be seen that specimens marked I. and II. broke at higher loads than those marked IV., and that the weakest of all were those marked III. TESTS OF THE TRANSVERSE STRENGTH OF WINDOW LINTELS. All the window lintels tested were of the form shown in the cut (Fig. 31), and all were supported at the ends and loaded in the middle, the span in every case being 52". From the cut it will be seen that the web varied in height, being 4 inches high above the flange in the centre, and decreasing to 2.5 inches at the ends over the supports. Inasin uch as the section, and hence the moment of inertia of the section varied, it became necessary to deduce a special approximate formula suitable to determine the modulus of elasticity from the observed deflections. In order to deduce this special formula, the moments of inertial were first determined at the following five sections, viz.: Hence this was used for I in the general deflection equation: A perusal of the results will show that the P's which in tension bore the least were in every case the ones which in the form of lintels stood the most. On the whole, the tensile and the compressive moduli of rupture compare very well with the tensile and the compressive strength of the iron respectively. SUMMARY OF THE EXPERIMENTS OF MESSRS. EASTMAN AND GERRISH. The object of this thesis was to determine the constants suitable to use in the formulæ for determining the strength of the arms of cast iron pulleys; and also, incidentally, to determine the holding power of keys and set screws. Some old pulleys which had been in use at the shops were employed for these tests. They were all about fifteen inches in diameter, and were bored for a shaft 1 inches in diameter. Inasmuch as this size of shaft would not bear the strain necesFig: 32 sary to break the arms, the hubs were bored out to a diameter of 11 inches diameter, and key-seated for a key one-half an inch square. In order to strengthen the hubs sufficiently, two wrought iron rings were shrunk on them, so as to make it a test of the arms and not of the hub. The machine used for applying the stress is shown in the cut (Fig. 32). The pulley under test is keyed to a shaft which, in its turn, is keyed to a pair of castings supported by two wrought iron I beams, resting upon a pair of jackscrews, by means of which the strain is applied. A wire rope is wound around the rim of the pulley, and leaves it in a tangential direction vertically. This rope is connected with the weighing lever of the machine, and weighs the load applied. The idea of the arrangement was to get a pull upon the rim of the pulley as nearly as possible like the belt pull, to which it is subjected in practice, and at the same time to have some means of weighing this pull. In practice there are two pulls upon the rim, that of the tight side, and that of the loose side of the belt, the sum of the two tending to produce a bending of the shaft and a compression of the rim and arms of the pulley, while the difference of the two causes a rotation of the pulley and a bending moment in all the arms. It will be seen in the arrangement used that while there is no tight side and loose side of a belt, yet there is a compression of both rim and arms, which must be very similar to that caused by a belt, and a bending moment in the arms such as occurs in practice. In a number of the experiments one arm gave way first, and then the unsupported part of the rim broke. The breaking load of the separate pulleys was, of course, determined, and then it was sought to compute from this the modulus of rupture of the cast iron, if so it can be called. The method commonly given for computing the strength of pulley arms is to consider them in one of two ways, viz., either as beams fixed in direction at one end and loaded at the other, or else to consider them as fixed in direction at both ends, thus making of each arm a pair of cantilevers, half as long as the arın, fixed at one end and loaded at the other. If we let I moment of inertia of section, n = number of arms, y = half depth of each arm distance from neutral axis to out side fibre, = length of each arm in a radial direction, x= Pbreaking load determined by experiment: Then we should have, for the outside fiber stress at fracture, 3123% fo 10408 XXX 5,600 17 x Hub cracked. Not a test of the arms. Not a test of the arms. 2 2,200 24,425 12,212 All the arms broke 5.12 3 133 51% × 1 × 614334 335 61 16 414 51 x 11 x 2,100 23,314 at the hub. One arm broke at Load subsequently increased to 8,000 when the rim rim and hub. at the hub. One arm broke at Load subsequently increased to 5,300 when the rim the hub. broke. One arm broke at Load subsequently increased to 2,200 when the rim One of the arms was broken in driving it on to the There was a bushing inside the hub keyed to shaft, One arm broke at the hub during test of keys. Average. 26,528 13,264 |