A Now for a point K between C and D. There is no B B choice of sides, suppose we take the left, then Κ = P. AC+ (P - W1)CK (1). If then we draw KL= MK (Fig. 200), and draw EM parallel to AB; we have KL=KM+LM=CE+LM. But CEP. AC, .'. LM=(P– W1)CK (from 1). Therefore LM/CK or LM/EM is constant for all positions of K, so that L lies on a straight line through E. Evidently exactly similar reasoning would prove that L must lie on a straight line through F. Hence we conclude that L lies on EF, and therefore AEFB is the diagram of B. M. Commencing at A, between A and C, F is + P, and the curve is RS parallel to AC. 1 Between C and D we have to the left of any point P – W1 upward; now we notice that in Fig. 200 we have drawn L above E, which assumes that P-W1 is positive; therefore F is + (P – W1) between Ĉ and D, and the curve is TV where CT=P-W1 or ST = W1. In BD we have P − (W1+ W2), which is negative, on the left, or +Q on the right; F then is Q, and we set off Q downwards, the curve then being UX. Thus RSTVUX is the full curve of shear. The student may for practice commence at B and work to the left; he should thus obtain the same curve. We have worked this question by the preceding method simply to show how, by keeping strictly to the definitions, we are able to find the value, and hence draw the curves, of B. M. and S. F. at every point. Generally, however, the simpler way will be to use a method which will be explained in the next chapter. EXAMPLES. 1. A plank is laid across an opening 12 feet wide. A man weighing 156 lbs. walks across it. Draw diagrams showing the S. F. and B. M. when he is 2, 4, and 6 ft. respectively from one end. Give the maximum values in each case. Ans. 130, 104, 78 lbs.; 260, 416, 468 lbs. -feet. 2. A similar plank to that of (1) is broken by the bending effect of 3 cwt. placed 3 ft. from the end. What is the greatest load the man could safely carry across? Ans. 96 lbs. 3. The air pump of an engine is driven by a lever or rocking arm, total length 8 feet. The air pump stroke is half the piston stroke. Find the greatest bending moment on the lever, when lifting the pump bucket 18 ins. diameter against an effective pressure of 16 lbs. to the square inch. Ans. 10,862 lbs. -feet. 4. The speed of periphery of a spur wheel is 20 f.s. The teeth are 1 in. long. Find the bending moment at the root of a tooth, for each H. P. transmitted, assuming only one tooth in gear at a time, and the whole pressure to come on the point of the tooth. Ans. 41 lbs. -inches. 5. A loaded truck weighing 10 tons rests on two axles. The axles are supported by the wheels 5 ft. apart, and the centres of the axle boxes are 4 ft. 4 ins. apart. Draw curves of B. M. and S. F., and give numerical values. Ans. B. M.-o at ends to 10 tons-inches at and between axle boxes; S. F.- ton from ends to boxes, o between boxes. 6. The horse-power of an engine is 100. Stroke, 4 feet. Revolutions, 50 per minute. Find the bending moment on the crank arm at its junction with the shaft, the diameter of the latter being 6 ins. Ans. 14,437.5 lbs. -feet. 7. A beam, 7 feet long between supports, overhangs a ft. and b ft. at the two ends respectively. At the first end a load W1 hangs, and at the second a load W2. Show from the definition that the bending moment midway between the supports is W1a+W2b and is thus independent of 7. Draw the curves of B. M. and S. F. 2 66 8. The top of a combustion chamber 2 ft. 8 ins. deep is supported by rows of three stays, each stay carrying a load of 6000 lbs., spaced at equal distances apart and from the front and back plates. Each row is supported by a girder" or dog," the ends of which rest on the front and back plates. Find the bending moment on a girder at each of the three points where it carries a stay. Ans. 6000, 8000, 6000 lbs. -feet. CHAPTER XV B. M. AND S. F. UNDER DISTRIBUTED LOADS-PRINCIPLE OF SUPERPOSITION IN the last chapter we considered the effects of loads acting at points, qualified by explaining that "at a point" really means over a small length. The work then applies to cases in which the load carried is concentrated on one or more small lengths of the beam, e.g. a locomotive on a bridge. Here the total weight is concentrated at the two points C and D, in proportions depending on the distribution C Fig. 201. of weights in the locomotive. In Fig. 201 the engine is of appreciable length compared to the bridge; but if the engine were, say, on one of the spans of the Forth Bridge, then, for all practical purposes, it would be quite sufficient to consider the engine as one weight only, concentrated at its C. G. But now take the case in which a bridge is covered by a densely packed crowd; then the points of application of the loads are so numerous that practically there is a continuous load at every point of the bridge. One kind of load we can see is perfectly continuous, viz. the weight of the beam itself. Such loads as we have just mentioned are called Distributed Loads; and if they be so distributed that the load on every unit length is the same, they are then said to be uniformly distributed. In the present work this latter is the only kind we shall consider. Intensity of Load. The intensity of the loading is generally estimated by the load on each foot length, or running foot, of the beam; so that it will be stated as so many lbs., cwts., or tons per foot run. For example, a bridge 20 ft. span, loaded with a uniformly distributed load of 10 tons, would be said to be under a loading of ton per foot run. The word span here used signifies the length between the supports. be denoted by or 2a-the latter being often useful These must be in feet. Then to avoid fractions. The total load on the beam=wl, This we generally denote by W, ... W=wl, or 2wa. or 2wa. These are First we must find the supporting forces. evidently equal, and each is, therefore, W/2 or wa. The uniform loading is represented by the small arrows, but its amount is not indicated in the figure. Take C the centre, and take any point K, K being defined by its distance CK from the centre. This dis tance we call x. Then MK = moment of forces to the left of K about K, This latter consists of an infinite number of small forces, but we can find its moment, because, by a principle of Statics, |