done when the pipes are of cast-iron. In the latter case the short straight pipes forming the sides of the polygon are often united by a double socket-joint, whose faces are inclined at a certain angle, as sketched in Fig. 17, while in other cases the straight lengths may meet each other with a mitre joint, and may be united by flanged ends bolted or riveted together, as sketched diagrammatically in Fig. 15, which may be regarded as the vertical section of an upright bend, and will serve to illustrate the question in its simplest form. The pipe AB in this polygonal bend will be a cylinder whose ends are formed by the inclined planes EG and FH; and it is obvious that the fluid pressure upon the side EF will have a greater magnitude than the opposite pressure upon the side GH, so that there will be a displacing force in the direction OD, which, in this particular case, would tend to lift the pipe vertically upwards. The magnitude of this lifting force will easily be found if we imagine the pipe to be divided into two semi-cylindrical troughs by a section along the centre line AB, for the upper trough will then contain all those elements of surface which can be subject to any upward force, and vice versa. Therefore, drawing a projection of the internal surfaces of these troughs on a plane at right angles to OD, we have the areas shown in Fig. 15a, the entire area EABF, etc., being the projection of the upper trough, or the positive area, while the smaller area GABH, etc., will be the projection of the lower trough, or the negative area. Subtracting the negative from the positive area, we have remaining the two ellipses, whose combined areas, multiplied by p, will give us the total lifting force exerted upon the pipe AB. If d is the internal diameter of the pipe, and if 0 denotes the angle EOF, which is bisected by OD, then the minor axis of each ellipse will be d tan and the area of each ellipse will be 0 0 2' d2 tan The displacing force in the direction OD will 4 therefore be 2pd2 tan; or, if we write P for the quantity pd3, we may express the magnitude of the displacing force by— 4' In this expression the quantity P is nothing else than the axial fluid stress in the cylindrical pipe of diameter d; and if we consider the fluid stress acting upon the two ends of the pipe AB as producing a pair of external forces which have to be equilibrated, we shall find that their resultant has the direction OD, and has exactly the magnitude already given by the above formula. Thus, regarding the fluid contents of the pipe EFGH in Fig. 15 as forming the voussoir of a fluid arch, the planes EG and FH, forming the faces of this voussoir, will evidently be of elliptical shape, the major axis EG or FH being equal to d sec, and the fluid stress upon these surfaces will therefore Ꮎ have the magnitude P sec 02 in each case. The lines of action of these forces will, of course, be at right angles to the surfaces EG and FH respectively, and will intersect on the line OD, while their resultant will have the direction OD and the mag 0 × 2 sin, so that the combination of these forces upon the familiar principles of the parallelogram gives for their resultant the same value It follows, then, that one method of equilibrating the displacing forces upon a polygonal bend would be to apply on each side of the polygon EF, FL, etc., an external force or resistance equal and opposite to the calculated force S. In the case of a buried pipe having such a bend in the horizontal plane, the lateral resistance might be furnished by the reaction of the earth filling or by the concrete surrounding the pipe; while, in the case of an upright bend, the force to be applied to the pipe AB would be simply so much dead weight, in which the weight of the pipe itself and its fluid contents would constitute part of the required load. It is also obvious that if we wish to avoid any bending stress in the pipe, the load or reaction ought, strictly speaking, to be applied in exact opposition to the unbalanced fluid pressure, or at two points near the ends of the pipe and opposite the centres of the two elliptical areas. Another method of preventing any displacement of the polygonal bend would be to apply a sufficient external force the angle ABC. Drawing the sections MN and HK normal to AB, and HG normal to BC, the angle KHG, or 0, will be the supplement of the angle ABC, and will be bisected by HF. The internal pressures on the opposite sides of the pipe, between the normal sections NM and KH, will exactly balance each other, but the pressures on all the parts of the internal surface KFGH will not balance each other. To find the displacing force acting on this surface in the direction HF, we may draw a projection of the surface on a plane at right angles to HF, as in Fig. 16a; and subtracting as before the negative from the positive projection, we have the two ellipses, whose minor axes KH and HG will each be equal to 0 d sin The displacing force will accordingly be― Exactly the same result will, of course, be reached if we regard the cylindrical columns AB and BC as members of a fluid polygonal frame under an axial compressive force in each member, equal to P or pd2 4 When the straight lengths of pipe are formed with plain ends united by double sockets, whose faces are inclined to each other at the angle 0, as sketched in Fig. 17, there will evidently be no unbalanced force resulting from the internal pressure on the sides of the straight pipes; while the socket itself will be subjected to a displacing force in the direction HF, whose magnitude will be independent of the length of the socket, but A Ꮎ will be proportional to sin as before. 2 It seems probable, however, that the fluid would find its way into the joint so far as to take effect over the annular surface at the ends of the plain pipe; and if their external diameter is d1, the area of each elliptical projection 0 will be d2 sin; or, if we denote by 1 4 P1 the axial pressure taken over the whole section of the pipe (including the annular surface), the displacing force will be FIG. 17 F EXAMPLE. Suppose a cast-iron pipe, having an internal diameter of 36 inches, and an external diameter of 39 inches, to be laid in straight lengths, whose centre lines form the tangents to a curve of 50 feet radius. And let the length of each side, AB, BC, etc., be 5.24 feet, so that the angle e subtended at the centre of the curve by each side of the polygon is 6°, the internal angles ABC, etc., being each equal to 174°. If this pipe were charged under a hydrostatic head h of 320 feet, the axial fluid stress would be-- Р = 320 × 62.4 x 32 × 0·7854 ÷ 2240 = 63 tons But, in the case of the socket-joint, the axial stress over the whole area of the pipe would be P1 = 320 × 62·4 × 3·252 × 0·7854 ÷ 2240 = 73.94 tons Also we have tan = 2 The displacing force at right angles to each side of the polygon, in such a case as Fig. 15, would then be S = 2 × 63 × 0·0524 = 6.7 tons which is equivalent to 1-26 tons per lineal foot of the pipe AB. Or, again, at each angle of the same polygon, the radial displacing force would be N = 2 x 63 x 0.05233 = 6.59 tons But, in the case of the double socket illustrated in Fig. 17, if we allow for the pressure upon the annular ends, we shall have— N1 = 2 x 73.94 × 0·05233 = 7.74 tons The displacing force would be equilibrated either by a force equal and opposite to S acting upon each side, or by a force equal and opposite to N acting upon each angle. In this example the difference between N and S is very slight, because ♦ is a small angle; but if were large the difference would be greater. Thus, if were 90°, we should have S = N√2. Art. 10. Displacing Forces in a Circular or Segmental Bend.— If a cast-iron pipe is formed as a curved bend, its centre line ADB lying in the arc of a circle as sketched in Fig. 18, it is again evident that the pressures on the opposite internal sides of the pipe will not balance each other; for if we conceive the pipe to be divided into two bent troughs by a diametral section passing along the centre line ADB, the areas of those trough surfaces will not be equal, nor will their projections be equal; and the sum of all the components of fluid pressure measured in the direction OD will be greater than the components measured in the opposite direction. |