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

2024
Views:

342

Pages:

1
Author:

diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 97. From Problem 9.79: Ix = Problem 9.67: I xy = π 8 a4 Iy = π 2 a4 1 4 a 2 The Mohr’s circle is defined by the diameter XY, where 1  π X  a4, a4  2  8 I ave = Now 1  π Y  a4 , − a4  2  2 and 1 1π π  I x + I y =  a 4 + a 4  = 0.98175a 4 2 2 8 2  ( ) and R= 2 1  2  2 I x − I y  + I xy =   ( ) 2  1  π 4 π 4  1 4   a − a  +  a  2  2  2 8 2 = 0.77264a 4 The Mohr’s circle is then drawn as shown. tan 2θ m = − =− 2I xy Ix − I y π 8 1  2  a4  2  a4 − π 2 a4 = 0.84883 2θ m = 40.326° or θ m = 20.2° and ∴ The principal axes are obtained by rotating the xy axes through 20.2° counterclockwise Now I max, min = I ave ± R = 0.98175a 4 ± 0.77264a 4 or I max = 1.754a 4 and I min = 0.209a 4 From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .

COSMOS Chapter 9 Solution 97

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