COSMOS Chapter 9 Solution 102 | Answer Key - Edubirdie

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

2020
Views:

289

Pages:

1
Author:

behend6abs

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 102. From Problems 9.75 and 9.82 I x = 0.70134 × 106 mm 4 , Now I ave = I y = 7.728 × 106 mm 4 , 1 1 I x + I y = ( 0.70134 + 7.728 ) × 106 mm 4 = 4.2147 × 106 mm 4 2 2 ( ) 2 and R= I xy = 1.5732 × 106 mm 4  Ix − I y  2   + I xy = 2   2  ( 0.70134 − 7.728 ) × 106  6   + 1.5732 × 10 2   ( ) 2 = 3.8495 × 106 mm 4 Define points X ( 0.70134, 15732 ) × 106 mm Y ( 7.728, − 1.5732 ) × 106 mm Also  −2 (1.5732 )  2θ m = tan −1   = 24.122°, θ m = 12.06°  0.70134 − 7.728  or θ m = 12.06° counterclockwise Then I max, min = I ave ± R = ( 4.2147 ± 3.8495 ) × 106 mm 4 or I max = 8.06 × 106 mm 4 and I min = 0.365 × 106 mm 4 Note: From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .

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