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)] mm 2
= (144 + 384 + 288) mm 2
= 816 mm 2
Now
I x = ( I x )1 + ( I x )2 + ( I x )3
where
1
(24 mm)(6 mm)3 + (144 mm 2 )(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 mm 2 )(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 mm 2
or
I x = 390 × 103 mm 4
or
k x = 21.9 mm
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1435
PROBLEM 9.39
Determine the shaded area and its moment of inertia with respect to the
centroidal axis parallel to AA′, knowing that d1 = 30 mm and d2 = 10 mm, and
that the moments of inertia with respect to AA′ and BB′ are 4.1 × 106 mm4
and 6.9 × 106 mm4, respectively.
SOLUTION
I AA′ = 4.1 × 106 mm 4 = I + A(30 mm)2
(1)
I BB′ = 6.9 × 106 mm 4 = I + A(40 mm) 2
I BB′ − I AA′ = (6.9 − 4.1) × 106 = A(402 − 302 )
2.8 × 106 = A(700)
Eq. (1):
4.1 × 106 = I + (4000)(30)2
A = 4000 mm 2
I = 500 × 103 mm 4
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1444
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.
1
Then
A, mm 2
x , mm
108 × 60 = 6480
54
349,920
46
–59,616
2
1
− × 72 × 36 = −1296
2
Σ
5184
290,304
X Σ A = Σ xA : X (5184 mm 2 ) = 290,304 mm3
or
X = 56.0 mm
Now
I x = ( I x )1 − ( I x )2
where
xA, mm3
1
(108 mm)(60 mm)3 = 1.944 × 106 mm 4
12
1
1
× 72 mm × 18 mm (6 mm) 2
( I x )2 = 2
(72 mm)(18 mm)3 +
36
2
( 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
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1446
PROBLEM 9.41 (Continued)
Also
where
I y = ( I y )1 − ( I y ) 2
1
(60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2
12
= (6, 298,560 + 25,920) mm 4 = 6.324 × 106 mm 4
( I y )1 =
1
(36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2
36
= (373, 248 + 129, 600) mm 4 = 0.502 × 106 mm 4
( I y )2 =
Then
I y = (6.324 − 0.502)106 mm 4
or I y = 5.82 × 106 mm 4
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1447
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.
Then
A, in 2
x , in.
1
5 × 8 = 40
2.5
2
−2 × 5 = −10
1.9
Σ
30
where
4
100
160
4.3
–19
–43
81
117
X = 2.70 in.
Y Σ A = Σ yA: Y (30 in 2 ) = 117 in 3
Y = 3.90 in.
or
Now
yA, in 3
X Σ A = Σ xA: X (30 in 2 ) = 81 in 3
or
and
xA, in 3
y , in.
I x = ( I x )1 − ( I x )2
1
(5 in.)(8 in.)3 + (40 in 2 )[(4 − 3.9) in.]2
12
= (213.33 + 0.4) in 4 = 213.73 in 4
( I x )1 =
1
(2 in.)(5 in.)3 + (10 in 2 )[(4.3 − 3.9) in.]2
12
= (20.83 + 1.60) = 22.43 in 4
( I x )2 =
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1450
PROBLEM 9.43 (Continued)
Then
I x = (213.73 − 22.43) in 4
Also
I y = ( I y )1 − ( I y ) 2
where
or I x = 191.3 in 4
1
(8 in.)(5 in.)3 + (40 in 2 )[(2.7 − 2.5) in.]2
12
= (83.333 + 1.6) in 4 = 84.933 in 4
( I y )1 =
1
(5 in.)(2 in.)3 + (10 in 2 )[(2.7 − 1.9) in.]2
12
= (3.333 + 6.4) in 4 = 9.733 in 4
( I y )2 =
Then
I y = (84.933 − 9.733) in 4
or
I y = 75.2 in 4
PROPRIETARY MATERIAL. © 2013 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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1451