The moment of inertia of an area which may be regarded as consisting of a larger area minus other areas, is computed by sub tracting from the moment of inertia of the large area those of the "minus areas." Examples. 1. Compute the moment of inertia of the built up section represented in Fig. 30 (in part same as Fig. 25) with respect to a horizontal axis passing through the center of gravity, it being given that the moment of inertia of each channel section with respect to a horizontal axis through its center of gravity is 128.1 inches', and its area 6.03 square inches.
The center of gravity of the whole section was found in the example of Art. 49 to be 8.30 inches from the bottom of the section; hence the distances from the inertia-axis to the centers of gravity of the channel section and the plate are 2.30 and 3.95 inches respectively (see Fig. 30).
The moment of inertia of one channel section with respect to the axis AA (see equation 5, Art. 53) is: 128.1+ 6.03 X inches'.
The moment of inertia of the plate section (rectangle) with re spect to the line ea" (see Art. 52) is: [14 X 02)1=0.15 inches"; and with respect to the axis AA (the area being 7 square inches) it is: 0 15+ 7 x inches'.
Therefore the moment of inertia of the whole section with re spect to AA is: 2 X 160.00 +109.37=429.37 i n 2. It is compute the moment of inertia of the " hollow rectangle " of Fig. 29 with 'respect to a line through the
center of gravity and parallel to the short side.
The amount of inertia of the large rectangle with respect to the named axis (see Art. 52) is: Fig. 31.
and the moment of inertia of the smaller one with respect to the same axis is: (4 X = 170g; hence the moment of inertia of the hollow section with respect to the axis is: 1. Compute the moment of inertia of the section repre sented in Fig. 31, a, about the axis AA, it being 3.08 inches from the top. Given also that the area of one angle section is 5.06 square inches, its center of gravity C (Fig. 31, inches from the top, and its moment of inertia with respect to the axis as 17.68 inches'. Ans. 145.8 inches'.
2. Compute the moment of inertia of the section of Fig. 31, a, with respect to the axis BB. Given that distance of the center of gravity of one angle from one side is 1.66 inches (see Fig. 31,b), and its moment of inertia with respect to Lb 17.68 inches.
Ans. 77.5 inches'.
55. Table of Centers of Gravity and floments of Inertia.
Column 2 in Table A below gives the formula for moment' of inertia with respect to the horizontal line through the center of gravity. The numbers in the third column are explained in Art. 62; and those in the fourth, in Art. 80.