L4. Pipes Materials and Loads

Transcription

1 The Islamic University of Gaza- Civil Engineering Department SanitaryEngineering- ECIV 4325 L4. Pipes Materials an Loas Base on Dr. Fahi Rabah lecture notes

2 Pipes Materials an loas Materials use for pipes: Sewers are mae from: 1- Concrete 2- Reinforce concrete 3- Vitrifie clay 4- Asbestos cement 5- Cast iron (line with cement). 6- Ductile iron (line with cement). 7- Steel (line with cement). 8- PVC, UPVC 9- GRB (fiber glass). Factors effecting the selection of materials: 1- Chemical characteristics of wastewater an egree of resistance to aci, base, solvents, 2- Resistance to scour an flow (friction coefficient). 3- External forces an internal pressures. 4- Soil conitions. 5- Type of backfill 6- Useful life 7- Strength an water tightness of joints an effective control of infiltration an inflow. 8- Availability in iameter, length, an ease of installation. 9- Cast of construction an maintenance.

3 Loas on pipes When pipes are burie, many forces affect them. The following are the main forces expecte to affect burie pipes: A- Loas ue to Back fill: Back fill loa on a pipe epens on: Trench with Depth of excavation Unit weight of the fill material Frictional characteristics of the backfill. These factors are formulate in the following formula: W where, = 9.8 C * ρ * B 2 W = loa on burie pipe as ue to backfill (Newton per linear meter) C = Coefficient base on the type of backfill an ratio of trench epth to with. ρ = Density of backfill, kg/m 3 B = With of trench at top of the pipe, (m)

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5 Example (1) Calculate the backfill loa on a 610 mm pipe given the following information: Backfill epth (H) = 3.6 m Trench with at the top of the pipe is 1.2 m Backfill is saturate clay, P = 1920 kg/m3 Solution H/B = 3.6/1.2 = 3 from the figure C=2.2 Or you can calculate C from the equation W = 9.8 C * ρ * B W = 9.8 x 2.2 * 1920 * (1.2 ) 2 = 6082 N/m 2

6 B. Wheel loas from trucks Wheel loas from trucks an vehicles transmit live loas to burie sewer lines. When the sewer is eep, only a small portion of the loa is transmitte to the sewers. Equations to compute live loas are very complex that s why esigners pre-calculate ata as illustrate by tables Table gives the highway truck loas transmitte to burie circular pipe in kn/m. The information neee are: Pipe iameter (). Height of fill (H) above pipe (m). Note: The loa is base on (71.17 kn ual-tire wheel loa). If the cover fill is less than ( 1m ) the value obtaine from table Shoul be multiplie by an impact factor obtaine from next table.

7 Example (2) Calculate the wheel loa on a 610 mm pipe that has a backfill cover of 0.8 m. Solution From the table the wheel loa is kn / m (by interpolation) Since H = 0.80 m an in the range (0.61 m 0.90 m) use table to fin the impact factor of 1.1 Truck loa = X 1.1 = KN/m.

8 C- Superficial loas on burie pipes: Superficial loas are prouce by builings an other structures crossing the trench or built along the trench. The proportion of superficial loas that reach the pipe is estimate in tables 1 an 2; the forces are ivie into two types: Long superficial Length of Application > trench With. Short superficial Length of Application trench With. To fin the portion of superficial loa transmitte to the pipes from tables 1 an 2 one shoul know : a. Depth of trench b. With of trench c. Soil type 1 For table 2, the minimum values are for L W 10 Note: [The truck loa can be also estimate as a superficial loa]. Trench Trench Structure L Structure L W [Short superficial loa] W [Long superficial loa]

9 Table 1. Proportion of long superficial loas reaching pipe in trench Table 2. Proportion of short superficial loas reaching pipe in trench Max when L=W Min when L=<0.1 W

10 Example (3) A concrete structure 0.91 m wie with a weight of 1340 kg/m crosses a trench 1.22 m wie in amp clay. The structure bears on the soil 1.83 m above the top of the pipe. Fin the loa transmitte to the pipe. Solution The loa applie by the structure is 1340X1.22= 1635 kg The pressure applie to the soil above the pipe is P= 1635/0.91= 1795 kg/m The ratio of epth to the with is 1.83/1.22= 1.5 From table 2, the maximum proportion of the loa reaching the pipe will be Therefore the loa reaching the pipe will be Trench P = 1795X0.51=915 kg/m Structure L= 0.91 m W=1.22 m

11 Strength of pipes: The crushing strength of sewer pipes is etermine by the three-ege bearing test. The pipe is stresse until failure occurs. Table (3) gives the minimum crushing strength for clay pipes. Strength requirements for reinforce concrete pipes are given in table 4, for this table the crushing force correspon to 0.25mm crack. The values in the table are pre mm iameter, that s why they are calle D-Loas (KN/m.mm) The pipe strength in supporting loas epens on the metho of pipe being : -Class (D) being support the three bearing loa only. -Class(C) support (1.5) the three ege loa (1.5 = Loa factor) -Class (B) being has a loa factor of Class (A) being has a loa factor of 2.3 to Other pipe material has similar tables to estimate their strength The three- ege bearing test

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13 Example (4) A 610 mm concrete pipe is subjecte to a loa of 40 KN/m (backfill an wheel loa). The D-loa of the pipe to prouce 0.25 mm crack is 38.3N/m.mm (The three ege test strength). The pipe being is Class A with a loa factor of 3.4. a) What is the strength of this pipe b) Is the strength enough to resist the applie loa if the minimum factor of safety is 1.5 c) What is the final factor of safety in this case Solution a) Strength of the pipe= the three ege test strength X being loa factor = (38.3X610)X3.4= N/m = 79.4 KN/m b) Factore applie stress = applie stress X loa factor = 40X 1.5= 60 KN/m. c) Since the strength is 79.4 KN/m > factore applie stress (60 KN/m), so the strength is enough. Factor of safety = pipe strength/ applie stress= the three ege test strength X being loa factor/ applie stress.