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

2024
Views:

287

Pages:

3
Author:

hacedoras1kxy

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 186. (a) From the solutions to Problem 9.150 I x = 14.32 × 10−3 lb ⋅ ft ⋅ s 2 I y = I z = 18.62 × 10−3 lb ⋅ ft ⋅ s 2 I xy = I zx = 4.297 × 10−3 lb ⋅ ft ⋅ s 2 , I yz = 0 From Problem 9.172 I y = Iz, I xy = I zx , I yz = 0 ( ) ( ) ( )  K − (14.32 × 10 )(18.62 × 10 ) ) ( Substituting into Eq. (9.56) and using ( ) ( ) ( ) 2 2 K 3 − I x + 2I y K 2 +  I x 2I y + I y2 − 2 I xy  K − I x I y2 − 2I y I xy =0   ) ( K 3 − 14.32 × 10−3 + 2 18.62 × 10−2  K 2 +  14.32 × 10−3 ( 2 ) 18.62 × 10−3 + 18.62 × 10−3    ( − 2 4.297 × 10−3 or 2 −3 −3 2 ( )( 2 ) 2 − 2 18.62 × 10 −3 4.297 × 10 −3  = 0  K 3 − 51.56 × 10−3 K 2 + 0.84305 × 10−3 K − 0.004277 × 10−3 = 0 Solving: K1 = 0.010022 lb ⋅ ft ⋅ s 2 or K1 = 10.02 × 10−3 lb ⋅ ft ⋅ s 2 K 2 = 0.018624 lb ⋅ ft ⋅ s 2 or K 2 = 18.62 × 10−3 lb ⋅ ft ⋅ s 2 K3 = 0.022914 lb ⋅ ft ⋅ s 2 or K3 = 22.9 × 10−3 lb ⋅ ft ⋅ s 2 (b) To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Equations (9.54) and Equation (9.57). Then K1 : Begin with Equations (9.54b) and (9.54c): 0 − I xy (λx )1 + ( I y − K1)(λ y )1 − I yz (λz )1 = 0 − I zx (λx )1 − I yz (λ y )1 + ( I z − K1) (λz )1 = 0 or ) ( + (18.62 × 10 ) ) (λ ) − 4.297 × 10−3 (λx )1 + 18.62 × 10−3 − 10.02 × 10−3 (λ y )1 = 0 − 4.297 × 10−3 (λx )1 −3 − 10.02 × 10−3 z 1 =0 continued COSMOS: Complete Online Solutions Manual Organization System ( λ y )1 = ( λz )1 = 0.49965 ( λx )1 ( λx )12 + 2 0.49965 ( λx )1  2 =1 ( λx )1 = 0.81669 ( λ y )1 = ( λz )1 = 0.40806 (θ x )1 = 35.2°; (θ y )1 = (θ z )1 = 65.9° K2: Begin with Equations (9.54a) and (9.54b): ( I x − K 2 )( λx )2 − I xy (λ y )2 − I zx (λz )2 = 0 ( )( )2 − I yz (0λz )2 = 0 − I xy ( λx )2 + I y − K 2 λ y Substituting: (14.32 × 10 −3 ) − 18.62 × 10−3 ( λx )2 − 4.297 × 10−3 (λ y ) 2 − (λz ) 2  = 0 ( )( ) −4.297 × 10−3 ( λx )2 + 18.62 × 10−3 − 18.62 × 10−3 λ y ( λx )2 From (ii) 2 =0 (i) (ii) =0 ( λ y )2 = − ( λz )2 From (i) Substituting: ( λx )22 + ( λ y )2 +  − ( λz )2  2 ( λ y )2 = 2 =1 1 2 (θ x )2 ( )2 = 45.0°, (θ z )2 = 135.0° = 90.0°, θ y K3: Begin with Equations (9.54b) and (9.54c) − I xy ( λx )3 + I y − K3 λ y ( )( )3 + I yz ( λz )3 = 0 − I zx ( λx )3 − I yz 0( λ y ) + ( I z − K 3 ) ( λz )3 = 0 Substituting: ( + (18.62 × 10 ) )(λ ) − 4.297 × 10−3 ( λx )3 + 18.62 × 10−3 − 22.9 × 10−3 ( λz )3 = 0 − 4.297 × 10−3 ( λx )3 −3 − 22.9 × 10−3 ( λ y )3 = ( λ z ) 3 = − ( λ x )3 Simplifying: ( λx )23 2 + 2  − ( λx )3  = 1 ⇒ ( λx )3 = 1 3 and z 3 =0 ( λ y )3 = ( λz )3 = − (θ x )3 1 3 ( )3 = (θ z )3 = 125.3° = 54.7°, θ y continued COSMOS: Complete Online Solutions Manual Organization System Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set to obtain the direction cosines corresponding to the labeled axis, the negative root of Equation (i) must be chosen; that is: ( λ x )3 Then: =− 1 3 (θ x )3 = 125.3° (θ y )3 = (θ z )3 = 54.7°