Fig. 15. The analysis of the forces acting on the drum is simple, but its theoretical design is more complicated. It is evident that the drum acts as a beam of hollow circular cross section, and that its worst case of loading is when the rope is at or near the middle of the drum length. At the same time the metal of this circular cross section is in a state of torsion between the free end of the rope and the driving gear, due to the load on the gear tooth and the reaction of the rope. Also the wrapping of the rope around the drum tends to crush the metal of the section beneath it, the maximum effect of this action being near the free end of the rope where its tension has not been reduced by friction on the drum surface.

Now the "mechanics" to solve the problem of these three combined actions is rather complicated. It can be at least approximately solved, however, for it satisfies fairly well the case of combined compression and shear. But on a further study of this particular case, it is seen at once that the diameter of the drum is relatively large with respect to its length, which means that the thickness of the metal may be very small and yet give a large resisting area, or value of " I," both in direct bending as well as torsion; also it is so short that the external bending moment will be small. The practical condition now comes in, that the drum can be safely cast only when the thickness of the metal is at a minimum limit, for the core may be out of round, not set centrally, or by some other variation produce thin spots or even develop holes reaching out into the rope groove, discovered only when the latter is turned in the lathe.

Hence it seems reasonable and safe in this case to make the thickness of the drum depend simply upon the crushing caused by the wrapping of the rope around it, and we shall take the coil nearest the free end of the rope, assuming that it carries the full load of 5;000 pounds throughout one complete wrap around the drum.

The area resisting the crushing action may be considered to be that of the cross section of a ring, of width equal to the pitch of the groove. Assuming that § inch is the least thickness which can be safely allowed under the groove for casting purposes, let us figure the crushing fibre stress to see if this is sufficiently strong. Disregarding the small amount of metal existing above the bottom of the groove, this gives the area to resist the crushing 5/8 X ¾ X 15/32, or .47 inch. Since there are two of these sections and the rope acts on both sides, the equation of forces is: 5,000 X 2 = S X.47 X 2

S = 5,000 X 2 /.47 X 2 = 1,062 pounds per square inch.

This, for cast iron, is very low, and allows plenty of margin for the extra bending and torsional stress, which for such a considerable thickness would be slight.

The above case indicates a method of reasoning much used in designing machinery, which while following out the specified routine of thought as previously given in these pages, stops short of elaborate and minute theoretical calculation when such is obviously unnecessary. If a drum of great length were to be designed, and of small diameter, the same method of reasoning would deduce the fact that the design should be based on the bending and the torsional moments, the thickness in such a case being so great to withstand these that the intensity of the crushing due to wrap of the rope becomes of inappreciable value.

The remaining points of design of the drum are determined from practical considerations and judgment of appearance. The ribs behind the arms are put in to give lateral stiffness and guard against endwise collapse. The arms are subject to the same bend-ing as those of the gear, but as they are equally heavy it is not necessary to calculate them. The flange at the driving end is of course matched to that already designed for the gear. The rope is intended to be brought through the right-hand end with an easy bend and the standard form of button wedged on to prevent its pulling through.

This drum would probably be cast with its axis vertical, and the driving flange down to secure sound metal at that point. Heavy risers would be left at the other end to secure soundness where the rope is fastened. Drums are often cast with the axis horizontal, but the vertical method is more certain to produce a sound casting. The grooves should be turned from the solid metal, partly because it is a difficult matter to cast them, but principally because the rope should run on as smooth surface as possible to avoid undue wear. On drums which carry chain instead of wire rope the grooves are sometimes cast with success, although even in this case the turned groove is generally preferable.

The brake consists of a wrought-iron band to which are fastened wooden blocks, the iron band giving the requisite strength while the blocks give frictional grip on the drum surface and can be easily replaced when worn. As in the designing of a belt the object in view is the grip on the pulley surface by the leather to enable power to be transmitted from the belt to the pulley, so in the case of the brake if we put the proper tension in the strap it can be made to grip the brake drum so tightly that motion between it and the drum cannot occur. The latter case is really the reverse of the first, if the driven pulley be considered, but is identical with the case of the driving pulley, in which the power is transmitted from the pulley to the belt. Of course in the case of the brake no power is transmitted, as when the brake holds no motion occurs, but the principle of the relative tensions in the strap is the same as for the belt.

Since the brake drum surface is 28 inches in diameter, the load at that surface which the brake must hold is:

P = 5,000 X 27 / 14 X 2 = Pounds.

We have then the following calculation corresponding exactly to that of the belt given in Fig. 3.

Log. Tn / To = 2.729 X u X n Let u = .25

Tn - T° = P = 4,821 " n = .75 log. Tn / To = 2.729 X .25 X .75 = 0.512 (for which the natural number is 3.25).

Then Tn /To = 3.25 T0 = Tn / 3.25

Tn-T0 = 4,821 Tn- Tn / 3.25 =2.25 x Tn /3.25 = 4,821

Tn = 4,821 x 3.25 / 2.25 = 6,963 pounds (say 7,000) T0 = 6,963 - 4,821 = 2,142 pounds (say 2,200)

The tight end of the strap must then be capable of carrying a load of 7,000 pounds, and since the width has already been taken at 4½ inches, the problem is to find the necessary thickness. Equating the external load to the internal resistance we have:

7,000 = A X S Let t = thickness.

" S = fibre stress = 12,000 7,000 = 4.5 Xt X 12,000.

t = 7,000 / 4.5 X 12.000 = 13 inch.

This, however, can be but a preliminary figure, for the riveting of the strap will take out some of the effective area, and the thickness will have to be increased to allow for this. Suppose on the basis of this figure we assume the thickness at a slightly increased value, say 3/16 inch, and proceed to calculate the rivets.

A group of five rivets will work in well for this case, which gives 7,000 / 5 = 1,400 pounds per rivet. A safe shearing fibre stress is 6,000, hence the area necessary per rivet is 1,400 / 6,000 = .23 square inch. This comes nearest to the area 9/16 diameter, but for the sake of using the more general size of rivet (5/8 inch) the latter is chosen, for which the area is .30.

We must now try these rivets in a 3/16 -inch plate for their safe bearing value. The projected area of a 5/8-inch hole in a 3/16-inch plate is 5/8 x 3/16 =.117 square inch. 1,400 / .117 = 11,905 (15,000 would be safe)

Taking out two 5/8-inch rivets from the full width of 4½ inches leaves 4½ - (2 X 5/8)= 3.25, and makes the net area of strap to take stress 3.25 X 3/16 = .6l square inches. Re-calculating the fibre stress for this area gives 7,000 = .61X S.

S = 7,000 / .61 = 11,475 (which approximates the previous value of 12,000).

The slack end of the strap has to take but 2,200 pounds, hence a different calculation might be made for this end giving smaller rivets; but as it is impractical to change the thickness of the strap to meet this reduced load, it is well to maintain the same proportion of joint as at the tight end. The spacing of the rivets in both cases follows the ordinary rule allowing at least three times the diameter of the rivet as center distance, and one-half this value to the edge of the plate.

The threaded end of the forging on the strap also has to carry the load of 2,200 pounds, for which a size smaller than 1 inch would suffice. It is natural, however, for the sake of general proportion to make the bolt as strong as the strap, and a 1-inch bolt gives an area of .52 square inch, nearly equalling the value of .61 net area of strap noted above.
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