Shadows- 6 Determine the product c © Chegg Study | Guided S × V- (2) YouTube X.

Question

Shadows- 6 Determine the product c © Chegg Study | Guided S × V- (2) YouTube X. www.webassign.net/web/Student/Assignment-Responses/last?dep 17002074 2. -20 points NorEngStatics1 8.P.046 0 My Notes Ask You The figure below shows a thin-walled cold rolled steel cross-section. The thickness of all segments of the member is 0.450 in. 8 in 8 In 7 in 7 in Determine the moments of inertia of the cross-section I, Iy, and Jc in int with respect to the centroidal axes x' and y' and the corresponding radii of gyration in inches nns in in in 2-20 PM Type here to search

Explanation / Answer

moment of inertia of the given figure can be found by finding moment of inertia of individual parts about their center of mass and then applying parallel axis theorem and then adding resultant algebraically
hence

we have to find location of centroidal axis
hence let the location be at x from the left end
then
x = [8*0.45*4 + (14 - 0.45)*0.45*(8 - 0.45/2) + 2*8*0.45*(8 + 4)]/[8*0.45 + (14 - 0.45)*0.45 + 2*8*0.45]
x = 8.771 in

hence
Iy' = (8*0.45*8^2/12 + 8*0.45*(8.771 - 4)^2 + (14 - 0.45)*0.45^3/12 + (14 - 0.45)*0.45*((8.771 - 8 + 0.45/2)^2 + 2*8*0.45*8^2/12 + 2*8*0.45(12 - 8.771)^2)
Iy' = 220.7668756725 m^4

Ix' = 2*8*0.45*8^2/12 + 2*8*0.45*(7 - 0.45/2)^2 + 8*0.45^3/12 + (14 - 0.45)^3*0.45/12 + (14 - 0.45)*0.45*(7 - (14 - 0.45)/2)^2
Ix' = 462.54695625 m^4

[8*0.45 + (14 - 0.45)*0.45 + 2*8*0.45]*kx^2 = 462.54695625
kx = 5.2319 in

[8*0.45 + (14 - 0.45)*0.45 + 2*8*0.45]*kx^2 = 220.7668756725
ky = 3.61456 in

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