CHAPTER 9
PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At x  0, y  b : b  k  0  a 
k
b
a2
then
2
y
b
2
x  a  ; dA  ydx
2 
a
dI y  x 2 dA  x 2 y dx
Iy 
a b
2
  a x  x  a  dx
0
2
2

  a  x  2ax  a x dx

b 1 5 1 4 1 2 3 
x  ax  a x 
2
3
a 2  5
0
a b
0
4
3
2 2

2
a
Iy 
1 3
a b 
30
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PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
y  kx1/3
For x  a :
b  ka1/3
k  b /a1/3
Thus:
y
b 1/3
x
a1/3
dI y  x 2 dA  x 2 ydx
b 1/3
b
x dx  1/3 x7/3 dx
1/3
a
a
a
b
b  3

Iy  dI y  1/3
x 7/3 dx  1/3  a10/3 
0
a
a  10

dIy  x 2


Iy 
3 3
a b 
10
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PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
By observation
Now
y
h
x
b
dI y  x 2
dA  x 2 [(h  y )dx]
x

 hx 2 1   dx
 b
Then

I y  dI y 
b
2
x
 hx 1  b  dx
0
b
1
x4 
 h  x3  
4b  0
3
or
Iy 
1 3
b h 
12
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PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At x  0, y  b : b  ka3
k
b
a3
then
y
b 3
x ; dA  ydx
a3
dI y  x 2 dA  x 2  b  y  dx
Iy 

a
x3  
2
bx 1  3  dx
0

 a  
a
1
x6 
 b  x3  3 
6a  0
3
Iy 
1 3
a b 
6
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PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At x  0, y  b : b  k  0  a 
k
b
a2
then y 
b
2
x  a
2 
a
2
or

x  a 1 


y
 , take the negative root so that y  b when x  0.
b 
dI x  y 2 dA  y 2 x dy
Ix 
b

1
 a  y  b
0
2
12

y 5 2  dy

b
1

2 1 
 a  y3   1 2  y7 2 
3
7
b 

 0
Ix 
1 3
ab 
21
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PROBLEM 9.6
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
y
b 1/3
x
a1/3
3
1
1 b
1 b3

dI x  y 3 dx   1/3 x1/3  dx 
xdx
3
3 a
3 a


I x  dI x 
a 1 b3
 3a
0
xdx 
1 b3 a 2
3 a 2
Ix 
1 3
ab 
6
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PROBLEM 9.7
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
By observation
y
h
x
b
or
x
b
y
h
Now
dI x  y 2 dA  y 2 ( x dy )
b

 y 2  y dy 
h

b 3
 y dy
h
Then

I x  dI x 
hb
 h y dy
3
0
h
b 1 
  y4 
h  4 0
or
Ix 
1 3
bh 
4
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PROBLEM 9.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At x  a, y  b : b  ka3
k
b
a3
then
y
b 3
a
x ; x  1 3 y1 3
3
a
b
 a

dI x  y 2 dA  y 2  1 3 y1 3  dy
b

Ix 
b
a
0
13
  b

y 7 3 dy

b
3 a 

   1 3  y10 3 
 10  b 
 0
Ix 
3 3
ab 
10
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PROBLEM 9.9
Determine by direct integration the moment of inertia of the
shaded area with respect to the x axis.
SOLUTION
At x  3a,
y  b:
b  k (3a  a)3
or
k
b
8a3
Then
y
b
( x  a )3
3
8a
Now
dI x 
1 3
y dx
3
3
Then

1 b

( x  a)3  dx

3
3  8a


b3
( x  a )9 dx
1536a 9
3a
b3
b3  1

9
I x  dI x 
(
x

a
)
dx

( x  a)10 
9 
a 1536a 9
1536a 10
a



3a
b3
[(3a  a )10  0]
15,360a9
or
Ix 
1 3
ab 
15
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PROBLEM 9.10
Determine by direct integration the moment of inertia of the
shaded area with respect to the x axis.
SOLUTION
At x  0, y  b :
b  c(1  0)
or
cb
At x  a, y  0:
0  b(1  ka1/ 2 )
or k 
Then

x1/ 2 
y  b 1  1/2 
 a 
1
a
1/ 2
3
Now
1
1 
x1/ 2  
dI x  y 3 dx  b 1  1/ 2   dx
3
3   a  
or
1 
x1/2
x x3/ 2 
dI x  b3 1  3 1/ 2  3  3/ 2  dx
3 
a a 
a
Then
I x  dI x  2



x1/2
x x3/ 2 
b3 1  3 1/ 2  3  3/ 2  dx
0 3
a a 
a

a1
a
2 
x3/ 2 3 x 2 2 x5/ 2 
 b3  x  2 1/ 2 


3 
2 a 5 a 3/ 2  0
a
or
Ix 
1 3
ab 
15
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PROBLEM 9.11
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At x  a, y  b :
b  ke a/a
Then
y
Now
dI x 
Then
I x  dI x 
or k 
b
e
b x/a
e  be x/a 1
e
1 3
1
y dx  (be x/a 1 )3 dx
3
3
1 3 3( x/a 1)
 be
dx
3

a1
 3
0
b3e3( x/a 1) dx 
1
 ab3 (1  e 3 )
9
a
b3  a 3( x/a 1) 
e

3  3
0
or I x  0.1056ab3 
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PROBLEM 9.12
Determine by direct integration the moment of inertia of the
shaded area with respect to the y axis.
SOLUTION
At x  3a,
y  b:
b  k (3a  a)3
or
k
b
8a3
Then
y
b
( x  a )3
8a 3
Now
Then
 b

dI y  x 2 dA  x 2 ( y dx)  x 2  3 ( x  a)3 dx 
 8a

b
 3 x 2 ( x3  3x 2 a  3xa 2  a3 )dx
8a

3a
b
I y  dI y 
 8a ( x  3x a  3x a  a x )dx

b 1 6 3 5 3 2 4 1 3 3
x  ax  a x  a x 
5
4
3
8a 3  6
a

b  1
3
3
1

(3a)6  a (3a )5  a 2 (3a) 4  a 3 (3a )3 
3 
5
4
3
8a   6

a
5
4
3 2
3 2
3
3a
3
3
1
1

  ( a ) 6  a ( a )5  a 2 ( a ) 4  a 3 ( a )3  
5
4
3
6

or I y  3.43a3b 
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PROBLEM 9.13
Determine by direct integration the moment of inertia of the
shaded area with respect to the y axis.
SOLUTION
At x  0,
y  b:
b  c(1  0)
or c  b
x  a,
y  0:
0  c(1  ka1/ 2 )
or k 
1
a
1/ 2

x1/ 2 
y  b 1  1/2 
 a 
 
x1 2  
dI y  x 2 dA  x 2  ydx   x 2 b  1  1 2  dx 
  a  

I y  dI y  2
 x1 2 
bx 2 1  1 2  dx
0
 a 

a
a
1
2 x7 2 
I y  2b  x3 

7 a1 2  0
3
Then
Iy 
2 3
a b 
21
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PROBLEM 9.14
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At x  a,
y  b:
b  ke a/a
or
k
b
e
Then
y
b x/a
e  be x/a 1
e
dI y  x 2 dA  x 2 ( ydx )
Now
 x 2 (be x /a 1dx )

I y  dI y 
Then
a
 bx e
2 x/a 1
0
dx
Now use integration by parts with
a
 xe
Then
0
u  x2
dv  e x /a 1dx
du  2 xdx
v  ae x /a 1
2 x /a 1
a
dx   x 2 ae x /a 1  
0
 a 3  2a
a
 xe
a
 (ae
0
x /a 1
0
x /a 1
)2 xdx
dx
Using integration by parts with
Then
ux
dv  e x/a 1dx
du  dx
v  ae x/a 1
a



I y  b a3  2a ( xae x/a 1 ) |0a  (ae x/a 1 )dx  
0






 b a  2a  a  (a  a e ) 
 b a 3  2a  a 2  (a 2 e x/a 1 ) |0a 
3
2
2
2 1
or I y  0.264a3b 
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PROBLEM 9.15
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
For
y1  k1 x 2
y2  k2 x1/ 2
x  0 and
y1  y2  b
b  k1a 2
b  k2 a1/ 2
k1 
b
a2
k2 
b
a
y1 
b 2
x
a2
y2 
1/ 2
Thus,
b 1/ 2
x
a
1/ 2
dA  ( y2  y1 )dx
A
a
b
0
1/ 2
  a
x1/ 2 
b 2
x dx
a 2 
2 ba 3/ 2 ba 3

3 a1/ 2 3a 2
1
A  ab
3
A
1 3
1
y2 dx  y13 dx
3
3
3
1 b
1 b3 6

x3/ 2 dx 
x dx
3/ 2
3a
3 a6
dI x 

b3 a 3/ 2
b3 a 6
x
dx
x dx

3a3/ 2 0
3a 6 0
b3 a5/ 2 b3 a 7  2 1  3
3 3
 3/ 2 5  6
    ab I x 
ab 
35
7
15
21
3a


 2  3a

I x  dI x 
I
k x2  x
A
 ab 


3
3
35
ab
b
kx  b
9

35
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.16
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At x  2a,
h  k  2a 
y  h:
2
or
k
h
4a 2
Then
y
h 2
x
4a 2
dA  ydx 
Now

A  dA 
h
4a 2
2a
h 2
x dx
4a 2
 x dx
2
a
2a
3
3
a3  7
 h  x 
 h   8a
A   2     2 
   ah
3  12
 4a   3  a  4a   3
dI x 
1 3
1 h  6
y dx  
 x dx
3
3  64a 6 

I x  dI x 
Then
h3
192a 6
2a
 x dx
6
0
2a
 h3  x 7
h3  128a 7  a 7 
Ix  


6 
6 
7
 192a  7 a 192a 

Ix 
127
ah3  0.094494ah3
 7 192 
I x  0.0945ah3 
 k x 2 


I x 0.094494ah3

 0.161983h 2 
7
A
ah
12
k x  0.402h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.17
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 9.15.
1
A  ab
3
a
dI y  x 2 dA  x 2 ( y2  y1 )dx
2
b
b 2
b
x  dx  1/ 2
2
a
a

Iy 
 x  a
Iy 
b b7/2 b a5  2 1  3



   a b
a1/ 2  72  a 2 5  7 5 
k y2 
1/ 2
0
Iy
A

x1/2 
 a b
a
b
a
2
0
 x dx  a  x dx
0
5/ 2
4
Iy 
3 3
a b
35
3
3
35
ab
3
ky  a
9

35
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.18
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 9.16
y
h 2
x
4a 2
h
Iy  2
4a
Iy 
k y2 

A
2a
0
7
 h

ah dA  ydx dI y  x 2 dA  x 2  2 x 2  dx
12
 4a

2a
h  x5 
x dx  2  
4a  5  a
4
h  32a5 a5  31 3
  
ha

5  20
4a 2  5
Iy
A
 ha   93 a

31
20
Iy 
31 3
ha 
20
3
7
ah
12
35
2
ky  a
93

35
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.19
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
y1 :
At x  2a,
y  0:
0  c sin k (2a)
2ak  
y2 :
or k 

At x  a,
y  h:
h  c sin
At x  a,
y  2h :
2h  ma  b
At x  2a,
y  0:
0  m(2a )  b
2a

2a
(a ) or
ch
2h
, b  4h
a
Solving yields
m
Then
y1  h sin
Now
 
 2h
dA  ( y2  y1 )dx   ( x  2a )  h sin
x dx
2a 
a
Then
A  dA 


2a
x
2a
2h
x  4h
a
2h

(  x  2a )
a
y2  
2


 h  a ( x  2a)  sin 2a x  dx
a
2a
2a
 
 1
cos
x
 h   (  x  2a ) 2 

2a  a
 a
 2 a  1

2

 h     ( a  2a )2   ah  1  
a







 0.36338ah
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.19 (Continued)
Find:
I x and k x
We have
3
3
1 
1   2h
  
1
 
dI x   y2  y1  dx    ( x  2a)    h sin
x   dx
3 
3   a
2a  
3
 

Then
Now
Then
h3  8
 
( x  2a)3  sin 3
x

3
3 a
2a 

I x  dI x 

 
(  x  2a )3  sin 3
x dx

3
3 a
2a 
2 a h3  8
a
sin 3   sin  (1  cos2  )  sin   sin  cos2 
Ix 
h3
3
2a  8





  a ( x  2a)   sin 2a x  sin 2a x cos 2a x  dx
a
3
2
3
2a

2a
2a

 
h3  2
cos
cos3
 3 (  x  2a ) 4 
x
x

3  a
2a
3
2a  a


h 3  2 a 2 a  2
4

 
  3 (  a  2a ) 
3   3  a


2 3
2 
ah 1 
3
3
 

I x  0.52520ah3
and
k x2 
I x 0.52520ah3

A 0.36338ah
or I x  0.525ah3 
or
k x  1.202h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.20
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
y1 :
At x  2a,
y  0:
0  c sin k (2a)
2ak  
y2 :
or k 

At x  a,
y  h:
h  c sin
At x  a,
y  2h :
2h  ma  b
At x  2a,
y  0:
0  m(2a)  b
2a

2a
( a ) or
ch
2h
, b  4h
a
Solving yields
m
Then
y1  h sin
Now
 
 2h
dA  ( y2  y1 )dx   ( x  2a)  h sin
x dx
2a 
a
Then
A  dA 


2a
x
2a
2h
x  4h
a
2h

(  x  2a )
a
y2  
2


 h  a ( x  2a)  sin 2a x  dx
a
2a
2a
 
 1
cos x 
 h   (  x  2a ) 2 
a
2
a a


 2 a  1

2

 h     ( a  2a )2   ah 1  
 
   a

 0.36338ah
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.20 (Continued)
Find: I y and k y
  2h
  
dI y  x 2 dA  x 2   (  x  2a )  h sin
x dx 
a
2
a  

 
2
x dx
 h  (  x3  2ax 2 )  x 2 sin
2a 
a
We have

Then
2a
2


I y  dI y 
 h  a ( x  2ax )  x sin 2a x  dx
u  x2
dv  sin
du  2 x dx
v
3
2
2
a
Now using integration by parts with

2

x dx
cos


2a
x

 2a

 x sin 2a x dx  x    cos 2a x       cos 2a x (2 x dx)
2
Then
ux
dv  cos
du  dx
v
Now let
Then
2a
2a

2a


2a

2a


2a
sin
x dx

2a
4a   2a
x


 2a



 x sin 2a x dx    x cos 2a x    x   sin 2a x      sin 2a x  dx 
2
2
2 1
2

I y  h    x 4  ax3 
4
3
a

 
2a
 2a 2
8a 2
16a3


 
  
x cos
x  2 x sin
x  3 cos
x 
2a
2a
2a  


 
a
 2  1
2
16a3 
 2a
 h    (2a)4  a(2a )3  
(2a) 2  3 
3
 
 
 a  4
2
 2  1

2
 8a
 h    ( a ) 4  a ( a )3   2 ( a ) 
3
 
 a  4

 0.61345a 3 h
and
k y2 
Iy
A

0.61345a 3 h
0.36338ah
or
or
I y  0.613a3h 
k y  1.299a 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.21
Determine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point P.
SOLUTION
We have
dI x  y 2 dA  y 2 [( x2  x1 )dy ]
Then
 a
I x  2  y 2 (2a  a)dy 
 2

2a

 y (2a  0)dy 
2
a
2a
  1  a
 1  
 2  a  y 3   2a  y 3  
 3  a 
  3  0
 1

 2
 2 a  (a )3   a (2a )3  (a)3  
 3
 3

 10a 4
Also
dI y  x 2 dA  x 2 [( y2  y1 )dx]
Then
 a
I y  2  x 2 (2a  a)dx 
 0

2a

 x (2a  0)dx 
2
a
 10a 4
Now
JP  I x  I y  10a 4  10a 4
and
k P2 

or
J P  20a 4 
or
k P  1.826a 
JP
20a 4

A (4a )(2a )  (2a )( a )
10 2
a
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.22
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown with respect to point P.
SOLUTION
By observation
First note
Now
y
b
x
a
dA  ( y  2b)dx
b
 ( x  2a )dx
a
1 3
1 3

 ybottom
dI x   ytop
 dx
3
3

3

1  b 
  x   (2b)3  dx
3  a 


Then
1 b3 3
( x  8a 3 )dx
3
3a

I x  dI x 
a 1 b3
 3 a ( x  8a )dx
a
2
3
3
a

1 b3  1 4

x  8a3 x 
3 
3 a 4
a

1 b3   1 4
 1
  16
(a)  8a3 (a)    (a)4  8a3 (a)    ab3
3 
3 a  4
 4
 3
Also
b

dI y  x 2 dA  x 2  ( x  2a) dx 
a

Then
I y  dI y 

a b
 a x ( x  2a)dx
2
a
a

b 1 4 2 3
x  ax 
3
a  4
 a

2
4 3
b  1 4 2
3 1
4
3 
 (a)  a (a)    (a)  a (a)    a b
3
4
3
3
a   4
 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.22 (Continued)
Now
JP  Ix  I y 
16 3 4 3
ab  a b
3
3
JP 
and
kP2 
4
ab(a 2  4b 2 )
3

4
ab(a 2  4b2 )
JP
3

A (2a)(3b)  12 (2a)(2b)
1
 (a 2  4b2 ) or
3
kP 
a 2  4b 2

3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.23
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown with respect to Point P.
SOLUTION
y1 :
At x  2a,
y  2a :
2a  k1 (2a) 2
y2 :
At x  0,
y  a:
At x  2a,
y  2a :
or k1 
ac
2a  a  k2 (2a)2
Then
y1 
1 2
x
2a
or
k2 
y2  a 
1 2
x
4a

Now
Then
Now
1
2a
1
4a
1
(4a 2  x 2 )
4a
1 2
1
dA  ( y2  y1 )dx   (4a 2  x 2 ) 
x dx
2a 
 4a
1

(4a 2  x 2 )dx
4a

A  dA  2

2a
0
2a
1
1  2
1 
8
(4a 2  x 2 )dx 
4a x  x 3   a 2

4a
2a 
3 0
3
3
3
1 
1   1
1
 1
 
dI x   y23  y13  dx    (4a 2  x 2 )    x 2   dx
3 
3   4a
3
  2a  
1 1
1

(64a 6  48a 4 x 2  12a 2 x 4  x 6 )  3 x6  dx
 
3
3  64a
8a

1
(64a 6  48a 4 x 2  12a 2 x 4  7 x 6 )dx

192a 3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.23 (Continued)
Then

I x  dI x  2
2a
1
 192a (64a  48a x  12a x  7 x )dx
0
6
4 2
2 4
6
3
2a

1 
12

64a 6 x  16a 4 x3  a 2 x5  x7 
3 
5
96a 
0

1 
12

64a 6 (2a)  16a 4 (2a)3  a 2 (2a)5  (2a)7 
3 
5
96a 


1 4
12
 32 4
a 128  128   32  128  
a
96 
5
 15
Also
1

dI y  x 2 dA  x 2  (4a 2  x 2 )dx 
a
4


Then
I y  dI y  2



2a
0
2a
1 2
1 4 2 3 1 5
x (4a 2  x 2 )dx 
a x  x 
4a
2a  3
5 0
1 4 2
1
 32  1 1  32
a (2a)3  (2a)5   a 4     a 4
2a  3
5
 2
 3 5  15
Now
JP  I x  I y 
and
kP2 
32 4 32 4
a 
a
15
15
64 4
a
J P 15
8

 a2
2
8
A
5
a
3
64 4
a 
15
or
JP 
or
k P  1.265a 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.24
Determine the polar moment of inertia and the polar radius of gyration of
the shaded area shown with respect to point P.
SOLUTION
The equation of the circle is
x2  y 2  r 2
So that
x  r2  y2
Now
dA  xdy  r 2  y 2 dy
Then
A  dA  2
Let
y  r sin  ; dy  r cos  d
Then
A2

2
 /2

 /6


r
 r/ 2
r 2  y 2 dy
r 2  (r sin  ) 2 r cos d
 /2
  sin 2 
r 2 cos 2  d  2r 2  
 /6
4   /6
2
 /2
     sin  3  

3
2 
 2r 2  2   6 

   2r  

4  
8 
3
 2  2
 2.5274r 2
 r  y dy 
Now
dI x  y 2 dA  y 2
Then
I x  dI x  2
Let
y  r sin  ; dy  r cos  d
Then
Ix  2

2
Now

2
r
 r/ 2
 /2
2
y 2 r 2  y 2 dy
  (r sin  ) r  (r sin  ) r cos d
2
2
2
 /6
 /2
  r sin  (r cos )r cos d
2
2
 /6
sin 2  2sin  cos   sin 2  cos 2  
1
sin 2
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.24 (Continued)
Then
Ix  2

 /2
r 4   sin 4 
1

r 4  sin 2 2  d   
 /6
2 2
8   /6
4

 /2
1 3
1
x dy 
3
3

r 4  2  6 sin  23  
  

2  2  2
8  

r4  
3
 

2  3 16 
 r  y  dy
2
2
3
Also
dI y 
Then
I y  dI y  2
Let
y  r sin  ; dy  r cos  d
Then
Iy 
2
3
  [r  (r sin  ) ] r cos d
Iy 
2
3
  (r cos  )r cos d
Now
Then

 /2

r
1
 r/ 2 3
2
(r 2  y 2 )3/ 2 dy
2 3/ 2
 /6
 /2
3
3
 /6
1
cos 4   cos 2  (1  sin 2  )  cos 2   sin 2 2
4
Iy 
2
3
 /2
4
1

  r  cos   4 sin 2  d
2
2
 /6
 /2
2   sin 2  1   sin 4  
 r 4  



3  2
4  4  2
8    /6
2 4   2 1  2     6 sin  3 1   6 sin  23   

 

r      
 
3   2 4  2    2
4
4 2
8   


2    1  3  
1  3 
 r4     

 


3  4 16 12 4  2  48 32  2  


2 4 9 3 
r  

3  4
64 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.24 (Continued)
Now
JP  Ix  I y 
r4  
3  2 4 9 3 
 
 r  

2  3 16  3  4
64 

3
 r4  
 1.15545r 4
 3 16 


and
k P2 
J P 1.15545r 4

A
2.5274r 2
or
J P  1.155r 4 
or
k P  0.676 r 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.25
(a) Determine by direct integration the polar moment of inertia of the annular
area shown with respect to Point O. (b) Using the result of Part a, determine
the moment of inertia of the given area with respect to the x axis.
SOLUTION
dA  2 u du
(a)
dJ O  u 2 dA  u 2 (2 u du )
 2 u 3 du

R2
J O  dJ O  2
 u du
 2
1 4 2
u
4 R1
3
R1
JO 
 4
Ix 
 4
R
(b)
From Eq. (9.4):
(Note by symmetry.)
2
R2  R14  
Ix  I y
JO  I x  I y  2I x
Ix 
1

J O   R24  R14 
2
4
4
R2  R14  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.26
(a) Show that the polar radius of gyration kO of the annular area shown is
approximately equal to the mean radius Rm  ( R1  R2 ) 2 for small values of
the thickness t  R2  R1 . (b) Determine the percentage error introduced by
using Rm in place of kO for the following values of t/Rm: 1, 12 , and 101 .
SOLUTION
(a)
From Problem 9.23:
JO 
 4
R2  R14 

2
4
4

J O 2  R2  R1 
2
kO 

A  R22  R12


kO2 
1 2
 R2  R12 

2
Mean radius:
Rm 
1
( R1  R2 )
2
Thus,
1
1
R1  Rm  t and R2  Rm  t
2
2
Thickness: t  R2  R1
kO2 
2
2
1 
1  
1  
1
 Rm  t    Rm  t    Rm2  t 2
2 
2  
2  
4
For t small compared to Rm : kO2  Rm2
(b)
Percentage error 
kO  Rm 
(exact value)  (approximation value)
100
(exact value)
P.E.  
Rm2  14 t 2  Rm
Rm2  14 t 2
(100)  
   1100
1  
1  14
t
Rm
1
4
2
t
Rm
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.26 (Continued)
For
t
 1:
Rm
t
1
 :
For
Rm 2
t
1
 :
For
Rm 10
P.E. 
1  14  1
1  14
(100)  10.56%
1  14  12   1

2
P.E.  
1  14
 
1
2
2
1  14  101   1
(100)  2.99%

(100)  0.1250%

2
P.E. 
 
1  14 101
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.27
Determine the polar moment of inertia and the polar radius of gyration of
the shaded area shown with respect to Point O.
SOLUTION
At    , R  2a :
2a  a  k ( )
or
k
Then
Ra
Now
dA  (dr )(r d )
 rdr d
Then
A  dA 
a

a




  a 1 

 

a (1 / )
0
0
 
rdr d 

 1
a (1 / )

r2 

0 2
0
d

1     
A
a  1   d  a 2   1   
0 2
2  3    
 

 1
2
2
3

0
  
 7
1
  a 2 1    (1)3    a 2
6
  
 6
3
Now
dJ O  r 2 dA  r 2 (rdr d )
Then
J O  dJ O 



 1

a (1 / )
0
0
 
r 3 dr d
a (1 / )

r4 
0 
4

0
d 

4
 
a 4  1   d
0 4
 
 1

5
   5

1     
1
 a 4  1      a 4 1    (1)5 
4  5    
20
  

JO 
or
kO  1.153a 
0
and
31
 a 4 93
J
kO2  O  20 2  a 2
7
A
70
a
6
31 4
a 
20
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.28
Determine the polar moment of inertia and the polar radius of gyration of
the isosceles triangle shown with respect to point O.
SOLUTION
By observation:
h
y b x
2
b
y
2h
or
x
Now
 b 
dA  xdy  
y  dy
 2h 
and
dI x  y 2 dA 
Then
I x  dI x  2

b 3
y dy
2h
h
b
 2h y dy
3
0
h
b y4
1

 bh3
h 4 0 4
2h
x
b
From above:
y
Now
2h 

dA  (h  y )dx   h 
x  dx
b 


and
dI y  x 2 dA  x 2
h
(b  2 x)dx
b
h
(b  2 x)dx
b
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.28 (Continued)
Then

I y  dI y  2

b/ 2 h
0
b
x 2 (b  2 x)dx
b/ 2
h 1
1 
 2  bx3  x 4 
b 3
2 0
3
4
h b  b  1  b   1 3
2       
b h
b  3  2  2  2   48
Now
JO  I x  I y 
1 3 1 3
bh  b h
4
48
or
and
(12h2  b 2 ) 1
J O bh
2
48
kO 

 (12h2  b 2 )
1
A
bh
24
2
or
JO 
bh
(12h 2  b 2 ) 
48
kO 
12h 2  b 2

24
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.29*
Using the polar moment of inertia of the isosceles triangle of Prob. 9.28,
show that the centroidal polar moment of inertia of a circular area of
radius r is  r 4 /2. (Hint: As a circular area is divided into an increasing
number of equal circular sectors, what is the approximate shape of each
circular sector?)
PROBLEM 9.28 Determine the polar moment of inertia and the polar
radius of gyration of the isosceles triangle shown with respect to Point O.
SOLUTION
First the circular area is divided into an increasing number of identical circular sectors. The sectors can be
approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of
the isosceles triangles are as shown.
For an isosceles triangle (see Problem 9.28):
JO 
bh
(12h 2  b 2 )
48
b  r 
Then with
( J O )sector 

dJ O sector
Now
d
and h  r
1
(r  )(r )[12r 2  (r  ) 2 ]
48
1 4
r  (12   2 ) 
48
  J O sector 
1 4
2 
 lim 
  lim  r [12  ( ) ]
 0
 0  48







( J O )circle  dJ O sector 
Then

2 1
0
4
1 4
r
4
r 4 d 
1 4 2
r  0
4
or
( J O )circle 

2
*Advanced or specialty topic
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
r 4 
PROBLEM 9.30*
Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 /2 . (Hint:
Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same
area and the same centroid.)
SOLUTION
From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is
( J C )cir 

2
r4
The area of the circle is
Acir   r 2
So that
[ J C ( A)]cir 
A2
2
Two methods of solution will be presented. However, both methods depend upon the observation that as a
given element of area dA is moved closer to some Point C. The value of J C will be decreased
( J C   r 2 dA; as r decreases, so must J C ).
Solution 1
Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so
that its area is equal to A. To minimize the value of ( J C ) A , the area would have to be distributed as
closely as possible about C. This is accomplished by winding the strip into a tightly wound roll with C as
its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus
increasing the value of ( J C ) A . (The process is analogous to rewinding a length of tape back into a roll.)
Since the shape of the roll is circular, with the centroid of its area at C, it follows that
( JC ) A 
A2
Q.E.D. 
2



where the equality applies when the original area is circular.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.30* (Continued)
Solution 2
Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid)
at Point C. Without loss of generality, assume that
A1  A2
A3  A4
It then follows that
( J C ) A  ( J C ) cir  [ J C ( A1 )  J C ( A2 )  J C ( A3 )  J C ( A4 )]
Now observe that
J C ( A1 )  J C ( A2 )  0
J C ( A3 )  J C ( A4 )  0
Since as a given area is moved farther away from C, its polar moment of inertia with respect to C must
increase.
( J C ) A  ( J C )cir
or ( J C ) A 
A2
Q.E.D. 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A  A1  A2  A3
 [(24)(6)  (8)(48)  (48)(6)] mm 2
 (144  384  288) mm 2
 816 mm 2
I x  ( I x )1  ( I x ) 2  ( I x )3
Now
where
1
(24 mm)(6 mm)3  (144 mm 2 )(27 mm) 2
12
 (432  104,976) mm 4
( I x )1 
 105, 408 mm 4
1
(8 mm)(48 mm)3  73,728 mm 4
12
1
( I x )3  (48 mm)(6 mm)3  (288 mm 2 )(27 mm)2
12
 (864  209,952) mm 4  210,816 mm 4
( I x )2 
Then
I x  (105, 408  73,728  210,816) mm4
 389,952 mm4
and
k x2 
I x 389,952 mm 4

A
816 mm 2
or I x  390  103 mm 4 
or
k x  21.9 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.32
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 − A2 − A3
= [(5)(6) − (4)(2) − (4)(1)] cm2
= (30 − 8 − 4) cm2
= 18 cm2
I x = ( I x )1 − ( I x ) 2 − ( I x )3
Now
where
( I x )1 =
1
(5 cm)(6 cm)3 = 90 cm4
12
1
(4 cm)(2 cm)3 + (8 cm2)(2 cm)2
12
2
= 34 cm4
3
( I x )2 =
1
3
(4 cm)(1 cm)3 + (4 cm2)
cm
12
2
1
= 9 cm4
3
2
( I x )3 =
Then
2
1
I x = 90 − 34 − 9
cm4
3
3
and
k x2 =
I x 46.0 cm4
=
A
18 cm4
or
I x = 460 × 103 mm4
or k x = 16 mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.33
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A  A1  A2  A3
 [(24  6)  (8)(48)  (48)(6)] mm 2
 (144  384  288) mm 2
 816 mm 2
I y  ( I y )1  ( I y )2  ( I y )3
Now
where
1
(6 mm)(24 mm)3  6912 mm 4
12
1
( I y ) 2  (48 mm)(8 mm)3  2048 mm 4
12
1
( I y )3  (6 mm)(48 mm)3  55, 296 mm 4
12
( I y )1 
Then
I y  (6912  2048  55, 296) mm 4  64, 256 mm 4
I y  64.3  103 mm 4 
or
and
k y2 
Iy
A

64, 256 mm 4
816 mm 2
or
k y  8.87 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.34
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 − A2 − A3
= [(5)(6) − (4)(2) − (4)(1)] cm2
= (30 − 8 − 4) cm2
= 18 cm2
I y = ( I y )1 − ( I y ) 2 − ( I y )3
Now
where
( I y )1 =
1
(6 cm)(5 cm)3 = 62.5 cm4
12
( I y )2 =
1
2
(2 cm)(4 cm)3 = 10 cm4
3
12
( I y )3 =
1
1
(1 cm)(4 cm)3 = 5 cm4
12
3
Then
2
1
I y = 62.5 − 10 − 5 cm4
3
3
and
k y2 =
Iy
A
=
46.5 cm4
18 cm2
or
I y = 465 × 103 mm4
or k y = 16.1 mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.35
Determine the moments of inertia of the shaded area shown with respect to the x and
y axes.
SOLUTION
1
(2a)(2a)3
12
4
 a4
3
I x  ( I x )1  ( I x )2   I x 3 where
We have
( I AA )2   I BB 3 

a4
8
   Ad
 I AA 2  I x2
   
I x2
2
 I x3
( I x )1 
3
2
2

2

8  4
   4a   
a 4   a 2 
 

a
8
 2  3   8 9 
2
 8

4a
4 5
 I x 2  I x2 2  Ad     a 4    a 2  a       a 4
3   3 8 
 8 9 
2 
 
2
2
 8

4a
4 5
 I x 3  I x3 3  Ad     a 4    a 2  a        a 4
3   3 8 
 8 9 
2 
 
2
Finally, I x 
 3
I y  ( I y )1  ( I y ) 2  I y
Also
1
16
3
2
2
 2a  2a    2a   a   a 4
12
3
where
( I y )1 
but
( I y )2  I y
Then
4 4  4 5  4  4 5  4
4
a  
a   
 a I x  4a 
3
3 8 
 3 8 
 3
Iy 
16 4
a 
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.36
Determine the moments of inertia of the shaded area shown with respect to the
x and y axes.
SOLUTION
I x  ( I x )1  ( I x )2   I x 3
We have
1
2
2
(4a )(4a)3   4a   2a 
12
256 4

a
3
( I x )1 
where
( I AA )2   I BB 3 

 I AA 2   I x 2  Ad 2
8
a4
2
2
8  4
   4a   
I x2  I x3  a   a 2     
a
2
3
8
 2  3   8 9 
   

4
2
 8

5
4a
10 13
 I x 2  I x2 2  Ad     a 4    a 2  a       a 4
3   3
4 
 8 9 
 2 2
 
2
 I x 3   I x 3  Ad 2  
3

8
2

8  4   2 3
4a  
5  4
a   a  a 
a
  2 


9 
3  
4 
 2 2
Finally, I x 
Also
256 4  10 13  4 
5  4
4
a  
 a   2 
 a I x  69.9a 
3
3
4
4




 3
I y  ( I y )1  ( I y ) 2  I y
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PROBLEM 9.36 (Continued)
( I y )1 
1
256 4
a
 4a  4a 3   4a 2  2a 2 
12
3
 3  8 a 4   2 a2   2a 2  178  a4
( I y )2  I y
Then
Iy 
256 4
 17

a  2   a4 
3
8

I y  72.0a 4 
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PROBLEM 9.37
The centroidal polar moment of inertia J C of the 24-mm 2 shaded
area is 600 mm 4. Determine the polar moments of inertia JB and JD
of the shaded area knowing that JD = 2JB and d = 5 mm.
SOLUTION
A  24 mm2 J D  2 J B
Given:
J B  J C  A(dCB )2
J C  600 mm 4
(1)
J D  J C  A(dCD )2 (2)

J C  A  dCD   2 J C  A  dCB 
2
2

 dCB 2  a 2  d 2 and  dCD 2   2a 2  d 2



A 4a 2  d 2  J C  2 A a 2  d 2


1 J
a2   C  d 2 
2 A

1  600
2
a2  
  5 
2  24

a  5 mm
Then from Eqn. (1)
Then from Eqn. (2)


J  600  24 10  5 
J B  600  24 52  52
2
D
2
J B  1800 mm4 
J D  3600 mm4 
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PROBLEM 9.38
Determine the centroidal polar moment of inertia J C of the 25-mm2
shaded area knowing that the polar moments of inertia of the
area with respect to points, A, B and D are, respectively,
JA = 281 mm4, JB = 810 mm4, and JD = 1578 mm4.
SOLUTION
J A  J C  A  dCA 
2
 dCA   a 2
2
J B  J C  A(a)2
J B  J C  A  dCB 
(1)
2
 dCB   a 2  d 2
2
J B  J C  A(a 2  d 2 )
J D  J C  A  dCD 
 dCD   4a 2  d 2
2

J D  J C  A 4a 2  d 2
(2)
2

(3)
Eqn. (3) – Eqn. (2):
J D  J B  3 Aa 2
Eqn. (4) – 3[Eqn. (1)]
J D  J B  3J A  3J C
J C  281 
1
1578  810 
3
(4)
J C  25.0 mm4 
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PROBLEM 9.39
Determine the shaded area and its moment of inertia with respect to the
centroidal axis parallel to AA knowing that d1 = 25 mm and d2 = 10 mm
and that its moments of inertia with respect to AA and BB are 2.2  106
mm4 and 4  106 mm4, respectively.
SOLUTION
I AA  2.2  106 mm4  I  A(25 mm)2
(1)
I BB  4  106 mm4  I  A(35 mm)2
I BB  I AA  (4  2.2)  106  A(352  252 )
1.8  106  A(600)
Eq. (1):
2.2  106  I  (3000)(25)2
A  3000 mm 2 
I  325  103 mm 4 
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PROBLEM 9.40
Knowing that the shaded area is equal to 6000 mm2 and that its moment of
inertia with respect to AA is 18  106 mm4, determine its moment of inertia
with respect to BB for d1 = 50 mm and d2 = 10 mm.
SOLUTION
Given:
A  6000 mm2
I AA  I  Ad12 ; 18  106 mm2  I  (6000 mm2 )(50 mm)2
I  3  106 mm4
I BB  I  Ad 2  3  106 mm4  (6000 mm2 )(50 mm  10 mm)2
 3  106  6000(60)2  24.6  106 mm4
I BB  24.6  106 mm4 
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PROBLEM 9.41
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to side
AB.
SOLUTION
First locate centroid C of the area.
Symmetry implies X  0
A, mm 2
y , mm
yA, mm3
Flng.
180  40  7200
20
144,000
Web
80  60  4800
80
384,000

12,000
528,000
Then
Y  A   yA : X (12,000 mm2 )  528,000 mm3
or
Y  44.0 mm (measured from top of flange)
Now
I x  ( I x ) F  ( I x )W
where




1
2
(180 mm)(40 mm)3  7200 mm 2  44 mm  20 mm   5.170  106 mm 4
12
1

( I x )W   (60 mm)(80 mm)3  4800 mm 2 (80 mm  44 mm)2   8.78  106 mm 4
12

(I x )F 
Then
I x  (5.107  8.78)  106 mm 4
or
Also
I y  ( I y ) F  ( I y )W
Then
1
3
3
1
I y    40 180    80  60   mm 4
12
12

I x  13.89  106 mm 4 
or I y  20.9  106 mm 4 
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PROBLEM 9.42
Determine the moments of inertia I x and I y of the area shown with respect to
centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate C of the area.
Symmetry implies X  18 mm.
A, mm 2
y , mm
yA, mm3
1
36  28  1008
14
14,112
2
1
 36  42  756
2
42
31,752

Dimensions in mm
1764
45,864
Y  A   yA: Y (1764 mm 2 )  45,864 mm3
Then
or
Y  26.0 mm
I x  ( I x )1  ( I x )2
Now
where
1
(36 mm)(28 mm)3  (1008 mm 2 )[(26  14) mm]2
12
 (65,856  145,152) mm 4  211.008  103 mm 4
( I x )1 
1
(36 mm)(42 mm)3  (756 mm 2 )[(42  26) mm]2
36
 (74, 088  193,536) mm 4  267.624  103 mm 4
( I x )2 
Then
I x  (211.008  267.624)  103 mm4
or
I x  479  103 mm 4 

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PROBLEM 9.42 (Continued)
I y  ( I y )1  ( I y )2
Also
where
1
(28 mm)(36 mm)3  108.864  103 mm 4
12
1

1

( I y )2  2  (42 mm)(18 mm)3    18 mm  42 mm  (6 mm) 2 
36
2




( I y )1 
 2(6804  13,608) mm 4  40.824  103 mm 4
[( I y )2 is obtained by dividing A2 into
Then
]
I y  (108.864  40.824  103 mm 4
or I y  149.7  103 mm 4 
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PROBLEM 9.43
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid C of the area.
A, cm2
x , cm
y , cm
xA, cm3
yA, cm3
1
5 × 8 = 40
2.5
4
100
160
2
−2 × 5 = −10
1.9
4.3
–19
–43
Σ
30
81
117
X ΣA = Σ xA: X (30 cm2) = 81 cm3
Then
or
X = 2.70 cm
and
Y ΣA = Σ yA: Y (30 cm2) = 117 cm3
or
Y = 3.90 cm
I x = ( I x )1 − ( I x )2
Now
where
1
(5 cm)(8 cm)3 + (40 cm2)[(4 − 3.9)cm]2
12
= (213.33 + 0.4)cm4 = 213.73 cm4
( I x )1 =
1
(2 cm)(5 cm)3 + (10 cm2)[(4.3 − 3.9)cm]2
12
= (20.83 + 1.60) = 22.43 cm4
( I x )2 =
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PROBLEM 9.43 (Continued)
Then
I x = (213.73 − 22.43) cm4
Also
I y = ( I y )1 − ( I y ) 2
where
or I x = 1.91 × 106 mm4
1
(8 cm)(5 cm)3 + (40 cm2)[(2.7 − 2.5)cm]2
12
= (83.333 + 1.6) cm4 = 84.933 cm4
( I y )1 =
1
(5 cm)(2 cm)3 + (10 cm2)[(2.7 − 1.9)cm]2
12
= (3.333 + 6.4)cm4 = 9.733 cm4
( I y )2 =
Then
I y = (84.933 − 9.733) cm4
or
I y = 752 × 103 mm4
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PROBLEM 9.44
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
First locate centroid C of the area.
Then
A, cm2
x , cm
y , cm
xA, cm3
yA, cm3
1
3.6 × 0.5 = 1.8
1.8
0.25
3.24
0.45
2
0.5 × 3.8 = 1.9
0.25
2.4
0.475
4.56
3
1.3 × 1 = 1.3
0.65
4.8
0.845
6.24
Σ
5.0
4.560
11.25
X ΣA = Σ xA: X (5 cm2) = 4.560 cm3
or
X = 0.912 cm
and
Y ΣA = Σ yA: Y (5 cm2) = 11.25 cm3
or
Y = 2.25 cm
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PROBLEM 9.44 (Continued)
I x = ( I x )1 + ( I x )2 + ( I x )3
Now
where
1
(3.6 cm)(0.5 cm)3 + (1.8 cm2)[(2.25 − 0.25)cm]2
12
= (0.0375 + 7.20) cm4 = 7.2375 cm4
( I x )1 =
1
(0.5 cm)(3.8 cm)3 + (1.9 cm2)[(2.4 − 2.25)cm]2
12
= (2.2863 + 0.0428) cm4 = 2.3291 cm4
( I x )2 =
1
(1.3 cm)(1 cm)3 + (1.3 cm2)[(4.8 − 2.25)cm]2
12
= (0.1083 + 8.4533) cm4 = 8.5616 cm4
( I x )3 =
I x = (7.2375 + 2.3291 + 8.5616) cm4 = 18.1282 cm4
Then
or I x = 181.3 × 103 mm4
I y = ( I y )1 + ( I y ) 2 + ( I y )3
Also
where
1
(0.5 cm)(3.6 cm)3 + (1.8 cm2)[(1.8 − 0.912)cm]2
12
= (1.9440 + 1.4194) cm4 = 3.3634 cm4
( I y )1 =
1
(3.8 cm)(0.5 cm)3 + (1.9 cm2)[(0.912 − 0.25)cm]2
12
= (0.0396 + 0.8327) cm4 = 0.8723 cm4
( I y )2 =
1
(1 cm)(1.3 cm)3 + (1.3 cm2)[(0.912 − 0.65)cm]2
12
= (0.1831 + 0.0892) cm4 = 0.2723 cm4
( I y )3 =
Then
I y = (3.3634 + 0.8723 + 0.2723) cm4 = 4.5080 cm4
or
I y = 45.1 × 103 mm4
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PROBLEM 9.45
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
A, cm2
1
π
2
8
(6)2 = 56.5487
2
1
− (12)(4.5) = −27
2
Σ
29.5487
π
y , cm
yA, cm3
= 2.5465
144
1.5
–40.5
103.5
Y ΣA = Σ yA: Y (29.5487 cm2) = 103.5 cm3
Then
Y = 3.5027 cm
or
J O = ( J O )1 − ( J O ) 2
(a)
where
( J O )1 =
π
4
( J O ) 2 = ( I x′ ) 2 + ( I y ′ ) 2
1
(12 cm)(4.5 cm)3 = 91.125 cm4
12
Now
( I x′ ) 2 =
and
( I y′ ) 2 = 2
[Note: ( I y ′ ) 2 is obtained using
(6 cm)4 = 107.876 cm4
1
(4.5 cm)(6 cm)3 = 162.0 cm4
12
]
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PROBLEM 9.45 (Continued)
Then
Finally
( J O ) 2 = (91.125 + 162.0) cm4
= 253.125 cm4
J O = (1017.876 − 253.125) cm4 = 764.751 cm4
or
JO = 7.65 × 106 mm4
J O = J C + AY 2
(b)
or
J C = 764.751 cm4 − (29.5487 cm2)(3.5027 cm)2
or
JC = 4.02 × 106 mm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.46
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.
SOLUTION
Determination of centroid C of entire section:
Area, cm2
x , cm
xA, cm3
(4) 2  4
16
3
21.333
2
1
(4)(4)  8
2

4
3
10.6667
3
1
(4)(4)  8
2
4
3
10.6667

28.566
Section
1

4
21.333
X  A   xA: X (28.566 cm2)  21.333 cm3
X  0.74680 cm; by symmetry, Y  X  0.74680 cm
(a)
(b)
For section :
Ix 
11 4 1
4
4
  r    (4)  50.265 cm
44
 16
For sections ,:
Ix 
1 3 1
bh  (4)(4)3  21.333 cm 4
12
12
For total area,
I x  50.265  2(21.333)  92.931 cm 4
By symmetry,
I y  I x ; J O  I x  I y  185.862 cm4
Parallel axis theorem:
J C  J O  A(OC )2  J O  A( X 2  Y 2 )
J C  185.862  (28.566)(0.746802  0.746802 )
J O  185.9 cm4 
J C  154.0 cm4 
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PROBLEM 9.47
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
Symmetry: X  0
Dimensions in mm
Determination of centroid C of entire section
Y  A   yA
Y (4000 mm 2 )  122.67  103 mm3
Y  30.667 mm
Area mm2
(a)
1
1
(160)(80)  6400
2
2
1
 (80)(60)  2400
2

4000
y , mm
yA, mm3
80
3
170.67  103
20
48  103
122.67  103
Polar moment of inertia J O :
Section :
Ix 
1
(160)(80)3  6.8267  106 mm 4
12
For Iy consider the following two triangles
1

I y  2  (80)(80)3   6.8267  106 mm 4
12


J O  I x  I y  (6.8267  6.8267)106  13.653  106 mm4
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PROBLEM 9.47 (Continued)
(b)
Section :
Ix 
1
(80)(60)3  1.44  106 mm 4
12
For Iy consider the following two triangles
1

I y  2  (60)(40)3   0.640  106 mm4
12

J O  I x  I y  (1.44  0.640)106  2.08  106 mm4
Entire section
J O  ( J O )1  ( J O )2  13.653  106  2.08  106
J O  11.573  106 mm 4
(c)
J O  11.57  106 mm4 
Polar moment of inertia J O of intire area
J O  J C  AY 2
11.573  106  J C  (4000)(30.667)2
J C  7.811  106 mm4
J C  7.81  106 mm4 
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PROBLEM 9.48
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
Note that symmetry implies Y  0.
Dimensions in mm
A, mm 2

1
2
(84)(42)  5541.77

2
2

3

2

4

X , mm

2

56
(42) 2  2770.88

(54)(27)  2290.22

112

72

36
(27)2  1145.11

XA, mm3
 35.6507
197,568
 17.8254
49,392
 22.9183
52,488
 11.4592
–13,122
4877.32
–108,810
X A   xA: X (4877.32 mm 2 )  108,810 mm3
Then
X  22.3094
or
J O  ( J O )1  ( J O ) 2  ( J O )3  ( J O ) 4
(a)
where

(84 mm)(42 mm)[(84 mm) 2  (42 mm) 2 ]
8
 12.21960  106 mm 4
( J O )1 
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PROBLEM 9.48 (Continued)

(42 mm) 4
4
 2.44392  106 mm 4

( J O )3  (54 mm)(27 mm)[(54 mm)2  (27 mm)2 ]
8
 2.08696  106 mm 4

( J O )4  (27 mm)4
4
 0.41739  106 mm 4
( J O )2 
Then
J O  (12.21960  2.44392  2.08696  0.41739)  106 mm4
 12.15917  106 mm4
or
J O  12.16  106 mm4 
J O  J C  AX 2
(b)
or
J C  12.15917  106 mm4  (4877.32 mm2 )(22.3094 mm)2
or
J C  9.73  106 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.49
Two channels and two plates are used to form the column
section shown. For b  200 mm, determine the moments of
inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
From Figure 9.13B
For C250  22.8
A  2890 mm 2
I x  28.0  106 mm 4
I y  0.945  106 mm 4
Total area
A  2[2890 mm 4  (10 mm)(375 mm)]  13.28  103 mm 2
Given b 200 mm:
1

I x  2[28.0  106 mm 4 ]  2  (375 mm)(10 mm)3  (375 mm)(10 mm)(132 mm)2 
12


 186.743  106
k x2 
I x 186.743  106

 14.0620  103 mm 2
A
13.28  103
I x  186.7  106 mm 4 
k x  118.6 mm 
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PROBLEM 9.49 (Continued)
Channel
Plate
1

I y  ( I y  Ad 2 )  2[ I y  Ad 2 ]  2  (10 mm)(375 mm)3 
12


2

 200 mm
 
 2 0.945  106 mm 4  (2890 mm 2 ) 
 16.10 mm    87.891  106 mm 4
2

 

 2[0.945  106  38.955  106 ]  87.891  106
 167.691  106 mm4
k y2 
Iy
A

167.691  106
 12.6273  103
3
13.28  10
I y  167.7  106 mm 4 
k y  112.4 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9. 50
Two L152 ¥ 102 ¥ 12.7. mm angles are welded together to form the section shown.
Determine the moments of inertia and the radii of gyration of the comb ined section
with respect to the centroidal x and y axes.
12.7 mm
SOLUTION
Based on L152 × 102 × 12.7
From Fig. 9.13A
9.13B
2
4
.000645
m
Area = A = 3060
4.75 ¥mm
-7
6
4
7.2 ×¥10
4.16mm
¥ 10 m4
I x ′ = 17.4
152 mm
-7
¥ 104 m4
6.27 ¥× 4.16
I y ′ = 2.59
106 mm
50.3 mm
12.7 mm
102 mm
24.9 mm
(
)
4
+ +4.75
− 1.990
I x = 2  IIx¢x ′ + Ad
Ad22  == 22 717.4
in2)(76
200in000
(3060
− 50.3
)2 )2 in2 
(3.00

-7
4
17.4×in
= 18.44
4.8455
in 4  = 185 ¥ 10 m4
104 6+mm

Ix = I18.4
× 10
mm
¥610
m44 t
x = 185
-5
k x2 =
mt
= .054
kxk=x 54.9
mm
46
-3 22
I x 18.44
44.491×in10
3.02 ¥10mm
m
=
= 3013.7
2
A 2 24.75
(3060in)
(
)
(
)
2 2 2
− 0.987
I y = 2  I y′ + Ad 2  = 22 2.59
6.27 in
in2 ()(2.25
× 410+6 4.75
+ (3060
57.35
− 24.9
)in 
)

-5
46
4
= 11.62
2 6.27×in10
+mm
7.5771
in 4  = 1.15 ¥ 10 m4
m4 t
Iy =I 11.6
× 10¥610
mm
y = 1.15
-5
k y2 =
Iy
A
=
11.62
1046
-3 22
27.694× in
f 10mm
m
= 1.88
1898.7
2
(3060in)
2 24.75
(
)
kyk=y 43.6
mt
= .043mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.51
Four L76 × 76 × 6.4-mm angles are welded together to a rolled
W section as shown. Determine the moments of inertia and
the radii of gyration of the combined section with respect to
the centroidal x and y axes .
SOLUTION
W section:
A  5880 mm 2
I x  45.8 × 106 mm4
I y  15.4 3 106 mm4
A  929 mm2 I x  0.512 3 106 mm4
Angle:
I y  0.512 3 106 mm4
Atotal  AW  4AA
 5880  4(929)  9.596 3 103 mm2
I x  ( I x )W  4( I x )A
Now
where
( I x ) A  I xA  Ad 2
 0.512 3 106 mm4  (929 mm2)(101.5 mm  21.2 mm) 2  14.498 3 106 mm4
Then
I x  (45.8 + 4 × 14.498) 3 106 mm4 = 103.792 3 106 mm4
and
k x2 
Also
I y  ( I y ) W  4( I y ) A
where
I x  103.8 3 106 mm4 
Ix
103.792 3 106 mm4

9.596 3 103 mm2
Atotal
kx  104 mm 
 I y A  I y  Ad 2
A
 (0.512 3 10 6  929(125 2 21.2)2)mm4 5 10.521 3 106 mm4
I y  (15.4 + 4 3 10.521) 3 106 mm4 5 57.484 3 106 mm4
k y2 
Iy
 57.484 3 103 mm2
9.596 3 10 mm
Atotal
6
4
I y  57.5 3 106 mm4 
k y  77.4 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.52
Two 20-mm steel plates are welded to a rolled S section as shown. Determine
the moments of inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
A  6010 mm 2
S section:
I x  90.3  106 mm 4
I y  3.88  106 mm 4
Atotal  AS  2 Aplate
Note:
 6010 mm2  2(160 mm)(20 mm)
 12, 410 mm2
I x  ( I x )S  2( I x ) plate
Now
( I x ) plate  I xplate  Ad 2
where
1
(160 mm)(20 mm)3  (3200 mm 2 )[(152.5  10) mm]2
12
 84.6067  106 mm 4

I x  (90.3  2  84.6067)  106 mm4
Then
 259.5134  106 mm 4
k x2 
and
Also
Ix
259.5134  106 mm 4

Atotal
12410 mm 2
or I x  260  106 mm 4 
or
kx  144.6 mm 
I y  ( I y )S  2( I y ) plate
1

 3.88  106 mm 4  2  (20 mm)(160 mm)3 
12


 17.5333  106 mm 4
and
k y2 
Iy
Atotal

17.5333  106 mm 4
12, 410 mm 2
or I y  17.53  106 mm  
or
k y  37.6 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.53
C200 × 17.1
12 mm
300 mm
A channel and a plate are welded together as shown to form
a section that is symmetrical with respect to the y axis.
Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.
SOLUTION
57.4 + 12 − 14.5 = 48.9 mm
2
14.5 mm
57.4 mm
12 mm
300 mm
14.5 mm
From Figure 9.13B
B = 2 1 7 0 m m2
6
For C200 × 17.1:
I x''′ = 0.545 × 10 mm4
(Note change of axes)
I y = 13.5 × 10 mm
6
4
Location of centroid
Y Σ A = Σ y A:
Y [2170 mm2 + (300 mm)(12 mm)] = (2170 mm2 )(48.9 mm)
Y = 18.39 mm
Moment of inertia with respect to x axis.
1
(300 mm)(12 mm)3 + (300mm)(12 mm)(18.39)2
12
= 259200 + 1217491 = 1476691 mm4
Plate:
I x = I x + AY 2 =
Channel:
I x = I x′ + AY02 = 0.545 × 10 mm4 + (2170 mm )(48.9 mm − 18.39)2
6
2
= 2.564996 mm4
Entire section:
I x = 1476691 mm4 + 2564996 mm 4 = 4041687 mm4
6
I x = 4.04 × 10 mm4 
Moment of inertia with respect to y axis.
1
6
(12 mm)(300 mm)3 = 27 ×10 mm4
12
Plate:
Iy =
Channel:
I y = 13 .5 × 106 mm4
Entire section:
I y = 10 ( 27 + 13.5) = 40.5 × 10 mm4
6
6
I y = 40.5 mm4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.54
The strength of the rolled W section shown is increased by welding a channel to
its upper flange. Determine the moments of inertia of the combined section with
respect to its centroidal x and y axes.
SOLUTION
A  14, 400 mm 2
W section:
I x  554  106 mm 4
I y  63.3  106 mm 4
A  2890 mm2
Channel:
I x  0.945  106 mm4
I y  28.0  106 mm 4
First locate centroid C of the section.
A, mm2
y , mm
yA, mm3
W Section
14,400
231
33,26,400
Channel
2,890
49.9
1,44,211

17,290
31,82,189
Y  A   y A: Y (17, 290 mm 2 )  3,182,189 mm3
Then
Y  184.047 mm
or
I x  ( I x ) W  ( I x )C
Now
where
( I x ) W  I x  Ad 2
 554  106 mm 4  (14, 400 mm 2 )(231  184.047)2 mm 2
 585.75  106 mm 4
( I x )C  I x  Ad 2
 0.945  106 mm 4  (2,890 mm 2 )(49.9  184.047)2 mm 2
 159.12  106 mm 4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.54 (Continued)
Then
also
I x  (585.75  159.12)  106 mm4
or
I x  745  106 mm4 
or
I y  91.3  106 mm 4 
I y  ( I y ) W  ( I y )C
 (63.3  28.0)  106 mm 4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.55
Two L76  76  6.4-mm angles are welded to a C250  22.8 channel.
Determine the moments of inertia of the combined section with respect to
centroidal axes respectively parallel and perpendicular to the web of the
channel.
SOLUTION
A  929 mm 2
Angle:
I x  I y  0.512  106 mm 4
A  2890 mm2
Channel:
I x  0.945  106 mm4
I y  28.0  106 mm4
First locate centroid C of the section
A, mm 2
y , mm
yA, mm3
Angle
2(929)  1858
21.2
39,389.6
Channel
2890
16.1
46,529

4748
7139.4
Y  A   y A: Y (4748 mm2 )  7139.4 mm3
Then
Y  1.50366 mm
or
I x  2( I x ) L  ( I x )C
Now
where
( I x ) L  I x  Ad   0.512  106 mm 4  (929 mm 2 )[(21.2  1.50366) mm]2
 0.990859  106 mm 4
( I x )C  I x  Ad 2  0.949  106 mm 4  (2890 mm 2 )[(16.1  1.50366) mm]2
 1.56472  106 mm 4
Then
I x  [2(0.990859)  1.56472  106 ] mm 4
or
I x  3.55  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.55 (Continued)
I y  2( I y ) L  ( I y )C
Also
where
( I y ) L  I y  Ad 2  0.512  106 mm 4  (929 mm 2 )[(127  21.2) mm]2
 10.9109  106 mm 4
( I y )C  I y
Then
I y  [2(10.9109)  28.0]  106 mm 4
or
I y  49.8  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.56
Two steel plates are welded to a rolled W section as indicated.
Knowing that the centroidal moments of inertia I x and I y of
the combined section are equal, determine (a) the distance a,
(b) the moments of inertia with respect to the centroidal x and y
axes.
SOLUTION
A  7230 mm2, I x  160 3 106 mm4,
For W-section:
I y  11.1 3 106 mm4
Section
Area, mm2
x mm
xA, mm3
Plate
2(25)(650)5 32500
a
32500a
W-section
7230



39730
32500a
X A  xA :
X (39730 mm2) 5 32500a mm3
x  0.81802 a mm
Entire section:
I x  ( I xW  2 I xPL );
I xPL  I x  Ad 2
1
(650 mm) (25 mm)3 + (16250 mm2)(179 mm + 12.5 mm)2
12
= 596.77 3 106 mm4
I x  (160 + 2(596.77)) 3 106 5 1353.54 3 106 mm4 I x  755.9 3 10 6 

Ix  I y
I y  ( I yW  2 I yPL );
I y  755.9 3 10 6 
I yPL  I y  Ad 2

1
(25 mm)(650 mm)3 + (650 mm2)(a 2 0.81802a mm)2
12
= (572.135 3 106 + 21.526a 2) mm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.56 (Continued)
I yW  I yW  Ad 2
 11.1 3 106 mm4 + (7230 mm2)(0.81802a mm)2
 (11.1 3 106 + 4838 a 2) mm4
I y  (11.1 3 106 + 4838 a2) mm4 + 2(572.135 3 106 + 21.526a2 5 1353.54 3 106)
a  202 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.57
The panel shown forms the end of a trough that is filled with water to the
line AA. Referring to section 9.2, determine the depth of the point of
application of the resultant of the hydrostatic forces acting on the panel (the
center of pressure).
SOLUTION
From section 9.2:


R   ydA, M AA   y 2 dA
Let yP  distance of center of pressure from AA. We must have:


 y dA I
M
 AA
RyP  M AA : yP  AA 
R
 ydA y A
2
(1)
For triangular panel:
I AA 
yP 
1 3
ah
12
1
y h
3
1
ah3
I AA
 12
yA  13 h   12 ah 
A
1
ah
2
yP 
1
h
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.58
The panel shown forms the end of a trough that is filled with water
to the line AA. Referring to Section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 506 of the text:
yP 
I AA
yA
YA   yA
Now
h
4 1

(2b  h)  h   2b  h 
2
3 2

7
 bh 2
3

I AA  ( I AA )1  ( I AA ) 2
and
where
1
2
( I AA )1  (2b)(h)3  bh3
3
3
1
1
 4 
(2b)(h)3    2b  h  h 
36
2
 3 
11
 bh3
6
2
( I AA )2  I x  Ad 2 
2 3 11 3 5 3
bh  bh  bh
3
6
2
Then
I AA 
Finally,
5 3
bh
yP  2
7 2
bh
3
or
yP 
15
h
14
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.59
The panel shown forms the end of a trough that is filled with water to the line
AA. Referring to section 9.2, determine the depth of the point of application of
the resultant of the hydrostatic forces acting on the panel (the center of
pressure).
SOLUTION
Using the equation developed on page 506 of the text have:
yP 
For a quarter circle:
Then
I AA 
I AA
yA

16
4r
y
3
yP 
r4
A
 r4
16
4r
 4 r 2
3
or

4
r2
or
yP 
3
r
16
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.60*
The panel shown forms the end of a trough that is filled with water
to the line AA. Referring to section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 506 of the text:
For a parabola:
yP 
I AA
yA
y
2
h
5
A
4
ah
3
1
dI AA  (h  y )3 dx
3
Now
By observation
y
h 2
x
a2
3
So that
1
h
1 h3 2

dI AA   h  2 x 2  dx 
(a  x 2 )3 dx
3
3 a6
a

1 h 6

( a  3a 4 x 2  3a 2 x 4  x 6 )dx
6
3a
I AA  2
Then
a 1 h3
 3 a (a  3a x  3a x  x )dx
0
6
4 2
2 4
6
6
a
Finally,

2 h3  6
3
1 
a x  a 4 x3  a 2 x5  x 7 
6 
3a 
5
7 0

32 3
ah
105
32
ah3
y p  2 105 4
h  3 ah
5
or
yP 
4
h
7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.61
A vertical trapezoidal gate that is used as an automatic valve is held
shut by two springs attached to hinges located along edge AB.
Knowing that each spring exerts a couple of magnitude 1470 N · m,
determine the depth d of water for which the gate will open.
SOLUTION
From section 9.2:
R   yA
yP 
I SS 
yA
where R is the resultant of the hydrostatic forces acting on the gate
and yP is the depth to the point of application of R. Now
1

1

yA   yA  [(h  0.17) m]   1.2 m  0.51 m   [( h  0.34) m]   0.84  0.51 m 
2

2

3
 (0.5202h  0.124848) m
Recalling that   py, we have
R  (103 kg/m3  9.81 m/s 2 )(0.5202h  0.124848) m3
 5103.162(h  0.24) N
I SS   ( I SS  )1  ( I SS  ) 2
Also
where
( I SS  )1  I x  Ad 2

1
1

(1.2 m)(0.51 m)3    1.2 m  0.51 m  [(h  0.17) m]2
36
2


 [0.0044217  0.306(h  0.17)2 ] m 4
 (0.306h 2  0.10404h  0.0132651) m 4
( I SS  ) 2  I X 2  Ad 2

1
1

(0.84 m)(0.51 m)3    0.84 m  0.51 m  [( h  0.34) m]2
36
2

 [0.0030952  0.2142( h  0.34) 2 ] m 4
 (0.2142h 2  0.145656 h  0.0278567) m 4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.61 (Continued)
I SS   ( I SS  )1  ( I SS  ) 2
Then
 (0.5202 h 2  0.249696h  0.0411218) m 4
yP 
and

(0.5202h 2  0.244696h  0.0411218) m 4
(0.5202h  0.124848) m3
h 2  0.48h  0.07905
m
h  0.24
For the gate to open, require that
M AB : M open  ( yP  h) R
Substituting
or
 h2  0.48h  0.07905

2940 N  m  
 h  m  5103.162(h  0.24) N
h  0.24


5103.162(0.24h  0.07905)  2940
or
h  2.0711 m
Then
d  (2.0711  0.79) m
or
d  2.86 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.62
The cover for a 0.5-m-diameter access hole in a water storage
tank is attached to the tank with four equally spaced bolts as
shown. Determine the additional force on each bolt due to the
water pressure when the center of the cover is located 1.4 m
below the water surface.
SOLUTION
From section 9.2:
R   yA
yP 
I AA
yA
where R is the resultant of the hydrostatic forces acting on the cover
and yP is the depth to the point of application of R.
Recalling that   p  y, we have
R  (103 kg/m3  9.81 m/s 2 )(1.4 m)[ (0.25 m) 2 ]
 2696.67 N

(0.25 m)4  [ (0.25 m)2 ](1.4 m)2
4
 0.387913 m 4
Also
I AA  I x  Ay 2 
Then
yP 
.387913 m 4
 1.41116 m
(1.4 m)[ (0.25 m)2 ]
Now note that symmetry implies
FA  FB
FC  FD
Next consider the free-body of the cover.
We have
M CD  0: [2(0.32 m) sin 45](2 FA )
 [0.32 sin 45  (1.41116  1.4)] m
 (2696.67 N)  0
or
FA  640.92 N
Then
Fz  0: 2(640.92 N)  2 FC  2696.67 N  0
or
FC  707.42 N
FA  FB  641 N 
and
FC  FD  707 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.63*
Determine the x coordinate of the centroid of the volume shown.
(Hint: The height y of the volume is proportional to the x coordinate;
consider an analogy between this height and the water pressure on a
submerged surface.)
SOLUTION
y
First note that


x dV  xEL dV
Now
where
Then
h
x
a
h 
xEL  x dV  ydA   x  dA
a 
 x  xdA   x dA  (I )
x
 xdA  xdA ( xA)
h
a
2
h
a
z A
A
where ( I z ) A and ( x A ) A pertain to area.
OABC: ( I z ) A is the moment of inertia of the area with respect to the z axis, x A is the x coordinate
of the centroid of the area, and A is the area of OABC. Then
( I z ) A  ( I z ) A1  ( I z ) A2
1
1
 (b)(a )3  (b)(a )3
3
12
5 3
 ab
12
and
( xA) A   xA
 a 
  a   1

   ( a  b )       a  b  

 2 
  3   2
2
 a 2b
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.63* (Continued)
Finally,
5 3
ab
x  12 2
2
ab
3
or
5
x  a 
8
Analogy with hydrostatic pressure on a submerged plate:
Recalling that P   y , it follows that the following analogies can be established.
Height y ~ P
dV  ydA ~ pdA  dF
xdV  x( ydA) ~ ydF  dM
Recalling that

dM 
Mx 

yP 

R 
dF 


It can then be concluded that
x ~ yP


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.64*
Determine the x coordinate of the centroid of the volume shown;
this volume was obtained by intersecting an elliptic cylinder with
an oblique plane. (Hint: The height y of the volume is
proportional to the x coordinate; consider an analogy between this
height and the water pressure on a submerged surface.)
SOLUTION
Following the “Hint,” it can be shown that (see solution to Problem 9.63)
x
(I z ) A
x
( xA) A
where ( I z ) A and ( xA) A are the moment of inertia and the first moment of the area, respectively, of the
elliptical area of the base of the volume with respect to the z axis. Then
( I z ) A  I z  Ad 2

(39 mm)(64 mm)3  [ (64 mm)(39 mm)](64 mm)2
4
 12.779520  106 mm 4

( xA) A  (64 mm)[ (64 mm)(39 mm)]
 0.159744  106 mm 3
Finally
x
12.779520  106 mm 4
0.159744  106 mm 4
or
x  80.0 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.65*
Show that the system of hydrostatic forces acting on a
submerged plane area A can be reduced to a force P at the
centroid C of the area and two couples. The force P is
perpendicular to the area and is of magnitude P   Ay sin  ,
where  is the specific weight of the liquid, and the couples are
M x  ( I x sin  )i
and
M y   ( I xy  sin  ) j, where
I xy    x y  dA (see section 9.8). Note that the couples are
independent of the depth at which the area is submerged.
SOLUTION
The pressure p at an arbitrary depth ( y sin  ) is
p   ( y sin  )
so that the hydrostatic force dF exerted on an infinitesimal area dA is
dF  ( y sin  )dA
Equivalence of the force P and the system of infinitesimal forces dF requires



F : P  dF   y sin  dA   sin  y dA
or
P   Ay sin  
Equivalence of the force and couple (P, M x  M y ) and the system of infinitesimal hydrostatic
forces requires

M x :  yP  M x  ( ydF )
Now



 ydF   y ( y sin  )dA   sin  y 2 dA
 ( sin  ) I x
Then
or
 yP  M x  ( sin  ) I x
M x  ( sin  ) I x  y ( Ay sin  )
  sin  ( I x  Ay 2 )
or
M x   I x sin  

M y : xP  M y  x dF
Now
 x dF   x( y sin  )dA   sin   xy dA
 ( sin  ) I xy
(Equation 9.12)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.65* (Continued)
Then
or
xP  M y  ( sin  ) I xy
M y  ( sin  ) I xy  x ( Ay sin  )
  sin  ( I xy  Ax y )
or
M y    I x y  sin  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.66*
Show that the resultant of the hydrostatic forces acting on a
submerged plane area A is a force P perpendicular to the area
and of magnitude P   Ay sin   pA, where  is the specific
weight of the liquid and p is the pressure at the centroid C
of the area. Show that P is applied at a Point CP, called the
center of pressure, whose coordinates are xP  I xy /Ay and
y P  I x /Ay , where I xy   xydA (see section 9.8). Show also
that the difference of ordinates y P  y is equal to k x2 / y and
thus depends upon the depth at which the area is submerged.
SOLUTION
The pressure P at an arbitrary depth ( y sin  ) is
P   ( y sin  )
so that the hydrostatic force dP exerted on an infinitesimal area dA is
dP  ( y sin  )dA
The magnitude P of the resultant force acting on the plane area is then



P  dP   y sin  dA   sin  ydA
Now
  sin  ( yA)
p   y sin 
P  pA 
Next observe that the resultant P is equivalent to the system of infinitesimal forces d P . Equivalence then
requires

M x :  yP P   ydP
Now
 ydP   y( y sin  )dA   sin   y dA
2
 ( sin  ) I x
Then
y P P  ( sin  ) I x
or
yP 
( sin  ) I x
 sin  ( yA)
or
yP 
Ix

Ay
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.66* (Continued)

M y : xP P  xdP
 xdP   x( y sin  )dA   sin   xydA
Now
 ( sin  ) I xy
Then
xP P  ( sin  ) I xy
or
xP 
(Equation 9.12)
( sin  ) I xy
 sin  ( yA)
or
Now
I x  I x  Ay 2
From above
I x  ( A y ) yP
By definition
I x  k x A
Substituting
( Ay ) yP  k x2 A  Ay 2
xP 
I xy
Ay

2
yP  y 
Rearranging yields
k x2

y
Although k x is not a function of the depth of the area (it depends only on the shape of A), y is dependent
on the depth.
( y P  y )  f (depth) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.67
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.
SOLUTION
y  a 1
First note

x2
4a 2
1
4a 2  x 2
2
We have
dI xy  dI xy   xEL y EL dA
where
dI xy   0 (symmetry)
y EL 
1
1
y
4a 2  x 2
2
4
dA  ydx 
Then
xEL  x
1
4a 2  x 2 dx
2

I xy  dI xy 
2a
1
2  1
2 
 x  4 4a  x  2 4a  x  dx
2
2
0
2a
1 2a
1
1 

(4a 2 x  x3 )dx   2a 2 x 2  x 4 
0
8
8
4 0


1
a4 

2(2)2  (2)4 

8 
4

or
I xy 
1 4
a 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.68
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.
SOLUTION
dI xy  dI xy   xEL y EL dA
But
dI xy  0 (by symmetry) xEL  x
yEL 
1
h
2
dA  hd x

I xy  dI xy 

b
1 
1
b
 x  2 h  hdx  2 h  xdx
0
2
0
2
1 2b
h
2
2
I xy 
1 2 2
b h 
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.69
Determine by direct integration the product of inertia of the given area with
respect to the x and y axes.
SOLUTION
For
x  a, b  ka1 2
Thus,
y
b
a1 2
or k 
b 12
x
a1 2
dI xy  dI xy   xEL y EL dA, But
with
xEL  x
yEL 

I xy  dI xy 
1 b2

2 a
y
2
dA  ydx
1
 y
x   ydx 
0  2
2

dI xy   0 (by symmetry)
a

2
 b

x  1 2 x1 2  dx
0 a

a
a
b 2  x3 
x dx 
 
0
2a  3  0

a
2
I xy  a 2 b 2 /6 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.70
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.
SOLUTION
For
x  a,
Thus,
y
y   a;  a 
k
a
xEL  x

k  a 2
a2
x
dI xy  dI xy   xEL y EL dA, But
with
or
y
1  a2 
    dA  ydx
2
2 x 
y EL 
I xy  dI xy 
dI xy   0 (by symmetry)
 1 a2   a2 
a4
x 
   dx  
a
2
 2 x  x 

a

2a 1
a
x
dx
a4
2a
ln x a

2
a 4 2a
  ln
a
2

I xy  0.347 a 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.71
Using the parallel-axis theorem, determine the product of
inertia of the area shown with respect to the centroidal x
and y axes.
SOLUTION
Dimensions in mm
I xy  ( I xy )1  ( I xy ) 2  ( I xy )3
We have
Now symmetry implies
( I xy ) 2  0
and for the other rectangles
I xy  I x y   x yA
where
I xy  0 (symmetry)
Thus
I xy  ( x yA)1  ( x yA)3
A, mm2
x , mm
y , mm
x yA, mm4
1
10  80  800
55
20
880,000
3
10  80  800
55
20
880,000

1,760,000
I xy  1.760  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.72
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.
SOLUTION
Given area  Rectangle  (Two triangles)
For rectangle I xy  0, by symmetry
For each triangle:
Compare these triangles with the triangle of sample Problem 9.6, where I xy  
1 2 2
b h .
72
For orientation of axes of this problem,
I xy  
1 2 2 1
b h 
(60 mm)2 (40 mm)2  80  103 mm 4
72
72
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.72 (Continued)
Area, mm2
x , mm
y , mm
x yA, mm 4
1
1
(60)(40)  1200
2
40
26.67
1.280  106
2
1
(60)(40)  1200
2
40
26.67
1.280  106
 x y A  2.56  106 mm4
For two triangles:
I xy  ( I xy  x y A)  2(80  103 )  2.56  106
 2.40  106 mm 4
Since we must subtract triangles,
I xy  2.40  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.73
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2
For each semicircle
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
Thus
I xy = Σ x yA
1
2
Σ
π
2
π
2
A, cm2
x , cm
y , cm
(6) 2 = 18π
−3
−
(6) 2 = 18π
3
8
π
8
π
xyA
432
432
864
I xy = 8.64 × 106
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.74
24.9 mm
4
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
24.9
4
4
24.9
I xy = ( I xy )1 + ( I xy ) 2
We have
For each rectangle
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
Thus
I xy = Σ x yA
A, mm2
x , mm
y , mm
x yA, mm4
1
76 × 6.4 = 486.4
−13.1
9.2
−58620.9
2
6.4 × 44.6 = 285.44
21.7
−16.3
−100963.0
−159583.9
Σ
I xy = −159.6 × 103 mm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1478
PROBLEM 9.75
Using the parallel-axis theorem, determine the product of
inertia of the area shown with respect to the centroidal x and
y axes.
SOLUTION
I xy  ( I xy )1  ( I xy ) 2  ( I xy )3
We have
( I xy )1  0
Now symmetry implies
and for the other rectangles
I xy  I x y   x yA
where
I x y   0 (symmetry)
Thus
I xy  ( x yA) 2  ( x yA)3
A, mm 2
X , mm
y , mm
x yA, mm4
2
8  32  256
–46
–20
235,520
3
8  32  256
46
20
235,520

471,040
I xy  471  103 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.76
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3
We have
Now, symmetry implies
( I xy )1 = 0
I xy = I x′y′ + x yA
and for each triangle
where, using the results of Sample Problem 9.6, I x′y′ = − 72 b 2 h 2 for both triangles. Note that the sign of I x′y′
1
is unchanged because the angles of rotation are 0° and 180° for triangles 2 and 3, respectively.
Now
A, cm2
x , cm
y , cm
x yA, cm4
2
1
(9)(15) = 67.5
2
–9
7
–4252.5
3
1
(9)(15) = 67.5
2
9
–7
–4252.5
Σ
Then
–8505
I xy = 2 −
1
(9 cm)2 (15 cm)2 − 8505 cm4
72
= −9011.25 cm4
or
I xy = −90.1 × 106 mm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.77
1.3 cm
1.0 cm
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
y
0.412 cm
5.3 cm
C
x
2.25 cm
0.5 cm
3.6 cm
0.5 cm
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3
For each rectangle
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
Thus
I xy = Σ x yA
A, cm2
x , cm
y , cm
x yA, cm4
1
3.6 × 0.5 = 1.8
0.888
–2.00
–3.1968
2
0.5 × 3.8 = 1.9
–0.662
0.15
–0.18867
3
1.3 × 1.0 = 1.3
–0.262
2.55
–0.86853
Σ
–4.254
I xy = −42.5 × 103 mm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.78
Using the parallel-axis theorem, determine the product of
inertia of the area shown with respect to the centroidal x
and y axes.
SOLUTION
We have
I xy  ( I xy )1  ( I xy ) 2
For each rectangle
I xy  I x y   x yA
and
I x y   0 (symmetry)
Thus
I xy   x yA
1
2

A, mm 2
x , mm
y , mm
x yA, mm4
139.3  12.7  1769.11
+32.05
+18.55
1051.78  103
-43.95
-26.1
1485.95  103
12.7  (102)  1295.4
2537.7  103
I xy  2.54  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.79
Determine for the quarter ellipse of Problem 9.67 the moments of
inertia and the product of inertia with respect to new axes obtained by
rotating the x and y axes about O (a) through 45 counterclockwise,
(b) through 30 clockwise.
SOLUTION
From Figure 9.12:
Ix 

Iy 

From Problem 9.67:
First note

16

8
a4

16

2
I xy 
(2a )(a )3
(2a )3 (a )
a4
1 4
a
2
1
1
  5
(I x  I y )   a4  a4    a4
2
2 8
2  16
 
1
1
3
(I x  I y )   a4  a4     a4
2
2 8
2 
16
Now use Equations (9.18), (9.19), and (9.20).
Equation (9.18):
I x 

Equation (9.19):
I y 

Equation (9.20):
I xy 
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
5
3
1
 a 4   a 4 cos 2  a 4 sin 2
16
16
2
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
5
3
1
 a 4   a 4 cos 2  a 4 sin 2
16
16
2
1
( I x  I y )sin 2  I xy cos 2
2

3
1
 a 4 sin 2  a 4 cos 2
16
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.79 (Continued)
(a)
  45:
I x 
I y 
5
3
1
 a 4   a 4 cos 90  a 4 sin 90
16
16
2
or
I x  0.482a 4 
or
I y  1.482a 4 
5
3
1
   a 4 cos 90  a 4
16
16
2
I xy  
3
1
 a 4 sin 90  a 4 cos90
16
2
or I xy  0.589a 4 
(b)
  30:
I x 
5
3
1
 a 4   a 4 cos(60)  a 4 sin(60)
16
16
2
or
I y 
I x  1.120a 4 
5
3
1
 a 4   a 4 cos(60)  a 4 sin(60)
16
16
2
I xy   
or
I y  0.843a 4 
or
I xy  0.760a 4 
3
1
 a 4 sin(60)  a 4 cos(60)
16
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.80
Determine the moments of inertia and the product of inertia of the
area of Problem 9.72 with respect to new centroidal axes obtained
by rotating the x and y axes 30 counterclockwise.
SOLUTION
From Problem 9.72:
I xy  2.40  106 mm 4
Now, with two rectangles and two triangles:
1

1

I x  2  (60 mm)(40 mm)3   2  (60 mm)(40 mm)3 
3
12




I x  3.20  106 mm 4
1

1

I y  2  (40 mm)(60 mm)3   2  (40 mm)(60 mm)3   7.20  106 mm 4
3
12




1
1
( I x  I y )  (3.20  7.20)  106  5.20  106 mm 4
2
2
1
1
( I x  I y )  (3.20  7.20)  106  2.00  106 mm 4
2
2
Now
Using Eqs. (9.18), (9.19), (9.20):
Eq. (9.18):
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
 [5.20  ( 2.00) cos 60  (2.40) sin 60]  106 mm 4
I x 
or
I x  2.12  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.80 (Continued)
Eq. (9.19):
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
 [5.20  (2.00) cos 60  (2.40)sin 60]  106 mm 4
I y 
or
Eq. (9.20):
I y  8.28  106 mm 4 
1
( I x  I y )sin 2  I xy sin 2
2
 [(2.00)sin 60  (2.40) cos 60]  106 mm 4
I xy 
or
I xy  0.532  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.81
Determine the moments of inertia and the product of inertia of the
area of Problem 9.73 with respect to new centroidal axes obtained
by rotating the x and y axes 60° counterclockwise.
SOLUTION
From Problem 9.73:
I xy = 864 cm4
I x = ( I x )1 + ( I x ) 2
Now
( I x )1 = ( I x ) 2 =
π
(6 cm)4
8
= 162π cm4
where
Then
I x = 2(162π cm4) = 324π cm4
Also
I y = ( I y )1 + ( I y ) 2
( I y )1 = ( I y ) 2 =
where
π
8
(6 cm)4 +
π
2
(6 cm)2 (3 cm)2 = 324π cm4
I y = 2(324π cm4) = 648 π cm4
Then
1
1
( I x + I y ) = (324π + 648π ) = 486π cm4
2
2
1
1
( I x − I y ) = (324π − 648π ) = −162π cm4
2
2
Now
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [486π + (−162π ) cos120° − 864sin120°] cm4
I x′ =
or
I x′ = 10.3 × 106 mm4
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PROBLEM 9.81 (Continued)
Eq. (9.19):
Eq. (9.20):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
= [486π − (−162π ) cos120° + 864sin120°] cm4
I y′ =
or
I y′ = 20.2 × 106 mm4
or
I x′y′ = −8.73 × 106 mm4
1
( I x − I y )sin 2θ + I xy cos 2θ
2
= [(−162π ) sin120° + 864cos120°] cm4
I x′y′ =
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.82
Determine the moments of inertia and the product of inertia of the
area of Problem 9.75 with respect to new centroidal axes obtained
by rotating the x and y axes 45 clockwise.
SOLUTION
From Problem 9.75:
I xy  471, 040 mm 4
I x  ( I x )1  ( I x )2  ( I x )3
Now
where
1
(100 mm)(8 mm)3
12
 4266.67 mm 4
( I x )1 
1
(8 mm)(32 mm)3  [(8 mm)(32 mm)](20 mm)2
12
 124, 245.33
( I x ) 2  ( I x )3 
Then
I x  [4266.67  2(124, 245.33)] mm 4  252,757 mm4
Also
I y  ( I y )1  ( I y ) 2  ( I y )3
where
1
(8 mm)(100 mm)3  666,666.7 mm 4
12
1
( I y )2  ( I y )3  (32 mm)(8 mm)3  [(8 mm)(32 mm)](46 mm) 2
12
 543,061.3 mm 4
( I y )1 
I y  [666, 666.7  2(543, 061.3)] mm 4  1, 752, 789 mm 4
Then
1
1
( I x  I y )  (252, 757  1, 752, 789) mm 4  1, 002, 773 mm 4
2
2
1
1
( I x  I y )  (252, 757  1, 752, 789) mm 4  750, 016 mm 4
2
2
Now
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):

1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
 [1, 002,773  (750,016) cos(90)  471,040sin(90)]
I x 
or
I x  1.474  106 mm4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.82 (Continued)
Eq. (9.19):
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
 [1,002, 773  (750,016) cos(90)  471,040sin(90)]
I y 
or
Eq. (9.20):
I y  0.532  106 mm 4 
1
( I x  I y )sin 2  I xy cos 2
2
 [(750,016)sin(90)  471,040cos(90)]
I xy 
or
I xy  0.750  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.83
Determine the moments of inertia and the product of inertia of
the L76 × 51 × 6.4-mm angle cross section of Problem 9.74 with
respect to new centroidal axes obtained by rotating the x and
y axes 30° clockwise.
SOLUTION
From Figure 9.13:
I x = 0.162 × 106 mm4
I y = 0.454 × 106 mm4
From Problem 9.74:
I xy = −159.6 × 103 mm4
1
1
( I x + I y ) = (0.454 + 0.162) 106 mm4 = 0.308 × 106 mm 4
2
2
1
1
( I − I y ) = (0.162 − 0.454) 106 mm4 = −0.146 × 106 mm4
2 x
2
Now
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [0.308 × 10 6 + (−0.146 × 106 ) cos (−60°) − (−0.1596 × 106 )sin(−60°)]
I x′ =
or
Eq. (9.19):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
6
= [0.308 × 10 − (−0.146 × 106) cos (−60°) + (−0.1596 × 106 )sin(−60°)]
I y′ =
or
Eq. (9.20):
I x′ = 96.78 × 103 mm4
I y′ = 519.22 × 103 mm4
1
( I x − I y )sin 2θ + I xy cos 2θ
2
= [(−0.146 × 106) sin(−60°) + (−0.1596 × 106)cos(−60°)]
I x′y′ =
or I x′y′ = 46.64 × 103 mm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.84
Determine the moments of inertia and the product of inertia of the
L152  102  12.7-mm angle cross section of Prob. 9.78 with
respect to new centroidal axes obtained by rotating the x and y axes
30 clockwise.
SOLUTION
From Figure 9.13B:
Note the axes are reversed in sketch.
I x  2.59  106 mm 4
I y  7.20  106 mm 4
From Problem 9.78:
I xy  2.54  106 mm 4
1
1
( I x  I y )  (2.59  7.20)  106 mm 4
2
2
 4.895  106 mm 4
1
1
( I x  I y )  (2.59  7.20)  106 mm 4  2.305  106 mm 4
2
2
Now
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
 [4.895  2.305cos(60)  2.54sin(60)]  106 mm 4
I x 
or
Eq. (9.19):
1
1
( I x  I y )  ( I x  I y ) cos 2  I xy sin 2
2
2
 [4.895  2.305cos( 60)  2.54sin( 60)]  106 mm 4
I y 
or
Eq. (9.20):
I x  5.94  106 mm 4 
I y  3.85  106 mm 4 
1
( I x  I y )sin 2  I xy cos 2
2
 [(2.305sin(60)  2.54cos(60)]  106 mm 4
I xy 
or
I xy  3.27  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.85
For the quarter ellipse of Problem 9.67, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
SOLUTION

From Problem 9.79:
Ix 
Problem 9.67:
I xy 
Now Eq. (9.25):
tan2 m  
8
a4
Iy 

a4
2
1 4
a
2
2 I xy
Ix  I y


2 12 a 4

 a4   a4
8
2
8

 0.84883
3
2 m  40.326 and
Then
220.326
 m  20.2 and 110.2 
or
Also Eq. (9.27):
I max,min 

Ix  I y
2
2
 Ix  I y 
2
 
  I xy
 2 
1 4  4
a  a 
2  8
2 
2
1  
  1 
   a4  a4    a4 
2  2 
2  8
2
 (0.981,748  0.772,644) a 4
or
I max  1.754a 4 
and
I min  0.209a 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.86
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments of
inertia.
Area of Problem 9.72.
SOLUTION
From Problem 9.80:
I x  3.20  106 mm4
I y  7.20  106 mm4
From Problem 9.72:
Now Eq. (9.25):
I xy  2.40  106 mm 4
tan2 m  
2I xy
Ix  Iy

2(2.40  106 )
 1.200
(3.20  7.20)  106
2 m  50.194 and
Then
230.194
 m  25.1 and 115.1 
or
Ix  I y
2
 Ix  I y 
2
 
  I xy
 2 


Also Eq. (9.27):
I max, min 
Then
2
 3.20  7.20

 3.20  7.20 
2
I max,min  
 

(2.40)
 106 mm 4



2
2




2
 (5.20  3.1241)  106 mm 4
or
I max  8.32  106 mm 4 
and
I min  2.08  106 mm4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.87
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments of
inertia.
Area of Problem 9.73.
SOLUTION
From Problem 9.81:
I x = 324π cm4
Problem 9.73:
I xy = 864 cm4
Now Eq. (9.25):
tan2θ m = −
2 I xy
Ix − I y
=−
I y = 648π cm4
2(864)
324π − 648π
= 1.69765
2θ m = 59.500° and 239.500°
Then
θ m = 29.7° and 119.7°
or
Ix + Iy
Ix − I y
Also Eq. (9.27):
I max, min =
Then
324π + 648π
±
I max,min =
2
2
±
2
2
+ I xy2
324π − 648π
2
2
+ 8642
= (1526.81 ± 1002.75) cm4
or
I max = 25.3 × 106 mm4
and
I min = 5.24 × 106 mm4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.88
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments
of inertia.
Area of Problem 9.75.
SOLUTION
From Problem 9.82:
I x  252,757 mm 4
I y  1, 752, 789 mm 4
Problem 9.75:
Now Eq. (9.25):
I xy  471, 040 mm 4
tan 2 m  
2 I xy
Ix  I y
2(471, 040)
252, 757  1, 752, 789
 0.62804

2 m  32.130 and
Then
212.130°
 m  16.07 and 106.1 
or
Also Eq. (9.27):
I max, min 
Then
I max, min 
Ix  I y
2
2
 Ix  I y 
 
  I xy2
 2 


2
252, 757  1, 752, 789
 252, 757  1, 752, 789 
2
 
  471040
2
2


 (1, 002, 773  885, 665) mm 4
or
I max  1.888  106 mm4 
and
I min  0.1171  106 mm 4 
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
51 mm
PROBLEM 9.89
24.9 mm
6.4 mm
C
For the angle cross section indicated, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
12.4 mm
x
L76 × 51 × 6.4
The L76 × 51 × 6.4-mm angle cross section of Problem 9.74.
6.4 mm
76 mm
SOLUTION
From Problem 9.83:
I x = 0.162 × 106 mm4
Problem 9.74:
I xy = −0.1596 × 10 6 mm4
Now Eq. (9.25):
tan 2θ m = −
2 I xy
=−
Ix − I y
I y = 0.454 × 106 mm4
2(−0.1596)
0.162 − 0.454
= −1.09315
2
2θ m = −47.55° and 132.45°
Then
or
Also Eq. (9.27):
I max, min =
Then
I max, min =
Ix + I y
2
±
Ix − I y
θ m = −23.78° and 66.22°
2
2
+ I xy2
(0.162 + 0.454)106
±
2
0.162 × 106 − 0.454 × 106
2
2
+ (−0.1596 × 106)2
= (308000 ± 216305) mm4
or
I max = 524.3 × 103 mm4
and
I min = 91.7 × 103 mm4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.90
For the angle cross section indicated, determine the
orientation of the principal axes at the origin and the
corresponding values of the moments of inertia.
The L152  102  12.7-mm angle
Problem 9.78.
cross
section
of
SOLUTION
From Figure 9.13B:
I x  2.59  106 mm4
I y  7.20  106 mm 4
Problem 9.78:
Now Eq. (9.25):
I xy  2.54  106 mm 4
tan 2 m  
2 I xy
Ix  I y

2(2.54  106 )
(2.59  7.20)  106
 1.10195
2 m  47.777 and 227.777°
Then
or
Ix  Iy
 m  23.9 and 113.9° 
2
 Ix  I y 
 
  I xy2
 2 


Also Eq. (9.27):
I max, min 
Then
2


2.59  7.20
 2.59  7.20 
2
I max, min  
 

2.54
 106 mm 4



2
2




2
 (4.895  3.4300)  106 mm 4
or
I max  8.33  106 mm 4 
and
I min  1.465  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.91
Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67
the moments of inertia and the product of inertia with respect to new
axes obtained by rotating the x and y axes about O (a) through 45
counterclockwise, (b) through 30 clockwise.
SOLUTION
Ix 
From Problem 9.79:

8

Iy 
2
a4
1 4
a
2
I xy 
Problem 9.67:
a4
The Mohr’s circle is defined by the diameter XY, where
1 
1 


X  a 4, a 4  and Y  a 4,  a 4 
8
2
2
2 



Now
I ave 
1
1
  5
( I x  I y )   a 4  a 4    a 4  0.98175a 4
2
2 8
2  16
2
and
2
2
 Ix  I y 
 1   4  4   1 4 
2
R 
  I xy    a  a     a 
2   2 
2  8
 2 
 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2 m  

2 I xy
Ix  I y

2 12 a 4

8
4


a  2 a4
 0.84883
or
2 m  40.326
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.91 (Continued)
  90  40.326
Then
 49.674
  180  (40.326  60)
 79.674
(a)
  45:
I x  I ave  R cos   0.98175a 4  0.77264a 4 cos 49.674
I x  0.482a 4 
or
I y  I ave  R cos   0.98175a 4  0.77264a 4 cos 49.674
or
I y  1.482a 4 
or
I xy  0.589a 4 
I xy   R sin   0.77264a 4 sin 49.674
(b)
  30:
I x  I ave  R cos   0.98175a 4  0.77264a 4 cos 79.674
or
I x  1.120a 4 
I y  I ave  R cos   0.98175a 4  0.77264a 4 cos 79.674
or
I y  0.843a 4 
or
I xy  0.760a 4 
I xy  R sin   0.77264a 4 sin 79.674
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.92
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the area of Problem 9.72 with respect to new
centroidal axes obtained by rotating the x and y axes 30
counterclockwise.
SOLUTION
I x  3.20  106 mm4
From Problem 9.80:
I y  7.20  106 mm 4
I xy  2.40  106 mm 4
From Problem 9.72:
The Mohr’s circle is defined by the diameter XY, where X (3.20  106 , 2.40  106 )
and Y (7.20  106,  2.40  106 ).
Now
and
I ave 
1
1
( I x  I y )  (3.20  7.20)  106  5.20  106 mm 4
2
2
2
2
 1

 

1


2
R   ( I x  I y )   I xy    (3.20  7.20)   (2.40)2   106 mm 4
2


  2

 3.1241  106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2 m  
2 I xy
Ix  I y
2(2.40  106 )
(3.20  7.20)  106
 1.200

or
Then
2 m  50.1944
  60  50.1944  9.8056
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PROBLEM 9.92 (Continued)
Then
I x  I ave  R cos   (5.20  3.1241cos 9.8056)  106
or
I x  2.12  106 mm4 
I y  I ave  R cos   (5.20  3.1241cos 9.8056)  106
or
I y  8.28  106 mm 4 
or
I xy  0.532  106 mm 4 
I xy   R sin   (3.1241  106 )sin 9.8056
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.93
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the area of Problem 9.73 with respect to new centroidal
axes obtained by rotating the x and y axes 60° counterclockwise.
SOLUTION
I x = 324π cm4
From Problem 9.81:
I y = 648π cm4
I xy = 864 cm4
Problem 9.73:
The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864).
Now
I ave =
and
R=
1
1
( I x + I y ) = (324π + 648π ) = 1526.81 cm4
2
2
1
( I x − I y ) 2 + I xy2 =
2
1
(324π − 648π )
2
2
+ 8642
= 1002.75 cm4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(864)
324π − 648π
= 1.69765
=−
or
Then
2θ m = 59.500°
α = 120° − 59.500°
= 60.500°
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PROBLEM 9.93 (Continued)
Then
I x′ = I ave − R cos α = 1526.81 − 1002.75cos 60.500°
or
I x′ = 10.3 × 106 mm4
or
I y′ = 20.2 × 106 mm4
or
I x′y′ = −8.73 × 106 mm4
I y = I ave + R cos α = 1526.81 + 1002.75cos 60.500°
I x′y′ = − R sin α = −1002.75sin 60.500°
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1504
PROBLEM 9.94
Using Mohr’s circle, determine the moments of inertia and
the product of inertia of the area of Problem 9.75 with
respect to new centroidal axes obtained by rotating the x and
y axes 45 clockwise.
SOLUTION
From Problem 9.82:
I x  252,757 mm 4
I y  1, 752, 789 mm 4
Problem 9.75:
I xy  471, 040 mm 4
The Mohr’s circle is defined by the diameter XY, where X (252,757;
471,040) and Y (1,752,789; 471,040).
Now
1
1
( I x  I y )  (252, 757  1, 752, 789)
2
2
 1, 002, 773 mm 4
I ave 
2
and
1

R   ( I x  I y )   I xy2
2

1

  (252,757  1,752,789)2   471,0402
2

 885,665 mm 4
The Mohr’s circle is then drawn as shown.
tan 2 m  
2 I xy
Ix  I y
2(471,040)
252,757  1,752,789
 0.62804

or
Then
2 m  32.130
  180  (32.130  90)
 57.870
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PROBLEM 9.94 (Continued)
Then
I x  Iave  R cos   1,002,773  885,665cos57.870
or
I x  1.474  106 mm 4 
I y   I ave  R cos   1, 002, 773  885, 665 cos 57.870
or
I y  0.532  106 mm 4 
or
I xy  0.750  106 mm 4 
I xy   R sin   885, 665sin 57.870
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.95
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the L76 × 51 × 6.4-mm angle cross section of Problem 9.74
with respect to new centroidal axes obtained by rotating the x and y axes
30° clockwise.
SOLUTION
From Problem 9.83:
I x = 0.162 × 106 mm4
I y = 0.454 × 106 mm4
Problem 9.74:
I xy = −0.1596 × 106 mm4
The Mohr’s circle is defined by the diameter XY, where X(0.162 × 106, −0.1596 × 106) and Y(0.454 × 106,
0.1596 × 106).
1
I ave = ( I x + I y )
Now
2
1
= (0.162 + 0.454)106
2
= 0.308 × 106 mm4
R=
and
=
1
(I x − I y )
2
+ I xy2
2
1
(0.162 × 106 − 0.454 × 106) + (−0.1596 × 106)2
2
= 0.2163 × 106 mm4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−0.1596)
0.162 − 0.454
= −1.09315
=−
or
Then
2θ m = −47.55°
α = 60°− 47.55° = 12.45°
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PROBLEM 9.95 (Continued)
Then
I x′ = I ave − R cos α = (0.308 − 0.2163 cos 12.45°)106
or
I x′ = 96.79 × 103 mm4
or
I y = 519.21 × 103 mm4
I y′ = I ave + R cos α = (0.30 8 + 0.2163 cos 12.45°)106
I x′y′ = R sin α = 0.2163 × 106 sin 12.45°
or
I x′y′ = 46.63 × 103 mm4
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1508
PROBLEM 9.96
Using Mohr’s circle, determine the moments of inertia and
the product of inertia of the L152  102  12.7-mm angle
cross section of Problem 9.78 with respect to new
centroidal axes obtained by rotating the x and y axes 45
counterclockwise.
SOLUTION
From Problem 9.84:
I x  2.59  106 mm4
I y  7.20  106 mm 4
Problem 9.78:
I xy  2.54  106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (2.59  106 , 2.54  106 ),
Y (7.20  106 , 2.54  106 ).
Now
I ave 
1
1
( I x  I y )  (2.59  7.20)  106  4.895  106 mm 4
2
2
and
R
 I avg  I x   I xy2

2


4.895  2.54   2.542  106 mm 4
2
 3.43  106 mm 4
The Mohr’s circle is then drawn as shown.
DX
DC
2.54  106

2.305  106
 1.10195
tan  
or
  42.777
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PROBLEM 9.96 (Continued)
Then
  180  60  47.777
 72.223
and
I x  I ave  R cos   (4.895  3.43cos 72.223)  106
or
I x  5.94  106 mm4 
I y  I ave  R cos   (4.895  3.43cos 72.223)  106
or
I y  3.85  106 mm 4 
or
I xy  3.27  106 mm 4 
I xy  R sin   (3.43  106 )sin 72.223
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.97
For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine
the orientation of the principal axes at the origin and the corresponding
values of the moments of inertia.
SOLUTION
From Problem 9.79:
Ix 
Problem 9.67:
I xy 

8
a4
Iy 

2
a4
1 4
a
2
The Mohr’s circle is defined by the diameter XY, where
1 
1 


X  a 4, a 4  and Y  a 4,  a 4 
2 
2 
8
2
Now
I ave 
1
1
 
( I x  I y )   a 4  a 4   0.98175a 4
2
2 8
2 
and
2
1

2
R   ( I x  I y )   I xy
2

2
1  
   1 
   a4  a4    a4 
2   2 
2  8
2
 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2 m  

2 I xy
Ix  I y

2 12 a 4

 a4   a4
8
2
 0.84883
or
2 m  40.326
and
 m  20.2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.97 (Continued)
The principal axes are obtained by rotating the xy axes through
20.2 counterclockwise 

about O.

Now
I max,min  I ave  R  0.98175a 4  0.77264a 4
or
I max  1.754 a 4 
and
I min  0.209 a 4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.98
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes and the
corresponding values of the moments of inertia.
Area of Problem 9.78.
SOLUTION
From Fig. 9.13B:
I x  2.59  106 mm4
I y  7.20  106 mm 4
From Problem 9.78:
I xy  2.54  106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (2.59  106 , 2.54  106 )
and Y (7.20  106 ,  2.54  106 ).
1
(I x  I y )
2
1
 (2.59  7.20)  106 mm4
2
 4.895  106 mm 4
Now
I ave 
and
1

R   ( I x  I y )   I xy2
2


2
2
 1

 


   (2.59  7.20)   (2.54) 2   106 mm 4

  2

 3.430  106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2 m  

2 I xy
Ix  I y
2(2.54  106 )
 1.10195
(2.59  7.20)  106
or
2 m  47.777
and
m  23.9
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PROBLEM 9.98 (Continued)
The principal axes are obtained by rotating the xy axes through
23.9° counterclockwise 
about C.
Now
I max,min  I ave  R  (4.895  3.43) 106
or
I max  8.33  106 mm 4 
and
I min  1.465  106 mm 4 
From the Mohr’s circle it is seen that the x’ axis corresponds to I min and the y’ axis corresponds to I max . 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.99
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.76.
SOLUTION
From Problem 9.76:
I xy = −9011.25 cm4
I x = ( I x )1 + ( I x )2 + ( I x )3
Now
where
( I x )1 =
1
(24 cm)(4 cm)3 = 128 cm4
12
( I x ) 2 = ( I x )3 =
1
1
(9 cm)(15 cm)3 +
(9 cm)(15 cm) (7 cm)2
36
2
= 4151.25 cm4
I x = [128 + 2(4151.25)] cm4
Then
= 8430.5 cm4
I y = ( I y )1 + ( I y ) 2 + ( I y )3
Also
where
( I y )1 =
1
(4 cm)(24 cm)3= 4608 cm4
12
( I y ) 2 = ( I y )3 =
1
1
(15 cm)(9 cm)3 +
(9 cm)(15 cm) (9 cm)2
2
36
= 5771.25 cm4
Then
I y =[4608 + 2(5771.25)] cm4 = 16150.5 cm4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1515
PROBLEM 9.99 (Continued)
The Mohr’s circle is defined by the diameter XY, where X (8430.5, − 9011.25) and Y (16150.5, 9011.25).
Now
I ave =
and
R=
=
1
1
( I x + I y ) = (8430.5 + 16150.5) = 12290.5 cm4
2
2
1
(I x − I y )
2
2
+ I xy2
1
(8430.5 − 16150.5)
2
2
+ (−9011.25) 2
= 9803.17 cm4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − I y
2( −9011.25)
8430.5 − 16150.5
= −2.33452
=−
or
2θ m = −66.812°
and
θ m = −33.4°
The principal axes are obtained by rotating the xy axes through
33.4° clockwise
about C.
Now
I max, min = I ave ± R = 12290.5 ± 9803.17
or
and
I max = 221 × 106 mm4
I min = 24.9 × 106 mm4
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.100
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes and the corresponding
values of the moments of inertia.
Area of Problem 9.73
SOLUTION
From Problem 9.81:
I x  324 cm4
Problem 9.73:
I xy  864 cm4
I y  648 cm4
The Mohr’s circle is defined by the diameter XY, where X (324 , 864) and Y (648 ,  864).
Now
and
I ave 
1
1
( I x  I y )  (324  648 )  1526.81 cm4
2
2
1
 2
R   ( I x  I y )2   I xy
2


2
1

  (324  648 )   8642
2

 1002.75 cm4
The Mohr’s circle is then drawn as shown.
tan 2 m  
2 I xy
Ix  I y
2(864)
324  648
 1.69765

or
2 m  59.4998
and
m  29.7
The principal axes are obtained by rotating the xy axes through
29.7° counterclockwise 
about C.
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PROBLEM 9.100 (Continued)
Now
I max, min  I ave  R  1526.81  1002.75
or
I max  25.3 × 106 mm4 
and
I min  5.24 × 106 mm4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.101
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.74.
SOLUTION
I x = 0.162 × 106 mm4
From Problem 9.83:
I y = 0.454 × 106 mm4
I xy = −0.1596 × 106 mm4
Problem 9.74:
The Mohr’s circle is defined by the diameter XY, where X(0.162 × 106, −0.1596 × 106) and Y(0.454 × 106,
+0.1596 × 106).
1
1
Now
I ave = ( I x + I y ) = (0.162 + 0.454)106 = 0.308 × 10 6 mm4
2
2
and
R=
=
1
(I x − I y )
2
2
+ I xy2
2
1
(0.162 × 106 − 0.454 × 106) + (−0.1596 × 106)2
2
= 0.2163 × 106 mm4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−0.1596)
0.166 − 0.454
= −1.09315
=−
Then
2θ m = −47.55°
and
θ m = −23.78°
The principal axes are obtained by rotating the xy axes through
23.8° clockwise
about C.
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PROBLEM 9.101 (Continued)
Now
I max, min = I ave ± R = (0.308 ± 0.2163)106
or
I max = 0.524 × 106 mm4
and
I min = 0.0917 × 106 mm4
8
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1520
PROBLEM 9.102
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.77
(The moments of inertia I x and I y of the area of Problem 9.102
were determined in Problem 9.44).
SOLUTION
From Problem 9.44:
I x = 18.1282 cm4
I y = 4.5080 cm4
Problem 9.77:
I xy = −4.25320 cm4
The Mohr’s circle is defined by the diameter XY, where X (18.1282, − 4.25320) and Y (4.5080, 4.25320).
Now
I ave =
and
R=
1
1
( I x + I y ) = (18.1282 + 4.5080) = 11.3181 cm4
2
2
=
1
(I x − I y )
2
2
+ I xy2
1
(18.1282 − 4.5080)
2
2
+ ( −4.25320)2
= 8.02915 cm4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−4.25320)
18.1282 − 4.5080
= 0.62454
=−
or
2θ m = 31.986°
and
θ m = 15.99°
The principal axes are obtained by rotating the xy axes through
15.99° counterclockwise
about C.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1521
PROBLEM 9.102 (Continued)
Now
I max, min = I ave ± R = 11.3181 ± 8.02915
or
I max = 193.5 × 103 mm4
and
I min = 32.9 × 103 mm4
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.103
The moments and product of inertia of an L127 × 76 × 12.7-mm angle cross section with respect to
two rectangular axes x and y through C are, respectively, Iy = 3.93 × 106 mm4, Ix = 1.06 × 106 mm4, and
Ixy ,0, with the minimum value of the moment of inertia of the area with respect to any axis through C
being Imin = 0.647 × 106 mm4. Using Mohr’s circle, determine (a) the product of inertia Ixy of the area, (b)
the orientation of the principal axes, (c) the value of Imax.
SOLUTION
(Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when
the 127-mm leg of the angle is parallel to the x axis. Further, for I xy 0, the angle must be oriented as shown.)
1
1
( I x + I y ) = (3.93 + 1.06)106 = 2.495 × 106 mm4
2
2
Now
I ave =
and
I min = I ave − R or
R = (2.495 − 0.647)106 mm4
= 1.848 × 106 mm4
Using I ave and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, X (3.93 × 106, I xy )
and Y (1.06 × 106, | I xy |).
(a)
1
(I x − I y )
2
2
We have
R2 =
+ I xy2
or
I xy2 = (1.848 × 10 6) −
2
1
(1.06 × 106 − 3.93 × 106
2
Solving for I xy and taking the negative root (since I xy
2
0) yields I xy = −1.164 × 106 mm4
I xy = −1.164 × 106 mm4
(b)
We have
tan 2θ m = −
2 I xy
Ix − Iy
=−
2(−1.164)
1.06 − 3.93
= −0.8111
or
2θ m = −39.05°
θ m = −19.53°
The principal axes are obtained by rotating the xy axes through
19.53° clockwise
about C.
(c)
We have
I max = I ave + R = (2.495 + 1.848)106 mm4
or
I max = 4.343 × 106 mm4
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PROBLEM 9.104
Using Mohr’s circle, determine for the cross section of the
rolled-steel angle shown the orientation of the principal
centroidal axes and the corresponding values of the moments of
inertia. (Properties of the cross sections are given in Figure 9.13.)
SOLUTION
From Figure 9.13B:
I x  0.162  106 mm 4
I y  0.454  106 mm 4
We have
I xy  ( I xy )1  ( I xy ) 2
For each rectangle
I xy  I x y   x yA
and
I xy   0
I xy   x yA
(symmetry)
A, mm 2
x , mm
y , mm
x yA, mm4
1
76  6.4  486.4
 13.1
9.2
58620.93
2
6.4  (51  6.4)  285.44
21.7
–16.3
–100962.98

–159583.91
I xy  159584 mm 4
The Mohr’s circle is defined by the diameter XY where X (0.162  106 ,  0.159584  106 )
and Y (0.454  106 , 0.159584  106 )
Now
1
1
( I x  I y )  (0.162  0.454)  106
2
2
6
 0.3080  10 mm 4
I ave 
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PROBLEM 9.104 (Continued)
and
2
2


 1

1


2
R   ( I x  I y )   I xy
   (0.162  0.454)   (0.159584) 2   106
2
2



 

 0.21629  106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2 m  
2 I xy
Ix  I y
2(0.159584  106 )
(0.162  0.454)  106
 1.09304

or
2 m  47.545
and
 m  23.8
The principal axes are obtained by rotating the xy axes through
23.8° clockwise 
About C.
Now
I max, min  I ave  R  (0.3080  0.21629)  106
or
I max  0.524  106 mm 4 
and
I min  0.0917  106 mm 4 
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.105
Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes
and the corresponding values of the moments of inertia.
(Properties of the cross sections are given in Figure 9.13.)
SOLUTION
I x  3.93  106 mm 4
From Figure 9.13B:
I y  1.06  106 mm 4
Determine I xy :
 1   I xy 2 and for each rectangle,
I xy  I xy
I xy  I x ' y '  xyA
1
2

and by symmetry I x ' y '  0 thus I xy  xyA
A, mm 2
x , mm
y , mm
xyA, mm 4
76  12.7  965.2
19.1
-37.85
-697,777
-12.55
25.65
-467,284
12.7 127  12.7   1451.61
-1,165,061
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PROBLEM 9.105 (Continued)
I xy  1.16506  106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (3.93  106 ,  1.165061  106 )
and Y (1.06  106 , 1.165061  106 ).
Now
1
(I x  I y )
2
1
 (3.93  1.06)  106
2
 2.495  106 mm 4
I ave 
2
1

R   ( I x  I y )   I xy2
2

and
2
 1

 


   (3.93  1.06)   (1.165061) 2   106  1.84840  106 mm 4

  2

The Mohr’s circle is then drawn as shown.
tan 2 m  
2 I xy
Ix  I y

2(1.165061  106 )
(3.93  1.06)  106
 0.81189
or
2 m  39.073
and
m  19.54
The principal axes are obtained by rotating the xy axes through
19.54° counterclockwise 
about C.
Now
I max, min  I ave  R  (2.495  1.84840)  106
or
I max  4.34  106 mm 4 
6
4
and I min  0.647  10 mm 
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.106*
For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are
Ix = 1.2 × 106 mm4 and Iy = 0.3 × 106 mm4, respectively. Knowing that after rotating the x and y axes about
the centroid 30° counterclockwise, the moment of inertia relative to the rotated x axis is 1.45 × 106 mm4, use
Mohr’s circle to determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.
SOLUTION
I ave =
We have
Now observe that I x
I xy
I ave , I x′
0 and (for convenience) I x′y′
1
1
( I x + I y ) = (1200 + 300)103 = 750 × 103 mm4
2
2
I x , and 2θ = +60°. This is possible only if I xy < 0. Therefore, assume
0. Mohr’s circle is then drawn as shown.
2θ m + α = 60°
We have
Now using ABD:
R=
I x − I ave (1200 − 750) 103
=
cos 2θm
cos 2θm
=
Using AEF:
R=
450 × 103
(mm4)
cos 2θ m
I x′ − I ave (1450 − 750) 103
=
cos α
cos α
700 × 103
=
(mm4)
cos α
Then
450
700
=
cos 2θ m cos α
or
9cos (60° − 2θ m ) = 14 cos 2θ m
Expanding:
9(cos 60° cos 2θ m + sin 60° sin 2θ m ) = 14 cos 2θ m
tan 2θ m =
or
14 − 9 cos 60°
= 1.21885
9 sin 60°
2θ m = 50.633° and θ m = 25.3°
or
(Note: 2θ m
Finally,
(a)
α = 60° − 2θ m
60° implies assumption I x′y′
R=
0 is correct.)
450 × 103
= 709.46 × 103 mm4
cos 50.633°
From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given centroidal x
and y axes through θ m about the centroid C or
25.3° counterclockwise.
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1528
PROBLEM 9.106* (Continued)
(b)
We have
I max, min = I ave ± R = (750 ± 709.46)103 mm4
or
I max = 1. 459 × 10 6 mm4
and
I min = 40.5 × 103 mm4
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.107
It is known that for a given area I y  48  106 mm4 and I xy  –20  106 mm4, where the x and y axes are
rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by
rotating the x axis 67.5 counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia
I x of the area, (b) the principal centroidal moments of inertia.
SOLUTION
First assume I x  I y and then draw the Mohr’s circle as shown. (Note: Assuming I x  I y is not consistent
with the requirement that the axis corresponding to ( I xy )max is obtained after rotating the x axis through 67.5°
CCW.)
From the Mohr’s circle we have
2 m  2(67.5)  90  45
(a)
From the Mohr’s circle we have
Ix  I y  2
| I xy |
tan 2 m
 48  106  2
20  106
tan 45
or
(b)
We have
and
Now
I x  88.0  106 mm 4 
1
1
( I x  I y )  (88.0  48)  106
2
2
6
 68.0  10 mm 4
I ave 
R
| I xy |
sin 2 m

20  106
 28.284  106 mm 4
sin 45
I max, min  I ave  R  (68.0  28.284)  106
or
I max  96.3  106 mm 4 
and
I min  39.7  106 mm 4 
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PROBLEM 9.108
Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia
with respect to every axis through the centroid is the same, (b) the product of inertia with respect to every
pair of rectangular axes through the centroid is zero.
SOLUTION
Consider the regular pentagon shown, with centroidal axes x and y.
Because the y axis is an axis of symmetry, it follows that I xy  0. Since I xy  0, the x and y axes must be
principal axes. Assuming I x  I max and I y  I min , the Mohr’s circle is then drawn as shown.
Now rotate the coordinate axes through an angle  as shown; the resulting moments of inertia, I x and I y ,
and product of inertia, I xy , are indicated on the Mohr’s circle. However, the x  axis is an axis of
symmetry, which implies I xy  0. For this to be possible on the Mohr’s circle, the radius R must be equal
to zero (thus, the circle degenerates into a point). With R  0, it immediately follows that
(a)
I x  I y  I x  I y  Iave (for all moments of inertia with respect to an axis through C )
(b)
I xy  I xy  0 (for all products of inertia with respect to all pairs of rectangular axes with
origin at C )
Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 9.109
Using Mohr’s circle, prove that the expression I x I y  I x2y  is independent of the orientation of the x and
y axes, where Ix, Iy, and Ixy represent the moments and product of inertia, respectively, of a given area
with respect to a pair of rectangular axes x and y through a given Point O. Also show that the given
expression is equal to the square of the length of the tangent drawn from the origin of the coordinate
system to Mohr’s circle.
SOLUTION
First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave
and R are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the
moments of inertia, I x and I y , and the product of inertia, I xy , having been computed for an arbitrary
orientation of the x y  axes.
From the Mohr’s circle
I x  I ave  R cos 2
I y  I ave  R cos 2
I xy  R sin 2
Then, forming the expression
I x I y  I x2y 
I x I y  I x2y  ( I ave  R cos 2 )( I ave  R cos 2 )  ( R sin 2 ) 2


2
 I ave
 R 2 cos 2 2  ( R 2 sin 2 2 )
2
 I ave
 R2
which is a constant
I x I y  I x2y  is independent of the orientation of the coordinate axes Q.E.D. 
Shown is a Mohr’s circle, with line OA, of length L, the required tangent.
Noting that
OAC is a right angle, it follows that
2
L2  I ave
 R2
or
L2  I x I y   I x2y Q.E.D. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.110
Using the invariance property established in the preceding problem,
express the product of inertia Ixy of an area A with respect to a pair of
rectangular axes through O in terms of the moments of inertia Ix and Iy
of A and the principal moments of inertia Imin and Imax of A about O.
Use the formula obtained to calculate the product of inertia Ixy of the
L76 × 51 × 6.4-mm angle cross section shown in Figure 9.13B, knowing
that its maximum moment of inertia is 5.232 × 106 mm4 .
SOLUTION
Consider the following two sets of moments and products of inertia, which correspond to two different
orientations of the coordinate axes whole origin is at Point O.
Case 1:
I x′ = I x , I y ′ = I y , I x′y′ = I xy
Case 2:
I x′ = I max , I y′ = I min , I x′y′ = 0
The invariance property then requires
2
I x I y − I xy
= I max I min
From Figure 9.13B:
or I xy = ± I x I y − I max I min
I x = 454 × 103 mm4
I y = 162 × 103 mm4
Using Eq. (9.21):
Substituting
I x + I y = I max + I min
103(454 + 162) = 523.2 × 103 + Imin
or
I min = 92.8 × 103 mm4
Then
I xy = [(162)(454) − (523.2)(92.8)]103
= ±158.1 × 103 mm4
The two roots correspond to the following two orientations of the cross section.
For
I xy = −158.1 × 103 mm4
and for
I xy = 158.1 × 103 mm4
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PROBLEM 9.111
A thin plate of mass m is cut in the shape of an equilateral triangle of side a.
Determine the mass moment of inertia of the plate with respect to (a) the
centroidal axes AA and BB, (b) the centroidal axis CC that is perpendicular
to the plate.
SOLUTION
Area 
1  3 
3 2
a
a 
a


2  2  4
Mass  m  V   tA
I mass   t I area 
(a)
Axis AA:
m
I area
A
 1  3   a 3 
3 4
I AA,area  2  
a    
a


12  2   2   96
 3 4 1
m
I AA, area  3 a 2 
a 
ma 2
 96  24
A
4


m
I AA, mass 

3
Axis BB:
1  3 
3 4
I BB, area 
a
a 
a
36  2 
96
(we check that I AA  I BB )
 3 4 1
m
I BB, area  3 a 2 
a 
ma 2
 96  24
A
4


m
I BB, mass 
(b)
Eq. (9.38):
 1

I CC   IAA  I BB  2  ma 2 
 24


I CC  
1
ma 2 
12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.112
Determine the mass moment of inertia of a ring of mass m, cut from a
thin uniform plate, with respect to (a) the axis AA, (b) the centroidal axis
CC that is perpendicular to the plane of the ring.
SOLUTION
Mass  m  V   tA
I mass   t I area 
m
I area
A
We first determine

A   r22   r12   r22  r12
I AA,area 
I AA, mass 
(a)
(b)

4
r24 

4
r14 

4

r  r 
4
2
4
1
m
m
 4 4 1
I AA, area 
r2  r1  m r22  r12
2
2
A
4
 r2  r1 4



By symmetry:
I BB  IAA
Eq. (9.38):
I CC   IAA  I BB  2 IAA







I AA 
1
m r12  r22 
4
I CC  
1
m r12  r22 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.113
A thin, semielliptical plate has a mass m. Determine the mass moment of
inertia of the plate with respect to (a) the centroidal axis BB, (b) the
centroidal axis CC that is perpendicular to the plate.
SOLUTION
mass  m   tA
I mass   t I area 
2m
I
 ab area
1
A   a2
2
Area:
I x ,area  I BB ',area  Ay 2
   4b 
ab3   ab 

8
 2  3 
64 


I BB ',area  ab3 1  2 
8
 9 
I BB ',area 

I BB,mass 
(a)

(b)
Fig. (9.12): I AA ', area 

8
a 3 b,
then
I CC ,mass  I AA, mass  I BB,mass 
2
2m   3 
64 
ab 1  2 

 ab  8
 9 
1 2
64 
mb 1  2 
4
 9 
I BB  0.0699mb 2 
 2m   3  1 2
I AA ', mass  
 a b   ma
  ab  8
 4
1
1
64 

ma 2  mb 2 1  2 
4
4
 9 
ICC  
1
m(a 2  0.279b2 ) 
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.114
The parabolic spandrel shown was cut from a thin, uniform plate.
Denoting the mass of the spandrel by m, determine its mass
moment of inertia with respect to (a) the axis BB, (b) the axis
DD that is perpendicular to the spandrel. (Hint: See Sample
Problem 9.3.)
SOLUTION
First note
mass  m  V   tA
 
  t  ab 
2 
I mass   t I area
Also

(a)
2m
I area
 ab
We have
I x ,area  I BB,area  Ay 2
or
(a)
I BB,area 

   4b 
ab3   ab 

8
 2  3 
2
From Sample Problem 9.3:
1 3
ab
21
3m 1 3
I BB,mass 
 ab
ab 21
I BB,area 
Then
or
(b)
I BB 
1 2
mb 
7
From Sample Problem 9.3:
1
I AA,area  a 3b
5
Now
3 
I AA,area  I11,area  A  a 
4 
2
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PROBLEM 9.114 (Continued)
and
Then
1 
I 22,area  I11,area  A  a 
4 
2
 1  2  3  2 
I 22,area  I AA,area  A  a    a  
 4   4  
1
1  1
9

 a 3b  ab  a 2  a 2 
5
3  16
16 
1 3

ab
30
3m 1 3
 ab
ab 30
1
 ma 2
10
and
I 22,mass 
Finally,
I DD,mass  I BB,mass  I 22,mass

1 2 1
mb  ma 2
7
10
or
I DD 
1
m(7a 2  10b 2 ) 
70
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PROBLEM 9.115
A piece of thin, uniform sheet metal is cut to form the machine
component shown. Denoting the mass of the component by m,
determine its mass moment of inertia with respect to (a) the x axis,
(b) the y axis.
SOLUTION
First note
mass  m  V   tA

1
 a 
  t (2a )(a )  (2a )   
2
 2 

3
  ta 2
2
I mass   tI area
Also

(a)
Now
Then
2m
I area
3a 2
I x ,area  ( I x )1,area  2( I x )2,area

 1 a

1
(a )(2a )3  2    (a )3 
12
12  2 


7 4
a
12
I x, mass 
2m 7 4
 a
3a 2 12
or
(b)
We have
Ix 
7
ma 2 
18
I z ,area  ( I z )1,area  2( I z )2,area
3
1
1
a 
 (2a )(a )3  2  (a )   
3
 2  
12

Then
31 4
a
48
2m 31 4
 a
3a 2 48
31 2

ma
72
I z ,mass 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.115 (Continued)
Finally,
I y ,mass  I x,mass  I z ,mass
7
31
ma 2  ma 2
18
72
59 2

ma
72

or
I y  0.819ma 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.116
A piece of thin, uniform sheet metal is cut to form the machine
component shown. Denoting the mass of the component by m,
determine its mass moment of inertia with respect to (a) the axis
AA, (b) the axis BB, where the AA and BB axes are parallel to the
x axis and lie in a plane parallel to and at a distance a above the xz
plane.
SOLUTION
First note that the x axis is a centroidal axis so that
I  I x,mass  md 2
and that from the solution to Problem 9.115,
(a)
(b)
We have
We have
I x , mass 
7
ma 2
18
I AA,mass 
7
ma 2  m(a ) 2
18
I BB, mass 
or
I AA  1.389ma 2 
or
I BB  2.39ma 2 
7
ma 2  m(a 2 )2
18
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.117
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the x
axis, (b) the y axis.
SOLUTION
First note
mass  m  V   tA
1


  t  (2a )(a )  (2a )( a)   3 ta 2
2


I mass   tI area 
Also
(a)
Now
m
I area
3a 2
I x ,area  ( I x )1,area  ( I x ) 2,area
1
1
 (2a )(a)3  (2a )(a )3
3
12
5
 a4
6
Then
(b)
I x ,mass 
m 5 4
 a
3a 2 6
or
I x , mass 
5
ma 2 
18
We have
I z ,area  ( I z )1,area  ( I z ) 2, area
2
1
1
1
 1

 
  (a)(2a)3    (a)(2a)3  (2a)(a)  2a   2a    10a 4
2
3
3
  36

 
m
10
 10a 4  ma 2
2
3
3a
Then
I x ,mass 
Finally,
I y , mass  I x , mass  I z ,mass

5
10
ma 2  ma 2
18
3

65 2
ma
18
or I y ,mass  3.61ma 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.118
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the
centroidal axis CC that is perpendicular to the plate, (b) the axis
AA that is parallel to the x axis and is located at a distance 1.5a
from the plate.
SOLUTION
First locate the centroid C.
1


X  A   xA: X (2a 2  a 2 )  a(2a 2 )   2a   2a  (a 2 )
3


X
or
14
a
9
1 
1 
Z  A   zA: Z (2a 2  a 2 )   a  (2a 2 )   a  (a 2 )
2 
3 
Z
or
(a)
We have
4
a
9
I y , mass  I CC, mass  m( X 2  Z 2 )
From the solution to Problem 9.117:
Then
I y , mass 
65 2
ma
18
I cc, mass 
 14  2  4 2 
65 2
ma  m  a    a  
18
 9   9  
or
(b)
We have
I cc  0.994ma 2 
I x,mass  I BB,mass  m( Z )2
and
I AA, mass  I BB, mass  m(1.5a)2
Then
2

4  
I AA, mass  I x , mass  m  (1.5a ) 2   a  
 9  

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.118 (Continued)
From the solution to Problem 9.193:
I x ,mass 
Then
5
ma 2
18
2

5
4  
2
2
I AA, mass  ma  m  (1.5a )   a  
18
 9  

or
I AA  2.33ma 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.119
Determine by direct integration the mass moment of inertia
with respect to the z axis of the right circular cylinder shown,
assuming that it has a uniform density and a mass m.
SOLUTION
For the cylinder:
m  V   a 2 L
For the element shown:
dm   a 2 dx

and
dI z  dI z  x 2 dm

Then
m
dx
L
1 2
a dm  x 2 dm
4

I z  dI z 
L 1
2  m

  4 a  x  L dx 
2
0
L

1 
m 1 2
a x  x3 

3 0
L 4
Iz 
1
m(3a 2  4 L2 ) 
12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.120
The area shown is revolved about the x axis to form a
homogeneous solid of revolution of mass m. Using direct
integration, express the mass moment of inertia of the solid
with respect to the x axis in terms of m and h.
SOLUTION
We have
y
2h  h
xh
a
so that
r
h
( x  a)
a
For the element shown:
dm   r 2 dx dI x 
1 2
r dm
2
2
h

   ( x  a)  dx
a

h2
1
h2
2



(
x
a
)
dx
[( x  a )3 ]0a
0
3
a2
a2
1
h2
7
  2 (8a3  a 3 )   ah 2
3
3
a


a

Then
m  dm 
Now
ah
1 2
1

I x  dI x 
r (  r 2 dx)    ( x  a )  dx
0
2
2
a




4
1
1 h4
1
h4
5 a


 

[(
x
a
)
]
(32a5  a5 )
0
4
4
2
5a
10
a
31
  ah 4
10

From above:
Then
3
7
 ah 2  m
Ix 
31  3  2 93 2
mh
 mh 
10  7 
70
or
I x  1.329 mh 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.121
The area shown is revolved about the x axis to form a
homogeneous solid of revolution of mass m. Determine by
direct integration the mass moment of inertia of the solid with
respect to (a) the x axis, (b) the y axis. Express your answers in
terms of m and the dimensions of the solid.
SOLUTION
We have at
(a, h): h 
k
a
k  ah
or
For the element shown:
r4
dm   r 2 dx
2
 ah 
    dx
 x 

m  dm 
Then

3a
a
2
 ah 
 dx
 x 
 
2 2
3a
1
  a h   
 x a
 1  1  2
  a 2 h 2          ah 2
 3a  a   3
(a)
For the element:
Then
dI x 
1 2
1
r dm   r 4 dx
2
2

I x  dI x 

a
3a
4
1
 ah 
 1 1
   dx   a 4 h 4   3 
2
2
 x 
 3 x a
3a 1
 1 3  1 3  1 26
1
 ah 4
   a 4 h 4        
6
3
a
a
6
27







1 2
13
13 2
mh
  ah 2  h 2 
6 3
9
54
or
I x  0.241 mh 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.121 (Continued)
(b)
For the element:
dI y  dI y  x 2 dm

Then
1 2
r dm  x 2 dm
4

I y  dI y 

3a  1  ah 

 ah 
    x 2     dx
a 4  x 
 x 


  a h
2 2
2
2
3a  1 a 2 h 2
  4 x
a
4
3a

 1 a2 h2

 1 dx   a 2 h 2  
 x
3

 12 x
a
2 2
  1 a2 h2
 
 3    1 a h
  m  a 

 a
a
3
  
3
3
 2    12 (3a )
  12 (a )
 

 1 26 h2

3
 13 2

ma  
h  3a 2 
 2a   m 
a
2
12
27
108




or
I y  m(3a 2  0.1204h2 ) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.122
Determine by direct integration the mass moment of inertia with
respect to the x axis of the tetrahedron shown, assuming that it has
a uniform density and a mass m.
SOLUTION
We have
x
a
y

y  a  a 1  
h
h

and
z
b
y

y  b  b 1  
h
h


2
For the element shown:
y
1
 1

dm    xz dy    ab 1   dy
2
2
h




Then
m  dm 


2
1
y

 ab 1   dy
h 2
h

0
3
1
h 
y 
  ab  1   
2
3 
h  
0
h
1
  abh (1)3  (1  1)3 
6
1
  abh
6
Now, for the element:
Then
I AA,area 
1 3 1 3
y
xz  ab 1  
36
36
h


4
4
1 3 
y 
dI AA,mass   tI AA,area   (dy )  ab  1   
h  

 3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.122 (Continued)
Now
 2  1 2 
dI x  dI AA, mass   y   z   dm
 3  

4

1
y

 ab3 1   dy
36
h

2
2


1 
y     1
y

2
  y   b 1       ab 1   dy 
h     2
h


3 


4

1
1
y
y3 y 4 
3
  ab 1   dy   ab  y 2  2

 dy
12
2
h
h h 2 


Now
m
1
 abh
6
Then
4
 1 b2 
y
3m  2
y3 y 4 

dI x   m  1   
 y  2
  dy
h
h 
h h 2  
 2 h 
and
I x  dI x 


4

m  2
y
y3 y 4 

b 1    6  y 2  2
 dy
 h 2h 
h
h h 2  

 
0
5
1
1 y4
m  2 h
y
y5 

 2  
b  1    6  y 3 
2h 
5
2 h 5h  
h
3
0
h
1
1
m 1 2
1

5
3
(  h ) 4  2 (  h )5  

 b h(1)  6  (h) 
2h  5
2h
5h
3

or
Ix 
1
m(b 2  h 2 ) 
10
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.123
Determine by direct integration the mass moment of inertia with
respect to the y axis of the tetrahedron shown, assuming that it has
a uniform density and a mass m.
SOLUTION
We have
x
a
y

y  a  a 1  
h
h

and
z
b
y

y  b  b 1  
h
h

2
For the element shown:
y
1
 1

dm    xz dy    ab 1   dy
h
2
 2

Then
m  dm 

0 1
 2
h


 ab 1 
2
y
dy
h 
3
1
h 
y 
  ab  1   
2
3 
h  
0
h
1
  abh (1)3  (1  1)3 
6
1
  abh
6
I BB,area 
Also
Then, using
we have
1 3
xz
12
I DD,area 
1 3
zx
12
I mass   tI area
 1

dI BB, mass   (dy )  xz 3 
 12

 1

dI DD, mass   (dy )  zx3 
 12

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.123 (Continued)
Now
dI y  dI BB, mass  dI DD, mass
1
 xz ( x 2  z 2 )dy
12
2
2
1
y  2
y 

2 
  ab 1   (a  b ) 1    dy
12
h  
h  



4
We have
m
1
m
y

 abh  dI y  (a 2  b2 ) 1   dy
6
2h
h

Then

I y  dI y 

4
m 2
y

(a  b 2 ) 1   dy
 h 2h
h

0
5
m 2
h 
y 
a  b 2   1   

h  
2h
5 



0
h

m 2
a  b 2 (1)5  (1  1)5 

10
or
Iy 
1
m(a 2  b 2 ) 
10
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.124
Determine by direct integration the mass moment of inertia and
the radius of gyration with respect to the x axis of the paraboloid
shown, assuming that it has a uniform density and a mass m.
SOLUTION
r 2  y 2  z 2  kx:
at x  h, r  a; a 2  kh; k 
Thus: r 2 
a2
x
h
dm   r 2 dx  
m

h
0
a2
h
dm  
1 
dI x   r 2  dm
2 
h
1
I x  dI x 
0
2

a2
xdy
h
a2
h
1
h
 xdx; m  2  a h
  r   r dx   2   r dx
h
2
1
2
0
h
4
0
2

2
0
h  a2 
1
 a 4
  x  dx 
0
2
2h 2
 h 

h
 x dx
2
0
Ix 
Recall: m 
1
 a 2 h;
2
1
 a 4 h 
6
1
 1 
I x    a 2 h   a 2
2
 3 
1
I x  ma 2 
3
1 2
ma 
1 2
I 3
  k x    
  a 
m  m  3




2
kx 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
a
3

PROBLEM 9.125
A thin rectangular plate of mass m is welded to a vertical shaft AB as
shown. Knowing that the plate forms an angle  with the y axis,
determine by direct integration the mass moment of inertia of the
plate with respect to (a) the y axis, (b) the z axis.
SOLUTION
Projection on yz plane
Mass of plate: m   tab
(a)
For element shown:
dm   btdv
dI y  dI y  (v sin  ) 2 dm 
1 2
b dm  v 2 sin 2  dm
12
 1

  b2  v 2 sin 2    btdv
 12

a 1

I y  dI y   bt  b 2  v 2 sin 2   dv
0  12



a
 1

1 2
v3
a3
b v  sin    bt  ab2  sin 2  
  bt
12
3
3
 12

0
 (  tab)
1 2
(b  4a 2 sin 2  )
12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.125 (Continued)
Iy 
(b)
dI z  dI z  (v cos  )2 dm 
1
m(b 2  4a 2 sin 2  ) 
12
1 2
b dm  v 2 cos2  dm
12
 1

  b2  v 2 cos2    btdv
 12

a 1

I z  dI z   bt  b2  v 2 cos 2   dv
0  12



a
 1

1 2
v3
a3
b v  cos 2    bt  ab 2  cos 2  
  bt
12
3
3
 12

0
 (  tab)
1 2
(b  4a 2 cos 2  )
12
Iz 
1
m(b 2  4a 2 cos 2  ) 
12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.126*
A thin steel wire is bent into the shape shown. Denoting the mass per
unit length of the wire by m, determine by direct integration the mass
moment of inertia of the wire with respect to each of the coordinate
axes.
SOLUTION
dy
  x 1/3 (a 2/3  x 2/3 )1/ 2
dx
First note
2
 dy 
1     1  x 2/3 (a 2/3  x 2/3 )
 dx 
Then
a
 
x
2/3
2
 dy 
dm  mdL  m 1    dx
 dx 
For the element shown:
1/3
a
 m  
x
dx
a
a1/3
3
3
m 1/3 dx  ma1/3  x 2/3   ma
0
0
2
2
x
a



 a1/3 
(a 2/3  x 2/3 )3  m 1/3 dx 
0
 x

2
a a

 ma1/3  1/3  3a 4/3 x1/3  3a 2/3 x  x5/3  dx
0 x


Then
m  dm 
Now
I x  y 2 dm 

a

a
9
3
3
3

 ma1/3  a 2 x 2/3  a 4/3 x 4/3  a 2/3 x 2  x8/3 
4
2
8
2
0
 3 9 3 3 3
 ma 3       ma 3
 2 4 2 8 8
or
Symmetry implies

Ix 
1
ma 2 
4
Iy 
1
ma 2 
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.126* (Continued)
Alternative solution:

a
 a1/3 
x 2  m 1/3 dx   ma1/3 x5/3 dx
0
0
 x

a
3
3
 ma1/3   x8/3   ma3
0
8
8
1 2
 ma
4
I y  x 2 dm 
Also

a


I z  ( x 2  y 2 ) dm  I y  I x
or
Iz 
1
ma 2 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.127
Shown is the cross section of an idler roller. Determine its mass
moment of inertia and its radius of gyration with respect to the
axis AA′. (The density of bronze is 8580 kg/m3; of aluminum,
2770 kg/m3; and of neoprene, 1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that
m = ρV = ρ t ×
π
4
( d − d ) = π4 ρt ( d − d )
2
2
2
1
2
2
2
1
and, using Figure 9.28, that
I AA′ =
2
d
1
m2 2
2
2
=
1
8
=
1 π
ρt
8 4
ρt ×
−
π
4
d
1
m1 1
2
2
d 22 d 22 − ρ t ×
4
2
4
d12 d12
4
1
(
(
π
(d − d )
1 π
ρ t d 22 − d12
8 4
1
= m d12 + d 22
8
=
2
)( d + d )
2
2
2
1
)
Now treat the roller as three concentric rings and, working from the bronze outward, we have
m=
π
4
{( 8580 kg/m3)(0.02 m)[(0.008)2 − (0.006)2] m2
+ ( 2770 kg/m3)(0.016 m)[(0.012 m)2 − (0.008)2] m2
+ ( 1250 kg/m3)(0.016 m)[(0.028)2 − (0.012)2] m2}
= (3.774 + 2.785 + 10.053) × 10–3 kg
= 16.612 × 10–3 kg
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.127 (Continued)
and
I AA′ =
1
{( 3.774)[(0.006)2 + (0.008)2] + (2.785)[(0.008)2 + (0.012)2]
8
+( 10.053)[(0.028)2 + (0.012)2]} × 10–3 kg ⋅ m2
= (47.175 + 72.41 + 1166.15) × 10–9 kg ⋅ m2
= 1.286 × 10–6 kg ⋅ m2
or
I AA′ = 1.286 × 10–6 kg ⋅ m2
I AA′ 1.286 × 10–6 kg ⋅ m2
=
= 77.413 × 10–6 m2
–3
m
16.612 × 10 kg
Now
2
k AA
′ =
Then
k AA′ = 8.799 × 10–3 m
or
k AA′ = 8.8 mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.128
Shown is the cross section of a molded flat-belt
pulley. Determine its mass moment of inertia and its
radius of gyration with respect to the axis AA. (The
density of brass is 8650 kg/m3 and the density of the
fiber-reinforced polycarbonate used is 1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that
m  V   t 

4
d  d 
2
2
2
1
and, using Figure 9.28, that
2
I AA 

1  d2 
1 d 
m2    m1  1 
2  2 
2  2
2
 2 2 
 2  2
1 
  t  d 2  d 2    t  d1  d1 
8 
4 
4  

1 
   t  d 24  d14
8 4 
1 
   t  d 22  d12
8 4 
1
 m d12  d 22
8




 d  d 
2
2
2
1

Now treat the pulley as four concentric rings and, working from the brass outward, we have
m


8650 kg/m 3  (0.0175 m)  (0.0112  0.0052 ) m 2
4
 1250 kg/m3 [(0.0175 m)  (0.017 2  0.0112 ) m 2
 (0.002 m)  (0.0222  0.017 2 ) m 2
 (0.0095 m)  (0.0282  0.0222 ) m 2 ]}
 (11.4134  2.8863  0.38288  2.7980)  103 kg
 17.4806  103 kg
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.128 (Continued)
and
1
I AA  [(11.4134)(0.0052  0.0112 )  (2.8863)(0.0112  0.017 2 )
8
 (0.38288)(0.017 2  0.0222 )
 (2.7980)(0.0222  0.0282 )]  103 kg  m 2
 (208.29  147.92  37.00  443.48)  109 kg  m 2
 836.69  109 kg  m 2
or
Now
2
k AA
 
I AA  837  109 kg  m2 
I AA 836.69  109 kg  m 2

 47.864  106 m 2
3
m
17.4806  10 kg
or
k AA  6.92 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.129
The machine part shown is formed by machining a conical
surface into a circular cylinder. For b  12 h, determine the
mass moment of inertia and the radius of gyration of the
machine part with respect to the y axis.
SOLUTION
Mass:
mcyl   a 2 h
1
mcyl a 2
2
1
  a 4 h
2
I y : I cyl 
mcone 
1
h 1
 a 2   a 2 h
3
2 6
3
mcone a 2
10
1

 a 4 h
20
I cone 
For entire machine part:
1
5
 a 2 h   a 2 h
6
6
1
1
9
I y  I cyl  I cone   a 4 h 
 a 4 h   a 4 h
2
20
20
m  mcyl  mcone   a 2 h 
or
5
 6  9 
I y    a 2 h    a 2
6
 5  20 
Then
k y2 
I 27 2

a
m 50
Iy 
27
ma 2 
50
k y  0.735a 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.130
Knowing that the thin hemispherical shell shown has a mass m and thickness t,
determine the mass moment of inertia and the radius of gyration of the shell with
respect to the x axis. (Hint: Consider the shell as formed by removing a hemisphere
of radius r from a hemisphere of radius r + t; then neglect the terms containing t2
and t3 and keep those terms containing t.)
SOLUTION
Consider shell to be formed by removing hemisphere of radius r from hemisphere of radius r  t.
For a hemisphere,


2 2
1
mr ; A  4 r 2  2 r 2
5
2
1 4 3 2
m  V     r    r 3
2 3
 3
I
22

 r 3  r 2
5  3

4
  r 5
15
I 
Then
Now following the hint, for a hemispherical shell,
4
5
  r  t   r 5 


15
4
I    r 5  5r 4 t  10r 3t 2  ...  r 5 
15
I
Neglect terms with powers of t  1. I 


4
2
 r 4 t  2  r 2 r 2t
3
3
Mass of shell,


m  V   tA   t 2 r 2  2  r 2 t
Substituting with expression for m to simplify,
I
Radius of gyration,
k2 
2 2
mr 
3
I 2 2
 r
m 3
k  0.816r 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.131
A square hole is centered in and extends through the aluminum
machine component shown. Determine (a) the value of a for which
the mass moment of inertia of the component with respect to the axis
AA′, which bisects the top surface of the hole, is maximum, (b) the
corresponding values of the mass moment of inertia and the radius of
gyration with respect to the axis AA′. (The density of aluminum is
2770 kg/m3.)
SOLUTION
First note
m1 = ρV1 = ρ b 2 L
and
m2 = ρV2 = ρ a 2 L
(a)
Using Figure 9.28 and the parallel-axis theorem, we have
I AA′ = ( I AA′ )1 − ( I AA′ ) 2
=
1
a
m1 (b 2 + b 2 ) + m1
2
12
2
1
a
−
m2 (a 2 + a 2 ) + m2
2
12
= ( ρ b 2 L)
=
ρL
12
5 2
1 2 1 2
a
b + a − ( ρ a 2 L)
6
4
12
(2b 4 + 3b 2 a 2 − 5a 4 )
Then
dI AA′ ρ L
=
(6b 2 a − 20a 3 ) = 0
da
12
or
a=0
Also
d 2 I AA′
da
a=b
and
=
ρL
12
d I AA′
3
10
(6b 2 − 60a 2 ) =
1
ρ L(b2 − 10a 2 )
2
2
Now for a = 0,
and for a = b
2
2
da 2
3
,
10
d 2 I AA′
da 2
0
0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.131 (Continued)
3
10
( I AA′ )max occurs when
a=b
Then
a = (84 mm)
3
10
a = 46 mm
or
(b)
From Part a:
( I AA′ ) max =
=
ρL
2b + 3b
4
12
2
3
b
10
2
3
−5 b
10
where
2
k AA
′ =
=
49
ρ Lb 4
240
49
× 2770 kg/m3 × (0.3 m)(0.084 m)4
240
or
Now
4
(I AA′ )max = 8.447 × 10–3 kg ⋅ m2
( I AA′ ) max
m
m = m1 − m2 = ρ L(b 2 − a 2 )
3
= ρL b − b
10
2
2
=
7
ρ Lb2
10
49
Then
2
240
k AA
′ =
ρ Lb 4
7
ρ Lb 2
10
=
7 2 7
(84 mm) 2
b =
24
24
or
k AA′ = 45.4 mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1570
PROBLEM 9.132
The cups and the arms of an anemometer are fabricated from
a material of density . Knowing that the mass moment of
inertia of a thin, hemispherical shell of mass m and thickness
t with respect to its centroidal axis GG is 5ma 2 /12,
determine (a) the mass moment of inertia of the anemometer
with respect to the axis AA, (b) the ratio of a to l for which
the centroidal moment of inertia of the cups is equal to 1
percent of the moment of inertia of the cups with respect to
the axis AA.
SOLUTION
(a)
First note
marm  Varm   
and
dmcup   dVcup

4
d 2l
  [(2 a cos  )(t )(ad )]
Then

mcup  dmcup 

 /2
0
2 a 2 t cos  d
 2 a 2 t[sin  ]0 / 2
 2 a 2 t
Now
( I AA )anem  ( I AA )cups  ( I AA )arms
Using the parallel-axis theorem and assuming the arms are slender rods, we have
2 


( I AA )anem  3 ( I GG )cup  mcup d AG
  3  I arm  marm d AGarm 
2
2
 5

1
 a   
l 
2
2
2
 3  mcup a  mcup (l  a )       3  marm l  marm   
12
 2   
 2  

 2

5

 3mcup  a 2  2la  l 2   marm l 2
3

5
 

 3(2 a 2t )  a 2  2la  l 2     d 2 l  (l 2 )
3
4

 

or

 5 a2
 d 2l 
a
( I AA )anem   l 2  6a 2 t  2  2  1 
 
l
3 l
 4 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.132 (Continued)
(b)
We have
or
( I GG )cup
( I AA )cup
 0.01
5
5

mcup a 2  0.01mcup  a 2  2la  l 2  (from part a )
12
3

a
Now let   .
l
Then
5

5 2  0.12   2  2  1
3

or
40 2  2  1  0
2  (2)2  4(40)(1)
2(40)
Then
 
or
  0.1851 and  =  0.1351
a
 0.1851 
l
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.133
After a period of use, one of the blades of a shredder has been worn to
the shape shown and is of mass 0.18 kg. Knowing that the mass
moments of inertia of the blade with respect to the AA and BB axes are
0.320 g  m2 and 0.680 g  m2 , respectively, determine (a) the location
of the centroidal axis GG, (b) the radius of gyration with respect to axis
GG.
SOLUTION
(a)
d B  (0.08  d A ) m
We have
and, using the parallel axis
I AA  I GG   md A2
I BB  I GG   md B2
Then

I BB  I AA  m d B2  d A2

 m (0.08  d A ) 2  d A2 
 m(0.0064  0.16d A )
Substituting:
(0.68  0.32)  103 kg  m 2  0.18 kg(0.0064  0.16d A ) m 2
or
d A  27.5 mm 
to the right of A
(b)
We have
I AA  I GG  md A2
or
I GG  0.32  103 kg  m 2  0.18 kg  (0.0275 m) 2
 0.183875  103 kg  m 2
Then
2
kGG
 
I GG  0.183875  103 kg  m 2

 1.02153  103 m 2
0.18 kg
m
or
kGG  32.0 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.134
Determine the mass moment of inertia of the 0.5 kg machine component shown
with respect to the axis AA′.
SOLUTION
First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger
cone as indicated.
h
h + 60
=
10
30
Now
or h = 30 mm
The weight of the body is given by
m = (m1 + m2 − m3) = r (V1 + V2 − V3)
or
0.5 kg = r
p (0.02)2 (0.06) +
p
p
(0.005)2 (0.03) − (0.015)2 (0.09) m3
3
3
= r (75.398 + 0.785 − 21.205) × 10–6 m3
or
r = 9095 kg/m3
Then
m1 = (9095)(75.398 × 10–6) = 0.686 kg
m2 = (9095)(0.785 × 10–6) = 0.007 kg
m3 = (9095)(21.205 × 10–6) = 0.193 kg
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.134 (Continued)
Finally, using Figure 9.28, we have
I AA′ = ( I AA′ )1 + ( I AA′ ) 2 − ( I AA′ )3
=
1
3
3
m1a12 + m2 a22 − m3 a32
2
10
10
=
1
3
3
(0.686)(0.02)2 +
(0.007)(0.005)2 −
(0.193)(0.015)2 kg ⋅ m2
2
10
10
= (137.2 + 0.052 − 13.03) × 10–6 kg ⋅ m2
or
I AA′ = 124.2 × 10–6 kg ⋅ m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
To the instructor:
The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are
derived at this time for use in the solutions of Problems 9.135 through 9.140.
Thin rectangular plate
( I x ) m  ( I x ) m  md 2

 b  2  h  2 
1
m(b 2  h 2 )  m      
12
 2   2  
1
 m(b 2  h 2 )
3
( I y )m  ( I y )m  md 2
2
1
1
b
 mb 2  m    mb 2
12
3
2
I z  ( I z  ) m  md 2
2

1
1
h
mh 2  m    mh 2
12
2
3
 
Thin triangular plate
We have
1

m  V    bht 
2

1 3
bh
36
and
I z ,area 
Then
I z ,mass   tI z ,area
 t 
1 3
bh
36

1
mh2
18
Similarly,
I y ,mass 
1
mb 2
18
Now
I x,mass  I y , mass  I z , mass 
1
m(b 2  h 2 )
18
Thin semicircular plate
We have
and


m  V    a 2t 
2

I y ,area  I z ,area 

8
a4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(Continued)
I y , mass  I z ,mass   tI y ,area
Then

 t 

8
a4
1 2
ma
4
1
ma 2
2
Now
I x ,mass  I y , mass  I z , mass 
Also
I x, mass  I x,mass  my 2
 1 16 
or I x, mass  m   2  a 2
 2 9 
and
I z , mass  I z , mass  my 2
or
yz 
 1 16 
I z , mass  m   2  a 2
 4 9 
4a
3
Thin Quarter-Circular Plate
We have


m  V    a 2t 
4


a4
and
I y ,area  I z ,area 
Then
I y ,mass  I z , mass   tI y ,area
16
 t 


16
a4
1 2
ma
4
1
ma 2
2
Now
I x , mass  I y ,mass  I z , mass 
Also
I x,mass  I x,mass  m( y 2  z 2 )
or
 1 32 
I x, mass  m   2  a 2
 2 9 
and
I y ,mass  I y,mass  mz 2
or
 1 16 
I y, mass  m   2  a 2 
 4 9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.135
A 2-mm thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass moment of inertia of the component with
respect to each of the coordinate axes.
SOLUTION
m1  V  7850 kg/m3 (0.120 m)(0.200 m)(0.002 m)  0.3768 kg
m2  V  7850 kg/m3 (0.120 m)(0.100 m)(0.002 m)  0.1884 kg
Panel 1:
I x  I x  md 2 
1
 0.3768 kg  0.12 m 2   0.3768 kg   0.1 m 2   0.06 m 2 
12
I x  5.577  103 kg  m 2
I y  I y  md 2 
1
 0.3768 kg  0.2 m 2   0.3768 kg   0.1 m 2   0.1 m 2 
12
I y  8.792  103 kg  m 2
I z  I z  md 2 
1
 0.3768 kg   0.2 m 2   0.12 m 2    0.3768 kg   0.1 m 2   0.06 m 2 
12
I z  6.833  103 kg  m 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.135 (Continued)
Panel 2:
I x  I x  md 2 
1
 0.1884 kg   0.12 m 2   0.1 m 2    0.1884 kg   0.05 m 2   0.06 m 2 
12
I x  1.532  103 kg  m 2
I y  I y  md 2 
1
 0.1884 kg  0.1 m 2   0.1884 kg   0.05 m 2   0.2 m 2 
12
I y  8.164  103 kg  m 2
I z  I z  md 2 
1
 0.1884 kg  0.12 m 2   0.1884 kg   0.06 m 2   0.2 m 2 
12
I z  8.440  103 kg  m2
Total Mass Moments of Inertia:
IT  I1  I 2
I x  7.11  103 kg  m2 
I y  16.96  103 kg  m2 
I z  15.27  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.136
A 2-mm thick piece of sheet steel is cut and bent into the
machine component shown. Knowing that the density of steel is
7850 kg/m3, determine the mass moment of inertia of the
component with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m  STV  ST tA
Then
m1  (7850 kg/m3 )(0.002 m)(0.35  0.39) m 2
 2.14305 kg


m2  (7850 kg/m3 )(0.002 m)   0.1952  m 2  0.93775 kg
2


1

m3  (7850 kg/m3 )(0.002 m)   0.39  0.15  m 2  0.45923 kg
2

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have
I x  ( I x )1  ( I x )2  ( I x )3
2
1
 0.39  
2
  (2.14305 kg)(0.39 m)  (2.14305 kg) 
m 
 2
 
 12
 4  0.195 2

 1 16 

   2  (0.93775 kg)(0.195 m) 2  (0.93775 kg) 
 (0.195) 2  m 2 


 3

 2 9 

 1
 0.39 2  0.15 2  2 
  (0.45923 kg)[(0.39)2  (0.15)2 ] m 2  (0.45923 kg) 
 
 m 
 3   3  
18

 [(0.027163  0.081489)  (0.011406  0.042081)  (0.004455  0.008909)] kg  m2
 (0.108652  0.053487  0.013364) kg  m 2
 0.175503 kg  m 2
or
I x  175.5  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.136 (Continued)
I y  ( I y )1  ( I y ) 2  ( I y )3
 1
 0.35  2  0.39  2  2 
  (2.14305 kg)[(0.35) 2  (0.39) 2 ] m 2  (2.14305 kg) 
 
  m 
 2   2  
12

1

  (0.93775 kg)(0.195 m) 2  (0.93775 kg)(0.195 m)2 
4

2
 1

 0.39   2 
  (0.45923 kg)(0.39 m) 2  (0.45923 kg) (0.35) 2  
  m 
 3  

18

 [(0.049040  0.147120)  (0.008914  0.035658)
 (0.003880  0.064017)] kg  m 2
 (0.196160  0.044572  0.067897) kg  m2
 0.308629 kg  m2
or
I y  309  103 kg  m2 
I z  ( I z )1  ( I z )2  ( I z )3
2
1
 0.35  
  (2.14305 kg)(0.35 m)2  (2.14305 kg) 
m 
 2
 
12
1

  (0.93775 kg)(0.195 m) 2 
4


2
 1

 0.15   2 
  (0.45923 kg)(0.15 m) 2  (0.45923 kg) (0.35)2  
  m 
 3  

18

 [(0.021877  0.065631)  0.008914)  (0.000574  0.057404)] kg  m 2
 (0.087508  0.008914  0.057978) kg  m2
 0.154400 kg  m2
or
I z  154.4  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.137
A subassembly for a model airplane is fabricated from three pieces of 1.5mm plywood. Neglecting the mass of the adhesive used to assemble the
three pieces, determine the mass moment of inertia of the subassembly
with respect to each of the coordinate axes. (The density of the plywood
is 780 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m  V   tA
Then
1

m1  m2  (780 kg/m3 )(0.0015 m)   0.3  0.12  m 2
2

 21.0600  103 kg


m3  (780 kg/m3 )(0.0015 m)   0.122  m 2  13.2324  103 kg
4

Using the equations derived above and the parallel-axis theorem, we have
( I x )1  ( I x )2
I x  ( I x )1  ( I x )2  ( I x )3
2
1
 0.12  
 2  (21.0600  103 kg)(0.12 m)2  (21.0600  103 kg) 
m 
 3
 
18
1

  (13.2324  103 kg)(0.12 m) 2 
2


 [2(16.8480  33.6960)  (95.2733)]  106 kg  m 2
 [2(50.5440)  (95.2733)]  106 kg  m 2
or
I x  196.4  106 kg  m 2 






Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.137 (Continued)

I y  ( I y )1  ( I y ) 2  ( I y )3


2
1
 0.3  
  (21.0600  103 kg)(0.3 m)2  (21.0600 kg) 
m 
 3  
18
1
  (21.0600  103 kg)[(0.3)2  (0.12)2 ] m 2 
18

 0.3 2  0.12 2  2 
 (21.0600  103 kg) 
 
 m 
 3   3  

1

  (13.2324  103 kg)(0.12 m)2  
4

 [(105.300  210.600)  (122.148  244.296) 


 (47.637)]  106 kg  m 2
 (315.900  366.444  47.637)  106 kg  m 2 

or
I y  730  106 kg  m2 
Symmetry implies I y  I z
I z  730  106 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.138
A section of sheet steel 0.3 mm thick is cut and bent into the
sheet metal machine component shown. Determine the mass
moment of inertia of the component with respect to each of the
coordinate axes. (The density of steel is 7850 kg/m3 .)
SOLUTION
First consider one triangular plate with local axes u-v. We have
m  V   tA  (7850 kg/m3)
1
(0.0003 m)(0.12 m)(0.045 m) = 0.00636 kg
2
I mass   tI area 
m
I area
A
Then
 m  1
  bh3   1 mh 2 ,
Iu  
 1 bh   36
 18
 2 
Iu 
1
(0.00636)(0.045)2 = 0.716 3 1026 kg  m 2
18
Iv 
1
(0.00636)(0.12)2 = 5.088 3 1026 kg  m 2
18
Iv 
1
mb 2
18
For composite area:
I x  2  I u  md 2   2 0.716 3 1026 + 0.00636(0.015)2


I x  4.294 3 1026 kg  m 2 
For I y , consider plates separately.
Plate 1 in x-y plane:
I y1  I v  md 2  5.088 3 1026 + 0.00636(0.02)2 = 7.632 3 1026 kg  m 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.138 (Continued)
Plate 2 in x-z plane:
Note: By Eqn. 9.38,
I y '  I u  I v  0.716 3 1026 + 5.088 3 1026
I y '  5.804 3 1026 kg  m 2
Now applying the Parallel Axis Theorem,
I y2  I y '  md 2

2
2 
I y2  5.804 3 1026 + (0.00636) (0.02) + (0.015) 


I y2  9.779 3 1026 kg  m 2
Finally, total for both plates 1 and 2 gives,
I y  I y1  I y2  7.632 3 1026 + 9.779 3 1026
I y  17.41 3 1026 kg  m 2 
By symmetry, I y  I z
2
I z  17.41 3 1026 kg  m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.139
A framing anchor is formed of 2-mm-thick galvanized steel. Determine the
mass moment of inertia of the anchor with respect to each of the coordinate
axes. (The density of galvanized steel is 7530 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = rV = rtA
Then
m1 = 7530 kg/m3 × 0.002 m × (0.045 × 0.07) m2
= 47439 × 10–6 kg
m2 = 7530 kg/m3 × 0.002 m × (0.045 × 0.02) m2
= 13554 × 10–6 kg
m3 = 7530 kg/m3 × 0.002 m ×
= 28614 × 10–6 kg
1
× 0.04 × 0.095 m2
2
Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
I x = ( I x )1 + ( I x ) 2 + ( I x )3
=
1
(47439 × 10–6 kg)(0.07 m)2
12
+ (47439 × 10–6 kg)(0.035 m2)
+
1
(13554 × 10–6 kg)(0.02 m)2
12
+ (13554 × 10–6 kg)[ (0.07)2 + (0.01)2 ] m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.139 (Continued)
1
(28614 × 10–6 kg)[(0.095)2 + (0.04)2] m2
18
+
+ (28614 × 10–6 kg)
2
2
1
× 0.095 +
× 0.04
3
3
2
m2
= [(19.37 + 58.11) + (0.45 + 67.77)
+ (16.89 + 119.86)] × 10–6 kg · m2
= (77.48 + 68.22 + 136.75) × 10–6 kg · m2
I x = 282.5 × 10–6 kg · m2
or
I y = ( I y )1 + ( I y ) 2 + ( I y )3
=
1
(47439 × 10–6 kg)(0.045 m)2
12
+ (47439 × 10–6 kg)(0.0225 m)2
+
1
(13554 × 10–6 kg)[(0.045)2 + (0.02)2] m2
12
+ (13554 × 10–6 kg)[(0.0225)2 + (0.01)2] m2
+
1
(28614 × 10–6 kg)(0.04 m)2
18
+ (28614 × 10–6 kg) (0.045)2 +
1
× 0.04
3
2
m2
= [(8 + 24.02) + (2.74 + 8.22)
+ (2.54 + 63.03)] × 10–6 kg · m2
= (32.02 + 10.96 + 65.57) × 10–6 kg · m2
or
I y = 108.6 × 10–6 kg · m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.139 (Continued)
I z = ( I z )1 + ( I z ) 2 + ( I z )3
1
(47439 × 10–6 kg)[(0.045)2 + (0.07)2] m2
12
=
+ (47439 × 10–6 kg)[(0.0225)2 + (0.035)2] m2
+
1
(13554 × 10–6 kg)(0.045 m)2
12
+ (13554 × 10–6 kg)[(0.0225)2 + (0.07)2] m2
+
1
(28614 × 10–6 kg)(0.095 m)2
18
+ (28614 × 10–6 kg) (0.045)2 +
2
× 0.095
3
2
m2
= [(27.38 + 82.13) + (2.29 + 73.28)
+ (14.35 + 172.72)] × 10–6 kg · m2
= (109.51 + 75.57 + 187.07) × 10–6 kg · m2
or
I z = 372.2 × 10–6 kg · m2
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1584
PROBLEM 9.140*
A farmer constructs a trough by welding a rectangular piece
of 2-mm-thick sheet steel to half of a steel drum. Knowing
that the density of steel is 7850 kg/m3 and that the thickness
of the walls of the drum is 1.8 mm, determine the mass
moment of inertia of the trough with respect to each of the
coordinate axes. Neglect the mass of the welds.
SOLUTION
First compute the mass of each component. We have
m  STV  ST tA
Then
m1  (7850 kg/m3 )(0.002 m)(0.84  0.21) m 2
 2.76948 kg
m2  (7850 kg/m3 )(0.0018 m)(  0.285  0.84) m 2
 10.62713 kg


m3  m4  (7850 kg/m 3 )(0.0018 m)   0.2852  m 2
2

 1.80282 kg
Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have
( I x )3  ( I x ) 4
I x  ( I x )1  ( I x )2  ( I x )3  ( I x ) 4
2
1
0.21  2 

2
  (2.76948 kg)(0.21 m)  (2.76948 kg)  0.285 
m 
2 

12

1

 [(10.62713 kg)(0.285 m)2 ]  2  (1.80282 kg)(0.285 m)2 
2


 [(0.01018  0.08973)  (0.86319)  2(0.07322)] kg  m 2
 [(0.09991  0.86319  2(0.07322)] kg  m 2
or
I x  1.110 kg  m2 



Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.140* (Continued)
I y  ( I y )1  ( I y )2  ( I y )3  ( I y ) 4
1
  (2.76948 kg)[(0.84)2  (0.21)2 ] m 2
12
2
 0.84 2 
0.21   2 
0.285
 (2.76948 kg) 


 
 m 
2  

 2  
2
 1
 0.84  
  (10.62713 kg)[(0.84) 2  6(0.285) 2 ] m 2  (10.62713 kg) 
m 
 2
 
12
1

  (1.80282 kg)(0.285 m)2 
4


1

  (1.80282 kg)(0.285 m) 2  (1.80282 kg)(0.84 m)2 
4

 [(0.17302  0.57827)  (1.05647  1.87463)
 (0.03661)  (0.03661  1.27207)] kg  m 2
 (0.75129  2.93110  0.03661  1.30868) kg  m 2
or
I y  5.03 kg  m2 
I z  ( I z )1  ( I z ) 2  ( I z )3  ( I z )4
2
1
 0.84  
2
  (2.76948 kg)(0.84 m)  (2.76948 kg) 
m 
 2
 
 12
2
 1
 0.84  
  (10.62713 kg)[(0.84) 2  6(0.285) 2 ] m 2  (10.62713 kg) 
m 
 2
 
12
1

  (1.80282 kg)(0.285 m)2 
4

1

  (1.80282 kg)(0.285 m) 2  (1.80282 kg)(0.84 m)2 
4

 [(0.16285  0.48854)  (1.05647  1.87463)
 (0.03661)  (0.03661  1.27207)] kg  m 2
 (0.65139  2.93110  0.03661  1.30868) kg  m 2
or
I z  4.93 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.141
The machine element shown is fabricated from steel.
Determine the mass moment of inertia of the assembly with
respect to (a) the x axis, (b) the y axis, (c) the z axis. (The
density of steel is 7850 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m  STV
Then
m1  (7850 kg/m3 )( (0.08 m) 2 (0.04 m)]
 6.31334 kg
m2  (7850 kg/m3 )[ (0.02 m)2 (0.06 m)]  0.59188 kg
m3  (7850 kg/m3 )[ (0.02 m)2 (0.04 m)]  0.39458 kg
Using Figure 9.28 and the parallel-axis theorem, we have
(a)
I x  ( I x )1  ( I x ) 2  ( I x )3
1

  (6.31334 kg)[3(0.08) 2  (0.04) 2 ] m 2  (6.31334 kg)(0.02 m) 2 
12

1

  (0.59188 kg)[3(0.02)2  (0.06)2 ] m 2  (0.59188 kg)(0.03 m)2 
12


1

  (0.39458 kg)[3(0.02) 2  (0.04) 2 ] m 2  (0.39458 kg)(0.02 m) 2 
12

 [(10.94312  2.52534)  (0.23675  0.53269)
 (0.09207  0.15783)]  103 kg  m 2
 (13.46846  0.76944  0.24990)  103 kg  m 2
 13.98800  103 kg  m 2
or
I x  13.99  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.141 (Continued)
(b)
I y  ( I y )1  ( I y ) 2  ( I y )3
1

  (6.31334 kg)(0.08 m)2 
2


1


  (0.59188 kg)(0.02 m)2  (0.59188 kg)(0.04 m)2 
2

1

  (0.39458 kg)(0.02 m 2 )  (0.39458 kg)(0.04 m)2 
2

 [(20.20269)  (0.11838  0.94701)
 (0.07892  0.63133)]  103 kg  m 2
 (20.20269  1.06539  0.71025)  103 kg  m2
 20.55783  103 kg  m2
or
(c)
I y  20.6  103 kg  m 2 
I z  ( I z )1  ( I z ) 2  ( I z )3
1

  (6.31334 kg)[3(0.08) 2  (0.04) 2 ] m 2  (6.31334 kg)(0.02 m) 2 
12

1

  (0.59188 kg)[3(0.02)2  (0.06)2 ] m2  (0.59188 kg)[(0.04)2  (0.03) 2 ] m 2 
12

1

  (0.39458 kg)[3(0.02)2  (0.04)2 ] m 2  (0.03958 kg)[(0.04)2  (0.02)2 ] m 2 
12

 [(10.94312  2.52534)  (0.23675  1.47970)
 (0.09207  0.78916)]  103 kg  m 2
 (13.46846  1.71645  0.88123)  103 kg  m2
 14.30368  103 kg  m 2
or
I z  14.30  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
To the Instructor:
The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the
solutions of Problems 9.142 through 9.145.
From Figure 9.28:
Cylinder
( I x ) cyl 
1
mcyl a 2
2
( I y )cyl  ( I z )cyl 
1
mcyl (3a 2  L2 )
12
Symmetry and the definition of the mass moment of inertia ( I   r 2 dm) imply
( I )semicylinder 
( I x )sc 
1
( I )cylinder
2
11
2
 mcyl a 
22

( I y )sc  ( I z )sc 
and
msc 
1
mcyl
2
Thus,
( I x )sc 
1
msc a 2
2
and
( I y )sc  ( I z )sc 
However,
11

mcyl (3a 2  L2 ) 
2 12

1
msc (3a 2  L2 )
12
Also, using the parallel axis theorem find
 1 16 
I x  msc   2  a 2
 2 9 
 1 16 
1 
I z  msc   2  a 2  L2 
12 
 4 9 
where x and z  are centroidal axes.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.142
Determine the mass moments of inertia and the radii of
gyration of the steel machine element shown with respect
to the x and y axes. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m  STV
Then
m1  (7850 kg/m3 )(0.24  0.04  0.14) m3
 10.5504 kg


m2  m3  (7850 kg/m3 )  (0.07)2  0.04  m3  2.41683 kg
2


m4  m5  (7850 kg/m3 )[ (0.044) 2  (0.04)] m3  1.90979 kg
Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to
Problem 9.144) for a semicylinder, we have
I x  ( I x )1  ( I x ) 2  ( I x )3  ( I x ) 4  ( I x )5
where ( I x ) 2  ( I x )3
1

1

  (10.5504 kg)(0.042  0.142 ) m 2   2  (2.41683 kg)[3(0.07 m)2  (0.04 m) 2 ]
12

12

1


  (1.90979 kg)[3(0.044 m) 2  (0.04 m) 2 ]  (1.90979 kg)(0.04 m) 2 
12


1

  (1.90979 kg)[3(0.044 m) 2  (0.04 m) 2 ]
 12

 [(0.0186390)  2(0.0032829)  (0.0011790  0.0030557)  (0.0011790)] kg  m 2
 0.0282605 kg  m 2
or
I x  28.3  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.142 (Continued)
I y  ( I y )1  ( I y )2  ( I y )3  ( I y )4  ( I y )5
( I y )2  ( I y )3 (I y )4  |( I y )5 |
where
Then
1

I y   (10.5504 kg)(0.242  0.142 ) m 2 
12

2

4  0.07  2 
 1 16 

m 
 2 (2.41683 kg)   2  (0.07 m 2 )  (2.41683 kg)  0.12 
3 
 2 9 



 [(0.0678742)  2(0.0037881  0.0541678)] kg  m 2
 0.1837860 kg  m 2
or
Also
I y  183.8  103 kg  m2 
m  m1  m2  m3  m4  m5 where m2  m3 , m4  |m5 |
 (10.5504  2  2.41683) kg  15.38406 kg
Then
and
k x2 
k y2 
I x 0.0282605 kg  m 2

m
15.38406 kg
Iy
m

or
k x  42.9 mm 
or
k y  109.3 mm 
0.1837860 kg  m 2
15.38406 kg
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.143
Determine the mass moment of inertia of the steel machine
element shown with respect to the x axis. (The density of
steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
Then
m1 = (7850 kg/m3 )(0.09 × 0.03 × 0.15) m3
= 3.17925 kg
π

m2 = (7850 kg/m3 )  (0.045) 2 × 0.04  m3
2


= 0.998791 kg
m3 = (7850 kg/m3 )[π (0.038) 2 × 0.03] m3 = 1.06834 kg
Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to
Problem 9.142) for a semicylinder, we have
I x = ( I x )1 + ( I x )2 − ( I x )3
1

=  (3.17925 kg)(0.032 + 0.152 ) m 2 + (3.17925 kg)(0.075 m) 2 
12

 1 16 

1

+ (0.998791 kg)  − 2  (0.045 m) 2 + (0.04 m) 2 
12

 4 9π 

2

 4 × 0.045
 

+ (0.998791 kg) (0.13)2 + 
+ 0.015   m 2 
 3π
 


1

−  (1.06834 kg)[3(0.038 m) 2 + (0.03 m)2 ] + (1.06834 kg)(0.06 m)2 
12

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.143 (Continued)
= [(0.0061995 + 0.0178833) + (0.0002745 + 0.0180409)
− (0.0004658 + 0.0038460)] kg ⋅ m 2
= (0.0240828 + 0.0183154 − 0.0043118) kg ⋅ m 2
= 0.0380864 kg ⋅ m 2
or
I x = 38.1 × 10−3 kg ⋅ m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.144
40 mm
50 mm
60mm
45 mm
Determine the mass moment of inertia of the steel machine
element shown with respect to the y axis. (The density of
steel is 7850 kg/m3.)
38 mm
15 mm
15 mm
45 mm
45 mm
SOLUTION
First compute the mass of each component. We have
m   stV
m1  (7850 kg/m3)(0.09 3 0.03 3 0.15) m3
Then
 3.17925 kg


m2  (7850 kg/m3)  (0.045)2 3 0.04 m3

 2
 0.998791 kg
m3  (7850 kg/m3)[ (0.038)2 3 0.03] m3 = 1.06834 kg
Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to
Problem 9.142) for a semicylinder, we have
I y  ( I y )1  ( I y ) 2  ( I y )3
1
2

2
I y1  
3.17925 (0.09) + (0.15)   (3.17925) (0.075)2

12






I y1  25.99 3 1023 kg m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.144 (Continued)
I y2  I y2 '  md 2
I y2 
2
0.998791 3 (0.045)2 + (0.04)2 
  0.998791(0.13)



12
I y2  17.52 3 1023 kg m2
I y3 
2
2
1.06834
(0.038)  1.06834 (0.06)
2
I y3  4.62 3 1023 kg m2
Now combining all components and substituting for  ,
I y  (25.99 + 17.52 + 4.62) 3 1023
or
I y  48.13 3 1023 kg m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.145
Determine the mass moment of inertia of the steel fixture
shown with respect to (a) the x axis, (b) the y axis, (c) the
z axis. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m  STV
m1  7850 kg/m3  (0.08  0.05  0.160) m3
Then
 5.02400 kg
m2  7850 kg/m3  (0.08  0.038  0.07) m3  1.67048 kg


m3  7850 kg/m3    0.0242  0.04  m3  0.28410 kg
2

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
(a) I x  ( I x )1  ( I x ) 2  ( I x )3
 1
 0.05 2  0.16 2  2 
  (5.02400 kg)[(0.05)2  (0.16) 2 ] m 2  (5.02400 kg) 
 
 m 
12

 2   2  
2
2
 1

0.038  
0.07   2 
  (1.67048 kg)[(0.038)2  (0.07) 2 ] m 2  (1.67048 kg)  0.05 

0.05

m 
2  
2  

12


 1 16 

1
 (0.28410 kg)   2  (0.024)2  (0.04)2  m 2
12
 4 9 


2
2

4  0.024  
0.04   2 
0.16
 (0.28410 kg)  0.05 


m 
 
3
2  
 


 [(11.7645  35.2936)  (0.8831  13.6745)  (0.0493  6.0187)]  103 kg  m2
 (47.0581  14.5576  6.0680)  103 kg  m 2
 26.4325  103 kg  m 2
or I x  26.4  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.145 (Continued)
(b)
I y  ( I y )1  ( I y )2  ( I y )3
 1
 0.08 2  0.16 2  2 
  (5.02400 kg)[(0.08) 2  (0.16)2 ] m 2  (5.02400 kg) 
 
 m 
12

 2   2  
2
 1
 0.08 2 
0.07   2 
  (1.67048 kg)[(0.08)2  (0.07)2 ] m 2  (1.67048 kg) 

0.05

m 
 
2  
 2  
12

2
 1
 0.08 2 
0.04   2 
2
2
2
  (0.28410 kg)[3(0.024)  (0.04) ] m  (0.28410 kg) 
   0.16  2   m 
 
 2  
12

 [(13.3973  40.1920)  (1.5730  14.7420)  (0.0788  6.0229)]  103 kg  m2
 (53.5893  16.3150  6.1017)  103 kg  m2
 31.1726  103 kg  m2
(c)
or I y  31.2  103 kg  m2 
I z  ( I z )1  ( I z )2  ( I z )3
 1
 0.08 2  0.05 2  2 
2
2
2
  (5.02400 kg)[(0.08)  (0.05) ] m  (5.02400 kg) 
 
 m 
12
 2   2  


2
 1
 0.08 2 
0.038   2 
  (1.67048 kg)[(0.08)2  (0.038) 2 ] m 2  (1.67048 kg) 

0.05

m 
 
2  
 2  
12

2

 0.08 2 
4  0.024   2 
 1 16 
0.05
 (0.28410 kg)   2  (0.024 m) 2  (0.28410 kg) 


 
 m 
3
 2 9 
 
 2  


 [(3.7261  11.1784)  (1.0919  4.2781)  (0.0523  0.9049)]  103 kg  m 2
 (14.9045  5.3700  0.9572)  103 kg  m2
 8.5773  103 kg  m 2
or I z  8.58  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
To the instructor:
The following formulas for the mass moment of inertia of wires are derived or summarized at this time
for use in the solutions of Problems 9.146 through 9.148.
Slender Rod
Ix  0
I y  I z  
1
mL2 (Figure 9.28)
12
1
I y  I z  mL2 (Sample Problem 9.9)
3
Circle

We have
I y  r 2 dm  ma 2
Now
I y  Ix  Iz
And symmetry implies
Ix  Iz
Ix  Iz 
1 2
ma
2
Semicircle
Following the above arguments for a circle, We have
Ix  Iz 
1
ma 2
2
I y  ma 2
Using the parallel-axis theorem
I z  I z   mx 2
or
x
2a

1 4 
I z  m   2  a 2 
2  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.146
Aluminum wire with a mass per unit length of 0.049 kg/m is used
to form the circle and the straight members of the figure shown.
Determine the mass moment of inertia of the assembly with respect
to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m=
m
L
L
m1 = 0.049 kg/m × (2p × 0.32 m)
Then
= 0.0985 kg
m2 = m3 = m4 = m5 = 0.049 kg/m × 0.16 m
= 0.00784 kg
Using the equations given above and the parallel-axis theorem, we have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 + ( I x )5
=
1
(0.0985 kg)(0.32 m)2
2
+
1
(0.00784 kg)(0.16 m)2
3
+ [0 + (0.00784 kg)(0.16 m)2]
+
1
(0.00784 kg)(0.16 m)2 + (0.00784 kg)(0.082 + 0.322)m2
12
+
1
(0.00784 kg)(0.16 m)2 + (0.00784 kg)(0.162 + 0.242)m2
12
= [(5.043) + (0.067) + (0.201) + (0.017 + 0.853) + (0.017 + 0.652)] × 10−3 kg ⋅ m2
= 6.85 × 10−3 kg ⋅ m2
or
I x = 6.85 × 10−3 kg ⋅ m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.146 (Continued)
( I y )2 = ( I y )4 ,
( I y )3 = ( I y )5
I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5
= [(0.0985 kg)(0.32 m)2]
+ 2[0 + (0.00784 kg)(0.32 m)2]
+2
1
(0.00784 kg)(0.16 m)2 + (0.00784 kg)(0.24 m)2
12
= [(10.086) + 2(0.803) + 2(0.017 + 0.452)] × 10−3 kg ⋅ m2
= 12.63 × 10−3 kg ⋅ m2
or
Symmetry implies
Ix = Iz
I y = 12.63 × 10−3 kg ⋅ m2
I z = 6.85 × 10−3 kg ⋅ m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
PROBLEM 9.147
360 mm
The figure shown is formed of 3-mm-diameter steel wire. Knowing that the
density of the steel is 7850 kg/m3, determine the mass moment of inertia of the
wire with respect to each of the coordinate axes.
360 mm
360 mm
x
z
SOLUTION
First compute the mass of each component. We have
m = ρSTV = rAL
Then
m1 = m2 = 7850 kg/m3 ×
π
3 4 (0.003 m) 4 × (p × 0.36 m)
2
= 62.756 × 10−3 kg
m3 = m4 = 7850 kg/m3 ×
π
3 4 (0.003 m) 4 × 0.36 m
2
= 19.976 × 10−3 kg
Using the equations given above and the parallel-axis theorem, we have
( I x )3 = ( I x ) 4
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4
=
1
(0.062756 kg)(0.36 m)2
2
+
+2
1
(0.062756 kg)(0.36 m)2 + (0.062756 kg)(0.36 m)2
2
1
(0.019976 kg)(0.36 m)2 + (0.019976 kg)(0.182 + 0.362)m2
12
= [(4.067) + (4.067 + 8.133) + 2(0.216 + 3.236)] × 10−3
= 23.17 × 10−3 kg ⋅ m2
or
I x = 23.2 × 10−3 kg ⋅ m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.147 (Continued)
( I y )1 = ( I y ) 2 , ( I y )3 = ( I y ) 4
I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4
= 2[(0.062756 kg)(0.36 m)2]
+ 2[(0.019976 kg)(0.36 m)2]
= [2(8.133) + 2(2.589)] × 10−3 kg ⋅ m2
= 21.444 × 10−3 kg ⋅ m2
or
I y = 21.4 × 10−3 kg ⋅ m2
( I z )3 = ( I z ) 4
I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4
=
1
(0.062756 kg)(0.36 m)2
2
+ (0.062756 kg)
+ (0.062756 kg)
+2
1 4
−
(0.36 m)2
2 π2
2 × 0.36 2
π
+ (0.36)2 m2
1
(0.019976 kg)(0.36 m)2
3
= [(4.067) + (0.77 + 11.429) + 2(0.863)] × 10−3 kg ⋅ m2
= 17.99 × 10−3 kg ⋅ m2
or
I z = 18 × 10−3 kg ⋅ m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1604
PROBLEM 9.148
A homogeneous wire with a mass per unit length of 0.056 kg/m is
used to form the figure shown. Determine the mass moment of
inertia of the wire with respect to each of the coordinate axes.
SOLUTION
First compute the mass m of each component. We have
m  (m/L) L
 0.056 kg/m  1.2 m
 0.0672 kg
Using the equations given above and the parallel-axis theorem, we have
I x  ( I x )1  ( I x )2  ( I x )3  ( I x ) 4  ( I x )5  ( I x )6
( I x )1  ( I x )3  ( I x )4  ( I x )6
Now
and ( I x )2  ( I x )5
1

I x  4  (0.0672 kg)(1.2 m)2   2[0  (0.0672 kg)(1.2 m) 2 ]
3


Then
 [4(0.03226)  2(0.09677)] kg  m 2
 0.32258 kg  m2
or
I x  0.323 kg  m 2 
I y  ( I y )1  ( I y )2  ( I y )3  ( I y )4  ( I y )5  ( I y )6
( I y )1  0, ( I y )2  ( I y )6 , and ( I y )4  ( I y )5
Now
1

I y  2  (0.0672 kg)(1.2 m) 2   [0  (0.0672 kg)(1.2 m) 2 ]
3

1

 2  (0.0672 kg)(1.2 m) 2  (0.0672 kg)(1.22  0.62 ) m 2 
12


Then
 [2(0.03226)  (0.09677)  2(0.00806  0.12096)] kg  m 2
 [2(0.03226)  (0.09677)  2(0.12902)] kg  m 2
 0.41933 kg  m 2
Symmetry implies
I y  Iz
or
I y  0.419 kg  m 2 
I z  0.419 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.149
Determine the mass products of inertia Ixy, Iyz, and Izx of the
steel fixture shown. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m  V
m1  7850 kg/m3  (0.08  0.05  0.16) m3  5.02400 kg
Then
m2  7850 kg/m3  (0.08  0.038  0.07) m3  1.67048 kg


m3  7850 kg/m3    0.0242  0.04  m3  0.28410 kg
2

Now observe that the centroidal products of inertia, I xy , I yz , and I zx , of each component are zero
because of symmetry. Now
0.038 

y2   0.05 
m  0.031 m
2 

4  0.024 

y3   0.05 
 m  0.039814 m
3


and then
m, kg
x, m
y, m
z, m
mx y , kg  m2
my z , kg  m2
mz x , kg  m2
1
5.02400
0.04
0.025
0.08
5.0240  103
10.0480  103
16.0768  103
2
1.67048
0.04
0.031
0.085
2.0714  103
4.4017  103
5.6796  103
3
0.28410
0.04
0.039814
0.14
0.4524  103
1.5836  103
1.5910  103
Finally,
I xy  ( I xy )1  ( I xy )2  ( I xy )3  [(0  5.0240)  (0  2.0714)  (0  0.4524)]  103
 2.5002  103 kg  m 2
or
I xy  2.50  103 kg  m2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.149 (Continued)
I yz  ( I yz )1  ( I yz )2  ( I yz )3  [(0  10.0480)  (0  4.4017)  (0  1.5836)]  103
 4.0627  103 kg  m 2
or
I yz  4.06  103 kg  m2 
I zx  ( I zx )1  ( I zx )2  ( I zx )3  [(0  16.0768)  (0  5.6796)  (0  1.5910)]  103
 8.8062  103 kg  m 2
or
I zx  8.81  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.150
Determine the mass products of inertia Ixy, Iyz, and Izx
of the steel machine element shown. (The density of
steel is 7850 kg/m3.)
SOLUTION
Since the machine element is symmetrical with respect to the xy plane, Iyz = Izx= 0

I xy  0
Also,
for all components except the cylinder, since these are symmetrical with respect to the xz plane.
For the cylinder
m  V  (7850 kg/m 3 ) (0.022 m) 2 (0.02 m)  0.2387 kg
  I xy   mx y  0  (0.2387 kg)(0.06 m)(0.02 m)  286.4  106 kg  m 2
I xy
I xy  286  106 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.151
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast
aluminum machine component shown. (The density of aluminum
is 2770 kg/m3)
SOLUTION
First compute the mass of each component. We have
m = ρ ALV
Then
m1 = 2770 kg/m3 × (0.118 × 0.036 × 0.044)m3
= 0.518 kg
m2 = 2770 kg/m3 ×
π
2
(0.022)2 × 0.036 m3
= 0.0758 kg
m3 = 2770 kg/m3 × (0.028 × 0.022 × 0.024)m3
= 0.041 kg
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry. Now
x2 = − 0.118 +
4 × 0.022
m
3p
= −0.127 m
y3 = 0.036 −
0.022
m
2
= 0.025 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.151 (Continued)
and then
m, kg
x, m
y, m
z, m
mxy, kg · m2
myz , kg · m2
mz x , kg · m2
1
0.518
−0.059
0.018
0.022
−550.12 × 10−6
205.13 × 10−6
−672.36 × 10−6
2
0.0758
−0.127
0.018
0.022
−173.28 × 10−6
30.02 × 10−6
−211.79 × 10−6
3
0.041
−0.014
0.025
0.026
−14.35 × 10−6
26.65 × 10−6
−14.92 × 10−6
Finally,
I xy = ( I xy )1 + ( I xy ) 2 − ( I xy )3
= [(0 − 550.12) + (0 − 173.28) − (0 −14.35)] × 10−6
or I xy = −709.1 × 10−6 kg · m2
I yz = ( I yz )1 + ( I yz ) 2 − ( I yz )3
= [(0 + 205.13) + (0 + 30.02) − (0 + 26.65)] × 10−6
or I yz = 208.5 × 10−6 kg · m2
I zx = ( I zx )1 + ( I zx )2 − ( I zx )3
= [(0 − 672.36) + (0 − 211.79) − (0 −14.92)] × 10−6
or I zx = −869.2 × 10−6 kg · m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.152
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast
aluminum machine component shown. (The density of aluminum
is 2770 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = ρ ALV
m1 = 2770 kg/m3 × (0.156 × 0.016 × 0.032)m3
Then
= 0.221 kg
m2 = 2770 kg/m3 × (0.048 × 0.016 × 0.04)m3
= 0.0851 kg
m3 = 2770 kg/m3 ×
π
2
(0.016)2 × 0.016 m3 = 0.0178 kg
m4 = 2770 kg/m3 × [p (0.012)2 × 0.012]m3 = 0.015 kg
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry. Now
x3 = − 0.156 +
4 × 0.016
m = −0.163 m
3p
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.152 (Continued)
and then
m, kg
x, m
y, m z, m
mxy, kg · m2
myz , kg · m2
mzx , kg · m2
1
0.221
−0.078
0.008 −0.016 −137.904 × 10−6
−28.288 × 10−6
275.808 × 10−6
2
0.0851
−0.024
0.008 −0.052 −16.339 × 10−6
−35.401 × 10−6
106.204 × 10−6
3
0.0178
−0.163
0.008 −0.016 −23.211 × 10−6
−2.278 × 10−6
46.422 × 10−6
4
0.015
−0.156
0.022 −0.016
−51.48 × 10−6
−5.28 × 10−6
37.44 × 10−6
−228.934 × 10−6
−71.247 × 10−6
465.874 × 10−6
Σ
Then
0
I xy = Σ( I x′y′ + mx y )
0
I yz = Σ( I y ′z ′ + m y z )
0
I zx = Σ( I z ′x′ + mz x )
or
I xy = −228.9 × 10−6 kg · m2
or
I yz = −71.2 × 10−6 kg · m2
or
I zx = 465.9 × 10−6 kg · m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1610
PROBLEM 9.153
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of steel
is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz,
and Izx of the component.
SOLUTION
First compute the mass of each component.
We have
m  STV  ST tA
Then
m1  (7850 kg/m3 )[(0.002)(0.4)(0.45)]m3
 2.8260 kg

1

m2  7850 kg/m3 (0.002)   0.45  0.18   m3
2


 0.63585 kg
Now observe that
( I xy )1  ( I yz )1  ( I zx )1  0
( I y z ) 2  ( I z x ) 2  0
From Sample Problem 9.6:
( I xy  ) 2,area  
1 2 2
b2 h2
72
Then
1
 1

( I xy )2  ST t ( I xy ) 2,area  ST t   b22 h22    m2 b2 h2
36
 72

Also
x1  y1  z2  0
0.45 

x2    0.225 
 m   0.075 m
3 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.153 (Continued)
Finally,
 1
I xy  ( I xy  mx y )  (0  0)    (0.63585 kg)(0.45 m)(0.18 m)
 36
 0.18 m  
 (0.63585 kg)( 0.075 m) 

 3 
 (1.43066  103  2.8613  103 ) kg  m 2
and
or
I xy   4.29  103 kg  m2 
or
I yz  0 
or
I zx  0 
I yz  ( I yz  m y z )  (0  0)  (0  0)  0
I zx  ( I zx  m z x )  (0  0)  (0  0)  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.154
A section of sheet steel 2 mm thick is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass products of inertia Ixy, Iyz, and Izx of the
component.
SOLUTION
m1  V  7850 kg/m3 (0.120 m)(0.200 m)(0.002 m)  0.3768 kg
m2  V  7850 kg/m3 (0.120 m)(0.100 m)(0.002 m)  0.1884 kg
For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes.
m, kg
x, m
y, m
z, m
mx y
myz
mz x

0.3768
0.1
0.06
0.1
2.261  103
2.261  103
3.768  103

0.1884
0.2
0.06
0.05
2.261  103
0.565  103
1.884  103
4.522  103
2.826  103
5.652  103

I xy  4.52  103 kg  m2 
I yz  2.83  103 kg  m 2 
I zx  5.65  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.155
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of
steel is 7850 kg/m3, determine the mass products of inertia
Ixy, Iyz, and Izx of the component.
SOLUTION
First compute the mass of each component. We have
m  STV  ST tA
Then
m1  (7850 kg/m3 )(0.002 m)(0.35  0.39) m2
 2.14305 kg


m2  (7850 kg/m3 )(0.002 m)   0.1952  m2
2

 0.93775 kg
1

m3  (7850 kg/m3 )(0.002 m)   0.39  0.15  m2
2


 0.45923 kg
Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero
and
( I xy )3  ( I zx )3  0
Also
( I yz )3,mass  STt ( I yz )3,area
Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to
a 90° rotation, we have
( I y z  )3,area 
Then
Also
1 2 2
b3 h3
72
 1
 1
( I yz  )3  ST t  b32 h32  
m3b3 h3
 72
 36
y1  x2  0
y2 
4  0.195
m  0.082761 m
3
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PROBLEM 9.155 (Continued)
Finally,
I xy  ( I xy  mx y )

 0.15  
m 
 (0  0)  (0  0)  0  (0.45923 kg)(0.35 m)  
3



  8.0365  103 kg  m2
or I xy   8.04  103 kg  m 2 
I yz  ( I yz   my z )
 (0  0)  [0  (0.93775 kg)(0.082761 m)(0.195 m)]
1
 0.15  0.39  
  (0.45923 kg)(0.39 m)(0.15 m)  (0.45923 kg)  
m 
m 
36
3

 3


 [(15.1338)  (0.7462  2.9850)]  103 kg  m 2
 (15.1338  2.2388)  103 kg  m 2
 12.8950  103 kg  m2
or I yz  12.90  103 kg  m2 
I zx  ( I zx  mz x )
 [0  (2.14305 kg)(0.175 m)(0.195 m)]  (0  0)


 0.39 
 0  (0.45923 kg) 
m  (0.35 m) 
 3



 (73.1316  20.8950)  103 kg  m 2
 94.0266  103 kg  m2
or I zx  94.0  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.156
A section of sheet steel 2 mm thick is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass products of inertia Ixy, Iyz, and Izx of the
component.
SOLUTION
First compute the mass of each component. We have
m  STV  ST tA
Then
1

m1  m3  (7850 kg/m3 )(0.002 m)   0.225  0.135  m 2  0.23844 kg
2



m2  (7850 kg/m3 )(0.002 m)   0.2252  m 2  0.62424 kg
4





m4  (7850 kg/m3 )(0.002 m)   0.1352  m 2  0.22473 kg
4

Now observe that the following centroidal products of inertia are zero because of symmetry.
( I xy )1  ( I yz  )1  0
( I xy ) 2  ( I yz  ) 2  0
( I xy )3  ( I z x )3  0
( I xy ) 4  ( I zx )  0
Also
y1  y2  0
x3  x4  0
Now
I xy  ( I xy  mx y )
I xy  0 
so that
Using the results of Sample Problem 9.6, we have
1 2 2
b h
24
I uv , mass  ST t I uv ,area
I uv ,area 
Now
 1

 ST t  b 2 h 2 
24


1
 mbh
12
Thus,
( I zx )1 
1
m1b1h1
12
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PROBLEM 9.156 (Continued)
While
( I yz )3  
1
m3b3 h3
12
because of a 90° rotation of the coordinate axes.
To determine I uv for a quarter circle, we have
0
dI uv  d I uv  uEL vEL dm
Where
uEL  u vEL 
1
1 2
v
a  u2
2
2
dm  ST t dA  ST tv du  ST t a 2  u 2 du

I uv  dI uv 
Then
 (u)  2 a  u    t a  u du 
a
1
2 
2
ST
0
2
2

a
1
ST t u (a 2  u 2 ) du
0
2

1
1 
1
1
1
ma 4
ST t  a 2 u 2  u 4   ST t a 4 
2
2
4
8
2


0

a
( I zx )2  
Thus
1
m2 a22
2
because of a 90° rotation of the coordinate axes. Also
( I yz ) 4 
Finally,
1
m4 a42
2
0
0
I yz  ( I yz )  [( I yz )  m1 y1 z1 ]  [( I yz  )  m2 y2 z2 ]  ( I yz )3  ( I yz ) 4
 1
  1

(0.22473 kg)(0.135 m) 2 
   (0.23844 kg)(0.225 m)(0.135 m)   
 12
  2

 ( 0.60355  0.65185)  103 kg  m 2
I yz  48.3  106 kg  m2 
0
0
I zx  ( I zx )  ( I zx )1  ( I zx ) 2  [( I zx )3  m3 z3 x3 ]  [( I z x ) 4  m4 z4 x4 ]
or
1
  1

  (0.23844 kg)(0.225 m)(0.135 m)    
(0.62424 kg)(0.225 m) 2 
12
2


 

 (0.60355  5.02964)  103 kg  m 2
or
I zx   4.43  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.157
The figure shown is formed of 1.5-mm-diameter
aluminum wire. Knowing that the density of aluminum is
2800 kg/m3, determine the mass products of inertia Ixy, Iyz,
and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m   ALV   AL AL
Then


m1  m4  (2800 kg/m3 )  (0.0015 m) 2  (0.25 m)
4


 1.23700  103 kg


m2  m5  (2800 kg/m3 )  (0.0015 m) 2  (0.18 m)
4

 0.89064  103 kg


m3  m6  (2800 kg/m3 )  (0.0015 m) 2  (0.3 m)
4

 1.48440  103 kg
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.157 (Continued)
Now observe that the centroidal products of inertia, I xy , I yz , and I zx , of each component are zero
because of symmetry.
m, kg
x, m
y, m
z, m
mx y , kg  m2
my z , kg  m2
mz x , kg  m2
1
1.23700  103
0.18
0.125
0
27.8325  106
0
0
2
0.89064  103
0.09
0.25
0
20.0394  106
0
0
3
1.48440  103
0
0.25
0.15
0
55.6650  106
0
4
1.23700  103
0
0.125
0.3
0
46.3875  106
0
5
0.89064  103
0.09
0
0.3
0
0
24.0473  106
6
1.48440  103
0.18
0
0.15
0
0
40.0788  106
47.8719  106
102.0525  106
64.1261  106

Then
0
I xy  ( I xy  mx y )
0
I yz  ( I yz  my z )
0
I zx  ( I zx  mz x )
or
I xy  47.9  106 kg  m2 
or
I yz  102.1  106 kg  m 2 
or
I zx  64.1  106 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.158
Thin aluminum wire of uniform diameter is used to form the figure shown.
Denoting by m the mass per unit length of the wire, determine the mass
products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m
m    L  mL
L
  
m1  m5  m  R1   mR1
2  2
m2  m4  m( R2  R1 )
Then

 
m3  m  R2   mR2
2

 2
Now observe that because of symmetry the centroidal products of inertia, I xy , I yz , and I zx , of
components 2 and 4 are zero and
( I xy )1  ( I z x )1  0
( I xy )3  ( I yz  )3  0
( I yz )5  ( I z x )5  0
x1  x2  0
Also
y2  y3  y4  0
z4  z5  0
Using the parallel-axis theorem [Equations (9.47)], it follows that I xy  I yz  I zx for components 2 and 4.
To determine I uv for one quarter of a circular arc, we have dI uv  uvdm
where
u  a cos 
v  a sin 
and
dm   dV   [ A(ad )]
where A is the cross-sectional area of the wire. Now
 
 
m  m  a    A  a 
2 
2 
so that
dm  mad
and
dI uv  (a cos  )(a sin  )(mad )
 ma 3 sin  cos  d
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.158 (Continued)

I uv  dI uv 
Then

 /2
0
ma 3 sin  cos  d
 /2
1

 ma3  sin 2  
2
0

1
ma3
2
1
mR13
2
Thus,
( I yz )1 
and
1
( I zx )3   mR23
2
1
( I xy )5   mR13
2
because of 90 rotations of the coordinate axes. Finally,
0
0
I xy  ( I xy )  [( I xy )1  m1 x1 y1 ]  [( I xy )3  m3 x3 y3 ]  ( I xy )5
0
or
1
I xy   mR13 
2
or
I yz 
or
1
I zx   mR23 
2
0
I yz  ( I yz )  ( I yz )1  [( I yz )3  m3 y3 z3 ]  [( I yz )5  m5 y5 z5 ]
0
0
I zx  ( I zx )  [( I zx )1  m1 z1 x1 ]  ( I zx )3  [( I zx )5  m5 z5 x5 ]
1
mR13 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.159
Brass wire with a weight per unit length w is used to form the figure shown.
Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m
W 1
 wL
g g
W
w
(2  a )  2 a
g
g
w
w
m2  (a )  a
g
g
w
w
m3  (2a )  2 a
g
g
w
w
3 
m4   2  a   3 a
g
g
2 
m1 
Then
Now observe that the centroidal products of inertia, I xy , I yz , and I zx , of each component are zero
because of symmetry.
m
x
y
z
mx y
w
a
g
2a
a
a
4
2
w
a
g
2a
1
a
2
0
w 3
a
g
0
3
2
w
a
g
2a
0
a
0
0
4
4
3
w
a
g
2a
3
 a
2
2a
 9
12
1

2
w 3
a
g
my z
mz x
w 3
a
g
2
 4
w 3
a
g
0
w 3
a
g
 9
w 3
a
g
w
(1  5 ) a 3
g
11
w 3
a
g
4
w 3
a
g
w 3
a
g
w
(1  2 ) a 3
g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.159 (Continued)
Then
0
I xy  ( I xy  mx y )
or
0
I yz  ( I yz  my z )
or
0
I zx  ( I zx  mz x )
or
I xy 
w 3
a (1  5 ) 
g
I yz  11
I zx  4
w 3
a 
g
w 3
a (1  2 ) 
g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.160
Brass wire with a weight per unit length w is used to form the figure
shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire
figure.
SOLUTION
First compute the mass of each component. We have
m
W 1
 wL
g g
W
3  3 w
  a   a

g 
2  2 g
w
w
m2  (3a )  3 a
g
g
w
w
m3  (  a )   a
g
g
m1 
Then
Now observe that the centroidal products of inertia, I xy , I yz , and I zx , of each component are zero
because of symmetry.
x
y
z
mx y
23 
a
  2 
2a
1
a
2
9
0
1
a
2
2a
(a )
a
a
m
1
3 w
 a
2 g
2
3
w
a
g
3

w
a
g


2

my z
w 3
a
g
3 w 3
 a
2 g
0
3
2
w 3
a
g

11
w 3
a
g
mz x

9w 3
a
4g
w 3
a
g
0
w 3
a
g
2
w
 3
  3 a
g2


w 3
a
g
1w 3
a
4g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.160 (Continued)
Then
0
I xy  ( I xy  mx y )
or
0
I yz  ( I yz  my z )
or
0
I zx  ( I zx  mz x )
or
I xy  11
I yz 
w 3
a 
g
1w 3
a (  6) 
2g
I zx  
1w 3
a 
4g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.161
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of
inertia.
SOLUTION
We have

I xy  xydm
and
x  x  x
Consider
I xy  xydm

I yz  yzdm
y  y  y

I zx  zxdm
(9.45)
z  z  z
(9.31)

Substituting for x and for y

  xy dm  y  xdm  x  y dm  x y  dm
I xy  ( x  x )( y   y ) dm
By definition
and

I xy   xy dm
 xdm  mx 
 ydm  my 
However, the origin of the primed coordinate system coincides with the mass center G, so that
x  y  0
I xy  I xy  mx y Q.E.D. 
The expressions for I yz and I zx are obtained in a similar manner.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.162
For the homogeneous tetrahedron of mass m shown, (a) determine by
direct integration the mass product of inertia Izx, (b) deduce Iyz and Ixy from
the result obtained in part a.
SOLUTION
(a)
First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown.
a
z

z  a  a 1  
c
c


Now
x
and
b
z

y   z  b  b 1  
c
 c
The mass dm of the slab is
2
z
1
 1

dm   dV    xydz    ab 1   dz
2
2
c




Then
2
z

m  dm 
 ab 1   dz
0 2
 c


c1
c
3
 c 
1
z 
1
  ab   1      abc
2
6
 3  c  
0
Now
dI zx  dI zx  zEL xEL dm
where
dI zx  0 (symmetry)
and
zEL  z
Then
2

1 
z   1
z

I zx  dI zx  z  a 1      ab 1   dz 
0
 c

 3  c    2
xEL 



1
1 
z
x  a 1  
3
3  c
c
1 2 c
z2
z3 z 4 
 a b  z  3  3 2  3  dz
0
6
c
c
c 


c
m 1
z3 3 z 4 1 z5 
 a  z2  


c 2
c 4 c 2 5 c3  0
or
I zx 
1
mac 
20
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.162 (Continued)
(b)
Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular
permutation of ( x, y , z ) and (a, b, c) .
Thus,

I xy 
1
mab 
20
I yz 
1
mbc 
20
Alternative solution for part a:
First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown.
Now
x
a
y

y  a  a 1  
b
b


and
z
c
y

y  c  c 1  
b
 b
The mass dm of the slab is
2
y
1
 1

dm   dV    xzdy    ac 1   dy
2
2
b




Now
where
and dI zx,area 
Then
dI zx   tdI zx ,area
t  dy
1 2 2
x z from the results of Sample Problem 9.6.
24
2
2
 1  
y   
y   
dIzx   (dy )   a  1     c  1    
24 
b   
b  
 

4

Finally,
4
1
y
1m 
y

 a 2 c 2 1   dy 
ac  1   dy
24
b
4 b 
b


I zx  dI zx 

4
y

ac  1   dy
0 4 b
b

b1 m
b
5
1 m  b 
y 

ac    1   
4 b  5 
b  0
or
I zx 
1
mac 
20
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.162 (Continued)
Alternative solution for part a:
The equation of the included face of the tetrahedron is
x y z
  1
a b c
x z

y  b 1   
a
c

so that
For an infinitesimal element of sides dx, dy, and dz :
dm   dV   dydxdz
z

x  a 1  
 c
From part a
Now

I zx  zxdm 
c
a (1 z/c )
0
0


b (1 x/a  z/c )
 zx(  dydxdz )
0
c
a (1 z/c )
0
0

zx b 1  ax  cz   dxdz
a (1 z/c )
1
1 x3 1 z 2 
 b z  x2 

x 
0
3 a 2 c 0
2

 b

c
dz
2
3
2
1 
1
z
z 1 z 2
z 

z  a 2 1    a 3 1   
a 1    dz
0 2
 c  3a  c  2 c  c  

c
3
z

a 2 z 1   dz
0 6
c


2
c
1
z
z3 z 4 
  a 2 b  z  3  3 2  3  dz
0
6
c
c
c 

 b

c1

c

m  1 2 z3 3 z 4 1 z5 
a z  


c 2
c 4 c 2 5 c3  0
or I zx 
1
mac 
20
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.163
The homogeneous circular cone shown has a mass m.
Determine the mass moment of inertia of the cone with respect
to the line joining the origin O and Point A.
SOLUTION
2
First note that
9
3 
d OA   a   ( 3a ) 2  (3a ) 2  a
2
2


Then
OA  9  ai  3aj  3ak   (i  2 j  2k )
a 2
 3
2 
1 3

1
For a rectangular coordinate system with origin at Point A and axes aligned with the given x, y , z axes,
we have (using Figure 9.28)
3 1

I x  I z  m  a 2  (3a) 2 
5 4


Also, symmetry implies
Iy 
3
ma 2
10
111 2
ma
20
I xy  I yz  I zx  0
With the mass products of inertia equal to zero, Equation (9.46) reduces to
I OA  I x x2  I y y2  I z z2
2
2

111 2  1 
3
 2  111 2  2 
ma    ma 2    
ma  
20
3
10
20
 
 3
3

193 2
ma
60
2
or I OA  3.22ma 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.164
The homogeneous circular cylinder shown has a mass m. Determine the mass
moment of inertia of the cylinder with respect to the line joining the origin O and
Point A that is located on the perimeter of the top surface of the cylinder.
SOLUTION
Iy 
From Figure 9.28:
1
ma 2
2
and using the parallel-axis theorem
2
Ix  Iz 
Symmetry implies
1
1
h
m(3a 2  h 2 )  m    m(3a 2  4h2 )
12
 2  12
I xy  I yz  I zx  0
For convenience, let Point A lie in the yz plane. Then
OA 
1
2
h  a2
( hj  ak )
With the mass products of inertia equal to zero, Equation (9.46) reduces to
0
I OA  I x x2  I y y2  I z z2
2




1
h
1
a
 ma 2 
  m(3a 2  4h 2 ) 

 h2  a 2  12
 h2  a 2 
2




or
I OA 
2
1
10h 2  3a 2
ma 2

12
h2  a 2
Note: For Point A located at an arbitrary point on the perimeter of the top surface, OA is given by
OA 
1
h2  a 2
( a cos  i  hj  a sin  k )
which results in the same expression for I OA . 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.165
Shown is the machine element of Problem 9.141. Determine its
mass moment of inertia with respect to the line joining the
origin O and Point A.
SOLUTION
First compute the mass of each component.
We have
m  STV  (7850 kg/m 3 )
Then
m1  (7850 kg/m3 )[ (0.08 m) 2 (0.04 m)]  6.31334 kg
m2  (7850 kg/m3 )[ (0.02 m) 2 (0.06 m)]  0.59188 kg
m3  (7850 kg/m3 )[ (0.02 m) 2 (0.04 m)]  0.39458 kg
Symmetry implies
and
Now
I yz  I zx  0
( I xy )1  0
( I xy )2  ( I xy )3  0
I xy  ( I xy   mx y )  m2 x2 y2  m3 x3 y3
 [0.59188 kg (0.04 m)(0.03 m)]  [0.39458 kg ( 0.04 m)( 0.02 m)]
 (0.71026  0.31566)  103 kg  m 2
 0.39460  103 kg  m 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.165 (Continued)
From the solution to Problem 9.141, we have
I x  13.98800  103 kg  m 2
I y  20.55783  103 kg  m 2
I z  14.30368  103 kg  m 2
By observation
Substituting into Eq. (9.46)
λ OA 
1
13
(2i  3 j)
0
0
0
IOA  I x x2  I y y2  I z z2  2 I xy x y  2 I yz  y z  2 I zx z x
2
2

 2 
 3 
  (13.98800) 
  (20.55783) 


 13 
 13 
 2  2  
3
2
 2(0.39460) 

   10 kg  m
13
13



 (4.30400  14.23234  0.36425)  103 kg  m 2
or
I OA  18.17  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.166
Determine the mass moment of inertia of the steel fixture of
Problems 9.145 and 9.149 with respect to the axis through
the origin that forms equal angles with the x, y, and z axes.
SOLUTION
From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
I x  26.4325  103 kg  m 2
I y  31.1726  103 kg  m 2
I z  8.5773  103 kg  m 2
Problem 9.149:
I xy  2.5002  103 kg  m 2
I yz  4.0627  103 kg  m 2
I zx  8.8062  103 kg  m 2
From the problem statement it follows that
x   y   z
x2   y2  z2  1  3x2  1
Now
x   y  z 
or
1
3
Substituting into Eq. (9.46)
I OL  I x x2  I y  y2  I z z2  2 I xy x y  2 I yz  y z  2 I zx z x
Noting that
We have
x2   y2  z2  x  y   y z  z x 
1
3
1
I OL  [26.4325  31.1726  8.5773
3
 2(2.5002  4.0627  8.8062)]  103 kg  m 2
or I OL  11.81  103 kg  m 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.167
The thin bent plate shown is of uniform density and weight W. Determine its
mass moment of inertia with respect to the line joining the origin O and
Point A.
SOLUTION
First note that
m1  m2 
and that
OA 
1
3
1W
2 g
(i  j  k )
Using Figure 9.28 and the parallel-axis theorem, we have
I x  ( I x )1  ( I x ) 2
 1  1 W  2 1 W  a 2 
 

a 
2 g  2  
12  2 g 
2
2
 1  1 W  2
1 W  a   a   
2
 



(
a
a
)


    
2 g  2   2   
12  2 g 


1 W  1 1  2  1 1  2  1 W 2
a
   a     a  
2 g  12 4 
6 2  2 g
I y  ( I y )1  ( I y ) 2
2
2
 1  1 W 
1 W  a   a   
2
2
 



(
a
a
)



2 g  2   2   
12  2 g 

2 

 1  1 W  2 1 W
a 
 
(a )2     
a 
2 g 
 2   
12  2 g 

1 W  1 1  2  1 5  2  W 2
a    a   a
 
2 g  6 2 
 12 4   g
I z  ( I z )1  ( I z ) 2
 1  1 W  2 1 W  a 2 
 

a 
2 g  2  
12  2 g 
2
 1  1 W  2 1 W  2  a   
 
(a )     
a 
2 g 
 2   
12  2 g 
1 W  1 1  2  1 5  2  5 W 2


a    a  
a

2 g  12 4 
 12 4   6 g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.167 (Continued)
Now observe that the centroidal products of inertia, I xy , I yz , and I zx , of both components are zero
because of symmetry. Also, y1  0
0
1W
a 1W 2
(a )   
I xy  ( I xy  mx y )  m2 x2 y2 
a
Then
2 g
2 4 g
0
1 W  a  a  1 W 2
I yz  ( I yz   my z )  m2 y2 z2 
a
   
2 g  2  2  8 g
0
I zx  ( I zx  mz x )  m1z1 x1  m2 z2 x2

1 W  a  a  1 W  a 
3W 2
a
  
  (a ) 
2 g  2  2  2 g  2 
8 g
Substituting into Equation (9.46)
IOA  I x x2  I y y2  I z z2  2 I xy x y  2 I yz  y z  2 I zx z x
Noting that
x2   y2  z2  x  y   y z  z x 
1
3
We have
 1 W 2 1 W 2 3 W 2 
1 1 W 2 W 2 5 W 2
I OA  
a  a 
a  2
a 
a 
a 
3 2 g
g
6 g
8 g
8 g
4 g

1 14
 3  W
   2   a 2
3 6
 4  g
or
I OA 
5 W 2
a 
18 g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.168
A piece of sheet steel of thickness t and specific weight  is cut and
bent into the machine component shown. Determine the mass
moment of inertia of the component with respect to the joining the
origin O and Point A.
SOLUTION
First note that
1
λ OA 
6
(i  2 j  k )
Next compute the mass of each component. We have
m  V 
Then
m1 
m2 

g

g
(tA)
t (2a  2a )  4
t
g
a2
 
  t 2
t   a2  
a
g 2
 2 g
Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem
9.134) for a semicircular plate for component 2, we have
I x  ( I x )1  ( I x )2
 1   t 

t
   4 a 2  [(2a )2  (2a )2 ]  4 a 2 (a 2  a 2 ) 
g
12  g 

 1    t 2  2   t 2

 
a a 
a [(2a)2  (a)2 ]
2 g
 4  2 g 

 t 2  1
2


a2   2  a2 
a   5  a2
3
2
g
4




t 4
 18.91335 a
g
4
t
g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.168 (Continued)
I y  ( I y )1  ( I y )2
 1  t 

t
   4 a 2  (2a) 2  4 a 2 (a 2 ) 
g

12  g

2
   t  1 16 
 
  t 2  4a 
a2   2  a2 
a 

 (a ) 2  

2 g  2 9 
2 g
 3 
 

16
t 1 
  t 2  1 16

a
 4 a 2   1 a 2 


 1 a 2
g 3 
2 g  2 9 2 9 2

t 4
 7.68953 a
g
I z  ( I z )1  ( I z ) 2
 1  t 

t
   4 a 2  (2a) 2  4 a 2 (a 2 ) 
g

12  g

2
   t  1 16 

  t 2  4 a 
2 
a2   2  a2 
a 

  (2a)  
2 g
 3 
 
 2 g  4 9 
16
t  1 
  t 2  1 16

a  
 4 a 2   1 a 2 

 4  a2
g 3 
2 g  4 9 2 9 2

t 4
 12.00922 a
g
Now observe that the centroidal products of inertia, I xy , I yz , and I zx , of both components are zero
because of symmetry. Also x1  0.
0
  t 2  4a 
Then
I xy  ( I xy  mx y )  m2 x2 y2 
a
(2a )
2 g  3 
t
 1.33333 a 4
g
0
I yz  ( I yz  my z )  m1 y1 z1  m2 y2 z2
4
t
g
a 2 (a)(a ) 
 7.14159
t
g
 t
2 g
a 2 (2a )(a)
a4
0
I zx  ( I zx  mz x )  m2 z2 x2 
 0.66667
t
g
 t
2 g
 4a 
a 2 (a ) 

 3 
a4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.168 (Continued)
Substituting into Eq. (9.46)
I OA  I x x2  I y  y2  I z z2  2 I xy x  y  2 I yz  y z  2 I zx z x
 18.91335
2
t
 1 
t 4  2 
a4 
  7.68953 a 

g  6
g  6
 12.00922
t
2
2

 1 
 t 4   1  2 
a4 
  2 1.33333 a  


g  6
g

  6  6 

 t   2  2  
 t 4   1  1 
 2  7.14159 a 4  

  2  0.66667 a  


g
g

  6  6 

  6  6 
t
 (3.15223  5.12635  2.00154  0.88889  4.76106  0.22222) a 4
g
or
I OA  4.41
t
g
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a 4 
PROBLEM 9.169
Determine the mass moment of inertia of the machine
component of Problems 9.136 and 9.155 with respect to
the axis through the origin characterized by the unit vector
λ  ( 4i  8j  k )/9.
SOLUTION
From the solutions to Problems 9.136 and 9.155. We have
Problem 9.136:
I x  175.503  103 kg  m 2
I y  308.629  103 kg  m 2
I z  154.400  103 kg  m 2
Problem 9.155:
I xy  8.0365  103 kg  m 2
I yz  12.8950  103 kg  m 2
I zx  94.0266  103 kg  m 2
Substituting into Eq. (9.46)
IOL  I x x2  I y  y2  I z z2  2 I xy x  y  2 I yz  y z  2 I zx z x
2
2
2

 4
8
1
 175.503     308.629    154.400  
 9
9
9

 4  8 
 8  1 
 2(8.0365)      2(12.8950)   
9
9

 
 9  9 
 1  4  
 2(94.0266)       103 kg  m 2
 9  9  
 (34.6673  243.855  1.906  6.350
 2.547  9.287)  103 kg  m 2
or
I OL  281  103 kg  m 2 
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PROBLEM 9.170
For the wire figure of Problem 9.148, determine the mass
moment of inertia of the figure with respect to the axis
through the origin characterized by the unit vector
  (3i  6j  2k )/7.
SOLUTION
First compute the mass of each component. We have
m
m    L  0.056 kg/m  1.2 m
L
 0.0672 kg
Now observe that the centroidal products of inertia, I xy , I yz , and I zx , for each
component are zero because of symmetry.
Also
x1  x6  0
Then
y4  y5  y6  0
z1  z2  z3  0
0
I xy  ( I xy  mx y )  m2 x2 y2  m3 x3 y3
 (0.0672 kg)(0.6 m)(1.2 m)  (0.0672 kg)(1.2 m)(0.6 m)
 0.096768 kg  m 2
0
I yz  ( I yz  my z )  0
0
I zx  ( I zx  mz x )  m4 z4 x4  m5 z5 x5
 (0.0672 kg)(0.6 m)(1.2 m)  (0.0672 kg)(1.2 m)(0.6 m)
 0.096768 kg  m 2
From the solution to Problem 9.148, we have
I x  0.32258 kg  m 2
I y  I z  0.41933 kg  m 2
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PROBLEM 9.170 (Continued)
Substituting into Eq. (9.46)
0
I OL  I x x2  I y  y2  I z z2  2 I xy x  y  2 I yz  y z  2 I zx z x

 3
 6
2
 0.32258     0.41933     0.41933  
 7
 7
7

2
2
2
 3  6 
 2  3  
 2(0.096768)       2(0.096768)      kg  m 2
 7  7 
 7  7  
I OL  (0.059249  0.30808  0.034231  0.071095  0.023698) kg  m 2
or
IOL  0.354 kg  m2 
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y
PROBLEM 9.171
360 mm
For the wire figure of Problem 9.147, determine the mass moment of inertia of
the figure with respect to the axis through the origin characterized by the unit
vector λ = (−3i − 6j + 2k )/7.
360 mm
360 mm
x
z
SOLUTION
First compute the mass of each component. We have
m = ρSTV = rAL
Then
m1 = m2 = 7850 kg/m3 ×
π
4
(0.003 m)2 × (p × 0.36 m)
= 62.756 × 10−3 kg
m3 = m4 = 7850 kg/m3 ×
π
4
(0.003 m)2 × 0.36 m
= 0.019776 kg
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each component are zero because
of symmetry.
Also
x3 = x4 = 0
y1 = 0
Then
0
I xy = Σ( I x′y′ + mx y ) = m2 x2 y2
= (0.062756 kg) −
z1 = z2 = 0
2 × 0.36
m (0.36 m)
p
= −5.178 × 10−3 kg · m2
0
I yz = Σ( I y ′z′ + my z ) = m3 y3 z3 + m4 y4 z4
Now
m3 = m4 ,
y3 = y4 ,
z 4 = − z3
0
I zx = Σ( I z′x′ + mz x ) or
I yz = 0
I zx = 0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.171 (Continued)
From the solution to Problem 9.147, we have
I x = 0.0232 kg · m2
I y = 0.0214 kg · m2
I z = 0.018 kg · m2
Substituting into Eq. (9.46)
0
0
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
3
= 0.0232 −
7
2
6
+ 0.0214 −
7
− 2(−0.005178) −
3
7
−
6
7
2
2
+ 0.018
7
2
kg · m2
= (4.261 + 15.722 + 1.469 + 3.804) × 10−3 kg · m2
or
I OL = 25.3 × 10−3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.172
For the wire figure of Problem 9.146, determine the mass moment of
inertia of the figure with respect to the axis through the origin
characterized by the unit vector λ = (−3i − 6j + 2k )/7.
SOLUTION
First compute the mass of each component. We have
m=1
M
2L
L
Then
m1 = (0.049 kg/m)(2p × 0.32 m)
= 0.0985 kg
m2 = m3 = m4 = m5 = (0.049 kg/m)(0.32 m)
= 0.00784 kg
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry. Also
x1 = x4 = x5 = 0
y1 = 0
z1 = z2 = z3 = 0
0
I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3
Then
= 0.00784 kg (0.32 m)(0.08 m)
+ 0.00784 kg (0.24 m)(0.16 m)
= 0.502 × 10−3 kg · m2
Symmetry implies
I yz = I xy
I yz = 0.502 × 10−3 kg · m2
0
I zx = Σ( I z ′x′ + mz x ) = 0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.172 (Continued)
From the solution to Problem 9.146, we have
I x = I z = 6.85 × 10−3 kg · m2
I y = 12.63 × 10−3 kg · m2
Substituting into Eq. (9.46)
0
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
= 6.85 −
3
7
2
+ 12.63 −
− 2(0.502) −
3
7
−
6
7
2
+ 6.85
2
7
2
6
6
− 2(0.502) −
7
7
2
7
× 10−3 kg · m2
= (1.258 + 9.279 + 0.559 − 0.369 + 0.246) × 10−3 kg · m2
or
I OL = 10.97 × 10−3 kg · m2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1648
PROBLEM 9.173
For the homogeneous circular cylinder shown, of radius a and length L,
determine the value of the ratio a/L for which the ellipsoid of inertia of
the cylinder is a sphere when computed (a) at the centroid of the cylinder,
(b) at Point A.
SOLUTION
(a)
From Figure 9.28:
1 2
ma
2
1
I y  I z  m(3a 2  L2 )
12
Ix 
Now observe that symmetry implies
I xy  I yz  I zx  0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x x 2  I y y 2  I z z 2  1:
1 2 2 1
1
ma x  m(3a 2  L2 ) y 2  m(3a 2  L2 )  1
2
12
12
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore
1 2 1
ma  m(3a 2  L2 )
2
12
(b)
a
1


L
3
or
Using Figure 9.28 and the parallel-axis theorem, we have
I x 
1
ma 2
2
2
I y  I z 
1
7 2
L
1
m(3a 2  L2 )  m    m  a 2 
L
12
48 
4
4
Now observe that symmetry implies
I xy  I yz  I zx  0
From Part a it then immediately follows that
1 2
7 2
1
ma  m  a 2 
L 
2
4
48


or
a
7


L
12
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.174
For the rectangular prism shown, determine the values of the
ratios b/a and c/a so that the ellipsoid of inertia of the prism is a
sphere when computed (a) at Point A, (b) at Point B.
SOLUTION
(a)
Using Figure 9.28 and the parallel-axis theorem, we have
at Point A
1
I x  m(b 2  c 2 )
12
1
a
m( a 2  c 2 )  m  
12
2
1
 m(4a 2  c 2 )
12
2
I y 
2
1
1
a
I z   m(a 2  b 2 )  m    m(4a 2  b 2 )
12
 2  12
Now observe that symmetry implies
I xy  I yz  I zx  0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x x 2  I y y 2  I z z 2  1:
1
1
1
m(b 2  c 2 ) x 2  m(4a 2  c 2 ) y 2  m(4a 2  b 2 ) z 2  1
12
12
12
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore
1
1
m(b 2  c 2 )  m(4a 2  c 2 )
12
12
1
 m(4a 2  b2 )
12
Then
b 2  c 2  4a 2  c 2
or
b
 2
a
and
b 2  c 2  4a 2  b 2
or
c
 2
a
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.174 (Continued)
(b)
Using Figure 9.28 and the parallel-axis theorem, we have at Point B
2
I x 
1
1
c
m(b 2  c 2 )  m    m(b 2  4c 2 )
12
 2  12
2
1
1
c
I y  m(a 2  c 2 )  m    m(a 2  4c 2 )
12
 2  12
1
I z  m(a 2  b2 )
12
Now observe that symmetry implies
I xy  I yz  I zx  0
From part a it then immediately follows that
1
1
1
m(b 2  4c 2 )  m(a 2  4c 2 )  m(a 2  b 2 )
12
12
12
Then
b 2  4c 2  a 2  4c 2
or
b
 1
a
and
b 2  4c 2  a 2  b 2
or
c 1
 
a 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.175
For the right circular cone of Sample Problem 9.11, determine the
value of the ratio a/h for which the ellipsoid of inertia of the cone is a
sphere when computed (a) at the apex of the cone, (b) at the center of
the base of the cone.
SOLUTION
(a)
From Sample Problem 9.11, we have at the apex A
3
ma 2
10
3 1

I y  I z  m  a 2  h2 
5 4

Ix 
I xy  I yz  I zx  0
Now observe that symmetry implies
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x x 2  I y y 2  I z z 2  1:
3
3 1
3 1


ma 2 x 2  m  a 2  h 2  y 2  m  a 2  h 2  z 2  1
10
5 4
5 4


For the ellipsoid to be a sphere, the coefficients must be equal. Therefore,
3
3 1

ma 2  m  a 2  h 2 
10
5 4

(b)
or
a
 2
h
From Sample Problem 9.11, we have
and at the centroid C
I x 
3
ma 2
10
I y 
3  2 1 2
ma  h 
20 
4 
2
Then
I y  I z 
3  2 1 2
1
h
m a  h   m  
m(3a 2  2h2 )
20 
4 
20
4
Now observe that symmetry implies
I xy  I yz  I zx  0
From Part a it then immediately follows that
3
1
ma 2 
m(3a 2  2h 2 )
10
20
or
a
2


h
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.176
Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of
the body with respect to any one of the three axes cannot be larger than the sum of the mass moments of
inertia of the body with respect to the other two axes. That is, prove that the inequality I x  I y  I z and
the two similar inequalities are satisfied. Further, prove that I y  12 I x if the body is a homogeneous solid
of revolution, where x is the axis of revolution and y is a transverse axis.
SOLUTION
(i)
I y  Iz  Ix
To prove

I   ( x  y )dm
I y  ( z 2  x 2 )dm
By definition
2
2
z


I y  I z  ( z 2  x 2 ) dm  ( x 2  y 2 ) dm
Then


 ( y 2  z 2 )dm  2 x 2 dm
Now
 ( y  z )dm  I
2
2
x
and
 x dm  0
2
I y  Iz  Ix
Q.E.D.
The proofs of the other two inequalities follow similar steps.
(ii)
If the x axis is the axis of revolution, then
I y  Iz
and from part (i)
or
or
I y  Iz  Ix
2I y  I x
Iy 
1
Ix
2
Q.E.D. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.177
Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a
sphere, and use this property to determine the moment of inertia of the cube with respect to one of its
diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of
revolution, and determine the principal moments of inertia of the cube at that point.
SOLUTION
(a)
At the center of the cube have (using Figure 9.28)
1
1
I x  I y  I z  m(a 2  a 2 )  ma 2
12
6
Now observe that symmetry implies
I xy  I yz  I zx  0
Using Equation (9.48), the equation of the ellipsoid of inertia is
1 2 2 1 2 2 1 2 2
 ma  x   ma  y   ma  z  1
6

6

6

or
x2  y 2  z 2 
6
( R 2 ) 
ma 2
which is the equation of a sphere.
Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through
the center O of the cube must always

1
1 
I OL  ma 2 
be the same  R 
.


6
I OL 

(b)
The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes
through corner A. For a rectangular coordinate system at A and with one of the coordinate axes
aligned with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then
rotated 120 about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid
will also remain unchanged after it is rotated. As noted at the end of Section 9.17, this is possible
only if the ellipsoid is an ellipsoid of revolution, where the diagonal is both the axis of revolution
and a principal axis.
1
It then follows that
I x  I OL  ma 2 
6
In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal
and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallelaxis theorem between the center of the cube and corner A for any perpendicular axis
 3 
1
I y   I z  ma 2  m 
a
 2 
6


2
or
I y  I z  
11 2
ma 
12
(Note: Part b can also be solved using the method of Section 9.18.)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.177 (Continued)
First note that at corner A
2 2
ma
3
1
I xy  I yz  I zx  ma 2
4
Ix  I y  Iz 
Substituting into Equation (9.56) yields
k 3  2ma 2 k 2 
55 2 6
121 3 9
m a k
m a 0
48
864
For which the roots are
1 2
ma
6

11 2
k2  k3  ma
12
k1 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.178
Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with
origin at O, prove that the sum Ix + Iy + Iz of the mass moments of inertia of the body cannot be smaller
than the similar sum computed for a sphere of the same mass and the same material centered at O.
Further, using the result of Problem 9.176, prove that if the body is a solid of revolution, where x is the
axis of revolution, its mass moment of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10,
where a is the radius of the sphere of the same mass and the same material.
SOLUTION
(i)
Using Equation (9.30), we have



I x  I y  I z  ( y 2  z 2 ) dm  ( z 2  x 2 ) dm  ( x 2  y 2 )dm

 2 ( x 2  y 2  z 2 ) dm

 2 r 2 dm
where r is the distance from the origin O to the element of mass dm. Now assume that the given
body can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of
mass m and radius a which is centered at O; it then follows that
m1  m2 (mbody  msphere  m)
Then
( I x  I y  I z )body  ( I x  I y  I z )sphere  ( I x  I y  I z )V1
 ( I x  I y  I z )V2
or
( I x  I y  I z )body  ( I x  I y  I z )sphere  2
 r dm  2 r dm
2
m1
2
m2
Now, m1  m2 and r1  r2 for all elements of mass dm in volumes 1 and 2.
 r dm   r dm  0
2
m1
so that
2
m2
( I x  I y  I z )body  ( I x  I y  I z )sphere
Q.E.D.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.178 (Continued)
(ii)
First note from Figure 9.28 that for a sphere
Ix  I y  Iz 
Thus
( I x  I y  I z )sphere 
2 2
ma
5
6 2
ma
5
For a solid of revolution, where the x axis is the axis of revolution, we have
I y  Iz
Then, using the results of part (i)
( I x  2 I y ) body 
From Problem 9.178 we have
or
Iy 
6 2
ma
5
1
Ix
2
(2 I y  I x )body  0
Adding the last two inequalities yields
(4 I y ) body 
or
6 2
ma
5
( I y )body 
3
ma 2
10
Q.E.D.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.179*
The homogeneous circular cylinder shown has a mass m, and
the diameter OB of its top surface forms 45 angles with the x
and z axes. (a) Determine the principal mass moments of
inertia of the cylinder at the origin O. (b) Compute the angles
that the principal axes of inertia at O form with the coordinate
axes. (c) Sketch the cylinder, and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
First compute the moments of inertia using Figure 9.28 and the
parallel-axis theorem.
 a 2  a 2  13
1
2
m(3a 2  a 2 )  m 
      ma
12
2
12
 2    
1
3
I y  ma 2  m(a) 2  ma 2
2
2
Ix  Iz 
Next observe that the centroidal products of inertia are zero because of symmetry. Then
0
 a  a 
1
I xy  I xy  mx y  m 
ma 2
    
2 2
 2  2 
0
1
 a  a 
I yz  I yz  my z  m    
ma 2

2
2 2

 2 
0
 a  a  1 2
I zx  I z x  mz x  m 

  ma
 2  2  2
Substituting into Equation (9.56)
 13 3 13 
K 3      ma 2 K 2
 12 2 12 
2
2
2
 13 3   3 13   13 13  
1  
1  1 
2 2
              



 
    (ma ) K
12
2
2
12
12
12
2
 
 
  2 2   2 2    

2
 13 3 13   13  
1   3  1 
        
    
 12 2 12   12   2 2   2  2 
2
2

1 
1 
1   1 
 13  
2 3
   
  2 
 
    (ma )  0
 12   2 2 
 2 2  2 2   2  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.179* (Continued)
Simplifying and letting K  ma 2 yields
3
11 2 565
95
 

0
3
144
96
Solving yields
 1  0.363383
2 
19
12
 3  1.71995
The principal moments of inertia are then
K1  0.363ma 2 
K 2  1.583ma 2 
K3  1.720ma 2 
(b)
To determine the direction cosines x ,  y , z of each principal axis, we use two of the equations
of Equations (9.54) and (9.57).
Thus,
( I x  K )x  I xy  y  I zx z  0
(9.54a)
 I zx x  I yz  y  ( I z  K )z  0
(9.54c)
x2   y2  z2  1
(9.57)
(Note: Since I xy  I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of
 y during the solution process.)
Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)


1
 13 2

1

ma 2   y   ma 2  z  0
 12 ma  K  x   


2

 2 2



1
1

 13

ma 2   y   ma 2  K  z  0
  ma 2  x   
2

 12

 2 2

or
1
1
 13

 y  z  0
    x 
2
2 2
 12

(i)
and
1
1
 13

 y      z  0
 x 
2
12
2 2


(ii)
Observe that these equations will be identical, so that one will need to be replaced, if
13
1
  
12
2
or
 
19
12
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PROBLEM 9.179* (Continued)
Thus, a third independent equation will be needed when the direction cosines associated with K 2
are determined. Then for K1 and K3
Eq. (i) through Eq. (ii):
 13
 1  13
 1 

         x          z  0
 2 

12
 2  12
z  x
or
Substituting into Eq. (i):
1
1
 13

 y  x  0
    x 
2
2 2
 12


y  2 2   
or

7
 x
12 
Substituting into Equation (9.57):

x2   2

or
2
7 

2     x   ( x ) 2  1
12  

2

7   2

 2  8      x  1
12  


(iii)
K1: Substituting the value of 1 into Eq. (iii):
2

7  

 2  8  0.363383    (x )12  1
12  


or
and then
(x )1  (z )1  0.647249
7 

( y )1  2 2  0.363383   (0.647249)
12 

 0.402662
( x )1  ( z )1  49.7 ( y )1  113.7 
K 3 : Substituting the value of  3 into Eq. (iii):
2

7  

 2  8  1.71995    (x )32  1
12  


or
and then
(x )3  (z )3  0.284726
7 

( y )3  2 2 1.71995   (0.284726)
12 

 0.915348
( x )3  ( z )3  73.5
( y )3  23.7 
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PROBLEM 9.179* (Continued)

K2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b),
and (9.57).
Now
 I xy x  ( I y  K ) y  I yz z  0
(9.54b)
Substituting for the moments and products of inertia.




1
1
3

ma 2  x   ma 2  K   y   
ma 2  z  0
 
2

 2 2

 2 2

1
or
2 2
3



x       y 
2
1
2 2
z  0
(iv)
Substituting the value of 2 into Eqs. (i) and (iv):
1
1
 13 19 
( y ) 2  (z ) 2  0
 12  12  (x ) 2 
2
2 2


1
1
 3 19 
(x ) 2     ( y ) 2 
(z ) 2  0
2 2
2 2
 2 12 
or
( x ) 2 
and
1
2
( x ) 2 
( y ) 2  ( z ) 2  0
2
( y ) 2  ( z ) 2  0
6
Adding yields
( y ) 2  0
and then
( y ) 2  ( x ) 2
Substituting into Equation (9.57)
0
(x )22  ( y )22  (x )22  1
or
(x ) 2 
1
2
and (z ) 2  
1
2
( x )2  45.0 ( y )2  90.0 ( z )2  135.0 
(c)
Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B.
Principal axis 2 lies in the xz plane.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.180
For the component described in Problem 9.165, determine
(a) the principal mass moments of inertia at the origin, (b) the
principal axes of inertia at the origin. Sketch the body and show
the orientation of the principal axes of inertia relative to the x, y,
and z axes.
SOLUTION
(a)
From the solutions to Problems 9.141 and 9.165 we have
I x  13.98800  103 kg  m 2
Problem 9.141:
I y  20.55783  103 kg  m 2
I z  14.30368  103 kg  m 2
I yz  I zx  0
Problem 9.165:
I xy  0.39460  103 kg  m 2
Eq. (9.55) then becomes
Thus
Substituting:
Ix  K
 I xy
0
 I xy
Iy  K
0
0
0
Iz  K
2
 0 or ( I x  K )( I y  K )( I z  K )  ( I z  K ) I xy
0
Iz  K  0
2
I x I y  ( I x  I y ) K  K 2  I xy
0
K1  14.30368  103 kg  m 2
or
K1  14.30  103 kg  m 2 
and
(13.98800  103 )(20.55783  103 )  (13.98800  20.55783)(103 ) K  K 2  (0.39460  103 )2  0
or
Solving yields
and
K 2  (34.54583  103 ) K  287.4072  106  0
K 2  13.96438  103 kg  m 2
K3  20.58145  103 kg  m 2
or
K 2  13.96  103 kg  m 2 
or
K3  20.6  103 kg  m2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.180 (Continued)
(b)
To determine the direction cosines x ,  y , z of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
K1:
Begin with Eqs. (9.54a) and (9.54b) with I yz  I zx  0.
( I x  K1 )(x )1  I xy ( y )1  0
 I xy (x )1  ( I y  K1 )( y )1  0
Substituting:
[(13.98800  14.30368)  103 ](x )1  (0.39460  103 )( y )1  0
(0.39460  103 )(x )1  [(20.55783  14.30368)  103 ]( y )1  0
Adding yields
Then using Eq. (9.57)
(x )1  ( y )1  0
0
0
2
2
(x )1  (y )1  (z )12  1
(z )1  1
or
( x )1  90.0 ( y )1  90.0 ( z )1  0 
K2: Begin with Eqs. (9.54b) and (9.54c) with I yz  I zx  0.
 I xy (x ) 2  ( I y  K 2 )( y )2  0
(i)
( I z  K 2 )(z )2  0
I z  K 2  (z ) 2  0
Now
Substituting into Eq. (i):
(0.39460  103 )(x ) z  [(20.55783  13.96438)  103 ]( y )2  0
or
( y )2  0.059847(x )2
Using Eq. (9.57):
(x )22  [0.05984](x )2 ]2  (z )22  1
or
(x ) 2  0.998214
and
( y )2  0.059740
( x )2  3.4
0
( y )2  86.6
( z )2  90.0 
K3: Begin with Eqs. (9.54b) and (9.54c) with I yz  I zx  0.
 I xy (x )3  ( I y  K3 )( y )3  0
( I z  K3 )(z )3  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(ii)
PROBLEM 9.180 (Continued)
I z  K 3  (  z )3  0
Now
Substituting into Eq. (ii):
(0.39460  103 )(x )3  [(20.55783  20.58145)  103 ]( y )3  0
or
Using Eq. (9.57):
( y )3  16.70618(x )3
0
(x )32  [16.70618(x )3 ]2  (z )32  1
or
(x )3  0.059751 (axes right-handed set  "_")
and
( y )3  0.998211
( x )3  93.4
( y )3  3.43
( z )3  90.0 
Note: Since the principal axes are orthogonal and ( z ) 2  ( z )3  90, it follows that
|(x )2 |  |( y )3 |
|( y )2 |  | (z )3 |
The differences in the above values are due to round-off errors.
(c)
Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.181*
For the component described in Problems 9.145 and 9.149,
determine (a) the principal mass moments of inertia at the
origin, (b) the principal axes of inertia at the origin. Sketch
the body and show the orientation of the principal axes of
inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
I x  26.4325  103 kg  m 2
I xy  2.5002  103 kg  m 2
Problem 9.149:
I y  31.1726  103 kg  m2
I yz  4.0627  103 kg  m2
I z  8.5773  103 kg  m 2
I zx  8.8062  103 kg  m 2
Substituting into Eq. (9.56):
K 3  [(26.4325  31.1726  8.5773)(103 )]K 2
 [(26.4325)(31.1726)  (31.1726)(8.5773)  (8.5773)(26.4325)
 (2.5002)2  (4.0627) 2  (8.8062)2 ](106 ) K
 [(26.4325)(31.1726)(8.5773)  (26.4325)(4.0627) 2
 (31.1726)(8.8062) 2  (8.5773)(2.5002)2
 2(2.5002)(4.0627)(8.8062)](109 )  0
or
K 3  (66.1824  103 ) K 2  (1217.76  106 ) K  (3981.23  109 )  0
Solving yields

K1  4.1443  103 kg  m 2
or K1  4.14  103 kg  m2 
K 2  29.7840  103 kg  m2
or K 2  29.8  103 kg  m2 
K3  32.2541  103 kg  m2 
or K3  32.3  103 kg  m2 
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PROBLEM 9.181* (Continued)
(b)
To determine the direction cosines x ,  y , z of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K1: Begin with Eqs. (9.54a) and (9.54b).
( I x  K1 )(x )1  I xy ( y )1  I zx (z )1  0
I xy (x )1  ( I y  K1 )( y )1  I yz (z )1  0
Substituting:
[(26.4325  4.1443)(103 )](x )1  (2.5002  103 )( y )1  (8.8062  103 )(z )1  0
(2.5002  103 )(x )1  [(31.1726  4.1443)(103 )]( y )1  (4.0627  103 )(z )1  0
Simplifying
8.9146(x )1  ( y )1  3.5222(z )1  0
0.0925(x )1  ( y )1  0.1503(z )1  0
Adding and solving for (z )1:
(z )1  2.4022(x )1
( y )1  [8.9146  3.5222(2.4022)](x )1
and then
 0.45357(x )1
Now substitute into Eq. (9.57):
(x )12  [0.45357(x )1 ]2  [2.4022(x )1 ]2  1
(x )1  0.37861
or
and
( y )1  0.17173
(z )1  0.90950
( x )1  67.8 ( y )1  80.1 ( z )1  24.6 
K 2 : Begin with Eqs. (9.54a) and (9.54b).
( I x  K 2 )(x )2  I xy ( y )2  I zx (z )2  0
 I xy (x )2  ( I y  K 2 )( y )2  I yz (z )2  0
Substituting:
[(26.4325  29.7840)(103 )](x )2  (2.5002  103 )( y )2  (8.8062  103 )(z )2  0
(2.5002  103 )(x )2  [(31.1726  29.7840)(103 )](y )2  (4.0627  103 ) z )2  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.181* (Continued)
Simplifying
1.3405(x )2  ( y )2  3.5222(z )2  0
1.8005(x )2  ( y )2  2.9258(z )2  0
Adding and solving for (z ) 2 :
(z ) 2  0.48713(x ) 2
( y )2  [1.3405  3.5222(0.48713)](x ) 2
and then
 0.37527(x )2
Now substitute into Eq. (9.57):
(x )22  [0.37527(x )2 ]2  [0.48713(x )2 ]2  1
(x ) 2  0.85184
or
( y )2  0.31967
and
(z ) 2  0.41496
( x )1  31.6 ( y )2  71.4 ( z )2  114.5 
K3 :
Begin with Eqs. (9.54a) and (9.54b).
( I x  K3 )(x )3  I xy ( y )3  I zx (z )3  0
 I xy (x )3  ( I y  K3 )( y )3  I yz (z )3  0
Substituting:
[(26.4325  32.2541)(103 )](x )3  (2.5002  103 )( y )3  (8.8062  103 )(z )3  0
(2.5002  103 )(x )3  [(31.1726  32.2541)(103 )](y )3  (4.0627  103 )(z )3  0
Simplifying
2.3285(x )3  ( y )3  3.5222(z )3  0
2.3118(x )3  ( y )3  3.7565(z )3  0
Adding and solving for (z )3 :
(z )3  0.071276(x )3
and then
( y )3  [2.3285  3.5222(0.071276)](x )3
 2.5795(x )3
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PROBLEM 9.181* (Continued)
Now substitute into Eq. (9.57):
(x )32  [2.5795(x )3 ]2  [0.071276(x )3 ]2  1
(x )3  0.36134
or
and
(i)
( y )3  0.93208
(z )3  0.025755
( x )3  68.8 ( y )3  158.8 ( z )3  88.5 
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
that is, (x )3   0.36134.
Then
( x )3  111.2 ( y )3  21.2 ( z )3  91.5 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.182*
For the component described in Problem 9.167, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes of
inertia at the origin. Sketch the body and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solution of Problem 9.167, we have
1W 2
a
2 g
W
I y  a2
g
5W 2
Iz 
a
6 g
Ix 
1W 2
a
4 g
1W 2
I yz 
a
8 g
3W 2
I zx 
a
8 g
I xy 
Substituting into Eq. (9.56):
 1
5   W 
K 3    1    a 2   K 2
6  g

 2
2
2
2
2
 1 
 5   5  1   1   1   3    W 2 
   (1)  (1)                  a  K
 6   6  2   4   8   8    g

 2 
3
2
2
2
 1   5   1  1 
 3   5  1 
 1  1   3    W 
   (1)        (1)        2        a 2   0
 8   6  4 
 4  8   8    g
 2   6   2  8 

Simplifying and letting K 
W 2
a K yields
g
K 3  2.33333K 2  1.53125 K  0.192708  0
Solving yields
K1  0.163917 K 2  1.05402 K3  1.11539
The principal moments of inertia are then
K1  0.1639
W 2
a 
g
K 2  1.054
W 2
a 
g
K 3  1.115
W 2
a 
g
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PROBLEM 9.182* (Continued)
(b)
To determine the direction cosines x ,  y , z of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
K1:
Begin with Eqs. (9.54a) and (9.54b).
( I x  K1 )(x )1  I xy ( y )1  I zx (z )1  0
 I xy (x )1  ( I y  K 2 )( y )1  I yz (z )1  0
Substituting
 1
1W 2
3W 2
  W 2 
a  ( y )1  
a  (z )1  0
  0.163917   a   (x )1  
  g 
4 g 
8 g 
 2

1W 2
 W 
1W 2

a  (x )1  (1  0.163917)  a 2   ( y )1  
a  (z )1  0
4 g 
 g 
8 g 

Simplifying
1.34433(x )1  ( y )1  1.5(z )1  0
0.299013(x )1  ( y )1  0.149507(z )1  0
Adding and solving for (z )1:
(z )1  0.633715(x )1
( y )1  [1.34433  1.5(0.633715)](x )1
and then
 0.393758(x )1
Now substitute into Eq. (9.57):
(x )12  [0.393758(x )1 ]2  (0.633715(x )1 ]2  1
(x )1  0.801504 
or
and
( y )1  0.315599
(z )1  0.507925
( x )1  36.7 ( y )1  71.6 ( z )1  59.5 
K2:
Begin with Eqs. (9.54a) and (9.54b):
( I x  k2 )(x )2  I xy ( y )2  I zx (z )2  0
 I xy (x )2  ( I y  k2 )( y )2  I yz (z )2  0
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PROBLEM 9.182* (Continued)
Substituting
 1
1W 2
3W 2
  W 2 
a  ( y ) 2  
a  ( z ) 2  0
  1.05402   a   (x )2  
 g

4 g

8 g

 2

1W 2
 W 
1W 2
a  (x )2  (1  1.05402)  a 2   ( y )2  
a  ( z ) 2  0

4 g

 g

8 g


Simplifying
2.21608(x )2  ( y )2  1.5(z )2  0
4.62792(x )2  ( y )2  2.31396(z )2  0
Adding and solving for (z ) 2
(z ) 2  2.96309(x ) 2
( y )2  [2.21608  1.5(2.96309)](x )2
and then
 2.22856(x ) 2
Now substitute into Eq. (9.57):
(x )22  [2.22856(x )2 ]2  [2.96309(x )2 ]2  1
(x ) 2  0.260410
or
( y )2  0.580339
and
(z )2  0.771618
( x )2  74.9 ( y )2  54.5 ( z )2  140.5 

K3 :
Begin with Eqs. (9.54a) and (9.54b):
( I x  K3 )(x )3  I xy ( y )3  I zx (z )3  0
 I xy (x )3  ( I y  K3 )( y )3  I yz (z )3  0
Substituting
 1
1W 2
3W 2
  W 2 
a  ( y )3  
a  ( z )3  0
  1.11539   a   (x )3  
 g

4 g

8 g

 2

1W 2
 W 
1W 2

a  (x )3  (1  1.11539)  a 2   ( y )3  
a  ( z )3  0
4 g

 g

8 g


Simplifying
2.46156(x )3  ( y )3  1.5(z )3  0
2.16657(x )3  ( y )3  1.08328(z )3  0
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PROBLEM 9.182* (Continued)
Adding and solving for (z )3
(z )3  0.707885(x )3
and then
( y )3  [2.46156  1.5(0.707885)](x )3
 1.39973(x )3
Now substitute into Eq. (9.57):
(x )32  [1.39973(x )3 ]2  [0.707885(x )3 ]2  1
or
(x )3  0.537577
and
( y )3  0.752463
(i)
(z )3  0.380543
( x )3  57.5 ( y )3  138.8 ( z )3  112.4 
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
that is, (x )3  0.537577.
Then
( x )3  122.5 ( y )3  41.2 ( z )3  67.6 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.183*
For the component described in Problem 9.168, determine (a) the
principal mass moments of inertia at the origin, (b) the principal
axes of inertia at the origin. Sketch the body and show the
orientation of the principal axes of inertia relative to the x, y, and z
axes.
SOLUTION
(a)
From the solution to Problem 9.168, we have
I x  18.91335
t
a4
t
a4
g
t
I yz  7.14159 a 4
g
t
I zx  0.66667 a 4
g
I xy  1.33333
g
t
I y  7.68953 a 4
g
t
I z  12.00922 a 4
g
Substituting into Eq. (9.56):

  t 
K 3  (18.91335  7.68953  12.00922)  a 4   K 2
 g 

 [(18.91335)(7.68953)  (7.68953)(12.00922)  (12.00922)(18.91335)
2
t 
 (1.33333)  (7.14159)  (0.66667) ]  a 4  K
g

2
2
2
 [(18.91335)(7.68953)(12.00922)  (18.91335)(7.14159) 2
 (7.68953)(0.66667)2  (12.00922)(1.33333)2
3
t 
 2(1.33333)(7.14159)(0.66667)]  a 4   0
g 
Simplifying and letting
K
t
g
a4 K
yields
K 3  38.61210 K 2  411.69009 K  744.47027  0
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PROBLEM 9.183* (Continued)
Solving yields
K1  2.25890 K 2  17.27274 K3  19.08046
K1  2.26
The principal moments of inertia are then
K 2  17.27
K 3  19.08
(b)
t
g
t
g
t
g
a4 
a4 
a4 
To determine the direction cosines x ,  y , z of each principal axis, use two of the equations of
Eq. (9.54) and Eq. (9.57). Then
K1 : Begin with Eqs. (9.54a) and (9.54b):
( I x  K1 )(x )1  I xy ( y )1  I zx (z )1  0
 I xy (x )1  ( I y  K 2 )( y )1  I yz (z )1  0
Substituting

  t 4 

t 4 
t 4 

(18.91335  2.25890)  a   (x )1  1.33333 a  ( y )1   0.66667 a  (z )1  0
a 
g 

 g 




  t 

t 
t 
 1.33333 a 4  (x )1  (7.68953  2.25890)  a 4   ( y )1   7.14159 a 4  (z )1  0
g 
g 

 g 


Simplifying
12.49087(x )1  ( y )1  0.5(z )1  0
0.24552(x )1  ( y )1  1.31506(z )1  0
Adding and solving for (z )1
(z )1  6.74653(x )1
and then
( y )1  [12.49087  (0.5)(6.74653)](x )1
 9.11761(x )1
Now substitute into Eq. (9.57):
(x )12  [9.11761(x )1 ]2  [6.74653(x )1 ]2  1
or
(x )1  0.087825
and
( y )1  0.80075
(z )1  0.59251
( x )1  85.0 ( y )1  36.8 ( z )1  53.7 
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PROBLEM 9.183* (Continued)
K 2 : Begin with Eqs. (9.54a) and (9.54b):
( I x  K 2 )(x )2  I xy ( y )2  I zx (z )2  0
 I xy (x )2  ( I y  K 2 )( y )2  I yz (z )2  0
Substituting
t 


t 
t 
[(18.91335  17.27274)  a 4  (x )2  1.33333 a 4  ( y )2   0.66667 a 4  (z )2  0
g
g
g








t 

t 
t 
 1.33333 a 4  (x )2  (7.68953  17.27274)  a 4   ( y ) 2   7.14159 a 4  (z )2  0
g
g


g




Simplifying
1.23046(x )2  ( y )2  0.5(z )2  0
0.13913(x )2  ( y )2  0.74522(z )2  0
Adding and solving for (z ) 2
(z ) 2  5.58515(x ) 2
and then
( y )2  [1.23046  (0.5)(5.58515)](x )2
 4.02304 (x )2
Now substitute into Eq. (9.57):
(x )22  [4.02304(x )2 ]2  [5.58515(x )2 ]2  1
or
(x ) 2  0.14377
and
( y )2  0.57839
(z )2  0.80298
( x )2  81.7 ( y )2  54.7 ( z )2  143.4 
K3 :
Begin with Eqs. (9.54a) and (9.54b):
( I x  K3 )(x )3  I xy ( y )3  I zx (z )3  0
 I xy (x )3  ( I y  K3 )( y )3  I yz ( y )3  0
Substituting
t 


t 
t 
[(18.91335  19.08046)  a 4  (x )3  1.33333 a 4  ( y )3   0.66667 a 4  (z )3  0
g
g
g







  t 

t 
t 
 1.33333 a 4  (x )3  (7.68953  19.08046)  a 4   ( y )3   7.14159 a 4  (z )3  0
g 
g 

 g 


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.183* (Continued)
Simplifying
0.12533(x )3  ( y )3  0.5(z )3  0
0.11705(x )3  ( y )3  0.62695(z )3  0
Adding and solving for (z )3
(z )3  0.06522(x )3
and then
( y )3  [0.12533  (0.5)(0.06522)](x )3
 0.15794(x )3
Now substitute into Eq. (9.57):
(x )32  [0.15794(x )3 ]2  [0.06522(x ]3 ]2  1
or
(x )3  0.98571
and
( y )3  0.15568
(i)
(z )3  0.06429
( x )3  9.7 ( y )3  99.0 ( z )3  86.3 
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
that is, (x )3  0.98571.
Then
( x )3  170.3 ( y )3  81.0 ( z )3  93.7 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.184*
For the component described in Problems 9.148 and 9.170,
determine (a) the principal mass moments of inertia at the origin,
(b) the principal axes of inertia at the origin. Sketch the body and
show the orientation of the principal axes of inertia relative to the
x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.148 and 9.170. We have
I x  0.32258 kg  m 2
I y  I z  0.41933 kg  m 2
I xy  I zx  0.096768 kg  m 2
I yz  0
Substituting into Eq. (9.56) and using
I y  Iz
I xy  I zx
I yz  0
K 3  [0.32258  2(0.41933)]K 2
 [2(0.32258)(0.41933)  (0.41933) 2  2(0.096768) 2 ]K
 [(0.32258)(0.41933) 2  2(0.41933)(0.096768) 2 ]  0
Simplifying
K 3  1.16124 K 2  0.42764 K  0.048869  0
Solving yields
(b)
K1  0.22583 kg  m 2
or K1  0.226 kg  m 2 
K 2  0.41920 kg  m2
or
K 2  0.419 kg  m 2 
K3  0.51621 kg  m 2
or
K3  0.516 kg  m 2 
To determine the direction cosines x ,  y , z of each principal axis, use two of the equations of
Eqs. (9.54) and (9.57). Then
K1 : Begin with Eqs. (9.54b) and (9.54c):
0
 I xy (x )1  ( I y  K1 )( y )1  I yz (z )3  0
0
 I zx (x )1  I yz (y )1  ( I z  K1 )(z )3  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.184* (Continued)
Substituting
(0.096768)(x )1  (0.41933  0.22583)( y )1  0
(0.096768)(x )1  (0.41933  0.22583)(z )1  0
Simplifying yields
( y )1  (z )1  0.50009(x )1
Now substitute into Eq. (9.54):
(x )2  2[0.50009(x )1 ]2  1
or
(x )1  0.81645
and
( y )1  (z )1  0.40830
( x )1  35.3 ( y )1  ( z )1  65.9 
K 2 : Begin with Eqs. (9.54a) and (9.54b):
( I x  K 2 )(x )2  I xy ( y )2  I zx (z )2  0
0
 I xy (x )2  ( I y  K 2 )( y )2  I yz (z )2  0
Substituting
(0.32258  0.41920)(x )2  (0.096768)( y )2  (0.096768)(z )2  0
(i)
(0.96768)(x )2  (0.41933  0.41920)( y )2  0
(ii)
Eq. (ii)  (x )2  0
and then Eq. (i)  (z )2  ( y )2
Now substitute into Eq. (9.57):
0
(x )22  ( y )22  [( y )2 ]2  1
1
or
( y ) 2 
and
( z ) 2  
2
1
2
( x )2  90.0 ( y ) 2  45.0 ( z )2  135.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.184* (Continued)
K3 :
Begin with Eqs. (9.54b) and (9.54c):
0
I xy (x )3  ( I y  K3 )( y )3  I yz (z )3  0
0
 I zx (x )3  I yz ( y )3  ( I z  K3 )(z )3  0
Substituting
(0.096768)(x )3  (0.41933  0.51621)( y )3  0
(0.096768)(x )3  (0.41933  0.51621)(z )3  0
Simplifying yields
( y )3  (  z )3   ( x )3
Now substitute into Eq. (9.57):
(x )32  2[(x )3 ]2  1
(i)
1
or
( x )3 
and
( y )3  ( z )  
3
1
3
( x )3  54.7 ( y )3  ( z )3  125.3 
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain
the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
That is,
Then
( x ) 3  
1
3
( x )3  125.3
( y )3  ( z )3  54.7 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.185
Determine by direct integration the moments of inertia of the
shaded area with respect to the x and y axes.
SOLUTION
 x x2 
y  4h   2 
a a 
3
1
1   x x2 
dI x  y 3 dx   4h   2   dx
3
3   a a  

I x  dI x 
64h3
3

a  x3
x4
x5 x6 
 3  3 4  3 5  6  dx
0 a
a
a
a 

a

64h3 1 x 4 3 x5 1 x 6 1 x 7



3 4 a3 5 a 4 2 a5 7 a 6 0

64h3  1 3 1 1  64h3  3 
a     
a

3
3
4 5 2 7
 420 
Ix 
16 3
ah 
105
 x x2 
y  4h   2 
a a 
dA  ydx
 x x2 
dI y  x 2 dA  4hx 2   2  dx
a a 
a  x3
x4 
I y  4h   2  dx
0
 a a 

a
 x4
 a3 a3 
x5 
I y  4h   2   4h   
5 
 4 a 5a  0
 4
Iy 
1 3
ha 
5
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.186
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
x2 y 2

1
a 2 b2
x2
a2
y  b 1
dA  2 ydx
dI y  x 2 dA  2 x 2 ydx

I y  dI y 

x  a sin 
Set:
I y  2b
a
0
2 x 2 ydx  2b

a
0
x2 1 
x2
dx
a2
dx  a cos  d
 /2
 a sin  1  sin  a cos d
2
2
2
0
 2 a 3b

 /2
0
sin 2  cos 2  d  2a3b

 /2 1
0
4
sin 2 2 d
 /2
 /2 1
1
1
1
 a 3b
(1  cos 4 )d  a3b   sin 4
0
2
2
4
4
0


From solution of Problem 9.16:
1 3 
 
a b   0   a 3b
4
2
 8
1
I y   a 3b 
8
1
A   ab
2
Thus:
I y  k y2 A k y2 
Iy
A
 a 3b 1 2
 a
 ab 4
2
1
 81
ky 
1
a
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.187
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
At x  a1, y1  y2  b :
y1: b  ka 2
or k 
y2 : b  2b  ca 2
b
a2
or c 
b
a2

x2 
y2  b  2  2 
a 

b 2
x
a2
Then
y1 
Now
 

x2  b
2b
dA  ( y2  y1 ) dx  b  2  2   2 x 2  dx  2 (a 2  x 2 )dx
a
a
a

 

Then
2b 
1 
4
A  dA 
(a  x )dx  2  a 2 x  x3   ab
2
0 a
3 0 3
a 
Now
3
3

1  
x2   b 2  
1 3 1 3
dI x   y2  y1  dx   b  2  2     2 x   dx
3 
3   
a    a
3
 


Then


a
a 2b
2
2

1 b3
(8a 6  12a 4 x 2  6a 2 x 4  x6  x6 )dx
6
3a

2 b3
(4a 6  6a 4 x 2  3a 2 x 4  x6 )dx
3 a6

I x  dI x 
a 2 b3
 3 a (4a  6a x  3a x  x )dx
6
4 2
2 4
6
6
0
a
2 b3  6
3
1 

4a x  2a 4 x 3  a 6 x5  x 7 
6 
3a 
5
7 0

172 3
ab
105
or I x  1.638ab3 
3
and
k x2 
ab
I x 172
43
 1054
 b2
35
A
ab
3
or
k x  1.108b 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.188
Determine the moments of inertia I x and I y of the area shown
with respect to centroidal axes respectively parallel and
perpendicular to side AB.
SOLUTION
Dimensions in mm
First locate centroid C of the area.
Symmetry implies Y  30 mm.
A, mm 2
x , mm
xA, mm3
1
108  60  6480
54
349,920
2
1
  72  36  1296
2
46
–59,616

5184
290,304
X  A   xA : X (5184 mm 2 )  290,304 mm3
Then
or
X  56.0 mm
Now
I x  ( I x )1  ( I x )2
where
1
(108 mm)(60 mm)3  1.944  106 mm 4
12
1

1

( I x ) 2  2  (72 mm)(18 mm)3    72 mm  18 mm  (6 mm)2 
2

 36

( I x )1 
 2(11,664  23,328) mm 4  69.984  103 mm 4
[( I x ) 2 is obtained by dividing A2 into
Then
]
I x  (1.944  0.069984)  106 mm 4
or
I x  1.874  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.188 (Continued)
I y  ( I y )1  ( I y )2
Also
where
1
(60 mm)(108 mm)3  (6480 mm 2 )[(56.54) mm]2
12
 (6, 298,560  25,920) mm 4  6.324  106 mm 4
( I y )1 
1
(36 mm)(72 mm)3  (1296 mm 2 )[(56  46) mm]2
36
 (373, 248  129,600) mm 4  0.502  106 mm 4
( I y )2 
Then
I y  (6.324  0.502)106 mm4
or I y  5.82  106 mm4 
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PROBLEM 9.189
Determine the polar moment of inertia of the area shown with
respect to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the area.
A, mm 2
1
2


2

(54)(36)  3053.6

2
(18) 2  508.9
48

24

y , mm
yA, mm3
 15.2789
46,656
 7.6394
3888
42,768
2544.7
Y  A   y A: Y (2544.7 mm 2 )  42, 768 mm3
Then
or
Y  16.8067 mm 
J O  ( J O )1  ( J O ) 2
(a)


(54 mm)(36 mm)[(54 mm)2  (36 mm)2 ] 
8
 (3.2155  106  0.0824  106 ) mm 4

4
(18 mm)4
 3.1331  106 mm 4
or
J O  3.13  106 mm 4 
J O  J C  A(Y )2
(b)
or
J C  3.1331  106 mm 4  (2544.7 mm 2 )(16.8067 mm)2
or
J C  2.41  106 mm 4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.190
Two L102 × 102 × 12.7-mm angles are welded to a steel plate as
shown. Determine the moments of inertia of the combined section
with respect to centroidal axes respectively parallel and perpendicular to the plate.
SOLUTION
For L102 × 102 × 12.7-mm angle:
A = 2430 mm2,
Ix = Iy = 2.34 × 106 mm4
YA = Σ y A
Y (6300 mm2) = 237060 mm3
Y = 37.6 mm
Section
Area, mm2
y mm
yA, mm3
Plate
(12)(120) = 1440
60
86400
Two angles
2(2430) = 4860
31
150660
Σ
6300
237060
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PROBLEM 9.190 (Continued)
Entire section:
I x = Σ( I x′ + Ad 2 ) =
1
(12)(120)3 + (12)(120)(22.4)2 + 2[2.34 × 106 + (2430)(6.6)2]
12
= 1.728 × 106 + 0.723 × 106 + 4.68 × 106 + 0.212 × 106
Iy =
I x = 7.343 × 106 mm4
1
(120)(12)3 + 2[2.34 × 106 + (2430)(37)2]
12
= 17280 + 4.68 × 106 + 6.653 × 106 = 11.347 × 106 mm4
I y = 11.347 × 106 mm4
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1451
PROBLEM 9.191
Using the parallel-axis theorem, determine the product of inertia of
the L127 × 76 × 12.7-mm angle cross section shown with respect to
the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2
For each rectangle:
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
Thus
I xy = Σ x yA
A, mm2
1 76 × 12.7 = 965.2
x, mm
y, mm
19.1
−37.85 −697.78 × 10−3
2 114.3 × 12.7 = 1451.6 −12.55
25.65
xy A,mm4
−467.28 × 10−3
−1165.06 × 10−3
Σ
I xy = −1.165 × 106 mm4
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1689
PROBLEM 9.192
For the L127 × 76 × 12.7-mm angle cross section shown, use Mohr’s
circle to determine (a) the moments of inertia and the product of inertia
with respect to new centroidal axes obtained by rotating the x and y
axes 30° clockwise, (b) the orientation of the principal axes through
the centroid and the corresponding values of the moments of inertia.
SOLUTION
From Figure 9.13B:
Ix = 3.93 × 106 mm4
Iy = 1.06 × 106 mm4
From the solution to Problem 9.191
I xy = −1.165 × 106 mm4
The Mohr’s circle is defined by the diameter XY, where
X(3.93 × 106, −1.165 × 106 ),
Now
I ave =
and
R=
Y(1.06 × 106, 1.165 × 106)
1
1
( I x + I y ) = (3.93 + 1.06)106 = 2.495 × 106 mm4
2
2
1
(I x − I y )
2
2
2
+ I xy2 =
1
(3.93 − 1.06)106 + (−1.165 × 106)2
2
= 1.848 × 106 mm4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−1.165 × 106)
(3.93 − 1.06)106
= 0.81185
=−
or
2θ m = 39.07°
and
θ m = 19.54°
Then
α = 180° − (39.07° + 60°)
= 80.93°
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PROBLEM 9.192 (Continued)
(a)
We have
I x′ = I ave − R cos α = (2.495 − 1.848cos 80.93°) 106
I x′ = 2.204 × 106 mm4
or
I y′ = I ave + R cos α = (2.495 + 1.848cos 80.93°) 106
or
I y ′ = 2.786 × 106 mm4
or
I x′y′ = −1.825 × 106 mm4
I x′y′ = − R sin α = (−1.848 sin 80.93°) 106
First observe that the principal axes are obtained by rotating the xy axes through
(b)
19.54° counterclockwise
about C.
Now
I max, min = I ave ± R = (2.495 ± 1.848) 106
or
I max = 4.343 × 106 mm4
I min = 0.647 × 106 mm4
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 9.193
A thin plate of mass m was cut in the shape of a parallelogram as
shown. Determine the mass moment of inertia of the plate with
respect to (a) the x axis, (b) the axis BB, which is perpendicular to
the plate.
SOLUTION
mass  m   t A
I mass   t I area 
(a)
m
I area
A
Consider parallelogram as made of horizontal strips and slide strips to form a square since distance
from each strip to x axis is unchanged.
1
I x ,area  a 4
3
I x, mass 
(b)
m
m 1 
I x,area  2  a 4 
A
a 3 
1
I x  ma 2 
3
For centroidal axis y:
1  1
I y,area  2  a 4   a 4
12  6
m
m 1  1
I y,mass  I y,area  2  a 4   ma 2
A
a 6  6
1
7
I y  I y  ma 2  ma 2  ma 2  ma 2
6
6
For axis BB  to plate, Eq. (9.38):
1
7
I BB  I x  I y  ma 2  ma 2
3
6
I BB 
3 2
ma 
2
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PROBLEM 9.194
A thin plate of mass m was cut in the shape of a parallelogram as
shown. Determine the mass moment of inertia of the plate with
respect to (a) the y axis, (b) the axis AA, which is perpendicular to
the plate.
SOLUTION
See sketch of solution of Problem 9.117.
(a)
From part b of solution of Problem 9.117:
I y 
1
ma 2
6
I y  I y  ma 2 
(b)
1 2
ma  ma 2
6
Iy 
7
ma 2 
6
I AA 
1
ma 2 
2
From solution of Problem 9.115:
I y 
Eq. (9.38):
1 2
ma
6
1
and I x  ma 2
3
I AA  I y  I x

1 2 1 2
ma  ma
6
3
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PROBLEM 9.195
A 2-mm thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass moment of inertia of the component with respect
to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m  STV  ST t A
Then
m1  (7850 kg/m3 )(0.002 m)(0.76  0.48) m 2
 5.72736 kg
1

m2  (7850 kg/m3 )(0.002 m)   0.76  0.48  m 2
2


 2.86368 kg


m3  (7850 kg/m3 )(0.002 m)   0.482  m 2
4


 2.84101 kg
Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have
I x  ( I x )1  ( I x ) 2  ( I x )3
2
1
 0.48  
m 
  (5.72736 kg)(0.48 m)2  (5.72736 kg) 
 2
 
12
2
1
 0.48    1

m     (2.84101 kg)(0.48 m)2 
  (2.86368 kg)(0.48 m)2  (2.86368 kg) 
18
3
2

  


 [(0.109965  0.329896)  (0.036655  0.073310)  (0.327284)] kg  m 2
 (0.439861  0.109965  0.327284) kg  m 2
or
I x  0.877 kg  m2 






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PROBLEM 9.195 (Continued)
I y  ( I y )1  ( I y ) 2  ( I y )3
 1
 0.76 2  0.48 2  2 
  (5.72736 kg)(0.762  0.482 ) m 2  (5.72736 kg) 
 
 m 
 2   2  
12

2
1
 0.76    1

m     (2.84101 kg)(0.48 m)2 
  (2.86368 kg)(0.76 m) 2  (2.86368 kg) 
18
3
4

  


 [(0.385642  1.156927)  (0.091892  0.183785)  (0.163642)] kg  m 2
 (1.542590  0.275677  0.163642) kg  m 2
or
I y  1.982 kg  m2 
I z  ( I z )1  ( I z ) 2  ( I z )3
2
1
 0.76  
  (5.72736 kg)(0.76 m) 2  (5.72736 kg) 
m 
 2
 
12
 1
 0.76 2  0.48 2  2 
2
2
2
  (2.86368 kg)(0.76  0.48 ) m  (2.86368 kg) 
 
 m 
18
 3   3  


1

  (2.84101 kg)(0.48 m)2 
4

I z  [(0.275677  0.827031)  (0.128548  0.257095)
 (0.163642)] kg  m 2
 (1.102708  0.385643  0.163642) kg  m 2
or
I z  1.652 kg  m2 
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PROBLEM 9.196
Determine the mass moment of inertia of the steel
machine element shown with respect to the z axis.
(The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
Then
m1 = 7850 kg/m3 × (0.054 × 0.074 × 0.18) m3
= 5.646 kg
m2 = 7850 kg/m3 ×
π
2
× 0.0272 × 0.028 m3
= 0.252 kg
m3 = 7850 kg/m3 × (p × 0.0122 × 0.024) m3
= 0.0852 kg
m4 = 7850 kg/m3 × (0.018 × 0.012 × 0.18) m3 = 0.305 kg
The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations
derived above (component 2).
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PROBLEM 9.196 (Continued)
Find: I z
We have
I z = ( I z )1 + ( I z ) 2 + ( I z )3 − ( I z ) 4
=
1
(5.646 kg)[(0.054)2 + (0.074)2] m2
12
+ (5.646 kg)[(0.027)2 + (0.037)2] m2
+ (0.252 kg)
1 16
−
(0.027 m)2
2 9p 2
+ (0.252 kg) (0.027)2 + 0.074 +
+
4 × 0.027
3p
2
m2
1
(0.0852 kg)(0.012)2
12
+ (0.0852 kg)[(0.027)2 + (0.074)2] m2
−
1
(0.305 kg)[(0.018)2 + (0.012)2] m2
12
+ (0.305 kg)[(0.027)2 + (0.006)2] m2
I z = [(3.95 + 11.85) + (0.059 + 2.02)
+ (0.001 + 0.53) − (0.012 + 0.23)] × 10–3 kg · m2
or
I z = 18.2 × 10–3 kg · m2
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