Considering the segment ADB as forming either the entire bend, or one of a series of similar segments constituting the bend, let denote the angle subtended by the arc, and let the segment be terminated at A and B by the radial plane sections OA and OB. Then, to find the magnitude of the displacing force which acts in the direction OD at right angles to the chord AB, draw a projection of the trough surfaces on a plane parallel to AB, and, subtracting the negative area as before, we have the two ellipses of Fig. 18a, in each of which the minor axis is

0

2

The combined area of these two ellipses will be 2d2 sin The displacing force in the direction OD, or the resultant of the unbalanced pressures on the internal surfaces of the bend AB, will therefore have the magnitude

If we proceed to subdivide the segmental arc ADB into a number of indefinitely small and equal arcs, it is evident that by repeating the same calculation we shall find upon every one of those little arcs a displacing force acting upon the pipe in a direction radially outwards (from the centre of curvature 0), or at right angles to the curved centre line of the pipe; and on each of these equal elements of the arc, the displacing force will have the same magnitude, so that we have really to consider the force as a uniformly distributed pressure acting normally to the arc at all points; and the intensity of this pressure per lineal foot of the pipe is easily to be found.

If 0 is indefinitely small, the chord is sensibly equal to the length of the arc, and from the formula (56) above given, it is evident that the distributed radial force or pressure must have for its intensity per lineal unit of arc the value—

so that the pipe-bend would be perfectly equilibrated at all

points if, along with the internal fluid pressure, it were subjected to an external pressure uniformly distributed along the arc, and acting normally to the curve in a direction towards the

Р
Ꭱ R

centre O, with the intensity per unit of arc-length.

Exactly the same result will be reached if we regard the curved column of water as a fluid arch of segmental form in which the compressive stress is known to have the value Ppd2; for if the compressive stress in such an arch is to

π

4

follow the curved "line of pressure," of radius R, we know very well, by the ordinary principles of the equilibrium of arches, that it must be combined with a uniform external pressure (normal to the arch), whose intensity per lineal unit of arclength must be n = as before.

P
R'

EXAMPLE.—In a cast-iron pipe, whose internal diameter is everywhere 36 inches, let the segmental bend ADB, subtending an angle of 45°, be formed with a radius 40 of 20 feet; and let the pipe be subjected to a hydrostatic head of 320 feet.

Here the axial fluid stress P will have the value already found in a previous example, or—

while sin

P = 320 × 62.4 × 32 × 0.7854 2240 = 63 tons

Therefore the bend will be subjected to a uniformly distributed radial displacing force or pressure, whose intensity per unit of arc-length will be—

Upon the whole bend, the resultant of this distributed force will have the direction OD, and the magnitude

N = 2P sin = 2 × 63 × 0.38268 48.2 tons

Art. 11. Axial Stress in the Fluid and in the Pipe.-In the preceding articles we have found the displacing forces due to the unbalanced fluid pressures by a direct measurement of those pressures upon the actual internal surfaces of the pipe; but in each case it has been shown that the same result might be more simply obtained by considering the axial stress in the cylindrical column of fluid which fills the pipe. For in every equilibrated arch or polygonal frame there must be a certain adjustment between the so-called "thrust," or compressive axial stress, and

the external forces applied to the arch or to the joints of the frame. Of course the same adjustment must always subsist whether this axial stress takes effect wholly in the fluid contents of the pipe, or partly or wholly in the metal of the pipe itself; and therefore if we know in any given case that the pipe is itself subjected to a longitudinal compressive force Q, the effective axial stress P+Q must be substituted for P in calculating the external loads or lateral forces that are required to maintain the equilibrium of the arch or frame.

The formulæ already found for the polygonal bend will then become

while for the segmental bend or arch we shall have

or for the uniformly distributed pressure per lineal unit of arc-length

EXAMPLE. Suppose the segmental bend ADB already illustrated in Fig. 18 to be fitted at each end with an expansion joint, in which the plain cylindrical end of the pipe slides in the gland of a stuffing-box, as sketched in Fig. 19. Then, if the internal diameter d is uniform throughout the pipe, it is evident that the gross or effective axial stress will be somewhat greater than pd2 on account 4 of the fluid pressure upon the annular end of the pipe within the expansion joint.

If d1 denotes the external diameter of the sliding neck, or the bore of the gland, the area of annular surface will be (d,2 — d2), and the total axial stress P + Q

=

45°, as = 63

For example, let d = 36 inches, d1 39 inches, R = 20 feet, and ℗ = in the previous example; then for a head of 320 feet, we have as before P tons, P+Q = 73-94 tons; and the total lifting-force exerted in the direction OD will be

N = 2 × 73.94 × 0.38268 = 56.59 tons

or the distributed normal pressure exerted uniformly all along the arc will be

The compressive force Q transmitted through the metal of the pipe will be

FIG. 19

10.94 tons, and the stress per square inch of metal section will, at each point, depend on the thickness of the pipe. At the ends it will obviously be equal to p.

Art. 12. The Displacing Force resisted by Longitudinal Stress.The longitudinal stress Q referred to in the last article was a compressive stress of the same character as P; but it needs no demonstration to show that in all cases the effective axial stress, which is really concerned in the equilibrium of the bend, will always be the algebraical sum P+Q, whether Q be positive or negative.

Hence it is easy to see that the displacing forces, which result from unbalanced internal pressures on the sides of the pipe, will be exactly balanced and practically annulled if the pipe itself is subjected to a longitudinal pull - Q, equal and opposite to the axial fluid stress P, so that the sum P+Q of all the stresses over the whole cross-section shall be zero.

In this case the pipe-bend will of course need no lateral support, or load, or reaction, to maintain its equilibrium.

If the displacing forces are to be resisted wholly by this means, it is worthy of note that the requisite tensile stress or force Q will be independent of the radius R and of the angle 0, and must always be equal to P, whether the curvature of the

pipe be great or small; also that the metal of the pipe will then be subject to a pair of conjugate stresses, viz. the "hoop-tension"

and at right angles to this a longitudinal stress, whose

f

SO 2'

intensity q is always one-half of the hoop-tension, or q=2, that a pipe which is strong enough to resist the hoop-tension is always more than strong enough to resist the longitudinal stress q.

These simple considerations are sufficient in themselves to explain the fact that, in a large number of practical cases, the unbalanced internal pressures exhibit no tendency to produce any lateral displacement of the pipe; but in considering the equilibrium of any pipe-line, it is necessary to inquire whether the tensile stress -Q, equal and opposite to P, does or can take place in the metal of the pipe.

It is hardly necessary to remark that, if the bend were closed at both ends, the pressure upon the covers would at once subject the metal to the necessary longitudinal stress, nor to observe that in such a closed vessel the resultant of all the internal pressures must be zero.

Again, in the case of a flexible pipe which is not closed at the ends, a little consideration will show that the conditions are present under which the longitudinal tensile stress could be, and automatically would be, adjusted to an exact equality with the axial fluid stress; and in no other way could the equilibrium of such a charged pipe be accounted for.

But there are a good many other cases in which the unbalanced internal pressures give rise to very large displacing forces, which cannot possibly be resisted in this way; there are yet other cases in which the magnitude of the tensile stress Q is definitely determined by independent external conditions, or is limited by the tensile strength of the joints; while again there are some others in which it is difficult to estimate the value of that stress with any certainty.

Art. 13. Expansion Joints. If we introduce a sliding expansion joint at any point in a line of pipe, it is obviously impossible, so long as the joint is really free to slide, that any tensile stress whatever can, at this point, be transmitted through the pipe in a longitudinal direction; and therefore no part of the displacing forces can be resisted by this means.

D