Jawaban Quiz 13

March 11, 2024 | Author: Anonymous | Category: N/A
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PROBLEM 9.142 Determine the mass moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. (The density of steel is 7850 kg/m3.)

SOLUTION

First compute the mass of each component. We have m = ρSTV

Then

m1 = (7850 kg/m3 )(0.24 × 0.04 × 0.14) m3 = 10.5504 kg

π  m2 = m3 = (7850 kg/m3 )  (0.07)2 × 0.04  m3 = 2.41683 kg 2 

m4 = m5 = (7850 kg/m3 )[π (0.044) 2 × (0.04)] m3 = 1.90979 kg

Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to Problem 9.144) for a semicylinder, we have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 − ( I x )5

where ( I x ) 2 = ( I x )3

1  1  =  (10.5504 kg)(0.042 + 0.142 ) m 2  + 2  (2.41683 kg)[3(0.07 m) 2 + (0.04 m) 2 ] 12 12     1  +  (1.90979 kg)[3(0.044 m) 2 + (0.04 m) 2 ] + (1.90979 kg)(0.04 m) 2  12  1  −  (1.90979 kg)[3(0.044 m)2 + (0.04 m)2 ]  12  = [(0.0186390) + 2(0.0032829) + (0.0011790 + 0.0030557) − (0.0011790)] kg ⋅ m 2 = 0.0282605 kg ⋅ m 2

or

I x = 28.3 × 10−3 kg ⋅ m 2 

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. 1604

PROBLEM 9.142 (Continued)

I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 − ( I y )5 ( I y ) 2 = ( I y )3

where Then

(I y )4 = |( I y )5 |

1  I y =  (10.5504 kg)(0.242 + 0.142 ) m 2  12     1 16 + 2 (2.41683 kg)  − 2  2 9π 

2 4 × 0.07  2    2 + + (0.07 m ) (2.41683 kg) 0.12 m    3π    

= [(0.0678742) + 2(0.0037881 + 0.0541678)] kg ⋅ m 2 = 0.1837860 kg ⋅ m 2

or Also

I y = 183.8 × 10−3 kg ⋅ m 2 

m = m1 + m2 + m3 + m4 − m5 where m2 = m3 , m4 = |m5 | = (10.5504 + 2 × 2.41683) kg = 15.38406 kg

Then

and

k x2 =

k y2 =

I x 0.0282605 kg ⋅ m 2 = m 15.38406 kg

Iy m

=

or

k x = 42.9 mm 

or

k y = 109.3 mm 

0.1837860 kg ⋅ m 2 15.38406 kg

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. 1605

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