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Unformatted text preview: The Islamic University of Gaza Civil Engineering Department
Sanitary Engineering ECIV 4325 L4. Pipes Materials and Loads Based on Dr. Fahid Rabah lecture notes Pipes Materials and loads Materials used for pipes: Factors effecting the selection of materials: Sewers are made from:
1 Concrete
2 Reinforced concrete
3 Vitrified clay
4 Asbestos cement
5 Cast iron (lined with cement).
6 Ductile iron (lined with cement).
7 Steel (lined with cement).
8 PVC, UPVC
9 GRB (fiber glass). 1 Chemical characteristics of wastewater and degree of
resistance to acid, base, solvents,
2 Resistance to scour and flow (friction coefficient).
3 External forces and internal pressures.
4 Soil conditions.
5 Type of backfill
6 Useful life
7 Strength and water tightness of joints and effective
control of infiltration and inflow.
8 Availability in diameter, length, and ease of
installation.
9 Cast of construction and maintenance. Loads on pipes
When pipes are buried, many forces affect them. The following are the main forces expected
to affect buried pipes:
A Loads due to Back fill:
Back fill load on a pipe depends on:
• Trench width
• Depth of excavation
• Unit weight of the fill material
• Frictional characteristics of the backfill.
These factors are formulated in the following formula: W d = 9 .8 C * ρ *B 2
d
d where,
Wd = load on buried pipe as due to backfill (Newton per linear meter)
Cd = Coefficient based on the type of backfill and ratio of trench depth
to width. ρ = Density of backfill, kg/m3
Bd = Width of trench at top of the pipe, (m) Example (1)
Calculate the backfill load on a 610 mm pipe given the following information:
• Backfill depth (H) = 3.6 m
• Trench width at the top of the pipe is 1.2 m
• Backfill is saturated clay, P = 1920 kg/m3 Solution
H/Bd = 3.6/1.2 = 3 from the figure Cd=2.2
Or you can calculate Cd from the equation W W d d = 9 .8 C *ρ *B 2
d
d = 9 . 8 x 2 . 2 * 1920 * (1 . 2 ) 2 = 6082 N/m B. Wheel loads from trucks
Wheel loads from trucks and vehicles transmit live loads to buried sewer lines.
When the sewer is deep, only a small portion of the load is transmitted to the sewers.
Equations to compute live loads are very complex that’s why designers precalculated data as illustrated by
tables
Table gives the highway truck loads transmitted to buried circular pipe in kN/m. The information needed are:
Pipe diameter (d). Height of fill (H) above pipe (m).
Note: The load is based on (71.17 kN dualtire wheel load).
If the cover fill is less than ( 1m ) the value obtained from table
Should be multiplied by an impact factor obtained from next table. Example (2)
Calculate the wheel load on a 610 mm pipe that has a backfill cover of 0.8 m.
Solution
From the table the wheel load is 14.15 kN / m (by interpolation)
Since H = 0.80 m and in the range (0.61 m – 0.90 m) use table to find the impact factor
of 1.1
Truck load = 14.15 X 1.1 = 15.57 KN/m. C Superficial loads on buried pipes:
Superficial loads are produced by buildings and other structures crossing the trench or built
along the trench.
The proportion of superficial loads that reach the pipe is estimated in tables 1 and 2; the
forces are divided into two types:
Long superficial → Length of Application > trench Width.
Short superficial → Length of Application ≤ trench Width.
To find the portion of superficial load transmitted to the pipes from tables 1 and 2 one should
know :
a. Depth of trench
b. Width of trench
c. Soil type
1
L≤
W
For table 2, the minimum values are for →
10
Note: [The truck load can be also estimated as a superficial load].
Trench Structure Trench L W [Short superficial load] Structure L
W [Long superficial load] Table 1. Proportion of long superficial loads
reaching pipe in trench Table 2. Proportion of short superficial
loads reaching pipe in trench
Max when L=W
Min when L=<0.1 W Example (3)
A concrete structure 0.91 m wide with a weight of 1340 kg/m crosses a trench
1.22 m wide in damp clay. The structure bears on the soil 1.83 m above the top
of the pipe. Find the load transmitted to the pipe.
Solution
The load applied by the structure is 1340X1.22= 1635 kg
The pressure applied to the soil above the pipe is
P= 1635/0.91= 1795 kg/m
The ratio of depth to the width is 1.83/1.22= 1.5 From table 2, the maximum
proportion of the load reaching the pipe will be 0.51.
Therefore the load reaching the pipe will be Trench P = 1795X0.51=915 kg/m
Structure L= 0.91 m W =1.22 m Strength of pipes:
The crushing strength of sewer pipes is determined by the threeedge bearing test. The pipe
is stressed until failure occurs. Table (3) gives the minimum crushing strength for clay
pipes.
Strength requirements for reinforced concrete pipes are given in table 4, for this table the
crushing force correspond to 0.25mm crack. The values in the table are pre mm diameter,
that’s why they are called DLoads (KN/m.mm)
The pipe strength in supporting loads depends –on the method of pipe bedding :
 Class (D) bedding support the three bearing load only.
 Class(C) support (1.5) the three edge load (1.5 = Load factor)
 Class (B) bedding has a load factor of 1.9.
 Class (A) bedding has a load factor of 2.3 to 3.4.
 Other pipe material has similar tables to estimate their strength The three edge bearing test 3 4 Example (4)
A 610 mm concrete pipe is subjected to a load of 40 KN/m (backfill and wheel load). The
Dload of the pipe to produce 0.25 mm crack is 38.3N/m.mm (The three edge test
strength). The pipe bedding is Class A with a load factor of 3.4.
a) What is the strength of this pipe
b) Is the strength enough to resist the applied load 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 edge test strength X bedding load factor =
(38.3X610)X3.4= 79434.2 N/m = 79.4 KN/m
b) Factored applied stress = applied stress X load factor = 40X 1.5= 60 KN/m.
c) Since the strength is 79.4 KN/m > factored applied stress (60 KN/m), so the strength
is enough.
Factor of safety = pipe strength/ applied stress= the three edge test strength X bedding
load factor/ applied stress. ...
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 Spring '10
 Dr.Adib

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