y
PROBLEM 9.2
y = kx 1/3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
b
x
a
SOLUTION
y kx1/ 3
For x a:
b ka1/ 3
k b/a1/ 3
Thus:
y
b 1/ 3
x
a1/ 3
dI y x 2 dA x 2 ydx
dIy x 2
Iy ¨
b 1/ 3
b
x dx 1/ 3 x 7 / 3 dx
a1/ 3
a
a
¬
b
b 3
dI y 1/ 3
x 7 / 3 dx 1/ 3 žž a10 / 3 ­­­
®
0
a
a Ÿ10
¨
Iy 3 3
a b◀
10
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1422
PROBLEM 9.34
Determine the moment of inertia and the radius of gyration
of the shaded area with respect to the y axis.
SOLUTION
Assign area 1 to be the flanges and area 2 to be the web.
First note that
A 2 A1
A2
[2(3)(0.5)
(5)(0.5)] in 2
5.50 in2
Now
I y 2( I y )1
( I y )2
where
(I y )1 1
(0.5 in.)(3 in.)3
12
( I y )2 1
(5 in.)(0.5 in.)3 0.052083 in4
12
Then
I y 2(3.4688)
and
k y2 Iy
A
(0.5 in.)(3 in.)(1.25 in.)2 3.4688 in4
0.052083 in 4 6.9897 in 4
6.9897 in 4
5.50 in 2
1.27085 in 2
or I y 6.99 in4 ◀
or ky 1.127 in. ◀
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1463
A
B
PROBLEM 9.43
1.2 in.
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.
5.0 in.
1.8 in.
0.9 in.
2.0 in.
2.1 in.
SOLUTION
First locate centroid C of the area.
A, in2
x, in.
y, in.
xA, in3
yA, in3
1
5q8 40
2.5
4
100
160
2
2 q 5 10
1.9
4.3
–19
–43
4
30
81
117
Then
or
X4 A 4xA: X (30 in2 ) 81 in3
X 2.70 in.
and
or
Y 4 A 4 yA: Y (30 in2 ) 117 in3
Y 3.90 in.
Now
where
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
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1474
PROBLEM 9.43 (Continued)
Then
I x (213.73 22.43) in4
Also
Iy (Iy )1 (Iy )2
where
4
or I x 191.3 in ◀
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) in4
or
I y 75.2 in4 ◀
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1475
y
PROBLEM 9.142
120
44
Determine the mass moments of inertia and the radii of
gyration of the steel machine element shown with respect
3
to the x and y axes. (The density of steel is 7850 kg/m .)
120
70
44
70
40
20
20 x
z
Dimensions in mm
SOLUTION
First compute the mass of each component. We have
m SSTV
m1 (7850 kg/m3 )(0.24 q 0.04 q 0.14)m3
Then
10.5504 kg
Q
¯
m2 m3 (7850 kg/m3 ) ¡ (0.07)2 q 0.04° m3 2.41683kg
¢¡ 2
±°
m4 m5 (7850 kg/m3 )[Q(0.044)2 q (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
\
^
¯
1
1
¡ (10.5504 kg)(0.04 2 0.14 2 )m 2 ° 2
(2.41683 kg)[3(0.07 m)2 (0.04 m)2 ]
12
¢¡ 12
±°
1
(1.90979 kg)[3(0.044 m)2 (0.04 m)2 ] (1.90979 kg)(0.04 m)2
12
1
(1.90979 kg)[3(0.044 m)2 (0.04 m)2 ]
12
[(0.0186390) 2(0.0032839) (0.0011790 0.0030557) (0.0011790)]kg ¸ m 2
\
\
^
^
0.0282605 kg ¸ m 2
or
I x 28.3q103 kg ¸ m2 ◀
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1628
PROBLEM 9.142 (Continued)
I y ( I y )1
( I y )2 ( I y )3
where
( I y )2
( I y )4 ( I y )5
( I y )4 ( I y )5
¯
0.14 2 )m 2 °
°±
1
I y ¡ (10.5504 kg)(0.24 2
¡¢ 12
Then
( I y )3
 1 16 ¬

2 ¡¡(2.41683 kg) žž 2 ­­­ (0.07 m 2 ) (2.41683 kg) žž0.12
Ÿ 2 9Q ®
Ÿ
¢
[(0.0678742) 2(0.0037881 0.0541678)] kg ¸ m 2
4 q 0.07 ­¬2 2 ¯°
­ m °
3Q ­®
±
0.1837860 kg ¸ m 2
or
Also
m m1
m2
(10.5504
Then
and
k x2 k y2 m3
m 4 m5
I y 183.8q103 kg ¸ m2 ◀
where m2 m3 , m4 m5
2 q 2.41683) kg 15.38406 kg
I x 0.0282605 kg ¸ m2
m
15.38406 kg
Iy
m
or
kx 42.9 mm ◀
or
ky 109.3 mm ◀
0.1837860 kg ¸ m 2
15.38406 kg
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1629