COSMOS Chapter 9 Solution 92 | Answer Key - Edubirdie

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

2023
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443

Pages:

1
Author:

strofamay58

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 92. From the solution to Problem 9.72: I xy = 501.1875 in 4 Problem 9.80: I x = 865.6875 in 4 I y = 4758.75 in 4 1 I x + I y = 2812.21875 in 4 2 ( Now ) 1 I x − I y = −1946.53125 in 4 2 ( ) The Mohr’s circle is defined by the points X and Y where ( I x , I xy ) X: I ave = Now Y: ( I y , −I xy ) 1 I x + I y = 2812.2 in 4 2 ( ) 2 and R=  Ix − I y  2   + I xy = 2   2 1946.53125 ) + 501.18752 in 4 = 2010.0 in 4 Also, tan 2θ m = I xy Ix − I y = 501.1875 = 0.2575 1946.53125 2 or 2θ m = 14.4387° Then α = 180° − (14.4387° + 90° ) = 75.561° Then I x′ , I y′ = I ave ± R cos α = 2812.2 ± 2010.0cos 75.561° or I x′ = 3.31 × 103 in 4 and I y′ = 2.31 × 103 in 4 and I x′y′ = R sin α = 2010.0sin 75.561° or I x′y′ = 1.947 × 103 in 4

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