The various forces exerted on a sewer are explained as follows: 1. Internal Pressure of Sewage 2. Temperature Stresses 3. Forces due to External Load.
1. Internal Pressure of Sewage:
Most of the sewers are designed to run partially full in which case sewage flows through sewers under gravity and hence no internal pressure is exerted by the sewage on the sewers. However, due to overflow of sewers or due to chancy surcharge the sewers may run full in which case internal pressure may be exerted by the sewage on the sewers.
Similarly the internal pressure may be exerted by the sewage on outfall sewers which have to flow full under pressure. The internal pressure is invariably exerted by the sewage on the sewers which carry sewage under pressure, i.e., in the case of pressure mains or pressure sewers.
The internal pressure exerted on a sewer by the sewage flowing through it induces circumferential tensile stress called hoop stress in the wall of the sewer which is given by the following expression-
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σ = pD / 2t … (5.1)
In which
σ = hoop stress or circumferential tensile stress;
p = internal pressure of sewage;
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D = diameter of sewer; and
t = thickness of sewer wall.
Hence the pressure mains or pressure sewers must be designed to withstand the above stress.
2. Temperature Stresses:
The temperature stresses may be developed in sewers if they are laid on the ground surface and exposed to atmosphere, and are therefore subjected to temperature changes. The sewers expand during day time and contract at night.
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If this expansion or contraction of the sewer is prevented due to the fixed ends or any other similar cause, stress is induced in the sewer wall in the longitudinal direction which is known as temperature stress or longitudinal stress.
The magnitude of temperature stress is given by the expression-
f = E α T … (5.2)
In which
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f = temperature stress;
E = Young’s modulus of elasticity of the sewer material;
α = coefficient of linear expansion of the sewer material; and
T = change in temperature.
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In order to counteract temperature stresses expansion joints may be provided at suitable intervals (say 20 m to 30 m or so). However, most of the sewers are buried underground and hence they are not subjected to temperature stresses.
3. Forces due to External Loads:
Sewers are mostly buried under the ground surface and placed in trenches which are backfilled, and hence they are subjected to external loads. The external loads exerted on the buried sewers are of two categories, viz., load due to backfill material known as backfill load and superimposed load which again is of two types viz.- concentrated load and distributed load. Moving loads (such as traffic load) may be considered as equivalent to uniformly distributed load. By the external loads exerted on the sewers stresses are induced in the walls of the sewers.
The load imposed on the sewer and the stresses induced in it depends on whether the pipe used as sewer is rigid or flexible. Sewer lines are mostly constructed of stoneware, concrete or cast iron pipes which are considered as rigid pipes, while steel pipes, if used, are not considered as rigid pipes.
The methods of determining the external loads exerted on a buried sewer and the stresses induced in it are discussed below:
(a) Loads on Sewers due to Backfill:
For determining the loads on sewers due to backfill the methods developed by A. Marston for determining the vertical load on buried conduits due to gravity earth forces in commonly encountered conditions are generally accepted as the most suitable and reliable for computation.
Theoretically stated, the load on a buried conduit is equal to the weight of the prism of earth directly over the conduit, called the interior prism of earth, plus or minus the frictional shearing forces transferred to the prism by the adjacent prism of earth.
The general form of Marston’s formula is-
W = CwB2 … (5.3)
Where W = vertical load per unit length acting on the conduit due to gravity earth loads (or backfill);
w = unit weight of earth;
B = width of trench or conduit depending upon the type of installation condition; and
C = dimensionless coefficient.
The coefficient C measures the effect of:
(a) Ratio of height of fill to width of trench or conduit,
(b) Shearing forces between interior and adjacent earth prisms, and
(c) Direction and amount of relative settlement between interior and adjacent earth prisms for embankment conditions.
(b) Superimposed Loads on Sewers:
The superimposed loads which are generally encountered in the case of buried sewers (or conduits) may be classified as:
(a) Concentrated load and
(b) Uniformly distributed load.
These loads may be computed as indicated below:
(a) Concentrated Load:
The load exerted on the buried conduit due to superimposed concentrated load (such as a truck wheel) is given by the following formula:
In which
Wsc = load on the conduit per unit length;
P = concentrated load acting on the ground surface and vertically centred over the conduit; Is = impact factor, the value of which may be taken as 1.0 for air field runways, 1.5 for highway traffic and air field taxi ways and 1.75 for railway traffic;
Le = effective length of the conduit to which the load is transmitted; and
Cs = load coefficient
The load coefficient Cs is a function of (BC/2H) and (Le/2H)
Where Bc = outside width (diameter) of the conduit;
H = height of ground surface above the top of the conduit; and
Le = effective length of the conduit.
The effective length of the conduit is defined as the length over which the average load due to surface traffic units produces the same stress in the conduit wall as does the actual load which varies in intensity from point to point. This is generally taken as 1 m or the actual length of the conduit if it is less than 1 m.
The values of Cs for various values of (Bc/2H) and (Le/2H) are given in Table 5.5.
(b) Uniformly Distributed Load:
The load exerted on the buried conduit due to superimposed uniformly distributed load is given by the following formula:
Wsd = Csp lsBc … (5.15)
In which
Wsd = load on the conduit per unit length;
p = intensity of uniformly distributed load per unit area acting on the ground surface on an area of length M (measured parallel to the conduit) and width N (measured transverse to M), and vertically centred over the conduit;
ls = impact factor;
Bc = outside width (diameter) of the conduit; and
Cs = load coefficient
The load coefficient Cs is a function of (N/2H) and (M/2H), where
N and M are same as indicated earlier; and
H = height of ground surface above the top of the conduit.
The values of Cs for various values of (N/2H) and (M/2M) are given in Table 5.5.
Conduits under Highway:
For class AA loading of IRC, in the critical case of wheel load of 6.25 tonnes, the intensity of distributed load with wheel area 300 mm x 150 mm is given by
Conduits under Railway Track:
In which;
P = axle load (22.5 tonnes for Broad gauge);
ls = impact factor (1.75 for railroad);
2A = length of the sleeper (2.7 m for broad gauge);
2B = distance between the two axles (1.84 m for broad gauge);
Wt = weight of the track structure per unit length (0.3 tonnes/m for broad gauge);
Bc = outside width (diameter) of the conduit; and
Cs = load coefficient which depends on the height of the top of sleeper from the top of the conduit.
For broad gauge track the formula for the load on the conduit per unit length reduces to
W = 32.147 CsBc … (5.18)