141 150

 

PROBLEM 9.141

A 2-mm-thick piece of sheet steel is cut and bent into the machine 3 component shown. Knowing that the density of steel is 7850 kg/m , determine the moment of inertia of the component with respect to each of the coordinate axes.

SOLUTION

m =  ρstV = ρ sttA  

Have Then

m1  = 7850 kg kg/m3 × 0.002 m × ( 0.320 × 0.240 ) m 2  

= 1.20576 kg   m2  = 7850 kg kg/m3 × 0.002 m × ( 0.320 × 0.180 ) m 2  

= 0.90432 kg  

 

1  × 0.180 × 0.240  m 2   2 

m3 = 7850 kg kg/m3 × 0.002 m × 

= 0.33912 kg     2  π  m4 = 7850 kg kg/m3 × 0.002 m ×  ( 0.100 m )    2 

= 0.24662 kg   Using Fig. 9-2B for components 1 and 2 and the equations derived above for components 3 and 4, have

=  Now where

 I x

+

+

−

( I x )1 ( I x )2 ( I x )3 ( I x )4   1

2 1.20 2057 576 6 kg kg ) ( 0. 0.24 240 0 m)   ( I  x ) 1 =  (1.

3

= 23.151 × 10−3 kg ⋅ m 2   1

2 0.90432 0432 kg ) ( 0. 0.18 180m 0m )   ( I  x ) 2 =  ( 0.9

3

= 9.7667 × 10−3 kg ⋅ m2   1

2 2 ( I  x )3 = ( 0. 33912 kg ) ( 0.180 ) + ( 0.240 )  m2  

6

−3

2

= 5.0868 × 10 kg ⋅ m  

 

 

PROBLEM 9.141 CONTINUED

1

2 .246 466 62 kg kg ) ( 0.1 .100 00 m )   ( I  x ) 4 = ( 0.2

4

= 0.61655 × 10−3 kg ⋅ m 2   Then

 I  x

= + +  − × −3  ⋅ 2 ( 23.151 9.7667 5.0868 0.61655 ) 10  kg m   or  I  x = 37 .4 × 10−3 kg ⋅ m 2  

( )1 + ( I y )2 + ( I y )3 − ( I y )4  

 I y = I y

And

1

( I  y )1 =  3 (1. 20576 kg ) ( 0.320)2 + ( 0.240 )2  m2  

where

= 64.307 × 10−3 kg ⋅ m2   1

2 0.90 9043 432 2 kg kg ) ( 0. 0.32 320 0 m)   ( I  y ) 2 =  3 ( 0.

= 30.867 × 10−3 kg ⋅ m2   1

2 0.33 3391 912 2 kg kg ) ( 0.2 .240 40 m )   ( I  y ) 3 =  6 ( 0.

= 3.2556 × 10−3 kg ⋅ m2    1



( I  y )4  = ( 0.24662 kg )     2 − 9π 2  ( 0.100 m )2    16

 



2  2  4 × 0.100    2  + ( 0.160 ) +    m     3π         

= 7.5466 × 10−3 kg ⋅ m2   Then

 I  y = ( 64.307 + 30.067  + 3.2556 − 7.5466 )10−3  kg ⋅ m 2  





or  I  y = 90 .9 × 10−3 kg ⋅ m 2  And  where

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

2 ( I  z )1 = (1.20576 kkgg ) ( 0. 320 m ) =

3

41.157 × 10−3 kg ⋅ m 2  

1

2 2 kg ) ( 0.320 ) + ( 0.180 )  m 2   ( I  z )2 = ( 0. 90432 kg  

3

= 40.634 × 10−3 kg ⋅ m2   2 ( I  z )3 = 1 ( 0.33912 kkgg ) ( 0. 180 m ) = 1.83125 × 10−3 kg ⋅ m2  

6

 

 

PROBLEM 9.141 CONTINUED

 

 1  4



2 2  ( I  z )4 = ( 0.24662 kg )  ( 0.100 m )  +   ( 0.160 m )    





= 6.9300 × 10−3 kg ⋅ m 2   Then

( I  z ) z  = ( 41.157 + 40.634  + 1.83125 − 6.9300 ) × 10−3  kg ⋅ m 2   or  I  z  = 76 .7 × 10−3 kg ⋅ m 2  

 

 

PROBLEM 9.142

The piece of roof flashing shown is formed from sheet copper that is 3 0.032 in. thick. Knowing that the specific weight of copper is 558 lb/ft , determine the mass moment of inertia of the flashing with respect to each of the coordinate axes.

SOLUTION

Have

m =  ρ copper V   

= Then

γ copper   g 

tA  

m1 = m2   3

558 lb/ft 3

 1 ft  2 0 . 0 3 2 i n . 4 8 6 i n = 32. ×( × ) ×  12 in.    32.2 2 ft/ ft/ss × 2

.09 92422 2422 lb ⋅ s 2 /ft   = 0.0 Using Fig. 9-2B for components 1 and 2, have  Now

Then

 I x = ( I x )1 + ( I x )2  

and

( I x )1 = ( I x )2  

2 1 6   2  I  x = 2  ( 0.092422 lb ⋅ s /ft )  ft   = 1.54037 lb ⋅ ft ⋅ s 2    12    3

or  I  x = 1.5  4 × 10−3 lb ⋅ ft ⋅ s 2    I y = I y

( )1 + ( I y )2  

where

( I  y )1

 48 2  6 2  2 =  ( 0.092422 lb ⋅ s /ft )   +    ft   3  12   12   1

2

= 500.62 × 10−3 lb ⋅ ft ⋅ s 2   and

( I y )2 = ∫ ry2dm  

where

r y2 = x 2 + z 2   2

=  x 2 +  (ζ  cos30)  

 

 

PROBLEM 9.142 CONTINUED

Then

γ copper 

dm =  ρ dV =

and

( I y )2 = =

=

=

γ copper 

  g

γ copper   g

1 3

tdζ dx  

(

)

t ∫ L0 ∫ 0a x 2 + ζ 2 cos 2 30° dζ dx  

 

t ∫ L0  ax 2 +

1 γ copper  3

g

g

1 3

 

a3 cos 2 30°  dx  

(

t aL3 + a3 L cos 2 30°

(

)

A = aL  

)

m2 L2 + a 2 cos 2 30°  

2  2 1   2 = ( 0.092422 lb ⋅ s /ft )   ( 4 ) +  cos 30°   ft   3 2   

1

2

= 498.69 × 10−3 lb ⋅ ft ⋅ s 2   Finally,

 I  y = ( 500.62 +  498.69 ) × 10−3 lb ⋅ ft ⋅ s 2   or  I  y = 99  9 × 10−3 lb ⋅ ft ⋅ s 2   I z = ( I z )1 + ( I z )2  

where

2

 48  ( I  z )1 =  ( 0.092422 llbb ⋅ s / ft )  ft    3  12  1

 

2

−3

= 492.92 × 10 and

2

lb ⋅ ft ⋅ s  

( I z )2 = ∫ rz 2dm  

where

r z 2 = x 2 + y 2  

and

 y =  ζ  sin30°  

Then

2 ( I z )2 = ∫  x 2 + (ζ  sin30° )  dm  

 

 

PROBLEM 9.142 CONTINUED

( )2  

Similarly, as  I  y

1

( I z )2 =  

3

(

)

m2 L2 + a 2 sin 2 30°  

 2  1 2 2  / ft )   ( 4 ) +   sin 30°  ft 2   = ( 0.092422 lb ⋅ s /f 3 2   1

2

= 494.84 × 10−3 lb ⋅ ft ⋅ s 2   Then

 I  x = ( 492.92 +  494.84 ) × 10−3 lb ⋅ ft ⋅ s 2   or  I  z  = 98  8 × 10−3 lb ⋅ ft ⋅ s2  

 

 

PROBLEM 9.143

The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a (a) the x the  x axis,  axis, (b (b) the y the  y   3 axis, (c (c) the z  the z  axis.  axis. (The specific weight of steel is 0.284 lb / i n . )

SOLUTION

m =  ρ V =

Have

γ 

V =  g 

0.284 lb/in 3 32.2 ft/s 2

×V  

 

= ( 0.0088199 lb⋅ s 2/ft ⋅ in 3 )V    Then m1 = 0.0088199  lb ⋅ s 2 /ft ⋅ in 3 π  4 2 2 in 3 = 0.88667 lb lb ⋅ s 2 /ft     ( ) ( )

(

)

2 m2 = 0.0088199  lb ⋅ s 2 /ft ⋅ in 3 π  lb ⋅ s 2 /ft     (1) (3) in 3 = 0.083126 lb

(

) 



2 m3 = 0.0088199  lb ⋅ s 2 /ft ⋅ in 3 π  lb ⋅ s 2 /ft     (1) ( 2 )  in 3 = 0.055417 lb

(

) 



Using Fig. 9-28 and the parallel theorem, theorem, have (a)  I x = ( I x )1 + ( I x )2 − ( I x )3   2 2 2 1  1 ft 2   = ( 0.88667 lb ⋅ s 2 /ft )    3 ( 4 ) + ( 2 )  + (1)  in 2  ×    12    12 in.  2

2 2 2 1  1 ft  + ( 0.083126 lb ⋅ s //fft )    3 (1) + ( 3)  + (1.5 )  in 2  ×       12    12 in.  2

2

2 2 2 1  1 ft  − ( 0.055417 lb ⋅ s /ft )    3 (1) + ( 2 )  + (1)  in.  ×       12    12 in.  2

= 0.034106 lb ⋅ ft ⋅ s 2   or  I  x  = 0. 0.0341lb ⋅ ft ⋅ s 2  

 

 

PROBLEM 9.143 CONTINUED

( b)

( )1 + ( I y )2 − ( I y )3  

 I y = I y

2

 1 2  1 ft  = ( 0.88667 lb ⋅ s /ft )    ( 4 )  in 2 ×     2   12 i n.  2

2

2 1 2  1 ft  / ft )    (1) + ( 2 )  in 2 ×  + ( 0.083126 lb ⋅ s /f    2   12 in.  2

2

2 1 2  1 ft  / ft )   (1) + ( 2 )  in 2 ×  − ( 0.055417 lb ⋅ s /f    2   12 in.  2

= 5.0125 × 10−2 lb ⋅ ft ⋅ s2   or  I  y  = 0. 0.0501lb ⋅ ft ⋅ s 2  (c )

 I z = ( I z )1 + ( I z )2 − ( I z )3   2

2 2 2 1  1 ft  = ( 0.88667 lb ⋅ s /ft )    3 ( 4 ) + ( 2 )  + (1)  in 2  ×       12    12 in.  2

2

2 2 2 2  1  1 ft  + ( 0.083126 lb ⋅ s /ft )    3 (1) + ( 3)  + ( 2 ) + (1.5 )   in 2  ×         12   12 in.  2

2

2 2 2 2  1  1 ft  − ( 0.055417 lb ⋅ s /ft )    3 (1) + ( 2 )  + ( 2 ) + (1)   in 2  ×         12   12 in.  2

0.0 03487 34876 6 llb b ⋅ ft ⋅ s 2   = 0. 2

or  I  z   = 0.0349 lb ⋅ ft ⋅ s  

 

 

To 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 9.144–9.147. From Figure 9.28 Cylinder:

( I x ) cyl =

1 2

mcyla 2  

1

( I y )cyl   = ( I z )cyl = 12   mcyl (3a2 + L2 )  

(

)

Symmetry and the definition of the mass moment of inertia  I = ∫ r 2dm   imply 1

( I )semicylinder = ( I )cylinder    2

11  ∴   ( I x )sc =    mcyla 2    2 2  and

( I y )sc  = ( I z )sc = 12  121 mcyl (3a 2 + L2 )   

1

However,

msc =

Thus,

( I x )sc =

and

2



mcyl   1 2

msc a 2  

1

( I y )sc  = ( I z )sc = 12   msc (3a 2 + L2 )  

Also, using the parallel axis theorem find

 1 − 16  a 2   2   2 9π  

 I x′ = msc 

 1 16  1  − 2  a 2 + L2    12   4 9π  

 I z ′ = msc 

where  x′ and ′ are centroidal axes.

 

 

PROBLEM 9.144

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

SOLUTION

m =  ρ steel V   

Have

m1  = 7850 kg/m3 ( 0.200 × 0.120 × 0.600 ) m3  

Then

= 113.040 kg   m   = 7850 kg kg/m3 × 0.200 × 0.080 × 0.360 m3   2

= 45.216 kg  

(

)

  2  π   m3 = 7850 kg/m3 ×  ( 0.100 ) ( 0.120 )  m3   2 

= 14.7969 kg   2 m4 = 7850  kg/m3 × π  ( 0.050 ) ( 0.120 )  m3  





= 7.3985 kg   Using Figure 9.28 for components 1 and 2 and the equations derived above for components 3 and 4, have

( )1 + ( I y )2 + ( I y )3  

 I y = I y

 Now

where

( )1  I  y

 0.600 2  0.200 2   2 2 2  1  1 1 3 . 0 4 0 k g 0 . 6 0 0 0 . 2 0 0 = + +   ( )  ( ) ( )    +  2  m   1 2 2           15.072 720 0 kg ⋅ m 2   = 15.0

 

 

PROBLEM 9.144 CONTINUED

( I  y )2

 1  0.360 2  0.200 2   2 2 2   = ( 45.216 kg )  ( 0.360 ) + ( 0.200 )  +   +  2   m   1 2 2            .55 562 kg ⋅ m 2   = 2.5

( I  y )3

2   16  4 × 0.100    2 2      1   m2    = (14.7969 kg )   − 2  ( 0.100 )  + ( 0.100 ) +  0.600 +  2 9π   3π         

= 6.3024 kg ⋅ m 2    

 1 



( I  y )4  = ( 7.3985 kg )  2 ( 0.050 )2  +   ( 0.100) 2 + ( 0.600 )2   m2   



= 2.7467 kg ⋅ m2   Then

 I  y  = (15.0720 + 2.5562 + 6.3024 − 2.7467 ) kg ⋅ m 2   21.183 839 9 kg ⋅ m 2   = 21.1 or  I  y  = 21.2 kg ⋅ m 2  

 

 

PROBLEM 9.145

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

SOLUTION

See machine elements shown in Problem 9.145 m1  = 113 113.04 .040 0 kg  

Also note

m2  = 45. 45.216 216 kg   m3  = 14. 14.796 7969 9 kg   m4  = 7.3 7.3985 985 kg   Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3 and 4, have  Now

where

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

 0.200 2  0.120 2   2 2 2  1  + + 0 . 2 0 0 0 . 1 2 0 ) ( )    +  2  m   ( 1 2 2          

( I  z )1 = (113.040 kg ) 

= 2.0498 kg ⋅ m2  

   ( I  z )2 = ( 45.216 kg ) 

1  2 2 2 2  0.200 ) + ( 0.080 )  + ( 0.100 ) + ( 0.160 )   m 2   (        12 

1.78 7845 453 3 kg⋅ m 2   = 1. 2 2 2 2    1 ( I  z )3 = (14.7969 kg )  3 ( 00..100 ) + ( 0.120 )  + ( 0.100 ) + ( 0.060 )   m2  

12



.255 5599 99 kg⋅ m 2   = 0.2

   ( I  z )4 = ( 7.3985 kg ) 

1  2 2 2 2  3( 0 0..050 ) + ( 0.120 )  + ( 0.100 ) + ( 0.060 )   m 2       12 

.114 1412 122 2 kg ⋅ m 2   = 0.1 Then

 I  z   = ( 2.0498 + 1.78453 + 0.25599 − 0.114122 ) kg ⋅ m 2   3.97 9762 629 9 kg ⋅ m 2   = 3. or  I  z   = 3.98 kg ⋅ m 2  

 

 

PROBLEM 9.146

An aluminum casting has the shape shown. Knowing that the specific weight of aluminum is 0.1 0.100 00 lb / in3 , det determ ermine ine the momen momentt of ine inerti rtiaa o off the casting with respect to the z  the z  axis.  axis.

SOLUTION

m =  ρ V =

Have

0.10 0.1 0 lb/in lb/in3 V = V   g  32.2 ft/s 32.2 ft/s 2 γ 

 

.00 031056 lb ⋅ s 2 /ft ⋅ in 3 ) V    = ( 0.0 Then

(

)

m1  = 0.0031056 lb l b s2 /ft ⋅ in 3 (12 in i n.) ( 3 in in. )( i n. )   )( 4.8 in

= 0.53665 lb ⋅ s2 /ft   m2  = 0.0031056 lb l b ⋅ s 2 /ft ⋅ in 3 (1.5 in in.) ( 4.8 in in.) ( 3 in in.) = 0.06708 lb lb ⋅ s 2 /ft  

(

)

  2  π   m3 = 0.0031056 lb lb ⋅ s 2/ft ⋅ in 3    ( 2.4 in in.) × (1.5 in in.)    2 

(

)

.0421 148lb ⋅ s 2 /ft   = 0.042   2  π   m4 = m5 = 0.0031056 llb b ⋅ s 2/ft ⋅ in 3    ( 2.75 in in.) × (1.0 iin n. )    2 

(

)

.03689 6892 lb ⋅ s 2 /ft   = 0.03

 

 

PROBLEM 9.146 CONTINUED

Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3, 4, and 5, have  I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4 + ( I z )5  

where

( I z )4 = ( I z )5  

2 2 2  1 2 2  12 in.   3 in.    1 ft    I  z  = 0.53665 lb ⋅s /ft 1 2 (12 in.) + ( 3 in.)  +   2  +  2   ×  12 in.    ( )   2

2

2 2 2 2 1  1 ft  + ( 0.06708 lb ⋅s /ft )    ( 3 in.) + (1.5 in.)  + (1.5 in.) + ( 4.5 in. − 0.75 in.)  ×        12    12 i n.  2

(

 1 16    )  4 −  4π   ( 2.4 in.)2 + 112 (1.5 in.)2   

+ 0.042148 lb ⋅ s2 //fft 

2   1 ft  2 4 × 2.4 in.  2    3 i n . 4 . 5 i n . 0 . 7 5 i n . + + − ) ×  +(    3π  1 2 i n .     

2 2 2 2  1  1 ft 2   + 2 ( 0.036892 lb ⋅s 2 /ft )    3 ( 2.75 in.) + (1.0 in.)  + (12 in − 2.75 in ) + ( 0.5 in )   ×        12 in.   12 

= 0.252096 lb ⋅ ft ⋅ s 2   or  I  z   = 0.252 lb ⋅ ft ⋅ s 2  

 

 

PROBLEM 9.147

Determine the moment of inertia of the steel machine element shown with respect to (a (a) the  x  x axis,  axis, (b (b) the y the  y axis,  axis, (c (c) the  z  axis.   axis. (The specific 3 weight of steel is 490 lb/ft .)

SOLUTION

m =  ρ ST V =

Have

δ ST   g 

V    3

 1 ft  m1 = 3 1 4 in i n × × × × ( )      in..  32.2 32. 2 ft/ ft/ss 2  12 in 490 lb/ft 3

Then

3

= 105.676 × 10−3 lb ⋅ s 2 /ft   3

 1 ft  −3 2 m2 = in ×  × (1.5 × 1 × 2 ) in  = 26.419 × 10 lb⋅ s /ft   2 in..  32.2 32. 2 ft/s ft/s  12 in 490 lb/ft 3

  3

490 l  b/ft 3

3

2  π    1 ft  m3 = ×  ( 0.5 ) × 1.5 iinn 3 ×  = 5.1874 × 103 lb ⋅ s 2 /ft    2 32.2 32. 2 ft/s ft/s 2   12 in.  3

   3  1 ft  2  π  m4 = 1 . 4 0 .4  in ×  × × ( )    32.2 32. 2 ft/s ft/s2  2     12 in.  490  lb/ft 3

= 10.8491 × 10−3 lb ⋅ s 2/ft   (a) Using Fig. 9.28 for components 1 and and 2 and the equations equations derived above for com components ponents 3 and 4, have  I x = ( I x )1 + ( I x ) 2 + ( I x )3 − ( I x )4  

where

 1 2  4 2  2   1 ft 2  2  1  2 ( I  x )1  = (105.676 × 1100 llbb ⋅ s /ft )    (1) + ( 4 )  +   +    in ×     1 2 2 2 1 2 i n .                  

−3

 

2

= 4.1585 × 10−3 lb ⋅ ft ⋅ s 2   2

 1 ft  2 2  2 2  1 ( I  x )2  = ( 26.419 × 10 lb ⋅ s /ft )    (1) + ( 2 )  + ( 0.5) + ( 5 )   in 2  ×     12     12 in.   

−3

2

= 4.7089 × 10−3 lb ⋅ ft ⋅ s 2  

 

 

PROBLEM 9.147 CONTINUED

 

2 2  16  4 × 0.5    2 2    1   2  1 ft   10 lb l b ⋅ s /ft )   −  0.5 )  + ( 0.5) +  6 + ( I  x )3  = ( 5.1874 × 10    i n   ×  12 i n.    2 ( π  2 3 π  9               

−3

2

= 1.40209 × 10−3 lb ⋅ ft ⋅ s 2    2

2 2  1     4 0 . 5 1 f t × 2 2    2 ( I  x )4  = (10.8451 × 10 lb ⋅ s /ft )    3 (1.4 ) + ( 0.4 )  +  6 ×   i n   ×  12 i n.    1 2 3 π           

−3

= 0.38736 × 10−3 lb ⋅ ft ⋅ s 2    I  x = ( 4.1585 + 4.7089 +  1.40209 − 0.38736 ) × 10−3  lb ⋅ ft ⋅ s2  

Then





= 9.8821 × 10−3 lb ⋅ ft ⋅ s 2   or  I  x = 9.8  8 × 10−3 lb ⋅ ft ⋅ s 2 

( )1 + ( I y ) 2 + ( I y )3 − ( I y ) 4  

 I y = I y

(b) Have where

( I  y )1

 1  3 2  4 2    2  1 ft 2 2 2  10 lb l b ⋅ s /ft )    ( 3) + ( 4 ) +   +     in   ×   = (105.676 × 10         2    1 2 2 1 2 i n .             

 

−3

2

= 6.1155 × 10−3 lb ⋅ ft ⋅ s 2   2

    ( I  y )2  = ( 26.419 × 10 lb ⋅ s /ft ) 11 2 (1.5)2 + ( 2)2  + ( 0.75)2 + ( 5)2   in 2  ×  112 fitn.         

−3

2

= 4.7854 × 10−5 lb ⋅ ft ⋅ s 2  

( I  y )3

 

  1

2

16 





4.05   1 0 lb ⋅ s /ft   = (5.1874 × 10 )  4 −  9π 2  ( 0.5) + 12 (1.5)  + ( 0.75) +  6  + 3π   −3

2

1

2

2

2

   2

 1 ft 

  in   ×  12 in.   

= 1.4178 s × 10−3 lb ⋅ ft ⋅ s 2  

( I  y )4

2   1    4 × 1.4 2   1 ft  16  2 2   2  = (10.8451 × 10 llbb ⋅s /ft )  2 −  9π 2  (1.4 )  +  3 − 3π   + ( 2 )   in   ×  12 in.             

−3

 

2

= 0.78438 × 10−3 lb ⋅ ft ⋅ s 2  

 

2

 

PROBLEM 9.147 CONTINUED

 I  y = (16.1155 + 4.7854 +  1.41785 − 0.78438) × 10−3  lb⋅ ft ⋅ s 2  

Then





= 11.5344 × 10−3 lb ⋅ ft ⋅ s 2   or  I  y = 11. 53 × 10−3 lb ⋅ ft ⋅ s 2    I z = ( I z )1 + ( I z ) 2 + ( I z )3 − ( I z )4  

(c) Have where

 3 2  1 2    2  1 ft 2 2  1  2 10 lb l b ⋅ s /ft )    ( 3) + (1) +   +     in   ×  ( I  z )1  = (105.676 × 10         2    1 2 2 1 2 i n .             

 

−3

2

= 2.4462 × 10−3 lb ⋅ ft ⋅ s 2    1.5 2  1 2    2  1 ft 2 2 2  1  10 lb ⋅ s /ft )    (1.5) + (1) +  ( I  z )2  = ( 26.419 × 10      2  +  2    in   ×  12 in.      1 2          

 

−3

2

= 0.198754 × 10−3 lb ⋅ ft ⋅ s 2   2

 2  1 ft  2 2 2  1 ( I  z )3  = ( 5.1874 × 10 lb ⋅ s /ft )    3 ( 0.5) + (1.5)  + ( 0.75) + ( 0.5 )   in 2  ×     12     12 in.   

−3

2

= 0.038275 × 10−3 lb ⋅ ft ⋅ s 2    

2   1    4 × 1.4 2   1 ft  16  1 2 2 2   2 10 lb ⋅ s /ft )   −  1. 4 ) + ( I  z )4  = (10.8451 × 10 ( 0.4 )  +  3 −  + ( 0.8)   in   ×  12 in.    2 ( π  4 1 2 3 π  9               

−3

2

= 0.49543 × 10−3 lb ⋅ ft ⋅ s 2   Then

 I  z  = ( 2.4462 + 0.198754 +  0.038275 − 0.49543) × 10−3  lb ⋅ft ⋅ s 2  



= 2.1878 × 10−3 lb ⋅ ft ⋅ s2  



or  I  z  = 2.1  9 × 10−3 lb ⋅ ft ⋅ s2  

 

 

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 9.148–9.150 Slender Rod  I x = 0

I y′ = I z ′ =

1

mL2 (Fig. 9.28)  

12 1

 I y = I z  =

3

mL2  (Sample Problem 9.9)  

Circle Have

 I y = ∫ r 2dm = ma 2  

 Now

 I y = I x + I z     I x = I z   

And symmetry implies

∴    I x = I z  =

1

ma 2  

2 Semicircle Following the above arguments for a circle, have  I x = I z =

1 2

ma 2  

I y = ma 2  

Using the parallel-axis parallel-axis theorem  I z  = I z ′ + mx 2  

 I z ′

or

 

x =

2a π 

4  2 1 = m  2 − π 2  a  

 

 

PROBLEM 9.148

Aluminum wire with a mass per unit length of 0.049 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.

SOLUTION

First compute the mass of each component. Have

m =  ρ L  

Then

m1  = ( 0.049 kg k g/m )  2π ( 0.32 m )    0.0985 9852 2 kg   = 0.0 m2 = m3 = m4 = m5  

= ( 0.049 kkgg/m ) ( 0.160 m ) = 0.00784 kg kg   Using the equation given above and the parallel axis theorem, have  I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x )5  

 1   1  2 2 = ( 0.09852 kg )   ( 0.32 m )  + ( 0.00784 kg )   ( 0.160 m )     2    3   2 + ( 0.00784 kkgg ) 0 + ( 0.160 m)     

 1  2 2 2 + ( 0.00784 kg )   ( 0.16 m ) + ( 0.08 m ) + ( 0.32 m )        12  2 2 2 1 + ( 0.00784 kg )  ( 0.16 m ) + ( 0.16 m ) + ( 0.32 m − 0.08 m )       12

= ( 5.0442 + 0.06690 + 0.2007 + 0.86972 + 0.66901) × 10−3  kg ⋅ m2   = 6.8505 × 10−3 kg ⋅ m2  

or  I  x = 6.8  5 × 10−3 kg ⋅ m 2  

 

 

PROBLEM 9.148 CONTINUED

Have

 I y = ( I y ) + ( I y ) + ( I y ) + ( I y ) + ( I y )  

where

( I ) = ( I )

Then

1

 y

2

y

2

4

3

and

4

5

( I ) = ( I  ) y

3

y

5

 

2 2  I  y = ( 0.09  852 kg ) ( 0.32 m )  + 2 ( 0.00784 kg )  0 + ( 0.32 m )   









2 2 1 + 2 ( 0.00784 kg )  ( 0.16 m ) + ( 0.24 m )    12 

= 10.088 + 2 ( 0.80282 ) + 2 ( 0.46831)  × 10−3 kg ⋅ m2   = 12.6303 × 10−3 kg ⋅ m 2   or  I  y = 12.6  3 × 10−3 kg ⋅ m 2   By symmetry

 

 I z = I x  

or  I  z  = 6.8  5 × 10−3 kg ⋅ m 2  

 

PROBLEM 9.149

The figure shown is formed of 3-mm-diameter steel wire. Knowing that the density of the steel is 785 7850 0 kg kg/m /m3 , determ determine ine the the mass mass mome moment nt of inertia of the wire with respect to each of the coordinate axes.

SOLUTION

m =  ρV = ρ AL  

Have Then

2 m1 = m2 = 7850  kg/m3 π   ( 0.0015 m )  × (π  × 0.36 m )  

(

) 



m2 = m1 = 0.0 0.0627 62756 56 kg   2 m3 = m4 = 7850  kg/m3 π    ( 0.0015 m )  × ( 0.36 m )  

(

) 



0.0199 19976 76 kg   = 0.0 Using the equations given above and the parallel axis theorem, have  I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4  

( I x )3 = ( I x )4  

where Then

 

1

 I  x = ( 0.062756 kg ) 

2 2 2  ( 0.36 m )  + ( 0.062756 )  ( 0.36 m ) + ( 0.36 m ) 

1

2

2





2 2 2 1 + 2 ( 0.019976 kg )  ( 0.36 m ) + ( 0.18 m ) + ( 0.36 m )    12 

 

=  4.06659 + 12.19977 + 2 ( 3.45185 )  × 10−3 kg ⋅ m 2  

= 23.1701 × 10−3 kg ⋅ m2   or  I  x = 23 .2 × 10−3 kg ⋅ m 2   Have where

( )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4  

 I y = I y

( I y )1 = ( I y )2  

and

( I y )3 = ( I y ) 4  

 

 

`

PROBLEM 9.149 CONTINUED

Then 2 2  I  y = 2 ( 0.0 62756 kg kg ) ( 0.36 m )  + 2 ( 0.019976 kg kg )  0 + ( 0.36 m )   









= 2 (8.13318 × 10−3 kg ⋅ m 2 ) + 2 ( 2.58889 × 10−3 kg ⋅ m 2 )   = 21.44414 × 10−3 kg ⋅ m2   or  I  y = 21 .4 × 10−3 kg ⋅ m 2   Have  I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4  

( I z )3 = ( I z )4  

where Then

 

2 1 ( 0.36 m )    2 

 I  z  = ( 0.062756 k kg g) 

 1 4  2 + ( 0.062756 kg )  − 2  ( 0.36  m )    2 π  

 2 × 0.36 m2  2 0 . 3 6 m +  + ( )    π    

2 1 kg )  ( 0.36 m )    + 2 ( 0.019976 kg 3 

= 4.06659 + 12.1998 + 2 ( 0.86296 )  10−3 kg ⋅ m 2  

= 17.9923 × 10−3 kg ⋅ m2   or  I  z  = 17. 99 × 10−3 kg ⋅ m 2  

 

 

PROBLEM 9.150

A homogeneous wire with a weight per unit length of 0.041 lb/ft is used to form the figure shown. Determine the moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION

First compute the mass of each component. Mass of each component is indentical m =

( m/L )

L 

 g 

=

Have

.04 41lb/ 1lb/ft ) (1.5ft ) ( 0.0 32.2 32. 2 ft/s ft/s 2

 

.001 0190 909 994 lb ⋅ s 2 /ft   = 0.0 Using the equations given above and the parallel axis theorem, have

( I x )1 = ( I x )3 + ( I x )4 = ( I x )6 Then

and

( I x )2 = ( I x )5  

 I x = 4 ( I x )1 + 2 ( I x )2    I  x = 4 0.0019 0994 lb lb ⋅ s 2/ft  1 ( 5 fftt )2    3 

(

)

2 + 2 ( 0.00190994 lb lb ⋅ s 2 /ft )  0  + (1.5 fftt )     

.01 143246 lb ⋅ ft ⋅ s 2   = 0.0 or  I  x = 14. 32 × 10−3 lb ⋅ ft ⋅ s 2    Now

( I y )1 = 0

( I y ) 2 = ( I y )6

( I y )4 = ( I y )5  

 

 

`

PROBLEM 9.150 CONTINUED

Then

( )2 + ( I y )3 + 2 ( I y )4  

 I y = 2 I y

    1  2 2 = ( 0.0019094 lb ⋅ s 2/ft )   2   (1.5 ft )  +  0 + (1.5 ft )         3   2 2 2  1 + 2  (1.5 ft ) + (1.5 ft ) + ( 0.75  ft )     12 

lb ⋅ ft ⋅ s 2   = 0.0019094 (1.5 + 2.25 + 6 ) lb ⋅ ft ⋅ s 2 = 0.0186219 lb  I  y = 18. 62 × 10−3 lb ⋅ ft ⋅ s 2   By symmetry

 I z = I y    I  z  = 18. 62 × 10−3 lb ⋅ ft ⋅ s 2 