 + Í(7.68953 - 19.08046) Á a 4 ˜ ˙ ( l y )3 - Á 7.14159 a 4 ˜ ( l z )3 = 0
Ë g ¯˚
Ë
Ë
g ¯
g ¯
Î
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298
PROBLEM 9.183* (Continued)
Simplifying
-0.12533( l x )3 - ( l y )3 - 0.5( l z )3 = 0
0.11705( l x )3 + ( l y )3 + 0.62695( l z )3 = 0
Adding and solving for ( l z )3
( l z )3 = 0.06522( l x )3
and then
( l y )3 = [-0.12533 - (0.5)(0.06522)]( l x )3
= -0.15794( l x )3
Now substitute into Eq. (9.57):
( l x )32 + [-0.15794( l x )3 ]2 + [0.06522( l x )3 ]2 = 1
or
( l x )3 = 0.98571
and
( l y )3 = -0.15568
(i)
( l z )3 = 0.06429
(q x )3 = 9.7∞ (q y )3 = 99.0∞ (q z )3 = 86.3∞ 
(c)Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,
( l x )3 = -0.98571.
(q x )3 = 170.3∞ (q y )3 = 81.0∞ (q z )3 = 93.7∞ 
Then
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299
PROBLEM 9.184*
For the component described in Problems 9.148 and 9.170, determine
(a) the principal mass moments of inertia at the origin, (b) the principal
axes of inertia at the origin. Sketch the body and show the orientation of
the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.148 and 9.170. We have
I x = 0.32258 kg ◊ m2
I y = I z = 0.41933 kg ◊ m2
I xy = I zx = 0.096768 kg ◊ m2
I yz = 0
Substituting into Eq. (9.56) and using
I y = Iz
I xy = I zx
I yz = 0
K 3 - [0.32258 + 2(0.41933)]K 2
+ [2(0.32258)(0.41933) + (0.41933)2 - 2(0.096768)2 ]K
- [(0.32258)(0.41933)2 - 2(0.41933)(0.096768)2 ] = 0
Simplifying
K 3 - 1.16124 K 2 + 0.42764 K - 0.048869 = 0
Solving yields
K1 = 0.22583 kg ◊ m2 or
K1 = 0.226 kg ◊ m2 
K 2 = 0.41920 kg ◊ m2 or
K 2 = 0.419 kg ◊ m2 
K3 = 0.51621 kg ◊ m2 or
K3 = 0.516 kg ◊ m2 
(b)To determine the direction cosines l x , l y , l z of each principal axis, use two of the equations of
Eqs. (9.54) and (9.57). Then
K1 :
Begin with Eqs. (9.54b) and (9.54c):
0
- I xy ( l x )1 + ( I y - K1 )( l y )1 - I yz ( l z )3 = 0
0
- I zx ( l x )1 - I yz ( l y )1 + ( I z - K1 )( l z )3 = 0
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300
Problem 9.184* (continued)
Substituting
-(0.096768)( l x )1 + (0.41933 - 0.22583)( l y )1 = 0
-(0.096768)( l x )1 + (0.41933 - 0.22583)( l z )1 = 0
Simplifying yields
( l y )1 = ( l z )1 = 0.50009( l x )1
Now substitute into Eq. (9.54):
( l x )2 + 2[0.50009( l x )1 ]2 = 1
or
( l x )1 = 0.81645
and
( l y )1 = ( l z )1 = 0.40830
(q x )1 = 35.3∞ (q y )1 = (q z )1 = 65.9∞ 
K2 :
Begin with Eqs. (9.54a) and (9.54b)
( I x - K 2 )( l x )2 - I xy ( l y )2 - I zx ( l z )2 = 0
0
- I xy ( l x )2 + ( I y - K 2 )( l y )2 - I yz ( l z )2 = 0
Substituting
(0.32258 - 0.41920)( l x )2 - (0.096768)( l y )2 - (0.096768)( l z )2 = 0 (i)
-(0.96768)( l x )2 + (0.41933 - 0.41920)( l y )2 = 0 (ii)
Eq. (ii) fi ( l x )2 = 0
and then Eq. (i) fi ( l z )2 = -( l y )2
Now substitute into Eq. (9.57):
0
( l x )22
+ ( l y )22 + [-( l y )2 ]2 = 1
1
2
1
( l z )2 = 2
( l y )2 =
or
and
(q x )2 = 90∞ (q y )2 = 45∞ (q z )2 = 135∞ 
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301
PROBLEM 9.184* (Continued)
K3 :
Begin with Eqs. (9.54b) and (9.54c):
0
I xy ( l x )3 + ( I y - K3 )( l y )3 + I yz ( l z )3 = 0
0
- I zx ( l x )3 - I yz ( l y )3 + ( I z - K3 )( l z )3 = 0
Substituting
-(0.096768)( l x )3 + (0.41933 - 0.51621)( l y )3 = 0
-(0.096768)( l x )3 + (0.41933 - 0.51621)( l z )3 = 0
Simplifying yields
( l y )3 = ( l z )3 = - ( l x )3
Now substitute into Eq. (9.57):
( l x )32 + 2[-( l x )3 ]2 = 1
(i)
1
3
or
( l x )3 =
and
( l y )3 = ( l z ) = -
1
3
(q x )3 = 54.7∞ (q y )3 = (q z )3 = 125.3∞ 
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
That is,
( l x )3 = -
1
3
(q x )3 = 125.3∞ (q y )3 = (q z )3 = 54.7∞ 
Then
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302
PROBLEM 9.185
Determine by direct integration the moments of inertia of the shaded area
with respect to the x and y axes.
SOLUTION
x1 = a,
At
y1 :
y2 :
Then
b = ka2
Then
or k =
b
a2
b
b = ma or m =
a
y1 =
y2 =
Now
y1 = y2 = b
b
a2
x2
b
x
a
(
)
1 3
y2 - y13 dx
3
b3 ˆ
1 Ê b3
= Á 3 x 3 - 6 x 6 ˜ dx
3Ë a
a
¯
dI x =
I x = Ú dI x = Ú
a b3
0
Ê 1 3 1 6ˆ
Á x - 6 x ˜¯ dx
3 Ë a3
a
a
1
b3 È 1
˘
= Í 3 x4 - 6 x7 ˙
3 Î 4a
7a
˚0
Also
Then
or
Ix =
1 3
ab 
28
or
Iy =
1 3
ab
20
b ˆ ˘
ÈÊ b
dI y = x 2 dA = x 2 [( y2 - y1 )dx = x 2 ÍÁ x - 2 x 2 ˜ dx ˙
¯ ˚
Ë
a
a
Î
a Ê1
1 ˆ
I y = Ú dI y = Ú b Á x 3 - 2 x 4 ˜ dx
0 Ëa
¯
a
a
1
È1
˘
= b Í x 4 - 2 x5 ˙
5a
Î 4a
˚0
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303
PROBLEM 9.186
Determine the moments of inertia and the radii of gyration of the shaded
area shown with respect to the x and y axes.
SOLUTION
We have
Then
È Ê x ˆ 1/ 2 ˘
dA = ydx = b Í1 - Á ˜ ˙ dx
ÍÎ Ë a ¯ ˙˚
È Ê x ˆ 1/ 2 ˘
A = Ú dA = Ú b Í1 - Á ˜ ˙ dx
0
ÍÎ Ë a ¯ ˙˚
a
a
2 3/ 2 ˘
1
È
= b Íx x ˙ = ab
3 a
Î
˚0 3
3
Now
Then
1/ 2
1
1 ÔÏ È Ê x ˆ ˘ Ô¸
dI x = y 3 dx = Ìb Í1 - Á ˜ ˙ ˝ dx
3
3 Ô ÍÎ Ë a ¯ ˙˚ Ô
Ó
˛
1
/
2
3/ 2
b3 È
Ê xˆ
Ê xˆ Ê xˆ ˘
= Í1 - 3 Á ˜ + 3 Á ˜ - Á ˜ ˙ dx
Ë a¯ Ë a¯ ˙
Ë a¯
3 ÍÎ
˚
I x = Ú dI x = Ú
0
=
b3
3
1/ 2
3/ 2
È
Ê xˆ Ê xˆ ˘
Ê xˆ
Í1 - 3 Á ˜ + 3 Á ˜ - Á ˜ ˙ dx
Ë a¯ Ë a¯ ˙
Ë a¯
3 ÍÎ
˚
a b3
a
2 3/ 2 3 2
2
È
˘
x +
x + 3/ 2 x 5 / 2 ˙
Íx 2a
a
5a
Î
˚0
or
and
k x2 =
Ix
=
A
1 3
ab 
30
3
1
30 ab
1
3 ab
or
Also
Ix =
kx =
b

10
ÏÔ È Ê x ˆ 1/ 2 ˘ ¸Ô
dI y = x dA = x ( ydx ) = x Ìb Í1 - Á ˜ ˙ dx ˝
ÔÓ ÍÎ Ë a ¯ ˙˚ Ô˛
2
2
2
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304
PROBLEM 9.186 (Continued)
a
Then
a Ê
1 5/ 2 ˆ
2 7/ 2 ˘
È1
I y = Ú dI y = Ú b Á x 2 x ˜ dx = b Í x 3 x ˙
0 Ë
¯
a
7 a
Î3
˚0
or
and
k y2 =
Iy
A
=
Iy =
1 3
ab
21
1 3
21 a b
1
3 ab
or
ky =
a

7
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305
PROBLEM 9.187
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
At x = a,
y1 = y2 = b :
y1 : b = ka2
or k =
y2 : b = 2b - ca2
Then
Now
b
a2
or c =
b
a2
b 2
x
a2
Ê
x2 ˆ
y2 = b Á 2 - 2 ˜
a ¯
Ë
y1 =
dA = ( y2 - y1 ) dx
È Ê
x2 ˆ b ˘
= Íb Á 2 - 2 ˜ - 2 x 2 ˙ dx
a ¯ a
ÍÎ Ë
˚˙
2b
= 2 ( a2 - x 2 ) dx
a
Then
A = Ú dA = Ú
=
=
Now
a 2b
0
a
2
( a2 - x 2 )dx
2b È 2
1 ˘
a x - x3 ˙
2 Í
3 ˚
a Î
a
0
4
ab
3
È 2b
˘
dI y = x 2 dA = x 2 Í 2 ( a2 - x 2 ) dx ˙
Îa
˚
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306
PROBLEM 9.187 (Continued)
Then
I y = Ú dI y = Ú
a 2b
0
a2
x 2 ( a2 - x 2 )dx
a
=
2b È 1 2 3 1 5 ˘
a x - x ˙
5 ˚0
a2 ÍÎ 3
or
and
k y2 =
Iy
A
=
4 3
15 a b
4
3 ab
Iy =
4 3
ab
15
1
= a2
5
or
ky =
a

5
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307
PROBLEM 9.188
Determine the moments of inertia of the shaded area shown with respect to the x and
y axes.
SOLUTION
We have
where
I x = ( I x )1 + ( I x )2 - ( I x )3
( I x )1 =
4
1
(2a)(2a)3 = a 4
12
3
p 4
a
8
Now
( I AA )2 = ( I BB )3 =
and
( I AA )2 = ( I x2 )2 + Ad 2
or
( I x2 )2 = ( I x3 )3 =
2
Then
p 4 Ê p 2 ˆ Ê 4a ˆ
8ˆ
Êp
a - Á a ˜ Á ˜ = Á - ˜ a14
Ë 8 9p ¯
Ë 2 ¯ Ë 3p ¯
8
8ˆ
4a ˆ
Êp ˆÊ
Êp
( I x )2 = ( I x2 )2 + Ad22 = Á - ˜ a 4 + Á a2 ˜ Á a + ˜
Ë 2 ¯Ë
Ë 8 9p ¯
3p ¯
Ê 4 5p ˆ
= Á + ˜ a4
Ë3 8 ¯
2
4a ˆ
8ˆ
Êp ˆÊ
Êp
( I x )3 = ( I x3 )3 + Ad32 = Á - ˜ a 4 + Á a2 ˜ Á a - ˜
Ë 2 ¯Ë
Ë 8 9p ¯
3p ¯
Ê 4 5p ˆ
= Á - + ˜ a4
Ë 3 8¯
2
4 4 ÈÊ 4 5p ˆ 4 ˘ ÈÊ 4 5p ˆ 4 ˘
a + ÍÁ + ˜ a ˙ - ÍÁ - + ˜ a ˙ 3
ÎË 3 8 ¯ ˚ ÎË 3 8 ¯ ˚
Finally
Ix =
Also
I y = ( I y )1 + ( I y )2 - ( I y )3
where
( I y )1 =
I x = 4a4 
16
1
(2a)(2a)3 + (2a)2 ( a)2 = a 4
12
3
È p
˘
Êp ˆ
( I y ) 2 = ( I y )3 Í = a 4 + Á a 2 ˜ ( a ) 2 ˙
¯
Ë
2
Î 8
˚
Iy =
16 4
a 
3
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308
PROBLEM 9.189
Determine the polar moment of inertia of the area shown with
respect to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the area.
A, mm2
1
p
(54)(36) = 3053.6
2
2
-
S
p
(18)2 = -508.9
2
y, mm
yA, mm3
48
= 15.2789
p
46, 656
24
= 7.6394
p
-3888
2544.7
42, 768
Y S A = S y A: Y (2544.7 mm2 ) = 42,768 mm3
Then
or
Y = 16.8067 mm 
J O = ( J O )1 - ( J O )2
(a)
p
p
(54 mm)(36 mm)[(54 mm)2 + (36 mm)2 ] - (18 mm)4
8
4
6
6
4
= (3.2155 ¥ 10 - 0.0824 ¥ 10 ) mm
=
= 3.1331 ¥ 106 mm 4
or
J O = 3.13 ¥ 106 mm 4 
J O = J C + A(Y )2
(b)
or
J C = 3.1331 ¥ 106 mm 4 - (2544.7 mm2 )(16.8067 mm)2
or
J C = 2.41 ¥ 106 mm 4 
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309
PROBLEM 9.190
To form an unsymmetrical girder, two L76 ¥ 76 ¥ 6.4 mm angles and
two L152 ¥ 102 ¥ 12.7-mm angles are welded to a 16-mm steel plate
as shown. Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.
SOLUTION
A = 929 mm2
L76 ¥ 76 ¥ 6.4 :
I x ¥ I y = 0.512 ¥ 106 mm 4
A = 3060 mm2
L152 ¥ 102 ¥ 12.7 :
I x = 2.59 ¥ 106 mm 4
I y = 7.20 ¥ 106 mm 4
First locate centroid C of the section:
Then
or
A, mm2
y
yA, mm3
1
2(3060) = 6120
24.9
152,388
2
(16)(540) = 8640
270
2,332,800
3
2(929) = 1858
518.8
963,930
S
16,618
3,449,118
Y SA= SyA
Y (16, 618 mm2 ) = 3, 449,118
or
Y = 207.553 mm
Now
I x = 2( I x )1 + ( I x )2 + 2( I x )3
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310
PROBLEM 9.190 (Continued)
where
( I x )1 = I x + Ad 2 = 2.59 ¥ 106 mm 4 + (3060 mm2 )[(207.553 - 24.9) mm]2
= 104.678 ¥ 106 mm 4
1
(16 mm)(540 mm)3 + (16 mm)(540 mm)[(270 - 207.553) mm]2
12
= 243.645 ¥ 106 mm 4
( I x )2 = I x + Ad 2 =
( I x )3 = I x + Ad 2 = 0.512 ¥ 106 mm 4 + (929 mm2 [(518.8 - 207.553) mm]2
= 90.5086 ¥ 106 mm 4
Then
I x = [2(104.678) + 243.645 + 2(90.5086)] ¥ 106 mm 4
or
Also
where
I x = 634 ¥ 106 mm 4 
I y = 2( I y )1 + ( I y )2 + 2( I y )3
( I y )1 = I y + Ad 2 = 7.20 ¥ 106 mm 4 + (3060 mm2 )[(8 + 50.3) mm]2
= 17.6006 ¥ 106 mm 4
1
(540 mm)(16 mm)3
12
= 0.1843 ¥ 106 mm 4
( I y )2 =
( I y )3 = I y + Ad 2 = 0.512 ¥ 106 mm 4 + (929 mm2 )[(8 + 21.2) mm]2
= 1.3041 ¥ 106 mm 4
Then
I y = [2(17.6006) + 0.1843 + 2(1.3041)] ¥ 106 mm 4
or
I y = 38.0 ¥ 106 mm 4 
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311
PROBLEM 9.191
Using the parallel-axis theorem, determine the product of inertia of
the L127 ¥ 76 ¥ 12.7 mm angle cross section shown with respect
to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy )2
For each rectangle:
I xy = I x ¢y ¢ + x yA
I x ¢y ¢ = 0 (symmetry)
and
I xy = S x yA
Thus
A, mm2
x, mm
y, mm
x yA, mm 4
1
76 ¥ 12.7 = 965.2
19.1
-37.85
-697.777 ¥ 103
2
114.3 ¥ 12.7 = 1451.61
-12.55
25.65
-467.284 ¥ 103
-1165.061 ¥ 103
S
I xy = -1.165 ¥ 106 mm 4 
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312
PROBLEM 9.192
For the L127 ¥ 76 ¥ 12.7 mm angle cross section shown, use
Mohr’s circle to determine (a) the moments of inertia and the
product of inertia with respect to new centroidal axes obtained by
rotating the x and y axes 30∞ clockwise, (b) the orientation of the
principal axes through the centroid and the corresponding values
of the moments of inertia.
SOLUTION
From Figure 9.13a:
I x = 3.93 ¥ 106 mm 4
I y = 1.06 ¥ 106 mm 4
From the solution to Problem 9.191
I xy = -1.16506 ¥ 106 mm 4
The Mohr’s circle is defined by the diameter XY, where
X (3.93 ¥ 106 , - 1.1651 ¥ 106 ),
Now
1
1
I ave = ( I x + I y ) = (3.93 + 1.06) ¥ 106 = 2.495 ¥ 106 mm 4
2
2
2
2
and
Y (1.06 ¥ 106 , 1.1651 ¥ 106 )
2
È1
˘
È1
˘
R = Í ( I x - I y ) ˙ + I xy = Í (3.93 - 1.06) ¥ 106 ˙ + ( -1.1651 ¥ 106 )2
Î2
˚
Î2
˚
= 1.84843 ¥ 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2q m = -
2 I xy
Ix - I y
-2( -1.1651 ¥ 106 )
(3.93 - 1.06) ¥ 106
= 0.81192
=
or
and
Then
2q m = 39.074∞
q m = 19.537∞
a = 180∞ - (39.074∞ + 60∞)
= 80.926∞
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313
PROBLEM 9.192 (Continued)
(a)
We have
I x¢ = I ave - R cos a = 2.495 ¥ 106 - 1.84843 ¥ 106 cos 80.926∞
= 2.2035 ¥ 106 mm 4
I x¢ = 2.20 ¥ 106 mm 4 
or
I y¢ = I ave + R cos a = 2.495 ¥ 106 + 1.84843 ¥ 106 cos(80.926∞)
= 2.7865 ¥ 106 mm 4
or
I y¢ = 2.79 ¥ 106 mm 4 
or
I x ¢y ¢ = -1.825 ¥ 106 mm 4 
I x ¢y ¢ = - R sin a = -1.84843 ¥ 106 sin 80.926∞
= -1.82530 ¥ 106 mm 4
(b)
First observe that the principal axes are obtained by rotating the xy axes through
19.54° counterclockwise 
about C.
Now
I max, min = I ave ± R = (2.495 ± 1.84843) ¥ 106 mm 4
or I max = 4.34 ¥ 106 mm 4 
I min = 0.647 ¥ 106 mm 4 
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
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314
PROBLEM 9.193
A piece of thin, uniform sheet metal is cut to form the machine component
shown. Denoting the mass of the component by m, determine its mass
moment of inertia with respect to (a) the x axis, (b) the y axis.
SOLUTION
First note
mass = m = rV = rtA
1
È
Ê aˆ ˘
= rt Í(2a)( a) - (2a) Á ˜ ˙
Ë 2¯ ˚
2
Î
3
= rta2
2
I mass = rtI area
Also
=
(a)
Now
Then
2m
I area
3a2
I x , area = ( I x )1, area - 2( I x )2, area
=
1
È 1 Ê aˆ
˘
( a)(2a)3 - 2 Í Á ˜ ( a)3 ˙
¯
Ë
12
Î12 2
˚
=
7 4
a
12
I x, mass =
2m
3a
2
¥
7 4
a
12
or
(b)
We have
I z , area = ( I z )1, area - 2( I z )2, area
Ix =
7
ma2 
18
3
È1
1
Ê aˆ ˘
= (2a)( a)3 - 2 Í ( a) Á ˜ ˙
3
ÍÎ12 Ë 2 ¯ ˙˚
=
Then
31 4
a
48
2m 31 4
¥ a
3a2 48
31 2
=
ma
72
I z, mass =
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315
PROBLEM 9.193 (Continued)
Finally,
I y , mass = I x , mass + I z , mass
=
7
31
ma2 + ma2
18
72
=
59 2
ma
72
or
I y = 0.819ma2 
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316
PROBLEM 9.194
A piece of thin, uniform sheet metal is cut to form the machine component
shown. Denoting the mass of the component by m, determine its mass
moment of inertia with respect to (a) the axis AA¢, (b) the axis BB¢, where
the AA¢ and BB¢ axes are parallel to the x axis and lie in a plane parallel to
and at a distance a above the xz plane.
SOLUTION
First note that the x axis is a centroidal axis so that
I = I x , mass + md 2
and that from the solution to Problem 9.115,
I x, mass =
(a)
(b)
We have
We have
I AA¢ , mass =
I BB¢ , mass =
7
ma2
18
7
ma2 + m( a)2
18
or
I AA¢ = 1.389ma2 
or
I BB¢ = 2.39ma2 
7
ma2 + m( a 2 )2
18
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317
PROBLEM 9.195
A 2 mm thick piece of sheet steel is cut and bent into the machine component
shown. Knowing that the density of steel is 7850 kg/m3, determine the mass
moment of inertia of the component with respect to each of the coordinate
axes.
SOLUTION
First compute the mass of each component. We have
m = rSTV = rST t A
Then
m1 = (7850 kg/m3 )(0.002 m)(0.76 ¥ 0.48) m2
= 5.72736 kg
ˆ
Ê1
m2 = (7850 kg/m3 )(0.002 m) Á ¥ 0.76 ¥ 0.48˜ m2
¯
Ë2
= 2.86368 kg
ˆ
Êp
m3 = (7850 kg/m3 )(0.002 m) Á ¥ 0.482 ˜ m2
¯
Ë4
= 2.84101 kg
Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have
I x = ( I x )1 + ( I x )2 + ( I x )3
2
È1
Ê 0.48 ˆ ˘
= Í (5.72736 kg)(0.48 m)2 + (5.72736 kg) Á
m˜ ˙
¯ ˙
Ë 2
ÍÎ12
˚
2
È1
˘
Ê 0.48 ˆ ˘ È 1
+ Í (2.86368 kg)(0.48 m)2 + (2.86368 kg) Á
m˜ ˙ + Í (2.84101 kg)(0.48 m)2 ˙
¯
Ë
3
2
18
Î
˚
ÍÎ
˙˚
= [(0.109965 + 0.329896) + (0.036655 + 0.073310) + (0.327284)] kg ◊ m2
= (0.439861 + 0.109965 + 0.327284) kg ◊ m2
or
I x = 0.877 kg ◊ m2 
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318
PROBLEM 9.195 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3
ÏÔ 1
ÈÊ 0.76 ˆ 2 Ê 0.48 ˆ 2 ˘ 2 ¸Ô
= Ì (5.72736 kg)(0.762 + 0.482 ) m2 + (5.72736 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
2 ¯ ˙˚
ÍÎË 2 ¯
ÔÓ12
Ô˛
2
È1
Ê 0.76 ˆ ˘ È 1
˘
+ Í (2.86368 kg)(0.76 m)2 + (2.86368 kg) Á
m˜ ˙ + Í (2.84101 kg)(0.48 m)2 ˙
¯
Ë
3
4
18
Î
˚
ÍÎ
˙˚
= [(0.385642 + 1.156927) + (0.091892 + 0.183785) + (0.163642)] kg ◊ m 2
= (1.542590 + 0.275677 + 0.163642) kg ◊ m 2
or
I y = 1.982 kg ◊ m2 
I z = ( I z )1 + ( I z )2 + ( I z )3
2
È1
Ê 0.76 ˆ ˘
= Í (5.72736 kg)(0.76 m)2 + (5.72736 kg) Á
m˜ ˙
¯ ˙
Ë 2
ÍÎ12
˚
ÏÔ 1
ÈÊ 0.76 ˆ 2 Ê 0.48 ˆ 2 ˘ 2 ¸Ô
+ Ì (2.86368 kg)(0.762 + 0.482 ) m2 + (2.86368 kg) ÍÁ
˜ ˙m ˝
˜ + ÁË
3 ¯ ˙˚
ÍÎË 3 ¯
ÔÓ18
Ô˛
È1
˘
+ Í (2.84101 kg)(0.48 m)2 ˙
Î4
˚
I z = [(0.275677 + 0.827031) + (0.128548 + 0.257095)
+ (0.163642)] kg ◊ m2
= (1.102708 + 0.385643 + 0.163642) kg ◊ m2
or
I z = 1.652 kg ◊ m2 
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319
PROBLEM 9.196
Determine the mass moments of inertia and the radii of gyration of
the steel machine element shown with respect to the x and y axes.
(The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = rSTV
Then
m1 = (7850 kg/m3 )(0.24 ¥ 0.04 ¥ 0.14) m3
= 10.5504 kg
Èp
˘
m2 = m3 = (7850 kg/m3 ) Í (0.07)2 ¥ 0.04 ˙ m3 = 2.41683 kg
2
Î
˚
m4 = m5 = (7850 kg/m3 )[p (0.044)2 ¥ (0.04)] m3 = 1.90979 kg
Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to
Problem 9.144) for a semicylinder, we have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 - ( I x )5
where ( I x )2 = ( I x )3
È1
Ï1
¸
˘
= Í (10.5504 kg)(0.042 + 0.142 ) m2 ˙ + 2 Ì (2.41683 kg) ÈÎ3(0.07 m)2 + (0.04 m)2 ˘˚ ˝
Î12
˚
Ó12
˛
Ï1
¸
+ Ì (1.90979 kg) ÈÎ3(0.044 m)2 + (0.04 m)2 ˚˘ + (1.90979 kg)(0.04 m)2 ˝
Ó12
˛
Ï1
¸
- Ì (1.90979 kg) ÈÎ3(0.044 m)2 + (0.04 m)2 ˘˚ ˝
Ó12
˛
= [(0.0186390) + 2(0.0032829) + (0.0011790 + 0.0030557) - (0.0011790)] kg ◊ m2
= 0.0282605 kg ◊ m2
or
I x = 28.3 ¥ 10 -3 kg ◊ m2 
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320
PROBLEM 9.196 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 - ( I y )5
( I y ) 2 = ( I y )3
where
Then
(I y ) 4 = |( I y )5 |
È1
˘
I y = Í (10.5504 kg)(0.242 + 0.142 ) m2 ˙
Î12
˚
2
˘
4 ¥ 0.07 ˆ
È
Ê 1 16 ˆ
Ê
m2 ˙
+ 2 Í(2.41683 kg) Á - 2 ˜ (0.07 m2 ) + (2.41683 kg) Á 0.12 +
˜
Ë
Ë 2 9p ¯
3p ¯
Î
˙˚
= [(0.0678742) + 2(0.0037881 + 0.0541678)] kg ◊ m2
= 0.1837860 kg ◊ m2
Also
or
I y = 183.8 ¥ 10 -3 kg ◊ m2 
m = m1 + m2 + m3 + m4 - m5 where m2 = m3 , m4 = m5
= (10.5504 + 2 ¥ 2.41683) kg = 15.38406 kg
Then
k x2 =
I x 0.0282605 kg ◊ m2
=
m
15.38406 kg
or k x = 42.9 mm 
and
k y2 =
Iy
m
=
0.1837860 kg ◊ m2
15.38406 kg
or k y = 109.3 mm 
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321