An axis of symmetry is any line so drawn that, if part of the figure on one side of the line is folded on this line, it will coincide exactly with the other part, point for point and line for line. Thus, in Fig. 1, if the part a b is folded on the line A B, the upper half will coincide exactly with the lower half; also, if bc is folded on the line CD, the right-hand half will coincide exactly with the left-hand half. Hence, the point O where A B and C D intersect is the center of gravity of the rectangle abc d. If the figure has one axis of symmetry, the center of gravity may be found as follows: Let Fig. 2. The center m n be an axis of symmetry of the area in of gravity will lie somewhere on this line. perpendicular to mn. Divide the area into squares, rectangles, triangles, parallelograms, circles, etc., whose centers of gravity are easily found, and measure the perpendicular distances of these centers of gravity from, the line AB. Add the sum of the products obtained by multiplying each area by the distance of its center of gravity from the line A B, and divide by the area of the entire figure; the result is the distance of the center of gravity from A B measured on mn, or the point F. If the figure has no axis of symmetry, as in Fig. 3, draw any line, as A B, and find the distance x of the center of gravity from A B, and through a draw fg parallel to A B. Choose any other line, CD, and find the distance y of the center of gravity from CD by the same method, and through y draw mn parallel to CD. The point of intersection o of f g and m n is the center of gravity. Thus, suppose that the area of the triangle, Fig. 3, is A sq. in., and the distance of its center of gravity from A Bis a in., and from CD, a1 in.; that the area of the small rectangle is B sq. in., and the distance of its center of gravity from A B is b in., and from CD is b1 in.; that the area of the large rectangle is C sq. in., and the distance of its center of gravity from A B is c in., and from CD is c1 in.; then, and x = y = (A × a) + (B × b) + (C× c), (A× a1) + (B× b1) + (C×¢1) To find the center of gravity mechanically, suspend the object from a point near its edge and mark on it the direction of a plumb-line from that point; then suspend it from another point and again mark the direction of a plumb-line. The intersection of these two lines will be directly over the center of gravity. The center of gravity of a body having parallel sides may be found by drawing the outline of one of the sides upon heavy paper, and cutting out the exact shape of the figure. Then suspend the paper from the two points and find the center of gravity, as in the last case. The center of gravity of a triangle lies on a line drawn from a vertex to the middle point of the opposite side, and at a distance from that side equal to one-third of the length of the line. Or, draw a line from another vertex to the middle point of the side opposite, and the intersection of the two lines will be the center of gravity. For a parallelogram, the center of gravity is at the intersection of the two diagonals. For an irregular four-sided figure, draw a diagonal, dividing it into two triangles. Draw a line joining these centers of gravity. Draw the other diagonal, dividing the figure into two other triangles, and join the centers of gravity by a straight line. The intersection of these lines is the center of gravity of the figure. For a figure having more than four sides, find the center of gravity by the general method explained in connection with Fig. 3. For an arc of a circle, the center of gravity lies on the radius drawn to the middle point of the arc (an axis of symmetry) and at a distance from the center equal to the length of the chord multiplied by the radius and divided by the length of the arc. For a semicircle, the distance from the center .6366 r, when r the radius. = 2 T For the area included in a half circle, the distance of the center of gravity from the center 4 r = = .4244 r. 3 п For circular sector, the distance of the center of gravity from the center equals two-thirds of the length of the chord multiplied by the radius and divided by the length of the arc. For a circular segment, let A be its area and C the length of its chord; then the distance of the center of gravity from C3 the center of the circle is equal to 12 A For a solid having three axes of symmetry, all perpendicular to each other, like a sphere, cube, right parallelopiped, etc., their point of intersection is the center of gravity. For a cone or pyramid, draw a line from the apex to the center of gravity of the base; the required center of gravity is one-fourth the length of this line from the base, measured on the line. For two bodies, the larger weighing W lb., and the smaller P lb., the center of gravity will lie on the line joining the centers of gravity of the two bodies and at a distance from the larger body equal to Ρα where a is the distance between the centers of gravity of the two bodies. For any number of bodies, first find the center of gravity of two of them as above, and consider them as one weight whose center of gravity is at the point just found. Find the center of gravity of this combined weight and a third body. So continue for the rest of the bodies, and the last center of gravity will be the center of gravity of the whole system of bodies. MOMENT OF INERTIA. The moment of inertia of a body or section is a mathematical expression that is much used in computations relating to rotating bodies and to the strength of materials. It may be defined as follows: The moment of inertia of a body, rotating about a given axis, is the sum of the products obtained by multiplying the weights of the elementary particles of which it is composed by the square of their distances from the axis. It is often desirable to use the moment of inertia for a plane section; but as a plane surface has no weight, it is apparent that the above definition does not correctly apply. The following definition applies to plane surfaces: The moment of inertia of a plane surface about a given axis is the sum of the products obtained by multiplying each elementary areas into which the surface may be conceived to be divided by the square of its distance from the axis. The axis about which the body or surface rotates, or is assumed to rotate, i. e., the axis from which the distance to each area or particle is measured, is called the axis of rotation. The least moment of inertia is that value of the moment of inertia of a body or section when the axis of rotation passes through the center of gravity, since its value is less for that position of the axis than for any other. To find the moment of inertia of a body about a given axis: Divide the body or section into many small parts and multiply the weight or area of each part by the square of the distance from its center of gravity to the axis of rotation; the sum of these products will be the moment of inertia. NOTE. The results obtained by the above rules are really only approximate; for practically it is impossible to divide a body or surface into parts sufficiently small for absolute accuracy. The smaller the parts the more accurate will be the result; but the results obtained by these rules will always be slightly too small. The moment of inertia is usually designated by the letter I. Formulas for the values of I about an axis of rotation passing through the center of gravity of the section are given for various forms of sections in Table V, page 153. The moment of inertia about an axis of rotation not passing through the center of gravity is equal to the moment of inertia about a parallel axis through the center of gravity plus the product of the entire weight of the body (or area of the section) multiplied by the square of the distance between the two axes. 8.5 EXAMPLE.-It is desired to find the moment of inertia of a 6" I-beam of the dimensions shown in Fig. 1 about an axis zy perpendicular to the web of the beam at the center. SOLUTION. Since the axis about which the moment of inertia is to be found is an axis of symmetry of the beam, it is necessary to make the computations only for the half section of the beam lying at one side of the axis, and multiply the result by 2. As stated before, the smaller the parts into which the area is divided, the more accurate will be the result. .375 FIG. 1. It will be sufficiently accurate for present purposes to divide the section in the manner shown in Fig. 2. The operations are given at the side of the figure, and will be readily understood. The sum of the products is the approximate value of the moment of inertia of this half of the section about the axis xy, and when multiplied by 2 is the approximate value of I for the entire section. It is found to equal 23.444. |