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CHAPTER 3CHAPTER 3CHAPTER 9CHAPTER 9

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

PROBLEM 9.1

Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.

SOLUTION

By observation yh

bx=

Now dI x dA x h y dx

hxx

bdx

y = = -

= -ÊËÁ

ˆ¯

2 2

2 1

[( ) ]

Then I dI hxx

bdxy y

b= = -Ê

ËÁˆ¯Ú Ú 2

01

= -È

ÎÍ

˘

˚˙h x

x

b

b1

3 43

4

0

or I b hy = 1

123 b

4


PROBLEM 9.2


SOLUTION

At x a y a= =, : ak

a= or k a= 2

Then ya

x=

2

Now dI x dA x y dx

xa

xdx a x d

y = =

=Ê

ËÁˆ

¯=

2 2

22

2

( )

x

Then I dI a x dx a xa

a ay y a

a

a

a

= = = ÈÎÍ

˘˚

= -Ú Ú 22 2 22 2

2 21

2 22[( ) ( ) ]

or I ay = 3

24 b

5


PROBLEM 9.3


SOLUTION

y kx= 2

For x a= : b ka= 2

kb

a=

2

Thus: yb

ax=

22

dA b y dx= -( )

dI x dA x b y dx x bb

ax dx

I dI bxb

ax

y

y y

= = - = -ÊËÁ

ˆ¯

= = -ÊËÁ

ˆ¯Ú

2 2 22

2

22

4

( )

˜ = -Ú dx a b a ba 1

3

1

53 3

0 I a by = 2

153 b

6


PROBLEM 9.4


SOLUTION

y hx

a

x

a= -

Ê

ËÁˆ

¯4

2

2

dA y dx

dI x dA hxx

a

x

adx

I hx

a

x

adx

y

y

a

=

= = -Ê

ËÁˆ

¯

= -Ê

ËÁˆ

¯Ú

2 22

2

3 4

20

4

4

I hx

a

x

ah

a ay

a

= -È

ÎÍ

˘

˚˙ = -

Ê

ËÁˆ

¯4

4 54

4 5

4 5

20

3 3

I hay = 1

53 b

7


PROBLEM 9.5

Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.

SOLUTION

By observation yh

bx=

or xb

hy=

Now dI y dA y x dy

yb

hy dy

b

hy dy

x = =

= ÊËÁ

ˆ¯

=

2 2

2

3

( )

Then I dIb

hy dyx x

h= = ÚÚ 3

0

= ÈÎÍ

˘˚

b

hy

h1

44

0

or I bhx = 1

43 b

8


PROBLEM 9.6


SOLUTION

At x a y a= =, : ak

a= or k a= 2

Then ya

x=

2

Now dI y dxa

xdx

a

xdx

x = =Ê

ËÁˆ

¯

=

1

3

1

3

1

3

32 3

6

3

Then I dIa

xdx a

xx x

a

a

a

a= = = -È

ÎÍ˘˚ÚÚ

1

3

1

3

1

2

16

36

2

2

= - -È

ÎÍ

˘

˚˙

1

6

1

2

162 2

aa a( ) ( )

or I ax = 1

84 b

9


PROBLEM 9.7


SOLUTION

See figure of solution of Problem 9.3.

yb

ax=

2

dI b dx y dx b dxb

ax dx

I dI bb

ax

x

x x

= - = -

= = -Ê

ËÁ

1

3

1

3

1

3

1

3

1

3

1

3

3 3 33

66

33

66 ˆ

¯

= - = -ÊËÁ

ˆ¯

= -ÊËÁ

ˆ¯

ÚÚ dx

b ab

a

aab

ab

a

0

33

6

731

3

1

3 7

1

3

1

21

7

21

1

2133 36

21= ab I abx = 2

73 b

10


PROBLEM 9.8


SOLUTION

See figure of solution of Problem 9.4.

y h h hx

a

dI y dxx

= + -

=

1 2 1

31

3

( )

I dI h h hx

adx

h h hx

a

a

x y

a= = + -È

ÎÍ˘˚

= + -ÈÎÍ

˘˚

ÚÚ1

3

1

12

1 2 1

3

0

1 2 1

4

( )

( )hh h

a

h hh h

a h h h h h

a

2 1 0

2 124

14 2

212

2 1 2

12 12

-ÊËÁ

ˆ¯

=-

-( ) = ◊+( ) +

( )

( )( ---

h

h h1

2 1

)

Ia

h h h hx = +( ) +12 1

222

1 2( ) b

11


PROBLEM 9.9


SOLUTION

At x y b= =0, : b c= -( )1 0 or c b=

At x a y= =, :0 0 1 1 2= -b ka( )/ or/

ka

= 11 2

Then y bx

a= -

Ê

ËÁˆ

¯1

1 2

1 2

/

/

Now dI y dx bx

adxx = = -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

1

3

1

313

1 2

1 2

3/

/

or dI bx

a

x

a

x

adxx = - + -

Ê

ËÁˆ

¯1

31 3 33

1 2

1 2

3 2

3 2

/

/

/

/

Then I dI bx

a

x

a

x

adxx x

a= = - + -

Ê

ËÁˆ

¯Ú Ú21

31 3 33

1 2

1 2

3 2

3 20

/

/

/

/

= - + -È

ÎÍ

˘

˚˙

2

32

3

2

2

53

3 2

1 2

2 5 2

3 20

b xx

a

x

a

x

a

a/

/

/

/ or I abx = 1

153 b

12


PROBLEM 9.10


SOLUTION

At x a y b= =, : b ke kb

ea a= =/ or

Then yb

ee bex a x a= = -/ / 1

Now dI y dx be dx

b e dx

xx a

x a

= =

=

-

-

1

3

1

31

3

3 1 3

3 3 1

( )

( )

/

/

Then I dI b e dxb a

ex xx aa x a

a

= = = ÈÎÍ

˘˚Ú Ú - -1

3 3 33 3 1

0

33 1

0

( ) ( )/ /

= - -1

913 3ab e( ) or I abx = 0 1056 3. b

13


PROBLEM 9.11


SOLUTION

At x a y b= =3 , : b k a a= -( )3 3

or kb

a=

8 3

Then yb

ax a= -

8 33( )

Now dI y dx

b

ax a dx

b

ax a dx

x =

= -ÈÎÍ

˘˚

= -

1

3

1

3 8

1536

3

33

3

3

99

( )

( )

Then I dIb

ax a dx

b

ax a

b

x x a

a

a

a

= = - = -ÈÎÍ

˘˚

=

Ú Ú3

993 3

910

3

1536 1536

1

10( ) ( )

33

910

15 3603 0

,[( ) ]

aa a- -

or I abx = 1

153 b

14


PROBLEM 9.12


SOLUTION

At x y b= =0, : b c= -( )1 0 or c b=

x a y= =, :0 0 1 1 2= -c ka( )/ or/

ka

= 11 2

Then y bx

a= -

Ê

ËÁˆ

¯1

1 2

1 2

/

/ b

15


PROBLEM 9.13


SOLUTION

At x a y b= =, : b kea a= /

or kb

e=

Then yb

ee bex a x a= = -/ / 1

Now dI x dA x y dx

x be dx

y

x a

= =

= -

2 2

2 1

( )

( )/

Then I dI bx e dxy y

a x a= = ÚÚ -2

0

1/

Now use integration by parts with

u x dv e dx

du x dx v ae

x a

x a

= =

= =

-

-

2 1

12

/

/

Then x e dx x ae ae x dx

a a xe

a x a x a a x aa

x a

2

0

1 2 1

0

1

0

3

2

2

Ú Ú- - -

-

= ÈÎ ˘ -

= -

/ / /

/

( )

11

0

adxÚ

Using integration by parts with

u x dv e dx

du dx v ae

x a

x a

= =

= =

-

-

/

/

1

1

Then I b a a xae ae dx

b a a a a

yx a a x aa

= - -ÈÎÍ

˘˚{ }

= - -

- -Ú3 10

1

0

3 2 2

2

2

( ) | ( )

(

/ /

eex a a/ -ÈÎ ˘{ }10) |

= - - -ÈÎ ˘{ }-b a a a a a e3 2 2 2 12 ( ) or I a by = 0 264 3. b

16


PROBLEM 9.14


SOLUTION

At x a y b= =3 , : b k a a= -( )3 3

or kb

a=

8 3

Then yb

ax a= -

8 33( )

Now dI x dA x y dx xb

ax a dx

b

ax x x a xa

y = = = -ÈÎÍ

˘˚

= - + -

2 2 23

3

32 3 2 2

8

83 3

( ) ( )

( aa dx3)

Then I dIb

ax x a x a a x dx

b

ax ax a

y y a

a= = - + -

= - +

Ú Ú 83 3

8

1

6

3

5

3

4

3

3 5 4 3 2 3 2

36 5 2

( )

xx a x

b

aa a a a a a a

a

a4 3 3

3

36 5 2 4 3

1

3

8

1

63

3

53

3

43

1

33

-ÈÎÍ

˘˚

= - + -( ) ( ) ( ) ( ))

( ) ( ) ( ) ( )

3

6 5 2 4 3 31

6

3

5

3

4

1

3

ÈÎÍ

˘˚

ÏÌÓ

- - + -ÈÎÍ

˘˚¸˝˛

a a a a a a a

or I a by = 3 43 3. b

17


PROBLEM 9.15

Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.

SOLUTION

At x a y y b= = =, :1 2 y b ma mb

a

y b ka kb

a

1

23

3

:

:

= =

= =

or

or

Then yb

ax1 =

or xa

by1 =

and yb

ax2 3

3=

or xa

by2 1 3

1 3=/

/

Now dA x x dy

a

by

a

by dy

= -

= -ÊËÁ

ˆ¯

( )2 1

1 31 3

//

Then A dA ay

b by dy

ab

yb

y

b= = -

Ê

ËÁˆ

¯

= -ÈÎÍ

˘˚

Ú Ú21

23

4

1 1

2

1 3

1 30

1 34 3 2

0

/

/

//

bb

ab= 1

2

Now dI y dA ya

by

a

by dyx = = -Ê

ËÁˆ¯

ÈÎÍ

˘˚

2 21 3

1 3/

/

18


PROBLEM 9.15 (Continued)

Then I ab

yb

y dyx

b= -Ê

ËÁˆ¯Ú2

1 11 3

7 3 3

0 //

= -ÈÎÍ

˘˚

23

10

1 1

41 310 3 4

0

ab

yb

yb

// or I abx = 1

103 b

and kI

A

ab

abbx

x21

103

12

21

5= = = or k

bx =

5 b

19


PROBLEM 9.16


SOLUTION

y k x y k x1 12

2 21 2= = /

For x y y b= = =0 1 2and

b k a b k a

kb

ak

b

a

= =

= =

12

21 2

1 2 2 1 2

/

/

Thus,

yb

ax y

b

ax1 2

22 1 2

1 2= =/

/

dA y y dx

Ab

ax

b

ax dx

Aba

a

ba

a

= -

= -ÈÎÍ

˘˚

= -

Ú

( )2 1

1 21 2

22

0

3 2

1 2

32

3 3

//

/

/ aa

A ab

2

1

3=

dI y dx y dx

b

ax dx

b

ax dx

x = -

= -

1

3

1

3

1

3

1

3

23

13

3

3 23 2

3

66

//

I dIb

ax dx

b

ax dx

b

a

a b

a

x x

a a= = -

= ( ) -

Ú Ú Ú3

3 23 2

3

60

6

0

3

3 2

5 2

52

3

3 3

3 3

//

/

/

66

73

7

2

15

1

21

aab= -Ê

ËÁˆ¯ I abx = 3

353 b

kI

A

abx

xabb

2335

3

= =( )

k bx = 9

35 b

20


PROBLEM 9.17

Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.

SOLUTION

At x a y y b= = =, :1 2 y b ka kb

a

y b ma mb

a

13

3

2

:

:

= =

= =

or

or

Then yb

ax

yb

ax

1 33

2

=

=

Now dA y y dxb

ax

b

ax dx= - = -Ê

ËÁˆ¯

( )2 1 33

Then A dAb

ax

ax dx

b

ax

ax ab

a

a

= = -ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚

=

Ú Ú21

21

2

1

4

1

2

23

0

22

4

0

Now dI x dA xb

ax

b

ax dxy = = -Ê

ËÁˆ¯

ÈÎÍ

˘˚

2 23

3

Then I dIb

ax x

ax dxy y

a= = - 1Ê

ËÁˆ¯ÚÚ 2 2

23

0

= -ÈÎÍ

˘˚

21

4

1

6

142

6

0

b

ax

ax

a

or I a by = 1

63 b

and kI

A

a b

abay

y216

3

12

21

3= = = or k

ay =

3 b

21


PROBLEM 9.18


SOLUTION

See figure of solution on Problem 9.16.

A ab dI x dA x y y dx

I xb

ax

b

ax dx

y

y

a

= = = -

= -ÊËÁ

ˆ¯

=Ú

1

32 2

2 1

2

0 1 21 2

22

( )

// bb

ax dx

b

ax dx

aa

1 25 2

24

00// - ÚÚ

Ib

a

b b

a

aa by = ◊ ( ) - ◊ = -Ê

ËÁˆ¯1 2

7 2

72

2

53

5

2

7

1

5/

/

I a by = 3

353 b

kI

A

a by

yab

2335

3

3

= =( )

k ay = 9

35 b

22


PROBLEM 9.19


SOLUTION

y1 : At x a y= =2 0, : 0 2= c k asin ( )

22

ak ka

= =p por

At x a y h= =, : h ca

a= sin ( )p2

or c h=

y2 : At x a y h= =, :2 2h ma b= -

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

Solving yields mh

ab h= - =2

4,

Then y ha

x yh

ax h

h

ax a

1 22

24

22

= = - +

= - +

sin

( )

p

Now dA y y dxh

ax a h

ax dx= - = - + -È

ÎÍ˘˚

( ) ( ) sin2 12

22

p

Then A dA ha

x aa

x dxa

a= = - + -È

ÎÍ˘˚ÚÚ

22

2

2( ) sin

p

= - - + +ÈÎÍ

˘˚

ha

x aa

ax

a

a12

2

22

2

( ) cosp

p

= -Ê

ËÁˆ¯

+ - +ÈÎÍ

˘˚

= -ÊËÁ

ˆ¯

=

ha

aa a ah

ah

2 12 1

2

0 36338

2

p p( )

.

23



Find: I kx x and

We have dI y y dxh

ax a h

axx = -Ê

ËÁˆ¯

= - +ÈÎÍ

˘˚

- ÊËÁ

ˆ¯

Ï1

3

1

3

1

3

22

22 1

3 3

( ) sinp

ÌÌÔ

ÓÔ

¸˝Ô

Ô

= - + -ÈÎÍ

˘˚

dx

h

ax a

ax

3

33 3

3

82

2( ) sin

p

Then I dIh

ax a

ax dxx x a

a= = - + -È

ÎÍ˘˚ÚÚ

3

33 32

3

82

2( ) sin

p

Now sin sin ( cos ) sin sin cos3 2 21q q q q q q= - = -

Then Ih

ax a

ax

ax

ax dx a

a= - + - -Ê

ËÁˆ¯

ÈÎÍ

˘˚Ú

3

33 22

3

82

2 2 2( ) sin sin cos

p p pxx

h

ax a

a

ax

a

ax

h a

a

a

= - - + + -ÈÎÍ

˘˚

= -

3

34 3

2

3

3

22

2

2

2

3 2

3

2

( ) cos cosp

pp

p

pp p

p

+ÊËÁ

ˆ¯

+ - +ÈÎÍ

˘˚

= -ÊËÁ

ˆ¯

2

3

22

2

31

2

3

34

3

a

aa a

ah

( )

I ahx = 0 52520 3. or I ahx = 0 525 3. b

and kI

A

ah

ahxx2

30 52520

0 36338= = .

. or k hx = 1 202. b

24


PROBLEM 9.20


SOLUTION

y1 : At x a y= =2 0, : 0 2= c k asin ( )

22

ak ka

= =p por

At x a y h= =, : h ca

a= sin ( )p2

or c h=

y2 : At x a y h= =, :2 2h ma b= -

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

Solving yields mh

ab h= - =2

4,

Then y ha

x yh

ax h

h

ax a

1 22

24

22

= = - +

= - +

sin

( )

p

Now dA y y dxh

ax a h

ax dx= - = - + -È

ÎÍ˘˚

( ) ( ) sin2 12

22

p

Then A dA ha

x aa

x dxa

a= = - + -È

ÎÍ˘˚ÚÚ

22

2

2( ) sin

p

= - - + +ÈÎÍ

˘˚

ha

x aa

ax

a

a12

2

22

2

( ) cosp

p

= -Ê

ËÁˆ¯

+ - +ÈÎÍ

˘˚

= -ÊËÁ

ˆ¯

=

ha

aa a ah

ah

2 12 1

2

0 36338

2

p p( )

.

25



Find: I ky y and

We have dI x dA xh

ax a h

ax dx

ha

x ax

y = = - + -ÈÎÍ

˘˚

ÏÌÓ

¸˝˛

= - +

2 2

3 2

22

2

22

( ) sin

( )

p

--ÈÎÍ

˘˚

xa

x dx2

2sin

p

Then I dI ha

x ax xa

x dxy y a

a= = - + -È

ÎÍ˘˚ÚÚ

22

23 2 22

( ) sinp

Now using integration by parts with

u x dva

x dx= =2

2sin

p

du x dx va

ax= = -2

2

2pp

cos

Then xa

x dx xa

ax

a

ax x dx2 2

2

2

2

2

22sin cos cos ( )

pp

pp

pÚ Ú= -Ê

ËÁˆ¯

- -ÊËÁ

ˆ¯

Now let u x dva

x dx= = cosp2

du dx va

ax= = 2

2pp

sin

Then xa

x dxa

xa

xa

xa

ax

a

ax2 2

2

2

2

4 2

2

2

2sin cos sin sin

pp

pp p

pp

p= - + ÊËÁ

ˆ¯

-Ú ÊÊËÁ

ˆ¯

ÈÎÍ

˘˚Ú dx

I ha

x ax

ax

ax

ax

ax

a

y = - +ÊËÁ

ˆ¯

ÈÎÍ

- - + +

2 1

4

2

3

2

2

8

2

16

4 3

22

2pp

pp

cos sin33

3

2

2pp

cosa

xa

aÊ

ËÁˆ

¯

˘

˚˙˙

= - +ÈÎÍ

˘˚

- +ÏÌÔ

ÓÔ

¸˝ÔÔ

- -

ha

a a aa

aa

ha

2 1

42

2

32

22

16

2

4 3 23

3( ) ( ) ( )

p p

11

4

2

3

8

0 61345

4 32

2

3

( ) ( ) ( )

.

a a aa

a

a h

+ÈÎÍ

˘˚

-ÏÌÔ

ÓÔ

¸˝ÔÔ

=

p

or I a hy = 0 613 3. b

and kI

A

a h

ahyy2

30 61345

0 36338= = .

. or k ay = 1 299. b

26


PROBLEM 9.21

Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.

SOLUTION

We have dI y dA y x x dyx = = -2 22 1[( ) ]

Then I y a a dy y a dy

a y a

x a

aa

a

= - + -ÈÎÍ

˘˚

= ÈÎÍ

˘˚

+

ÚÚ2 2 2 0

21

32

1

3

2 22

2

3

0

( ) ( )

yy

a a a a a

a

a3

2

3 3 321

3

2

32

ÈÎÍ

˘˚

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= ÈÎÍ

˘˚

+ -ÈÎ ˘ÏÌÓ

( ) ( ) ( )¸˝˛

= 10 4a

Also dI x dA x y y dxy = = -2 22 1[( ) ]

Then I x a a dx x a dx

a

y a

aa= - + -È

ÎÍ˘˚

=

ÚÚ2 2 2 0

10

2 22

0

4

( ) ( )

Now J I I a aP x y= + = +10 104 4 or J aP = 20 4 b

and kJ

A

a

a a a aPP2

420

4 2 2= =

-( )( ) ( )( )

= 10

32a or k aP = 1 826. b

27


PROBLEM 9.22


SOLUTION

xa a y

aa y= + = +

2 2

1

2( )

dA x dy a y dy= = +1

2( )

1

2

1

2

1

2 2

1

2 2

3

40

2

00

22

2A dA a y dy ayy

aa

aa a

a

= = + = + = +Ê

ËÁˆ

¯=Ú Ú ( ) A a= 3

22 v

1

2

1

2

1

2

1

2 3 4

1

2

1

2 2 2 3

00

3 4

0

I y dA y a y dy ay y dy

ay y

x

aa

a

= = + = +

= + =

ÚÚÚ ( ) ( )

33

1

4

1

2

7

124 4+Ê

ËÁˆ¯

=a a I ax = 7

124 v

1

2

1

3

1

3

1

23

00

3

I x dy a y dyy

aa= = +Ê

ËÁˆ¯ÚÚ ( )

1

2

1

24

1

24

1

4

1

962

15

963

0

4

0

4 4 4I a y dy a y a a ay

aa

= + = + = -ÈÎ ˘ =Ú ( ) ( ) ( )

1

2

5

324I ay = I ay = 5

164 v

From Eq. (9.4): J I I a a aO x y= + = + = +ÊËÁ

ˆ¯

7

12

5

16

28 15

484 4 4 J aO = 43

484 b

J k A kJ

A

a

aa k aO O O

OO= = = = =2 2

4348

4

32

2243

72

43

72 k aO = 0 773. b

28


PROBLEM 9.23

(a) Determine by direct integration the polar moment of inertia of the annular area shown with respect to Point O. (b) Using the result of Part a, determine the moment of inertia of the given area with respect to the x axis.

SOLUTION

(a) dA u du= 2p

dJ u dA u u du

u du

O = =

=

2 2

3

2

2

( )p

p

J dJ u du

u

O O R

R

R

R

= =

=

Ú Ú2

21

4

3

4

1

2

1

2

p

p J R RO = -ÈÎ ˘p2 2

414 b

(b) From Eq. (9.4): (Note by symmetry.) I Ix y=

J I I I

I J R R

O x y x

x O

= + =

= = -ÈÎ ˘

2

1

2 4 24

14p

I R Rx = -ÈÎ ˘p4 2

414 b

29


PROBLEM 9.24

(a) Show that the polar radius of gyration kO of the annular area shown is approximately equal to the mean radius R R Rm = +( )1 2 2 for small values of the thickness t R R= -2 1. (b) Determine the percentage error introduced by using Rm in place of kO for the following values of t/Rm: 1, 1

2 , and 110 .

SOLUTION

(a) From Problem 9.23:

J R RO = -ÈÎ ˘p2 2

414

kJ

A

R R

R RO

O2 2 24

14

22

12

= =-ÈÎ ˘

-( )p

p

k R RO2

22

121

2= +ÈÎ ˘

Thickness: t R R= -2 1

Mean radius: R R Rm = +1

2 1 2( )

Thus, R R t R R tm m11

2

1

2= - = +and 2

k R t R t R tO m m m2

2 22 21

2

1

2

1

2

1

4= +Ê

ËÁˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +

For t small compared to Rm : k RO m2 2ª k RO mª b

(b) Percentage errorexact value) approximation value)

exact= -( (

( value)100

P E. . ( )= -+ -

+= -

+ ( ) -

+ ( )R t R

R t

m m

m

tR

tR

m

m

2 14

2

2 14

2

14

2

14

2100

1 1

1

100

30



For t

Rm

= 1: P.E. =+ -

+= -

1 1

1100 10 56

14

14

( ) . % b

For t

Rm

= 1

2: P.E. = -

+ ( ) -

+ ( )= -

1 1

1100 2 99

14

12

2

14

12

2( ) . % b

For t

Rm

= 1

10: P.E. =

+ ( ) -

+ ( )= -

1 1

1100 0 125

14

110

2

14

110

2( ) . % b

31


PROBLEM 9.25

Detetmine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.

SOLUTION

y1 : At x a y a= =2 2, :

2 21

212

1a k a ka

= =( ) or

y2 : At x y a= =0, : a c=

At x a y a= =2 2, :

2 222a a k a= + ( ) or k

a21

4=

Then ya

x y aa

x

aa x

12

22

2 2

1

2

1

41

44

= = +

= +( )

Now dA y y dxa

a xa

x dx

aa x dx

= - = + -ÈÎÍ

˘˚

= -

( ) ( )

( )

2 12 2 2

2 2

1

44

1

2

1

44

Then A dAa

a x dxa

a x x aa

a

= = - = -ÈÎÍ

˘˚

=ÚÚ 21

44

1

24

1

3

8

32 2

0

2 2 3

0

22( )

Now dI y y dxa

a xa

xx = -ÊËÁ

ˆ¯

= +ÈÎÍ

˘˚

- ÈÎÍ

˘˚

Ï1

3

1

3

1

3

1

44

1

223

13 2 2

32

3

( )ÌÌÔ

ÓÔ

¸˝Ô

Ô

= + + + -ÈÎÍ

˘˚

dx

aa a x a x x

ax dx

1

3

1

6464 48 12

1

836 4 2 2 4 6

36( )

== + + -1

19264 48 12 7

36 4 2 2 4 6

aa a x a x x dx( )

32



Then I dIa

a a x a x x dxx x

a= = + + -ÚÚ 2

1

19264 48 12 7

36 4 2 2 4 6

0

2( )

= + + -ÈÎÍ

˘˚

= +

1

9664 16

12

5

1

9664 2 16

36 4 3 2 5 7

0

2

36 4

aa x a x a x x

aa a a

a

( ) (2212

52 2

1

96128 128

12

532 128

3 2 5 7

4

a a a a

a

) ( ) ( )+ -ÈÎÍ

˘˚

= + + ¥ -ÊËÁ

ˆ¯ == 32

154a

Also dI x dA xa

a x dxy = = -ÈÎÍ

˘˚

2 2 2 21

44( )

Then I dIa

x a x dxa

a x x

a

y y

aa

= = - = -ÈÎÍ

˘˚

=

ÚÚ 21

44

1

2

4

3

1

5

1

2

4

3

2 2 2 2 3 5

0

2

0

2

( )

aa a a a a2 3 5 4 421

52

32

2

1

3

1

5

32

15( ) ( )-È

ÎÍ˘˚

= -ÊËÁ

ˆ¯

=

Now J I I a aP x y= + = +32

15

32

154 4 or J aP = 64

154 b

and kJ

A

a

aaP

P26415

4

83

228

5= = = or k aP = 1 265. b

33


PROBLEM 9.26


SOLUTION

The equation of the circle is

x y r2 2 2+ =

So that x r y= -2 2

Now dA x dy r y dy= = -2 2

Then A dA r y dyr

r= = -Ú Ú-

2 2 2

2/

Let y r dy r d= =sin ; cosq q q

Then A r r r d

r d r

= -

= = +

-

-

Ú

Ú

2

2 22

2

2

6

2

2 2

6

2 2

2

/

/

/

/

( sin ) cos

cossin

q q

q q q

p

p

p

p qq

p

p

p

p p p

4

22 2 4

23

3

6

2

2 2 6 3 2

ÈÎÍ

˘˚

= --

+-Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= +

- /

/

r rsin

88

2 5274 2

Ê

ËÁˆ

¯

= . r

Now dI y dA y r y dyx = = -( )2 2 2 2

Then I dI y r y dyx x r

r= = -

-ÚÚ 2 2

2

2 2

/


Then I r r r r d

r r r

x = -

=

-Ú2

2

2 2 2

6

2

2 2

(/

/sin ) ( sin ) cos

sin ( cos ) cos

q q q q

q q

p

p

qq qp

pd

-Ú /

/

6

2

Now sin sin cos sin cos sin2 21

422 2q q q q q q= fi =

34



Then I r dr

r

x = ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚

=

--Ú21

42

2 2

4

84 2

4

6

2

6

2sin

sinq q q q

p

p

p

p

/

/

/

/

442 6

23

4

2 2 2 8

2 3

3

16

p p p

p

- --Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= -Ê

ËÁˆ

¯

sin

r

Also dI x dy r y dyy = = -( )1

3

1

33 2 2

3

Then I dI r y dyy y r

r= = -Ú Ú-

21

32 2 3 2

2( ) /

/


Then I r r r dy = --Ú

2

32 2 3 2

6

2[ ( sin ) ] cosq q q

p

p /

/

/

I r r dy =-Ú

2

33 3

6

2( cos ) cosq q q

p

p

/

/

Now cos cos ( sin ) cos sin4 2 2 2 211

42q q q q q= - = -

Then I r d

r

y = -ÊËÁ

ˆ¯

= +ÊËÁ

ˆ¯

-

-Ú2

3

1

42

2

3 2

2

4

4 2 2

6

2

4

cos sin

sin

q q q

q q

p

p

/

/

11

4 2

4

8 6

2q q

p

p

-ÊËÁ

ˆ¯

ÈÎÍ

˘˚-

sin

/

/

= -Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

--

+-

--

--Ê2

3 2

1

4 2 2 4

1

4 2 84 2 2 6 3 6

23r

p p p p p psin sin

ËËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= - + +Ê

ËÁˆ

¯- +2

3 4 16 12

1

4

3

2 48

1

324r

p p p p 33

2

2

3 4

9 3

644

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= +Ê

ËÁˆ

¯r

p

35



Now J I Ir

rP x y= + = -Ê

ËÁˆ

¯+ +

Ê

ËÁˆ

¯

44

2 3

3

16

2

3 4

9 3

64

p p

= +Ê

ËÁˆ

¯=r r4 4

3

3

161 15545

p. or J rP = 1 155 4. b

and kJ

A

r

rP

P24

2

1 15545

2 5274= = .

. or k rP = 0 676. b

36


PROBLEM 9.27

Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point O.

SOLUTION

At q p= =, :R a2 2a a k= + ( )p

or ka=p

Then R aa

a= + = +ÊËÁ

ˆ¯p

q qp

1

Now dA dr r d

rdr d

==

( )( )qq

Then A dA rdr d r da

a

= = = ÈÎÍ

˘˚

+ +

ÚÚÚ Ú0

1

0

2

0

1

0

1

2

( )( )q pp q pp

q q/

/

A a d a

a

= +ÊËÁ

ˆ¯

= +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +

Ú1

21

1

2 31

1

61

22

23

00

2

qp

q p qp

p

pp

ppp

pÊËÁ

ˆ¯

-È

ÎÍÍ

˘

˚˙˙

=3

3 217

6( ) a

Now dJ r dA r rdr dO = =2 2 ( )q

Then J dJ r dr dO O

a= =

+ÚÚÚ 3

0

1

0

( )q ppq

/

= ÈÎÍ

˘˚

= +ÊËÁ

ˆ¯

= +Ê

+

ÚÚ1

4

1

41

1

4 51

4

0

14

4

00

4

r d a d

a

a( )q p ppq q

pq

p qp

/

ËËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

= +ÊËÁ

ˆ¯

-È

ÎÍÍ

˘

˚˙˙

5

0

45

51

201 1

p

p pp

a ( ) or J aO = 31

204p b

and kJ

A

a

aaO

O23120

4

76

2293

70= = =

pp

or k aO = 1 153. b

37


PROBLEM 9.28

Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.

SOLUTION

By observation: yh

xb

=2

or xb

hy=

2

Now dA xdyb

hy dy= = Ê

ËÁˆ¯2

and dI y dAb

hy dyx = =2 3

2

Then I dIb

hy dyx x

h= = ÚÚ 2

23

0

= =b

h

ybh

h4

0

3

4

1

4

From above: yh

bx= 2

Now dA h y dx hh

bx dx= - = -Ê

ËÁˆ¯

( )2

= -h

bb x dx( )2

and dI x dA xh

bb x dxy = = -2 2 2( )

38



Then I dIh

bx b x dxy y

b= = -ÚÚ 2 22

0

2( )

/

= -ÈÎÍ

˘˚

21

3

1

23 4

0

2h

bbx x

b/

= ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=23 2

1

2 2

1

48

3 43h

b

b b bb h

Now J I I bh b hO x y= + = +1

4

1

483 3 or J

bhh bO = +

4812 2 2( ) b

and kJ

A

h b

bhh bO

Obh

2 482 2

12

2 212 1

2412= =

+= +

( )( ) or k

h bO = +12

24

2 2

b

39


PROBLEM 9.29*

Using the polar moment of inertia of the isosceles triangle of Problem 9.28, show that the centroidal polar moment of inertia of a circular area of radius r ispr4 /2. (Hint: As a circular area is divided into an increasing number of equal circular sectors, what is the approximate shape of each circular sector?)

PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.

SOLUTION

First the circular area is divided into an increasing number of identical circular sectors. The sectors can be approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of the isosceles triangles are as shown.

For an isosceles triangle (see Problem 9.28):

Jbh

h bO = +48

12 2 2( )

Then with b r h r= =Dq and

( ) ( )( ) ( )D D DJ r r r rO sector ;1

4812 2 2q q+ÈÎ ˘

= +ÈÎ ˘1

48124 2r D Dq q( )

Now dJ

d

Jr

O Osector sector

q qq

q q=

DD

ÊËÁ

ˆ¯

= + DÈD Æ D Ælim lim ( )

0 0

4 21

4812ÎÎ ˘Ï

ÌÓ

¸˝˛

= 1

44r

Then ( )J dJ r d rO Ocircle sector= = = [ ]ÚÚ1

4

1

44 4

02

0

2q q pp

or ( )J rO circle = p2

4 b

40


PROBLEM 9.30*

Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 2/ p . (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.)

SOLUTION

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

( )J rC cir = p2

4

The area of the circle is

A rcir = p 2

So that [ ( )]J AA

C cir =2

2pTwo 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 JC will be decreased ( ;J r dAC = Ú 2 as r decreases, so must JC ).

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 ( ) ,JC 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 ( ) .JC 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

( )JA

C A ≥2

2pQ.E.D. b

where the equality applies when the original area is circular.

41


PROBLEM 9.30* (Continued)

Solution 2

Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid) at Point C. Without loss of generality, assume that

A A A A1 2 3 4= =

It then follows that

( ) ( ) [ ( ) ( ) ( ) ( )]J J J A J A J A J AC A C C C C C= + - + -cir 1 2 3 4

Now observe that

J A J AC C( ) ( )1 2 0- �

J A J AC C( ) ( )3 4 0- �

since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase.

( ) ( )J JC A C� cir or Q.E.D.( )JA

C A �2

2p b

42


PROBLEM 9.31

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

SOLUTION

First note that

A A A A= + +

= + +

= + +

=

1 2 3

24 6 8 48 48 6

144 384 288

8

[( )( ) ( )( ) ( )( )]

( )

mm

mm

2

2

116 mm2

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

where

( ) ( ( )(

( , )

I x 11

1224 144 27

432 104 976

= +

= +

mm)(6 mm) mm mm)

mm

3 2 2

44

4 mm= 105 408,

( ) ( ,

( ) (

I

I

x

x

2

3

1

128 73 728

1

12

= =

= +

mm)(48 mm) mm

48 mm)(6 mm)

3 4

3 (( )(

( , ) ,

288 27

864 209 952 210 816

mm mm)

mm mm

2 2

4 4= + =

Then I x = + +( , , , )105 408 73 728 210 816 mm4

= 389 952, mm4 or mm4I x = ¥390 103 b

and kI

Axx2 389 952

816= = , mm

mm

4

2 or kx = 21 9. mm b

43


PROBLEM 9.32

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

SOLUTION

First note that

A A A A= - -

= - -

= - -

1 2 3

120 144 48 96 96 24

17280 4608 2

[( )( ) ( )( ) ( )( )]

(

mm2

3304

10368

) mm

mm

2

2=

Now I I I Ix x x x= - -( ) ( ) ( )1 2 3

where ( ) (I x 11

12120 144= = ¥mm)( mm) 29.8594 10 mm3 6 4

( ) ( ( ) (

.

I x 2

6

1

1296 4608

11 501568 10

= +

= ¥

mm)(48 mm) mm 48 mm)

m

3 2 2

mm4

( ) ( ( )

.

I x 32

6

1

1296 2304 36

3 096576 10

= + ( )

= ¥

mm)(24 mm) mm mm

mm

3 2

44

Then I x = - -( ) ¥29 8594 11 501568 3 096576 106 4. . . mm

=15.261256 ´ 106 mm4 or mmI x = ¥15 26 106 4. b

and kI

Axx2

6

2

15 261256 10

10368= = ¥. mm

mm

4

or kx = 38.4 mm b

44


PROBLEM 9.33

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.

SOLUTION

First note that

A A A A= + +

= ¥ + +

= + +

=

1 2 3

2

24 6 8 48 48 6

144 384 288

8

[( ) ( )( ) ( )( )]

( )

mm

mm

2

116 mm2

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

where

( ) (

( ) (

I

I

y

y

1

2

1

126 6912

1

1248 20

= =

= =

mm)(24 mm) mm

mm)(8 mm)

3 4

3 448

1

126 55 2963

mm

mm)(48 mm) mm

4

3 4( ) ( ,I y = =

Then I y = + + =( , ) ,6912 2048 55 296 64 2564mm mm4

or I y = ¥64 3. 10 mm3 4 b

and kI

Ayy2 64 256

816= = , mm

mm

4

2 or ky = 8 87. mm b

45


PROBLEM 9.34

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.

SOLUTION

First note that

A A A A= - -

= - -

= - -

1 2 3

120 144 48 96 96 24

17280 4608 2

[( )( ) ( )( ) ( )( )]

(

mm2

3304

10368

) mm

mm

2

2=

Now I I I Iy y y y= - -( ) ( ) ( )1 2 3

where

( ) ( .I y 161

12144 20 736 10= = ¥ mm)(120 mm) mm3 4

( ) .I y 261

123 538944 10= = ¥(48 mm)(96 mm) mm3 4

( ) ( .I y 361

1224 1 769472 10= = ¥mm)(96 mm) mm3 4

Then I y = - -

=

( . . . )

.

20 736 3 53844 1 769472 10

15 428088 10

6

6

¥

¥

mm4

or mm4I y = ¥15 43 106. b

and kI

Ayy2

615 428088 10

10368= = ¥. mm

mm

4

2 or mmky = 38 6. b

46


PROBLEM 9.35

Determine the moments of inertia of the shaded area shown with respect to the x andy axes when a = 20 mm.

SOLUTION

We have I I Ix x x= +( ) ( )1 22

where ( ) (

.

I x 1

3

1

1240

213 33 10

=

= ¥

mm)(40 mm)

mm

3

4

( ) ( (I x 2

2

820

220

4 20

3= - ¥Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

p pp

mm) mm) mm4 2

+ ¥ +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

pp2

204 20

320

2

( mm) mm2

= ¥527 49 103. mm4

Then I x = + ¥[ . ( . )]213 33 2 527 49 103 mm4

or I x = ¥1 268 106. mm4 b

Also I I Iy y y= +( ) ( )1 22

where ( ) ( .I y 131

1240 213 33 10= = ¥ mm)(40 mm) mm3 4

( ) ( .I y 23

820 62 83 10= = ¥p

mm) mm4 4

Then I y = + ¥[ . ( . )]213 33 2 62 83 103 mm4

or I y = ¥339 103 mm4 b

47


PROBLEM 9.36

Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.

SOLUTION

Given: Area Square Semicircles)= - 2(

For a = 20 mm, we have

Square: I Ix y= = = ¥1

1260 1080 104 3( ) mm4

Semicircle :

I x = = ¥p8

20 62 83 104 3( ) . mm4

I I Ad IAA y y¢ ¢ ¢= + = + ÊËÁ

ˆ¯

2 4 2 2

820

220 8 488; ( ) ( ) ( . )

p p

I y¢ = ¥ - ¥62 83 10 45 27 103 3. .

I y¢ = ¥17 56 103. mm4

I I Ay y= + - = ¥ +¢ ( . ) . ( ) ( . )30 8 488 17 56 102

20 21 5122 3 2 2p

I y = ¥308 3 103. mm4

48



Semicircle : Same as semicircle .

Entire Area:

I x = ¥ - ¥

= ¥

1080 10 2 62 83 10

954 3 10

3 3

3

( . )

. mm4 I x = ¥954 103 mm4 b

I y = ¥ - ¥

= ¥

1080 10 2 308 3 10

463 3 10

3 3

3

( . )

. mm4 I y = ¥463 103 mm4 b

49


PROBLEM 9.37

For the 4000-mm2 shaded area shown, determine the distance d2 and the moment of inertia with respect to the centroidal axis parallel to AA¢ knowing that the moments of inertia with respect to AA¢ and BB¢ are 12 10 mm6 4¥ and 23 9 106. ¥ mm4, respectively, and that d1 25= mm.

SOLUTION

We have I I AdAA¢ = + 12 (1)

and I I AdBB¢ = + 22

Subtracting I I A d dAA BB¢ ¢- = -( )12

22

or d dI I

AAA BB

22

12= -

-¢ ¢

= --

=

(( . )

2512 23 9 10

4000

3600

6

mm)mm

mm

mm

24

2

2 or d2 60 0= . mm

Using Eq. (1): I = ¥ -12 10 4000 256 mm mm mm)4 2 2( )( or mm4I = ¥9 50 106. b

50


PROBLEM 9.38

Determine for the shaded region the area and the moment of inertia with respect to the centroidal axis parallel to BB¢, knowing that d1 = 25 mm and d2 15= mm and that the moments of inertia with respect to AA¢ and BB¢ are 7 84 106. ¥ mm4 and 5 20 106. ¥ mm4, respectively.

SOLUTION

We have I I AdAA¢ = + 12 (1)

and I I AdBB¢ = + 22

Subtracting I I A d dAA BB¢ ¢- = -( )12

22

or AI I

d dAA BB=

--

=-

-¢ ¢

12

22

67 84 5 20 10

25 15

( . . )

( (

mm

mm) mm)

4

2 2

or A = 6600 mm2 b

Using Eq. (1): I = ¥ -

= ¥

7 84 10 6600 25

3 715

6. ( )(

.

mm mm mm)

10 mm

4 2 2

6 4

or I = ¥3 72 106. mm4 b

51


PROBLEM 9.39

The shaded area is equal to 3.125 ¥ 104 mm2. Determine its centroidal moments of inertia I x and I y , knowing that I Iy x= 2 and that the polar moment

of inertia of the area about Point A is JA = 878 9 106. .¥ mm4

SOLUTION

Given: A I I Jy x A= ¥ = = ¥3 125 10 2 878 104 6 4. , mm mm2

J J AA C= + (150 mm)2

878 9 10 3 125 106 4. ( .¥ = + ¥ mm mm )(150 mm)4 2 2JC

JC = ¥175 775 106. mm4

J I I I IC x y y x= + =with 2

175 775 10 26. ¥ = + mm4 I Ix x I x = ¥58 6 106. mm4 b

I Iy x= = ¥2 117 2 106. mm4 b

52


PROBLEM 9.40

The polar moments of inertia of the shaded area with respect to Points A, B, and D are, respectively, J JA B= =1125 mm , 2625 10 mm ,4 6 4¥10 ¥6 and JD = 1781.25 ¥ 106 mm4. Determine the shaded area, its centroidal moment of inertia JC, and the distance d from C to D.

SOLUTION

See figure at solution of Problem 9.39.

Given: J J JA B D= ¥ = ¥ = ¥1125 10 2625 10 1781 25 106 6 6mm mm mm4 4 4, , .

J J A CB J A dB C C= + ¥ = + +( ) ; ( )2 2 22625 15010 mm6 4 (1)

J J A CD J AdD C C= + ¥ = +( ) ; .2 6 21781 25 10 mm4 (2)

Eq. (1) subtracted by Eq. (2): J J AB D- = ¥ =843 75 10 1506 2. ( )mm4 A = 37500 mm2 b

J J A AC JA C C= + ¥ = +( ) ; ( )(2 61125 10 37500 150mm mm mm)4 2 2

J

d

C =

¥ = ¥ +

281 25 10

1781 25 10 281 25 10 37500

6 4

6 6 2

.

. . ( )

¥ mm

mm mm mm4 4 2

JC = ¥281 106 mm4 b

Eq. (2):

J

d

C =

¥ = ¥ +

281 25 10

1781 25 10 281 25 10 37500

6 4

6 6 2

.

. . ( )

¥ mm

mm mm mm4 4 2 d = 200 mm b

53


PROBLEM 9.41

Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION

Dimensions in mm

First locate centroid C of the area.

Symmetry implies Y = 30 mm.

A, mm2 x , mm xA, mm3

1 108 60 6480¥ = 54 349,920

2 - ¥ ¥ = -1

272 36 1296 46 –59,616

S 5184 290,304

Then X A xA XS S= =: ( ) ,5184 290 3042 3mm mm

or X = 56 0. mm

Now I I Ix x x= -( ) ( )1 2

where ( ) ( )( ) .

( ) ( )( )

I

I

x

x

13 6

2

1

12108 60 1 944 10

21

3672 18

= = ¥

=

mm mm mm

mm mm

4

33 2

4

1

272 6

2 11 664 23 328 69 9

+ ¥ ¥ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= + =

mm 18 mm mm

mm

( )

( , , ) . 884 103 4¥ mm

[( )I x 2 is obtained by dividing A2 into ]

Then I x = - ¥( . . )1 944 0 069984 106 4mm

or I x = ¥1 874 106 4. mm b

54



Also I I Iy y y= -( ) ( )1 2

where ( ) ( )( ) ( )[( . ) ]

( , ,

I y 13 21

1260 108 6480 56 54

6 298 560 2

= +

= +

mm mm mm mm2

55 920 6 324 104 6 4, ) .mm mm= ¥

( ) ( )( ) ( )[( ) ]

( , ,

I y 23 21

3636 72 1296 56 46

373 248 129

= + -

= +

mm mm mm mm2

6600 0 502 106 4) .mm mm4 = ¥

Then I y = -( . . )6 324 0 502 106 4mm

or I y = ¥5 82 106 4. mm b

55


PROBLEM 9.42


SOLUTION

First locate C of the area:

Symmetry implies X = 12 mm.

A, mm2 y, mm yA, mm3

1 12 22 264¥ = 11 2904

21

224 18 216( )( ) = 28 6048

S 480 8952

Then Y A yA YS S= =: ( )480 89522 3mm mm

Y = 18 65. mm

Now I I Ix x x= +( ) ( )1 2

where ( ) ( )( ) ( )[( . ) ]

,

(

I

I

x 13 2

4

1

1212 22 264 18 65 11

26 098

= + -

=

mm mm mm mm

mm

2

xx ) ( )( ) ( )[( . ) ]

,

23 2

4

1

3624 18 216 28 18 65

22 771

= + -

=

mm mm mm mm

mm

2

Then I x = + ¥( . . )26 098 22 771 103 4mm

or I x = ¥48 9 103 4. mm b

56



Also I I Iy y y= +( ) ( )1 2

where ( ) ( )( )

( ) ( )( )

I

I

y

y

13 4

23

1

1222 12 3168

21

3618 12

1

21

= =

= + ¥

mm mm mm

mm mm 88 4

5184

2

4

mm 12 mm mm

mm

¥ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

( )

[( )I y 2 is obtained by dividing A2 into ]

Then I y = +( )3168 5184 4mm or I y = ¥8 35 103 4. mm b

57


PROBLEM 9.43


SOLUTION


A, mm2 x , mm y, mm xA, mm3 yA, mm3

1 100 ¥ 160 = 16 ¥ 103 50 80 800 ¥ 103 1280 ¥ 103

2 – 100 ¥ 40 = – 4 ¥ 103 38 86 –152 ¥ 103 –344 ¥ 103

S 12 ¥ 103 648 ¥ 103 936 ¥ 103

Then X A xA XS S= ¥ = ¥: ( )12 10 648 103 2 3 3mm mm

or X = 54 0. mm

and Y A yA YS S= ¥ = ¥: ( )12 10 936 103 2 3 3mm mm

or Y = 78 0. mm

Now I I Ix x x= -( ) ( )1 2

where ( ) ( )( ) ( )[( ) ]

( .

I x 13 3 21

12100 160 16 10 80 78

34 133 0

= + ¥ -

= +

mm mm mm mm2

.. )

.

( ) ( )( ) (

064 10

34 19733 10

1

1240 100 4 10

6 4

6

23

¥

= ¥

= + ¥

mm

mm

mm mm

4

I x33 2

6 6 4

86 78

3 3333 0 256 10 3 58933 10

mm mm

mm

2 )[( ) ]

( . . ) ( . )

-

= + ¥ = ¥

58



Then I x = - ¥

= ¥

( . . )

.

34 19733 3 58933

30 608 10

4

6

10 mm

mm

6

4 or I x = ¥30 6 106 4. mm b

Also I I Iy y y= -( ) ( )1 2

where ( ) ( )( ) ( )[( ) ]

( .

I y 13 3 2 21

12160 100 16 10 54 50

13 33333

= + ¥ -

=

mm mm mm mm

++ ¥ =

= + ¥

0 256 10 13 58933

1

12100 40 4 10

6 4 4

23 3

. ) .

( ) ( )( ) (

mm mm

mm mm mI y mm mm

mm mm

2 2

6 4 6 4

54 38

0 53333 1 024 10 1 55733 10

)[( ) ]

( . . ) .

-

= + ¥ = ¥

Then I y = - ¥

= ¥

( . . )

.

13 58933 1 55733 10

12 032 10

6 4

6

mm

mm4 or I y = ¥12 03 106 4. mm b

59


PROBLEM 9.44


SOLUTION


A, mm2 x , mm y, mm xA, mm3 yA, mm3

1 72 10 720¥ = 36 5 25.92 ¥ 103 3.6 ¥ 103

2 10 76 760¥ = 5 mm 48 mm 3.8 ¥ 103 36.48 ¥ 103

3 26 20 520¥ = 13 mm 96 mm 676 ¥ 103 49.92 ¥ 103

S 2000 36.48 ¥ 103 90 ¥ 103

Then X A xA XS S= = ¥: ( ) .2000 36 48 102 3mm mm3

or X = 18 24. mm

and Y A yA YS S= = ¥: ( )2000 90 102 3 3mm mm

or Y = 45 mm

60



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

where ( ) ( )( ) ( )[( ) ]

( )

I x 13 2

3 4

1

1272 10 720 45 5

6 1152 10

= + -

= + ¥

mm mm mm mm

mm

2

== ¥

= + -

=

1158 10

1

1210 76 760 48 45

3 4

23 2

mm

mm mm mm mm2( ) ( )( ) ( )[( ) ]

(

I x

3365 8133 6 84 10 372 6533 103 4 3 4. . ) ( . )+ ¥ = ¥mm mm

( ) ( )( ) ( )[( ) )]

( . )

I x 33 2 21

1226 20 520 96 45

17 3333 135

= + -

= +

mm mm mm mm

22 52 10 1369 8533 103 4 3 4. ) .¥ = ¥mm mm

Then I x = + + ¥ = ¥( . . .1158 372 6533 1369 8533 2900 5066 104 3 4) 10 mm mm3

or I x = ¥2 900 106 4. mm b

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

where ( ) ( )( ) ( )[( . ) ]

( .

I y 13 21

1210 72 720 36 18 24

311 04 227

= + -

= +

mm mm mm mm2

.. ) .1 10 538 14 103 4 3 4¥ = ¥mm mm

( ) ( )( ) ( )[( . ) ]

( . .

I y 23 2 21

1276 10 760 18 24 5

6 3333 133

= + -

= +

mm mm mm mm

22262 10 139 5595 10

1

1220 26 520

3 3 4

33

) .

( ) ( )( ) (

¥ =

= +

mm mm

mm mm mm

4 ¥

I y22 2

3 4 3 4

18 24 13

29 2933 14 2780 10 43 5713 10

)[( . ) ]

( . . ) .

-

= + ¥ = ¥

mm

mm mm

Then I y = + + ¥ = ¥( . . . ) .538 14 139 5595 43 5713 10 721 2708 103 4 3 4mm mm

or I y = ¥0 721 106 4. mm b

61


PROBLEM 9.45

Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.

SOLUTION

Dimensions in mm Symmetry: X Y=

Determination of centroid, C, of entire section.

Section Area mm, 2 y, mm yA, mm3

1 p4

100 7 854 102 3( ) .= ¥ 42.44 333 3 103. ¥

2 ( )( )50 100 5 103= ¥ 50 250 103¥

3 ( )( )100 50 5 103= ¥ –25 - ¥125 103

S 17 854 103. ¥ 458 3 103. ¥

Y A yA YS S= ¥ = ¥: ( . ) .17 854 10 458 3 103 3mm mm2 3

Y X Y= = =25 67 25 67. .mm mm

Distance O to C: OC Y= = =2 2 25 67 36 30( . ) . mm

(a) Section 1: JO = = ¥p8

100 39 27 104 6 4( ) . mm

Section 2: J J A ODO = + = + + ÊËÁ

ˆ¯

+( ) ( )( )[ ] ( )( )2 2 221

1250 100 50 100 50 100

50

2

100

22

5 208 10 15 625 10 20 83 10

2

6 6 6

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= ¥ + ¥ = ¥JO . . . mm4

62



Section 3: Same as Section 2; JO = ¥20 83 106. mm4

Entire section:

JO = ¥ + ¥

= ¥

39 27 10 2 20 83 10

80 94 10

6 6

6

. ( . )

. JO = ¥80 9 106. mm4 b

(b) Recall that, OC A= = ¥36 30 17 854 103. .mm and mm2

J J A OC

J

O C

C

= +

¥ = + ¥

( )

. ( . )( . )

2

6 4 3 280 94 10 17 854 10 36 30mm mm mm2

JC = ¥57 41 106 4. mm JC = ¥57 4 106 4. mm b

63


PROBLEM 9.46


SOLUTION

First locate centroid C of the figure.

Note that symmetry implies Y = 0.

Dimensions in mm

A, mm2 X , mm XA, mm3

1p2

84 42 5541 77( )( ) .= - = -11235 6507

p. -197 568,

2p2

42 2770 882( ) .= 5617 8254

p= . 49,392

3 - = -p2

54 27 2290 22( )( ) . - = -7222 9183

p. 52,488

4 - = -p2

27 1145 112( ) .36

11 4592p

= . –13,122

S 4877.32 –108,810

Then X A xA XS S= = -: ( . ) ,4877 32 108 810mm mm2 3

or X = -22 3094.

(a) J J J J JO O O O O= + - -( ) ( ) ( ) ( )1 2 3 4

where ( ) ( )( )[( ) ( ) ]

.

JO 12 2

6 4

884 42 84 42

12 21960 10

= +

= ¥

pmm mm mm mm

mm

64



( ) ( )

.

( ) ( )( )[( )

J

J

O

O

24

6 4

32

442

2 44392 10

854 27 54

=

= ¥

=

p

p

mm

mm

mm mm mm ++

= ¥

=

= ¥

( ) ]

.

( ) ( )

.

27

2 08696 10

427

0 41739 10

2

6

44

6 4

mm

mm

mm

mm

4

JOp

Then JO = + - - ¥

= ¥

( . . . . )

.

12 21960 2 44392 2 08696 0 41739 10

12 15917 10

6

6

mm

m

4

mm4

or JO = ¥12 16 106. mm4 b

(b) J J AXO C= + 2

or JC = ¥ - -12 15917 10 4877 32 22 30946 4 2. ( . )( . )mm mm mm2

or JC = ¥ 49 73 106. mm b

65


PROBLEM 9.47

Determine the polar moment of inertia of the area shown with respect to (a) Point O,(b) the centroid of the area.

SOLUTION

Section Area mm, 2 x , mm xA, mm3

1 p4

90 6 3617 102 3( ) .= ¥ 38.1972 243.000 ¥ 103

2 - = - ¥p4

60 2 8274 102 3( ) . 25.4648 –72.0000 ¥ 103

S 3.5343 ¥ 103 171 ¥ 103

Then XA xA X

X

= ¥ = ¥=

S : ( . )

.

3 5343 10 171 10

48 3830

23 3 3mm mm

mm

Then JO = - = ¥p p8

908

60 20 6756 104 4 6 4( ) ( ) .mm mm mm JO = ¥20 7 106. mm4 b

OC X

J J A OC

J

O C

C

= = =

= +

¥ = +

2 2 48 383 68 4239

20 6756 10

2

6 4

( . ) .

( ) :

.

mm mm

mm (( . )( . )

.

3 5343 68 4239

4 1286 10

2

6

mm mm

mm

2

4JC = ¥ JC = ¥4 13 106 4. mm b

66


PROBLEM 9.48


SOLUTION

First locate centroid C of the figure.


1 p2

120 22 61947 102 3( ) .= ¥ 16050 9296

p= . 1152 ¥ 103

2 - = - ¥1

2240 90 10 8 103( )( ) . 30 –324 ¥ 103

S 11.81947 ¥ 103 828 ¥ 103

Then Y A yA YS S= ¥ = ¥: ( . )11 81947 10 828 103 3mm mm2 3

or Y = 70 0539. mm

(a) J J JO O O= -( ) ( )1 2

where ( ) ( ) .

( ) ( ) ( )

J

J I I

O

O x y

14 6 4

2 2 2

4120 162 86016 10= = ¥

= +¢ ¢

pmm mm

Now ( ) ( )( ) .I x¢ = = ¥23 6 41

12240 90 14 58 10mm mm mm

and ( ) ( )( ) .I y¢ = ÈÎÍ

˘˚

= ¥23 6 42

1

1290 120 25 92 10mm mm mm

[Note: ( )I y¢ 2 is obtained using ]

67



Then ( ) ( . . )

.

JO 26 4

6 4

14 58 25 92 10

40 5 10

= + ¥

= ¥

mm

mm

Finally JO = - ¥ = ¥( . . ) .162 86016 40 5 10 122 36016 106 4 6 4mm mm

or JO = ¥122 4 106 4. mm b

(b) J J AYO C= + 2

or JC = ¥ - ¥

= ¥

122 36016 10 11 81947 10 70 0539

64 3555

6 4 3 2 2. ( . )( . )

.

mm mm mm

1106 mm4 or JC = ¥64 4 106 4. mm b

68


PROBLEM 9.49

Two 20-mm steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.

SOLUTION

S section: A

I

I

x

y

=

= ¥

= ¥

6010

90 3 10

3 88 10

2

6 4

6

mm

mm

mm4

.

.

Note: A A Atotal S plate

2

mm mm)(20 mm)

mm

= +

= +

=

2

6010 2 160

12 410

2 (

,

Now I I Ix x x= +( ) ( )S plate2

where ( )

( )( ) ( )[( .

I I Adx xplate

2

plate

mm mm mm

= +

= + +

2

31

12160 20 3200 152 5 100

84 6067 10

2

6 4

) ]

.

mm

mm= ¥

Then I x = + ¥ ¥( . . )90 3 2 84 6067 106 4mm

= ¥259 5134 106 4. mm or I x = ¥260 106 4mm b

and kI

Axx2

6 4

2

259 5134 10

12410= =

¥

total

mm

mm

. or kx = 144 6. mm b

69



Also I I Iy y y= +( ) ( )S plate2

= ¥ + ÈÎÍ

˘˚

= ¥

3 88 10 21

1220 160

17 5333 10

6 3

6 4

. ( )( )

.

mm mm mm

mm

4

or I y = ¥ 417 53 106. mm b

and kI

Ayy2

6 4

2

17 5333 10

12 410= =

¥

total

mm

mm

.

, or ky = 37 6. mm b

70


PROBLEM 9.50

Two channels are welded to a rolled W section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.

SOLUTION

W section: A

I

I

x

y

=

= ¥

= ¥

5880

45 8 10

15 4 10

2

6 4

6

mm

mm

mm4

.

.

Channel: A

I

I

x

y

=

= ¥

= ¥

2170

13 5 10

0 545 10

2

6 4

6

mm

mm

mm4

.

.

A A Atotal W chan

mm

= +

= + =

2

5880 2 2170 10220 2( )

Now I I Ix x x= +( ) ( )W chan2

where ( )

. ( )( . ) .

I I Adx xchan chan

mm mm mm

= +

= ¥ + =

2

6 4 2 213 5 10 2170 114 5 41 94922 106 4¥ mm

Then I x = + ¥ ¥ = ¥[( . ( . )] .45 8 2 41 9492 10 129 6984 106 6 4mm mm4 I x = ¥129 7 106 4. mm b

and kI

Axx2

6 4

2

129 6984 10

10220= = ¥

total

mm

mm

. kx = 112 7. mm b

Also I I Iy y y= +( ) ( )W chan2

= + ¥ = ¥[ . ( . )] .15 4 2 0 545 10 16 49 106 4 6 4mm mm I y = ¥16 49 106 4. mm b

and kI

Ayy2

6 4

2

16 49 10

10220= =

¥

total

mm

mm

. ky = 40 2. mm b

71


PROBLEM 9.51

To form a reinforced box section, two rolled W sections and two plates are welded together. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal axes shown.

SOLUTION

(Note: x and y interchanged as w.r.t. Fig. (19.3A)

W section: A

I

I

x

y

=

= ¥

= ¥

5880

15 4 10

45 8 10

6

6

mm

mm

mm

2

4

4

.

.

Note: A A Atotal W plate

2

2

mm mm mm

mm

= +

= +

=

2 2

2 5880 2 203 6

14 196

( ) ( )( )

,

Now I I Ix x x= +2 2( ) ( )W plate

where ( ) . ( )

.

I I Adx xW4 2mm mm mm= + = ¥ + Ê

ËÁˆ¯

= ¥

2 62

15 4 10 5880203

2

75 9772 1066

2

31

12203 6 1218 203

mm

mm mm mm

4

plate

2

plate( )

( )( ) ( )[(

I I Adx x= +

= + ++

= ¥

3

51 6907 10

2

6

) ]

.

mm

mm4

Then I x = + ¥

= ¥

[ ( . ) ( . )]

.

2 75 9772 2 51 6907 10

255 336 10

6

6

mm

mm

4

4

or I x = ¥255 106 mm4

and kI

Axx2

6255 336 10

14 196= =

¥

total

4

2

mm

mm

.

,

or kx = 134 1. mm

72



also I I Iy y y= +2 2( ) ( )W plate

where ( )

( ) ( )( ) .

I I

I

y y

y

W

plate4mm mm mm

=

= = ¥1

126 203 4 1827 103 6

Then I y = + ¥

= ¥

[ ( . ) ( . )]

.

2 45 8 2 4 1827 10

99 9654 10

6

6

mm

mm

4

4

or I y = ¥100 0 106. mm4

and kI

Ayy2

699 9654 10

14 196= =

¥

total

4

2

mm

mm

.

, or ky = 83 9. mm

73


PROBLEM 9.52

Two channels are welded to a d ¥ 300 mm steel plate as shown. Determine the width d for which the ratio I Ix y/ of the centroidal moments of inertia of the section is 16.

SOLUTION

Channel: A

I

I

x

y

=

= ¥

= ¥

2890

28 0 10

0 945 10

2

6

6

mm

mm

mm

4

4

.

.

Now I I I

d

x x C x= +

= ¥ +

= ¥

2

2 28 0 101

12300

56 10

6

6

( ) ( )

( . ) (

(

plate

4 3

4

mm mm)

mm ++ ¥2 25 106. )d

and I I Iy y C y= +2( ) ( )plate

where ( )I I Ady C y= + 2

= ¥ + +ÊËÁ

ˆ¯

= ¥ +

0 945 10 28902

16 1

0 945 10 2890

62

6

. ( ) .

( . ( )

mm mm mm4 2 d

d222

2 6

416 1 16 1

722 5 46529 1 69412 10

+ +Ê

ËÁˆ

¯

= + + ¥

. .

( . . )

( )

d

d d

I y platee mm= =1

12300 253 3( )d d

I d d d

d d d

y = + + + ¥

= + + +

25 2 722 5 46529 1 69412 10

25 1445 93058

3 2 6

3 2

( . . )

( 33388240

I I d d d dx y= fi ¥ + ¥ = + + + ¥16 56 10 2 25 10 16 25 1445 93058 3 38824 106 6 3 2 6. ( . ))

400 23120 761 072 10 1 78816 10 03 2 3 6d d d+ - ¥ - ¥ =. .

Solving yields d = 25 0999. mm (the other roots are negative)

d = 25 10. mm

74


PROBLEM 9.53

Two L76 ¥ 76 ¥ 6.4-mm angles are welded to a C250 ¥ 22.8 channel. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the web of the channel.

SOLUTION

Angle: A

I Ix y

=

= = ¥

929

0 512 106

mm

mm

2

4.

Channel: A

I

I

x

y

=

= ¥

= ¥

2890

0 945 10

28 0 10

6

6

mm

mm

mm

2

4

4

.

.

First locate centroid C of the section


Angle 2 929 1858( ) = 21.2 39,389.6

Channel 2890 -16.1 -46,529

S 4748 -7139.4

Then Y A y A YS S= = -: ( ) .4748 7139 4mm mm2 3

or Y = -1 50366. mm

Now I I Ix x L x C= +2( ) ( )

where ( ) . ( )[( . . ) ]

.

I I Adx L x= + = ¥ + +

=

2 0 512 10 929 21 2 1 50366

0 99

6 2mm mm mm4 2

00859 10

0 949 10 2890 16 1 1 502 6

¥

= + = ¥ + -

6 4

4 2

mm

mm mm( ) . ( )[( . .I I Adx C x 3366

1 56472 10

2

6

) ]

.

mm

mm4= ¥

Then I x = + ¥[ ( . ) . ]2 0 990859 1 56472 106 4mm

or I x = ¥3 55 106. mm4

75



Also I I Iy y L y C= +2( ) ( )

where ( ) . ( )[( . ) ]

.

I I Ady L y= + = ¥ + -

= ¥

2 6 20 512 10 929 127 21 2

10 9109

mm mm mm4 2

1106 mm4

( )I Iy C y=

Then I y = + ¥[ ( . ) . ]2 10 9109 28 0 106 mm4

or I y = ¥49 8 106. mm4

76


PROBLEM 9.54

Two L102 ¥ 102 ¥ 12.7-mm angles are welded to a steel plate as shown. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the plate.

SOLUTION

For 102 102 12 7¥ ¥ . mm A I Ix y= = = ¥2420 2 30 106mm mm2 4, .

Section Area, mm2 y, mm yA, mm3

Plate ( )( )12 240 2880= 120 345.6 ¥ 103

Two angles 2 2420 4840( ) = 30 145.2 ¥ 103

S 7720 490.8 ¥ 103

YA y A

Y

Y

=

= ¥=

S

( ) .

.

7720 490 8 10

63 5751

3mm mm

mm

2 3

77



Entire section:

I I Adx x= + = +ÈÎÍ

˘˚

+¢S( ) ( )( ) ( )( )( . ) [ .2 3 21

1212 240 12 240 56 4249 2 2 330 10 2420 33 57516 2¥ + ( )( . ) ]

= ¥ + ¥ + ¥ + ¥

= ¥

13 824 10 9 1693 10 4 6 10 2 7280 10

30 3213 10

6 6 6 6

6

. . . .

. mm4 I x = ¥30 3 106. mm4

I y = + ¥ +1

12240 12 2 2 30 10 2420 363 6 2( )( ) [ . ( )( ) ]

= ¥ + ¥ + ¥

= ¥

( . . . )

.

34 56 10 4 6 10 6 27264 10

10 9072 10

3 6 6

6 4mm I y = ¥10 91 106. mm4

78


PROBLEM 9.55

The strength of the rolled W section shown is increased by welding a channel to its upper flange. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.

SOLUTION

W Section: A

I

I

x

y

=

= ¥

= ¥

14 400

554 10

63 3 10

6

6

,

.

mm

mm

mm

2

4

4

Channel: A

I

I

x

y

=

= ¥

= ¥

2890

0 945 10

28 0 10

6

6

mm

mm

mm

2

4

4

.

.

First locate centroid C of the section.


W Section 14400 -231 -3326400

Channel 2890 49.9 144211

S 17290 -3182189

Then Y A y A YS S= = -: ( , ) , ,17 290 3 182 189mm mm2 3

or Y = -184 047. mm

Now I I Ix x x= +( ) ( )W C

where ( )

( , )( . )

.

I I Adx xW

4 2 2mm mm mm

= +

= ¥ + -

=

2

6 2554 10 14 400 231 184 047

585 755 10

0 945 10 2890 49 9 184 047

6

2

6

¥

= -

= ¥ + +

mm

mm mm

4

C

4 2

( )

. ( )( . . )

I I Adx x

22

6159 12 10

mm

mm

2

4= ¥.

79



Then I x = + ¥( . . )585 75 159 12 106 mm4

or I x = ¥745 106 mm4

also I I Iy y y= +

= + ¥

( ) ( )

( . . )

W C

4mm63 3 28 0 106

or I y = ¥91 3 106. mm4

80


PROBLEM 9.56

Two L127 ¥ 76 ¥ 12.7-mm angles are welded to a 12 mm steel plate. Determine the distance b and the centroidal moments of inertia I x and I y of the combined section, knowing that I y = 4 I x .

SOLUTION

Angle: A

I

I

x

y

=

=

=

93 75

235 75

63 75

.

.

.

mm

mm

mm

2

4

4

First locate centroid C of the section.

Area, mm2 y, mm yA, mm3

Angle 2 93 75 187 50( . ) .= 43.50 / 326.25 13.05

Plate ( )( . )10 12 5 125= - 6.25 - 31.25

S 312.50 29.50

Then Y A yA YS S= =: ( ) .312 29 50mm mm2 3

or Y = 23 6. mm

Now I I Ix x x= +2( ) ( )angle plate

where ( ) . ( . )[( . . ) ]I I Adx xangle4 2mm mm mm= + = + -

=

2 2235 75 93 75 43 50 23 6

295..

( ) ( )( . ) ( )[( .

15

1

12250 12 5 125 6 252 3

mm

mm mm mm

4

plate2I I Adx x= + = + ++

=

23 6

180 8

2. ) ]

.

mm

mm4

Then I x = + =[ ( . ) . ] .2 295 15 180 8 771 1125mm mm4 4

or I x = 771 1. mm 4

81



We have I Iy x= = =4 4 771 1125 3084 45( . ) .mm mm4 4

or I y = 3084 5. mm4

Now I I Iy y y= +2( ) ( )angle plate

where ( ) . ( . )[( . ) ]

( )

I I Ad b

I

y y

y

angle4 2

plat

mm mm mm= + = + -2 263 75 93 75 18 65

ee4mm mm mm= =1

1212 5 250 1041 66753( . )( ) .

Then 3084 45 2 63 75 93 75 18 65 1041 66752 4. [ . . ( . ) ] .mm mm mm4 4= + - +b

or b = 98 5. mm

82


PROBLEM 9.57

The panel shown forms the end of a trough that is filled with water to the line AA¢. Referring to Section 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).

SOLUTION

From Section 9.2: R y dA M y dAAA= =¢Ú Úg g, 2

Let yP = distance of center of pressure from AA¢. We must have

Ry M yM

R

y dA

y dA

I

y AP AA PAA AA= = = =¢

¢ ¢ÚÚ

:g

g

2

(1)

For semicircular panel:

I r yr

A rAA¢ = = =pp

p8

4

3 24 2

yI

yA

r

rr

PAA= =

ÊËÁ

ˆ¯

¢

p

pp

843 2

4

2 y rP = 3

16

p

83


PROBLEM 9.58


SOLUTION

See solution of Problem 9.57 for derivation of Eq. (1):

yI

yAPAA= ¢ (1)

For quarter ellipse panel:

I ab yb

A abAA¢ = = =pp

p16

4

3 43

yI

yA

ab

bab

PAA= =

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

¢

p

pp

1643 4

3

y bP = 3

16

p

84


PROBLEM 9.59


SOLUTION

Using the equation developed on page 491 of the text:

yI

yAPAA= ¢

Now YA yA

hb h h b h

bh

=

= ¥ + ¥ ¥ÊËÁ

ˆ¯

=

S

22

4

3

1

22

7

32

( )

and I I IAA AA AA¢ ¢ ¢= +( ) ( )1 2

where

( ) ( )( )

( ) ( )( )

I b h bh

I I Ad b h b h

AA

AA x

¢

¢

= =

= + = + ¥ ¥

13 3

22 3

1

32

2

3

1

362

1

22ÊÊ

ËÁˆ¯

ÊËÁ

ˆ¯

=

4

3

11

6

2

3

h

bh

Then I bh bh bhAA¢ = + =2

3

11

6

5

23 3 3

Finally, ybh

bhP =

5273

3

2

or y hP = 15

14

85


PROBLEM 9.60*


SOLUTION

Using the equation developed on page 491 of the text:

yI

yAPAA= ¢

For a parabola: y h A ah= =2

5

4

3

Now dI h y dxAA¢ = -1

33( )

By observation yh

ax=

22

So that dI hh

ax dx

h

aa x dx

h

aa a x

AA¢ = -ÊËÁ

ˆ¯

= -

= - +

1

3

1

3

1

33 3

22

3 3

62 2 3

66 4 2

( )

( aa x x dx2 4 6- )

Then Ih

aa a x a x x dx

h

aa x a x a x

AA

a

¢ = - + -

= - +

Ú21

33 3

2

3

3

5

0

3

66 4 2 2 4 6

3

66 4 3 2 5

( )

--ÈÎÍ

˘˚

=

1

7

32

105

7

0

3

x

ah

a

Finally, yah

h ahp =

¥

32105

25

43

3

or y hp = 4

7

86


PROBLEM 9.61

The cover for a 0.5 m-diameter access hole in a water storage tank is attached to the tank with four equally spaced bolts as shown. Determine the additional force on each bolt due to the water pressure when the center of the cover is located 1.4 m below the water surface.

SOLUTION

From Section 9.2:

R yA yI

yAPAA= = ¢g

where R is the resultant of the hydrostatic forces acting on the cover

and yP is the depth to the point of application of R.

Recalling that g r= g, we have

R = ¥

=( . )( . )[ ( . ) ]

.

10 9 81 1 4 0 25

2696 67

3 2kg/m m/s m m

N

3 2 p

Also I I AyAA x¢ = + = +

=

2 4 2 2

40 25 0 25 1 4

0 387913

p p( . ) [ ( . ) ]( . )

.

m m m

m4

Then yP =0

=.

( . )[ ( . ) ].

387913

1 4 0 251 41116

2

m

m mm

4

pNow note that symmetry implies

F F F FA B C D= =

Next consider the free-body of the cover.

We have + SM FCD A= ∞- ∞ - -¥

0 2 0 32 45 2

0 32 45 1 41116 1 4

: [ ( . ) sin ]( )

[ . sin ( . . )]

(

m

m

22696 67 0. )N =or FA = 640 92. NThen + SF Fz C= + - =0 2 640 92 2 2696 67 0: ( . ) .N N

or FC = 707 42. N

F FA B= = 641 N

and F FC D= = 707 N

87


PROBLEM 9.62

A vertical trapezoidal gate that is used as an automatic valve is held shut by two springs attached to hinges located along edge AB. Knowing that each spring exerts a couple of magnitude 1470 N · m, determine the depth d of water for which the gate will open.

SOLUTION

From Section 9.2:

R yA yI

yAPSS= = ¢g

where R is the resultant of the hydrostatic forces acting on the gate

and yP is the depth to the point of application of R. Now

yA yA h h= = + ¥ ¥Ê

ËÁˆ¯

+ + ¥ ¥S [( . ) ] . . [( . ) ] .0 171

21 2 0 51 0 34

1

20 84 0m m m m ..

( . . )

51

0 5202 0 124848

m

m3

ÊËÁ

ˆ¯

= +h

Recalling that g r= g, we have

R h

h

= ¥ += +

( . )( . . )

. ( . )

10 9 81 0 5202 0 124848

5103 162 0 24

3 kg/m m/s m3 2 3

NN

Also I I ISS SS SS¢ ¢ ¢= +( ) ( )1 2

where ( )

( . )( . ) . [( .

I I Ad

h

SS x¢ = +

= + ¥ ¥ÊËÁ

ˆ¯

+

12

31

361 2 0 51

1

21 2 0m m m 0.51 m 117

0 0044217 0 306 0 17

0 306 0 10404 0 0

2

2 4

2

) ]

[ . . ( . ) ]

( . . .

m

m= + +

= + +

h

h h 1132651) m4

( )

( . )( . ) . [(

I I Ad

h

SS X¢ = +

= + ¥ ¥ÊËÁ

ˆ¯

22

3

2

1

360 84 0 51

1

20 84m m m 0.51 m ++

= + +

= +

0 34

0 0030952 0 2142 0 34

0 2142 0 14565

2

2

2

. ) ]

[ . . ( . ) ]

( . .

m

m4h

h 66 0 0278567h + . ) m4

88



Then I I I

h h

SS SS SS¢ ¢ ¢= +

= + +

( ) ( )

( . . . )

1 2

20 5202 0 249696 0 0411218 m4

and yh h

h

h

P =+ +

+

=

( . . . )

( . . )

0 5202 0 244696 0 0411218

0 5202 0 124848

2

2

m

m

4

3

++ ++

0 48 0 07905

0 24

. .

.

h

hm

For the gate to open, require that

SM M y h RAB P: ( )open = -

Substituting 29400 48 0 07905

0 245103 162 0 24

2

N m m N◊ = + ++

-Ê

ËÁˆ

¯¥ +h h

hh h

. .

.. ( . )

or 5103 162 0 24 0 07905 2940. ( . . )h + =

or h = 2 0711. m

Then d = +( . . )2 0711 0 79 m

or d = 2 86. m

89


PROBLEM 9.63*

Determine the x coordinate of the centroid of the volume shown. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)

SOLUTION

First note that yh

ax=

Now x dV x dVEL= ÚÚ

where x x dV y dAh

ax dAEL = = = Ê

ËÁˆ¯

Then xx x dA

x dA

x dA

x dA

I

xA

ha

ha

z A

A

=( )

= =ÚÚ

ÚÚ

2( )

( )

where ( )I z A and ( )xA A pertain to area.

OABC: ( )I z A is the moment of inertia of the area with respect to the z axis, xA is the x coordinate of the centroid of the area, and A is the area of OABC. Then

( ) ( ) ( )

( )( ) ( )( )

I I I

b a b a

a b

z A z A z A= +

= +

=

1 2

1

3

1

125

12

3 3

3

and ( )

( )

xA xA

aa b

aa b

A =

= ÊËÁ

ˆ¯

¥ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯

¥ ¥ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

S

2 3

1

2

2

332a b

90



Finally, xa b

a b=

512

3

23

2

or x a= 5

8

Analogy with hydrostatic pressure on a submerged plate:

Recalling that P y= g , it follows that the following analogies can be established.

Height y P~

dV y dA pdA dF

xdV x y dA y dF dM

= == =

~

( ) ~

Recalling that yM

R

dM

dFP

x= =Ê

ËÁÁ

ˆ

¯˜˜

ÚÚ

It can then be concluded that x yP~

91


PROBLEM 9.64*

Determine the x coordinate of the centroid of the volume shown; this volume was obtained by intersecting an elliptic cylinder with an oblique plane. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)

SOLUTION

Following the “Hint,” it can be shown that (see solution to Problem 9.63)

xI

xAz A

A

=( )

( )x

where ( )I z A and ( )xA A are the moment of inertia and the first moment of the area, respectively, of the elliptical area of the base of the volume with respect to the z axis. Then

( )

( [ (

I I Adz A z= +

= +

=

2

439 64

12

p p mm)(64 mm) mm)(39 mm)](64 mm)3 2

..779520 106p ¥ mm4

( ) (xA A =

= ¥

64 mm)[ (64 mm)(39 mm)]

0.159744 10 mm6 3

p

p

Finally x =¥¥

12 779520 10

0 159744 10

6

6

.

.

pp

mm

mm

4

4

or x = 80 0. mm

92


PROBLEM 9.65*

Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a force P at the centroid C of the area and two couples. The force P is perpendicular to the area and is of magnitude P = g gAy sin ,q where g is the density of the liquid, and the couples are M ¢ ¢=x xgI( sin )g q i and M j¢ ¢ ¢=y x ygI( sin ) ,g q where I x y dAx y¢ ¢ = Ú ¢ ¢ (see Section 9.8). Note that the couples are independent of the depth at which the area is submerged.

SOLUTION

Let g be the specific weight = rg

The pressure p at an arbitrary depth ( sin )y q is

p y= g ( sin )q

so that the hydrostatic force dF exerted on an infinitesimal area dA is

dF y dA= ( sin )g q

Equivalence of the force P and the system of infinitesimal forces dF requires

SF P dF y dA y dA: = = =Ú Úg gsin sinq q

or P Ay= g sinq

Equivalence of the force and couple ( , )P M M ¢ ¢+x y and the system of infinitesimal hydrostatic forces requires

SM yP M y dFx x: - - = -¢ Ú ( )

Now - = - = -Ú ÚÚ y dF y y dA y dA( sin ) sing gq q 2

= -( sin )g q I x

Then - - = -¢yP M Ix x( sin )g q

or M I y Ayx x¢ = -( sin ) ( sin )g gq q

= -g sin ( )q I Ayx2

or M Ix x¢ ¢= g sinq

93



SM xP M x dFy y: + =¢ ÚNow x dF x y dA xy dA= = ÚÚÚ ( sin ) sing gq q

= ( sin )g q I xy ( )Equation 9.12

Then xP M Iy xy+ =¢ ( sin )g q

or M I x Ayy xy¢ = -( sin ) ( sin )g gq q

= -g sin ( )q I Ax yxy

or M Iy x y¢ ¢ ¢= g sinq

94


PROBLEM 9.66*

Show that the resultant of the hydrostatic forces acting on a submerged plane area A is a force P perpendicular to the area and of magnitude P = =g gAy pAsin ,q where g is the density of the liquid and p is the pressure at the centroid C of the area. Show that P is applied at a point CP, called the center of pressure, whose coordinates are x I AyP xy= / and y I AyP x= / , where I xy = Ú xy dA (see Section. 9.8). Show also that the difference of ordinates y yP - is equal to k yx¢

2 / and thus depends upon the depth at which the area is submerged.

SOLUTION


The pressure P at an arbitrary depth ( sin )y q is

P y= g ( sin )q

so that the hydrostatic force dP exerted on an infinitesimal area dA is

dP y dA= ( sin )g q

The magnitude P of the resultant force acting on the plane area is then

P dP y dA y dA

yA

= = =

=ÚÚÚ g g

g

sin sin

sin ( )

q q

qNow p y= g sinq P pA=

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

SM y P y dPx P: - = -ÚNow y dP y y dA y dA= =Ú ÚÚ ( sin ) sing gq q 2

= ( sin )g q I x

Then y P IP x= ( sin )g q

or yI

yAPx=

( sin )

sin ( )

gg

qq

or yI

AyPx=

95



SM x P x dPy P: = ÚNow x dP x y dA xy dA= = ÚÚÚ ( sin ) sing gq q

= ( sin )g q I xy (Equation 9.12)

Then x P IP xy= ( sin )g q

or xI

yAPxy=

( sin )

sin ( )

gg

qq

or xI

AyPxy=

Now I I Ayx x= +¢2

From above I Ay yx P= ( )

By definition I k Ax x¢ ¢=2

Substituting ( )Ay y k A AyP x= +¢2 2

Rearranging yields y yk

yPx- = ¢2

Although kx¢ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on the depth.

( ) (y y fP - = depth)

96


PROBLEM 9.67

Determine by direct integration the product of inertia of the given area with respect to the x and y axes.

SOLUTION

First note y ax

a= -1

4

2

2

= -1

24 2 2a x

We have dI dI x y dAxy x y EL EL= +¢ ¢

where dI x xx y EL¢ ¢ = =0 (symmetry)

y y a xEL = = -1

2

1

44 2 2

dA ydx a x dx= = -1

24 2 2

Then I dI x a x a x dxxy xy

a= = -Ê

ËÁˆ¯

-ÊËÁ

ˆ¯ÚÚ

1

44

1

242 2 2 2

0

2

= - = -ÈÎÍ

˘˚Ú

1

84

1

82

1

42 3 2 2 4

0

2

0

2( )a x x dx a x x

aa

= -ÈÎÍ

˘˚

a42 4

82 2

1

42( ) ( )

or I axy = 1

24

97


PROBLEM 9.68


SOLUTION

At x a y b a kb= = =, : 2

or ka

b=

2

Then xa

by=

22

We have dI dI x y dAxy x y EL EL= +¢ ¢

where dI x xa

byx y EL¢ ¢ = = =0

1

2 2 22(symmetry)

y y dA x dya

by dyEL = = =

22

Then I dIa

by y

a

by dyxy xy

b= = Ê

ËÁˆ¯

ÊËÁ

ˆ¯ÚÚ 2 2

22

2

0( )

= = ÈÎÍ

˘˚Ú

a

by dy

a

by

bb2

45

2

46

002 2

1

6

or I a bxy = 1

122 2

98


PROBLEM 9.69


SOLUTION

First note that

yh

bx= -

Now dI dI x y dAxy x y EL EL= +¢ ¢

where dIx y¢ ¢ = 0 (symmetry)

x x y yh

bxEL EL= = = -1

2

1

2

dA y dxh

bx dx= = -

Then I dI xh

bx

h

bx dxxy xy b

= = -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯-ÚÚ

1

2

0

= = ÈÎÍ

˘˚--Ú

1

2

1

2

1

4

2

23

2

24

00h

bx dx

h

bx

bb

or I b hxy = - 1

82 2

99


PROBLEM 9.70


SOLUTION

y h h hx

a= + -1 2 1( )

dI dI x y dAxy x y EL EL= +¢ ¢

dIx y¢ ¢ = 0 (by symmetry)

x x y y dA y dxEL EL= = =1

2

I dI x y y dx xy dx

x h h hx

a

xy xy

a a= - Ê

ËÁˆ¯

=

= + -ÈÎÍ

˘˚

Ú ÚÚ1

2

1

2

1

2

0

2

0

1 2 1( ) ˙

= + - + -È

ÎÍ

˘

˚˙

Ú

Ú

0

2

12

1 2 1

2

2 12

3

20

1

22

a

a

dx

h x h h hx

ah h

x

adx( ) ( )

= + - + -È

ÎÍ

˘

˚˙

= +

1

2 22

3 4

1

2 2

2

3

12

2

1 2 1

3

2 12

4

2

12

2

1

ha

h h ha

ah h

a

a

ha

h

( ) ( )

hh a h a h a h h a h a

ah

22

12 2

22 2

2 1 12 2

2

12

2

3

1

4

1

2

1

4

2

1

2

2

3

1

4

- + - +È

ÎÍ

˘

˚˙

= - +ÊÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= ÊËÁ

ˆ¯

+

h h h

ah

1 2 22

2

12

2

3

1

2

1

4

2

1

12hh h h1 2 2

21

6

1

4ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

ÈÎÍ

˘˚

Ia

h h h hxy = + +( )2

12

1 2 22

242 3

100



The following table is provided for the convenience of the instructor, as many problems in this and the next lesson are related.

Type of Problem

Compute

and I Ix yFigure 9.12 Figure 9.13B Figure 9.13A

Compute

I xy9.67 9.72 9.73 9.74 9.75 9.78

I I Ix y x y¢ ¢ ¢ ¢, ,

by equations9.79 9.80 9.81 9.83 9.82 9.84

Principal axes by equations 9.85 9.86 9.87 9.89 9.88 9.90

I I Ix y x y¢ ¢ ¢ ¢, ,

by

Mohr’s circle

9.91 9.92 9.93 9.95 9.94 9.96

Principal axes by Mohr’s circle 9.97 9.98 9.100 9.101 9.103 9.106

101


PROBLEM 9.71

Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.

SOLUTION

Dimensions in mm

We have I I I Ixy xy xy xy= + +( ) ( ) ( )1 2 3

Now symmetry implies

( )I xy 2 0=

and for the other rectangles

I I x yAxy x y= +¢ ¢

where I x y¢ ¢ = 0 ( )symmetry

Thus I x yA x yAxy = +( ) ( )1 3

A, mm2 x , mm y, mm x yA, mm4

1 10 80 800¥ = -55 20 -880 000,

3 10 80 800¥ = 55 -20 -880 000,

S -1 760 000, ,

I xy = - ¥1 760 106. mm4

102


PROBLEM 9.72


SOLUTION

Dimensions in mm

We have I I I Ixy xy xy xy= - -( ) ( ) ( )1 2 3

Now symmetry implies ( )I xy 1 0=

and for each triangle I I x yAxy x y= +¢ ¢

where, using the results of Sample Problem 9.6, I b hx y¢ ¢ = - 172

2 2 for both triangles. Note that the sign of I x y¢ ¢ is unchanged because the angles of rotation are 0∞ and 180∞ for triangles 2 and 3, respectively. Now


21

2120 60 3600( )( ) = -80 -40 11 520 106. ¥

31

2120 60 3600( )( ) = 80 40 11 520 106. ¥

S 23 040 106. ¥

Then I xy = - -ÈÎÍ

˘˚

+ ¥ÏÌÓ

¸˝˛

21

72120 60 23 040 106( ( . mm) mm) mm2 2 4

or I xy = - ¥21 6 106. mm4

103


PROBLEM 9.73


SOLUTION

We have I I Ixy xy xy= +( ) ( )1 2

For each semicircle I I x yAxy x y= +¢ ¢

and I x y¢ ¢ = 0 ( )symmetry

Thus I x yAxy = S

A, mm2 x , mm y, mm xyA, mm4

1p p2

150 112502( ) = -75 - 200

p168.75 ¥ 106

2p p2

150 112502( ) = 75 200

p168.75 ¥ 106

S 337.5 ¥ 106

I xy = ¥338 106 mm4

104


PROBLEM 9.74


SOLUTION


For each rectangle


and I x y¢ ¢ = 0 ( )symmetry

Thus I x yAxy = S


1 7 6 6 4 486 4. . .¥ = -13 7. 9.2 -58620 928.

2 6 4 44 6 285 44. . .¥ = 21.7 -16.3 -100962.982

S -159583.91

I xy = ¥159 6 103 4. mm

105


PROBLEM 9.75


SOLUTION


Now symmetry implies ( )I xy 1 0=

and for the other rectangles


where I x y¢ ¢ = 0 (symmetry)

Thus I x yA x yAxy = +( ) ( )2 3

A, mm2 X , mm y, mm x yA, mm4

2 8 32 256¥ = – 46 -20 235,520

3 8 32 256¥ = 46 20 235,520

S 471,040

I xy = ¥471 103 mm4

106


PROBLEM 9.76


SOLUTION


Now, symmetry implies

( )I xy 1 0=

and for each triangle I I x yAxy x y= +¢ ¢

where, using the results of Sample Problem 9.6, I b hx y¢ ¢ = - 172

2 2 for both triangles. Note that the sign of I x y¢ ¢

is unchanged because the angles of rotation are 0∞ and 180∞ for triangles 2 and 3, respectively.

Now


21

2228 380 43 32 103( )( ) .= ¥ -228

533

31.75481 ¥ 109

31

2228 381 43 434 103( )( ) ( . )= ¥ 228 -178 1.76273 ¥ 109

S 3.51754 ¥ 109

Then I xy = - ¥ ¥ÏÌÓ

¸˝˛

- ÏÌÓ

¸˝˛

-1

72228 380

1

72228 381 3 517542 2 2 2( ) ( ) ( ) ( ) . ¥¥

= - ¥

10

3 7266 10

9

9.

or I xy = - ¥3 73 9. 10 mm4

107


SOLUTION


For each rectangle I I x yAxy x y= +¢ ¢

and I x y¢ ¢ = 0 (symmetry)

Thus I x yAxy = S


1 91 13 1183¥ = 22.5 -50.5 -1.3442 ¥ 106

2 13 96 1248¥ = -16.5 4 -82.368 ¥ 103

3 33 25 825¥ = -6.5 64.5 -345.88125 ¥ 103

S -1.7724 ¥ 106

I xy = - ¥1 772 106 4. mm

PROBLEM 9.77


108


SOLUTION




Thus I x yAxy = S


1 76 12 7 965 2¥ =. . –19.1 –37.85 697,777

2 12 7 127 12 7 1451 61. ( . ) .¥ - = 12.55 25.65 467,284

S 1,165,061

I xy = ¥1 165 106. mm4

PROBLEM 9.78


109


PROBLEM 9.79

Determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45∞ counterclockwise, (b) through 30∞ clockwise.

SOLUTION

From Figure 9.12: I a a

a

I a a

a

x

y

=

=

=

=

p

p

p

p

162

8

162

2

3

4

3

4

( )( )

( ) ( )

From Problem 9.67: I axy = 1

24

First note 1

2

1

2 8 2

5

16

1

2

1

2 8 2

4 4 4

4 4

( )

( )

I I a a a

I I a a

x y

x y

+ = +ÊËÁ

ˆ¯

=

- = -ÊËÁ

p p p

p p ˆ¯

= - 3

164pa

Now use Equations (9.18), (9.19), and (9.20).

Equation (9.18): I I I I I Ix x y x y xy¢ = + + - -1

2

1

22 2( ) ( )cos sinq q

= - -5

16

3

162

1

224 4 4p p q qa a acos sin

Equation (9.19): I I I I I Iy x y x y xy¢ = + - - +1

2

1

22 2( ) ( )cos sinq q

= + +5

16

3

162

1

224 4 4p p q qa a acos sin

Equation (9.20): I I I Ix y x y xy¢ ¢ = - +1

22 2( )sin cosq q

= - +3

162

1

224 4p q qa asin cos

110



(a) q = + ∞45 : I a a ax¢ = - ∞ - ∞5

16

3

1690

1

2904 4 4p p cos sin

or I ax¢ = 0 482 4.

I a ay¢ = + ∞ +5

16

3

1690

1

24 4p p cos

or I ay¢ = 1 482 4.

I a ax y¢ ¢ = - ∞ + ∞3

1690

1

2904 4p sin cos

or I ax y¢ ¢ = -0 589 4.

(b) q = - ∞30 : I a a ax¢ = - - ∞ - - ∞5

16

3

1660

1

2604 4 4p p cos( ) sin( )

or I ax¢ = 1 120 4.

I a a ay¢ = + - ∞ + - ∞5

16

3

1660

1

2604 4 4p p cos( ) sin( )

or I ay¢ = 0 843 4.

I a ax y¢ ¢ = - - ∞ + - ∞3

1660

1

2604 4p sin( ) cos( )

or I ax y¢ ¢ = 0 760 4.

111


PROBLEM 9.80

Determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30∞ counterclockwise.

SOLUTION

From Problem 9.72: I xy = - ¥21 6 106 4. mm

Now I I I Ix x x x= - -( ) ( ) ( )1 2 3

where

( ) ( )( )

.

I x 13

6 4

1

12240 160

81 920 10

=

= ¥

mm mm

mm

( ) ( ) ( )( )

( )( ) ( )

I Ix x2 33

2

1

36120 60

1

2120 60 40

= =

+ ÈÎÍ

˘˚

mm mm

mm mm mm

== ¥6 480 106 4. mm

Then I x = - ¥ = ¥[ . ( . )] .81 920 2 6 480 10 68 96 106 4 6 4mm mm

Also I I I Iy y y y= - -( ) ( ) ( )1 2 3

where ( ) ( )( ) .

( ) ( ) (

I

I I

y

y y

13 6 4

2 3

1

12160 240 184 320 10

1

3660

= = ¥

= =

mm mm mm

mmm mm

mm mm mm

mm

)( )

( )( ) ( )

.

120

1

2120 60 80

25 920 10

3

2

6 4

+ ÈÎÍ

˘˚

= ¥

Then I y = - ¥ = ¥[ . ( . )] .184 320 2 25 920 10 132 48 106 4 6 4mm mm

112



Now 1

2

1

268 96 132 48 10 100 72 10

1

2

1

26

6 6 4( ) ( . . ) .

( ) (

I I

I I

x y

x y

+ = + ¥ = ¥

- =

mm

88 96 132 48 10 31 76 106 6. . ) .- ¥ = - ¥ mm4

Using Eqs. (9.18), (9.19), and (9.20):

Eq. (9.18): I I I I I Ix x y x y xy= + + - -

= + -

1

2

1

22 2

100 72 31 76 60

( ) ( )cos sin

[ . ( . )cos

q q

∞∞ - - ∞ ¥( . )sin ]21 6 60 106 4mm

or I x¢ = ¥103 5 106 4. mm

Eq. (9.19): I I I I I Iy x y x y xy¢ = + - - +

= - -

1

2

1

22 2

100 72 31 76 6

( ) ( )cos sin

[ . ( . )cos

q q

00 21 6 60 106 4∞ + - ∞ ¥( . )sin ] mm

or I y¢ = ¥97 9 106 4. mm

Eq. (9.20): I I I Ix y x y xy¢ ¢ = - +

= - ∞ + - ∞

1

22 2

31 76 60 21 6 60

( )sin sin

[ . sin ( . )cos ]

q q

¥¥106 4mm

or I x y¢ ¢ = - ¥38 3 106 4. mm

113


SOLUTION

From Problem 9.73:

I xy = ¥337 5 106 4. mm

Now I I Ix x x= +( ) ( )1 2

where ( ) ( ) ( )

( . )

I Ix x1 24

6

8150

63 28125 10

= =

= ¥

p

p

mm

mm4

Then I Ix xj= = ¥2 126 5625 106( . )p mm4

Also I I Iy y y= +( ) ( )1 2

where ( ) ( ) ( ) ( ) ( ) ( .I Iy y1 24 2 2

8150

2150 75 126 5625 1= = + È

ÎÍ˘˚

= ¥p pmm mm mm 006 4)p mm

Then I y = ¥ = ¥2 126 5625 10 253 125 106 6( . ) ( . )p pmm mm4 4

Now 1

2 2126 5625 10 253 125 10 189 84375 10

1

2

6 6 6( ) ( . . ) .I Ix y+ = ¥ + ¥ = ¥p p mm4

(( ) ( . . ) .I Ix y- = ¥ - ¥ - ¥p p2

126 5625 10 253 125 10 63 28125 106 6 6 mm4

Using Eqs. (9.18), (9.19), and (9.20):

Eq. (9.18): I I I I I Ix x y x y xy¢ = + + - -

= + -

1

2

1

22 2

189 843751 63 28

( ) ( )cos sin

[{ . ( .

q q

1125 120 337 5 120 10

403 53 10

6 4

6

)cos } . (sin )]

.

∞ - ∞ ¥

= ¥

p mm

mm4

or I x¢ = ¥404 106 mm4

PROBLEM 9.81

Determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60∞ counterclockwise.

114




= - -

1

2

1

22 2

189 84375 63 281

( ) ( )cos sin

[{ . ( .

q q

225 120 337 5 120 10 789 29 106 4 6 4)cos } . sin ] .∞ + ∞ ¥ = ¥p mm mm

or I y¢ = ¥789 106 4mm

Eq. (9.20): I I I Ix y x y xy¢ ¢ = - +

= - ∞+

1

22 2

63 28125 120 337 5

( )sin cos

[( . )sin . co

q q

p ss ]

.

120 10

340 919 10

6

6

∞ ¥

= - ¥

mm

mm

4

4

or I x y¢ ¢ = - ¥341 106 4mm

115


PROBLEM 9.82

Determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise.

SOLUTION

From Problem 9.75:

I xy = 471 4,040 mm

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

where ( ) ( )( )

.

I x 13

4

1

12100 8

4266 67

=

=

mm mm

mm

( ) ( ) ( )( ) [( )( )]( )

, .

I Ix x2 33 21

128 32 8 32 20

124 245 3

= = +

=

mm mm mm mm mm

33

Then I x = + =[ . ( , . )] ,4266 67 2 124 245 33 252 7574 4mm mm

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

where ( ) ( )( ) , .

( ) ( ) ( )(

I

I I

y

y y

13 4

2 3

1

128 100 666 666 7

1

1232 8

= =

= =

mm mm mm

mm mmm mm mm mm

mm

) [( )( )]( )

, .

3 2

4

8 32 46

543 061 3

+

=

Then I y = + =[ , . ( , . )] , ,666 666 7 2 543 061 3 1 752 7894 4mm mm

Now 1

2

1

2252 757 1 752 789 1 002 773

1

2

1

4 4( ) ( , , , ) , ,

( )

I I

I I

x y

x y

+ = + =

- =

mm mm

22252 757 1 752 789 750 0164( , , , ) ,- = -mm mm4

Using Eqs. (9.18), (9.19), and (9.20):


= + -

1

2

1

22 2

1 002 773 750 016

( ) ( )cos sin

[ , , ( ,

q q

))cos( ) , sin( )]- ∞ - - ∞90 471 040 90

or I x¢ = ¥1 474 106 4. mm

116




= - -

1

2

1

22 2

1 002 773 750 016

( ) ( )cos sin

[ , , ( ,

q q

))cos( ) , sin( )]- ∞ + - ∞90 471 040 90

or I y¢ = ¥0 532 106 4. mm

Eq. (9.20): I I I Ix y x y xy¢ ¢ = - +

= - - ∞ +

1

22 2

750 016 90 471 040

( )sin cos

[( , )sin( ) ,

q q

ccos( )]- ∞90

or I x y¢ ¢ = ¥0 750 106 4. mm

117


PROBLEM 9.83

Determine the moments of inertia and the product of inertia of the L mm76 51 6 4¥ ¥ . angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.

SOLUTION

From Figure 9.13:

I

I

x

y

= ¥

= ¥

0 162

0 454 10

4

6

.

.

10 mm

mm

6

4

From Problem 9.74:

I xy = - ¥0 159584 106 4. mm

Now 1

2

1

20 162 0 454 10 0 308 10

1

2

1

2

6 6 4( ) ( . . ) .

( ) (

I I

I I

x y

x y

+ = + ¥ = ¥

- =

mm mm4

00 162 0 454 10 0 146 106 4 6 4. . ) .- ¥ = - ¥mm mm

Using Eqs. (9.18), (9.19), and (9.20):


= + - -

1

2

1

22 2

0 308 0 146

( ) ( )cos sin

[ . ( . )cos(

q q

660 0 159584 60 106∞ - - - ∞ ¥) ( . )sin( ) mm4

or I x¢ = ¥96 8 103 4. mm


= - - -

1

2

1

22 2

0 308 0 146

( ) ( )cos sin

[ . ( . )cos(

q q

660 0 159584 60 106∞ + - - ∞ ¥) ( . )sin( )] mm4

or I y¢ = ¥519 103 4mm

Eq. (9.20): I I I Ix y x y xy¢ ¢ = - +

= - - ∞ + -

1

22 2

0 146 60 0 15958

( )sin cos

[( . )sin( ) ( .

q q

44 60 106 4)cos( )]- ∞ ¥ mm

or I x y¢ ¢ = ¥46 6 103 4. mm

118


PROBLEM 9.84

Determine the moments of inertia and the product of inertia of the L127 ¥ 76 ¥ 12.7 mm angle cross section of Problem 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 45∞ counterclockwise.

SOLUTION

From Figure 9.13:

I

I

x

y

= ¥

= ¥

3 93 10

1 06 10

6 4

6 4

.

.

mm

mm

From Problem 9.78:

I xy = ¥1 165061 106 4. mm

Now 1

2

1

23 93 1 06 10

2 495 10

1

2

1

23

6 4

6 4

( ) ( . . )

.

( ) ( .

I I

I I

x y

x y

+ = + ¥

= ¥

- =

mm

mm

993 1 06 10 1 435 106 6 4- ¥ = ¥. ) .mm mm4

Using Eqs. (9.18), (9.19), and (9.20):


= + ∞ -

1

2

1

22 2

2 495 1 435 90 1

( ) ( )cos sin

[ . . cos

q q

.. sin ]165061 90 106 4∞ ¥ mm

or I x¢ = ¥1 330 106 4. mm


= - ∞ +

1

2

1

22 2

2 495 1 435 90 1

( ) ( )cos sin

[ . . cos

q q

.. sin ]165061 90 106 4∞ ¥ mm

or I y¢ = ¥3 66 106 4. mm

Eq. (9.20): I I I Ix y x y xy¢ ¢ = - +

= ∞ + ∞

1

22 2

1 435 90 1 165061 90

( )sin cos

[( . sin . cos

q q

]] ¥106 4mm

or I x y¢ ¢ = ¥1 435 106 4. mm

119


SOLUTION

From Problem 9.79: I a I ax y= =p p8 2

4 4

Problem 9.67: I axy = 1

24

Now Eq. (9.25): tan22 2

8

30 84883

12

4

84

24

q

p

p pmxy

x y

I

I I

a

a a= -

-= -

( )-

= = .

Then 2 40 326 220 326qm = ∞ ∞. .and

or qm = ∞ ∞20 2 110 2. .and

Also Eq. (9.27): II I I I

I

a a

a

x y x yxymax, min =

+±

-ÊËÁ

ˆ¯

+

= +ÊËÁ

ˆ¯

±

2 2

1

2 8 2

1

2 8

22

4 4p p

p 44 42

42

4

2

1

2

0 981 748 0 772 644

-ÊËÁ

ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯

= ±

pa a

a( . , . , )

or I amax .= 1 754 4

and I amin .= 0 209 4

By inspection, the a axis corresponds to Imin and the b axis corresponds to Imax .

PROBLEM 9.85

For the quarter ellipse of Problem 9.67, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.

120


PROBLEM 9.86

For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.

Area of Problem 9.72.

SOLUTION

From Problem 9.80: I

I

x

y

= ¥

= ¥

68 96 10

132 48 10

6 4

6 4

.

.

mm

mm

Problem 9.72: I xy = - ¥21 6 106 4. mm

Now Eq. (9.25): tan22 2 21 6 10

68 96 132 48 10

0 68010

6

6qm

xy

x y

I

I I= -

-= - - ¥

- ¥= -

( . )

( . . )

.

Then 2 34 220 145 780qm = - ∞ ∞. .and

or qm = - ∞ ∞17 11 72 9. .and


Ix y x y

xymax, min =+

±-Ê

ËÁˆ

¯+

2 2

2

2

Then Imax, min. . . .

( . )= + ± -ÊËÁ

ˆ¯

+ -È

ÎÍÍ

˘

˚

68 96 132 48

2

68 96 132 48

221 6

22 ˙

˙¥

= ± ¥

10

100 72 38 409 10

6

6 4( . . ) mm

or Imax .= ¥139 1 106 4mm

and Imin .= ¥62 3 106 4mm


121


PROBLEM 9.87



SOLUTION

From Problem 9.81: I Ix y= ¥ = ¥( . ) ( . )126 5625 10 253 125 106 6p pmm mm4 4

Problem 9.73: I xy = ¥337 5 106.

Now Eq. (9.25): tan22 2 337 5 10

126 5625 253 125 10

1 6

6

6q

pmxy

x y

I

I I= -

-= - ¥

- ¥=

( . )

( . . )

. 99765

Then 2 59 500 239 500qm = ∞ ∞. .and

or qm = ∞ ∞29 7 119 7. .and


Ix y x y

xymax, min =+

±-Ê

ËÁˆ

¯+

2 2

2

2

Then Imax, min( . . ) ( . . )

=+ ¥

±- ¥126 5625 253 125 10

2

126 5625 253 125 10

2

6 6p pÏÏÌÔ

ÓÔ

¸˝ÔÔ

+ ¥

= ¥ ± ¥

( . )

( . . )

337 5 10

596 412 10 391 700 10

6 2

6 6 mm4

or Imax = ¥988 106 mm4

and Imin = ¥205 106 mm4


122


PROBLEM 9.88



SOLUTION

From Problem 9.82: I x = 252 757, mm4

I y = 1 752 789, , mm4

Problem 9.75: I xy = 471 040, mm4

Now Eq. (9.25): tan

( , )

, , ,

.

22

2 471 040

252 757 1 752 789

0 62804

qmxy

x y

I

I I= -

-

= --

=

Then 2 32 130qm = ∞ ∞. and 212.130

or qm = ∞ ∞16 07 106 1. .and


Ix y x y

xymax, min =-

±-Ê

ËÁˆ

¯+

2 2

2

2

Then Imax,, , , , , ,

min =+

±-Ê

ËÁˆ¯

+252 757 1 752 789

2

252 757 1 752 789

2471

2

0040

1 002 773 885 665

2

= ±( , , , ) mm4

or Imax .= ¥1 888 106 mm4

and Imin .= ¥0 1171 106 mm4


123


PROBLEM 9.89

For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.

The L mm76 51 6 4¥ ¥ . angle cross section of Problem 9.74.

SOLUTION

From Problem 9.83: I Ix y= ¥ = ¥0 162 10 0 454 106 6 4. .mm mm4

Problem 9.74: I xy = - ¥0 159584 106. mm4

Now Eq. (9.25): tan( . )

( . . )

.

22 2 0 159584 10

0 162 0 454 10

1 09

6

6qm

xy

x y

I

I I= -

-= -

- ¥- ¥

= - 3304

Then 2 47 545qm = - ∞ ∞. and 132.455

or qm = - ∞ ∞23 8. and 66.2


Ix y x y

xymax, min =+

±-Ê

ËÁˆ

¯+

2 2

2

2

Then Imax,( . . ) ( . . )

min =+ ¥

±- ¥Ï

ÌÔ

ÓÔ

¸˝ÔÔ

0 162 0 454 10

2

0 162 0 454 10

2

6 6 2

++ - ¥

= ¥ ± ¥

( . )

( . . )

0 159584 10

0 308 10 0 21629 10

6 2

6 6 4mm

or Imax .= ¥0 524 410 mm6

and Imin .= ¥0 0917 410 mm6


124


PROBLEM 9.90

For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.

The L127 76 12.7-mm ¥ ¥ angle cross section of Problem 9.78.

SOLUTION

From Problem 9.84: I x = ¥3 93 106. mm4

I y = ¥1 06 106. mm4

Problem 9.78: I xy = ¥1 165061 106. mm4

Now Eq. (9.25): tan( . )

( . . )

.

22 2 1 165061 10

3 93 1 06 10

0 81189

6

6qm

xy

x y

I

I I= -

-= - ¥

- ¥= -

Then 2 39 073qm = - ∞ ∞. and 140.927

or qm = - ∞ ∞19 54. and 70.5


Ix y x y

xymax, min =+

±-Ê

ËÁˆ

¯+

2 2

2

2

Then Imax,. . . .

. min = + ± -ÊËÁ

ˆ¯

+È

ÎÍÍ

˘

˚˙˙

¥3 93 1 06

2

3 93 1 06

21 165061 1

22 00

2 495 1 84840 10

6 4

6

mm

mm4= ± ¥( . . )

or Imax .= ¥4 34 106 mm4

and Imin .= ¥0 647 106 mm4

By inspection, the a axis corresponds to Imax and the b axis corresponds to Imin.

125


PROBLEM 9.91

Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30∞ clockwise.

SOLUTION

From Problem 9.79: I ax = p8

4

I ay = p2

4


24

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

X a a Y a ap p8

1

2 2

1

24 4 4 4, ,Ê

ËÁˆ¯

-ÊËÁ

ˆ¯

and

Now I I I a a a ax yave = + = +ÊËÁ

ˆ¯

= =1

2

1

2 8 2

5

160 981754 4 4 4( ) .

p p p

and RI I

I a a ax y

xy=-Ê

ËÁˆ¯

+ = -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯2

1

2 8 2

1

2

22 4 4

24

2p p

= 0 77264 4. a

The Mohr’s circle is then drawn as shown.

tan22

qmxy

x y

I

I I= -

-

= -( )

-

2 12

4

84

24

a

a ap p

= 0 84883.

or 2 40 326qm = ∞.

126



Then a = ∞ - ∞= ∞

90 40 326

49 674

.

.

b = ∞ - ∞ + ∞= ∞

180 40 326 60

79 674

( . )

.

(a) q = + ∞45 : I I R a ax¢ = - = - ∞ave cos . . cos .a 0 98175 0 77264 49 6744 4

or I ax¢ = 0 482 4.

I I R a ay¢ = + = + ∞ave cos . . cos .a 0 98175 0 77264 49 6744 4

or I ay¢ = 1 482 4.

I R ax y¢ ¢ = - = - ∞sin . sin .a 0 77264 49 6744

or I ax y¢ ¢ = -0 589 4.

(b) q = - ∞30 : I I R a ax¢ = + = + ∞ave cos . . cos .b 0 98175 0 77264 79 6744 4

or I ax¢ = 1 120 4.

I I R a ay¢ = - = - ∞ave cos . . cos .b 0 98175 0 77264 79 6744 4

or I ay¢ = 0 843 4.

I R ax y¢ ¢ = = ∞sin . sin .b 0 77264 79 6744

or I ax y¢ ¢ = 0 760 4.

127


PROBLEM 9.92

Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30∞ counterclockwise.

SOLUTION


I y = ¥132 48 106. mm4

Problem 9.72: I xy = - ¥21 6 106. mm4

The Mohr’s circle is defined by the diameter XY, where X ( . , . )68 96 10 21 6 106 6¥ - ¥ and Y ( . ,132 48 106¥ 21.6 106¥ ).

Now I I Ix yave4 mm= - = + ¥ = ¥1

2

1

268 96 132 48 10 100 72 106 6( ) ( . . ) .

and R I I Ix y xy= -ÈÎÍ

˘˚

+ = -ÈÎÍ

˘˚

+ -ÏÌ

1

2

1

268 96 132 48 21 6

22

22( ) ( . . ) ( . )

ÔÔ

ÓÔ

¸˝Ô

Ô¥

= ¥

10

38 409 10

6 4

6

mm

mm4.


tan

( . )

( . . )

.

22

2 21 6 10

68 96 132 48 10

0 68010

6

6

qmxy

x y

I

I I= -

-

= - - ¥- ¥

= -

or 2 34 220qm = - ∞.

Then a = ∞ - ∞ + ∞= ∞

180 34 220 60

85 780

( . )

.

128



Then I I Rx¢ = + = + ∞ ¥ave cos ( . . cos . )a 100 72 38 409 85 780 106

or I x¢ = ¥103 5 106. mm4

I I Ry¢ = - = - ∞ ¥ave cos ( . . cos . )a 100 72 38 409 85 780 106

or I y¢ = ¥97 9 106. mm4

I Rx y¢ ¢ = - = - ¥ ∞sin ( . )sin .a 38 409 10 85 7806

or I x y¢ ¢ = - ¥38 3 106. mm4

129


SOLUTION

From Problem 9.81: I x = ¥( . )126 5625 106 p mm4

I y = ¥( . )253 125 106 p mm4

Problem 9.73: I xy = ¥337 5 106 4. mm

The Mohr’s circle is defined by the diameter XY, where X Y( . , . ( . ,126 5625 10 337 5 10 253 125 106 6 6¥ ¥ ¥p p) and - ¥337 5 106. ).

Now I I Ix yave

4

mm

mm

= + = ¥

= ¥

1

2189 84375 10

596 412 10

6 4

6

( ) ( . )

.

p

and R I I Ix y xy= - +ÈÎÍ

= ¥

1

2

391 700 10

2 2

6

( )

. mm4

(from solution of Problem 9.87)


tan

( . )

( . . )

22

2 337 5 10

126 5625 253 125 10

1

6

6

qmxy

x y

I

I I= -

-

= -- ¥

- ¥ ¥=

p..69765

or 2 59 500qm = ∞.

Then a = ∞ - ∞= ∞

120 59 500

60 500

.

.

PROBLEM 9.93

Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60∞ counterclockwise.

130



Then I I Rx¢ = - = ¥ - ¥ ∞ = ¥ave cos . . cos . .a 596 412 10 391 700 10 60 500 403 53 106 6 66

or I x¢ = ¥404 106 mm4

I I Ry = + = ¥ + ¥ = ¥ave mmcos . . cos( . ) .a 596 412 10 391 7 10 60 5 789 294 106 6 6 44

or I y¢ = ¥789 106 mm4

I Rx y¢ ¢ = - = - ¥ ∞sin . sin( . )a 391 700 10 60 56

or I x y¢ ¢ = - ¥341 106 mm4

131


PROBLEM 9.94

Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45∞ clockwise.

SOLUTION

From Problem 9.82: I x = 252 757, mm4

I y = 1 752 789, , mm4

Problem 9.75: I xy = 471 040, mm4

The Mohr’s circle is defined by the diameter XY, where X (252,757; 471,040) and Y (1,752,789; -471,040).

Now I I Ix yave

4 mm

= + = +

=

1

2

1

2252 757 1 752 789

1 002 773

( ) ( , , , )

, ,


˘˚

+

= -ÈÎÍ

˘˚

+

1

2

1

2252 757 1 752 789 471 0

22

2

( )

( , , , ) , 440

885 665

2

= , mm4


tan

( , )

, , ,

.

22

2 471 040

252 757 1 752 789

0 62804

qmxy

x y

I

I I= -

-

= --

=

or 2 32 130qm = ∞.

Then a = ∞ - + ∞= ∞

180 32 130 90

57 870

( . )

.

132



Then I I Rx¢ = + = + ∞ave cos , , , cos .a 1 002 773 885 665 57 870

or I x¢ = ¥1 474 106. mm4

I I Ry¢ = - = - ∞ave cos , , , cos .a 1 002 773 885 665 57 870

or I y¢ = ¥0 532 106. mm4

I Rx y¢ ¢ = = ∞sin , sin .a 885 665 57 870

or I x y¢ ¢ = ¥0 750 106. mm4

133


PROBLEM 9.95

Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L mm76 51 6 4¥ ¥ . angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.

SOLUTION


I y = ¥0 454 106. mm4

Problem 9.74:

I xy = - ¥0 159584 106. mm4

The Mohr’s circle is defined by the diameter XY, where X (0.162 ´ 106,- 0.159584 ´ 106 ) and Y (0.454 ´ 106,0.159584 ´ 106 ).

Now I I Ix yave

4mm

= +

= ¥

1

2

0 308 106

( )

.


˘˚

+

= ¥

1

2

0 21629 10

22

6

( )

. mm4

(from Problem 9.89)


tan

( . )

( . . )

.

22

2 0 159584 10

0 162 0 454 10

1 09

6

6

qmxy

x y

I

I I= -

-

= -- ¥

- ¥= - 3304

or 2 47 545qm = - ∞.

Then a = ∞ - ∞ = ∞60 47 545 12 455. .

134



Then I I Rx¢ = - = ¥ - ¥ ∞ = ¥ave mcos . . cos . .a 0 308 10 0 21629 10 12 455 0 0968 106 6 6 mm4

or I x¢ = ¥0 0968 106. mm4

I I Ry¢ = + = ¥ + ¥ ∞ = ¥ave mcos . . cos . .a 0 308 10 0 21629 10 12 455 0 5192 106 6 6 mm4

or I y = ¥0 519 106. mm4

I Rx y¢ ¢ = = ¥ ∞sin . sin( . )a 0 21629 10 12 4556

or I x y¢ ¢ = ¥0 0466 106. mm4

135


PROBLEM 9.96

Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L127 76 12.7-mm¥ ¥ angle cross section of Problem 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 45∞ counterclockwise.

SOLUTION


I y = ¥1 06 106. mm4

Problem 9.78: I xy = ¥1 165061 106. mm4

The Mohr’s circle is defined by the diameter XY, where X ( . , ),3 93 106¥ ¥ 1.165061 106

(1.06 06Y ¥1 , - ¥1.165061 106 ).

Now I I Ix yave4mm= + = + ¥ = ¥1

2

1

23 93 1 06 10 2 495 106 6( ) ( . . ) .


˘˚

+

= -ÈÎÍ

˘˚

+ÏÌÔ

Ó

1

2

1

23 93 1 06 1 165061

22

22

( )

( . . ) .ÔÔ

¸˝Ô

Ô¥

= ¥

10

1 84840 10

6 4

6

mm

mm4.


tan

( . )

( . . )

.

22

2 1 165061 10

3 93 1 06 10

0 81189

6

6

qmxy

x y

I

I I= -

-

= - ¥- ¥

= -

or 2 39 073qm = - ∞.

136



Then a = ∞ - ∞ + ∞= ∞

180 39 073 90

50 927

( . )

.

Then I I Rx¢ = - = - ∞ ¥ave cos ( . . cos . )a 2 495 1 84840 50 927 106

or I x¢ = ¥1 330 106. mm4

I I Ry¢ = + = + ∞ ¥ave cos ( . . cos . )a 2 495 1 84840 50 927 106

or I y¢ = ¥3 66 106 4. mm

I Rx y¢ ¢ = = ¥ ∞sin ( . )sin .a 1 84840 10 50 9276

or I x y¢ ¢ = ¥1 435 106. mm4

137


PROBLEM 9.97

For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine the orientation the principal axes at the origin and the corresponding values of the moments of inertia.

SOLUTION

From Problem 9.79: I a I ax y= =p p8 2

4 4


24


X a a Y a ap p8

1

2 2

1

24 4 4 4, , and Ê

ËÁˆ¯

-ÊËÁ

ˆ¯

Now I I I a a ax yave = + = +ÊËÁ

ˆ¯

=1

2

1

2 8 20 981754 4 4( ) .

p p

and

R I I I

a a a

x y xy= -ÈÎÍ

˘˚

+

= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯

1

2

1

2 8 2

1

2

22

4 42

4

( )

p p˜

=

2

40 77264. a


tan

.

22

2

0 84883

12

4

84

24

q

p p

mxy

x y

I

I I

a

a a

= --

= -( )

-=

or 2 40 326qm = ∞.

and qm = ∞20 2.

138



The Principal axes are obtained by rotating the xy axes through

20 2. ∞ counterclockwise

about O.

Now I I R a amax, min . .= ± = ±ave 0 98175 0 772644 4

or I amax .= 1 754 4

and I amin .= 0 209 4

From the Mohr’s circle it is seen that the a axis corresponds to Imin and the b axis corresponds to Imax .

139


PROBLEM 9.98

Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia.


SOLUTION


I y = ¥132 48 106. mm4

Problem 9.72: I xy = - ¥21 6 106. mm4

The Mohr’s circle is defined by the diameter XY, where X ( . , )68 96 106¥ - ¥21.6 106

and Y ( . , )132 48 106¥ ¥21.6 106

Now I I Ix yave

4mm

= +

= + ¥

= ¥

1

21

268 96 132 48 10

100 72 10

6

6

( )

( . . )

.


˘˚

+

= -ÈÎÍ

˘˚

+ -ÏÌ

1

2

1

268 96 132 48 21 6

22

22

( )

( . . ) ( . )ÔÔ

ÓÔ

¸˝Ô

Ô¥

= ¥

10

38 409 10

6

6. mm4


tan

( . )

( . . )

.

22

2 21 6 10

68 96 132 48 10

0 68010

6

6

qmxy

x y

I

I I= -

-

= - - ¥- ¥

= -

or 2 34 220qm = - ∞.

and qm = - ∞17 11.

140



The principal axes are obtained by rotating the xy axes through

17.11° clockwise

about C.

Now I I Rmax, ( . . ) min ave= ± = + ¥100 72 38 409 106

or Imax4mm= ¥139 1 106.

and Imin .= ¥62 3 106 mm4


141


SOLUTION

From Problem 9.76:

I xy = - ¥3 7266 109. mm4

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

where ( ) ( .I x 161

12608 53 767872 10= = ¥mm)(102mm) mm3 4

( ) ( (I x 21

36228

1

2228

533

3= + È

ÎÍ˘˚ÊËÁ

ˆ¯

mm)(380 mm) mm)(380 mm) mm3˜

= ¥

= +

2

4

3

mm

mm)(381mm) mm)(381mm

1 71494 10

1

36228

1

2228

9

3

.

( ) ( (I x )) mm

mm

2

4

ÈÎÍ

˘˚( )

= ¥

178

1 72644 109.

Then I x = + + ¥

= ¥

[ . . . )

.

0 053768 1 71494 1 72644 10

3 495148 10

9

9

mm

mm

4

4

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

where ( ) ( .I y 191

12102 1 91042 10= = ¥mm)(608mm) mm3 4

( ) ( ( ( .I y 21

36380

1

2228 228 2= +

È

ÎÍ

˘

˚˙ =mm)(228mm) mm)(380 mm) mm)3 2 337706 10

1

36381

1

2228

9

3

¥

= +È

ÎÍ

˘

mm

mm)(228mm) mm)(381mm)

4

3( ) ( (I y˚˙ = ¥( .228 2 38331 109mm) mm2 4

Then I y = + + ¥ = ¥( . . . ) .1 91042 2 37706 2 38331 10 6 67079 109 9mm mm4 4

PROBLEM 9.99



142



The Mohr’s circle is defined by the diameter XY, where X ( . ,3 495148 10 109 9¥ - ¥ 3.7266 ) and

Y ( . ,6 67079 10 109 9¥ ¥ 3.7266 )

Now I I Ix yave = + = + ¥ = ¥1

2

1

23 495148 6 67079 10 5 082969 109 9( ) ( . . ) .


˘˚

+

= - ¥ÈÎÍ

˘˚

+ -

1

2

1

23 495148 6 67079 10

22

92

( )

( . . ) ( 33 7266 10

4 05077 10

9 2

9

. )

.

¥

= ¥ mm4


tan

( . )

( . . )

22

2 3 7266 10

3 495148 6 67079 10

2

9

9

qmxy

x y

I

I I= -

-

= -- ¥

- ¥= - ..34699

or 2 66 922qm = - ∞.

and qm = - ∞33 5.


33 5. ∞ clockwise

about C.

Now I I Rmax, ( . . ) min ave mm= ± = ± ¥5 082969 4 05077 109 4

or Imax .= ¥9 13 109 mm4

and Imin .= ¥1 032 109 mm4


143


PROBLEM 9.100



SOLUTION

From Problem 9.81: I Ix y= ¥ = ¥( . ) ( . )126 5625 10 253 125 106 6p pmm mm4 4

Problem 9.73: I xy = ¥337 5 106. mm4

The Mohr’s circle is defined by the diameter XY, where X ( . ,126 5625 10 106 6¥ ¥p 337.5 ) and Y ( . ,253 125 106¥ p - ¥337 5 106. ).

Now I I Ix yave4mm= + = ¥1

2596 412 106( ) .


˘˚

+

= ¥

1

2

391 700 10

2 2

6

( )

. mm4 (from Solution of Problem 9.87)


tan

( . )

( . . )

.

22

2 337 5 10

126 5625 253 125 10

1 6

6

6

q

p

mxy

x y

I

I I= -

-

= -¥

- ¥= 99765

or 2 59 4998qm = ∞.

and qm = ∞29 7.


29.7° counterclockwise

about C.

144



Now I I Rmax, . .min ave= ± = ¥ ± ¥596 412 10 391 700 106 6

or Imax = ¥988 106 4mm

and Imin = ¥205 106 4mm


145


PROBLEM 9.101



SOLUTION


I

x

y

=

=

0 162 10

0 454 10

6 4

6

.

.

¥

¥

mm

mm4

Problem 9.74: I xy = -0 159584 106 4. ¥ mm

The Mohr’s circle is defined by the diameter XY, where X Y( . , . ) ( . ,0 162 10 0 159584 10 0 454 106 6 6¥ ¥ ¥- and

0 159584 106. ).¥

Now I I Ix yave mm= + =1

20 308 106 4( ) . ¥


˘˚

+

=

1

2

0 21629 10

22

6 4

( )

. ¥ mm (from Solution of Problem 9.89)


tan

( . )

( . . )

.

22

2 0 159584 10

0 162 0 454 10

1 09

6

6

qmxy

x y

I

I I= -

-

= - --

= -

¥¥

3304

Then 2 47 545qm = - .

and qm = -23 8. ∞


23.8° clockwise b

about C.

146



Now I I Rmax, min . .= ± = ±ave 0 308 10 0 21629 106 6¥ ¥

or Imax .= 0 524 106 4¥ mm b

and Imin .= 0 0917 106 4¥ mm b


147


SOLUTION


I

x

y

=

=

2 90051 10

0 721271 10

6 4

6 4

.

.

¥

¥

mm

mm

Problem 9.77: I xy = -0 680632 106 4. ¥ mm

The Mohr’s circle is defined by the diameter XY, where X Y( . , . ) ( . ,2 90051 10 0 680632 10 0 72127 106 6 6¥ ¥ ¥- and

0 680632 106. ).¥

Now I I Ix yave mm= + =1

21 81089 106 4( ) . ¥


˘˚

+

= -ÈÎÍ

˘˚

+ -

1

2

1

22 90051 0 721271 10

22

62

( )

( . . ) (¥ 00 680632 10

1 28473 10

6 2

6 4

. )

.

¥

¥= mmThe Mohr’s circle is then drawn as shown.

tan

( . )

( . . )

22

2 0 680632 10

2 90051 0 721271 10

6

6

qmxy

x y

I

I I= -

-

= - --

=

¥¥

00 6246.or 2 31 99qm = . ∞and qm = 16 00. ∞

The principal axes are obtained by rotating the xy axes through 16.00° counterclockwise b about C.

PROBLEM 9.102


Area of Problem 9.77

(The moments of inertia I x and I y of the area of Problem 9.102 were determined in Problem 9.44).

148



Now I I Rmax, min ( . . )= ± = ±ave 1 81089 1 28473 106¥

or Imax .= 3 10 4¥106 mm b

and Imin .= ¥0 526 4106 mm b

From the Mohr’s circle it is seen that the a axis corresponds to Imax and the b axis corresponds to Imin .

149


PROBLEM 9.103

The moments and product of inertia of an L102 ¥ 102 ¥ 12.7-mm angle cross section with respect to two rectangular axes x and y through C are, respectively, I x= 2.30 ¥ 106 mm4, I y = 2.30 ¥ 106 mm4, and I xy � 0, with the minimum value of the moment of inertia of the area with respect to any axis through C beingI min = 1.15 ¥ 106 mm4. Using Mohr’s circle, determine (a) the product of inertia I xy of the area, (b) the orientation of the principal axes, (c) the value of I max .

SOLUTION

(Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when the 104-mm leg of the angle is parallel to the x axis. Further, for I xy � 0, the angle must be oriented as shown.)

Now I I Ix yave610 mm= + = =1

2

1

22 30 10 2 2 306 4( ) ( . ) .¥ ¥ ¥

and I I R Rmin . .= - = -ave or 2 30 10 1 15 106 6¥ ¥

= 1 15 106 4. ¥ mm

Using Iave and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, X Ixy( . , )2 30 106¥ and Y Ixy( . , ).2 30 106¥ | |

(a) We have R I I I I I

R I

x y xy x y

xy

22

21

2= -È

ÎÍ˘˚

+ =

=

( ) ; Since

or I xy = 1 15 106 4. ¥ mm

Solving for I xy and taking the negative root (since I xy � 0) yields

I xy = -1 15 4. ¥10 mm6

I xy = -1 150 4. ¥10 mm6 b

(b) We have from formula or otherwise clearly

or 2 90

45

qq

m

m

= -= -

∞∞


45° clockwise b

about C.

(c) We have I I Rmax . .

.

= + = +

=ave

4mm

2 30 10 1 15 10

3 45 10

6 6

6

¥ ¥

¥ or Imax .= 3 45 106 4¥ mm b

150


SOLUTION

From Figure 9.13B: I

I

x

y

= ¥

= ¥

0 162 10

0 454 10

6 4

6 4

.

.

mm

mm



and I x y¢ ¢ = 0 (symmetry) I x yAxy = S


1 76 6 4 486 4¥ =. . -13 1. 9.2 -58620 93.

2 6 4 51 6 4 285 44. ( . ) .¥ - = 21.7 –16.3 –100962.98

Â –159583.91

I xy = -159584 4 mm

The Mohr’s circle is defined by the diameter XY where X ( . , . )0 162 10 0 159584 106 6¥ - ¥and Y ( . , . )0 454 10 0 159584 106 6¥ ¥

Now I I Ix yave

mm

= + = + ¥

= ¥

1

2

1

20 162 0 454 10

0 3080 10

6

6 4

( ) ( . . )

.

PROBLEM 9.104

Using Mohr’s circle, determine for the cross section of the rolled-steel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Figure 9.13.)

151




˘˚

+ = -ÈÎÍ

˘˚

+ -1

2

1

20 162 0 454 0 159584

22

2

( ) ( . . ) ( . )22 6

6 4

10

0 21629 10

ÏÌÔ

ÓÔ

¸˝Ô

Ô¥

= ¥. mm


tan

( . )

( . . )

.

22

2 0 159584 10

0 162 0 454 10

1 09

6

6

qmxy

x y

I

I I= -

-

= - - ¥- ¥

= - 3304

or 2 47 545qm = - .

and qm = - ∞23 8.


23.8° clockwise b

About C.

Now I I Rmax, min ( . . )= ± = ± ¥ave 0 3080 0 21629 106

or Imax .= ¥0 524 106 4mm b

and Imin .= ¥0 0917 106 4mm b


152


SOLUTION

From Figure 9.13B: I

I

x

y

= ¥

= ¥

3 93 10

1 06 10

6 4

6 4

.

.

mm

mm

Problem 9.7B: I xy = - ¥1 165061 106 4. mm

(Note that the figure of Problem 9.105 is obtained by replacing x with –x in the figure of Problem 9.78; thus the change in sign of I xy .)

The Mohr’s circle is defined by the diameter XY, where X Y( . , . ) ( . ,3 93 10 1 165061 10 1 06 106 6 6¥ - ¥ ¥ and 1 165061 106. ).¥

Now I I Ix yave

mm

= +

= + ¥

= ¥

1

21

23 93 1 06 10

2 495 10

6

6 4

( )

( . . )

.


˘˚

+

= -ÈÎÍ

˘˚

+ -Ï

1

2

1

23 93 1 06 1 165061

22

22

( )

( . . ) ( . )ÌÌÔ

ÓÔ

¸˝Ô

Ô¥

= ¥

10

1 84840 10

6

6 4. mm

PROBLEM 9.105

Using Mohr’s circle, determine for the cross section of the rolled-steel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Figure 9.13.)

153




tan

( . )

( . . )

.

22

2 1 165061 10

3 93 1 06 10

0 81189

6

6

qmxy

x y

I

I I= -

-

= - - ¥- ¥

=

or 2 39 073qm = ∞.

and qm = ∞19 54.


19.54° counterclockwise b

about C.


or Imax .= ¥4 34 106 4mm b

and Imin .= ¥0 647 106 4mm b


154


PROBLEM 9.106*

For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are I x= 12 × 106 mm4 and I y = 3 × 106 mm4, respectively. Knowing that after rotating the x and y axes about the centroid 30∞ counterclockwise, the moment of inertia relative to the rotated x axis is 14.5 × 106 mm4, use Mohr’s circle to determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.

SOLUTION

We have I I Ix yave mm= + = + ¥ = ¥1

2

1

212 3 10 7 5 106 6 4( ) ( ) .

Now observe that I Ix � ave , I Ix x¢ � , and 2 60q = + ∞. This is possible only if I xy < 0. Therefore, assume

I xy � 0 and (for convenience) I x y¢ ¢ � 0. Mohr’s circle is then drawn as shown.

We have 2 60q am + = ∞

Now using DABD: RI Ix

m m

m

=-

= -

=

ave

mm

cos

.

cos

.

cos( )

2

12 10 7 5 10

2

4 5 10

2

6 6

64

q q

q

¥ ¥

¥

Using DAEF: RI Ix=

-= -

=

¢ ave

mm

cos

. .

cos

cos( )

a a

a

14 5 10 7 5 10

7 0

6 6

64

¥ ¥

¥1

Then 4 5 10

2

7 060 2

6 6.

cos cos

¥ ¥1 ∞q a

a qm

m= = -

or 9 60 2 14 2cos( ) cos∞ - =q qm m

Expanding: 9 60 2 60 2 14 2(cos cos sin sin ) cos∞ ∞q q qm m m+ =

or tancos

sin.2

14 9 60

9 601 21885qm =

-=

∞∞

or 2 50 633 25 3q qm m= =. .∞ ∞and

(Note: 2 60qm � ∞ implies assumption I x y¢ ¢ � 0 is correct.)

Finally, R = =4 5 10

50 6337 0946

64.

cos ..

¥∞

¥10 mm6

(a) From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given centroidal x and y axes through qm about the centroid C or

25.3° counterclockwise b

155



(b) We have I I Rmax, min ( . . )= ± = ±ave 7 5 7 0946 106¥

or Imax .= 14 59 4¥10 mm6 b

and Imin .= 0 405 4¥10 mm6 b


156


PROBLEM 9.107

It is known that for a given area I y = 48 × 106 mm4 and I xy = –20 × 106 mm4, where the x and y axes are rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating the x axis 67.5∞ counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia I x of the area, (b) the principal centroidal moments of inertia.

SOLUTION

First assume I Ix y� and then draw the Mohr’s circle as shown. (Note: Assuming I Ix y� is not consistent with the requirement that the axis corresponding to ( )maxI xy is obtained after rotating the x axis through 67.5° CCW.)

From the Mohr’s circle we have

2 2 67 5 90 45qm = ∞ - ∞ = ∞( . )

(a) From the Mohr’s circle we have

I II

x yxy

m

= + = ¥ + ¥∞

22

48 10 220 10

456

6| |

tan tanq

or I x = ¥88 0 106 4. mm b

(b) We have I I Ix yave

mm

= + = + ¥

= ¥

1

2

1

288 0 48 10

68 0 10

6

6 4

( ) ( . )

.

and RIxy

m

= = ¥∞

= ¥| |

mmsin sin

.2

20 10

4528 284 10

66 4

q


or Imax .= ¥96 3 106 4mm b

and Imin .= ¥39 7 106 4mm b

157


PROBLEM 9.108

Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of rectangular axes through the centroid is zero.

SOLUTION

Consider the regular pentagon shown, with centroidal axes x and y.

Because the y axis is an axis of symmetry, it follows that I xy = 0. Since I xy = 0, the x and y axes must be principal axes. Assuming I I I Ix y= =max min ,and the Mohr’s circle is then drawn as shown.

Now rotate the coordinate axes through an angle a 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 I I I Ix y x y= = = =¢ ¢ ave (for all moments of inertia with respect to an axis through C ) b

(b) I Ixy x y= =¢ ¢ 0 (for all products of inertia with respect to all pairs of rectangular axes with origin at C ) b

158


PROBLEM 9.109

Using Mohr’s circle, prove that the expression I I Ix y x y¢ ¢ ¢ ¢- 2 is independent of the orientation of the x¢ and y¢ axes, where Ix¢, Iy¢, and Ix¢y¢ represent the moments and product of inertia, respectively, of a given area with respect to a pair of rectangular axes x¢ and y¢ through a given Point O. Also show that the given expression is equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr’s circle.

SOLUTION

First observe that for a given area A and origin O of a rectangular coordinate system, the values of Iave 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 I R

I I R

I R

x

y

x y

¢

¢

¢ ¢

= += -

=

ave

ave

cos

cos

sin

2

2

2

qq

q

Then, forming the expression I I Ix y x y¢ ¢ ¢ ¢- 2

I I I I R I R R

I R

x y x y¢ ¢ ¢ ¢- = + - -

= -

2 22 2 2( cos )( cos ) ( sin )ave ave

ave2

q q q22 2 2 2

2

2 2cos ( sin )q q( ) -

= -

R

I Rave2 which is a constant

I I Ix y x y¢ ¢ ¢ ¢- 2 is independent of the orientation of the coordinate axes Q.E.D. b

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

L I R2 2= -ave2

or L I I Ix y x y2 2= -¢ ¢ ¢ ¢ Q.E.D. b

159


PROBLEM 9.110

Using the invariance property established in the preceding problem, express the product of inertia Ixy of an area A with respect to a pair of rectangular axes through O in terms of the moments of inertia Ix and Iy of A and the principal moments of inertia Imin and Imax of A about O. Use the formula obtained to calculate the product of inertia Ixy of the L76 × 51 × 6.4 mm angle cross section shown in Figure 9.13A, knowing that its maximum moment of inertia is 0.6 × 106 mm4.

SOLUTION

Consider the following two sets of moments and products of inertia, which correspond to two different orientations of the coordinate axes whole origin is at Point O.

Case 1: I I I I I Ix x y y x y xy¢ ¢ ¢ ¢= = =, ,

Case 2: I I I I Ix y x y¢ ¢ ¢ ¢= = =max min, , 0

The invariance property then requires

I I I I Ix y xy- =2max min or I I I I Ixy x y= ± - max min b

From Figure 9.13A: I

I

x

y

=

=

0 454 10

0 162 10

6 4

6 4

.

.

¥

¥

mm

mm

Using Eq. (9.21): I I I Ix y+ = +max min

Substituting ( . . ) . min0 454 0 162 10 0 6 106 6+ = +¥ ¥ I

or Imin .= 0 016 106 4¥ mm

Then I xy = ± -

= ±

( . )( . ) ( . )( . )

.

0 454 10 0 162 10 0 6 10 0 016 10

0 25288

6 6 6 6¥ ¥ ¥ ¥

¥1106 4mmThe two roots correspond to the following two orientations of the cross section.

For I xy = -0 253 106 4. ¥ mm b

and for I xy = 0 253 106 4. ¥ mm b

160


PROBLEM 9.111

Determine the mass moment of inertia of a ring of mass m, cut from a thin uniform plate, with respect to (a) the axis AA¢, (b) the centroidal axis CC¢ that is perpendicular to the plane of the ring.

SOLUTION

Mass

mass area area

= = =

= =

m V tA

I t Im

AI

r r

r

We first determine

A r r r r

I r r r rAA

= - = -( )= - = -( )¢

p p p

p p p22

12

22

12

24

14

24

14

4 4 4,area

(a) Im

AI

m

r rr r m r rAA AA¢ ¢= =

-( ) -( ) = +( ), ,mass areap

p

22

12 2

414

22

12

4

1

4

I m r rAA¢ = +( )1

4 12

22 b

(b) By symmetry: I IBB AA¢ ¢=

Eq. (9.38): I I I ICC AA BB AA¢ ¢ ¢ ¢= + = 2 I m r rCC ¢ = +( )1

2 12

22 b

161


PROBLEM 9.112

A thin semicircular plate has a radius a and a mass m. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis B¢, (b) the centroidal axis CC¢ that is perpendicular to the plate.

SOLUTION

mass

mass area area

= =

= =

m tA

I t Im

AI

r

r

Area: A a= 1

22p

I I a a

I I

AA DD

AA DD

¢ ¢

¢ ¢

= = ÊËÁ

ˆ¯

=

= =

, ,

, ,

area area

mass mass

1

2 4

1

84 4p p

mm

AI

m

aa maAA¢ = Ê

ËÁˆ¯

=, area 12

24 21

8

1

4pp

(a) I I m AC ma ma

BB DD¢ ¢= - = - ÊËÁ

ˆ¯

( )2 221

4

4

3p

= -( . . )0 25 0 1801 2ma I maBB¢ = 0 0699 2. b

(b) Eq. (9.38): I I I ma maCC AA BB¢ ¢ ¢= + = +1

40 06992 2.

I maCC ¢ = 0 320 2. b

162


PROBLEM 9.113

The quarter ring shown has a mass m and was cut from a thin, uniform plate. Knowing that r1 = 3

4 r2, determine the mass moment of inertia of the quarter ring with respect to (a) the axis AA¢, (b) the centroidal axis CC¢ that is perpendicular to the plane of the quarter ring.

SOLUTION

First note mass = = =

= -( )m V tA

t r r

r r

r p4 2

212

Also I t I

m

r rI

mass area

area

=

=-( )

r

p4 2

212

(a) Using Figure 9.12, I r rAA¢ = -( ), areap16 2

414

Then Im

r rr r

mr r

AA¢ =-( )

¥ -( )

= +( )

, mass pp

416

4

22

12

24

14

22

12

= + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

mr r

4

3

422

2

2

or I mrAA¢ = 25

64 22 b

(b) Symmetry implies

I IBB AA¢ ¢=, ,mass mass

Then I I I

mr

mr

DD AA BB¢ ¢ ¢= +

= ÊËÁ

ˆ¯

=

225

64

25

32

22

22

163



Now locate centroid C.

X A x AS S=

or X r rr

rr

rp p

pp

pp

4 4

4

3 4

4

3 422

12 2

22 1

12-Ê

ËÁˆ¯

= ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

or Xr r

r r=

--

4

323

13

22

12p

Now r X

r r

r r

r

=

=- Ê

ËÁˆ¯

- ÊËÁ

ˆ¯

=

2

4 2

3

34

34

37 2

21

23

2

3

22

2

2

2

p

p

Finally, I I mrDD CC¢ ¢= + 2

or 25

32

37 2

2122

2

2

mr I m rCC= +Ê

ËÁˆ

¯¢ p or I mrCC ¢ = 0 1522 2

2. b

164


SOLUTION


= ÊËÁ

ˆ¯

m V tA

t ab

r r

r p2

Also I t I

m

abI

mass area

area

=

=

r

p2

(a) We have

or

I I Ay

I ab abb

x BB

BB

, ,

,

area area

area

= +

= - ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

¢

¢

2

32

8 2

4

3

p pp

(a) From Sample Problem 9.3:

Then

I ab

Im

abab

BB

BB

¢

¢

=

= ¥

,

,

area

mass

1

21

3 1

21

3

3

or I mbBB¢ = 1

72 b

(b) From Sample Problem 9.3:

Now

I a b

I I A a

AA

AA

¢

¢ ¢

=

= + ÊËÁ

ˆ¯

,

, ,

area

area area

1

5

3

4

3

11

2

PROBLEM 9.114

The parabolic spandrel shown was cut from a thin, uniform plate. Denoting the mass of the spandrel by m, determine its mass moment of inertia with respect to (a) the axis BB ¢, (b) the axis DD¢ that is perpendicular to the spandrel. (Hint: See Sample Problem 9.3.)

165



and I I A a22 11

21

4¢ ¢= + ÊËÁ

ˆ¯, ,area area

Then I I A a a

a b

AA22

2 2

3

1

4

3

4

1

5

1

¢ ¢= + ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +

, ,area area

33

1

16

9

16

1

30

2 2

3

ab a a

a b

-ÊËÁ

ˆ¯

=

and Im

aba b

ma

223

2

3 1

301

10

¢ = ¥

=

, mass

Finally, I I IDD BB¢ ¢ ¢= +, , ,mass mass mass22

= +1

7

1

102 2mb ma or I m a bDD¢ = +1

707 102 2( ) b

166


PROBLEM 9.115

A thin plate of mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the axis B¢, which is perpendicular to the plate.

SOLUTION

mass

mass area area

= =

= =

m t A

I t Im

AI

r

r

(a) Consider parallelogram as made of horizontal strips and slide strips to form a square since distance from each strip to x axis is unchanged.

I ax, area = 1

34

Im

AI

m

aax x, ,mass area= = Ê

ËÁˆ¯2

41

3 I max = 1

32 b

(b) For centroidal axis y¢:

I a a

Im

AI

m

aa

y

y y

¢

¢ ¢

= ÈÎÍ

˘˚

=

= = ÊËÁ

ˆ¯

,

, ,

area

mass area

21

12

1

6

1

6

4 4

24˜ =

= + = + =¢¢ ¢

1

6

1

6

7

6

2

2 2 2 2

ma

I I ma ma ma may y

For axis BB¢ ^ to plate, Eq. (9.38):

I I I ma maBB x y¢ ¢¢= + = +1

3

7

62 2 I maBB¢ = 3

22 b

167


SOLUTION

See sketch of the solution of Problem 9.115.

(a) From Part b of solution of Problem 9.115:

I may¢ = 1

62

I I ma ma may y= + = +¢2 2 21

6 I may = 7

62 b

(b) From solution of Problem 9.115:

I ma I may x¢ = =1

6

1

32 2and

Eq. (9.38): I I IAA y x¢ ¢= +

= +1

6

1

32 2ma ma I maAA¢ = 1

22 b

PROBLEM 9.116

A thin plate of mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the y axis, (b) the axis AA¢, which is perpendicular to the plate.

168


SOLUTION


= +ÈÎÍ

˘˚

=

m V tA

t a a a a ta

r r

r r( )( ) ( )( )21

22 3 2

Also I tIm

aImass area area= =r

3 2

(a) Now I I I

a a a a

a

x x x, , ,( ) ( )

( )( ) ( )( )

area area area= +

= +

=

1 2

3 3

4

1

32

1

122

5

6

Then Im

aax, mass = ¥

3

5

624 or massI max, = 5

182 b

(b) We have

I I I

a a a a

z z z, , ,( ) ( )

( )( ) ( )( )

area area area= +

= ÈÎÍ

˘˚

+

1 2

3 31

32

1

362 ++ + ¥Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

=1

22 2

1

32 10

24( )( )a a a a a

Then Im

aa max, mass = ¥ =

310

10

324 2

Finally, I I I

ma ma

y x z, , ,mass mass mass= +

= +5

18

10

32 2

= 65

182ma or massI may, .= 3 61 2 b

PROBLEM 9.117

A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis.

169


SOLUTION

First locate the centroid C.

X A xA X a a a a a a aS S= + = + + ¥ÊËÁ

ˆ¯

: ( ) ( ) ( )2 2 21

322 2 2 2

or X a= 14

9

Z zA Z a a a a a aS SA = + = ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

: ( ) ( ) ( )21

22

1

32 2 2 2

or Z a= 4

9

(a) We have I I m X Zy CC, , ( )mass mass= + +¢2 2

From the solution to Problem 9.117:

I may, mass = 65

182

Then I ma m a acc¢ = - ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

, mass65

18

14

9

4

92

2 2

or 2I macc¢ = 0 994. b

(b) We have I I m Zx BB, , ( )mass mass= +¢2

and I I m aAA BB¢ ¢= +, , ( . )mass mass 1 5 2

Then I I m a aAA x¢ = + - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

, , ( . )mass mass 1 54

92

2

PROBLEM 9.118

A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis CC¢ that is perpendicular to the plate, (b) the axis AA¢ that is parallel to the x axis and is located at a distance 1.5a from the plate.

170



From the solution to Problem 9.117:

I max, mass = 5

182

Then I ma m a aAA¢ = + - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

, ( . )mass5

181 5

4

92 2

2

or I maAA¢ = 2 33 2. b

171


PROBLEM 9.119

The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Using direct integration, express the mass moment of inertia of the solid with respect to the x axis in terms of m and h.

SOLUTION

We have yh h

ax h= - +2

so that rh

ax a= +( )

For the element shown:

dm r dx dI r dm

h

ax a dx

x= =

= +ÈÎÍ

˘˚

rp

rp

2 2

2

1

2

( )

Then m dmh

ax a dx

h

ax a

h

aa a

aa= = + = +

= -

ÚÚ rp rp

rp

2

22

2

23

00

2

23 3

1

3

1

38

( ) [( ) ]

( )) = 7

32rpah

Now I dI r r dxh

ax a dx

h

a

x x

a= = = +È

ÎÍ˘˚

= ¥

ÚÚÚ1

2

1

2

1

2

1

5

2 24

0

4

4

( ) ( )

[

rp rp

rp (( ) ] ( )x ah

aa a

ah

a+ = -

=

50

4

45 5

4

1

1032

31

10

rp

rp

From above: rpah m2 3

7=

Then I m h mhx = ÊËÁ

ˆ¯

=31

10

3

7

93

702 2

or I mhx = 1 329 2. b

172


PROBLEM 9.120

Determine by direct integration the mass moment of inertia with respect to the y axis of the right circular cylinder shown, assuming that it has a uniform density and a mass m.

SOLUTION

For element shown: dm dV a dx= =r rp 2

dI dI x dm

a dm x dm

a x a dx

I dI a a

y y

y y

= +

= +

= +ÊËÁ

ˆ¯

= =

2

2 2

2 2 2

2

1

4

1

4

1

4

rp

rp 22 2

2

2

2 23

2

22 21

4 32

1

4 2

+ÊËÁ

ˆ¯

= + = +

-

+

-

+

ÚÚ x dx

a a xx

a aL

L

L

L

L

/

/

/

/

rp rp 11

3 2

3LÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

I a L a Ly = +1

1232 2 2rp ( )

But for whole cylinder: m V a L= =r rp 2

Thus, I m a Ly = +1

123 2 2( ) b

173


PROBLEM 9.121

The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Determine by direct integration the mass moment of inertia of the solid with respect to (a) the x axis, (b) the y axis. Express your answers in terms of m and the dimensions of the solid.

SOLUTION

We have at ( , ):a h hk

a=

or k ah=

For the element shown:

r

dm r dx

ah

xdx

=

=

= ÊËÁ

ˆ¯

42

2

rp

rp

Then m dmah

xdx

a hx

a ha a

a

a

a

a

= = ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚

= - - -

ÚÚ rp

rp

rp

23

2 23

2 2

1

1

3

1ÊÊËÁ

ˆ¯

ÈÎÍ

˘˚

= 2

32rpah

(a) For the element: dI r dm r dxx = =1

2

1

22 4rp

Then I dIah

xdx a h

xx x a

a

a

a

= = ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚

= -

ÚÚ1

2

1

2

1

3

1

1

6

44 4

3

33

rp rp

rpaa ha a

ah

ah

4 43 3

4

2

1

3

1 1

6

26

27

1

6

2

3

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= ¥

= ¥ ¥

rp

rp 113

9

13

542 2h mh=

or I mhx = 0 241 2. b

174



(b) For the element: dI dI x dm

r dm x dm

y y= +

= +

2

2 21

4

Then I dIah

xx

ah

xdx

a h

y y a

a= = Ê

ËÁˆ¯

+È

ÎÍÍ

˘

˚˙˙

ÊËÁ

ˆ¯

=

Ú Ú1

4

1

4

22

23

2 2

rp

rp aa h

xdx a h

a h

xx

m

a

a

a

a 2 2

42 2

2 2

3

33

11

12

3

2

+Ê

ËÁˆ

¯= - +

È

ÎÍ

˘

˚˙

= ÊËÁ

ˆ¯

Ú rp

˜ - +È

ÎÍ

˘

˚˙ - - +

È

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝ÔÔ

aa h

aa

a h

aa

1

12 33

1

12

2 2

3

2 2

3( ) ( )

= ¥ +Ê

ËÁˆ

¯= +Ê

ËÁˆ¯

3

2

1

12

26

272

13

1083

22 2ma

h

aa m h a

or I m a hy = +( . )3 0 12042 2 b

175


PROBLEM 9.122

Determine by direct integration the mass moment of inertia with respect to the x axis of the tetrahedron shown, assuming that it has a uniform density and a mass m.

SOLUTION

We have xa

hy a a

y

h= + = +Ê

ËÁˆ¯

1

and zb

hy b b

y

h= + = +Ê

ËÁˆ¯

1

For the element shown: dm xz dy aby

hdy= Ê

ËÁˆ¯

= +ÊËÁ

ˆ¯

r r1

2

1

21

2

Then m dm aby

hdy

abh y

h

h

h

= = +ÊËÁ

ˆ¯

= ¥ +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

-

-

ÚÚ1

21

1

2 31

20

3

r

r00

3 31

61 1 1

1

6

= - -ÈÎ ˘

=

r

r

abh

abh

( ) ( )

Now, for the element: I xz aby

hAA¢ = = +ÊËÁ

ˆ¯, area

1

36

1

3613 3

4

Then dI tI dy aby

hAA AA¢ ¢= = +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

, , ( )mass arear r 1

313

4

176



Now

dI dI y z dm

aby

hd

x AA= + + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +ÊËÁ

ˆ¯

¢, mass2

2

34

1

3

1

361r yy

y by

hab

y

hdy+ + +Ê

ËÁˆ¯

ÈÎÍ

˘˚

ÏÌÔ

ÓÔ

¸˝Ô

Ô+Ê

ËÁˆ¯

È

ÎÍÍ

22 21

31

1

21r

˘

˚˙˙

= +ÊËÁ

ˆ¯

+ + +Ê

ËÁˆ

¯1

121

1

223

42

3 4

2r rab

y

hdy ab y

y

h

y

hdy

Now m abh= 1

6r

Then dI mb

h

y

h

m

hy

y

h

y

hdyx = +Ê

ËÁˆ¯

+ + +Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

1

21

32

2 42

3 4

2

and I dIm

hb

y

hy

y

h

y

hdyx x h

= = +ÊËÁ

ˆ¯

+ + +Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

Ú Ú- 21 6 22

42

3 4

2

0

== ¥ +ÊËÁ

ˆ¯

+ + +Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=

-

m

hb

h y

hy

y

h

y

h

m

h2 5

1 61

3

1

2 52

53

4 5

2

0

22

1

51 6

1

3

1

2

1

52 5 3 4

25

hb h h

hh

hh( ) ( ) ( ) ( )- - + - + -È

ÎÍ˘˚

ÏÌÓ

¸˝˛

or I m b hx = +1

102 2( ) b

177


PROBLEM 9.123

Determine by direct integration the mass moment of inertia with respect to the y axis of the tetrahedron shown, assuming that it has a uniform density and a mass m.

SOLUTION

We have xa

hy a a

y

h= + = +Ê

ËÁˆ¯

1

and zb

hy b b

y

h= + = +Ê

ËÁˆ¯

1

For the element shown: dm xz dy aby

hdy= Ê

ËÁˆ¯

= +ÊËÁ

ˆ¯

r r1

2

1

21

2

Then m dm aby

hdy

abh y

h

h

h

= = +ÊËÁ

ˆ¯

= ¥ +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

-

-

ÚÚ1

21

1

2 31

20

3

r

r00

3 31

61 1 1

1

6

= - -ÈÎ ˘

=

r

r

abh

abh

( ) ( )

Also I xz I zxBB DD¢ ¢= =, ,area area 1

12

1

123 3

Then, using I tImass area= r

we have dI dy xz dI dy zxBB DD¢ ¢= ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯, ,( ) ( )mass massr r1

12

1

123 3

178



Now

dI dI dI

xz x z dy

aby

h

y BB DD= +

= +

= +ÊËÁ

ˆ¯

¢ ¢, ,

( )

mass mass

1

12

1

121

2 2r

r22

2 22

1( )a by

hdy+ +Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

We have m abh dIm

ha b

y

hdyy= = + +Ê

ËÁˆ¯

1

6 212 2

4

r fi ( )

Then

I dIm

ha b

y

hdy

m

ha b

h y

h

y y h= = + +Ê

ËÁˆ¯

= +( ) ¥ +ÊËÁ

ˆ¯

Ú Ú- 21

2 51

2 24

0

2 2

( )

˜È

ÎÍÍ

˘

˚˙˙

= +( ) - -ÈÎ ˘

-

5 0

2 2 5 5

101 1 1

h

ma b ( ) ( )

or I m a by = +1

102 2( ) b

179


PROBLEM 9.124*

Determine by direct integration the mass moment of inertia with respect to the z axis of the semiellipsoid shown, assuming that it has a uniform density and a mass m.

SOLUTION

First note that when

z y bx

a= = -

Ê

ËÁˆ

¯0 1

2

2

1 2

:

/

y z cx

a= = -

Ê

ËÁˆ

¯0 1

2

2

1 2

:

/

For the element shown: dm yzdx bcx

adx= = -

Ê

ËÁˆ

¯r p pr( ) 1

2

2

Then m dm bcx

adx

bc xa

x abc

a

a

= = -Ê

ËÁˆ

¯

= -ÈÎÍ

˘˚

=

Ú Ú pr

pr pr

0

2

2

23

0

1

1

3

2

3

For the element: I zyAA¢ =, areap4

3

Then dI tI dx zyAA AA¢ ¢= = ÊËÁ

ˆ¯, , ( )mass arear r p

43

Now dI dI x dm

b cx

adx x bc

x

a

z AA= +

= -Ê

ËÁˆ

¯+ -

Ê

ËÁˆ

¢, mass2

32

2

22

2

241 1

p r pr¯

È

ÎÍÍ

˘

˚˙˙

= - +Ê

ËÁˆ

¯+ -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

dx

m

a

b x

a

x

ax

x

a

3

2 41 2

2 2

2

4

42

4

2˚˙˙dx

180



Finally, I dI

m

a

b x

a

x

ax

x

a

z z

a

=

= - +Ê

ËÁˆ

¯+ -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

Ú

Ú3

2 41 2

2 2

2

4

42

4

20ddx

m

a

bx

x

a

x

ax

x

a= - +

Ê

ËÁˆ

¯+ -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

3

2 4

2

3

1

5

1

3

1

5

2 3

2

5

43

5

200

a

= - +ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙

3

2 41

2

3

1

5

1

3

1

5

22m

ba

or I m a bz = +( )1

52 2 b

181


PROBLEM 9.125*

A thin steel wire is bent into the shape shown. Denoting the mass per unit length of the wire by m¢, determine by direct integration the mass moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION

First note dy

dxx a x= - --1 3 2 3 2 3 1 2/ / / /( )

Then 1 12

2 3 2 3 2 3

2 3

+ ÊËÁ

ˆ¯

= + -

= ÊËÁ

ˆ¯

-dy

dxx a x

a

x

/ / /

/

( )

For the element shown: dm m dL m

dy

dxdx

ma

xdx

= ¢ = ¢ + ÊËÁ

ˆ¯

= ¢ ÊËÁˆ¯

12

1 3/

Then m dm ma

xdx m a x m a

aa= = ¢ = ¢ ÈÎ ˘ = ¢Ú Ú

1 3

1 31 3 2 3

00

3

2

3

2

/

// /

Now I y dm a x ma

xdx

m aa

xa

x

a= = - ¢

Ê

ËÁˆ

¯

= ¢ -

Ú Ú2 2 3 2 3 3

0

1 3

1 3

1 32

1 33

( )/ //

/

//

44 3 1 3 2 3 5 3

0

1 3 2 2 3 4 3 4 3

3

3

2

9

4

/ / / /

/ / / /

x a x x dy

m a a x a x

a+ -

Ê

ËÁˆ

¯

= ¢ - +

Ú33

2

3

8

3

2

9

4

3

2

3

8

3

8

2 3 2 8 3

0

3 3

a x x

m a m a

a/ /-È

ÎÍ˘˚

= ¢ - + -ÊËÁ

ˆ¯

= ¢

or I max = 1

42 b

Symmetry implies I may = 1

42 b

182



Alternative solution:

I x dm x ma

xdx m a x dx

m a

y

aa= = ¢

Ê

ËÁˆ

¯= ¢

= ¢ ¥

Ú ÚÚ2 21 3

1 31 3 5 3

00

1 3 3

8

/

// /

/ xx m a

ma

a8 3

0

3

2

3

81

4

/ÈÎ ˘ = ¢

=

Also I x y dm I Iz y x= +( ) = +Ú 2 2

or I maz = 1

22 b

183


SOLUTION

For line BC: z = - +

= -

ha

x h

h

aa x

2

2( )

Also m V t ah

tah

= = ÊËÁ

ˆ¯

=

r r

r

1

2

1

2

(a) We have dI dm dm

dm

x = ¢ + ÊËÁ

ˆ¯ ¢

= ¢

1

12 2

1

3

22

2

z z

z

where dm t dx¢ = r z

Then I dI t dx

th

aa x dx

x x

a

a

= =

= -ÈÎÍ

˘˚

ÚÚ

Ú

21

3

2

32

2

0

2

3

0

2

z r z

r

( )

( )

/

/

= ¥ -ÊËÁ

ˆ¯

-ÈÎ ˘

= - - -

2

3

1

4

1

22

1

12

3

34

0

2

3

34 4

r

r

th

aa x

th

aa a a

a( )

( ) ( )

/

ÈÈÎ ˘

= 1

123rtah

or I mhx = 1

62 b

PROBLEM 9.126

A thin triangular plate of mass m is welded along its base AB to a block as shown. Knowing that the plate forms an angle q with the y axis, determine by direct integration the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis, (c) the z axis.

184



Now I x dmz = Ú 2

and I x dm x t dxa

z r z= ¢ =Ú Ú2 2

0

22 ( )

/

= -ÈÎÍ

˘˚

= -ÈÎÍ

˘˚

=

Ú2 2

23

1

4

2

2

0

2

3 4

0

2

r

r

r

t xh

aa x dx

th

a

ax x

th

a

a

a

( )/

/

aa a a

ta h ma

3 2

1

4 2

1

48

1

24

3 4

3 2

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= =r

(b) We have I r dm x dmy y= = +ÈÎ ˘Ú Ú2 2 2( sin )z q

= + ÚÚ x dm dm2 2 2sin q z

Now I dm I I Ix y x= fi = +Úz qz2 2sin

= +1

24

1

62 2 2ma mh sin q

or Im

a hy = +24

42 2 2( sin )q b

(c) We have I r dm x y dmz z= = +Ú Ú2 2 2( )

= +ÈÎ ˘

= +

= +

= +

ÚÚÚ

x dm

x dm dm

I I

ma m

x

2 2

2 2 2

2

21

24

1

6

( cos )

cos

cos

z q

q z

qz

hh2 2cos q

or Im

a hz = +24

42 2 2( cos )q b

185


SOLUTION

First note for the cylindrical ring shown that

m V t d d= = ¥ -( )r r p4 2

212

and, using Figure 9.28, that

I md

md

t d d t

AA¢ = ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

= ¥ÊËÁ

ˆ¯

- ¥

1

2 2

1

2 2

1

8 4

22

2

11

2

22

22r p r pp

p r

p r

4

1

8 4

1

8 4

12

12

24

14

2

d d

t d d

t d

ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= ÊËÁ

ˆ¯

-( )

= ÊËÁ

ˆ¯

2212

22

12

12

221

8

-( ) +( )= +( )

d d d

m d d

Now treat the pulley as four concentric rings and, working from the brass outward, we have

m = ¥ ¥ -{+

p4

8650 0 0175 0 011 0 005

1250 0 017

3 2 2 2

3

kg/m m m

kg/m

( . ) ( . . )

( . 55 0 017 0 011

0 002 0 022 0 017

0 009

2 2 2

2 2 2

m m

m m

) ( . . )

( . ) ( . . )

( .

¥ -ÈÎ+ ¥ -+ 55 0 028 0 0222 2 2m m) ( . . )¥ - ˘}

= + + + ¥

= ¥

-

-

( . . . . )11 4134 2 8863 0 38288 2 7980 10 3 kg

17.4806 10 kg3

PROBLEM 9.127

Shown is the cross section of a molded flat-belt pulley. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA¢. (The density of brass is 8650 kg/ m3 and the density of the fiber-reinforced polycarbonate used is 1250 kg/m3.)

186



and I AA¢ = + + +ÈÎ

+

1

811 4134 0 005 0 011 2 8863 0 011 0 0172 2 2 2( . )( . . ) ( . )( . . )

(( . )( . . )

( . )( . . )

0 38288 0 017 0 022

2 7980 0 022 0 028 10

2 2

2 2 3

+

+ + ˘ ¥ - kg ◊◊ m2

= + + + ¥ ◊

= ¥ ◊

-

-

( . . . . )

.

208 29 147 92 37 00 443 48 10

836 69 10

9 2

9 2

kg m

kg m

or I AA¢-= ¥ ◊837 10 9 2kg m b

Now kI

mAAAA

¢¢

-

--= =

¥ ◊¥

= ¥29

36 2836 69 10

17 4806 1047 864 10

.

..

kg m

kgm

2

or kAA¢ = 6 92. mm b

187


SOLUTION

First note for the cylindrical ring shown that

m V t d d t d d= = ¥ -( ) = -( )r r p p r4 42

212

22

12

and, using Figure 9.28, that

I md

md

t d d t

AA¢ = ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

= ¥ÊËÁ

ˆ¯

- ¥

1

2 2

1

2 2

1

8 4

22

2

11

2

22

22r p r pp

p r

p r

4

1

8 4

1

8 4

12

12

24

14

2

d d

t d d

t d

ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= ÊËÁ

ˆ¯

-( )= Ê

ËÁˆ¯

2212

22

12

12

221

8

-( ) +( )= +( )

d d d

m d d

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

Have m = ¥ ( ) -ÈÎ ˘{+

p4

8580 0 0195 0 009 0 006

2770

3 2 2( ) . ( . ) ( . )kg/m m m m

kg/m33 2 2

3

0 0165 0 012 0 009

1250 0 0165

( )( ) -ÈÎ ˘

+ ( )( ). ( . ) ( . )

. (

m m m

kg/m m 00 027 0 0122 2. ) ( . )m m-ÈÎ ˘}

= + +[ ]= +

-

-

p4

7 52895 2 87942 12 06563 10

5 9132 10 2 26149

3

3

. . .

. .

¥

¥

kg

kg ¥¥ ¥

¥

10 9 47632 10

17 6510 10

3 3

3

- -

-

+

=

kg kg

kg

.

.

PROBLEM 9.128

Shown is the cross section of an idler roller. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA¢. (The specific weight of bronze is 8580 kg/m3; of aluminum, 2770 kg/m3; and of neoprene, 1250 kg/m3.)

188



And I AA¢-= ¥ +ÈÎ ˘

+ ¥

1

85 9132 10 0 006 0 009

2 26149 10

3 2 2 2{( . ) ( . ) ( . )

( .

kg m

--

-

+ÈÎ ˘

+ ¥ +

3 2 2 2

3 2

0 009 0 012

9 47632 10 0 012 0

kg) ( . ) ( . ) m

( . kg) ( . ) ( .0027

1

8691 844 508 835 8272 827 10

1 1841

2 2

9 2

) m }

( . . . ) kg m

.

ÈÎ ˘

= + + ◊

=

-

99 10 6 2¥ ◊- kg m

or I AA¢-= ¥ ◊1 184 10 6 2. kg m b

Now kI

mAAAA

¢¢

-

-

-

= =¥ ◊

¥

= ¥

26 2

3

6

1 18419 10

17 6510 10

67 08902 10

. kg m

. kg

. m22

38 19079 10kAA¢-= ¥. m

or kAA¢ = 8 19. mm b

189


PROBLEM 9.129

Knowing that the thin cylindrical shell shown is of mass m, thickness t, and height h, determine the mass moment of inertia of the shell with respect to the x axis. (Hint: Consider the shell as formed by removing a cylinder of radius a and height h from a cylinder of radius a + t and height h; then neglect terms containing t2 and t3 and keep those terms containing t.)

SOLUTION

From Figure 9.28 for a solid cylinder (change orientation of axes):

I m a h

I I mh

m a h mh

m a

x

x x

¢

¢

= +

= + ÊËÁ

ˆ¯

= + +

=

1

123

2

1

123

4

3

12

2 2

2

2 22

2

( )

( )

++Ê

ËÁˆ

¯h2

3 I m a hx = +1

123 42 2( )

Shell = Cylinder – Cylinder

I m a t h m a h

a t h a t

x = + + - +

= + +

1

123 4

1

123 4

1

123

12 2

22 2

2

[ ( ) ] ( )

[ ( ) ][ ( )rp 22 2 2 2 241

123 4+ - +h a h a h] ( )( )rp

190



Expand ( )a t+ factors and neglecting terms in t t t2 3 4, , : and

Ih

a at t a at t h a a h

ha

x = + +( ) + + +( )ÈÎ

˘˚ - -

=

rp

rp12

2 3 6 3 4 3 4

123

2 2 2 2 2 4 2 2

44 3 2 2 2 2 3 2 4 2 2

3

6 3 4 6 8 3 4

1212 8

+ + + + + - -( )= +

a t a t a h a t ath a a h

ha t ath

rp 22

2 2

1212 8

( )= +( )rphat

a h

Mass of shell: m ath hat= =r p rp( )2 2

Ihat

a h

ma h

x = +( )= +( )

2

2412 8

4

243 2

2 2

2 2

rp

or Im

a hx = +( )6

3 22 2 b

Checks:

a I mh

h I ma

x

x

= = -

= = -

¸

˝ÔÔ

˛ÔÔ

01

3

01

2

2

2

;

;

slender rod

thin ringOK

191


SOLUTION

Mass: m a h m ah

a hcyl cone= = =rp rp rp2 2 21

3 2

1

6

I I m a I m a

a h a h

y : cyl cyl cone cone= =

= =

1

2

3

101

2

1

20

2 2

4 4rp rp

For entire machine part:

m m m a h a h a h

I I I a hy

= - = - =

= - = -

cyl cone

cyl cone

rp rp rp

rp

2 2 2

4

1

6

5

61

2

1

200

9

204 4rp rpa h a h=

or I a h ay = ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

5

6

6

5

9

202 2rp I may = 27

502 b

Then kI

may

2 227

50= = k ay = 0 735. b

PROBLEM 9.130

The machine part shown is formed by machining a conical surface into a circular cylinder. For b h= 1

2 , determine the mass moment of inertia and the radius of gyration of the machine part with respect to the y axis.

192


SOLUTION

(a) We have d dB A= -( . )0 08 m

and, using the parallel axis

I I md

I I md

AA GG A

BB GG B

¢ ¢

¢ ¢

= +

= +

2

2

Then I I m d d

m d d

m d

BB AA B A

A A

A

¢ ¢- = -( )= - -ÈÎ ˘

= -

2 2

2 20 08

0 0064 0 16

( . )

( . . )

Substituting: ( . . ) . ( . . )0 68 0 32 10 0 18 0 0064 0 163 2- ¥ = -- kg m kg m2◊ dA

or dA = 27 5. mm b

(b) We have I I mdAA GG A¢ ¢= + 2

or IGG¢-

-

= ¥ -

= ¥

0 32 10 0 18

0 183875 10

3 2

3 2

. .

.

kg m kg (0.0275 m)

kg m

2◊ ◊

◊

Then kI

mGGGG

¢¢

--= =

¥= ¥2

330 183875 10

0 181 02153 10

.

..

kg m

kgm

22◊

or kGG¢ = 32 0. mm b

PROBLEM 9.131

After a period of use, one of the blades of a shredder has been worn to the shape shown and is of mass 0.18 kg. Knowing that the mass moments of inertia of the blade with respect to the AA¢ and B¢ axes are 0 320. g m2◊and 0 680. ,g m2◊ respectively, determine (a) the location of the centroidal axis GG¢, (b) the radius of gyration with respect to axis GG¢.

193


SOLUTION

First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger cone as indicated.

Now h h

h10

60

3030= + =or mm

The weight of the body is given by

m m m m V V V= + - = + -( ) ( )1 2 3 1 2 3r

or 0 4 0 02 0 063

0 005 0 033

0 015 0 092 2 2. ( . ) ( . ) ( . ) ( . ) ( . ) ( . )kg = + -ÈÎÍ

r p p p ˘˚m3

= + - ¥ -r( . . . )7 54 0 079 2 12 10 5 m3

or r = 7274 kg/m3

Then m

m

15

27

7274 7 54 10 0 5485

7274 7 9 10 0 00575

= ¥ =

= =

-

-

( )( . ) .

( )( . ) .

kg

kg¥

mm357274 2 12 10 0 1542= ¥ =-( )( . ) . kg

Finally, using Figure 9.28 have,

I I I I

m a m a m a

AA AA AA AA¢ ¢ ¢ ¢= + -

= + -

( ) ( ) ( )1 2 3

1 12

2 22

3 321

2

3

10

3

10

= + -1

20 5485 0 02

3

100 00575 0 005

3

100 1542 0 0152 2( . )( . ) ( . )( . ) ( . )( . )22È

ÎÍ˘˚

= + - ¥ -( . . . )10 97 0 00431 1 042 10 5 2 kg m◊ or I AA¢

-= ¥9 93 10 5 2. kg m◊ b

PROBLEM 9.132

Determine the mass moment of inertia of the 400-g machine component shown with respect to the axis AA¢.

194


SOLUTION

First note m V b L1 12= =r r

And m V a L2 22= =r r

(a) Using Figure 9.28 and the parallel-axis theorem, have

I I I

m b b ma

m

AA AA AA¢ ¢ ¢= -

= + + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

-

( ) ( )

( )

1 2

12 2

1

21

12 2

1

12 222 2

2

2

2 2 2 2

2

1

6

1

4

( )

( ) ( )

a a ma

b L b a a L

+ + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +ÊËÁ

ˆ¯

-r r 55

12

122 3 5

2

4 2 2 4

a

Lb b a a

ÊËÁ

ˆ¯

= + -r( )

Then dI

da

Lb a aAA¢ = - =r

126 20 02 3( )

or a a b= =03

10and

Also d I

da

Lb a L b aAA

2

22 2 2 2

126 60

1

210¢ = - = -r r( ) ( )

Now, for

a = 0, d I

daAA

2

20¢ � and for a b= 3

10,

d I

daAA

2

20¢ �

\ ( )maxI AA¢ occurs when a b= 3

10

a = =1053

1057 511. mm

or a = 57 5. mm b

PROBLEM 9.133

A square hole is centered in and extends through the aluminum machine component shown. Determine (a) the value of a for which the mass moment of inertia of the component with respect to the axis AA¢, which bisects the top surface of the hole, is maximum, (b) the corresponding values of the mass moment of inertia and the radius of gyration with respect to the axis AA¢. (The density of aluminum is 2770 kg/ m3.)

195



(b) From part (a)

( )massIL

b b b bAA¢ = +Ê

ËÁˆ

¯-

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=

r12

2 33

105

3

10

49

2

4 22 4

440

49

2402770 0 375 0 105

25 778 10

4 3 4

3 2

rLb =

= ¥ ◊-

( kg/m )( . m)( . m)

. kg m

or (I AA¢-= ¥ ◊) . kg mmass 25 8 10 3 2 b

and kI

mAAAA

¢¢=2 ( )mass

where

m m m L b a L b b Lb= - = - = -Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=1 22 2 2

223

10

7

10r r r( )

Then

kLb

LbbAA¢ = = = =2

4

2

2 2 2

492407

10

7

24

7

24105 3215 625

r

r( ) .mm mm

kAA¢ = 56 706. mm

or mmkAA¢ = 56 7. b

196


SOLUTION

(a) First note m V d larm arm= = ¥r r p4

2

and dm dV

a t ad

cup cup=

=

r

r p q q[( cos )( )( )]2

Then m dm a t d

a t

a t

cup cup

/

/

= =

=

=

Ú Ú 2

2

2

2

0

2

20

2

2

pr q q

pr q

pr

p

p

cos

[sin ]

Now ( ) ( ) ( )I I IAA AA AA¢ ¢ ¢= +anem cups arms

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

( ) ( )I I m d I m dAA GG AG AG¢ ¢= +ÈÎ ˘ + +ÈÎ ˘

=

anem cup cup arm arm arm3 3

35

2

112 23

1

22 2

22m a m l a

am lcup cup arm+ + + Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô+( ) ++ Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

= + +ÊËÁ

ˆ¯

+

=

ml

m a la l m l

arm

cup arm

2

35

32

3 2

2

2 2 2

( prr p ra t a la l d l l2 2 2 2 25

32

4) ( )+ +ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

or ( )I l a ta

l

a

l

d lAA¢ = + +

Ê

ËÁˆ

¯+

È

ÎÍÍ

˘

˚˙˙

anem pr 2 22

2

2

65

32 1

4 b

PROBLEM 9.134

The cups and the arms of an anemometer are fabricated from a material of density r. Knowing that the mass moment of inertia of a thin, hemispherical shell of mass m and thickness t with respect to its centroidal axis GG¢ is 5 122ma / , determine (a) the mass moment of inertia of the anemometer with respect to the axis AA¢, (b) the ratio of a to l for which the centroidal moment of inertia of the cups is equal to 1 percent of the moment of inertia of the cups with respect to the axis AA¢.

197



(b) We have ( )

( ).

I

IGG

AA

¢

¢=cup

cup

0 01

or 5

120 01

5

322 2 2m a m a la l acup cup (from Part )= + +Ê

ËÁˆ¯

.

Now let z = a

l.

Then 5 0 125

32 12 2z z z= + +Ê

ËÁˆ¯

.

or 40 2 1 02z z- - =

Then z =± - - -2 2 4 40 1

2 40

2( ) ( )( )

( )

or z z= -0 1851. and = 0.1351

a

l= 0 1851. b

To the instructor:

The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at this time for use in the solutions of Problems 9.135 through 9.140.

Thin rectangular plate

( ) ( )

( )

I I md

m b h mb h

x m x m= +

= + + ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=

¢2

2 22 21

12 2 2

11

32 2m b h( )+

( ) ( )I I md

mb mb

mb

y m y m= +

= + ÊËÁ

ˆ¯

=

¢2

22

21

12 2

1

3

I I md

mh mh

mh

z z m= +

= + ÊËÁ

ˆ¯

=

¢( ) 2

22

21

12 2

1

3

198



Thin triangular plate

We have m V bht= = ÊËÁ

ˆ¯

r r 1

2

and I bhz, area = 1

363

Then I tI

t bh

mh

z z, ,mass area=

= ¥

=

r

r 1

361

18

3

2

Similarly, I mby, mass = 1

182

Now I I I m b hx y z, , , ( )mass mass mass= + = +1

182 2

Thin semicircular plate

We have m V a t= = ÊËÁ

ˆ¯

r r p2

2

and I I ay z, ,area area= = p8

4

Then I I tI

t a

ma

y z y, , ,mass mass area= =

= ¥

=

r

r p8

1

4

4

2

Now I I I max y z, , ,mass mass mass= + = 1

22

Also I I my I m ax x x, , ,mass mass massor= + = -ÊËÁ

ˆ¯¢ ¢

22

21

2

16

9p

and I I my I m az z z, , ,mass mass massor= + = -ÊËÁ

ˆ¯¢ ¢

22

21

4

16

9p

199



y za= = 4

3p

Thin Quarter-Circular Plate

We have m V a t= = ÊËÁ

ˆ¯

r r p4

2

and I I ay z, ,area area= = p16

4

Then I I tI

t a

ma

y z y, , ,mass mass area= =

= ¥

=

r

r p16

1

4

4

2

Now I I I max y z, , ,mass mass mass= + = 1

22

Also I I m y zx x, , ( )mass mass= + +¢2 2

or I m ax¢ = -ÊËÁ

ˆ¯, mass

1

2

32

9 22

p

and I I mzy y, ,mass mass= +¢2

or I m ay¢ = -ÊËÁ

ˆ¯, mass

1

4

16

9 22

p

200


SOLUTION

First compute the mass of each component. We have

m V tA= =r rST ST

Then

m1 7850 0 002

1 413

==

( )( .

.

kg/m m)(0.3 m)

kg

3 2

m m2 3 7850 0 002

0 2826

= = ¥ ¥=

( )( .

.

kg/m m) (0.15 0.12) m

kg

3 2

Using Figure 9.28 and the parallel-axis theorem, we have

I I Ix x x= +

= ÈÎÍ

˘˚

+

( ) ( )

( .

( .

1 22

1

121 413

21

120 2826

kg)(0.3 m)

kg

2

))(0.12 m) kg)(0.15 m2 2+ +ÈÎÍ

˘˚

= +

( . . )

[( . )

0 2828 0 06

0 0105975

2 2

22 0 0003391 0 0073759( . . )]+ kg m2◊

= +[( . ) ( . )]0 0105975 2 0 0077150 kg m2◊

or I x = ¥ -26 0 10 3. kg m2◊ b

I I Iy y y= +

= +ÈÎÍ

˘˚

+

( ) ( )

( . . )

( .

1 2

2 2

2

1

121 413 0 3

21

120 28

kg)(0.3 m2

226 0 2826 0 15

0 021

2 2 kg)(0.15 m) kg)(0.075 m2 2+ +ÈÎÍ

˘˚

=

( . . )

[( . 11950 2 0 0005299 0 0079481

0 0211950 2 0 0084

) ( . . )]

[( . ) ( .

+ +

= +

kg m2◊

7780)] kg m2◊ or I y = ¥ -38 2 10 3. kg m2◊ b

PROBLEM 9.135

A 2 mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is7850 kg/m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.

201



I I Iz z z= +

= ÈÎÍ

˘˚

+

( ) ( )

( .

( .

1 22

1

121 413

21

120 2826

kg)(0.3 m)

kg

2

))(0.15 m kg)(0.075 m22 2 2 2 20 12 0 2826 0 06

0 01

+ + +ÈÎÍ

˘˚

=

. ) ( . . )

[( . 005975 2 0 0008690 0 0026070

0 0105975 2 0 0034

) ( . . )]

[( . ) ( .

+ +

= +

kg m2◊

7760)] kg m2◊

or I z = ¥ -17 55 10 3. kg m2◊ b

202


PROBLEM 9.136

A 2 mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/ m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.

SOLUTION


m V tA= =r rST ST

Then

m1 7850 0 002

2 14305

= ¥=

( )( .

.

kg/m m)(0.35 0.39) m

kg

3 2

m22 27850 0 002 0 195 0 93775= ¥Ê

ËÁˆ¯

=( )( . . . kg/m m)2

m kg3 p

m3 7850 0 0022

0 39 0 15 0 45923= ¥ ¥ÊËÁ

ˆ¯

=( )( . . . . kg/m m)1

m kg3 2

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have

I I I Ix x x x= + +

= +

( ) ( ) ( )

( . ( .

1 2 3

1

122 14305 2 14305 kg)(0.39 m) kg)2 00.39

2 m

kg)(0.195 m

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ -ÊËÁ

ˆ¯

2

2

1

2

16

90 93775

p( . )) kg)

4 0.195m2 + ¥Ê

ËÁˆ¯

+È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝( . ( . )0 93775

30 195

22 2

pÔÔ

Ô

+ + + ÊËÁ

ˆ¯

1

180 45923 0 15 0 459232 2( . ( . ) ] ( . kg)[(0.39) m kg)

0.39

32

˜ + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

2 20 15

3

.m2

= + + + + +[( . . ) ( . . ) ( . .0 027163 0 081489 0 011406 0 042081 0 004455 0 0089009)] kg m2◊

= + +( . . . )0 108652 0 053487 0 013364 kg m2◊

= 0 175503. kg m2◊

or I x = ¥ -175 5 10 3. kg m2◊ b

203



I I I Iy y y y= + +

= + +

( ) ( ) ( )

( . ( . ) ] (

1 2 3

21

122 14305 0 39 kg)[(0.35) m2 2 22 14305

0 39

2

2 22.

. kg)

0.35

2mÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

+ +ÈÎÍ

˘˚

+

1

40 93775 0 93775

1

18

( . ( . kg)(0.195 m) kg)(0.195 m)2 2

(( . ( ..

0 45923 0 459230 39

3

2

kg)(0.39 m) kg) (0.35)2 2+ + ÊËÁ

ˆ¯

È

ÎÍÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô= + + +

m2

0 049040 0 147120 0 008914 0 035658[( . . ) ( . . ))

( . . )]+ +0 003880 0 064017 kg m2◊

= + +( . . . )0 196160 0 044572 0 067897 kg m2◊

= 0 308629. kg m2◊

or I y = ¥ -309 10 3 kg m2◊ b

I I I Iz z z z= + +

= +

( ) ( ) ( )

( . ( .

1 2 3

1

122 14305 2 14305 kg)(0.35 m) kg)2 00.35

2 mÊ

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

2

+ ÈÎÍ

˘˚

+ +

1

40 93775

1

180 45923

( .

( . (

kg)(0.195 m)

kg)(0.15 m)

2

2 00 459230 15

3

0 0

22.

.

[( .

kg) (0.35) m2 + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= 221877 0 065631 0 008914 0 000574 0 057404+ + + +. ) . ) ( . . )] kg m2◊

= + +

=

( . . . )

.

0 087508 0 008914 0 057978

0 154400

2kg m

kg m2

◊

◊

or I z = ¥ -154 4 10 3. kg m2◊ b

204


SOLUTION

Have m V= r

Now m1 2770 0 00125 0 060 0 075

0 015581

==

( / )( . )( . )( . )

.

kg m m m m

kg

3

m2 2770 0 00125 0 075 0 155

0 040252

==

( / )( . )( . )( . )

.

kg m m m m

kg

3

m3 2770 0 00125 0 060 0 155

0 032201

==

( / )( . )( . )( . )

.

kg m m m m

kg

3

Using Figure 9.28 and the parallel-axis theorem, have

( ) ( ) ( ) ( ) ( ) ( ) ( )

( . ) ( .

I I I I I I Ix x x x x x x= + + + =

=

1 2 3 4 3 4

0 015581 0

and

kg 0061

12

1

40 040252 0 155

1

12

1

42 2) ( . ) ( . )+Ê

ËÁˆ¯

ÈÎÍ

˘˚

+ +ÊËÁ

ˆ¯

m kg2 ÈÈÎÍ

˘˚

+ +ÈÎ ˘ +ÊËÁ

ˆ¯

ÏÌÓ

m

kg)

2

2 0 032201 0 06 0 1551

12

1

42 2( . ( . ) ( . )

¸˝˛

m2

= + +- - -( . . . )1 86972 10 3 2235 10 5 9304 105 4 4¥ ¥ ¥ ◊kg m2

= -934 087 10 6 2. ¥ ◊kg m

or I x = -934 10 6 2¥ ◊kg m b

PROBLEM 9.137

The cover for an electronic device is formed from sheet aluminum that is 1.25 mm thick. Determine the mass moment of inertia of the cover with respect to each of the coordinate axes. (The density of aluminum is 2770 kg/m3.)

205



I I I I Iy y y y y= + + +

= ÈÎÍ

˘˚

+

( ) ( ) ( ) ( )

( . ) ( . )

1 2 3 4

20 0155811

30 075kg m2 (( . ) ( . ) ( . )

( .

0 040252 0 075 0 1551

3

0 0322

2 2kg m2+ÈÎ ˘ ÊËÁ

ˆ¯

ÏÌÓ

¸˝˛

+ 001 0 1551

30 032201

1

20 155 0 02 2kg m kg2) ( . ) ( . ) ( . ) ( .Ê

ËÁˆ¯

ÈÎÍ

˘˚

+ + 7751

40 1552 2) ( . )+È

ÎÍ˘˚m2

= + + + =- - - -( . . . . )2 9214 10 3 9782 10 2 5788 10 4 3901 10 115 4 4 4 2¥ ¥ ¥ ¥ ◊kg m 223 924 10 6. ¥ ◊- kg m2

or I y = ¥ -1124 10 6 kg m2◊ b

I I I I Iz z z z z= + + +

= +ÈÎ ˘

( ) ( ) ( ) ( )

( . ) ( . ) ( . )

1 2 3 4

2 20 015581 0 075 0 06kg ˚ÊËÁ

ˆ¯

ÏÌÓ

¸˝˛

+ ÊËÁ

ˆ¯

ÈÎÍ

˘˚

+

1

30 040252 0 075

1

3

0

2m kg m2 2( . ) ( . )

( .. ) ( . ) ( . ) ( .032201 0 061

30 032201

1

302kg m kg2Ê

ËÁˆ¯

ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯

006 0 0752 2) ( . )+ÈÎÍ

˘˚m2

= + + +

=

- - - -( . . . . )4 7912 10 7 5473 10 3 8641 10 2 1977 10

38

5 5 5 4 2¥ ¥ ¥ ¥ ◊kg m

11 796 10 6. ¥ ◊- kg m2

or I z = ¥ -382 10 6 kg m2◊ b

206


SOLUTION

First compute the mass of each component.

Have m V At= =r r

Then

m137530 0 002 0 045 0 070

0 047439

==

( kg/m )( . m)( . m)( . m)

. kg

m237530 0 002 0 045 0 045

0 013554

==

( kg/m )( . m)( . m)( . m)

. kg

m337530 0 002

1

20 04 0 095

0 028614

= ¥=

( kg/m )( . m) ( . m)( . m)

. kg

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, have

I I I Ix x x x= + +

= ÈÎÍ

˘˚

+

( ) ( ) ( )

( . kg) ( . ) m ( .

1 2 3

2 20 0474391

30 07 0 0135554

1

120 020 0 07 0 01

0 0286141

1

2 2 2 2kg) ( . ) [( . ) ( . ) ] m

( . kg)

+ +ÏÌÓ

¸˝˛

+88

0 095 0 041

92 0 095 0 040

0 0

2 2 2 2 2[( . ) ( . ) ] [( . ) ( . ) ] m

[( .

+ + ¥ +ÏÌÓ

¸˝˛

= 777484 0 068222 0 136751 10 0282457 103 2 3 2+ + ¥ ◊ = ¥ ◊- -. . ) ]kg m kg m

or I x = ¥ ◊-0 2825 10 3 2. kg m b

PROBLEM 9.138

A framing anchor is formed of 2-mm-thick galvanized steel. Determine the mass moment of inertia of the anchor with respect to each of the coordinate axes. (The density of galvanized steel is 7530 kg/m .)3

207



I I I Iy y y y= + +

= ÈÎÍ

˘˚

+

( ) ( ) ( )

( . kg) ( . ) m ( .

1 2 3

2 20 0474391

30 045 0 0135554 0 045 0 02

1

3

0 0286141

18

2 2 2kg) [( . ) ( . ) ] m

( . kg) (

+ ÊËÁ

ˆ¯

ÏÌÓ

¸˝˛

+ 00 04 0 0451

90 04

0 03202 0 010956 0 065

2 2 2 2. ) ( . ) ( . ) m

[( . . .

+ÈÎÍ

˘˚

= + + 5574 10

0 10855 10

3 2

3 2

) ]kg m

. kg m

¥ ◊

= ¥ ◊

-

or I y = ¥ ◊0 1086 103 2. kg m b

I I I Iz z z z= + +

= + ÊËÁ

ˆ

( ) ( ) ( )

( . kg) [( . ) ( . ) ]

1 2 3

2 20 047439 0 045 0 0701

3¯ÏÌÓ

¸˝˛

+ ÊËÁ

ˆ¯

+

+

m ( . kg)[( . )

( . ) ]m

( .

2 2

2 2

0 013554 0 0451

3

0 070

0 02286141

180 095 0 045

2

30 095

0

2 22

2kg) ( . ) ( . ) . m

[(

+ + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= .. . . ) ]kg m

. kg m

0109505 0 075564 0 187064 10

0 37213 10

3 3

3 2

+ + ¥ ◊

= ¥ ◊

-

-

or I z = ¥ ◊-0 372 10 3 2. kg m b

208


PROBLEM 9.139

A subassembly for a model airplane is fabricated from three pieces of 1.5 mm plywood. Neglecting the mass of the adhesive used to assemble the three pieces, determine the mass moment of inertia of the subassembly with respect to each of the coordinate axes. (The density of the plywood is 780 kg/m .)3

SOLUTION


m V tA= =r r

Then

m m1 22780 0 0015 0 3 0 12

21 0600 10

= = ¥ ¥ÊËÁ

ˆ¯

= ¥ -

( )( . . .

.

kg/m m)1

2m3

33 kg

m32 2 3780 0 0015

40 12 13 2324 10= ¥Ê

ËÁˆ¯

= ¥ -( )( . . . kg/m m) m kg3 p

Using the equations derived above and the parallel-axis theorem, we have

( ) ( )

( ) ( ) ( )

( .

I I

I I I Ix x

x x x x

1 2

1 2 3

321

1821 0600 10

== + +

= ¥ - kg)(0.12 mm) kg)0.12

m2 + ¥ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ ¥

-

-

( .

( .

21 0600 103

1

213 2324 10

32

3 kkg)(0.12 m)

kg

2ÈÎÍ

˘˚

= + + ¥ -[ ( . . ) ( . )]2 16 8480 33 6960 95 2733 10 6 ◊ mm

kg m

2

2= + ¥ -[ ( . ) ( . )]2 50 5440 95 2733 10 6 ◊

or I x = ¥ -196 4 10 6. kg m2◊ b

209



I I I Iy y y y= + +

= ¥ +-

( ) ( ) ( )

( . ( .

1 2 3

31

1821 0600 10 21 0600kg)(0.3 m) 2 kkg)

0.3

3mÊ

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

2

+ ÏÌÓ

¥ +

+ ¥

-

-

1

1821 0600 10 0 12

21 0600 10

3 2 2

3

( . ( . ) ]

( .

kg)[(0.3) m

kg)

2

00.3

3mÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô

2 220 12

3

.

+ ¥ÈÎÍ

˘˚

-1

413 2324 10 3( . kg)(0.12 m)2

= + + +

+ ¥ -

[( . . ) ( . . )

( . )]

105 300 210 600 122 148 244 296

47 637 10 6 kg m2◊

= + + ¥ -( . . . )315 900 366 444 47 637 10 6 kg m2◊

or I y = ¥ -730 10 6 kg m2◊ b

Symmetry implies I Iy z=

I z = ¥ -730 10 6 kg m2◊ b

210


SOLUTION


m V tA= =r rST ST

Then

m1 7850 0 002

2 76948

= ¥=

( )( .

.

kg/m m)(0.84 0.21) m

kg

3 2

m2 7850 0 0018

10 62713

= ¥ ¥=

( )( .

.

kg/m m)( 0.285 0.84) m

kg

3 2p

m m3 427850 0 0018 0 285

1 80282

= = ¥ÊËÁ

ˆ¯

=

( )( . .

.

kg/m m)2

m

kg

3 2p

Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have

( ) ( )

( ) ( ) ( ) ( )

( .

I I

I I I I Ix x

x x x x x

3 4

1 2 3 4

1

122 76948

== + + +

= kg)(0.21 m) kg) 0.2850.21

2m2 2+ -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

( .2 769482

+ + ÈÎÍ

˘˚

[( . ) ] ( .10 62713 21

21 802822 kg)(0.285 m kg)(0.285 m)2

== + + +

=

[( . . ) ( . ) ( . )]

[( .

0 01018 0 08973 0 86319 2 0 07322

0 09991

kg m2◊

++ +0 86319 2 0 07322. ( . )] kg m2◊

or I x = 1 110. kg m2◊ b

PROBLEM 9.140*

A farmer constructs a trough by welding a rectangular piece of 2 mm-thick sheet steel to half of a steel drum. Knowing that the density of steel is 7850 kg/m3 and that the thickness of the walls of the drum is 1.8 mm, determine the mass moment of inertia of the trough with respect to each of the coordinate axes. Neglect the mass of the welds.

211



I I I I Iy y y y y= + + +

= +

( ) ( ) ( ) ( )

( . ( . )

1 2 3 4

21

122 76948 0 21 kg)[(0.84)2 ]]

( . ..

m

kg)0.84

2

2

2 2

2 76948 0 2850 21

2

ÏÌÓ

+ ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ôm 2

+ + +1

1210 62713 6 0 285 10 627132 2( . ( . ) ] ( . kg)[(0.84) m kg)

0.84

2 2 mmÊ

ËÁˆ¯

ÏÌÔ

ÓÔ

¸˝Ô

Ô

2

+ ÈÎÍ

˘˚

1

41 80282( . kg)(0.285 m)2

+ +ÈÎÍ

˘˚

1

41 80282 1 80282( . ( . kg)(0.285 m) kg)(0.84 m)2 2

= + + +

+ + +

[( . . ) ( . . )

( . ) ( .

0 17302 0 57827 1 05647 1 87463

0 03661 0 03661 1.. )]27207 kg m2◊

= + + +( . . . . )0 75129 2 93110 0 03661 1 30868 kg m2◊

or I y = 5 03. kg m2◊ b

I I I I Iz z z z z= + + +

= +

( ) ( ) ( ) ( )

( . ( .

1 2 3 4

1

122 76948 2 769 kg)(0.84 m)2 448

1

1210 62713 6 0

2

kg)0.84

2 m

kg)[(0.84)2

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ +( . ( .. ) ] ( .285 10 6271322

m kg)0.84

2 m2 + Ê

ËÁˆ¯

ÏÌÔ

ÓÔ

¸˝Ô

Ô

+ ÈÎÍ

˘˚

+ +

1

41 80282

1

41 80282 2

( .

( . ) (

kg)(0.285 m)

kg)(0.285 m

2

11 80282. kg)(0.84 m)2ÈÎÍ

˘˚

= + + +

+ + +

[( . . ) ( . . )

( . ) ( .

0 16285 0 48854 1 05647 1 87463

0 03661 0 03661 1.. )]27207 kg m2◊

= + + +( . . . . )0 65139 2 93110 0 03661 1 30868 kg m2◊

or I z = 4 93. kg m2◊ b

212


SOLUTION


m V= rST

Then

m1 7850 0 08 0 04==

( )[( ( . ( . kg/m m) m)]

6.31334 kg

3 2p

m2 7850 0 02 0 06= =( )[ ( . ( . kg/m m) m)] 0.59188 kg3 2p

m3 7850 0 02 0 04 0 39458= =( )[ ( . ( . . kg/m m) m)] kg3 2p


(a) I I I Ix x x x= + -

= + +

( ) ( ) ( )

( . ( . ) ]

1 2 3

21

126 31334 0 04 kg)[3(0.08) m2 2 (( .6 31334 kg)(0.02 m)2Ï

ÌÓ

¸˝˛

+ + +1

120 59188 0 06 0 591882( . ( . ) ] ( . kg)[3(0.02) m kg)(0.03 m)2 2 2ÏÏ

ÌÓ

¸˝˛

- + +1

120 39458 0 04 0 394582( . ( . ) ( . kg)[3(0.02) ] m kg)(0.02 m)2 2 22Ï

ÌÓ

¸˝˛

= + + +

- + ¥

[( . . ) ( . . )

( . . )]

10 94312 2 52534 0 23675 0 53269

0 09207 0 15783 110 3- kg m2◊

= + - ¥ -( . . . )13 46846 0 76944 0 24990 10 3 kg m2◊

= ¥ -13 98800 10 3. kg m2◊

or I x = ¥ -13 99 10 3. kg m2◊ b

PROBLEM 9.141

The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m .)3

213



(b) I I I Iy y y y= + -

= ÈÎÍ

˘˚

+

( ) ( ) ( )

( .

( .

1 2 3

1

26 31334

1

20 59

kg)(0.08 m)2

1188 0 59188

1

20 39458

kg)(0.02 m) kg)(0.04 m)

k

2 2+ÈÎÍ

˘˚

-

( .

( . gg)(0.02 m kg)(0.04 m)2 2) ( .+ÈÎÍ

˘˚

0 39458

= + +

- + ¥ -

[( . ) ( . . )

( . . )]

20 20269 0 11838 0 94701

0 07892 0 63133 10 3 kg m◊ 22

= + - ¥ -( . . . )20 20269 1 06539 0 71025 10 3 kg m2◊

= ¥ -20 55783 10 3. kg m2◊

or I y = ¥ -20 6 10 3. kg m2◊ b

(c) I I I Iz z z z= + -

= + +

( ) ( ) ( )

( . ( . ) ]

1 2 3

21

126 31334 0 04 kg)[3(0.08) m2 2 (( .6 31334 kg)(0.02 m)2Ï

ÌÓ

¸˝˛

+ + +1

120 59188 0 06 0 591882( . ( . ) ] ( . kg)[3(0.02) m kg)[(0.04)2 2 2 ++Ï

ÌÓ

¸˝˛

( . ) ]0 03 2 m2

- + +1

120 39458 0 04 0 039582( . ( . ) ] ( . kg)[3(0.02) m kg)[(0.04)2 2 2 ++Ï

ÌÓ

¸˝˛

( . ) ]0 02 2 m2

= + + +

- + ¥

[( . . ) ( . . )

( . . )]

10 94312 2 52534 0 23675 1 47970

0 09207 0 78916 110 3- kg m2◊

= + - ¥ -( . . . )13 46846 1 71645 0 88123 10 3 kg m2◊

= ¥ -14 30368 10 3. kg m2◊

or I z = ¥ -14 30 10 3. kg m2◊ bTo the Instructor:

The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.142 through 9.145.

From Figure 9.28:

Cylinder ( )I m ax cyl cyl= 1

22

( ) ( ) ( )I I m a Ly zcyl cyl cyl= = +1

123 2 2

214



Symmetry and the definition of the mass moment of inertia ( )I r dm= Ú 2 imply

( ) ( )I Isemicylinder cylinder= 1

2

( )I m ax sc cyl= ÊËÁ

ˆ¯

1

2

1

22

and ( ) ( ) ( )I I m a Ly zsc sc cyl= = +ÈÎÍ

˘˚

1

2

1

123 2 2

However, m msc cyl= 1

2

Thus, ( )I m ax sc sc= 1

22

and ( ) ( ) ( )I I m a Ly zsc sc sc= = +1

123 2 2

Also, using the parallel axis theorem find

I m ax¢ = -ÊËÁ

ˆ¯sc

1

2

16

9 22

p

I m a Lz¢ = -ÊËÁ

ˆ¯

+ÈÎÍ

˘˚˙sc

1

4

16

9

1

1222 2

p

where ¢x and ¢z are centroidal axes.

215


SOLUTION


m V= rST

Then m137850 0 18 0 074 0 054

5 6463

= ¥ ¥ ¥=

kg m m

kg

3/ ( . . . )

.

m2

3 2 378502

0 027 0 028

0 25170

= ¥ÈÎÍ

˘˚

=

kg m m

kg

/ ( . ) ( . )

.

p ¥

m33 2 37850 0 012 0 024

0 085230

= ¥ ¥ÈÎ ˘

=

kg m m

kg

/ ( . ) ( . )

.

p ¥

m4

3 37850 0 018 0 18 0 012

0 30521

= ¥ ¥ ¥=

kg m m

kg

/ ( . . . )

.

The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations derived above (component 2).

PROBLEM 9.142

Determine the mass moment of inertia of the steel machine element shown with respect to the y axis. (The density of steel is 7850 kg/ m3.)

216



Find: Iy

We have I I I I Iy y y y y= + + -

= ÏÌÓ

+

( ) ( ) ( ) ( )

( . ( . ) ( . )

1 2 3 4

21

125 6463 0 054 0 18kg) 22

2 2

5 64630 18

2

ÈÎ ˘

+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝

m

kg)0.054

2 m

2

2( .. ÔÔ

Ô

+ +ÏÌÓ

+

1

120 25170 0 028

0 25170

2( . ( . )

( .

kg)[3(0.027) ] m

kg) (0.027

2 2

)) m2 2+ -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô0 18

0 028

2

2

..

+ ÏÌÓ

+

+

1

120 08523 0 024

0 08523

2( . ( . ) ]

( .

kg)[3(0.012) m

kg)[(0.02

2 2

77) m2 2+ +ÊËÁ

ˆ¯

¸˝Ô

Ô0 18

0 024

2

2

..

- +ÏÌÓ

+

1

120 30521 0 18

0 30521

2( . ( . ) ]

( .

kg)[(0.018) m

kg) (0.027)

2 2

22 2 m+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô

0 180

2

2.

= + + +

+ + -

[( . . ) ( . . )

( . . ) ( .

16617 1 49851 2 62 317 7119 3

7 1593 3204 1 832 311 2694 7 10 6 2+ ¥ -. )] kg m◊

= -73 334 10 3. ¥ ◊kg m2

or I y = -73 3 10 3. ¥ ◊kg m2 b

217


PROBLEM 9.143

Determine the mass moment of inertia of the steel machine element shown with respect to the z axis. (The density of steel is 7850 kg/m3.)

SOLUTION


m V= rST

Then m13 37850 0 18 0 074 0 054

5 6463

= ¥ ¥ ¥=

kg/m m

kg

( . . . )

.

m2

3 2 378502

0 027 0 028

0 25170

= ¥ÈÎÍ

˘˚

=

kg/m m

kg

p( . ) ( . )

.

¥

m33 2 37850 0 012 0 024

0 085230

= ¥ ¥ÈÎ ˘

=

kg/m m

kg

p ( . ) ( . )

.

¥

m4

3 37850 0 018 0 18 0 012

0 30521

= ¥ ¥ ¥=

kg/m m

kg

( . . . )

.

The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations derived above (component 2).

218



Find: I z

We have I I I I Iz z z z z= + + -

= +È

( ) ( ) ( ) ( )

( . ) ( . ) ( . )

1 2 3 4

2 21

125 6463 0 054 0 074kg ÎÎ ˘Ï

ÌÓ

+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

m

kg)0.054

2 m

2

2( ..

5 64630 074

2

2 2 ¸˝Ô

Ô

+ -ÊËÁ

ˆ¯

ÏÌÓ

+ +

( . ) ( . )

( .

0 25171

2

16

90 027

0 2517

22kg m

kg) (0.027)

2

2

p

00 0744 0 027

3

2

..+ ¥Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ôp m2

+ Ï

ÌÓ

+ +

1

20 08523

0 08523 0 074 2

( .

( . ( . ) ]

kg)(0.012 m)

kg)[(0.027)

2

2 mm2}

- +ÏÌÓ

+

1

20 30521 0 018 0 012

0 30521

2 2( . )[( . ) ( . ) ]

( .

kg m

kg) (0.027)

2

2 ++ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô

0 012

2

. 22 m

I z = + + +

+ + -

[( . . ) ( . . )

( . . ) ( .

3948 6 11845 9 58 693 2021 7

6 1366 528 85 11 9003 233 49 10 6+ ¥ -. )] kg m2◊

= -18 1645 10 3 2. ¥ ◊kg m

or I z = -18 16 10 3 2. ¥ ◊kg m b

219


SOLUTION


m V= rST

Then m1 7850 0 09 0 03 0 15

3 17925

= ¥ ¥=

( )( . . . )

.

kg/m m

kg

3 3

m2

278502

0 045 0 04

0 998791

= ¥ÈÎÍ

˘˚

=

( ) ( . ) .

.

kg/m m

kg

3 3p

m327850 0 038 0 03 1 06834= ¥ =( )[ ( . ) . ] . kg/m m kg3 3p

Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to Problem 9.142) for a semicylinder, we have

I I I Ix x x x= + -

= + +

( ) ( ) ( )

( . . ) ( .

1 2 3

21

123 17925 0 15 3 17 kg)(0.03 m2 2 9925 kg)(0.075 m)2È

ÎÍ˘˚

+ -ÊËÁ

ˆ¯

+ÈÎÍ

˘˚˙

Ï( . ( . ( .0 998791

1

4

16

90 045

1

120 04

2 kg) m) m)2 2

pÌÌÓ

+ + ¥ +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

( ..

.0 9987914 0 045

30 015

2

kg) (0.13) m2 2

p

¸˝Ô

Ô

- + +1

121 06834 0 03 1 06834( . ( . ] ( . kg)[3(0.038 m) m) kg)(0.06 2 2 mm)2Ï

ÌÓ

¸˝˛

PROBLEM 9.144

Determine the mass moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m .)3

220



= + + +

- +

[( . . ) ( . . )

( .

0 0061995 0 0178833 0 0002745 0 0180409

0 0004658 0.. )]0038460 kg m2◊

= + - ◊( . . . )0 0240828 0 0183154 0 0043118 kg m2

= ◊0 0380864. kg m2

or I x = ¥ ◊-38 1 10 3. kg m2 b

Now m m m m= + - = + -=

1 2 3 3 17925 0 998791 1 06834( . . . ) kg

3.10970 kg

and kI

mxx2 0 0380864

3 10970= = ◊.

.

kg m

kg

2

or k x = 110 7. mm b

221


PROBLEM 9.145

Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3.)

SOLUTION

First compute the mass of each component. we have

m V= rST

Then m13 37850 0 08 0 05 0 160

5 02400

= ¥ ¥ ¥=

kg/m m

kg

( . . . )

.

m

m

23 3

33

7850 0 08 0 038 0 07 1 67048

7850

= ¥ ¥ ¥ =

= ¥

kg/m m kg

kg/m

( . . . ) .

pp2

0 024 0 04 0 284102 3¥ ¥ÊËÁ

ˆ¯

=. . .m kg

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have

(a) I I I Ix x x x= - -

= + +

( ) ( ) ( )

( . )[( . ) ( . ) ] (

1 2 3

2 2 21

125 02400 0 05 0 16 5 kg m .. )

. .02400

0 05

2

0 16

2

2 22 kg mÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

- + + -1

121 67048 1 67048 0 05

0 03( . ( ] ( . ) .

. kg)[(0.038) 0.07) m kg2 2 2 88

20 05

0 07

2

2 22Ê

ËÁˆ¯

+ +ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô.

.m

- -ÊËÁ

ˆ¯

+ÈÎÍ

˘˚˙

ÏÌÓ( . ) ( . ) ( . )0 28410

1

4

16

90 024

1

120 04

22 2 2 kg m

p

+ - ¥ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚( . ) .

..

.0 28410 0 05

4 0 024

30 16

0 04

2

2 2

kgp

˙˙

¸˝Ô

Ôm2

= + - + - + ¥ -[( . . ) ( . . ) ( . . )]11 7645 35 2936 0 8831 13 6745 0 0493 6 0187 10 3 kkg m

kg m

2

2

◊

= - - ¥ ◊-( . . . )47 0581 14 5576 6 0680 10 3

= ¥ ◊-26 4325 10 3. kg m2 or kg m2I x = ¥ ◊-26 4 10 3. b

222



(b) I I I Iy y y y= - -

= + +

( ) ( ) ( )

( . )[( . ) ( . ) ] (

1 2 3

2 2 21

125 02400 0 08 0 16 5 kg m .. )

. .02400

0 08

2

0 16

2

2 22 kg mÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

- + + ÊËÁ

ˆ¯

1

121 67048 0 08 0 07 1 67048

0 08

22 2 2( . )[( . ) ( . ) ] ( . )

. kg m kg ˜ + +Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

2 220 05

0 07

2.

.m

- + + ÊËÁ

1

120 28410 3 0 024 0 04 0 28410

0 08

22 2 2( . )[ ( . ) ( . ) ] ( . )

. kg m kg ˆ

¯+ -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

2 220 16

0 04

2.

.m

= + - + - + ¥ -[( . . ) ( . . ) ( . . )]13 3973 40 1920 1 5730 14 7420 0 0788 6 0229 10 3 kkg m

kg m

2

2

◊

= - - ¥ ◊-( . . . )53 5893 16 3150 6 1017 10 3

= ¥ ◊-31 1726 10 3. kg m2 or kg m2I y = ¥ ◊-31 2 10 3. b

(c) I I I Iz z z z= - -

= + +

( ) ( ) ( )

( . )[( . ) ( . ) ] (

1 2 3

2 2 21

125 02400 0 08 0 05 5 kg m .. )

. .02400

0 08

2

0 05

2

2 22 kg mÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

- + + ÊËÁ

ˆ1

121 67048 0 08 0 038 1 67048

0 08

22 2 2( . )[( . ) ( . ) ] ( . )

. kg m kg

¯+ -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

2 220 05

0 038

2.

.m

- -ÊËÁ

ˆ¯

+ ÊËÁ

( . ) ( . ) ( . ).

0 284101

2

16

90 024 0 28410

0 08

222 kg m kg

pˆ¯

+ - ¥ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

2 220 05

4 0 024

3.

.

pm

= + - + - + ¥ -[( . . ) ( . . ) ( . . )]3 7261 11 1784 1 0919 4 2781 0 0523 0 9049 10 3 kg ◊◊

= - - ¥ ◊-

m

kg m

2

2( . . . )14 9045 5 3700 0 9572 10 3

= ¥ ◊-8 5773 10 3. kg m2 or kg m2I z = ¥ ◊-8 58 10 3. b

To the instructor:

The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use in the solutions of Problems 9.146 through 9.148.

Slender Rod

I I I mLx y z= = =¢ ¢01

122 (Figure 9.28)

I I mLy z= = 1

32 ( )Sample Problem 9.9

223



Circle

We have I r dm may = =Ú 2 2

Now I I Iy x z= +

And symmetry implies I Ix z=

I I max z= = 1

22

Semicircle

Following the above arguments for a circle, we have

I I ma I max z y= = =1

22 2

Using the parallel-axis theorem

I I mx xa

z z= + =¢2 2

p

or I m az¢ = -ÊËÁ

ˆ¯

1

2

42

2

p

224


SOLUTION


Have m L= r

Then m

m m m m

1

2 3 4 5

0 05

0 125664

0 05

=

=

= = =

=

( .

.

( .

kg/m)[2 (0.4 m)]

kg

kg/m)(0

p

..2m) 0.01kg=

Using the equations given above and the parallel-axis theorem, have

I I I I I Ix x x x x x= + + + +

= ÊËÁ

ˆ¯

( ) ( ) ( ) ( ) ( )

( . ) ( .

1 2 3 4 5

0 1256641

20 4kg m)) ( . ( . )2 20 01

1

30 2

ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯

ÈÎÍ

˘˚

kg) m

+ +ÈÎ ˘

+ ÊËÁ

ˆ¯

+ +

( . ( .

( . ( . ( . (

0 01 0 0 2

0 011

120 2 0 1 0

2

2 2

kg) m)

kg) m) m) ..

( . ( . ( . ( .

4

0 011

120 2 0 2 0 4

2

2 2 2

m)

kg) m) m) m 0.1m)

ÈÎÍ

˘˚

+ ÊËÁ

ˆ¯

+ + -ÈÈÎÍ

˘˚

= + + + + ¥ ◊

=

-[( . . . . . ) ]

.

10 053 0 13333 0 4 1 7333 1 3333 10

13

3 2kg m

66529 10 3 2¥ ◊- kg m

or kg m2I x = ¥ ◊-13 65 10 3. b

PROBLEM 9.146

Aluminum wire with a weight per unit length of 0.05 kg/m is used to form the circle and the straight members of the figure shown. Determine the mass moment of inertia of the assembly with respect to each of the coordinate axes.

225



Have I I I I I Iy y y y y y= + + + +( ) ( ) ( ) ( ) ( )1 2 3 4 5

where ( ) ( ) ( ) ( )I I I Iy y y y2 4 3 5= =and

Then I y = + +( . ] ( . ]0 125664 2 0 012 2kg)[(0.4 m) kg)[0 (0.4 m)

+ +ÈÎÍ

˘˚

= + +

2 0 011

120 2 0 3

20 106 2 1 6 2 0 9333

2 2( . ( . ) ( .

[ . ( . ) ( .

kg) m m)

))]

.

¥ ◊

= ¥ ◊

-

-

10

25 1726 10

3 2

3 2

kg m

kg m

or I y = ¥ ◊-25 2 10 3 2. kg m b

By symmetry I Iz x= or I z = ¥ ◊-13 65 10 3 2. kg m b

226


PROBLEM 9.147

The figure shown is formed of 3-mm steel wire. Knowing that the density of the steel is 7850 kg/m3 determine the mass moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION

Have m V AL= =r r

Then

m m1 23 27850 0 0015 0 45= = ÈÎ ˘ ¥ ¥( ) ( . ( .kg/m m) m)p p

m m2 1 0 078445= = . kg

m m3 43 27850 0 0015 0 45= = ÈÎ ˘ ¥( ) ( . ( .kg/m m) m)p

= 0 024970. kg

Using the equations given above and the parallel-axis theorem, have

I I I I Ix x x x x= + + +( ) ( ) ( ) ( )1 2 3 4

where ( ) ( )I Ix x3 4=

Then

I x = ÈÎÍ

˘˚

+ +( . ( . ( . ( . ( .0 0784451

20 45 0 078445

1

20 45 02 2kg) m) kg) m) 445

2 0 024971

120 45 0 225 0 45

2

2 2 2

m)

m) m) m)

ÈÎÍ

˘˚

+ + +ÈÎÍ

˘˚

( . ) ( . ( . ( . ˙

= + + ¥ ◊

= ¥ ◊

-

-

[ . ( . )]

.

794256 23 8277 2 6 7419 10

45 254 10

3 2

3 2

kg m

kg m

or kg mI x = ¥ ◊-45 3 10 3 2. b

Have I I I I Iy y y y y= + + +( ) ( ) ( ) ( )1 2 3 4

where ( ) ( ) ( ) ( )I I I Iy y y y1 2 3 4= =and

227



Then

I y = ÈÎ ˘ + +

=

2 0 078445 0 45 2 0 02497 0 0 45

2 15

2 2( . ( . ( . [ ( . ) ]

(

kg) m) kg) m

.. ) ( . )

.

8851 10 2 5 0564 10

41 883 10

3 2 3 2

3 2

¥ ◊ + ¥ ◊

= ¥ ◊

- -

-

kg m kg m

kg m

or kg mI y = ¥ ◊-41 9 10 3 2. b

Have

I I I I Iz z z z z= + + +( ) ( ) ( ) ( )1 2 3 4

where ( ) ( )I Iz z3 4=

Then

I z = ÈÎÍ

˘˚

+ -ÊËÁ

ˆ¯

( . ( .

( . (

0 0784451

20 45

0 0784451

2

40

2

2

kg) m)

kg)p

.. ).

( .

( . ) (

452 0 45

0 45

2 0 024971

30

22

2mm

m)

kg

+¥Ê

ËÁˆ¯

+È

ÎÍÍ

˘

˚˙˙

+ +

p

..

[ . . ( . )]

.

45

7 94256 23 8277 2 1 6855 10

35 141

2

3 2

m)

kg m

ÈÎÍ

˘˚

= + + ¥ ◊

=

-

¥¥ ◊-10 3 2kg m

or kg mI z = ¥ ◊-35 1 10 3 2. b

228


SOLUTION

First compute the mass m of each component. We have

m m L L== ¥=

( )

. .

.

/

kg/m m

kg

0 056 1 2

0 0672

Using the equations given above and the parallel-axis theorem, we have

I I I I I I Ix x x x x x x= + + + + +( ) ( ) ( ) ( ) ( ) ( )1 2 3 4 5 6

Now ( ) ( ) ( ) ( ) ( ) ( )I I I I I Ix x x x x x1 3 4 6 2 5= = = =and

Then I x = ÈÎÍ

˘˚

+ +

=

41

30 0672 1 2 2 0 0 0672 1 22 2( . )( . ) [ ( . )( . ) ]

[

kg m kg m

44 0 03226 2 0 09677( . ) ( . )]+ ◊ kg m2

= ◊0 32258. kg m2 or I x = ◊0 323. kg m2 b

I I I I I I Iy y y y y y y= + + + + +( ) ( ) ( ) ( ) ( ) ( )1 2 3 4 5 6

Now ( ) , ( ) ( ) , ( ) ( )I I I I Iy y y y y1 2 6 4 50= = =and

Then I y = ÈÎÍ

˘˚

+ +

+

21

30 0672 1 2 0 0 0672 1 2

21

2 2( . )( . ) [ ( . )( . ) ] kg m kg m

1120 0672 1 2 0 0672 1 2 0 6

2 0

2 2 2 2( . )( . ) ( . )( . . )

[ (

kg m kg m+ +ÈÎÍ

˘˚

= .. ) ( . ) ( . . )]

[ ( . ) (

03226 0 09677 2 0 00806 0 12096

2 0 03226

+ + + ◊

= +

kg m2

00 09677 2 0 12902. ) ( . )]+ ◊ kg m2

= ◊0 41933. kg m2 or I y = ◊0 419. kg m2 b

Symmetry implies I Iy z= I z = ◊0 419. kg m2 b

PROBLEM 9.148

A homogeneous wire with a mass per unit length of 0.056 kg/m is used to form the figure shown. Determine the mass moment of inertia of the wire with respect to each of the coordinate axes.

229


SOLUTION


m V= r

Then m

m

13

2

7850 0 08 0 05 0 16 5 02400

7850

= ¥ ¥ ¥ =

= ¥

kg/m m kg

kg/m

3

3

( . . . ) .

(00 08 0 038 0 07 1 67048

78502

0 024 0 04

3

32

. . . ) .

. .

¥ ¥ =

= ¥ ¥ ¥

m kg

kg/m3mpÊÊ

ËÁˆ¯

=m kg3 0 28410.

Now observe that the centroidal products of inertia, I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢, , ,and of each component are zero because of symmetry. Now

y

y

2

3

0 050 038

20 031

0 054 0 024

30 03

= -ÊËÁ

ˆ¯

=

= - ¥ÊËÁ

ˆ¯

=

..

.

..

.

m m

mp

99814 m

and then

m, kg x , m y, m z , m mx y, kg m2◊ my z , kg m2◊ mz x , kg m2◊1 5.02400 0.04 0.025 0.08 5 0240 10 3. ¥ - 10 0480 10 3. ¥ - 16 0768 10 3. ¥ -

2 1.67048 0.04 0.031 0.085 2 0714 10 3. ¥ - 4 4017 10 3. ¥ - 5 6796 10 3. ¥ -

3 0.28410 0.04 0.039814 0.14 0 4524 10 3. ¥ - 1 5836 10 3. ¥ - 1 5910 10 3. ¥ -

Finally,

I I I Ixy xy xy xy= - - = + - + - +( ) ( ) ( ) [( . ) ( . ) ( . )1 2 3 0 5 0240 0 2 0714 0 0 4524 ]] ¥ -10 3

= ¥ ◊-2 5002 10 3. kg m2 or I xy = ¥ ◊-2 50 10 3. kg m2 b

PROBLEM 9.149

Determine the mass products of inertia Ixy, Iyz, and Izx of the steel fixture shown. (The density of steel is 7850 kg/m3.)

230



I I I Iyz yz yz yz= - - = + - + - +( ) ( ) ( ) [( . ) ( . ) ( .1 2 3 0 10 0480 0 4 4017 0 1 5836))] ¥ -10 3

= ¥ ◊-4 0627 10 3. kg m2

or I yz = ¥ ◊-4 06 10 3. kg m2 b

I I I Izx zx zx zx= - - = + - + - +( ) ( ) ( ) [( . ) ( . ) ( .1 2 3 0 16 0768 0 5 6796 0 1 5910))] ¥ -10 3

= ¥ ◊-8 8062 10 3. kg m2

or I zx = ¥ ◊-8 81 10 3. kg m2 b

231


SOLUTION


m V= rST

Then m

m

13

2

7850 0 1 0 018 0 09 1 27170

7850

= ¥ ¥ ¥ =

= ¥

kg/m m kg

kg/m

3

3

( . . . ) .

(00 016 0 06 0 05 0 37680

7850 0 012 0 01

3

32

. . . ) .

( . .

¥ ¥ =

= ¥ ¥ ¥

m kg

kg/m3m p )) .m kg3 0 03551=


m, kg x , m y, m z , m mx y, kg m2◊ my z , kg m2◊ mz x , kg m2◊

1 1.27170 0.05 0.009 0.045 0 57227 10 3. ¥ - 0 51504 10 3. ¥ - 2 86133 10 3. ¥ -

2 0.37680 0.092 0.048 0.045 1 66395 10 3. ¥ - 0 81389 10 3. ¥ - 1 55995 10 3. ¥ -

3 0.03551 0.105 0.054 0.045 0 20134 10 3. ¥ - 0 08629 10 3. ¥ - 0 16778 10 3. ¥ -

Â 2 43756 10 3. ¥ - 1 41522 10 3. ¥ - 4 58906 10 3. ¥ -

Then I I mx yxy x y= +¢ ¢S( ) or I xy = ¥ ◊-2 44 10 3. kg m2 b

I I my zyz y z= +¢ ¢S ( ) or I yz = ¥ ◊-1 415 10 3. kg m2 b

I I mzxzx z x= +¢ ¢S( ) or I zx = ¥ ◊-4 59 10 3. kg m2 b

PROBLEM 9.150

Determine the mass products of inertia Ixy, Iyz, and Izx of the steel machine element shown. (The density of steel is 7850 kg/m3.)

0

0

0

232


SOLUTION


m V= rAL

Then m

m

1 3

3

2

2770 0 156 0 016 0 032

221 245 10

2770

= ¥ ¥ ¥

= ¥

=

-

kg

mm

kg

kg

m

3( . . . )

.

33

3

0 048 0 016 0 04

85 0944 10

¥ ¥ ¥

= ¥ -

( . . . )

.

m

kg

3

m

m

3 32 3

4

27702

0 016 0 016 17 8221 10

277

= ¥ ¥ ¥ÈÎÍ

˘˚

= ¥

=

-kg

mm kg3p

( . ) . .

00 0 011 0 012 12 6356 103

2 3kg

mm kg3¥ ¥ ¥ÈÎ ˘ = ¥ -p ( . ) . .


x3 156

4 16

3162 791

0 162791

= - + ¥ÊËÁ

ˆ¯

= -

= -p

mm mm

m

.

.

PROBLEM 9.151

Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The specific weight of aluminum is 2770 kg/m3)

233



and then

m, kg x , m y, m z , m mx y, kg m2◊ my z , kg m2◊ mz x , kg m2◊

1 221 245 10 3. ¥ - -0 078. 0 008. -0 016. - ¥ -138 057 10 6. - ¥ -28 3194 10 6. 276 114 10 6. ¥ -

2 85 0944 10 3. ¥ - -0 024. 0 008. -0 052. - ¥ -16 3381 10 6. - ¥ -35 3993 10 6. 106 198 10 6. ¥ -

3 17 8221 10 3. ¥ - -0 1628. 0 008. -0 016. - ¥ -23 2115 10 6. - ¥ -2 28123 10 6. 46 423 10 6. ¥ -

4 12 6356 10 3. ¥ - -0 156. 0 022. -0 016. - ¥ -43 3654 10 6. - ¥ -4 44773 10 6. 31 5385 10 6. ¥ -

Â - ¥ -220 972 10 6. - ¥ -70 448 10 6. 460 274 10 6. ¥ -

Then I I mx yxy x y= +¢ ¢S( ) or I xy = - ¥ ◊-221 10 6 kg m2

I I my zyz y z= +¢ ¢S( ) or I yz = - ¥ ◊-70 4 10 6. kg m2

I I mz xzx z x= +¢ ¢S( ) or I zx = ¥ ◊-460 10 6 kg m2

0

0

0

234


PROBLEM 9.152

Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The density of aluminum is 2770 kg/m3.)

SOLUTION

Have m V= r

Then m

m

1

2

2770 0 118 0 036 0 044 0 51775

2770

= ¥ ¥ =

=

( )( . . . ) .

(

kg/m m kg

kg/m

3 3

3)) ( . ) . .

( )( .

p2

0 022 0 036 0 075814

2770 0 028

2

3

¥ÈÎÍ

˘˚

=

= ¥

m kg

kg/m

3

3m 00 022 0 024 0 04095. . ) .¥ =m kg3

Now observe that I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢, , and are zero because of symmetry

Now

x

y

2

3

0 1184 0 023

30 12734

0 0360 062

2

= - + ¥ÊËÁ

ˆ¯

= -

= -ÊËÁ

ˆ¯

..

.

..

pm m

m m= 0 025.

m kg, x , m y, m z , m mx y, kg m2◊ my z , kg m2◊ mz x , kg m2◊1 0 51775. -0 059. 0 018. 0 022. - ¥ -0 54985 10 3. 0 20503 10 3. ¥ - - ¥ -0 67204 10 3.

2 0 075814. -0 12734. 0 018. 0 022. - ¥ -0 17377 10 3. 0 03002 10 3. ¥ - - ¥ -0 21239 10 3.

3 0 04095. kg -0 041. 0 025. 0 026. - ¥ -0 014198 10 3. 0 02662 10 3. ¥ - - ¥ -0 04365 10 3.

235



And

I I mx y

I I my z

I I mx z

xy x y

yz y z

zx z x

= +

= +

= +

¢ ¢

¢ ¢

¢ ¢

S

S

S

( )

( )

( )

Finally, I I I Ixy xy xy xy= + - = - ¥ ◊-( ) ( ) ( ) .1 2 330 70942 10 kg m2

or I xy = - ¥ ◊-0 709 10 3. kg m2

I I I Iyz yz yz yz= + - = ¥ ◊-( ) ( ) ( ) .1 2 330 20843 10 kg m2

or I yz = ¥ ◊-0 208 10 3. kg m2

I I I Izx zx zx zx= + - = - ¥ ◊-( ) ( ) ( ) .1 2 330 84078 10 kg m2

or I zx = - ¥ ◊-0 841 10 3. kg m2

0

0

0

236


PROBLEM 9.153

A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the component.

SOLUTION

m V

m V

1

2

7850 0 2 0 3 0 002 0 942

7850

= = =

= =

r

r

kg/m m m m kg

kg/m

3

3

( . )( . )( . ) .

(( . )( . )( . ) .0 4 0 3 0 002 1 884m m m kg=

For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes.

m, kg x , m y, m z , m mx y my z mz x

0 942. 0.15 0.1 0.4 + ¥ -14 13 10 3. + ¥ -37 68 10 3. + ¥ -56 52 10 3.

1 884. 0.15 0 0.2 0 0 + ¥ -56 52 10 3.

Â + ¥ -14 13 10 3. + ¥ -37 68 10 3. + ¥ -113 02 10 3.

I xy = + ¥ ◊-14 13 10 3. kg m2 I xy = + ◊14 13. g m2

I yz = + ¥ ◊-37 68 10 3. kg m2 I yz = + ◊37 7. g m2

I zx = + ¥ ◊-113 02 10 3. kg m2 I zx = + ◊113 0. g m2

237


PROBLEM 9.154

A section of sheet steel 2-mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the component.

SOLUTION


We have m V tA= =r rST ST

Then m

m

13

2

7850 0 002 0 4 0 45

2 8260

7850

==

=

( )[( . )( . )( . )]

.

(

kg/m m

kg

kg/m

3

3 00 0021

20 45 0 18

0 63585

3. ) . .

.

¥ ¥ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

m

kgNow observe that

( ) ( ) ( )

( ) ( )

I I I

I I

x y y z z x

y z z x

¢ ¢ ¢ ¢ ¢ ¢

¢ ¢ ¢ ¢

= = =

= =1 1 1

2 2

0

0

From Sample Problem 9.6: ( )I b hx y¢ ¢ = -2,area1

72 22

22

Then ( ) ( ) ,I t I t b h m b hx y x y¢ ¢ ¢ ¢= = -ÊËÁ

ˆ¯

= -2 2 22

22

2 2 21

72

1

36r rST area ST

Also x y z x1 1 2 20 0 2250 45

30 075= = = = - +Ê

ËÁˆ¯

= -..

.m m

238



Finally, I I mx yxy xy= + = + + -ÈÎÍ

+

S( ) ( ) ( . )( . )( . )

( .

0 01

360 63585 0 45 0 18

0 6

kg m m

33585 0 0750 18

3

1 43066 10 2 8613 103

kg mm

)( . ).

( . .

- ÊËÁ

ˆ¯

˘˚

= - ¥ - ¥- -33 2) kg m◊

or I xy = - ¥ ◊-4 29 10 3 2. kg m

and I I m y zyz y z= + = + + + =¢ ¢S( ) ( ) ( )0 0 0 0 0

or I yz = 0

I I m z xzx z x= + = + + + =¢ ¢S( ) ( ) ( )0 0 0 0 0

or I zx = 0

239


PROBLEM 9.155


SOLUTION


m V tA= =r rST ST

Then m

m

1

2

7850 0 002 0 35 0 39

2 14305

7850

= ¥=

=

( )( . )( . . )

.

(

kg/m m m

kg

kg/m

3 2

3))( . ) .

.

( )( .

0 0022

0 195

0 93775

7850 0 002

2 2

3

m m

kg

kg/m m3

p ¥ÊËÁ

ˆ¯

=

=m )) . .

.

1

20 39 0 15

0 45923

¥ ¥ÊËÁ

ˆ¯

=

m

kg

2

Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero and

( ) ( )I Ix y z x¢ ¢ ¢ ¢= =3 3 0

Also ( ) ( ), ,I t Iy z y z¢ ¢ ¢ ¢=3 3mass ST arear

Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds toa 90° rotation, we have

( ) ,I b hy z¢ ¢ =3 32

321

72area

Then ( )I t b h m b hy z¢ ¢ = ÊËÁ

ˆ¯

=3 32

32

3 3 31

72

1

36rST

Also y x

y

1 2

2

0

4 0 195

30 082761

= =

= ¥ =..

pm m

240



Finally,

I I mx yxy x y= +

= + + + + + -ÊË

¢ ¢S( )

( ) ( ) ( . )( . ).

0 0 0 0 0 0 45923 0 350 15

3kg m mÁÁ

ˆ¯

ÈÎÍ

˘˚

= - ¥ ◊-8 0365 10 3. kg m2 or kg m2I xy = - ¥ ◊-8 04 10 3.

I I my zyz y z= +

= + + +

+

¢ ¢S( )

( ) [ ( . )( . )( . )]0 0 0 0 93775 0 082761 0 195kg m m

11

360 45923 0 39 0 15 0 45923

0 15

3

0 39( . )( . )( . ) ( . )

. .kg m m kg m+ -Ê

ËÁˆ¯ 33

15 1338 0 7462 2 9850 10

15

3

m

kg m2

ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= + - ¥ ◊

=

-[( . ) ( . . )]

( .11338 2 2388 10 3- ¥ ◊-. ) kg m2

= ¥ ◊-12 8950 10 3. kg m2 or kg m2I yz = ¥ ◊-12 90 10 3.

I I mz xzx z x= += + + +

+ +

¢ ¢S( )

[ ( . )( . )( . )] ( )

(

0 2 14305 0 175 0 195 0 0

0

kg m m

00 459230 39

30 35

73 1316 20 8950 10

. ).

( . )

( . . )

kg m mÊËÁ

ˆ¯

ÈÎÍ

˘˚

= + ¥ -33 kg m2◊

= ¥ ◊-94 0266 10 3. kg m2 or kg m2I zx = ¥ ◊-94 0 10 3.

241


PROBLEM 9.156


SOLUTION


m V tA= =r rST ST

Then

m m

m

1 3 7850 0 0021

20 225 0 135 0 23844= = ¥ ¥Ê

ËÁˆ¯

=( )( . ) . . .kg/m m m kg3 2

222 2

4

7850 0 0024

0 225 0 62424

7850

= ¥ÊËÁ

ˆ¯

=

=

( )( . ) . .

(

kg/m m m kg

k

3 p

m gg/m m m kg3)( . ) . .0 0024

0 135 0 224732 2p ¥ÊËÁ

ˆ¯

=

Now observe that the following centroidal products of inertia are zero because of symmetry.

( ) ( ) ( ) ( )

( ) ( ) (

I I I I

I I

x y y z x y y z

x y z x

¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢

¢ ¢ ¢ ¢

= = = =

= =1 1 2 2

3 3

0 0

0 II Ix y z x¢ ¢ ¢ ¢= =) ( )4 0

Also y y1 2 0= = x x3 4 0= =

Now I I mx yxy x y= +¢ ¢S( )

so that I xy = 0

Using the results of Sample Problem 9.6, we have

I b huv, area = 1

242 2

Now I t I

t b h

mbh

uv uv, mass ST ,area

ST

=

= ÊËÁ

ˆ¯

=

r

r 1

24

1

12

2 2

Thus, ( )I m b hzx 1 1 1 11

12=

242



while ( )I m b hyz 3 3 3 31

12= -

because of a 90° rotation of the coordinate axes.

To determine Iuv for a quarter circle, we have

dI d I u v dmuv u v= +¢ ¢ EL EL

where u u v v a u

dm t dA tv du t a u du

EL EL

ST ST ST

= = = -

= = = -

1

2

1

22 2

2 2r r r

Then I dI u a u t a u du

t u a u d

uv uv

a= = -Ê

ËÁˆ¯ -( )

= -

ÚÚ ( )

( )

1

2

1

2

2 2 2 20

2 2

r

r

ST

ST uu

t a u u t a ma

a

a

0

2 2 4

0

4 41

2

1

2

1

4

1

8

1

2

Ú

= -ÈÎÍ

˘˚

= =r rpST ST

Thus ( )I m azx 2 2 221

2= -

p

because of a 90° rotation of the coordinate axes. Also

( )I m ayz 4 4 421

2=

p

Finally,

I I I m y z I m y z I Iyz yz y z y z yz yz= = + + + + +¢ ¢ ¢ ¢S( ) [( ) ] [( ) ] ( ) ( )1 1 1 2 2 2 3 4

== -ÈÎÍ

˘˚

+1

120 23844 0 225 0 135

1

20 22473 0 1( . )( . )( . ) ( . )( .kg m m kg

p335

0 60355 0 65185 10 3

m

kg m

2

2

)

( . . )

ÈÎÍ

˘˚

= - + ¥ ◊-

or I yz = ¥ ◊-48 3 10 6. kg m2

I I I I I m z x I m z xzx zx zx zx z x z x= = + + + + +¢ ¢ ¢ ¢S( ) ( ) ( ) [( ) ] [( )1 2 3 3 3 3 4 4 4 44

1

120 23844 0 225 0 135

1

20 62424 0

]

( . )( . )( . ) ( . )(= ÈÎÍ

˘˚

+ -kg m m kgp

.. )

( . . )

225

0 60355 5 02964 10

2

3

m

kg m2

ÈÎÍ

˘˚

= - ¥ ◊-

or I zx = - ¥ ◊-4 43 10 3. kg m2

0 0

0 0

0

243


PROBLEM 9.157

Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION


mW

g gwL= = 1

Then mW

ga

w

ga

mw

ga

w

ga

mw

ga

w

ga

mw

ga

1

2

3

4

2 2

2 2

23

2

= ¥ =

= =

= =

= ¥ÊËÁ

ˆ

( )

( )

( )

p p

p¯

= 3p w

ga

Now observe that the centroidal products of inertia, I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢, , ,and of each component are zero because of symmetry.

m x y z mx y my z mz x

1 2p w

ga 2a a -a 4 3p w

ga -2 3p w

ga - 4 3p w

ga

2w

ga 2a

1

2a 0

w

ga3 0 0

3 2w

ga 2a 0 a 0 0 4 3w

ga

4 3p w

ga 2a - 3

2a 2a -9 3p w

ga -9 3p w

ga 12 3p w

ga

Âw

ga( )1 5 3- p -11 3p w

ga 4 1 2 3w

ga( )+ p

244



Then I I mx yxy x y= +¢ ¢S( ) or Iw

gaxy = -3 1 5( )p

I I my zyz y z= +¢ ¢S( ) or Iw

gayz = -11 3p

I I mz xzx z x= +¢ ¢S( ) or Iw

gazx = +4 1 23( )p

0

0

0

245


PROBLEM 9.158

Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION


mW

g gwL= = 1

Then mW

ga

w

ga

mw

ga

w

ga

mw

ga

w

ga

1

2

3

3

2

3

2

3 3

= ¥ÊËÁ

ˆ¯

=

= =

= ¥ =

p p

p p

( )

( )


m x y z mx y my z mz x

13

2p w

ga - Ê

ËÁˆ¯

2 3

2pa 2a

1

2a -9 3w

ga

3

23p w

ga - 9

43w

ga

2 3w

ga 0

1

2a 2a 0 3 3w

ga 0

3 p w

ga

2

p( )a -a a -2 3w

ga -p w

ga3 2 3w

ga

Â -11 3w

ga

w

ga

p2

3 3+ÊËÁ

ˆ¯

- 1

43w

ga

246



Then I I mx yxy x y= +¢ ¢S( ) or Iw

gaxy = -11 3

I I my zyz y z= +¢ ¢S( ) or Iw

ga byz = +1

23( )p

I I mz xzx z x= +¢ ¢S( ) or Iw

gazx = - 1

43

0

0

0

247


PROBLEM 9.159

The figure shown is formed of 1.5-mm-diameter aluminum wire. Knowing that the density of aluminum is 2800 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION


m V AL= =r rAL AL

Thenm m

m

1 42

3

28004

0 0015 0 25

1 23700 10

= = ÈÎÍ

˘˚

= ¥ -

( ) ( . ) ( . )

.

kg/m m m

kg

3 p

22 52

3

28004

0 0015 0 18

0 89064 10

= = ÈÎÍ

˘˚

= ¥ -

m

m

( ) ( . ) ( . )

.

kg/m m m

kg

3 p

33 62

3

28004

0 0015 0 3

1 48440 10

= = ÈÎÍ

˘˚

= ¥ -

m ( ) ( . ) ( . )

.

kg/m m m

kg

3 p

248




m, kg x , m y, m z , m mx y, kg m2◊ my z , kg m2◊ mz x , kg m2◊1 1 23700 10 3. ¥ - 0.18 0.125 0 27 8325 10 6. ¥ - 0 0

2 0 89064 10 3. ¥ - 0.09 0.25 0 20 0394 10 6. ¥ - 0 0

3 1 48440 10 3. ¥ - 0 0.25 0.15 0 55 6650 10 6. ¥ - 0

4 1 23700 10 3. ¥ - 0 0.125 0.3 0 46 3875 10 6. ¥ - 0

5 0 89064 10 3. ¥ - 0.09 0 0.3 0 0 24 0473 10 6. ¥ -

6 1 48440 10 3. ¥ - 0.18 0 0.15 0 0 40 0788 10 6. ¥ -

Â 47 8719 10 6. ¥ - 102 0525 10 6. ¥ - 64 1261 10 6. ¥ -

Then I I mx yxy x y= +¢ ¢S( ) or I xy = ¥ ◊-47 9 10 6. kg m2

I I my zyz y z= +¢ ¢S( ) or I yz = ¥ ◊-102 1 10 6. kg m2

I I mz xzx z x= +¢ ¢S( ) or I zx = ¥ ◊-64 1 10 6. kg m2

0

0

0

249


PROBLEM 9.160

Thin aluminum wire of uniform diameter is used to form the figure shown. Denoting by m¢ the mass per unit length of the wire, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION


mm

LL m L= Ê

ËÁˆ¯

= ¢

Then m m m R m R

m m m R R

m m R

1 5 1 1

2 4 2 1

3 2

2 2

2 2

= = ¢ ÊËÁˆ¯

= ¢

= = ¢ -

= ¢ ÊËÁˆ¯

=

p p

p p( )

¢¢m R2

Now observe that because of symmetry the centroidal products of inertia, I Ix y y z¢ ¢ ¢ ¢, , and I z x¢ ¢ , of components 2 and 4 are zero and

( ) ( ) ( ) ( )

( ) ( )

I I I I

I I

x y z x x y y z

y z z x

¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢

¢ ¢ ¢ ¢

= = = =

= =1 1 3 3

5 5

0 0

0

Also x x y y y z z1 2 2 3 4 4 50 0 0= = = = = = =

Using the parallel-axis theorem [Equations (9.47)], it follows that I I Ixy yz zx= = for components 2 and 4.

To determine Iuv for one quarter of a circular arc, we have dI uvdmuv =

where u a v a= =cos sinq q

and dm dV A ad= =r r q[ ( )]

where A is the cross-sectional area of the wire. Now

m m a A a= ¢ ÊËÁˆ¯

= ÊËÁ

ˆ¯

p r p2 2

so that dm m ad= ¢ q

and dI a a m ad

m a d

uv = ¢

= ¢

( cos )( sin )( )

sin cos

q q q

q q q3

250



Then I dI m a d

m a m a

uv uv= = ¢

= ¢ ÈÎÍ

˘˚

= ¢

ÚÚ 3

0

2

3 2

0

231

2

1

2

sin cos

sin

q q q

q

p

p

/

/

Thus, ( )I m Ryz 1 131

2= ¢

and ( ) ( )I m R I m Rzx xy3 23

5 131

2

1

2= - ¢ = - ¢

because of 90∞ rotations of the coordinate axes. Finally,

I I I m x y I m x y Ixy xy x y x y xy= = + + + +¢ ¢ ¢ ¢S( ) [( ) ] [( ) ] ( )1 1 1 1 3 3 3 3 5

or I m Rxy = - ¢1

2 13

I I I I m y z I m y zyz yz yz y z y z= = + + + +¢ ¢ ¢ ¢S( ) ( ) [( ) ] [( ) ]1 3 3 3 3 5 5 5 5

or I m Ryz = ¢1

2 13

I I I m z x I I m z xzx zx z x zx z x= = + + + +¢ ¢ ¢ ¢S( ) [( ) ] ( ) [( ) ]1 1 1 1 3 5 5 5 5

or I m Rzx = - ¢1

2 23

0 0

00

00

251


PROBLEM 9.161

Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of inertia.

SOLUTION

We have I xydm I yzdm I zxdmxy yz zx= = =Ú Ú Ú (9.45)

and x x x y y y z z z= ¢ + = ¢ + = ¢ + (9.31)

Consider I xydmxy = ÚSubstituting for x and for y

I x x y y dm

x y dm y x dm x y dm x y dm

xy = ¢ + ¢ +

= ¢ ¢ + ¢ + ¢ +

ÚÚ Ú Ú Ú

( )( )

By definition I x y dmx y¢ ¢ = ¢ ¢Úand ¢ = ¢

¢ = ¢

ÚÚ

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 I mx yxy x y= +¢ ¢ Q.E.D.

The expressions for I yz and I zx are obtained in a similar manner.

252


PROBLEM 9.162

For the homogeneous tetrahedron of mass m shown, (a) determine by direct integration the mass product of inertia Izx, (b) deduce Iyz and Ixy from the result obtained in Part a.

SOLUTION

(a) First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown.

Now xa

cz a a

z

c= - + = -Ê

ËÁˆ¯

1

and yb

cz b b

z

c= - + = -Ê

ËÁˆ¯

1

The mass dm of the slab is

dm dV xydz abz

cdz= = Ê

ËÁˆ¯

= -ÊËÁ

ˆ¯

r r r1

2

1

21

2

Then m dm abz

cdz

abc z

c

c= = -Ê

ËÁˆ¯

= -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

È

ÎÍÍ

˘

Ú Ú1

21

1

2 31

2

0

3

r

r˚˙˙

=0

1

6

c

abcr

Now dI dI z x dmzx z x= +¢ ¢ EL EL

where dIz x¢ ¢ = 0 ( )symmetry

and z z x x az

cEL EL= = = -ÊËÁ

ˆ¯

1

3

1

31

Then I dI z az

cab

z

cdzzx zx

c= = -Ê

ËÁˆ¯

ÈÎÍ

˘˚

-ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙Ú Ú

1

31

1

21

0

2

r˙

= - + -Ê

ËÁˆ

¯

= - + -

Ú1

63 3

1

2

3

4

1

22 3

2

4

30

23 4

2

ra b zz

c

z

c

z

cdz

m

ca z

z

c

z

c

c

55

5

30

z

c

cÈ

ÎÍ

˘

˚˙

or I maczx = 1

20

253



(b) Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular permutation of ( , , )x y z and ( , , )a b c .

Thus,

I mabxy = 1

20

I mbcyz = 1

20

Alternative solution for Part a:

First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown.

Now xa

by a a

y

b= - + = -Ê

ËÁˆ¯

1

and zc

by c c

y

b= - + = -Ê

ËÁˆ¯

1

The mass dm of the slab is

dm dV xzdy acy

bdy= = Ê

ËÁˆ¯

= -ÊËÁ

ˆ¯

r r r1

2

1

21

2

Now dI tdIzx zx= r , area

where t dy=

and dI x zzx, area = 1

242 2 from the results of Sample Problem 9.6.

Then dIzx dy ay

bc

y

b= -Ê

ËÁˆ¯

È

ÎÍ

˘

˚˙ -Ê

ËÁˆ¯

ÈÎÍ

˘˚

ÏÌÔ

ÓÔ

¸˝Ô

˛r( )

1

241 1

2 2

ÔÔ

= -ÊËÁ

ˆ¯ = -Ê

ËÁˆ¯

1

241

1

412 2

4 4

ra cy

bdy

m

bac

y

bdy

Finally, I dIm

bac

y

bdyzx zx

b= = -Ê

ËÁˆ¯ÚÚ

1

41

4

0

= -ÊËÁ

ˆ¯ -Ê

ËÁˆ¯

È

ÎÍ

˘

˚˙

1

4 51

5

0

m

bac

b y

b

b

or I maczx = 1

20

254



Alternative solution for Part a:

The equation of the included face of the tetrahedron is

x

a

y

b

z

c+ + = 1

so that y bx

a

z

c= - -Ê

ËÁˆ¯

1

For an infinitesimal element of sides dx dy, , and dz :

dm dV dydxdz= =r r

From Part a x az

c= -Ê

ËÁˆ¯

1

Now

I zxdm zx dydxdzzx

b x a z ca z cc= =Ú ÚÚÚ

- --( )

( )( )r

0

1

0

1

0

/ //

= - -( )ÈÎ ˘

= - -È

Î

-ÚÚr

r

zx b dxdz

b z xx

a

z

cx

xa

zc

a z cc1

1

2

1

3

1

2

0

1

0

23

2

( )/

ÍÍ˘

˚˙

= -ÊËÁ

ˆ¯

- -ÊËÁ

ˆ¯

-

-

Ú0

1

0

22

331

21

1

31

1

2

a z cc

dz

b z az

c aa

z

c

( )/

r zz

ca

z

cdz

b a zz

cdz

a

c

c

22

0

23

0

1

1

61

1

6

-ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= -ÊËÁ

ˆ¯

=

Ú

Úr

r 222 3

2

4

30

23 4

2

5

3

3 3

1

2

3

4

1

5

b zz

c

z

c

z

cdz

m

ca z

z

c

z

c

z

c

c- + -

Ê

ËÁˆ

¯

= - + -È

Ú

ÎÎÍ

˘

˚˙

0

c

or I maczx = 1

20

255


PROBLEM 9.163

The homogeneous circular cylinder shown has a mass m. Determine the mass moment of inertia of the cylinder with respect to the line joining the origin O and Point A that is located on the perimeter of the top surface of the cylinder.

SOLUTION

From Figure 9.28: I may = 1

22

and using the parallel-axis theorem

I I m a h mh

m a hx z= = + + ÊËÁ

ˆ¯

= +1

123

2

1

123 42 2

22 2( ) ( )

Symmetry implies I I Ixy yz zx= = = 0

For convenience, let Point A lie in the yz plane. Then

lOAh a

h a=+

+12 2

( )j k

With the mass products of inertia equal to zero, Equation (9.46) reduces to

I I I IOA x x y y z z= + +l l l2 2 2

=+

Ê

ËÁ

ˆ

¯˜ + +

+

Ê

ËÁ

ˆ

¯˜

1

2

1

123 42

2 2

2

2 2

2 2

2

mah

h am a h

a

h a( )

or I mah a

h aOA = +

+1

12

10 322 2

2 2

Note: For Point A located at an arbitrary point on the perimeter of the top surface,lOA is given by

l f fOAh a

a h a=+

+ +12 2

( cos sin )i j k

which results in the same expression for IOA.

0

256


PROBLEM 9.164

The homogeneous circular cone shown has a mass m. Determine the mass moment of inertia of the cone with respect to the line joining the origin O and Point A.

SOLUTION

First note that d a a a aOA = ÊËÁ

ˆ¯

+ - + =3

23 3

9

2

22 2( ) ( )

Then lOA aa a a= - +Ê

ËÁˆ¯

= - +1 3

23 3

1

32 2

92

i j k i j k( )

For a rectangular coordinate system with origin at Point A and axes aligned with the given x y z, , axes,we have (using Figure 9.28)

I I m a a I max z y= = +ÈÎÍ

˘˚

=3

5

1

43

3

102 2 2( )

= 111

202ma

Also, symmetry implies I I Ixy yz zx= = = 0

With the mass products of inertia equal to zero, Equation (9.46) reduces to

I I I IOA x x y y z z= + +l l l2 2 2

= ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

111

20

1

3

3

10

2

3

111

20

2

32

22

22

2

ma ma ma

= 193

602ma or I maOA = 3 22 2.

257


PROBLEM 9.165

Shown is the machine element of Problem 9.141. Determine its mass moment of inertia with respect to the line joining the origin O and Point A.

SOLUTION


We have m V= rST

Then

m

m

13 2

2

7850 0 08 0 04 6 31334

7850

= =

=

( )[ ( . ) ( . )] .

( )[

kg/m m m kg

kg/m3

p

p (( . ) ( . )] .

( )[ ( . ) ( . )

0 02 0 06 0 59188

7850 0 02 0 04

2

2

m m kg

kg/m m m33

=

=m p ]] .= 0 39458 kg

Symmetry implies I I Iyz zx xy= = =0 01( )

and ( ) ( )I Ix y x y¢ ¢ ¢ ¢= =2 3 0

Now I I mx y m x y m x yxy x y= + = -

= -¢ ¢S( )

[ . ]2 2 2 3 3 3

0 59188 kg (0.04 m)(0.03 m) [[ . . ]

( . . )

0 39458 0 04

0 71026 0 31566 10 3

kg ( m)( 0.02 m)

kg m2

- -

= - ¥ ◊

=

-

00 39460 10 3 2. ¥ ◊- kg m

258



From the solution to Problem 9.141, we have

I

I

I

x

y

z

= ¥ ◊

= ¥ ◊

= ¥

-

-

-

13 98800 10

20 55783 10

14 30368 10

3

3

3

.

.

.

kg m

kg m

2

2

kkg m2◊

By observation lOA = +1

132 3( )i j

Substituting into Eq. (9.46)

I I I I I I IOA x x y y z z xy x y yz y z zx z x= + + - - -l l l l l l l l l2 2 2 2 2 2

= Ê

ËÁˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

-

( . ) ( . )

( . )

13 988002

1320 55783

3

13

2 0 394602

2 2

113

2

1310

4 30400 14 23234 0 36425

3ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

˘

˚˙ ¥ ◊

= + -

- kg m2

( . . . ) ¥¥ ◊-10 3 kg m2

or kg m2IOA = ¥ ◊-18 17 10 3.

0 0 0

259


PROBLEM 9.166

Determine the mass moment of inertia of the steel fixture of Problems 9.145 and 9.149 with respect to the axis through the origin that forms equal angles with the x, y, and z axes.

SOLUTION

From the solutions to Problems 9.145 and 9.149, we have

Problem 9.145: I

I

I

x

y

z

= ¥ ◊

= ¥ ◊

= ¥ ◊

-

-

-

26 4325 10

31 1726 10

8 5773 10

3

3

3

.

.

.

kg m

kg m

kg m

2

2

22

Problem 9.149: I

I

I

xy

yz

zx

= ¥ ◊

= ¥ ◊

= ¥ ◊

-

-

-

2 5002 10

4 0627 10

8 8062 10

3

3

3

.

.

.

kg m

kg m

kg

2

2

mm2

From the problem statement it follows that

l l lx y z= =

Now l l l lx y z x2 2 2 21 3 1+ + = fi =

or l l lx y z= = = 1

3Substituting into Eq. (9.46)

I I I I I I IOL x x y y z z xy x y yz y z zx z x= + + - - -l l l l l l l l l2 2 2 2 2 2

Noting that l l l l l l l l lx y z x y y z z x2 2 2 1

3= = = = = =

We have IOL = + +

- + + ¥ -

1

326 4325 31 1726 8 5773

2 2 5002 4 0627 8 8062 10

[ . . .

( . . . )] 33 2kg m◊

or kg m2IOL = ¥ ◊-11 81 10 3.

260


PROBLEM 9.167

The thin bent plate shown is of uniform density and weight W. Determine its mass moment of inertia with respect to the line joining the origin O and Point A.

SOLUTION

First note that m mW

g1 21

2= =

and that lOA = + +1

3( )i j k


I I I

W

ga

W

g

a

x x x= +

=ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+

( ) ( )1 2

221

12

1

2

1

2 2

1

12

1

2

WW

ga a

W

g

a aÊËÁ

ˆ¯

ÏÌÓÔ

+ + ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô

=

( )2 22 21

2 2 2

11

2

1

12

1

4

1

6

1

2

1

22 2 2W

ga a

W

ga+Ê

ËÁˆ¯

+ +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

I I I

W

ga a

W

g

a a

y y y= +

=ÊËÁ

ˆ¯

+ + ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

( ) ( )

( )

1 2

2 221

12

1

2

1

2 2 2

22

2 21

12

1

2

1

2 2

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

+ÊËÁ

ˆ¯

ÏÌÓÔ

+ + ÊËÁ

ˆ¯

W

ga

W

ga

a( )

22

2 21

2

1

6

1

2

1

12

5

4

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô

= +ÊËÁ

ˆ¯

+ +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=W

ga a

W

ga22

I I I

W

ga

W

g

a

z z z= +

=ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+

( ) ( )1 2

221

12

1

2

1

2 2

1

12

1

2

WW

ga

W

ga

a

W

g

ÊËÁ

ˆ¯

ÏÌÓÔ

+ + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

¸˝Ô

Ô

= +

2 221

2 2

1

2

1

12

1

4

( )

ÊÊËÁ

ˆ¯

+ +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=a aW

ga2 2 21

12

5

4

5

6

261



Now observe that the centroidal products of inertia, I Ix y y z¢ ¢ ¢ ¢, , and I z x¢ ¢ , of both components are zero because of symmetry. Also, y1 0=

Then I I mx y m x yW

ga

a W

gaxy x y= + = = Ê

ËÁˆ¯

=¢ ¢S( ) ( )2 2 221

2 2

1

4

I I my z m y zW

g

a a W

gayz y z= + = = Ê

ËÁˆ¯

ÊËÁ

ˆ¯

=¢ ¢S( ) 2 2 221

2 2 2

1

8

I I mz x m z x m z xzx z x= + = +¢ ¢S( ) 1 1 1 2 2 2

= ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯ ( ) =1

2 2 2

1

2 2

3

82W

g

a a W

g

aa

W

ga

Substituting into Equation (9.46)

I I I I I I IOA x x y y z z xy x y yz y z zx z x= + + - - -l l l l l l l l l2 2 2 2 2 2

Noting that

l l l l l l l l lx y z x y y z z x2 2 2 1

3= = = = = =

We have

IW

ga

W

ga

W

ga

W

ga

W

ga

W

gaOA = + + - + +

ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙

1

3

1

2

5

62

1

4

1

8

3

82 2 2 2 2 2

== - ÊËÁ

ˆ¯

ÈÎÍ

˘˚

1

3

14

62

3

42W

ga or I

W

gaOA = 5

182

0

0

0

262


PROBLEM 9.168

A piece of sheet steel of thickness t and density r is cut and bent into the machine component shown. Determine the mass moment of inertia of the component with respect to the line joining the origin O and Point A.

SOLUTION


First note that

lOA = + +1

62( )i j k

Next compute the mass of each component. We have

m Vg

tA= =r g( )

Then mg

t a at

ga

mg

t at

ga

12

22 2

2 2 4

2 2

= ¥ =

= ¥ÊËÁ

ˆ¯

=

g g

g p p g

( )

Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem 9.134) for a semicircular plate for component 2, we have

I I I

t

ga a a

t

ga a a

x x x= +

=ÊËÁ

ˆ¯

+ + +Ï

( ) ( )

[( ) ( ) ] ( )

1 2

2 2 2 2 2 21

124 2 2 4

g gÌÌÓÔ

¸˝Ô

+ÊËÁ

ˆ¯

+ +ÏÌÓÔ

¸˝Ô

=

1

4 2 22

4

2 2 2 2 2p g p g

g

t

ga a

t

ga a a

t

[( ) ( ) ]

gga a

t

ga a

t

ga

2 2 2 2

4

2

32

2

1

45

18 91335

+ÊËÁ

ˆ¯

+ +ÊËÁ

ˆ¯

=

p g

g.

263



I I I

t

ga a

t

ga a

t

g

y y y= +

=ÊËÁ

ˆ¯

+È

ÎÍ

˘

˚˙

+

( ) ( )

( ) ( )

1 2

2 2 2 21

124 2 4

2

g g

p gaa a

t

ga

aa2

22 2

221

2

16

9 2

4

3-Ê

ËÁˆ¯

+ ÊËÁ

ˆ¯

+È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝

pp g

p( )

ÔÔ

Ô

= +ÊËÁ

ˆ¯

+ - + +ÊËÁ

ˆ¯

=

41

31

2

1

2

16

9

16

91

7

2 2 22 2

2g p gp p

t

ga a

t

ga a

.668953 4g t

ga

I I I

t

ga a

t

ga a

t

g

z z z= +

=ÊËÁ

ˆ¯

+È

ÎÍ

˘

˚˙

+

( ) ( )

( ) ( )

1 2

2 2 2 21

124 2 4

2

g g

p gaa a

t

ga

aa2

22 2

221

4

16

9 2

4

32-Ê

ËÁˆ¯

+ ÊËÁ

ˆ¯

+È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸

pp g

p( ) ˝

Ô

Ô

= +ÊËÁ

ˆ¯

+ - + +ÊËÁ

ˆ¯

=

41

31

2

1

4

16

9

16

94

1

2 2 22 2

2g p gp p

t

ga a

t

ga a

22 00922 4.g t

ga

Now observe that the centroidal products of inertia, I Ix y y z¢ ¢ ¢ ¢, , and I z x¢ ¢ , of both components are zero because of symmetry. Also x1 0= .

Then I I mx y m x yt

ga

aa

t

ga

xy x y= + = = ÊËÁ

ˆ¯

=

¢ ¢S( ) ( )

.

2 2 22

4

2

4

32

1 33333

p gp

g

I I my z m y z m y z

t

ga a a

t

ga a a

yz y z= + = +

= +

¢ ¢S( )

( )( ) ( )( )

1 1 1 2 2 2

2 242

2g p g

== 7 14159 4.g t

ga

I I mz x m z x

t

ga a

a

t

ga

zx z x= + = = ÊËÁ

ˆ¯

=

¢ ¢S( ) ( )

.

2 2 22

4

2

4

3

0 66667

p gp

g

0

0

0

264




I I I I I I IOA x x y y z z xy x y yz y z zx z x= + + - - -

=

l l l l l l l l l

g

2 2 2 2 2 2

18 91335.tt

ga

t

ga4

24

21

67 68953

2

6

ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

.g

+ ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

12 009221

62 1 33333

1

6

2

64

24. .

g gt

ga

t

ga ˜

-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

2 7 141592

6

2

62 0 666674 4. .

g gt

ga

t

ga

11

6

1

6

3 15223 5 12635 2 00154 0 88889 4 76106

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

= + + - -( . . . . . -- 0 22222 4. )g t

ga

or I t aOA = 4 41 4. r

265


PROBLEM 9.169

Determine the mass moment of inertia of the machine component of Problems 9.136 and 9.155 with respect to the axis through the origin characterized by the unit vector l = - + +( ) / .4 8 9i j k

SOLUTION

From the solutions to Problems 9.136 and 9.155, we have

Problem 9.136: I

I

I

x

y

z

= ¥ ◊

= ¥ ◊

= ¥ ◊

-

-

-

175 503 10

308 629 10

154 400 10

3

3

3

.

.

.

kg m

kg m

kg

2

2

mm2

Problem 9.155: I

I

I

xy

yz

zx

= - ¥ ◊

= ¥ ◊

= ¥

-

-

-

8 0365 10

12 8950 10

94 0266 10

3

3

3

.

.

.

kg m

kg m

2

2

kkg m2◊


I I I I I I IOL x x y y z z xy x y yz y z zx z x= + + - - -l l l l l l l l l2 2 2 2 2 2

= -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

-

175 5034

9308 629

8

9154 400

1

9

2

2 2 2

. . .

(( . ) ( . )

(

- -ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

-

8 03654

9

8

92 12 8950

8

9

1

9

2 944 02661

9

4

910

34 6673 243 855 1

3 2. )

( . . .

ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

˘˚

¥ ◊

= + +

- kg m

9906 6 350

2 547 9 287 10 3 2

-

- + ¥ ◊-

.

. . ) kg m

or IOL = ¥ ◊-281 10 3 kg m2

266


PROBLEM 9.170

For the wire figure of Problem 9.148, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector l = - - +( ) / .3 6 2 7i j k

SOLUTION


m

m

LL= Ê

ËÁˆ¯

= ¥

=

0 056

0 0672

.

.

kg/m 1.2 m

kg

Now observe that the centroidal products of inertia, I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢, , ,and for eachcomponent are zero because of symmetry.

Also x x y y y z z z1 6 4 5 6 1 2 30 0 0= = = = = = = =

Then I I mx y m x y m x yxy x y= + = +¢ ¢S( ) 2 2 2 3 3 3

= +

= ◊

( . )( . )( . ) ( . )( . )( . )

.

0 0672 0 6 1 2 0 0672 1 2 0 6

0 096768

kg m m kg m m

kg mm2

I I my zyz y z= + =¢ ¢S( ) 0

I I mz x m z x m z xzx z x= + = +

= +

¢ ¢S( )

( . )( . )( . ) ( .

4 4 4 5 5 5

0 0672 0 6 1 2 0 0kg m m 6672 1 2 0 6

0 096768 2

kg m m

kg m

)( . )( . )

.= ◊

From the solution to Problem 9.148, we have

I

I I

x

y z

= ◊

= = ◊

0 32258

0 41933 2

.

.

kg m

kg m

2

0

0

0

267




I I I I I I IOL x x y y z z xy x y yz y z zx z x= + + - - -

= -

l l l l l l l l l2 2 2 2 2 2

0 322583

.77

0 419336

70 41933

2

7

2 0 09676

2 2 2ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

-

. .

( . 883

7

6

72 0 096768

2

7

3

72) ( . )-Ê

ËÁˆ¯

-ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

˘˚

◊kg m

IOL = + + - + ◊( . . . . . )0 059249 0 30808 0 034231 0 071095 0 023698 2kg m

or IOL = ◊0 354. kg m2

0

268


PROBLEM 9.171

For the wire figure of the problem indicated, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector l = - - +( 3 6 2 )/7.i j k

SOLUTION

Have

m V AL= =r r

Then

m m1 23 27850 0 0015 0 45= = ÈÎ ˘ ¥ ¥( ) ( . ( .kg/m m) m)p p

m m2 1 0 078445= = . kg

m m3 43 27850 0 0015 0 45= = ÈÎ ˘ ¥( ) ( . ( .kg/m m) m)p

= 0 024970. kg

Now observe that the centroidal products of inertia, I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢= = = 0 for each component.

Also x x y z z3 4 1 1 20 0 0= = = = =, ,

Then I I mx y m x yxy x y= + =¢ ¢S( ) 2 2 2

= -¥Ê

ËÁˆ¯

= - ¥ ◊-

( . ).

( . )

.

0 0784452 0 45

0 45

10 1128 10 3

kgm

m

kg m2

p

I I my z m y z m y zyz y z= + = +¢ ¢S( ) 3 3 3 4 4 4

where m m y y z z I yz3 4 3 4 4 3 0= = = - =, , , so that

I I mz x m z x m z xzx z x= + = + =¢ ¢S( ) 1 1 1 2 2 2 0

0

0

269



From the solution to Problem 9.147

I

I

I

x

y

z

= ¥ ◊

= ¥ ◊

= ¥ ◊

-

-

-

45 254 10

41 883 10

35 141 10

3

3

3

.

.

.

kg m

kg m

kg m

2

2

2

Now lOL = - - +1

73 6 2( )i j k

Have

I I I I I I IOL x x y y z z xy x y yz y z zx z x= + + - - -l l l l l l l l l2 2 2 2 2 2 [Eq. (9.46))]

= -ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

-

45 2543

741 883

6

735 141

2

7

2 2 2

. . .

22 10 11283

7

6

7

8 3120 30 771

2( . )

( . .

- -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

˘˚

¥ ◊

= +

-10 kg m3

22 2 8687 7 4298 10

49 3817 10

3 2

3 2

+ + ¥ ◊

= ¥ ◊

-

-

. . )

.

kg m

kg m

or IOL = ¥ ◊-49 4 10 3 2. kg m

0 0

270


PROBLEM 9.172

For the wire figure of Problem indicated, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector l = - - +( 3 6 2 )/7.i j k

SOLUTION


Have m L= r

Then m

m m m m

1

2 3 4 5

0 05

0 125664

0 05

=

=

= = =

=

( .

.

( .

kg/m)[2 (0.4 m)]

kg

kg/m)(0

p

..2m) 0.01kg=

Now observe that I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢= = = 0 for each component.

Also, x x x y z z z1 4 5 1 1 2 30 0 0= = = = = = =, ,

Then I I mx y m x y m x yxy x y= + = +¢ ¢S( ) 2 2 2 3 3 3

=

= ¥ ◊-

( .

.

0 01

1 000 10 3

kg)[(0.40 m)(0.1m) + (0.3m)(0.2m)]

kg m2

By symmetry I I Iyz xy yz= = ¥ ◊-1 000 10 3 2. kg m

Now I I mz xzx z x= + =¢ ¢S( ) 0


I I

I

x z

y

= = ¥ ◊

= ¥ ◊

-

-

13 6529 10

25 1726 10

3

3

.

.

kg m

kg m

2

2

0

271



Now lOL = - - +1

73 6 2( )i j k

Have

I I I I I I IOL x x y y z z xy x y yz y z zx z x= + + - - -l l l l l l l l l2 2 2 2 2 2 [Eq.(9.46)]]

kg= -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô¥ ◊-( . )13 6529

3

7

2

710

2 23 mm

10 kg m

2

3+ { -ÊËÁ

ˆ¯

ÏÌÔ

ÓÔ¥ ◊

- -ÊËÁ

ˆ¯

-

-( . )

( . )

25 17266

7

2 1 0003

7

22

66

7

6

7

2

7

3 6222 18

2ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ÈÎÍ

˘˚

¸˝˛

¥ ◊

= +

-10 kg m3

( . .44942 0 2449

21 8715

2

2

- ¥ ◊

= ¥ ◊

-

-

. )

.

10 kg m

10 kg m

3

3

or IOL = ¥ ◊-21 9 2. 10 kg m3

0

272


PROBLEM 9.173

For the rectangular prism shown, determine the values of the ratios b/a and c/a so that the ellipsoid of inertia of the prism is a sphere when computed (a) at Point A, (b) at Point B.

SOLUTION

(a) Using Figure 9.28 and the parallel-axis theorem, we have at Point A

I m b c

I m a c ma

m a c

I

x

y

z

¢

¢

¢

= +

= + + ÊËÁ

ˆ¯

= +

=

1

12

1

12 2

1

124

2 2

2 22

2 2

( )

( )

( )

11

12 2

1

1242 2

22 2m a b m

am a b( ) ( )+ + Ê

ËÁˆ¯

= +

Now observe that symmetry implies

I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢= = = 0

Using Eq. (9.48), the equation of the ellipsoid of inertia is then

I x I y I z m b c x m a c y m ax y z¢ ¢ ¢+ + = + + + + +2 2 2 2 2 2 2 2 2 211

12

1

124

1

124: ( ) ( ) ( bb z2 2 1) =

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore

1

12

1

124

1

124

2 2 2 2

2 2

m b c m a c

m a b

( ) ( )

( )

+ = +

= +

Then b c a c2 2 2 24+ = + orb

a= 2

and b c a b2 2 2 24+ = + orc

a= 2

273



(b) Using Figure 9.28 and the parallel-axis theorem, we have at Point B

I m b c mc

m b c

I m a c m

x

y

¢¢

¢¢

= + + ÊËÁ

ˆ¯

= +

= + +

1

12 2

1

124

1

12

2 22

2 2

2 2

( ) ( )

( )cc

m a c

I m a bz

2

1

124

1

12

22 2

2 2

ÊËÁ

ˆ¯

= +

= +¢¢

( )

( )


I I Ix y y z z x¢¢ ¢¢ ¢¢ ¢¢ ¢¢ ¢¢= = = 0

From Part a it then immediately follows that

1

124

1

124

1

122 2 2 2 2 2m b c m a c m a b( ) ( ) ( )+ = + = +

Then b c a c2 2 2 24 4+ = + or b

a= 1

and b c a b2 2 2 24+ = + or c

a= 1

2

274


PROBLEM 9.174

For the right circular cone of Sample Problem 9.11, determine the value of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere when computed (a) at the apex of the cone, (b) at the center of the base of the cone.

SOLUTION

(a) From Sample Problem 9.11, we have at the apex A

I ma

I I m a h

x

y z

=

= = +ÊËÁ

ˆ¯

3

103

5

1

4

2

2 2

Now observe that symmetry implies I I Ixy yz zx= = = 0


I x I y I z ma x m a h y m a hx y z2 2 2 2 2 2 2 2 2 21

3

10

3

5

1

4

3

5

1

4+ + = + +Ê

ËÁˆ¯

+ +ÊËÁ

: ˆ¯

=z2 1

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore,

3

10

3

5

1

42 2 2ma m a h= +Ê

ËÁˆ¯ or

a

h= 2

(b) From Sample Problem 9.11, we have

I max¢ = 3

102

and at the centroid C I m a hy¢¢ = +ÊËÁ

ˆ¯

3

20

1

42 2

Then I I m a h mh

m a hy z¢ ¢= = +ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

= +3

20

1

4 4

1

203 22 2

22 2( )


I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢= = = 0


3

10

1

203 22 2 2ma m a h= +( ) or

a

h= 2

3

275


PROBLEM 9.175

For the homogeneous circular cylinder shown, of radius a and length L, determine the value of the ratio a/L for which the ellipsoid of inertia of the cylinder is a sphere when computed (a) at the centroid of the cylinder, (b) at Point A.

SOLUTION

(a) From Figure 9.28:

I ma

I I m a L

x

y z

=

= = +

1

21

123

2

2 2( )

Now observe that symmetry implies I I Ixy yz zx= = = 0


I x I y I z ma x m a L y m a Lx y z2 2 2 2 2 2 2 2 2 21

1

2

1

123

1

123 1+ + = + + + + =: ( ) ( )

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore

1

2

1

1232 2 2ma m a L= +( ) or

a

L= 1

3

(b) Using Figure 9.28 and the parallel-axis theorem, we have

I ma

I I m a L mL

m a L

x

y z

¢

¢ ¢

=

= = + + ÊËÁ

ˆ¯

= +ÊËÁ

ˆ¯

1

2

1

123

4

1

4

7

48

2

2 22

2 2( )


I I Ix y y z z x¢ ¢ ¢ ¢ ¢ ¢= = = 0


1

2

1

4

7

482 2 2ma m a L= +Ê

ËÁˆ¯ or

a

L= 7

12

276


PROBLEM 9.176

Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of the body with respect to any one of the three axes cannot be larger than the sum of the mass moments of inertia of the body with respect to the other two axes. That is, prove that the inequality I I Ix y z� + and the two similar inequalities are satisfied. Further, prove that I Iy x� 1

2 if the body is a homogeneous solid of revolution, where x is the axis of revolution and y is a transverse axis.

SOLUTION

(i) To prove I I Iy z x+ �

By definition I z x dm

I x y dm

y

z

= +

= +

ÚÚ( )

( )

2 2

2 2

Then I I z x dm x y dmy z+ = + + +ÚÚ ( ) ( )2 2 2 2

= + + ÚÚ ( )y z dm x dm2 2 22

Now ( )y z dm I x dmx2 2 2 0+ =Ú Ú and �

I I Iy z x+ � Q.E.D.

The proofs of the other two inequalities follow similar steps.

(ii) If the x axis is the axis of revolution, then

I Iy z=

and from Part (i) I I Iy z x+ �

or 2I Iy x�

or I Iy x�1

2 Q.E.D.

277


PROBLEM 9.177

Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a sphere, and use this property to determine the moment of inertia of the cube with respect to one of its diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution, and determine the principal moments of inertia of the cube at that point.

SOLUTION

(a) At the center of the cube have (using Figure 9.28)

I I I m a a max y z= = = + =1

12

1

62 2 2( )


I I Ixy yz zx= = = 0

Using Equation (9.48), the equation of the ellipsoid of inertia is

1

6

1

6

1

612 2 2 2 2 2ma x ma y ma zÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

=

or x y zma

R2 2 22

26+ + = =( )

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

be the same RIOL

=Ê

ËÁ

ˆ

¯˜

1. I maOL = 1

62

(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 I max OL¢ = = 1

62

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 parallel-axis theorem between the center of the cube and corner A for any perpendicular axis

I I ma m ay z¢ ¢= = +Ê

ËÁˆ

¯1

6

3

22

2

or I I may z¢ ¢= = 11

122

(Note: Part b can also be solved using the method of Section 9.18.)

278



First note that at corner A

I I I ma

I I I ma

x y z

xy yz zx

= = =

= = =

2

31

4

2

2

Substituting into Equation (9.56) yields

k ma k m a k m a3 2 2 2 6 3 9255

48

121

8640- + - =

For which the roots are

k ma

k k ma

12

2 32

1

611

12

=

= =

279


PROBLEM 9.178

Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin at O, prove that the sum Ix + Iy + Iz of the mass moments of inertia of the body cannot be smaller than the similar sum computed for a sphere of the same mass and the same material centered at O. Further, using the result of Problem 9.176, prove that if the body is a solid of revolution, where x is the axis of revolution, its mass moment of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10, where a is the radius of the sphere of the same mass and the same material.

SOLUTION

(i) Using Equation (9.30), we have

I I I y z dm z x dm x y dmx y z+ + = + + + + +Ú Ú Ú( ) ( ) ( )2 2 2 2 2 2

= + +Ú2 2 2 2( )x y z dm

= Ú2 2r dm

where r is the distance from the origin O to the element of mass dm. Now assume that the given body can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m and radius a which is centered at O; it then follows that

m m m m m1 2= = =( )body sphere

Then

( ) ( ) ( )I I I I I I I I Ix y z x y z x y z V+ + = + + + + +body sphere 1

- + +( )I I Ix y z V2

or ( ) ( )I I I I I I r dm r dmx y z x y z m m+ + = + + + -Ú Úbody sphere 2 22 2

1 2

Now, m m1 2= and r r1 2� for all elements of mass dm in volumes 1 and 2.

r dm r dmm m

2 2

1 2

0Ú Ú- �

so that ( ) ( )I I I I I Ix y z x y z+ + ≥ + +body sphere Q.E.D.

280



(ii) First note from Figure 9.28 that for a sphere

I I I max y z= = = 2

52

Thus ( )I I I max y z+ + =sphere6

52

For a solid of revolution, where the x axis is the axis of revolution, we have

I Iy z=

Then, using the results of Part (i)

( )I I max y+ 26

52

body �

From Problem 9.178 we have I Iy x�1

2

or ( )2 0I Iy x- body �

Adding the last two inequalities yields

( )46

52I may body �

or ( )I may body Q.E.D.�3

102

281


PROBLEM 9.179*

The homogeneous circular cylinder shown has a mass m, and the diameter OB of its top surface forms 45∞ angles with the x andz axes. (a) Determine the principal mass moments of inertia of the cylinder at the origin O. (b) Compute the angles that the principal axes of inertia at O form with the coordinate axes. (c) Sketch the cylinder, and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION

(a) First compute the moments of inertia using Figure 9.28 and the

parallel-axis theorem.

I I m a a m

a ama

I

x z

y

= = + + ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=

=

1

123

2 2

13

122 2

2 22( )

11

2

3

22 2 2ma m a ma+ =( )

Next observe that the centroidal products of inertia are zero because of symmetry. Then

I I mx y ma a

maxy x y= + = ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

= -¢ ¢2 2

1

2 22

I I my z ma a

mayz y z= + = -ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

= -¢ ¢2 2

1

2 22

I I mz x ma a

mazx z x= + = ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

=¢ ¢2 2

1

22


K ma K3 2 213

12

3

2

13

12- + +Ê

ËÁˆ¯

+ ¥ÊËÁ

ˆ¯

+ ¥ÊËÁ

ˆ¯

+ ¥ÊËÁ

ˆ¯

ÈÎÍ

- -ÊËÁ

ˆ¯

13

12

3

2

3

2

13

12

13

12

13

12

1

2 2

22 2 22 21

2 2

1

2- -Ê

ËÁˆ¯

- ÊËÁ

ˆ¯

˘

˚˙˙( )ma K

- ¥ ¥ÊËÁ

ˆ¯

ÈÎÍ

- ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

13

12

3

2

13

12

13

12

1

2 2

3

2

1

2

2ˆ¯

- ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

- -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

˘

2

213

12

1

2 22

1

2 2

1

2 2

1

2 ˚˙ =( )ma2 3 0

0

0

0

282



Simplifying and letting K ma= 2z yields

z z z3 211

3

565

144

95

960- + - =

Solving yields

z z z1 2 30 36338319

121 71995= = =. .

The principal moments of inertia are then

K ma120 363= .

K ma221 583= .

K ma321 720= .

(b) To determine the direction cosines l l lx y z, , of each principal axis, we use two of the equations of Equations (9.54) and (9.57).

Thus,

( )I K I Ix x xy y zx z- - - =l l l 0 (9.54a)

- - + - =I I I Kzx x yz y z zl l l( ) 0 (9.54c)

l l lx y z2 2 2 1+ + = (9.57)

(Note: Since I Ixy yz= , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of ly during the solution process.)

Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)

13

12

1

2 2

1

202 2 2ma K ma max y z-Ê

ËÁˆ¯

- -ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

=l l l

-ÊËÁ

ˆ¯

- -ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

=1

2

1

2 2

13

1202 2 2ma ma ma Kx y zl l l

or 13

12

1

2 2

1

20-Ê

ËÁˆ¯

+ - =z l l lx y z (i)

and - + + -ÊËÁ

ˆ¯

=1

2

1

2 2

13

120l l z lx y z (ii)

Observe that these equations will be identical, so that one will need to be replaced, if

13

12

1

2

19

12- = - =z z or

283



Thus, a third independent equation will be needed when the direction cosines associated with K2 are determined. Then for K1 and K3

Eq. (i) through Eq. (ii): 13

12

1

2

1

2

13

120- - -Ê

ËÁˆ¯

ÈÎÍ

˘˚

+ - - -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=z l z lx z

or l lz x=

Substituting into Eq. (i): 13

12

1

2 2

1

20-Ê

ËÁˆ¯

+ - =z l l lx y x

or l z ly x= -ÊËÁ

ˆ¯

2 27

12

Substituting into Equation (9.57):

l z l lx x x2

222 2

7

121+ -Ê

ËÁˆ¯

ÈÎÍ

˘˚

+ =( )

or 2 87

121

22+ -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

=z lx (iii)

K1 : Substituting the value of z1 into Eq. (iii):

2 8 0 3633837

121

2

1

2+ -ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙( ) =. lx

or ( ) ( ) .l lx z1 1 0 647249= =

and then ( ) . ( . )ly 1 2 2 0 3633837

120 647249= -Ê

ËÁˆ¯

= -0 402662.

( ) ( ) . .q q qx z y1 1 149 7 113 7= = ∞ = ∞ ( )

K3 : Substituting the value of z3 into Eq. (iii):

2 8 1 719957

121

2

32+ -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

=. ( )lx

or ( ) ( ) .l lx z3 3 0 284726= =

and then ( ) . ( . )ly 3 2 2 1 719957

120 284726= -Ê

ËÁˆ¯

= 0 915348. ( ) ( ) . .q q qx z y3 3 373 5 23 7= = ∞ = ∞ ( )

284



K2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57). Now

- + - - =I I K Ixy x y y yz zl l l( ) 0 (9.54b)

Substituting for the moments and products of inertia.

- -ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

- -ÊËÁ

ˆ¯

=1

2 2

3

2

1

2 202 2 2ma ma K max y zl l l

or 1

2 2

3

2

1

2 20l x l lx y z+ -Ê

ËÁˆ¯

+ = (iv)

Substituting the value of x2 into Eqs. (i) and (iv):

13

12

19

12

1

2 2

1

20

1

2 2

3

2

19

12

2 2 2

2

-ÊËÁ

ˆ¯

+ - =

+ -ÊËÁ

( ) ( ) ( )

( )

l l l

l

x y z

xˆ¯

+ =( ) ( )l ly z2 21

2 20

or - + - =( ) ( ) ( )l l lx y z2 2 21

20

and ( ) ( ) ( )l l lx y z2 2 22

60- + =

Adding yields ( )ly 2 0=

and then ( ) ( )l ly x2 2= -


( ) ( ) ( )l l lx y x22

22

22 1+ + - =

or ( ) ( )l lx z2 21

2

1

2= = -and

( ) . ( ) . ( ) .q q qx y z2 2 245 0 90 0 135 0= ∞ = ∞ = ∞

(c) Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B. Principal axis 2 lies in the xz plane.

0

285


PROBLEM 9.180

For the component described in Problem 9.165, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION

(a) From the solutions to Problems 9.141 and 9.165 we have

Problem 9.141: I

I

I

x

y

z

= ¥ ◊

= ¥ ◊

= ¥

-

-

-

13 98800 10

20 55783 10

14 30368 10

3

3

3

.

.

.

kg m

kg m

2

2

kkg m2◊

Problem 9.165: I I

I

yz zx

xy

= =

= ¥ ◊-

0

0 39460 10 3. kg m2

Eq. (9.55) then becomes

I K I

I I K

I K

I K I K I K I K Ix xy

xy y

z

x y z z xy

- -- -

-= - - - - - =

0

0

0 0

0 2or ( )( )( ) ( ) 00

Thus I K I I I I K K Iz x y x y xy- = - + + - =0 02 2( )

Substituting: K1314 30368 10= ¥ ◊-. kg m2

or K1314 30 10= ¥ ◊-. kg m2

and ( . )( . ) ( . . )( )13 98800 10 20 55783 10 13 98800 20 55783 103 3 3¥ ¥ - + +- - - K KK 2 3 20 39460 10 0- ¥ =-( . )

or K K2 3 634 54583 10 287 4072 10 0- ¥ + ¥ =- -( . ) .

Solving yields K2313 96438 10= ¥ ◊-. kg m2

or K2313 96 10= ¥ ◊-. kg m2

and K3320 58145 10= ¥ ◊-. kg m2 or K3

320 6 10= ¥ ◊-. kg m2

286



(b) To determine the direction cosines l l lx y z, , of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then

K1: Begin with Eqs. (9.54a) and (9.54b) with I Iyz zx= = 0.

( )( ) ( )

( ) ( )( )

I K I

I I K

x x xy y

xy x y y

- - =

- + - =1 1 1

1 1 1

0

0

l l

l l Substituting:

[( . . ) ]( ) ( . )( )

( .

13 98800 14 30368 10 0 39460 10 0

0 3

31

31- ¥ - ¥ =

-

- -l lx y

99460 10 20 55783 14 30368 10 031

31¥ + - ¥ =- -)( ) [( . . ) ]( )l lx y

Adding yields ( ) ( )l lx y1 1 0= =

Then using Eq. (9.57) ( ) ( ) ( )l l lx y z12

12

12 1+ + =

or ( )lz 1 1=

( ) . ( ) . ( )q q qx y z1 1 190 0 90 0 0= ∞ = ∞ = ∞

K2: Begin with Eqs. (9.54b) and (9.54c) with I Iyz zx= = 0.

- + - =

- =

I I K

I K

xy x y y

z z

( ) ( )( )

( )( )

l l

l2 2 2

2 2

0

0

(i)

Now I Kz zπ fi =2 2 0( )l

Substituting into Eq. (i):

- ¥ + - ¥ =- -( . )( ) [( . . ) ]( )0 39460 10 20 55783 13 96438 10 03 32l lx z y

or ( ) . ( )l ly x2 20 059847=

Using Eq. (9.57): ( ) [ . ]( ) ] ( )l l lx x z22

22

220 05984 1+ + =

or ( ) .lx 2 0 998214=

and ( ) .ly 2 0 059740=

( ) . ( ) . ( ) .q q qx y z2 2 23 4 86 6 90 0= ∞ = ∞ = ∞

K3: Begin with Eqs. (9.54b) and (9.54c) with I Iyz zx= = 0.

- + - =

- =

I I K

I K

xy x y y

z z

( ) ( )( )

( )( )

l l

l3 3 3

3 3

0

0

(ii)

0 0

0

287



Now I Kz zπ fi =3 3 0( )l

Substituting into Eq. (ii):

- ¥ + - ¥ =- -( . )( ) [( . . ) ]( )0 39460 10 20 55783 20 58145 10 033

33l lx y

or ( ) . ( )l ly x3 316 70618= -

Using Eq. (9.57): ( ) [ . ( ) ] ( )l l lx x z32

32

3216 70618 1+ - + =

or ( ) . "_"lx 3 0 059751= - fi(axes right-handed set )

and ( ) .ly 3 0 998211=

( ) . ( ) . ( ) .q q qx y z3 3 393 4 3 43 90 0= ∞ = ∞ = ∞

Note: Since the principal axes are orthogonal and ( ) ( ) ,q qz z2 3 90= = ∞ it follows that

| | |( ) | | | |3( ) ( ) | ( )l l l lx y y z2 2 3= =

The differences in the above values are due to round-off errors.

(c) Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.

0

288


PROBLEM 9.181*

For the component described in Problems 9.145 and 9.149, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION

(a) From the solutions to Problems 9.145 and 9.149, we have

Problem 9.145: I x = ¥ ◊-26 4325 10 3. kg m2 Problem 9.149: I xy = ¥ ◊-2 5002 10 3. kg m2

I y = ¥ ◊-31 1726 10 3. kg m2 I yz = ¥ ◊-4 0627 10 3. kg m2

I z = ¥ ◊-8 5773 10 3. kg m2 I zx = ¥ ◊-8 8062 10 3. kg m2

Substituting into Eq. (9.56):

K K3 3 226 4325 31 1726 8 5773 10

26 4325 31 1726

- + ++ +

-[( . . . )( )]

[( . )( . ) (( . )( . ) ( . )( . )

( . ) ( . ) (

31 1726 8 5773 8 5773 26 4325

2 5002 4 06272 2

+

- - - 88 8062 10

26 4325 31 1726 8 5773 26 4325 4

2 6. ) ]( )

[( . )( . )( . ) ( . )( .

-

- -

K

00627

31 1726 8 8062 8 5773 2 5002

2 2 5002 4 0

2

2 2

)

( . )( . ) ( . )( . )

( . )( .

- -

- 6627 8 8062 10 09)( . )]( )- =

or K K K3 3 2 6 966 1824 10 1217 76 10 3981 23 10 0- ¥ + ¥ - ¥ =- - -( . ) ( . ) ( . )

Solving yields

K134 1443 10= ¥ ◊-. kg m2 or kg m2K1

34 14 10= ¥ ◊-.

K2329 7840 10= ¥ ◊-. kg m2 or kg m2K2

329 8 10= ¥ ◊-.

K3332 2541 10= ¥ ◊-. kg m2 or kg m2K3

332 3 10= ¥ ◊-.

289



(b) To determine the direction cosines l l lx y z, , of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then

K1: Begin with Eqs. (9.54a) and (9.54b).

( )( ) ( ) ( )

( ) ( )( ) (

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

+ - -1 1 1 1

1 1 1

0l l l

l l llz )1 0= Substituting:

[( . . )( )]( ) ( . )( ) ( .26 4325 4 1443 10 2 5002 10 8 8062 131

31- - ¥ - ¥- -l lx y 00 03

1- =)( )lz

- ¥ + - - ¥- -( . )( ) [( . . )( )]( ) ( .2 5002 10 31 1726 4 1443 10 4 062731

31l lx y 110 03

1- =)( )lz

Simplifying

8 9146 3 5222 0

0 0925 0 15031 1 1

1 1

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l l lx y z

x y

- - =

- + - zz )1 0=

Adding and solving for ( ) :lz 1

( ) . ( )l lz x1 12 4022=

and then ( ) [ . . ( . )]( )

. ( )

l l

ly x

x

1 1

1

8 9146 3 5222 2 4022

0 45357

= -

=

Now substitute into Eq. (9.57):

( ) [ . ( ) ] [ . ( ) ]l l lx x x12

12

120 45357 2 4022 1+ + =

or ( ) .lx 1 0 37861=

and ( ) .ly 1 0 17173= ( ) .lz 1 0 90950=

( ) . ( ) . ( ) .q q qx y z1 1 167 8 80 1 24 6= ∞ = ∞ = ∞

K 2: Begin with Eqs. (9.54a) and (9.54b).

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -2 2 2 2

2 2 2

0l l l

l l (( )lz 2 0=

Substituting:

[( . . )( )]( ) ( . )( ) ( .26 4325 29 7840 10 2 5002 10 8 806232

32- - ¥ - ¥- -l lx y 110 03

2- =)( )lz

- ¥ + - -- -( . )( ) [( . . )( )]( ) ( .2 5002 10 31 1726 29 7840 10 4 062732

32l lx y ¥¥ ( =-10 03

2) )lz

290



Simplifying

- - - =

- + -

1 3405 3 5222 0

1 8005 2 92582 2 2

2 2

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l lx y z

x y llz )2 0=


( ) . ( )l lz x2 20 48713= -

and then ( ) [ . . ( . )]( )

. ( )

l l

ly x

x

2 2

2

1 3405 3 5222 0 48713

0 37527

= - - -

=


( ) [ . ( ) ] [ . ( ) ]l l lx x x22

22

220 37527 0 48713 1+ + - =

or ( ) .lx 2 0 85184=

and ( ) .ly 2 0 31967= ( ) .lz 2 0 41496= -

( ) . ( ) . ( ) .q q qx y z1 2 231 6 71 4 114 5= ∞ = ∞ = ∞

K 3: Begin with Eqs. (9.54a) and (9.54b).

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -3 3 3 3

3 3 3

0l l l

l l (( )lz 3 0=

Substituting:

[( . . )( )]( ) ( . )( ) ( .26 4325 32 2541 10 2 5002 10 8 806233

33- - ¥ - ¥- -l lx y 110 03

3- =)( )lz

- ¥ + - -- -( . )( ) [( . . )( )]( ) ( .2 5002 10 31 1726 32 2541 10 4 062733

33l lx y ¥¥ =-10 03

3)( )lz

Simplifying

- - - =

+ +

2 3285 3 5222 0

2 3118 3 75653 3 3

3 3

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l l lx y z

x y zz )3 0=


( ) . ( )l lz x3 30 071276=

and then ( ) [ . . ( . )]( )

. ( )

l l

ly x

x

3 3

3

2 3285 3 5222 0 071276

2 5795

= - -

= -

291




( ) [ . ( ) ] [ . ( ) ]l l lx x x32

32

322 5795 0 071276 1+ - + = (i)

or ( ) .lx 3 0 36134=

and ( ) .ly 3 0 93208= ( ) .lz 3 0 025755=

( ) . ( ) . ( ) .q q qx y z3 3 368 8 158 8 88 5= ∞ = ∞ = ∞

(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is, ( ) . .lx 3 0 36134= -

Then ( ) . ( ) . ( ) .q q qx y z3 3 3111 2 21 2 91 5= ∞ = ∞ = ∞

292


PROBLEM 9.182*


SOLUTION

(a) From the solution of Problem 9.167, we have

IW

ga I

W

ga

IW

ga I

W

ga

IW

ga I

W

ga

x xy

y yz

z zx

= =

= =

= =

1

2

1

4

1

8

5

6

3

8

2 2

2 2

2 2


KW

ga K3 2 21

21

5

6

1

21 1

5

6

- + +ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙

+ ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

( ) ( ) ++ ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

5

6

1

2

1

4

1

8

3

8

2 2 2 W

gaa K2

2

21

21

5

6

1

2

1

81

3

8

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

- ÊËÁ

( ) ( ) ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

2 25

6

1

42

1

4

1

8

3

8

W

gga2

3

0ÊËÁ

ˆ¯

=

Simplifying and letting KW

ga K= 2 yields

K K K3 22 33333 1 53125 0 192708 0- + - =. . .

Solving yields

K K K1 2 30 163917 1 05402 1 11539= = =. . .

The principal moments of inertia are then KW

ga1

20 1639= .

KW

ga2

21 054= .

KW

ga3

21 115= .

293



(b) To determine the direction cosines l l lx y z, , of each principal axis, use two of the equations ofEqs. (9.54) and Eq. (9.57). Then

K1: Begin with Eqs. (9.54a) and (9.54b).

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -1 1 1 1

1 2 1

0l l l

l l (( )lz 1 0=

Substituting

1

20 163917

1

4

321

21-Ê

ËÁˆ¯

ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙ -

ÊËÁ

ˆ¯

-. ( ) ( )W

ga

W

gax yl l

8802

1W

ga z

ÊËÁ

ˆ¯

=( )l

-ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙ -1

41 0 163917

1

82

12

1W

ga

W

ga

W

gax y( ) ( . ) ( )l l 22

1 0ÊËÁ

ˆ¯

=( )lz

Simplifying

1 34433 1 5 01 1 1. ( ) ( ) . ( )l l lx y z- - =

- + - =0 299013 0 149507 01 1 1. ( ) ( ) . ( )l l lx y z


( ) . ( )l lz x1 10 633715=

and then ( ) [ . . ( . )]( )

. ( )

l l

ly x

x

1 1

1

1 34433 1 5 0 633715

0 393758

= -

=


( ) [ . ( ) ] ( . ( ) ]l l lx x x12

12

120 393758 0 633715 1+ + =

or ( ) .lx 1 0 801504=

and ( ) .ly 1 0 315599= ( ) .lz 1 0 507925=

( ) . ( ) . ( ) .q q qx y z1 1 136 7 71 6 59 5= ∞ = ∞ = ∞

K 2: Begin with Eqs. (9.54a) and (9.54b):

( )( ) ( ) ( )

( ) ( )( )

I k I I

I I k I

x x xy y zx z

xy x y y yz

- - - =

- + - -2 2 2 2

2 2 2

0l l l

l l (( )lz 2 0=

294



Substituting

1

21 05402

1

4

3

82

22

2-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙ -

ÊËÁ

ˆ¯

-. ( ) ( )W

ga

W

gax yl l WW

ga

W

ga

W

ga

z

x

22

22

2

0

1

41 1 05402

ÊËÁ

ˆ¯

=

-ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

( )

( ) ( . )

l

lÈÈ

ÎÍ

˘

˚˙ -

ÊËÁ

ˆ¯

=( ) ( )l ly zW

ga2

22

1

80

Simplifying

- - - =2 21608 1 5 02 2 2. ( ) ( ) . ( )l l lx y z

4 62792 2 31396 02 2 2. ( ) ( ) . ( )l l lx y z+ + =

Adding and solving for ( )lz 2

( ) . ( )l lz x2 22 96309= -

and then ( ) [ . . ( . )]( )

. )

l l

ly x

x

2 2

2

2 21608 1 5 2 96309

2 22856

= - - -

= (


( ) [ . ( ) ] [ . ( ) ]l l lx x x22

22

222 22856 2 96309 1+ + - =

or ( ) .lx 2 0 260410=

and ( ) .ly 2 0 580339= ( ) .lz 2 0 771618= -

( ) . ( ) . ( ) .q q qx y z2 2 274 9 54 5 140 5= ∞ = ∞ = ∞

K 3: Begin with Eqs. (9.54a) and (9.54b):

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -3 3 3 3

3 3 3

0l l l

l l (( )lz 3 0= Substituting

1

21 11539

1

4

3

82

32

3-ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙ -

ÊËÁ

ˆ¯

-. ( ) ( )W

ga

W

gax yl l WW

ga

W

ga

W

ga

z

x

23

23

2

0

1

41 1 11539

ÊËÁ

ˆ¯

=

-ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

( )

( ) ( . )

l

lÈÈ

ÎÍ

˘

˚˙ -

ÊËÁ

ˆ¯

=( ) ( )l ly zW

ga3

23

1

80

Simplifying

- - - =

+ +

2 46156 1 5 0

2 16657 1 083283 3 3

3 3

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l l lx y z

x y zz )3 0=

295




( ) . ( )l lz x3 30 707885= -

and then ( ) [ . . ( . )]( )

. ( )

l l

ly x

x

3 3

3

2 46156 1 5 0 707885

1 39973

= - - -

= -


( ) [ . ( ) ] [ . ( ) ]l l lx x x32

32

321 39973 0 707885 1+ - + - = (i)

or ( ) .lx 3 0 537577=

and ( ) . ( ) .l ly z3 30 752463 0 380543= - = -

( ) . ( ) . ( ) .q q qx y z3 3 357 5 138 8 112 4= ∞ = ∞ = ∞

(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,( ) .lx 3 0 537577= -

Then ( ) . ( ) . ( ) .q q qx y z3 3 3122 5 41 2 67 6= ∞ = ∞ = ∞

296


PROBLEM 9.183*


SOLUTION

(a) From the solution to Problem 9.168, we have

It

ga

It

ga

It

ga

x

y

z

=

=

=

18 91335

7 68953

12 00922

4

4

4

.

.

.

g

g

g

It

ga

It

ga

It

ga

xy

yz

zx

=

=

=

1 33333

7 14159

0 66667

4

4

4

.

.

.

g

g

g


Kt

ga K3 4 218 91335 7 68953 12 00922

18 913

- + +ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙

+

( . . . )

[( .

g

335 7 68953 7 68953 12 00922 12 00922 18 91335

1

)( . ) ( . )( . ) ( . )( . )

( .

+ +

- 333333 7 14159 0 666672 2 2 42

) ( . ) ( . ) ]- +ÊËÁ

ˆ¯

g t

ga K

- -

-

[( . )( . )( . ) ( . )( . )

( .

18 91335 7 68953 12 00922 18 91335 7 14159

7 68

2

9953 0 66667 12 00922 1 33333

2 1 33333 7 14159 0

2 2)( . ) ( . )( . )

( . )( . )(

-

- .. )]66667 043

g t

ga

ÊËÁ

ˆ¯

=

Simplifying and letting

Kt

ga K= g 4 yields

K K K3 238 61210 411 69009 744 47027 0- + - =. . .

297



Solving yields

K K K1 2 32 25890 17 27274 19 08046= = =. . .

The principal moments of inertia are then Kt

ga1

42 26= .g

Kt

ga2

417 27= .g

Kt

ga3

419 08= .g

(b) To determine the direction cosines l l lx y z, , of each principal axis, use two of the equations ofEqs. (9.54) and Eq. (9.57). Then

K1 : Begin with Eqs. (9.54a) and (9.54b):

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -1 1 1 1

1 2 1

0l l l

l l (( )lz 1 0= Substituting

( . . ) ( ) .18 91335 2 25890 1 3333341

4-ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙ - Ê

ËÁˆ¯

g l gt

ga

t

aax ˜ -

ÊËÁ

ˆ¯

=( ) . ( )l g ly zt

ga1

410 66667 0

-ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍ

˘

˚1 33333 7 68953 2 258904

14. ( ) ( . . )

g l gt

ga

t

gax ˙ -

ÊËÁ

ˆ¯

=( ) . ( )l g ly zt

ga1

417 14159 0

Simplifying

12 49087 0 5 0

0 24552 1 315061 1 1

1 1

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l lx y z

x y

- - =

- + - llz )1 0=


( ) . ( )l lz x1 16 74653=

and then ( ) [ . ( . )( . )]( )

. ( )

l l

ly x

x

1 1

1

12 49087 0 5 6 74653

9 11761

= -

=

Now substitute into Eq. (9.57): ( ) [ . ( ) ] [ . ( ) ]l l lx x x12

12

129 11761 6 74653 1+ + =

or ( ) .lx 1 0 087825=

and ( ) . ( ) .l ly z1 10 80075 0 59251= =

( ) . ( ) . ( ) .q q qx y z1 1 185 0 36 8 53 7= ∞ = ∞ = ∞

298



K 2 : Begin with Eqs. (9.54a) and (9.54b):

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -2 2 2 2

2 2 2

0l l l

l l (( )lz 2 0=

Substituting

[( . . ) ( ) . (18 91335 17 27274 1 3333342

4-ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

g l g lt

ga

t

gax y )) . ( )

. ( ) ( .

24

2

42

0 66667 0

1 33333 7 6

-ÊËÁ

ˆ¯

=

-ÊËÁ

ˆ¯

+

g l

g l

t

ga

t

ga

z

x 88953 17 27274 7 1415942

4-ÊËÁ

ˆ¯

È

ÎÍ

˘

˚˙ -

ÊËÁ

ˆ¯

. ) ( ) . (g l g lt

ga

t

gay zz )2 0=

Simplifying

1 23046 0 5 0

0 13913 0 745222 2 2

2 2

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l l lx y z

x y z

- - =

+ + ))2 0=


( ) . ( )l lz x2 25 58515= -

and then ( ) [ . ( . )( . )]( )

. )

l l

ly x

x

2 2

2

1 23046 0 5 5 58515

4 02304

= - -

= (


( ) [ . ( ) ] [ . ( ) ]l l lx x x22

22

224 02304 5 58515 1+ + - =

or ( ) .lx 2 0 14377=

and ( ) . ( ) .l ly z2 20 57839 0 80298= = -

( ) . ( ) . ( ) .q q qx y z2 2 281 7 54 7 143 4= ∞ = ∞ = ∞

K 3 : Begin with Eqs. (9.54a) and (9.54b):

( )( ) ( ) ( )

( ) ( )( )

I K I I

I I K I

x x xy y zx z

xy x y y yz

- - - =

- + - -3 3 3 3

3 3 3

0l l l

l l (( )ly 3 0= Substituting

[( . . ) ( ) . (18 91335 19 08046 1 3333343

4-ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

g l g lt

ga

t

gax y )) . ( )3

430 66667 0-

ÊËÁ

ˆ¯

=g lt

ga z

-ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

È

ÎÍ

˘1 33333 7 68953 19 080464

34. ( ) ( . . )

g l gt

ga

t

gax

˚˙ -

ÊËÁ

ˆ¯

=( ) . ( )l g ly zt

ga3

437 14159 0

299



Simplifying

- - - =

+ +

0 12533 0 5 0

0 11705 0 626953 3 3

3 3

. ( ) ( ) . ( )

. ( ) ( ) . (

l l l

l l lx y z

x y zz )3 0=


( ) . ( )l lz x3 30 06522=

and then ( ) [ . ( . )( . )]( )

. ( )

l l

ly x

x

3 3

3

0 12533 0 5 0 06522

0 15794

= - -

= -


( ) [ . ( ) ] [ . ( ) ]l l lx x x32

32

320 15794 0 06522 1+ - + = (i)

or ( ) .lx 3 0 98571=

and ( ) . ( ) .l ly z3 30 15568 0 06429= - =

( ) . ( ) . ( ) .q q qx y z3 3 39 7 99 0 86 3= ∞ = ∞ = ∞

(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,( ) . .lx 3 0 98571= -

Then ( ) . ( ) . ( ) .q q qx y z3 3 3170 3 81 0 93 7= ∞ = ∞ = ∞

300


PROBLEM 9.184*

For the component described in Problems 9.148 and 9.170, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION

(a) From the solutions to Problems 9.148 and 9.170. We have

I I I

I I I

x y z

xy zx

= ◊ = = ◊

= = ◊

0 32258 0 41933

0 096768

. .

.

kg m kg m

kg m

2 2

2yyz = 0

Substituting into Eq. (9.56) and using

I I I I Iy z xy zx yz= = = 0

K K3 2

2

0 32258 2 0 41933

2 0 32258 0 41933 0 41933

- +

+ + -

[ . ( . )]

[ ( . )( . ) ( . ) 22 0 096768

0 32258 0 41933 2 0 41933 0 096768

2

2 2

( . ) ]

[( . )( . ) ( . )( . )

K

- - ]] = 0

Simplifying

K K K3 21 16124 0 42764 0 048869 0- + - =. . .

Solving yields

K1 0 22583= ◊. kg m2 or kg m2K1 0 226= ◊.

K2 0 41920= ◊. kg m2 or kg m22K = ◊0 419.

K3 0 51621= ◊. kg m2 or kg m2K3 0 516= ◊.

(b) To determine the direction cosines l l lx y z, , of each principal axis, use two of the equations ofEqs. (9.54) and (9.57). Then

K1 : Begin with Eqs. (9.54b) and (9.54c):

- + - - =I I K Ixy x y y yz z( ) ( )( ) ( )l l l1 1 1 3 0

- - + - =I I I Kzx x yz y z z( ) ( ) ( )( )l l l1 1 1 3 0

0

0

301


Problem 9.184* (continued)

Substituting

- + - =

- +

( . )( ) ( . . )( )

( . )( )

0 096768 0 41933 0 22583 0

0 0967681 1

1

l l

lx y

x (( . . )( )0 41933 0 22583 01- =lz

Simplifying yields

( ) ( ) . ( )l l ly z x1 1 10 50009= =


( ) [ . ( ) ]l lx x2

122 0 50009 1+ =

or ( ) .lx 1 0 81645=

and ( ) ( ) .l ly z1 1 0 40830= =

( ) . ( ) ( ) .q q qx y z1 1 135 3 65 9= ∞ = = ∞

K 2 : Begin with Eqs. (9.54a) and (9.54b)

( )( ) ( ) ( )I K I Ix x xy y zx z- - - =2 2 2 2 0l l l

- + - - =I I K Ixy x y y yz z( ) ( )( ) ( )l l l2 2 2 2 0

Substituting

( . . )( ) ( . )( ) ( . )( )0 32258 0 41920 0 096768 0 096768 02 2 2- - - =l l lx y z (i)

- + - =( . )( ) ( . . )( )0 96768 0 41933 0 41920 02 2l lx y (ii)

Eq. (ii) fi ( )lx 2 0=

and then Eq. (i) fi ( ) ( )l lz y2 2= -


( ) ( ) [ ( ) ]l l lx y y22

22

22 1+ + - =

or ( )ly 21

2=

and ( )lz 21

2= -

( ) ( ) ( )q q qx y z2 2 290 45 135= ∞ = ∞ = ∞

0

0

302



K 3 : Begin with Eqs. (9.54b) and (9.54c):

I I K Ixy x y y yz z( ) ( )( ) ( )l l l3 3 3 3 0+ - + =

- - + - =I I I Kzx x yz y z z( ) ( ) ( )( )l l l3 3 3 3 0

Substituting

- + - =( . )( ) ( . . )( )0 096768 0 41933 0 51621 03 3l lx y

- + - =( . )( ) ( . . )( )0 096768 0 41933 0 51621 03 3l lx z

Simplifying yields

( ) ( ) ( )l l ly z x3 3 3= = -


( ) [ ( ) ]l lx x32

322 1+ - = (i)

or ( )lx 31

3=

and ( ) ( )l ly z31

3= = -

( ) . ( ) ( ) .q q qx y z3 3 354 7 125 3= ∞ = = ∞

(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;

That is, ( )lx 31

3= -

Then ( ) .qx 3 125 3= ∞ ( ) ( ) .q qy z3 3 54 7= = ∞

0

0

303


PROBLEM 9.185

Determine by direct integration the moments of inertia of the shaded area with respect to the x and y axes.

SOLUTION

At x a y y b1 1 2= = =,

y1 : b ka kb

a= =2

2or

y2 : b ma mb

a= =or

Then yb

ax1 2

2=

yb

ax2 =

Now dI y y dx

b

ax

b

ax dx

x = -( )= -

Ê

ËÁˆ

¯

1

3

1

3

23

13

3

33

3

66

Then I dIb

ax

ax dx

b

ax

ax

x x

a

a

= = -ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚

Ú Ú3

33

66

0

3

34

67

0

3

1 1

3

1

4

1

7

or I abx = 1

283

Also dI x dA x y y dx xb

ax

b

ax dxy = = - = -Ê

ËÁˆ¯

ÈÎÍ

˘˚

2 22 1

22

2[( )

Then I dI ba

xa

x dx

ba

xa

x

y y

a

a

= = -ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚

Ú Ú1 1

1

4

1

5

32

4

0

42

5

0

or I a by = 1

203

304


SOLUTION

We have dA ydx bx

adx= = - Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

11 2/

Then A dA bx

adx

b xa

x ab

a

a

= = - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= -ÈÎÍ

˘˚

=

Ú Ú 1

2

3

1

3

1 2

0

3 2

0

/

/

Now dI y dx bx

adx

b x

a

x = = - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= - Ê

1

3

1

31

31 3

31 2

3

3

/

ËËÁˆ¯

+ ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

1 2 3 2

3/ /x

a

x

adx

Then I dIb x

a

x

a

x

adxx x= = - Ê

ËÁˆ¯

+ ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

Ú3 1 2 3 2

0 31 3 3

/ /aa

ab

xa

xa

xa

x

Ú

= - + +ÈÎÍ

˘˚

33 2 2

3 25 2

03

2 3

2

2

5/

//

or I abx = 1

303

and kI

A

ab

abxx2

130

3

13

= =

or kb

x =10

Also dI x dA x ydx x bx

adxy = = = - Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô2 2 2

1 2

1( )/

PROBLEM 9.186

Determine the moments of inertia and the radii of gyration of the shaded area shown with respect to the x and y axes.

305



Then I dI b xa

x dx b xa

xy y

aa

= = -ÊËÁ

ˆ¯

= -ÈÎÍ

˘˚ÚÚ 2 5 2 3 7 2

00

1 1

3

2

7/ /

or I a by = 1

213

and kI

A

a b

abyy2

121

3

13

= =

or ka

y =7

306


SOLUTION

At x a y y b= = =, :1 2 y b ka kb

a

y b b ca cb

a

12

2

22

22

:

:

= =

= - =

or

or

Then yb

ax

y bx

a

1 22

2

2

22

=

= -Ê

ËÁˆ

¯

Now dA y y dx

bx

a

b

ax dx

b

aa x dx

= -

= -Ê

ËÁˆ

¯-

È

ÎÍÍ

˘

˚˙˙

= -

( )

( )

2 1

2

2 22

22 2

2

2

Then A dAb

aa x dx

b

aa x x

ab

a

a

= = -

= -ÈÎÍ

˘˚

=

ÚÚ2

2 1

3

4

3

22 2

0

22 3

0

( )

Now dI x dA xb

aa x dxy = = -È

ÎÍ˘˚

2 22

2 22( )

PROBLEM 9.187


307



Then I dIb

ax a x dx

b

aa x x

y y

a

a

= = -

= -ÈÎÍ

˘˚

Ú Ú2

2 1

3

1

5

22 2 2

0

22 3 5

0

( )

or I a by = 4

153

and kI

A

a b

abay

y24

153

43

21

5= = =

or ka

y =5

308


SOLUTION

We have I I I Ix x x x= + -( ) ( ) ( )1 2 3

where ( ) ( )( )I a a ax 13 41

122 2

4

3= =

Now ( ) ( )I I aAA BB2 34

8= = p

and ( ) ( )I I AdAA x2 22

2= +

or ( ) ( )I I a aa

ax x2 32 34 2

2

14

8 2

4

3 8

8

9= = - Ê

ËÁˆ¯

ÊËÁ

ˆ¯

= -ÊËÁ

ˆ¯

p pp

pp

Then ( ) ( )I I Ad a a aa

x x2 2 22 4 2

2

2 8

8

9 2

4

3

4

= + = -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

+ÊËÁ

ˆ¯

=

pp

pp

33

5

8

8

8

9 2

4

3 3 32 4 2

3

+ÊËÁ

ˆ¯

= + = -ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

p

pp

p

a

I I Ad a a ax x( ) ( ) --ÊËÁ

ˆ¯

= - +ÊËÁ

ˆ¯

4

3

4

3

5

8

2

4

a

a

pp

Finally I a a ax = + +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

- - +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

4

3

4

3

5

8

4

3

5

84 4 4p p

I ax = 4 4

Also I I I Iy y y y= + -( ) ( ) ( )1 2 3

where ( ) ( )( ) ( ) ( )I a a a a ay 13 2 2 41

122 2 2

16

3= + =

( ) ( ) ( )I I a a ay y2 34 2 2

8 2= = + Ê

ËÁˆ¯

ÈÎÍ

˘˚

p p

I ay = 16

34

PROBLEM 9.188

Determine the moments of inertia of the shaded area shown with respect to the x andy axes.

309


SOLUTION



1p2

54 36 3053 6( )( ) .= 48

15 2789p

= . 46 656,

2 - = -p2

18 508 92( ) .24

7 6394p

= . -3888

S 2544 7. 42 768,

Then Y A y A YS S= =: ( . ) ,2544 7 42 768mm mm2 3

or Y = 16 8067. mm

(a) J J JO O O= -

= + -

=

( ) ( )

( )( )[( ) ( ) ] ( )

(

1 2

2 2 4

854 36 54 36

418

p pmm mm mm mm mm

33 2155 10 0 0824 10

3 1331 10

6 6

6

. . )

.

¥ - ¥

= ¥

mm

mm

4

4

or JO = ¥3 13 106. mm4

(b) J J A YO C= + ( )2

or JC = ¥ -3 1331 10 2544 7 16 80676 2. ( . )( . )mm mm mm4 2

or JC = ¥2 41 106. mm4

PROBLEM 9.189


310


SOLUTION

L76 76 6 4¥ ¥ . : A

I Ix y

=

¥ = ¥

929

0 512 106

mm

mm

2

4.

L152 102 12 7¥ ¥ . : A

I

I

x

y

=

= ¥

= ¥

3060

2 59 10

7 20 10

6

6

mm

mm

mm

2

4

4

.

.

First locate centroid C of the section:

A, mm2 y yA, mm3

1 2 3060 6120( ) = 24.9 152,388

2 ( )( )16 540 8640= 270 2,332,800

3 2 929 1858( ) = 518.8 963,930

S 16,618 3,449,118

Then Y A y AS S=

or Y ( , ) , ,16 618 3 449 118mm2 =

or Y = 207 553. mm

Now I I I Ix x x x= + +2 21 2 3( ) ( ) ( )

PROBLEM 9.190

To form an unsymmetrical girder, two L76 ¥ 76 ¥ 6.4 mm angles and two L152 ¥ 102 ¥ 12.7-mm angles are welded to a 16-mm steel plate as shown. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.

311



where ( ) . ( )[( . . ) ]

.

I I Adx x12 6 22 59 10 3060 207 553 24 9

104

= + = ¥ + -

=

mm mm mm4 2

6678 10

1

1216 540 16 540 2

6

22 3

¥

= + = +

mm

mm mm mm mm

4

( ) ( )( ) ( )( )[(I I Adx x 770 207 553

243 645 10

0 512 10

2

6

32 6 4

-

= ¥

= + = ¥ +

. ) ]

.

( ) . (

mm

mm

mm

4

I I Adx x 9929 518 8 207 553

90 5086 10

2

6

mm mm

mm

2

4

[( . . ) ]

.

-

= ¥

Then I x = + + ¥[ ( . ) . ( . )]2 104 678 243 645 2 90 5086 106 mm4

or I x = ¥634 106 mm4

Also I I I Iy y y y= + +2 21 2 3( ) ( ) ( )

where ( ) . ( )[( . ) ]

.

I I Ady y12 6 27 20 10 3060 8 50 3

17 6006 10

= + = ¥ + +

= ¥

mm mm mm4 2

66

23

6

32

1

12540 16

0 1843 10

0 5

mm

mm mm

mm

4

4

( ) ( )( )

.

( ) .

I

I I Ad

y

y y

=

= ¥

= + = 112 10 929 8 21 2

1 3041 10

6 2

6

¥ + +

= ¥

mm mm mm

mm

4 2

4

( )[( . ) ]

.

Then I y = + + ¥[ ( . ) . ( . )]2 17 6006 0 1843 2 1 3041 106 mm4

or I y = ¥38 0 106. mm4

312


SOLUTION

We have

I I Ixy xy xy= +( ) ( )1 2

For each rectangle:



Thus I x yAxy = S


1 76 12 7 965 2¥ =. . 19.1 -37 85. - ¥697 777 103.

2 114 3 12 7 1451 61. . .¥ = -12 55. 25.65 - ¥467 284 103.

S - ¥1165 061 103.

I xy = - ¥1 165 106 4. mm

PROBLEM 9.191

Using the parallel-axis theorem, determine the product of inertia of the L127 ¥ 76 ¥ 12.7 mm angle cross section shown with respect to the centroidal x and y axes.

313


SOLUTION

From Figure 9.13a:

I Ix y= ¥ = ¥3 93 10 1 06 106 6. .mm mm4 4


I xy = - ¥1 16506 106. mm4


X Y( . , . ), ( . , . )3 93 10 1 1651 10 1 06 10 1 1651 106 6 6 6¥ - ¥ ¥ ¥

Now I I Ix yave4mm= + = + ¥ = ¥1

2

1

23 93 1 06 10 2 495 106 6( ) ( . . ) .


˘˚

+ = - ¥ÈÎÍ

˘˚

+ - ¥1

2

1

23 93 1 06 10 1 1651

22 6

2

( ) ( . . ) ( . 110

1 84843 10

6 2

6

)

.= ¥ mm4


tan

( . )

( . . )

.

22

2 1 1651 10

3 93 1 06 10

0 81192

6

6

qmxy

x y

I

I I= -

-

= - - ¥- ¥

=

or 2 39 074qm = ∞.

and qm = ∞19 537.

Then a = ∞ - ∞ + ∞= ∞

180 39 074 60

80 926

( . )

.

PROBLEM 9.192

For the L127 ¥ 76 ¥ 12.7 mm angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30∞ clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia.

314



(a) We have I I Rx¢ = - = ¥ - ¥ ∞

= ¥ave

m

cos . . cos .

.

a 2 495 10 1 84843 10 80 926

2 2035 10

6 6

6 mm4

or I x¢ = ¥2 20 106. mm4

I I Ry¢ = + = ¥ + ¥ ∞

= ¥ave cos . . cos( . )

.

a 2 495 10 1 84843 10 80 926

2 7865 10

6 6

66 mm4

or I y¢ = ¥2 79 106. mm4

I Rx y¢ ¢ = - = - ¥ ∞

= - ¥

sin . sin .

.

a 1 84843 10 80 926

1 82530 10

6

6 mm4

or I x y¢ ¢ = - ¥1 825 106. mm4

(b) First observe that the principal axes are obtained by rotating the xy axes through

19.54° counterclockwise

about C.

Now I I Rmax, ( . . )min ave4mm= ± = ± ¥2 495 1 84843 106

or Imax .= ¥4 34 106 mm4

Imin .= ¥0 647 106 mm4


315


SOLUTION


= - ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

m V tA

t a a aa

ta

r r

r

r

( )( ) ( )21

22

2

3

22

Also I tI

m

aI

mass area

area

=

=

r2

3 2

(a) Now I I I

a aa

a

x x x, , ,( ) ( )

( )( ) ( )

area area area= -

= - ÊËÁ

ˆ¯

1 2

3

2

1

122 2

1

12 233

47

12

ÈÎÍ

˘˚

= a

Then Im

aax, mass = ¥2

3

7

1224

or I max = 7

182

(b) We have I I I

a a aa

z z z, , ,( ) ( )

( )( ) ( )

area area area= -

= - ÊËÁ

ˆ¯

1 2

33

2

1

32 2

1

12 2

ÈÈ

ÎÍÍ

˘

˚˙˙

= 31

484a

Then Im

aa

ma

z, mass = ¥

=

2

3

31

4831

72

24

2

PROBLEM 9.193

A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the x axis, (b) the y axis.

316



Finally, I I I

ma ma

ma

y x z, , ,mass mass mass= +

= +

=

7

18

31

72

59

72

2 2

2

or I may = 0 819. 2

317


SOLUTION

First note that the x axis is a centroidal axis so that

I I mdx= +, mass2

and that from the solution to Problem 9.115,

I max, mass = 7

182

(a) We have I ma m aAA¢ = +, ( )mass7

182 2

or I maAA¢ = 1 389. 2

(b) We have I ma m aBB¢ = +, ( )mass7

1822 2

or I maBB¢ = 2 39. 2

PROBLEM 9.194

A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the axis AA¢, (b) the axis BB¢, where the AA¢ and BB¢ axes are parallel to the x axis and lie in a plane parallel to and at a distance a above the xz plane.

318


SOLUTION


m V t A= =r rST ST

Then m

m

1

2

7850 0 002 0 76 0 48

5 72736

7850

= ¥=

=

( )( . )( . . )

.

(

kg/m m m

kg

kg/m

3 2

3))( . ) . .

.

( )( .

0 0021

20 76 0 48

2 86368

7850 0 03

m m

kg

kg/m

2

3

¥ ¥ÊËÁ

ˆ¯

=

=m 0024

0 48

2 84101

2m m

kg

2) .

.

p ¥ÊËÁ

ˆ¯

=

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have

I I I Ix x x x= + +

= +

( ) ( ) ( )

( . )( . ) ( . ).

1 2 3

21

125 72736 0 48 5 72736

0 4kg m kg

88

2

1

182 86368 0 48 2 86368

0 48

2

2

m

kg m kg

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ +( . )( . ) ( . ).

33

1

22 84101 0 48

0 109965

22m kg mÊ

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

+ ÈÎÍ

˘˚

= +

( . )( . )

[( . 00 329896 0 036655 0 073310 0 327284

0 439861

. ) ( . . ) ( . )]

( .

+ + + ◊

= +

kg m2

00 109965 0 327284. . )+ ◊kg m2

or I x = ◊0 877. kg m2

PROBLEM 9.195

A 2 mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.

319



I I I Iy y y y= + +

= + +

( ) ( ) ( )

( . )( . . ) ( .

1 2 3

2 21

125 72736 0 76 0 48 5 7273kg m2 66

0 76

2

0 48

2

1

182 8

2 2

kg m2). .

( .

ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

+ 66368 0 76 2 863680 76

3

1

42 8412

2

kg m kg m)( . ) ( . ).

( .+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ 001 0 48

0 385642 1 156927 0 091892 0 1837

2kg m)( . )

[( . . ) ( . .

ÈÎÍ

˘˚

= + + + 885 0 163642

1 542590 0 275677 0 163642

) ( . )]

( . . . )

+ ◊

= + + ◊

kg m

kg m

2

2

or I y = ◊1 982. kg m2

I I I Iz z z z= + +

= +

( ) ( ) ( )

( . )( . ) ( . ).

1 2 3

21

125 72736 0 76 5 72736

0 7kg m kg

66

2

1

182 86368 0 76 0 48 2 86368

2

2 2

m

kg m2

ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

+ + +( . )( . . ) ( . kkg m2). .

( .

0 76

3

0 48

3

1

42 841

2 2ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

+ 001 0 48 2kg m)( . )ÈÎÍ

˘˚

I z = + + +

+ ◊

[( . . ) ( . . )

( . )]

0 275677 0 827031 0 128548 0 257095

0 163642 kg mm

kg m

2

2= + + ◊( . . . )1 102708 0 385643 0 163642

or I z = ◊1 652. kg m2

320


SOLUTION


m V= rST

Then m

m m

1

2 3

7850 0 24 0 04 0 14

10 5504

7850

= ¥ ¥=

= =

( )( . . . )

.

(

kg/m m

kg

kg

3 3

//m m kg

kg/m

3

3

) ( . ) . .

( )[

p

p

20 07 0 04 2 41683

7850

2 3

4 5

¥ÈÎÍ

˘˚

=

= =m m (( . ) ( . )] .0 044 0 04 1 909792 3¥ =m kg

Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to Problem 9.144) for a semicylinder, we have

I I I I I I I Ix x x x x x x x= + + + - =

=

( ) ( ) ( ) ( ) ( ) ( ) ( )

( .

1 2 3 4 5 2 3

1

1210 55

where

004 0 04 0 14 21

122 41683 3 0 07 02 2 2 2 kg m kg m)( . . ) ( . ) ( . ) (+È

ÎÍ˘˚

+ + .. )

( . ) ( . ) ( . )

04

1

121 90979 3 0 044 0 04

2

2 2

m

kg m m

ÈÎ ˘ÏÌÓ

¸˝˛

+ +ÈÎ ˘ +ÏÌÓ

¸˝˛

- +

( . )( . )

( . ) ( . )

1 90979 0 04

1

121 90979 3 0 044

2

2

kg m

kg m (( . )

[( . ) ( . ) ( .

0 04

0 0186390 2 0 0032829 0 0011790

2 mÈÎ ˘ÏÌÓ

¸˝˛

= + + + 00 0030557 0 0011790

0 0282605

2. ) ( . )]

.

- ◊

= ◊

kg m

kg m2

or I x = ¥ ◊-28 3 10 3. kg m2

PROBLEM 9.196

Determine the mass moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. (The density of steel is 7850 kg/m3.)

321



I I I I I Iy y y y y y= + + + -( ) ( ) ( ) ( ) ( )1 2 3 4 5

where ( ) ( ) |( ) |I I I Iy y y y2 3 4 5= =( )

Then I y = +ÈÎÍ

˘˚

+ -

1

1210 5504 0 24 0 14

2 2 416831

2

16

2 2 2( . )( . . )

( . )

kg m

kg99

0 07 2 41683 0 124 0 07

32

22

p pÊËÁ

ˆ¯

ÈÎÍ + + ¥Ê

ËÁˆ¯

˘( . ) ( . ) .

. m kg m2

˚˙˙

= + + ◊

=

[( . ) ( . . )]

.

0 0678742 2 0 0037881 0 0541678

0 1837860

kg m

k

2

gg m2◊

or I y = ¥ ◊-183 8 10 3. kg m2

Also m m m m m m m m m m= + + + - = == + ¥

1 2 3 4 5 2 3 4 5

10 5504 2 2 41683

where

k

,

( . . ) gg kg= 15 38406.

Then kI

mxx2 0 0282605

15 38406= =

◊.

.

kg m

kg

2

or kx = 42 9. mm

and kI

myy2 0 1837860

15 38406= =

◊.

.

kg m

kg

2

or ky = 109 3. mm

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