The areas of the parts are 6 and 3 square inches, and their arms with respect to Or are 4 inches and j inch respectively, and with respect to OL" inch and 1 inches. Hence the equations of moments with respect to OL' and OL" (the whole area being 9 square inches) are: 9Xa= 25.5, 9xb = =7.5.
Hence, ' a = 25.5÷.9 = 2.83 inches, =-- 7.5=9=0.83 " .
2. It is .required to locate the center of gravity of Fig. 22, b, the width being uniformly one inch.
The figure can be divided up into three rectangles. Let CI, and be the centers of gravity of such parts, C the center of gravity of the whole; and let a denote the (unknown) distance of C from the base. The areas of the parts are 4, 10 and 4 square inches, and their " arms " with respect to the base are 2, and 2 inches respectively; hence the equation of moments with re spect to the base (the entire area being 18 square inches) is: .18 xa =4x 2+10x i±4x 2 = 21.
Hence, a = 21֥18 = 1.17 inches.
From the symmetry of the area it is plain that the center of gravity is midway be tween the sides.
1. Locate the center of gravity of Fig. 23.
Ans. 2.6 inches above the base.
49. Center of Gravity of Sections. In Fig. 24 there are represented cross-section's of various kinds of rolled steel, called "shape steel," which is used extensively in steel construction. Manufacturers of this material publish "handbooks" giving full information in regard thereto, among other things, the position of the center of gravity of each cross section. With such a handbook available, it is therefore not necessary actually to compute the tion of the center of gravity of any section, as we did in the ceding article; but sometimes several shapes are riveted together to make a "built-up" section (see Fig. 25), and then it may be sary to compute the position of the center of gravity of the section. Example. It is desired to locate the center of gravity of the section of a built-up beam represented in Fig. 25. The beam con sista of two channels and a plate, the area of the cross-section of a channel being 6.03 square inches.
Evidently the center of gravity of each channel section is 6 inches. and that of the plate section is 12i inches, from the bottom. Let c denote the dis tance of the center of gravity of the whole section from the bot toin ; then since the area of the plate section is 7 square inches, and that of the whole sec tion is 19.06, 1. Locate the center of gravity of the built-up section of Fig. 26, a, the area of each "angle" being 5.00 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. 20, a, and locate the center of gravity of the remainder.
Ans. 5 Distance from top, 3.65 inches, " " left side, 1.19 inches.
so. foment of Inertia. If a plane area be divided into an infinite number of infinitesimal parts, then the sum of the prod. ucts obtained by multiplying the area of each part by the square of its distance from some line is sailed 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 infini tesimal), 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 dis tances from the axis to the centers of gravity of the parts are and 11 inches. For the four parts lying nearest the axis the product (area times distance squared) is: I X( if=i; and for the other parts it is 1 X Hence the approximate value of the moment of inertia of the area with respect to the axis, is 4(i)±4(1)=----10.
If the area is divided into 32 parts, as shown in Fig. 27, b, the area of each part is I- 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 1/ inches; for the next eight it is for the next eight /; and for the remainder 1- inch. Hence an approximate value of the moment of inertia of the rectangle with respect to the axis is: 8 xix xix If we divide the rectangle into still smaller parts and form the products (small area) X and add the products just as we have done, we shall get a larger answer than 101. The smaller the parts into which the tangle is divided, the larger will be the answer, but it will never be larger than 103. This 10g- is the 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.