shearing stress, or R=sA, the number n of rivets required to transfer a stress T by single shear is T T In Fig. 2, the rivet is subjected to shear on two sections, d and e, and it is Isaid to be in double shear. The amount of stress that one rivet can carry in double shear is twice that of one in single shear, and, using the preceding notation, n= T 2R The bearing value of a rivet is the compressive stress induced by the rivet in bearing on the plate, and is also calculated by the simple-stress formula in which P=SA P= value of rivet in bearing; s=unit working stress in bearing; A bearing area. It is customary to assume that the bearing area A is the thickness of the plate multiplied by the diameter of the rivet. In calculating the required number of rivets, both the shearing and the bearing value of one rivet are determined and the critical value (the smaller) used. The accompanying table gives the shearing and bearing values of rivets, in pounds, for different values of the working stress. 3. Strength of Cylindrical Shells and Pipes With Thin Walls.-When a cylinder is subjected to internal pressure, the tensile stress developed in the walls or shell of the cylinder is called circumferential stress, or hoop tension. Let s=intensity of this stress; d=internal diameter of cylinder; p-intensity of pressure on inner surface of the cylinder; The first formula serves to compute the thickness when p, d, and s (working stress) are given; and the second one is used to compute the intensity of stress when the intensity of pressure p and the dimensions of the cylinder are given. EXAMPLE. What should be the thickness of walls of a cast-iron water pipe, inside diameter 24 in., to resist a water pressure of 200 lb. per sq. in., using a unit working stress of 2,000 lb.? SOLUTION. Here, d=24, p =200, and s=2,000. Substituting in the formula for t, 200 X 24 t= = 1.2 in. 2X 2000 4. Temperature Stresses.—If a bar subjected to change of temperature is constrained so that it can neither expand nor contract, the constraint exerts on it a force sufficient to prevent the deformation. This causes in the bar a corresponding stress called temperature stress. It is compressive when the change of temperature is a rise, and tensile when a fall. Let T be the stress induced in a bar, whose area is a, by a rise or fall of t°; also let c be the coefficient of expansion and E the modulus of elasticity of the material. Then, T= ctaE The coefficient of expansion for a number of substances is given in the section on Heat, Fuels, Etc. EXAMPLE.-A wrought-iron bar 1.5 in. square has its ends fastened to firm supports. What is the stress produced in it by a change of 50° in its temperature? SOLUTION. Here, E=25,000,000; a=1.5X1.5=2.25 sq. in., and t=50; and, c=.00000686. Substituting in the formula, T=,00000686×50×2.25 X25,000,000 = 19,294 lb. BEAMS A beam is a body resting upon supports and liable to transverse stress. Beams are designated by the number and location of the supports, and may be simple, cantilever, fixed, or continuous, A simple beam is one that is supported at each end, the distance between its supports being the span. A cantilever is a beam that has one or both ends overhanging the support; or a beam that has one end firmly fixed and the other end free. A fixed beam is one that has both ends firmly secured. A continuous beam is one which rests upon more than two supports. Reactions. The loads acting on a beam are balanced by the reactions or supporting forces; their sum must therefore be equal to the sum of the loads. To find any reaction, as R2, at B, Fig. 1, take moments of all the external forces about the other support A and divide their sum by the span. With reference to Fig. 1, Wia+W2b+W3C R2= The reaction R1 can be found in a similar manner by taking moments about the support B. Their sum R1+R2 must be equal to the sum of loads W1+ W2 +W3. EXAMPLE. Find the reactions of a cantilever bridge loaded as shown in Fig. 2. SOLUTION.-Substituting given values in the formula and noting that the moment of P1 about B is of opposite sign to the moments of the other loads, 10,000×120+8,000×90+15,000×40-20,000×30 R1 = and R2= 150 10,000×30+8,000 × 60+15,000×110+20,000 × 180 150 = = 12,800 lb. 40,200 lb. The sum of the loads is 10,000+8,000+15,000+20,000=53,000. The sum of the reactions is 40,200+12,800=53,000. External Shear and Bending Moment.-The forces acting on a beam tend, on the one hand, to shear its fibers vertically and, on the other hand, to bend it, producing compressional stresses in the fibers on one side of the neutral axis and tensional on the other side. The tendency to shear the fibers vertically is determined by the external shear, and that of bending by the bending moment. For brevity, external shear is often called simply shear, but it must not be confused with shearing stress at the section. Forces acting upwards are considered positive, and those acting downwards, negative. The external shear at any section of a beam is the algebraic sum of all the external forces (loads and reactions) on one side of the section. It is equal to either reaction minus the sum of the loads between that reaction and the section considered. The maximum shear is always equal to the greater reaction. For a simple beam with a uniformly distributed load, the maximum shear is at the supports, and is equal to one-half the load, or to the reaction; the shear changes at every point of the loaded length, the minimum shear being zero at the center of the span. The maximum shear in a simple beam having a single load concentrated at the center is equal to one-half the load, and is uniform throughout the beam. Where a beam supports several concentrated loads, changes in the amount of shear occur only at the points where the loads are applied. V The external shear is resisted by the internal shear, or shearing stress, of the beam, which is numerically equal to the external shear. If the external shear is denoted by V, and the area of the cross-section by A, the average intensity of shearing stress in the section is This shearing stress is not uniformly distributed, and in beams of rectangular cross-section, the maximum intensity of shearing stress is Hence, a rectangular beam must be so designed that this value will not exceed the working shearing strength of the material. 3V In metallic beams with thin webs (plate girders), the shearing stress may be considered as uniformly distributed over the cross-section of the web. There is, also, at every horizontal or longitudinal section of the beam, a horizontal shearing stress the intensity of which at any point is equal to the intensity of the vertical shearing stress at that point. Although the maximum intensity of shearing stress, both horizontal and vertical, in wooden beams is usually small, the shearing strength of wood along the grain is also small. As the horizontal external shear usually acts along the grain, the safe load for a wooden beam may depend on its shearing strength and not on its bending strength. For instance, the safe load for a beam 4 in. X 12 in. and 4 ft. long is 16,000 lb., uniformly distributed, when based on a fiber strength of 1,000 lb. per sq. in. Such a load will produce a shearing stress per unit of area equal to 3X8,000 250 lb. per sq. in., which exceeds the working shearing stress for the wood along the grain by about 100 lb. per sq. in. The bending moment at any section of a loaded beam is equal to the algebraic sum of the moments of all the external forces (loads and reactions) to the right or FIG. 3 W left of the section about that section. For example, the bending moments at several points on the beam shown in Fig. 3 are as follows: At Wi Ria; at W2 R1(a+b) — W1b; at W3=R1(a+b+c) -[W2c+W1(b+c)], or Red. The bending moment varies, depending on the shear, and attains a maximum value at the point R1 E2where the shear changes sign. If the loads are concentrated at several points, the maximum bending moment will be under the load at which the sum of all the loads between one support up to and including the load in question first becomes equal to, or greater than, the reaction at the support. Hence, to find the maximum bending moment in any simple beam: Rule. Compute the reactions and determine the point where the shear changes sign. Calculate the moment about this point of either reaction, and of each load between the reaction and the point, and subtract the sum of the latter moments from the former. EXAMPLE.-What is the maximum bending moment of the beam loaded ås shown in Fig. 4? 25ft. ·6ft 7ft 000'9 4,000 P 9ft -Load-e ning foot -12ft R2 FIG. 4 SOLUTION.-The reactions due to the uniform load are equal to one-half of the load; those due to the concentrated loads are computed by the principle given under Reactions. Both added give R1 = 18,170 lb. and R2 = 14,330 lb. Beginning at R1 and subtracting the loads in succession, it is found that the shear just to the left of the load d is 18,170-16,500; and just to right of the load d it becomes negative. Hence, the shear changes sign under the load d and the bending moment is maximum at that point. It is equal to 18,170X13-10,000 × 7— 132 X 500 = 123,960 ft.-lb. Formulas for the maximum bending moments and shears for beams loaded and supported in different ways are given in the following table. For a beam supporting moving loads, the maximum bending moment occurs: 1. For a single load, when the load is at the middle of the span. 2. For two equal loads, under either load, when the two loads are on opposite sides of the center, and one of the loads is at a distance from the center equal to one-fourth the distance between the loads. 3. For two unequal loads, under the heavier load, when that load and the center of gravity of the two loads are equidistant from the center of gravity of the beam. EXAMPLE. A beam 24 ft. long supports two moving loads 6 ft. apart. The left-hand load is 8,000 lb., and the right-hand load is 4,000 lb. Find the maximum bending moment. SOLUTION. The center of gravity of the loads is 2 ft. from the left-hand load. The maximum bending moment occurs under the heavy load, and obtains when the latter is 1 ft. to the left of the center of the beam. The left • 12,000 X 11 reaction is, then, = 5,500 lb., and the maximum bending moment is 24 5,500X 11=60,500 ft.-lb. Designing of Beams. In every section of a carrying beam there is induced an internal moment called the moment of resistance, which is equal to the bending |