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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, 18-inch space by Table 14: then the area of plate between two rivet-holes

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1 × 1 8125 square inch, giving in a single-riveted joint by col. 7 of Table 5, 34110 × ·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 1 inch diameter for inch plate, showing that in a single-riveted joint the principle of equality between the strains on the rivet and plate is complied with.

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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 × 8125 = 33962 lbs. from the plate requiring 33962 ÷ 49280 = 689 square inch, area of two rivets, or 345 square inch each, = say, inch diameter, instead of inch, as for single-riveting; but by most practical men rivets would be deemed too light for 1-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, hence 1 × = square inch of metal, which in a single-riveted joint would give 34100 x 17055 lbs., requiring 17055÷ 49280 346 square inch of rivet = say1inch 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).

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(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 15651386 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 384.

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.

CHAPTER III.

COHESION APPLIED TO PIPES.

(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

of 25° = a, b

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0.4142, hence with radius, as in our case, we have 0.2071, and a, c = 0·4142 inch. The pressure on a, c will act direct as a tendency to rupture on the points B, B, the force being 0·4142 x 100 = 41.42 lbs.; but the pressure on c, e and a, g will act obliquely. Thus the strain on c, e will of course be 41.42 lbs., as on a, c, but it will act in the direction of the arrow x, and must be resolved into two equivalent forces, one in the direction of the arrow y, which being at right angles to B, B, will tend to rupture on those points; the other in the direction of the arrow z being parallel to B, B will have no effect. By the well-known parallelogram of forces, Fig. 20 making the diagonal D = 41.42 lbs., we have two equivalent strains, the direction and force of which are given by the two sides of the parallelogram E, F, each 29.29 lbs., F being a direct tensile strain on the points B, B in Fig. 19. Of course the side a, g will give 29.29 lbs. also, and the combined strain will be from a, g 29.29; a, c = 41.42; and c, e = 29.29 lbs., or 29 29+ 41.4229.29 100 lbs., being precisely the same as with the tube, Fig. 18, 1 inch square.

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Calculating in this way with a polygon of any number of sides we should obtain the same result, and a circle being regarded as a polygon with an infinite number of sides, we thus find that the strain on a cylindrical tube is the same as on a square one of the same dimensions. From this it follows that the strength of a cylinder of thin plate, such as an ordinary boiler, is simply and directly proportional to the thickness, and inversely as the diameter.

(57.) "Lap-welded Tubes."-Say that we require the strength of a small boiler 24 inches diameter, 1-inch plate, with welded joint, made of Staffordshire plates whose tensile strength, namely, that of a solid plate, is 20 tons per square inch. By Mr. Bertram's experiments at Woolwich the strength of a lap-welded joint may be taken at 65 per cent. of that of the solid plate: hence 20 × ⚫65 13 tons, or 29,120 lbs. per square inch. In our case rupture strains a square inch (or inch at each side); hence 29120 × = 14560 lbs., which on 24 inches gives 1456024 :

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607 lbs. per square inch

bursting pressure; with 6 for the Factor of safety (78) we obtain 6076 = say 100 lbs. per square inch, safe or working pressure. From this we have the general rules :

(58.)

(59.)

For welded boilers: P = 58200 x t÷d.

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101, say

In which t = thickness of plate in inches; d= inside diameter in inches; P the bursting pressure, and Р = the safe working pressure in lbs. per square inch: thus for the 24-inch boiler we have considered, the rule gives 9700 × 1÷24 100 lbs. per square inch working pressure, as before. (60.) "Steam-boilers with Riveted Joints."-Staffordshire plates are now so extensively used for boilers, that it will be expedient to take them as a basis for general rules, although, as shown by Table 5, their strength is inferior to the mean of British plateiron, and still more inferior to Yorkshire iron.

We have shown in (44) that the pitch of rivets, and thereby the general proportions of joints in steam-boilers, is governed by the pressure of steam as affecting the tendency to leakage, irrespective of strength to resist bursting.

(61.) For the purpose of fixing general proportions, we may take as a "standard" case the working pressure of 50 lbs. per square inch, the proportions due to which will suffice for all lower pressures; and also with sufficient accuracy for practical purposes up to say 70 or 80 lbs. per square inch. The proportions for higher pressures should be found by special calculation (68) (76).

We have first to find the space between rivet-holes with the different thicknesses of plate for 50-lb. steam by Table 13; taking the nearest pressures in that Table we obtain col. 6 in Table 14. Thus, for 3-inch plate we have for 50-lb. steam the space 5 Table 14 gives rivets, as in col. 3;

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hence the pitch 2 inches, col. 4; the ratio of the metal between holes to the solid part of the plate = 1÷ 2, or 2132 656, as in col. 7. The apparent strength in

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single-riveted joints of Staffordshire plates being 34,110 lbs. per square inch by col. 7 of Table 5, that on the solid part of the plate = 34110 x 656 22380 lbs., as in col. 8 of Table 14.

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of 25° = 0·4142, hence with radius, as in our case, we have a, b = 0.2071, and a, c = 0·4142 inch. The pressure on a, c will act direct as a tendency to rupture on the points B, B, the force being 0·4142 x 100 = 41.42 lbs.; but the pressure on c, e and a, g will act obliquely. Thus the strain on c, e will of course be 41-42 lbs., as on a, c, but it will act in the direction of the arrow x, and must be resolved into two equivalent forces, one in the direction of the arrow y, which being at right angles to B, B, will tend to rupture on those points; the other in the direction of the arrow z being parallel to B, B will have no effect. By the well-known parallelogram of forces, Fig. 20 making the diagonal D 41.42 lbs., we have two equivalent strains, the direction and force of which are given by the two sides of the parallelogram E, F, each 29.29 lbs., F being a direct tensile strain on the points B, B in Fig. 19. Of course the side a, g will give 29 29 lbs. also, and the combined strain will be from a, g 29.29; a, c = 41.42; and c, e = 29.29 lbs., or 29.2941.42+29.29 100 lbs., being precisely the same as with the tube, Fig. 18,

1 inch square.

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Calculating in this way with a polygon of any number of sides we should obtain the same result, and a circle being regarded as a polygon with an infinite number of sides, we thus find that the strain on a cylindrical tube is the same as on a square one of the same dimensions. From this it follows that the strength of a cylinder of thin plate, such as an ordinary boiler, is simply and directly proportional to the thickness, and inversely as the diameter.

(57.) "Lap-welded Tubes."-Say that we require the strength of a small boiler 24 inches diameter, 1-inch plate, with welded joint, made of Staffordshire plates whose tensile strength, namely, that of a solid plate, is 20 tons per square inch. By Mr. Bertram's experiments at Woolwich the strength of a lap-welded joint may be taken at 65 per cent. of that of the solid plate: hence 20 x 65 13 tons, or 29,120 lbs. per square inch. In our case rupture strains a square inch (or inch at each side); hence 29120 × 14560 lbs., which on 24 inches gives 1456024 607 lbs. per square inch

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