Solucionario de Problemas de Momento de Inercia

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 = A1 + A2 + A3 = [(2 (24 4)(6) + (8)(4 (48 8) + (4 (48 8)(6 (6))] mm =

(144 + 384 + 288) mm mm

= 816

 Now

2

2

mm 2

 I x = ( I x )1 + ( I x )2 + ( I x )3

where ( I  x )1 =

1

3

2

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

2

12 4 = (4 (432 32 + 104 04,97 ,976) 6) mm = 105 105,, 408

( I  x )2 = ( I  x )3 =

1 12 1

mm

4

3

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

4

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

12 4 4 = (864 + 209, 95 952) mm = 210, 81 816 mm mm

Then

 I  x = (105,40 ,408 8 + 73,7 ,72 28 + 210,8 ,81 16) mm 4 = 389,952 mm

and

k  x2 =

 I  x  A

=

4

or  I  x = 390 × 103 m mm m4 

389,, 952 mm 4 389 816 mm

2

or

k  x = 21.9 mm 

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1416

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 = A1 − A2 − A3 2

(6)) − (4 (4))(2 (2)) − (4 (4))(1)] in. = [(5)(6 = (30 − 8 − 4) = 18

 Now where

2

in. in

in.2

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

1 12 1 12

(5 in.)(6 in.)3 = 90 in.4 3

2

(4 in.)(2 in.) + (8 in. )(2 in.)

= 34

2 3

2

in.4

3  ( I  x )3 = (4 in.)(1 in.) + (4 in. )  in.  12 2  1

=9

1 3

3

4

in.

 

Then

 I  x =  90 − 34

and

k  x2 =

 I  x  A

2

2

=

2 3

−9

1

4

 in. 3

46.0 in.4 18 in.4

or

 I  x = 46.0 in.4 

or

k  x = 1.599 in. 

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1417

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 = A1 + A2 + A3 = [(24 × 6) + (8)(48) + (48)(6)] = (144 + 384 + 288) mm = 816

 Now

mm 2

2

mm 2

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

where ( I  y )1 = ( I  y ) 2 = ( I  y )3 = Then

and

1 12 1 12 1 12

(6 mm)(24 mm)3 = 6912 mm 4 (48 mm)(8 mm)3 = 2048 mm 4 (6 mm)(48 mm) 3 = 55, 296 mm 4

 I  y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4

k  y2 =

 I  y  A

=

64,256 mm 816 mm

or 

 I  y = 64.3 × 103 mm 4 

or

k  y = 8.87 mm 

4

2

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1418

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 = A1 − A2 − A3 = [(5)(6) − (4)(2) − (4)(1)] in. = (30 − 8 − 4) = 18

 Now

2

2

in.

in.2

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

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

1 12 1 12

(6 in.)(5 in.)3 = 62.5 in.4 (2 in.)(4 in.)3 = 10

2 3

in.4

1

1 (1 in.)(4 in.)3 = 5 in.4 12 3

 

Then

 I  y =  62.5 − 10

and

k  y2 =

 I  y  A

=

2 3

−5

1

4

 in. 3

46.5 in.4

4 or  I  y = 46.5 in. 

or

18 in.2

k  y = 1.607 in. 

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1419

PROBLEM 9.35 Determine the moments of inertia of the shaded area shown with respect to the  x and y axes when a  = 20 mm.

SOLUTION We have where

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

1

(40 mm)(40 mm)3

12

213.33 × 103 mm 4

π ( I  x )2 =  (20 mm) 4  8 +

2

 4 × 20   (20 mm)  + 20  mm  2   3π   

π  

2

2

= 527.49 × 10

Then

−

2  4 × 20   (20 mm)  mm   2  3π    

π  

3

mm 4

 I  x = [213.33 + 2(527.49)] × 103 mm 4 or   I  x = 1.268 × 106 mm 4 

Also where

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

Then

1 12 π  

8

(40 mm)(40 mm) 3 = 213.33 × 10 3 mm 4

(20 mm)4 = 62.83 × 103 mm 4

 I  y = [213.33 + 2(62.83)] × 103 mm 4 or 

 I  y = 339 × 103 mm 4 

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1420

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 − 2(Semicircles)

For a  = 20 mm,  we have

Square:

 I x = I y =

1 12

(60) 4 = 1080 ×103 mm 4

Semicircle :  I  x =

π  

8

(20) 4 = 62.83 × 103 mm 4

 I AA′ = I y′ + Ad 2 ;

π

8

 π    (20) 2 (8.488) 2  2

(20)4 = I y′ + 

 I  y′ = 62.83 × 103 − 45.27 ×103  I  y′ = 17.56 × 103 mm 4  I y = I y′ + A(30 − 8.488)2 = 17.56 ×103 + 3

 I  y = 308.3 × 10 mm

π  

2

(20) 2 (21.512) 2

4

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1421

PROBLEM 9.36 (Continued)

Semicircle : Same as semicircle . Entire Area:  I  x = 1080 × 103 − 2(62.83 ×103 ) = 954.3 × 10

3

mm 4

 I  x = 954 × 103 mm 4 

 I  y = 1080 × 103 − 2(308.3 ×103 ) =

463.3 × 103 mm 4

 I  y = 463 ×103 mm 4 

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1422

PROBLEM 9.37 For the 4000-mm2 shaded area shown, determine the distance d 2 and the moment of inertia with respect to the centroidal axis  parallel to  AA′ knowing that the moments of inertia with respect to 4  AA′ and BB′ are 12 × 106 mm 4  and 23.9 × 106 mm , respectively, and that d 1 = 25 mm.

SOLUTION We have

 I AA′ = I + Ad 12  

and

 I BB′ = I + Ad 22

Subtracting

(1)

(

 I AA′ − I BB′ = A d12 − d 22

or 

d 22 = d 12 −

)

 I AA′ − I BB′  A 6

=

= 3600

Using Eq. (1):

2

(25 mm) − mm

(12 − 23.9)10 mm 4000 mm

4

2

2

or  d 2 = 60.0 mm

 I  = 12 × 106 mm 4 − (4000 mm 2 )(25 mm) 2

or  I  = 9.50 × 106 mm 4 

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1423

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 d 1 = 25 mm  and d 2 = 15 mm  and that the moments of inertia with 4 respect to  AA′  and  BB′  are 7.84 × 106 mm 4   and 5.20 × 106 mm , respectively.

SOLUTION 2

We have

 I AA′ = I + Ad 1  

and

 I BB′ = I + Ad 22

Subtracting or 

(

 I AA′ − I BB′ = A d12 − d 22  A =

(1)

) 6

 I AA′ − I BB′

=

d12 − d 22

(7.84 − 5.20)10 mm

4

(25 mm) 2 − (15 mm)2 or 

Using Eq. (1):

6

4

2

 I  = 7.84 × 10 mm − (6600 mm )(25 mm) 6

= 3.715 × 10

mm

 A = 6600 mm 2 

2

4

or   I  = 3.72 × 106 mm 4 

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1424

PROBLEM 9.39 2

The shaded area is equal to 50 in. . Determine its centroidal moments of inertia  I  x and  I  y , knowing that  I y = 2 I x  and that the polar moment 4 of inertia of the area about Point  A is  J   A = 2250 in. .

SOLUTION Given:

2

 A = 50 in.

 I y = 2 I x ,

4

J A = 2250 in.

 J A = J C  + A(6 in.) 4

2

2

2

with

I y = 2 I x

2250 in. =  J C  + (50 in. )(6 in.)  J C  = 450 in.4 C =  I x +

Iy

450 in.4 = I x + 2 I x

 I  x = 150.0 in.4  4  I y = 2 I x = 300 in. 

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1425

PROBLEM 9.40 The polar moments of inertia of the shaded area with respect to Points  A,  B, and  D are, respectively,  J A = 2880 in.4, J B = 6720 in.4, 4 and J  D  =  4560 in. . Determine the shaded area, its centroidal moment of inertia C ,  and the distance d  from C  to D.

SOLUTION See figure at solution of Problem 9.39.  J A = 2880 in.4 ,

Given:

J B = 6720 in.4 ,

J D = 4560 in.4

 B =  J C +

A(CB) 2 ; 6720 in.4 = J C  + A(62 + d 2 )  

(1)

 D =  J C +

A(CD) 2 ; 4560 in.4 = J C  + Ad 2  

(2)

Eq. (1) subtracted by Eq. (2):  J B − J D = 2160 in.4 = A(6) 2  J A = J C + A( AC ) 2 ; 2880 in.4 = J C  + (60 in.2 )(6 in.) 2 Eq. (2):

4560 in.4 = 720 in.4 + (60 in.2 )d 2

 A = 60.0 in.2   J C  = 720 in.4  d  = 8.00 in. 

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1426

PROBLEM 9.41 Determine the moments of inertia  I  x  and  I   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.

1 −

2

1 2

 A, mm2

 x , mm

108 × 60 = 6480

54

349,920

× 72 × 36 = −1296

46

–59,616

5184

Σ

Then

 xA, mm3

290,304 2

 X ΣA = Σ xA :

X (5184 mm ) = 290, 304 mm

or 

 X  = 56.0 mm

 Now

 I x = ( I x )1 − ( I x )2

where

( I  x )1 =

1 12

 36

=

3

4

1  2 × 72 mm ×18 mm  (6 mm)  2  

(72 mm)(18 mm) +  4

3

2(11, 664 + 23, 328) mm = 69.984 × 10 mm

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

6

(108 mm)(60 mm) = 1.944 ×10 mm

1

( I  x )2 = 2 

3

3

4

]

 I  x = (1.944 − 0.069984) × 106 mm4 or

 I  x = 1.874 × 106 mm 4 

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1427

PROBLEM 9.41 (Continued)

Also where

 I y = ( I y )1 − ( I y ) 2 ( I  y )1 =

1 12

3

2

(60 mm)(108 mm) + (6480 mm )[(56.54) mm] 4

= (6, 298, 560 + 25, 920) mm = 6.324 × 10

( I  y ) 2 =

1 36

3

mm 4

2

(36 mm)(72 mm) + (1296 mm )[(56 − 46) mm] 4

= (373, 248 + 129, 600) mm = 0.502 × 10

Then

6

2

6

 I  y = (6.324 − 0.502)10 mm

6

mm

2

4

4

6

4

or   I  y = 5.82 × 10 mm 

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1428

PROBLEM 9.42 Determine the moments of inertia  I  x  and  I   of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION First locate C  of the area: Symmetry implies  X  = 12 mm.

1 2

 A, mm2

 y , mm

 yA, mm3

12 × 22 = 264

11

2904

28

6048

1 2

Σ

Then

(24)(18) = 216 480

8952

Y ΣA = Σ yA: Y (480   mm 2 ) = 8952 mm 3 Y  = 18.65 mm

 Now where

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

( I  x )2 = =

Then

1 12

(12 mm)(22 mm) 3 + (264 mm 2 )[(18.65 − 11) mm] 2

26,098 mm4 1 36

(24 mm)(18 mm) 3 + (216 mm 2 )[(28 − 18.65) mm] 2

22, 771mm 4

 I  x = (26.098 + 22.771) × 103 mm 4 or

 I  x = 48.9 × 103 mm 4 

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1429

PROBLEM 9.42 (Continued)

Also where

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

1 12

(22 mm)(12 mm) 3 = 3168 mm 4

1

( I  y ) 2 = 2 

 36

1  2 × 18 mm ×12 mm  (4 mm)  2  

(18 mm)(12 mm) 3 +  4

= 5184 mm

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

]

 I  y = (3168 + 5184)mm 4

or   I  y = 8.35 × 103 mm 4 

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1430

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.  A, in.2

 x , in.

1

5 × 8 = 40

2.5

2

−2 × 5 = −10

1.9

Σ

30

Then

 X ΣA = Σ xA:

where

4

100

160

4.3

–19

–43

81

117

 X  = 2.70 in. Y ΣA = Σ yA: Y (30   in.2 ) = 117 in.3 Y  = 3.90 in.

or  Now

 yA, in.3

X (30 in.2 ) = 81 in.3

or and

 xA, in.3

 y , in.

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

1 12

(5 in.)(8 in.)3 + (40 in.2 )[(4 − 3.9) in.]2 4

= (213.33 + 0.4) in. =

( I  x )2 =

1 12

213.73 in.4

(2 in.)(5 in.)3 + (10 in.2 )[(4.3 − 3.9) in.]2

= (20.83 + 1.60) =

22.43 in.4

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1431

PROBLEM 9.43 (Continued)

Then

 I  x = (213.73 − 22.43) in.4

Also

 I y = ( I y )1 − ( I y ) 2

where

( I  y )1 =

1 12

or   I  x = 191.3 in.4 

(8 in.)(5 in.)3 + (40 in.2 )[(2.7 − 2.5) in.]2 4

4

= (83.333 + 1.6) in. = 84.933 i n.

( I  y ) 2 =

1 12

(5 in.)(2 in.)3 + (10 in.2 )[(2.7 − 1.9)in.]2 4

4

= (3.333 + 6.4) in. = 9.733 in.

Then

 I  y = (84.933 − 9.733)in.4

or   I  y = 75.2 in.4 

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1432

PROBLEM 9.44 Determine the moments of inertia  I  x  and  I   of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION

First locate centroid C  of the area.  A, in.2

 x , in.

 y , in.

 xA, in.3

 yA, in.3

1

3.6 × 0.5 = 1.8

1.8

0.25

3.24

0.45

2

0.5 × 3.8 = 1.9

0.25

2.4

0.475

4.56

3

1.3 × 1 = 1.3

0.65

4.8

0.845

6.24

Σ

5.0

4.560

11.25

Then or and or

 X ΣA = Σ xA:

X (5 in.2 ) = 4.560 in.3  X  = 0.912 in.

Y ΣA = Σ yA: Y (5  in.2 ) = 11.25 in.3 Y  = 2.25 in.

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1433

PROBLEM 9.44 (Continued)

 Now where

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

1 12

(3.6 in.)(0.5 in.)3 + (1.8 in.2 )[(2.25 − 0.25)in.]2 4

4

= (0.0375 + 7.20) in. = 7.2375 i n.

( I  x ) 2 =

1 12

(0.5 in.)(3.8 in.)3 + (1.9 in.2 )[(2.4 − 2.25)in.]2 4

= (2.2863 + 0.0428) in. =

( I  x )3 =

1 12

2.3291 in.4

(1.3 in.)(1 in.)3 + (1.3 in.2 )[(4.8 − 2.25 in.)]2 4

4

= (0.1083) + 8.4533)in. = 8.5616 in.

Then

 I  x = (7.2375 + 2.3291 + 8.5616) in.4 = 18.1282 in.4  I  x = 18.13 in.4 

or Also where

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

1 12

3

2

2

(0.5 in.)(3.6 in.) + (1.8 in. )[(1.8 − 0.912)in.] 4

4

= (1.9440 + 1.4194)in. = 3.3634 in.

( I  y ) 2 =

1 12

(3.8 in.)(0.5 in.)3 + (1.9 in.2 )[(0.912 − 0.25) in.]2 4

4

= (0.0396 + 0.8327) in. = 0.8723 i n.

( I  y )3 =

1 12

(1 in.)(1.3 in.)3 + (1.3 in.2 )[(0.912 − 0.65) in.]2 4

4

= (0.1831 + 0.0892) in. = 0.2723 i n.

Then

 I  y = (3.3634 + 0.8723 + 0.2723) in.4 = 4.5080 i n.4 or

4  I  y = 4.51in. 

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1434

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 Symmetry:  X = Y 

Dimensions in mm

Determination of centroid, C , of entire section. Area, mm2

Section π  

1

4

 y , mm

 yA, mm3

42.44

333.3 × 103 250 × 103

(100) 2 = 7.854 × 103

2

(50)(100) = 5 × 103

50

3

(100)(50) = 5 × 103

 –25

Σ

17.854 × 103

−125 × 10

3

458.3 × 103

3 2 3 3 Y ΣA = Σ yA: Y (17.854   × 10 mm ) = 458.3 × 10 mm

Y = 25.67 mm

X = Y  = 25.67 mm

Distance O to C :

OC = 2Y  = 2(25.67) = 36.30 mm

(a)

Section 1:

 J O =

Section 2:

 50  2  100  2   JO = J + A(OD) = (50)(100)[50 + 100 ] + (50)(100)   +   12  2   2  

π  

8

(100) 4 = 39.27 × 106 mm 4

2

1

2

2

 J O = 5.208 × 106 + 15.625 × 106 = 20.83 × 10 6 mm 4

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1435

PROBLEM 9.45 (Continued)

Section 3: Same as Section 2;

 J O = 20.83 × 106 mm 4

Entire section:  J O = 39.27 × 106 + 2(20.83 × 106 ) = 80.94 × 10

(b)

Recall that,

6

OC = 36.30 mm

 J O = 80.9 × 106 mm 4  and

A = 17.854 × 103 mm 2

 J O = J C  + A(OC ) 2 80.94 × 106 mm 4 =  J C  + (17.854 × 103 mm 2 )(36.30 mm) 2  J C  = 57.41 × 106 mm 4

 J C  = 57.4 × 106 mm 4 

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1436

PROBLEM 9.46 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.

SOLUTION First locate centroid C  of the figure.  Note that symmetry implies Y  = 0.

Dimensions in mm

 A, mm2 π  

1

2

(84)(42) = 5541.77

2 −

3

π  

2 −

4

= −35.6507

−197,568

56

= 17.8254

49,392

= −22.9183

52,488

π  

(54)(27) = −2290.22

−

72 π  

π  

2

36

(27) 2 = −1145.11

= 11.4592

4877.32  X ΣA = Σ xA:

 –13,122

π  

–108,810

X (4877.32 mm 2 ) = −108,810 mm 3  X  = −22.3094

or

 J O = ( J O )1 + ( J O ) 2 − ( JO )3 − ( J O )4

(a)  

112

(42) 2 = 2770.88

Σ

Then

−

 XA, mm3

π  

π  

2

 X , mm

where

( J O )1 =

π  

8

2

2

(84 mm)(42 mm)[(84 mm) + (42 mm) ]

= 12.21960 × 10

6

mm 4

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1437

PROBLEM 9.46 (Continued)

( J O ) 2 =

π  

(42 mm) 4

4 6 4 = 2.44392 × 10 mm

( J O )3 = =

( J O ) 4 =

π  

8

(54 mm)(27 mm)[(54 mm) 2 + (27 mm) 2 ] 6

2.08696 × 10 mm π  

4

(27 mm) 4

4 6 4 = 0.41739 × 10 mm

 

Then

 J O = (12.21960 + 2.44392 − 2.08696 − 0.41739) × 10 6 mm 4 = 12.15917 × 10

6

mm 4 or

 J O = 12.16 × 106 mm 4 

 J O = J C  + AX 2

(b) or

 J C  = 12.15917 × 106 mm 4 − (4877.32 mm 2 )( −22.3094 mm) 2 or

 J C  = 9.73 × 106 mm 4 

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1438