My question comes from Mechanics of Materials (5th) Ch.4 P51 I have a question a

Question

My question comes from Mechanics of Materials (5th) Ch.4 P51 I have a question about the calculation of the Moment of Inertia for the entire transformed section.

In the solution posted it says I = I 1 + I2 + I3 I = [(30 x 53)/12 + 150(2.7252)] + [(12 x .2253)/3] + [25.133 x 16.2752] I = [312.5 + 1113.8] + [.0456] + [6657.0] = 8083.4 in4 I underdstand how I1 and I2 were calculated but shouldn't I3 also have it's centriodal moment of inertia because it is not lying on the z axis?
Can someone explain to me why I3  isnt:    4[(p/4)(.5)(43) + 6.283(16.2752)] = The solution given here on the site is the solution also given in the book. Can someone please help me out here. Thanks in advance. Here is a sketch of the original cross section.

Explanation / Answer

I do not have the book, but based on the sketch shown on the soulution that 25.133 is a given number. therefore, the solution book simply use that number to calculate. If the question give you the re-bar size and you use the formular to do, your appraoch is correct also

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