=

square inch bursting pressure of steam, and the ordinary working pressure being 120 lbs., the factor is 414 120 3.45: occasionally the pressure is 150 lbs., or even more, and the factor becomes 414 150

=

2.76.

From all this we may admit that in ordinary cases the factor should be 6, but for exceptional cases it may be 4, as in (68), &c., or even 3 with comparatively new and sound boilers: but this is a matter which must be left to the judgment of the engineer.

STRENGTH OF THICK PIPES.

(79.) The strength of a pipe in resisting internal pressure is not simply proportional to the thickness of the metal; the material stretches under a tensile strain, the result being that the metal inside is more strained than that outside, and that thick pipes are weaker in proportion to their thickness than thin

ones.

To illustrate this, let Fig. 22 be a 10-inch pipe, 5 inches thick, therefore 20 inches outside, and let an internal pressure be exerted until the inside diameter becomes 101 inches: now if the metal at A were strained in the same proportion as at B, it would be extended or stretched in the same proportion, and the outside diameter would become 20 inches. But obviously the cross-sectional area must be the same in all cases, 20 inches being 314-16, and 10 inches = €78.54, the area of the annulus must be 314 16 - 78.54 235 62 square inches, therefore the outside diameter in Fig. 23 will be found by adding the area of 101, or 80·516 to 235 62, and we thus obtain 80.516235.62 = 316·136, the diameter due to which

=

201 inches instead of 201 inches, and if we admit that the strains are proportional to the extensions, the metal at A is strained to only of that at B: for instance, if the strain at A = 4 tons per square inch, that at B will be 1 ton only, and between A and B we have an infinite series of strains progressively diminishing from 4 to 1 ton per square inch.

(80.) It will now be seen that the strain is inversely proportional to the square of the distance from the centre: in our case the strain at B being 4, that at A will be 4 × 5o÷102

= 1 ton, &c. Let Fig. 24 be the section of a 10-inch pipe with various thicknesses up to 10 inches: we will assume that the strain at C, where it is a maximum, is 7 tons per square inch, this being nearly the breaking weight for ordinary cast iron (4), the extension due to which by rule (605) is :

E (00015 x 7) + (0000122 x 72)=0016487.

=

This being at 5 inches from the centre, that at D, or 6 inches, will be 0016487 x 5262 .001144, the strain due to which

by the rule (606) becomes:

=

per square inch, as per col. 3 of Table 18; hence the mean strain on the ring A is (7+5·32) ÷ 2 = 6.16 tons as in col. 4, and as we have 2 square inches of metal per inch run (namely 1 inch at each side) we obtain 6·16 × 2 = 12.32 ton bursting pressure on the whole diameter, or 12.32 ÷ 1.232 ton per square inch as in col. 5. Calculating in this way we obtain the strains and pressures in cols. 4, 5 of Table 18: thus for 10-inch pipes, 5 inches thick, the mean strain throughout the section becomes

10

=

(6·16 +4.75 +3·775 +3·07 +2·54) ÷ 5 = 4·059 tons

per square inch of metal as in col. 4. Then as we have 10 square inches of metal per inch run (or 5 inches at each side) we have 4.059 × 10 = 40.59 tons on the whole diameter, or 40·59 ÷ 10 = 4.059 tons internal pressure per square inch, col. 5. If the whole cross-sectional area had yielded the maximum strain of 7 tons per square inch, we should have had 7 x 10 = 70 tons on the whole diameter, or 70 ÷ 10 7 tons pressure per square

inch instead of 4·059 as per col. 5.

=

We should obtain nearly the same results by the following rules :

In which

=

the internal pressure per square inch in tons, lbs., &c., dependent on the value of S.

S = the maximum tensile strain, or that at the inside of the pipe, in tons, lbs., &c., per square inch.

R = the external, and r = the internal radius of the pipe in inches.

=

Thus for example, with a 10-inch pipe 5 inches thick, R = 10, and r = 5 inches: taking S 7 tons, which is nearly the ultimate or breaking tensile strength of ordinary cast iron, we get 7 × (102 — 52) = 4.2 tons per square inch: calculating

p =

102 +52

in this way we obtain col. 6 of Table 18.

Again: say that with a cylinder 12 inches bore, 5 inches thick, and an internal pressure of 2 tons per square inch, we require the maximum strain on the metal or that at the inside of the cylinder. Then R being 11, r = 6, rule (82) becomes 2 × (112 + 62)

S

=

112-62

= 3.7 tons per square inch of metal.

(83.) The ordinary proportions adopted almost universally by practical engineers for hydraulic-press cylinders, is to make the thickness equal to the internal radius, and it is supposed that those proportions will allow a working pressure of 4 tons per square inch, or say 3 tons per circular inch. But Table 18 shows by cols. 5 or 6, that with ordinary cast iron these are really bursting pressures. It is shown in (883) that cast iron will sustain for years a strain very nearly equal to the breaking weight, but it is not safe to trust to that fact, and in most cases the working pressure should not exceed say half the ultimate pressure, or in our case 2 tons per square inch with ordinary iron. Many presses, however, may be found which seem to bear much heavier pressures than that as shown by the safety-valve, but as usually constructed a safety-valve is a very unreliable indicator of pressure, the breadth of the conical seat being great, and the acting or effective area uncertain. A better form is shown by Fig. 202: the valve V is of hardened steel formed like a hollow punch, the cutting edge imbeds itself in the hard gun

=

metal seat and forms it own bed, giving a precise area, and thereby a certain pressure. Thus for 3 inch diameter, the area = ·11 inch, requiring for say 1 ton per square inch 2240 ×·11 246 lbs. strain, and with a leverage of say 20 to 1, we have 24620 12.3 lbs. weight on the lever per ton pressure. The knife-edges at A and B, also the key K must be of hardened steel, and in order to adjust the level of the lever and compensate for wear (which is a practical necessity) the upper edge of the key should be wedge-shaped, and at an angle adapted to its seat in the slotted recess prepared for it.

(84.) The actual load on the ram of a hydraulic press is not often known with accuracy, but in the presses used for raising the Conway and Britannia bridges we have more precise information. For the Conway tube, a ram 183 inches diameter, or 265 square inches area, was used at each end; the gross weight of the tube, &c., was about 1300 tons, or 650 tons at each end hence we have 650265 = 2.45 tons per square inch. The cylinder was 20 inches diameter internally, and 10 inches thick hence R 20, r = 10, and the rule (82) gives 2.45 x (2010)

S

=

202-102

=

= 4.08 tons tensile strain per square

inch of metal; this being the maximum strain, or that at the inside of the cylinder (80).

(85.) With the Britannia tube, the two Conway presses were used at one end and a large one with 20-inch ram at the other. The gross weight of the tube, &c., was about 1640 tons, or 820 tons at each end: then the 20-inch ram being 314 square inches area, we have 820314 2.61 tons pressure per square inch. The cylinder was 22 inches internal diameter and 11 inches thick, hence R 22, and r = 11, and rule (82) gives 2.61 x (2211) 222 · 112

S =

per square inch of metal.

=

=

= 4.35 tons maximum tensile strain

It is probable that this pressure, 2.61, and strain 4.35 tons per square inch, were very nearly the breaking weights, indeed one cylinder failed by the bottom blowing off, which would have led to most disastrous results, but for the wise precaution taken of blocking up the tube inch by inch as it was raised by the