Subido por Guillermo Ortigas

# Solucionario Cap. 8 - Momentos de Inercia

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```Problem 8.1 Use the method described in Active
Example 8.1 to determine IY and ky for the rectangular
area.
Solution: The height of the vertical strip of width dx is 0.6 m, so
the area is dA D (0.6 m) dx.
We can use this expression to determine Iy .
y
x 2 dx D (0.6 m)
Iy D
0.4 m
x 2 dx
0.2m
A
D (0.6 m)
x3
3
0.4
m
D 0.0416 m4
0.2 m
0.6 m
ky D
Iy
D
A
0.0416 m4
D 0.416 m
(0.4 m) (0.6 m)
x
0.2 m
Iy D 0.0416 m4 , ky D 0.416 m
0.4 m
Problem 8.2 Use the method described in Active
Example 8.1 to determine Ix and kx for the rectangular
area.
y
Solution: It was shown in Active Example 8.1 that the moment
of inertia about the x axis of a vertical strip of width dx and height
f(x) is
1
Ix strip D [fx]3 dx.
3
For the rectangular strip, fx D 0.6 m. Integrating to determine Ix
for the rectangular area.,
0.4m
Ix D
0.2m
1
1
4
0.6 m3 dx D 0.6 m3 [x]0.4m
0.2m D 0.0288 m
3
3
0.6 m
kx D
Ix
D
A
0.0288 m4
D 0.346 m
(0.4 m)(0.6 m)
x
0.2 m
Ix D 0.0288 m4 , ky D 0.346 m
0.4 m
Problem 8.3 In Active Example 8.1, suppose that the
triangular area is reoriented as shown. Use integration
to determine Iy and ky .
Solution: The height of a vertical strip of width dx is h h/bx,
so the area
dA D
h
h
x
b
y
h
dx.
We can use this expression to determine Iy :
b
x 2 dA D
Iy D
x2 h 0
A
h
x
b
dx D h
x3
3
x4
4b
x
b
D
0
1 3
hb
12
b
ky D
Iy D
Iy
D
A
1 3
hb
b
12
D p
1
6
hb
2
hb3
b
, ky D p .
12
6
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601
Problem 8.4 (a) Determine the moment of inertia Iy
of the beam’s rectangular cross section about the y axis.
Solution:
(b) Determine the moment of inertia Iy 0 of the beam’s
values, show that Iy D Iy 0 C d2x A, where A is the area of
the cross section.
(a)
40 mm
0
x 2 dydx D 1.28 &eth; 106 mm4
0
(b)
60 mm
Iy D
20 mm
30 mm
Iy 0 D
20 mm
x 2 dydx D 3.2 &eth; 105 mm4
30 mm
y
Iy D Iy 0 C dx 2 A
y
dx
1.28 &eth; 106 mm4
D 3.2 &eth; 105 mm4 C 20 mm2 [40 mm60 mm]
60 mm
x
O
dy
x
O
40 mm
Problem 8.5 (a) Determine the polar moment of
inertia JO of the beam’s rectangular cross section about
the origin O.
(b) Determine the polar moment of inertia JO0 of the
beam’s cross section about the origin O0 . Using your
numerical values, show that JO D JO0 C d2x C d2y A,
where A is the area of the cross section.
Solution:
(a)
40 mm
0
(b)
60 mm
JO D
x 2 C y 2 dydx D 4.16 &eth; 106 mm4
0
20 mm
30 mm
JO 0 D
20 mm
x 2 C y 2 dydx D 1.04 &eth; 106 mm4
30 mm
JO D JO0 C dx 2 C dy 2 A
(c)
4.16 &eth; 106 mm4 D 1.04 &eth; 106 mm4 C [20 mm2
C 30 mm2 ][40 mm60 mm]
Problem 8.6 Determine Iy and ky .
y
Solution:
A D 0.3 m1 m C
1 m 0.3 mC0.3x
Iy D
0
0.6 m
0.3 m
ky D
x
1
0.3 m1 m D 0.45 m2
2
0
Iy
D
A
x 2 dydx D 0.175 m4
0.175 m4
D 0.624 m
0.45 m2
1m
602
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Problem 8.7 Determine JO and kO .
Solution:
A D 0.3 m1 m C
1m
0.3 mC0.3x
JO D
0
1
0.3 m1 m D 0.45 m2
2
x 2 C y 2 dydx D 0.209 m4
0
0.209 m4
D 0.681 m
0.45 m2
kO D
Problem 8.8 Determine Ixy .
Solution:
1m
0.3 mC0.3x
Ixy D
0
xydydx D 0.0638 m4
0
Problem 8.9 Determine Iy .
y
Solution: The height of a vertical strip of width dx is 2 x2 , so
y 2 x2
the area is
dA D 2 x2 dx.
We can use this expression to determine Iy :
x 2 dA D
Iy D
1
x 2 2 x2 dx D
0
A
2x3
x5
3
5
1
0
D 0.467.
Iy D 0.467.
1
Problem 8.10 Determine Ix .
Solution: It was shown in Active Example 8.1 that the moment
of inertia about the x axis of a vertical strip of width dx and height
fx is
1
Ix strip D [fx]3 dx.
3
x
y
y 2 x2
In this problem fx D 2 x3 . Integrating to determine Ix for the area,
1
Ix D
0
D
D
1
3
1
3
1
2 x2 3 dx
3
1
8 12x2 C 6x4 x 6 dx
0
1
12x3
6x5
x7
8x C
D 1.69.
3
5
7 0
1
x
Ix D 1.69.
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603
Problem 8.11 Determine JO .
y
y 2 x2
Solution: See the solutions to Problems 8.9 and 8.10. The polar
moment of inertia is
JO D Ix C Iy D 1.69 C 0.467 D 2.15.
JO D 2.15.
1
Problem 8.12 Determine Ixy .
x
y
y 2 x2
Solution: It was shown in Active Example 8.1 that the product of
inertia of a vertical strip of width dx and height f(x) is
Ixy strip D
1
[fx]2 xdx.
2
In this problem, fx D 2 x2 . Integrating to determine Ixy for the
area,
1
1
2 x2 2 xdx
2
Ixy D
0
D
D
1
2
1
4x 4x3 C x 5 x
0
1
4x4
x6
1 4x2
C
D 0.583.
2 2
4
6 0
1
x
Ixy D 0.583.
Problem 8.13 Determine Iy and ky .
y
y
1 2
x 4x 7
4
Solution: First we need to locate the points where the curve intersects the x axis.
4 š
1
x 2 C 4x 7 D 0 ) x D
4
Now
14
x
x2 /4C4x7
dydx D 72
2
0
14
x2 /4C4x7
Iy D
2
ky D
604
p
16 41/47
D 2,14
21/4
0
Iy
D
A
x 2 dydx D 5126
5126
D 8.44
72
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Problem 8.14 Determine Ix and kx .
Solution: See Solution to Problem 8.13
14
x2 /4C4x7
Ix D
2
0
Ix
D
A
kx D
y 2 dydx D 1333
1333
D 4.30
72
Problem 8.15 Determine JO and kO .
Solution: See Solution to 8.13 and 8.14
JO D Ix C Iy D 1333 C 5126 D 6459
JO
6459
D
D 9.47
kO D
A
72
Problem 8.16 Determine Ixy .
Solution:
14
x2 /4C4x7
xydydx D 2074
Ixy D
2
0
Problem 8.17 Determine Iy and ky .
y
y
1 2
x 4x 7
4
y5
Solution: First we need to locate the points where the curve intersects the line.
4 š
1 2
x C 4x 7 D 5 ) x D
4
12
p
16 41/412
D 4,12
21/4
x
x2 /4C4x7
dydx D 21.33
4
5
12
x2 /4C4x7
Iy D
4
ky D
5
Iy
D
A
x 2 dydx D 1434
1434
D 8.20
21.33
Problem 8.18 Determine Ix and kx .
Solution: See Solution to Problem 8.17
12
x2 /4C4x7
Ix D
4
kx D
5
Ix
D
A
y 2 dydx D 953
953
D 6.68
21.33
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605
Problem 8.19 (a) Determine Iy and ky by letting dA
be a vertical strip of width dx.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO D 12 R4 . Explain how you
y
x
R
Solution:
The equation of the circle is x2 C y 2 D R2 , from which
p
y D š pR2 x 2 . The strip dx wide and y long has the elemental area
dA D 2 R2 x 2 dx. The area of the semicircle is
R2
Iy D
2
R
x 2 dA D 2
x2
R2 x 2 dx
0
A
R
R2 xR2 x 2 1/2
R4
x
xR2 x 2 3/2
C
C
sin1
D2 4
8
8
R 0
(b) If the integration were done for a circular area with the center at
the origin, the limits of integration for the variable x would be from
R to R, doubling the result. Hence, doubling the answer above,
Iy D
R4
.
4
By symmetry, Ix D Iy , and the polar moment would be
JO D 2Iy D
D
R4
8
ky D
R4
,
2
which is indeed the case. Also, since kx D ky by symmetry for the full
circular area,
R
Iy
D
A
2
kO D
Iy
Ix
C
D
A
A
2
Iy
D
A
JO
A
as required by the definition. Thus the result checks.
Problem 8.20 (a) Determine Ix and kx for the area
in Problem 8.19 by letting dA be a horizontal strip of
height dy.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO D 12 R4 . Explain how you
Solution: Use the results of the solution to Problem 8.19, A D
R2
. The equation for the circle is x2 C y 2 D R2 , from which x D
2
š R2 y 2 . The horizontal strip is from 0 to R, hence the element of
area is
CR
y 2 dA D
Ix D
A
y2
D
kx D
606
and JO D 2Ix D
R2 y 2 dy
R
R4
R4
C
D
8 2
8 2
R4
.
4
R
y
R2 yR2 y 2 1/2
R4
yR2 y 2 3/2
C
C
sin1
D 4
8
8
R R
Ix D
By symmetry Iy D Ix ,
R2 y 2 dy.
dA D
(b) If the area were circular, the strip would be twice as long, and the
moment of inertia would be doubled:
R4
R4
,
2
which is indeed the result. Since kx D ky by symmetry for the full
circular area, the
kO D
Iy
Ix
C
D
A
A
2
Ix
D
A
JO
A
8
as required by the definition. This checks the answer.
R
Ix
D .
A
2
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Problem 8.21 Use the procedure described in
Example 8.2 to determine the moment of inertia Ix
and Iy for the annular ring.
y
Ro
x
Ri
Solution: We first determine the polar moment of inertia JO by
integrating in terms of polar coordinates. Because of symmetry and
the relation JO D Ix C Iy , we know that Ix and Iy each equal 12 JO .
Integrating as in Example 8.2, the polar moment of inertia for the
annular ring is
r 2 dA D
JO D
A
Therefore Ix D Iy D
Ro
r 2 2rdr D
Ri
1
Ro4 Ri4 2
1
Ro4 Ri4 4
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607
y
Problem 8.22 What are the values of Iy and ky for the
elliptical area of the airplane’s wing?
y2
x2
21
b
a2
Solution:
a
x 2 dA D
Iy D
A
0
a
y
y
x 2 dy dx
y
0
a
Iy D 2
0
5m
x2
b1 2 1/2
a
[x2 y]0
y
dx
0
1/2
x2
x2 b 1 2
dx
a
a
x
2m
x 2 dy dx
Iy D 2
0
a
Iy D 2
x2 + y2 = 1
a2 b2
2b
x
x2
Iy D 2b
1
0
x2
a2
dx
a
Rewriting
Iy D
2b
a
a
x2
y = b 1– x2
a2
a2 x 2 dx
0
p
xa2 x 2 3/2
a2 x a2 x 2
2b
Iy D
C
a
4
8
x
a4
C sin1
8
a
a
0
D
a
D

0
p 0

2b  aa2 a2 3/2 a3 a2 a2
Iy D
C

a 
4
8



a  
a4
sin1

8
a
x2
b 1
dx
a
2b
a
a
a2 x 2 1/2 dx
0
p
a
x
x a2 x 2
a2
C
sin1
2
2
a
0
(from the integral tables)
C
2b
a
0
p
a2
a
2b
a 0
C
sin1
D
a
2
2
a
0
0a2 3/2
p
a2
0
0 a
C
sin1
2
2
a
4

0p

a2 &ETH; 0 a2 a4 01 0 
C
C sin


8
8
a

ab
2b a2 D
a 2 2
2
Evaluating, we get
A D 7.85 m2
Iy D
a4
2b
a 8 2
Finally
Iy
D
A
2a3 b
Iy D
8
ky D
Evaluating, we get
ky D 2.5 m
49.09
7.85
Iy D 49.09 m4
The area of the ellipse (half ellipse) is
a
x2
b 1 a
0
608
1/2
dy dx
0
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Problem 8.23 What are the values of Ix and kx for the
elliptical area of the airplane’s wing in Problem 8.22?
Solution:
y
y 2 dA D 2
Ix D
a
Ix D 2
0
b
3 a
y
3
p
b
yD a
y2
x2
—
+ —2 = 1
a2
b
p
a2 x2
y 2 dy dx
0
0
A
2b
a
y = b a2 – x2
a
dx
0
b3 2
a x 2 3/2 dx
3a3
Ix D 2
0
2b3
3a3
p
3a2 x a2 x 2
xa2 x 2 3/2
C
4
8
3 4 1 x
a sin
8
a
C
2b3
Ix D 3
3a
3
a4
8
2
Ix D
2b3
&ETH;
3a3
Ix D
3ab3 ab3 D
3.8
8
a
Evaluating (a D 5, b D 1)
0
p
3a3 0
3 a0
C
C a4
4
8
8 2
Ix D
p
3a2 &ETH; 0 a2
0a2 C0
4
8
5
D 1.96 m4
8
From Problem 8.22, the area of the wing is A D 7.85 m2
x
a2 x2
a
Ix D
a
kx D
Ix
D
A
1.96
7.85
Problem 8.24 Determine Iy and ky .
kx D 0.500 m
y
Solution: The straight line and curve intersect where x D x2 20.
Solving this equation for x, we obtain
xD
1š
y = x 2 – 20
p
1 C 80
D 4, 5.
2
y=x
If we use a vertical strip: the area
x
dA D [x x2 20] dx.
Therefore
A
D
5
x 2 dA D
Iy D
x 2 x x 2 C 20 dx
4
x4
x5
20x3
C
4
5
3
5
D 522.
The area is
y = x2 – 20
A
ky D
y=x
x x 2 C 20 dx
x
4
x2
x3
C 20x
2
3
So
5
dA D
D
y
4
Iy
D
A
dA
5
D 122.
4
dx
522
D 2.07.
122
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609
Problem 8.25 Determine Ix and kx for the area in
Problem 8.24.
Solution: Let us determine the moment of inertia about the x axis
of a vertical strip holding x and dx fixed:
y 2 dAs D
Ix strip D
As
x
y
y 2 dx dy D dx
x2 20
y3
3
x
y=x
y=
x2 20
x2 –
20
dAs
x
dx
x 6 C 60x4 C x 3 1200x2 C 8000.
3
D
Integrating this value from x D 4 to x D 5 (see the solution to
Problem 8.24), we obtain Ix for the entire area:
5
x
1
x 6 C 60x4 C x 3 1200x2 C 8000 dx
3
Ix D
4
dx
7
5
x
x4
400x3
8000x
D D 10,900.
C 4x5 C
C
21
12
3
3
4
From the solution to Problem 8.24, A D 122 so
kx D
Ix
D
A
10,900
D 9.45.
122
Problem 8.26 A vertical plate of area A is beneath the
surface of a stationary body of water. The pressure of
the water subjects each element dA of the surface of the
plate to a force p0 C y dA, where p0 is the pressure
at the surface of the water and is the weight density
of the water. Show that the magnitude of the moment
about the x axis due to the pressure on the front face of
the plate is
Mx
x
A
y
D p0 yA C Ix ,
axis
where y is the y coordinate of the centroid of A and Ix
is the moment of inertia of A about the x axis.
Solution: The moment about the x axis is dM D yp0 C y dA
integrating over the surface of the plate:
p0 C yy dA.
MD
A
Noting that p0 and are constants over the area,
y dA C M D p0
y 2 dA.
A
By definition,
y dA
A
yD
A
y 2 dA,
and Ix D
A
then M D p0 yA C IX , which demonstrates the result.
610
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Problem 8.27 Using the procedure described in Active
Example 8.3, determine Ix and kx for the composite area
by dividing it into rectangles 1 and 2 as shown.
Solution: Using results from Appendix B and applying the
parallel-axis theorem, the moment on inertia about the x axis for area
1 is
Ix 1 D Ix C d2y A D
y
1m
1
4m
2
1m
1
(1 m)(3 m)3 C (2.5 m)[(1 m)(3 m)]
12
x
3m
D 21.0 m4
The moment of inertia about the x axis for area 2 is
Ix 2 D
1
(3 m)(1 m)3 D 1 m4 .
3
The moment of inertia about the x axis for the composite area is
Ix D Ix 1 C Ix 2 D 22.0 m4 .
kx D
Ix
D
A
22.0 m4
D 1.91 m
6 m2
Ix D 22.0 m4 , kx D 1.91 m.
Problem 8.28 Determine Iy and ky for the composite
area by dividing it into rectangles 1 and 2 as shown.
Solution: Using results from Appendix B, the moment of inertia
about the y axis for area 1 is
Iy 1 D
1m
1
4m
1
(3 m)(1 m)3 D 1 m4 .
3
The moment of inertia about the y axis for area 2 is
Iy 2 D
y
2
1m
3m
x
1
(1 m)(3 m)3 D 9 m4 .
3
The moment of inertia about the y axis for the composite area is
Iy D Iy 1 C Iy 2 D 10 m4 .
ky D
Iy
D
A
10 m4
D 1.29 m
6 m2
Iy D 10 m4 , kx D 1.29 m.
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611
Problem 8.29 Determine Ix and kx .
y
Solution: Break into 3 rectangles
Ix D
1
1
0.60.23 C
0.20.63 C 0.20.60.52
3
12
0.8 m
1
0.80.23 C 0.80.20.92 D 0.1653 m4
12
C
0.2 m
A D 0.20.6 C 0.60.2 C 0.80.2 D 0.4 m2
0.6 m
kx D
Ix
D
A
0.1653 m4
D 0.643 m
0.4 m2
0.2 m
x
Ix D 0.1653 m4
)
0.2 m
kx D 0.643 m
0.6 m
Problem 8.30 In Example 8.4, determine Ix and kx for
the composite area.
y
Solution: The area is divided into a rectangular area without the
20 mm
cutout (part 1), a semicircular areas without the cutout (part 2), and
the circular cutout (part 3).
x
Using the results from Appendix B, the moment of inertia of part 1
Ix 1 D
1
120 mm80 mm3 D 5.12 &eth; 106 mm4 ,
12
40 mm
120 mm
the moment of inertia of part 2 is
Ix 2 D
1
40 mm4 D 1.01 &eth; 106 mm4 ,
8
and the moment of inertia of part 3 is
Ix 3 D
1
20 mm4 D 1.26 &eth; 105 mm4 .
4
The moment of inertia of the composite area is
Ix D Ix 1 C Ix 2 C Ix 3 D 6.00 &eth; 106 mm4 .
From Example 8.4, the composite area is A D 1.086 &eth; 104 mm4 , so
kx D
Ix
D
A
6.00 &eth; 106 mm4
D 23.5 mm.
1.086 &eth; 104 mm2
Ix D 6.00 &eth; 106 mm4 , kx D 23.5 mm.
612
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Problem 8.31 Determine Ix and kx .
y
0.8 m
0.2 m
x
0.6 m
0.2 m
0.2 m
0.6 m
Solution: Break into 3 rectangles — See 8.29
First locate the centroid
dD
0.60.20.1 C 0.20.60.5 C 0.80.20.9
D 0.54 m
0.60.2 C 0.20.6 C 0.80.2
1
0.20.63 C 0.20.6d 0.52
12
1
0.60.23 C 0.60.2d 0.12
C
12
1
0.80.23 C 0.80.20.9 d2 D 0.0487 m4
C
12
Ix
0.0487 m4
D
kx D
D 0.349 m
A
0.4 m2
Ix D
d
Problem 8.32 Determine Iy and ky .
Solution: Break into 3 rectangles — See 8.29
Iy D
1
1
1
0.60.23 C
0.20.63 C
0.20.83
12
12
12
D 0.01253 m4
Iy
0.01253 m4
ky D
D 0.1770 m
D
A
0.4 m2
Problem 8.33 Determine JO and kO .
Solution: See 8.29, 8.31 and 8.32
JO D Ix C Iy D 0.0612 m4
JO
0.0612 m4
D
KO D
D 0.391 m
A
0.4 m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
613
y
Problem 8.34 If you design the beam cross section so
that Ix D 6.4 &eth; 105 mm4 , what are the resulting values
of Iy and JO ?
h
Solution: The area moment of inertia for a triangle about the
x
base is
Ix D
1
12
bh3 ,
h
from which Ix D 2
1
12
60h3 D 10h3 mm4 ,
30
mm
30
mm
Ix D 10h3 D 6.4 &eth; 105 mm4 ,
from which h D 40 mm.
Iy D 2
1
12
2h303 D
from which Iy D
1
h303 3
1
40303 D 3.6 &eth; 105 mm4
3
and JO D Ix C Iy D 3.6 &eth; 105 C 6.4 &eth; 105 D 1 &eth; 106 mm4
614
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.35 Determine Iy and ky .
160
mm
Solution: Divide the area into three parts:
Part (1): The top rectangle.
A1 D 16040 D 6.4 &eth; 103 mm2 ,
dx1
160
D 80 mm,
D
2
Iyy1 D
1
12
40 mm
200
mm
40
mm
40 mm
x
120
mm
401603 D 1.3653 &eth; 107 mm4 .
From which
Iy1 D d2x1 A1 C Iyy1 D 5.4613 &eth; 107 mm4 .
Part (2): The middle rectangle:
A2 D 200 8040 D 4.8 &eth; 103 mm2 ,
dx2 D 20 mm,
Iyy2 D
1
12
120403 D 6.4 &eth; 105 mm4 .
From which,
Iy2 D d2x2 A2 C Iyy2 D 2.56 &eth; 106 mm4 .
Part (3) The bottom rectangle:
A3 D 12040 D 4.8 &eth; 103 mm2 ,
dx3 D
120
D 60 mm,
2
Iyy3 D
1
12
401203 D 5.76 &eth; 106 mm4
From which
Iy3 D d2X3 A3 C Iyy3 D 2.304 &eth; 107 mm4
The composite:
Iy D Iy1 C Iy2 C Iy3 D 8.0213 &eth; 107 mm4
ky D
Iy
D 70.8 mm.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
615
Problem 8.36 Determine Ix and kx .
Solution: Use the solution to Problem 8.35. Divide the area into
from which
three parts:
Part (1): The top rectangle.
A1 D 6.4 &eth; 103 mm2 ,
Ix2 D d2y2 A2 C Ixx2 D 5.376 &eth; 107 mm4
Part (3) The bottom rectangle:
A3 D 4.8 &eth; 103 mm2 ,
dy1 D 200 20 D 180 mm,
Ixx1 D
1
12
dy3 D 20 mm,
160403 D 8.533 &eth; 105 mm4 .
Ixx3 D
From which
Ix1 D d2y1 A1 C Ixx1 D 2.082 &eth; 108 mm4
Part (2): The middle rectangle:
A2 D 4.8 &eth; 103 mm2 ,
dy2
120
C 40 D 100 mm,
D
2
Ixx2 D
1
12
1
12
120403 D 6.4 &eth; 105 mm4
and Ix3 D d2y3 A3 C Ixx3 D 2.56 &eth; 106 mm4 .
The composite:
Ix D Ix1 C Ix2 C Ix3 D 2.645 &eth; 108 mm4
kx D
Ix
D 128.6 mm
A1 C A2 C A3 401203 D 5.76 &eth; 106 mm4
Problem 8.37 Determine Ixy .
Solution: (See figure in Problem 8.35). Use the solutions in
from which
Problems 8.35 and 8.36. Divide the area into three parts:
Part (1): A1 D 16040 D 6.4 &eth; 103 mm2 ,
dx1
160
D 80 mm,
D
2
dy1 D 200 20 D 180 mm,
Ixy2 D dx2 dy2 A2 D 9.6 &eth; 106 mm4 .
Part (3): A3 D 12040 D 4.8 &eth; 103 mm2 ,
dx3 D
120
D 60 mm,
2
dy3 D 20 mm,
Ixxyy1 D 0,
from which
from which
Ixy1 D dx1 dy1 A1 C Ixxyy1 D 9.216 &eth; 107 mm4 .
Ixy3 D dx3 dy3 A3 D 5.76 &eth; 106 .
The composite:
Part (2) A2 D 200 8040 D 4.8 &eth; 103 mm2 ,
Ixy D Ixy1 C Ixy2 C Ixy3 D 1.0752 &eth; 108 mm4
dx2 D 20 mm,
dy2 D
616
120
C 40 D 100 mm,
2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.38 Determine Ix and kx .
y
Solution: The strategy is to use the relationship Ix D d2 A C Ixc ,
160
mm
where Ixc is the area moment of inertia about the centroid. From this
Ixc D d2 A C Ix . Use the solutions to Problems 8.35, 8.36, and 8.37.
Divide the area into three parts and locate the centroid relative to the
coordinate system in the Problems 8.35, 8.36, and 8.37.
40 mm
x
200
mm
40
mm
Part (1) A1 D 6.4 &eth; 103 mm2 ,
40 mm
120
mm
dy1 D 200 20 D 180 mm.
Part (2) A2 D 200 8040 D 4.8 &eth; 103 mm2 ,
dx1 D
160
D 80 mm,
2
dy2 D
120
C 40 D 100 mm,
2
dx2 D 20 mm,
Part (3) A3 D 12040 D 4.8 &eth; 103 mm2 ,
dx3 D
120
D 60 mm,
2
dy3 D 20 mm.
The centroid coordinates are
xD
A1 dx1 C A2 dx2 C A3 dx3
D 56 mm,
A
yD
A1 dy1 C A2 dy2 C A3 dy3
D 108 mm
A
from which
Ixc D y2 A C Ix D 1.866 &eth; 108 C 2.645 &eth; 108
D 7.788 &eth; 107 mm4
The total area is
A D A1 C A2 C A3 D 1.6 &eth; 104 mm2 .
kxc D
Ixc
D 69.77 mm
A
Problem 8.39 Determine Iy and ky .
Solution: The strategy is to use the relationship Iy D d2 A C Iyc ,
where Iyc is the area moment of inertia about the centroid. From
this Iyc D d2 A C Iy . Use the solution to Problem 8.38. The centroid
coordinates are x D 56 mm, y D 108 mm, from which
Iyc D x2 A C Iy D 5.0176 &eth; 107 C 8.0213 &eth; 107
D 3.0 &eth; 107 mm4 ,
kyc D
Iyc
D 43.33 mm
A
Problem 8.40 Determine Ixy .
Solution: Use the solution to Problem 8.37. The centroid
coordinates are
x D 56 mm,
y D 108 mm,
from which Ixyc D xyA C Ixy D 9.6768 &eth; 107 C 1.0752 &eth; 108
D 1.0752 &eth; 107 mm4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
617
Problem 8.41 Determine Ix and kx .
Solution: Divide the area into two parts:
y
3 ft
4 ft
Part (1): a triangle and Part (2): a rectangle. The area moment of inertia
for a triangle about the base is
Ix D
1
12
3 ft
x
bh3 .
The area moment of inertia about the base for a rectangle is
Ix D
1
bh3 .
3
Part (1) Ix1 D
Part (2) Ix2 D
1
12
433 D 9 ft2 .
1
333 D 27.
3
The composite: Ix D Ix1 C Ix2 D 36 ft4 . The area:
1
43 C 33 D 15 ft2 .
2
kx D
Ix
D 1.549 ft.
A
Problem 8.42 Determine JO and kO .
Solution: (See Figure in Problem 8.41.) Use the solution to
from which
Problem 8.41.
Part (1): The area moment of inertia about the centroidal axis parallel
to the base for a triangle is
Iyc D
1
36
bh3 D
1
36
343 D 5.3333 ft4 ,
from which
Iy1
2
8
D
A1 C Iyc D 48 ft4 .
3
Iy2 D 5.52 A2 C Iyc D 279 ft4 ,
where A2 D 9 ft2 .
The composite: Iy D Iy1 C Iy2 D 327 ft4 , from which, using a result
from Problem 8.41,
JO D Ix C Iy D 327 C 36 D 363 ft4
and kO D
JO
D 4.92 ft
A
where A1 D 6 ft2 .
Part (2): The area moment of inertia about a centroid parallel to the
base for a rectangle is
Iyc D
618
1
12
bh3 D
1
12
333 D 6.75 ft4 ,
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.43 Determine Ixy .
Solution: (See Figure in Problem 8.41.) Use the results of the
solutions to Problems 8.41 and 8.42. The area cross product of the
moment of inertia about centroidal axes parallel to the bases for a
1 2 2
b h , and for a rectangle it is zero. Therefore:
triangle is Ix0 y 0 D
72
Ixy1 D
1
72
42 32 C
8
3
A1 D 18 ft4
3
3
and Ixy2 D 1.55.5A2 D 74.25 ft4 ,
Ixy D Ix0 y 0 1 C Ixy2 D 92.25 ft4
y
Problem 8.44 Determine Ix and kx .
Solution: Use the results of Problems 8.41, 8.42, and 8.43. The
strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate
yD
A1 1 C A2 1.5
D 1.3 ft.
A
4 ft
3 ft
3 ft
x
From which
Ixc D y2 A C Ix D 10.65 ft4
and kxc D
Ixc
D 0.843 ft
A
Problem 8.45 Determine JO and kO .
Solution: Use the results of Problems 8.41, 8.42, and 8.43. The
strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate:
A1
xD
8
C A2 5.5
3
D 4.3667 ft,
A
from which
IYC D x2 A C IY D 40.98 ft4 .
Using a result from Problem 8.44,
JO D IXC C IYC D 10.65 C 40.98 D 51.63 ft4
and kO D
JO
D 1.855 ft
A
Problem 8.46 Determine Ixy .
Solution: Use the results of Problems 8.41–8.45. The strategy is
to use the parallel axis theorem and solve for the area moment of inertia
about the centroidal axis. Using the centroidal coordinates determined
in Problems 8.44 and 8.45,
Ixy D xyA C Ixy D 85.15 C 92.25 D 7.1 ft4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
619
Problem 8.47 Determine Ix and kx .
y
120
mm
20
mm
80
mm
x
40 mm
80 mm
Solution: Let Part 1 be the entire rectangular solid without the
y
hole and let part 2 be the hole.
m
20 m
Area D hb
R2
D 80120 40 mm
R2
y′
Ix1 D 13 bh3
where b D 80 mm
h D 120 mm
Ix1 D 13 801203 D 4.608 &eth; 107 mm4
40 mm
120 mm
Part 2
x′
For Part 2,
Ix0 2 D 14 R4 D 14 204 mm4
Ix0 2 D 1.257
&eth; 105
dy = 80 mm
mm4
Ix2 D Ix0 2 C d2y A
Part 1
x
where A D R2 D 1257 mm2
80 mm
d D 80 mm
Ix2 D 1.257 &eth; 105 C 202 802
Ix2 D 0.126 &eth; 106 C 8.042 &eth; 106 mm4
D 8.168 &eth; 106 mm4 D 0.817 &eth; 107 mm4
Ix D Ix1 Ix2 D 3.79 &eth; 107 mm4
Area D 8343 mm2
kx D
620
Ix
D 67.4 mm
Area
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.48 Determine JO and kO .
Solution: For the rectangle,
y
40 mm
JO1 D Ix1 C Iy1 D 13 bh3 C 13 hb3
JO1 D 4.608 &eth; 107 C 2.048 &eth; 107 mm4
y′
R = 20 mm
JO1 D 6.656 &eth; 107 mm4
x′
A1 D bh D 9600 mm2
120 mm
A2
For the circular cutout about x0 y 0
J0O2 D Ix0 2 C Iy 0 D 14 R4 C 14 R4
2
(h) 80 mm
A1
J0O2 D 1.257 &eth; 105 C 1.257 &eth; 105 mm4
J0O2 D 2.513 &eth; 105 mm2
x
Using the parallel axis theorem to determine JO2 (about x, y)
80 mm
(b)
JO2 D J002 C d2x C d2y A2
A2 D R2 D 1257 mm2
JO2 D 1.030 &eth; 107 mm4
JO D 5.63 &eth; 107 mm4
kO D
JO D JO1 JO2
JO
D
Area
JO
A1 A2
kO D 82.1 mm
JO D 6.656 &eth; 107 1.030 &eth; 107 mm4
Problem 8.49 Determine Ixy .
Solution:
A1 D 80120 D 9600
y
80 mm
mm2
A2 D R2 D 202 D 1257 mm2
R = 20 mm
y′
For the rectangle A1 x′
Ixy1 D 14 b2 h2 D 14 802 1202
Ixy1 D 2.304 &eth; 107 mm2
For the cutout
A2
A1
120 mm
A2
dy = 80 mm
A1
Ix 0 y 0 2 D 0
and by the parallel axis theorem
x
dx = 40 mm
Ixy2 D Ix0 y 0 2 C A2 dx dy Ixy2 D 0 C 12574080
Ixy2 D 4.021 &eth; 106 mm4
Ixy D Ixy1 Ixy2
Ixy D 2.304 &eth; 107 0.402 &eth; 107 mm4
Ixy D 1.90 &eth; 107 mm4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
621
y
Problem 8.50 Determine Ix and kx .
20
mm
120
mm
x
80
mm
40 mm
80 mm
Solution: We must first find the location of the centroid of the total
y
area. Let us use the coordinates XY to do this. Let A1 be the rectangle
and A2 be the circular cutout. Note that by symmetry Xc D 40 mm
Y
80 mm
Rectangle1
Circle2
Area
Xc
Yc
9600 mm2
40 mm
60 mm
mm2
40 mm
80 mm
1257
R = 20 mm
120 mm
A1 D 9600 mm2
A2 D 1257 mm2
x
80 mm
For the composite,
Xc D
A1 Xc1 A2 Xc2
D 40 mm
A1 A2
Yc D
A1 Yc1 A2 Yc2
D 57.0 mm
A1 A2
Now let us determine Ix and kx about the centroid of the
composite body.
Rectangle about its centroid (40, 60) mm
Ix1 D
1 3
1
bh D
801203
12
12
Ix1 D 1.152
&eth; 107
mm3 ,
X
40 mm
40 mm
Now to C ! dy2 D 80 57 D 23 mm
Ixc2 D Ix2 C dy2 2 A2
Ixc2 D 7.91 &eth; 105 mm4
For the composite about the centroid
Ix D Ixc1 Ixc2
Now to C
Ix D 1.08 &eth; 107 mm4
Ixc1 D Ix1 C 60
Yc 2 A1
The composite Area D 9600 1257 mm2
Ixc1 D 1.161 &eth; 107 mm4
D 8343 mm2
Circular cut out about its centroid
A2 D
R2
D
202 D 1257
mm2
kx D
Ix
D 36.0 mm
A
Ix2 D 14 R4 D 204 /4
Ix2 D 1.26 &eth; 105 mm4
622
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.51 Determine Iy and ky .
Solution: From the solution to Problem 8.50, the centroid of the
y
composite area is located at (40, 57.0) mm.
The area of the rectangle, A1 , is 9600 mm2 .
The area of the cutout, A2 , is 1257 mm2 .
The area of the composite is 8343 mm2 .
(1)
Rectangle about its centroid (40, 60) mm.
Iy1 D
1 3
1
hb D
120803
12
12
+
x
80 mm
Iy1 D 5.12 &eth; 106 mm4
dx1 D 0
(2)
40 mm
Circular cutout about its centroid (40, 80)
80 mm
Iy2 D R4 /4 D 1.26 &eth; 105 mm4
dx2 D 0
Since dx1 and dx2 are zero. (no translation of axes in the xdirection), we get
Iy D Iy1 Iy2
Iy D 4.99 &eth; 106 mm4
Finally,
ky D
Iy
D
A1 A2
4.99 &eth; 106
8343
ky D 24.5 mm
Problem 8.52 Determine JO and kO .
y
Solution: From the solutions to Problems 8.51 and 8.52,
Ix D 1.07 &eth; 107 mm4
Iy D 4.99 &eth; 106 mm4
and A D 8343 mm2
20
mm
JO D Ix C Iy D 1.57 &eth; 107 mm4
kO D
JO
D 43.4 mm
A
x
120 mm
80 mm
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
623
y
Problem 8.53 Determine Iy and ky .
12 in
x
20 in
Solution: Treat the area as a circular area with a half-circular
y
y
y
cutout: From Appendix B,
1
Iy 1 D 14 204 in4
and Iy 2 D 18 124 in4 ,
12
in.
20 in.
2
x
x
2 in.
x
20 in.
so Iy D 14 204 18 124 D 1.18 &eth; 105 in4 .
The area is A D 202 12 122 D 1030 in2
so,
ky D
Iy
D
A
1.18 &eth; 105
1.03 &eth; 103
D 10.7 in
Problem 8.54 Determine JO and kO .
Solution: Treating the area as a circular area with a half-circular
cutout as shown in the solution of Problem 8.53, from Appendix B,
JO 1 D Ix 1 C Iy 1 D 12 204 in4
and JO 2 D Ix 2 C Iy 2 D 14 124 in4 .
Therefore JO D 12 204 14 124
D 2.35 &eth; 105 in4 .
From the solution of Problem 8.53,
A D 1030 in2 Ro D
D
624
JO
A
2.35 &eth; 105
D 15.1 in.
1.03 &eth; 103
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.55 Determine Iy and ky if h D 3 m.
y
Solution: Break the composite into two parts, a rectangle and a
semi-circle.
1.2 m
For the semi-circle
Ix 0 c D
9
8
8
R4
h
Iy 0 c D
1 4
4R
R d D
8
3
y′
x
y
d
x′
1.2 m
AC
d = 4R
3π
AR
To get moments about the x and y axes, the dxc , dyc for the semicircle are
dxc D 0,
dyc D 3 m C
3m = h
4R
3
x
and Ac D R2 /2 D 2.26 m2
Iy 0 c D
To get moments of area about the x, y axes, dxR D 0, dyR D 1.5 m
1 4
R
8
and Iyc D Iy 0 c C d2xc A
0
IyR D Iy 0 R C dxR 2 bh
dx D 0
1
32, 43 m4
12
Iyc D Iy 0 c D 1.24 /8
IyR D Iy 0 R D
Iyc D 0.814 m4
IyR D 3.456 m2
For the Rectangle
Ix 0 R D
1 3
bh
12
Iy 0 R D
1 3
hb
12
AR D bh D 7.2 m2
Iy D Iyc C IyR
Iy D 4.27 m2
To find ky , we need the total area, A D AR C Ac
AR D bh
A D 7.20 C 2.26 m2
y′ y
A D 9.46 m2
2.4 m
ky D
h
3m
b
Iy
D 0.672 m
A
x′
x
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
625
Problem 8.56 Determine Ix and kx if h D 3 m.
Solution: Break the composite into two parts, the semi-circle and
the rectangle. From the solution to Problem 8.55,
Ix 0 c D
9
8
8
dyc D
3C
4R
3
y
Ac
R4
m
Ac D 2.26 m2
R = 1.2 m
h=3m
b = 2.4 m
AR
3m
h
Ixc D Ix0 c C Ac d2yc
Substituting in numbers, we get
dyc D 3.509 m
x
2.4 m
Ix0 c D 0.0717 m4
yc′
and Ixc D Ix0 c C Ac d2y
Ixc D 27.928 m2
xc′
R
For the Rectangle h D 3 m, b D 2.4 m
Area: AR D bh D 7.20 m2
Ix 0 R D
1 3
bh , dyR D 1.5 m
12
4R
3π
IxR D Ix0 R C d2yR AR
Substituting, we get
Ix0 R D 5.40 m4
IxR D 21.6 m4
For the composite,
Ix D IxR C Ixc
Ix D 49.5 m4
Also kx D
Ix
D 2.29 m
AR C Ac
kx D 2.29 m
626
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.57 If Iy D 5 m4 , what is the dimension
of h?
Solution: From the solution to Problem 8.55, we have:
y c′
y
xc′
For the semicircle
Iy 0 c D Iy D 1.24/8 D 0.814 m2
h
For the rectangle
Iy 0 R D IyR D
1.2
y ′R
m
h
x ′R
b
1
h2.43 m4
12
2.4 m
x
Also, we know IyR C Iyc D 5 m4 .
Hence 0.814 C
1
h2.43 D 5
12
Solving, h D 3.63 m
Problem 8.58 Determine Iy and ky .
Solution: Let the area be divided into parts as shown. The areas
and the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
A3 D
1
302 D 707 in2 ,
4
x 3 D 20 C
430
D 32.7 in,
3
y 3 D 40 C
430
D 52.7 in.
3
Using the results from Appendix B, the moments of inertia of the parts
Iy 1 D
1
40 in50 in3 D 167 &eth; 104 in4 ,
3
Iy 2 D
1
30 in20 in3 D 8.00 &eth; 104 in4 ,
3
Iy 3 D
4
16
9
&quot;
430 in 2 ! 30 in4 C 20 in C
30 in2
3
4
D 80.2 &eth; 104 in4 .
The moment of inertia of the composite area about the y axis is
Iy D Iy 1 C Iy 2 C Iy 3 D 2.55 &eth; 106 in4 .
2
The composite area is A D A1 C A2 C A3 D 3310
in .
Iy
D 27.8 in.
A
Iy D 2.55 &eth; 106 in4 , ky D 27.8 in.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
627
Problem 8.59 Determine Ix and kx .
Solution: See the solution to Problem 8.58.
Let the area be divided into parts as shown. The areas and the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
A3 D
1
302 D 707 in2 ,
4
y 3 D 40 C
430
D 52.7 in.
3
Using the results from Appendix B, the moments of inertia of the parts
Ix 1 D
1
50 in40 in3 D 1.07 &eth; 106 in4 ,
3
Ix 2 D
1
20 in30 in3 C 55 in2 600 in2 D 1.86 &eth; 106 in4 ,
12
Ix 3 D
4
4
9
&quot;
430 in 2 ! 30 in4 C 40 in C
30 in2 D 2.01 &eth; 106 in4 .
3
4
The moment of inertia of the composite area about the x axis is
Ix D Ix 1 C Ix 2 C Ix 3 D 4.94 &eth; 106 in4 .
2
The composite area is A D A1 C A2 C A3 D 3310
in .
Ix
D 38.6 in.
A
Ix D 4.94 &eth; 106 in4 , kx D 38.6 in
628
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.60 Determine Ixy .
Solution: See the solution to Problem 8.58.
Let the area be divided into parts as shown. The areas and the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
A3 D
1
302 D 707 in2 ,
4
y 3 D 40 C
430
D 52.7 in.
3
Using the results from Appendix B, the products of inertia of the parts
Ixy 1 D
1
50 in2 40 in2 D 10.0 &eth; 105 in4 ,
4
Ixy 2 D 10 in55 in600 in2 D 3.30 &eth; 105 in4 ,
Ixy 3 D
4
1
8
9
430 in
430 in
30 in4 C 20 in C
40 in C
[707 in2 ]
3
3
D 12.1 &eth; 105 in4 .
The product of inertia of the composite area is
Ixy D Ixy 1 C Ixy 2 C Ixy 3 D 2.54 &eth; 106 in4 .
Ixy D 2.54 &eth; 106 in4 .
y
Problem 8.61 Determine Iy and ky .
Solution: See the solution to Problem 8.58.
In terms of the coordinate system used in Problem 8.58, the areas and
the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
x 1 D 25 in,
y 1 D 20 in,
A2 D 2030 D 600 in2 ,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
A3 D
1
302 D 707 in2 ,
4
30 in
40 in
x
430
D 52.7 in.
y 3 D 40 C
3
20 in
The composite area is A D A1 C A2 C A3 D 3310 in2 .
x 1 A1 C x 2 A2 C x 3 A3
The x coordinate of its centroid is x D
D 23.9 in
A
The moment of inertia about the y axis in terms of the coordinate
system used in Problem 8.58 is Iy D 2.55 &eth; 106 in4 . Applying the
parallel axis theorem, the moment of inertia about the y axis through
the centroid of the area is
Iy D 2.55 &eth; 106 in4 23.9 in2 3310 in2 D 6.55 &eth; 105 in4 .
Iy
D 14.1 in.
A
Iy D 6.55 &eth; 105 in4 , ky D 14.1 in.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
629
y
Problem 8.62 Determine Ix and kx .
Solution: See the solution to Problem 8.59.
In terms of the coordinate system used in Problem 8.59, the areas and
the coordinates of their centroids are
A1 D 4050 D 2000 in2 ,
A2 D 2030 D 600 in2 ,
A3 D
1
302 D 707 in2 ,
4
x 1 D 25 in,
y 1 D 20 in,
x 2 D 10 in,
y 2 D 55 in,
x 3 D 20 C
430
D 32.7 in,
3
30 in
40 in
x
430
D 52.7 in.
3
y 3 D 40 C
20 in
2
The composite area is A D A1 C A2 C A3 D 3310 in .
y A1 C y 2 A2 C y 3 A3
The x coordinate of its centroid is y D 1
D 33.3 in
A
The moment of inertia about the x axis in terms of the coordinate
system used in Problem 8.59 is Ix D 4.94 &eth; 106 in4 . Applying the
parallel axis theorem, the moment of inertia about the x axis through
the centroid of the area is
Ix D 4.94 &eth; 106 in4 33.3 in2 3310 in2 D 1.26 &eth; 106 in4 .
Ix
D 19.5 in.
A
Ix D 1.26 &eth; 106 in4 , kx D 19.5 in.
Problem 8.63 Determine Ixy .
Solution: See the solution to Problem 8.60.
Ixy D [0 C 1.0 m0.8 mdx 0.5 mdy 0.4 m]
[0 C 0.2 m2 dx 0.4 mdy 0.3 m]
C
1
1
0.8 m2 0.6 m2 0.8 m0.6 m0.2 m
24
2
C
Solving:
0.8 m
3
0.8 m
1
0.8 m0.6 mdx 1.2 m dy 2
3
Ixy D 0.0230 m4
Check using the noncentroidal product of inertia from Problem 8.60
we have
Ixy D Ixy 0 Adx dy D 0.2185 m4 0.914 m2 0.697 m0.379 m
D 0.0230 m4
630
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.64 Determine Iy and ky .
y
18 in
x
6 in
6 in
Solution: Divide the area into three parts:
The composite area:
Part (1) The rectangle 18 by 18 inches; Part (2) The triangle with base
6 in and altitude 18 in; Part (3) The semicircle of 9 in radius.
3
#
6 in
Ai D 397.23 in2 .
1
Part (1): A1 D 1818 D 324 in2 ,
x1 D 9 in,
Iy D x21 A1 C Iyy1 x22 A2 Iyy2 C x23 A3 C Iyy3 ,
y1 D 9 in,
Ixx1 D
Iyy1 D
1
12
1
12
The area moment of inertia:
Iy D 4.347 &eth; 104 in4 ,
18183 D 8748 in4 ,
ky D
Iy
D 10.461 in
A
18183 D 8748 in4 .
1
186 D 54 in2 ,
2
Part (2): A2 D
x2 D 9 in,
y2 D
1
18 D 6 in,
3
Ixx2 D
1
36
6183 D 972 in4 ,
Iyy2 D 1/181833 D 27 in4 .
Part (3) A3 D
92 D 127.23 in2 ,
2
x3 D 9 in,
y3 D 18 C
Ixx3 D
Iyy3 D
49
3
D 21.82 in,
1
49 2
A3 D 720.1 in4 ,
94 8
3
1
94 D 2576.5 in4 .
8
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
631
Problem 8.65 Determine Ix and kx .
Solution: Use the results of the solution to Problem 8.64.
IX D y21 A1 C IXX1 y22 A2 IXX2 C y23 A3 C IXX3 ,
Ix D 9.338 &eth; 104 in4 ,
kx D
Ix
D 15.33 in
A
Problem 8.66 Determine Ixy .
Solution: Use the results of the solutions to Problems 8.63
and 8.64.
Ixy D x1 y1 A1 x2 y2 A2 C x3 y3 A3
Ixy D 4.8313 &eth; 104 in4
Problem 8.67 Determine Iy and ky .
y
Solution: We divide the composite area into a triangle (1), rect-
6 in
angle (2), half-circle (3), and circular cutout (4):
2 in
Triangle:
Iy 1 D 14 1283 D 1536 in4
x
Rectangle:
Iy 2 D
8 in
1
1283 C 122 812 D 14,336 in4 .
12
8 in
y
Half-Circle:
Iy 3 D
8
8
9
2
46 2 1
62 D 19,593 in4
64 C 16 C
3
2
3
4
1
x
Circular cutout:
Iy 4 D 14 24 C 162 22 D 3230 in4 .
y
8 in.
y
Therefore
Iy D Iy 1 C Iy 2 C Iy 3 Iy 4 D 3.224 &eth; 104 in4 .
12 in.
2
1
12 in.
The area is
x
A D A1 C A2 C A3 A4
D
1
1
128 C 812 C 62 22 D 188 in2 ,
2
2
so ky D
Iy
D
A
x
8 in.
12 in.
y
y
3
6 in.
4
3.224 &eth; 104
D 13.1 in.
188
6 in.
x
16 + 4(6)
in.
3π
632
2 in.
x
16 in.
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.68 Determine JO and kO .
Solution: Iy is determined in the solution to Problem 8.67. We
will determine Ix and use the relation JO D Ix C Iy . Using the figures
in the solution to Problem 8.67,
Triangle:
Ix 1 D
1
8123 D 1152 in4 .
12
Rectangle:
Ix 2 D 13 8123 D 4608 in4 .
Half Circle:
Ix 3 D 18 64 C 62 21 62 D 2545 in4 .
Circular Cutout:
Ix 4 D 14 24 C 62 22 D 465 in4 .
Therefore
Ix D Ix 1 C Ix 2 C Ix 3 Ix 4 D 7840 in4 .
Using the solution of Problem 8.67,
JO D Ix C Iy D 0.784 &eth; 104 C 3.224 &eth; 104 D 4.01 &eth; 104 in4 .
From the solution of Problem 8.67, A D 188 in2 , so
R0 D
JO
D
A
4.01 &eth; 104
D 14.6 in.
188
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
633
Problem 8.69 Determine Iy and ky .
y
4 in
Solution: Divide the area into four parts: Part (1) The rectangle
8 in by 16 in. Part (2): The rectangle 4 in by 8 in. Part (3) The semicircle of radius 4 in, and Part (4) The circle of radius 2 in.
Part (1): A1 D 168 D 128 in2 ,
x1 D 8 in,
2 in
4 in
8 in
y1 D 4 in,
Ixx1 D
Iyy1 D
1
12
1
12
x
1683 D 682.67 in4 ,
16 in
8163 D 2730.7 in4 .
Part (2): A2 D 48 D 32 in2 ,
Iyy2 D
1
12
1
12
843 D 42.667 in4 ,
42 D 25.133 in2 ,
2
Ixx3 D
24 D 12.566 in4 ,
3
#
Ai A4 D 172.566 in2 .
1
The area moment of inertia:
Iy D x21 A1 C Iyy1 C x22 A2 C Iyy2 C x23 A3 C Iyy3 x24 A4 Iyy4
44
3
D 13.698 in.
The area moments of inertia about the centroid of the semicircle are
Iyy3 D
4
Iyy4 D Ixx4 D 12.566 in4 .
483 D 170.667 in4 .
x3 D 12 in.
\$1%
The composite area:
Part (3): A3 D
y3 D 12 C
x4 D 12 in,
Ixx4 D
y2 D 10 in,
Part (4): A4 D 22 D 12.566 in2 ,
y4 D 12 in,
x2 D 12 in,
Ixx2 D
12 in
1
44 D 100.53 in4 ,
8
Iy D 1.76 &eth; 104 in4 ,
ky D
Iy
D 10.1 in
A
1
44 2
A3 D 28.1 in4 .
44 8
3
Check:
Ixx3 D 0.1098R4 D 28.1 in4 .
check.
Problem 8.70 Determine Ix and kx .
Solution: Use the results in the solution to Problem 8.69.
Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 y24 A4 Ixx4
Ix D 8.89 &eth; 103 in4
kx D
634
Ix
D 7.18 in
A
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.71 Determine Ixy .
Solution: Use the results in the solution to Problem 8.69.
Ixy D x1 y1 A1 C x2 y2 A2 C x3 y3 A3 x4 y4 A4 ,
Ixy D 1.0257 &eth; 104 in4
Problem 8.72 Determine Iy and ky .
y
4 in
2 in
4 in
Solution: Use the results in the solutions to Problems 8.69 to 8.71.
The centroid is
xD
x1 A1 C x2 A2 C x3 A3 x4 A4
A
1024 C 384 C 301.6 150.8
D 9.033 in,
D
172.567
x
8 in
12 in
16 in
from which
Iyc D x2 A C Iy D 1.408 &eth; 104 C 1.7598 &eth; 104 D 3518.2 in4
kyc D
Iyc
D 4.52 in
A
Problem 8.73 Determine Ix and kx .
Solution: Use the results in the solutions to Problems 8.69 to 8.71.
The centroid is
yD
y1 A1 C y2 A2 C y3 A3 y4 A4
D 5.942 in,
A
from which
Ixc D y2 A C Ix D 6092.9 C 8894 D 2801 in4
kxc D
Ixc
D 4.03 in
A
Problem 8.74 Determine Ixy .
Solution: Use the results in the solutions to Problems 8.69–8.71.
Ixyc D xyA C Ixy D 9.263 &eth; 103 C 1.0257 &eth; 104 D 994.5 in4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
635
Problem 8.75 Determine Iy and ky .
y
Solution: We divide the area into parts as shown:
Iy 1 D
5 mm
1
50 C 15 C 15303 D 180,000 mm4
12
Iy 2 D Iy 3 D Iy 4 D
15 mm
50 mm
1
30103 C 202 1030
12
5 mm
5 mm
D 122,500 mm4
Iy 5 D Iy 6 D Iy 7 D
8
8
9
x
15 mm
15 mm
154
10 15 15 10
mm mm mm mm
415 2 1
152 D 353,274 mm4
C 25 C
3
2
y
1
Iy 8 D Iy 9 D Iy 10
2 5
8
1
D 54 C 252 52 D 49,578 mm4 .
4
Therefore,
50 mm
7 4
10
Iy D Iy 1 C 3Iy 2 C 3Iy 5 3Iy 8 D 1.46
&eth; 106
3 6
9
mm4 .
x
10
15
mm mm
The area is
A D A1 C 3A2 C 3A5 3A8
D 3080 C 31030 C 3
1
152 352
2
D 4125 mm2
so ky D
Iy
D
A
1.46 &eth; 106
D 18.8 mm
4125
Problem 8.76 Determine JO and kO .
Solution: Iy is determined in the solution to Problem 8.75. We
will determine Ix and use the relation JO D Ix C Iy . Dividing the area
as shown in the solution to Problem 8.75, we obtain
Ix 1 D
1
30803 C 252 3080 D 2, 780, 000 mm4
12
Ix 2 D
1
10303 C 502 1030 D 772, 500 mm4
12
Ix 3 D Ix 4 D
Ix 5 D
1
10303 D 22, 500 mm4
12
Therefore
Ix D Ix 1 C Ix 2 C 2Ix 3 C Ix 5 C 2Ix 6 Ix 8 2Ix 9
D 4.34 &eth; 106 mm4
and JO D Ix C Iy D 5.80 &eth; 106 mm4 .
From the solution to Problem 8.75, A D 4125 mm2
so kO D
1
1
154 C 502 152 D 903,453 mm4
8
2
D
Ix 6 D Ix 7 D
Ix 8 D
5.80 &eth; 106
4125
D 37.5 mm.
1
54 C 52 502 ,
4
Ix 9 D Ix 10 D
636
1
154 D 19,880 mm4 ,
8
JO
A
1
54 D 491 mm4 .
4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.77 Determine Ix and Iy for the beam’s
cross section.
5 in
2 in
Solution: Use the symmetry of the object
1
1
Ix
D 3 in.8 in3 C
3 in3 in3 C 3 in2 11.5 in2
2
3
12
C
5 in4
5 in2
16
4
C
4[5 in]
3
2
5 in2
4
8 in
4[5 in] 2
8 in C
3
x
3 in
in4
2
16
in2
2
4
C
4[2 in]
3
5 in
3 in
5 in
2
2 in2
4
4[2 in] 2
8 in C
3
Solving we find
Ix D 7016 in4
Iy
1
1
D 3 in3 in3 C
8 in3 in3 C 8 in3 in6.5 in2
2
3
12
C
5 in4
5 in2
16
4
4[5 in]
3
2
5 in2
C
4
2 in2
2 in4
16
4
C
4[2 in]
3
4[5 in]
3 in C
3
2 2
2 in2
4
3 in C
4[2 in]
3
2 Solving we find
Iy D 3122 in4
y
Problem 8.78 Determine Ix and Iy for the beam’s
cross section.
5 in
2 in
x
Solution: Use Solution 8.77 and 7.39. From Problem 7.39 we
know that
y D 7.48 in, A D 98.987 in2
8 in
Ix D 7016 in4 Ay 2 D 1471 in4
Iy D 3122 in4 A02 D 3122 in4
3 in
5 in
5 in
3 in
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
637
Problem 8.79 The area A D 2 &eth; 104 mm2 . Its moment
of inertia about the y axis is Iy D 3.2 &eth; 108 mm4 . Determine its moment of inertia about the yO axis.
ŷ
y
Solution: Use the parallel axis theorem. The moment of inertia
A
about the centroid of the figure is
Iyc D x 2 A C Iy D 1202 2 &eth; 104 C 3.2 &eth; 108
D 3.20 &eth; 107 mm4 .
x, x̂
The moment of inertia about the yO axis is
100 mm
120 mm
IyO D x2 A C Iyc
IyO D 2202 2 &eth; 104 C 3.2 &eth; 107
D 1 &eth; 109 mm4
Problem 8.80 The area A D 100 in2 and it is
symmetric about the x 0 axis. The moments of inertia
Ix0 D 420 in4 , Iy 0 D 580 in4 , JO D 11000 in4 , and Ixy D
4800 in4 . What are Ix and Iy ?
y
y'
A
x'
O'
O
x
Solution: The basic relationships:
(1)
(2)
(3)
(4)
(5)
(6)
Ix D y 2 A C Ixc ,
Iy D x2 A C Iyc ,
JO D Ar 2 C Jc ,
JO D Ix C Iy ,
Jc D Ixc C Iyc , and
Ixy D Axy C Ixyc ,
where the subscript c applies to the primed axes, and the others to the
unprimed axes. The x, y values are the displacement of the primed
axes from the unprimed axes. The steps in the demonstration are:
From symmetry about the xc axis, the product of inertia Ixyc D 0.
JO J c
D 100 in2 , from which r 2 D x2 C
From (3): r 2 D
A
y 2 D 100 in2
Ixy
(iii) From (6) and Ixyc D 0, y D
, from which x2 r 2 D x4 C
Ax
2
Ixy
. From which: x4 100x2 C 2304 D 0.
A
(iv) The roots: x12 D 64, and x22 D 36. The corresponding values of y
p
are found from y D r 2 x 2 from which x1 , y1 D 8, 6, and
x2 , y2 D 6, 8.
(v) Substitute these pairs to obtain the possible values of the area
moments of inertia:
(i)
(ii)
Ix1 D Ay12 C Ixc D 4020 in4 ,
Iy1 D Ax12 C Iyc D 6980 in4
Ix2 D Ay22 C Ixc D 6820 in4 ,
Iy2 D Ax22 C Iyc D 4180 in4
638
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.81 Determine the moment of inertia of the
with the moment of inertia of a solid square cross section
of equal area. (See Example 8.5.)
y
20 mm
x
Solution: We first need to find the location of the centroid of the
composite. Break the area into two parts. Use X, Y coords.
160 mm
y
2
100
20
A1 = 2000 mm2
XC = 0
160 mm
1
A2 = 3200 mm2
XC2 = 0
20 mm
YC2 = 80 mm
100 mm
1
YC1 = 170 mm
x
100 mm
y Y
20 mm
Xc D
Xc1 A1 C Xc2 A2
D0
A1 C A2
Yc D
Yc1 A1 C Yc2 A2
A1 C A2
C1
1
For the composite
20 mm
x
160 mm
2
C2
Substituting, we get
Xc D 0 mm
X
Yc D 114.6 mm
We now find Ix for each part about its center and use the parallel axis
theorem to find Ix about C.
20 mm
Part (1): b1 D 100 mm, h1 D 20 mm
Finally, Ix D Ix1 C Ix2
1
1
b1 h13 D
100203 mm4
12
12
Ix D 1.686 &eth; 107 mm4
Ix 0 1 D
for our composite shape.
Ix0 1 D 6.667 &eth; 104 mm4
dy1 D Yc1 Yc D 55.38 mm
Ix1 D Ix0 1 C dy1 2 A1 Now for the comparison. For the solid square with the same total area
A1 C A2 D 5200 mm2 , we get a side of length
l2 D 5200:
l D 72.11 mm
And for this solid section
Ix1 D 6.20 &eth; 106 mm4
Part (2) b2 D 20 mm, h2 D 160 mm
Ix 0 2 D
1
1
b2 h2 3 D
201603 mm4
12
12
IxSQ D
1 3
1 4
bh D
l
12
12
IxSQ D 2.253 &eth; 106 mm4
1.686 &eth; 107
2.253 &eth; 106
Ix0 2 D 6.827 &eth; 106 mm4
Ratio D Ix /IxSQ D
dy2 D Yc2 Yc D 34.61 mm
Ratio D 7.48
Ix2 D Ix0 2 C dy2 A2
This matches the value in Example 8.5.
Ix2 D 1.066 &eth; 107 mm4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
639
Problem 8.82 The area of the beam cross section is
5200 mm2 . Determine the moment of inertia of the beam
the moment of inertia of a solid square cross section of
equal area. (See Example 8.5.)
y
x
20 mm
Solution: Let the outside dimension be b mm, then the inside
dimension is b 40 mm. The cross section is A D b2 b 402 D
5200 mm2 . Solve: b D 85 mm. Divide the beam cross section into two
parts: the inner and outer squares. Part (1)
A1 D 852 D 7225 mm2 ,
Ixx1 D
1
12
85853 D 4.35 &eth; 106 .
Part (2)
A2 D 452 D 2025 mm2 .
Ixx2 D
1
12
45453 D 3.417 &eth; 105 .
The composite moment of inertia about the centroid is
Ix D Ixx1 Ixx2 D 4.008 &eth; 106 mm4 .
For a square cross section of the same area, h D
p
5200 D 72.111 mm.
The area moment of inertia is
Ixb D
1
12
72.11172.1113 D 2.253 &eth; 106 in4 .
The ratio:
RD
4.008 &eth; 106
D 1.7788 D 1.78
2.253 &eth; 106
which confirms the value given in Example 8.5.
640
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.83 If the beam in Fig. a is subjected to
couples of magnitude M about the x axis (Fig. b), the
beam’s longitudinal axis bends into a circular are whose
y
y
z
x
EIx
RD
,
M
where Ix is the moment of inertia of the beam’s cross
section about the x axis. The value of the term E, which
is called the modulus of elasticity, depends on the material of which the beam is constructed. Suppose that a
beam with the cross section shown in Fig. c is subjected
to couples of magnitude M D 180 N-m. As a result,
the beam’s axis bends into a circular arc with radius
R D 3 m. What is the modulus of elasticity of the beam’s
material? (See Example 8.5.)
Solution: The moment of inertia of the beam’s cross section about
the x axis is
&amp;
'
1
1
mm4
393 C 2
933 C 62 93
Ix D
12
12
D 2170 mm4 D 2.17 &eth; 109 m4 .
The modulus of elasticity is
ED
RM
3 m180 N-m
D
D 2.49 &eth; 1011 N/m2
Ix
2.17 &eth; 109 m4
M
R
M
(b) Subjected to couples at the ends.
y
3 mm
x
9 mm
3 mm
3
mm
9 mm
(c) Beam cross section.
E D 2.49 &eth; 1011 N/m2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
641
Problem 8.84 Suppose that you want to design a beam
made of material whose density is 8000 kg/m3 . The
beam is to be 4 m in length and have a mass of
320 kg. Design a cross section for the beam so that
Ix D 3 &eth; 105 m4 . (See Example 8.5.)
Solution: The strategy is to determine the cross sectional area,
and then use the ratios given in Figure 8.14 to design a beam. The
volume of the beam is V D AL D 4A m3 . The mass of the beam is m D
V8000 D 32000A D 320 kg, from which A D 0.01 m2 . The moment
of inertia for a beam of square cross section with this area is
Ixxb D
1
12
h
0.10.13 D 8.333 &eth; 106 m4 .
The ratio is R D
b
&eth; 105
3
D 3.6.
8.333 &eth; 106
From Figure 8.6, this ratio suggests an I-beam of the form shown in
the sketch. Choose an I-beam made up of three equal area rectangles,
of dimensions b by hm in section. The moment of inertia about the
centroid is Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 .
Since all areas are equal, A1 D A2 D A3 D bh, and y1 D
0, and y3 D y1 , this reduces to
Ix D
bCh
, y2 D
2
1
bCh 2
1
bh3 C 2
hb3 .
hb C
6
2
12
A
, where A is the known total cross section area. These
3
are two equations in two unknowns. Plot the function
Note that bh D
fb D
1
bCh 2
1
bh C
bh3 C 2
hb3 Ix
6
2
12
Problem 8.85 The area in Fig. (a) is a C230&eth;30
American Standard Channel beam cross section. Its cross
sectional area is A D 3790 mm2 and its moments of
inertia about the x and y axes are Ix D 25.3 &eth; 106 mm4
and Iy D 1 &eth; 106 mm4 . Suppose that two beams with
C230&eth;30 cross sections are riveted together to obtain a
composite beam with the cross section shown in Fig. (b).
What are the moments of inertia about the x and y axes
of the composite beam?
f
(
b
)
*
E
+
5
I - beam Flange dimension
1
.8
.6
.4
.2
0
–.2
–.4
–.6
–.8
–1
.03 .04 .05 .06 .07 .08 .09 .1
b
A
.The function was graphed using
3
TK Solver Plus. The graph crosses the zero axis at approximately
b D 0.0395 m. and b D 0.09 m. The lower value is an allowable value
for h and the greater value corresponds to an allowable value of b.
Thus the I beam design has the flange dimensions, b D 90 mm and
h D 39.5 mm.
subject to the condition that hb D
y
y
x
x
14.8 mm
(a)
(b)
Solution:
Ix D 225.3 &eth; 106 mm4 D 50.6 &eth; 106 mm4
Iy D 2106 mm4 C [3790 mm2 ][14.8 mm]2 D 3.66 &eth; 106 mm4
642
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.86 The area in Fig. (a) is an L152&eth;102&eth;
12.7 Angle beam cross section. Its cross sectional
area is A D 3060 mm2 and its moments of inertia
about the x and y axes are Ix D 7.24 &eth; 106 mm4 and
Iy D 2.61 &eth; 106 mm4 . Suppose that four beams with
L152&eth;102&eth;12.7 cross sections are riveted together to
obtain a composite beam with the cross section shown
in Fig. (b). What are the moments of inertia about the x
and y axes of the composite beam?
y
y
24.9
mm
x
Solution:
x
50.2 mm
(a)
Ix D 47.24 &eth; 106 mm4 C [3060 mm2 ][50.2 mm]2 D 59.8 &eth; 106 mm4
Iy D 42.61 &eth; 106 mm4 C [3060 mm2 ][24.9 mm]2 D 18.0 &eth; 106 mm4
(b)
Problem 8.87 In Active Example 8.6, suppose that the
vertical 3-m dimension of the triangular area is increased
to 4 m. Determine a set of principal axes and the corresponding principal moments of inertia.
y
Solution: From Appendix B, the moments and products of inertia
of the area are
Ix D
1
4 m4 m3 D 21.3 m4 ,
12
Iy D
1
4 m3 4 m D 64 m4 ,
4
Ixy D
1
4 m2 4 m2 D 32 m4 .
8
3m
x
4m
From Eq. (8.26),
tan2p D
2Ixy
232
D 1.50 ) p D 28.2&deg; .
D
Iy Ix
64 21.3
From Eqs. (8.23) and (8.24), the principal moments of inertia are
Ix ‘ D
Ix C Iy
Ix Iy
C
cos 2p Ixy sin 2p
2
2
D
21.3 C 64
2
C
21.3 64
2
cos2[28.2&deg; ] 32 sin2[28.2&deg; ]
D 4.21 m4
Iy ‘ D
Ix Iy
Ix C Iy
cos 2p C Ixy sin 2p
2
2
D
21.3 C 64
2
21.3 64
2
cos2[28.2&deg; ] C 32 sin2[28.2&deg; ]
D 81.8 m4
p D 28.2&deg; , principal moments of inertia are
4.21 m4 , 81.1 m4
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
643
y
Problem 8.88 In Example 8.7, suppose that the area is
reoriented as shown. Determine the moments of inertia
Ix0 , Iy 0 and Ix0 y 0 if D 30o .
Solution: Based on Example 8.7, the moments and product of
1 ft
3 ft
inertia of the reoriented area are
1 ft
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
x
4 ft
Applying Eqs. (8.23)–(8.25),
Ix 0 D
D
Iy 0 D
D
Ix 0 y 0 D
D
Ix Iy
Ix C Iy
C
cos 2 Ixy sin 2
2
2
10 C 22
10 22
C
cos 60&deg; 6 sin 60&deg; D 7.80 ft4 ,
2
2
Ix Iy
Ix C Iy
C
cos 2 C Ixy sin 2
2
2
10 C 22
10 22
cos 60&deg; C 6 sin 60&deg; D 24.2 ft4 ,
2
2
Ix Iy
sin 2 C Ixy cos 2
2
12 22
sin 60&deg; C 6 cos 60&deg; D 2.20 ft4
2
Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix0 y 0 D 2.20 ft4 .
y
Problem 8.89 In Example 8.7, suppose that the area
is reoriented as shown. Determine a set of principal
axes and the corresponding principal moments of inertia.
Based on the result of Example 8.7, can you predict a
value of p without using Eq. (8.26)?
Solution: Based on Example 8.7, the moments and product of
1 ft
3 ft
inertia of the reoriented area are
1 ft
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
x
2Ixy
26
D
D 1 ) p D 22.5&deg;
Ix Iy
22 10
This value could have been anticipated from Example 8.7 by reorienting the axes.
Substituting the angle into Eqs. (8.23) and (8.24), the principal
moments of inertia are
From Eq. (8.26), tan 2p D
Ix 0 D
D
Iy 0 D
D
4 ft
Ix Iy
Ix C Iy
C
cos 2p Ixy sin 2p
2
2
10 C 22
10 22
C
cos 45&deg; 6 sin 45&deg; D 7.51 ft4 ,
2
2
Ix Iy
Ix C Iy
cos 2p C Ixy sin 2p
2
2
10 22
10 C 22
cos 45&deg; C 6 sin 45&deg; D 24.5 ft4 ,
2
2
p D 22.5&deg; , principal moments of inertia are
7.51 ft4 , 24.5 ft4 .
644
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.90 The moment of inertia of the area are
y
y
Ix D 1.26 &eth; 106 in4 ,
Iy D 6.55 &eth; 105 in4 ,
Ixy D 1.02 &eth; 105 in4
Determine the moments of inertia of the area Ix0 , Iy 0 and
Ix0 y 0 if D 30&deg; .
x
u
x
Solution:
Applying Eqs. (8.23)–(8.25),
Ix 0 D
Ix Iy
Ix C Iy
C
cos 2 Ixy sin 2&ETH;
2
2
D
1.26 C 0.655
1.26 0.655
C
cos 60&deg; 0.102 sin 60&deg; &eth; 106 in4
2
2
D 1.20 &eth; 106 in4
Iy 0 D
Ix Iy
Ix C Iy
cos 2 C Ixy sin 2
2
2
D
1.26 0.655
1.26 C 0.655
cos 60&deg; C 0.102 sin 60&deg; &eth; 106 in4
2
2
D 7.18 &eth; 105 in4
Ix 0 y 0 D
Ix Iy
sin 2 C Ixy cos 2
2
D
1.26 0.655
sin 60&deg; C 0.102 cos 60&deg; &eth; 106 in4
2
D 2.11 &eth; 105 in4
Ix0 D 1.20 &eth; 106 in4 , Iy 0 D 7.18 &eth; 105 in4 , Ix0 y 0 D 2.11 &eth; 105 in4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
645
Problem 8.91 The moment of inertia of the area are
y
y
Ix D 1.26 &eth; 106 in4 ,
Iy D 6.55 &eth; 105 in4 ,
Ixy D 1.02 &eth; 105 in4
x
u
Determine a set of principal axes and the corresponding
principal moments of inertia.
x
Solution: From Eq. (8.26),
tan 2p D
2Ixy
2.102
D
D 0.337
Iy Ix
0.655 1.26
) p D 9.32&deg;
Substituting this angle into Eqs. (8.23) and (8.24), the principal
moments of inertia are
Ix 0 D
Ix Iy
Ix C Iy
C
cos 2p Ixy sin 2p
2
2
D
1.26 0.655
1.26 C 0.655
C
cos 18.63&deg; 0.102 sin 18.63&deg;
2
2
&eth; 106 in4 D 1.28 &eth; 106 in4
Iy 0 D
Ix C Iy
Ix Iy
cos 2p C Ixy sin 2p
2
2
D
1.26 0.655
1.26 C 0.655
cos 18.63&deg; C 0.102 sin 18.63&deg;
2
2
&eth; 106 in4 D 6.38 &eth; 105 in4
p D 9.32&deg; , principal moments of inertia are
1.28 &eth; 106 in4 , 6.38 &eth; 105 in4 .
646
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.92* Determine a set of principal axes and
the corresponding principal moments of inertia.
160 mm
Solution: We divide the area into 3 rectangles as shown: In terms
of the xO , yO coordinate system, the position of the centroid is
x̂ D
40 mm
x̂1 A1 C x̂2 A1 C x̂3 A3
A1 C A2 C A3
x
D
ŷ D
2040200 C 10012040 C 808040
D 56 mm,
40200 C 12040 C 8040
200 mm
40 mm
ŷ1 A1 C ŷ2 A1 C ŷ3 A3
A1 C A2 C A3
40 mm
10040200 C 18012040 C 208040
D
D 108 mm.
40200 C 12040 C 8040
120 mm
The moments and products of inertia in terms of the xO , yO system are
y
y^
(Ix D (Ix 1 C (Ix 2 C (Ix 3
D
160 mm
2
1
1
402003 C
120403 C 1802 12040
3
12
C
1
80403 D 26.5 &eth; 107 mm4 ,
3
200
mm
x
40
mm
(Iy D (Iy 1 C (Iy 2 C (Iy 3
40 mm
3
120 mm
D
1
1
200403 C
401203 C 1002 12040
3
12
C
40 mm
1
x^
y y′
1
40803 C 802 8040 D 8.02 &eth; 107 mm4 ,
12
(Ixy D (Ixy 1 C (Ixy 2 C (Ixy 3
x
12.1&deg;
x′
D 2010040200 C 10018040120
C 20804080 D 10.75 &eth; 107 mm.
The moments and product of inertia in terms of the xO , yO system are
Ix D (Ix yO 2 A D 77.91 &eth; 106 mm4 ,
Iy D (Iy xO 2 A D 30.04 &eth; 106 mm4 ,
Ixy D (Ixy x̂ŷA D 10.75 &eth; 106 mm4 ,
from Equation (8.26),
tan 2p D
2Ixy
210.75 &eth; 106 ,
D
Iy Ix
30.04 &eth; 106 77.91 &eth; 106 we obtain p D 12.1&deg; . We can orient the principal axes as shown:
Substituting the values of Ix , Iy and Ixy into Equations (8.23)
and (8.24) and setting D 12.1&deg; , we obtain
Ix1 D 80.2 &eth; 106 mm4
Iy 1 D 27.7 &eth; 106 mm4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
647
Problem 8.93 Solve Problem 8.87 by using Mohr’s
Circle.
y
3m
x
4m
Solution: The vertical 3-m dimension is increased to 4 m. From
Problem 8.87, the moments and product of inertia for the unrotated
system are
Ix D
1
4 m4 m3 D 21.3 m4 ,
12
Iy D
1
4 m3 4 m D 64 m4 ,
4
Ixy D
1
4 m2 4 m2 D 32 m4 .
8
Mohr’s circle (shown) has a center and radius given by
CD
21.3 C 64
D 42.7 m4
2
RD
21.3 64
2
2
C 322 D 38.5 m4
The angle and principal moments are now
tan2p D
32
) p D 28.2&deg; ,
64 42.7
I1 D C C R D 81.1 m4 , I2 D C R D 4.21 m4 .
p D 28.2&deg; , principal moments of inertia are 4.21 m4 , 81.1 m4
648
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.94 Solve Problem 8.88 by using Mohr’s Circle.
1 ft
3 ft
1 ft
x
4 ft
Solution: Based on Example 8.7, the moments and product of
inertia of the reoriented area are
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
For Mohr’s circle we have the center, radius, and angle
CD
10 C 22
D 16 ft4 ,
2
RD
p D
22 10
2
1
tan1
2
2
C 62 D 8.49 ft4 ,
6
22 16
D 22.5&deg;
Now we can calculate the new inertias
Ix D C R cos 15&deg; D 7.80 ft4
Iy D C C R cos 15&deg; D 24.2 ft4
Ixy D R sin 15&deg; D 2.20 ft4
Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix‘y‘ D 2.20 ft4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
649
y
Problem 8.95 Solve Problem 8.89 by using Mohr’s Circle.
1 ft
3 ft
1 ft
x
4 ft
Solution: Based on Example 8.7, the moments and product of
inertia of the reoriented area are
Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 .
For Mohr’s circle we have the center, radius, and angle
CD
10 C 22
D 16 ft4 ,
2
RD
p D
22 10
2
1
tan1
2
2
C 62 D 8.49 ft4 ,
6
22 16
D 22.5&deg;
Now we can calculate the principal moments of inertias
I1 D C C R D 24.5 ft4
I2 D C R D 7.51 ft4
p D 22.5&deg; , principal moments of inertia are 7.51 ft4 , 24.5 ft4 .
650
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.96 Solve Problem 8.90 by using Mohr’s Circle. y
y
x
u
x
Solution: For Mohr’s circle we have the center, radius, and angle
CD
12.6 C 6.55
2
RD
p D
12.6 6.55
2
1
tan1
2
&eth; 105 D 9.58 &eth; 105 in4 ,
2
C 1.022 &eth; 105 D 3.19 in4 ,
1.02
12.6 9.58
D 9.32&deg;
Now we can calculate the new inertias
Ix0 D C C R cos 22.7&deg; D 12.0 &eth; 105 in4 ,
Iy 0 D C R cos 22.7&deg; D 7.18 &eth; 105 in4 ,
Ix0 y 0 D R sin 22.7&deg; D 2.11 &eth; 105 in4 .
Ix0 D 1.20 &eth; 106 in4 , Iy 0 D 7.18 &eth; 105 in4 , Ix0 y 0 D 2.11 &eth; 105 in4 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
651
Problem 8.97 Solve Problem 8.91 by using Mohr’s Circle. y
y
x
u
x
Solution: For Mohr’s circle we have the center, radius, and angle
CD
12.6 C 6.55
2
RD
p D
12.6 6.55
2
1
tan1
2
&eth; 105 D 9.58 &eth; 105 in4 ,
2
C 1.022 &eth; 105 D 3.19 in4 ,
1.02
12.6 9.58
D 9.32&deg;
Now we can calculate the principal inertias
I1 D C C R D 12.8 &eth; 105 in4 ,
Iy 0 D C R D 6.38 &eth; 105 in4 ,
p D 9.32&deg; , principal moments of inertia are
1.28 &eth; 106 in4 , 6.38 &eth; 105 in4 .
652
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.98* Solve Problem 8.92 by using Mohr’s
circle.
Solution: The moments and product of inertia are derived in terms
of the xy coordinate system in the solution of Problem 8.92:
50
◊106
(+)
Ix D 77.91 &eth; 106 mm4
Iy D 30.04 &eth; 106 mm4
1
2θp
Ixy D 10.75 &eth; 106 mm4 .
The Mohr’s circle is: Measuring the 2p, angle we estimate that p D
12&deg; , and the principle moments of inertia are approximately 81 &eth;
106 mm4 and 28 &eth; 106 mm4 the orientation of the principal axes is
shown in the solution of Problem 8.92.
2
(30.0, –10.8)
(77.9,10.8)
(+)
100
6
◊10
- 50
◊106
Problem 8.99 Derive Eq. (8.22) for the product of
inertia by using the same procedure we used to derive
Eqs. (8.20) and (8.21).
Solution: Suppose that the area moments of inertia of the area A
x′
are known in the coordinate system x, y,
y
2
y dA,
Ix D
y′
A
A
0
x 2 dA,
Iy D
x
A
and Ixy D
xyA.
A
The objective is to find the product of inertia in the new coordinate
system x 0 , y 0 in terms of the known moments of inertia. The new
x 0 , y 0 system is formed from the old x, y system by rotation about
the origin through a counterclockwise angle .
By definition,
Substitute into the definition:
Ix0 y 0 D cos2 sin2 xy dA
A
A
0 0
x y dA.
Ix 0 y 0 D
x 2 dA ,
y 2 dA C cos sin A
from which
A
From geometry,
x 0 D x cos C y sin ,
Ix0 y 0 D cos2 sin2 Ixy C Ix Iy sin cos ,
which is the expression required.
and y 0 D x sin C y cos .
The product is
x0 y 0 D xy cos2 xy sin2 C y 2 cos sin x2 cos sin .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
653
Problem 8.100 The axis LO is perpendicular to both
segments of the L-shaped slender bar. The mass of the
bar is 6 kg and the material is homogeneous. Use the
method described in Example 8.10 to determine is moment
Solution: Use Example 8.10 as a model for this solution.
1m
LO
2m
Introduce the coordinate system shown and divide the bar into two
parts as shown
y=1
y
dy
2
r
dx
1
0
2
r 2 dm D
I0 1 D
Ax 2 dx D A
0
I0 1 D
x
2
x
x3
3
2
0
8
A
3
However m1 D Al1 D 2A.
Since part 1 is 2/3 of the length, its mass is 2/36 kg D 4 kg. Part 2
has mass 2 kg.
For part 2, dm D A dy and
22 C y 2
rD
1
r 2 dm D
I0 2 D
I0 2 D A4y C
I0TOTAL D
A22 C y 2 dy
0
m2
y3
3
1
D A
0
13
3
13
8
21
A C A D
A
3
3
3
I0 TOTAL D 7A
The total mass D 3A D 6 kg
I0TOTAL D
654
7
6 kg &ETH; m2 D 7 kg m2
6
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.101 Two homogenous slender bars, each
of mass m and length l, are welded together to form
the T-shaped object. Use integration to determine the
moment of inertia of the object about the axis through
point O that is perpendicular to the bars.
l
O
l
Solution: Divide the object into two pieces, each corresponding
to a slender bar of mass m; the first parallel to the y axis, the second
to the x axis. By definition
l
r 2 dm C
ID
0
r 2 dm.
m
For the first bar, the differential mass is dm D A dr. Assume that
the second bar is very slender, so that the mass is concentrated at a
distance l from O. Thus dm D A dx, where x lies between the limits
l
l
x .
2
2
The distance to a differential dx is r D
becomes
l
I D A
r 2 dr C A
l
2
0
D A
r3
3
D ml2
l
2
p
l2 C x 2 . Thus the definition
l2 C x 2 dx I
l/2
x3
C A l2 x C
3 l/2
0
l
1
1
C1C
3
12
D
17 2
ml
12
Problem 8.102 The slender bar lies in the x –y plane.
Its mass is 6 kg and the material is homogeneous. Use
integration to determine its moment of inertia about the
z axis.
y
2m
50
x
1m
Solution:
1
m
2
Iz D
6 kg
D 2 kg/m
The density is D
3m
x 2 dx
y
0
C
m
2m
[1 m C s cos 50&deg; 2 C s sin 50&deg; 2 ]ds
0
50&deg;
Iz D 15.14 kg m2
Problem 8.103 Use integration to determine the
moment of inertia of the slender bar in Problem 8.102
x
1m
Solution: See solution for 8.102
Iy D
0
1
m
2
x 2 dx C
m
1 m C s cos 50&deg; 2 ds D 12.01 kg m2
0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
655
y
Problem 8.104 The homogeneous thin plate has mass
m D 12 kg and dimensions b D 2 m and h D 1 m. Use
the procedure described in Active Example 8.9 to
determine the moments of inertia of the plate about the
x and y axes.
h
x
Solution: From Appendix B, the moments of inertia about the x
and y axes are
1 3
1 3
bh , Iy D
hb .
Ix D
36
36
b
Therefore the moments of inertia of the plate about the x and y axes
are
1 3
m
m
1
1
b D
mh2 D 12 kg1 m2 D 0.667 kg-m2
Ixaxis D Ix 1
A
36
18
8
bh
2
Iyaxis D
m
m
Iy D
1
A
bh
2
1 3
hb
36
D
1
1
mb2 D
12 kg2 m2 D 2.67 kg-m2
18
18
Ixaxis D 0.667 kg-m2 , Iyaxis D 2.67 kg-m2 .
y
Problem 8.105 The homogenous thin plate is of
uniform thickness and mass m.
(a)
(b)
Determine its moments of inertia about the x and z
axes.
Let Ri D 0, and compare your results with the
values given in Appendix C for a thin circular
plate.
Ro
Ri
x
Solution:
(a)
(b)
The area moments of inertia for a circular area are Ix D Iy D
R4
. For the plate with a circular cutout, Ix D Ro4 Ri4 . The
4
4
m
area mass density is , thus for the plate with a circular cut,
A
m
m
,
from
which the moments of inertia
D
2
A
Ro2 Ri Ri
Ro
m
mRo4 Ri4 D Ro2 C Ri2 4
4Ro2 Ri2 Ix
axis
D
Iz
axis
D 2Ix
axis
D
m 2
R C Ri2 .
2 o
Let Ri D 0, to obtain
Ix
axis
D
m 2
R ,
4 o
Iz
axis
D
m 2
R ,
2 o
which agrees with table entries.
656
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.106 The homogenous thin plate is of
uniform thickness and weighs 20 lb. Determine its
moment of inertia about the y axis.
1
y = 4 – – x 2 ft
4
Solution:
y D4
1 2
x ft
4
x
y
The plate’s area is
4
4
4
1 2
x
4
dx D 21.3 ft2 .
y=4–
1 2
x ft
4
The plate’s density per unit area is
υ D 20/32.2/21.3 D 0.0291 slug/ft2 .
x
–4 ft
x
The moment of inertia about the y axis is
Iy
axis
D
4 ft
dx
1
x 2 υ 4 x 2 dx
4
4
4
D 1.99 slug-ft2 .
Problem 8.107 Determine the moment of inertia of the
plate in Problem 8.106 about the x axis.
Solution: See the solution of Problem 8.106. The mass of the strip
so the moment of inertia of the plate about the x axis is
element is
mstrip
1
D υ 4 x2
4
dx.
Ix
axis
4
D
4
3
1
1
dx D 2.27 slug-ft2 .
υ 4 x2
3
4
The moment of inertia of the strip about the x axis is
Istrip D
D
2
1
1
mstrip 4 x 2
3
4
3
1
1
υ 4 x2
dx,
3
4
Problem 8.108 The mass of the object is 10 kg. Its
moment of inertia about L1 is 10 kg-m2 . What is its
moment of inertia about L2 ? (The three axes lie in the
same plane.)
Solution: The strategy is to use the data to find the moment of
inertia about L, from which the moment of inertia about L2 can be
determined.
IL D 0.62 10 C 10 D 6.4 m2 ,
0.6 m
L
from which IL2 D 1.22 10 C 6.4 D 20.8 m2
0.6 m
L1
L2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
657
Problem 8.109 An engineer gathering data for the
design of a maneuvering unit determines that the
astronaut’s center of mass is at x D 1.01 m, y D 0.16 m
and that her moment of inertia about the z axis is
105.6 kg-m2 . Her mass is 81.6 kg. What is her moment
of inertia about the z0 axis through her center of mass?
y
y
x
x
Solution: The distance d from the z axis to the z0 axis is
1.012 C 0.162
dD
D 1.0226 m.
From the parallel-axis theorem,
D Iz0
axis
C d2 m :
105.6 D Iz0
axis
C 1.02262 81.6.
Iz
axis
Solving, we obtain
Iz0
D 20.27 kg-m2 .
axis
Problem 8.110 Two homogenous slender bars, each of
mass m and length l, are welded together to form the Tshaped object. Use the parallel axis theorem to determine
the moment of inertia of the object about the axis through
point O that is perpendicular to the bars.
l
O
l
Solution: Divide the object into two pieces, each corresponding
to a bar of mass m. By definition
l
ID
r 2 dm.
0
For the first bar, the differential mass is dm D A dr, from which the
moment of inertia about one end is
l
I1 D A
r 2 dr D A
0
r3
3
l
D
0
ml2
.
3
For the second bar
I2 D A
l
2
l
2
r3
r dr D A
3
2
2l
l
2
D
ml2
12
is the moment of inertia about the center of the bar. From the parallel
axis theorem, the moment of inertia about O is
Io D
658
17 2
ml2
ml2
C l2 m C
D
ml .
3
12
12
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.111 Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object in
Problem 8.110 about the axis through the center of mass
of the object that is perpendicular to the two bars. (See
Active Example 8.11.)
Solution:
The location of the center of mass of the object is x D
\$ %
l
2
C lm
3
D l. Use the results of Problem 8.110 for the moment
2m
4
of inertia of a bar about its center. For the first bar,
m
I1 D
2
l
ml2
7 2
mC
D
ml .
4
12
48
For the second bar,
I2 D
2
l
ml2
7 2
mC
D
ml .
4
12
48
The composite:
Ic D I1 C I2 D
7 2
ml
24
y
Problem 8.112 The mass of the homogenous slender
bar is 20 kg. Determine its moment of inertia about the
z axis.
y'
x'
Solution: Divide the object into three segments. Part (1) is the 1 m
bar on the left, Part (2) is the 1.5 m horizontal segment, and Part (3)
is the segment on the far right. The mass density per unit length is
1m
x
20
m
p D 5.11 kg/m.
D
D
L
1 C 1.5 C 2
1.5 m
1m
The moments of inertia about the centers of mass and the distances to
the centers of mass from the z axis are:
Part (1)
I1 D l31
12
D m1
l21
D 0.426 kg-m2 ,
12
m1 D 5.11 kg,
d1 D 0.5 m,
Part (2),
I2 D l2
l32
D m2 2 D 1.437 kg- m2 ,
12
12
m2 D 7.66 kg,
d2 D
Part (3)
p
0.752 C 12 D 1.25 m
I3 D p
22
l33
D m3
D 1.204 kg-m2 ,
12
12
m3 D 7.23 kg,
d3 D
p
22 C 0.52 D 2.062 m.
The composite:
I D d21 m1 C I1 C d22 m2 C I2 C d23 m3 C I3 D 47.02 kg-m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
659
Problem 8.113 Determine the moment of inertia of the
bar in Problem 8.112 about the z0 axis through its center
of mass.
Solution: The center of mass:
xD
D
yD
D
x 1 m 1 C x2 m 2 C x3 m 3
20
0 C 0.757.66 C 27.23
D 1.01 m.
20
0.5m1 C 1m2 C 0.5m3
20
0.55.11 C 17.66 C 0.57.23
D 0.692 m.
20
The distance from the z axis to the center of mass is d D x2 C y2 D
1.224 m. The moment of inertia about the center o mass:
Ic D d2 20 C Io
D 17.1 kg-m2
Problem 8.114 The homogeneous slender bar weighs
5 lb. Determine its moment of inertia about the z axis.
y
y'
Solution: The Bar’s mass is m D 5/32.2 slugs. Its length is
L D L1 C L 2 C L 3 D 8 C
4 in
p
82 C 82 C 4 D 31.9 in.
The masses of the parts are therefore,
L1
m1 D
mD
L
m2 D
L2
mD
L
m3 D
L3
mD
L
8
31.9
5
32.2
x'
D 0.0390 slugs,
p
264
5
D 0.0551 slugs,
31.9
32.2
4
31.9
5
32.2
x
8 in
D 0.0612 slugs.
y′
y
The center of mass of part 3 is located to the right of its center C a
distance 2R/ D 24/ D 2.55 in. The moment of inertia of part 3
2
C
r 2 dm D m3 r 2 D 0.061242 D 0.979 slug-in2 .
m3
1
The moment of inertia of part 3 about the center of mass of part 3 is
therefore
4 in
x′
3
x
8 in
I3 D 0.979 m3 2.552 D 0.582 slug-in2 .
The moment of inertia of the bar about the z axis is
Iz
axis
D 13 m1 L12 C 13 m2 L22 C I3 C m3 [8 C 2.552 C 42 ]
D 11.6 slug-in2 D 0.0802 slug-ft2 .
660
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.115 Determine the moment of inertia of the
bar in Problem 8.114 about the z0 axis through its center
of mass.
Solution: In the solution of Problem 8.114, it is shown that the
moment of inertia of the bar about the z axis is Iz axis D 11.6 slug-in2 .
The x and y coordinates of the center of mass coincide with the
centroid of the axis:
xD
x1 L1 C x2 L2 C x3 L3
L1 C L2 C L3
p
24
4
48 C 4 82 C 82 C 8 C
p
D
D 6.58 in,
8 C 82 C 82 C 4
yD
D
y1 L1 C y2 L2 C y3 L3
L1 C L2 C L3
p
0 C 4 82 C 82 C 44
p
D 3.00 in.
8 C 82 C 82 C 4
The moment of inertia about the z axis is
Iz0
axis
D Iz
axis
x2 C y2 5
32.2
D 3.43 slug-in2 .
y
Problem 8.116 The rocket is used for atmospheric
research. Its weight and its moment of inertia about the
z axis through its center of mass (including its fuel) are
10 kip and 10,200 slug-ft2 , respectively. The rocket’s
fuel weighs 6000 lb, its center of mass is located at
x D 3 ft, y D 0, z D 0, and the moment of inertia of
the fuel about the axis through the fuel’s center of mass
parallel to z is 2200 slug-ft2 . When the fuel is exhausted,
what is the rocket’s moment of inertia about the axis
through its new center of mass parallel to z?
Solution: Denote the moment of inertia of the empty rocket as IE
about a center of mass xE , and the moment of inertia of the fuel as IF
about a mass center xF . Using the parallel axis theorem, the moment
of inertia of the filled rocket is
x
From which
xE D 186.335
124.224
3 D 4.5 ft
2m ,
IR D IE C xE2 mE C IF C xF
F
is the new location of the center of mass. Substitute:
about a mass center at the origin (xR D 0). Solve:
2m
IE D IR xE2 mE IF xF
F
2m .
IE D IR xE2 mE IF xF
F
D 10200 2515.5 2200 1677.01
The objective is to determine values for the terms on the right from
the data given. Since the filled rocket has a mass center at the origin,
the mass center of the empty rocket is found from 0 D mE xE C mF xF ,
from which
D 3807.5 slug-ft2
xE D mF
mE
xF .
Using a value of g D 32.2 ft/s2 ,
mF D
WF
6000
D
D 186.34 slug,
g
32.2
mE D
10000 6000
WR WF D
D 124.23 slug.
g
32.2
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661
Problem 8.117 The mass of the homogenous thin plate
is 36 kg. Determine its moment of inertia about the x
axis.
y
0.4 m
0.4 m
Solution: Divide the plate into two areas: the rectangle 0.4 m by
0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The
m
mass density is D . The area is
A
0.3 m
A D 0.40.6 C 0.40.3 D 0.36 m2 ,
0.3 m
x
from which
D
36
D 100 kg/m2 .
0.36
The moment of inertia about the x axis is
Ix
axis
D
\$1%
3
0.40.63 C \$1%
3
0.40.33 D 3.24 kg-m2
Problem 8.118 Determine the moment of inertia of the
plate in Problem 8.117 about the z axis.
Solution: The basic relation to use is
Iz
axis
D Ix
axis
C Iy
axis .
The value of Ix axis is given in the solution of Problem 8.117. The
moment of inertia about the y axis using the same divisions as in
Problem 8.117 and the parallel axis theorem is
Iy
axis
D
1
1
0.60.43 C 0.30.43
3
12
C 0.62 0.30.4 D 5.76 kg-m2 ,
from which
Iz
axis
D Ix
axis
C Iy
axis
D 3.24 C 5.76 D 9 kg-m2.
y
Problem 8.119 The homogenous thin plate weighs
10 lb. Determine its moment of inertia about the x axis.
5 in
5 in
Solution: Divide the area into two parts: the lower rectangle 5 in
by 10 in and the upper triangle 5 in base and 5 in altitude. The mass
W
. The area is
density is D
gA
A D 510 C
\$1%
2
Using g D 32 ft/s2 , the mass density is
D
10 in
55 D 62.5 in2 .
W
D 0.005 slug/in2 .
gA
5 in
x
Using the parallel axis theorem, the moment of inertia about the x axis
is
Ix
axis
D
1
1
1053 C 553 3
36
5 2 1
C 5C
55 D 4.948 slug-in2
3
2
Ix
662
axis
D 0.03436 slug-ft2
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Problem 8.120 Determine the moment of inertia of the
plate in Problem 8.119 about the y axis.
Solution: Use the results of the solution in Problem 8.119 for the
area and the mass density.
Iy
axis
D
1
1
5103 C 553 3
36
10 2 1
C 5C
55
3
2
D 12.76 slug-in2 D 0.0886 slug-ft2
Problem 8.121 The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a
homogenous, thin rectangular plate. Its mass is 5 slugs.
Determine its moment of inertia about the x, y, and
z axes.
y
3 ft
6 ft
3 ft
Solution: The area is A D 93 D 27 ft2 . The mass density is
D
5
m
D
D 0.1852 slugs/ft2 .
A
27
2 ft
x
The moment of inertia about the centroid of the rectangle is
Ixc D Iyc D 1
12
1
12
933 D 3.75 slug-ft2 ,
393 D 33.75 slug-ft2 .
Use the parallel axis theorem:
Ix
axis
D A2 C 1.52 C Ixc D 65 slug-ft2 ,
Iy
axis
D A4.5 32 C Iyc D 45 slug-ft2 .
Iz
axis
D Ix
axis
C Iy
axis
D 110 slug-ft2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
663
Problem 8.122 The homogeneous cylinder has mass m,
length l, and radius R. Use integration as described in
Example 8.13 to determine its moment of inertia about
the x axis.
y
x
Solution: The volume of the disk element is R2 dz and its mass is
dm D R2 dz, where is the density of the cylinder. From Appendix
C, the moment of inertia of the disk element about the x0 axis is
dIx0 axis D
1
1
dmR2 D R2 dzR2 .
4
4
R
l
Applying the parallel-axis theorem, the moment of inertia of the disk
element about the x axis is
dIxaxis
z
1
D dIx0 axis C z2 dm D R2 dzR2 C z2 R2 dz
4
Integrating this expression from z D 0 to z D l gives the moment of
inertia of the cylinder about the x axis.
l
Ixaxis D
0
1
1
1
R4 C R2 z2 dz D R4 l C R2 l3 .
4
4
3
In terms of the mass of the cylinder m D R2 l,
Ixaxis D
1 2 1 2
mR C ml
4
3
Problem 8.123 The homogenous cone is of mass m.
Determine its moment of inertia about the z axis, and
compare your result with the value given in Appendix C.
(See Example 8.13.)
y
x
Solution: The differential mass
dm D
R
3m
m
r 2 dz D 2 r 2 dz.
V
R h
The moment of inertia of this disk about the z axis is 12 mr 2 . The radius
varies with z,
rD
h
z
R
z,
h
from which
Iz
664
axis
D
3mR2
2h5
h
0
z4 dz D
3mR2
2h5
z5
5
h
D
0
3mR2
10
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Problem 8.124 Determine the moments of inertia of
the homogenous cone in Problem 8.123 about the x and
y axes, and compare your results with the values given
in Appendix C.
3m
m
D
. The differential
V
R2 h
element of mass is dm D r 2 dz. The moment of inertia of this
elemental disk about an axis through its center of mass, parallel to
the x- and y-axes, is
Solution: The mass density is D
\$1%
dIx D
4
r 2 dm.
Use the parallel axis theorem,
\$1%
Ix D
m
4
r 2 dm C
z2 dm.
m
Noting that r D
R
z, then
h
r 2 dm D and z2 dm D R4
h4
R2
h2
z4 dz,
z4 dz.
Substitute:
Ix D R4
4h4
h
z4 dz C 0
R2
h2
h
z4 dz.
0
Integrating and collecting terms:
Ix D
3mR2
3m
C 3
4h5
h
z5
5
h
D m
0
3 2 3 2
R C h .
20
5
By symmetry, Iy D Ix
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665
Problem 8.125 The mass of the homogeneous wedge
is m. Use integration as described in Example 8.13 to
determine its moment of inertia about the z axis. (Your
answer should be in terms of m, a, b, and h.)
y
x
Solution: Consider a triangular element of the wedge of thickness
dz. The mass of the element is the product of the density of the
material and the volume of the element, dm D 12 bhdz. The moments
of inertia of the triangular element about the x’ and y’ axes are given
by Eqs. (8.30) and (8.31) in terms of the mass of the element, its
triangular area, and the moments of inertia of the triangular area:
dIx0 axis
1
bhdz 1
dm 0
1
2
I D
rbh3 dz,
bh3 D
D
1
A x
12
12
bh
2
dIy 0 axis
1
bhdz 1
dm 0
1
I D 2
hb3 D rhb3 dz,
D
1
A y
4
4
bh
2
h
a
z
b
The moment of inertia of this thin plate about the z axis is
dIzaxis D dIx0 axis C dIy 0 axis D
1
1
bh3 dz C hb3 dz.
12
4
Integrating this expression from z D 0 to z D a gives the moment of
inertia of the wedge about the z axis:
a
Izaxis D
0
1
1
1
1
bh3 C hb3 dz D
bh3 a C hb3 a.
12
4
12
4
In terms of the mass m D 12 bha,
Izaxis D
666
1 2 1 2
mh C mb .
6
2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.126 The mass of the homogeneous wedge
is m. Use integration as described in Example 8.13 to
determine its moment of inertia about the x axis. (Your
answer should be in terms of m, a, b, and h.)
y
x
Solution: Consider a triangular element of the wedge of thickness
dz. The mass of the element is the product of the density of the
material and the volume of the element, dm D 12 bhdz. The moments
of inertia of the triangular element about the x’ axis is given by Eq.
(8.30) in terms of the mass of the element, its triangular area, and the
moments of inertia of the triangular area:
dIx0 axis
h
a
z
b
1
bhdz 1
dm
1
Ix 0 D 2
rbh3 dz,
bh3 D
D
1
A
36
36
bh
2
Applying the parallel-axis theorem, the moment of inertia of the
triangular element about the x axis is
1
dIxaxis D dIx0 axis C z2 C h2 dm
3
D
1
1
1
1
1
bh3 dz C [z2 C h2 ] bhdz D
bh3 dz C bhz2 dz
36
3
2
12
2
Integrating this expression from z D 0 to z D a gives the moment of
inertia of the wedge about the x axis:
a
Ixaxis D
0
1
1
1
1
bh3 C bhz2 dz D
bh3 a C bha3 .
12
2
12
6
In terms of the mass m D 12 bha,
Ixaxis D
1 2 1 2
mh C ma .
6
3
Problem 8.127 In Example 8.12, suppose that part of
the 3-kg bar is sawed off so that the bar is 0.4 m long
and its mass is 2 kg. Determine the moment of inertia
of the composite object about the perpendicular axis L
through the center of mass of the modified object.
Solution: The mass of the disk is 2 kg. Measuring from the left
0.2 m
0.6 m
L
end of the rod, we locate the center of mass
xD
2 kg0.2 m C 2 kg0.6 m
D 0.4 m.
2 kg C 2 kg
The center of mass is located at the point where the rod and disk are
connected. The moment of inertia is
'
&amp;
1
1
I D 2 kg0.4 m2 C
2 kg0.2 m2 C 2 kg0.2 m2
3
2
I D 0.227 kg-m2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
667
y
Problem 8.128 The L-shaped machine part is composed of two homogeneous bars. Bar 1 is tungsten alloy
with mass density 14,000 kg/m3 , and bar 2 is steel
with mass density 7800 kg/m3 . Determine its moment
of inertia about the x axis.
240 mm
1
40 mm
2
80 mm
Solution: The masses of the bars are
m1 D 14,0000.240.080.04 D 10.75 kg
80 mm
z
240 mm
m2 D 78000.240.080.04 D 5.99 kg.
Using Appendix C and the parallel axis theorem the moments of inertia
of the parts about the x axis are
Ix
axis1
D
1
m1 [0.042 C 0.242 ] C m1 0.122 D 0.2079 kg-m2 ,
12
Ix
axis2
D
1
m2 [0.042 C 0.082 ] C m2 0.042 D 0.0136 kg-m2 .
12
x
Therefore
Ix
668
axis
D Ix
axis1
C Ix
axis2
D 0.221 kg-m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.129 The homogeneous object is a cone
with a conical hole. The dimensions R1 D 2 in, R2 D
1 in, h1 D 6 in, and h2 D 3 in. It consists of an
aluminum alloy with a density of 5 slug/ft3 . Determine
its moment of inertia about the x axis.
y
x
R1
Solution: The density of the material is
D 5 slug/ft3
1 ft
12 in
3
D 0.00289 slug/in3 .
The volume of the conical object without the conical hole is
V1 D
R2
h1
z
h2
1 2
1
R h1 D 2 in2 6 in D 25.1 in3 .
3 1
3
The mass of the conical object without the conical hole is m1 D V1 D
0.0727 slug. From Appendix C, the moment of inertia of the conical
object without the conical hole about the x axis is
Ix 1 D m1
3 2
3 2
h C
R
5 1 20 1
D 0.0727 slug
3
3
6 in2 C
2 in2 D 1.61 slug-in2
5
20
The volume of the conical hole is
V2 D
1 2
1
R h2 D 1 in2 3 in D 3.14 in3 .
3 2
3
The mass of the material that would occupy the conical hole is m2 D
V2 D 0.00909 slug. The z coordinate of the center of mass of the
material that would occupy the conical hole is
z D h1 h2 C
3
3
h2 D 6 in 3 in C 3 in D 5.25 in.
4
4
Using Appendix C and applying the parallel-axis theorem, the moment
of inertia about the x axis of the material that would occupy the conical
hole is
Ix 2 D m2 3 2
3 2
h C
R C z2 m2 D 0.255 slug-in2 .
80 2 20 2
The moment of inertia of the conical object with the conical hole is
Ix D Ix 1 Ix 2 D 1.36 slug-in2 .
Ix D 1.36 slug-in2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
669
Problem 8.130 The circular cylinder is made of
aluminum (Al) with density 2700 kg/m3 and iron (Fe)
with density 7860 kg/m3 . Determine its moments of
inertia about the x 0 and y 0 axes.
y
y
Al
z
Fe
600 mm
200 mm
z
Solution: We have Al D 2700 kg/m3 , Fe D 7860 kg/m3
600 mm
x, x
We first locate the center of mass
xD
Al [0.1 m]2 [0.6 m]0.3 m C Fe [0.1 m]2 [0.6 m]0.9 m
Al [0.1 m]2 [0.6 m] C Fe [0.1 m]2 [0.6 m]
D 0.747 m
We also have the masses
mAl D Al 0.1 m2 0.6 m, mFe D Fe 0.1 m2 0.6 m
Now find the moments of inertia
1
1
mAl 0.1 m2 C mFe 0.1 m2 D 0.995 kg m2
2
2
[0.1 m]2
[0.6 m]2
C
D mAl
C mAl x 0.3 m2
12
4
[0.1 m]2
[0.6 m]2
C mFe
C
C mFe 0.9 m x2
12
4
D 20.1 kg m2
Ix 0 D
Iy 0
Problem 8.131 The homogenous half-cylinder is of
mass m. Determine its moment of inertia about the axis
L through its center of mass.
Solution:
The centroid of the half cylinder is located a distance
R
L
T
4R
from the edge diameter. The strategy is to use the parallel
3
axis theorem to treat the moment of inertia of a complete cylinder as
the sum of the moments of inertia for the two half cylinders. From
Problem 8.118, the moment of inertia about the geometric axis for a
cylinder is IcL D mR2 , where m is one half the mass of the cylinder.
of
By the parallel axis theorem,
IcL D 2
4R
3
2
m C IhL
.
Solve
IhL D
IcL
2
D mR2
D mR2
670
4R
3
2 2 16
mR
mR2
m D
2
2
9
16
1
2
92
16
1
2
92
D 0.31987 mR2 D 0.32 mR2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.132 The homogeneous machine part is
made of aluminum alloy with density D 2800 kg/m3 .
Determine its moment of inertia about the z axis.
y
y
20 mm
x
z
40 mm
120 mm
Solution: We divide the machine part into the 3 parts shown: (The
dimension into the page is 0.04 m). The masses of the parts are
y
m1 D 28000.120.080.04 D 1.075 kg,
m2 D 2800 12 0.042 0.04 D 0.281 kg,
40
mm
0.12
m
1
0.08
m
x
m3 D 28000.022 0.04 D 0.141 kg.
y
Iz
axis1
0.12 m
C 2
0.12 m
1
m1 [0.082 C 0.122 ] C m1 0.062
D
12
The moment of inertia of part 2 about the axis through the center C
that is parallel to the z axis is
x
0.02 m
Iz
axis2
D 0.000144 C m2 0.12 C 0.0172 D 0.00543 kg-m2 .
The moment of inertia of the material that would occupy the hole 3
D 12 m2 0.042 .
The distance along the x axis from C to the center of mass of part 2 is
40.04/3 D 0.0170 m.
Iz
axis3
Therefore Iz
Therefore, the moment of inertia of part 2 about the z axis through its
center of mass that is parallel to the axis is
1
2
2 m2 0.04
3
x –
0.04 m
Using this result, the moment of inertia of part 2 about the z axis is
D 0.00573 kg-m2 .
1
2
2 m2 R
y
+
Using Appendix C and the parallel axis theorem the moment of inertia
of part 1 about the z axis is
axis
D 12 m3 0.022 C m3 0.122 D 0.00205 kg-m2 .
D Iz
axis1
C Iz
axis2
Iz
axis3
D 0.00911 kg-m2 .
m2 0.01702 D 0.000144 kg-m2 .
Problem 8.133 Determine the moment of inertia of the
machine part in Problem 8.132 about the x axis.
Solution: We divide the machine part into the 3 parts shown in
the solution to Problem 8.132. Using Appendix C and the parallel axis
theorem, the moments of inertia of the parts about the x axis are:
1
m1 [0.082 C 0.042 ] D 0.0007168 kg-m2
12
Ix
axis1
D
Ix
axis2
D m2
1
1
0.042 C 0.042
12
4
D 0.0001501 kg-m2
Ix
axis3
D m3
1
1
0.042 C 0.022
12
4
D 0.0000328 kg-m2 .
Therefore, Ix
axis
D Ix
axis1
C Ix
axis2
Ix
axis3
D 0.000834 kg-m2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
671
Problem 8.134 The object consists of steel of density
D 7800 kg/m3 . Determine its moment of inertia about
the axis LO .
20 mm
O
100 mm
Solution: Divide the object into four parts: Part (1) The semicylinder of radius R D 0.02 m, height h1 D 0.01 m.
Part (2): The rectangular solid L D 0.1 m by h2 D 0.01 m by w D
0.04 m. Part (3): The semi-cylinder of radius R D 0.02 m, h1 D
0.01 m. Part (4) The cylinder of radius R D 0.02 m, height h D
0.03 m.
10 mm
30 mm
LO
Part (1)
m1 D
I1 D
R2 h1
D 0.049 kg,
2
m1 R 2
D 4.9 &eth; 106 kg-m2 ,
4
Part (2):
m2 D wLh2 D 0.312 kg,
I2 D
1
12
m2 L 2 C w2 C m2
2
L
D 0.00108 kg-m2 .
2
Part (3)
m3 D m1 D 0.049 kg,
I3 D 4R
3
2
4R 2
m2 C I 1 C m 3 L D 0.00041179 kg m2 .
3
Part (4)
m4 D R2 h D 0.294 kg,
I4 D
\$1%
2
m4 R2 C m4 L 2 D 0.003 kg m2 .
The composite:
ILo D I1 C I2 I3 C I4 D 0.003674 kg m2
Problem 8.135 Determine the moment of inertia of
the object in Problem 8.134 about the axis through the
center of mass of the object parallel to LO .
Solution: The center of mass is located relative to LO
4R
4R
m1 C m2 0.05 m3 0.1 C m4 0.1
3
3
xD
m1 C m 2 m 3 C m 4
D 0.066 m,
Ic D x2 m C ILo D 0.00265 C 0.00367 D 0.00102 kg m2
672
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.136 The thick plate consists of steel of
density D 15 slug/ft3 . Determine its moment of inertia
y
y
4 in
2 in
2 in
x
Solution: Divide the object into three parts: Part (1) the rectangle
8 in by 16 in, Parts (2) &amp; (3) the cylindrical cut outs. Part (1):
m1 D 8164 D 4.444 slugs.
I1 D
1
12
z
4 in
4 in
8 in
4 in
4 in
m1 162 C 82 D 118.52 slug in2 .
Part (2):
m2 D 22 4 D 0.4363 slug,
I2 D
m2 22 C m2 42 D 7.854 slug in2 .
2
Part (3):
m3 D m2 D 0.4363 slugs,
I3 D I2 D 7.854 slug-in2 .
The composite:
Iz
axis
D I1 2I2 D 102.81 slug-in2
Iz
axis
D 0.714 slug-ft2
Problem 8.137 Determine the moment of inertia of the
plate in Problem 8.136 about the x axis.
Solution: Use
Problem 8.136.
the
same
divisions
of
the
object
as
in
Part (1):
I1x
axis
D
1
12
m1 82 C 42 D 29.63 slug-in2 ,
Part (2):
I2x
axis
D
1
12
m2 322 C 42 D 1.018 slug-in2 .
The composite:
Ix
axis
D I1x
axis
2I2x
axis
D 27.59 slug in2
D 0.1916 slug ft2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
673
Problem 8.138 Determine Iy and ky .
y
(1, 1)
Solution:
1
dA D dx dyA D
dx
Iy
D
A
1
x 2 dx D .
3
x2
x 2 dx
1
dy D
0
0
A
ky D
0
1
x 2 dA D
Iy D
1
dy D
0
0
y = x2
x2
x 4 dx D
0
x5
5
1
D
0
1
5
x
3
5
Problem 8.139 Determine Ix and kx .
Solution: (See figure in Problem 8.138.) dA D dx dy,
kx D
[x7 ]10 D
y 2 dy D
0
0
1
21
x2
dx
A
D
1
y 2 dA D
Ix D
1
3
1
x 6 dx
0
1
21
Ix
1
D p
A
7
Problem 8.140 Determine JO and kO .
Solution: (See figure in Problem 8.138.)
JO D Ix C Iy D
kO D
1
26
1
C
D
,
5
21
105
kx2 C ky2 D
1
3
C D
5
7
26
35
Problem 8.141 Determine Ixy .
Solution: (See figure in Problem 8.138.) dA D dx dy
D
674
1
2
0
1
x2
x dx
0
A
1
xy dA D
Ixy D
x 5 dx D
y dy
0
1 6 1
1
[x ]0 D
12
12
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.142 Determine Iy and ky .
Solution: By definition,
y
y = x – 1– x 2
4
x 2 dA.
Iy D
A
The element of area is dA D dx dy. The limits on the variable x are
0 x 4. The area is
4
dy D
0
0
4
Iy D
x3
x2
2
12
xx2 /4
x 2 dx
4
0
x5
x4
4
20
4
D 2.6667
0
x
dy D
0
D
xx2 /4
dx
x
0
x2
4
x 2 dx
4
D 12.8
0
from which
ky D
Iy
D 2.19
A
Problem 8.143 Determine Ix and kx .
Solution: By definition,
y 2 dA,
Ix D
A
from which
4
0
Ix D
xx2 /4
dx
Ix D
y 2 dy D
0
4 3
1
x2
dx
x
3
4
0
4
4
1
3 6
x7
x
3 5
D 0.6095.
x C
x 3
4
20
96
448 0
From Problem 8.142,
A D 2.667, kx D
Ix
D 0.4781
A
Problem 8.144 Determine Ixy .
Solution:
Ixy D
xy dA,
A
4
x dx
D
0
xx2 /4
y dy
0
D
4 2
1
x2
x dx
x
2
4
0
D
4
4
x
x5
x6
1
C
D 2.1333
2
4
10
96 0
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
675
Problem 8.145 Determine Iy 0 and ky 0 .
Solution: The limits on the variable x are 0 x 4. By definition,
y dA D
Ay D
4
y dy
0
4 2
1
x2
D
dx
x
2
4
0
D
y'
y = x – 1– x 2
4
xx2 /4
dx
0
A
y
x'
x
3
4
x4
x5
1
x
C
D 1.06667.
2
3
8
80 0
From Problem 8.142 the area is A D 2.667, from which y D 0.3999 D
0.4. Similarly,
4
XX2 /4
x dx
Ax D
0
dy
0
4
D
0
3
4
x2
x
x4
x x
D 5.3333,
dx D
4
3
16 0
from which x D 1.9999 D 2. The area moment of inertia is Iyy D
x2 A C Iy . Using the result of Problem 8.142, Iy D 12.8, from which
the area moment of inertia about the centroid is
Iy 0 D 10.6666 C 12.8 D 2.133
and ky 0 D
Iy 0
D 0.8944
A
Problem 8.146 Determine Ix0 and kx0 .
Solution: Using the results of Problems 8.143 and 8.145, Ix D
0.6095 and y D 0.4. The area moment of inertia about the centroid is
Ix0 D y2 A C Ix D 0.1828
and k D
x0
Ix 0
D 0.2618
A
Problem 8.147 Determine Ix0 y 0 .
Solution: From Problems 8.143 and 8.144, Ixy D 2.133 and x D
2, y D 0.4. The product of the moment of inertia about the centroid is
Ix0 y 0 D xyA C Ixy D 2.133 C 2.133 D 0
676
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
y
Problem 8.148 Determine Iy and ky .
Solution: Divide the section into two parts: Part (1) is the upper
rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm
by 40 mm.
Part (1) A1 D 0.0400.200 D 0.008 m2 ,
y1 D 0.180 m
40
mm
160
mm
x1 D 0,
Iy1 D
1
12
0.040.23 D 2.6667 &eth; 105 m4 .
x
80
mm
40
mm
80
mm
Part (2): A2 D 0.040.16 D 0.0064 m2 ,
y2 D 0.08 m,
x2 D 0,
Iy2 D
1
12
0.160.043 D 8.5 &eth; 107 m4 .
The composite:
A D A1 C A2 D 0.0144 m2 ,
Iy D Iy1 C Iy2 ,
Iy D 2.752 &eth; 105 m4 D 2.752 &eth; 107 mm4 ,
and ky D
Iy
D 0.0437 m D 43.7 mm
A
Problem 8.149 Determine Ix and kx for the area in
Problem 8.148.
Solution: Use the results in the solution to Problem 8.148. Part (1)
A1 D 0.0400.200 D 0.008 m2 ,
y1 D 0.180 m,
Ix1 D
1
12
0.20.043 C 0.182 A1 D 2.603 &eth; 104 m4 .
Part (2):
A2 D 0.040.16 D 0.0064 m2 ,
y2 D 0.08 m,
Ix2 D
1
12
0.040.163 C 0.082 A2 D 5.461 &eth; 105 m4 .
The composite: A D A1 C A2 D 0.0144 m2 , The area moment of
inertia about the x axis is
Ix D Ix1 C Ix2 D 3.15 &eth; 104 m4 D 3.15 &eth; 108 mm4 ,
and kx D
Ix
D 0.1479 m D 147.9 mm
A
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
677
y
Problem 8.150 Determine Ix and kx .
Solution: Use the results of the solutions to Problems 8.148–
40
mm
8.149. The centroid is located relative to the base at
xc D
x1 A1 C x2 A2
D 0,
A
yc D
y1 A1 C y2 A2
D 0.1356 m.
A
x
160
mm
The moment of inertia about the x axis is
80
mm
Ixc D y2C A C IX D 5.028 &eth; 107 mm4
and kxc D
40
mm
80
mm
Ixc
D 59.1 mm
A
Problem 8.151 Determine JO and kO for the area in
Problem 8.150.
Solution: Use the results of the solutions to Problems 8.148–
8.149. The area moments of inertia about the centroid are
Ixc D 5.028 &eth; 105 m4
and Iyc D Iy D 2.752 &eth; 105 m4 ,
from which
JO D Ixc C Iyc D 7.78 &eth; 105 m4 D 7.78 &eth; 107 mm4
and kO D
JO
D 0.0735 m
A
D 73.5 mm
Problem 8.152 Determine Iy and ky .
y
Solution: For a semicircle about a diameter:
Iyy D Ixx D
Iy D
1
1
44 24 D 44 24 D 94.25 ft4 ,
8
8
8
ky D
678
1
R4 ,
8
2 ft
x
4 ft
2Iy
D 2.236 ft
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.153 Determine JO and kO . for the area in
Problem 8.152.
Solution: For a semicircle:
Iyy D Ixx D
1
R4 .
8
4
4 24 D 94.248 ft4 .
8
Ix D
kx D
2Ix
D 2.236 ft.
42 22 Also use the solution to Problem 8.152.
JO D Ix C Iy D 294.248 D 188.5 ft4
kO D
2JO
D 3.16 ft
42 22 Problem 8.154 Determine Ix and kx .
Solution: Break the area into three parts: Part (1) The rectangle
with base 2a and altitude h; Part (2) The triangle on the right with
base b a and altitude h, and Part (3) The triangle on the left with
base b a and altitude h. Part (1) The area is
y
3 ft
3 ft
6 ft
A1 D 2ah D 24 ft2 .
The centroid is
x
x1 D 0
2 ft
h
and y1 D D 3 ft.
2
2 ft
y
b
The area moment of inertia about the centroid is
Ixc1 D
1
12
2ah3 D
Part (2): A2 D
x2 D a C
y2 D
ba
D 2.3333 ft,
3
1
36
h
1
hb a D 3 ft2 ,
2
2
h D 4 ft,
3
Ixc2 D
1
ah3 D 72 ft4 .
6
a
x
The composite moment of inertia
Ix D y1 2 A1 C Ixc1 C y2 2 A2 C Ixc2 C y3 2 A3 C Ixc3 ,
Ix D 396 ft4
b ah3 D 6 ft4 .
kx D
Ix
D
A
396
D 3.633 ft
30
Part (3): A3 D A2 ,
x3 D x2 , y3 D y2 , Ixc3 D Ixc2 .
The composite area is
A D A1 C A2 C A2 D 30 ft2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
679
Problem 8.155 Determine Iy and ky for the area in
Problem 8.154.
Solution: Divide the area as in the solution to Problem 8.154.
Part (1) The area is A1 D 2ah D 24 ft2 . The centroid is x1 D 0 and
h
y1 D D 3 ft. The area moment of inertia about the centroid is
2
Iyc1 D
1
12
h2a3 D
Part (2): A2 D
x2 D a C
y2 D
2
ha3 D 32 ft4
3
1
hb a D 3 ft2 ,
2
ba
D 2.3333 ft,
3
2
h D 4 ft,
3
Iyc2 D
1
36
hb a3 D 0.1667 ft4 .
Part (3): A3 D A2 ,
x3 D x2 , y3 D y2 , Iyc3 D Iyc2 .
The composite area is
A D A1 C A2 C A2 D 30 ft2 .
The composite moment of inertia,
Iy D x21 A1 C Iyc1 C x22 A2 C Iyc2 C x23 A3 C Iyc3 ,
Iy D 65 ft4
and ky D
Iy
D 1.472 ft
A
Problem 8.156 The moments of inertia of the area are
Ix D 36 m4 , Iy D 145 m4 , and Ixy D 44.25 m4 . Determine a set of principal axes and the principal moment
of inertia.
Solution: The principal angle is
D
1
2Ixy
tan1
D 19.54&deg; .
2
Iy Ix
y
3m
4m
3m
x
The principal moments of inertia are
IxP D Ix cos2 2Ixy sin cos C Iy sin2 D 20.298 D 20.3 m4
IyP D Ix sin2 C 2Ixy sin cos C Iy cos2 D 160.70 m4
680
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.157 The moment of inertia of the 31-oz bat
about a perpendicular axis through point B is 0.093 slugft2 . What is the bat’s moment of inertia about a perpendicular axis through point A? (Point A is the bat’s “instantaneous center,” or center of rotation, at the instant shown.)
31
D 0.06023 slugs.
1632.17
Use the parallel axis theorem to obtain the moment of inertia about the
center of mass C, and then use the parallel axis theorem to translate
to the point A.
Solution: The mass of the bat is m D
IC D IA D
12
12
2
12 C 14
12
m C 0.093 D 0.0328 slug-ft2
2
m C 0.0328 D 0.3155 slug-ft2
C
12 in
B
14 in
A
Problem 8.158 The mass of the thin homogenous plate
is 4 kg. Determine its moment of inertia about the y axis.
Solution: Divide the object into two parts: Part (1) is the semicircle of radius 100 mm, and Part (2) is the rectangle 200 mm by
280 mm. The area of Part (1)
A1 D
y
100 mm
140 mm
x
R2
D 15708 mm2 .
2
140 mm
The area of Part (2) is
200 mm
A2 D 280200 D 56000 mm2 .
The composite area is A D A2 A1 D 40292 mm2 . The area mass
density is
D
4
D 9.9275 &eth; 105 kg/mm2 .
A
For Part (1) x1 D y1 D 0,
Iy1 D 1
R4 D 3898.5 kg-mm2 .
8
For Part (2) x2 D 100 mm.
Iy2 D x22 A2 C 1
12
2802003 D 74125.5 kg-mm2 .
The composite:
Iy D Iy2 Iy1 D 70226 kg-mm2 D 0.070226 kg-m2
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
681
Problem 8.159 Determine the moment of inertia of the
plate in Problem 8.158 about the z axis.
Solution: Use the same division of the parts and the results of the
solution to Problem 8.158. For Part (1),
Ix1 D 1
R4 D 3898.5 kg-mm2 .
8
For Part (2)
Ix2 D 1
12
2002803 D 36321.5 kg-mm2 .
The composite: Ix D Ix2 Ix1 D 32423 kg-mm2 ,
from which, using the result of the solution to Problem 8.158
Iz D Ix C Iy D 32422 C 70226 D 102649 kg-mm2
D 0.10265 kg-m2
y
Problem 8.160 The homogenous pyramid is of mass m.
Determine its moment of inertia about the z axis.
Solution: The mass density is
D
x
3m
m
D 2 .
V
w h
h
The differential mass is dm D ω2 dz. The moment of inertia of this
element about the z axis is
z
1
dIZ D
ω2 dm.
6
Noting that ω D
dIz D w4
6h4
z4 dz D
y
y
wz
, then
h
mw2 4
z dz.
2h5
Integrating:
Iz
axis
D
x
w
z
w
h
2
mw
2h5
h
z4 dz D
0
1
mw2
10
Problem 8.161 Determine the moment of inertia of the
homogenous pyramid in Problem 8.160 about the x and
y axes.
Solution: Use the results of the solution of Problem 8.160 for the
Noting that ω D
mass density. The elemental disk is dm D ω2 dz. The moment of
inertia about an axis through its center of mass parallel to the x axis is
dIX D
1
12
Ix
axis
D
w4
12h4
ω2 dm.
Ix
682
axis
D
1
12
h
z4 dz C
0
w2
h2
h
z4 dz.
0
Integrating and collecting terms
Use the parallel axis theorem:
w
z, the integral is
h
Ix
ω2 dm C
m
axis
Dm
1 2 3 2
w C h .
20
5
z2 dm.
m
By symmetry, Iy
axis
D Ix
axis
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 8.162 The homogenous object weighs
400 lb. Determine its moment of inertia about the x axis.
y
y
Solution: The volumes are
Vcyl D
4692
z
6 in
9 in
3
D 11,706 in ,
Vcone D 13 62 36 D 1357in3 ,
36 in
x
36 in
46 in
x
46 in
so V D Vcyl Vcone D 10,348 in3 .
The masses of the solid cylinder and the material that would occupy
the conical hole are
mcyl D
mcone D
Vcyl
V
Vcone
V
400
32.2
400
32.2
D 14.052 slug,
D 1.629 slug.
Using results from Appendix C,
Ix
axis
D
1
3
mcyl 92 mcone 62
2
10
D 551 slug-in2
D 3.83 slug-ft2
Problem 8.163 Determine the moments of inertia of
the object in Problem 8.162 about the y and z axes.
Solution: See the solution of Problem 8.162. The position of the
center of mass of the material that would occupy the conical hole is
3
36 D 37 in.
4
x D 46 36 C
6 in
x , x′
From Appendix C,
X
Iy 0
axiscone
y′
y
D mcone
3
3
362 C
62
80
20
36 in
D 87.97 slug-in2 .
The moment of inertia about the y axis for the composite object is
Iy
axis
D mcyl
)1
2
3 46
\$
Iy 0
C 14 92
axiscone
*
C x2 mcone
%
D 7877 slug-in2
D 54.7 slug-ft2 .
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
683
Problem 8.164 Determine the moment of inertia of the
14-kg flywheel about the axis L.
50 mm
70 mm
120 mm
L
100 mm
440 mm
500 mm
150 mm
Solution: The flywheel can be treated as a composite of the objects
shown:
The volumes are
=
–
+
–
+
4
V1 D 1502502 D 294.5 &eth; 105 mm3 ,
1
2
–
5
6
3
V2 D 1502202 D 228.08 &eth; 105 mm3 ,
V3 D 502202 D 76.03 &eth; 105 mm3 ,
V4 D 50602 D 5.65 &eth; 105 mm3 ,
V5 D 100602 D 11.31 &eth; 105 mm3 ,
V6 D 100352 D 3.85 &eth; 105 mm3 .
The volume
V D V1 V2 C V3 V4 C V5 V6
D 144.3 &eth; 105 mm3 ,
so the density is
υD
14
D 9.704 &eth; 107 kg/mm3 .
V
The moment of inertia is
IL D 12 υV1 2502 12 υV2 2202
C 12 υV3 2202 12 υV4 602
C 12 υV5 602 12 υV6 352
D 536,800 kg-mm2
D 0.5368 kg-m2 .
684