bottom angles. The distances of the centers of gravity of the subdivisions of this section from the axis XX, Fig. 2, are: For the cover-plate 15.25-9.63.... = 5.62 = 2.13 For the 3"X3"X" L's, 15.00-1.06-9.63. = 4.31 = 8.45 The moments of inertia of the respective parts about their own neutral axes parallel to XX are: From a steel manufacturer's handbook, the value of I for a 3"x3"XL is found to be 3.64; and for a 4"X4"X4" L it is 5.56. Applying principle II, the moment of inertia of the entire section is, I=.25+24X5.622+281.25+2×15×× 2.132+2 3.64 + 2 × 3.25 X 4.312 +2 X 5.56 +2×3.75X8.452 = 1,403.22. RADIUS OF GYRATION Let I denote the moment of inertia of any section and a its area; then, the relation between I and a is expressed in the formula, I = ar2, in which r is a constant depending on the shape of the section and is called the radius of gyration of the section referred to the same axis as I. Then, 1= a EXAMPLE 1.-What is the radius of gyration of the section shown in Fig. 1 about the axis XX? SOLUTION. The moment of inertia of this section has been found to be 618.6 and its area is 2X12X+10.5×}=23.25 sq. in. Substituting in the formula, EXAMPLE 2.-Determine the distance b in the strut made up of two latticed channels, as shown in Fig. 3, so that the radii of gyration about the axes XX and YY will be equal. SOLUTION.-Let Ix, rx, Iy, ry be, respectively, the moments of inertia and radii of gyration of a single about the axes Y FIG. 3 A practical rule giving good approximate results for a channel column or strut is to subtract ry from r; the result is b. Applying this rule for the 15" E of 33 lb. column or strut, b = 5.62 -.912-4.708. The expression —, с AND MOMENT OF RESISTANCE in which I is the moment of inertia and c the distance of the outermost fiber of the section from the neutral axis, is called the section modulus. For a given material, this quantity is a measure of the capacity of the section to resist bending. Multiplied by the unit stress to which the outermost fibers are subjected under given loads, the product gives the amount of bending moment the section is resisting, and is therefore called moment of resistance. If f is the unit stress that certain loads develop in the outermost fibers of the section, the moment of resistance is I C EXAMPLE 1.-What is the section modulus of a 20-in. I beam at 75 lb. whose moment of inertia is 1,268.9? SOLUTION. Since the neutral axis passes through the center of the section, the distance c is in this case equal to one-half the depth; that is, 40=10. The section modulus is therefore EXAMPLE 2.—When subjected to loads perpendicular to the cover-plates the outermost fibers of the section shown in Fig. 2, are stressed to 16,000 lb. per sq. in., What is the resisting moment of the section? SOLUTION.-The moment of inertia of the section has been found to be 1,403.22 and the outermost fibers are 9.63 in. from the neutral axis; hence, the section modulus is equal to 1,403.22 9.63 = 145.7; this multiplied by 16,000 gives 2,331,200 in.-lb. Formulas for obtaining directly the section moduli of sections frequently used are given in the table of Moments of Inertia, etc. For rolled-steel sections, they are given in manufacturers' handbooks. FRICTION Friction is the resistance that a body meets from the surface on which it moves. It depends on the degree of roughness of the surfaces in contact, and is directly proportional to the perpendicular pressure between the surfaces. It is independent of the extent of the surfaces in contact, so long as the normal pressure remains the same. It is generally greater between surfaces of the same material than between those of different materials, and greater between soft bodies than hard ones. Coefficient of Friction.-The ratio between the resistance to the motion of a body due to friction and the perpendicular pressure between the surfaces is called the coefficient of friction. When the coefficient of friction between two surfaces is known, the frictional resistance is obtained by multiplying the normal pressure by the coefficient. EXAMPLE. What is the resistance per linear foot of a retaining wall against sliding when the normal pressure on the foundation is 10,000 lb. per lin. ft. of wall and the coefficient of friction of the masonry on the foundation is .65? SOLUTION.-The frictional resistance is 10,000X.65=6,500 lb. The coefficient of friction of the wheels of suddenly stopping engines and cars on the rails is usually assumed at .20. The rails on bridges or trestles will transfer to the bridge or trestle tower the frictional forces produced by the brakes in order to stop the cars, causing stresses that have to be provided for in the structure. EXAMPLE.-What is the longitudinal force on a bridge caused by the sudden stopping of a car weighing 60,000 lb.? SOLUTION.- 60,000X.20=12,000 lb. Angle of Friction.-When a body, as B in the accompanying illustration, weighing W lb. is placed on an inclined plane making an angle a with the horizontal, the normal pressure is N=W cos a; and, if the coefficient of friction is denoted by f, the frictional resistance against sliding down of the body is F=ƒN=ƒW cos a. This force acts in a direction opposite to that of the force P = W sin a. When the angle a is such that F just balances, or is equal to, P, so that the slight N W est force will cause the body to slide, the angle is then called The tangent of that angle is equal to the angle of friction. the coefficient of friction, or f= tan a. Angle of Repose.-On a sloping bank of loose material, such as sand, earth, etc., when the angle of slope is such that the particles are on the point of moving, the angle is called the angle of repose. It is the same as the angle of friction of the material on itself. The slope is then called the slope of repose, or the natural slope of the material, for it is the slope that the material will assume when subject to gravity only. EXAMPLE. The coefficient of friction of dry sand on itself is .65; what is its angle of repose? SOLUTION. The angle of repose is the same as the angle of friction, whose tangent equals the coefficient of friction; consequently, .65=tan a, and from a table of natural tangents a=33°. The accompanying tables give coefficients of friction and angles of repose of a number of materials. COEFFICIENTS OF FRICTION AND ANGLES OF REPOSE FOR MASONRY MATERIALS Rolling Friction.-The friction between the circumference of a rolling body and the surface upon which it rolls is known as rolling friction. It is due to the compressibility of substances, the weight of the rolling body causing a small depression in the supporting surface and a flattening of the roller. Its magnitude .50 to .60 .50 27 to 31 19 to 24 19 |