This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

Fig. 25.

Evidently the center of gravity of each channel section is 6 inches, and that of the plate section is 12¼ inches, from the bottom.

Let c denote the distance of the center of gravity of the whole section from the bottom; then since the area of the plate section is 7 square inches, and that of the whole section is 19.00,

19.06 x c = 6.03 X 6 + 6.03 X6 + 7 X 12½ = 158.11. Hence, c = 158.11 ÷ 19.06 = 8.30 inches, (about).

1. Locate the center of gravity of the built-up section of

Fig. 26.

Fig. 26, a, the area of each "angle" being 5.0G square inches, and the center of gravity of each being as shown in Fig. 26, b.

Ans. Distance from top, 3.08 inches. 2. Omit the left-hand angle in Fig. 26, a, and locate the center of gravity of the remainder.

Ans. | Distance from top, 3.65 inches, |

" " left side, 1.19 inches. |

50. Moment of Inertia. If a plane area be divided into an infinite number of infinitesimal parts, then the sum of the products obtained by multiplying the area of each part by the square of its distance from some line is called the moment of inertia of the area with respect to the line. The line to which the distances are measured is called the inertia-axis; it may be taken anywhere in the plane of the area. In the subject of beams (where we have sometimes to compute the moment of inertia of the cross-section of a beam), the inertia-axis is taken through the center of gravity of the section and horizontal.

An approximate value of the moment of inertia of an area can be obtained by dividing the area into small parts (not infinitesimal), and adding the products obtained by multiplying the area of each part by the square of the distance from its center to the inertia-axis.

Example. If the rectangle of Fig. 27, a, is divided into 8 parts as shown, the area of each is one square inch, and the distances from the axis to the centers of gravity of the parts are ½ and 1½ inches. For the four parts lying nearest the axis the product (area times distance squared) is: l X ( ½) = ¼; and for the other parts it is

1 x (1½)2 = 9/4.

Hence the approximate value of the moment of inertia of the area with respect to the axis, is

4(¼) + 4 (9/4) = 10.

If the area is divided into 32 parts, as shown in Fig. 27, b, the area of each part is ¼ square inch. For the eight of the little squares farthest away from the axis, the distance from their centers of gravity to the axis is 1¾ inches; for the next eight it is 1¼; for the next eight ¾; and for the remainder ¼ inch. Hence an approximate value of the moment of inertia of the rectangle with respect to the axis is:

8 x ¼ X (1¾)2 + 8 X ¼ X (1¼)2 + 8 X ¼ X (¾)2 + 8 X ¼ X (¼)2 =10½,

If we divide the rectangle into still smaller parts and form the products

(small area) X (distance)2, and add the products just as we have done, we shall get a larger answer than 10½. The smaller the parts into which the rectangle is divided, the larger will be the answer, but it will never be larger than 102/3. This 10| is the sum corresponding to a division of the rectangle into an infinitely large number of parts (infinitely small) and it is the exact value of the moment of inertia of the rectangle with re-spect to the axis selected.

There are short methods of computing the exact values of the moments of inertia of simple figures (rectangles, circles, etc.,), but they cannot be given here since they involve the use of difficult mathematics. The foregoing method to obtain approximate values of moments of inertia is used especially when the area is quite irregular in shape, but it is given here to explain to the student the meaning of the moment of inertia of an area. He should understand now that the moment of inertia of an area is simply a name for such sums as we have just computed. The name is not a fitting; one, since the sum has nothing; whatever to do with inertia. It was first used in this connection because the sum is very similar to certain other sums which had previously been called moments of inertia.

51. Unit of Moment of Inertia. The product (area X distance2) is really the product of four lengths, two in each factor ; and since a moment of inertia is the sum of such products, a moment of inertia is also the product of four lengths. Now the product of two lengths is an area, the product of three is a volume, and the product of four is moment of inertia-unthinkable in the way in which we can think of an area or volume, and therefore the source of much difficulty to the student. The units of these quantities (area, volume, and moment of inertia) are respectively : the square inch, square foot, etc.,

Fig. 27.

" cubic " , cubic " " , " biquadratic inch, biquadratic foot, etc.; but the biquadratic inch is almost exclusively used in this connection; that is, the inch is used to compute values of moments of inertia, as in the preceding illustration. It is often written thus: Inches4.

52. Moment of Inertia of a Rectangle. Let b denote the base of a rectangle, and a its altitude; then by higher mathematics it can be shown that the moment of inertia of the rectangle with respect to a line through its center of gravity and parallel to its base, is 1/12 ba3.

Example. Compute the value of the moment of inertia of a rectangle 4 x12 inches with respect to a line through its center of gravity and parallel to the long side.

Here b =12, and a = 4 inches ; hence the moment of inertia desired equals

1/12 (12 X43) = 64 inches4.

1. Compute the moment of inertia of a rectangle 4 x12 inches with respect to a line through its center of gravity and parallel to the short side. Ans. 576 inches4.

53. Reduction Formula. In the previously mentioned "handbooks" there can be found tables of moments of inertia of all the cross-sections of the kinds and sizes of rolled shapes made. The inertia-axes in those tables are always taken through the center of gravity of the section, and usually parallel to some edge of the section. Sometimes it is necessary to compute the moment of inertia of a "rolled section" with respect to some other axis, and if the two axes (that is, the one given in the tables, and the other) are parallel, then the desired moment of inertia can be easily computed from the one given in the tables by the following rule:

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