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**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

+

=

=

816 mm

2

*A*

3

=

*I*

1

+

*I*

2

+

*I*

3

2

2

*I*

1

=

1

12

=

(24 mm)(6 mm)

3

=

105, 408 mm

4

4

+

*I*

3

2

=

=

1

12

1

12

=

(8 mm)(48 mm)

3

(48 mm)(6 mm)

3

4

=

73,728 mm

4

+

=

210,816 mm

4

4

*I x*

=

=

389,952 mm

4

*k*

2

*x*

=

*I x*

*A*

=

389,952 mm

4

816 mm

2 or or

2

2

*I x*

=

4

*k x*

=

21.9 mm

*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*

′

=

×

×

*I*

*BB*

′

−

*I*

*AA*

′

=

4

4

*A*

(30 mm)

2

*A*

(40 mm)

2

6

=

*A*

(40

2

−

× 6 =

*A*

(700)

×

6

(4000)(30)

2

(1)

*I*

=

*A*

=

4000 mm

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. *

**1444 **

**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*

.

First locate centroid

*C*

of the area.

Symmetry implies

*Y*

=

30 mm.

*A*

, mm

2

*x*

, mm

1

54

*xA*

, mm

3

349,920

2

Σ

2

72 36

= −

1296

5184

46

Then

Then

= Σ

*xA*

:

*X*

or

Now where

*X*

=

56.0 mm

*I x*

=

*I*

1

−

*I*

2

*I*

*I*

1

=

1

12

(108 mm)(60 mm)

3 =

2

=

2

1

36

(72 mm)(18 mm)

3 +

=

4

=

1

2

×

×

*I*

2

is obtained by dividing

*A*

2 into ]

*I x*

=

4

3

–59,616

4

290,304

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

−

2

*I*

1

=

1

12

=

(60 mm)(108 mm)

3

+

4 =

*I y*

2

=

=

1

36

(36 mm)(72 mm)

3

+

+

4

=

=

4

2

−

4

4

2 or

*I y*

=

4

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

**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

= −

10

30

Then or and or

Now where

*x*

, in.

2.5

1.9

*y*

, in.

4

4.3

= Σ

*xA*

:

*X*

=

3

*X*

=

2.70 in.

= Σ

*yA*

:

*Y*

3

*Y*

=

3.90 in.

*I x*

=

*I*

1

−

2

*I*

*I*

1

=

1

12

=

(5 in.)(8 in.)

3

+

+

4

−

=

213.73 in

4

2

=

1

12

=

(2 in.)(5 in.)

3

+

4

*xA*

, in

3

100

–19

81

2

2

*yA*

, in

3

160

–43

117

*PROPRIETARY MATERIAL.*

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

**1450 **

Then

Also where

Then

*I x*

=

*I y*

=

*I*

1

−

2

4

*I*

1

=

1

12

(8 in.)(5 in.)

3

=

+

4 =

84.933 in

4

2

=

1

12

(5 in.)(2 in.)

3

= 4

+

=

9.733 in

4

*I y*

=

4 or

*I x*

=

191.3 in

4

2

2 or

*I y*

=

75.2 in

4

*PROPRIETARY MATERIAL.*

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

**1451 **