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

2020
Views:

389

Pages:

2
Author:

strofamay57

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 45. Dimensions in mm Determination of centroid, C: Par Area mm 2 t y mm yA mm3 170.67 × 103 1 1 (160 )(80 ) = 6400 2 80 3 2 − 1 (80 )( 60 ) = − 24 2 20 Σ ΣyA ΣΑ 122.67 × 103 mm 3 = 4000 mm 2 = 30.667 mm y = − 48.0 × 103 122.67 × 103 4000 (a) Polar moment of inertia with respect to point O, J O : Part : Ix = 1 (160 mm )(80 mm )3 = 6.8267 × 106 mm 4 12 3 1 I y = 2  ( 80 mm )( 80 mm )  = 6.8267 × 106 mm 4 12   J O = I x + I y = ( 6.8267 + 6.8267 ) × 106 = 13.653 × 106 mm 4 Part : Ix = 1 (80 mm )( 60 mm )3 = 1.440 × 106 mm 4 12 3 1 I y = 2  ( 60 mm )( 40 mm )  = 0.640 × 106 mm 4 12   J O = I x + I y = (1.440 + 0.640 ) × 106 = 2.080 × 106 mm 4 continued COSMOS: Complete Online Solutions Manual Organization System Entire Section: ( ) ( ) J O = J O1 − J O2 = 13.653 × 106 − 2.080 × 106  mm 4 = 11.573 × 106 mm 4 J O = 11.57 × 106 mm 4 (b) Polar moment of inertia with respect to centroid, J C : J O = J C + Ay 2 ( ) 11.573 × 106 mm 4 = J C + 4000 mm 2 ( 30.667 mm ) 2 J C = 7.8116 × 106 mm 4 J C = 7.81 × 106 mm 4

COSMOS Chapter 9 Solution 45

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