  99. Stresses in Buried Pipe.—The stresses which sewer pipe should be designed to resist are: internal bursting pressure, for sewers flowing under pressure; stresses due to handling, for precast pipe; temperature stresses; and external loads. The latter is by far the most important and frequently is the only stress considered in design. The thickness of a pipe to resist internal stress should be PR ft,
The derivation of this expression is simple. The stresses due to handling cannot be computed and are cared for by a thickness of material dictated by experience. These thicknesses are given for vitrified clay and cement pipe in the specifications in the preceding chapter. Temperature stresses are not allowed for in the design of the pipe ring, but allowance must be made for them in long rigid pipe lines exposed to wide variations in temperature. Such a condition seldom exists in sewerage works. The external forces are ordinarily the controlling features in the design of sewer rings. The simplest problems arise in the design of a circular pipe. If the external loading is uniform about the circumference of the pipe the internal stresses will all be compression. Almost all other forms of loading will cause bending moments resulting in tension and compression in different parts of the pipe. The maximum bending is caused by two concentrated loads diametrically opposed. As such a condition is extreme it is not cared for in ordinary design, but a loading between 100. Design of Steel Pipe.—The stresses which may occur in steel sewer pipes are commonly caused by the internal or bursting pressure of the contained liquid. Occasionally a steel pipe may be used as a bridge or as a stressed member of a bridge, but steel pipes should not be used to withstand compression normal to the axis. In order to avoid such stresses the bursting tensile stresses should exceed the external compressive stresses. Such a condition in design requires that buried pipes shall never be emptied, a condition that cannot always be fulfilled. Precaution should be taken, by the installation of proper valves, to prevent the emptying of the pipe at so rapid a rate that a vacuum is created resulting in the collapse of the pipe. Steel pipes are ordinarily made of plates curved to the proper diameter, the edges being held together by rivets. The design of the pipe consists in the determination of the thickness of the plate and the design of the riveted joint. The longitudinal joint and the thickness of the plate are first designed. The design of the joint consists in determining the diameter and pitch of the rivets and the thickness of the plate so that the full strength of the uncut metal shall be developed as nearly as possible under bearing, tearing, and shearing. This is done by making the efficiency of the joint the same under all stresses. The efficiency of the joint is the ratio of the strength of the joint under any kind of stress to the strength in tension of the unpunched plate. Properties of riveted joints are given in Table 41. The diameter of the rivet holes should be computed as 1 The transverse joint must be designed empirically as the stresses in it are indeterminate. The common form of joint for pipes less than 48 inches in diameter is a single-riveted lap joint, and for larger pipes or for pipes exposed to unusual stresses, a double riveted lap joint is used. The same size rivets are used as
Fig. 78.—Lock Bar Pipe. Lock Bar Pipe is a steel pipe with a special form of joint made by the East Jersey Pipe Corporation. It is arranged as shown in Fig. 78 and has the advantage of developing the full strength of the plate. It is equivalent to a joint with 100 per cent efficiency, which permits the use of thinner plates. 101. Design of Wood Stave Pipe.—In the design of wood stave pipe S = Cr(R + t)
The preceding expression can be derived easily by the application of the laws of mechanics, and from it the p = S PR + kt
Fig. 79.—Shoe for Wood Stave Pipe. Transverse joints between staves are closed by inserting metal strips between them, or by shaping the edges irregularly so that they fit closely together with an irregular joint. Transverse joints between all staves at any one point are avoided by splitting the joints between staves. Longitudinal joints between staves are usually made smooth and are closed by steel bands which are drawn tight about the pipe by inserting the ends in coupling shoes as shown in Fig. 79. Fig. 80.—B in Formula W = CwB2 102. External Loads on Buried Pipe.—Prof. Anston Marston and H. C. Anderson published
For example, let it be desired to determine the load on a 72–inch concrete sewer with a 9–inch shell under the following conditions: depth of backfill over the top of the pipe, 15 feet; character of backfill, saturated yellow clay; superimposed load, pile of building brick 6 feet high. The ratio of the depth of backfill to the width of the trench is 15 ÷ 9 or 1.67. The coefficient in the expression CwB2 is 1.39, from Table 42. The weight of saturated yellow clay is 130 pounds per cubic foot, from Table 43. Therefore the load per foot length of the sewer due to the backfill is: W = CwB2 = 1.39 × 130 × 81 = 14,600 pounds.
The pressure of the pile of brick per square foot of trench area is, from Table 46, 120 × 6 = 720 pounds per square foot. The value of C from Table 44, is about 0.70. Therefore Lp is 0.7 × 9 × 720 = 4536 pounds. The equivalent depth of backfill weighing 130 pounds per cubic foot is 4536
103. Stresses in Circular Ring—In Fig. 81a the loads shown indicate the distribution ordinarily assumed in sewer design, the forces being uniformly distributed across the diameter. To find the bending moment in the pipe caused by this loading, let ab in Fig. 81b represent a section of a pipe loaded with equally distributed horizontal and vertical forces. Then the vertical component on a strip of differential length ds is wds cos T and the horizontal component is wds sin T and resolving, the resultant normal to the surface is wds, in which w is the intensity per unit length of the horizontal and vertical forces and T is the angle which the tangent to ds makes with the horizontal. Thus the loading of the nature shown in Fig. 81b is equivalent to a loading of equally distributed normal forces which give no moment in the ring. Fig. 81.—Distribution of Stresses on Buried Pipe. Considering a ring subjected to vertical forces only, the moments will be as shown in Fig. 81c and if loaded with horizontal forces only, the moments will be as shown in Fig. 81d. Because of the symmetry of the figure, moment (1) equals moment (4) but is opposite in direction and moment (2) equals moment (3) but is opposite in direction. When the horizontal and vertical forces are combined on the same ring as in Fig. 81b these moments cancel each other as has been proven. Therefore moment (1) equals moment (2) and moment (3) equals moment (4). Then in Fig. 81e, Ma = Mb. Now ?M = O for conditions of equilibrium, therefore Ma + Mb + (W
104. Analysis of Sewer Arches.—The preceding method for the determination of the stresses in a sewer ring has referred only to a circular pipe uniformly loaded. Other methods must be used if the pipe is not circular or the load is not uniformly distributed. The simplest method, is the static or so-called vouissoir method. In this method the arch is assumed to be fixed at both ends, presumably at the springing line or line of intersection between the inside face of the arch and the abutment, and it is so designed that the resultant of all the forces acting on any section shall lie within the middle third of that section. Fig. 82.—Voussoir Arch Analysis. Fig. 83.—Force Polygon for Voussoir Arch Analysis. To design an unreinforced sewer arch by the vouissoir method, a desired arch is drawn to scale in apparently good proportions for the loadings anticipated. The arch is then divided into any number of sections of equal or approximately equal length called vouissoirs, and the line of action of the resultant load, including the weight of the vouissoir is drawn above each vouissoir as shown in Fig. 82. The forces are assumed to act as shown in the figure. In symmetrically loaded sewer arches there is no vertical reaction at the crown. The resultant R is assumed to act at the lower middle third of the skewback, which is the inclined joint between the arch and the abutment. The upper horizontal force H is assumed to act at the upper middle third of the middle or crown section. The magnitude of H is computed by equating the sum of the moments of all forces about the point of application of R at the skewback to zero, and solving. The force polygon is then drawn as shown in Fig. 83, and the equilibrium polygon is completed in Fig. 82 with its rays parallel to the corresponding strings drawn from the end of H as origin in Fig. 83. If the equilibrium polygon line, called the resistance line, lies wholly within the middle third of each vouissoir, the arch is satisfactory to support the assumed load without reinforcement. If any portion of the resistance line lies outside of the middle third, an attempt should be made to find a resistance line which lies wholly within the middle third. The true resistance line is that which deviates the least from the neutral axis of the arch. To approximate more nearly the true resistance line find two points at which the resistance line already drawn deviates the most from the neutral axis of the arch. Select points M and N on these joints, M being nearer the crown than N. Then let W1 and W2 be the sum of all the loads between the crown and M and N respectively, y represent the vertical distance from the crown to N, and y' represent the vertical distance between M and N, and x1 and x2 represent the horizontal distance from W1 and W2 to M and N respectively. Then the horizontal thrust, H, and a, the distance from the crown to the point of application of H, are, H = (W2x2 - W1x1) y', Fig. 84.—Method for Dividing Arch into Proportion I The elastic method for the design of arches locates the true line of resistance without approximations and is more accurate though not so simple to apply as the static or vouissoir method. In this method a desired form of arch is drawn as in the static method and subdivided into vouissoirs so that the distance S along the neutral axis between joints is such that the ratio I Fig. 85.—Elastic Arch Analysis. The following steps in the procedure are taken from the second edition of the American Civil Engineers Pocket Book, p. 634: In Fig. 85 let the middle points of the joints be marked 1, 2, 3, etc. and the coordinates x and y from the crown be found for each by computation or measurement. For a load W placed at one of these points, let z denote the distance from it, toward the nearest skewback, to another middle point. Let ?zx be the sum of the products of all the values of z by the corresponding x, and ?zy be the sum of all the products of z by the corresponding y; that is, each z in the last two summations is multiplied by the x or y of the point back of W which corresponds to z. For a single load W on the left semi-arch of Fig. 85 the following formulas are deduced from the elastic theory, Horizontal thrust, H = (W 2)n?zy - ?y·?z n?y2 - (?y)2 (1) Moment at Crown, M0 = ½W?z - H?y n (2) Shear at Crown, V0 = ½W?zx ?x2 (3) For symmetrical loading such as W on the left and W on the right the horizontal thrust and crown moment due to both loads are double those found by the above formulas, while the crown shear V0 is zero. For several loads unsymmetrically placed the formulas are to be applied to each in succession and the results added algebraically, the value of V0 being taken as negative for the left semi-arch and positive for the right semi-arch. For any joint whose middle point is at a distance x from the crown M = M0 + Hy + V0x - ?Wz, V = V0 - ?W, where ?W is the sum of all the loads between the joint and the crown and ?Wz is the sum of the moments of those loads with respect to the middle of the joint. The components of the resultant thrust normal and parallel to the joints are, N = H cos ? - V sin ?, F = H sin ? + V cos ?, in which ? is the angle which the plane of the joint makes with the vertical. The distances from the neutral axis to the resistance line are, at the crown, e0 = M0 H, at the joint, e = M N. The resistance line should be located as in the vouissoir method and if not within the middle third a new design should be studied. I = Ic + nIs, in which I is the moment of inertia to be employed, Ic is the moment of inertia of the concrete, Is is the moment of inertia of the steel, and n is the ratio of their moduli of elasticity, generally taken as 15. All of the moments of inertia are referred to the neutral axis of the beam. The reinforcement called for in precast circular pipes is given in Table 39. Sewers cast in place are ordinarily designed to avoid reinforcement, except where the depth of cover is small and the sewer may be subjected to superimposed loads. Concrete sewers are sometimes reinforced longitudinally, with expansion joints from 30 to 50 feet apart. This reinforcement is to reduce the size of expansion and contraction cracks by distributing them over the length of a section. The pipe is divided into sections to concentrate motion due to expansion or contraction at definite points where it can be cared for. The amount of longitudinal reinforcement to be used is a matter of judgment. It varies in practice from 0.1 to 0.4 per cent of the area of the section. Since the coefficients of expansion of concrete and of steel are nearly the same, movements of the structure are as important as the stresses due to changes in temperature. Because of the uncertain and difficult conditions under which concrete sewers are frequently constructed it is advisable to specify the best grade of concrete and not to stress the concrete over 450 pounds per square inch in compression, with no allowable stress in tension. The concrete covering of reinforcing steel should be thicker than is ordinarily used for concrete building design, because of the possibility of poor concrete allowing the sewage to gain access to the steel, resulting in more rapid deterioration than would be caused by exposure to the atmosphere. A minimum covering of about 2 inches is advisable, except in very thin sections not in contact with the sewage. A minimum thickness of concrete of about 9 inches is frequently used in design, although crown thicknesses of 4½ inches have been used with Brick linings are often provided for the invert where moderately high velocities of about 10 feet per second when flowing full are to be expected. For velocities in the neighborhood of 20 feet per second the invert should be lined with the best quality vitrified brick. Although concrete may erode no faster than brick under the same conditions, brick linings are more easily replaced and at a smaller expense. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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