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intersecting cylindrical shells (fig. 171). Fig. 170 is a section through the line ced.
Assuming that the larger cylinder is cut up into numerous rings of the thickness è, and that internal pressure is p, then if the ring under consideration were severed at the point C it would have to be retained
D in its position by a pull Q =
2 Now the vertical component of this pull could be transmitted to the straight side of the cylinder d, but a stay would have to be fitted
Fig. 171 to take up the horizontal componenta. A little consideration will show that it bears the same relation to Q as the distance ce bears to half the diameter D. The same arguments show that the horizontal component of the narrow strip cF is also proportional to its projected
This shows that the greatest pull is found at G and the smallest at H, and instead of fitting the chief stays at H, as is often proposed, they should be fitted at c and G, unless the latter point is well supported by the flanges.
Stresses in Cylindrical Flues. The same formula as for cylindrical shells may be applied here, but, when experimenting, buckles will show themselves long before even the limit of elasticity for compression has been reached, and the pressure cannot then be increased, because the flue grows weaker and weaker the more it alters its shape. It is important to know at what pressure this change takes place. Let the elliptical dotted line (fig. 172) be the shape of the originally cylindrical fue, and let the black line be the curve of thrusts. This is a nearly circular line, and can be found by graphic construction, as is done in the case of arches, &c. The intensity of the thrust is D -p, where p is the external pressure. To find the bending moment
at any position (say, at the angle a) it is only necessary to multiply this thrust into the distance h between the two lines. At the
major and minor axes the moments are maxima m, =- m2 =
2.2 For any other point
m = m; cos 2a. This equation is practically identical with that obtained for a
D. straight strut of the length with loose ends and loaded with the
4 above force. If accidentally bent, such a strut would straighten itself while supporting its load K, provided it does not exceed
D2 Substituting the value for K, and also for I=
the moment of inertia, we have
D 16. E. t3
from which it follows that р
is limited : 8
= 8.107. 3 D3
per sq. in.
As will be seen, p in this formula depends only on the modulus of elasticity, and not on the working strength of the material ; or, in other words, the strength of a flue to resist a collapsing or buckling pressure depends on its rigidity, whereas its strength to resist a crushing pressure depends on the strength of the material. The two values are equal when
that is, when
S 4 t
E 3' D2 Here S is the plastic limit of the material. This formula expresses the conditions under which a flue would give way simultaneously both by crushing and collapsing. Assuming that É S = 1200, we have
1 = 1.000
P = 32
This means, that as long as the ratio of t to D is greater than to, a perfectly cylindrical tube, unsupported at its ends, will give way by crushing and not by collapsing. Boiler tubes are not supported at their ends, which is the same thing as if they were of infinite length; yet, on account of their relative great thickness, it is not wrong to estimate their strength by the formula
2.ts p =
D Corrugated Flues.—By corrugating a flue, I, the moment of
t3 t3 inertia of one inch of its section is increased from to
12 where h is the depth of the corrugations, and we have
t. h3 D3 12 8
t3 Let h 1} in., then can be neglected, and we find that the crush
12 ing and collapsing pressures are equal when
t.s 4 . t . ? . E . D
h2 that is, when
This shows that even very slight corru2.E gations will make ordinary boiler flues independent of their end supports (see Dr. F. Grashoff, 1866, pp. 232, 235).
Influence of Rings and Furnace Ends.—The previous formulæ take no account of the strengthening effect of rings or furnace ends. One way of dealing with this question is to regard the shell of a furnace as if it were a wide flat column, supported at its ends by two solid guides (fig. 173).
Let L be one-eighth of the furnace circumference, whose diameter is D, and let l equal half its unsupported length, while q is the circumferential thrust per inch of length.
Then, if we take a narrow vertical strip (fig. 173A), the bending moment would be q. y. This would produce a curvature
q .y P d.x2
If, on the other hand, we divide the shell into horizontal strips, they will be bent as shown in fig. 174. The external forces producing this deflection are due to certain horizontal shearing forces, which when
summed up can be represented by an imaginary horizontal pressure p, which is balanced by P at either end. The slight extra relief due to the action of force q on the radius p may be neglected (fig. 175).
We then have
= (1-21) dzi?
It is evident that to work out these formulæ would require more space than can be spared, and as the result, like the previous ones, is 9
only applicable to perfectly circular furnaces, it will not be carried further. In practice the problem is made more complicated by the back end being irregularly supported.
This is also shown in fig. 173, which represents the case of © a furnace whose saddle is indicated by a vertical dotted line
and is supported by the tube plate, while the bottom extends
the whole length as far as the combustion chamber back '9 plate.
Ribbed and Flanged Furnaces.-Another way of looking Fig. 175
at the problem, and one which is intimately mixed up with the above, is the relief afforded to the circumferential stresses by the end plates or stiffening rings. Thus in fig. 176 the line AB represents the original position of the cylindrical part of the flue, whose section is shown above. CD is its position when the diameter has decreased under pressure
= Pi. D2
if rings are fitted to the ends, while EF would be its position if no rings were fitted. The circumferential stresses are, of course, proportional to the compressions ò, and , showing that the relief afforded at the centre is
In this fig. 176 y, and
y, have the same meaning as in fig. 169, A
B p. 199, and could, if required, C
D be calculated by the formulæ E
F to be found there. When
Fig. 176 doing this it must not be
dy forgotten that = 0 both when x= 0 and when x= l, and further
dx that approximately
4. E. t 2.E.a' where a is the excess sectional area of the strengthening ring over and above that of the cylindrical part, and Q has the same value as on p. 199.
Oval Furnaces. The same reasoning which led (fig. 172) to a limiting pressure for a perfectly cylindrical furnace has to be modified if the furnace was originally elliptical. Let 2. A be the difference between the major and minor axes before straining, and let 2.0 be the difference after the external pressure has been applied; then it can easily be shown that the furnace still remains elliptical, and that 16.E.I
32. E.I 16. E. I-K. D2 32. E. I-p. D3 Example: Let A = in., t = } in., D = 40 in., p = 100 lbs.,
s E= 30,000,000 pounds; then
= 2.78 1 – 0:64
2 =14 ins. This is of course only true for long furnaces where the supporting influence of the ends can be negleeted. In a corrugated flue whose moment of inertia of section is about ten times greater than the above half-inch plate, the ratio would be
1 0.064 This shows how enormously rigid these furnaces are, even although they, like long plain ones, have no effective end support.
When the furnaces are short, or when strengthening rings are fitted, the problem grows to be a very complicated one, every deformation producing secondary strains and stresses. The approximate formulæ which the author has worked out are far too complicated to be of practical value. It may, however, be pointed out that the amount of deflection is a fair measure of the stresses. Thus, in the above
င် example of a plain furnace, where = 2.78 for an infinite length, it