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Let A, B, C, Fig. 17, be three beams, all of the same depth (or thickness) and breadth, but varying in length in the ratio 1, 2, 4; then by analogy these may be regarded as three steam joints having distances between insides of rivet-holes in the ratio 1, 2, 3, &c. Now, if we admit that a crack of any measurable amount will cause leakage, that amount will be the same in all three cases, so that the problem becomes this; to find what the respective loads must be, to give one and the same deflection in all the three cases. By the laws of deflection in (662) it is

d° X b X shown that W

In our case do, b, d, and Care

L' x 0 constant, therefore W will be inversely proportional to L simply, hence the lengths C, B, A, being in our case 1, 2, 4, the loads will be in the ratio 4, 23, 1, or 64, 8, 1. But in our case, the surfaces over which these loads are spread are also in the ratio 1, 2, 4, and our special object is to find the pressure or load per square inch; with A we have a load of 1 spread over a length of 4, hence 1 : 4 = ) per unit of length; with B, a load of 8 spread over a length of 2, or 8 = 2 = 4 per unit of length; and with C, a load of 64 spread over a length of 1, or 64 per unit of length. Thus, with lengths 4, 2, 1, we obtain pressures 1, 4, 64; or in the ratio 1, 16, 256, which are inversely as the fourth power of the lengths, for 14, 2*, 4* are 1, 16, 256, and we thus find that with constant thickness, the pressure tending to produce leakage of steam will be inversely proportional to Sʻ, or the fourth power of the space, or distance between the insides of the rivet-holes.

The formula in (662) shows that W is directly proportional to t’, hence we have the rules :(46.)

p = ML XtS.

= (47.)

S = y Mix t = p. (48.)

M = St X pt. In which s

between insides of rivet-holes in inches. t = thickness of plate in ths of an inch. p = working pressure of steam in lbs. per square inch. = a constant from practice = 5.5 for wrought

iron; 6.2 for steel plate.

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To find the value of Mi we may take a standard case, say

plate, it rivets, 2 inches pitch, 116 space, and 50 lbs. per square inch; these are common proportions, and have been proved to be satisfactory by universal experience. We may find 12

R* by the square of the square of that number; thus 1,8% = 1.723, and 1.7232 = 2.969, which is the fourth power of 118; then the rule M, = 8+ X p ; to, becomes 2.969 x 50 : 27 = 5.48, say 5.5, the value of ML.

Mr. (49.) Again: to find S for say 150-1b. steam with l-inch plate, the rule (47) becomes S = 35.5 x 43 = 150 = 1.238, or say

= 14 inch. For example, 5.5 x 64 = 150 = 2.35; then we may obtain the 4th root of 2.35 by finding the square root of the square root of that number: thus V 2.35 1.533, and Ņ1.533 = 1.238 inch as before, this being the 4th root of 2:35. We should obtain the same result direct by the use of logarithms: thus the log. of 2.35 or •371068 4 = .092767, the natural number due to which 1.238 inch as before.

Again: to find p for say it plate with rivets, 11-inch space, therefore { + 14 = 21 pitch, the rule (46) gives p =

+11 5.5 x 3.1 ; 118 = 80-lb. steam. Thus 1,0 = 1.72, and

? 1.722 = 2.96, which is the 4th power of 11. Then 3. being

42.87, we obtain p = 5.5 X 42.87 = 2.96 = 80-lb. steam as before.

The London and North-Western Railway Co. at Crewe, for their 4-foot locomotive boilers, use 13-inch plates, rivets, 1; pitch, therefore l-inch



р 5.5 x 318 = 14 = 189-1b. steam; the actual ordinary working pressure is 120 lbs.; occasionally 150 lbs. per square inch.

(50.) Table 13 has been calculated by rule (46). It should be understood that these rules are approximate only, giving a fair working pressure. Possibly a pressure double or even treble that given by the rule, would not cause the joint to leak instantly, but in all probability it would eventually do so, and as it is essential that boilers should be perfectly steam-tight, it will be advisable that the working pressure should not much exceed those given by the rules, and Table 13. When a steam joint or anything else is overstrained, failure is always more or less a question of time.

5 2 16

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Joints, as governed by the Space between Rivet-holes.

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(51.) We have admitted in (28) that the diameter of the rivets shall be governed by the thickness of the plate alone, irrespective of the pressure of steam or other considerations ; and in (44) we have allowed that the space between rivet-holes (and thereby the pitch of the rivets) shall be dominated by the pressure of the steam. But under these two conditions it is impossible to secure that equality between the shearing strain on the rivets and the tensile strain on the plate, which is an essential principle in riveting, as shown in (29). For instance, in Table 14, the pitch is allowed to be the same whether the joints are single or double-riveted: but obviously if the area of the rivets is properly proportioned for the

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former, they must have an excess of strength for the latter, because in one case each space is matched by one rivet, and in the other case by two.

(52.) This anomaly might be avoided if we allow that the diameter may be varied so as to adapt it to the strain, irrespective of the mere thickness of the plate. Say that we take }-inch plate for 50-lb. steam, = 15-inch space by Table 14: then the area of plate between two rivet-holes 1 x .8125 square inch, giving in a single-riveted joint by col. 7 of Table 5, 34110 X •8125 = 27710 lbs., and as rivets yield 49,280 lbs. per square inch (19), we have 27710 = 49280 = -562 square inch of rivet = say bare 3-inch diameter, agreeing nearly with col. 2 of Table 6, which gives 12 inch diameter for l-inch plate, showing that in a single-riveted joint the principle of equality between the strains on the rivet and plate is complied with.

But with double-riveted joints we have two rivets to each space, and 41,800 lbs. per square inch of plate by col. 7 of Table 5: then, we have 41800 X .8125 33962 lbs. from the plate requiring 33962 -- 49280 = .689 square inch, area of two rivets, or · 345 square inch each, say, 16 inch diameter, instead of } inch, as for single-riveting; but by most practical men 1 rivets would be deemed too light for l-inch plates.

(53.) Besides, there is this anomaly, that the higher the pressure of steam, the smaller the rivets become, this being due to the reduced space between rivets. Thus, for 350-lb. steam, and -inch plates, the space = 1 inch by Table 13, 'i hence 1x} = } square inch of metal, which in a single-riveted joint would give 34100 x } = 17055 lbs., requiring 17055 ; 49280 = 346 square inch of rivet = say it inch diameter for 350-lb. steam, whereas for 50-lb. steam we obtained ( inch.

These calculated proportions are no doubt correct so far as the strains on the rivet and plate are concerned, but there are other considerations which render it inexpedient that they should be followed, and we must admit the practical dictum (28) that the diameter of the rivet shall be proportional to the thickness of the plate, as given by the rule (27).

(54.) Space between Rivets with Steel Plates.By col. 4 of

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Table 105, steel is stiffer than wrought iron, the difference being 1565 ; 1386 = 1.13 or 13 per cent., and the pressure of steam would be greater in that ratio; hence the rule (46) becomes :

p = 6.2 x 3 = S4. But the difference of 13 per cent. is so small, and the rule such an approximate one, that we may safely admit that the working pressure with steel will be the same as for wrought iron as given by Tables 13, &c.

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(55.) It is necessary to consider this subject under two different heads; namely, thin and thick pipes; the former being usually of wrought sheet metal, such as ordinary steamboilers, and the latter of cast metals, such as strong waterpipes, hydraulic-press cylinders, &c. The strains in these two cases differ considerably from one another, the latter being much more complex than the former.

Thin Tubes.”—Let Fig. 18 be a tube 1 inch square, and for the sake of illustration, say 1 inch deep, subjected to an internal fluid pressure of 100 lbs. per square inch, acting, of course, in all directions. Now the surface c, d having an area of 1 square inch, will exert a force of 100 lbs. in the direction of the arrow a, and will be resisted by a similar force on the surface e, f, acting in the direction of b; hence we have a tensile strain of 50 lbs. on each of the sides c, e and d, f, tending to produce rupture say on the line B, B.

(56.) Let Fig. 19 be an octagonal tube 1 inch inside, and 1 inch deep as before: we have first to find the dimensions of the sides of the polygon. The half-side a, b is evidently the tangent of the angle a, d, b, which being the sixteenth part of the circle will be 360 = 16 = 25°: then by any table of natural tangents we find that with radius 1:0, the tangent

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