Solution manual vector mechanics engineers dynamics 8th beer chapter 09

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Chapter 9, Solution 1.

First note:
Have

y=

b2 − b1
x + b1
a

I y = ∫ x 2dA
b2 − b1
x + b1
a
a
x 2d ydx
0 0

=∫ ∫

a
 b − b1

= ∫ 0 x2  2
x + b1  dx
 a

a

 1 b − b1 4 1 3 
= 2
x + b1x 
3
4 a
0
=

1 3
a ( b1 + 3b2 )
12
Iy =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

1 3
a ( b1 + 3b2 )
12


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 2.

At

5

x = a, y = b :

∴ y =

b
a

5
2

5

b = ka 2

x2

or

dI y =

1 3
x dy
3

=

Then

Iy =

x=

or

a
b

b
5

a2

2

2
5

y5

1 a3 65
y dy
3 b 65
1 a3 b 65
∫ y dy
3 b 65 0

1 5 a3 115
=
y

3 11 b 65
=

k =

b

0

5 a3 115
b
33 b 65
or I y =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

5 3
ab
33


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 3.
At x = 0:

First note:

At x = a :

a = k ( 2b − b )
k =
∴ x=

Have

2

c = −b

or

or

0 = k (b + c)

2

a
b2
a
2
y − b)
2(
b

I y = ∫ x 2dA
=∫

a
2
y − b)
2b b 2 (
0
0

∫

x 2dxdy
3

=

1 2b  a
2
y − b )  dy
∫
0  2(
3
b

2b

1 a3 1
7
=
× ( y − b)
6

3b
7
b
=

1 3
ab
21
Iy =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

1 3
ab
21


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 4.

y = kx 2 + c

Have

b = k ( 0) + c

x = 0, y = b :

At

c=b

or
At

x = 2a, y = 0:

or

k =−
y =−

Then

=
Then

I y = ∫ x 2dA,
2a

I y = ∫ a x 2dA =

2

0 = k ( 2a ) + b

b

4a 2

b 2
x +b
4a 2

b
4a 2 − x 2
4a 2

(

dA = ydx =

)
(

)

b
4a 2 − x 2 dx
4a 2

(

)

b 2a 2
2
2

∫ x 4a − x dx
4a 2 a
2a

b  2 x3 x5 
=
− 
 4a
3
5 a
4a 2 
=

b
b
8a3 − a3 −
32a5 − a5
2
3
20a

=

7a3b 31a3b
−
3
20

(

)

(

)

Iy =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

47 3
ab
60


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 5.

First note:
Have

y =

b2 − b1
x + b1
a

I x = ∫ y 2dA
=

b2 − b1
x + b1
a
a
0 0

∫ ∫

y 2d ydx
3

1 a  b − b1

= ∫0  2
x + b1  dx
3  a

4
1  1 a   b2 − b1

= ×
x + b1 

3  4 b2 − b1   a


a

0

=

1
a
b24 − b14
12 b2 − b1

=

1
a
( b2 + b1 )( b2 − b1 ) b22 + b12
12 b2 − b1

=

1
a ( b1 + b2 ) b12 + b22
12

(

)
(

(

)

)
Ix =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

1
a ( b1 + b2 ) b12 + b22
12

(

)


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 6.
SOLUTION

5

At x = a, y = b : b = ka 2
or k =

∴y =

b
5

a2

b
a

5

x2

5
2

I x = ∫ y 2dA
=

b

a


2
y 5 dy 
 b


2
∫0 y 

a

dA = xdy

2
5

5 175
= 2 ×
y
b 5 17

b

0

17

5a b 5
=
17 b 52

or I x =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

5 3

ab
17


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 7.
At x = 0: 0 = k ( b + c )

First note:
or

2

c = −b
At x = a : a = k ( 2b − b )

Have

or

k =

a
b2

∴

x=

a
2
y − b)
2(
b

2

I x = ∫ y 2dA

=∫

a
2
y − b)
2b b 2 (
0
b

∫

y 2dxdy

=

a 2b 2
2
y ( y − b ) dy
2 ∫b
b

=

a 2b 4
y − 2by 3 + b 2 y 2 dy
2 ∫b
b

=

a 1 5 1 4 1 2 3
 y − by + b y 
2
3
b2  5
b

=

a  1
1
1 2
1 2 3 
5
4
3
1 5 1
4
 ( 2b ) − ( 2b ) + b ( 2b )  −  b − b b + b b  
2

3
5
2
3
b 2   5
 


(

)

2b

( )

( )

8 1 1 1
 32
= ab3 
−8+ − + − 
3 5 2 3
 5
=

31 3
ab
30
Ix =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

31 3
ab
30


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 8.

Have

y = kx 2 + c

At

x = 0, y = b: b = k (0) + c

or

c=b

At

x = 2a, y = 0: 0 = k (2a) 2 + b

or

k =−

Then

y =

Now

dI x =
=

b
4a 2

b
4a 2 − x 2
4a 2

(

)

1 3
y dx
3
3
1 b3
4a 2 − x 2 dx

6
3 64a

(

)

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Then

I x = ∫ dI x
=

3
1 b3 2 a
4a 2 − x 2 dx
6 ∫a
3 64a

=

b3

2a
64a 6 − 48a 4 x 2 + 12a 2 x 4 − x 6 dx
6 ∫a
192a

(

)

(

)

2a

b3 
12 2 5 x 7 
6
4 3
64
a
x
16
a
x
a x −
=
−
+



5
7 a
192a 6 
=

b3 
64a 7( 2 − 1) − 16a 7 ( 8 − 1)
192a 6 

+

=

12 7
1

a ( 32 − 1) − (128 − 1) 
5
7


ab3 
372 127 
3
−
 64 − 112 +
 = 0.043006ab
192 
5

7 

I x = 0.0430ab3

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 9.

x2
y2
+
=1
a 2 b2

x = a 1−

y2
b2

dA = xdy
dI x = y 2dA = y 2 xdy
b

b

I x = ∫ dI x = ∫ − b xy 2dy = a ∫ − b y 2 1 −
Set: y = b sin θ

y2
dy
b2

dy = b cosθ dθ

π

I x = a ∫ 2π b 2 sin 2 θ 1 − sin 2 θ b cosθ dθ
−

2

π

π

−

−

= ab3 ∫ 2π sin 2 θ cos 2 θ dθ = ab3 ∫ 2π
2

2

1 2

sin 2θ dθ
4
π

π

1
1
1
1

2
= ab3 ∫ 2π (1 − cos 4θ ) dθ = ab3 θ − sin 4θ 
− 2
4
8
4

−π
2

=

2

1 3  π  π  π 3
ab  −  −   = ab
8
 2  2  8

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

Ix =

1
π ab3
8


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 10.

At

2a = kb3

x = 2a, y = b:

or

k =

2a
b3

Then

x=

2a 3
y
b3

or

y =

Now

dI x =

b

( 2a )

1

1
3

x3

1 3
1 b3
y dx =
xdx
3

3 2a
2a

Then

1 b3 2 a
1 b3 1 2
I x = ∫ dI x =
xdx
x
=
∫
3 2a a
6 a 2 a
=

b3
4a 2 − a 2
12a

(

)
Ix =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

1 3

ab
4


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 11.
−a
At x = a : b = k 1 − e a 



First note:
or
Have

k =

b
1 − e−1

I x = ∫ y 2dA
b
a 1− e
0 0

=∫ ∫

−x


 1− e a


−1 

3






y 2dydx
3

=

−x
1  b  a
1 − e a  dx

−1  ∫ 0 
31 − e 



=

−x
−2 x

−3 x
1  b  a
1 − 3e a + 3e a − e a  dx

−1  ∫ 0 
31 − e 



3

3

a

−x
1 b  
 a  −2 x  a −3x  
x − 3( − a ) e a + 3 −  e a −  − e a 
= 
−1  
31 − e  
 2
 3
 0

3

1  b  
1

1 
 
= 
a + 3ae−1 − 1.5ae−2 + ae −3  −  3a − 1.5a + a  
−1  
3  1 − e  
3
3 
 

=

1
ab3
3 1 − e −1

(

)

3

11 

1.91723 − 
6


= 0.1107ab3

I x = 0.1107ab3

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 12.
x2
y2
+
=1
a 2 b2

y = b 1−

x2
a2

dA = 2 ydx
dI y = x 2dA = 2 x 2 ydx
a

a

I y = ∫ dI y = ∫ 0 2 x 2 ydx = 2b ∫ 0 x 2 1 −
x = a sin θ

Set:

x2
dx
a2

dx = a cosθ dθ

π

I y = 2b∫ 02 a 2 sin 2 θ 1 − sin 2 θ a cosθ dθ
3

π

2

2

3

π

= 2a b∫ sin θ cos θ dθ = 2a b∫ 02
2
0

π

1 2

sin 2θ dθ
4
π

1
1
1
1

2
= a3b∫ 02 (1 − cos 4θ ) dθ = a3b θ − sin 4θ 
2
2
4
4

0

=

1 3 π
 π
a b  − 0  = a3b
4
2
 8
Iy =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.

1 3
πa b
8


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 13.

At

x = 2a, y = b : 2a = kb3
2a 3
y
b3

Then

x=

or

y =

Now

I y = ∫ x 2dA

Then

I y = ∫ a x2

b

( 2a )

=

=

=

1

b

( 2a )

1
3

x 3 dx

7

b

( 2a )

x3

dA = ydx

2a

=

1

1
3

1
3

b
1

( 2a ) 3

2a
∫ a x 3 dx

3 103
x
10

3b

10 ( 2a )

1
3

2a

a

 2a 103 − a 103 
( )



10
3ba3  103
2 − 1 3 
1 

10 ( 2 ) 3 

= 2.1619a3b
or I y = 2.16a3b

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 14.
First note:
or
Have

−a
At x = a : b = k 1 − e a 



k =

b
1 − e−1

I y = ∫ x 2dA
b
a 1− e
0 0

−x

1− e a


−1 




 2
 x d ydx

=

∫ ∫

=

∫ 0 x 1 − e−1 1 − e a   dx



a 2



b



−x



a



−x


2
 
b 1 3
e a  1  2
 1
x
x
2
x
2
=
−
−
−
−
+






3 
1 − e−1  3
 a
 
 1   a 
− 



 a

0
=

2

b   1 3
a
3 −1  a
3
+
a
a
e

 2 + 2 + 2   − a × 2
−1 
a
1 − e   3
a

 

=

a3b  1

+ 5e −1 − 2 

−1 
1− e 3


(



)


= 0.273a3b

I y = 0.273 a3b

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 15.

k1 =

or
Then

y1 =

and

x1 =

b 4
x
a4
a
b

1
4

b = k2a 4

b
a4

k2 =
b

y2 =

1
4

b
1

a4

1

x4

a
a
x2 = 4 y 4
b

1

y4

A = ∫ ( y2 − y1 ) dx = b∫

Now

1

b = k1a 4

x = a, y = b:

At

 x 14

a


0

a

1
4

−

x 4 
dx
a4 


a

 4 x 54 1 x5 
 = 3 ab
= b
−
5
 5 a 14 5 a 4 

0
I x = ∫ y 2dA

Then

dA = ( x1 − x2 ) dy

 a 1

a
b
I x = ∫ 0 y 2  1 y 4 − 4 y 4  dy
 4

b
b

b

 4 y 134
1 y 7 
= a
−
7 b4 
13 b 14

0

1
 4
= ab3  − 
 13 7 
or I x =

Now

kx =

Ix
=
A

15 3
ab
91
=
3
ab
5

15 3
ab
91

25 2
b = 0.52414b
91
or k x = 0.524b

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 16.
First note:

At x = a :

2b = k a

or
Straight line:
Now:

k=

2b
a

y1 =

b
x
a

∴ y2 =

2b
a

x

b 

a  2b 12
A = 2∫ 0 
x − x  dx
a 
 a
a

 4 x 23
1 2 
= 2b 
−
x
 3 a 2a 

0
=
Have

5
ab
3

I x = ∫ y 2dA
1

2b 2
x
a
a
0 bx

a

= 2∫ ∫

=

y 2dydx

2 a  8b3 32 b3 3 
x − 3 x  dx
∫ 

3 0  a 23
a

a


2b3  2
8 5
1
=
 × 3 x 2 − 3 x 4 
3  5 a2
4a
0

Ix =
And

kx =

=

59 3
ab
30

Ix
A
59 3
ab
30
5
ab
3

= b 1.18

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

k x = 1.086 b


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 17.
At

x = a, y = b:

k1 =

or
Then

Now

y1 =

1

b = k1a 4

b = k2a 4

b
a4

b

b 4
x
a4

k2 =

y2 =

and

A = ∫ ( y2 − y1 ) dx = b∫

1

a4

 x 14

a

0

a

1
4

−

b
a

1
4

1

x4

x 4 
dx
a4 


a

 4 x 54 1 x5 
 = 3 ab
= b
−
1
4
5
5 a4 5 a 

0

dA = ( y2 − y1 ) dx

Now

I y = ∫ x 2dA

Then

 b 1


b
a
I y = ∫ 0 x 2  1 x 4 − 4 x 4  dx
 4

a
a

= b∫

 x 94

a

0

a

1
4

−

x 6 
dx
a4 

a

 4 x 134

1 x 7 
= b
−
7 a4 
13 a 14

0

1 
 4
= b  a3 − a3 
7 
 13

or I y =

Now

ky =

Iy
A

=

15 3
ab
91
=
3

ab
5

15 3
ab
91

25
a = 0.52414a
91
or k y = 0.524a

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 18.
First note:

At x = a :

2b = k a

or

Straight line:

Now:

k=

2b
a

y1 =

b
x
a

∴ y2 =

2b
a

x

b 
a  2b 12
A = 2∫ 0 
x − x  dx
a 
 a
a

 4 x 23
1 2 

= 2b 
−
x
 3 a 2a 

0

=
Have

5
ab
3

I y = ∫ x 2dA
1

2b 2
x
a
a
b
0
x
a

= 2∫ ∫

x 2dydx

 2b 12 b 
a
x − x  dx
= 2∫ 0 x 2 
a 
 a
7

2 x2
1 x 4 

= 2b 2 ×
−

7 a 4 a 



=

a

0

9 3
ab
14

Iy =

9 3
ab
14
continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

And

ky =

=

=a

Iy
A

9 3
ab
14
5
ab
3

27
70

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

k y = 0.621 a


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 19.
First note:

At x = 0: b = c cos ( 0 )

c=b

or

At x = 2a : b = b sin k ( 2a )
2ka =

or

k=
Then

π

2

π
4a

π
π 
2a 
A = ∫ a  b sin
x − b cos
x  dx
4a
4a 

2a

π
4a
π 
 4a
= b  − cos
x−
sin
x
4a
π
4a  a
 π
=−
=

Have

4ab 
1 
 1
+
(1) − 

π 
2 
 2

4ab

π

(

)

2 −1

I x = ∫ y 2dA
π

2 a b sin 4a x

= ∫a ∫

b cos

=

π

4a

x

y 2dydx

1 2a  3 3 π

3
3 π
∫  b sin 4a x − b cos 4a x  dx
3 a 

2a

b3  4a
π
1 4a
π   4a
π
1 4a 3 π  
x+
cos3
x  −  sin
x−

sin
x 
=
 − cos
3  π
4a
3π
4a   π
4a
3π
4a   a
=

4ab3  
 −1 +
3π  


3
3
1  1
1 1 
1
1  1   


−
−
+
−

+
3   2 3  2 
3  2   
2



=

4ab3  5
2
2− 

3π  6
3
continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System

Ix =

2
ab3 5 2 − 4
9π

(

)

= 0.217 ab3
And

kx =

=

I x = 0.217 ab3

Ix
A
2
ab3 5 2 − 4
9π
4
ab 2 − 1

π

(
(

)

)

= 0.642b

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

k x = 0.642 b


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 20.
At x = 0: b = c cos ( 0 )

First note:

c=b

or

At x = 2a : b = b sin k ( 2a )
2ka =

or

k=

π
2

π
4a

π
π 
2a 
A = ∫ a  b sin
x − b cos
x  dx
4a
4a 


Then

2a

π
4a
π 
 4a
= b  − cos
x−
sin
x
4a
π
4a  a
 π
=−

=
Have

4ab 
1 
 1
+
(1) − 

π 
2 
 2

4ab

π

(

)

2 −1

I y = ∫ x 2dA
π

2 a b sin 4a x 2
x dydx
a b cos π x

4a

=

∫ ∫

=

2a 2 

∫ a x  b sin 4a x − b cos 4a x  dx

π



π



continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.


COSMOS: Complete Online Solutions Manual Organization System






2
π
2
π
x
π 
 2x
= b 
+
−
sin
x
cos
x
cos
x
2
3

π 
4a
4a
4a 
π 
  π 








   4a 
 4a 

 4a 

2a





2
2x
π
2
π
x
π 
cos
x
sin
x
sin
x
−

−
+

3
  π 2
π 
4a
4a
4a  
π 
 



 

 4a 
 4a 
  4a 
  a

2a

64a3b 
π
π  π  
π
π 
π 2 2 
Iy =

sin
x
cos
x
x
sin
x
cos
x
2
x 
−
+
+
−







4a  2a  
4a
4a  
π 3  4a
16a 2  
a

=


64a3b  
π 2    1
1 
π 2   
1
π
1
2
2
+
−
−
+
−
(
)(
)
(
)






 





4    2
16   
π 3  
2  



I y = 1.482a3b

= 1.48228a3b
And

ky =

=

Iy
A
1.48228a3b
4ab
2 −1

π

(

)

= 1.676a

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

k y = 1.676a


COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 21.

dI x =

(a)

Ix =

1 3
a dx
3
1 3 a
a3
2
a ∫ − a dx =
( 2a ) = a 4
3
3
3

dI y = x 2dA = x 2adx

I y = a∫

a 2
x dx
−a

JO = I x + I y =

JO =

kO2 A

a

 x3 
2
= a   = a4
 3  −a 3

2 4 2 4
a + a
3
3

4 4
a
JO

2
k =
= 3 2 = a2
A
3
2a
2

JO =

4 4
a
3

kO = a

2
3

(b)

dI x =
Ix =

1 3
a dx
12
a 3 2a
a 3 2a 1 4
dx

=
[ x] = 6 a
∫
12 0
12 0

dI y = x 2dA = x 2 ( adx )
I y = a∫

2a 2
x dx
0

JO = I x + I y =

J O = kO2 A

2a

 x3 
8
= a   = a4
3
 3 0

1 4 8 4 17 4
a + a =
a
6
3

6

17 4
a
J
17 2
kO2 = O = 6 2 =
a
A
12
2a

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

JO =

17 4
a
6

kO = a

17
12