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
2)(d
4) = O and solving Ma = 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.
| TABLE 48 | |||||||
|---|---|---|---|---|---|---|---|
| Maximum Stress in Flexible Rings Due to Different Loadings | |||||||
| (From Marston) | |||||||
| Symmetrical Vertical Loadings | Moment at Crown of Sewer | Moment at End of Horizontal Diameter | Compressive Thrust at Crown | Compressive Thrust at End of Horizontal Diameter | Shear at Crown | Shear at End of Horizontal Diameter | |
| Character | Width | ||||||
| Concentrated | 0° | + .318RW 12 | - .182RW 12 | 0.000 | + .500W 12 | 0.500W 12 | 0.000 |
| Uniform | 60° | + .207RW 12 | - .168RW 12 | 0.000 | + .500W 12 | 0.000W 12 | 0.000 |
| Uniform | 90° | + .169RW 12 | - .154RW 12 | 0.000 | + .500W 12 | 0.000W 12 | 0.000 |
| Uniform | 180° | + .125RW 12 | - .125RW 12 | 0.000 | + .500W 12 | 0.000W 12 | 0.000 |
| R = the radius of the pipe, W = total weight of ditch filling and superimposed load plus ⅝ of the weight of the pipe itself (usually neglected), expressed in pounds per foot length of pipe. Moments are inch-pounds per inch length of pipe. Shears and thrusts are in pounds per inch length of pipe. | |||||||
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′,