Divide the cross-section into simple parts, and find the moment of inertia of each simple part around its own neutral axis (parallel to main neutral axis); then, if we call the moment of inertia of the whole cross-section i, and that of each part i1, i11, i111, i1111, etc., and, further, it we call the area of each part a1, + a11, + a111, etc., and the distance of the centre of gravity of each part from the main neutral axis of the whole cross-section, d1, + d11, + d111,1111, etc., we have: i = (d12a1+i1) + (d112a11+i11) + (d1112a111+i111) + (d11112a1111+i1111) + , etc

Referring back to Figure 1, we should have for Part I: - i1 =11/14 x14. (See Table I., column 8.)

For Part II we should have: i11 = (b11h113) /12

And for Part III: (i111 = b111h1113 )/12

For the distances of individual centres of gravity from main centre of gravity we should have for Part I: d,-d.

For Part II: d11-d.

And for Part III: d-d111.

Howtofind moment of inertia of any cross-section.

Therefore the moment of inertia, i, of the whole deck-beam would be: -

But a1=22/7 r12

Further, a11= b11 h11,

And a111, = b111 h111, which, inserted above, gives for

The following table (I) gives the values for the moment of inertia (i), moment of resistance (r), area (a), square of radius of gyration (Q2), etc., for nearly every cross-section likely to be used in building. Those not given can be found from Table I by dividing the cross-section into several simpler parts, for which examples can be found in the table. Note, that it makes a great difference whether the neutral axis is located through the centre of gravity (of the part), or elsewhere. When making calculations we must, of course, insert in the different formulae in place of i, r, a, Q2, their values (for the respective cross-section), as given in Table I.