Cap 09 by Cristian Manrique

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 1.

First note: Have

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.

x = a, y = b :

∴ y =

b a

5 2

b = ka 2

dI y =

1 3 x dy 3

Then

Iy =

a b

b 5

2 5

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 =

5 a3 115 b 33 b 65 or I y =

5 3 a b 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

c = −b

0 = k (b + c)

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 =

1 3 a b 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 :

c=b

or At

x = 2a, y = 0:

k =− y =−

Then

= Then

I y = ∫ x 2dA, 2a

I y = ∫ a x 2dA =

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 =

47 3 a b 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 

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 =

1 a ( b1 + b2 ) b12 + b22 12

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 6. SOLUTION

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

∴y =

b 5

b a

5 2

I x = ∫ y 2dA =

a

 2 y 5 dy   b 

2 ∫0 y 

dA = xdy

2 5

5 175 = 2 × y b 5 17

5a b 5 = 17 b 52

or I x =

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

c = −b At x = a : a = k ( 2b − b )

Have

k =

a b2

∴

a 2 y − b) 2( b

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   

(

)

( )

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

31 3 ab 30 Ix =

31 3 ab 30

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 8.

Have

y = kx 2 + c

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

c=b

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

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

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

(

)

(

)

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

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

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

y2 dy b2

dy = b cosθ dθ

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

−

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

1 2 sin 2θ dθ 4 π

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

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

Ix =

1 π ab3 8

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 10.

2a = kb3

x = 2a, y = b:

k =

2a b3

Then

2a 3 y b3

y =

Now

dI x =

( 2a )

1 3

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 =

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 

   

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   

−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

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

(

)

11   1.91723 −  6 

= 0.1107ab3

I x = 0.1107ab3

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

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

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

= 2a b∫ sin θ cos θ dθ = 2a b∫ 02 2 0

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 =

1 3 πa b 8

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 13.

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

Then

y =

Now

I y = ∫ x 2dA

Then

I y = ∫ a x2

( 2a )

1 3

x 3 dx

( 2a )

dA = ydx

1 3

b 1

( 2a ) 3

2a ∫ a x 3 dx

3 103 x 10

3b 10 ( 2a )

1 3

 2a 103 − a 103  ( )  

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

= 2.1619a3b or I y = 2.16a3b

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





−x



  −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

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

a a x2 = 4 y 4 b

A = ∫ ( y2 − y1 ) dx = b∫

Now

b = k1a 4

x = a, y = b:

 x 14

a  0

a

1 4

−

x 4  dx a4  

 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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 16. First note:

At x = a :

2b = k a

or Straight line: Now:

2b a

y1 =

b x a

∴ y2 =

2b a

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

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 =

b = k1a 4

b = k2a 4

b a4

b 4 x a4

k2 =

y2 =

and

A = ∫ ( y2 − y1 ) dx = b∫

 x 14

a  0

a

1 4

−

b a

1 4

x 4  dx a4  

 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 a b 91

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 18. First note:

At x = a :

2b = k a

Straight line:

Now:

2b a

y1 =

b x a

∴ y2 =

2b a

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   

9 3 ab 14

Iy =

9 3 a b 14 continued

COSMOS: Complete Online Solutions Manual Organization System

And

ky =

Iy A

9 3 ab 14 5 ab 3

27 70

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

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

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

(

)

2 −1

I x = ∫ y 2dA π

2 a b sin 4a x

= ∫a ∫

b cos

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

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

( (

)

= 0.642b

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

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

π 2

π 4a

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

Then

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

= Have

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

4ab

(

)

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





continued

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

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

(

)

= 1.676a

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

 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

 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

JO =

17 4 a 6

kO = a

17 12

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 22. Have: A = ( 2a )( 3b ) −

1 ( 2a )( b ) 2

= 5ab

Have:

JP = Ix + I y

Where:

I x = ∫ y 2dA a 2b b − x a

= 2∫ 0 ∫

y 2dydx

3 2 a 3  b   − − 2 b x  ∫ ( )  a   dx 3 0    

2  x4  = b3  8x + 3  3  4a  = And

11 3 ab 2

I y = ∫ x 2dA a 2b 2 b x dydx − x a

= 2∫ 0 ∫

 a  b  = 2∫ 0 x 2  2b −  − x   dx  a   a

1 4 2 = 2b  x3 + x  4a  0 3 =

11 3 ab 6 continued

COSMOS: Complete Online Solutions Manual Organization System

Then:

Also:

JP =

kP =

11 3 11 3 ab + a b 2 6 JP =

11 ab a 2 + 3b 2 6

kP =

11 2 a + 3b 2 30

(

)

JP A 11 ab a 2 + 3b 2 6 5ab

(

)

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 23.

y1 :

x = a,

y = 2b: 2b = ma

2b a

Then

y1 =

2b x a

y2 :

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

x = a, y = 2b: 2b = ka 2 + b

k =

y2 =

Then Now

c=b

b a2

b 2 b x + b = 2 x2 + a2 a2 a

(

)

2b  a b A = ∫ ( y2 − y1 ) dx = ∫ 0  2 x 2 + a 2 − x dx a  a

(

)

 b 1  b  =  2  x3 + a 2 x  − x 2   a 0 a  3 = Now

1 b 1 3  b a + a3  − a 2 = ab 2 3 a 3  a a

I y = ∫ x 2dA = ∫ 0 x 2 ( y2 − y1 ) dx =

2b  x  dx 

a 2 2 2 ∫ 0 x  a2 ( x + a ) − a 

b  1 1  2b x 4  = 2  x5 + a 2 x3  −  3 a  5  a 4 0 =

b  a5 1 5  2b a 4 1 3 + a  − = ab 2  3  a 4 30 a  5 continued

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And

1  1 I x = ∫ dI x = ∫  y23 − y13  dx 3 3   =

1 a  b3 2 x + a2 ∫ 0  6 3 a

1 b3 a  1 6  x + 3x 4a 2 + 3x 2a 4 + a 6 − 8x3  dx ∫ 3 a3 0  a3 

(

)

−

8b3 3  x  dx a3 

(

)

1 b3  1  x 7 3 2 5 3 4 3 8 4 6  Ix = a x a x a x + + + −  3  3  4x  3 a  a  7 5 3  0  =

Finally

  1 b3  1  a 7 3 7 26 3 + a + a 7 + a 7  − 2a 4  = ab 3  3  3 a  a  7 5   105 26 3 1 3 ab + ab 105 30

JP = Ix + I y =

or J P =

And

kP =

JP = A

ab 7a 2 + 52b 2 210

(

ab 7a 2 + 52b 2 210 1 ab 3 or k P =

(

)

) 7a 2 + 52b 2 70

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 24. x = r 2 − y2

First note:

JP = Ix + I y r

I x = ∫ y 2dA = 2∫ r ∫ 0

r 2 − y2

y 2dxdy

= 2∫ r y 2 r 2 − y 2 dy 2

Let Then Thus

y = r sin θ dy = r cosθ dθ ;

y = r:

θ =

π 2

;

r : 2

θ =

π 6

I x = 2∫ π2 r 2 sin 2 θ ( r cosθ )( r cosθ dθ ) 6

Now

sin 2θ = 2sin θ cosθ Ix =

thus

1 2 sin 2θ 4

1 4 π2 2 r ∫ π sin 2θ dθ 2 6

1  θ sin 4θ  = r4  −  2 2 8 

π 2

π 6

1 4π π 1 3 + × r  −  2  4 12 8 2 

3 r4  π  +  4 3 8  r

I y = ∫ x 2dA = 2∫ r ∫0

r 2 − y2

sin 2 θ cos 2 θ =

x 2dxdy

2 r 2 2 ∫ r r − y 2 dy 3 2

(

)

continued

COSMOS: Complete Online Solutions Manual Organization System

Let Then

Now

Thus

y = r sin θ Iy =

2 π2 3 3 ∫ π r cos θ 3 6

(

) ( r cosθ dθ )

(

)

cos 4 θ = cos 2 θ 1 − sin 2 θ = cos 2 θ −

Iy =

1 2 sin 2θ 4

2 4 π2  2 1  r ∫ π  cos θ − sin 2 2θ  dθ 3 4 6   π

2  θ sin 2θ  1  θ sin 4θ   2 = r 4  + −  −  3  2 4  4 2 8  π 6

Then

1 4  π 1 π   π 1 3 1 π 1 1 3  r  − ×  −  + × − × + × ×  3  2 4 2   6 2 2 4 6 4 4 2  

r4  π 3 3   −  4  3 8 

JP =

r4  π 3  r4  π 3 3   + +  −  43 8  4  3 8 

r4 8π − 3 3 48

(

) J P = 0.415r 4

Now

A = 2∫ r xdy 2

= ∫ r r 2 − y 2 dy 2

continued

COSMOS: Complete Online Solutions Manual Organization System

Let

y = r sin θ π

A = 2∫π2 ( r cosθ )( r cosθ dθ ) 6

2 θ

sin 2θ  2 = 2r  +  4 π 2 6

π π 1 3 = r 2  − − ×  2  2 6 2

= Have

kP =

r2 4π − 3 3 12

(

)

JP A r4 8π − 3 3 48 r2 4π − 3 3 12

( (

) ) k P = 0.822 r

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 25.

J O ≡ I 0 = ∫ r 2dA

(a) Have

dA =

Where

Then

R  3π J O = ∫R 2 r 2  1  2

3π rdr 2

 r  dr 

3π 4 2 3π 4 r = R2 − R14 8 8 R1

(

)

JO =

I x = 3 ( I x )1

so that

( )1 + ( I y )2 + ( I y )3 = 3 ( I y )1

Iy = Iy

( I x )1 = ( I y )1

Symmetry implies

Ix = I y

Then

)

( I x )1 = ( I x )2 = ( I x )3

By inspection

Now

(

I x = ( I x )1 + ( I x )2 + ( I x )3

(b) Now

Similarly,

3π 4 R2 − R14 8

JO = I x + I y Ix = I y =

JO 3π 4 = R2 − R14 2 16

(

)

or I x = I y =

3π 4 R2 − R14 16

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 26.

( (

3π 3π 2 rdr = R2 − R12 2 4

(

) ( )

Then

R2 = Rm + kO2 =

t 2

)(

(

1 ( R1 + R2 ) 2

Rm =

For

) )

3π 4 R2 − R14 R22 + R12 R22 − R12 J 1 2 2 O 8 kO = = = = R2 + R12 2 2 3π 2 A 2 2 R − R 2 2 1 R2 − R1 4

Now

And

(

A = ∫ dA = ∫R 2 1

And

Then

3π 4 R2 − R14 8

JO =

(a) From Problem 9.25

R1 = Rm −

and 2

kO2

(

t 2

2

1  = Rm2 + t 2 4 

Rm2 kO

(b) Have

)

t = R2 − R1

and

1  t t   Rm +  +  Rm −  2  2 2 

R1, R2

)

R − kO % error = m × 100% = kO

1− 1+

Rm2 +

Rm −

Rm2 +

1 t    4  Rm 

1 t  1+   4  Rm 

1 2 t 4 × 100%

1 2 t 4

× 100%

continued

COSMOS: Complete Online Solutions Manual Organization System

Then

t = 1: % error = Rm

1− 1+ 1+

1 4

1 4 × 100%

or % error = −10.56%

t 1 = : % error = Rm 4

1− 1+

11   44

11 1+   4 4

× 100%

or % error = −0.772%

1 t = : % error = Rm 16

1 1  1− 1+   4  16  1+

1 1    4  16 

× 100%

or % error = −0.0488%

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Chapter 9, Solution 27. π

Have:

a cos 2θ

A = 2∫ 04 ∫ 0

rdrdθ

= ∫ 04 a 2 cos 2 2θ dθ π

2 θ

sin 4θ  4 =a  +  8 0 2 = Have:

π 8

J O = ∫ r 2dA π

a cos 2θ

= 2∫ 04 ∫ 0

r 2 ( rdrdθ )

1 = ∫ 04 a 4 cos 4 2θ dθ 2 Now:

(

cos 4 2θ = cos 2 2θ 1 − sin 2 2θ

= cos 2 2θ −

)

1 2 sin 4θ 4

Then:

1 1   J O = a 4 ∫ 04  cos 2 2θ − sin 2 4θ  dθ 2 4   1  θ sin 4θ  1  θ sin 8θ = a 4  + −  − 2  2 8  4 2 16

 4   0

1 4π 1 π  a  − ×  4 4 4 4

JO =

3π 4 a 64 continued

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And:

kO =

JO A

3π 4 a 64

a2 kO =

a 6 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 28.

y =

By observation

b y 2h

 b dA = xdy =   2h

Now

dI x = y 2dA =

Then

I x = ∫ dI x = 2∫ 0

b 3 y dy 2h

1 3 bh 4

= 0

y =

From above Now

 y  dy 

b 3 y dy 2h

And

b y4 = h 4

h x b 2

2h x b

2h   dA = ( h − y ) dx =  h − x  dx b   =

h ( b − 2 x ) dx b

And

dI y = x 2dA = x 2

Then

I y = ∫ dI y = 2∫ 02

h ( b − 2 x ) dx b h 2 x ( b − 2 x ) dx b b

h 1 1 2 = 2  bx3 − x 4  b 3 2 0 3 4 h b  b  1b  1 3 bh =2    −    = b  3  2  2  2   48

continued

COSMOS: Complete Online Solutions Manual Organization System

Now

JO = I x + I y =

1 3 1 3 bh + bh 4 48

or J O =

And

bh 12h 2 + b 2 J 1 2 48 O kO = 12h 2 + b 2 = = 1 A 24 bh 2

(

)

(

bh 12h 2 + b 2 48

(

)

) or kO =

12h 2 + b 2 24

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 29.

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 =

b = r ∆θ

Then with

dJ O sector dθ

and

)

h=r

1 4  r ∆θ 12 + ∆θ 2    48

(

)

 ∆ J O sector  2  1 4 = lim   = lim  r 12 + ( ∆θ )   ∆θ → 0   ∆θ  ∆θ → 0  48 =

Then

(

1 ( r ∆θ )( r ) 12r 2 + ( r ∆θ )2  48

( ∆ J O )sector

Now

bh 12h 2 + b 2 48

1 4 r 4

( J O )circle

= ∫ dJ O sector =

2π 2π 4 4 ∫ 0 4 r dθ = 4 r [θ ]0

or ( J O )circle =

π 2

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Chapter 9, Solution 30.

From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is

( J C )cir

π 2

The area of the circle is

Acir = π r 2 So that

A2  J C ( A )  = cir 2π

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 2dA ; 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. 2π

where the equality applies when the original area is circular. continued

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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

( JC ) A

≥

A2 Q.E.D. 2π

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 31.

Area: A = 2 (10 mm )( 40 mm )  + ( 90 mm )(10 mm ) = 1700 mm 2 Part :

I x = I x + Ad 2 =

1 (10 mm )( 40 mm )3 + (10 mm )( 40 mm )( 25 mm )2 12

= 303.3 × 103 mm 4 1 ( 90 mm )(10 mm )3 = 7.50 × 103 mm 4 12

Part :

Ix = Ix =

Part :

(Same as Part ) I x = 303.3 × 103 mm 4

Thus for entire area:

I x = ( 303.3 + 7.50 + 303.3) × 103 = 614.2 × 103 mm 4

I x = 614 × 103 mm 4 k x2 =

Ix 614.2 × 103 mm 4 = = 361.27 mm 2 2 A 1700 mm k x = 19.01 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 32.

Have

A = A1 − A2 − A3

= (12 )( 8 ) − ( 5 )( 4 ) − ( 2 )( 6 )  in 2 = 64 in 2

Have

I x = ( I x )1 − ( I x )2 − ( I x )3

 1 3 3 3 1 1 =   (12 )( 8 )  −  ( 5 )( 4 )  −  ( 2 )( 6 )   in 4 12 12 12      

= ( 512 − 26.667 − 36 ) in 4 = 449.33 in 4 I x = 449 in 4 And

kx =

Ix A

449.33 in 4 64 in 2

= 2.65 in. k x = 2.65 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 33.

Area:

A = 2 (10 mm )( 40 mm )  + ( 90 mm )(10 mm ) = 1700 mm 2

Part : I y = I y + Ad 2 =

1 ( 40 mm )(10 mm )3 + ( 40 mm )(10 mm )( 40 mm )2 12

= 643.3 × 103 mm 4 Part : I y = I y =

1 (10 mm )( 90 mm )3 = 607.5 × 103 mm 4 12

Part : (Same as Part ) I y = 643.3 × 103 mm 4 Thus for entire area:

I y = ( 643.3 + 607.5 + 643.3) × 103 mm 4

= 1.894 × 106 mm 4 k y2 =

Iy A

I y = 1.894 × 106 mm 4

1.894 × 106 mm 4 = 1114.2 mm 2 2 1700 mm

k y = 33.4 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 34.

Have

A = A1 − A2 − A3

= (12 )( 8 ) − ( 5 )( 4 ) − ( 2 )( 6 )  in 2 = 64 in 2 Have

( )1 − ( I y )2 − ( I y )3

Iy = Iy

2   1  1 3 3 3 2  1   1 =   ( 8 )(12 )  −  ( 4 )( 5 ) + 20    −  ( 6 )( 2 ) + 12 ( 4 )   in 4  12  2    12     12

= (1152 ) − ( 41.667 + 5 ) − ( 4 + 192 )  in 4 = 909.33 in 4 And

ky =

I y = 909 in 4

Iy A

909.33 in 4 64 in 2 = 3.77 in.

k y = 3.77 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 35.

Have

I x = ( I x )1 + ( I x )2 + ( I x )3 3 3 1 1 =  ( 2a )( 4a )  +  ( a )( 3a )  3 3     2 2   π π  4a   π  4a   +   a 4 − a 2    + a 2  3a +   16 4  3π   4  3π     

4 9π 4  4  128 4   27 4   π a + a + a = − + +2+ 3 3 16 9 π 4 9 π        161 37π  4 4 = +  a = 60.9316a 16   3 or I x = 60.9a 4 Also

( )1 + ( I y )2 + ( I y )3

Iy = Iy

3 3 1 1 π  =  ( 4a )( 2a )  +  ( 3a )( a )  +  a 4  3  3  16 

π   32 = + 1 +  a 4 = 11.8630a 4 16   3 or I y = 11.86a 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 36.

Have

I x = ( I x )1 − ( I x )2 − ( I x )3 3 1 π  π  =  ( 3a )( 2a )  −  a 4  −  a 4  12   8  8 

π π π   =  2 − −  a4 =  2 −  a4 8 8 4   or I x = 1.215a 4 Also

( )1 − ( I y )2 − ( I y )3

Iy = Iy

2 1 3 a  =  ( 2a )( 3a ) + ( 3a )( 2a )     2   12 2 2   π π  4a   π  4a   −   a 4 − a 2    + a 2  2a −   2  3π   2  3π     8 

2 2   π π  4a   π  4a   −  a4 − a2    + a2  a −   2  3π   2  3π     8 

8 8 8  4 9 3 π =  +  a4 −  − + 2π − + a 3 9π  2 2  8 9π 8 π 4 8 π − − + − + 2 3 9π  8 9π

11π  4   a =  10 − 4  

 4 a 

= 1.3606a 4 or I y = 1.361a 4

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Chapter 9, Solution 37.

I AA′ = 2.2 × 106 mm 4 = I + A ( 25 mm ) I BB′ = 4 × 106 mm 4 = I + A ( 35 mm )

(

I BB′ − I AA′ = ( 4 − 2.2 ) × 106 = A 352 − 252

) 1.8 × 106 = A ( 600 )

Then

(

)

I AA′ = 2.2 × 106 mm 4 = I + 3000 mm 2 ( 25 mm )

A = 3000 mm 2

I = 325 × 103 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 38. Given: A = 6000 mm 2

(

)

I AA′ = 18 × 106 mm 4 = I + 6000 mm 2 ( 50 mm )

I = 3 × 106 mm 4

(

)

I BB′ = I + Ad 2 = 3 × 106 mm 4 + 6000 mm 2 ( 60 mm )

= 24.6 × 106 mm 4

I BB′ = 24.6 × 106 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 39. Have

J A = J C + Ad12

( = A ( 3a

d1 = 2a

J B = J C + A d 22 + a 2 Then

J A − JB

Substituting

)

− d 22

) (

)

2 256 in 4 − 190 in 4 = 24 in 2 3a 2 − ( 2 in.)   

a = 1.500 in. And

(

)

2 256 in 4 = J C + 24 in 2  4 (1.500 in.)   

J C = 40.0 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 40. J A = J C + Ad12

Have

(

J B = J C + A d 22 + a 2

)

J B = 3 J A; d1 = d 2 = 2.5 in.

(a) Then

(

)

a2 = 2

(

3 J C + Ad12 = J C + A d12 + a 2

)

JC + 2d12 A

 52.5 in 4 2 = 2 + 2.5 in. ( )  2  30 in 

a = 4.00 in. (b)

Have

(

)

2 2 J B = 52.5 in 4 + 30 in 2 ( 2.5 in ) + ( 4.00 in.)   

J B = 720 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 41. Determination of centroid: y = 0 by symmetry. Part 1 2 3

Σ x =

Area

( ) 2

3 ( 0.75 ) = 2.25

6 ( 0.75 ) = 4.50

3 ( 0.75 ) = 2.25 9.00

x ( in.) 1.5

( )

xA in 3 3.375

0.375

1.6875

1.5

3.375 8.4375

ΣxA 8.4375 in 3 = = 0.9375 in. ΣΑ 9.00 in 2

Determination of I x : Part : I x =

1 ( 3 in.)( 0.75 in.)3 + ( 3 in.)( 0.75 in.)( 3.375 in.)2 = 25.734 in 4 12 1 Part : I x = ( 0.75 in.)( 6 in.)3 = 13.50 in 4 12 Part : (Same as Part ) I x = 25.734 in 4

Entire Section: I x = ( 25.734 + 13.50 + 25.734 ) in 4

= 64.97 in 4

I x = 65.0 in 4

Determination of I y : Part : I y =

2 1 ( 0.75 in.)( 3 in.)3 + ( 0.75 in.)( 3 in.) (1.5 − 0.9375) in. 12

= 2.3994 in 4 Part : I y =

2 1 ( 6 in.)( 0.75 in.)3 + ( 6 in.)( 0.75 in.) ( 0.9375 − 0.375) in. 12

= 1.6348 in 4 Part : (Same as Part ) I y = 2.3994 in 4 Entire Section: I y = ( 2.3994 + 1.6348 + 2.3994 ) in 4

= 6.434 in 4

I y = 6.43 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 42. Locate centroid

y1 = 52.5 mm A1 = ( 84 mm )(105 mm )

= 8820 mm 2 y2 = 90 mm A2 = −

1 ( 42 mm )( 45 mm ) 2

= − 945 mm 2 y =

Then

Σyi Ai ΣAi

( 52.5 mm ) (8820 mm 2 ) + ( 90 mm ) ( − 945 mm 2 ) 8820 mm 2 − 945 mm 2

= 48.0 mm I x = ( I x )1 − ( I x )2

Have

3 2 1 =  ( 84 mm )(105 mm ) + 8820 mm 2 ( 52.5 mm − 48.0 mm )  12 

(

)

3 2 1 −  ( 42 mm )( 45 mm ) + 945 mm 2 ( 90.0 mm − 48.0 mm )   36 

(

)

= ( 8103375 + 178605 ) − (106312.5 + 1666980 )  mm 4

I x = 6.51 × 106 mm 4

( )1 − ( I y )2

And I y = I y

3 3 1 1 =  (105 mm )( 84 mm )  − 2  ( 45 mm )( 21 mm )  12  12 

= 5186160 − 2 ( 34728.75 )  mm 4 I y = 5.12 × 106 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 43. Locate centroid:

x1 = 8 in. y1 = 3 in. A1 = (16 )( 6 ) = 96 in 2 x2 = −1.5 in. y2 = 2 in. A2 = ( 3)( 4 ) = 12 in 2 x3 = 14 in. y3 = − 5.25 in. A3 = ( 4 )(10.5 ) = 42 in 2

x =

ΣxA ΣΑ

8 ( 96 ) − 1.5 (12 ) + 14 ( 42 )  in 3 = = 8.92 in. ( 96 + 12 + 42 ) in 2

y =

( ) ( ) ( )

ΣyA ΣA

3 ( 96 ) + 2 (12 ) − 5.25 ( 42 )  in 3 = = 0.610 in. ( 96 + 12 + 42 ) in 2

I x = I x1 + I x2 + I x3

3 2 3 2 1 1 =  (16 )( 6 ) + ( 96 )( 3 − 0.610 )  in 4 +  ( 3)( 4 ) + (12 )( 2 − 0.610 )  in 4 12 12     3 2 1 +  ( 4 )(10.5 ) + ( 42 )( − 5.25 − 0.610 )  in 4 12  

= ( 288 + 548.36 ) + (16 + 23.185 ) + ( 385.875 + 1442.26 )  in 4 = 2703.7 in 4

I x = 2700 in 4

( ) ( ) ( )

I y = I y1 + I y2 + I y3

3 2 3 2 1 1 =  ( 6 )(16 ) + ( 96 )( 8 − 8.92 )  in 4 +  ( 4 )( 3) + (12 )( −1.5 − 8.92 )  in 4 12 12     3 2 1 +  (10.5 )( 4 ) + ( 42 )(14 − 8.92 )  in 4 12 

= ( 2048 + 81.254 ) + ( 9 + 1302.92 ) + ( 56 + 1083.87 )  in 4 = 4581.0 in 4

I y = 4580 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 44. Locate centroid:

x1 = 20 mm y1 = 45 mm A1 = ( 40 mm )( 90 mm ) = 3600 mm 2

x2 = 50 mm y2 = 51 mm A2 =

1 ( 48 mm )( 30 mm ) 2

= 720 mm 2

Then

x = =

Σxi Ai ΣΑi

( 20 mm ) ( 3600 mm 2 ) + ( 50 mm ) ( 720 mm 2 ) 3600 mm 2 + 720 mm 2

= 25.0 mm And

y =

Σyi Ai ΣAi

( 45 mm ) ( 3600 mm 2 ) + ( 51 mm ) ( 720 mm 2 ) 3600 mm 2 + 720 mm 2

= 46.0 mm Now

I x = ( I x )1 + ( I x )2 3 2 1 =  ( 40 mm )( 90 mm ) + 3600 mm 2 (1 mm )  12  

(

)

1 3 2 1 +  ( 30 mm )( 24 mm ) + 720 mm 2 ( 59 mm − 46.0 mm )  2  36 

(

)

1 3 2 1 +  ( 30 mm )( 24 mm ) + 720 mm 2 ( 46.0 mm − 43 mm )  2  36 

(

) (

)

) (

)

=  2.430 × 106 + 3600 + 11.520 × 103 + 60.840 × 103 + 11.520 × 103 + 3240  mm 4  

I x = 2.52 × 106 mm 4 continued

COSMOS: Complete Online Solutions Manual Organization System

And

( )1 + ( I y )2

Iy = Iy

3 2 1 =  ( 90 mm )( 40 mm ) + 3600 mm 2 ( 5 mm )  12  

(

)

3 2 1 +  ( 48 mm )( 30 mm ) + 720 mm 2 ( 25 mm )   36 

(

) (

)

=  480 × 103 + 90 × 103 + 36 × 103 + 450 × 103  mm 4  

I y = 1.056 × 106 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 45.

Dimensions in mm Determination of centroid, C: Par

Area mm 2

y mm

yA mm3 170.67 × 103

1 (160 )(80 ) = 6400 2

80 3

−

1 (80 )( 60 ) = − 24 2

ΣyA ΣΑ 122.67 × 103 mm 3 = 4000 mm 2 = 30.667 mm

y =

− 48.0 × 103

122.67 × 103

4000

(a) Polar moment of inertia with respect to point O, J O : Part :

Ix =

1 (160 mm )(80 mm )3 = 6.8267 × 106 mm 4 12

3 1 I y = 2  ( 80 mm )( 80 mm )  = 6.8267 × 106 mm 4 12  

J O = I x + I y = ( 6.8267 + 6.8267 ) × 106 = 13.653 × 106 mm 4 Part :

Ix =

1 (80 mm )( 60 mm )3 = 1.440 × 106 mm 4 12

3 1 I y = 2  ( 60 mm )( 40 mm )  = 0.640 × 106 mm 4 12  

J O = I x + I y = (1.440 + 0.640 ) × 106 = 2.080 × 106 mm 4

continued

COSMOS: Complete Online Solutions Manual Organization System

Entire Section:

( ) ( )

J O = J O1 − J O2 = 13.653 × 106 − 2.080 × 106  mm 4 = 11.573 × 106 mm 4

J O = 11.57 × 106 mm 4

(b) Polar moment of inertia with respect to centroid, J C :

J O = J C + Ay 2

(

)

11.573 × 106 mm 4 = J C + 4000 mm 2 ( 30.667 mm )

J C = 7.8116 × 106 mm 4

J C = 7.81 × 106 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 46. Locate centroid: By symmetry, y = 0

A1 = π (100 mm )( 60 mm ) = 6000π mm 2

x1 = 0

x2 =

4 ( 45 mm ) 3π

A2 = − x3 =

All dimensions in mm

π 2

3π

π 2

( 45mm )2 = −1012.5π

− 4 ( 30 mm )

A3 = −

=−

mm 2

( 30 mm )2 = − 450π

mm 2

60 40  0 ( 6000π ) + ( −1012.5π ) − ( − 450π ) mm3 ΣxA  π π  x= = = − 2.9990 mm ΣA ( 6000π − 1012.5π − 450π ) mm 2 (a)

Have

J O = ( J O )1 − ( J O )2 − ( J O )3 4 4 1  1 1 =  π (100 )( 60 ) 1002 + 602  mm 4 −  π ( 45 )  mm 4 −  π ( 30 )  mm 4 4  4  4 

(

)

(

)

= π 20 400 000 − 1 025 156 − 202 500 mm 4 = 60232000 mm 4

(b)

Have With

J O = 60.2 × 106 mm 4

J C = 60.1 × 106 mm 4

J O = J C + Ax 2 A = ( 6000π − 1012.5π − 450π ) mm 2

= 4537.5π mm 2 Then

(

)

J C = 60 232 000 mm 4 − 4537.5π mm 2 ( − 2.9990 mm )

= 60 232 000 mm 4 − 128209 mm 4 = 60 104 000 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 47. Locate centroid:

(a)

Have

x1 = 4 in.

y1 = −1.15 in.

A1 = (16 )( 2.3) = 36.8 in 2

x2 = 4 in.

y2 = 3.2 in.

A2 =

x3 = 2 in.

y3 = 1.6 in.

1 (12 )( 9.6 ) = 57.6 in 2 2 1 A3 = ( 6 )( 4.8 ) = −14.4 in 2 2

3 ΣxA  4 ( 36.8 ) + 4 ( 57.6 ) + 2 ( −14.4 )  in 348.8 in 3 = = − = 4.360 in. ΣA 80 in 2 ( 36.8 + 57.6 − 14.4 ) in 2

3 ΣyA  −1.15 ( 36.8 ) + 3.2 ( 57.6 ) + 1.6 ( −14.4 )  in 118.96 in 3 = =− = 1.487 in. 2 ΣA 80 in 80 in 2

I x = ( I x )1 + ( I x )2 − ( I x )3 =

1 1 1 (16 )( 2.3)3 in 4 + (12 )( 9.6 )3 in 4 − ( 6 )( 4.8)3 in 4 3 12 12

= 64.891 in 4 + 884.736 in 4 − 55.296 in 4 = 894.33 in 4

( )1 + ( I y )2 − ( I y )3

Iy = Iy

1 1 3 2 3 3 1 =  ( 2.3)(16 ) + 2.3 (16 )( 4 )  in 4 + ( 9.6 )(12 ) in 4 − ( 4.8 )( 6 ) in 4 12 12 12  = 1373.87 in 4 + 1382.4 in 4 − 86.4 in 4 = 2669.9 in 4 Now

J O = I x + I y = 894.33 in 4 + 2669.9 in 4 = 3564.2 in 4

(b)

J O = 3560 in 4

Have

J O = J C + Ad 2

Then

2 2 J C = 3564.2 in 4 − 80 in 2 ( 4.360 in.) + (1.487 in.)   

where

(

d2 = x2 + y2

)

= 3564.2 in 4 − 1697.66 in 4 = 1866.54 in 4

J C = 1867 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 48.

Locate centroid:

4 (12 in.)

y1 =

3π

A1 =

y2 = 4 in.

(12 in )2

= 72π in 2

in.

A2 = − (12 in.)( 8 in.) = − 96 in 2

Then

y =

Σ yi Ai ΣΑi

 16  2 2  in.  72π in + ( 4 in.) − 96 in π   = 72π in 2 − 96 in 2

(

)

(

)

= 5.8989 in. (a)

Have

JO = I x + I y

Where

I x = ( I x )1 − ( I x )2

1 4 3 π =  (12 in.) − (12 in.)( 8 in.)  3 8  = 6095.0 in 4 And

( )1 − ( I y )2

Iy = Iy

1 4 3 π =  (12 in.) − ( 8 in.)(12 in.)  12 8  = 6991.0 in 4

Then

J O = ( 6095.0 + 6991.0 ) in 4

= 13086.0 in 4 (b)

Have

J O = 13.09 × 103 in 4

J O = J C + Ay 2 2 13086.0 in 4 = J C + ( 72π − 96 ) in 2  ( 5.8989 in.)

J C = 8.555 × 103 in 4 J C = 8.56 × 103 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 49. From Fig. 9.13A

Area = A = 4.75 in 2 I x′ = 17.4 in 4 I y′ = 6.27 in 4

(

)

2 I x = 2  I x′ + Ad 2  = 2 17.4 in 4 + 4.75 in 2 ( 3.00 − 1.990 ) in 2   

= 2 17.4 in 4 + 4.8455 in 4  = 44.491 in 4

I x = 44.5 in 4 k x2 =

Ix 44.491 in 4 = = 4.683 in 2 A 2 4.75 in 2

(

k x = 2.16 in.

)

(

)

2 I y = 2  I y′ + Ad 2  = 2 6.27 in 4 + 4.75 in 2 ( 2.25 − 0.987 ) in 2   

= 2 6.27 in 4 + 7.5771 in 4  = 27.694 in 4

I y = 27.7 in 4 k y2 =

Iy A

27.694 in 4

(

2 4.75 in 2

)

= 2.9152 in 2

k y = 1.707 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 50.

Data for C 250 × 30:

A = 3780 mm 2 I x = 32.6 × 106 mm 4 I y = 1.14 × 106 mm 4

Dimensions in mm

A = 2 3780 mm 2 + (10 mm )( 375 mm )  = 15.06 × 103 mm 2

Total Area,

3 2 1 I x = 2 32.6 × 106 mm 4  + 2  ( 375 mm )(10 mm ) + ( 375 mm )(10 mm )(132 mm )  12  

= 65.2 × 106 mm 4 + 130.74 × 106 mm 4 = 195.94 × 106 mm 4 k x2 =

I x 195.94 × 106 mm 4 = = 13.01 × 103 mm 2 3 2 A 15.06 × 10 mm

I x = 195.9 × 106 mm 4

k x = 114.0 mm

2 3 1 I y = 2 1.14 × 106 mm 4 + 3780 mm 2 (115.3 mm )  + 2  (10 mm )( 375 mm )    12  

(

)

= 102.783 × 106 mm 4 + 87.891 × 106 mm 4 = 190.674 × 106 mm 4

k y2 =

Iy A

190.674 × 106 mm 4 = 12.661 × 103 mm 2 15.06 × 103 mm 2

I y = 190.7 × 106 mm 4

k y = 112.5 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 51.

Shape Data:

Fig. 9.13A

A = 10.3 in 2

S10 × 35:

A = 5.88 in 2

C10 × 20:

I x = 147 in 4

I x = 78.9 in 4

I y = 8.36 in 4

I y = 2.81 in 4

Combined section:

(

)

A = AS + 2 AC = 10.3 in 2 + 2 5.88 in 2 = 22.06 in 2

(

)

I x = ( I x )S + 2 ( I x )C = 147 in 4 + 2 78.9 in 4 = 304.8 in 4 or

I x = 305 in 4

( )S + 2 ( I y )C + ACd 2 

Iy = Iy

2    4 2  4.944 = 8.36 in + 2  2.81 in + 5.88 in  in. + 2.739 in. − 0.606 in.    2   

(

)

= ( 8.36 + 5.62 + 249.38 ) in 4 = 263.36 in 4

kx = ky =

I y = 263 in 4

Ix = A

304.8 in 4 22.06 in 2

k x = 3.72 in.

263.36 in 4 22.06 in 2

k y = 3.46 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 52.

A = 3780 mm 2

Channel:

I x = 32.6 × 106 mm 4 I y = 1.14 × 106 mm 4

( )C + ( I x )plate

Ix = 2 Ix

Now

(

)

= 2 32.6 × 106 mm 4 +

(

d ( 300 mm )3 12

)

= 65.2 × 106 + 2.25d × 106 mm 4 And

( )channel + ( I y )plate

Iy = 2 Iy

2 3  d   ( 300 mm ) d = 2 1.14 × 106 mm 4 + 3780 mm 2  + 15.3 mm   + 12 2   

(

)

(

)

=  2.28 × 106 + 1890d 2 + 115.668 × 103 d + 1.7697 × 106 + 25d 3  mm 4  

(

)

= 25d 3 + 1890d 2 + 115.67 × 103 d + 4.0497 × 106 mm 4 I x = 16I y

Given Then

65.2 × 106 + 2.25d × 106

(

= 16 25d 3 + 1890d 2 + 115.67 × 103 d + 4.0497 × 106 or Solving numerically

)

25d 3 + 1890d 2 − 24955d − 25300 = 0

d = 12.2935 mm or d = 12.29 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 53.

Locate centroid:

x1 = 15.3 mm y1 = 127 mm

A1 = 3780 mm 2

x2 = − 21.4 mm y2 = 76 mm − 21.4 mm

A2 = 932 mm 2

= 54.6 mm x3 = − 21.4 mm y3 = 76 mm + 21.4 mm

A3 = 932 mm 2

= 97.4 mm Then

x =

Σxi Ai ΣΑi

(15.3 mm ) ( 3780 mm 2 ) + 2 ( − 21.4 mm ) ( 932 mm 2 )

(

3780 mm 2 + 2 932 mm 2

)

= 3.1794 mm And

Σyi Ai ΣΑi

(127 mm ) ( 3780 mm 2 ) + ( 76 mm ) ( 2 × 932 mm 2 )

(

3780 mm 2 + 2 932 mm 2

)

= 110.157 mm Now

I x = ( I x )1 + ( I x )2 + ( I x )3

(

)

2 = 32.6 × 106 mm 4 + 3780 mm 2 (127 mm − 110.157 mm )   

( ) + ( 932 mm ) (110.157 mm − 97.4 mm )  

2 + 0.517 × 106 mm 4 + 932 mm 2 (110.157 mm − 54.6 mm )   

+ 0.517 × 106 mm 4 

continued

COSMOS: Complete Online Solutions Manual Organization System

(

) ( + 0.151675 × 10 )  mm 

=  32.6 × 106 + 1.07234 × 106 + 0.517 × 106 + 2.8767 × 106 

(

+ 0.517 × 106

)

I x = 37.7 × 106 mm 4 And

( )1 + ( I y )2 + ( I y )3

Iy = Iy

(

)

2 = 1.14 × 106 mm 4 + 3780 mm 2 (15.3 mm − 3.1794 mm )   

(

)

2 + 2 0.517 × 106 mm 4 + 932 mm 4 ( 3.1794 mm + 21.4 mm )   

(

) (

)

=  1.14 × 106 + 0.55532 × 106 + 2 0.517 × 106 + 0.56306 × 106  mm 4  

I y = 3.86 × 106 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 54.

Angle:

L3 × 3 ×

1 : 4

A = 1.44 in 2 L6 × 4 ×

1 : 2

A = 4.75 in 2 Plate:

I x = I y = 1.24 in 4

I x = 6.27 in 4

I y = 17.4 in 4

A = ( 27 in.)( 0.8 in.) = 21.6 in 2

Ix =

1 ( 0.8 in.)( 27 in.)3 = 1312.2 in 4 12

Iy =

1 ( 27 in.)( 0.8 in.)3 = 1.152 in 4 12 continued

COSMOS: Complete Online Solutions Manual Organization System

X =0

Centroid:

Y =

(

)

(

)

2  1.44 in 2 ( 27 in. − 0.84 in.)  + 2  4.75 in 2 ( 0.987 in.)  + 21.6 in 2 (13.5 in.)    Y =  2 1.44 in 2 + 4.75 in 2 + 21.6 in 2

= Now

)

ΣAy ΣA

(

)

376.31 in 3 = 11.0745 in. 33.98 in 2

I x = 2 ( I x )1 + 2 ( I x )3 + ( I x )2 2 2 = 2 6.27 + 4.75 (11.075 − 0.987 )  in 4 + 2 1.24 + 1.44 ( 27 − 0.842 − 11.075 )  in 4     2 + 1312.2 + 21.6 (13.5 − 11.075 )  in 4  

= 2 ( 489.67 ) in 4 + 2 ( 328.84 ) in 4 + 1439.22 in 4 = 3076.24 in 4

or I x = 3076 in 4 Also

( I y ) = 2 ( I y )1 + 2 ( I y )3 + ( I y )2 2 2 = 2 17.4 + 4.75 ( 0.4 + 1.99 )  in 4 + 2 1.24 + 1.44 ( 0.4 + 0.842 )  in 4 + 1.152 in 4    

= 2 ( 44.532 ) in 4 + 2 ( 3.4613) in 4 + 1.152 in 4 = 97.139 in 4 or I y = 97.1 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 55.

A = 2420 mm 2

Angle:

I x = 3.93 × 106 mm 4

I y = 1.06 × 106 mm 4 A = ( 200 mm )(10 mm ) = 2000 mm 2

Plate:

Ix =

1 ( 200 mm )(10 mm )3 = 0.01667 × 106 mm 4 12

Iy =

1 (10 mm )( 200 mm )3 = 6.6667 × 106 mm 4 12

X =0

Centroid

Y = or

Y =

(

)

ΣAy ΣA

2 2420 mm 2 ( 44.4 mm ) + 2000 mm 2 ( −5 mm )  2 ( 2420 ) + 2000  mm 2

204896 mm 3 6840 mm 2

= 29.9556 mm Now

I x = 2 ( I x )angle + ( I x )plate 2 = 2 3.93 × 106 + ( 2420 )( 44.4 − 29.9556 )  mm 4   2 + 0.01667 × 106 + ( 2000 )( 29.9556 + 5 )  mm 4  

(

)

= 2 4.4349 × 106 mm 4 + 2.4605 × 106 mm 4 = 11.3303 × 106 mm 4 or I x = 11.33 × 106 mm 4

continued

COSMOS: Complete Online Solutions Manual Organization System

Also Where

( )angle + ( I y )plate

Iy = 2 Iy

( I y )angle = 1.06 × 106 mm 4 + ( 2420 mm 2 ) ( b − 19.0 mm )2

(

)

= 1.06 × 106 + ( 2420 ) b 2 − 38b + 361  mm 4  

=  2420b 2 − 91960b + 1.93362 × 106  mm 4 and Now Then or

( I y )plate = 6.6667 × 106 mm 4 ( )

I y = 3 Ix

(

)

2  2420b 2 − 91960b + 1.93362 × 106  mm 4 + 6.6667 × 106 mm 4 = 3  11.33 × 106 mm 4    2420b 2 − 91960b + 1.93362 × 106 − 13.662 × 106 = 0 b 2 − 38.0b − 4846.5 = 0

b = 91.16 mm or b = 91.2 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 56.

(a)

Using shape data from Fig. 9.13A

xA = 1.78 in.

AA = 8.44 in 2

xC = − 0.499 in.

AC = 3.09 in 2

a  xP =  6 −  in. 2 

AP = ( 0.75a ) in 2

From the condition

ΣxA =0 ΣΑ

ΣxA = 0



(1.78 in.) (8.44 in 2 ) − ( 0.499 in.) ( 3.09 in 2 ) +  6 − 

a 2 − 12a − 35.950 = 0

a  2  in. 0.75a in = 0 2 

(

)

a = 14.4823 in. a = 14.48 in.

or and (b)

AP = ( 0.75 in.)(14.4823 in.) = 10.8617 in 2

Locate centroid

ΣyA = ΣΑ

6  in.  3.09 in 2 − ( 0.375 in.) 10.8617 in 2 2  (8.44 + 3.09 + 10.8617 ) in 2

(1.78 in.) (8.44 in 2 ) + 

(

)

(

)

= 0.90302 in.

continued

COSMOS: Complete Online Solutions Manual Organization System

I x = ( I x ) A + ( I x )C + ( I x ) P

Now

(

)

2 =  28.2 in 4 + 8.44 in 2 (1.78 in. − 0.90302 in.)   

(

)

2 + 15.2 in 4 + 3.09 in 2 ( 3.0 in. − 0.90302 in.)   

3 2 1 +  (14.4823 in.)( 0.75 in.) + 10.8617 in 2 ( 0.375 in. + 0.90302 in.)  12 

(

)

= ( 28.2 + 6.4912 ) + (15.2 + 13.5877 ) + ( 0.5091 + 17.7408 )  in 4

= 81.729 in 4 I x = 81.7 in 4

or and

( ) A + ( I y )C + ( I y ) P

Iy = Iy

(

)

(

)

2 2 =  28.2 in 4 + 8.44 in 2 (1.78 in.)  + 0.866 in 4 + 3.09 in 2 ( 0.499 in.)      2 1 14.4823 in.   3  +  ( 0.75 in.)(14.4823 in.) + 10.8617 in 2  6 in. −   2    12

(

)

= ( 28.2 + 26.741) + ( 0.866 + 0.7694 ) + (189.842 + 16.7319 )  in 4 = 263.15 in 4 or

I y = 263 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 57.

R = γ ∫ ydA

From Sec. 9.2:

M AA′ = γ ∫ y 2dA Let yP = Distance of center of pressure from AA′. We must have

RyP = M AA ' :

yP =

2 M AA′ γ ∫ y dA I AA′ = = γ ∫ ydA R yA

For a triangular panel:

I AA′ =

Thus

1 3 ah 12

1 h 3

1 ah 2

1 3 ah 1 12 yP = = h  1  1  2  h  ah   3  2  yP =

1 h 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 58.

From Sec. 9.2:

R = γ ∫ ydA M AA′ = γ ∫ y 2dA

Let yP = Distance of center of pressure from AA′. We must have RyP = M AA ' :

For a semiellipse

yP = I AA′ = y =

2 M AA′ γ ∫ y dA I AA′ = = R yA γ ∫ ydA

π 8

ab3

4b π , A = ab 3π 2

π Then

yP =

ab3

8  4b  π    ab   3π  2  or yP =

3π b 16

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Chapter 9, Solution 59.

R = γ ∫ ydA

From Sec. 9.2:

M AA′ = γ ∫ y 2dA Let yP = Distance of center of pressure from AA′ : We must have RyP = M AA ' :

yP =

2 M AA′ γ ∫ y dA I AA′ = = R yA γ ∫ ydA

Divide Trapezoid as shown:

I AA′ =

1 1 1 1 ( a − b ) h3 + bh3 = ah3 + bh3 12 3 12 4

yA = y1 A1 + y2 A2 =

1 1 1 1  1 h  ( a − b ) h  + h ( bh ) = ah 2 + bh 2 3 2 6 3  2

1 3 1 3 I AA′ 12 ah + 4 bh yP = = 1 2 1 2 yA ah + bh 6 3

yP =

a + 3b h 2a + 4b

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Chapter 9, Solution 60.

From Sec. 9.2:

R = γ ∫ ydA M AA′ = γ ∫ y 2dA

Let yP = Distance of center of pressure from AA′. We must have

RyP = M AA ' : where

yP =

2 M AA′ γ ∫ y dA I AA′ = = R yA γ ∫ ydA

I AA′ = ( I AA′ )1 + ( I AA′ )2 2 2   π π  4r   π 2  4r   3 1 r r =  ( 2r )( r )  +   r 4 − r 2  + +      2  3π   2  3π   3    8 

= And

2 4 π 8 π 4 9 r + − + + + 3 2 3 8π  8 9π

5π  4  4  r = 2 + r 8   

4r  π 2   r   YA = ΣyA =  ( 2r × r )  +  r +  r   3π  2   2  

π 2  5 π  = 1 + +  r 3 =  +  r 3 2 3  3 2

Then

5π  4  2 + r 8  yP =  = 1.2242r 5 π  3  + r 3 2

or yP = 1.224r

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Chapter 9, Solution 61.

Using equation developed on page 491 of text:

yP =

R = 920

Then

Ix I = AA′ yA yA

R = γ yA = ρ gyA

kg m × 9.81 2 × 3 m × ( 0.55 × 0.25 ) m 2 3 m s

= 3722.9 N I AA′ =

and

1 ( 0.55 m )( 0.25 m )3 + ( 0.55 m )( 0.25 m )( 3 m )2 12

= 1.238216 m 4

FBD Cover:

and

yA = ( 3 m )( 0.55 m )( 0.25 m ) = 0.4125 m 3

Then

yP =

Symmetry implies

1.238216 m 4 = 3.001736 m 0.4125 m 3 FA = FB

and

FC = FD .

Equilibrium:

ΣM CD = 0:

− ( 0.125 − 0.001736 ) m × ( 3722.9 N ) + ( 0.25 m )( 2FA ) = 0 FA = 917.80 N

ΣFx = 0:

or FA = FB = 918 N

− 2 ( 917.80 N ) + 3722.9 N − 2 FC = 0 FC = 943.65 N

or FC = FD = 944 N

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 62. Using equations developed on page 491 of text: yP =

Now

Ix I = ss′ yA yA

R = γ yA

yA = Σ yA 1  = ( h + 17 in.)  × 120 in. × 51 in.  2  1  + ( h + 34 in.)  × 84 in. × 51 in.  2  

d = ( h + 79 in.)

= ( 5202h + 124848 ) in 3 = ( 36.125h + 72.25 ) ft 3 Then,

(

)

R = γ yA = 62.4 lb/ft 3 ( 36.125h + 72.75 ) ft 3

= 2254.2 ( h + 2 ) lb Also,

I ss′ = ( I ss′ )1 + ( I ss′ )2 2  1  120  51 3  1  120  51    17  2  =  ft  ft  +   ft  ft    h +  ft  12   2  12  12     36  12  12   2  1  84  51 3  1  84  51    34  2  +  ft  ft  +   ft  ft    h +  ft  12   2  12  12     36  12  12  

 34 289  1156   4   2 68 h+ h+ =  21.324 + 21.25  h 2 +  + 14.9266 + 14.875  h +   ft 12 144  12 144     

(

)

= 36.125h 2 + 144.5h + 198.311 ft 4

continued

COSMOS: Complete Online Solutions Manual Organization System

(

Then

= FBD of Gate:

)

36.125h 2 + 144.5h + 198.311 ft 4 I ss′ yP = = yA ( 36.125h + 72.25 ) ft 3

h 2 + 4h + 5.4896 ft h+2

For gate to open:

ΣM AB = 0: − M open + ( yP − h ) R = 0

 h 2 + 4h + 5.4896   2 ( 8000 lb ⋅ ft ) −  − h  ft   2254.2 ( h + 2 ) lb  = 0 h+2    or Thus

2h − 1.60826 = 0

d =h+

h = 0.80413 ft

79 79   ft =  0.80413 +  ft 12 12  

= 7.3875 ft

or d = 7.39 ft

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 63.

x dV x = ∫ EL ∫ dV

Have

dV = ydA

where

and

xEL = x

60 1 x= x 300 5

Now

y =

Then

1    x = 1  ∫  5 x  dA  

∫ x  5 x  dA

∫ x dA = ( I z ) A ∫ xdA ( xA) A 2

where ( I z ) A is the moment of inertia of the area with respect to the z axis, and x is analogous to y p Now

( Iz )A

1 1 ( 240 mm )( 300 mm )3 + ( 240 mm )( 300 mm ) ( 200 mm )2 36 2

= 1.620 × 109 mm 4 and

Then

1  xA = ( 200 mm )  ( 240 mm )( 300 mm )  = 7.20 × 106 mm 3 2  x =

1.620 × 109 mm 4 7.20 × 106 mm 3 or x = 225 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 64.

xv of the volume is defined by

xvV = ∫ xEL dV

Selecting the element of volume shown

dV = ydA = kxdA V = k ∫ xdA = kx A A Where x A = coordinate of the centroid of area A

xvV = ∫ xEL dV = ∫ x ( kx dA ) = k ∫ x 2dA = kI z Where

I z = moment of inertia of area with respect to z-axis.

Thus

xv =

kI z I = z which is the same as for center of pressure. kx A A xA A

For circular area:

I z = I z′ + Aa 2 =

1 4 5 π a + π a2 a2 = π a4 4 4

( )

xA = a A = π a2

Thus

5 4 πa Iz = 4 2 xv = xA A a π a

( )

xv =

5 a 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 65. The pressure p at an arbitrary depth ( y sin θ ) is

p = γ ( y sin θ ) so that the hydrostatic force dF exerted on an infinitesimal area dA is

dF = (γ y sin θ ) dA Equivalence of the force P and the system of infinitesimal forces dF requires

ΣF: P = ∫ dF = ∫ γ y sin θ dA = γ sin θ ∫ ydA or P = γ Ay sin θ Equivalence of the force and couple

( P, M x′ + M y′ ) and

the system of infinitesimal hydrostatic forces

requires ΣM x: − yP − M x′ = Now

∫ ( − ydF )

−∫ ydF = −∫ y (γ y sin θ ) dA = − γ sin θ ∫ y 2dA = − (γ sin θ ) I x

Then or

− yP − M x′ = − (γ sin θ ) I x

M x′ = (γ sin θ ) I x − y (γ Ay sin θ )

(

= γ sin θ I x − Ay 2

) or M x′ = γ I x′ sin θ

ΣM y : xP + M y′ = ∫ xdF Now

∫ xdF = ∫ x (γ y sin θ ) dA = γ sin θ ∫ xydA = (γ sin θ ) I xy ( Equation 9.12 ) continued

COSMOS: Complete Online Solutions Manual Organization System

Then or

xP + M y′ = (γ sin θ ) I xy M y′ = (γ sin θ ) I xy − x (γ Ay sin θ )

(

= γ sin θ I xy − Ax y

)

or, using Equation 9.13,

M y′ = γ I x′y′ sin θ

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 66. The pressure p at an arbitrary depth ( y sin θ ) is

p = γ ( y sin θ ) so that the hydrostatic force dP exerted on an infinitesimal area dA is

dP = (γ y sin θ ) dA The magnitude P of the resultant force acting on the plane area is then

P = ∫ dP = ∫ γ y sin θ dA = γ sin θ ∫ ydA = γ sin θ ( yA) Now

p = γ y sin θ

∴ P = pA

Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then requires

ΣM x: − yP P = −∫ ydP Now

2 ∫ ydP = ∫ y (γ y sin θ ) dA = γ sin θ ∫ y dA

= ( γ sin θ ) I x Then or

yP P = (γ sin θ ) I x yP =

(γ sin θ ) I x γ sin θ ( yA ) or yP =

ΣM y : xP P = ∫ xdP Now

Ix Ay

∫ xdP = ∫ x (γ y sin θ ) dA = γ sin θ ∫ xydA = (γ sin θ ) I xy ( Equation 9.12 )

continued

COSMOS: Complete Online Solutions Manual Organization System

Then

xP P = (γ sin θ ) I xy

xP =

(γ sin θ ) I xy γ sin θ ( yA ) or xP =

Now

I x = I x′ + Ay 2

From above

I x = ( Ay ) yP

By definition

I x′ = k x2′ A

Substituting

( Ay ) yP

I xy Ay

= k x2′ A + 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.

∴ ( yP − y ) = f ( depth )

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Chapter 9, Solution 67.

y = a 1−

First note

1 4a 2 − x 2 2

dI xy = dI x′y′ + xEL yEL dA

Have

dI x′y′ = 0

where

yEL =

( symmetry )

xEL = x

1 1 y = 4a 2 − x 2 2 4

dA = ydx = Then

x2 4a 2

1 4a 2 − x 2 dx 2

2a  1  1  4a 2 − x 2  4a 2 − x 2  dx I xy = ∫ dI xy = ∫ 0 x  4  2  2a

1 2a 1 1  = ∫ 0 4a 2 x − x3 dx =  2a 2 x 2 − x 4  8 8 4 0

(

a4 8

)

1 2 4  2 ( 2 ) − 4 ( 2 )    or I xy =

1 4 a 2

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Chapter 9, Solution 68. First note:

At x = a : or k =

Now

b = ka3

b b ; y = 3 x3 3 a a

I xy = ∫ xydA a b b 3 x a3

= ∫0 ∫

xydydx

1 a  2 b2 6  ∫ x  b − a 6 x  dx 2 0   a

b2  1 2 1 8 =  x − 6x  2 2 8a 0 =

3 2 2 ab 16

I xy =

3 2 2 ab 16

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Chapter 9, Solution 69.

Have

I xy = ∫ xydA =

∫ − b ∫ h x xydydx b

1 0  h2 2  ∫ x  − x  dx 2 − b  b 2  0

=− =

1 h2 4 x 8 b 2 −b

1 2 2 b h 8 I xy =

1 2 2 b h 8

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 70. a

At x = a : b = ke a

First note:

or k = Now

b b x ; y = ea e e

I xy = ∫ xydA b ax a ee 0 0

=∫ ∫ =

xydydx

1 b 2 a 2ax ∫ xe dx 2 e2 0 a

  2x  2  a 1b  e 2  =  x − 1  2 e2   2 2  a      0   a  =

1 b2  a 2   2   e (1) − (1)( −1)  2 e2  4  

a 2b 2 2 e +1 8e 2

(

) I xy = 0.1419a 2b 2

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Chapter 9, Solution 71.

I xy = I xy

Where

( I xy )1 = 0

and Now

x2 = −15 mm

( )1 − ( I xy )2 − ( I xy )3

Have

y2 = 10 mm

I xy = I x′y′ + A x y for areas and A2 = ( 50 mm )( 20 mm ) = 1000 mm 2

x3 = 15 mm

y3 = −10 mm A3 = ( 50 mm )( 20 mm ) = 1000 mm 2

Then

(

)

I xy = − 1000 mm 2 ( −15 mm )(10 mm )

− (1000 mm )(15 mm )( −10 mm ) = 300 × 103 mm 4 I xy = 300 × 103 mm 4

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Chapter 9, Solution 72.

Note: Orientation of A3 corresponding to a 180° rotation of the axes. Equation 9.20 then yields

I x′y′ = I xy •

Symmetry implies Using Sample Problem 9.6 and Similarly, and

( I xy )1 = 0 ( I x′y′ )2 = − 721 ( 9 in.)2 ( 4.5 in.)2 = − 22.78125 in 4

X 2 = 9 in.

Y2 = 1.5 in.

A2 =

1 ( 9 in.)( 4.5 in.) = 20.25 in 2 2

( I x′y′ )3 = − 721 ( 9 in.)2 ( 4.5 in.)2 = − 22.78125 in 4 X 3 = − 9 in.

Y2 = −1.5 in.

A3 =

( )1 + ( I xy )2 + ( I xy )3

1 ( 9 in.)( 4.5 in.) = 20.25 in 2 2

Then

I xy = I xy

and

I xy = I x′y′ + x y A

Therefore,

I xy = 2  − 22.78125 + ( 9 )(1.5 )( 20.25 )  in 4

with

( I xy )2 = ( I xy )3

= 501.1875 in 4

or I xy = 501 in 4

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Chapter 9, Solution 73.

( )1 + ( I xy )2

Have

I xy = I xy

For each semicircle

I xy = I x′y′ + x y A

Thus

I xy = Σx y A

A, mm 2 1

π 2

x , mm

(120 )2

= 7200π

− 60

(120 )2

= 7200π

I x′y′ = 0 (symmetry)

and

y , mm

−

160

Ax y , mm 4 69.12 × 106

69.12 × 106 138.24 × 106

or I xy = 138.2 × 106 mm 4

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Chapter 9, Solution 74.

( )1 + ( I xy )2

Have

I xy = I xy

For each rectangle

I xy = I x′y′ + Ax y

Thus

I xy = Σ x y A

and

I x′y′ = 0 (symmetry)

A, mm 2

x , mm

y , mm

Ax y , mm 4

76 ( 6.4 ) = 486.4

−12.9

9.4

− 58 980.86

44.6 ( 6.4 ) = 285.44

21.9

−16.1

−100 643.29

−159 624.15 or I xy = − 0.1596 × 106 mm 4

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Chapter 9, Solution 75.

Have Now symmetry implies

( )1 + ( I xy )2 + ( I xy )3

I xy = I xy

( I xy )1 = 0

and for the other rectangles

I xy = I x′y′ + x y A

Thus

I xy = ( x y A )2 + ( x y A)3

where

I x′y′ = 0 (symmetry)

= ( − 69 mm )( − 25 mm ) (12 mm )( 38 mm )  + ( 69 mm )( 25 mm ) (12 mm )( 38 mm )  = ( 786 600 + 786 600 ) mm 4 = 1 573 200 mm 4

or I xy = 1.573 × 106 mm 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 76.

( I xy )1 = 0

Symmetry implies

Using Sample Problem 9.6 and Equation 9.20, note that the orientation of A2 corresponds to a 1 2 2 90° rotation of the axes; thus I x′y′ = b h 2 72

( )

( I x′y′ )3 = 721 b2h2

Also, the orientation of A3 corresponds to a 270° rotation of the axes; thus

( I x′y′ )2 = 721 ( 9 in.)2 ( 6 in.)2 = 40.5 in 4

Then and

x2 = − 6 in.,

y2 = 2 in.,

x3 = 6 in.,

and

Then

1 ( 9 in.)( 6 in.) = 27 in 2 2

( I x′y′ )3 = ( I x′y′ )2 = 40.5 in 4

Also

Now

A2 =

y3 = − 2 in.,

( )1 − ( I xy )2 − ( I xy )3

I xy = I xy

and

A3 = A2 = 27 in 2 I xy = I x′y′ + x y A

(

( I xy )2 = ( I xy )3

)

I xy = −2  40.5 in 4 + ( 6 in.)( −2 in.) 27 in 2   

= −2 ( 40.5 − 324 ) in 4

or I xy = 567 in 4

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Chapter 9, Solution 77.

Have

I xy = I x′y′ + x yA

Where

I x′y′ = 0 for each rectangle

Then

I xy = I xy

Section :

( )1 + ( I xy )2 + ( I xy )3 = Σ x yA x1 = − ( 8.92 in. − 8 in.) = − 0.92 in. y1 = 3 in. − 0.61 in. = 2.39 in.

A1 = (16 in.)( 6 in.) = 96 in 2 Section :

x2 = ( − 8.92 in. − 1.5 in.) = −10.42 in. y2 = 2 in. − 0.61 in. = 1.39 in.

A2 = ( 3 in.)( 4 in.) = 12 in 2 Section :

x3 = (16 in. − 8.92 in.) − 2 in. = 5.08 in. y3 = − ( 0.61 in. + 5.25 in.) = − 5.86 in. A3 = ( 4 in.)(10.5 in.) = 42 in 2

Then

(

)

(

)

I xy = ( − 0.92 in.)( 2.39 in.) 96 in 2  + ( −10.42in.)(1.39 in.) 12 in 2     

(

)

+ ( 5.08 in.)( − 5.86 in.) 42 in 2   

= − ( 211.08 + 173.806 + 1250.29 ) in 4 = −1635.18 in 4

I xy = −1635 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 78.

( )1 + ( I xy )2

I xy = I xy

Have For each rectangle

I xy = I x′y′ + x yA Then

and

I x′y′ = 0 (symmetry)

I xy = Σ x yA = ( −0.75 in.)( −1.5 in.) ( 3 in.)( 0.5 in.)  + ( 0.5 in.)(1.00 in.) ( 4.5 in.)( 0.5 in.)  = (1.6875 + 1.125 ) in 4 = 2.8125 in 4

or I xy = 2.81 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 79.

Ix =

From Figure 9.12:

( 2a )( a )3

Iy =

π 2

( 2a )3 ( a )

I xy =

From Problem 9.67:

1 4 a 2

1 1 π π  5 I x + I y =  a4 + a4  = π a4 2 2 8 2  16

(

First note

)

1 1π π  3 I x − I y =  a4 − a4  = − π 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 Ix + I y + I x − I y cos 2θ − I xy sin 2θ 2 2

(

)

(

)

5 3 1 π a 4 − π a 4 cos 2θ − a 4 sin 2θ 16 16 2 1 1 Ix + I y − I x − I y cos 2θ + I xy sin 2θ 2 2

(

)

(

)

5 3 1 π a 4 + π a 4 cos 2θ + a 4 sin 2θ 16 16 2 1 I x − I y sin 2θ + I xy cos 2θ 2

=−

(

)

3 1 π a 4 sin 2θ + a 4 cos 2θ 16 2

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Chapter 9, Solution 80.

From the solution to Problem 9.72

I xy = 501.1875 in 4

A2 = A3 = 20.25 in 2 First compute the moment of inertia

I x = ( I x )1 + ( I x )2 + ( I x )3

( I x )2 = ( I x )3

with

3 3 1 1 =  (12 in.)( 9 in.)  + 2  ( 9 in.)( 4.5 in.)  12 12    

= ( 729 + 136.6875 ) in 4 = 865.6875 in 4 and

( )1 + ( I y )2 + ( I y )3

Iy = Iy

with

( I y ) 2 = ( I y )3

3 3 2 1 1 =  ( 9 in.)(12 in.)  + 2  ( 4.5 in.)( 9 in.) + 20.25 in 2 ( 9 in.)  12   36 

(

)

= (1296 + 182.25 + 3280.5 ) in 4 = 4758.75 in 4 From Equation 9.18

I x′ = =

Ix + I y 2

Ix − I y 2

cos 2θ − I xy sin 2θ

865.6875 in 4 + 4758.75 in 4 865.6875 in 4 − 4758.75 in 4 + cos  2 ( −45° )  2 2 −501.1875 in 4 sin  2 ( −45° ) 

= ( 2812.21875 + 501.1875 ) in 4 = 3313.4063 in 4

or I x′ = 3.31 × 103 in 4

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Similarly

I y′ =

Ix + I y 2

−

Ix − I y 2

cos 2θ + I xy sin 2θ

= ( 2812.21875 − 501.1875 ) in 4 = 2311.0313 in 4

or I y′ = 2.31 × 103 in 4 and

I x′y′ = =

Ix − I y 2

sin 2θ + I xy cos 2θ

865.6875 in 4 − 4758.75 in 4 sin  2 ( −45° )  2 + 501.1875cos  2 ( −45° ) 

= ( −1946.53125 )( −1) in 4 = 1946.53125 in 4

or I x′y′ = 1.947 × 103 in 4

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Chapter 9, Solution 81.

From Problem 9.73,

I xy = 138.24 × 106 mm 4 I x = ( I x )1 + ( I x )2

( I x )1 = ( I x )2

4 π = 2  (120 mm )  8 

= 51.84π × 106 mm 4

( )1 + ( I y )2

( I y )1 = ( I y )2

Iy = Iy

π 4 2 2 π = 2  (120 mm ) + (120 mm ) ( 60 mm )  2 8  = 103.68π × 106 mm 4

( ) = 2 ( 51.84π × 10 ) = 103.68π × 10

Have

I x = 2 25.92π × 106 = 51.84π × 106 mm 4

and

Then

1 I x + I y = 77.76π × 106 mm 4 2

and

(

mm 4

)

1 I x − I y = −25.92π × 106 mm 4 2

(

)

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Now, from Equations 9.18, 9.19, and 9.20 Equation 9.18:

I x′ =

1 1 Ix + I y + I x − I y cos 2θ − I xy sin 2θ 2 2

(

)

(

)

= 77.76π × 106 − 25.92π × 106 cos ( −60° ) − 138.24 × 106 sin ( −60° )  mm 4 = 323.29 × 106 mm 4

or I x = 323 × 106 mm 4 Equation 9.19:

I y′ =

1 1 Ix + I y − I x − I y cos 2θ + I xy sin 2θ 2 2

(

)

(

)

= 77.76π × 106 + 25.92π × 106 cos ( −60° ) + 138.24 × 106 sin ( −60° )  mm 4 = 165.29 × 106 mm 4

or I y′ = 165.29 × 106 mm 4 Equation 9.20:

I x′y′ =

1 I x − I y sin 2θ + I xy cos 2θ 2

(

)

=  −25.92π × 106 sin ( −60° ) + 138.24 × 106 cos ( −60° )  mm 4 = 139.64 × 106 mm 4

or I x′y′ = 139.6 × 104 mm 4

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Chapter 9, Solution 82.

From Problem 9.75

I xy = 1.5732 × 106 mm 4

Now

I x = ( I x )1 + ( I x )2 + ( I x )3

( I x )1 =

where and

( I x )2 = ( I x )3

1 (150 mm )(12 mm )3 = 21 600 mm 4 12

1 (12 mm )( 38 mm )3 + (12 mm )( 38 mm ) ( 25 mm )2 12

= 339 872 mm 4 Then

I x =  21 600 + 2 ( 339 872 )  mm 4 = 701 344 mm 4 = 0.70134 × 106 mm 4

( )1 + ( I y )2 + ( I y )3

Iy = Iy

Also where and

( I y )1 = 121 (12 mm )(150 mm )3 = 3.375 × 106 mm 4 ( I y )2 = ( I y )3 = 121 ( 38 mm )(12 mm )3 + (12 mm )( 38 mm ) ( 69 mm )2 = 2.1765 × 106 mm 4

Then

I y = ( 3.375 + 2 ( 2.1765 )  × 106 mm 4 = 7.728 × 106 mm 4

Now

1 I x + I y = 4.2146 × 106 mm 4 2

and

1 I x − I y = −3.5133 × 106 mm 4 2

( (

)

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Using Equations 9.18, 9.19, and 9.20 From Equation 9.18:

I x′ =

Ix + I y 2

Ix − I y

cos 2θ − I xy sin 2θ

(

)

=  4.2147 × 106 + −3.5133 × 106 cos (120° ) − 1.5732 × 106 sin (120° )  mm 4   = 4.6089 × 106 mm 4

or I x′ = 4.61 × 106 mm 4 From Equation 9.19:

I y′ =

Ix + I y 2

−

Ix − I y 2

cos 2θ + I xy sin 2θ

(

)

=  4.2147 × 106 − −3.5133 × 106 cos (120° ) + 1.5732 × 106 sin (120° )  mm 4   = 3.8205 × 106 mm 4 or I y′ = 3.82 × 106 mm 4 From Equation 9.20:

I x′y′ =

Ix − I y 2

sin 2θ + I xy cos 2θ

=  −3.5133 × 106 sin (120° ) + 1.5732 × 106 cos (120° )  mm 4 = −3.8292 × 106 mm 4

or I x′y′ = −3.83 × 106 mm 4

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Chapter 9, Solution 83.

From Problem 9.74

I xy = −0.1596 × 106 mm 4 From Figure 9.13

I x = 0.166 × 106 mm 4 I y = 0.453 × 106 mm 4 1 I x + I y = 0.3095 × 106 mm 4 2

(

Now

)

1 I x − I y = −0.1435 × 106 mm 4 2

(

)

Using Equations (9.18), (9.19), and (9.20) Equation (9.18):

I x′ =

Ix + I y 2

Ix − I y 2

(

cos 2θ − I xy sin 2θ

)

(

)

= 0.3095 × 106 + −0.1435 × 106 cos ( −90° ) − −0.1596 × 106 sin ( −90° )  mm 4   = 0.1499 × 106 mm 4 or I x′ = 0.1499 × 106 mm 4

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Equation (9.19):

I y′ =

Ix + I y 2

−

Ix − I y

cos 2θ + I xy sin 2θ

(

)

(

)

= 0.3095 × 106 − −0.1435 × 106 cos ( −90° ) + −0.1596 × 106 sin ( −90° )  mm 4  

= 0.4691 × 106 mm 4

or I y′ = 0.469 × 106 mm 4 Equation (9.20):

I x′y′ =

Ix − I y 2

sin 2θ + I xy cos 2θ

=  −0.1435 × 106 sin ( −90° ) + 0.1596 × 106 cos ( −90° )  mm 4 = 0.1435 × 106 mm 4

or I x′y′ = 0.1435 × 106 mm 4

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Chapter 9, Solution 84.

From Problem 9.78

I xy = 2.8125 in 4 From Figure 9.13

I x = 9.45 in 4 ,

I y = 2.58 in 4

1 I x + I y = 6.015 in 4 2

(

Now

)

1 I x − I y = 3.435 in 4 2

(

)

Using Equations (9.18), (9.19), and (9.20) Equation (9.18):

Ix + I y

I x′ =

Ix − I y 2

cos 2θ − I xy sin 2θ

= 6.015 + 3.435cos ( 60° ) − 2.8125sin ( 60° )  in 4 = 5.2968 in 4 or I x′ = 5.30 in 4 Equation (9.19):

I y′ =

Ix + I y 2

−

Ix − I y 2

cos 2θ + I xy sin 2θ

= 6.015 − 3.435cos ( 60° ) + 2.8125sin ( 60° )  in 4 = 6.7332 in 4 or I y′ = 6.73 in 4 Equation (9.20):

I x′y′ =

Ix − I y 2

sin 2θ + I xy cos 2θ

= 3.435sin ( 60° ) + 2.8125cos ( 60° )  in 4 = 4.3810 in 4

or I x′y′ = 4.38 in 4

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Chapter 9, Solution 85.

From Problem 9.79:

Ix =

tan 2θ m = −

= Then

Iy =

I xy =

Problem 9.67:

Now, Equation (9.25):

π 2

1 4 a 2

2I xy Ix − I y

=−

π 8

1  2  a4  2  a4 −

8 = 0.84883 3π

2θ m = 40.326°

and

220.326°

or θ m = 20.2° and 110.2°

Also, Equation (9.27):

I max, min =

Ix + I y 2

 Ix − I y  2 ±   + I xy 2  

1π 4 π 4  a + a  2 8 2  2

 1  π 4 π 4  1 4   a − a  +  a  2  2  2  8

= ( 0.981748 ± 0.772644 ) a 4

or I max = 1.754a 4 and I min = 0.209a 4 By inspection, the a axis corresponds to Imin and the b axis corresponds to Imax.

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Chapter 9, Solution 86. From the solutions to Problem 9.72 and 9.80

1 I x + I y = 2812.21875 in 4 2

(

I xy = 501.1875 in 4

)

1 I x − I y = −1946.53125 in 4 2

(

Then Equation (9.25): or

)

tan 2θ m = −

2I xy Ix − I y

2θ m = 14.4387°

=−

501.1875 = 0.257477 −1946.53125 194.4387°

and

or θ m = 7.22° and 97.2° Equation (9.27):

I max, min =

Ix + I y 2

 Ix − I y  ±  + I xy2  2   

= 2812.21875 ±

( −1946.53125 )2 + ( 501.1875 )2

= ( 2812.21875 ± 2010.0181) in 4

or I max = 4.82 × 103 in 4 and I min = 802 in 4

By inspection, the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 87. From Problems 9.73 and 9.81 I x = 51.84π × 106 mm 4 I y = 103.68π × 106 mm 4 I xy = 138.24 × 106 mm 4

tan 2θ m = −

Equation (9.25):

2 I xy Ix − I y

=−

(

2 138.24 × 106 6

)

51.84π × 10 − 103.68π × 106

= 1.69765

2θ m = 59.500°

and

239.500° or θ m = 29.7° and 119.7° !

Then

I max, min

 Ix − I y  1 2 = Ix + I y ±   + I xy  2  2 

(

)

( 51.84 + 103.68)π × 106 2

 ( 51.84 − 103.68 ) π × 106  6   + 138.24 × 10 2  

(

)

= ( 244.29 ± 160.44 ) × 106 mm 4 or I max = 405 × 106 mm 4 ! and I min = 83.9 × 106 mm 4 !

Note: By inspection the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 88. From Problems 9.75 and 9.82

I x = 0.70134 × 106 mm 4 I y = 7.728 × 106 mm 4

I xy = 1.5732 × 106 mm 4 Then

1 I x + I y = 4.2147 × 106 mm 4 2

(

)

1 I x − I y = −3.5133 × 106 mm 4 2

(

Equation (9.25):

)

tan 2θ = −

2I xy Ix − I y

=−

(

2 1.5732 × 106 6

)

0.70134 × 10 − 7.728 × 106

= 0.44778

Then

2θ m = 24.12°

204.12°

and

or θ m = 12.06° and 102.1° Also, Equation (9.27):

I max, min =

Ix + I y 2

 Ix − I y  2 I ±   2  xy  

= 4.2147 × 106 ±

( −3.5133 × 10 ) + (1.5732 × 10 ) 6

= ( 4.2147 ± 3.8494 ) × 106 mm 4

or I max = 8.06 × 106 mm 4 and I min = 0.365 × 106 mm 4

By inspection, the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 89. From Problems 9.74 and 9.83 I x = 0.166 × 106 mm 4 I y = 0.453 × 106 mm 4 I xy = −0.1596 × 106 mm 4

1 I x + I y = 0.3095 × 106 mm 4 2

(

Then

)

1 I x − I y = −0.1435 × 106 mm 4 2

(

Equation (9.25):

)

tan 2θ m = −

2 I xy Ix − I y

=−

)

( 0.166 − 0.453) × 106

2θ m = −48.041°

Then

(

2 −0.1596 × 106

and

= −1.1122

131.96°

θ m = −24.0° and 66.0° !

or Also, Equation (9.27):

I max, min

( Ix + I y ) ± = 2

 Ix − I y  2   + I xy 2  

= 0.3095 × 106 ±

( −0.1435 × 10 ) + ( −0.1596 × 10 ) 6

= ( 0.3095 ± 0.21463) × 106 mm 4 or

I max = 0.524 × 106 mm 4 ! I min = 0.0949 × 106 mm 4 !

By inspection, the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 90. From Problems 9.78 and 9.84 I xy = 2.81 in 4

I x = 9.45 in 4 I y = 2.58 in 4

1 I x + I y = 6.015 in 4 2

(

Then

)

1 I x − I y = 3.435 in 4 2

(

Equation (9.25):

)

tan 2θ m = −

2 I xy Ix − I y

2θ m = −39.2849°

Then

=−

and

2 ( 2.81) 9.45 − 2.58

= −0.8180

140.7151° or θ m = −19.64° and 70.36° !

Also, Equation (9.27):

I max, min

( Ix + I y ) ± = 2

 Ix − I y  2   + I xy 2  

= 6.015 ± 3.4352 − 2.812 = ( 6.015 ± 4.438 ) in 4 or I max = 10.45 in 4 ! and I min = 1.577 in 4 !

Note: By inspection, the a axis corresponds to I max and the b axis corresponds to I min .

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Chapter 9, Solution 91. π

Ix =

From Problem 9.79:

Iy =

1 4 a 2

I xy =

Problem 9.67:

The Mohr’s circle is defined by the diameter XY, where

1  π X  a4, a4  8 2   Now

I ave =

1 1π π  5 I x + I y =  a 4 + a 4  = π a 4 = 0.98175a 4 2 2 8 2  16

(

)

and

1  π Y  a4 , − a4  2 2  

and

 Ix − I y  2   + I xy =  2 

 1  π 4 π 4  1 4   a − a  +  a  2  2  2  8

= 0.77264a 4 The Mohr’s circle is then drawn as shown.

tan 2θ m = −

=−

2I xy Ix − I y

π 8

1  2  a4  2  a4 −

= 0.84883

2θ m = 40.326°

continued

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Then

α = 90° − 40.326° = 49.674°

β = 180° − ( 40.326° + 60° ) = 79.674°

(a)

I x′ = I ave − R cos α = 0.98175a 4 − 0.77264a 4 cos 49.674° or I x′ = 0.482a 4 I y′ = I ave + R cos α = 0.98175a 4 + 0.77264a 4 cos 49.674°

or I y′ = 1.482a 4 I x′y′ = − R sin α = −0.77264a 4 sin 49.674° or I x′y′ = −0.589a 4 (b)

I x′ = I ave + R cos β = 0.98175a 4 + 0.77264a 4 cos 79.674° or I x′ = 1.120a 4 I y′ = I ave − R cos β = 0.98175a 4 − 0.77264a 4 cos 79.674°

or I y′ = 0.843a 4 I x′y′ = R sin β = 0.77264a 4 sin 79.674° or I x′y′ = 0.760a 4

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Chapter 9, Solution 92.

From the solution to Problem 9.72:

I xy = 501.1875 in 4

Problem 9.80:

I x = 865.6875 in 4 I y = 4758.75 in 4

1 I x + I y = 2812.21875 in 4 2

(

Now

)

1 I x − I y = −1946.53125 in 4 2

(

)

The Mohr’s circle is defined by the points X and Y where

( I x , I xy )

I ave =

Now

( I y , −I xy )

1 I x + I y = 2812.2 in 4 2

(

and

 Ix − I y  2   + I xy = 2  

)

( −1946.53125)2 + 501.18752

in 4

= 2010.0 in 4 Also,

tan 2θ m =

I xy Ix − I y

501.1875 = 0.2575 1946.53125

2 or

2θ m = 14.4387°

Then

α = 180° − (14.4387° + 90° ) = 75.561°

Then

I x′ , I y′ = I ave ± R cos α = 2812.2 ± 2010.0cos 75.561°

or I x′ = 3.31 × 103 in 4 and I y′ = 2.31 × 103 in 4 and

I x′y′ = R sin α = 2010.0sin 75.561°

or I x′y′ = 1.947 × 103 in 4

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Chapter 9, Solution 93.

From Problems 9.73 and 9.81

I xy = 138.24 × 106 mm 4 I x = 51.84π × 106 mm 4 = 162.86 × 106 mm 4

I y = 103.68π × 106 mm 4 = 325.72 × 106 mm 4

I ave =

Now

1 Ix + I y 2

(

)

= 244.29 × 106 mm 4 2

 Ix − I y  2   + I xy 2  

= 160.4405 × 106 mm 4 2θ m = 59.5°

From Problem 9.87 Then Then

α = 180 − 60° − 2θ m = 60.5° I x′ = I ave + R cos α = ( 244.29 + 160.4405cos 60.5° ) × 106 = 323.29 × 106 mm 4

or I x′ = 323 × 106 mm 4 I y′ = I ave − R cos α = ( 244.24 − 160.4405cos 60.5° ) × 106 = 165.29 × 106 mm 4

or I y′ = 165.3 × 106 mm 4 I x′y′ = R sin α = 160.44 × 106 sin 60.5° = 139.6 × 106 mm 4

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Chapter 9, Solution 94.

From Problems 9.75 and 9.82

I x = 0.70134 × 106 mm 4 I y = 7.728 × 106 mm 4 I xy = 1.5732 × 106 mm 4 Now

I ave =

1 I x + I y = 4.2147 × 106 mm 4 2

(

)

and

Then and Then

 Ix − I y  2 6 4   + I xy = 3.8494 × 10 mm 2  

 −2 (1.5732 )  2θ m = tan −1   = 24.12°  0.70134 − 7.728 

α = 120° − 24.12° − 90° = 5.88° I x′ = I ave + R sin α = ( 4.2147 + 3.8494sin 5.88° ) × 106 mm 4 = 4.6091 × 106 mm 4 or I x′ = 4.61 × 106 mm 4

I y′ = I ave − R sin α = ( 4.2147 − 3.8494sin 5.88° ) × 106 mm 4 = 3.8203 × 106 mm 4

or I y′ = 3.82 × 106 mm 4

I x′y′ = − R cos α = −3.8494cos 5.88° = −3.8291 × 106 mm 4 or I x′y′ = −3.83 × 106 mm 4

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Chapter 9, Solution 95.

From Problems 9.74 and 9.83

I x = 0.166 × 106 mm 4 I y = 0.453 × 106 mm 4 I xy = −0.1596 × 106 mm 4 Now

I ave =

1 I x + I y = 0.3095 × 106 mm 4 2

(

)

and

 Ix − I y  2   + I xy 2  

= 0.21463 × 106 mm 4 Then and

 −2 ( −0.1596 )  2θ m = tan −1   = −48.04°  0.166 − 0.453 

α + 90° − 2θ = 90°; α = 2θ m

Then

I x′ = I ave − R sin α = ( 0.3095 − 0.21463sin 48.04° ) × 106 mm 4 = 0.14989 × 106 mm 4

or I x′ = 0.1499 × 106 mm 4 and

I y′ = I ave + R sin α = ( 0.3095 + 0.21463sin 48.04° ) × 106 mm 4 = 0.46910 × 106 mm 4

or I y′ = 0.4690 × 106 mm 4 and

I x′y′ = R cos α = 0.21463cos 48.04° = 0.1435 × 106 mm 4 or I x′y′ = 0.1435 × 106 mm 4

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Chapter 9, Solution 96. I x = 9.45 in 4

Have

I y = 2.58 in 4 I xy = 2.8125 in 4

From Problem 9.78

I ave =

Now 2

and

 Ix − I y    + I xy  2 

( )

Ix + I y 2

= 6.015 in 4

= 4.43952 in 4 Then

 −2 ( 2.8125 )  2θ m = tan −1   = −39.31°  9.45 − 2.58  2θ m + 60° + α = 180°,

Then

α = 80.69°

I x′ = I ave − R cos α = 6.015 in 4 − ( 4.43952 in 4 ) cos80.69° = 5.29679 in 4

or I x′ = 5.30 in 4 I y′ = I ave + R cos α = 6.015 in 4 + ( 4.43952 in 4 ) cos80.69°

= 6.73321 in 4

or I y′ = 6.73 in 4 I x′y′ = R sin α = ( 4.43952 in 4 ) sin 80.69° = 4.38104 in 4

or I x′y′ = 4.38 in 4

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Chapter 9, Solution 97.

From Problem 9.79:

Ix =

Problem 9.67:

I xy =

π 8

Iy =

π 2

1 4 a 2

The Mohr’s circle is defined by the diameter XY, where

1  π X  a4, a4  2  8

I ave =

Now

1  π Y  a4 , − a4  2  2

and

1 1π π  I x + I y =  a 4 + a 4  = 0.98175a 4 2 2 8 2 

(

)

and

1  2  2 I x − I y  + I xy =  

(

)

 1  π 4 π 4  1 4   a − a  +  a  2  2  2 8

= 0.77264a 4 The Mohr’s circle is then drawn as shown.

tan 2θ m = −

=−

2I xy Ix − I y

π 8

1  2  a4  2  a4 −

= 0.84883

2θ m = 40.326°

θ m = 20.2°

and

∴ The principal axes are obtained by rotating the xy axes through 20.2° counterclockwise Now

I max, min = I ave ± R = 0.98175a 4 ± 0.77264a 4

or I max = 1.754a 4 and I min = 0.209a 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 .

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Chapter 9, Solution 98.

From the solution to Problem 9.72:

I xy = 501.1875 in 4 From the solution to Problem 9.80:

I x = 865.6875 in 4 I y = 4758.75 in 4 1 I x + I y = 2812.21875 in 4 2

(

)

1 I x − I y = −1946.53125 in 4 2

(

)

The Mohr’s circle is defined by the point

I ave =

Now 2

and

 Ix − I y  2   + I xy = 2  

( I x , I xy ) ,

( I y,

− I xy

)

1 I x + I y = 2812.2 in 4 2

(

)

( −1946.53125 )2 + 501.18752

= 2010.0 in 4

continued

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tan 2θ m = −

I xy  Ix − I y     2 

=−

501.1875 = 0.2575, −1946.53125

2θ m = 14.4387°

or θ m = 7.22° counterclockwise

Then

I max, min = I ave ± R = ( 2812.2 ± 2010.0 ) in 4 or I max = 4.82 × 103 in 4 and I min = 802 in 4

Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 99.

I xy = 567 in 4

From the solution to Problem 9.76 Now

I x = ( I x )1 − ( I x )2 − ( I x )3 , where

π 4

( I x ) 2 = ( I x )3

1 12

(15 in.)4 − 2  ( 9 in.)( 6 in.)3  = ( 39761 − 324 ) in 4 

= 39, 437 in 4

( )1 − ( I y )2 − ( I y )3 ,

Iy = Iy

and

π 4

where

( I y )2 = ( I y )3

1  36

1 2

(15 in.)4 − 2  ( 6 in.)( 9 in.)3 + ( 9 in.)( 6 in.)( 6 in.)2  

= ( 39, 761 − 243 − 1944 ) in 4 = 37,574 in 4 The Mohr’s circle is defined by the point (X, Y) where X:

I ave =

Now

( I x , I xy )

( I y , −I xy )

1 1 I x + I y = ( 39, 437 + 37,574 ) in 4 = 38,506 in 4 2 2

(

)

and

 Ix − I y  2   + I xy = 2  

1  2 4  2 ( 39, 437 − 37,574 )  + 567 = 1090.5 in   continued

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tan 2θ m =

− I xy Ix − I y 2

−567 = −0.6087 1 ( 39, 437 − 37,574 ) 2 or θ m = −15.66° clockwise

Then

I max, min = I ave ± R = ( 38,506 ± 1090.50 ) in 4

or I max = 39.6 × 103 in 4 and I min = 37.4 × 103 in 4

Note: From the Mohr’s circle it is seen that the a axis corresponds to the I max and the b axis corresponds to

I min .

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Chapter 9, Solution 100.

I x = 162.86 × 106 mm 4

From Problems 9.73 and 9.81

I y = 325.72 × 106 mm 4 I xy = 138.24 × 106 mm 4 X (162.86,138.24 ) × 106 mm 4

Define points

I ave =

Now

Y ( 325.72, − 138.24 ) × 106 mm 4

1 1 I x + I y = (162.86 + 325.72 ) × 106 mm 4 2 2

(

)

= 244.29 × 106 mm 4 2

and

 Ix − I y  2   + I xy = 2  

 (162.86 − 325.72 )  × 106  + 138.24 × 106  2  

(

)

= 160.44 × 106 mm 4 and

 −2 (138.24 ) × 106  2θ m = tan −1  = 59.4999° 6  (162.86 − 325.72 ) × 10  or θ m = 29.7° counterclockwise

Then

(

)

I max, min = I ave ± R = 244.29 × 106 ± 160.44 × 106 mm 4

or I max = 405 × 106 mm 4 and I min = 83.9 × 106 mm 4 Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 101.

From Problems 9.74 and 9.83

I x = 0.166 × 106 mm 4 ,

I y = 0.453 × 106 mm 4 ,

X ( 0.166, −0.1596 ) × 106 mm 4

Define points

I ave =

Now

and

I xy = −0.1596 × 106 mm 4 Y ( 0.453, 0.1596 ) × 106 mm 4

1 1 I x + I y = ( 0.166 + 0.453) × 106 mm 4 2 2

(

)

= 0.3095 × 106 mm 4 2

and

 Ix − I y  2   + I xy = 2  

 ( 0.166 − 0.453) × 106  6   + −0.1596 × 10 2  

(

)

= 0.21463 × 106 mm 4 Also

 −2I xy 2θ m = tan −1   Ix − I y 

  −2 ( −0.1596 )   = tan −1   = −48.04°   0.166 − 0.453  

θ m = −24.02° or θ = −24.0° clockwise

Then

I max,

min

= I ave ± R = ( 0.3095 ± 0.21463) × 106 mm 4

or I max = 0.524 × 106 mm 4 and I min = 0.0949 × 106 mm 4 Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 102.

From Problems 9.75 and 9.82

I x = 0.70134 × 106 mm 4 , Now

I ave =

I y = 7.728 × 106 mm 4 ,

1 1 I x + I y = ( 0.70134 + 7.728 ) × 106 mm 4 = 4.2147 × 106 mm 4 2 2

(

)

and

I xy = 1.5732 × 106 mm 4

 Ix − I y  2   + I xy = 2  

 ( 0.70134 − 7.728 ) × 106  6   + 1.5732 × 10 2  

(

)

= 3.8495 × 106 mm 4

Define points

X ( 0.70134, 15732 ) × 106 mm Y ( 7.728, − 1.5732 ) × 106 mm

Also

 −2 (1.5732 )  2θ m = tan −1   = 24.122°, θ m = 12.06°  0.70134 − 7.728 

or θ m = 12.06° counterclockwise Then

I max,

min

= I ave ± R = ( 4.2147 ± 3.8495 ) × 106 mm 4

or I max = 8.06 × 106 mm 4 and I min = 0.365 × 106 mm 4 Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .

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Chapter 9, Solution 103. From the solution to Problem 9.71

I xy = 300 × 103 mm 4 Now

I x = ( I x )1 − ( I x )2 − ( I x )3 3 3 2 1 1 =  (120 mm )( 80 mm )  − 2  ( 50 mm )( 20 mm ) + 1000 mm 2 (10 mm )  12 12    

(

)

= 5.120 × 106 − 2 33.333 × 103 + 100 × 103  mm 4   = 4.8533 × 106 mm 4 and

( )1 − ( I y )2 − ( I y )3

Iy = Iy

3 3 2 1 1 =  ( 80 mm )(120 mm )  − 2  ( 20 mm )( 50 mm ) + 1000 mm 2 (15 mm )  12  12 

(

)

= 11.520 × 106 − 2 208.33 × 103 + 225 × 103  mm 4   = 10.6533 × 106 mm 4

(

x I x , I xy

)

(

y I y , − I xy

Center:

)

1 Ix + I y 2 1 = ( 4.8533 + 10.6533) × 106 mm 4 2

I ave =

(

)

= 7.7533 × 106 mm 4 2

Radius:

 Ix − I y  R=  + I xy2  2    2

2 1  = 106  ( 4.8533 − 10.6533)  + ( 0.300 ) mm 2 

= 2.9155 × 106 mm continued

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Chapter 9, Solution 104.

From Prob. 9.43 and 9.77

I y = 4581.0 in 4

I x = 2703.7 in 4 Define Points:

x ( 2703.7, −1635.18 ) in 4 y ( 4581.0, 1635.18 ) in 4

Then

I ave =

I xy = −1635.18 in 4

1 1 I x + I y = ( 2703.7 + 4581.0 ) in 4 = 3642.4 in 4 2 2

(

)

2  Ix − I y  2  2703.7 − 4581.0  4 + I xy2 =  R=    + ( −1635.18 ) = 1885.44 in  2  2    

tan 2θ m = −

2I xy Ix − I y

=−

2 ( −1635.18 ) = −1.74206 2703.7 − 4581.0

θ m = − 30.1° I max,min = I ave ± R = ( 3642.4 ± 1885.44 ) in 4

I max = 5530 in 4 I min = 1757 in 4

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Chapter 9, Solution 105.

I x = 0.166 × 106 mm 4 , I y = 0.453 × 106 mm 4 and I xy < 0

Given:

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 102 mm leg of the angle is parallel to the x axis. For I xy < 0 the angle must be oriented as shown. (a) Now

I ave =

1 1 I x + I y = ( 0.166 + 0.453) × 106 mm 4 2 2

(

)

= 0.3095 × 106 mm 4

I min = I ave − R

Now

R = I ave − I min

R = ( 0.3095 − 0.051) × 106 mm 4

Then

= 0.2585 × 106 mm 4 2

 Ix − I y  R = + I xy  2    2

From

I xy =

( )

2  2  0.166 − 0.453   6 4 ( 0.2585 ) −    × 10 mm 2    

I xy = ±0.21501 × 106 mm 4 Since (b)

I xy < 0,

I xy = −0.21501 × 106 mm 4

or I xy = −0.215 × 106 mm 4

 −2 ( −0.21501)  2θ m = tan −1   = −56.28°  0.166 − 0.453 

or θ m = −28.1° clockwise (c)

I max = I ave + R = ( 0.3095 + 0.2585 ) × 106 mm 4

or I max = 0.568 × 106 mm 4

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Chapter 9, Solution 106. I x = 9.45 in 4

From Figure 9.13

I y = 2.58 in 4 I xy = − 2.81 in 4

From Problem 9.78

The Mohr’s circle is defined by the diameter XY where

X ( 9.45, − 2.81) in 4 Y ( 2.58, 2.81) in 4 Now

I ave =

1 1 Ix + I y = 9.45 in 4 + 2.58 in 4 2 2

(

)

= 6.015 in 4

and

R= =

(

)

1 4 4  4  2 9.45 in − 2.58 in  + 2.81 in  

(

)

= 4.438 in 4

tan 2θ m =

1  2  2 I x − I y  + I xy  

−2I xy

Ix − I y

(

−2 −2.81 in 4 4

)

(

9.45 in − 2.58 in 4

)

= 0.81805

2θ m = 32.285°

or θ m = 19.64° counterclockwise Now

I max, min = I ave ± R = ( 6.015 ± 4.438 ) in 4

or I max = 10.45 in 4 and I min = 1.577 in 4 From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .

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Chapter 9, Solution 107.

I x = 7.20 × 106 mm 4 ,

From Figure 9.13B:

( )1 + ( I xy )2 ,

I xy = I xy

Have

x1 =

Now

where

102 − 25.3 = 25.7 mm, 2

I y = 2.64 × 106 mm 4 I xy = I x′y′ + x yA

y1 = 50.3 −

and

I x′y′ = 0

12.7 = 43.95 mm 2

A1 = 102 × 12.7 = 1295.4 mm 2 x2 = −25.3 +

12.7 = −18.95 mm 2

1  y2 = −  (152 − 12.7 ) − ( 50.3 − 12.7 )  = − 32.05 mm 2 

A2 = (12.7 )(152 − 12.7 ) = 1769.11 mm 2 Then

{

(

)

(

)}

I xy = ( 25.7 mm )( 43.95 mm ) 1295.4 mm 2  + ( −18.95 mm )( −32.05 mm ) 1769.11 mm 2  × 106     = (1.46317 + 1.07446 ) × 106 mm 4 = 2.5376 × 106 mm 4

The Mohr’s circle is defined by points X and Y, where

(

) (

X I x , I xy , Y I y , − I xy Now

I ave =

)

1 1 I x + I y = ( 7.20 + 2.64 ) × 106 mm 4 = 4.92 × 106 mm 4 2 2

(

)

continued

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and

2 2    Ix − I y  1  2 2 6 4   + I xy =   ( 7.20 − 2.64 )  + 2.5376  × 10 mm   2 2      

= 3.4114 × 106 mm 4

tan θ m = −

2I xy Ix − I y

=−

2 ( 2.5376 ) = −1.11298, ( 7.20 − 2.64 )

2θ = −48.0607°

θ = −24.0° clockwise

or Now

I max, min = I ave ± R = ( 4.92 ± 3.4114 ) × 106 mm 4

or and

I max = 8.33 × 106 mm 4 I min = 1.509 × 106 mm 4

Note: From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .

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Chapter 9, Solution 108. I ave =

Have

1 1 Ix + I y = 640 in 4 + 280 in 4 = 460 in 4 2 2

(

)

1 1 Ix − I y = 640 in 4 − 280 in 4 = 180 in 4 2 2

(

)

I x′y′ = −180 in 4 ,

Also have Letting the points

(

)

( I x , I xy ) and ( I x′, I x′y′ )

2θ = −120°,

Ix > I y

be denoted by X an X ′, respectively, three possible Mohr’s

circles can be constructed

Assume the first case applies Then Also

Ix − I y 2

= R cos 2θ m

I x′y′ = R cos α

R cos 2θ m = 180 in 4

or or

R cos α = 180 in 4

∴ α = ±2θ m

120° = 2θ m + ( 90° − α )

Also have

∴ α = −2θ m Note

and

2 ( 2θ m ) = 30°

2θ m − α = 30°

or or

2θ m = α = 15°

2θ m > 0   implies case 2 applies α < 0 continued

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(a) Therefore,

(b)

Have Then

θ m = 7.5° clockwise

R cos15° = 180

R = 186.35 in 4

I max, min = I ave ± R = 460 ± 186.350 or

I max = 646 in 4

and

I min = 274 in 4

Note: From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .

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Chapter 9, Solution 109.

First assume

Ix > I y

( )max is

(Note: Assuming I x < I y is not consistent with the requirement that the axis corresponding to the I xy obtained by rotating the x axis through 67.5° counterclockwise)

2θ m = 2 ( 67.5° ) − 90° = 45°

From Mohr’s circle have (a)

tan 2θ m =

From

Have I x = I y + 2

I xy tan 2θ m

= 300 in 4 + 2

2 I xy Ix − I y

125 in 4 = 550 in 4 tan 45°

or I x = 550 in 4 (b) Now

and

Then

I ave = R=

1 550 + 300 4 Ix + I y = in = 425 in 4 2 2

(

I xy sin 2θ m

)

125 in 4 = 176.78 in 4 sin 45°

I max, min = I ave ± R = ( 425 ± 176.76 ) in 4 = ( 601.78, 248.22 ) in 4

or I max = 602 in 4 and I min = 248 in 4

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Chapter 9, Solution 110.

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 α 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′ = I ave (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)

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Chapter 9, Solution 111.

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 2θ

I y′ = I ave − R cos 2θ I x′y′ = R sin 2θ I x′ I y′ − I x2′y′

Then, forming the expression

I x′ I y′ − I x2′y′ = ( I ave + R cos 2θ )( I ave − R cos 2θ ) − ( R sin 2θ )

(

) (

2 = I ave − R 2 cos 2 2θ − R 2 sin 2 2θ

)

2 = I ave − R 2 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.

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Chapter 9, Solution 112.

From Problem 9.111 have, I x′ I y′ − I x′y′ = constant Now consider the following two cases Case 1:

I x′ = I x ,

I y′ = I y ,

Case 2:

I x′ = I max ,

I x′y′ = I xy

I y′ = I min ,

2 I x I y − I xy = I max I min

Then

I xy = ± I x I y − I max I min

or From Figure 9.13B: Now

I x = 453 × 103 mm 4 I ave =

Finally

I y = 166 × 103 mm 4

1 1 Ix + I y = I max + I min 2 2

(

)

(

)

I max = 524 × 103 mm 4

With then

I x′y′ = 0

I min = ( 453 + 166 − 524 ) × 103 mm 4 = 95.0 × 103 mm 4 I xy = ±  

( 453)(166 ) − ( 524 )( 95.0 )  × 103 mm 4

= ±159.43 × 103 mm 4 The two roots corresponding to the following orientations of the cross section:

(a)

(a) I xy = −159.4 × 103 mm 4 (b) I xy = 159.4 × 103 mm 4

(b)

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Chapter 9, Solution 113.

Mass = m = ρ tA

I mass = ρ tI area = Area = A =

m I area A

1 2 πa 2

I AA′, area = I DD′, area = I AA′, mass = I DD′, mass =

(a)

I BB′ = I DD′ − m ( AC ) =

1π 4 1 4  a  = πa 2 4  8 m m 1 4 1 2 I AA′, area = π a  = ma 1 2  8 A  4 πa 2

1 2  4a  ma − m   4  3π 

= ( 0.25 − 0.1801) ma 2

I BB′ = 0.0699 ma 2

(b) Eq. ( 9.38 ) :

I CC ′ = I AA′ + I BB′

1 2 ma + 0.0699ma 2 4

I CC ′ = 0.320 ma 2

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Chapter 9, Solution 114. Mass = m = ρV = ρ tA

I mass = ρ tI area =

m I area A

(

Area = A = π r22 − π r12 = π r22 − r12

I AA′, area =

(a) I AA′, mass =

π 4

r24 −

π 4

r14 =

m m π 4 1 I AA′, area = r − r14 = m r22 + r12 2 2 4 2 4 A π r2 − r1

(b) By Symmetry: Eq. ( 9.38 ) :

(

)

(

)

(

4 2

)

− r14

)

) I AA′ =

1 m r22 + r12 4

I CC ′ =

1 m r22 + r12 2

(

)

I BB′ = I AA′ I CC ′ = I AA′ + I BB′ = 2 I AA′

(

)

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Chapter 9, Solution 115.

mass = m = ρV = ρ tA

First note

= ρt

π 4

2 2

− r12

)

I mass = ρ tI area

Also

π 4

I AA′, area =

(a) Using Figure 9.12, Then

I AA′, mass =

)

m 2 r2 + r12 4

)

m 2 1   r2 +  r2  4  2 

r22

−

(

(r 16 π

r12

(

4 2

(

− r14

m r22 − r12

)

2

m5   =  r22  44  

or I AA′ = (b) Symmetry implies Then,

5 mr22 16

I BB′, mass = I AA′, mass I OO′ = I AA′ + I BB′

 5  = 2  mr22  16  

5 2 mr2 8 continued

)

I area

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Now locate centroid C.

X ΣA = ΣxA or

π  4r  π  4r  π  π X  r22 − r12  = 2  r22  − 1  r12  4  3π  4  3π  4  4

X =

Now

4 r23 − r13 3π r22 − r12 r = X 2 3

1  r3 − r 4 2 2  2 2  14 2 = = r2 2 3π 2  1  9 π r2 −  r2  2  Finally or

I OO′ = I CC′ + mr 2  14 2  5 2 mr2 = I CC ′ + m  r2  8  9 π 

or ICC ′ = 0.1347mr22

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 116. Locate centroid:

y1 =

b 2

A1 = 3ab

y2 =

3 b 2

A2 = ab

y =

Then

Σyi Ai ΣΑi

b 3 ( 3ab ) + b ( ab ) 2 = 2 3ab + ab

Uniform thickness:

m is proportional to A

(a)

I AA′ = ( I AA′ )1 + ( I AA′ )2

3 b 4

∴ m1 =

3 m 4

m2 =

1 m 4

 1  3  2  3  b 2   1  1  2  1  3  2  =   m  b +  m    +   m  b +  m  b    4  2   12  4   4  2   12  4 

3 1 9 3 = mb 2  + + +   48 16 48 16  I AA′ = (b)

Have or

5 2 mb 6

I AA′ = I x + my 2

Ix = =

5 2 3  mb − m  b  6 4 

13 2 mb 48 continued

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 117. Mass = m = ρV = ρ tA m I mass = ρ tI area = I area A 1 A = bh 2

(a)

Axis AA′:

Axis BB′:

 1  b 3  1 3 I AA′, area = 2  h    = hb 12 2 48     m m 1 3 1 I AA′, mass = I AA′, area = ⋅ hb = mb 2 1 A 48 24 bh 2

1 mb 2 24

I BB′ =

1 mh 2 18

1 3 bh 36 m m 1 3 1 = I BB′, area = ⋅ bh = mh 2 1 A 36 18 bh 2

I BB′, area =

I BB′, mass

(b)

I AA′ =

Axis CC′: Eq. (9.38):

I CC ′ = I AA′ + I BB′ 1 1 = mb 2 + mh 2 24 18

I CC ′ =

m 3b 2 + 4h 2 72

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 118. From Prob. 9.117: 1 I AA′ = mb 2 24 1 I BB′ = mh 2 18

Note that AA′ and BB′ are centroidal axes. 1 mb 2 + md 2 Hence I DD′ = I AA′ + md 2 = 24

I EE ′ = I BB′ + md 2 =

)

(

)

m 2 b + 24d 2 24

I EE ′ =

m 2 h + 18d 2 18

1 mh 2 + md 2 18

(

I DD′ =

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 119. mass = m = ρV = ρ tA

First note

1   = ρ t ( 2a )( a ) + ( 2a )( a )  2  

= 3ρ ta 2 I mass = ρ tI area

Also

m I area 3a 2

I x, area = ( I x )1, area + ( I x )2, area

(a) Now

1 1 ( 2a )( a )3 + ( 2a )( a )3 3 12

5 4 a 6

I x, mass =

Then

m 5 × a4 2 6 3a or I x, mass =

5 ma 2 18

(b) Have

I z, area = ( I z )1, area + ( I z )2, area 2 1 1 1 3 3 1    =  ( a )( 2a )  +  ( a )( 2a ) + ( 2a )( a )  2a + × 2a   2 3 3   36   

= 10a 4 Then

I z, mass =

= Finally,

m × 10a 4 3a 2 10 2 ma 3

I y, mass = I x, mass + I z, mass

5 10 2 ma 2 + ma 18 3

65 2 ma 18

or I y, mass = 3.61ma 2 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 120. First locate the centroid C

1   X 2a 2 + a 2 = a 2a 2 +  2a + × 2a  a 2 3  

(

X ΣA = ΣxA:

)

( )

X =

Z ΣA = ΣzA:

14 a 9

1  1  Z 2a 2 + a 2 =  a  2a 2 +  a  a 2 2   3 

(

)

( )

Z =

or (a) Have

( )

4 a 9

(

I y, mass = I CC ′, mass + m X 2 + Z 2

)

From the solution to Problem 9.119

I y, mass =

Then

I cc′, mass =

65 2 ma 18

 14 2  4 2  65 2 ma − m  a  +  a   18  9    9 

or I cc′ = 0.994 ma 2 (b) Have and Then

( )

I x, mass = I BB′, mass + m Z

I AA′, mass = I BB′, mass + m (1.5a )

I AA′, mass = I x, mass

2  2 4   + m (1.5a ) −  a    9   

From the solution to Problem 9.119

I x, mass =

Then

I AA′, mass =

5 ma 2 18 2  5 2 4   ma 2 + m (1.5a ) −  a   18  9   

or I AA′ = 2.33ma 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 121.

b = ka 2

x = a, y = b:

y =

Then

k =

b 2 x a2

b a2

( )

dm = ρ π r 2 dx

Now

 b  = πρ  2 x 2  dx a  Then

m = πρ

b2 a 4 ∫ x dx a4 0 a

1 b2 = πρ 4 x5 5 a 0 =

Now

1 πρ ab 2 5

πρ =

5m ab 2

2 2  1 b 1     b  d I x =  r 2  dm =  2 x 2  πρ  2 x 2  dx  2 a 2     a  

5m 1 b 2 4 b 2 4 5 b2 8 × x × x dx = m x dx 2 a9 ab 2 2 a 4 a4 a

Then..

5 b2 a 5 b2 1 I x = m 9 ∫0 x8dx = m 9 × x9 2 a 2 a 9 0

or I x =

5 mb 2 18

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 122. ∴a = L

r1 = 2 r2

Now

m = ρV

or Now

Have Where

dm = ρ dV

(

= ρ π r 2dx ∴

π 2 2 π = ρ  ( 2L )( 2r2 ) − ( L )( r2 )  3 3  7π = ρ Lr22 3 3 m πρ = 7 Lr22 r −r r = 2 1 x + r1 L x  = r2  2 −  L   dI z = dI z′ + x 2dm 1 = ( dm ) r 2 + x 2dm 4

)

1  dI z = πρ  r 4 + r 2 x 2  dx 4   4 2  1     x  x    = πρ   r2  2 −   + x 2  r2  2 −    L  L      4   

Then

Iz =

4 3 m 2 L1 2  x x x2  2 2 4 4 r r − + x − +     dx 2 ∫0 2    7 Lr22 L L L2    4  

( )

5 3 m 1 2  1  x 4 x 4 1 x5   = +  r2   ( − L )  2 −  + x3 −  7 L  4  5  L 3 L 5 L2   0

Iz =

m 93r22 + 32L2 140

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 123. 3

2a = k ( 2a )

x = 2a

Then

k =

1 3 x 4a 2

(

dm = ρ π r 2dx

Now

1 4a 2

) 2

πρ 6  1  = πρ  2 x3  dx = x dx 16a 4  4a  πρ

Then

16a 4

or πρ =

2a 6 ∫a x dx

πρ 1 4

16a 7

= a

πρ  127 πρ a3 ( 2a )7 − ( a )7  = 4   112 112a

112m 127a3 2

(a) Now

1 1 1    πρ 6  d I x =  r 2  dm =  2 x3   x dx  4 2 2 4 a      16a  =

1 6 112m 7m x6 x dx = x12dx × × 4 3 4 11 32a 127a 16a 4064a

Then

Ix = =

7m 7m 1 13 2a 12 ∫ x dx = 4064a11 13 x 4064a11 a a

7m  57337 2 ma = 1.0853ma 2 ( 2a )13 − ( a )13  = 52832 52832a11 

or I x = 1.085ma 2

continued

COSMOS: Complete Online Solutions Manual Organization System

(b) Have

2 1  1  πρ 6 1   dI y =  r 2 + x 2  dm =   2 x3  + x 2  x dx 4 4    16a  4  4a

1 112m  1 12  x + x8  dx ×  4 16a 127a3  64a 4  2a

7m 2a  1 12 7m  1 1   x + x8  dx = x13 + x9  Then I y = 7 ∫a  4 7 4 9 a 127a 127a  832a  64a  =

7m  1 1 1 1 9 13 9 13 2a ) + ( 2a ) − a − (a)  7  4( 4( ) 9 9 127a  832a 832a 

7m  8191 9 511 9  a + a  = 3.67211ma 2 7  832 9 127a  

or I y = 3.67 ma 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 124.

For the element shown:

ab  y  y  dm = ρ dV = ρ  a  b  dy = ρ 2 y 2dy h h h   

dIx′ =

1  y 1 b 2 2  ab  2 1 ab3 4 y  ρ 2  y dy = y dy ρ  b  dm = 2 12  h  12 h 12 h 4  h 

Parallel-axis theorem

dI x = dI x′ + d 2dm =

1 y where d 2 = y 2 +  b  2 h

 1 ab3 4 1 y 2  ab ρ 4 y dy +  y 2 + b 2 2  ρ 2 y 2dy 12 h 4 h  h 

 ρ ab3 ab  =  + ρ 2  y 4dy 4 h  3 h  ρ ab3 ab  h ρ ab3h ρ abh3 I x = ∫ dI x =  + ρ 2  ∫ 0 y 4dy = + 4 15 5 h  3 h

1 ρ abh 3

For pyramid,

m = ρv =

Thus

1   b 2 3h 2  I x =  ρ abh   +  5  3  5

Ix =

1 m b 2 + 3h 2 5

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 125.

For the element shown:

ab  y  y  dm = ρ  b  a  dy = ρ 2 y 2dy h  h  h  For thin plate:

dI y = dI x′′ + dI z′′ 2

1 y  1 y  1 2 2 2  b  dm +  a  dm = 2 b + a y dm 3 h  3 h  3h

1  ab  ρ ab 2 a 2 + b 2 y 2  ρ 2 y 2dy  = a + b 2 y 4dy 2 4 3h  h  3h

(

m = ρV =

(

)

I y = ∫ dI y = For pyramid,

ρ ab 3h

)

(

+ b2

)∫

h 0

y 4dy =

ρ ab 15

)

+ b2 h

1 ρ abh 3

1 1 I y =  ρ abh  a 2 + b 2 3 5

(

)

Iy =

1 m a 2 + b2 5

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 126. r 2 = y 2 + z 2 = kx : x = h,

Thus, r 2 =

r = a;

= Now

dI y = dI y′ + x 2dm =

a2 xdx h h

∫ 0 dm = ρπ h ∫ 0 xdx

1 ρπ a 2h 2

 1 a2  1 2 r dm + x 2dm =  x + x 2  dm 4 4 h 

= ρπ

 a2 h  1 a2 2 x + x3  dx ∫ 0  h 4 h 

= ρπ

a 2  1 a 2 h3 h 4  +   h  4 h 3 4 

1 ρπ a 2h a 2 + 3h 2 12 1 m = ρπ a 2h 2

(

)

1 1 I y =  ρπ a 2h  a 2 + 3h 2 2 6

(

And

a2 h

2  a2 h h 1 a I y = ∫ 0 dI y = ∫ 0  x + x 2 ρπ xdx h 4 h 

Recall:

a2 x h

Have dm = ρπ r 2dx = ρπ Then m =

a 2 = kh;

k y2 =

Iy m

(

= a 2 + 3h 2

)

I y = m a 2 + 3h 2

)

ky =

)

(

+ 3h 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 127.

dy = dx

First note

 dy  1+    dx 

Then

2 2  2 −x  a 3 − x 3      2 2 2 − = 1 + x 3 a3 − x3      −

1 3

 a 3 =   x

 dy  dm = m′dL = m′ 1 +   dx  dx 

For the element shown

 a 3 = m′   dx x 1

Then

Now

1  2 3 3 m = ∫ dm = m′ 1 dx = m′a 3  x 3  = m′a 2 2   0 x3 1 3  2   2 3   a a 2 3 3 I x = ∫ y dm = ∫ 0  a − x   m′ 1 dx       x3   

∫ 0a

 2 1 4 1 2 5 a = m′a 3 ∫ 0a  1 − 3a 3 x 3 + 3a 3 x − x 3  dx    3  x  =

1 m′a 3

2 4 4 2 8 3 9 3 3 3 3 2 3 3 2 3  a x − a x + a x − x  4 2 8   2 0

 3 9 3 3 3 = m′a3  − + −  = m′a3  2 4 2 8 8 1 2 ma 4 1 I y = ma 2 4

or I x = Symmetry implies

continued

COSMOS: Complete Online Solutions Manual Organization System

Alternative Solution



I y = ∫ x dm =

= Also

1 m′a 3

∫ 0a

 1 5  a 3 3 ′ ′ x  m 1 dx  = m a ∫ 0 x dx  x3    2

8 3  3 ×  x 3  = m′a3 8  8  0

1 2 ma 4

(

)

I z = ∫ x 2 + y 2 dm = I y + I x

or I z =

1 2 ma 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 128.

ζ =−

For line BC

h x+h a 2

h ( a − 2x ) a

1  m = ρV = ρ t  ah  2 

Also

1 ρ tah 2 2

(a) Have

dI x = =

1 2 ζ dm′ 3

dm′ = ρ tζ dx

where Then

1 2 ζ  ζ dm′ +   dm′ 12 2

a 1 I x = ∫ dI x = 2 ∫ 02 ζ 2 ( ρ tζ dx ) 3

a h 2  ρ t ∫ 02  ( a − 2 x )  dx 3 a  

2 h3 1  1  4 = ρ t 3 ×  −  ( a − 2 x )  2  0 3 a 4 2 =− =

1 h3 4 4 ρ t 3 ( a − a ) − ( a )   12 a 

1 ρ tah3 12

or I x = continued

1 2 mh 6

COSMOS: Complete Online Solutions Manual Organization System

Iζ = ∫ x 2dm

Now and

Iζ = ∫ x 2dm′ = 2 ∫ 02 x 2 ( ρ tζ dx ) a h  = 2 ρ t ∫ 02 x 2  ( a − 2 x )  dx a  

h a 1 2 = 2 ρ t  x3 − x 4  a 3 4 0 3

h a  a  1a = 2ρ t    −   a  3  2  4 2

= (b) Have

4

 

1 1 ρ ta3h = ma 2 48 24

2 I y = ∫ ry2dm = ∫  x 2 + (ζ sin θ )  dm  

= ∫ x 2dm + sin 2 θ ∫ ζ 2dm Now

I x = ∫ ζ 2dm ⇒ I y = Iζ + I x sin 2 θ

1 1 ma 2 + mh 2 sin 2 θ 24 6

or I y = (c) Have

(

m 2 a + 4h 2 sin 2 θ 24

(

)

I z = ∫ rz2dm = ∫ x 2 + y 2 dm 2 = ∫  x 2 + (ζ cosθ )  dm  

= ∫ x 2dm + cos 2 θ ∫ ζ 2dm = Iζ + I x cos 2 θ =

1 1 ma 2 + mh 2 cos 2 θ 24 6 or I z =

m 2 a + 4h 2 cos 2 θ 24

(

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 129. Mass of cylindrical ring: m = ρV = =

γ π

π 2 2  d 2 − d1  t g4 4 

π γ 4 g

2 2

)

− d12 t

Now treat the wheel as a series of 4 concentric rings. (Note - the steel is treated as a large ring minus two smaller rings.) mwheel = ∑ mring =

π 0.310 lb/in 3 4 32.2 ft/s +

π 0.284 lb/in 3 4 32.2 ft/s 2

−2 × +

(

)

× 1.5 in. × 0.7 2 − 0.52 in 2

(

π 0.284 lb/in 3 4 32.2 ft/s 2

π 0.043 lb/in 3 4 32.2 ft/s 2

)

× 1.5 in. × 4.42 − 0.7 2 in 2 ×

1.1 in. × 42 − 1.22 in 2 2

(

)

× 1.5 in. × 52 − 4.42 in 2

(

mwheel = 2.7221 × 10−3 + 196.072 × 10−3 − 110.945 × 10−3 + 8.8730 × 10−3 = 96.722 × 10−3

) lbft⋅s

lb ⋅ s 2 ft continued

COSMOS: Complete Online Solutions Manual Organization System

For a cylindrical ring:

( I AA′ )ring =

1  d2  1 d  m2   − m1  1  2  2  2  2

1π γt 2 2 1π γt 2 2 d2  d2 −  d1  d1  8 4 g 8 4 g   π γt 4 d 2 − d14 = 32 g =

(

= =

π γt 32 g

)

2 2

)(

− d12 d 22 + d12

1 mring d 22 + d12 8

(

)

Then: I AA′ = ∑ ( I AA′ )ring

 1ft  1 lb ⋅ s 2  2 2 2 =  2.7221 × 10−3  0.7 + 0.5 in ×   8 ft   12 in. 

(

)

 1 ft  1 lb ⋅ s 2  2 2 2 + 196.072 × 10−3  4.4 + 0.7 in ×   8 ft   12 in. 

(

−

)

2  1 ft  1 2 2 2 −3 lb ⋅ s  × 110.945 10   4 + 1.2 in ×   8 ft   12 in. 

(

)

 1 ft  1 lb ⋅ s 2  2 2 2 +  8.8730 × 10−3  5 + 4.4 in ×   8 ft   12 in. 

(

)

(

)

= 1.74857 × 10−6 + 3.3785 × 10−3 − 1.67958 × 10−3 + 341.67 × 10−6 lb ⋅ ft ⋅ s 2 = 2.0423 × 10−3 lb ⋅ ft ⋅ s 2

I AA′ = 2.04 × 10−3 lb ⋅ ft ⋅ s 2 And

k AA′ =

I AA′ m

2.04 × 10−3 lb ⋅ ft ⋅ s 2 96.722 × 10−3 lb ⋅ s 2/ft

= 0.145311 ft k AA′ = 1.744 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 130.

First note for the cylindrical ring shown that

m = ρV = ρ t ×

(d 4

2 2

)

− d12 =

π 4

(

ρ t d 22 − d12

)

and, using Figure 9.28, that 2

I AA′

1 d  1 d  = m2  2  − m1  1  2  2  2  2

1  π 2 2  π 2  2  ρ t × d 2  d 2 −  ρ t × d1  d1  8  4  4   

1π  4 4  ρ t  d 2 − d1 8 4 

1π  2 2 2 2  ρ t  d 2 − d1 d 2 + d1 8 4 

1 m d12 + d 22 8

(

)

)(

)

Now treat the roller as three concentric rings and, working from the bronze outward, have Have

π 4

{(8580 kg/m ) ( 0.0195 m ) ( 0.009 m ) 3

(

)

(

)

2 − ( 0.006 m )  

2 2 + 2770 kg/m 3 ( 0.0165 m ) ( 0.012 m ) − ( 0.009 m )   

}

2 2 + 1250 kg/m 3 ( 0.0165 m ) ( 0.027 m ) − ( 0.012 m )   

[7.52895 + 2.87942 + 12.06563] × 10−3 kg 4 = 5.9132 × 10−3 kg + 2.26149 × 10 −3 kg =

+ 9.47632 × 10−3 kg = 17.6510 × 10−3 kg continued

COSMOS: Complete Online Solutions Manual Organization System

And

I AA′ =

{(

)

(

)

(

)

1 2 2 5.9132 × 10−3 kg ( 0.006 ) + ( 0.009 )  m 2   8 2 2 + 2.26149 × 10−3 kg ( 0.009 ) + ( 0.012 )  m 2  

2 2 + 9.47632 × 10−3 kg ( 0.012 ) + ( 0.027 )  m 2  

}

1 ( 691.844 + 508.835 + 8272.827 )10−9 kg ⋅ m 2 8

= 1.18419 × 10−6 kg ⋅ m 2

or I AA′ = 1.184 × 10−6 kg ⋅ m 2 Now

2 k AA ′ =

I AA′ 1.18419 × 10−6 kg m 2 = m 17.6510 × 10−3 kg

= 67.08902 × 10−6 m 2

k AA′ = 8.19079 × 10−3 m or k AA′ = 8.19 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 131. Consider shell to be formed by removing hemisphere of radius r hemisphere of radius r + t. For hemisphere:

from

I =

2 2 mr 5

m = ρV =

Area =

1 4π r 2 = 2π r 2 2

(

)

1 4 3 2 ρ π r = ρπ r 3 2 3 3 I =

Thus

2 2 4 3 2 ρπ r 5  ρπ r  r = 53 15 

For hemispherical shell: I =

4 4 5 ρπ ( r + t ) − r 5  = ρπ  r 5 + 5r 4t + 10r 3t 2 + ... − r 5    15 15

Neglect terms with powers of t > 1, I =

4 ρπ r 4t 3

Mass of shell:

(

I = 2 ρπ r 2t

(

)

m = ρV = ρ tA = ρ t 2π r 2 = 2 ρπ r 2t

) 23 r

2 2 mr 3

I =

2 2 mr 3

Radius of gyration: k2 =

I 2 = r2 m 3

k = 0.816r

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 132. For solid cylinder: I =

∴I =

1 2 ma 2

m = ρV = ρπ a 2h

1 ρπ a 4h 2

For ring: I AA′ =

1 1 1 ρπ a24h − ρπ a14h = ρπ h a24 − a14 2 2 2

(

m = ρπ h a22 − a12

)

(a) By parallel-axis theorem I BB′ = I AA′ + ma12 =

1 ρπ h a24 − a14 + ρπ h a22 − a12 a12 2

(

)

(

)

I BB′ =

1 ρπ h a24 + 2a22a12 − 3a14 2

(

)

(b) For Maximum I BB′ : dI BB′ 1 = ρπ h 4a22a1 − 12a13 = 0 da1 2

(

a1 = 0;

)

(

)

4 a22 − 3a12 = 0

a12 =

1 2 a2 3

a1 =

1 a2 3

2 4  1 a  a   ρπ h  a24 + 2a22  2  − 3  2   2   3  3  

1 2 1   ρπ h  a24 + a24 − a24  2 3 3  

1 4  ρπ h  a24  2 3 

I BB′ =

2 ρπ ha24 3

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 133.

First note and Now

L L − 240 = 120 80

L = 720 mm

m = ρ stV 2 1  π m1 = ρ st  π a12h1  = × 7850 kg/m 3 × ( 0.120 m ) ( 0.720 m ) 3 3  

= 85.230 kg 3 2  2 m2 = ρ st  π a23  = π × 7850 kg/m 3 × ( 0.090 m ) = 11.9855 kg 3  3 2 1  π m3 = ρ st  π a32h3  = × 7850 kg/m 3 × ( 0.080 m ) ( 0.720 − 0.240 ) m 3  3

= 25.253 kg Now

( )1 − ( I y )2 − ( I y )3

Iy = Iy

continued

COSMOS: Complete Online Solutions Manual Organization System

( IGG′ )cup ( I AA′ )cup

(b) Have

= 0.01

5 5  mcup a 2 = 0.01mcup  a 2 + 2la + l 2  12 3 

Now let ζ =

( From Part a )

a l

Then

5  5ζ 2 = 0.12  ζ 2 + 2ζ + 1 3 

40ζ 2 − 2ζ − 1 = 0

Then or

ζ =

2±

ζ = 0.1851

( −2 )2 − 4 ( 40 )( −1) 2 ( 40 ) and

ζ = − 0.1351 ∴

a = 0.1851 l

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 134.

Have

dB =

(a) and

4 − d A = ( 0.33333 − d A ) ft 12

I AA′ = I GG′ + md A2 I BB′ = I GG′ + md B2

Then

(

I BB′ − I AA′ = m d B2 − d A2

)

2 = m ( 0.33333 − d A ) − d A2   

= m ( 0.11111 − 0.66666d A ) Then

(1.26 − 0.6 ) × 10−3 lb ⋅ ft ⋅ s 2 =

0.40 lb ( 0.11111 − 0.66666d A ) ft 2 32.2 ft/s 2

d A = 0.08697 ft d A = 1.044 in.

I AA′ = I GG′ + md A2

(b) or

I GG′ = 0.6 × 10−3 lb ⋅ ft ⋅ s 2 −

0.4 lb 2 0.08697 ft ) 2( 32.2 ft /s

= 0.50604 × 10−3 lb ⋅ ft ⋅ s 2 Then

2 kGG ′ =

I GG′ 0.50604 × 10−3 lb ⋅ ft ⋅ s 2 = 0.4 lb m 32.2 ft/s 2

= 0.04074 ft 2 kGG′ = 0.20183 ft = 2.4219 in. or kGG′ = 2.42 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 135.

marm = ρVarm = ρ ×

(a) First note

π 4

d 2l

dmcup = ρ dVcup

and

= ρ ( 2π a cosθ )( t )( adθ )  π

mcup = ∫ dmcup = ∫ 02 2πρ a 2t cosθ dθ

Then

= 2πρ a 2t [sin θ ]02 = 2πρ a 2t

( I AA′ )anem. = ( I AA′ )cups + ( I AA′ )arms

Now

Using the parallel-axis theorem and assuming the arms are slender rods, have

( I AA′ )anem.

2  = 3 ( I GG′ )cup + mcup d AG  

2  + 3  I arm + marm d AG arm  2  5  2  a    = 3  mcup a 2 + mcup ( l + a ) +      2    12 

2 1 l  +3  marml 2 + marm     2    2

5  = 3mcup  a 2 + 2la + l 2  + marml 2 3  

5  π  = 3 2πρ a 2t  a 2 + 2la + l 2  +  ρ d 2l  l 2 3  4 

(

( I AA′ )anem

)

( )

  5 a2  d 2l  a = πρ l 2 6a 2t  2 + 2 + 1 +  l 4  3 l   continued

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 136.

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, have

I AA′ = ( I AA′ )1 − ( I AA′ )2 2 1 a  2 2 =  m1 b + b + m1     2   12

(

)

2 1 a  −  m2 a 2 + a 2 + m2     2   12

(

)

1  1  5  = ρ b2 L  b2 + a 2  − ρ a 2 L  a 2  4  6  12 

(

)

ρL

( 2b 12

(

+ 3b 2a 2 − 5a 4

)

dI AA′ ρL = 6b 2a − 20a3 = 0 12 da

(

Then

a=0

)

3 10

a=b

and

d 2 I AA′ ρL 1 6b 2 − 60a 2 = ρ L b 2 − 10a 2 = 2 12 2 da

Also

(

)

(

)

and for

a=b

3 d 2 I AA′ , <0 10 da 2

a=b

3 10

Now, for

a = 0, ∴

d 2 I AA′ >0 da 2

( I AA′ )max

occurs when

a = 84

3 = 46.009 mm 10

a = 46.0 mm continued

COSMOS: Complete Online Solutions Manual Organization System

(b) From part (a)

( I AA′ )mass

2 4  3  3   2b + 3b  b = − 5 b  10   10   12       

ρ L 

2

49 49 4 ρ Lb 4 = 2800 kg/m 3 ( 0.3 m )( 0.084 m ) 240 240

(

)

= 8.5385 × 10−3 kg ⋅ m 2

( I AA′ )mass

2 k AA ′ =

and

= 8.54 × 10−3 kg ⋅ m 2

( I AA′ )mass m

where

(

m = m1 − m2 = ρ L b − a

)

2     3 7 2 = ρ L b −  b ρ Lb 2   =    10 10    

Then 2 k AA ′

49 ρ Lb 4 7 2 7 240 b = = = (84 mm )2 = 2058 mm 2 7 24 24 2 ρ Lb 10

k AA′ = 45.3652 mm or k AA′ = 45.4 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 137. m = ρV =

m1 =

Then:

γ g

 1 ft  490 lb/ft 3 × 0.08 in. × ( 3.6 in.)(1.2 in.) ×   2 32.2 ft/s  12 in. 

lb ⋅ s 2 ft

= 3.0435 × 10−3

 1 ft  490 lb/ft 3 1 m2 = × 0.08 in. × (1.8 in.)(1.2 in.) ×   2 2 32.2 ft/s  12 in.  lb ⋅ s 2 ft

= 760.87 × 10−6

 1 ft  490 lb/ft 3 π 2 × 0.08 in. × (1.8 in.) ×  m3 =  2 2 32.2 ft/s  12 in.  = 3.5855 × 10−3

lb ⋅ s 2 ft

I x = ( I x )1 + ( I x )2 + ( I x )3

Now

2 2 1  2   1 ft  −3 lb ⋅ s  I x =   3.0435 × 10  (1.2 in.)    ft   3    12 in.  2 2 2 1   −6 lb ⋅ s  −6 lb ⋅ s  2 2 2 2 2 2  1 ft  +   760.87 × 10  1.8 + 1.2 in +  760.87 × 10  0.6 + 0.4 in    ft  ft  18    12 in.  

(

)

(

)

2 2  2   1 ft  −3 lb ⋅ s   1  +  3.5855 × 10    (1.8 in.)    ft   4     12 in. 

(

) (

)

=  10.1450 × 10−6 + 1.37379 × 10−6 + 2.7476 × 10−6 + 20.168 × 10−6  lb ⋅ ft ⋅ s 2  

= 34.435 × 10−6 lb ⋅ ft ⋅ s 2

I x = 34.4 lb ⋅ ft ⋅ s 2 continued

COSMOS: Complete Online Solutions Manual Organization System

( )1 + ( I y )2 + ( I y )3

Iy = Iy

2 1  lb ⋅ s 2  2   1 ft  =   3.0435 × 10−3 3.6 in. )  (  ft   3    12 in.  2 2 1   lb ⋅ s 2  2 −6 lb ⋅ s  2 2 2  1 ft  +   760.87 × 10−6 + × + 1.8 in. 760.87 10 1.8 0.6 in  )  (   ft  ft  18    12 in.  

(

)

2 2 2   2   1 ft 2  16  2  2 −3 lb ⋅ s   1 −3 lb ⋅ s   4 × 1.8  +  3.5855 × 10  −     (1.8 in.) +  3.5855 × 10  + 1.8  in  ft   2 9π 2  ft   3π      12 in.  

(

) (

)

=  91.305 × 10−6 + 0.95109 × 10−6 + 19.0218 × 10−6 + 25.805 × 10 −6 + 92.205 × 10−6  lb ⋅ ft ⋅ s 2   = 232 × 10−6 lb ⋅ ft ⋅ s 2

I y = 232 × 10−6 lb ⋅ ft ⋅ s 2 I z = ( I z )1 + ( I z )2 + ( I z )3 2 1   lb ⋅ s 2  2 2 2  1 ft  =   3.0435 × 10−3 + 3.6 1.2 in    ft   3    12 in. 

(

)

2 2 1   lb ⋅ s 2  2 −6 lb ⋅ s  2 2 2  1 ft  +   760.87 × 10−6 + × + 1.2 in. 760.87 10 1.8 0.4 in )   (   ft  ft  18    12 in.  

(

)

2 2 1   lb ⋅ s 2  2 2   1 ft  −3 lb ⋅ s  1.8 in. 3.5855 10 1.8 in. +   3.5855 × 10−3 + × )  )  ( (  ft  ft    4    12 in. 

(

) (

)

=  101.45 × 10−6 + 0.42271 × 10−6 + 17.9650 × 10−6 + 100.842 × 10−6  lb ⋅ ft ⋅ s 2  

= 221 × 10−6 lb ⋅ ft ⋅ s 2

I z = 221 × 10−6 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

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 9.137–9.142. Thin rectangular plate

( I x )m

( )m + md 2

= I x′ =

 b  2  h  2  1 m b 2 + h 2 + m   +    12  2    2 

1 m b2 + h2 3

(

)

(

)

( I y )m = ( I y′ )m + md 2 =

1 b mb 2 + m   12 2

1 2 mb 3

( )m + md 2

I z = I z′

1 h = mh 2 + m   12 2 =

1 2 mh 3

Thin triangular plate Have

1  m = ρV = ρ  bht  2   1 3 bh 36

and

I z, area =

Then

I z, mass = ρ tI z, area

= ρt × =

1 3 bh 36

1 mh 2 18 continued

COSMOS: Complete Online Solutions Manual Organization System

I y, mass =

Similarly, Now

1 mb 2 18

1 m b2 + h2 18

(

I x, mass = I y, mass + I z, mass =

)

Thin semicircular plate Have

π  m = ρV = ρ  a 2t  2 

And

I y, area = I z, area =

Then

π 8

I y, mass = I z , mass = ρ tI y, area

= ρt ×

π 8

1 2 ma 4

I x, mass = I y, mass + I z, mass =

Now

1 2 ma 2

Also

I x, mass = I x′, mass + my 2

1 I x′, mass = m  2

And

I z, mass = I z′, mass + my 2

1 I z′, mass = m  4

Thin quarter-circular plate

y = z =

Have

π  m = ρV = ρ  a 2t  4 

and

I y, area = I z, area =

Then

I y, mass = I z, mass = ρ tI y, area

π 16

= ρt × =

4a 3π

π 16

1 2 ma 4 continued

COSMOS: Complete Online Solutions Manual Organization System

1 2 ma 2

Now

I x, mass = I y, mass + I z, mass =

Also

I x, mass = I x′, mass + m y 2 + z 2

(

32  2 1 I x′, mass = m  − a 2  2 9π 

and

I y, mass = I y′, mass + mz 2

16  2 1 I y′, mass = m  − a 2  4 9π 

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 138.

First compute the mass of each component. Have m = ρ stV = ρ st tA

(

)

m1 = 7850 kg/m 3 ( 0.003 m )( 0.70 m )( 0.780 m ) = 12.858 kg  π  2 m2 = 7850 kg/m 3 ( 0.003 m )   ( 0.39 m )  = 5.6265 kg  2  

(

)

 1   m3 = 7850 kg/m 3 ( 0.003 m )   ( 0.780 m )( 0.3 m )  = 2.7554 kg 2   

(

)

Using Fig. 9.28 for component and the equations derived above for components and have I x = ( I x )1 + ( I x )2 + ( I x )3 2 1 2  0.78   2 = (12.858 kg )  ( 0.78 ) +   m  2   12 2   1 16  2  4 × 0.39  2  2 0.39 + ( 5.6265 kg )  − +  )   + ( 0.39 )   m 2 (  3π     2 9π 

 0.78 2  0.30 2   2 2 2 1 + ( 2.7554 kg )  ( 0.78 ) + ( 0.30 )  +  +   m   3  18   3    

= ( 2.6076 + 1.2836 + 0.3207 ) kg ⋅ m 2 = 4.2119 kg ⋅ m 2 or I x = 4.21 kg ⋅ m 2 continued

COSMOS: Complete Online Solutions Manual Organization System

And

( )1 + ( I y )2 + ( I y )3

Iy = Iy

1 2 2 2 2  1 = (12.858 kg )  ( 0.7 ) + ( 0.78 )  + ( 0.7 ) + ( 0.78)   m 2  4  12  2 2 1 + ( 5.6265 kg )  ( 0.39 ) + ( 0.39 )  m 2 4   2  1  2 2  0.78    4 + ( 2.7554 kg )  ( 0.78 ) + ( 0.7 ) +    m 18  3     

= ( 4.7077 + 1.0697 + 1.6295 ) kg ⋅ m 2 = 7.4069 kg ⋅ m 2 or I y = 7.41 kg ⋅ m 2 And

I z = ( I z )1 + ( I z )2 + ( I z )3

 1  2 1 = (12.858 kg ) ( 0.7 )  +  m 2  12 4    2 1 + ( 5.6265 kg )  ( 0.39 )  m 2 4   2  1  2 2  0.30    2 + ( 2.7554 kg )  ( 0.3) + ( 0.70 ) +    m 18  3     

= ( 2.1001 + 0.21395 + 1.39145 ) kg ⋅ m 2 = 3.7055 kg ⋅ m 2

or I z = 3.71 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 139. t = 2 mm

ρ = 7850 kg/m3

Part :

(

)

m1 = ρV = 7850 kg/m 3 ( 0.08 m )( 0.1 m )( 0.002 m ) = 0.1256 kg

I x = I x + md 2 =

1 ( 0.1256 kg )( 0.08 m )2 + ( 0.1256 kg ) ( 0.1 m )2 + ( 0.04 m )2  12

(

)

= 66.99 × 10−6 + 1.457 × 10−3 kg ⋅ m 2 = 1.524 × 10 −3 kg ⋅ m 2

I y = I y + md 2 =

1 ( 0.1256 kg )( 0.1 m )2 + ( 0.1256 kg ) ( 0.05 m )2 + ( 0.1 m )2  12

(

)

= 0.1047 × 10−3 + 1.57 × 10−3 kg ⋅ m 2 = 1.675 × 10−3 kg ⋅ m 2 I z = I z + md 2 =

1 ( 0.1256 kg ) ( 0.1 m )2 + ( 0.08 m )2  + ( 0.1256 kg ) ( 0.05 m )2 + ( 0.04 m )2  12

( ) = ρV = ( 7850 kg/m ) ( 0.2 m )( 0.2 m )( 0.002 m ) = 0.628 kg

= 0.1717 × 10−3 + 0.515 × 10−3 kg ⋅ m 2 = 0.687 × 10−3 kg ⋅ m 2

Part :

Ix =

1 1 ma 2 = ( 0.628 kg )( 0.2 m )2 = 2.093 × 10−3 kg ⋅ m 2 12 12 continued

COSMOS: Complete Online Solutions Manual Organization System

Iy =

1 1 m a2 + b2 = ( 0.628 kg ) ( 0.2 m )2 + ( 0.2 m )2  = 4.187 × 10−3 kg ⋅ m 2 12 12

Iz =

1 1 mb 2 = ( 0.628 kg )( 0.2 m )2 = 2.093 × 10−3 kg ⋅ m 2 12 12

(

)

Part : Same values as Part Total mass moment of inertia:

(

)

(

)

(

)

I x = 2 1.524 × 10−3 kg ⋅ m 2 + 2.093 × 10−3 kg ⋅ m 2

I y = 2 1.675 × 10−3 kg ⋅ m 2 + 4.187 × 10−3 kg ⋅ m 2 I z = 2 0.687 × 10−3 kg ⋅ m 2 + 2.093 × 10−3 kg ⋅ m 2

I x = 5.14 × 10−3 kg ⋅ m 2 I y = 7.54 × 10−3 kg ⋅ m 2

I z = 3.47 × 10−3 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 140.

First compute the mass of each component

m = ρV = ρ At

Have Now

( ) = ( 7530 kg/m ) ( 0.002 m )( 0.045 m )( 0.020 m ) = 0.013554 kg

m1 = 7530 kg/m 3 ( 0.002 m )( 0.045 m )( 0.070 m ) = 0.047439 kg m2

(

)

m3 = 7530 kg/m 3 ( 0.002 m ) ×

1 ( 0.04 m )( 0.095 m ) = 0.028614 kg 2

Using Fig. 9.28 for components and and the equations derived above for components , have I x = ( I x )1 + ( I x )2 + ( I x )3 2 2 2 2  1 1 = ( 0.047439 kg )  ( 0.07 )  m 2 + ( 0.013554 kg )  ( 0.020 ) + ( 0.07 ) + ( 0.01)   m 2   3  12

1 2 2 2 2  1 + ( 0.028614 kg )  ( 0.095 ) + ( 0.04 )  + ( 2 × 0.095) + ( 0.040 )   m 2     9 18

= ( 0.077484 + 0.068222 + 0.136751) × 10−3  kg ⋅ m 2 = 0.282457 × 10−3 kg ⋅ m 2 or I x = 0.2825 × 10−3 kg ⋅ m 2 continued

COSMOS: Complete Online Solutions Manual Organization System

continued

( )1 + ( I y )2 + ( I y )3

Iy = Iy

 2 2 2  1  1 = ( 0.047439 kg )  ( 0.045 )  m 2 + ( 0.013554 kg )  ( 0.045 ) + ( 0.02 )     m 2    3  3  

1 2 2 2 1 + ( 0.028614 kg )  ( 0.04 ) + ( 0.045 ) + ( 0.04 )  m 2 9  18 

= ( 0.03202 + 0.010956 + 0.065574 ) × 10−3  kg ⋅ m 2 = 0.10855 × 10−3 kg ⋅ m 2 or I y = 0.1086 × 10−3 kg ⋅ m 2 I z = ( I z )1 + ( I z )2 + ( I z )3

  2 2  1  21 2 = ( 0.047439 kg )  ( 0.045 ) + ( 0.070 )     m 2 + ( 0.013554 kg ) ( 0.045 )   + ( 0.070 )  m 2   3 3        2 1 2 2   + ( 0.028614 kg )  ( 0.095 ) + 0.0452 +  0.095   m 2 3   18

= ( 0.0109505 + 0.075564 + 0.187064 ) × 10−3  kg ⋅ m 3 = 0.37213 × 10−3 kg ⋅ m 2 or I z = 0.372 × 10−3 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 141. t = 0.1 in.

γ st = 490 lb/ft 3 Have m = ρstV =

γ st g

490 lb/ft 3  0.1  ×  ft × A 32.2 ft/s 2  12 

 lb ⋅ s 2 2  =  0.126812 ⋅ ft  A ft    lb ⋅ s 2 2   16 12  2 lb ⋅ s 2 m1 =  0.126812 ⋅ ft   ×  ft = 0.169083 ft ft    12 12 

Then

 lb ⋅ s 2 2   16 9  2 lb ⋅ s 2 m2 =  0.126812 ⋅ ft   ×  ft = 0.126812 ft ft    12 12   lb ⋅ s 2 2   1 12 9  2 lb ⋅ s 2 m3 =  0.126812 ⋅ ft   × ×  ft = 0.047555 ft ft    2 12 12  2  lb ⋅ s 2 2   π  5   2 lb ⋅ s 2 m4 =  0.126812 ⋅ ft      ft = 0.034583 ft ft    2  12  

Using Fig. 9.28 for components and and the equations derived previously for components and , have I x = ( I x )1 + ( I x )2 + ( I x )3 − ( I x )4

Where

( I x )1

1 lb ⋅ s 2   12  2 =  0.169083   ft  = 0.056361 lb ⋅ ft ⋅ s 3 ft   12 

( I x )2

( I x )3

2 2 2 2  lb ⋅ s 2   1  12   9    4   3    2 =  0.047555     +    +   +     ft ft  18  12   12    12   12     

1 lb ⋅ s 2   9  0.126812 ft  = 0.023777 lb ⋅ ft ⋅ s 2   3  ft   12 

continued

COSMOS: Complete Online Solutions Manual Organization System

= 0.0123841 lb ⋅ ft ⋅ s 2

( I x )4

1 lb ⋅ s 2   5  ft  = 0.00150100 lb ⋅ ft ⋅ s 2  0.034583  4 ft   12 

Then I x = ( 0.056361 + 0.023777 + 0.0123841 − 0.00150100 ) lb ⋅ ft ⋅ s 2

I x = 91.0 × 10−3 lb ⋅ ft ⋅ s 2

( )1 + ( I y )2 + ( I y )3 − ( I y )4

Iy = Iy Where

2 2 1 lb ⋅ s 2   16   12   2 2  0.169083    +    ft = 0.156558 lb ⋅ ft ⋅ s 3 ft   12  12   

( )1

( I y )2

1 lb ⋅ s 2   16  2 =  0.126812   ft  = 0.075148 lb ⋅ ft ⋅ s 3 ft   12 



( I y )3 =  0.047555 lbft⋅ s

  1  12 2  4 2  2 2     +    ft = 0.0079258 lb ⋅ ft ⋅ s 18 12 12        

2 2  8  2  4   1 16  5  5   2  + + × ft    −       ft  2   3π 12     12    2 9π  12 





( I y )4 =  0.034583 lbft⋅ s 

= 0.0183722 lb ⋅ ft ⋅ s 2 Then I y = ( 0.156558 + 0.075148 + 0.0079258 − 0.0183722 ) lb ⋅ ft ⋅ s 2

I y = 221 × 10−3 lb ⋅ ft ⋅ s 2 I z = ( I z )1 + ( I z )2 + ( I z )3 − ( I z )4 Where

( I z )1 =

1 lb ⋅ s 2   16  2  0.169083   ft  = 0.100197 lb ⋅ ft ⋅ s 3 ft   12 

( I z )2

2 2 1 lb ⋅ s 2   16   9  2 2  0.126812    +    ft = 0.098925 lb ⋅ ft ⋅ s 3 ft   12   12  

2 2  lb ⋅ s 2   1  9   3  2 2 0.047555 I = +  ( z )3        ft = 0.0044583 lb ⋅ ft ⋅ s ft 18 12 12        

( I z )4

2 2  lb ⋅ s 2   1  5   8   2 =  0.034583    ft  +  ft   = 0.0168712 lb ⋅ ft ⋅ s ft 4 12 12         

Then I z = ( 0.100197 + 0.098925 + 0.0044583 − 0.0168712 ) lb ⋅ ft ⋅ s 2

I z = 186.7 × 10−3 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 142.

Have Then

m = ρcuV = ρcu tA

(

)

m1 = m2 = 8940 kg/m 3 ( 0.0008 m )(1.2 m )( 0.15 m ) = 1.28736 kg

Using Fig. 9.28 for components and , have

I x = ( I x )1 + ( I x )2 and ( I x )1 = ( I x )2 Then

2 1 I x = 2  (1.28736 kg )( 0.15 m )  = 0.0193104 kg ⋅ m 2 3 

I x = 19.31 × 10−3 kg ⋅ m 2

( )1 + ( I y )2

Also

Iy = Iy

Where

( I y )1 = 13 (1.28736 kg ) (1.2 m )2 + ( 0.15 m )2  = 0.62759 kg ⋅ m 2

and

( I y )2 = ∫ ry2 dm ry2 = x 2 + z 2 = x 2 + (ζ cos 30° )

dm = ρ dV = ρ cu tdζ dx

Then

( I y )2 = ρcu t ∫ 0L ∫ 0a ( x2 + ζ 2 cos2 30°) dζ dx continued

COSMOS: Complete Online Solutions Manual Organization System

1 L  = ρcu t ∫ 0  ax 2 + a3 cos 2 30°  dx 3  

Thus

1 ρcu t aL3 + a3L cos2 30° 3

1 m2 L2 + a 2 cos 2 30° 3

1 (1.28736 kg ) (1.2 m )2 + ( 0.15 m )2 cos2 30° = 0.62517 kg ⋅ m 2 3

(

)

where V = aLt

)

I y = ( 0.62759 + 0.62517 ) kg ⋅ m 2 I y = 1.253 kg ⋅ m 2

Also

I z = ( I z )1 + ( I z )2

Where

( I z )1 = (1.28736 kg )(1.2 m )2

and

( I z )2

1 3

= 0.61793 kg ⋅ m 2

= ∫ rz2dm

rz2 = x 2 + y 2 = x 2 + (ζ sin 30° )

dm = ρ dV = ρ cu tdζ dx

Then

( I z )2

(

)

= ρcu t ∫ 0 ∫ 0 x 2 + ζ 2 sin 2 30° dζ dx 1 L  = ρcu t ∫ 0  ax 2 + a3 sin 2 30°  dx 3  

1 ρcu t aL3 + a3 L sin 2 30° 3

1 m2 L2 + a 2 sin 2 30° 3

1 (1.28736 kg ) (1.2 m )2 + ( 0.15 m )2 sin 2 30° 3

(

)

where V = aLt

)

= 0.62035 kg ⋅ m 2 Thus

I z = ( 0.61793 + 0.62035 ) kg ⋅ m 2

I z = 1.238 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 143.

m = ρV =

Have

γ g

V =

0.284 lb/in 3 ×V 32.2 ft/s 2

(

)

= 0.0088199 lb ⋅ s 2/ft ⋅ in 3 V

(

)

(

)

(

)

2 Then m1 = 0.0088199 lb ⋅ s 2 /ft ⋅ in 3 π ( 4 ) ( 2 )  in 3 = 0.88667 lb ⋅ s 2 /ft   2 m2 = 0.0088199 lb ⋅ s 2/ft ⋅ in 3 π (1) ( 3)  in 3 = 0.083126 lb ⋅ s 2/ft   2 m3 = 0.0088199 lb ⋅ s 2 /ft ⋅ in 3 π (1) ( 2 )  in 3 = 0.055417 lb ⋅ s 2/ft  

Using Fig. 9-28 and the parallel theorem, have (a)

I x = ( I x )1 + ( I x )2 − ( I x )3 2 2 2 1  1 ft  = 0.88667 lb ⋅ s 2 /ft  3 ( 4 ) + ( 2 )  + (1)  in 2 ×     12    12 in. 

(

)

2 2 2 1  1 ft  + 0.083126 lb ⋅ s 2 /ft  3 (1) + ( 3)  + (1.5 )  in 2 ×     12    12 in. 

(

)

2 2 2 1  1 ft  − 0.055417 lb ⋅ s 2 /ft  3 (1) + ( 2 )  + (1)  in. ×     12    12 in. 

(

)

= 0.034106 lb ⋅ ft ⋅ s 2 or I x = 0.0341 lb ⋅ ft ⋅ s 2 continued

COSMOS: Complete Online Solutions Manual Organization System

( )1 + ( I y )2 − ( I y )3

Iy = Iy

(b)

2 1  1 ft  = 0.88667 lb ⋅ s 2 /ft  ( 4 )  in 2 ×   2   12 in. 

(

)

2 1 2  1 ft  + 0.083126 lb ⋅ s 2 /ft  (1) + ( 2 )  in 2 ×   2   12 in. 

(

)

2 1 2  1 ft  − 0.055417 lb ⋅ s 2 /ft  (1) + ( 2 )  in 2 ×   2   12 in. 

(

)

= 5.0125 × 10−2 lb ⋅ ft ⋅ s 2 or I y = 0.0501 lb ⋅ ft ⋅ s 2 (c)

I z = ( I z )1 + ( I z )2 − ( I z )3 2 2 2 1  1 ft  = 0.88667 lb ⋅ s 2 /ft  3 ( 4 ) + ( 2 )  + (1)  in 2 ×    12    12 in. 

(

)

2 2 2 2  1  1 ft  + 0.083126 lb ⋅ s 2 /ft  3 (1) + ( 3)  + ( 2 ) + (1.5 )   in 2 ×      12   12 in. 

(

)

2 2 2 2  1  1 ft  − 0.055417 lb ⋅ s 2 /ft  3 (1) + ( 2 )  + ( 2 ) + (1)   in 2 ×      12   12 in. 

(

)

= 0.034876 lb ⋅ ft ⋅ s 2 or I z = 0.0349 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 144. m = ρsteel V

Have

m1 = 7850 kg/m 3 ( 0.200 × 0.120 × 0.600 ) m 3

Then

= 113.040 kg m2 = 7850 kg/m 3 × ( 0.200 × 0.080 × 0.360 ) m 3

= 45.216 kg 2 π  m3 = 7850 kg/m 3 ×  ( 0.100 ) ( 0.120 )  m 3 2 

= 14.7969 kg 2 m4 = 7850 kg/m 3 × π ( 0.050 ) ( 0.120 )  m 3  

= 7.3985 kg Using Figure 9.28 for components and and the equations derived above for components and , have

( )1 + ( I y )2 + ( I y )3

Iy = Iy

Now

where

( )1 Iy

 1  0.600 2  0.200 2   2 2 2 = (113.040 kg )  ( 0.600 ) + ( 0.200 )  +  +   m   2  2     12  

= 15.0720 kg ⋅ m 2

continued

COSMOS: Complete Online Solutions Manual Organization System

( )2 Iy

 1  0.360 2  0.200 2   2 2 2 = ( 45.216 kg )  ( 0.360 ) + ( 0.200 )  +  +   m   2  2     12  

= 2.5562 kg ⋅ m 2

( )3 Iy

2   1  16  4 × 0.100    2 2 2  = (14.7969 kg )   − + + + 0.100 0.100 0.600  ( ) ( )    m  2 3π         2 9π 

= 6.3024 kg ⋅ m 2

( I y )4 = ( 7.3985 kg )  12 ( 0.050 )2  + ( 0.100)2 + ( 0.600 )2   m 2 



= 2.7467 kg ⋅ m 2 Then

I y = (15.0720 + 2.5562 + 6.3024 − 2.7467 ) kg ⋅ m 2

= 21.1839 kg ⋅ m 2 or I y = 21.2 kg ⋅ m 2 continued

COSMOS: Complete Online Solutions Manual Organization System

To the Instructor: The following formulas for the mass moment of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.144–9.147. From Figure 9.28

( I x ) cyl

Cylinder:

1 mcyla 2 2

( I y )cyl = ( I z )cyl = 121 mcyl (3a2 + L2 )

(

Symmetry and the definition of the mass moment of inertia I = ∫ r 2dm imply

( I )semicylinder ∴

and

( I x )sc

1 (I ) 2 cylinder

11 2  mcyla  2 2 

( I y )sc = ( I z )sc = 12 121 mcyl ( 3a2 + L2 )

However,

msc =

1 mcyl 2

Thus,

( I x )sc

and

1 msca 2 2

( I y )sc = ( I z )sc = 121 msc ( 3a2 + L2 )

Also, using the parallel axis theorem find

16  2 1 I x′ = msc  − a 2 9 π2    1 16  2 1 2 I z′ = msc  − a + L 2 12   4 9π 

where x′ and z′ are centroidal axes.

)

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 145. See machine elements shown in Problem 9.144

m1 = 113.040 kg

Also note

m2 = 45.216 kg m3 = 14.7969 kg m4 = 7.3985 kg

Using Fig. 9.28 for components and and the equations derived above for components and , have

I z = ( I z )1 + ( I z )2 + ( I z )3 − ( I z )4

Now

where

2 2  1  ( 0.200 )2 + ( 0.120 )2  +  0.200  +  0.120    m 2        2    12   2 

( I z )1 = (113.040 kg )  = 2.0498 kg ⋅ m 2

( I z )2

2 2 2 2  1 = ( 45.216 kg )  ( 0.200 ) + ( 0.080 )  + ( 0.100 ) + ( 0.160 )   m 2     12

= 1.78453 kg ⋅ m 2 1  2 2 2 2  3 ( 0.100 ) + ( 0.120 )  + ( 0.100 ) + ( 0.060 )   m 2     12 

( I z )3 = (14.7969 kg )  = 0.25599 kg ⋅ m 2

( I z )4

2 2 2 2  1 = ( 7.3985 kg )  3 ( 0.050 ) + ( 0.120 )  + ( 0.100 ) + ( 0.060 )   m 2    12 

= 0.114122 kg ⋅ m 2 Then

I z = ( 2.0498 + 1.78453 + 0.25599 − 0.114122 ) kg ⋅ m 2

= 3.97629 kg ⋅ m 2 or I z = 3.98 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 146. m = ρV

= m1 =

γ g

0.100 lb/in 3 ( 4 in.)(1 in.)( 3 in.) 32.2 ft/s 2

= 37.267 × 10−3 m2 =

0.100 lb/in 3 ( 2 in.)(1.2 in.)( 3 in.) 32.2 ft/s 2

= 22.360 × 10−3 m3 =

lb ⋅ s 2 ft

0.100 lb/in 3 π 2 × ( 0.9 in.) ( 2 in.) 2 2 32.2 ft/s

= 7.9028 × 10−3 m4 =

lb ⋅ s 2 ft

0.100 lb/in 3 π 2 × ( 0.5 in.) ( 3 in.) 2 2 32.2 ft/s

= 3.6587 × 10−3

lb ⋅ s 2 ft

continued

COSMOS: Complete Online Solutions Manual Organization System

Have

I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4 2 2 2 1   2 2 2 2  1 ft  −3 lb ⋅ s  2 −3 lb ⋅ s  2 =   37.267 × 10  4 + 1 in +  37.267 × 10  2 + 0.5 in    ft  ft  12    12 in.  

(

)

(

)

2 2 1   lb ⋅ s 2  2 −3 lb ⋅ s  2 2 2 2 2  1 ft  +   22.360 × 10−3 + + × + 2 1.2 in 22.360 10 1 1.6 in      ft  ft  12    12 in.  

(

)

(

)

lb ⋅ s 2   1 16  1 2 2  0.9 in.) + ( 2 in.)  +  7.9028 × 10−3   − 2 ( ft   4 9π  12   2 2 2   4 × 0.9  2    1 ft  2  −3 lb ⋅ s  +  7.9028 × 10  (1 in.) +  2.2 +   in   ft   3π       12 in. 

lb ⋅ s 2   1 16  2  0.5 in.) +  3.6587 × 10−3   − 2 ( ft   2 9π   2 2 2   4 × 0.5  2    1 ft  2  −3 lb ⋅ s  +  3.6587 × 10  ( 3.5 in.) + 1 +   in   ft   3π      12 in.  

(

) (

=  1.46653 × 10−3 + 70.393 × 10−6 + 552.79 × 10−6 

(

)

) (

)

+ 21.400 × 10−6 + 420.75 × 10−6 + 2.0318 × 10−6 + 348.58 × 10−6  lb ⋅ ft ⋅ s 2  = 2.88 × 10−3 lb ⋅ ft ⋅ s 2 I z = 2.88 × 10−3 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 147. Have

m = ρst V =

Then

m1 =

δ st g

 1 ft  490 lb/ft 3 × ( 3 × 1 × 4 ) in 3 ×   2 32.2 ft/s  12 in. 

= 105.676 × 10−3 lb ⋅ s 2 /ft 3

 1 ft  490 lb/ft 3 2 −3 m2 = × (1.5 × 1 × 2 ) in 3 ×   = 26.419 × 10 lb ⋅ s /ft 2 12 in. 32.2 ft/s   3

m3 =

490 lb/ft 3  π 2   1 ft  2 −3 ×  ( 0.5 ) × 1.5 in 3 ×   = 5.1874 × 10 lb ⋅ s /ft 2 32.2 ft/s 2   12 in. 

 1 ft  490 lb/ft 3  π 2  m4 = ×  (1.4 ) × 0.4  in 3 ×   2 32.2 ft/s 2   12 in. 

= 10.8491 × 10−3 lb ⋅ s 2/ft (a)

Using Fig. 9.28 for components and and the equations derived above for components and , have

I x = ( I x )1 + ( I x )2 + ( I x )3 − ( I x )4 where

( I x )1 = (105.676 × 10

−3

 1  1 2  4 2  2  1 ft 2  2 2  lb ⋅ s /ft  (1) + ( 4 )  +   +    in ×      2  12   2    12 in.    

)

= 4.1585 × 10−3 lb ⋅ ft ⋅ s 2

( I x )2

 1 ft  2 2 2 2  1 = 26.419 × 10−3 lb ⋅ s 2 /ft  (1) + ( 2 )  + ( 0.5) + ( 5)   in 2 ×      12   12 in. 

(

)

= 4.7089 × 10−3 lb ⋅ ft ⋅ s 2 continued

COSMOS: Complete Online Solutions Manual Organization System

( I x )3

(

= 5.1874 × 10

2 2   1  16  4 × 0.5    2  1 ft  2 2  lb ⋅ s /ft   −   ( 0.5 )  + ( 0.5 ) +  6 +    in ×  2 9π 2  3π        12 in.    

−3

)

= 1.40209 × 10−3 lb ⋅ ft ⋅ s 2

( I x )4

 1 ft  2 2 1  = 10.8451 × 10−3 lb ⋅ s 2 /ft  3 (1.4 ) + ( 0.4 )  + 22 + 0.82  in 2 ×    12    12 in. 

(

)

(

)

= 0.38736 × 10−3 lb ⋅ ft ⋅ s 2 I x = ( 4.1585 + 4.7089 + 1.40209 − 0.38736 ) × 10−3  lb ⋅ ft ⋅ s 2

Then

= 9.8821 × 10−3 lb ⋅ ft ⋅ s 2 or I x = 9.88 × 10−3 lb ⋅ ft ⋅ s 2 (b) Have

( )1 + ( I y )2 + ( I y )3 − ( I y )4

Iy = Iy

where 2  1  3 2  4 2    1 ft  2 2 = 105.676 × 10−3 lb ⋅ s 2 /ft  ( 3) + ( 4 )  +   +     in 2 ×     2   2     12 in.  12  

( )1 ( Iy

)

= 6.1155 × 10−3 lb ⋅ ft ⋅ s 2

( I y )2 = ( 26.419 × 10

 1 ft  2 2 2 2  1 lb ⋅ s /ft  (1.5 ) + ( 2 )  + ( 0.75 ) + ( 5 )   in 2 ×      12   12 in. 

−3

)

= 4.7854 × 10−3 lb ⋅ ft ⋅ s 2

( I y )3 = (5.1874 × 10

−3

2 2   1  16  1 4 × 0.5    2  1 ft  2 2 2  lb ⋅ s /ft   −   ( 0.5 ) + (1.5 )  + ( 0.75 ) +  6 +    in ×  4 9π 2  12 3π        12 in.    

)

= 1.41785 × 10−3 lb ⋅ ft ⋅ s 2

( I y )4 = (10.8451 × 10

−3

2 2   1    1 ft  16  4 × 1.4  2 2  2 lb ⋅ s /ft   −   (1.4 )  +  3 −  + ( 2 )   in ×  2 9π 2  3π       12 in.    2

)

= 0.78438 × 10−3 lb ⋅ ft ⋅ s 2 continued

COSMOS: Complete Online Solutions Manual Organization System

I y = ( 6.1155 + 4.7854 + 1.41785 − 0.78438 ) × 10−3  lb ⋅ ft ⋅ s 2

Then

= 11.5344 × 10−3 lb ⋅ ft ⋅ s 2 or I y = 11.53 × 10−3 lb ⋅ ft ⋅ s 2

I z = ( I z )1 + ( I z )2 + ( I z )3 − ( I z )4

( I z )1 = (105.676 × 10

−3

 1  3 2  1 2   2  1 ft 2 2 2  lb ⋅ s /ft  ( 3) + (1)  +   +     in ×     2  12   2     12 in.   

)

= 2.4462 × 10−3 lb ⋅ ft ⋅ s 2  1  1.5 2  1 2   2  1 ft 2 2 2 −3 2  I = × ⋅ + + 26.419 10 lb s /ft 1.5 1 ( z )2 ( )    +     in ×   ( )   2     2   12 in.  12

(

)

= 0.198754 × 10−3 lb ⋅ ft ⋅ s 2

( I z )3

(

= 5.1874 × 10

−3

 1 ft  2 2 2 2  1 lb ⋅ s /ft  3 ( 0.5 ) + (1.5)  + ( 0.75) + ( 0.5 )   in 2 ×      12   12 in.  2

)

= 0.038275 × 10−3 lb ⋅ ft ⋅ s 2

( I z )4

(

= 10.8451 × 10

−3

2 2   1    1 ft  16  1 4 × 1.4  2 2 2  2 lb ⋅ s /ft   −   (1.4 ) + ( 0.4 )  +  3 −  + ( 0.8 )   in ×  4 9π 2  12 3π     12 in.      2

)

= 0.49543 × 10−3 lb ⋅ ft ⋅ s 2 Then

I z = ( 2.4462 + 0.198754 + 0.038275 − 0.49543) × 10 −3  lb ⋅ ft ⋅ s 2

= 2.1878 × 10−3 lb ⋅ ft ⋅ s 2

or I z = 2.19 × 10−3 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 148.

First compute the mass of each component.

m = ρL

Have Then

m1 = ( 0.049 kg/m )  2π ( 0.32 m ) 

= 0.09852 kg m2 = m3 = m4 = m5

= ( 0.049 kg/m )( 0.160 m ) = 0.00784 kg Using the equation given above and the parallel axis theorem, have

I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 + ( I x )5  1   1  2 2 = ( 0.09852 kg )   ( 0.32 m )  + ( 0.00784 kg )   ( 0.160 m )   2    3   2 + ( 0.00784 kg ) 0 + ( 0.160 m )   

 1  2 2 2 + ( 0.00784 kg )   ( 0.16 m ) + ( 0.08 m ) + ( 0.32 m )   12    2 2 2 1 + ( 0.00784 kg )  ( 0.16 m ) + ( 0.16 m ) + ( 0.32 m − 0.08 m )   12

= ( 5.0442 + 0.06690 + 0.2007 + 0.86972 + 0.66901) × 10−3  kg ⋅ m 2 = 6.8505 × 10−3 kg ⋅ m 2

or I x = 6.85 × 10−3 kg ⋅ m 2

continued

COSMOS: Complete Online Solutions Manual Organization System

Iy = (Iy ) + (Iy ) + (Iy ) + (Iy ) + (Iy )

Have

(I ) = (I )

where Then

y 2

y 4

and

(I ) = (I ) y 3

y 5

2 2 I y = ( 0.09852 kg ) ( 0.32 m )  + 2 ( 0.00784 kg ) 0 + ( 0.32 m )     

2 2 1 + 2 ( 0.00784 kg )  ( 0.16 m ) + ( 0.24 m )  12 

= 10.088 + 2 ( 0.80282 ) + 2 ( 0.46831)  × 10−3 kg ⋅ m 2 = 12.6303 × 10−3 kg ⋅ m 2 or I y = 12.63 × 10−3 kg ⋅ m 2 By symmetry

Iz = Ix

continued

or I z = 6.85 × 10−3 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

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 9.148–9.150. Slender Rod

Ix = 0

I y′ = I z′ =

1 mL2 (Fig. 9.28) 12

1 2 mL (Sample Problem 9.9) 3

I y = Iz = Circle Have

I y = ∫ r 2dm = ma 2

Now

Iy = Ix + Iz Ix = Iz

And symmetry implies

∴ Ix = Iz =

1 2 ma 2

Semicircle Following the above arguments for a circle, have

I x = Iz =

1 2 ma 2

I y = ma 2

Using the parallel-axis theorem

I z = I z′ + mx 2 or

4  1 I z′ = m  − 2  a 2 2 π 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 149.

m = ρV = ρ AL

Have Then

(

)

2 m1 = m2 = 7850 kg/m 3 π ( 0.0015 m )  × (π × 0.36 m )  

m2 = m1 = 0.062756 kg

(

)

2 m3 = m4 = 7850 kg/m 3 π ( 0.0015 m )  × ( 0.36 m )  

= 0.019976 kg Using the equations given above and the parallel axis theorem, have

I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 where

( I x )3 = ( I x ) 4

Then 2 2 2 1 1 I x = ( 0.062756 kg )  ( 0.36 m )  + ( 0.062756 )  ( 0.36 m ) + ( 0.36 m )  2  2 

2 2 2 1 + 2 ( 0.019976 kg )  ( 0.36 m ) + ( 0.18 m ) + ( 0.36 m )  12  

=  4.06659 + 12.19977 + 2 ( 3.45185 )  × 10−3 kg ⋅ m 2 = 23.1701 × 10−3 kg ⋅ m 2 or I x = 23.2 × 10−3 kg ⋅ m 2

Have where

( )1 + ( I y )2 + ( I y )3 + ( I y )4

Iy = Iy

( I y )1 = ( I y )2

and

( I y )3 = ( I y )4 continued

COSMOS: Complete Online Solutions Manual Organization System

Then 2 2 I y = 2 ( 0.062756 kg ) ( 0.36 m )  + 2 ( 0.019976 kg ) 0 + ( 0.36 m )     

(

) (

= 2 8.13318 × 10−3 kg ⋅ m 2 + 2 2.58889 × 10−3 kg ⋅ m 2

)

= 21.44414 × 10−3 kg ⋅ m 2 or I y = 21.4 × 10−3 kg ⋅ m 2 Have

I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4 where

( I z )3 = ( I z )4

Then 2 1 I z = ( 0.062756 kg )  ( 0.36 m )  2  2  1 4  2  2 × 0.36 m  2 0.36 m + ( 0.062756 kg )  − 2  ( 0.36 m ) +  +  ( )  π    2 π  

2 1 + 2 ( 0.019976 kg )  ( 0.36 m )  3  

=  4.06659 + 12.1998 + 2 ( 0.86296 )  10−3 kg ⋅ m 2 = 17.9923 × 10−3 kg ⋅ m 2 or I z = 17.99 × 10−3 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 150.

First compute the mass of each component. Mass of each component is indentical

Have

( m/L ) L g

( 0.041 lb/ft )(1.5 ft ) 32.2 ft/s 2

= 0.00190994 lb ⋅ s 2 /ft Using the equations given above and the parallel axis theorem, have

( I x )1 = ( I x )3 = ( I x )4 = ( I x )6

and

( I x ) 2 = ( I x )5

I x = 4 ( I x )1 + 2 ( I x )2

Then

2 1 I x = 4 0.00190994 lb ⋅ s 2 /ft  (1.5 ft )  3 

(

)

2 + 2 0.00190994 lb ⋅ s 2/ft 0 + (1.5 ft )   

= 0.0143246 lb ⋅ ft ⋅ s 2 or I x = 14.32 × 10−3 lb ⋅ ft ⋅ s 2 Now

( I y )1 = 0 ( I y )2 = ( I y )6

( I y )4 = ( I y )5 continued

COSMOS: Complete Online Solutions Manual Organization System

Then

( )2 + ( I y )3 + 2 ( I y )4

Iy = 2 Iy

   1  2 2 = 0.0019094 lb ⋅ s 2 /ft   2   (1.5 ft )  + 0 + (1.5 ft )    3     

(

)

2 2 2  1 + 2  (1.5 ft ) + (1.5 ft ) + ( 0.75 ft )   12  

= 0.0019094 (1.5 + 2.25 + 6 ) lb ⋅ ft ⋅ s 2 = 0.0186219 lb ⋅ ft ⋅ s 2 I y = 18.62 × 10−3 lb ⋅ ft ⋅ s 2 By symmetry

Iz = I y I z = 18.62 × 10−3 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 151. From the solution to Problem 9.147

m1 = 105.676 × 10−3 lb ⋅ s 2 /ft

m3 = 5.1874 × 10 −3 lb ⋅ s 2/ft

m2 = 26.419 × 10−3 lb ⋅ s 2/ft

m4 = 10.8451 × 10 −3 lb ⋅ s 2/ft

First note that symmetry implies I x′y′ = I y′z ′ = I z ′x′ = 0 for each component

I uv = I u′v′ + mu v = mu v

Now

( Iuv )body

so that

Then

= Σmu v

 1.5  0.5   0.75  0.5  I xy = Σmx y = 105.676 × 10−3 lb ⋅ s 2 /ft  ft  ft  + 26.419 × 10−3 lb ⋅ s 2 /ft  ft  ft   12  12   12  12 

(

)

(

)

 0.75  0.5  + 5.1874 × 10 −3 lb ⋅ s 2 /ft  ft  ft   12  12   4 × 1.4 in.   1 ft    0.8  − 10.8451 × 10−3 lb ⋅ s 2 /ft  3 in. − ft     3π   12 in.    12  

(

)

= ( 550.40 + 68.799 + 13.5089 − 144.952 ) × 10−6 lb ⋅ ft ⋅ s 2 = 487.76 × 10−6 lb ⋅ ft ⋅ s 2 or I xy = 0.488 × 10−3 lb ⋅ ft ⋅ s 2 continued

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 152. From the solution to Problem 9.146

m1 = 37.267 × 10−3

lb ⋅ s 2 ft

m2 = 22.360 × 10−3

lb ⋅ s 2 ft

m3 = 7.9028 × 10−3

lb ⋅ s 2 ft

m4 = 3.6587 × 10−3

lb ⋅ s 2 ft

I uv = I u′v′ + mu v

Have Symmetry implies

I x′y′ = I y′z ′ = I z ′x′ = 0

for each element.

∴

I uv = ∑ miui v i

Then:

I xy

  1 ft  lb ⋅ s 2  =  37.267 × 10−3  ( 2 in.)( 0.5 in.)   ft   12 in.     1 ft  lb ⋅ s 2  +  22.360 × 10−3  (1 in.)(1.6 in.)   ft   12 in.  

  1 ft  lb ⋅ s 2  4 × 0.9   +  7.9028 × 10−3  (1 in.)  2.2 +   in. ×  π ft 3    12 in.      1 ft  lb ⋅ s 2  4 × 0.5   +  3.6587 × 10−3  ( 3.5 in.) 1 +   in. ×  ft  3π    12 in.  

continued

COSMOS: Complete Online Solutions Manual Organization System

(

)

= 258.80 × 10−6 + 248.44 × 10−6 + 141.700 × 10−6 + 107.80 × 10−6 lb ⋅ ft ⋅ s 2 = 757 × 10−6 lb ⋅ ft ⋅ s 2

I xy = 757 × 10−6 lb ⋅ ft ⋅ s 2   1 ft  lb ⋅ s 2  I yz =  37.267 × 10−3  ( 0.5 in.)(1.5 in.)   ft   12 in.  

  1 ft  lb ⋅ s 2  +  22.360 × 10−3  (1.6 in.)(1.5 in.)   ft   12 in.  

  1 ft  lb ⋅ s 2   4 × 0.9  +  7.9028 × 10−3   2.2 +   in. × (1.5 in.)  ft   3π   12 in.  

  1 ft  lb ⋅ s 2   4 × 0.5  +  3.6587 × 10−3  1 +   in. × (1.5 in.) ×   ft   3π   12 in.  

(

)

= 194.099 × 10−6 + 372.67 × 10−6 + 212.55 × 10−6 + 46.199 × 10 −6 lb ⋅ ft ⋅ s 2 = 826 × 10

−6

lb ⋅ ft ⋅ s

I yz = 826 × 10−6 lb ⋅ ft ⋅ s 2

I zx

  1 ft  lb ⋅ s 2  =  37.267 × 10−3  (1.5 in.)( 2 in.)   ft   12 in.  

  1 ft  lb ⋅ s 2  +  22.360 × 10−3  (1.5 in.)(1 in.)   ft   12 in.  

  1 ft  lb ⋅ s 2  +  7.9028 × 10−3  (1.5 in.)(1 in.)   ft  12 in.   

  1 ft  lb ⋅ s 2  +  3.6587 × 10−3  (1.5 in.)( 3.5 in.)   ft   12 in.  

(

)

= 776.40 × 10−6 + 232.92 × 10−6 + 82.321 × 10−6 + 133.390 × 10−6 lb ⋅ ft ⋅ s 2 = 1.225 × 10−3 lb ⋅ ft ⋅ s 2

I zx = 1.225 × 10−3 lb ⋅ ft ⋅ s 2

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Chapter 9, Solution 153. Have

m = ρalV

Then

kg   m1 =  2700 3  ( 0.350 × 0.100 × 0.030 ) m 3 m   = 2.8350 kg

kg   m2 =  2700 3  ( 0.200 × 0.100 × 0.050 ) m 3 m   = 2.7000 kg

kg  2  m3 =  2700 3  π ( 0.025 ) × 0.100  m3   m   = 0.53014 kg First note that symmetry implies I x′y′ = I y′z ′ = I z ′x′ = 0 for each component Now

I uv = I u ′v′ + mu v

where

I u′v′ = 0

I xy = Σmx y = ( 2.8350 kg )( 0.175m )( 0.050 m ) + ( 2.7000 kg )( 0.100 m )( 0.050 m ) − ( 0.53014 kg )( 0.080 m )( 0.050 m ) = ( 24.806 + 13.500 − 2.1206 ) × 10 −3 kg ⋅ m 2 = 36.1854 × 10 −3 kg m 2 or I xy = 36.2 × 10−3 kg ⋅ m 2

I yz = Σmy z = ( 2.8350 kg )( 0.050 m )( 0.015m ) + ( 2.7000 kg )( 0.050 m )( 0.055m ) − ( 0.53014 kg )( 0.050 m )( 0.040 m ) = ( 2.1263 + 7.4250 − 1.06028 ) × 10−3 kg ⋅ m 2 = 8.49102 × 10 −3 kg ⋅ m 2 or I yz = 8.49 × 10−3 kg ⋅ m 2 continued

COSMOS: Complete Online Solutions Manual Organization System

I zx = Σmz x = ( 2.8350 kg )( 0.015m )( 0.175m ) + ( 2.7000 kg )( 0.055m )( 0.100 m ) − ( 0.53014 kg )( 0.040 m )( 0.080 m ) = ( 7.4419 + 14.850 − 1.69645 )10−3 kg ⋅ m 2 = 20.59545 × 10 −3 kg ⋅ m 2 or I zx = 20.6 × 10−3 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 154.

m = ρV

Have

(

)

m1 = 2700 kg/m 3 ( 0.118 × 0.036 × 0.044 ) m 3 = 0.50466 kg

Then

2 π  m2 = 2700 kg/m 3  ( 0.022 ) × 0.036  m 3 = 0.07389 kg 2 

(

)

(

)

m3 = 2700 kg/m 3 ( 0.028 × 0.022 × 0.024 ) m 3 = 0.03992 kg Now observe that I x′y′ , I y′z ′ and I z′x′ are zero because of symmetry

4 × 0.022   x2 = −  0.118 +  m = −0.12734 m 3π  

Now

0.022   y3 =  0.036 −  m = 0.025 m 2  

m, kg

x, m

y, m

z, m

mx y kg ⋅ m 2

my z kg ⋅ m 2

mz x kg ⋅ m 2

0.50466

−0.059

0.018

0.022

−0.53595 × 10−3

0.19985 × 10 −3

−0.65505 × 10−3

0.07389

−0.12734

0.018

0.022

−0.16932 × 10−3

0.02926 × 10 −3

−0.20695 × 10 −3

0.03992

− 0.014

0.025

0.026

−0.01397 × 10−3

0.02594 × 10 −3

−0.01453 × 10 −3

continued

COSMOS: Complete Online Solutions Manual Organization System

And

(

I xy = Σ I x′y′ + mx y

(

I yz = Σ I y′z′ + my z

) )

I zx = Σ ( I z′x′ + mx z ) Finally

( )1 + ( I xy )2 − ( I xy )3 = −0.6913 × 10−3 kg ⋅ m 2

I xy = I xy

or I xy = −0.691 × 10−3 kg ⋅ m 2

( )1 + ( I yz )2 − ( I yz )3 = 0.20317 × 10−3 kg ⋅ m 2

I yz = I yz

or I yz = 0.203 × 10−3 kg ⋅ m 2

I zx = ( I zx )1 + ( I zx )2 − ( I zx )3 = −0.84747 × 10−3 kg ⋅ m 2 or I zx = −0.848 × 10−3 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 155. m = ρV

Have

= ρ tA

kg   m1 =  7860 3  ( 0.003 m )( 0.2 × 0.090 ) m 2 m  

Then

= 424.44 × 10−3 kg

kg  π 2  m2 =  7860 3  ( 0.003 m ) × ( 0.045 m ) 2 m   = 75.005 × 10−3 kg

I uv = I u′v′ + m u v

Now

I x′y′ = I y′z ′ = I z ′x′ = 0

Symmetry implies

∴ Then

for both elements.

Iuv = ∑ miui v i

(

)

(

)

I xy = 424.44 × 10−3 kg ( 0.050 m )( 0.045 m ) + 75.005 × 10 −3 kg ( − 0.050 m )( 0.045 m )

(

)

= 954.99 × 10−6 − 168.761 × 10−6 kg ⋅ m 2 = 786 × 10−6 kg ⋅ m 2

I xy = 786 × 10−6 kg ⋅ m 2 continued

COSMOS: Complete Online Solutions Manual Organization System

 4 × 0.045  I yz = 424.44 × 10−3 kg ( 0.045 m )( 0 ) + 75.005 × 10−3 kg ( 0.045 m )  m 3π  

(

)

(

)

= 64.5 × 10−6 kg ⋅ m 2

I yz = 64.5 × 10−6 kg ⋅ m 2  4 × 0.045  I zx = 424.44 × 10−3 kg ( 0 )( 0.050 m ) + 75.005 × 10−3 kg  m  ( − 0.050 m ) 3π  

(

)

(

)

= − 71.6 × 10−6 kg ⋅ m 2

I zx = − 71.6 × 10−6 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 156.

First compute the mass of each component

m = ρstV = ρst tA

Have Then

m1 = ( 7860 kg/m 3 ) ( 0.003)( 0.08 )( 0.09 )  m 3 = 0.169776 kg  1  m2 = 7860 kg/m 3 ( 0.003)  × 0.09 × 0.036   m 3 2  

= 0.03820 kg Now observe that

(I ) = (I ) = (I ) x′y ′

y ′z ′

(I ) = (I ) y ′z ′

z ′x′

(I )

Also

x1 = y1 = z2 = 0

x′y ′ 2

( )

= ρ sT t I x′y′

( I x′y′ )2,area = − 721 b22 h22

From Sample Problem 9.6 Then

z ′x′

2,area

1  1  = ρst t  − b22 h22  = − m2 b2 h2 36  72 

0.09   x2 =  −0.045 +  m = −0.015 m 3   continued

COSMOS: Complete Online Solutions Manual Organization System

Finally

 1 I xy = Σ I xy + mx y = ( 0 + 0 ) +  − ( 0.03820 kg )( 0.09 m )( 0.036 m )  36

(

)

 0.036 m   + ( 0.03820 kg )( −0.015 m )   3  

(

)

= −3.4379 × 10−6 − 6.876 × 10−6 kg ⋅ m 2 = −10.3139 × 10−6 kg ⋅ m 2 or I xy = −10.31 × 10−6 kg ⋅ m 2 And

(

)

I yz = Σ I y′z ′ + m y z = ( 0 + 0 ) + ( 0 + 0 ) = 0

or I yz = 0

I zx = Σ ( I z ′x′ + m z x ) = ( 0 + 0 ) + ( 0 + 0 ) = 0

or I zx = 0

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 157.

First compute the mass of each component Have

m = ρstV = ρst tA

Then

m1 = ( 7860 kg/m3 ) ( 0.003)( 0.7 )( 0.78 )  m 3 = 12.875 kg  π  m2 = ( 7860 kg/m 3 ) ( 0.003)  × 0.392   m 3 = 5.6337 kg 2     1  m3 = ( 7860 kg/m 3 ) ( 0.003)  × 0.78 × 0.3   m 3 = 2.7589 kg 2  

Now observe that because of symmetry the centroidal products of inertia

( )3 = ( I z′x′ )3 = 0 ( I y′z′ )3,mass = ρstt ( I y′z′ )3,area

of components and are zero and I x′y′ Also

Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to a 90° rotation, have

( I y′z′ )3,area = 721 b32h32 Then

( I y′z′ )3 = ρstt  721 b32h32  = 361 m3b3h3 Also Finally

y1 = x2 = 0

(

y2 =

4 × 0.39 m = 0.16552 m 3π

)

I xy = Σ I x′y′ + mx y = ( 0 + 0 ) + ( 0 + 0 )

  − 0.3m   2 + 0 + ( 2.7589 kg ) ( 0.7 m )    = −0.19312 kg ⋅ m 3    or I xy = −0.1931 kg ⋅ m 2 continued

COSMOS: Complete Online Solutions Manual Organization System

(

I yz = Σ I y′z′ + my z

)

= ( 0 + 0 ) + 0 + ( 5.6337 kg ) ( 0.16552 m )( 0.39 m ) 

1  −0.3 m  0.78 m   + ( 2.7589 kg )  ( 0.78 m )( 0.3 m ) +     3  3    36 = ( 0.36367 + 0.017933 − 0.07173) kg ⋅ m 2 = 0.30987 kg ⋅ m 2

or I yz = 0.310 kg ⋅ m 2

(

)

I zx = Σ I z ′x′ + mz x =  0 + (12.875 kg ) ( 0.35 m )( 0.39 m ) 

   0.78 m  + ( 0 + 0 ) + 0 + ( 2.7589 kg )   ( 0.7 m )   3    = (1.75744 + 0.50212 ) kg ⋅ m 2 = 2.25956 kg ⋅ m 2 or I zx = 2.26 kg ⋅ m 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 158. γ st

First note

m = ρstV =

Then

 490 lb/ft 3   1ft  m1 =  0.08 in.) ( 6 × 3.6 ) in 2    2 (  12in.   32.2 ft/s 

tA 3

= 15.2174 × 10−3 lb ⋅ s 2 /ft  490 lb/ft 3  2   1 ft  π m2 =  0.08 in.)  (1.8 in.)    2 ( 2   12 in.   32.2 ft/s 

= 3.5855 × 10−3 lb ⋅ s 2 /ft  490 lb/ft 3  2   1 ft  π m3 =  0.08 in.)  ( 3.6 in.)    2 ( 4   12 in.   32.2 ft/s 

= 7.1710 × 10−3 lb ⋅ s 2 /ft Note that symmetry implies

( I x′y′ )1,2 = ( I y′z′ )1,2 = ( I z′x′ )1,2 = 0

(I ′ ′ ) = (I ′ ′ ) xy

Now

I uv = I u′v′ + mu v continued

COSMOS: Complete Online Solutions Manual Organization System

Thus

I xy = Σmx y

= m1x1 y1 − m2 x2 y2 + m3 x3 y3

 0.6  1.8  = (15.2174 × 10−3 lb ⋅ s 2 /ft )  − ft  ft   12  12   4 × 1.8  1 ft   1.8  ft  − ( 3.5855 × 10−3 lb ⋅ s 2/ft )  2.4 −   in. × 3π  12 in.   12  

= ( −114.131 − 73.326 ) × 10−6 lb ⋅ ft ⋅ s 2

or I xy = −187.5 × 10−6 lb ⋅ ft ⋅ s 2

I yz = Σmy z = m1 y1 z1 − m2 y2 z2 + m3 y3 z3

Now

I yz = 0

I zx = ( I zx )1 − ( I zx )2 + ( I zx )3

Also

= m1 z1x1 − m2 z2x2 + ( I zx )3 Now determine ( I zx )3

( dI zx )3 = ( dI z′x′ )3 + z x dm

Have

  x  γ = ( z )  −   st t x dz   2  g 

=−

γ st

( I zx )3

Therefore,

=−

I zx = −

 t  a32  g 4 

t =

4m3 π a32

(

)

1 m3a32 2π

1  3.6  7.1710 × 10−3 lb ⋅ s 2 /ft  ft  2π  12 

(

)

2m3 a 2 2m  1 1  a3 z − z 3 dz = − 23  a32 z 2 − z 4  2 ∫0 4 0 π a3 π a3  2

=−

Finally

(

γ st  π

m3 =

Now

1 γ st tz a32 − z 2 dz 2 g

)

or I zx = −102.7 × 10−6 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 159. Have

wL g

Then

m1 =

w 3  π × a  g 2 

3π wa 2 g

m2 =

w ( 2a ) g

=2 m3 =

w (π × a ) g

=π

wa g

I uv = I u′v′ + mu v

Now

I x′y′ = I y′z ′ = I z ′x′ = 0

Symmetry implies

∴ Then

wa g

for each element.

Iuv = ∑ miui v i  3π wa   3  2 3   wa  2  I xy =    a  × a  +  π  ( 3a )  × a  π   2 g   2  π 2   g  3 wa  27  = + 6  g  4 

I xy = 12.75

w 3 a g

I yz = 7

wa3 g

 3π wa   2 3   wa   2 × a  I yz =    × a  ( 2a ) +  π  (− a)  2 g  π 2   g  π 

wa3 (9 − 2) g

 3π wa   wa   3   wa  I zx =   ( 2a )  a  +  2  ( a )( 3a ) +  π  ( − a )( 3a ) 2   g   2 g   g 

wa3  9π  + 6 − 3π   g  2 

I zx = 1.5

w 3 a (π + 4 ) g

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 160.

First compute the mass of each component. Have

Then

W 1 = wL g g

m1 =

w w ( 2π × a ) = 2π a g g

m2 =

w w (a) = a g g

m3 =

w w ( 2a ) = 2 a g g

m4 =

w 3  w  2π × a  = 3π a g 2  g

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. continued

COSMOS: Complete Online Solutions Manual Organization System

2π

w a g

1 a 2 0

w a g

3π

w a g

mx y

−a 0

3 − a 2

−9π

w 3 a g

w (1 − 5π ) a3 g

(

I xy = Σ I x′y′ + mx y I yz = Σ I y′z0′ + my z

)

(

)

I zx = Σ I z′x0′ + mz x

w 3 a g

−11π

w 3 a g

−4π

w 3 a g 0

−9π

)

(

mz x

w 3 a g

−2π

w 3 a g

Then

w 3 a g

4π

my z

w 3 a g

12π 4

w 3 a g

w (1 + 2π ) a3 g

or I xy =

w 3 a (1 − 5π ) g

or I yz = −11π or I zx = 4

w 3 a g

w 3 a (1 + 2π ) g

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 161.

First note

m = ρV =

γ g

V =

γ g

Specific weight of aluminium = 0.10 lb/in 3 = 172.8 lb/ft 3 Then

2 172.8 lb/ft 3  π  0.075   ft    L 32.2 ft/s 2  4  12  

(

)

= 0.16464 × 10−3 L lb ⋅ s 2/ft 2 Now

L1 = L4 = 12.5 in. = 1.04167 ft

m1 = m4 = 0.17150 × 10−3 lb ⋅ s 2 /ft L2 = L5 = 9 in. = 0.75 ft

m2 = m5 = 0.12348 × 10−3 lb ⋅ s 2 /ft L3 = L6 = 15 in. = 1.25 ft

m3 = m6 = 0.20580 × 10−3 lb ⋅ s 2 /ft and

I x′y′ = I y′z ′ = I z ′x′ = 0 continued

COSMOS: Complete Online Solutions Manual Organization System

m, lb ⋅ s 2 /ft

x , ft

y , ft

z , ft

mx y, lb⋅ft ⋅s2

my z , lb ⋅ ft ⋅ s 2

m z x , lb ⋅ ft ⋅ s 2

0.17150 × 10−3

0.75

0.5208

0.06699 × 10−3

0.12348 × 10−3

0.375

1.04167

0.04823 × 10−3

0.20580 × 10−3

1.04167

0.625

0.13398 × 10−3

0.17150 × 10−3

0.5208

1.25

0.111646 × 10−3

0.12348 × 10−3

0.375

1.25

0.05788 × 10−3

0.20580 × 10−3

0.75

0.625

0.09647 × 10−3

0.11522 × 10−3

0.24563 × 10−3

0.15435 × 10−3

I xy = Σ ( I x′y′ + mx y ) = 0.115222 × 10−3 lb ⋅ ft ⋅ s 2

I xy = 0.1152 × 10−3 lb ⋅ ft ⋅ s 2

0 my z = 0.24563 × 10−3 lb ⋅ ft ⋅ s2 I yz = Σ ( I y′z ′ + ) or I yz = 0.246 × 10−3 lb ⋅ ft ⋅ s 2 I zx = Σ ( I z ′x′ +0 mz x ) = 0.15435 × 10−3 lb ⋅ ft ⋅ s 2

I zx = 0.1543 × 10−3 lb ⋅ ft ⋅ s 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 162. mring = 2π ( 0.5 m )(1.8 kg/m ) = 5.655 kg

mrod = ( 0.8 m )(1.8 kg/m ) = 1.44 kg

For each ring x = z = 0 and I x′y′ = I y′z′ = I z′x′ = 0, thus the mass product of inertia of the rings is zero with respect to each pair of coordinate axes. For each rod: Since each rod lies in the x-y plane, I yz = I zx = 0 I yz = I zx = 0

Thus for entire wire figure

(

I xy = ∑ I x′y′ + m x y

Hence

)

where I x′y′ = 0

I xy = ∑ mx y = (1.44 kg )( − 0.5 m )( 0.4 m ) + (1.44 kg )( 0.5 m )( − 0.4 m ) = − 0.576 kg ⋅ m 2

I xy = − 0.576 kg ⋅ m 2

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Chapter 9, Solution 163. Have and Consider

I xy = ∫ xydm x = x′ + x

I yz = ∫ yzdm y = y′ + y

I zx = ∫ zxdm z = z′ + z

(9.45) (9.31)

I xy = ∫ xydm

Substituting for x and for y

I xy = ∫ ( x′ + x )( y′ + y ) dm

= ∫ x′y′dm + y ∫ x′dm + x ∫ y′dm + x 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.

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Chapter 9, Solution 164. (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 = ρ dV = ρ  xydz  = ρ ab 1 −  dz c 2  2  c

2 3  c  1 z 1 z   m = ∫ dm = ∫ ρ ab 1 −  dz = ρ ab  − 1 −   c 2 c    02  3  0 c

Then

1 ρ abc 6

dI zx = dI z′x′ + z EL xEL dm

Now

dI z′x′ = 0

where

z EL = z

and

Then

xEL =

( symmetry ) 1 1  z x = a 1 −  3 3  c

2 c  1  z   1 z  I zx = ∫ dI zx = ∫ z  a 1 −    ρ ab  1 −  dz  c    2 c   0 3 

1 2 c z2 z3 z 4  ρ a b∫ 0  z − 3 + 3 2 − 3  dz 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

continued

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(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 1 I xy = 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 = ρ dV = ρ  xzdy  = ρ ac 1 −  dy b 2  2  Now

dI zx = ρ tdI zx, Area

where

t = dy

and dI zx, Area = Then

1 2 2 x z from the results of Sample Problem 9.6 24

2 2  1   y    y    dIzx = ρ ( dy )   a 1 −   c 1 −    b    b    24    4

1 y 1m  y  = ρ a 2c 2 1 −  dy = ac 1 −  dy 24 b 4 b  b 

Finally

I zx = ∫ dI zx

1m  y =∫ ac 1 −  dy 4b  b b 0

5 1 m  b  y  ac  −  1 −   = 4 b  5  b  0

or I zx =

1 mac 20

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.

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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 = ρ dV = ρ dydxdz z  x = a 1 −  c 

From part a Now I zx = ∫ zxdm = ∫ 0 ∫ 0 ( c

= ρ∫0 ∫0 ( c

a 1− cz

= ρ b∫

c 1 z 0 

a 1− cz

) b(1− ax − cz ) zx ( ρ dydxdz ) ∫0

) zx b (1 − 

x a

−

z c

) dxdz (

a 1− cz

1 x3 1 z 2  x − x  − 3 a 2 c 0 2 2

) dz

2 3 2 z 1 3 z 1 z 2 z  c 1  a 1 −  − a 1 −   dz = ρ b∫ 0 z  a 2 1 −  − c 3a  c 2c  c    2 

= ρ b∫ 0 =

1 2  z a z 1 −  dz 6 c 

1 2 c z2 z3 z 4  ρ a b∫ 0  z − 3 + 3 2 − 3  dz 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

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Chapter 9, Solution 165.

Iy =

From Figure 9.28

1 2 ma 2

and using the parallel-axis theorem

Ix = Iz =

1 1 h m 3a 2 + h 2 + m   = m 3a 2 + 4h 2 12 12 2

(

)

(

)

I xy = I yz = I zx = 0

Symmetry implies

For convenience, let point A lie in the yz plane. Then

λOA =

h + a2

( hj + a k )

With the mass products of inertia equal to zero, Equation (9.46) reduces to

I OA = I xλx2 + I yλ y2 + I z λz2  1 h = ma 2   2 2 2  h +a

 1 m 3a 2 + 4h 2  + 12 

(

)

 a  2 2  h +a

or I OA =

  

1 10h 2 + 3a 2 ma 2 2 12 h + a2

Note: For point A located at an arbitrary point on the perimeter of the top surface, λOA is given by

λOA =

1 h2 + a 2

( a cos φ i + hj + a sin φ k )

which results in the same expression for I OA

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Chapter 9, Solution 166. 2

dOA =

First note that

λ OA =

Then

9 2 2 3   a  + ( −3a ) + ( 3a ) = a 2 2  

1 3  1 ai − 3aj + 3ak  = ( i − 2 j + 2k ) 9  2  3 a 2

For a rectangular coordinate system with origin at point A and axes aligned with the given x, y, z axes, have (using Figure 9.28)

Ix = Iz = =

3 1 2 2 m  a + ( 3a )  5 4 

Iy =

3 ma 2 10

111 2 ma 20

I xy = I yz = I zx = 0

Also, symmetry implies

With the mass products of inertia equal to zero, Equation (9.46) reduces to

I OA = I xλx2 + I yλ y2 + I z λz2 2

111 2  1  3 111 2  2   2 ma   + ma 2  −  + ma   20 3 10 3 20     3

193 2 ma 60

or I OA = 3.22ma 2

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Chapter 9, Solution 167.

First compute the mass of each component

m = ρstV =

Have

0.284 lb/in 3 V = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 V 2 32.2 ft/s

(

)

Then 2 m1 = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 π ( 4 in.) ( 2 in.)  = 0.88667 lb ⋅ s 2 /ft   2 m2 = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 π (1 in.) ( 3 in.)  = 0.083125 lb ⋅ s 2 /ft  

m3 = 0.008819 lb ⋅ s 2 /ft ⋅ in 3 π (1 in.) ( 2 in.)  = 0.055417 lb ⋅ s 2 /ft   2

I yz = I zx = 0

Symmetry implies

( I xy )1 = 0

( I x′y′ )2 = ( I x′y′ )3 = 0

and Now

(

)

I xy = Σ I x′y′ + mx y = m2 x2 y2 − m3 x3 y3

 1 ft 2  = ( 0.083125 lb ⋅ s 2 /ft ) ( 2 in.)(1.5 in.)  ×  2   144 in   1 ft 2  − ( 0.055417 lb ⋅ s 2 /ft ) ( −2 in.)( −1 in.)  ×  2   144 in  = 0.96209 × 10−3 lb ⋅ ft ⋅ s 2 From the solution to Problem 9.143:

I x = 34.106 × 10−3 lb ⋅ ft ⋅ s 2 I y = 50.125 × 10−3 lb ⋅ ft ⋅ s 2

I z = 34.876 × 10−3 lb ⋅ ft ⋅ s 2 continued

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1 ( 2i + 3j) 13

λ OA =

By observation Then

I OA = I xλx2 + I y λ y2 + I z λz2 − 2I xy λxλ y − 2I yz λ y λz − 2I zxλz λx  2  = 34.106 × 10−3 lb ⋅ ft ⋅ s 2    13 

(

)

 3  + 50.125 × 10−3 lb ⋅ ft ⋅ s 2    13 

(

)

 2  3  − 2 0.96209 × 10−3 lb ⋅ ft ⋅ s 2     13  13 

(

)

= (10.4942 + 34.7019 − 0.8881) × 10−3 lb ⋅ ft ⋅ s 2 = 44.308 × 10−3 lb ⋅ ft ⋅ s 2 or I OA = 44.3 × 10−3 lb ⋅ ft ⋅ s 2

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Chapter 9, Solution 168. I x = 9.8821 × 10−3 lb ⋅ ft ⋅ s 2

From Problem 9.147:

I y = 11.5344 × 10−3 lb ⋅ ft ⋅ s 2

I z = 2.1878 × 10−3 lb ⋅ ft ⋅ s 2 I xy = 0.48776 × 10−3 lb ⋅ ft ⋅ s 2

Problem 9.151:

I yz = 1.18391 × 10−3 lb ⋅ ft ⋅ s 2

I zx = 2.6951 × 10−3 lb ⋅ ft ⋅ s 2 Now

λ x = λ y = λz

and

λx2 + λ y2 + λz2 = 1 3λx2 = 1

Therefore,

λx = λ y = λz =

1 3

Equation 9.46

I OL = I xλx2 + I y λ y2 + I z λz2 − 2 I xy λxλ y − 2I yz λ y λz − 2I zxλz λx 2 2 2   1   1   1   1  1  = 9.8821 + 11.5344 + 2.1878      − 2 ( 0.48776 )      3  3  3  3  3 

1  1   1  1   −3 2   − 2 ( 2.6951)     × 10 lb ⋅ ft ⋅ s 3 3 3 3       

− 2 (1.18391) 

= 4.95692 × 10−3 lb ⋅ ft ⋅ s 2 or I OL = 4.96 × 10−3 lb ⋅ ft ⋅ s 2

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Chapter 9, Solution 169.

m1 = m2 =

First note that

λ OA =

And that

1W 2 g

1 (i + j + k ) 3

Using Figure 9.28 and the parallel-axis theorem have

I x = ( I x )1 + ( I x )2

 1 1W =  12  2 g

 2 1W a + 2 g 

a   2

2

 

1W  1  1 W  2 2 +   a +a + 2 g 12  2 g 

(

)

 a 2  a 2     +      2     2 

1 W  1 1 1 1  1W 2 +  a2 +  +  a2  = a  2 g  12 4  6 2  2 g

( )1 + ( I y )2

Iy = Iy

 1  1 W  1W 2 2 =   a +a + 12 2 g 2 g   

(

 1  1 W +  12  2 g

)

 a 2  a 2     +      2     2 

2  2 1 W  2  a    a a + +  ( )     2 g   2    

1 W  1 1  2  1 5  W 2 a +  a2  =  +  a +  2 g  6 2  g  12 4   continued

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I z = ( I z )1 + ( I z )2

 1 1W =  12  2 g

 2 1W a + 2 g 

 1  1 W +  12  2 g

a   2

 2 1W a + 2 g 

2

 

 2  a 2   ( a ) +      2    

1 W  1 1 5  5W 2  1 a +  a2 +  +  a2  =  2 g  12 4   12 4   6 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, y1 = 0

(

)

I xy = Σ I x′y′ + mx y = m2 x2 y2 =

Then

1W a 1W 2 a ( a )   = 2 g 2   4 g

1W I yz = Σ I y′z′ +0my z = m2 y2z2 = 2 g

(

)

(

 a  a  1 W 2 a    =  2  2  8 g

)

I zx = Σ I z ′x′ + mz x = m1z1x1 + m2 z2 x2

1W 2 g

 a  a  1 W    +  2  2  2 g

3W 2 a a  (a) = 8 g 2

Substituting into Equation (9.46)

I OA = I xλx2 + I y λ y2 + I z λz2 − 2I xy λxλ y − 2I yz λ y λz − 2I zxλz λx Noting that

λx2 = λ y2 = λz2 = λxλ y = λ yλz = λz λx =

1 3

Have

I OA =

1 1 W 2 W 2 5 W 2 a + a + a  3 2 g 6 g g  1 W 2 1 W 2 3 W 2  a + a + a  − 2 8 g 8 g 4 g 

1 14  3  W 2  − 2   a 3 6  4  g or I OA =

5 W 2 a 18 g

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Chapter 9, Solution 170.

m = ρV = ρ tA

Have

m1 = ρ ta 2

Then

m2 =

1 ρ ta 2 2

Compute moments and moments of inertia with respect to point A Now

I x′′ = ( I x′′ )1 + ( I x′′ )2

 1  = ρ ta  a 2  + 12   2 

 1   a  2  2 2   2  2  1 2    + ( a )   + ρ ta   a  +  a    18   3     2    2   

19 ρ ta 4 12

( )1 + ( I y′′ )2

I y′′ = I y′′

  1   a 2  2  = ρ ta 2   a 2  +   + ( a )      12   2   a 2  2a 2   1 2 1  2 2 + ρ ta   a + a  +   +     2 18  3     3  =

5 ρ ta 4 3

I z ′′ = ( I z ′′ )1 + ( I z′′ )2

1  1 1  = ρ ta 2  a 2 + a 2  + ρ ta 2  a 2  3 2   6 

(

)

3 ρ ta 4 4

Now note symmetry implies

( I x′y′ )1 = ( I y′z′ )1 = ( I z′x′ )1 = 0 ( I x′y′ )2 = ( I y′z′ )2 = 0 continued

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I uv = I u′v′ + mu v

Now

0  a  a  1 I x′′y′′ = m1x1′′ y1′′ + m2 x2′′ y2′′ = ρ ta 2    = ρ ta 4  2  2  4

Therefore

0 1 a I y′′z′′ = m1 y ″1 z ″1 + m2 y2′′ z′′2 = ρ ta 2   ( −a ) = − ρ ta 4 2 2

( )2 + m2 z ″2 x ″2 

I z′′x′′ = m1z ″1 x ″1 +  I z ′x′ 

1 ( I z′x′ )  = − a4 2  area  72

From Sample Problem 9.6 Then

( I z′′x′′ )2 = ρ t ( I z′x′ )2  area

Then

a I z′′x′′ = ρ ta 2 ( − a )   2

=−

1 ρ ta 4 72

1  1  2  1   +  − ρ ta 4 + ρ ta 2  − a  a  2  72  3  3  

5 = − ρ ta 4 8 By observation

λ AB =

1 (i + j − k ) 3

Now, Equation 9.46

I AB = I x′′λx2′′ + I y′′λ y2′′ + I z′′λz2′′ − 2I x′′y′′λx′′y′′ − 2I y′′z′′λ y′′λz′′ − 2I z′′x′′λz′′λx′′  19  1 2 5  1 2 3  1 2 = ρ ta    +   + −  3 3  4 3  12  3  4

 1   1  1   1   1  1  − 2    − 2 −   −  4 3    3  3   2   3 

 5   1  1   − 2 −  −   3  3    8  or I AB =

5 ρ ta 4 12

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Chapter 9, Solution 171. I x = 4.212 kg ⋅ m 2

From Problem 9.138:

I y = 7.407 kg ⋅ m 2

I z = 3.7055 kg ⋅ m 2 I xy = −0.19312 kg ⋅ m 2

From Problem 9.157:

I yz = 0.30987 kg ⋅ m 2

I zx = 2.25956 kg ⋅ m 2 Now Eq. (9.46):

λ OL =

1 ( −4i + 8j + k ) 9

I OL = I xλx2 + I y λ y2 + I z λz2 − 2 I xy λxλ y − 2I yz λ y λz − 2I zxλz λx 2 2 2   4 8 1 =  4.212  −  + 7.407   + 3.7055    9 9 9 

 4  8   8  1  − 2 ( −0.19312 )  −   − 2 ( 0.3098 )     9  9   9  9   1  4   − 2 ( 2.25956 )   −  kg ⋅ m 2  9  9   = ( 0.832 + 5.85244 + 0.04575 − 0.15259 − 0.061195 + 0.22317 ) kg ⋅ m 2

= 6.73957 kg ⋅ m 2 I OL = 6.74 kg ⋅ m 2

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Chapter 9, Solution 172.

Mass of each leg is identical:

 W /L  m= L  g 

0.041 lb/ft (1.5 ft ) 32.2 ft/s

= 0.00190994 lb ⋅ s 2 /ft

Also, I x′y′ = I y′z′ = I z′x′ = 0 for each leg, and

x1 = x6 = 0

Now

I xy = Σ I x′y′ + mx y = m2 x2 y2 + m3x3 y3

(

y4 = y5 = y6 = 0

z1 = z2 = z3 = 0

)

(

)

= 0.00190994 lb ⋅ s 2 /ft ( 0.75 )(1.5 ) + (1.5 )( 0.75 )  ft 2 = 0.0042974 lb ⋅ ft ⋅ s 2 = 4.2974 × 10−3 lb ⋅ ft ⋅ s 2 I yz = 0 I zx = Σ ( I z′x′ + mz x ) = m4z4 x4 + m5z5 x5

(

)

= 0.00190994 lb ⋅ s 2 /ft ( 0.75 )(1.5 ) + (1.5 )( 0.75 )  ft 2 = 0.0042974 lb ⋅ ft ⋅ s 2 = 4.2974 × 10−3 lb ⋅ ft ⋅ s 2 continued

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I x = 14.32 × 10−3 lb ⋅ ft ⋅ s 2

From Problem 9.150

I y = I z = 18.62 × 10−3 lb ⋅ ft ⋅ s 2

λ OL =

Now

1 ( −3i − 6 j + 2k ) and then 7

 Eq. ( 9.46 ) 

I OL = I xλx2 + I y λ y2 + I z λz2 − 2I xy λxλ y − 2I yz λ y λz − 2I zxλz λx 2    3 =   14.32 × 10−3  −  + 18.62 × 10−3  7  

(

)

(

)

 −6 2  2 2    +    − 2 4.2974 × 10−3  7    7 

(

)  − 73  −76 

 2  −3    − 2 4.2974 × 10−3      lb ⋅ ft ⋅ s 2  7  7   

(

)

(

)

= 2.6302 × 10−3 + 15.20 × 10−3 − 3.1573 × 10−3 + 1.05242 × 10−3 lb ⋅ ft ⋅ s 2 = 15.725 × 10−3 lb ⋅ ft ⋅ s 2 or I OL = 15.73 × 10−3 lb ⋅ ft ⋅ s 2

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Chapter 9, Solution 173.

First compute the mass of each component

m = ρ stV = mAL

Have

(

)

2 = 7850 kg/m 3 π ( 0.0015 m )  L  

= ( 0.055488L ) kg/m m1 = m2 = 0.055488 kg/m (π × 0.36 m )

Then

= 0.062756 kg m3 = m4 = 0.055488 kg/m ( 0.36 m )

= 0.019976 kg Now observe that the centroidal products of inertia I x′y′ = I y′z ′

= I z′x′ = 0 for each component. x3 = x4 = 0,

Also

y1 = 0,

z1 = z2 = 0

Then

(

)

I xy = Σ I x′y′ + mx y = m2x2 y2

 2 × 0.36 m  −3 2 = ( 0.062756 kg )  −  ( 0.36 m ) = −5.1777 × 10 kg ⋅ m π  

(

)

I yz = Σ I y′z ′ + my z = m3 y3 z3 + m4 y4 z4 where

m3 = m4 ,

y3 = y4 , z4 = − z3 ,

so that

I yz = 0

I zx = Σ ( I z′x′ + mz x ) = m1z1x1 + m2 z2 x2 = 0 continued

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From the solution to Problem 9.149

I x = 23.170 × 10−3 kg ⋅ m 2 I y = 21.444 × 10−3 kg ⋅ m 2

I z = 17.992 × 10−3 kg ⋅ m 2

λ OL =

Now

1 ( −3i − 6 j + 2k ) 7

Have

I OL = I xλx2 + I yλ y2 + I z λz2 − 2I xy λxλ y − 2I yz λ yλz − 2I zxλz λx  Eq. ( 9.46 )  2 2 2   3  6 2 =  23.170  −  + 21.444  −  + 17.992    7  7 7 

 3  6   − 2 ( −5.1777 )  −  −   × 10−3 kg ⋅ m 2  7  7 

= ( 4.2557 + 15.755 + 1.4687 + 3.8040 ) × 10−3 kg ⋅ m = 25.283 × 10−3 kg ⋅ m 2 or I OL = 25.3 × 10−3 kg ⋅ m 2

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Chapter 9, Solution 174.

First compute the mass of each component.

m = ( m/L ) L

Have

= ( 0.049 kg/m ) L m1 = ( 0.049 kg/m )( 2π × 0.32 m )

Then

= 0.09852 kg m2 = m3 = m4 = m5 = ( 0.049 kg )( 0.160 m )

= 0.00784 kg I x′y′ = I y′z′ = I z′x′ = 0 for each component.

Now observe that

x1 = x4 = x5 = 0,

Also,

y1 = 0,

z1 = z2 = z3 = 0

Then

(

)

I xy = Σ I x′y′ + mx y = m2 x2 y2 + m3x3 y3

= ( 0.00784 kg ) ( 0.32 m )( 0.08 m ) + ( 0.24 m )( 0.16 m )  = 0.50176 × 10−3 kg ⋅ m 2 By symmetry Now

I yz = I xy = 0.50176 × 10−3 kg ⋅ m 2

I zx = Σ ( I z′x′ + mz x ) = 0

From the solution to Problem 9.148

I x = I z = 6.8505 × 10−3 kg ⋅ m 2 I y = 12.630 × 10−3 kg ⋅ m 2 continued

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λ OL =

Now

1 ( −3i − 6 j + 2k ) 7

Have

I OL = I xλx2 + I y λ y2 + I z λz2 − 2 I xy λxλ y − 2I yz λ y λz − 2I zxλz λx  Eq. ( 9.46 )    3 2  2 2   = ( 6.8505 )  −  +     × 10−3 kg ⋅ m 2  7      7  2    6 + (12.63)  −   × 10−3 kg ⋅ m 2  7   

 3  6   6  2    −2 ( 0.50176 )  −  −  +  −     × 10−3 kg ⋅ m 2  7  7   7  7  

= (1.8175 + 9.2792 − 0.12288 ) × 10−3 kg ⋅ m 2 = 10.9738 × 10−3 kg ⋅ m 2 or I OL = 10.97 × 10−3 kg ⋅ m 2

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Chapter 9, Solution 175.

(a) Using Figure 9.28 and the parallel-axis theorem have at point A.

I x′ =

I y′

1 m b2 + c2 12

(

) 2

1 1 a = m a2 + c2 + m   = m 4a 2 + c 2 12 12 2

(

I z′ =

)

(

1 1 a m a2 + b2 + m   = m 4a 2 + b 2 12 12 2

(

)

(

) )

Now observe that symmetry implies

I x′y′ = I y′z′ = I z′x′ = 0 Using Equation (9.48), the equation of the ellipsoid of inertia is then

I x′ x 2 + I y ′ y 2 + I z ′ z 2 = 1 or

1 1 1 m b2 + c 2 x2 + 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 1 m b2 + c 2 = m 4a 2 + c 2 = m 4a 2 + b 2 12 12 12

(

)

(

)

(

b 2 + c 2 = 4a 2 + c 2

Then

b 2 + c 2 = 4a 2 + b 2

and

)

b = 2 a

c = 2 a

(b) Using Figure 9.28 and the parallel-axis theorem, we have at point B

I x′′ =

I y′′

1 1 c m b2 + c 2 + m   = m b 2 + 4c 2 12 2 12  

(

)

(

1 1 c = m a2 + c2 + m   = m a 2 + 4c 2 12 12 2

I z′′ =

(

1 m a 2 + b2 12

(

)

(

)

) continued

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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 + b2 12 12 12

(

)

(

)

(

)

Then

b 2 + 4c 2 = a 2 + 4c 2

b = 1 a

and

b 2 + 4c 2 = a 2 + b 2

c 1 = a 2

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Chapter 9, Solution 176.

(a) From sample Problem 9.11, we have at the apex A

Ix =

3 ma 2 10

I y = Iz =

3 1 2  m  a + h2  5 4 

Now observe that symmetry implies I xy = I yz = I zx = 0 Using Equation (9.48), the equation of the ellipsoid of inertia is then

I x x2 + I y y 2 + I z z 2 = 1 or

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 

a = 2 h

(b) From Sample Problem 9.11, we have

I x′ =

I y′′ =

and at the centroid C

Then

I y′

3 ma 2 10

3  2 1 2 m a + h  20  4 

3  2 1 2 h m a + h  + m  = I z′ = 20  4  4 =

1 m 3a 2 + 2h 2 20

(

)

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

(

)

a = h

2 3

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Chapter 9, Solution 177. (a) From Figure 9.28

Ix =

1 2 ma 2

I y = Iz =

1 m 3a 2 + L2 12

(

)

Now observe that symmetry implies

I xy = I yz = I zx = 0 Using Equation (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

(

)

a L

1 3

(b) Using Fig. 9.28 and the parallel-axis theorem Have I x′ =

1 2 ma 2

I y′ = I z′ =

1 7 2 L 1 m 3a 2 + L2 + m   = m  a 2 + L  12 4 4 48    

(

)

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  

a L

7 12

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Chapter 9, Solution 178. I y + Iz ≥ Ix

(i) To prove By

(

)

I y = ∫ z + x dm Then

(

)

(

)

I z = ∫ x + y dm

(

)

I y + I z = ∫ z 2 + x 2 dm + ∫ x 2 + y 2 dm

(

)

= ∫ y 2 + z 2 dm + 2∫ x 2dm Now..

2 2 ∫ ( y + z ) dm = I x

and

∴ I y + Iz ≥ Ix

Q.E.D.

2 ∫ x dm ≥ 0

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) I y + I z ≥ I x or 2I y ≥ I x or I y ≥

1 Ix 2

Q.E.D.

definition

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Chapter 9, Solution 179.

(a) At the center of the cube have (using Figure 9.28)

Ix = I y = Iz =

1 1 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 = R2 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  be the same  R = .  I OL  

∴ I OL =

1 2 ma 6

(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. It then follows that

I x′ = I OL =

continued

1 2 ma 6

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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

I y′ = I z ′ =

 3  1 2 ma + m  a  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. First note that at corner A

Ix = I y = Iz =

2 2 ma 3

I xy = I yz = I zx =

1 2 ma 4

Substituting into Equation (9.56) yields

k 3 − 2ma 2k 2 +

55 2 6 121 3 9 m a k− ma =0 48 864

For which the roots are

k1 =

1 2 ma 6

k2 = k3 =

11 2 ma 12

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Chapter 9, Solution 180.

(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 2dm 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 )V

(

− Ix + I y + Iz

( I x + I y + I z )body = ( I x + I y + I z )sphere + 2∫ m r 2dm − 2∫ m r 2dm 1

Now, m1 = m2 and r1 ≥ r2 for all elements of mass dm in volumes 1 and 2.

∴ ∫ m r 2dm − ∫ m r 2dm ≥ 0 1 2 so that

( I x + I y + I z )body ≥ ( I x + I y + I z )sphere

Q.E.D.

continued

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(ii) First note from Figure 9.28 that for a sphere

Ix = I y = Iz = Thus,

2 2 ma 5

( I x + I y + I z )sphere = 65 ma2

For a solid of revolution, where the x axis is the axis of revolution, have

I y = Iz Then, using the results of part i

( I x + 2I y )body ≥ 65 ma2 Iy ≥

From Problem 9.178 have or

1 Ix 2

( 2I y − I x )body ≥ 0

Adding the last two inequalities yields

( 4I y )body ≥ 65 ma2 or

( I y )body ≥ 103 ma2

Q.E.D.

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Chapter 9, Solution 181.

(a) First compute the moments of inertia using Figure 9.28 and the parallel-axis theorem.

 a 2  a 2  13 1 2 2 2 Ix = Iz = m 3a + a + m   +  2   = 12 ma 12     2  1 3 2 I y = ma 2 + m ( a ) = ma 2 2 2

(

)

Next observe that the centroidal products of inertia are zero because of symmetry. Then

0 1  a  a  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  2 2 K3 −  + +  ma K  12 2 12  2 2 2  13 3   3 13   13 13   1  1   1  2 +  ×  +  ×  +  ×  −  − − − −    ma    12 2 2 12 12 12 2       2 2     2 2 

(

)

3  13 3 13   13   1   3   1   13   1  1  1   1   2 −  × ×  −    − −    −  − − 2− −    ma = 0       2 2   2 2   2   12 2 12   12   2 2   2   2   12   2 2 

(

Simplifying and letting K = ma 2ζ yields

ζ3 −

11 2 565 95 ζ + ζ − =0 3 144 96 continued

)

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Solving yields

ζ 1 = 0.363383

ζ2 =

19 12

ζ 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 λx , λ y , λz of each principal axis, we use two of the equations of Equations (9.54) and Equation (9.57). Thus (9.54a) ( I x − K ) λx − I xyλ y − I zxλz = 0

− I zxλx − I yz λ y + ( I z − K ) λz = 0

(9.54c)

λx2 + λ y2 + λz2 = 1

(9.57)

Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of λ 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  λ y −  ma 2  λz = 0  ma − K  λx −  −  12  2   2 2  1   1   13  −  ma 2  λx −  − ma 2  λ y +  ma 2 − K  λz = 0 2   12   2 2  or and

1 1  13  − ζ  λx + λ y − λz = 0  2 2 2  12 

(i)

1 1  13  − λx + λ y +  − ζ  λz = 0 2 12 2 2  

(ii)

Observe that these equations will be identical, so that one will need to be replaced, if

13 1 19 −ζ = − ζ = or 12 2 12 Thus, a third independent equation will be needed when the direction cosines associated with K 2 are determined. Then for K1 and K 3 continued

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 13  1  13  1   − ζ   λz = 0 Eq.(i) – Eq.(ii)  − ζ −  −   λx +  − −   2   12  2  12

λz = λx

1 1  13  − ζ  λx + λ y − λx = 0  2 2 2  12 

Substituting into Eq.(i)

7



λ y = 2 2  ζ −  λx 12  

Substituting into Equation (9.57)

λx2

 7  2  +  2 2  ζ −  λx  + ( λx ) = 1 12     2  7    2 + 8  ζ −   λx2 = 1 12    

(iii)

K1 : Substituting the value of ζ 1 into Eq.(iii) 2  7  2   2 + 8  0.363383 −   ( λx )1 = 1 12    

or and then

( λx )1 = ( λz )1 = 0.647249

( λ y )1 = 2

7  2  0.363383 −  ( 0.647249 ) 12  

= −0.402662

∴

(θ x )1 = (θ z )1 = 49.7°

(θ y )1 = 113.7°

K 3 : Substituting the value of ζ 3 into Eq.(iii) 2  7  2   2 + 8 1.71995 −   ( λx )3 = 1 12    

( λx )3 = ( λz )3

and then

( λ y )3 = 2

= 0.284726

7  2 1.71995 −  ( 0.284726 ) 12  

= 0.915348

∴

(θ x )3 = (θ z )3

= 73.5°

(θ y )3 = 23.7° continued

COSMOS: Complete Online Solutions Manual Organization System

K 2 : For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57). Now

(

)

− I xy λx + I y − K λ y − I yz λz = 0

(9.54b)

Substituting for the moments and products of inertia.

1 1     3  − − ma 2  λx +  ma 2 − K  λ y −  − ma 2  λz = 0 2 2 2 2 2       or

3



λx +  − ζ  λ y + λz = 0 2 2 2 2 2 

(iv)

Substituting the value of ζ 2 into Eqs.(i) and (iv)

1  13 19  −  ( λx )2 + λy  12 12 2 2  

( )2 − 12 ( λz )2 = 0

1 2 2

3 19  −  λy 2 12  

( )2 + 2 1 2 ( λz )2 = 0

( λx )2 + 

− ( λx )2 +

1 λy 2

( λx )2 −

2 λy 6

and

( )2 − ( λz )2 = 0 ( )2 + ( λz ) 2 = 0

Adding yields

( λ y )2 = 0

and then

( λz )2

= − ( λx )2

Substituting into Equation (9.57)

( λx )22 + ( λ y )2 + ( −λx )22 2

( λx )2 ∴

1 2

(θ x )2

and

( λz )2

=−

1 2

( )2 = 90.0° (θ z )2 = 135.0°

= 45.0° θ y

(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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 182. From the solution to Problem 9.143 and 9.167

I x = 34.106 × 10−3 lb ⋅ ft ⋅ s 2 I y = 50.125 × 10−3 lb ⋅ ft ⋅ s 2 I z = 34.876 × 10−3 lb ⋅ ft ⋅ s 2 I xy = 0.96211 × 10−3 lb ⋅ ft ⋅ s 2 I yz = I zx = 0 (a) From Equation 9.55

Ix − K I xy 0 0 I xy Iy − K =0 0 0 Iz − K

( Ix − K )( I y

) ( I z − K ) − ( I z − K ) I xy2 = 0 ( I z − K ) ( I x − K ) ( I y − K ) − I xy2  = 0 2 I z − K = 0 and I x I y − ( I x + I y ) K + K 2 − I xy =0

or or Then

−K

K1 = I z = 34.876 × 10−3 lb ⋅ ft ⋅ s 2

Now

or K1 = 34.9 × 10−3 lb ⋅ ft ⋅ s 2 and

( 34.106 × 10 )(50.125 × 10 ) − (34.106 × 10 + K − ( 0.96211 × 10 ) = 0 −3

or Solving yields

−3

)

+ 50.125 × 10 −3 K

1.70864 × 10−3 − 84.231 × 10−3 K + K 2 = 0

K 2 = 34.0486 × 10−3 K3 = 50.1824 × 103 or K 2 = 34.0 × 10−3 lb ⋅ ft ⋅s 2 and K3 = 50.2 × 10−3 lb ⋅ ft ⋅ s 2 continued

COSMOS: Complete Online Solutions Manual Organization System

(b) To determine the directions cosines λx , λ y and λz of each principal axis use two of the Equations 9.54 and Equation 9.57

K1 : Using Equation 9.54(a) and Equation 9.54(b) with I yz = I zx = 0 , we have

( I x − K1 )( λx )1 − I xy ( λ y )1 = 0

(

)( )1 = 0

− I xy ( λx )1 + I y − K1 λ y Substituting

( 34.106 × 10

−3

)

( )1 = 0

− 34.876 × 10−3 ( λx )1 − 0.96211 × 10−3 λ y

(

)( )

− 0.96211 × 10−3 ( λx )1 + 50.125 × 10−3 − 34.876 × 10−3 λ y

( )1 = 0

− 0.770 × 10−3 ( λx )1 − 0.96211 × 10−3 λ y

( )1 = 0

− 0.96211 × 10−3 ( λx )1 + 15.249 × 10−3 λ y Solving yields

From Equation 9.57 and

( λx )1 = ( λ y )1 = 0 02

( λx )12 + ( λ y )1 + ( λz )12

(θ x )1 = 90.0°,

= 1 or

(θ y )1 = 90.0°,

( λz )1 = 1 (θ z )1 = 0°

K 2 : Using Equation 9.54(b) and Equation 9.54(c) with I yz = I zx = 0

(

)( )2 = 0

− I xz ( λx )2 + I y − K 2 λ y

( Iz Now

− K 2 )( λz )2 = 0

I z ≠ K 2 ⇒ ( λz )2 = 0

Substituting

(

)( )

− 0.96211 × 10−3 ( λx )2 + 50.125 × 10−3 − 34.0486 × 10−3 λ y or

( λ y )2 = 0.05985 ( λx )2 continued

COSMOS: Complete Online Solutions Manual Organization System

( λx )22 + 0.05985 ( λx )2 

Then

( λx )2

+ ( λz )2 = 1

= 0.99821

( λ y )2 = 0.05974 (θ x )2

and

= 3.43°,

(θ y )2 = 86.6°, (θ z )2 = 90.0°

(

)( )3 = 0

− I xy ( λx )3 + I y − K3 λ y

K3 :

( Iz

− K3 ) ( λz )3 = 0

I z ≠ K3 ⇒ ( λz )3 = 0

Now Substituting

(

)( )

− 0.96211 × 10−3 ( λx )3 + 50.125 × 10−3 − 50.1824 × 10−3 λ y

( )3 = 0

− 0.96211 × 10−3 ( λx )3 − 0.0574 × 10−3 λ y

( λ y )3 = −16.7615 ( λx )3

or Have yields and

( λx )32 +  −16.7615 ( λx )3  ( λ x )3

= − 0.059555

(θ x )3

= 93.4°,

and

0 2

+ ( λz )3 = 1

( λ y )3 = 0.998231

(θ y )3 = 3.41°,

θ z = 90.0°

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Chapter 9, Solution 183. I x = 9.8821 × 10−3 lb ⋅ ft ⋅s 2

From Problem 9.147:

I y = 11.5344 × 10−3 lb ⋅ ft ⋅ s 2

I z = 2.1878 × 10−3 lb ⋅ ft ⋅ s 2 I xy = 0.48776 × 10−3 lb ⋅ ft ⋅ s 2

From Problem 9.151:

I yz = 1.18391 × 10−3 lb ⋅ ft ⋅ s 2

I zx = 2.6951 × 10−3 lb ⋅ ft ⋅ s 2 (a)

From Equation 9.56

(

)

2 2 2 K 3 − I x + I y + I z K 2 + I x I y + I y I z + I z I x − I xy − I yz − I zx K

(

)

2 2 2 − I x I y I z − I x I yz − I y I zx − I z I xy − 2I xy I yz I zx = 0

Substituting

{

K 3 − ( 9.8821 + 11.5344 + 2.1878 ) × 10−3  K 2 + ( 9.8821)(11.5344 ) + (11.5344 )( 2.1878 )

}

2 2 2 + ( 2.1878 )( 9.8821) − ( 0.48776 ) − (1.18391) − ( 2.6951)  × 10 −6 K  2 2 − ( 9.8821)(11.5344 )( 2.1878 ) − ( 9.8821)(1.18391) − (11.5344 )( 2.6951)  2 − ( 2.1878 )( 0.48776 ) − 2 ( 0.48776 )(1.18391)( 2.6951)  × 10 −9 = 0 

(

)

(

)

K 3 − 23.6043 × 10−3 K 2 + 151.9360 × 10−6 K − 148.1092 × 10−9 = 0

Solving numerically

K1 = 1.180481 × 10−3 lb ⋅ ft ⋅s 2

K1 = 1.180 × 10 −3 lb ⋅ ft ⋅s 2

K 2 = 10.72017 × 10−3 lb ⋅ ft ⋅ s 2

K 2 = 10.72 × 10 −3 lb ⋅ ft ⋅s 2

K3 = 11.70365 × 10−3 lb ⋅ ft ⋅ s 2

K3 = 11.70 × 10 −3 lb ⋅ ft ⋅s 2

continued

COSMOS: Complete Online Solutions Manual Organization System

(b)

From Equations 9.54(a) and 9.54(b)

( I x − K )( λx ) − I xy ( λ y ) − I zx ( λz ) = 0

(

)( )

− I xy ( λx ) + I y − K1 λ y − I yz ( λz ) = 0

( )1 and ( λz )1 .

K1: Substitute K1 and solve for λ to get ( λx )1 , λ y

( )1 − 2.6951( λz )1  × 10−3 = 0

( 9.8821 − 1.180481) ( λ ) − 0.48776 λ x 1 y 

( )1 − 1.18391( λz )1  × 10−3 = 0

 − 0.48776 ( λ ) + (11.5344 − 1.180481) λ x 1 y 

( )1 − 5.52546 ( λz )1 = 0

17.83996 ( λx )1 − λ y

( )1 − 0.11434 ( λz )1 = 0

− 0.0471( λx )1 + λ y Then

( λz )1 = 3.1549 ( λx )1

and

( λ y )1 = 0.40769 ( λx )1 ( λx )12 + ( λ y )1 + ( λz )12 2

Equation 9.57: Substituting

( λx )12 + 0.40769 ( λx )1 

=1 2

+ 3.1549 ( λx )1  = 1

( λx )1 = 0.29989

then

(θ x )1 = 72.5°

and

( λy )1 = 0.122262

then

(θ y )1 = 83.0°

( λz )1 = 0.94612

then

(θ z )1 = 18.89°

K 2: Substitute K 2 and solve for λ .

( )2 − 2.6951( λz )2  × 10−3 = 0

( 9.8821 − 10.72017 ) ( λ ) − 0.48776 λ x 2 y 

( )2 − 1.18391( λz )2  × 10−3 = 0

 − 0.48776 ( λ ) + (11.5344 − 10.72017 ) λ x 2 y 

continued

COSMOS: Complete Online Solutions Manual Organization System

( )2 − 5.52546 ( λz )2 = 0

−1.718202 ( λx )2 − λ y

( )2 − 1.45402 ( λz )2 = 0

− 0.599045 ( λx )2 + λ y

= −0.33201( λx )2

Then

( λz )2

and

( λ y )2 = 0.116306 ( λx )2 ( λx )22 + 0.116306 ( λx )2 

Then

( λx )2

= 0.94333

(θ x )2

then

And

( λ y )2 = 0.109715

( λz )2

= −0.31320

then

+  −0.33201( λx )2  = 1

= 19.38°

(θ y )2 = 83.7°

then

(θ z )2

= 108.3°

K 3: Substitute K 3 and solve for λ .

( )3 − 2.6951( λz )3  × 10−3 = 0

( 9.8821 − 11.70365 ) ( λ ) − 0.48776 λ x 3 y 

( )3 − 1.18391( λz )3  × 10−3 = 0

 − 0.48776 ( λ ) + (11.5344 − 11.70365 ) λ x 3 y 

( )3 − 5.52546 ( λz )3 = 0

−3.73452 ( λx )3 − λ y

( )3 + 6.99504 ( λz )3 = 0

2.88189 ( λx )3 + λ y

= 0.58019 ( λx )3

Then

( λz )3

and

( λ y )3 = −6.9403 ( λx )3 2

2 Then ( λx )3 +  − 6.9403 ( λx )3  + 0.58019 ( λx )3  =

( λx )3

= − 0.142128*

( λ y )3 = 0.98641 ( λz )3 *

= − 0.082461

then

(θ x )3

then

(θ y )3 = 9.46°

then

Note: the negative root of ( λx )3 is taken so that axes 1, 2, 3 form a right-handed set.

(θ z )3

= 98.2°

= 94.7°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 184. (a) From the solution of Problem 9.169 have

Ix =

1W 2 a 2 g

I xy =

1W 2 a 4 g

Iy =

W 2 a g

I yz =

1W 2 a 8 g

Iz =

5W 2 a 6 g

I zx =

3W 2 a 8 g

Substituting into Equation (9.56) 2 2 2 2  1 5   W 2   2  1   5   5  1   1  1  3   W 2  K −  + 1 +   a   K +   (1) + (1)   +    −   −   −     a  K 6   g   6   6  2   4  8  8    g   2   2 3

3 2 2  1   5   1  1 2  3  5  1   1  1  3    W 2  −   (1)   −    − (1)   −    − 2       a  = 0 8  6  4   4  8  8    g  2   6   2  8  

Simplifying and letting K =

W 2 a ζ yields g

ζ 3 − 2.33333ζ 2 + 1.53125ζ − 0.192708 = 0 Solving yields

ζ 1 = 0.163917

ζ 2 = 1.05402

ζ 3 = 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

continued

COSMOS: Complete Online Solutions Manual Organization System

(b) To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Equations (9.54) and (9.57). Then

K1 : Begin with Equations (9.54a) and (9.54b).

( I x − K1 )( λx )1 − I xy ( λ y )1 − I zx ( λz )1 = 0

(

)( )1 − I yz ( λz )1 = 0

− I xy ( λx )1 + I y − K 2 λ y Substituting

 1 1W 2   W 2  a  λy  − 0.163917   a   ( λx )1 −    g  4 g   2



 1W 2  W  − a  ( λx )1 + (1 − 0.163917 )  a 2   λ y 4 g   g  



( )1 −  83 Wg a2  ( λz )1 = 0 

( )1 −  18 Wg a2  ( λz )1 = 0 

Simplifying yields

( )1 − 1.5 ( λz )1 = 0

1.34433 ( λx )1 − λ y

( )1 − 0.149507 ( λz )1 = 0

− 0.299013 ( λx )1 + λ y Adding and solving for ( λz )1

( λz )1 = 0.633715 ( λx )1

( λ y )1 = 1.34433 − 1.5 ( 0.633715) ( λx )1

and then

= 0.393758 ( λx )1 Now substitute into Equation (9.57)

( λx )12 + 0.393758 ( λx )1 

+ 0.633715 ( λx )1  = 1

( λx )1 = 0.801504

( λ y )1 = 0.315599

and

∴

(θ x )1 = 36.7°

(θ y )1 = 71.6°

(θ z )1 = 59.5°

( λz )1 = 0.507925

K 2 : Begin with Equations (9.54a) and (9.54b).

( I x − K 2 )( λx )2 − I xy ( λ y )2 − I zx ( λz )2

(

)( )2 − I yz ( λz )2 = 0

− I xy ( λx )2 + I y − K 2 λ y

continued

COSMOS: Complete Online Solutions Manual Organization System

Substituting  1 1W 2   W 2  a  λy  − 1.05402   a   ( λx )2 −  2 g   4 g  



 1W 2  W  − a  ( λx )2 + (1 − 1.05402 )  a 2   λ y 4 g   g  



( )2 −  83 Wg a2  ( λz )2 = 0 

( )2 −  18 Wg a 2  ( λz )2 = 0 

Simplifying yields

( )2 − 1.5 ( λz )2 = 0

− 2.21608 ( λx )2 − λ y

( )2 + 2.31396 ( λz )2 = 0

4.62792 ( λx )2 + λ y Adding and solving for ( λz )2

( λz )2

= − 2.96309 ( λx )2

( λ y )2 = − 2.21608 − 1.5 ( − 2.96309 ) ( λx )2

and then

= 2.22856 ( λx )2 Now substitute into Equation (9.57)

( λx )22 + 2.22856 ( λx )2 

+  − 2.96309 ( λx )2  = 1

( λx )2

or and

= 0.260410

( λ y )2 = 0.580339 ∴

(θ x )2

( λz )2

= − 0.771618

(θ y )2 = 54.5°

= 74.9°

(θ z )2

= 140.5°

K 3 : Begin with Equations (9.54a) and (9.54b).

( I x − K3 )( λx )3 − I xy ( λ y )3 − I zx ( λz )3

(

)( )3 − I yz ( λz )3 = 0

− I xy ( λx )3 + I y − K 3 λ y Substituting

 1 1W 2   W 2  a  λy  − 1.11539   a   ( λx )3 −    g  4 g   2



 1W 2  W  a  ( λx )3 + (1 − 1.11539 )  a 2   λ y − 4 g   g  



( )3 −  83 Wg a2  ( λz )3 = 0 

( )3 −  81 Wg a2  ( λz )3 = 0 

continued

COSMOS: Complete Online Solutions Manual Organization System

Simplifying yields

( )3 − 1.5 ( λz )3 = 0

− 2.46156 ( λx )3 − λ y

( )3 + 1.08328 ( λz )3 = 0

2.16657 ( λx )3 + λ y Adding and solving for ( λz )3

( λz )3

= − 0.707885 ( λx )3

( λ y )3 = − 2.46156 − 1.5 ( − 0.707885) ( λx )3

and then

= −1.39973 ( λx )3 Now substitute into Equation (9.57)

( λx )32 +  −1.39973 ( λx )3 

+  − 0.707885 ( λx )3  = 1

( λ x )3

or and

(i)

= 0.537577

( λ y )3 = − 0.752463 ∴

(θ x )3

( λz )3 = 57.5°

= − 0.380543

(θ y )3 = 138.8°

(θ 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 Equation (i) must be chosen; that is, ( λx )3 = − 0.537577 Then (θ x )3 = 122.5°

(θ y )3 = 41.2°

(θ z )3

= 67.6°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 185.

m1 = ρ ta 2

From Problem 9.170

Now

I x = ( I x )1 + ( I x )2 =

m2 =

1 ρ ta 2 2

1 11 5  ρ ta 2 a 2 +  ρ ta 2  a 2 = ρ ta 4 3 6 2 12 

(

)

( )1 + ( I y )2 = 13 ( ρta2 ) a2 + 16  12 ρta2  ( a2 + a2 ) = 12 ρta4

Iy = Iy

I z = ( I z )1 + ( I z )2 = Now note that symmetry implies

1 11 3  ρ ta 2 a 2 + a 2 +  ρ ta 2  a 2 = ρ ta 4 3 6 2 4 

(

)(

)

( I x′y′ )1 = ( I y′z′ )1 = ( I z′x′ )1 = 0 ( I x′y′ )2 = ( I y′z′ )2 = 0

Have

I uv = I u′v′ + mu v continued

COSMOS: Complete Online Solutions Manual Organization System

 a  a  1 I xy = m1x1 y1 + m2 x2 y2 = ρ ta 2    = ρ ta 4  2  2  4

Then

I yx = m1 y1z1 +0 m2 y2 z02

( )2 + m2 z2 x2 

I zx = m1z1x1 +  I z ′x′ 

( I z′x′ )2 = − 721 ρ ta 4

From Problem 9.170

I zx = −

Then

1 1 1  1  1  ρ ta 4 +  ρ ta 2  a  a  = ρ ta 4 72 2 3 3 24    

(a) Equation 9.56

(

)

(

)

2 2 2 K 3 − I x + I y + I z K 2 + I x I y + I y I z + I z I x − I xy − I yz − I zx K

(

2 2 2 − I x I y I z − I x I yz − I y I zx − I z I xy − 2I xy I yz I zx = 0

Substituting 2   5  1   1  3   3  5   1 2  5  1 3  1   + +  ρ ta 4  K 2 +     +    +    −   − 0 −    ρ ta 4 K 3 −   24    12 2 4     12  2   2  4   4  12   4 

(

2 2   5  1  3    1  1   3  1  −      − 0 −    −    − 0  ρ ta 4 12 2 4  2  24   4  4       

(

Solving numerically...



3

)  = 0 

K = ρ ta 4ζ

Simplifying and letting yields

2

)  K

5 3

ζ3 − ζ2 + ζ 1 = 0.203032

479 125 ζ − =0 576 1152

K1 = 0.203ρ ta 4

ζ 2 = 0.698281

K 2 = 0.698ρ ta 4

ζ 3 = 0.765354

K3 = 0.765ρ ta 4

(b) Equations 9.54a and 9.54b

( I x − K )( λx ) − I xy ( λ y ) − I zx ( λz ) = 0

(

)( )

− I xy ( λx ) + I y − K λ y − I yz ( λz ) = 0 continued

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Substituting K1

 5 1   12 − 0.203032  ( λx )1 − 4 λ y  

( )1 − 241 ( λz )1  ρta4 = 0 

 1 1   − ( λx )1 +  − 0.203032  λ y 4 2   

( )1 − 0 ρ ta4 = 0 

( λ y )1 = 0.841842 ( λx )1

and

( λz )1 = 0.0761800 ( λx )1

Equation 9.57 Substituting

( λx )12 + ( λ y )1 + ( λz )12 2

( λx )12 + 0.841842 ( λx )1  or

=1 2

+ 0.0761800 ( λx )1  = 1

( λx )1 = 0.763715

then

(θ x )1 = 40.2°

( λ y )1 = 0.642927

then

(θ y )1 = 50.0°

( λz )1 = 0.0581798

then

(θ z )1 = 86.7°

Substituting K 2  5 1   12 − 0.698281 ( λx )2 − 4 λ y   

( )2 − 241 ( λz )2  ρta4 = 0 

 1 1   − ( λx )2 +  − 0.698281 λ y 2   4

( )2 − 0 ρ ta4 = 0

or and Then



( λ y )2 = −1.260837 ( λx )2 ( λz )2

= 0.806278 ( λx )2

( λx )22 +  −1.260837 ( λx )2  or

+ 0.806278 ( λx )2  = 1

( λx )2

= 0.555573

( λ y )2 = − 0.700487 ( λz )2

= 0.447946

then then

then

(θ x )2

= 56.2°

(θ y )2 = 134.5° (θ z )2

= 63.4°

continued

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Substituting K 3

 5 1   12 − 0.765354  ( λx )3 − 4 λ y  

( )3 − 241 ( λz )3  ρta4 = 0 

 1 1   − ( λx )3 +  − 0.765354  λ y 2   4

( )3 − 0 ρ ta4 = 0



( λ y )3 = − 0.942138 ( λx )3

And Then

( λz )3

= − 2.71567 ( λx )3

( λx )32 +  − 0.942138 ( λx )3  or

+  − 2.71567 ( λx )3  = 1

( λx )3

= 0.328576

then

(θ x )3

= 70.8°

( λ y )3 = − 0.309564

then

(θ y )3 = 108.0°

( λz )3

then

(θ z )3

= − 0.892304

(c)

= 153.2°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 186. (a) From the solutions to Problem 9.150

I x = 14.32 × 10−3 lb ⋅ ft ⋅ s 2 I y = I z = 18.62 × 10−3 lb ⋅ ft ⋅ s 2 I xy = I zx = 4.297 × 10−3 lb ⋅ ft ⋅ s 2 , I yz = 0

From Problem 9.172

I y = Iz,

I xy = I zx ,

I yz = 0

(

)

( ) ( )  K − (14.32 × 10 )(18.62 × 10 )

) (

Substituting into Eq. (9.56) and using

(

)

( )

2 2 K 3 − I x + 2I y K 2 +  I x 2I y + I y2 − 2 I xy  K − I x I y2 − 2I y I xy =0  

) (

K 3 − 14.32 × 10−3 + 2 18.62 × 10−2  K 2 +  14.32 × 10 −3 ( 2 ) 18.62 × 10−3 + 18.62 × 10 −3   

(

− 2 4.297 × 10−3 or

−3

(

)(

)

2 − 2 18.62 × 10 −3 4.297 × 10 −3  = 0 

K 3 − 51.56 × 10−3 K 2 + 0.84305 × 10−3 K − 0.004277 × 10−3 = 0 Solving:

K1 = 0.010022 lb ⋅ ft ⋅ s 2

or K1 = 10.02 × 10−3 lb ⋅ ft ⋅ s 2

K 2 = 0.018624 lb ⋅ ft ⋅ s 2

or K 2 = 18.62 × 10−3 lb ⋅ ft ⋅ s 2

K3 = 0.022914 lb ⋅ ft ⋅ s 2

or K3 = 22.9 × 10−3 lb ⋅ ft ⋅ s 2

(b) To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Equations (9.54) and Equation (9.57). Then

K1 : Begin with Equations (9.54b) and (9.54c):

− I xy (λx )1 + ( I y − K1)(λ y )1 − I yz (λz )1 = 0 − I zx (λx )1 − I yz (λ y )1 + ( I z − K1) (λz )1 = 0 or

)

( + (18.62 × 10

) ) (λ )

− 4.297 × 10−3 (λx )1 + 18.62 × 10−3 − 10.02 × 10−3 (λ y )1 = 0 − 4.297 × 10−3 (λx )1

−3

− 10.02 × 10−3

z 1

=0 continued

COSMOS: Complete Online Solutions Manual Organization System

( λ y )1 = ( λz )1 = 0.49965 ( λx )1 ( λx )12 + 2 0.49965 ( λx )1 

( λx )1 = 0.81669

( λ y )1 = ( λz )1 = 0.40806 (θ x )1 = 35.2°;

(θ y )1 = (θ z )1 = 65.9°

K2: Begin with Equations (9.54a) and (9.54b): ( I x − K 2 )( λx )2 − I xy (λ y )2 − I zx (λz )2 = 0

(

)( )2 − I yz (0λz )2 = 0

− I xy ( λx )2 + I y − K 2 λ y

Substituting:

(14.32 × 10

−3

)

− 18.62 × 10−3 ( λx )2 − 4.297 × 10−3 (λ y ) 2 − (λz ) 2  = 0

(

)( )

−4.297 × 10−3 ( λx )2 + 18.62 × 10−3 − 18.62 × 10−3 λ y

( λx )2

From (ii)

(i) (ii)

( λ y )2 = − ( λz )2

From (i) Substituting:

( λx )22 + ( λ y )2 +  − ( λz )2  2

( λ y )2 =

1 2

(θ x )2

( )2 = 45.0°, (θ z )2 = 135.0°

= 90.0°, θ y

K3: Begin with Equations (9.54b) and (9.54c) − I xy ( λx )3 + I y − K3 λ y

(

)( )3 + I yz ( λz )3 = 0 − I zx ( λx )3 − I yz 0( λ y ) + ( I z − K 3 ) ( λz )3 = 0

Substituting:

( + (18.62 × 10

) )(λ )

− 4.297 × 10−3 ( λx )3 + 18.62 × 10−3 − 22.9 × 10−3 ( λz )3 = 0 − 4.297 × 10−3 ( λx )3

−3

− 22.9 × 10−3

( λ y )3 = ( λ z ) 3 = − ( λ x )3

Simplifying:

( λx )23

+ 2  − ( λx )3  = 1 ⇒ ( λx )3 =

1 3

and

z 3

( λ y )3 = ( λz )3 = − (θ x )3

1 3

( )3 = (θ z )3 = 125.3°

= 54.7°, θ y

continued

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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 Equation (i) must be chosen; that is:

( λ x )3 Then:

=−

1 3

(θ x )3

= 125.3°

(θ y )3 = (θ z )3 = 54.7°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 187.

x = 0, y = 0: 0 = ka 2 + c or k = −

x = a,

y = b:

∴k = − y =−

=− Now

c a2

b=c

b a2

b 2 x − a) + b 2( a b 2 x − 2ax + a 2 + b a2

(

)

dI y = x 2dA = x 2 ( y2 − y1 ) dx b 2b   = x2  b + 2 x2 − x + b − b  dx a a   2b 3  b  =  2 x4 − x + bx 2  dx a a  2b 3 a  b  I y = ∫0  2 x 4 − x + bx 2  dx a a  a

b 4 b 3 1 b 5 = x − x + x  2 2a 3 0 5 a 1 1 1 = a3b  − +   5 2 3

1 3 ab 30 Iy =

1 3 ab 30

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Chapter 9, Solution 188.

x = 0, y = 0: 0 = ka 2 + c

k =−

c a2

x = a, y = b : k =−

Now

dI x = y 2dA = y 2 ( xdy )

( x − a )2

From above

a2 (b − y ) b

x = a 1−

and

x − a = a 1−

Then

b a2

b 2 x − a) 2( a

y =b−

Then

b=c

y b

y +a b

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

y3 =a 3

 y b + a ∫0 y 2  1 −  dy b  continued

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Chapter 9, Solution 189.

First note that

A = A1 − A2 − A3

= (100 mm )(120 mm ) − ( 80 mm )( 40 mm ) − ( 80 mm )( 20 mm ) = (12 000 − 3200 − 1600 ) mm 2 = 7200 mm 2 I x = ( I x )1 − ( I x )2 − ( I x )3

Now where

Then

( I x )1 =

1 (100 mm )(120 mm )3 = 14.4 × 106 mm 4 12

( I x )2

1 (80 mm )( 40 mm )3 + 3200 mm 2 ( 40 mm )2 = 5.5467 × 106 mm4 12

( I x )3

1 (80 mm )( 20 mm )2 + 1600 mm 2 ( 30 mm )2 = 1.4933 × 106 mm 4 12

(

)

I x = (14.4 − 5.5467 − 1.4933) × 106 mm 4 = 7.36 × 106 mm 4 or I x = 7.36 × 106 mm 4

and

k x2 =

Ix 7.36 × 106 = = 1022.2 mm 2 A 7200 or k x = 32.0 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 190.

First note that

A = A1 − A2 − A3

= (100 mm )(120 mm ) − ( 80 mm )( 40 mm ) − ( 80 mm )( 20 mm ) = (12 000 − 3200 − 1600)mm 2 = 7200 mm 2

( )1 − ( I y )2 − ( I y )3

Iy = Iy

Now

where

( I y )1 = 121 (120 mm )(100 mm )3 = 10 × 106 mm 4 ( I y )2 = 121 ( 40 mm )(80 mm )3 = 1.7067 × 106 mm 4 ( I y )3 = 121 ( 20 mm )(80 mm )3 = 0.8533 × 106 mm 4

Then

I y = (10 − 1.7067 − 0.8533) × 106 mm 4 = 7.44 × 106 mm 4 or I y = 7.44 × 106 mm 4

And

k y2 =

Iy A

7.44 × 106 mm 4 = 1033.33 mm 2 7200 mm 2

k = 32.14550 mm or k y = 32.1 mm

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Chapter 9, Solution 191.

y , mm

yA, mm3

50.9296

1.1520 × 106

− 0.324 × 106

A, mm 2

2 Σ

Now

−

(120 )2

= 22 619.5

1 ( 240 )( 90 ) = −10 800 2 11 819.5

Y =

0.828 × 106

0.828 × 106 mm3 ΣAY = = 70.054 mm ΣA 11819.5 mm 2

J O = ( J O )1 − ( J O )2

(a) where

and

( J O )1 =

π 4

(120 mm ) = 162.86 × 10

( J O ) 2 = ( I x′ ) 2 + ( I y ′ ) 2

mm 4

1 1 ( 240 mm )( 90 mm )3 + 2  ( 90 mm )(120 mm )3  12 12 

= 40.5 × 106 mm 4 Then

J O = (162.86 − 40.5 ) × 106 mm 4 = 122.36 × 106 mm 4 or J O = 122.4 × 106 mm 4 ! J O = J C + Ay 2

(b) or

(

)

J C = 122.36 × 106 mm 4 − 11 819.5 mm 2 ( 70.054 mm )

= (122.36 − 58.005 )106 mm 4 or J C = 64.4 × 106 mm 4 !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 192.

W section (fig. 9.13A):

A = 7.08 in 2 I x = 18.3 in 4 I y = 82.8 in 4

A = 2 AW + 2 Aplate = 2  7.08 in 2 + ( 7.93 in.)( 0.3 in.)  = 18.918 in 2 Now

( )W + 2 ( I x )plate

Ix = 2 Ix

2  4 2  6.495 in.  = 2 18.3 in + 7.08 in    2    

(

)

 ( 7.93 in.)( 0.3 in.)3  2 +2 + ( 7.93 in.)( 0.3 in.)  ( 6.495 in. + 0.15 in.)  12   = 2 92.967 in 4  + 2 105.07 in 4  = 396.07 in 4

I x = 396 in 4

or and

k x2 =

Ix 396.07 in 4 = = 20.936 in 2 A 18.918 in 2

k x = 4.58 in.

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Chapter 9, Solution 193.

Angle:

A = 1.44 in 2

I x = I y = 1.24 in 4

A = 5.88 in 2

Channel:

I x = 2.81 in 4

I y = 78.9 in 4

Locate the centroid

X =0

(

)

(

)

2  1.44 in 2 ( 0.842 in.)  + 5.88 in 2 ( − 0.606 in.) ΣAy  =  Y = ΣA 2 1.44 in 2 + 5.88 in 2

= Now

(

( 2.42496 − 3.5638) in 3 8.765 in 4

)

= − 0.12995 in.

( I x ) = 2 ( I x )L + ( I x )C = 2 1.24 in 4 + (1.44 in 2 ) ( 0.842 in. + 0.12995 in.)2 

(

)

2 +  2.81 in 4 + 5.88 in 2 ( 0.606 in. − 0.12995 in.)   

= 2 ( 2.6003) in 4 + 4.1425 in 4 = 9.3431 in 4 or I x = 9.34 in 4 Also

( I y ) = 2 ( I y )L + ( I y )C = 2 1.24 in 4 + 1.44 in 2 (5 in. − 0.842 in.)2  + 7.89 in 4 = 2 ( 26.136 ) in 4 + 78.9 in 4 = 131.17 in 4 or I y = 131.2 in 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 9, Solution 194. (a) Have

I AA′ = I GG′ + mra2

ra + rb = 290 mm = 0.29 m

and

I BB′ = I GG′ + mrb2

(

I BB′ − I AA′ = m rb2 − ra2

Subtracting

( 41 − 78) g ⋅ m 2

)

= (2000 g) ( rb + ra )( rb − ra )

− 37 = (2000) (0.29) ( rb − ra )

ra − rb = 63.793 × 10−3 m

now

ra + rb = 0.29 m

so that

2ra = 0.35379 m ra = 0.17689 m or ra = 176.9 mm

I AA′ = I GG′ + mra2

(b) Have Then

I GG′ = 78 g ⋅ m 2 − (2000 g) (0.17689 m) 2

= 15.420 g ⋅ m 2 Finally,

2 kGG ′ =

I GG′ 15.420 g ⋅ m 2 = = 0.007710 m 2 m 2000 g

kGG′ = 0.08781 m kGG′ = 87.8 mm

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Chapter 9, Solution 195.

( )1 + ( I xy )2 + ( I xy )3

Have

I xy = I xy

Symmetry implies

( I xy )2 = 0 I xy = I x′y′ + x yA

For the other rectangles

I x′y′ = 0

Where symmetry implies

A in 2

x , in.

y , in.

Ax y in 4

4 ( 0.5 ) = 2

− 2.75

1.0

− 5.5

4 ( 0.5 ) = 2

2.75

−1.0

− 5.5

−11.00 or I xy = −11.00 in 4

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Chapter 9, Solution 196.

I xy = −11.0 in 4

From Problem 9.195 Compute I x and I y for area of Problem 9.195 3

Ix =

5 in. × ( 0.5 in.) 12

 ( 0.5 in.)( 4 in.)3  2  +2 + ( 4 in. × 0.5 in.)(1.0 in.)  12  

= 9.38542 in 4  ( 0.5 in.)3 ( 4 in.)  0.5 in. × ( 5 in.)3 2 Iy = 2 + ( 4 in. × 0.5 in.)( 2.75 in.)  + 12 12  

= 35.54167 in 4 X ( 9.38542, −11) ,

Define points Now

I ave =

Ix + I y 2

Y ( 35.54167, 11)

9.38542 in 4 + 35.54167 in 4 = 22.46354 in 4 2 2

and

 Ix − I y    + I xy  2 

( )

2  9.38542 − 35.54167    + (11.0 ) 2  

= 17.08910 in 4 Also Then

  − 2 ( −11.0 ) 2θ m = tan −1   = − 40.067  9.38542 − 35.54167 

or θ m = − 20.0° clockwise

I max, min = I ave ± R = 22.46354 ± 17.08910 = 39.55264, 5.37444

Note: The a axis corresponds to I min and b axis corresponds to I max .

or I max = 39.6 in 4 I min = 5.37 in 4

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Chapter 9, Solution 197.

m = ρV = ρπ a 2 L

For the cylinder

dm = ρπ a 2dx

For the element shown

= dI z = dI z + x 2dm

and

Then

m dx L

1 2 a dm + x 2dm 4 L

1 2 1   m  m  1 I z = ∫ dI z = ∫ a + x 2  dx  =  a 2 x + x3  3 0 4  L  L  4 L 0 

m1 2 1 3  a L+ L L4 3  or I z =

1 m 3a 2 + 4L2 12

(

)

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Chapter 9, Solution 198.

First compute the mass of each component

Then

γ g

m1 =

0.284 lb/in 3 ( 5 in. × 4.5 in. × 0.9 in.) = 0.1786 lb ⋅ s2/ft 32.2 ft/s 2

m2 =

0.284 lb/in 3 ( 3 in. × 2.5 in. × 0.8 in.) = 0.05292 lb ⋅ s 2/ft 32.2 ft/s 2

m3 =

0.284 lb/in 3  2 π ( 0.6 in.) × 0.5 in. = 0.0049875 lb ⋅ s 2 /ft 2    32.2 ft/s

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 I uv = I u′v′ + muv so that ( Iuv )body = Σmu v .

m, lb ⋅ s 2 /ft

x , ft

y , ft

z , ft

mx y

my z

mz x

lb ⋅ ft ⋅ s 2

0.1786

0.2083 3

0.037 5

0.187 5

1.39531 × 10−3

1.25578 × 10−3

6.97656 × 10−3

0.05292

0.3833 3

0.20

0.187 5

4.0572 × 10−3

1.98451 × 10−3

3.80362 × 10−3

0.0049875

0.4375

0.225

0.187 5

0.49095 × 10−3

0.21041 × 10−3

0.40913 × 10−3

3.45069 × 10−3

11.18909 × 10−3

Then

5.94347 × 10−3

or I xy = 5.94 × 10−3 lb ⋅ ft ⋅ s 2 or I yz = 3.45 × 10−3 lb ⋅ ft ⋅ s 2 or I zx = 11.19 × 10−3 lb ⋅ ft ⋅ s 2