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

2019
Views:

483

Pages:

2
Author:

slypery5hy

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 82. From Problem 9.75 I xy = 1.5732 × 106 mm 4 Now I x = ( I x )1 + ( I x )2 + ( I x )3 ( I x )1 = where and ( I x )2 = ( I x )3 1 (150 mm )(12 mm )3 = 21 600 mm 4 12 1 (12 mm )( 38 mm )3 + (12 mm )( 38 mm ) ( 25 mm )2 12 = = 339 872 mm 4 Then I x =  21 600 + 2 ( 339 872 )  mm 4 = 701 344 mm 4 = 0.70134 × 106 mm 4 ( )1 + ( I y )2 + ( I y )3 Iy = Iy Also where and ( I y )1 = 121 (12 mm )(150 mm )3 = 3.375 × 106 mm 4 ( I y )2 = ( I y )3 = 121 ( 38 mm )(12 mm )3 + (12 mm )(38 mm ) ( 69 mm )2 = 2.1765 × 106 mm 4 Then I y = ( 3.375 + 2 ( 2.1765 )  × 106 mm 4 = 7.728 × 10 mm Now 1 I x + I y = 4.2146 × 106 mm 4 2 and 1 I x − I y = −3.5133 × 106 mm 4 2 ( ( ) ) COSMOS: Complete Online Solutions Manual Organization System Using Equations 9.18, 9.19, and 9.20 From Equation 9.18: I x′ = Ix + I y 2 Ix − I y + 2 cos 2θ − I xy sin 2θ ( ) =  4.2147 × 106 + −3.5133 × 106 cos (120° ) − 1.5732 × 106 sin (120° )  mm 4   = 4.6089 × 106 mm 4 or I x′ = 4.61 × 106 mm 4 From Equation 9.19: I y′ = Ix + I y 2 − Ix − I y 2 cos 2θ + I xy sin 2θ ( ) =  4.2147 × 106 − −3.5133 × 106 cos (120° ) + 1.5732 × 106 sin (120° )  mm 4   = 3.8205 × 106 mm 4 or I y′ = 3.82 × 106 mm 4 From Equation 9.20: I x′y′ = Ix − I y 2 sin 2θ + I xy cos 2θ =  −3.5133 × 106 sin (120° ) + 1.5732 × 106 cos (120° )  mm 4 = −3.8292 × 106 mm 4 or I x′y′ = −3.83 × 106 mm 4