(a) By considering the relationship between the supporting power and the length and size of the pile, the weight of the ram, height of fall, and the distance the pile was moved by the last blow. The pile is assumed to be driven under ordinary conditions; the head, if broomed or battered unreasonably, should be sawn off before striking the test blow; and it is also assumed that the pile has been sinking with a fair degree of regularity under the last few blows, and that the apparent uniformity of set is not deceptive. (b) By applying a load or direct pressure to each of a number of piles, observing the amount each will support, and expressing the result in terms of the depth driven, size of pile, and kind of soil. The former method may be expressed in a rational formula, the latter in an empyrical formula. The former method has been used more or less in the formulæ of Rankine, Weisbach, Sanders, Professor Baker, Trautwine, Crowell, Wellington, and others. = The energy accumulated in the ram when it strikes the pile head is Wh, where W weight of ram in pounds, and h = height of fall in feet. This energy is expended in compressing the ram and the head of the pile, in moving the ram as a whole against the resistance of the soil, in overcoming the inertia of the pile and soil. Only Wellington's formula will be here given, as it is a rational formula less complex than the others, and gives quite as good results. Let L denote the bearing pressure; s, the penetration under the last blow in inches; then 12Wh L = Wellington recommends a factor of safety of 6, so that the safe working pressure, L', is Fig. 112 shows a timber viaduct for 24-feet centres, designed to carry a single line of railway. In this example there are two main compound beams, each formed with two beams of ironbark timber 12 inches square, bolted together, with a space of two inches between them, with wedges inserted in corresponding notches cut in the beams, and resting upon corbels over the piers. An American deck of ironbark timber is shown similar to that of the 10-feet spans. In order that the compound beam should be equal in strength to a solid beam 24 inches in depth, it is necessary that the wedges and bolts should be capable of resisting the maximum horizontal and vertical shearing stresses. Since the bending moment increases as the square of the span, and the moment of resistance of the beam as the square of the depth, it follows that a span of 20 feet would give the same factor of safety as that found for the 10-feet spans, provided the equivalent distributed load remained the same. The equivalent distributed load on a 24-feet span, with consolidation engines having 12 tons on each of the four driving-axles, would be about 3.1 tons per foot run; hence the span may be increased from 3.75 20 feet to 20 x = 24 feet. The total depth of the compound 3.1 beam is 26 inches, but it will be taken as 24 inches solid throughout. The area is reduced in the centre by the bolthole, and to a slight extent by the wedges, but this is more than compensated for by the reduction of span due to the corbels over piers. The shearing stress over the piers is or 18.6 tons for each beam, which is distributed over the 24 inches by 12 inches in the manner explained in Chapter V., Fig. 106. The maximum shearing stress horizontally and vertically is, therefore S = 3 × 18.6 = 1.16 ton per lineal inch of beam The wedges immediately over the corbels are spaced 15 inches centre to centre, and will have to resist a horizontal shearing stress of 15 x 1·16 17·4 tons, and the corresponding bolts will have to resist a vertical shearing stress of the same amount. = Let x denote the width of the wedges, measured along the beam; the area exposed to shearing along the fibre (neglecting the portion of the wedge which projects beyond the beam for driving) is 12x. The safe intensity of shearing stress may be taken as 450 pounds per square inch in the timbers which are most suitable for this purpose (which require at least 2000 pounds per square inch to shear them along the grain), therefore the resistance of the wedge is 12 × 450 x x = 5400x = 17·4 × 2240 The wedges may therefore be made 7 inches wide by 6 inches deep. The area required in the bolts is square inches, or 2 inches in diameter. The working stress on the bolts is taken as 6 tons. The shearing stress at any other point may be calculated from the formula If x = 0, S1 = 4.65 tons; if x = 6, S1 = 10.5 tons; and these stresses are distributed over the section of the beam as before. The maximum shearing stress per lineal inch in the centre of the beam is, therefore The bolts and wedges are spaced 18 inches centre to centre, therefore the central bolts may be called upon to resist a stress of In a similar manner it may be shown that the bolts 6 feet from the centre of the beam will require to be 13 inch in diameter, and the width of the corresponding wedges 5400r 11.8 × 2240 .. 4.89 inches hence the section of the wedge may be made 5 inches by 4 inches. The sizes of the remaining bolts and wedges may be determined in a similar manner. The design of the piers and abutments is sufficiently illustrated in Figs. 112, 113, and 114. In the foregoing calculations, it has been assumed that each wedge and bolt resists the shearing stresses which are allotted to them according to their position in the span, the largest wedges and bolts occurring near the points of support. Some engineers make the wedges and bolts uniform in section throughout the span, while others omit the wedges in the centre third of the span. If heavier engines are used than those considered, as, for example, on the New South Wales railways, where the heaviest consolidation engine would produce a bending moment in the centre of a 24-feet span which is equivalent to a uniform load (allowing for dynamic effect) of 4 tons per foot run, in such a case it is found to be most convenient to arrange three compound beams, so that each carries of a ton per foot run for the live load, and about 0'4 ton per foot run for the dead load. The wedges are made 6 inches wide by 3 inches deep, and the bolts 1 inch in diameter throughout. We will now consider a compound beam consisting of two beams of spruce timber, each 12 inches by 12 inches, carrying a floor in a building over a span of 30 feet. The two beams are bolted together, and wedges are inserted between the beams. Let it be required to determine the sizes of the wedges and bolts, (a) for carrying a distributed load, and (b) for carrying a central load. The working stress on the extreme fibres of the spruce beam may be taken at 1000 lbs. per square inch. The shearing resistance of the keys or wedges, which should be made of suitable timber, may be taken at 200 lbs. per square inch. The bolts may be stressed up to 12,000 lbs. per square inch. (a) Distributed load. Hence the reactions at supports 12,800 lbs., and the = maximum shearing stress per lineal inch is— Let the wedges be the same length as the width of the beam, and assume that the beams are notched each 3 inches deep, and the depth of the wedges is 6 inches. Assume also that the wedges are spaced 3-feet centres at the ends, and let width of the wedges; then the If the wedges are spaced 24 inches centre to centre, they will only require to be 8 inches wide. Adopting this latter pitch for the wedges, the area of the end bolts will be— At a distance of 7 feet 6 inches from the supports, only half of the area will be necessary if the same pitch is adopted throughout, or the pitch may be doubled for the middle half of the beam, retaining the same sections in wedges and bolts throughout. The reduction in the moment of resistance in the centre of the beam may be neglected. (b) If the beam carries a load in the centre The shearing stress is uniform throughout the beam, excepting at the centre, where it is zero; hence Hence, if the bolts and wedges are spaced at 4-feet centres, they will have the same dimensions as the largest in the last example, namely, 8 inches wide by 6 inches deep. In order to test the accuracy of the foregoing theory of the compound beam, the author has made several large-size scale models, which, when tested, gave results in close agreement |