an ounce of platinum; but the ounce of platinum can have the same moment of inertia as the ounce of cork, if its center of gyration has the same position with respect to the axis of rotation. The center of gyration is not at the center of gravity, nor at the center of oscillation, but at some point in a straight line between those centers. The radius of gyration is the distance from the axis of rotation to the center of gyration. The square of the radius of gyration is the average of the squares of the distances from the axis of rotation to each elementary particle of the body, or to each elementary area of the section, as the case may be. But the sum of these squares of distances, multiplied by the weight or area of each elementary part, equals the moment of inertia; therefore, the moment of inertia divided by the weight of the body or area of the section equals the square of the radius of gyration; the square root of this quotient is the radius of gyration. But, according to the rule for finding the radius of oscillation, the quotient obtained by dividing the moment of inertia by the weight or area equals the product of the distance from the axis of rotation to the center of gravity, multiplied by the radius of oscillation; and, therefore, the radius of gyration is a mean proportional between these distances. If the distance from the axis of rotation to the center of gravity is known, and the radius of oscillation is known, the radius of gyration may be found by multiplying these two known distances together and extracting the square root of the product. In the example of the I-beam, Fig. 2, page 126, the sum of the areas of the half section of the beam is 1.917, and the area of the entire section is 3.834 sq. in. Therefore, the radius of gyration of this beam about an axis through the center of 23.44 gravity perpendicular to the web = V3.834 = 2.47 in. In the example of the iron bar 12 in. long (see figure, page 128), the distance from the axis of rotation to the center of gravity is 6 in., and the radius of oscillation was found to equal 8 in. Therefore, the radius of gyration about an axis perpendicular to the bar at one end = √6×8 6.93 in. Or, the moment of inertia of the bar = 161.586, and the weight of the bar = 3.372 lb. Therefore, the radius of gyra tion = 161.586 3.372 6.93 in., very nearly. The radius of gyration is used in determining the strength of columns. The axis must be taken in such a direction that the result will be the least radius of gyration of the column; this condition is usually obtained when the axis is perpendicular to the least diameter or side of the column. The various relations between these quantities may be concisely expressed by the following formulas, in which A area of section (or weight of body if the weight is used); g= distance from axis of rotation to center of gravity; = radius of gyration; G To find the radius of oscillation, radius of gyration, and moment of inertia, experimentally. The connecting-rod of an engine is represented in the figure. It is desired to find the moment of inertia of the rod about an axis of rotation through the center of the crosshead pin 4. This may be accomplished, experimentally, as follows: Suspend the rod from the crosshead pin in such a manner that it will swing freely; cause it to swing, or oscillate, and note the exact time of the vibrations. Remove the crosshead pin and reverse the rod, but, instead of suspending it by the crank pin, suspend it by a movable pin B, that can be clamped at any desired point upon the rod. C is another view of this pin. There will be a point on the rod from which it may be suspended by means of the movable pin, so that it will vibrate in exactly the same intervals of time as when suspended from the crosshead pin. This point is the center of oscillation, for the center of oscillation and the center of rotation are interchangeable; the point will be found at about one-third the length of the rod from the crankpin. Find this center of oscillation, experimentally, and carefully measure the distance from the center of the movable pin to the center of the crosshead-pin hole. This distance is the radius of oscillation =r. Next remove the movable pin, and find the center of gravity (lengthwise) of the rod by balancing it across a knife edge, and measure the distance from the center of gravity thus found to the center of the crosshead-pin hole; this distance g. Finally, weigh the rod. The product of the weight (= A), the radius of oscillation (= r.), and the distance from the center of crosshead pin (axis of rotation) to the center of gravity (= g) will be the moment of inertia. For, by the formula, I = A gr。. The radius of gyration G may be found by the formula MOMENT OF RESISTANCE. If the moment of inertia of the cross-section of a beam is divided by the distance from the neutral axis (see definition on next page) to the extreme fiber, i. e., the fiber that is farthest from the axis, the quotient will be the quantity known as the moment of resistance. It is evident that, if a beam is strained by a vertical load, the greatest stress will be in the extreme upper and lower fibers of the beam. The intensity of the stress that can be borne by the extreme fibers is the limit of the strength of the beam. The upper fibers are compressed and the lower fibers are stretched, but somewhere along or near the center of a vertical section of the beam, the fibers are neither extended nor compressed; the position of these fibers is called the neutral surface, and the line where this neutral surface intersects a right section of the beam is the neutral axis of the section. The neutral axis passes through the center of gravity of the section. If the moment of resistance is multiplied by the amount of stress that may be allowed per square inch upon the extreme fiber, the product will represent the efficiency of the beam to resist bending moment. EXAMPLE.-Referring to the 6" I-beam, Figs. 1 and 2, pages 126 and 127, for which the moment of inertia of the section has been found, it is desired to ascertain the load that a wrought-iron beam of the same dimensions as Fig. 1 will carry at the center of a span 8 ft. between supports. SOLUTION.-The moment of resistance for the section 23.48 3 = = 7.83. In Table II, page 151, the ultimate strength or fiber stress for wrought iron is given as 50,000 lb. per sq. in., and in Table I, page 151, the factor of safety given for wrought iron under a steady stress is 4; therefore, the safe S 50,000 fiber stress for wrought iron 12,500 lb. per = J 4 sq. in., and the moment of resistance multiplied by the safe 8 ft., or 96 in.; equating the bending moment for a load at fore, W = 4,078 lb., the load that can be safely supported at the center of the beam. |