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

2024
Views:

159

Pages:

2
Author:

garciavelascogr17

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 91. π Ix = From Problem 9.79: 8 π Iy = 2 a4 1 4 a 2 I xy = Problem 9.67: a4 The Mohr’s circle is defined by the diameter XY, where 1  π X  a4, a4  8 2   Now I ave = 1 1π π  5 I x + I y =  a 4 + a 4  = π a 4 = 0.98175a 4 2 2 8 2  16 ( ) 2 and R= 1  π Y  a4 , − a4  2 2   and  Ix − I y  2   + I xy =  2  2  1  π 4 π 4  1 4   a − a  +  a  2  2  2  8 = 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 or 2θ m = 40.326° continued 2 COSMOS: Complete Online Solutions Manual Organization System Then α = 90° − 40.326° = 49.674° β = 180° − ( 40.326° + 60° ) = 79.674° (a) I x′ = I ave − R cos α = 0.98175a 4 − 0.77264a 4 cos 49.674° or I x′ = 0.482a 4 I y′ = I ave + R cos α = 0.98175a 4 + 0.77264a 4 cos 49.674° or I y′ = 1.482a 4 I x′y′ = − R sin α = −0.77264a 4 sin 49.674° or I x′y′ = −0.589a 4 (b) I x′ = I ave + R cos β = 0.98175a 4 + 0.77264a 4 cos 79.674° or I x′ = 1.120a 4 I y′ = I ave − R cos β = 0.98175a 4 − 0.77264a 4 cos 79.674° or I y′ = 0.843a 4 I x′y′ = R sin β = 0.77264a 4 sin 79.674° or I x′y′ = 0.760a 4

COSMOS Chapter 9 Solution 91

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