**PROBLEM 9.31 **

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

Now where

First note that

Then and

*I x*

=

1

*+ A*

2

*+ A*

3

=

=

= 816 mm 2

*= I*

1

*+ I*

2

*+ I*

3

2

*I x*

=

*= 389,952 mm 4 k 2 x*

=

*I x*

*A*

=

389,952 mm 4

816 mm 2

2

( )

1

=

1

12

=

(24 mm)(6 mm) 3 + (144 mm )(27 mm) 2

4

= 105, 408 mm 4

*I*

3

2

=

=

1

12

1

12

=

(8 mm)(48 mm) 3 = 73,728 mm 4

(48 mm)(6 mm) 3 + (288 mm )(27 mm) 2

+ 4 = 210,816 mm 4

4 or or

*I x*

*= 390 10 mm k x*

= 21.9 mm

4

*PROPRIETARY MATERIAL.*

* © 2013 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. *

**1435 **

Eq. (1):

**PROBLEM 9.39 **

Determine the shaded area and its moment of inertia with respect to the centroidal axis parallel to AA ′ , knowing that d that the moments of inertia with respect to and 6.9 × 10 6 mm 4 , respectively.

1

*= 30 mm and d*

AA ′ and BB ′

2

= 10 mm, and

are 4.1 × 10 6 mm 4

*I*

AA ′

= ×

*I*

BB ′

= ×

4

4

= + A (30 mm) 2

= + A (40 mm) 2

*I*

BB ′

*− I*

AA ′

= (6.9 4.1) 10 6 = A (40 2 − 30 )

× 6 = A (700)

4.1 10 6

I (4000)(30) 2

(1)

A = 4000 mm 2

I = 500 10 mm 4

*PROPRIETARY MATERIAL.*

* © 2013 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. *

**1444 **

**PROBLEM 9.41 **

Determine the moments of inertia respect to centroidal axes respectively parallel and perpendicular to side AB .

*I x*

* and I y*

of the area shown with

First locate centroid C of the area.

Symmetry implies Y = 30 mm.

A , mm 2 x , mm xA , mm 3

1

2

1

2

72 36 = − 1296

5184

54

46

349,920

–59,616

Σ 290,304

Then X A = Σ xA : X = 3 or

Now where

X = 56.0 mm

*I x*

*= I*

1

*− I*

2

*I*

1

=

1

12

(108 mm)(60 mm) 3 = ×

*I*

2

= 2

1

36

(72 mm)(18 mm) 3 +

1

2

×

=

4

+ 4 = × 4

[( )

2

* is obtained by dividing A*

2 into ]

*Then I x*

*= − × 4 or I x*

=

2

4

*PROPRIETARY MATERIAL.*

* © 2013 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. *

**1446 **

Also where

Then

*I y*

*= I*

1

*− I*

2

( )

1

=

1

12

=

(60 mm)(108 mm) 3 + (6480 mm )[(56.54) mm] 2

+ 4 = × 4

*I*

2

=

=

1

36

(36 mm)(72 mm) 3 + −

+ 4 = × 4

*I y*

= − 4

*2 or I y*

= 4

*PROPRIETARY MATERIAL.*

**1447 **

**PROBLEM 9.43 **

*Determine the moments of inertia I x*

* and I y*

of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB .

First locate centroid C of the area.

1

2

Σ

A , in 2

5 8 40

2 5 = − 10

30

Then or x , in.

2.5

1.9 y , in.

4

4.3

X A = Σ xA : X = 3

X = 2.70 in.

and or

Now where xA , in

100

–19

81

3

Y A = Σ yA : Y = 3

Y = 3.90 in.

*I x*

*= I*

1

*− I*

2

*I*

*I*

1

=

1

12

=

(5 in.)(8 in.) 3 + − 2

+ 4 = 213.73 in 4

2

=

1

12

=

(2 in.)(5 in.) 3 + − 2

4 yA , in 3

160

–43

117

*PROPRIETARY MATERIAL.*

**1450 **

Then

Also where

Then

*I x*

=

*I y*

*= I*

1

*− I*

2

4 or

*I*

1

=

1

12

(8 in.)(5 in.)

3 + − 2

= + 4 = 84.933 in 4

2

=

1

12

(5 in.)(2 in.)

3 + − 2

= + 4 = 9.733 in 4

*I x*

= 191.3 in 4

*I y*

*= 4 or I y*

= 75.2 in

4

*PROPRIETARY MATERIAL.*

**1451 **