University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
Views:

259

Pages:

2
Author:

Adowj7b

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 141. t = 0.1 in. γ st = 490 lb/ft 3 Have m = ρstV = m= γ st g tA 490 lb/ft 3  0.1  ×  ft × A 32.2 ft/s 2  12   lb ⋅ s 2 2  =  0.126812 ⋅ ft  A ft    lb ⋅ s 2 2   16 12  2 lb ⋅ s 2 m1 =  0.126812 ⋅ ft   ×  ft = 0.169083 ft ft    12 12  Then  lb ⋅ s 2 2   16 9  2 lb ⋅ s 2 m2 =  0.126812 ⋅ ft   ×  ft = 0.126812 ft ft    12 12   lb ⋅ s 2 2   1 12 9  2 lb ⋅ s 2 m3 =  0.126812 ⋅ ft   × ×  ft = 0.047555 ft ft    2 12 12  2  lb ⋅ s 2 2   π  5   2 lb ⋅ s 2 ⋅ ft      ft = 0.034583 m4 =  0.126812 ft ft    2  12   Using Fig. 9.28 for components and and the equations derived previously for components and , have I x = ( I x )1 + ( I x )2 + ( I x )3 − ( I x )4 Where 2 ( I x )1 1 lb ⋅ s 2   12  2 =  0.169083   ft  = 0.056361 lb ⋅ ft ⋅ s 3 ft   12  ( I x )2 = ( I x )3 2 2 2 2  lb ⋅ s 2   1  12   9    4   3    2 =  0.047555     +    +   +     ft ft  18  12   12    12   12      2 1 lb ⋅ s 2   9  0.126812 ft  = 0.023777 lb ⋅ ft ⋅ s 2   3  ft   12  continued COSMOS: Complete Online Solutions Manual Organization System = 0.0123841 lb ⋅ ft ⋅ s 2 ( I x )4 2 1 lb ⋅ s 2   5  ft  = 0.00150100 lb ⋅ ft ⋅ s 2  0.034583  4 ft   12  = Then I x = ( 0.056361 + 0.023777 + 0.0123841 − 0.00150100 ) lb ⋅ ft ⋅ s 2 I x = 91.0 × 10−3 lb ⋅ ft ⋅ s 2 ( )1 + ( I y )2 + ( I y )3 − ( I y )4 Iy = Iy Where 2 2 1 lb ⋅ s 2   16   12   2 2  0.169083    +    ft = 0.156558 lb ⋅ ft ⋅ s 3 ft   12  12    ( )1 = ( I y )2 1 lb ⋅ s 2   16  2 =  0.126812   ft  = 0.075148 lb ⋅ ft ⋅ s 3 ft   12  Iy 2  ( I y )3 =  0.047555 lbft⋅ s 2   1  12 2  4 2  2 2     +    ft = 0.0079258 lb ⋅ ft ⋅ s 18 12 12         2 2 2  8  2  4   1 16  5  5   2  ft + + ×    −       ft  2   3π 12     12    2 9π  12    ( I y )4 =  0.034583 lbft⋅ s  = 0.0183722 lb ⋅ ft ⋅ s 2 Then I y = ( 0.156558 + 0.075148 + 0.0079258 − 0.0183722 ) lb ⋅ ft ⋅ s2 I y = 221 × 10−3 lb ⋅ ft ⋅ s 2 I z = ( I z )1 + ( I z )2 + ( I z )3 − ( I z )4 Where 2 ( I z )1 = 1 lb ⋅ s 2   16  2  0.169083   ft  = 0.100197 lb ⋅ ft ⋅ s 3 ft   12  ( I z )2 2 2 1 lb ⋅ s 2   16   9  2 2  0.126812    +    ft = 0.098925 lb ⋅ ft ⋅ s 3 ft   12   12   = 2 2  lb ⋅ s 2   1  9   3  2 2 = + I 0.047555  ( z )3        ft = 0.0044583 lb ⋅ ft ⋅ s ft 18 12 12         ( I z )4 2 2  lb ⋅ s 2   1  5   8   2 =  0.034583    ft  +  ft   = 0.0168712 lb ⋅ ft ⋅ s ft 4 12 12          Then I z = ( 0.100197 + 0.098925 + 0.0044583 − 0.0168712 ) lb ⋅ ft ⋅ s 2 I z = 186.7 × 10−3 lb ⋅ ft ⋅ s 2