COSMOS Chapter 9 Solution 183

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 183. I x = 9.8821 × 10−3 lb ⋅ ft ⋅s 2 From Problem 9.147: I y = 11.5344 × 10−3 lb ⋅ ft ⋅ s 2 I z = 2.1878 × 10−3 lb ⋅ ft ⋅ s 2 I xy = 0.48776 × 10−3 lb ⋅ ft ⋅ s 2 From Problem 9.151: I yz = 1.18391 × 10−3 lb ⋅ ft ⋅ s 2 I zx = 2.6951 × 10−3 lb ⋅ ft ⋅ s 2 (a) From Equation 9.56 ( ( ) ) 2 2 2 − I yz − I zx K 3 − I x + I y + I z K 2 + I x I y + I y I z + I z I x − I xy K ( ) 2 2 2 − I x I y I z − I x I yz − I y I zx − I z I xy − 2I xy I yz I zx = 0 Substituting { K 3 − ( 9.8821 + 11.5344 + 2.1878 ) × 10−3  K 2 + ( 9.8821)(11.5344 ) + (11.5344 )( 2.1878 ) } 2 2 2 + ( 2.1878 )( 9.8821) − ( 0.48776 ) − (1.18391) − ( 2.6951)  × 10 −6 K  2 2 − ( 9.8821)(11.5344 )( 2.1878 ) − ( 9.8821)(1.18391) − (11.5344 )( 2.6951)  2 − ( 2.1878 )( 0.48776 ) − 2 ( 0.48776 )(1.18391)( 2.6951)  × 10 −9 = 0  or ( ) ( ) K 3 − 23.6043 × 10−3 K 2 + 151.9360 × 10−6 K − 148.1092 × 10−9 = 0 Solving numerically K1 = 1.180481 × 10−3 lb ⋅ ft ⋅s 2 or K1 = 1.180 × 10−3 lb ⋅ ft ⋅s 2
K 2 = 10.72017 × 10−3 lb ⋅ ft ⋅ s 2 or K 2 = 10.72 × 10 −3 lb ⋅ ft ⋅s 2
K3 = 11.70365 × 10−3 lb ⋅ ft ⋅ s 2 or K3 = 11.70 × 10 −3 lb ⋅ ft ⋅s 2
continued COSMOS: Complete Online Solutions Manual Organization System (b) From Equations 9.54(a) and 9.54(b) ( I x − K )( λx ) − I xy ( λ y ) − I zx ( λz ) = 0 ( )( ) − I xy ( λx ) + I y − K1 λ y − I yz ( λz ) = 0 ( )1 and ( λz )1 . K1: Substitute K1 and solve for λ to get ( λx )1 , λ y ( )1 − 2.6951( λz )1  × 10−3 = 0 ( 9.8821 − 1.180481) ( λ ) − 0.48776 λ x 1 y  ( )1 − 1.18391( λz )1  × 10−3 = 0  − 0.48776 ( λ ) + (11.5344 − 1.180481) λ x 1 y  or ( )1 − 5.52546 ( λz )1 = 0 17.83996 ( λx )1 − λ y ( )1 − 0.11434 ( λz )1 = 0 − 0.0471( λx )1 + λ y Then ( λz )1 = 3.1549 ( λx )1 and ( λ y )1 = 0.40769 ( λx )1 ( λx )12 + ( λ y )1 + ( λz )12 2 Equation 9.57: Substituting ( λx )12 + 0.40769 ( λx )1  2 =1 2 + 3.1549 ( λx )1  = 1 or ( λx )1 = 0.29989 then (θ x )1 = 72.5° and ( λy )1 = 0.122262 then (θ y )1 = 83.0°
( λz )1 = 0.94612 then
(θ z )1 = 18.89°
K 2: Substitute K 2 and solve for λ . ( )2 − 2.6951( λz )2  × 10−3 = 0 ( 9.8821 − 10.72017 ) ( λ ) − 0.48776 λ x 2 y  ( )2 − 1.18391( λz )2  × 10−3 = 0  − 0.48776 ( λ ) + (11.5344 − 10.72017 ) λ x 2 y  continued COSMOS: Complete Online Solutions Manual Organization System ( )2 − 5.52546 ( λz )2 = 0 −1.718202 ( λx )2 − λ y or ( )2 − 1.45402 ( λz )2 = 0 − 0.599045 ( λx )2 + λ y = −0.33201( λx )2 Then ( λz )2 and ( λ y )2 = 0.116306 ( λx )2 2 ( λx )2 + 0.116306 ( λx )2  Then or ( λx )2 = 0.94333 (θ x )2 then And ( λ y )2 = 0.109715 ( λz )2 = −0.31320 then 2 2 +  −0.33201( λx )  = 1 = 19.38°
(θ y )2 = 83.7°
then (θ z )2 = 108.3°
K 3: Substitute K 3 and solve for λ . ( )3 − 2.6951( λz )3  × 10−3 = 0 ( 9.8821 − 11.70365 ) ( λ ) − 0.48776 λ x 3 y  ( )3 − 1.18391( λz )3  × 10−3 = 0  − 0.48776 ( λ ) + (11.5344 − 11.70365 ) λ x 3 y  ( )3 − 5.52546 ( λz )3 = 0 −3.73452 ( λx )3 − λ y or ( )3 + 6.99504 ( λz )3 = 0 2.88189 ( λx )3 + λ y Then ( λz )3 and ( λ y )3 = −6.9403 ( λx )3 = 0.58019 ( λx )3 2 2 2 Then ( λx )3 +  − 6.9403 ( λx )3  + 0.58019 ( λx )3  = or ( λx )3 = − 0.142128* ( λ y )3 = 0.98641 ( λz )3 * = − 0.082461 then (θ x )3 then (θ y )3 = 9.46°
then Note: the negative root of ( λx )3 is taken so that axes 1, 2, 3 form a right-handed set. (θ z )3 = 98.2°
= 94.7°