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

in which P =

the intensity of internal pressure;

R =

the radius of the inside of the pipe, and

f_t =

the unit-strength of the material in tension

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 this condition and perfect distribution is assumed, as explained in Art. 103.

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
16 of an inch larger than the diameter of the rivets. Rivets and plates should be designed for the nearest or next largest commercial size, and a generous allowance for corrosion should be made in determining the thickness of the plate. The distance from the edge of the plate to the side of the rivet should not be less than 1½ times the diameter of the rivet. The unit-strengths of the metal are given in the preceding chapter.

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 in the longitudinal joint. The maximum permissible distance between rivets should be used in the transverse joint.

Pipes used as compression members of a bridge are stiffened by riveting standard rolled steel sections longitudinally on the 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^[66] the entire bursting pressure is taken up by steel bands wrapped around the outside of wood staves which make up the shell of the pipe. The pipe is not designed to resist external loads except those which may be overcome by the internal pressure in the pipe. The thickness of the staves is fixed by experience. The sizes of staves and bands recommended by J. F. Partridge^[67] are given in Table 40. The size of the steel bands can be determined from the expression;

in which S =

the total stress in the band;

R =

the radius of the inside of the pipe;

t =

the thickness of the stave;

r =

the area of bearing per unit length of the band on the wood. For circular bands it is assumed as the radius of the band;

C =

the crushing strength of wood, usually taken at 650 pounds per sq. in.

The preceding expression can be derived easily by the application of the laws of mechanics, and from it the expression for the distance between bands follows logically. It is,

in which S =

the strength of the band;

p =

the distance between bands;

P =

the intensity of bursting pressure in the pipe;

R =

the radius of the inside of the pipe;

t =

the thickness of the staves;

k =

the swelling strength of wood, usually taken at 100 pounds per sq. in.

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.

102. External Loads on Buried Pipe.—Prof. Anston Marston and H. C. Anderson published^[68] the results of a series of experiments on the loads on buried pipes which are of extreme value in the design of sewer pipe. The load on the pipe is given by the empirical expression W = CwB², in which w is the weight of the backfilling material in pounds per cubic foot, B is the width of the trench in feet at the elevation of the end of a radius making an angle of 45 degrees upwards with the horizontal diameter of the pipe as illustrated in Fig. 80, and C is a coefficient dependent on the character of the backfill and the ratio of the width to the depth of the trench. Values of C are given in Table 42. The weights of various classes of backfilling are given in Table 43.

Where surface loads are to be carried on the sewer trench the proper proportion of the load to be carried by the sewer is determined by the expression L_p = CL, in which L_p is the equivalent backfill load per unit length of the trench, L is the surface load per unit length of the trench, and C is a coefficient in which allowance is made for the character of the backfilling, the ratio of depth to width of trench, and the character of the load, whether long or short. A long load is a load extending along the length of the trench such as a pile of building material. A short load is one extending across the trench and for only a short distance along it, such as that caused by a street car or road roller crossing the trench. Values of C are given in Table 44 for long loads, and in Table 45 for short loads. Values of long and short loads occasionally met in practice are given in Tables 46 and 47 respectively.

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:

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 L_p is 0.7 × 9 × 720 = 4536 pounds. The equivalent depth of backfill weighing 130 pounds per cubic foot is 4536
130 × 9 = 3.88 foot. The total equivalent depth of backfill is therefore 3.88 + 15 = 18.88 feet. The ratio of depth to width is 18.88
9 = 2.98. The coefficient C in the expression W = CwB² is 2.17. The total load per foot length of sewer is therefore W = 2.17 × 130 × 81 = 22,800 pounds.

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 Θ and the horizontal component is wds sin Θ 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 Θ 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.

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, M_a = M_b. Now ∑M = O for conditions of equilibrium, therefore M_a + M_b + (W
2)(d
4) = O and solving M_a = Wd
16. This moment occurs at the ends of the horizontal and vertical diameters and causes tension on the inside of the pipe at the top and on the outside at the ends of the horizontal diameter. There will also be compression at each end of the horizontal diameter equal to one-half of the total load on the pipe. If the material of the pipe is homogeneous, the maximum fiber stress f can be found through the expression f = My
I ± P
A in which M is the bending moment, y is the distance from the neutral axis to the extreme fiber of a cross-section of the shell of the pipe of unit length, I is the moment of inertia of this cross-section about its neutral axis, P is one-half the total load on the pipe, and A is the area of the cross-section. For reinforced concrete, the standard formulas should be used with this expression for M. The stresses in a circular ring subjected to other distributions of loads are shown in Table 48. An exhaustive study of the stresses in circular rings was published by Prof. A. N. Talbot in Bulletin No. 22 of the Engineering Experiment Station at the University of Illinois, 1908.

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.

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,

A resistance line should be drawn with this new horizontal thrust. If no resistance line can be found lying wholly within the middle third, new sections should be designed until a resistance line can be drawn lying wholly within the middle third—unless the arch is to be reinforced. A number of satisfactory arches should be designed and the easiest one to build should be selected. This method is limited in its application to sewer arches with rigid side walls and it cannot be extended to include the invert. Although an approximate method it is accurate within less than 10 per cent of the true stresses and is usually quite close.

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
S shall be the same for all vouissoirs. I is the average of the moments of inertia of the surfaces of the two limiting joints about the neutral axis. If the thickness of the arch is constant the distance between joints will be the same. The method for dividing the arch into sections such that the ratio I
S shall be a constant^[73] is as follows: divide the half arch axis into any number of equal parts; measure the radial depth at each point of division; lay off the length of the arch axis to scale on a straight line; divide this line into the same number of equal parts as the half arch, as shown in Fig. 84; at each point erect a perpendicular equal in length by scale to the moment of inertia at the corresponding point on the arch section; draw a smooth curve through the tops of these lines; draw a line ab at any slope from the center of the original straight line to the curve, and then a line bc back to the straight line to form an isosceles triangle abc; continue forming these triangles in a similar manner thus dividing the original straight line in the required ratio. The distance between joints is represented by the bases of the triangles. By construction the altitude of the triangle represents the average moment of inertia between the two limiting joints. The base of each isosceles triangle is S, and I
S = ½ tan α in which α is the base angle of all the isosceles triangles.