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For the cross-section of the angle shown below, use Mohrs Circle to determine the orientation of the principal axes with ori

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General case: = b x 3 I centodial Moment of Iy= nbs Inertia. k xx 2 Ixy = o IX = bh3 622 = h 63 Iny Ly Parallel Anis the ov(Iy), 18.9 x 25.2 = 100.81 X10 in 4 39 (Ixy) = 25.2 ? x 18.42 = 56.71x103 mi 4 4 For portion a = 6.3 + 18.9 - 15.75 y = 6.37Mohr circle: Iy, Iry) centre es 15.76 Xid + 34.648X103 C = (50.408x10% in , 0) Rodion R J CENIJ* + (pagga 115,76x103.34,648XImma = 37.641 x 03 nt Orientation Op tan 2 opa 2 Iny In-ly =219.46X163) 15-766103 - 34.648x16 tan 28P = -1.00 28p = tant(1.00

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