# PROBLEM 9.31

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 &times; 103 mm 4
or
k x = 21.9 mm
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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 &times; 106 mm4
and 6.9 &times; 106 mm4, respectively.
SOLUTION
I AA′ = 4.1 &times; 106 mm 4 = I + A(30 mm)2
(1)
I BB′ = 6.9 &times; 106 mm 4 = I + A(40 mm) 2
I BB′ − I AA′ = (6.9 − 4.1) &times; 106 = A(402 − 302 )
2.8 &times; 106 = A(700)
Eq. (1):
4.1 &times; 106 = I + (4000)(30)2
A = 4000 mm 2
I = 500 &times; 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 &times; 60 = 6480
54
349,920
46
–59,616
2
1
− &times; 72 &times; 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 &times; 106 mm 4
12
1
1
&times; 72 mm &times; 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 &times; 103 mm 4
[( I x ) 2 is obtained by dividing A2 into
Then
]
I x = (1.944 − 0.069984) &times; 106 mm 4
or
I x = 1.874 &times; 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 &times; 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 &times; 106 mm 4
( I y )2 =
Then
I y = (6.324 − 0.502)106 mm 4
or I y = 5.82 &times; 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 &times; 8 = 40
2.5
2
−2 &times; 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
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1451