COSMOS Chapter 9 Solution 182

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 182. From the solution to Problem 9.143 and 9.167 I x = 34.106 × 10−3 lb ⋅ ft ⋅ s 2 I y = 50.125 × 10−3 lb ⋅ ft ⋅ s 2 I z = 34.876 × 10−3 lb ⋅ ft ⋅ s 2 I xy = 0.96211 × 10−3 lb ⋅ ft ⋅ s 2 I yz = I zx = 0 (a) From Equation 9.55 Ix − K I xy 0 I xy Iy − K 0 =0 0 0 Iz − K ( Ix − K )( I y ) ( I z − K ) − ( I z − K ) I xy2 = 0 ( I z − K ) ( I x − K ) ( I y − K ) − I xy2  = 0 2 =0 I z − K = 0 and I x I y − ( I x + I y ) K + K 2 − I xy or or Then −K K1 = I z = 34.876 × 10−3 lb ⋅ ft ⋅ s 2 Now or K1 = 34.9 × 10−3 lb ⋅ ft ⋅ s 2
and ( 34.106 × 10 )(50.125 × 10 ) − (34.106 × 10 + K − ( 0.96211 × 10 ) = 0 −3 2 or Solving yields −3 −3 −3 ) + 50.125 × 10 −3 K 2 1.70864 × 10−3 − 84.231 × 10−3 K + K 2 = 0 K 2 = 34.0486 × 10−3 K3 = 50.1824 × 103 or K 2 = 34.0 × 10−3 lb ⋅ ft ⋅ s 2
and K3 = 50.2 × 10−3 lb ⋅ ft ⋅ s 2
continued COSMOS: Complete Online Solutions Manual Organization System (b) To determine the directions cosines λx , λ y and λz of each principal axis use two of the Equations 9.54 and Equation 9.57 K1 : Using Equation 9.54(a) and Equation 9.54(b) with I yz = I zx = 0 , we have ( I x − K1 )( λx )1 − I xy ( λ y )1 = 0 ( )( )1 = 0 − I xy ( λx )1 + I y − K1 λ y Substituting (34.106 × 10 −3 ) ( )1 = 0 − 34.876 × 10−3 ( λx )1 − 0.96211 × 10− λ y ( )( ) − 0.96211 × 10−3 ( λx )1 + 50.125 × 10−3 − 34.876 × 10−3 λ y 1 =0 or ( )1 = 0 − 0.770 × 10−3 ( λx )1 − 0.96211 × 10−3 λ y ( )1 = 0 − 0.96211 × 10−3 ( λx )1 + 15.249 × 10−3 λ y Solving yields 0 From Equation 9.57 and ( λx )1 = ( λ y )1 = 0 02 ( λx )12 + ( λ y )1 + ( λz )12 (θ x )1 = 90.0°, = 1 or (θ y )1 = 90.0°, ( λz )1 = 1 (θ z )1 = 0°
K 2 : Using Equation 9.54(b) and Equation 9.54(c) with I yz = I zx = 0 ( )( )2 = 0 − I xz ( λx )2 + I y − K 2 λ y ( Iz Now − K 2 )( λz )2 = 0 I z ≠ K 2 ⇒ ( λz )2 = 0 Substituting ( )( ) − 0.96211 × 10−3 ( λx )2 + 50.125 × 10−3 − 34.0486 × 10−3 λ y or ( λ y )2 = 0.05985 ( λx )2 continued 2 =0 COSMOS: Complete Online Solutions Manual Organization System ( λx )22 + 0.05985 ( λx )2  Then ( λx )2 2 + ( λz )2 = 1 = 0.99821 ( λ y )2 = 0.05974 (θ x )2 and = 3.43°, (θ y )2 = 86.6°, (θ z )2 = 90.0°
( )( )3 = 0 − I xy ( λx )3 + I y − K3 λ y K3 : ( Iz − K3 ) ( λz )3 = 0 I z ≠ K3 ⇒ ( λz )3 = 0 Now Substituting ( )( ) − 0.96211 × 10−3 ( λx )3 + 50.125 × 10−3 − 50.1824 × 10−3 λ y 3 =0 ( )3 = 0 − 0.96211 × 10−3 ( λx )3 − 0.0574 × 10−3 λ y ( λ y )3 = −16.7615 ( λx )3 or Have yields and ( λx )32 +  −16.7615 ( λx )3  ( λ x )3 = − 0.059555 (θ x )3 = 93.4°, 2 and 0 2 + ( λz )3 = 1 ( λ y )3 = 0.998231 (θ y )3 = 3.41°, θ z = 90.0°
(c) Principal axis 1 coincides with the z axis, while the principal axes 2 and 3 lie in the xy plane