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 Lp = CL, in which Lp 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.

TABLE 44
Ratio of Load on Pipe to Long Load on Trench[[70]]
Ratio of Depth to WidthSand and Damp Top SoilSaturated Top SoilDamp Yellow ClaySaturated Yellow Clay
0.01.001.001.001.00
0.50.850.860.880.89
1.00.720.750.770.80
1.50.610.640.670.72
2.00.520.530.590.64
2.50.440.480.520.57
3.00.370.410.450.51
4.00.270.310.350.41
5.00.190.230.270.33
6.00.140.170.200.26
8.00.070.090.120.17
10.00.040.050.070.11

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.

TABLE 45
Ratio of Load on Pipe to Short Load on Trench[[71]]
Ratio of Height to Width of TrenchSand and Damp Top SoilSaturated Top SoilDamp Yellow ClaySaturated Yellow Clay
Length of Load Equal to
Width of Trench⅒ Width of TrenchWidth of Trench⅒ Width of TrenchWidth of Trench⅒ Width of TrenchWidth of Trench⅒ Width of Trench
0.01.001.001.001.001.001.001.001.00
0.50.770.120.780.130.790.130.810.13
1.00.590.020.610.020.630.020.660.02
1.50.46 0.48 0.51 0.54
2.00.35 0.38 0.40 0.44
2.50.27 0.29 0.32 0.35
3.00.21 0.23 0.25 0.29
4.00 12 0.12 0.16 0.19
5.00.07 0.09 0.10 0.13
6.00.04 0.05 0.06 0.08
8.00.02 0.02 0.03 0.04
10.00.01 0.01 0.01 0.02
TABLE 46
Weights or Common Building Material When Piled for Storage. Pounds per Cubic Foot
Brick120
Cement90
Sand90
Broken stone150
Lumber35
Granite paving160
Coal50
Pig iron400

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
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.

TABLE 47
Weights of Short Loads on Sewer Trenches
(Adapted from Specifications of the American Bridge Company for Bridges)
Street railways, heavyA load of 24 tons on 2 axles on 10 foot centers.
Street railways, lightA load of 18 tons on 2 axles on 10 foot centers.
For city streets, heavy trafficA load of 24 tons on 2 axles 10 feet apart and 5 foot gage.
For city streets, moderate trafficA load of 12 tons on 2 axles 10 feet apart and 5 foot gage.
For city streets, light traffic or country roadsA load of 6 tons on 2 axles 10 feet apart and 5 foot gage.
Road rollersTotal weight 30,000 pounds. Weight on front wheel, 12,000 pounds, and on each of two rear wheels, 9,000 pounds. Width of front wheel, 4 feet and of each of two rear wheels 20 inches. Distance between front and rear axles 11 feet. Gage of rear wheels, 5 feet, c. to c.

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.