# PROBLEM 9.31 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

# SOLUTION

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.

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):

# SOLUTION

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.

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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

.

# SOLUTION

First locate centroid

C

of the area.

Symmetry implies

Y

=

30 mm.

### Dimensions in 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.

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1446

Also where

Then

# PROBLEM 9.41 (Continued)

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

### PROPRIETARY MATERIAL.

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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

.

## SOLUTION

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.

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1450

Then

Also where

Then

# PROBLEM 9.43 (Continued)

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.