When a beam is subjected to loading, as in Fig. 5, the fibers of the beam tend to resist the compression at the top and the tension at the bottom, each fiber exerting a force or moment directly proportional to the distance it is located from the neutral axis (provided the elastic limit of the material is not exceeded); hence, the topmost and bottommost fibers are of considerably more value than those located near the neutral axis. It is therefore necessary, before the strength of any section can be obtained, to ascertain the average value of all the fibers in the section; this value is called the moment of inertia.

The moment of inertia (usually designated by the letter I) of any body or figure is the sum of the products of each particle of the body or elementary area of the figure multiplied by the square of its distance from the axis around which the body would rotate. This axis is the neutral axis. For example, assuming that the section shown in Fig. 6 is 3 in. X 12 in., and that it is divided into 36 equal parts, each having an area of 1 sq. in., the approximate moment of inertia may be calculated as follows:

Moment Of Inertia 213

Fig. 5.

Fibers b, b, b, b, b,b =

6 sq. in. X 5.5 X 5.5



Fibers c, c, c, c, c, c =

6 sq. in. X 4.5 X 4.5



Fibers d, d, d, d,d,d =

6 sq. in. X 3.5 X 3.5



Fibers e, e, e, e, e, e =

6 sq. in. X 2.5 X 2.5



Fibers f, f, f, f,f,f =

6 sq. in. X 1.5 X 1.5



Fibers g, g, g, g, g, g =

6 sq. in. X .5 X .5



I, or moment of inertia = 429.00

In this manner the approximate moment of inertia for any section may be obtained. Table XII, page 83, gives convenient formulas by which the moment of inertia for usual sections may be determined. For instance, according to this table, the formula for the moment of inertia of any rectangular section is I =bd3 /12, in which which is nearly the same as the approximate result, 429, obtained in the previous calculation.

6 is the breadth of the beam, and d the depth.

Thus, the moment of inertia for the section shown in Fig. 6 may be found as follows:

I=(3x12x12x12) /12 = 432

Moment Of Inertia 214

Fig. 6.

Tables XIII to XIX, on pages 84 to 90, give the moment of inertia for rolled steel sections, and will be found useful in designing structural steel work.

As beam or column sections are often made up of several elementary sections, the moment of inertia is then found thus: Rule. - The moment of inertia is equal to the sum of the products of each elementary area multiplied by the square of its distance from the neutral axis, plus its moment of inertia about a parallel axis through its center of gravity.

If i represents the moment of inertia of each elementary figure, a the area of each, d2 the square of the perpendicular distance from the center of gravity of each elementary section to the neutral axis a b, Fig. 7, and I the required moment of inertia, the rule may he expressed thus: I=Σ(ad2+i).