COSMOS Chapter 9 Solution 108 | Michigan State University - Edubirdie

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Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2019
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2
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COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 108. I ave = Have 1 1 Ix + I y = 640 in 4 + 280 in 4 = 460 in 4 2 2 ( ( ) ) 1 1 Ix − I y = 640 in 4 − 280 in 4 = 180 in 4 2 2 ( ) I x′y′ = −180 in 4 , Also have Letting the points ( ) ( I x , I xy ) and ( I x′, I x′y′ ) 2θ = −120°, Ix > I y be denoted by X an X ′, respectively, three possible Mohr’s circles can be constructed Assume the first case applies Then Also Ix − I y 2 = R cos 2θ m I x′y′ = R cos α R cos 2θ m = 180 in 4 or or R cos α = 180 in 4 ∴ α = ±2θ m 120° = 2θ m + ( 90° − α ) Also have ∴ α = −2θ m Note and 2 ( 2θ m ) = 30° 2θ m − α = 30° or or 2θ m = α = 15° 2θ m > 0   implies case 2 applies α < 0 continued COSMOS: Complete Online Solutions Manual Organization System (a) Therefore, (b) Have Then θ m = 7.5° clockwise R cos15° = 180 or R = 186.35 in 4 I max, min = I ave ± R = 460 ± 186.350 or I max = 646 in 4 and I min = 274 in 4 Note: From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .

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