Chapter 6 - Design of PE Piping Systems

Chapter 6 155

Design of PE Piping Systems

Chapter 6

Design of PE Piping Systems Introduction Design of a PE piping system is essentially no different than the design undertaken with any ductile and flexible piping material. The design equations and relationships are well-established in the literature, and they can be employed in concert with the distinct performance properties of this material to create a piping system which will provide very many years of durable and reliable service for the intended application. In the pages which follow, the basic design methods covering the use of PE pipe in a variety of applications are discussed. The material is divided into four distinct sections as follows: Section 1 covers Design based on Working Pressure Requirements. Procedures are included for dealing with the effects of temperature, surge pressures, and the nature of the fluid being conveyed, on the sustained pressure capacity of the PE pipe. Section 2 deals with the hydraulic design of PE piping. It covers flow considerations for both pressure and non-pressure pipe. Section 3 focuses on burial design and flexible pipeline design theory. From this discussion, the designer will develop a clear understanding of the nature of pipe/soil interaction and the relative importance of trench design as it relates to the use of a flexible piping material. Finally, Section 4 deals with the response of PE pipe to temperature change. As with any construction material, PE expands and contracts in response to changes in temperature. Specific design methodologies will be presented in this section to address this very important aspect of pipeline design as it relates to the use of PE pipe.

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156 Chapter 6


This chapter concludes with a fairly extensive appendix which details the engineering and physical properties of the PE material as well as pertinent pipe characteristics such as dimensions of product produced in accordance with the various industry standards.

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Section 1 Design Based on Required Pressure Capacity Pressure Rating The methodology for arriving at the standard pressure rating, PR, for PE pipe is discussed in detail in Chapter 5. The terms pressure rating (PR), pressure class (PC), are used in various consensus standards from ASTM, AWWA, CSA and others to denote the pipe’s capacity for safely resisting sustained pressure, and typically is inclusive of the capacity to resist momentary pressure increases from pressure surges such as from sudden changes in water flow velocity. Consensus standards may treat pressure surge capacity or allowances differently. That treatment may vary from the information presented in this handbook. The reader is referred to the standards for that specific information. Equations 1-1 and 1-2 utilize the Hydrostatic Design Stress, HDS, at 73º F (23ºC) to establish the performance capability of the pipe at that temperature. HDS’s for various PE pipe materials are published in PPI TR-4, “PPI Listing of Hydrostatic Design Basis (HDB), Hydrostatic Design Stress (HDS), Strength Design Basis (SDB), Pressure Design Basis (PDB) and Minimum Required Strength (MRS) Ratings for Thermoplastic Piping Materials”. Materials that are suitable for use at temperatures above 100ºF (38ºC) will also have elevated temperature Hydrostatic Design Basis ratings that are published in PPI TR-4. The PR for a particular application can vary from the standard PR for water service. PR is reduced for pipelines operating above the base design temperatures, for pipelines transporting fluids that are known to have some adverse effect on PE, for pipelines operating under Codes or Regulations, or for unusual conditions. The PR may be reduced by application of a factor to the standard PR. For elevated temperature applications the PR is multiplied by a temperature factor, FT. For special fluids such as hydrocarbons, or regulated natural gas, an environmental application factor, AF, is applied. See Tables 1-2 and Appendix, Chapter 3. The reader is alerted to the fact that the form of the ISO equation presented in Equations 1-1 and 1-2 has changed from the form of the ISO equation published in the previous edition of the PPI PE Handbook. The change is to employ HDS rather than HDB, and is necessitated by the additional ratings available for high performance materials. In the earlier form of the ISO equation, PR is given as a function of the HDB, not the HDS as in Equations 1-1 and 1-2. This difference is significant and can result in considerable error if the reader uses the Environmental Applications Factors given in Table 1-2 as the “Design Factor” in the HDB form of the ISO equation.

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7, ASTM D2447, ASTM D3035, ASTM F714, AWWA C901, AWWA E.(3,4,5,6,7,8,9,10) The appendix provides specific dimensional ide diameter controlled polyethylene pipe and tubing made in 158 Chapter 6 of PE Piping Systems ected ASTM andDesign AWWA standards.

e used to determine an average inside diameter for OD-controlled ade to dimension ratio (DR) specifications in accordance with the d standards. In(1-1) these standards, pipe dimensions are specified as 2 HDS F T A F ameter and, typically, is specified as a minimum PR =wall thickness +12% tolerance is applied. (DR-1) Therefore, an average ID for flow may be determined by deducting twice the average wall thickness (1-2) 2 HDS F A F from the average outside ness plus half the wall PR =tolerance orT 6%)

(IDR+1)

WHERE §D · = Pressure DPR DO 2rating, .12¨psi O ¸ A © DRStress, ¹ psi (Table 1-1) HDS = Hydrostatic Design

Eq. 1-1

AF = Environmental Application Factor (Table 1-2) NOTE: The environmental application factors given in Table 1-2 are not to be confused with the Design Factor, DF,

used in previous editions of the PPI Handbook and in older standards. pipe average inside diameter, in FT = Service Temperature Design Factor (See Appendix to Chapter 3) pipe outside diameter, in DR = OD -Controlled Pipe Dimension Ratio dimension ratio

(1-3)

DR

DO t

Eq. 1-2

pipe minimum wall thickness, in

DO = OD-Controlled Pipe Outside Diameter, in. t = Pipe Minimum Wall Thickness, in. IDR = ID -Controlled Pipe Dimension Ratio

5

Controlled Pipe(1-4)

D IDR = I Eq. 1-6 t diameter controlled pipes provide average dimensions for the pipe are used for flow calculations. ID-controlled pipe standards include ID-Controlled Pipe Inside Diameter, in. DI = ID-Controlled Pipe Inside Diameter, in. M D2239, ASTM F894 and AWWA C901. (11,12,13)

Table 1-1 nd “IDR” are used with outside diameter controlled and inside Hydrostatic Design Stress and Service Temperatures drostatic Design Basis Ratings and Service Temperatures pipe respectively. Certain dimension ratios that meet an ASTM2708, PE 3608, ASTM eries are “standardized dimension ratios,” thatPEis2606,SDR orPE SIDR. y PE 3408 PE 2406 Property Standard PE2706 PE 3708, PE 4608

Standard

Hydrostatic Design Stress, D 2837 HDS at 73°F (23°C) 1600

23°C)

ed temperature ervice ed Temperature Service

psi

ASTM D2837 & (11.04PPIMPa) TR-3

Maximum recommended – temperature for140°F operating Pressure Service*

(60°C)*-

– recommended 180°F (82°C) Maximum operating temperature for materials are stress rated at temperatures Non-Pressure Service

PE 3710, PE 4710

630 psi (4.6 800 psi (5.5 MPa) 1250 psi MPa) (8.62 MPa)

1000 psi (6.9 MPa)

140°F 140°F (60°C)(60°C)140°F (60°C)

140°F (60°C)

180°F (82°C) 180°F (82°C)

180°F (82°C)

lene piping as high as 180° F. For n regarding these materials and their use, the reader is referred to PPI, TR-4

180°F (82°C)

* Some PE piping materials are stress rated at temperatures as high as 180°F. For more information regarding these materials and their use, the reader is referred to PPI, TR-4.

gth of thermoplastic pipe is based on regression analysis of stressd in accordance with ASTM D2837. Analysis of the data obtained s utilized to establish a stress intercept for the material under 0 hours. This intercept when obtained at 73° F is called the longength or LTHS. The LTHS typically falls within one of several hat are detailed in ASTM D2837. This categorization of the LTHS 154-264.indd 158 erial establishes its hydrostatic design basis or HDB. The HDB is

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Chapter 6 159


The Hydrostatic Design Stress, HDS, is the safe long-term circumferential stress that PE pipe can withstand. It is derived by applying an appropriate design factor, DF, to the Hydrostatic Design Basis, HDB. The method for establishing the Hydrostatic Design Stress for PE pipe is described in Chapters 3 and 5. At the time of this printing, AWWA is in the process of revising AWWA C906 to incorporate PE4710 material and to use the HDS values in Table 1-1. The version in effect at the time of this printing, AWWA C906-07, limits the maximum Hydrostatic Design Stress to 800 psi for HDPE and to 630 psi for MDPE. AWWA C901-08 has been revised to incorporate the materials listed in Table 1-1. The Environmental Application Factor is used to adjust the pressure rating of the pipe in environments where specific chemicals are known to have an effect on PE and therefore require derating as described in Chapter 3. Table 1-2 gives Environmental Applications Factors, AF, which should only be applied to pressure equations (see Equations 1-1 and 1-2) based on the HDS, not the HDB.

Table 1-2 PE Pipe Environmental Application Factors (AF)* Pipe Environment

Environmental Application Factor (AF) at 73ºF (23ºC)

Water: Aqueous solutions of salts, acids and bases; Sewage; Wastewater; Alcohols; Glycols (anti-freeze solutions)

1.0

Nitrogen; Carbon dioxide; Methane; Hydrogen sulfide; Non-Federally regulated applications involving dry natural gas or other non-reactive gases

1.0

Fluids such as solvating/permeating chemicals in pipe or soil (typically hydrocarbons) in 2% or greater concentration, natural or other fuel-gas liquids condensates, crude oil, fuel oil, gasoline, diesel, kerosene, hydrocarbon fuels, wet gas gathering, multiphase oilfield fluids, LVP liquid hydrocarbons, oilfield water containing >2% hydrocarbons.

0.5

* Certain codes and standards include prohibitions and/or strength reduction factors relating to the presence of certain constituents in the fluid being transported. In a code controlled application the designer must ensure compliance with all code requirements.

When choosing the environmental applications factor (AF), consideration must be given to Codes and Regulations, the fluid being transported, the external environment, and the uncertainty associated with the design conditions of internal pressure and external loads. The pressure rating (PR) for PE pipe in water at 73ºF over the range of typical DR’s is given in Tables 1-3 A and 1-3 B in this chapter.

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160 Chapter 6


Pressure Rating for Fuel Gas Pipe Compared to other common thermoplastic pipes, PE pipe can be used over a broader temperature range. For pressure applications, it has been successfully used from -40º F (-40º C) to 140ºF (60º C). In the case of buried non-pressure applications it has been used for conveying fluids that are at temperatures as high as 180°F (82°C). See Table 1-1. For pressure applications above 80ºF (27º C) the Service Temperature Design Factor is applied to determine the pressure rating. See Table A.2 in the Appendix to Chapter 3. The pressure rating for gas distribution and transmission pipe in US federally regulated applications is determined by Title 49, Transportation, of The Code of Federal Regulations. Part 192 of this code, which covers the transportation of natural and other gases, requires that the maximum pressure rating (PR) of a PE pipe be determined based on an HDS that is equal to the material’s HDB times a DF of 0.32. (See Chapter 5 for a discussion of the Design Factor, DF.) This is the equivalent of saying that for high density PE pipe meeting the requirements of ASTM D2513 the HDS is 500 psi at 73º F and for medium density PE pipe meeting D2513 the HDS is 400 psi at 73º F. There are additional restrictions imposed by this Code, such as the maximum pressure at which a PE pipe may be operated (which at the time of this writing is 125 psi for pipe 12-in and smaller and 100 psi for pipe larger than 12-in through 24-in.) and the acceptable range of operating temperatures. The temperature design factors for federally regulated pipes are different than those given in Table A.2 in the Appendix to Chapter 3. Consult with the Federal Regulations to obtain the correct temperature design factor for gas distribution piping. At the time of this writing, there is an effort underway to amend the US federal code to reflect changes already incorporated in ASTM F714 and D3035. When amended, these changes will increase the pressure rating (PR) of pipe made with high performance PR resins - those that meet the higher performance criteria listed in Chapter 5 (see “Determining the Appropriate Value of HDS”), to be 25% greater than pressure ratings of pipe made with ‘traditional’ resins. In Canada gas distribution pipe is regulated per CSA Z662-07. CSA allows a design factor of 0.4 to be applied to the HDB to obtain the HDS for gas distribution pipe. PE pipe meeting the requirements of ASTM D2513 may be used for the regulated distribution and transmission of liquefied petroleum gas (LPG). NFPA/ANSI 58 recommends a maximum operating pressure of 30 psig for LPG gas applications involving polyethylene pipe. This design limit is established in recognition of the higher condensation temperature for LPG as compared to that of natural gas and, thus, the maximum operating pressure is recommended to ensure that plastic pipe is not subjected to excessive exposure to LPG condensates. The Environmental Application Factor for LP Gas Vapors (propane, propylene, and butane) is 0.8 with

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a maximum HDS of 800 psi at 73º F for HDPE and 630 psi for MDPE. For further information the reader is referred to PPI’s TR-22, Polyethylene Piping Distribution Systems for Components of Liquid Petroleum Gases. The pressure rating for PE gas gathering lines in the US may differ depending upon the class location (population density) of the gathering line. Gas gathering lines in Class 2, 3 and 4 locations are regulated applications and subject to US federal codes the same as gas distribution and transmission lines. Gas gathering lines in Class 1 locations are not regulated in accordance with US federal codes, and may be operated at service pressures determined using Equation 1-1. Non-regulated gas gathering lines may use PE pipe meeting ASTM F2619 or API 15LE, and may be larger than 24” diameter. PE pipe meeting ASTM D2513 is not required for nonregulated gas gathering lines. In Canada, PE gas gathering lines are regulated in accordance with CSA Z662 Clause 13.3 and are required to meet API 15LE. PE gas gathering lines may be operated at service pressures equivalent to those determined using Equation 1-1. Pressure Rating for Liquid Flow Surge Pressure Surge pressure events, which give rise to a rapid and temporary increase in pressure in excess of the steady state condition, are the result of a very rapid change in velocity of a flowing liquid. Generally, it is the fast closing of valves and uncontrolled pump shutdowns that cause the most severe changes and oscillations in fluid velocity and, consequently in temporary major pressure oscillations. Sudden changes in demand can also lead to lesser but more frequent pressure oscillations. For many pipe materials repeated and frequent pressure oscillations can cause gradual and cumulative fatigue damage which necessitate specifying higher pressure class pipes than determined solely based on sustained pressure requirements. And, for those pipe materials a higher pressure class may also be required for avoiding pipe rupture under the effect of occasional but more severe high-pressure peaks. Two properties distinguish PE pipes from these other kinds of pipes. The first is that because of their lower stiffness the peak value of a surge pressures that is generated by a sudden change in velocity is significantly lower than for higher stiffness pipes such as metallic pipes. And, the second is that a higher pressure rating (PR), or pressure class (PC), is generally not required to cope with the effects of pressure surges. Research, backed by extensive actual experience, indicates that PE pipes can safely tolerate the commonly observed maximum peak temporary surge pressure of twice the steady state condition. Furthermore, the long-term strength of PE pipes is not adversely affected by repeated cyclic loading – that is, PE pipes are very fatigue resistant.

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162 Chapter 6


In the design of PE pipe, pressure surges are generally classified as Occasional pressure surges, Recurring pressure surges, and Negative pressures. • Occasional surge pressures are caused by emergency operations such as fire flow or as a result of a malfunction, such as a power failure or system component failure, which includes pump seize-up, valve stem failure and pressure relief valve failure. • Recurring surge pressures are inherent to the design and operation of a system. Recurring surge pressures can be caused by normal pump start up or shut down, normal valve opening and closing, and/or “background” pressure fluctuations associated with normal pipe operation. • Negative pressure may be created by a surge event and cause a localized collapse by buckling. (Negative pressure may also occur inside flowing pipelines due to improper hydraulic design.) In recognition of the performance behavior of PE pipes the following design principles have been adopted by AWWA for all PE pressure class (PC) rated pipes. These design principles, which are as follows, are also applicable to PE water pipes that are pressure rated (PR) in accordance with ASTM and CSA standards: 1. Resistance to Occasional Pressure Surges:

• The resultant total pressure – sustained plus surge – must not exceed 2.0 times the pipe’s temperature compensated pressure rating (PR). See Tables 1-3 A and 1-3 B for standard surge allowances when the pipe is operated at its full rated pressure. • In the rare case where the resultant total pressure exceeds 2.0 times the pipe’s temperature adjusted PR, the pipe must be operated at a reduced pressure so that the above criterion is satisfied. In this event the pipe’s reduced pressure rating is sometimes referred to as the pipe’s “working pressure rating” (WPR), meaning that for a specific set of operating conditions (temperature, velocity, and surge) this is the pipe’s pressure rating. AWWA uses the term WPR not just for a reduced pressure rating but for any pressure rating based on application specific conditions. Where the total pressure during surge does not exceed the standard allowance of 2.0 (occasional) and 1.5 (recurring) the WPR equals the temperature adjusted PR. • The maximum sustained pressure must never exceed the pipe’s temperature adjusted pressure rating (PR). Example:

A PE pipe has a DR = 17 and is made from a PE4710 material. Accordingly, its standard pressure rating (PR) for water, at 73°F is 125 psi (See Table A.1 in Appendix to Chapter 3). The maximum sustained water temperature shall remain below 73°F. Accordingly, no temperature compensation is required and therefore, the pipe’s initial WPR is equal to its standard PR or, 125 psi.

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Let us first assume that the maximum occasional surge pressure shall never exceed 120 psi. Since a WPR of 125 psi plus a surge of 120 psi is less than 2 times 125 psi the pipe’s initial WPR of 125 psi remains at that value. Now let us assume a second case in which the maximum occasional surge pressure can be as high as 150 psi. This pressure plus the pipe’s initial WPR of 125 psi result in a total momentary pressure of 275 psi, which is 25 psi above the limit of 2 x 125 psi = 250 psi. To accommodate this 25 psi excess it is necessary to reduce the pipe’s initial WPR of 125 to a final WPR of 100 psi. 2. Resistance to Recurring Pressure Surges:

• The resultant total momentary pressure – sustained plus surge – must not exceed 1.5 times the pipe’s temperature adjusted pressure rating (PR). See Tables 1-3 A and 1-3 B for standard surge allowance when the pipe is operated at its full rated pressure. • In the rare case where the resultant total pressure exceeds 1.5 times the pipe’s temperature adjusted PR the pressure rating must be reduced to the pipe’s WPR so that the above criterion is satisfied. • The maximum sustained pressure must never exceed the pipe’s temperature adjusted PR. 3. Resistance to Localized Buckling When Subjected to a Negative Pressure Generated by a Surge Event

A buried pipe’s resistance to localized buckling while under the combined effect of external pressure and a very temporary full vacuum should provide an adequate margin of safety. The design for achieving this objective is discussed in a later section of this chapter. It has been shown that a DR21 pipe can withstand a recurring negative pressure surge equal to a full vacuum at 73°F. Higher DR pipes may also be able to withstand a recurring negative surge equal to full vacuum if they are properly installed and have soil support. Their resistance may be calculated using Luscher’s Equation presented later in this chapter. Estimating the Magnitude of Pressure Surges

Regardless of the type of pipe being used surge or water hammer problems can be complex especially in interconnected water networks and they are best evaluated by conducting a formal surge analysis (See References 25 and 32). For all water networks, rising mains, trunk mains and special pump/valve circumstances a detailed surge analysis provides the best way of anticipating and designing for surge.

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surge pressures are caused by emergency operations. Occasional ures are usually the result of a malfunction, such as power failure or 164 Chapter 6 of PEincludes Piping Systems pump seize-up, valve stem failure and mponent failure, Design which ief valve failure.

WPR for a selected DR, the pressure surge must be calculated. The s may be used to estimate the pressure surge created in pressure Absent a formal surge analysis, an estimate of the magnitude of a surge pressure can ms. be made by evaluating the surge pressure that results from an anticipated sudden change in velocity in the water flowing inside a PE pipe.

in the velocity of a flowing liquid generates a pressure wave. The e may be determined using Equation 1-21.of a flowing liquid in a pipe generates a pressure An abrupt change in the velocity wave. The velocity of the wave may be determined using Equation 1-5. (1-5)

4660

a 1

Eq. 1-21

K BULK ( DR 2) Ed

16

Where

Wave velocity a(celerity), = Wave velocityft/sec (celerity), ft/sec Bulk modulus KofBULK fluid at working temperature = Bulk modulus of fluid at working temperature (typically 300,000 psi for water at 73˚F) ynamic instantaneous effective modulus of pipe material (typically 300,000 psi for water at 73oF)modulus Ed = Dynamic instantaneous effective of pipe material 150,000 psi for all PE pipe at 73˚F (23˚C)); see Appendix to Chapter 3 pically 150,000 psi(typically for PE pipe) DR = Pipe dimension ratio pe dimension ratio

The resultant transient surge pressure, Ps, may be calculated from the wave velocity,

t surge pressure, Ps,the may be change calculated from the ∆ wave velocity, a, a, and sudden in fluid velocity, V. d velocity, ∆v. (1-6)  ∆V  Eq. 1-22  Ps = a  2.31g  Where PS = Transient surge pressure, psig

a = Wave velocity (celerity), ft/sec ansient surge pressure, psig ∆V = Sudden velocity change, ft/sec ave velocity (celerity), ft/sec g = Constant of gravitational acceleration, 32.2 ft/sec dden velocity change, ft/sec Figure 1-1acceleration, represents the pressure surge2 curves for all PE pipes as calculated using nstant of gravitational 32.2 ft/sec 2

Equations 1-5 and 1-6 for Standard Dimension Ratios (SDR’s).

e pressure surge curves for PE3408 as calculated using Equations 1-21 imension Ratios (DR’s).

SDR 7.3

SDR 9 SDR 11 SDR 13.5 SDR 17 SDR 21 SDR 26 SDR 32.5

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Chapter 6 165

Pressure Surge, psig


Sudden Changes in Flow Rate, fps * A value of 150,000 psi and 300,000 psi were used for Ed and K, respectively. ** Calculated surge pressure values applicable to water at temperatures not exceeding 80ºF (27ºC).

Figure 1-1 Sudden Velocity Change vs. Pressure Surge for All PE Pipes

The surge pressure values in Figure 1-1 are based on a sudden change in velocity, which may more often be the case for events like a sudden pump shut-down or a rapid valve closure. A sudden shut-down or a rapid closure occurs faster than the “critical time” (the time it takes a pressure wave initiated at the beginning of a valve closing to return again to the valve). Under ordinary operations, during which valve closings and pump shut-downs are slower than the “critical time”, the actual pressure surge is smaller than that in Figure 1-1. The “critical time” is determined by means of the following relationship: (1-7)

T CR = 2L/a

WHERE TCR = critical time, seconds

L = distance within the pipeline that the pressure wave moves before it is reflected back by a boundary

condition, ft

a = wave velocity (celerity) of pressure wave for the particular pipe, ft/s. (See Equation 1-5)

Generally, PE pipe’s capacity for safely tolerating occasional and frequently occurring surges is such that seldom are surge pressures large enough to require a de-rating of the pipe’s static pressure rating. Tables 1-3 A and 1-3 B show the maximum allowable sudden changes in water flow velocity (ΔV) that are safely tolerated without the need

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166 Chapter 6


to de-rate the pressure rating (PR) or, the pressure class (PC), of a PE pipe. If sudden changes in velocity are expected to be greater than the values shown in these Tables, they then must be accommodated by lowering the pipe’s static pressure rating. As previously discussed, the new rating is called the working pressure rating (WPR).The procedure for establishing a WPR has been discussed earlier in this Section.

Table 1-3A Allowances for Momentary Surge Pressures Above PR or PC for Pipes Made From PE4710 and PE3710 Materials1. Standard Allowance for Momentary Surge Pressure Above the Pipe’s PR or PC

Pipe Standard Diameter Ratio (SDR)

Standard Static Allowance for Recurring Surge Pressure Rating (PR) or, Standard Resultant Pressure Class Allowable (PC) for water @ Allowable Surge Sudden Change 73°F, psig Pressure, psig in Velocity, fps

Allowance for Occasional Surge

Allowable Surge Pressure, psig

Resultant Allowable Sudden Change in Velocity, fps

32.5

63

32

4.0

63

8.0

26

80

40

4.5

80

9.0

21

100

50

5.0

100

10.0

17

125

63

5.6

125

11.2

13.5

160

80

6.2

160

12.4

11

200

100

7.0

200

14.0

9

250

125

7.7

250

15.4

7.3

320

160

8.7

320

17.4

1. AWWA C906-07 limits the maximum Pressure Class of PE pipe to the values shown in Table B. At the time of this printing C906 is being revised to allow PC values in Table A to be used for PE3710 and PE4710 materials. Check the latest version of C906

Table 1-3 B Allowances for Momentary Surge Pressures Above PR or PC for Pipes Made from PE 2708, PE3408, PE3608, PE3708 and PE4708 Materials. Standard Allowance for Momentary Surge Pressure Above the Pipe’s PR or PC

Pipe Standard Diameter Ratio (SDR)

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Standard Static Allowance for Recurring Surge Pressure Rating Resultant (PR) or, Standard Allowable Pressure Class (PC), for Water @ Allowable Surge Sudden Change Pressure, psig in Velocity, fps 73°F, psig

Allowance for Occasional Surge

Allowable Surge Pressure, psig

Resultant Allowable Sudden Change in Velocity, fps

32.5

50

25

3.1

50

6.2

26

63

32

3.6

63

7.2

21

80

40

4.0

80

8.0

17

100

50

4.4

100

8.8

13.5

125

63

4.9

125

9.8

11

160

80

5.6

160

11.2

9

200

100

6.2

200

12.4

7.3

250

125

6.8

250

13.6

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The surge pressure allowance in Table 1-3 A and 1-3 B are not the maximum surge limits that the pipe can safely withstand. Higher surge pressures can be tolerated in pipe where the working pressure rating (WPR) of the pipe is limited to a pressure less than the pressure rating (PR). This works because the combined total pressure for surge and for pumping pressure is limited to 1.5 times the PR (or PC) for recurring surge and 2.0 times the PR (or PC) for occasional surge. If the pumping pressure is less than the PR (or PC) then a higher surge than the standard allowance given in Table A and B is permitted. The maximum permitted surge pressure is equal to 1.5 x PR – WP for recurring surge and 2.0 x PR – WP for occasional surge, where WP is the pumping or working pressure of the pipeline. For example a DR21 PE4710 pipe with an operating pressure of 80 psi can tolerate a recurring surge pressure of 1.5 x 100 psi – 80 psi = 70 psi. Note that in all cases WP must be equal or less than PR. Controlling Surge Pressure Reactions Reducing the rate at which a change in flow velocity occurs is the major means by which surge pressure rises can be minimized. Although PE pipe is very tolerant of such rises, other non-PE components may not be as surge tolerant; therefore, the prudent approach is to minimize the magnitude of surge pressures by taking reasonable precautions to minimize shock. Hydrants, large valves, pumps, and all other hydraulic appurtenances that may suddenly change the velocity of a column of water should be operated slowly, particularly during the portion of travel near valve closing which has the larger effect on rate of flow. If the cause of a major surge can be attributable to pump performance – especially, in the case of an emergency stoppage – then, proper pressure relief mechanisms should be included. These can include traditional solutions such as by providing flywheels or by allowing the pumps to run backwards. In hilly regions, a liquid flow may separate at high points and cause surge pressures when the flow is suddenly rejoined. In such cases measures should be taken to keep the pipeline full at all times. These can consist of the reducing of the flow rate, of the use at high points of vacuum breakers or, of air relief valve. Also, potential surge pressure problems should be investigated in the design of pumping station piping, force mains, and long transmission lines. Proven and suitable means should be provided to reduce the effect of surges to a minimum that is practicable and economical. Although PE pipe is much more tolerant of the effect of sudden pressure increases traditional measures should be employed for the minimizing of the occurrence of such increases.

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168

Chapter 6


Section 2 Hydraulic Design of PE Pipe This section provides design information for determining the required flow diameter for PE pipe. It also covers the following topics: general fluid flows in pipe and fittings, liquid (water and water slurry) flow under pressure, non-pressure(gravity) liquid flow, and compressible gas flow under pressure. Network flow analysis and design is not addressed. (1,2) The procedure for piping system design is frequently an iterative process. For pressure liquid flows, initial choice of pipe flow diameter and resultant combinations of sustained internal pressure, surge pressure, and head loss pressure can affect pipe selection. For non-pressure systems, piping design typically requires selecting a pipe size that provides adequate reserve flow capacity and a wall thickness or profile design that sufficiently resists anticipated static and dynamic earthloads. This trial pipe is evaluated to determine if it is appropriate for the design requirements of the application. Evaluation may show that a different size or external load capacity may be required and, if so, a different pipe is selected then reevaluated. The Appendix to Chapter 3 provides engineering data for PE pipes made to industry standards that are discussed in this chapter and throughout this handbook. Pipe ID for Flow Calculations Thermoplastic pipes are generally produced in accordance with a dimension ratio (DR) system. The dimension ratio, DR or IDR, is the ratio of the pipe diameter to the respective minimum wall thickness, either OD or ID, respectively. As the diameter changes, the pressure rating remains constant for the same material, dimension ratio and application. The exception to this practice is production of thermoplastic pipe in accordance with the industry established SCH 40 and SCH 80 dimensions such as referenced in ASTM D 2447. Flow Diameter for Outside Diameter Controlled Pipe OD-controlled pipe is dimensioned by outside diameter and wall thickness. Several sizing systems are used including IPS, which specifies the same OD’s as iron pipe sized (IPS) pipe: DIPS pipe which specifies the same OD’s as ductile iron pipe; and CTS, which specifies the same OD’s as copper tubing sizes. For flow calculations, inside diameter is calculated by deducting twice the average wall thickness from the specified outside diameter. OD-controlled pipe standards include ASTM D2513, ASTM D2737, ASTM D2447, ASTM D3035, ASTM F714, AWWA C901, AWWA C906 and API 15LE.(3,4,5,6,7,8,9,10) The Appendix to this chapter provides specific dimensional information for outside diameter controlled PE pipe and tubing that is made to

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in PPI TR-4. Two design factors, DF and FT, are used to relate the outside diameter. OD-controlled pipeto standards include itions and service temperature conditions the product. SeeASTM D2737, ASTM D2447, ASTM D3035, ASTM F714, AWWA C901, AWWA If the HDB at an elevated temperature is known, that HDB value 15LE.(3,4,5,6,7,8,9,10) The appendix provides specific dimensional Chapter 6 quation 1-3 or 1-4, and the service temperature design factor, FTDesign , of PE Piping Systems outside diameter controlled polyethylene pipe and tubing made in he elevated HDB is not known, then FT should be used, but this will hower selected ASTM and AWWA standards. or more conservative pressure rating.

169

may be used to determine an average inside diameter for OD-controlled pe made to dimension ratio(DR) (DR) specifications in accordance with the ASTM, dimension ratio requirements in accordance with a number of different renced standards. In these standards, pipe dimensions are specified as AWWA, and API 2 HDBCSA x DF x FTstandards. P Eq. 1-3 de diameter and, typically, wall thickness is specified as a minimum DR 1 diameter for such pipes has been calculated using Equation 2-1. The average d a +12% tolerance is inside applied. Therefore, an average ID for flow Typically, wall thickness is specified as a minimum dimension, and a plus 12% poses may be determined by deducting twice the average wall thickness tolerance is applied. the average ID is determined by deducting 2 HDB DF tolerance x In FTthis equation, thickness plus half the xwall or 6%) from the average outside P twice Eq. 1-4 the average wall thickness (minimum wall thickness plus a tolerance of 6%) IDR 1 from the average outside diameter. (2-1)

 DO  2 . 12 D = D −   I A O Pressure rating, psi  DR  Hydrostatic Design Basis, psi Design Factor, Where from Table 1-2 DI = pipe average inside diameter, in Service Temperature Design Factor, = pipe average inside diameter, infrom Table 1-3 D = specified average value of pipe diameter, in 1.0 used. O temperature HDB is outside = if the pipeelevated outside diameter, in DR = dimension ratio OD = -Controlled dimension Pipe ratio Dimension Ratio (2-2)

DR

D D DRO = O t t

Eq. 1-1

Eq. 1-2 Eq. 1-5

t = pipe Outside minimum wallDiameter, thickness,in in in. = pipe minimum wall thickness, OD-Controlled Pipe Pipe Minimum Wall Thickness, in. Pipe Dimension Diameter for ID Controlled Pipe ID -Controlled Pipe Ratio

Standards for inside diameter controlled pipes provide average dimensions for

for ID Controlled thePipe pipe inside diameter that are used for flow calculations. ID-controlled pipe

standards include ASTM D2104, ASTM D2239, ASTM F894 and AWWA C901. (11,12,13)

nside diameter The controlled pipes provide average dimensions for the pipe terms “DR” and “IDR” identify the diameter to wall thickness dimension ratios that are used for flow calculations. ID-controlled pipe standards include for outside diameter controlled and inside diameter controlled pipe, respectively. ASTM D2239, ASTM F894 and AWWA C901. (11,12,13) When those ratios comply with standard values they are called “standard dimension ratios”, that is SDR or SIDR. A discussion of standard dimension ratios is included in Chapter 5. are used with outside diameter controlled and inside

R” and “IDR” olled pipe respectively. Certain dimension ratios that meet an ASTMber series are Fluid “standardized ratios,” that is SDR or SIDR. Flow in PEdimension Piping Head Loss in Pipes – Darcy-Weisbach/Colebrook/Moody

Viscous shear stresses within the liquid and friction along the pipe walls create resistance to flow within a pipe. This resistance results in a pressure drop, or loss of head in the piping system. The Darcy-Weisbach formula, Equation 2-3., and the Colebrook formula, Equation 2-6, are generally accepted methods for calculating friction losses due to liquids

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or Fanning Equation 1-7,aand Colebrook formula, within a pipe. formula, This resistance within pipethe results in a pressure drop, enerally for calculating friction due to stresses within themethods liquid and friction along the losses pipe walls create he pipingaccepted system. pipes.(15,16) These formulas recognize dependence on pipe boredrop, w within a pipe. This resistance within a pipe results in a pressure 170 Chapter 6 acteristics, liquid viscosity and flow velocity. Design of PE Piping Systems nach the or piping system. Fanning formula, Equation 1-7, and the Colebrook formula,

e generally accepted methods for calculating friction losses due to formula bach or is: FanningThese formula, Equation 1-7, and the Colebrook formula, ull pipes.(15,16) formulas recognize dependence on pipe bore are generally accepted methods for calculating friction losses due to haracteristics, liquid viscosity and flow velocity. 2 full pipes.(15,16) TheseL formulas recognize dependence on pipe bore V flowing in full pipes.(15,16) These formulas recognize dependence Eq. 1-7on pipe bore and hliquid f = fviscosity and flow velocity. characteristics, d ' 2 g ach formula is: pipe surface characteristics, liquid viscosity and flow velocity.

The Darcy-Weisbach formula is:

bach formula is:(2-3) LV 2 Eq. 1-7 f iction (head) loss, ft.h fof liquid d ' 2 g 2 ipeline length, ft. LV Eq. 1-7 hf f ipe inside diameter, ft. d ' 2g Where ow velocity, ft/sec. = friction (head) loss, ft. of liquid friction (head)hf loss, ft. of liquid L = pipeline length, ft. pipeline length, ft.0 . 4085 Q = pipe inside diameter, ft. Vd’ =loss, Eq. 1-8 friction (head) ft. liquid pipe inside diameter, ft. 2of V = flow velocity, ft/sec. D l pipeline length, ft. factor (dimensionless, but dependent upon pipe surface roughness and Reynolds number) f ft/sec. = friction flow velocity, g = constant of gravitational acceleration (32.2ft/sec pipe inside diameter, ft. onstant of gravitational acceleration (32.2ft/sec2)) flow velocity, 0 . 4085 Q ow rate, gpm Theft/sec. equation Vflow velocity 2may be computed by means of the following Eq. 1-8 ipe inside diameter, in D l (2-4) 0 . 4085 Q iction factor (dimensionless, but dependent upon pipe surface Eq. 1-8 V 2 constant of gravitational acceleration (32.2ft/sec2) Dl oughness and Reynolds number) flow rate, gpm 2 constant gravitational pipe insideofdiameter, in acceleration (32.2ft/sec ) Where flow rate, gpm Q(dimensionless, = flow rate, gpm friction factor but dependent upon pipe surface pipe inside diameter, DI of =Reynolds pipe insidein diameter, in and number) willroughness assume one three flow regimes. The flow regime may be friction factor (dimensionless, but dependent upon pipe surface in transition Liquid between laminar turbulent. Inflow laminar flow flow in pipes willand assume one of three regimes. The flow regime may be laminar, roughness and number) e, below 2000), theReynolds pipe’s roughness hasand noturbulent. effect and is turbulent or in surface transition between laminar In laminar flow (Reynolds number, Re, As such, the below friction factor, f , is calculated using Equation 1-9. 2000), the pipe’s surface roughness has no effect and is considered negligible. As such, the es will assume friction one offactor, three regimes. The flow ƒ, isflow calculated using Equation 2-5. regime may be or in transition between laminar and turbulent. In laminar flow (2-5) 64 pes will assume one of three flow regimes. The flow may be r, Re, below 2000), fthe effect and is = pipe’s surface roughness has noregime Eq. 1-9 Re factor, nt in such, transition between laminar and turbulent. In laminar ble.or As the friction f, is calculated using Equation 1-9.flow er, Re, below 2000), the pipe’s surface roughness has no effect and is Where Reynolds number, Re, above the friction f, 1-9. is gible. As such, the friction factor,4000), f, is calculated usingfactor, Equation 64 dimensionless = < 2000 for laminar flow, see Equation 2-7 Re = Reynolds number, actors, the Reynolds number and pipe surface roughness. Eq. The 1-9 f > 4000 for turbulent flow, see Figure 2-1 Re 64 For turbulent f flow (Reynolds number, Re, above 4000), the friction Eq. factor, 1-9 ƒ, is w (Reynolds number, Re, above 4000), thenumber friction factor, f, roughness. is Re dependent on two factors, the Reynolds and pipe surface The friction factor may determined fromroughness. Figure 2-1, the Moody o factors, the resultant Reynolds number andbepipe surface The Diagram. (17) ow (Reynolds This number, Re, to above theto all friction factor, f, Moody is factor applies all kinds4000), of PE’s and pipe sizes . In the Diagram, wo factors, therelative Reynolds number The roughness, ε/d (seeand Tablepipe 2-1 forsurface ε) is usedroughness. which is the ratio of absolute 2

roughness to the pipe inside diameter. The friction factor may also be determined using the Colebrook formula. The friction factor can also be read from the Moody diagram with enough accuracy for calculation.

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 ε 2Figure °.51H ½°2.51 2.51 ctor may1 be determined 1-1, the ½°Moody Diagram, which can be 1 1 ° H from Eq. 1-10 Eq. 1-10 Eq. 1-10 = − + 2 log 2 log 2 log ®  ® ¾  ¾ 10 10 10 various fpipe f materials ' 3.7Re d°¯sizes.(17) ' 3.f7Re d°¿ ' fRe f °¯ 3.7 dand f °¿ the Moody Diagram, relative  In s, İ/d’ (see Table 1-4 for ɽ) is used. The friction factor may then be determined Chapter 6 Design of PE Piping Systems Colebrook formula. The friction factor can also be read from the Moody 1-10, sand 1-91-10, terms and 1-10, terms are as terms are previously as are previously as previously defined, defined, and: defined, and: and: with enough accuracy for calculation. esolute roughness, absolute roughness, roughness, ft. ft. ft. eynolds ds number, Reynolds number, dimensionless number, dimensionless dimensionless brook formula is: '' U Vd Vd ' Vd Vd Vd The Colebrook formula is: '' ρ Vd ' U Eq. 1-11 Eq. 1-11 Eq. 1-11 Re Re = Re = (2-6) v Hg 1 vP g vµ g° P 2.51 ½° Eq. 1-10 2 log10 ® ¾ f °¯ 3.7 d ' Re f °¿ 3126 Q 3126 Q3126 Q Eq. 1-12 Eq. 1-12 Eq. 1-12 Re Re = Re DI are k Das k D k ulas 1-9 and 1-10, terms previously defined, and: I I

171

For Formulas 2-5 and 2-6, terms are as previously defined, and:

= absolute roughness, ft. ft. (see Table 2-1) ε = absolute roughness, = Reynolds2 Rnumber, dimensionless number, dimensionless (see Equation 2-5) e = Reynolds nematic tic viscosity, kinematic viscosity, ftviscosity, /sec ft2/secft2/sec ' Vd 'U Vd Liquid flow a pipe occurs turbulent Eq. 1-11 * g inRe Γ g * g in one of three flow regimes. It can be laminar, P g and turbulent. The nature Q transition ν = between Q v laminar Eq. 1-13 Eq. Eq.depends 1-13 on or in of the1-13 flow

U

ρ

U

the pipe diameter, the density and viscosity of the flowing fluid, and the velocity 3 3of flow. value Q of a dimensionless combination of these parameters is 3126 nsity, id density, fluid lb/ft density, lb/ft lb/ft3 The numerical Eq. 1-12 Re 2 2 2 known aslb-sec/ft the Reynolds number DI k and the resultant value of this number is a predictor cnamic viscosity, dynamic viscosity, lb-sec/ft viscosity, lb-sec/ft of the nature of the flow. One form of the equation for the computing of this number nematic tic viscosity, kinematic viscosity, centistokes viscosity, centistokes centistokes is as follows:

=

2 z Q z z 3160 kinematic (2-7) viscosity, kRe =k ft= /sec k s k Di s s

Eq. 1-14 Eq. 1-14 Eq. 1-14

*g Q Eq. 1-13 cnamic viscosity, dynamic viscosity, centipoises viscosity, centipoises centipoises 3 Where 3 3 U ensity, uidliquid density, gm/cm density, gm/cm gm/cm Q = rate of flow, gallons per minute r, k = kinematic = fluid density, lb/ft3 viscosity, in centistokes (See Table 2-3 for values for water) r, Re, 2 in Di = internal diameter of pipe, ,=the dynamic viscosity, lb-sec/ft through ction loss kinematic through loss one through size one pipe size oneis pipe size known, is pipe known, the is known, friction the friction the lossfriction through loss through loss another through another another = viscosity, centistokes iameter ent er may diameter be may found be may found by: bethe found by: by: When friction loss through one size pipe is known, the friction loss through another pipe of different size z may be found by: Eq. 1-14 k 5 5 s 5 (2-8) § d ' · d ' § d ' · Eq. 1-15 Eq. 1-15 Eq. 1-15 h f 1 h ff 12 ¨¨= h1f 12¸¸ h1f 2¨¨ 1 ¸¸ = dynamic viscosity, dcentipoises © ' 23¹ d ' 2 © d ' 2 ¹ = liquid density, gm/cm The subscripts 1 and 2 refer to the known and unknown pipes. Both pipes must have the same surface roughness, and the fluid must be the same viscosity and have the through one size pipe is known, the friction loss through another same flow rate.

friction loss ferent diameter may be found by:

hf1

154-264.indd 171

§ d' h f 2 ¨¨ 1 © d '2

· ¸¸ ¹

5

Eq. 1-15

1/16/09 9:56:55 AM

172

Chapter 6


Table 2-1 Surface Roughness for Various New Pipes ‘ε’ Absolute Roughness of Surface, ft Type of Pipe

Values for New Pipe Reported by Reference (18)

Values for New Pipe and Recommended Design Values Reported by Reference (19) Mean Value

Recommended Design Value

Riveted steel

0.03 - 0.003

–

–

Concrete

0.01 – 0.001

–

–

Wood stave Cast Iron – Uncoated Cast Iron – Coated Galvanized Iron

0.0003 – 0.0006

–

–

0.00085

0.00074

0.00083

–

0.00033

0.00042

0.00050

0.00033

0.00042

Cast Iron – Asphalt Dipped

0.0004

–

–

Commercial Steel or Wrought Iron

0.00015

–

–

0.000005 corresponds to “smooth pipe”

–

–

Drawn Tubing Uncoated Stee

–

0.00009

0.00013

Coated Steel

–

0.00018

0.00018

Uncoated Asbestos – Cement

–

Cement Mortar Relined Pipes (Tate Process)

–

0.00167

0.00167

Smooth Pipes (PE and other thermoplastics, Brass, Glass and Lead)

–

“smooth pipe” ( 0.000005 feet) (See Note)

“smooth pipe” (0.000005) (See Note)

Note: Pipes that have absolute roughness equal to or less than 0.000005 feet are considered to exhibit “smooth pipe” characteristics.

Pipe Deflection Effects Pipe flow formulas generally assume round pipe. Because of its flexibility, buried PE pipe will deform slightly under earth and other loads to assume somewhat of an elliptical shape having a slightly increased lateral diameter and a correspondingly reduced vertical diameter. Elliptical deformation slightly reduces the pipe’s flow area. Practically speaking, this phenomenon can be considered negligible as it relates to pipe flow capacity. Calculations reveal that an elliptical deformation which reduces the pipe’s vertical diameter by 7% results in a flow reduction of approximately 1%.

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Head Loss in Fittings Fluids flowing through a fitting or valve will experience a friction loss that can be directly Chapter 6 173 expressed using a resistance coefficient, K’, for the particular Design fitting.(20) As shown in of PE Piping Systems the discussion that follows, head loss through a fitting can be conveniently added into system flow calculations as an equivalent length of straight pipe having the same diameter as system piping. Table 1-5 presents K´ factors for various fittings. Wh ere a pip elin e con tain sa lar ge nu mb er of fitti ngs in clo se pro xim ity Note for the Moody Diagram: D = pipe inside diameter, ft to each other, this simplified method of predicting flow loss may not be adequate due to Figure 2-1 The Moody Diagram the cumulative systems effect. Where this is a design consideration, the designer Head Loss in Fittings Fluids flowing through a fitting or valve will experience a friction loss that can be directly expressed using a resistance coefficient, K’, which represents the loss in terms of an equivalent length of pipe of the same diameter. (20) As shown in the discussion that follows, this allows the loss through a fitting to be conveniently added into the system flow calculations. Table 2-2 presents K’ factors for various fittings. Where a pipeline contains a large number of fittings in close proximity to each other, this simplified method of predicting flow loss may not be adequate due to the cumulative systems effect. Where this is a design consideration, the designer should consider an additional frictional loss allowance, or a more thorough treatment of the fluid mechanics. The equivalent length of pipe to be used to estimate the friction loss due to fittings may be obtained by Eq. 2-9 where LEFF = Effective Pipeline length, ft; D is pipe bore diameter in ft.; and K’ is obtained from Table 2-2. (2-9)

154-264.indd 173

LEFF = K’D

1/16/09 9:56:55 AM

ping Component

K’ 40

174 Chapter 6


21 6

ow

32

ow

27 Table 2-2 Representative Fittings Factor, K’, To Determine Equivalent Length of Pipe

ow

Piping Component

ow

90° Molded Elbow

w

45° Molded elbow

w

90° Fabricated Elbow (3 or more miters)

15° Molded Elbow 90° Fabricated Elbow (2 miters)

e

90° Fabricated Elbow (1 miters)

Run/Branch


Run/Run


60° Fabricated Elbow (2 or more miters) 45° Fabricated Elbow (2 or more miters) 30° Fabricated Elbow (2 or more miters)

21

K’

1640 21

116

524 30 6060 25

6016 15

2012 8

entional, Fully Open 30° Fabricated Elbow (1 miters)

340 8


145 60

entional, Fully Open Equal Outlet Tee, Run/Branch

in , Fully Open

entional Swing

Equal Outlet Tee, Run/Run Globe Valve, Conventional, Fully Open Angle Valve, Conventional, Fully Open Butterfly Valve, >8”, Fully Open Check Valve, Conventional Swing

6

20 40 340 145 135 40

135

- K values are based on Crane Technical Paper No 410-C

vation Change - K value for Molded Elbows is based on a radius that is 1.5 times the diameter. - K value for Fabricated Elbows is based on a radius that is approximately 3 times the diameter.

e lost or gained from a change in elevation. For liquids, the Head Loss Due to Elevation Change evation change is given by:

E

Line pressure may be lost or gained from a change in elevation. For liquids, the pressure for a given elevation change is given by: (2-10)

=

hE = h2 − h1

Eq. 1-17

Elevation head, ft of liquid where hE = Elevation head, ft of liquid

h1 = Pipeline elevation at point 1, ft h2 = Pipeline elevation at point 2, ft

If a pipeline is subject to a uniform elevation rise or fall along its length, the two points would be the elevations at each end of the line. However, some pipelines may have several elevation changes as they traverse rolling or mountainous terrain. These pipelines may be evaluated by choosing appropriate points where the pipeline slope changes, then summing the individual elevation heads for an overall pipeline elevation head.

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vation changes as they traverse rolling or mountainous terrain. These e conveying liquids and running full, pressure in the pipe due to elevation may be evaluated by choosing appropriate points where the pipeline slope her or not liquid is flowing. At any low point in the line, internal pressure will hen summing the individual elevation heads for an overall pipeline elevationChapter 6 the height of the liquid above the point multiplied by the specific Design weight of Systems of PE Piping If liquid is flowing in the line, elevation head and head loss due to liquid flow are added to determine the pressure in the pipe at a given point in the e conveying liquids and running full, pressure in the pipe due to elevation her or not liquid is flowing. At any low point in the line, internal pressure will the height of the liquid above theliquids pointand multiplied by pressure the specific weight of In a pipeline conveying running full, in the pipe due to If liquid is flowing in the line, elevation head and head loss due to liquid flow elevation exists whether or not liquid is flowing. At any low point in the line, are of added determine thewill pressure the pipe at liquid a given in multiplied the pressure be equal tointhe height of the abovepoint the point low Waterto–internal Hazen-Williams

175

by the specific weight of the liquid. If liquid is flowing in the line, elevation head and head loss due to liquid flow in the pipe are added to determine the pressure in the method of flow resistance calculation may be applied to liquid and pipe at a given point in the pipeline.

Weisbach its solution can be complex. For many applications, empirical formulas are nd, when used within their limitations, reliable results are obtained with low of Water –Pressure Hazen-Williams Flow of and Water – Hazen-Williams venience. For example, Hazen Williams developed Equation an empirical formula of water in pipes 60° F. TheatDarcy-Weisbach method of flow resistance calculation may be applied to liquid Weisbach method of flow calculation be applied to liquid andformulas and gases, butresistance its solution can be complex.may For many applications, empirical its solution canarebeavailable complex. For many applications, empirical formulas are and, within their results obtained -Williams formula for water at when 60°Fused (16°C) can belimitations, applied reliable to water andare other nd, when usedwith within limitations, reliable results are obtained with 2 greatertheir convenience. For example, Hazen and Williams developed an empirical /sec), or ng the same viscosity of 1.130 centistokes (0.00001211 ft venience. Forkinematic example, Hazen and Williams developed an empirical formula formula for the flow of water in pipes at 60º F. The viscosity of water with temperature, so some error can occur at of water in pipes at 60°varies F. The Hazen-Williams formula for water at 60º F (16ºC) can be applied to water es other than 60°F (16°C). and other liquids having the same kinematic viscosity of 1.130 centistokes which

2 -Williams formula for 0.00001211 water at ft 60°F (16°C) can be applied to waterThe and other of /sec or 31.5 SSU (Saybolt Second Universal). viscosity equals 2 water varies with temperature, some error can occur at temperatures other ng the same kinematic viscosity of 1.130socentistokes (0.00001211 ft /sec), or ams viscosity formula for friction (head)with loss temperature, in feet: 60ºF (16ºC). The ofthan water varies so some error can occur at es other than 60°F (16°C).formula for friction (head) loss in feet of water head: Hazen-Williams

(2-11)

1.85

0.002083 L § 100 Q · hf ¸ ¨ 4.8655 DI © C ¹ ams formula for friction (head) loss in feet: ams formula forHazen-Williams friction (head) loss in(head) psi:loss in psi: formula for friction (2-12)

Eq. 1-18

1.85

0.0009015L 0.002083 L  100 Q  hpf =  1.85  4.8655 0.0009015 L§ 100 C Q· DI pf ¸ ¨ 4.8655 DI © C ¹ are as previously ams formula forTerms friction (head)defined, loss and: in psi:

Eq. 1-18

Eq. 1-19

hf = friction (head) loss, ft. of water. pf = friction (head) loss, psi 1.85 0.diameter, 0009015 DI = pipe inside in L  100 Q  Eq. 1-19 pf =   4.8655 C = Hazen-Williams c = 150-155 for PE , (not related to Darcy-Weisbach C  DFriction Factor,  dimensionless

friction factor, ƒ)I

Q = flow rate, gpm

The Hazen-Williams Friction Factor, C, for PE pipe was determined in a hydraulics laboratory using heat fusion joined lengths of pipe with the inner bead present. Other forms of these equations are prevalent throughout the literature.(21) The reader is referred to the references at the end of this chapter.

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176

Chapter 6


Table 2-3 Properties of Water Temperature, °F/°C

Specific Weight, lb/ft3

Kinematic Viscosity, Centistokes

32 / 0

62.41

1.79

60 / 15.6

62.37

1.13

75 / 23.9

62.27

0.90

100 / 37.8

62.00

0.69

120 / 48.9

61.71

0.57

140 / 60

61.38

0.47

14 14 14

14

Water flow through pipes having different Hazen-Williams factors and different

equation ng equation 1-21 1-21 and and C =1-21 C150, =C 150, Using Using equation equation 1-21 and and = 150, C = 150, flow diameters may be determined using the following equations: (2-13)

pf

1.85

1.85

1.85 1.85 0.0009015 0.0009015 (15000 ) §(15000 100 ) D 100 ((50 ·50) §(50 C .0009015 0(.15000 0009015 15000 )I2§)(100 100 (350 ).3· psi 2) ·= p f =p f %p0 flow 11 . 11 psi ¸  ¨  ! = 100 4D 11.3 psi ¸ 11.3 psi ¨ ¨ ¸ .86554.8655 150 3.938 3.f4938 I1  .8655 3.93834©..8655 938150 ¹ ¹ © ¹ 150 ©C 1150

ermine determine the elevation elevation head, head, assume assume point point 1 is 1atisthe atisthe bottom the ofbottom the To determine Tothe determine the elevation the elevation head, head, assume assume point point 1 at 1 bottom is theatof bottom the of theof the Where the subscripts 1 and 2 refer to the designated properties for two separate pipe on, vation, and and point point 2 is 2 at is the at the top. top. Using Using Equation Equation 1-17, 1-17, elevation, elevation, and profiles, point and point 2inisthis at 2 case, is theattop. the top. UsingUsing Equation Equation 1-17,1-17, the pipe inside diameter (D in inches) of the one pipe (1) versus I

that of the second pipe (2) and the Hazen-Williams factor for each respective profile.

hE h150 E 0 −150 0 = ft0of ft150 water h150 0of ftwater 150 of water ft of water E =h150 E 150

Pipe Flow Design Example

3 3 3is a is 3pressure eecific specific weight weight of water ofweight water 60°F at is 60°F 62.37 isat62.37 lb/ft , which pressure of 62.37 lb 62.37 The specific The specific weight ofatwater of 60°F water at 60°F islb/ft 62.37 is, which 62.37 lb/ft ,lb/ft which ,a which is a pressure isofa 62.37 pressure oflb62.37 of lb lb 2 2 A PE pipeline conveying water at 60°F is 15,000 feet long and is2 laid 2on a uniform er1over aft 1square ft square area, area, or a or pressure a pressure of 62.37 of 62.37 / 144 / 144 = 0.43 = 0.43 lb/in lb/in . Therefore, . Therefore, over a 1 ftasquare 1 ft square area,area, or a pressure or a pressure of 62.37 of 62.37 / 144/ =144 0.43 = 0.43 lb/in lb/in . Therefore, . Therefore,

grade that rises 150 feet. What is the friction head loss in 4” IPS DR 17 PE 3408 pipe 14 14 14 for a 50 gpm flow? What is the elevation head? What is the internal pressure at the hE of hE150 =hpipe 150 0 h0when −150 .43 0 0.water 43 64 .05.64 psi.5 psi 150 0640=..543 0flowing 43 64.5 psi bottom the ispsi uphill? When flowing downhill? When full E E not flowing? uation uation 1-21 1-21 and andbut C C= = 150, 150,

( )

Using equation 1-21 and C =head 150, water en water iswater flowing, is water flowing, elevation elevation and the the friction friction head are are added. added. Theadded. The The The When When isUsing flowing, is flowing, elevation head head and and the friction thehead friction head head are added. are equation 2-12 and Celevation =head 150and um ximum friction friction head head acts acts at the at the source source point, point, and and the the maximum maximum elevation elevation head head 1 . 85 1 1 . 85 . 85 maximum maximum friction friction headhead acts acts at theatsource the source point,1point, and the andmaximum the maximum elevation elevation headhead .85 00..0009015 0009015 (15000 ))when 100 ((50 50)))··§uphill, §§100 0(Therefore, .15000 0009015 (flowing 15000 100uphill, (flowing 50 ).3uphill, ·psi he owest point. Therefore, Therefore, when flowing the the pressure, pressure, P, at P, the at the bottom bottom at lowest theat lowest the lowest point. point. Therefore, when when flowing uphill, the pressure, the pressure, P, at P, the at bottom the bottom pppoint. 11 11 . 3 psi ¸ ¸ ¨ ¨ p f 444.8655 11.3 psi fff 8655 ¸ ¨ ..8655 8655 938 150 150 33the ..938 ¹¹© head ©©4.head ation levation head head plus plus the friction friction head because because thebecause the flow flow is from is from the the bottom bottom to the to the 150 3.938 is elevation is elevation head head plus plus the friction the friction head because the flow the flow is from is from the bottom the bottom to the to the ¹ . top. top. mine mine the the elevation elevation head, head, assume assume point 11 is ispoint at at the the1bottom bottom of ofbottom the the of the To determine the elevation head,point assume is at the To determine the elevation head, assume point 1 is at the bottom of the elevation, ,elevation, and and point pointand 22 is ispoint at at the the2top. top. Using Using Equation Equation 1-17, 1-17, is2 at the top. Using Equation 1-17, is at the Using Equation 2-10, PandP hpoint .h5ptop. 64 11 . 5 + . 3 11 . 75 3 = . 8 75 psi . 8 psi PEp f+hpPE64 p 64 . 5 64 11 . 5 . 3 11 75 . 3 . 8 psi 75.8 psi E=h f = Ef f hhEEE 150 150h00E 150 150 ft ft of of water water 150 0 150 ft of water flowing en flowing downhill, downhill, water water flows flows from from the the topfrom top to the to bottom. bottom. Friction Friction head head When When flowing flowing downhill, downhill, waterwater flows flows from the top thethe to topthe to bottom. the bottom. Friction Friction headhead 333 3 3 lb/ft cific The specific weight of water at 60°F is 62.37 (see Table 2-3), which, for each foot ific ific weight weight of of water water at at 60°F 60°F is is 62.37 62.37 lb/ft lb/ft , , which which is is a a pressure pressure of of 62.37 62.37 lb lb plies from from thefrom the source source point point atpoint the the top, top, so so the pressure pressure developed developed from from the the The specific weight of water at 60°F is lb/ftthe , which is a pressure of 62.37 lb the applies applies from the source the source point at the at62.37 top, the top, so so pressure the pressure developed developed from from the 222 2a pressure of 62.37/144 of head exerts a pressure of 62.37 lb over a 1 ft square area, or ft t square square area, area, or or a a pressure pressure of of 62.37 62.37 / / 144 144 = = 0.43 0.43 lb/in lb/in . . Therefore, Therefore, wnhill lldownhill flow flow applied is flow applied in the in athe opposite opposite direction asdirection the the elevation elevation head. head. over adownhill 1 is ft square pressure ofdirection 62.37 /direction 144 =as 0.43 lb/in . the Therefore, flow is area, applied is or applied in the in the opposite opposite as the as elevation elevation head. head. = 0.43 lb/in2. Therefore, for a 150 ft. head, erefore, ore, Therefore, Therefore, 150hE00 00..43 hhEEE 150 43 64 64 psi psig 150 0. .505.psi 43 64.5 psi P Ph E=P h Ep − pE 64 .11 564 −.f311 53 311 =..52.53 .253 psi h.p5E64 .5 . 64 3psi 11 .3 .2 psi 53 .2 psi f hP f = f p ater ater is is flowing, flowing, elevation head head and andhead the the friction friction head are arehead added. added. The When water is elevation flowing, elevation and thehead friction are The added. The m mmaximum friction friction head head acts acts at at the the source source point, point, and and the the maximum maximum elevation elevation head head acts at and the maximum elevation en he the pipe pipe is friction full, is but but water iswater not is the not flowing, flowing, no friction no friction head head develops. develops. When When the pipe thefull, pipe ishead full, iswater but full, but water is source notisflowing, notpoint, flowing, no friction no friction head head develops. develops.head west point. Therefore, when flowing flowing uphill, theuphill, pressure, P, at at the the bottom bottom estthe point. Therefore, when the pressure, P, at lowest point. Therefore, whenuphill, flowing the pressure, P, at the bottom on on head head plus plus the the friction friction head head because because the the flow flow is is from from the the bottom bottom to to the the is elevation head 154-264.indd 176 plus the friction head because the flow is from the bottom to the 1/16/09 9:56:56 AM P P h =hp + p 64=.564 .05 + 64 0 =.564 psi .5 psi

hE 150 0 0is .4362.37 64.5lb/ft psi3, which is a pressure of 62.37 lb pecific weight of water at 60°F 2 a 1 ft square area, orh a pressure 150 0 0of.4362.37 64.5/ 144 psi = 0.43 lb/in . Therefore, E water is flowing, elevation head and the friction head are added. The Chapter 6 Design of PE Piping Systems friction acts at and the maximum elevation numwater is head flowing, head are added.head The )0.43point, hEelevation = (the 150 source − 0head =and 64.5the psi friction owestfriction point. head Therefore, flowingpoint, uphill,and thethe pressure, P, at the bottom mum acts atwhen the source maximum elevation head head plus Therefore, the elevation friction when head because thefriction flow from bottom the eation lowest point. flowing the is pressure, P, at theto bottom water is flowing, head and uphill, the head the are added. The vation head head plus the because the the flowmaximum is from the bottom to the mum friction actsfriction at the head source point, and elevation head lowest point. Therefore, flowing uphill, pressure, P, at bottom When water iswhen flowing, elevation head the and the friction head arethe added. The Pmaximum hE friction pfriction 64 . 5 11 . 3 75 . 8 psi vation head plus the head because the flow is from the bottom to the head head acts at the source point, and the maximum elevation f atPthe h lowest point.64 Therefore, .5 11.3when 75.flowing 8 psi uphill, the pressure, P, at the bottom is E pf elevation plus from the friction because flow is from the bottom flowing downhill, waterhead flows the head top to the the bottom. Friction headto the top. from the source point at the top, so the pressure developed from the P = hwater = 64.5from + 11.3the = 75top .8 psig psi n flowing downhill, to the bottom. Friction head E + p f flows ll flow theat opposite direction as thedeveloped elevation from head. es from is theapplied source in point the top, so the pressure the ore, nhill flow downhill, is applied in flows the opposite direction the head. Whenwater flowing downhill, water theas top to the elevation bottom. Friction head flowing from theflows topfrom to the bottom. Friction head efore, appliespoint from the pointso at the so the pressure developed from the the es from the source at source the top, thetop,pressure developed from is applied in the opposite direction as the elevation head. Therefore, P downhill h E in pflow 64 . 5 11 . 3 53 . 2 psi hill flow is applied the opposite direction as the elevation head. f fore, P h E p f 64 .5 11 .3 53 .2 psig psi he pipe is full, but water is not flowing, no friction head develops. full, but water not no friction head develops. PWhen = h Ethe − pipe p f is =is not 64 .5flowing, − 11 .3 =isno 53 .friction 2flowing, psi head n the pipe is full, but water develops. P hE p f 64.5 0 64.5 psig psi the pipe is full, but water is not flowing, no P hE p f 64.5 0 64.5friction psi head develops. derations Pressure Flow of Liquid Slurries siderations Liquid P = hslurry + ppiping = 64systems .5 + 0 =transport 64.5 psisolid particles entrained in a liquid carrier.

177

m must be designed forE continuous operating pressure and for transient (surge) f Water application. is typically used as a liquid carrier,and and solid particles are commonly osed by the particular Surge allowance effects vary granular tem must be designed for continuous operating pressuretemperature and for transient (surge) iderations materials such as sand, conclusions fly-ash or coal.may Key design considerations involve the nature erial to pipe material, andapplication. erroneous be drawn when comparing posed by the particular Surge allowance and temperature effects vary lass (PC) of different pipe materials. of the solid material, it’s particle size and the carrier liquid. aterial to pipe material, and erroneous conclusions may be drawn when comparing em must be designed for continuous operating pressure and for transient (surge) Class by (PC) different pipe Turbulent flowmaterials. is preferred to ensure that particles are suspended in thevary liquid. posed theofparticular application. Surge allowance and temperature effects Turbulent flow also reduces pipeline wear because particles suspended in terial to pipe material, and erroneous conclusions may be drawn when comparing andle temporary pressure surges is a major advantage of polyethylene. Due tothe carrier Class (PC)ofofpolyethylene, different pipebounce materials. liquid will off thesystem pipe inside surface. pipe has viscoelastic properties chandle nature piping safelyPEwithstand momentarily temporary pressurea surges is a majorcan advantage of polyethylene. Due to that are combine with high molecular weight toughness to provide service life um pressures that significantly above the pipe’s PC. The strain from an that can stic nature of polyethylene, a piping system can safely withstand momentarily significantly exceed many metal piping materials. Flow velocity that is too low to ited load of short that duration met with anabove elasticthe response, whichThe is relieved uponan mum are is significantly pipe’s PC. strain from handlepressures temporary pressure surges is a major advantage of polyethylene. Due to maintain fully turbulent flow for a given particle size can allow solids to drift to the mited load of duration isa met withsystem an elastic response, which is relieved upon tic nature of short polyethylene, piping can safelyHowever, withstand momentarily bottom of the pipe and slide along the surface. compared to metals, PE is a mum pressures that are significantly above the pipe’s PC. The strain from an softer material. Under sliding bed and direct impingement conditions, PE may wear mited load of short duration is met with an elastic response, which is relieved upon appreciably. PE directional fittings are generally unsuitable for slurry applications because the change of flow direction in the fitting results in direct impingement. Directional fittings in liquid slurry applications should employ hard materials that are resistant to wear from direct impingement. Particle Size As a general recommendation, particle size should not exceed about 0.2 in (5 mm), but larger particles are occasionally acceptable if they are a small percentage of the solids in the slurry. With larger particle slurries such as fine sand and coarser particles, the viscosity of the slurry mixture will be approximately that of the carrying liquid. However, if particle size is very small, about 15 microns or less, the slurry viscosity will increase above that of the carrying liquid alone. The rheology

154-264.indd 177

1/16/09 9:56:57 AM

178

Chapter 6


of fine particle slurries should be analyzed for viscosity and specific gravity before determining flow friction losses. Inaccurate assumptions of a fluid’s rheological properties can lead to significant errors in flow resistance analysis. Examples of fine particle slurries are water slurries of fine silt, clay and kaolin clay. Slurries frequently do not have uniform particle size, and some particle size nonuniformity can aid in transporting larger particles. In slurries having a large proportion of smaller particles, the fine particle mixture acts as a more viscous carrying fluid that helps suspend larger particles. Flow analysis of non-uniform particle size slurries should include a rheological characterization of the fine particle mixture. Solids Concentration and Specific Gravity Equations 2-14 through 2-17 are useful in determining solids concentrations and 23 23 23 23specific mixture specific gravity. Tables 2-4, 2-5, and 2-6 provide information about gravity and particle size of some slurries.

SS −SS CCV V= CCMV M SSMLMLSSLL SVSS S−SSSSL LSS

Eq.1-31 1-31 Eq. Eq. 1-31 1-31 Eq.

CC SSS C SCVSS S VV V S CC CWW WW= C SSMM SS

Eq.1-32 1-32 Eq. Eq. 1-32 1-32 Eq.

(2-14)

SS

(2-15)

LL

MM

SCS SV−SSSSLSL)+ SSSSLL L SS L SMM= SC S(2-16) SC V V( SC M M

V

S

L

L

SSL L SS L (2-17) L SSMM= SS M M CC ( SS S−SSSSL L) SS C WW S C LL 11− 11 WW SS SSS S SS

Eq.1-33 1-33 Eq. Eq. 1-33 1-33 Eq. Eq.1-34 1-34 Eq. Eq. 1-34 1-34 Eq.

SS

:

Where

Sspecific liquid specific gravity liquid gravity carrier liquid specific gravity L = carrier =carrier carrier liquid specific gravity = carrier liquid specific gravity S = solids specific gravity Sgravity solids specific gravity solids specific = solids specific gravity = solids specific gravity Sspecific = slurry mixture specific gravity Mspecific mixture gravity slurry mixture gravity =slurry slurry mixture specific gravity = slurry mixture specific gravity Cconcentration = percent solids concentration by volume V concentration percent solids byvolume volume percent solids by = percent solids concentration by volume volume = percent solids concentration by CW = percent solids concentration by weight percent solids concentration by weight percent solids concentration by weight percent solids solids concentration concentration by by weight weight == percent

ty ty

Critical Velocity

As pointed out above, turbulent flow is preferred to maintain particle suspension. A turbulent flow regime avoids the formation of a sliding bed of solids, excessive pipeline wear and possible clogging. Reynolds numbers above 4000 will generally insure turbulent flow.

bove, turbulent flowflow ispreferred preferred maintain particle suspension. ove, turbulent flow is totomaintain particle suspension. AA AA ut above, turbulent flow is preferred preferred to maintain maintain particle suspension. ut above, turbulent is to particle suspension. ime avoids the formation of a sliding bed of solids, excessive pipeline me avoids the formation of a sliding bed of solids, excessive pipeline regime avoids avoids the the formation formation of of aa sliding sliding bed bed of of solids, solids, excessive excessive pipeline pipeline regime ble clogging. Reynolds numbers above 4000 will generally insure e clogging. Reynolds numbers above 4000 will generally insure ossible clogging. clogging. Reynolds Reynolds numbers numbers above above 4000 4000 will will generally generally insure insure ossible . 154-264.indd 178

1/16/09 9:56:58 AM

ut above, turbulent flow is preferred to maintain particle suspension. A regime avoids the formation of a sliding bed of solids, excessive pipeline Chapter 6 of PE Piping Systems ossible clogging. Reynolds numbers above 4000 will generallyDesign insure .

179

he flow velocity of a slurry at about 30% above the critical settlement ood practice. This insures that the particles will remain in suspension the flow velocity wear. of a slurry about 30% above the critical critical settlement ing the potentialMaintaining for excessive pipeline Forathorizontal pipes, velocity is a good1-35. practice. This insures that the particles will remain in suspension be estimated using Equation thereby avoiding the potential for excessive pipeline wear. For horizontal pipes,

perience with this equation relationships critical velocityvaries. may be Other estimated using Equationare 2-18.offered in the e Thompson and Aude (26). A test section may be installed to verify Individual experience with this equation varies. Other relationships are offered in f this equation for specific projects.

24 to verify the literature. See Thompson and Aude (26). A test section may be installed applicability of this equation for specific projects. (2-18)

FL 2 gd ' S S 1

VC

Eq. 1-35

inimum velocitydefined for fineand particle slurries (below 50 microns, 0.05 mm) are previously 24 terms previously definedA andguideline minimum velocity for ovided turbulentWhere flow isaremaintained. V = critical settlement velocity, ft/sec C ries (oversettlement 150 microns, 0.15 mm) critical velocity, ft/secis provided by Equation 1-36. FL = velocity coefficient (Tables 2-7 and 2-8) velocity coefficient (Tables 1-11 and 1-12) d’ = pipe inside diameter, ft pipe inside diameter, ft 14 d ' V Eq. 13650 microns, min particle minimum velocity for fine slurries (below inimum velocityAnforapproximate fine slurries (below 50particle microns, 0.05 mm)

mm)isismaintained. 4 to 7 ft/sec, provided turbulent flow is maintained. guideline ovided turbulent0.05 flow A guideline minimum velocityAfor minimum velocity for larger particle slurries (over 150 microns, 0.15 mm) is provided ies (over 150 microns, 0.15 mm) is provided by Equation 1-36. by Equation 2-19.

(2-19) velocity, ft/sec proximate minimum V = 14 d '

Eq. 1- 36

min

Where

minimum velocity, min = approximate t velocity and Vminimum velocity for ft/sec turbulent flow increases with Critical settlement velocity1-37 and minimum velocity for turbulent flow increases with ore. The relationship in Equation is derived from the Darcyproximate minimum velocity, ft/sec increasing pipe bore. The relationship in Equation 2-20 is derived from the Darcyn.

Weisbach equation. (Equation 2-3) (2-20)

d '2 velocity and minimum velocity for turbulent flow increases Eq. 1-with 37 V2 V1 ore. The relationship in Equation 1-37 is derived from the Darcyd '1 n. The subscripts 1 and 2 are for the two pipe diameters.

V2 =

d '2

V1 d '1 nd 2 are for the two pipe diameters.

Eq. 1- 37

Table 1-8: Scale of Particle Sizes U.S. Standard Mesh

Inches

nd 2 are for the two pipe 1.3 diameters. – 2.5 Table 1-8:

5 U.S. Standard 10 Mesh 18 35 60 154-264.indd

179

Microns

33,000 – 63,500 0.6 – 1.3 15,200 – 32,000 0.321 8,000 Scale of Particle Sizes 0.157 4,000 0.079 2,000 Inches Microns 0.039 1,000 1.3 – 2.5 33,000500 – 63,500 0.0197 0.6 – 1.3 15,200250 – 32,000 0.0098

Class Very coarse gravel Coarse gravel Medium gravel Fine gravel VeryClass fine gravel Very coarse sand Very coarsesand gravel Coarse Coarse Mediumgravel sand

1/16/09 9:56:58 AM

180

Chapter 6


Table 2-4 Scale of Particle Sizes Tyler Screen Mesh

U.S. Standard Mesh

Inches

Microns

Class Very coarse gravel

–

–

1.3 – 2.5

33,000 – 63,500

–

–

0.6 – 1.3

15,200 – 32,000

Coarse gravel

2.5

–

0.321

8,000

Medium gravel

5

5

0.157

4,000

Fine gravel

9

10

0.079

2,000

Very fine gravel Very coarse sand

16

18

0.039

1,000

32

35

0.0197

500

Coarse sand

60

60

0.0098

250

Medium sand

115

120

0.0049

125

Fine sand

250

230

0.0024

62

Very fine sand

400

–

0.0015

37

Coarse silt

–

–

0.0006 – 0.0012

16 – 31

Medium silt

–

–

–

8 – 13

Fine silt

–

–

-

4–8

Very fine silt

–

–

–

2–4

Coarse clay

–

–

–

1–2

Medium clay

–

–

–

0.5 - 1

Fine clay

Table 2-5 Typical Specific Gravity and Slurry Solids Concentration (Water Slurries)

154-264.indd 180

Typical Solids Concentration

Material

Specific Gravity

Gilsonite

1.05

40 –45

39 – 44

Coal

1.40

45 – 55

37 – 47

% by Weight

% by Volume

Sand

2.65

43 – 43

23 – 30

Limestone

2.70

60 – 65

36 – 41

Copper Concentrate

4.30

60 – 65

26 – 30

Iron Ore

4.90

–

–

Iron Sands

1.90

–

–

Magnetite

4.90

60 - 65

23 - 27

1/16/09 9:56:58 AM

Chapter 6 181


Table 2-6 Water-Base Slurry Specific Gravities Concentration by Weight Percent, CW

Solid Specific Gravity, SS 1.4

1.8

2.2

2.6

3.0

3.4

3.8

4.2

4.6

5.0

5

1.01

1.02

1.03

1.03

1.03

1.04

1.04

1.04

1.04

1.04

10

1.03

1.05

1.06

1.07

1.07

1.08

1.08

1.08

1.08

1.09

15

1.04

1.07

1.09

1.10

1.11

1.12

1.12

1.13

1.13

1.14

20

1.05

1.10

1.12

1.14

1.15

1.16

1.17

1.18

1.19

1.19

25

1.08

1.13

1.16

1.18

1.20

1.21

1.23

1.24

1.24

1.25

30

1.09

1.15

1.20

1.23

1.25

1.27

1.28

1.30

1.31

1.32

35

1.11

1.18

1.24

1.27

1.30

1.33

1.35

1.36

1.38

1.39

40

1.13

1.22

1.28

1.33

1.36

1.39

1.42

1.44

1.46

1.47

45

1.15

1.25

1.33

1.38

1.43

1.47

1.50

1.52

1.54

1.56

50

1.17

1.29

1.38

1.44

1.50

1.55

1.58

1.62

1.64

1.67

55

1.19

1.32

1.43

1.51

1.58

1.63

1.69

1.72

1.76

1.79

60

1.21

1.36

1.49

1.59

1.67

1.73

1.79

1.84

1.89

1.92

65

1.23

1.41

1.55

1.67

1.76

1.85

1.92

1.98

2.04

2.08

70

1.25

1.45

1.62

1.76

1.88

1.98

2.07

2.14

2.21

2.27

Table 2-7 Velocity Coefficient, FL (Uniform Particle Size) Velocity Coefficient, FL

Particle Size, mm

CV = 2%

CV = 5%

CV = 10%

CV = 15%

0.1

.76

0.92

0.94

0.96

0.2

0.94

1.08

1.20

1.28

0.4

1.08

1.26

1.41

1.46

0.6

1.15

1.35

1.46

1.50

0.8

1.21

1.39

1.45

1.48

1.0

1.24

1.04

1.42

1.44

1.2

1.27

1.38

1.40

1.40

1.4

1.29

1.36

1.67

1.37

1.6

1.30

1.35

1.35

1.35

1.8

1.32

1.34

1.34

1.34

2.0

1.33

1.34

1.34

1.34

2.2

1.34

1.34

1.34

1.34

2.4

1.34

1.34

1.34

1.34

2.6

1.35

1.35

1.35

1.35

2.8

1.36

1.36

1.36

1.36

≥ 3.0

1.36

1.36

1.36

1.36

154-264.indd 181

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182

Chapter 6


1-12: Velocity Coefficient, FL (50% Passing Particle Size) CV = 5% 0.48 0.58 0.70 0.77 0.83 0.85 0.97 1.09 1.15 1.21 1.24 1.33 1.36

Velocity Coefficient, FL CV = 10% CV = 20%

CV = 30%

0.48 0.48 Table 2-8 0.59 1.60 Velocity Coefficient, FL (50% Passing Particle Size) 0.72 0.79 0.86 CV = 5% 0.88 0.48 1.00 0.58 1.13 0.70 1.21 0.77 1.25 0.83 1.29 0.85 1.36 0.97 1.38 1.09

Particle Size, mm 0.01 0.02 0.04 0.06 0.08 0.10 0.20 0.40

0.74

Velocity Coefficient, 0.81 FL CV = 10% 0.48 0.59 0.72 0.79 0.86 0.88 1.00 1.13

C0.86 V = 20%

0.92 0.48 1.05 0.60 1.18 0.74 1.26 0.81 1.31 0.86 1.33 0.92 1.38 1.05 1.39 1.18

CV = 30% 0.48 0.61 0.76 0.83 0.91 0.95 1.08 1.23

0.60

1.15

1.21

1.26

1.30

0.80

1.21

1.25

1.31

1.33

1.0

1.24

1.29

1.33

1.35

2.0

1.33

1.36

1.38

1.40

3.0

1.36

1.38

1.39

1.40

0.48 0.61 0.76 0.83 0.91 0.95 0.18 1.23 1.30 1.33 1.35 1.40 1.40

arcy-Weisbach, and Equations 1-18 and 1-19, Hazen-Williams, may be 27 Equation and flows Equations 2-11 and 2-12, may be ne friction head loss 2-3, for Darcy-Weisbach, pressure slurry provided the Hazen-Williams, viscosity to determine friction Elevation head loss forhead pressure slurry flows provided equations are used taken into account. loss is increased by the viscosity limitations of the equations are taken into account. Elevation head loss is increased ity of the slurry mixture. Equations for High Pressure Gas Flow by the specific gravity of the slurry mixture.

(2-21) 27 h2 h1 equations used in industry hE areS Mempirical Eq. for 1- 38 s 1-39 through 1-42 pressure han 1 psig. (26) Calculated results may vary due to the assumptions inherent Compressible Gas Flow of the livation Equations forequation. High Pressure Gas Flow

Flow equations for smooth pipe may be used to estimate compressible gas flow

Gas Flow through PE pipe. squation 1-39 through 1-42 are empirical equations used in industry for pressure Equations for High Gas to Flow han 1 psig. (26)Empirical Calculated results mayPressure vary due the assumptions inherent or smooth pipe may be used to estimate compressible gas flow through 0 . 575 rivation of the equation. 2 empirical 2 2.725 Equations 2-22 through 2-25 are equations used in industry for pressure e. § p1 p 2 · 2826 DI ¸ vary due to the assumptions ¨ results may Eq.inherent 1-39 in Qhthan 1 psig.0.425 greater Calculated ¸ ¨ L S g ¹ © the derivation of the equation. Equation Mueller Equation (2-22)

Qh =

h Equation

2826 DI Sg

2.725

0.425

0.575

 p1 2 − p 2 2   L 

   

§ p1 2 p 2 2 ¨ ¨ L ©

· ¸ ¸ ¹

Eq. 1-39

Weymouth Equation (2-23)

Qh

th Equation

bution Equation 154-264.indd 182

Qh =

2034 DI Sg

2.667

0.5

2034 DI Sg

2.667

Eq. 1-40

0.5

 p1 2 − p 2 2      L  2  2 0.555 · §

2.667

0.5

0.5

Eq. 1-40 1/16/09 9:56:59 AM

th Equation

 p1 2 − p 2 2  Qh = 0.5 2  2 0L.5 2.667 S g§ p1 p 2 · 2034 DI ¨ ¸ 0.5 ¨ ¸ L Sg © ¹ 2034 DI

Qh

2.667

   

0.5

Eq. 1-40 Chapter 6 183

Eq. 1-40


T Distribution Equation

ibution Equation

 p1 2 − p 2 2  Qh = 0.444 2  2 0.555 2.667 S § 2679 DI g p1 p 2 · L ¸ ¨ 0.444 ¸ ¨ L Sg ¹ ©

IGT Distribution Equation (2-24)

Qh

2679 DI

2.667

   

0.555

Eq. 1-41 Eq. 1-41

itzglass Equation

s Equation

Spitzglass Equation (2-25)

  2 2 5   D I · 0.5 3410  p1 − p§ 2   Qh =   5 3.6 ¸ 20.5  2 0.5 ¨ 3410 §¨ Sp1g p 2 ·¸ L ¨  D + 0¸.03 D I 1+ I Qh DI 0.5 ¨ ¸  ¨ ¸ 3 . 6 L Sg © ¹ ¨1 0.03 D I ¸ DI Where: © ¹ Where 0.5

(Equations 2-22 through 2-25) 3

     

0.5

Eq. 1-42 Eq. 1-42 28

28

flow, standard ft /hour here: Qh = flow, standard ft 3/hour Sg = Qh =gas specific 3 gravity Sg standard = gas specific gravity Qh =p flow, ft /hourlb/in2 absolute = inlet pressure, 1 2 absolute p = inlet pressure, lb/in Sg =p gas gravity 2 1 = pspecific outlet pressure, lb/in absolute 2 pirical Equations for Low Pressure Gas Flow = outlet pressure, lb/in22 absolute absolute Equations Low Pressure Gas Flow p1 = L for inlet lb/in 2pressure, = L = length, length, ft 2 ft p2 =D outlet lb/in absolute = D =pressure, pipe inside diameter, in I inside diameter, in I pipe applications where internal pressures arethan less 1than psig, as such as landfill L = length, ft ations where internal pressures are less psig,1 such landfill gas gas hering or wastewater odor control, Equations 1-43 or 1-44 may be used. = pipe inside diameter, in D I or wastewater odor control, Equations or 1-44 Empirical Equations for Low1-43 Pressure Gasmay Flow be used.

eller Equation quation

For applications where internal pressures are less than 1 psig, such as landfill gas gathering or wastewater odor control, Equations 2-26 or 2-27 may be used. Mueller Equation 2.725

(2-26)

Qh

0.575 2.725 D 2971 2971 § hI1 h§¨2 h·1 h2 ·¸ Qh D I ¨0.425 © ¸ L ¹ 0.425 S g© L ¹ Sg

0.575

Eq. 1-43 Eq. 1-43

zglass Equation Equation

Spitzglass Equation (2-27)

Qh

0.5 § · § ¨ 0 . 5 5 3350h§ h·10.5¨h2 · ¨D 5 DI ¸ 3350 I 2 ¸ ¨ Qh §¨ h1 ¨ ¸ 0.5 ¸ L ¹ ¨ 3 . 6 0.5 S © ¨ 3 . 6 S g © g L ¹ 1 ¨ 10.03 D 0¸.03 DI ¨ I ¸ D I DI© ¹ ©

· ¸ ¸ ¸ ¸ ¹

0.5

Eq. 1-44 Eq. 1-44

Where terms are previously defined, and

h1are = inletpreviously pressure, in H2O defined, and terms ereWhere terms are previously defined, and h2 = outlet pressure, in H2O = pressure, inlet pressure, h1 =h1 inlet in H2Oin H2O = outlet pressure, h2 =h2 outlet pressure, in H2Oin H2O

seation Permeation

154-264.indd 183

1/16/09 9:57:00 AM

inlet pressure, in H2O outlet pressure, in H2O 184 Chapter 6


Permeation gasses may deliver slightly less gas due pelines carryingGas compressed on through the Long pipedistance wall. Permeation losses are small, but deliver it mayslightly be less gas pipelines carrying compressed gasses may inguish between losses andthe possible leakage. 1- but it duepermeation to gas permeation through pipe wall. PermeationEquation losses are small, ed to determine volumeto distinguish of a gasbetween that will permeate through maythe be necessary permeation losses and possible leakage. e of a given wallEquation thickness: 2-28 may be used to determine the volume of a gas that will permeate through PE pipe of a given wall thickness: (2-28)

qP =

K P As ΘPA t'

Eq. 1-45

Where 3 (gas at standard temperature and pressure) qP = permeated, volume of gas permeated, volume of gas cm3 cm (gas at standard temperature and Cm3 mil pressure) K P = permeability constant (Table 2-9); units: 100 in2 atm day wall area in units of 100 square inches permeabilityAconstant 1-13) s = pipe outside(Table 2 surface areaPAof the outside wall of the (1pipe, 100= in = pipe internal pressure, atmospheres atmosphere 14.7 lb/in2 ) pipe internalΘ pressure, atmospheres (1 atmosphere = 14.7 lb/in2 ) = elapsed time, days

t’ = wall thickness, mils

Table 2-9 Permeability Constants (28)

154-264.indd 184

Gas

KP

Methane

85

Carbon Monoxide

80

Hydrogen

425

1/16/09 9:57:00 AM

Chapter 6 185


Table 2-10 Physical Properties of Gases (Approx. Values at 14.7 psi & 68ºF) Gas Acetylene (ethylene) Air Ammonia

Chemical Formula

Molecular Weight

Weight Density, lb/ft 3

Specific Gravity, (Relative to Air) Sg

C2H2

26.0

0.0682

0.907

–

29.0

0.0752

1.000

NH3

17.0

0.0448

0.596

Argon

A

39.9

0.1037

1.379

Butane

C4H10

58.1

0.1554

2.067

Carbon Dioxide

CO2

44.0

0.1150

1.529

Carbon Monoxide

CO

28.0

0.0727

0.967

Ethane

C2H6

30.0

0.0789

1.049

Ethylene

C2H4

28.0

0.0733

0.975

Helium

He

4.0

0.0104

0.138

Hydrogen Chloride

HCl

36.5

0.0954

1.286

H

2.0

0.0052

0.070

Hydrogen Hydrogen Sulphide

H2S

34.1

0.0895

1.190

Methane

CH4

16.0

0.0417

0.554

Methyl Chloride

CH3Cl

50.5

0.1342

1.785

Natural Gas

–

19.5

0.0502

0.667

Nitric Oxide

NO

30.0

0.0708

1.037

Nitrogen

N2

28.0

0.0727

0.967

N2O

44.0

0.1151

1.530

Oxygen

O2

32.0

0.0831

1.105

Propane

C3H8

44.1

0.1175

1.562

Propene (Propylene)

C3H6

42.1

0.1091

1.451

Sulfur Dioxide

SO2

64.1

0.1703

2.264

Landfill Gas (approx. value)

–

–

–

1.00

Carbureted Water Gas

–

–

–

0.63

Nitrous Oxide

Coal Gas

–

–

–

0.42

Coke-Oven Gas

–

–

–

0.44

Refinery Oil Gas

–

–

–

0.99

Oil Gas (Pacific Coast)

–

–

–

0.47

“Wet” Gas (approximate value)

–

–

–

0.75

Gravity Flow of Liquids In a pressure pipeline, a pump of some sort, generally provides the energy required to move the fluid through the pipeline. Such pipelines can transport fluids across a level surface, uphill or downhill. Gravity flow lines, on the other hand, utilize the energy associated with the placement of the pipeline discharge below the inlet. Like pressure flow pipelines, friction loss in a gravity flow pipeline depends on viscous shear stresses within the liquid and friction along the wetted surface of the pipe bore.

154-264.indd 185

1/16/09 9:57:00 AM

placement of the pipeline discharge below the inlet. Like pressure friction nes, friction loss in loss a gravity in the a gravity flow pipeline flow pipeline depends depends on bore. viscous on viscous shearshear stresses stresses , and along surface of the on lossfriction in a gravity flowwetted pipeline depends on pipe viscous shear stresses d,liquid, and friction and friction alongalong the wetted the wetted surface surface of theofpipe the bore. pipe bore. d friction along the wetted surface of the pipe bore. 186 Chaptermay 6 ow piping systems become very complex, especially if the pipeline Design of PE Piping Systems vity ow piping flow piping systems systems may become may along become very complex, very complex, especially especially if the ifpipeline the pipeline ecause frictionmay loss become will vary with theespecially varying grade. Sections of iping systems very complex, if the pipeline ies, ecause because friction friction loss will loss vary will along vary along with the with varying the varying grade. grade. Sections Sections of of internal pressure, or vacuum, and may have varying liquid ay develop se friction loss will vary along with the varying grade. Sections of ne ay may develop develop internal internal pressure, pressure, or vacuum, or vacuum, and may and may have have varying varying liquidliquid e bore. evelop internal pressure, or vacuum, and may have varying liquid pe he bore. pipe bore. re.

Some gravity flow piping systems may become very complex, especially if the pipeline grade varies, because friction loss will vary along with the varying grade. nel water flow under ofmay constant and uniform channel Sectionsconditions of the pipeline develop grade, internal pressure, or vacuum, and may have nel channel water water flow under flow under conditions conditions of constant of constant grade, grade, and uniform and uniform channel channel varying liquid be levels in the(28,29,30) pipegrade, bore. and he Manning equation may Open channel flow exists water flow under conditions of used. constant uniform channel

ion, the Manning equation equation may be may used. beHazen-Williams used. (28,29,30) (28,29,30) Open Open channel channel flow exists flow exists nhe itManning runs partially full.be Like the formula, theexists Manning anning equation may used. (28,29,30) Open channel flow ned when ittoruns it runs partially partially full.with Like full. Like the Hazen-Williams the viscosity Hazen-Williams formula, formula, the Manning the Manning Manning Flow water or full. liquids aEquation kinematic equal tothe water. runs partially Like the Hazen-Williams formula, Manning ted s limited to water to water or liquids or liquids with a with kinematic a kinematic viscosity viscosity equal equal to water. to water. Forwith openachannel waterviscosity flow underequal conditions of constant grade, and uniform o water or liquids kinematic to water.

ion Equation ion

channel cross section, the Manning equation may be used.(29,30) Open channel flow exists in a pipe when it runs partially full. Like the Hazen-Williams formula, the Manning equation is applicable to water or liquids with a kinematic viscosity equal to water. Manning Equation (2-29)

V

1.486 2 / 3 1 / 2 V 1.486 1rH.486 S 2/3 2/3 1V .486 = Vn2 / 3 rH1 / 2 SrH1 / 2 S 1 / 2 rHn S n n

ere Where = flow velocity, ft/sec V = flow velocity, ft/sec = V flow = velocity, flow velocity, ft/secft/sec =flow velocity, roughness dimensionless ft/sec n coefficient, = roughness coefficient, dimensionless = n roughness = roughness coefficient, coefficient, dimensionless dimensionless =roughness hydraulic radius, ft coefficient, dimensionless rH = hydraulic radius, ft = = hydraulic radius, radius, ft ft rH hydraulic SH =fthydraulic slope, ft/ft hydraulic radius, AC rH A AC (2-30) rAHC = PrWHC rH PW PW W pipe bore, ft2 = cross-sectional areaPof A =C perimeter cross-sectional = cross-sectional area of area pipe bore, pipe bore, ft2 ft2 = flow, ftof area of flow C = cross-sectional cross-sectionalAwetted area ofby pipe bore, ft2bore, ft2 P =W hydraulic perimeter = perimeter wetted by wetted flow, by flow, ft ft Pwetted perimeter = slope, ft/ft W= perimeter wetted by flow, ft by flow, ft S =H hydraulic = hydraulic slope,slope, ft/ft ft/ft hydraulic slope, ft/ft (2-31) hf h h S H hUU − hhDDU hDf hf = hSDH L h f = L ShHU SH L L L L Lpipe elevation, = upstream hpipe elevation, ftL ft U = upstream h =U downstream upstream = upstream elevation, pipe elevation, ft ft ftft hpipe downstream pipe elevation, D =pipe = elevation, upstream pipe elevation, ft h =D to friction downstream = downstream pipe elevation, pipe elevation, ft with ft hf = friction (head) loss, ft of liquid ent combine the Manning equation = (head) loss, ft of downstream pipe elevation, ftliquid ombine Manning equation with L = length, ft loss, = hf friction = thefriction (head) (head) loss, ft of liquid ft of liquid friction (head) loss, ft of liquid It is convenient to combine the Manning equation with Q ACV (2-32) Q ACV

Eq. 1- 46 Eq. 1-Eq. 461- 46 Eq. 1- 46

Eq. 1- 47 Eq. 1-Eq. 471- 47 Eq. 1- 47

Eq. 1-48 Eq. 1-48 Eq. 1-48 Eq. 1-48 31 31

Eq. 1-49 Eq. 1-49

To obtain

1.486 A 2/3 1/ 2 1.486 Q AC 2 / 3 C r1H/ 2 S H Q rH n S H n are as defined above, and defined above, and = flow, ft3/sec flow, ft3/sec ular pipe is running full or half-full, e is running full or half-full, (2-33)

154-264.indd 186

d'

DI

Eq. 1-50 Eq. 1-50

1/16/09 9:57:01 AM

Q

ACV

Q = ACV

Eq. 1-49 Eq. 1-49

1.486 AC 2 / 3 1 / 2 rH S H n 1.486 AC 2 / 3 1 / 2 1.486QA=C 2 / n3 1 /r2H S H re as defined above, and rH S H Q n 3 rms as ftdefined /sec above, and = are flow, as defined above,3and /sec flow,Where ftfull rQpipe =is running or half-full, terms are as defined above, and flow, ft3/sec Q = flow, ft3/sec ircular pipe is running full or half-full, d ' DI pe is running full or half-full, rH When a circular pipe 4 is running ' orDhalf-full, d full 48 I r = H (2-34) d ' DI 4 = 48 rH 48 equation with onvenient to combine the 4Manning ere = pipe inside diameter, ft =d’ pipe diameter, in = inside pipewhere inside diameter, ft d’ = pipe inside diameter, ft Q ACV 3 = inside diameter, ft DIpipe pipe inside diameter, in insideestimated diameter, in using: n ft per secondDmay I = pipe be pipe inside in 3 thediameter, otain combine Manning equation with using: second may be estimated flow in ft per

Q

3

Chapter 6 Eq.Design 1-50 of PE Piping Systems Eq. 1-50 Eq. 1-50

187

Eq. 1-51 31 Eq. 1-51 Eq. 1-51 31

Eq. 1-49

8/3

Full pipe in ft 3 per second may be estimated S 1 / 2 using: per second may beflow estimated 4 DI using: 6 . 136 u 10 Eq. 1-52 Q(2-35) 1 . 486 A 28/ /33 11/ 2/ 2 FPS C Q Q ACV Eq. 1-49 Eq. 1-50 SSH − 4n rD HI ×810 Eq. 1-52 QFPS = 6.136 n/ 3 S 1 / 2 4 DI n 6may .136 ube 10 Eq. 1-52 Qminute FPS ne gallons peras terms are defined above,estimated and n using: Full pipe flow in gallons per minute may be estimated using: low in gallons per minute may be estimated using: 3 486 A Q = flow, 2 /83/ 3 11/ 2/ 2 (2-36) ft1./sec C rD SSHusing: Eq. 1-50 allons per minuteQmay be estimated H I ' 0 . 275 Eq. 1-53 Q n full or half-full, 8 / 3 1/ 2 a circular pipe is running DI S n 8 / 3 1/ 2 Eq. 1-53 S as defined above, and QD' =I 0.275 n ' 0 . 275 Eq. 1-53 Q DI more liquid ' carry ular pipes will carry more liquid than ad completely full pipe. When slightly than a completely full pipe. rH will Eq.When 1-51 flow, ft3/secNearly full circularnpipes slightly less than full, the perimeter wetted by flow is reduced, but the actual flow he hydraulic radius is significantly reduced, but the actual flow area is 4 48 l circular pipes will carry more liquid than a completely full pipe. When slightly pipe isthe running or half-full, area is only slightly lessened. results in pipe. abut larger hydraulic radius than when ssened. Maximum flow is achieved atThis about 93% of full pipe flow, and pipes will carryfull more liquid than a completely full When slightly full, hydraulic radius is significantly reduced, the actual flow area is where the pipe is running full. Maximum flow is achieved at about 93% of full pipe flow, city at about 78% of full pipe flow. hydraulic radius is significantly reduced, at butabout the actual flow tly lessened. Maximum flow is achieved 93% of fullarea pipe isflow, and DI at ' pipe d velocity d’Maximum = about pipe inside diameter, and maximum at ft about 78%93% of fullof pipe flow. Manning’s n is often assumed ned. flow is achieved about full pipe flow, velocity at 78% of full flow. rH Eq. and 1-51 DI = diameter, in Actually, n has been found to be slightly larger in to be constant 4with flow 48 depth. at about 78% pipe of fullinside pipe flow.

(

3

)

non-full flow.

ipe flow in ft per second may be estimated using: Table 1-15: Values of n for Use with Manning Equation pipe inside diameter, ft 8 / 3n, typical Table 1-15: Values Manning Table 2-11 of n for Use with S 1 / 2 Equation Surface 4 DI pipe inside diameter, in design 6.136Equation u 10 Eq. 1-52 QUse ble 1-15: Values ofofnn for for Use with Manning Equation Values with Manning n, typical FPS Surface n 0.009design 3 Polyethylene pipe n, typical ft per Uncoated secondSurface may estimated using: Surface n, typical0.013 design castPolyethylene orbe ductile ironpipe pipe design 0.009 PE pipe 0.009 pe flow in Polyethylene gallons per minute may be estimated using: Corrugated steel 0.024 0.013 Uncoated cast ductile iron8 /pipe pipeorpipe 0.009 Uncoated cast or ductile iron pipe 3 S 1 / 2 0.013 D Concrete pipe 0.013 0.024 4 I pipe Uncoated cast orCorrugated ductile iron pipe 0.013 6.136 usteel 10 Eq. 1-52 Q Corrugated steel pipe 8 / 3 1 / 2 0.024 FPS Vitrified clay pipe 0.013 0.013 Concrete pipe n DI S 0.024 Corrugated steel pipe pipe Concrete 0.013 ' 0.275 Eq. 1-53 Q Brick and cement mortar sewers 0.015 0.013 Vitrified pipe Concrete pipe Vitrifiedclay clay pipe 0.013 n 0.013 Brick Brick and cement mortar sewers 0.015 clay pipe 0.013 gallons perVitrified minute may be estimated using: and cement mortar sewers 0.015 Brick and cement mortar sewers 0.015 Wood stave y full circular pipes will carry more liquid than a 0.011 completely full pipe. When slightly 8 / 3 1/ 2 Rubble masonry 0.021 DI is significantly S han full, the hydraulic radius reduced, but the actual flow area is ' n-value 0.275of 0.009 Eq. 1-53 Note:Q The for PE pipe is for clear water applications. lightly lessened. Maximum flow is achieved at such about 93% of full pipe flow, and n utilized for applications An n-value of 0.010 is typically as sanitary sewer, etc. mum velocity at about 78% of full pipe flow. ar pipes will carry more liquid than a completely full pipe. When slightly e hydraulic radius is significantly reduced, but the actual flow area is ened. Maximum flow is achieved at about 93% of full pipe flow, and y at about 78%1-15: of full Values pipe flow. Table of n for Use with Manning Equation

154-264.indd 187

Surface

n, typical

1/16/09 9:57:02 AM

32 188

Chapter 6


Wood stave Rubble masonry

0.011 0.021

Note: The n-value of 0.009 for polyethylene pipe is for clear water applications. An n-value of 0.010 is typically utilized for applications such as sanitary sewer, etc.

tive Flows for Slipliners Comparative Flows for Slipliners Deteriorated gravity flow pipes may be rehabilitated by sliplining with PE pipe. This

ted gravity flowprocess pipesinvolves may be pipe.pipe therehabilitated installation of aby PE sliplining liner inside with of thepolyethylene deteriorated original cess involves the installation of a polyethylene liner deteriorated as described in subsequent chapters within this inside manual. of Forthe conventional sliplining, ipe as described in subsequent chapters within this manual. For conventional clearance between the liner outside diameter and the existing pipe bore is required clearance between liner thus outside diameter and thechannel existing pipe than borethat is of to installthe the liner; after rehabilitation, the flow is smaller to install the liner; thus after the flow channel is smaller theand the original pipe. rehabilitation, However, it is often possible to rehabilitate with a PEthan slipliner, pipe. However, regain it is often possible to rehabilitate with slipliner, all or most of the original flow capacity dueatopolyethylene the extremely smooth inside surface of the PE pipe its resistance or build-up. Because PE pipe in all or most of the original flowand capacity due totodeposition the extremely smooth inside is mostly joined means thisto results in no effective of of the polyethylene pipe byand itsof butt-fusion, resistance deposition or reduction build-up. flow diameter at joint locations flow capacities of circular pipes may be tive flow capacities of circular pipes mayComparative be determined by the following: determined by the following: (2-37)

 D I 18 / 3     n1  Q1  = 100  % flow = 100 8/3 Q2  DI 2     n2   

Eq. 1-54

Table 2-12 was developed using Equation 2-36 where D I1 = the inside diameter (ID) 16 was developed using Equation 1-54 where DI1 = the inside diameter (ID) of of the liner, and D I2 = the original inside diameter of the deteriorated host pipe. and DI2 = the original inside diameter of the deteriorated host pipe.

ine r OD, n.

.50 0 .50 0 .37 5 .62 5 .12 5 .62 5 0.7 50 2.7 50

Table 1-16: Comparative Flows for Slipliners

Liner DR 32.5 % flow % Line vs. flow r ID, concre vs. in.† te clay 3.27 84.5 97.5% 2 % 4.20 56.0 64.6% 6 % 5.02 90.0 103.8% 4 % 6.19 73.0 84.2% 3 % 6.66 88.6 102.2% 0 % 8.06 81.3 93.8% 2 % 10.0 90.0 103.8% 49 % 11.9 78.2 90.3% 18 %

154-264.indd 188

Liner DR 26 % flow % Line vs. flow r ID, concre vs. in.† te clay 3.21 80.6 93.0% 5 % 4.13 53.5 61.7% 3 % 4.93 85.9 99.1% 7 % 6.08 69.6 80.3% 5 % 6.54 84.5 97.5% 4 % 7.92 77.6 89.5% 2 % 9.87 85.9 99.1% 3 % 11.7 74.6 86.1% 10 %



1/16/09 9:57:02 AM

Chapter 6 189


Table 2-12 Comparative Flows for Slipliners Liner DR 32.5 Liner DR 26 Liner DR 21 Liner DR 17 Existing Liner % flow % flow % flow % flow Liner ID, % flow Liner ID, % flow Liner ID, % flow Liner ID, % flow vs. vs. vs. vs. Sewer OD, in.† vs. clay in.† vs. clay in.† vs. clay in.† vs. clay concrete concrete concrete concrete ID, in in. 4

3.500

3.272

97.5%

84.5%

3.215

93.0%

80.6%

3.147

87.9%

76.2%

3.064

81.8%

70.9%

6

4.500

4.206

64.6%

56.0%

4.133

61.7%

53.5%

4.046

58.3%

50.5%

3.939

54.3%

47.0%

6

5.375

5.024

103.8%

90.0%

4.937

99.1%

85.9%

4.832

93.6%

81.1%

4.705

87.1%

75.5%

8

6.625

6.193

84.2%

73.0%

6.085

80.3%

69.6%

5.956

75.9%

65.8%

5.799

70.7%

61.2%

8

7.125

6.660

102.2%

88.6%

6.544

97.5%

84.5%

6.406

92.1%

79.9%

6.236

85.8%

74.4%

10

8.625

8.062

93.8%

81.3%

7.922

89.5%

77.6%

7.754

84.6%

73.3%

7.549

78.8%

68.3%

12

10.750

10.049 103.8%

90.0%

9.873

99.1%

85.9%

9.665

93.6%

81.1%

9.409

87.1%

75.5%

15

12.750

11.918

78.2%

11.710

86.1%

74.6%

11.463

81.4%

70.5%

11.160

75.7%

65.6%

90.3%

15

13.375

12.503 102.5%

88.9%

12.284

97.8%

84.8%

12.025

92.4%

80.1%

11.707

86.1%

74.6%

16

14.000

13.087

84.5%

2.858

93.0%

80.6%

12.587

87.9%

76.2%

12.254

81.8%

70.9%

97.5%

18

16.000

14.956 101.7%

88.1%

14.695

97.0%

84.1%

14.385

91.7%

79.4%

14.005

85.3%

74.0%

21

18.000

16.826

92.3%

80.0%

16.532

88.1%

76.3%

16.183

83.2%

72.1%

15.755

77.5%

67.1%

24

20.000

18.695

85.6%

74.2%

18.369

81.7%

70.8%

17.981

77.2%

66.9%

17.506

71.9%

62.3%

24

22.000

20.565 110.4%

95.7%

20.206 105.3%

91.3%

19.779

99.5%

86.2%

19.256

92.6%

80.3%

88.1%

22.043

27

24.000

22.434 101.7%

97.0%

84.1%

21.577

91.7%

79.4%

21.007

85.3%

74.0%

30

28.000

26.174 115.8% 100.4% 25.717 110.5%

95.8%

25.173 104.4%

90.5%

24.508

97.2%

84.2%

33

30.000

28.043 108.0%

93.6%

27.554 103.0%

89.3%

26.971

97.3%

84.3%

26.259

90.6%

78.5%

36

32.000

29.913 101.7%

88.1%

29.391

84.1%

28.770

91.7%

79.4%

28.009

85.3%

74.0%

97.0%

36

34.000

31.782 119.5% 103.6% 31.228 114.1%

98.9%

30.568 107.7%

93.4%

29.760 100.3%

86.9%

42

36.000

33.652

92.3%

76.3%

32.366

72.1%

31.511

67.1%

80.0%

33.065

88.1%

83.2%

77.5%

48

42.000

39.260

97.5%

84.5%

38.575

93.0%

80.6%

37.760

87.9%

76.2%

36.762

81.8%

70.9%

54

48.000

44.869 101.7%

88.1%

44.086

97.0%

84.1%

43.154

91.7%

79.4%

42.014

85.3%

74.0%

60

54.000

50.478 105.1%

91.1%

49.597 100.3%

86.9%

48.549

94.8%

82.1%

47.266

88.2%

76.5%

† Liner ID calculated per Equation 2-1.

Flow Velocity Acceptable flow velocities in PE pipe depend on the specific details of the system. For water systems operating at rated pressures, velocities may be limited by surge allowance requirements. See Tables 1-3A and 1-3B. Where surge effects are reduced, higher velocities are acceptable, and if surge is not possible, such as in many gravity flow systems, water flow velocities exceeding 25 feet per second may be acceptable. Liquid flow velocity may be limited by the capabilities of pumps or elevation head to overcome friction (head) loss and deliver the flow and pressure required for the application. PE pipe is not eroded by water flow. Liquid slurry pipelines may be subject to critical minimum velocities that ensure turbulent flow and maintain particle suspension in the slurry. Gravity liquid flows of 2 fps (0.6 m/s) and higher can help prevent or reduce solids deposition in sewer lines. When running full, gravity flow pipelines are subject to the same velocity considerations as pressure pipelines.

154-264.indd 189

1/16/09 9:57:02 AM

190 Chapter 6


Flow velocity in compressible gas lines tends to be self-limiting. Compressible gas flows in PE pipes are typically laminar or transitional. Fully turbulent flows are possible in short pipelines, but difficult to achieve in longer transmission and distribution lines because the pressure ratings for PE pipe automatically limit flow capacity and, therefore, flow velocity. Pipe Surface Condition, Aging Aging acts to increase pipe surface roughness in most piping systems. This in turn increases flow resistance. PE pipe resists typical aging effects because PE does not rust, rot, corrode, tuberculate, or support biological growth, and it resists the adherence of scale and deposits. In some cases, moderate flow velocities are sufficient to prevent deposition, and where low velocities predominate, occasional high velocity flows will help to remove sediment and deposits. As a result, the initial design capabilities for pressure and gravity flow pipelines are retained as the pipeline ages. Where cleaning is needed to remove depositions in low flow rate gravity flow pipelines, water-jet cleaning or forcing a “soft” (plastic foam) pig through the pipeline are effective cleaning methods. Bucket, wire and scraper-type cleaning methods will damage PE pipe and must not be used.

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Section 3 Buried PE Pipe Design Introduction This section covers basic engineering information for calculating earth and live-load pressures on PE pipe, for finding the pipe’s response to these pressures taking into account the interaction between the pipe and its surrounding soil, and for judging that an adequate safety factor exists for a given application. Soil pressure results from the combination of soil weight and surface loads. As backfill is placed around and over a PE pipe, the soil pressure increases and the pipe deflects vertically and expands laterally into the surrounding soil. The lateral expansion mobilizes passive resistance in the soil which, in combination with the pipe’s inherent stiffness, resists further lateral expansion and consequently further vertical deflection. During backfilling, ring (or hoop) stress develops within the pipe wall. Ring bending stresses (tensile and compressive) occur as a consequence of deflection, and ring compressive stress occurs as a consequence of the compressive thrust created by soil compression around the pipe’s circumference. Except for shallow pipe subject to live load, the combined ring stress from bending and compression results in a net compressive stress. The magnitude of the deflection and the stress depends not only on the pipe’s properties but also on the properties of the surrounding soil. The magnitude of deflection and stress must be kept safely within PE pipe’s performance limits. Excessive deflection may cause loss of stability and flow restriction, while excessive compressive stress may cause wall crushing or ring buckling. Performance limits for PE pipe are given in Watkins, Szpak, and Allman(1) and illustrated in Figure 3-1. The design and construction requirements can vary somewhat, depending on whether the installation is for pressure or non-pressure service. These differences will be addressed later in this chapter and in Chapter 7, “Underground Installation of PE Pipe.”

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Calculations Section 3 describes how to calculate the soil pressure acting on PE pipe due to soil weight and surface loads, how to determine the resulting deflection based on pipe and soil properties, and how to calculate the allowable (safe) soil pressure for wall compression (crushing) and ring buckling for PE pipe. Detailed calculations are not always necessary to determine the suitability of a particular PE pipe for an application. Pressure pipes that fall within the Design Window given in AWWA M-55 “PE Pipe – Design and Installation” regarding pipe DR, installation, and burial depth meet specified deflection limits for PE pipe, have a safety factor of at least 2 against buckling, and do not exceed the allowable material compressive stress for PE. Thus, the designer need not perform extensive calculations for pipes that are sized and installed in accordance with the Design Window. AWWA M-55 Design Window

AWWA M-55, “PE Pipe – Design and Installation”, describes a Design Window. Applications that fall within this window require no calculations other than constrained buckling per Equation 3-15. It turns out that if pipe is limited to DR 21 or lower as in Table 3-1, the constrained buckling calculation has a safety factor of at least 2, and no calculations are required. The design protocol under these circumstances (those that fall within the AWWA Design Window) is thereby greatly simplified. The designer may choose to proceed with detailed analysis of the burial design and utilize the AWWA Design Window guidelines as a means of validation for his design calculations and commensurate safety factors. Alternatively, he may proceed with confidence that the burial design for these circumstances (those outlined within the AWWA Design Window) has already been analyzed in accordance with the guidelines presented in this chapter. The Design Window specifications are: • Pipe made from stress-rated PE material. • Essentially no dead surface load imposed over the pipe, no ground water above the surface, and provisions for preventing flotation of shallow cover pipe have been provided. • The embedment materials are coarse-grained, compacted to at least 85% Standard Proctor Density and have an E’ of at least 1000 psi. The native soil must be stable; in other words the native soil must have an E’ of at least 1000 psi. See Table 3-7. • The unit weight of the native soil does not exceed 120 pcf. • The pipe is installed in accordance with manufacturer’s recommendations for controlling shear and bending loads and minimum bending radius, and installed

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in accordance with ASTM D2774 for pressure pipes or ASTM D2321 for nonpressure pipes. • Minimum depth of cover is 2 ft (0.61 m); except when subject to AASHTO H20 truck loadings, in which case the minimum depth of cover is the greater of 3 ft (0.9 m) or one pipe diameter. • Maximum depth of cover is 25 ft (7.62 m). Table 3-1 AWWA M-55 Design Window Maximum and Minimum Depth of Cover Requiring No Calculations

DR

Min. Depth of Cover With H20 Load

Min. Depth of Cover Without H20 Load

Maximum Depth of Cover

7.3

3 ft

2 ft

25 ft

9

3 ft

2 ft

25 ft

11

3 ft

2 ft

25 ft

13.5

3 ft

2 ft

25 ft

17

3 ft

2 ft

25 ft

21

3 ft

2 ft

25 ft

* Limiting depths where no calculations are required. Pipes are suitable for deeper depth provided a sufficient E’ (1,000 psi or more) is accomplished during installations. Calculations would be required for depth greater than 25 ft.

Installation Categories For the purpose of calculation, buried installations of PE pipe can be separated into four categories depending on the depth of cover, surface loading, groundwater level and pipe diameter. Each category involves slightly different equations for determining the load on the pipe and the pipe’s response to the load. The boundaries between the categories are not definite, and engineering judgment is required to select the most appropriate category for a specific installation. The categories are: 1. Standard Installation-Trench or Embankment installation with a maximum

cover of 50 ft with or without traffic, rail, or surcharge loading. To be in this category, where live loads are present the pipe must have a minimum cover of at least one diameter or 18” whichever is greater. Earth pressure applied to the pipe is found using the prism load (geostatic soil stress). The Modified Iowa Formula is used for calculating deflection. Crush and buckling are performance limits as well. The Standard Installation section also presents the AWWA “Design Window.”

2. Shallow Cover Vehicular Loading Installation applies to pipes buried at a

depth of at least 18” but less than one pipe diameter. This installation category

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uses the same equations as the Standard Installation but with an additional equation relating wheel load to the pipe’s bending resistance and the soil’s supporting strength. 3. Deep Fill Installation applies to embankments with depths exceeding 50 ft.

The soil pressure calculation may be used for profile pipe in trenches less than 50 ft. The Deep Fill Installation equations differ from the Standard Installation equations by considering soil pressure based on armored, calculating deflection from the Watkins-Gaube Graph, and calculating buckling with the Moore-Selig Equation.

4. Shallow Cover Flotation Effects applies to applications where insufficient

cover is available to either prevent flotation or hydrostatic collapse. Hydrostatic buckling is introduced in this chapter because of its use in subsurface design.

Section 3 of the Design Chapter is limited to the design of PE pipes buried in trenches or embankments. The load and pipe reaction calculations presented may not apply to pipes installed using trenchless technologies such as pipe bursting and directional drilling. These pipes may not develop the same soil support as pipe installed in a trench. The purveyor of the trenchless technology should be consulted for piping design information. See the Chapter on “PE Pipe for Horizontal Directional Drilling” and ASTM F1962, Use of Maxi-Horizontal Directional Drilling (HDD) for Placement of Polyethylene Pipe or Conduit Under Obstacles, Including River Crossings for additional information on design of piping installed using directional drilling. 37

Figu re 21. Perf orm ance Limi ts for Buri ed PE Pipe

Figure 3-1 Performance Limits for Buried PE Pipe Design Process The interaction between pipe and soil, the variety of field-site soil conditions, and the Dimension Ratios make the design of buried pipe seem challenging. This section of the Design Chapter has been written with the intent of The interaction between pipe and soil, the variety ofapproaches field-sitefor soil conditions, and easing the designer’s task. While some very sophisticated design buried pipe systems may be justified in certain applications, the simpler, empirical the range of available pipe Dimension Ratios make the design methodologies presented herein have been proven by experience to provide reliableof buried pipe seem results for virtuallyThis all PEsection pipe installations. challenging. of the Design Chapter has been written with the intent range of Process available pipe Design

of easing the designer’s task. While some very sophisticated design approaches for The design process consists of the following steps: 1) Determine the vertical soil pressure acting at the crown of the pipe due to earth, live, and surcharge loads. 2) Select a trial pipe, which means selecting a trial dimension ratio (DR) or, in the case of profile pipe, a trial profile as well. 154-264.indd 194

3) Select an embedment material and degree of compaction. As will be described later, soil type and compaction are relatable to a specific modulus of soil

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buried pipe systems may be justified in certain applications, the simpler, empirical methodologies presented herein have been proven by experience to provide reliable results for virtually all PE pipe installations. The design process consists of the following steps: 1. Determine the vertical soil pressure acting at the crown of the pipe due to earth, live, and surcharge loads. 2. Select a trial pipe, which means selecting a trial dimension ratio (DR) or, in the case of profile pipe, a trial profile. 3. Select an embedment material and degree of compaction. As will be described later, soil type and compaction are relatable to a specific modulus of soil reaction value (E’) (Table 3-8). (As deflection is proportional to the combination of pipe and soil stiffness, pipe properties and embedment stiffness can be traded off to obtain an optimum design.) 4. For the trial pipe and trial modulus of soil reaction, calculate the deflection due to the vertical soil pressure. Compare the pipe deflection to the deflection limit. If deflection exceeds the limit, it is generally best to look at increasing the modulus of soil reaction rather than reducing the DR or changing to a heavier profile. Repeat step 4 for the new E’ and/or new trial pipe. 5. For the trial pipe and trial modulus of soil reaction, calculate the allowable soil pressure for wall crushing and for wall buckling. Compare the allowable soil pressure to the applied vertical pressure. If the allowable pressure is equal to or higher than the applied vertical pressure, the design is complete. If not, select a different pipe DR or heavier profile or different E’, and repeat step 5. Since design begins with calculating vertical soil pressure, it seems appropriate to discuss the different methods for finding the vertical soil pressure on a buried pipe before discussing the pipe’s response to load within the four installation categories. Earth, Live, and Surcharge Loads on Buried Pipe Vertical Soil Pressure The weight of the earth, as well as surface loads above the pipe, produce soil pressure on the pipe. The weight of the earth or “earth load” is often considered to be a “dead-load” whereas surface loads are referred to as “surcharge loads” and may be temporary or permanent. When surcharge loads are of short duration they are usually referred to as “live loads.” The most common live load is vehicular load. Other common surcharge loads include light structures, equipment, and piles of stored materials or debris. This section gives formulas for calculating the vertical

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soil pressure due to both earth and surcharge loads. The soil pressures are normally calculated at the depth of the pipe crown. The soil pressures for earth load and each surcharge load are added together to obtain the total vertical soil pressure which is then used for calculating deflection and for comparison with wall crush and wall buckling performance limits. Earth Load In a uniform, homogeneous soil mass, the soil load acting on a horizontal plane within the mass is equal to the weight of the soil directly above the plane. If the mass contains areas of varying stiffness, the weight of the mass will redistribute itself toward the stiffer areas due to internal shear resistance, and arching will occur. Arching results in a reduction in load on the less stiff areas. Flexible pipes including PE pipes are normally not as stiff as the surrounding soil, so the resulting earth pressure acting on PE pipe is reduced by arching and is less than the weight of soil above the pipe. (One minor exception to this is shallow cover pipe under dynamic loads.) For simplicity, engineers often ignore arching and assume that the earth load on the pipe is equal to the weight of soil above the pipe, which is referred to as the “prism load” or “geostatic stress.” Practically speaking, the prism load is a conservative loading for PE pipes. It may be safely used in virtually all designs. Equation 3-1 gives the vertical soil pressure due to the prism load. The depth of cover is the depth from the ground surface to the pipe crown. (3-1)

P E = wH

Where PE = vertical soil pressure due to earth load, psf

w = unit weight of soil, pcf H = depth of cover, ft

UNITS CONVENTION: To facilitate calculations for PE pipes, the convention used with rigid pipes for taking the load on the pipe as a line load along the longitudinal axis in units of lbs/lineal-ft of pipe length is not used here. Rather, the load is treated as a soil pressure acting on a horizontal plane at the pipe crown and is given in units of lbs/ft2 or psf.

Soil weight can vary substantially from site to site and within a site depending on composition, density and load history. Soil weights are often found in the construction site geotechnical report. The saturated unit weight of the soil is used when the pipe is below the groundwater level. For design purposes, the unit weight of dry soil is commonly assumed to be 120 pcf, when site-specific information is not available.

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Generally, the soil pressure on profile pipe and on DR pipe in deep fills is significantly less than the prism load due to arching. For these applications, soil pressure is best calculated using the calculations that account for arching in the 40 “Deep Fill Installation” section.

Figure 2-2. Figure 3-2 Prism LoadPrism

Load

Live Load

Even though wheel loadings from cars and other light vehicles may be frequent, these loads generally have little impact on subsurface piping compared to the less frequent but significantly heavier loads from trucks, trains, or other heavy vehicles. h wheel loadings cars andunder other lightand vehicles may these Forfrom design of pipes streets highways, onlybe thefrequent, loadings from these heavier ally have little impact on subsurface piping compared to the less frequent vehicles are considered. The pressure transmitted to a pipe by a vehicle depends ntly heavier loads from trucks, or other heavy vehicles. Forsize, design ofspeed, on the pipe’s depth,trains, the vehicle’s weight, the tire pressure and vehicle streets and highways, then, only the loadings from these heavier vehicles surface smoothness, the amount and type of paving, the soil, and the distance from red. The pressure transmitted toofaloading. pipe by vehicle depends onsuch theaspipe’s the pipe to the point Forathe more common cases, AASHTO, H20 vehicle's weight, the tire pressure and size, vehicle speed, surface HS20 truck traffic on paved roads and E-80 rail loading, this information has been , the amount and type ofand paving, soil, andand the3-5distance theFor pipe to cases, simplified put intothe Table 3-3, 3-4, to aid thefrom designer. special loading. For the more common cases, asvehicles, H20 (HS20) truck traffic on be used. such as mine trucks, cranes, or such off-road Equations 3-2 and 3-4 may

s and E-80 rail loading, this information has been simplified and put into load under a wheel occurs at the surface andmine diminishes with depth. 2-3, and 2-4 toThe aidmaximum the designer. For special cases, such as trucks, PE pipes should be installed a minimum of one diameter or 18”, whichever is greater, ff-road vehicles, Equations 2-2 and 2-4 may be used. beneath the road surface. At this depth, the pipe is far enough below the wheel load to significantly reduce soil pressure and the pipe can fully utilize the embedment a wheel occurs at the surface and diminishes with depth. soil for load resistance. Where design considerations do not permit installation with

um load under e pipes should be installed a minimum of one diameter or 18”, whichever is eath the road surface. At this depth, the pipe is far enough below the wheel ificantly reduce soil pressure and the pipe can fully utilize the embedment resistance. Where design considerations do not permit installation with at ameter of cover, additional calculations are required and are given in the cussing154-264.indd "Shallow Cover Vehicular Loading Installation." State highway 197

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at least one diameter of cover, additional calculations are required and are given in the section discussing “Shallow Cover Vehicular Loading Installation.” State highway departments often regulate minimum cover depth and may require 2.5 ft to 5 ft of cover depending on the particular roadway. During construction, both permanent and temporary underground pipelines may be subjected to heavy vehicle loading from construction equipment. It may be advisable to provide a designated vehicle crossing with special measures such as temporary pavement or concrete encasement, as well as vehicle speed controls to limit impact loads. The following information on AASHTO Loading and Impact Factor is not needed to use Tables 3-3 and 3-4. It is included to give the designer an understanding of the surface loads encountered and typical impact factors. If the designer decides to use Equations 3-2 or 3-4 rather than the tables, the information will be useful. AASHTO Vehicular Loading Vehicular loads are typically based on The American Association of State Highway and Transportation Officials (AASHTO) standard truck loadings. For calculating the soil pressure on flexible pipe, the loading is normally assumed to be an H20 (HS20) truck. A standard H20 truck has a total weight of 40,000 lbs (20 tons). The weight is distributed with 8,000 lbs on the front axle and 32,000 lbs on the rear axle. The HS20 truck is a tractor and trailer unit having the same axle loadings as the H20 truck but with two rear axles. See Figure 3-3. For these trucks, the maximum wheel load is found at the rear axle(s) and equals 40 percent of the total weight of the truck. The maximum wheel load may be used to represent the static load applied by either a single axle or tandem axles. Some states permit heavier loads. The heaviest tandem axle loads normally encountered on highways are around 40,000 lbs (20,000 lbs per wheel). Occasionally, vehicles may be permitted with loads up to 50 percent higher.

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4242 AASHTO H20 Wheel AASHTO H20 Load AASHTO H20Wheel WheelLoad Load

AASHTO Wheel AASHTO HS20 Wheel Load AASHTO HS20 HS20 Wheel Load Load

Figure 3-3 AASHTO H20 and HS20 Vehicle Loads

Figure 2-3.AASHTO AASHTO H20 HS20 VehicleLoads Loads Figure 2-3. H20 && HS20 Vehicle Impact Factor Road surfaces are rarely smooth or perfectly even. When vehicles strike bumps in the road, the impact causes an instantaneous increase in wheel loading. Impact load Impact Factor Impact Factor may be found by multiplying the static wheel load by an impact factor. The factor varies with depth. Table 3-2 gives impact factors for vehicles on paved roads. For Road surfaces rarely smooth or perfectly even.When When vehiclesstrike strikebumps bumpsininthe the Road surfaces areare rarely smooth even. unpaved roads, impact factorsor ofperfectly 2.0 or higher may occur, vehicles depending on the road road, the impact causes an instantaneous increase in wheel loading. Impact load may road, the impact causes an instantaneous increase in wheel loading. Impact load may surface.

found multiplying static wheel load impact factor.The Thefactor factorvaries varieswith with be be found by by multiplying thethe static wheel load byby anan impact factor. depth. Table 2-1 gives impact factors for vehicles on paved roads. For unpaved roads, depth. Table 2-1 gives impact factors for vehicles on paved roads. For unpaved roads, impact factors of 2.0 or higher may occur, depending on the road surface. impactTable factors3-2 of 2.0 or higher may occur, depending on the road surface. Typical Impact Factors for Paved Roads Cover Depth, ft 1 2

Impact Factor, If

Table1.35 2-1: Typical Impact Factors for Paved Roads Table 2-1: Typical Impact Factors for Paved Roads 1.30

3

1.25Cover

4

1.20

Depth, ft Cover Depth, ft

6

1.10

8

1.00

1

Impact Factor, If Impact Factor, If

1

Derived from Illinois DOT dynamic load2formula (1996).

2 3

3

Vehicle Loading through Highway Pavement (Rigid)

1.35 1.35 1.30 1.30 1.25 1.25

1.20A stiff, rigid pavement Pavement reduces the live load4 pressure reaching a pipe. 4 1.20 spreads load out over a large subgrade area thus significantly reducing the vertical 6 1.10 6 1.10

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200 Chapter 6


soil pressure. Table 3-3 gives the vertical soil pressure underneath an H20 (HS20) truck traveling on a paved highway (12-inch thick concrete). An impact factor is incorporated. For use with heavier trucks, the pressures in Table 3-3 can be adjusted proportionally to the increased weight as long as the truck has the same tire area as an HS20 truck.

Table 3-3 Soil Pressure under H20 Load (12” Thick Pavement) Depth of cover, ft.

Soil Pressure, lb/ft2

1

1800

1.5

1400

2

800

3

600

4

400

5

250

6

200

7

175

8

100

Over 8

Neglect

Note: For reference see ASTM F7906. Based on axle load equally distributed over two 18 by 20 inch areas, spaced 72 inches apart. Impact factor included.

Vehicle Loading through Flexible Pavement or Unpaved Surface

Flexible pavements (or unpaved surfaces) do not have the bridging ability of rigid pavement and thus transmit more pressure through the soil to the pipe than given by Table 3-3. In many cases, the wheel loads from two vehicles passing combine to create a higher soil pressure than a single dual-tire wheel load. The maximum pressure may occur directly under the wheels of one vehicle or somewhere in between the wheels of the two vehicles depending on the cover depth. Table 3-4 gives the largest of the maximum pressure for two passing H20 trucks on an unpaved surface. No impact factor is included. The loading in Table 3-3 is conservative and about 10% higher than loads found by the method given in AASHTO Section 3, LRFD Bridge Specifications Manual based on assuming a single dual-tire contact area of 20 x 10 inches and using the equivalent area method of load distribution.

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Chapter 6 201


Table 3-4 Soil Pressure Under H20 Load (Unpaved or Flexible Pavement) Depth of cover, ft.

Soil Pressure, lb/ft2

1.5

2000

2.0

1340

2.5

1000

3.0

710

3.5

560

4.0

500

6.0

310

8.0

200

10.0

140

Note: Based on integrating the Boussinesq equation for two H20 loads spaced 4 feet apart or one H20 load centered over pipe. No pavement effects or impact factor included.

Off-Highway Vehicles

Off-highway vehicles such as mine trucks and construction equipment may be considerably heavier than H20 trucks. These vehicles frequently operate on unpaved construction or mine roads which may have very uneven surfaces. Thus, except for slow traffic, an impact factor of 2.0 to 3.0 should be considered. For off-highway vehicles, it is generally necessary to calculate live load pressure from information supplied by the vehicle manufacturer regarding the vehicle weight or wheel load, tire footprint (contact area) and wheel spacing. The location of the vehicle’s wheels relative to the pipe is also an important factor in determining how much load is transmitted to the pipe. Soil pressure under a point load at the surface is dispersed through the soil in both depth and expanse. Wheel loads not located directly above a pipe may apply pressure to the pipe, and this pressure can be significant. The load from two wheels straddling a pipe may produce a higher pressure on a pipe than from a single wheel directly above it. For pipe installed within a few feet of the surface, the maximum soil pressure will occur when a single wheel (single or dual tire) is directly over the pipe. For deeper pipes, the maximum case often occurs when vehicles traveling above the pipe pass within a few feet of each other while straddling the pipe, or in the case of off-highway vehicles when they have closely space axles. The minimum spacing between the centerlines of the wheel loads of passing vehicles is assumed to be four feet. At this spacing for H20 loading, the pressure on a pipe centered midway between the two passing vehicles is greater than a single wheel load on a pipe at or below a depth of about four feet. For design, the soil pressure on the pipe is calculated based on the vehicle location (wheel load locations) relative to the pipe that produces the maximum pressure. This

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46

PL =

202 Chapter 6


PL =

· I f W w §¨ H ¸ 1 1.5 2 aC ¨© ( r T + H 2 ) ¸¹ 3

3 · I f W w §¨ H ¸ 1 2 ¨ 2 1.5 ¸ aC © ( r T + H ) ¹Where:

46 Eq. 2-2

PLunder = vertical pressure to live generally involves comparing the pressure a singlesoil wheel with that due occurring with two wheels straddling the pipe. TheITimoshenko = impact Equation factor can be used to find Where: 3 ·under a single wheelf load, whereas the Boussinesq Equation § pressureHdirectly I f W wthe ¸ ¨1 Eq. 2-2 2 PL = =towheel load, lb W ¸ pressure ¨ be( used 2 1.5the P2L+=to vertical soil pressure due live load, lb/ft w not find from wheels directly above the pipe. a can ) H C

Where:

rT ¹ If = impact factor Timoshenko’s Equation ©

aC = contact area, ft2

load, lb/ft2

46

The Timoshenko Equation gives at a point directly radius,under ft a rT = equivalent load, lbthe soil pressure Ww = wheel distributed surface load, neglecting 2 any pavement.

H = depth of cover, ft aC = contact area, ft PL(3-2) = vertical soil pressure due to· live load, lb/ft2 3 I f W w §¨ H Eq. 2-2 1radius, ft¸ P L r=T = equivalent 2 ¨ If = impact factor aC © ( r T + H 2 )1.5 ¸¹ The equivalent H = depth of cover, ftradius is given by: Ww = wheel load, lb Where aC 2 to live load, lb/ft 2 vertical soilarea, pressureft due aCP=L =contact rT = Where: S If =radius impact factor The equivalent is given by: ft soil pressure due to live load, lb/ft2 rT = Wwequivalent = wheel load, P lb Lradius, = vertical a contact area, ft 2 Ha=C =depth of Icover, = ft C factor Eq. 2-3 r Tftimpact f = rT = equivalent radius, S For standard H2O and HS20 highway vehicle loading, the contact area H = depth of cover, ft load, lb Wtaken for dual wheels, that is, 16,000 lb over 10 in. by 20 in. area. w = wheel uivalent radius is given by: is given by: The equivalent radius aC = contact area, ft2 For standard H2O and HS20 highway vehicle loading, the contact area is normally (3-3) a C = equivalent radius, r Calculation taken for dual wheels, is, 16,000 lbExample over 10 ftin. by 20 in. area. Eq. 2-3 r T = thatTTimoshenko

S

HFind = depth of cover, ft the vertical pressure on a 24" polyethylene pipe buried 3 ft ben For standard H2O andwhen HS20 highway vehicle thethe contact areaThe is normally road an R-50 truckloading, is over pipe. manufacturer lists the Timoshenko Example Calculation taken for dual wheels, that is, 16,000 lb over a 10 in. by 20 in. area. weight ofby: 183,540 21X35 E3 is tires, each having a 30,590 lb loa ndard H2O and HS20 radius highway vehicle loading,lbs theon contact area normally The equivalent is given 2 Find the vertical pressure on a 24" polyethylene pipe buried 3 ft beneath an unpaved area of 370 or dual wheels, that is, 16,000 lb over 10inin.. by 20 in. area. Timoshenko Example Calculation road when an R-50 truck is over the pipe. The manufacturer lists the truck with a gross aC Use Equations 2-2 and an 2-3. Sinceroad the vehicleEq. is operatin Find the verticalSOLUTION: pressure on a 24” PE pipe buried 3 ft beneath unpaved 2-3 r T = weight of 183,540 lbs on 21X35 E3 tires, each having a 30,590 lb load over an imprint S when an R-50 off-road truck is over the pipe. The manufacturer lists the truck with road, an impact factor of 2.0 is appropriate. 2 enko . areaExample of 370 inCalculation a gross weight of 183,540 lbs on 21X35 E3 tires, each having a 30,590 lb load over an

imprint area of 370 2-2 in 2. and 2-3. Since the vehicle is operating on an unpaved Use Equations e SOLUTION: vertical pressure on a 24" polyethylene pipe buried 3 ft beneath an unpaved For H2O and HS20 highway vehicle loading, the contact area is normally road, an standard impact factor 2.0Equations is appropriate. 370 /the 144 SOLUTION: 3-2 and 3-3. Since vehicle is operating on an unpaved hen an R-50 truck is overofUse the pipe. The manufacturer lists the truck with a gross = over 10 in. = r T lb taken for road, dualan wheels, that is, 16,000 by0.90ft 20 in. area. impact factor of 2.0 is appropriate. S of 183,540 lbs on 21X35 E3 tires, each having a 30,590 lb load over an imprint 2 370 in . 3 · (2.0)(30,590) § 370 / 144 3 1= = = 0.90ft ¨ P L vehicle 47 r T Example 1.5 ¸ ION: Timoshenko Use Equations 2-2SandCalculation 2-3. Since the is operating on an unpaved 2 2 370 © ( 0.90 + 3 ) ¹ n impact factor of 2.0 is appropriate. 144 pipe buried 3 ft beneath an unpaved Find the vertical pressure on a 24" 3 polyethylene · (2.0)(30,590) 2 § 3 / ft is road whenP LLan R-50 truck pipe. The manufacturer lists the truck with a gross = 2890lb ¨ 1-over the 2 2 1.5 ¸ 370 ( + © ¹ ) 0.90 3 weight of 183,540 lbs on 21X35 E3 tires, each having a 30,590 lb load over an imprint 370 / 144 144 area r T = of 370 in2. = 0.90ft S Boussinesq Equation SOLUTION: Use Equations 2-2 and 2-3. Since the vehicle is operating on an unpaved 3 § · (2.0)(30,590)factor 3 is appropriate. road, 2.0 ¨ 1- ofgives P L = an impact The Boussinesq Equation the 1.5 ¸ pressure at any point in a soil mass under a 2 2 370 154-264.indd 202 1/16/09 9:57:04 AM © ( 0.90 + 3 ) ¹

sq Equation gives the pressure at any point in a soil mass under a ft 2 L = 2890lb urface Pload. The /Boussinesq Equation may be used to find the pressure m a wheel load to a point that is not along the line of action of the load. Chapter 6 Design of PE Piping Systems cts are neglected. esq Equation

203

3 the pressure at any point in a soil mass under a ussinesq Equation gives 3I f W w H Eq. = rated surface The Boussinesq Equation may be used to find the2-4 pressure P Lload. Boussinesq 5 Equation 2 S r ted from a wheel to a Equation point that is the notpressure along atthe of the load. The load Boussinesq gives anyline pointofinaction a soil mass under a ent effects are neglected. concentrated surface load. The Boussinesq Equation may be used to find the pressure

transmitted from a wheel load to a point that is not along the line of action of the load. Pavement effects are neglected.

3 = vertical soil (3-4) pressure3I due to live load lb/ft2 f Ww H PL = Ww = wheel load, lb 2S r 5

L

Eq. 2-4

H = vertical depth to pipe crown, ft Where

= impact factorPL = vertical soil pressure due to live load lb/ft Where: =

2

Ww = wheel load, lb = vertical depthof to pipe crown, ftto live load distance fromH the point load application to pipe soil pressure due lb/ft2crown, PL = vertical If = impact factor Ww = wheel load,from lb the point of load application to pipe crown, ft r = distance (3-5) H = vertical depth to2 pipe r= X + H 2 crown, ft

ft 4848

Eq. 2-5

If = impact factor r = distance from the point of load application to pipe crown, ft

r = X 2+ H2

Eq. 2-5

Figure 3-4 Illustration of Boussinesq Point Loading

Figure Figure2-4: 2-4:Illustration IllustrationofofBoussinesq BoussinesqPoint PointLoading Loading Example Using Boussinesq Point Loading Technique

Example Example Using Using Boussinesq Boussinesq Point Point Loading Loading Technique Technique Determine the vertical soil pressure applied to a 12” pipe located 4 ft deep under a

dirt road when two vehicles traveling over the pipe and in opposite lanes pass each

Determine Determine the thevertical verticalsoil soilpressure pressureapplied appliedtotoa a12" 12"pipe pipelocated located4 4ft ftdeep deepunder undera adirt dirt other. Assume center lines of wheel loads are at a distance of 4 feet. Assume a wheel road roadwhen whentwo twovehicles vehiclestraveling travelingover overthe thepipe pipeand andininopposite oppositelanes lanespass passeach eachother. other. load of 16,000 lb. Assume Assume center centerlines linesofofwheel wheelloads loadsare areatata adistance distanceofof4 4feet. feet. Assume Assumea awheel wheelload loadofof 16,000 16,000lb.lb. SOLUTION: SOLUTION:Use UseEquation Equation2-4, 2-4,and andsince sincethe thewheels wheelsare aretraveling, traveling,a a2.0 2.0impact impactfactor factor isisapplied. applied.The Themaximum maximumload loadwill willbebeatatthe thecenter centerbetween betweenthe thetwo twowheels, wheels,sosoXX= =2.0 2.0 ft.ft.Determine Determiner from r fromEquation Equation2-5. 2-5.

2 2 r= r = 424+2 + .47ft ft 2.20.0==4.447

154-264.indd 203

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when two vehicles traveling over the pipe and in opposite lanes pass each other. me center lines of wheel loads are at a distance of 4 feet. Assume a wheel load of mine 0 lb. the vertical soil pressure applied to a 12" pipe located 4 ft deep under a dirt when two vehicles traveling over the pipe and in opposite lanes pass each other. 204 Chapter 6 Design of PE Piping Systems me center lines of wheel loads are at a distance of 4 feet. Assume a wheel load of UTION: Use Equation 2-4, and since the wheels are traveling, a 2.0 impact factor 0 lb. plied. The maximum load will be at the center between the two wheels, so X = 2.0 etermine r from Equation 2-5. UTION: Use Equation 2-4, and since the wheels are traveling, a 2.0 impact factor plied. The maximum load will be at the center between the two wheels, so X = 2.0 etermine r from SOLUTION: Equation 2-5. Use Equation 3-4, and since the wheels are traveling, a 2.0 impact 2 rfactor = is + 2.02 =The 4.47 ft 4 applied. maximum load will be at the center between the two wheels, so X = 2.0 ft. Determine r from Equation 3-5.

2 2 = for = load 4.47 ftdue to a single wheel. 4 +P2L,.0the solve Equation r2-4

Then solve Equation 3-4 for PL, the load due to a single wheel.

solve Equation 2-4 for PL, the load due 3 to a single wheel. 3( 2.0)( 16,000 )( 4 ) PL = 2S ( 4.47 )5 3 3( 2.0)( 16,2000 )( 4 ) = lb / 548 ft PL 2π ( 4.47 )5

49

The load on the pipe crown 2 is from both wheels, so

/ ft both wheels, so L = 548 oad on the pipe Pcrown islbfrom 2 P L = 2( 548 ) = 1096 lb / ft 2

load is calculated in this exampleso is higher than that given in Table 3-4 for a oad on the pipeThe crown from both wheels,

comparable depth even after correcting for the impact factor. Both the Timoshenko

oad calculatedand inBoussinesq this example is higher than that given in Table 2-3 for a Equations give the pressure applied at a point in the soil. In solving arable depth even after correcting for the impact factor. Both the Timoshenko and for pipe reactions it is assumed that this point pressure is applied across the entire sinesq Equations give pressure at a the point in applied the soil. In solving foraway pipe surface ofthe a unit length ofapplied pipe, whereas actual pressure decreases ons it is assumed thisofpoint applied across the entire surfaceover of a fromthat the line action pressure of the wheelisload. Methods that integrate this pressure ength of pipe, whereas the actual applied pressure decreases away from the line of the pipe surface such as used in deriving Table 3-4 gives more accurate loading n of the wheel values. load. However, Methods that integrate this pressure over the pipe surface the error in the point pressure equations is slight and conservative, as used in deriving Table 2-3 giveequations more accurate so they are still effective for design. loading values. However, the in the point pressure equations is slight and conservative, so they are still effective ions for design.Railroad Loads

Table

The live loading configuration used for pipes under railroads is the Cooper E-80 loading, which is an 80,000 lb load that is uniformly applied over three 2 ft by 8 ft areas on 5 ft centers. The area represents the 8 ft width of standard railroad ties and the standard spacing between locomotive drive wheels. Live loads are based on 2-4: Live Load Pressure forbyE-80 Railroad and Loading the axle weight exerted on the track two locomotives their tenders coupled together in doubleheader fashion. See Table 3-5. Commercial railroads frequently Depth of cover, ft. Soil Pressure, lb/ft2 require casings for pressure pipes if they are within 25 feet of the tracks, primarily for safety reasons in the event of a washout. Based upon design and permitting 2.0 3800 requirements, the designer should determine whether or not a casing is required.

154-264.indd 204

5.0

2400

8.0

1600

10.0

1100

12.0

800

15.0

600

1/16/09 9:57:06 AM

der fashion. See Table 2-4. Commercial railroads frequently require casings e pipes if they are within 25 feet of the tracks, primarily for safety reasons in f a washout. Based upon design and permitting requirements, the6 designer Chapter 205 Design of PE Piping Systems ermine whether or not a casing is required. Load Table 3-5 Live Load Pressure for E-80 Railroad Loading

loads may be distributed loads, such as a footing, foundation, or an ash pile, Depth of cover, ft. Soil Pressure*, lb/ft concentrated loads,2.0such as vehicle 3800 wheels. The load will be dispersed e soil such that there5.0is a reduction in 2400 pressure with an increase in depth or 8.0 1600 distance from the surcharged area. Surcharge loads not directly over the 10.0 1100 12.0 800 exert pressure on the pipe as well. The pressure at a point beneath a oad depends on the15.0 load magnitude 600 and the surface area over which the 20.0 300 100 s applied. Methods 30.0 for calculating vertical pressure on a pipe either located Over 30.0 Neglect eath a surcharge or located near a surcharge are given below. 2

For referecne see ASTM A796. * The values shown for soil pressure include impact.

Surcharge Load

loads may be distributed loads, such as a footing, foundation, or an ash y Beneath aSurcharge Surcharge Load

pile, or may be concentrated loads, such as vehicle wheels. The load will be dispersed through the soil such that there is a reduction in pressure with an increase in depth or horizontal from the surcharged area. Surcharge loadsanot directly over area with n method is for findingdistance the vertical soil pressure under rectangular the pipe may exert pressure on the pipe as well. The pressure at a point beneath a distributed surcharge load. This may be used in place of Tables 2-2 to 2-4 surcharge load depends on the load magnitude and the surface area over which ons 2-2 and the 2-4surcharge to calculate vertical soil pressure due to wheel loads. To do is applied. Methods for calculating vertical pressure on a pipe either s knowledgelocated of the tire beneath imprint area and impact factor. are given below. directly a surcharge or located near a surcharge Pipe Directly Beneath a Surcharge Load

pressure onThisthe pipe atis for depth, H,vertical is found byunder dividing thearearectangular design method finding the soil pressure a rectangular with a uniformly distributed surcharge load. This may be used in place of Tables area (ABCD) into four sub-area rectangles (a, b, c, and d) which have a 3-3 to 3-5 and Equations 3-3 and 3-5 to calculate vertical soil pressure due to wheel orner, E, in the surcharge area, and over the pipe. The surcharge pressure, loads. This requires knowledge of the tire imprint area and impact factor. int directly under E is the sum of the pressure due to each of the four subThe point pressure on the pipe at depth, H, is found by dividing the rectangular Refer to Figure 2-5 A. surcharge area (ABCD) into four sub-area rectangles (a, b, c, and d) which have a

common corner, E, in the surcharge area, and over the pipe. The surcharge pressure, PL, at a point directly under E is the sum of the pressure due to each of the four re due to each sub-area is calculated by multiplying the surcharge pressure sub-area loads. Refer to Figure 3-5 A. ce by an Influence Value, IV. Influence Values are proportionality constants The pressure due to each sub-area is calculated by multiplying the surcharge re what portion of a surface load reaches the subsurface point in question. pressure at the surface by an Influence Value, IV. Influence Values are proportionality derived usingconstants the Boussinesq Equation and are giventheinsubsurface Table point 2-5.in that measure what portion of a surface load reaches question. They were derived using the Boussinesq Equation and are given in Table 3-6. (3-6)

Where: 154-264.indd

205

P L = p a + pb + pc + p d

Eq. 2-6

1/16/09 9:57:06 AM

51

pa = pressure to sub-area a, lb/ft2 206 Chapterdue 6 Design of PE Piping Systems pb = pressure due to sub-area b, lb/ft2 pc = pressure due to sub-area c, lb/ft2 pd = pressure due to sub-area d, lb/ft2

51

pa = pressure due to to the sub-area a, lb/ft2 equals: o the surcharge applied i-th sub-area Where pb = pressure due to sub-area b, lb/ft2 PL = vertical soil pressure due to surcharge 2pressure, lb/ft 2 pc = pressure due to sub-area c, lb/ft pa = pressure due to sub-area a, lb/ft 2 pd = pressure due to tosub-area d,2 lb/ft2 pb = pressure sub-area b, lb/ft pi = Idue V wS

o the

pc = pressure due to sub-area c, lb/ft 2 surchargepdapplied to tothe i-th d,sub-area = pressure due sub-area lb/ft 2

Where:

Eq. 2-7

equals:

Pressure due to the surcharge applied to the i-th sub-area equals:

(3-7) Eq. 2-7 pValue IV = Influence wS Table 2-5 i = I V from wS = distributed pressure of surcharge load at ground surface, lb/ft2 Where IV = Influence Value from Table 3-6

Where: areas are equivalent, thenpressure Equation 12 may be simplified wS = distributed of surcharge load at ground surface, lb/ft 2 to IV = Influence Value from Table 2-5 If the four sub-areas are equivalent, then Equationload 3-7 may simplified to: wS = distributed pressure of surcharge atbeground surface, lb/ft2 (3-8) Eq. 2-8 P L = 4I V wS

The influence is dependent upon dimensions areas are equivalent, thenvalue Equation 12 may bethe simplified toof the rectangular area and upon the depth to the pipe crown, H. Table 3-6 Influence Value terms depicted in Figure 3-6, are as: dependent upon thedefined dimensions of the rectangular area and upon

value is H = Table depth of ftInfluence Value terms depicted in Figure e pipe crown, H. 2-5 4I Eq.2-5, 2-8 P L =cover, V wS M = horizontal distance, normal to the pipe centerline, from the center of the load to the load edge, ft :

N = horizontal distance, parallel to the pipe centerline, from the center of the load to the load edge, ft

value is dependent upon the the rectangular area and Interpolation maydimensions be used to findofvalues not given in Table 3-6. Theupon influence value th of cover, ft e pipe crown, H. Table 2-5 Influence Value depicted Figure gives the portion (or influence) of the terms load that reaches a in given depth2-5, beneath the normal the pipe :izontal distance,corner of thetoloaded area. centerline, from the center of the load he load edge, ft zontal distance, parallel to the pipe centerline, from the center of the load he edge, th load of cover, ft ft

5252

izontal distance, normal to the pipe centerline, from the center of the load he load edge, ft may be used to find values not given in Table 2-5. The influence value zontal parallel the pipe center of the load ion (ordistance, influence) of thetoload that centerline, reaches a from giventhedepth beneath the he load edge, ft oaded area.

may be used to find values not given in Table 2-5. The influence value ion (or influence) of the load that reaches a given depth beneath the oaded area. Figure 3-5 Illustration of Distributed Loads

Figure Figure2-5: 2-5: Illustration IllustrationofofDistributed DistributedLoads Loads

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Chapter 6 207


Table 3-6 Influence Values, IV for Distributed Loads* N/H M/H

0.062

0.089

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

∞

0.1

0.005 0.009 0.013 0.017 0.020 0.022 0.024 0.026 0.027 0.028 0.029 0.030 0.031 0.032

0.2

0.009 0.018 0.026 0.033 0.039 0.043 0.047 0.050 0.053 0.055 0.057 0.060 0.061 0.062

0.3

0.013 0.026 0.037 0.047 0.056 0.063 0.069 0.073 0.077 0.079 0.083 0.086 0.089 0.090

0.4

0.017 0.033 0.047 0.060 0.071 0.080 0.087 0.093 0.098 0.101 0.106 0.110 0.113 0.115

0.5

0.020 0.039 0.056 0.071 0.084 0.095 0.103 0.110 0.116 0.120 0.126 0.131 0.135 0.137

0.6

0.022 0.043 0.063 0.080 0.095 0.107 0.117 0.125 0.131 0.136 0.143 0.149 0.153 0.156

0.7

0.024 0.047 0.069 0.087 0.103 0.117 0.128 0.137 0.144 0.149 0.157 0.164 0.169 0.172

0.8

0.026 0.050 0.073 0.093 0.110 0.125 0.137 0.146 0.154 0.160 0.168 0.176 0.181 0.185

0.9

0.027 0.053 0.077 0.098 0.116 0.131 0.144 0.154 0.162 0.168 0.178 0.186 0.192 0.196

1.0

0.028 0.055 0.079 0.101 0.120 0.136 0.149 0.160 0.168 0.175 0.185 0.194 0.200 0.205

1.2

0.029 0.057 0.083 0.106 0.126 0.143 0.157 0.168 0.178 0.185 0.196 0.205 0.209 0.212

1.5

0.030 0.060 0.086 0.110 0.131 0.149 0.164 0.176 0.186 0.194 0.205 0.211 0.216 0.223

2.0

0.031 0.061 0.088 0.113 0.135 0.153 0.169 0.181 0.192 0.200 0.209 0.216 0.232 0.240

53

0.032 0.062 0.156 0.089 0.116 0.137 0.185 0.156 0.172 0.185 0.196 0.212 0.223 0.250 0.116 0.137 0.172 0.196 0.2050.2050.212 0.2230.2400.240 ∞

0.250

* H, M, and N are per Figure 3-5.

Vertical Surcharge Example # 1 al Surcharge Example #1

Find the vertical surcharge load for the 4’ x 6’, 2000 lb/ft2 footing shown below.

he vertical surcharge load for the 4' x 6', 2000 lb/ft2 footing shown below.

SOLUTION: Use equations 3-6 and 3-7, Table 3-6, and Figure 3-5. The 4 ft x 6 ft footing is divided into four sub-areas, such that the common corner of the sub-areas equations 2-6the and 2-7, Figure The 4 ft xa 6 is directly over pipe. SinceTable the pipe2-5, is notand centered under2-5. the load, sub-areas

TION: Use ft g is divided into four sub-areas, such that the common corner of the sub-areas is y over the pipe. Since the pipe is not centered under the load, sub-areas a and b dimensions of 3 ft x 2.5 ft, and sub-areas c and d have dimensions of 3 ft x 1.5 ft. 154-264.indd 207

1/16/09 9:57:07 AM

208 Chapter 6


and b have dimensions of 3 ft x 2.5 ft, and sub-areas c and d have dimensions of 3 ft x 1.5 ft. 54

pi

Determine sub-area dimensions for M, N, and H, then calculate M/H and N/H. Find , pb, pc, pd, and the 190 Influence Value from190 Table 3-6, then solve126 for each sub area, pa126 sum for PL. Sub-area

re: PL = 632 lbs/ft

a

b

c

d

2M

2.5

2.5

1.5

1.5

N

3.0

3.0

3.0

3.0

M/H

0.5

0.5

0.3

0.3

N/H

0.6

0.6

0.6

0.6

pi

190

190

126

126

jacent to, but Not Directly Beneath, a0.095 Surcharge IV 0.095 0.063 Load 0.063

sign method may be used find the surcharge load on buried pipes near, but 2 Therefore: PL = 632 to lbs/ft ctly below, uniformly distributed loads such as concrete slabs, footings and r other rectangular area loads, including wheel loads that are not directly over Pipe Adjacent to, but Not Directly Beneath, a Surcharge Load .

This design method may be used to find the surcharge load on buried pipes near, but not directly below, uniformly distributed loads such as concrete slabs, footings and tical pressure is found by first adding an imaginary loaded area that covers the floors, or other rectangular area loads, including wheel loads that are not directly en determiningover thethesurcharge pressure due to the overall load (actual and pipe.

ry) based on the previous section, and finally by deducting the pressure due to vertical is foundload. by first adding an imaginary loaded area that covers ginary load fromThe that due pressure to the overall

the pipe, then determining the surcharge pressure due to the overall load (actual and imaginary) based on the previous section, and finally by deducting the pressure due Since there is no surcharge directly above the pipe centerline, an to the imaginary load from that due to the overall load.

Figure 2-5 B. ry surcharge load, having the same pressure per unit area as the actual load, is B. Since there is no surcharge directly above theand pipe b+c centerline, to sub-areas c Refer and tod.Figure The3-5 surcharge pressure for sub-areas a+d are an imaginary surcharge the same pressure perdunit as the actual ned, then the surcharge loads fromload, the having imaginary areas c and arearea deducted to load, is applied to sub-areas c and d. The surcharge pressure for sub-areas a+d and ne the surcharge pressure on the pipe.

b+c are determined, then the surcharge loads from the imaginary areas c and d are deducted to determine the surcharge pressure on the pipe. (3-9)

P L = p a+d + pb+c - p d - p c

Eq. 2-9

Where terms are as previously defined above, and 2 Pa+dare = surcharge load of combined sub-areas above, a and d, lb/ftand where terms as previously defined

Pb+c = surcharge load of combined sub-areas b and c, lb/ft 2

pa+d = surcharge load of combined sub-areas a and d, lb/ft2 pb+c = surcharge load of combined sub-areas b and c, lb/ft2

Surcharge Example # 2 154-264.indd 208

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Chapter 6 209


55

nd the vertical surcharge pressure for the 6' x 10', 2000 lb/ft2 slab shown below. Vertical Surcharge Example # 2

Find the vertical surcharge pressure for the 6’ x 10’, 2000 lb/ft2 slab shown below.

SOLUTION: Use Equations 3-7 and 3-9, Table 3-6, and Figure 3-5 B. The surcharge area is divided into two sub-areas, a and b. The area between the surcharge and the lineEquations of the pipe crown divided two sub-areas, c and d, as well. TheB. imaginary OLUTION: Use 2-7 isand 2-9,into Table 2-5, and Figure 2-5 The surcharge load is applied to sub-areas c and d. Next, the four sub-areas are treated as a single ea is divided into two sub-areas, a and b. The area between the surcharge and the surcharge area. Unlike the previous example, the pipe is located under the edge e of the pipe crown is divided into two sub-areas, c and d, as well. The imaginary of the surcharge area rather than the center. So, the surcharge pressures for the ad is applied tocombined sub-areas c and d. Next, the four sub-areas are treated as a single sub-areas a+d and b+c are determined, and then for the sub-areas c and rcharge area. d.Unlike the previous example, the pipe is located under the edge of the The surcharge pressure is the sum of the surcharge pressure due to the surcharge rcharge area rather thea+d center. the surcharge pressures for the combined acting onthan sub-areas and b+c, So, less the imaginary pressure due to the imaginary b-areas a+d and b+c are determined, and then for the sub-areas c and d. The surcharge acting on sub-areas c and d.

rcharge pressure is the sum of the surcharge pressure due to the surcharge acting on Sub-area b-areas a+d and b+c, less the imaginary pressure due to the imaginary surcharge a+d b+c c d ting on sub-areas Mc and d.10 10 4 4 N

5

5

5

5

M/H

2.0

2.0

0.8

0.8

N/H

1.0

1.0

1.0

IV

0.200

0.200

0.160

400

400

pi

a+d

Therefore PL = 160 lb/ft 2

(320)

b+c

1.0

Sub-area 0.160 (320)

c

d

10

10

4

4

N

5

5

5

5

M/H

2.0

2.0

0.8

0.8

N/H

1.0

1.0

1.0

1.0

IV

0.200

0.200

0.160

0.160

pi

400

400

(320)

(320)

M

154-264.indd 209

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210

Chapter 6


Installation Category 1: Standard Installation - Trench or Embankment Pipe Reaction to Earth, Live, and Surcharge Loads

Now might be a good time to review the “Design Process” that appeared earlier in Section 3. After calculating the vertical pressure applied to the pipe the next design step is to choose a trial pipe (DR or profile). Then, based on the Installation Category and the selected embedment and compaction, calculate the anticipated deflection and resistance to crush and buckling. The Standard Installation category applies to pipes that are installed between 18 inches and 50 feet of cover. Where surcharge, traffic, or rail load may occur, the pipe must have at least one full diameter of cover. If such cover is not available, then the application design must also consider limitations under the Shallow Cover Vehicular Loading Installation category. Where the cover depth exceeds 50 ft an alternate treatment for dead loads is given under the Deep Fill Installation category. Where ground water occurs above the pipe’s invert and the pipe has less than two diameters of cover, the potential for the occurrence of flotation or upward movement of the pipe may exist. See Shallow Cover Flotation Effects. While the Standard Installation is suitable for up to 50 feet of cover, it may be used for more cover. The 50 feet limit is based on A. Howard’s (3) recommended limit for use of E’ values. Above 50 feet, the E’ values given in Table B.1.1 in Chapter 3 Appendix are generally thought to be overly conservative as they are not corrected for the increase in embedment stiffness that occurs with depth as a result of the higher confinement pressure within the soil mass. In addition, significant arching occurs at depths greater than 50 feet. The Standard Installation, as well as the other design categories for buried PE pipe, looks at a ring or circumferential cross-section of pipe and neglects longitudinal loading, which is normally insignificant. They also ignore the re-rounding effect of internal pressurization. Since re-rounding reduces deflection and stress in the pipe, ignoring it is conservative. Ring Deflection Ring deflection is the normal response of flexible pipes to soil pressure. It is also a beneficial response in that it leads to the redistribution of soil stress and the initiation of arching. Ring deflection can be controlled within acceptable limits by the selection of appropriate pipe embedment materials, compaction levels, trench width and, in some cases, the pipe itself. The magnitude of ring deflection is inversely proportional to the combined stiffness of the pipe and the embedment soil. M. Spangler (4) characterized this relationship

154-264.indd 210

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57 approach for soil characterization, thus developing the Modified Iowa Formula. In 1964, Burns and Richards [6] published a closed-form solution for ring deflection and pipe stress based linear elasticity. In 1976 M. Katona et. al. [7] Formula. developed finite Chapter 6 211 approach for on soilclassical characterization, thus developing the Modified Iowa Ina1964, Design of PE Piping Systems element program called (Culvert Analysis and Design) is now available in Burns and Richards [6]CANDE published a closed-form solution for which ring deflection and pipe astress PC version and can be used to predict pipe deflection and stresses. based on classical linear elasticity. In 1976 M. Katona et. al. [7] developed a finite

element program called CANDE (Culvert Analysis and Design) which is now available in The recent maytomake better the Iowa Formula, but they a PCmore version andsolutions can be used predict pipepredictions deflection than and stresses. require detailed information on soil and pipe properties, e.g. more soil lab testing. Often (5) in the Iowa Formula in 1941. R. Watkins modified this equation to allow a simpler the improvement in precision is all but lost in construction variability. Therefore, the soil characterization, developing the Modified Formula. In The moreapproach recent for solutions may makethus better predictions than Iowa the Iowa Formula, but they (6) Modified Iowa Formula remains the most frequently used method of determining ring 1964, Burns and Richards closed-form solution for more ring deflection require detailed information on published soil and apipe properties, e.g. soil laband testing. Often (7) deflection. pipe stress based on classical elasticity. M. Katona et.variability. al. developed a the improvement in precision islinear all but lost In in 1976 construction Therefore, the program called CANDE (Culvert Analysis used and Design) whichofis determining now Modified finite Iowaelement Formula remains the most frequently method ring available in a PC version and can be used to predict pipe deflection and stresses. Spangler's deflection. Modified Iowa Formula can be written for use with conventionally extruded DR pipe as: The more recent solutions may make better predictions than the Iowa Formula,

butModified they require detailed information and pipe properties, e.g.conventionally more soil lab Spangler's Iowa Formula canon besoilwritten for use with extruded testing. Often the improvement in precision is all but lost in construction variability. DR pipe as: Therefore, the Modified Iowa Formula remains the most frequently used method of determining ring deflection.





 Iowa Formula can be writtenfor use with solid wall PE pipe as: Spangler’s Modified (3-10)

1  K BED LDL PE + K BED PL  ·  3 D M 144 §¨ 2E  1  ′ ¸  +0.061F SE  K P L-DL1 P 1 ¨ 3KBED ¸ 'X E BED L DR  ¸ ¨ = 3 D M 144 ¨ 2E § 1 · ¸ ¨ 3 ¨ DR - 1 ¸ + 0.061FS E c ¸ © ¹ ¹ ©

∆X

=

Eq. 2-10 Eq. 2-10

and for use ASTM F894F894 profile wall as: and with for use with ASTM profile wallpipe pipe as: (3-11)    F894 profile wall pipe as: and for use with ASTM K BED LDL ∆X P   = ·  1.24(RSC) § 144 DI + 0.061FS E ′ ¸ ¨  'X P ¨ D M K BED LDL ¸ = ¸ ¨ 1.24(RSC) 144 WhereD I + 0.061FS E c ¸ Where: ¨ in ∆ X = Horizontal deflection, DM ¹ ©

Eq. 2-11 Eq. 2-11

K BED = Bedding factor, typically 0.1 LDL = ǻX Deflection factor Where: = lag Horizontal deflection, in PE = Vertical soil pressure due to earth load, typically psf 0.1 KBED = Bedding factor, PL = Vertical soil pressure due to live load, psf = Deflection lag factor LǻX DL Horizontal deflection, in 2 E = Apparent=modulus of elasticity of pipe material, lb/in P = Vertical soil pressure due to earth load, psf E KBEDof Soil = Bedding E’ =Modulus reaction, psi factor, typically 0.1 P = Vertical soil pressure due to live load, psf L = Factor Deflection lag factor LDL FS = Soil Support ERing=Stiffness Apparent modulus of elasticity of pipe material, lb/in2 RSC = P Constant,soil lb/ft pressure due to earth load, psf E = Vertical E' Modulus of Soil reaction, psi PL== Vertical DR = Dimension Ratio, OD/t soil pressure due to live load, psf F = Soil Support Factorof elasticity of pipe material, lb/in2 DM = Mean (DI+2z or DOmodulus -t), in ES =diameter Apparent

RSC = Ring Stiffness Constant, z = Centroid section, in of Soil E' =of wall Modulus reaction,lb/ft psi DR Ratio, OD/t t = Minimum wall Dimension thickness, in FS == Soil Support Factor

D = =diameter, Mean (DI+2z or lb/ft DO-t), M DI = pipe inside indiameterConstant, RSC Ring Stiffness DO = pipe diameter, in DRoutside = Dimension Ratio, OD/t

in

DM = Mean diameter (DI+2z or DO-t), in

154-264.indd 211

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212

Chapter 6


Deflection is reported as a percent of the diameter which can be found by multiplying 100 times ∆X/DM or ∆X/DI. (When using RSC, the units of conversion are accounted for in Equation 3-11.) Apparent Modulus of Elasticity for Pipe Material, E

59

The apparent modulus of PE is dependent on load-rate or, duration of laoding and temperature.Apparent elastic modulus values for high and medium density PE may be found in Table B.1.1 in Chapter 3 Appendix. These values can be used in Spangler’s Iowa Formula. It has long been an industry practice to use the short-term modulus in the Iowa Formula for thermoplastic pipe. This is based on the idea that, in granular embedment soil, deformation is a series of instantaneous deformations consisting rearrangement and fracturing of grains while bending stress in 2-6: Design Values forofApparent Modulus of Elasticity, E @the73°F the pipe wall is decreasing due to stress relaxation. Use of the short-term modulus has proven10 effective 100 and reliable for corrugated uration Short1000 1 yearand profile 10 wall pipes. 50 These pipes typically have pipe stiffness values of 46 psi or less when measured per ASTM Term hours hours hours years years D2412. Conventional DR pipes starting with DR17 or lower have significantly higher stiffness and therefore they may carry a greater proportion of the earth and live load PE 110,000 57,500 51,200 43,700 38,000 31,600 28,200 than corrugated or profile pipe; so it is conservative to use the 50-year modulus for lus of DR pipes that have low DR values when determining deflection due to earth load.

ity, psi

PE lus of ity, psi

s Constant,

Vehicle loads are generally met with a higher modulus than earth loads, as load duration may be nearly instantaneous for moving vehicles. The deflection due 88,000 46,000 41,000 35,000 30,400 25,300 22,600 to a combination of vehicle or temporary loads and earth load may be found by separately calculating the deflection due to each load using the modulus appropriate for the expected load duration, then adding the resulting deflections together to get the total deflection. When doing the deflection calculation for vehicle load, the Lag Factor will be one. An alternate, but conservative, method for finding deflection for combined vehicle and earth load is to do one calculation using the 50-year modulus, RSC but separate the vertical soil pressure into an earth load component and a live load component and apply the Lag Factor only to the earth load component.

pes manufactured to ASTM F894, “Standard Specification for Polyethylene Diameter ProfileRing Wall Sewer and Drain Pipe,” are classified on the basis of Stiffness Constant, RSC ffness ConstantProfile (RSC). 2-12 gives the F894, RSC.“Standard Specification for wall Equation pipes manufactured to ASTM

Polyethylene (PE) Large Diameter Profile Wall Sewer and Drain Pipe,” are classified on the basis of their Ring Stiffness Constant (RSC). Equation 3-12 gives the RSC. (3-12)

RSC

=

6 . 44 EI D M2

Eq. 2-12

Where: E =

Apparent modulus of elasticity of pipe material (Short-term value Table 2-6) @73oF I = Pipe wall moment of inertia, in4/in (t3/12, if solid wall construction) z = Pipe wall centroid in 154-264.indd 212

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Chapter 6 213


Where E = Apparent modulus of elasticity of pipe material @73°F (See Chapter 3 Appendix)

I = Pipe wall moment of inertia, in4/in (t 3/12, if solid wall construction)

z = Pipe wall centroid in DI = Pipe inside diameter in DM = Mean diameter (DI + 2z or DO-t), in

t = Minimum wall thickness, in

Modulus of Soil Reaction, E’

The soil reaction modulus is proportional to the embedment soil’s resistance to the lateral expansion of the pipe. There are no convenient laboratory tests to determine the soil reaction modulus for a given soil. A. Howard (8) determined E’ values empirically from numerous field deflection measurements by substituting site parameters (i.e. depth of cover, soil weight) into Spangler’s equation and “backcalculating” E’. Howard developed a table for the Bureau of Reclamation relating E’ values to soil types and compaction efforts. See Table 3-7. In back-calculating E’, Howard assumed the prism load was applied to the pipe. Therefore, Table 3-7 E’ values indirectly include load reduction due to arching and are suitable for use only with the prism load. In 2006, Howard published a paper reviewing his original 1977 publication from which Table 3-7 is taken. For the most part the recent work indicates that the E’ values in Table 3-7 are conservative. Due to differences in construction procedures, soil texture and density, pipe placement, and insitu soil characteristics, pipe deflection varies along the length of a pipeline. Petroff (9) has shown that deflection measurements along a pipeline typically fit the Normal Distribution curve. To determine the anticipated maximum deflection using Eq. 3-10 or 3-11, variability may be accommodated by reducing the Table 3-7 E’ value by 25%, or by adding to the calculated deflection percentage the correction for ‘accuracy’ percentage given in Table 3-7. In shallow installations, the full value of the E’ given in Table 3-7 may not develop. This is due to the lack of “soil confining pressure” to hold individual soil grains tightly together and stiffen the embedment. Increased weight or equivalently, depth, increases the confining pressure and, thus, the E’. J. Hartley and J. Duncan (10) published recommended E’ values based on depth of cover. See Table 3-8. These are particularly useful for shallow installations. Chapter 7, “Underground Installation of PE Pipe” covers soil classification for pipe embedment materials and preferred methods of compaction and installation for selected embedment materials. Some of the materials shown in Table 3-7 may not be appropriate for all pipe installation. One example would be fine-grained soils in wet ground, which would not be appropriate embedment, under most circumstances, for either profile pipe or pipes with high DR’s. Such limitations are discussed in Chapter 7.

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214 Chapter 6


Table 3-7 Values of E’ for Pipe Embedment (See Howard (8)) E’ for Degree of Embedment Compaction, lb/in2 Slight, Dumped Soil Type-pipe Embedment Material (Unified Classification System)1 Fine-grained Soils (LL > 50)2 Soils with medium to high plasticity; CH, MH, CH-MH

70% Relative Density

No data available: consult a competent soils engineer, otherwise, use E’ = 0.

Fine-grained Soils (LL < 50) Soils with medium to no plasticity, CL, ML, MLCL, with less than 25% coarse grained particles.

50

200

400

1000

Fine-grained Soils (LL < 50) Soils with medium to no plasticity, CL, ML, ML-CL, with more than 25% coarse grained particles; Coarse-grained Soils with Fines, GM, GC, SM, SC3 containing more than 12% fines.

100

400

1000

2000

Coarse-grained soils with Little or No Fines GW, GP, SW, SP3 containing less than 12% fines

200

1000

2000

3000

Crushed Rock

1000

3000

3000

3000

Accuracy in Terms of Percentage Deflection4

± 2%

±2%

±1%

±0.5%

1

ASTM D-2487, USBR Designation E-3 LL = Liquid Limit 3 Or any borderline soil beginning with one of these symbols (i.e., GM-GC, GC-SC). 4 For ±1% accuracy and predicted deflection of 3%, actual deflection would be between 2% and 4%. 2

Note: Values applicable only for fills less than 50 ft (15 m). Table does not include any safety factor. For use in predicting initial deflections only; appropriate Deflection Lag Factor must be applied for long-term deflections. If embedment falls on the borderline between two compaction categories, select lower E’ value, or average the two values. Percentage Proctor based on laboratory maximum dry density from test standards using 12,500 ft-lb/cu ft (598,000 J/m2) (ASTM D-698, AASHTO T-99, USBR Designation E-11). 1 psi = 6.9 KPa.

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Chapter 6 215


Table 3-8 Values of E’ for Pipe Embedment (See Duncan and Hartley(10))

Type of Soil Fine-grained soils with less than 25% sand content (CL, ML, CL-ML)

Coarse-grained soils with fines (SM, SC)

Coarse-grained soils with little or no fines (SP, SW, GP, GW)

Depth of Cover, ft

E’ for Standard AASHTO Relative Compaction, lb/in2 85%

90%

95%

100%

0-5

500

700

1000

1500

5-10

600

1000

1400

2000

10-15

700

1200

1600

2300

15-20

800

1300

1800

2600

0-5

600

1000

1200

1900

5-10

900

1400

1800

2700

10-15

1000

1500

2100

3200

15-20

1100

1600

2400

3700

0-5

700

1000

1600

2500

5-10

1000

1500

2200

3300

10-15

1050

1600

2400

3600

15-20

1100

1700

2500

3800

Soil Support Factor, FS

Ring deflection and the accompanying horizontal diameter expansion create lateral earth pressure which is transmitted through the embedment soil and into the trench sidewall. This may cause the sidewall soil to compress. If the compression is significant, the embedment can move laterally, resulting in an increase in pipe deflection. Sidewall soil compression is of particular concern when the insitu soil is loose, soft, or highly compressible, such as marsh clay, peat, saturated organic soil, etc. The net effect of sidewall compressibility is a reduction in the soil-pipe system’s stiffness. The reverse case may occur as well if the insitu soil is stiffer than the embedment soil; e.g. the insitu soil may enhance the embedment giving it more resistance to deflection. The Soil Support Factor, FS, is a factor that may be applied to E’ to correct for the difference in stiffness between the insitu and embedment soils. Where the insitu soil is less stiff than the embedment, FS is a reduction factor. Where it is stiffer, FS is an enhancement factor, i.e. greater than one. The Soil Support Factor, FS, may be obtained from Tables 3-9 and 3-10 as follows: • Determine the ratio Bd/DO, where Bd equals the trench width at the pipe springline (inches), and DO equals the pipe outside diameter (inches). • Based on the native insitu soil properties, find the soil reaction modulus for the insitu soil, E’N in Table 3-9. • Determine the ratio E’N/E’. • Enter Table 3-10 with the ratios Bd/DO and E’N/E’ and find FS.

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216 Chapter 6


Table 3-9 Values of E’N, Native Soil Modulus of Soil Reaction, Howard (3) Native In Situ Soils Granular

Cohesive

Std. Pentration ASTM D1586 Blows/ft

Description

Unconfined Compressive Strength (TSF)

Description

>0-1

very, very loose

> 0 - 0.125

very, very soft

50

1-2

very loose

0.125 - 0.25

very soft

200

E’N (psi)

2-4

very loose

0.25 - 0.50

soft

700

4-8

loose

0.50 - 1.00

medium

1,500

8 - 15

slightly compact

1.00 - 2.00

stiff

3,000

15 - 30

compact

2.00 - 4.00

very stiff

5,000

30 - 50

dense

4.00 - 6.00

hard

10,000

> 50

very dense

> 6.00

very hard

20,000

Rock

–

–

–

50,000

Table 3-10 Soil Support Factor, FS E’N/E’

Bd/DO 1.5

Bd/DO 2.0

Bd/DO 2.5

Bd/DO 3.0

Bd/DO 4.0

Bd/DO 5.0

0.1

0.15

0.30

0.60

0.80

0.90

1.00

0.2

0.30

0.45

0.70

0.85

0.92

1.00

0.4

0.50

0.60

0.80

0.90

0.95

1.00

0.6

0.70

0.80

0.90

0.95

1.00

1.00

0.8

0.85

0.90

0.95

0.98

1.00

1.00

1.0

1.00

1.00

1.00

1.00

1.00

1.00

1.5

1.30

1.15

1.10

1.05

1.00

1.00

2.0

1.50

1.30

1.15

1.10

1.05

1.00

3.0

1.75

1.45

1.30

1.20

1.08

1.00

5.0

2.00

1.60

1.40

1.25

1.10

1.00

Lag Factor and Long-Term Deflection

Spangler observed an increase in ring deflection with time. Settlement of the backfill and consolidation of the embedment under the lateral pressure from the pipe continue to occur after initial installation. To account for this, he recommended applying a lag factor to the Iowa Formula in the range of from 1.25 to 1.5. Lag occurs in installations of both plastic and metal pipes. Howard (3, 11) has shown that the lag factor varies with the type of embedment and the degree of compaction. Many plastic pipe designers use a Lag Factor of 1.0 when using the prism load as it

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factor to the Formula in the range of fromfor 1.25 to 1.5. Lag occurs in applying installations of to occur afterIowa initial installation. To account this, he recommended a lag both plastic and metal pipes. Howard 11] has lag factor varies with factor to the Iowa Formula in the range [3, of from 1.25shown to 1.5.that Lagthe occurs in installations of the type of embedment and the degree[3, of 11] compaction. Many designers use both plastic and metal pipes. Howard has shown that plastic the lagpipe factor varies with Chapter 6 217 Design of PE Piping Systems a Lag of 1.0 when the prism load as it accounts for backfill settlement. the typeFactor of embedment and using the degree of compaction. Many plastic pipe designers use This makes even more sense when the Soil Support Factor is included in the a Lag Factor of 1.0 when using the prism load as it accounts for backfill settlement. calculation. This makes even more sense when the Soil Support Factor is included in the calculation. Vertical Deflection Example accounts for backfill settlement. This makes even more sense when the Soil Support Vertical Deflection Example Estimate the vertical deflection ofcalculation. a 24” diameter HDPE DR 26 pipe installed under 18 Factor is included in the feet of cover. The embedment material is a well-graded sandy gravel, compacted a Estimate the vertical deflection of a 24” diameter HDPE DR 26 pipe installed underto18 Vertical Deflection Example minimum 90 percent of Standardmaterial Proctorisdensity, and thesandy nativegravel, groundcompacted is a saturated, feet of cover. The embedment a well-graded to a Estimate the vertical deflection of a density, 24” diameter DRthe 26 pipe produced fromisa PE4710 soft clayey90soil. The anticipated trench width is 42”.and minimum percent of Standard Proctor native ground a saturated, material is installedtrench under 18 feet ofiscover. soft clayey soil. The that anticipated width 42”.The embedment material is a wellgraded sandy gravel, compacted to a minimum 90 percent of Standard Proctor Use theand prism load,ground Equation 2-1, Tables 2-7,soil. 2-9,The and 2-10, and Equation density, the native is a saturated, soft clayey anticipated trench widththe is 42”. 2-7 gives an E' for Equation a compacted sandy 2-7, gravel GW-SW soil Equation as 2000 Use prism load, 2-1, Tables 2-9,orand 2-10, and

SOLUTION: 2-10. Table SOLUTION: lb/in2. To estimate maximum due to variability, this or value will besoil reduced by 2-10. Table 2-7 gives an E' deflection for a compacted sandy gravel GW-SW as 2000 SOLUTION: 2 Use the prism load, Equation 3-1, Tables 3-7, 3-9, and 3-10, and Equation 25%,2. or 1500 lb/in . Table 2-9 gives an E’ of 700 psi for soft clay. Since B /D To to estimate maximum deflection due to variability, this value will be reduced by lb/in N d 3-10. Table23-7 gives an E’ for a compacted sandy gravel or GW-SW soil as 2000 lb/in 2. equals 1.75 and E’ /E’ equals 0.47, F is obtained by interpolation and equal 0.60. 25%, or to 1500 lb/in . Table 2-9Modulus gives anElasticity E’N offor700 psi material for softobtained clay. from Since Bd/D N S The Short-Term Apparent of PE 4710 equals 1.75 and E’ /E’ equals 0.47, F is obtained by interpolation and equal 0.60. N S Table B.2.1 equals 130,000 psi. To estimate maximum deflection due to variability, this value will be reduced by 25%, or to 1500 lb/in 2. Table 3-9 gives an E’ of 700

N The prism loadpsion the pipe is equal to: for soft clay. Since Bd/D equals 1.75 and E’N/E’ equals 0.47, FS is obtained by The prism loadinterpolation on the pipe is equal to: and equal 0.60.

The prism load on the pipe is equal to:

2

P E = (120)(18) = 2160lb / ft 2 P E = (120)(18) = 2160lb / ft

Substituting these values into into Equation 2-10 Substituting these values Equation 3-10 gives:gives: Substituting these values into Equation 2-10 gives:    ∆X 2160 § (0.1)(1.0 ) · = ¨ ¸ ' 0) 110,000) 1(0.1)(1. 144 ¨ 2((130,000)  3 DXM = 2160  2(110,000) ( 1 ) + (0.061)(0.60)(1500) ¸ 144 ¨ ¸ 3 DM ( 26 − 1) 3 + (0.061)(0.60)(1500) ¸ ¨ 3 26 1 © ¹

∆X = 0.025 = 2.5 % ' DXM = 0.025 = 2.5 % DM

Deflection Limits

The designer limits ring deflection in order to control geometric stability of the pipe, wall bending strain, pipeline hydraulic capacity and compatibility with cleaning equipment, and, for bell-and-spigot jointed pipe, its sealing capability. Only the limits for geometric stability and bending strain will be discussed here. Hydraulic capacity is not impaired at deflections less than 7.5%. Geometric stability is lost when the pipe crown flattens and loses its ability to support earth load. Crown flattening occurs with excessive deflection as the increase in horizontal diameter reduces crown curvature. At 25% to 30% deflection, the

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218

Chapter 6


crown may completely reverse its curvature inward and collapse. See Figure 3-1A. A deflection limit of 7.5% provides at least a 3 to 1 safety factor against reverse curvature. Bending strain occurs in the pipe wall as a result of ring deflection — outer-fiber tensile strain at the pipe springline and outer-fiber compressive strain at the crown and invert. While strain limits of 5% have been proposed, Jansen (12) reported that, on tests of PE pipe manufactured from pressure-rated resins and subjected to soil pressure only, “no upper limit from a practical design point of view seems to exist for the bending strain.” In other words, as deflection increases, the pipe’s performance limit will not be overstraining but reverse curvature collapse. Thus, for non-pressure applications, a 7.5 percent deflection limit provides a large safety factor against instability and strain and is considered a safe design deflection. Some engineers will design profile wall pipe and other non-pressure pipe applications to a 5% deflection limit, but allow spot deflections up to 7.5% during field inspection. The deflection limits for pressurized pipe are generally lower than for nonpressurized pipe. This is primarily due to strain considerations. Hoop strain from pressurization adds to the outer-fiber tensile strain. But the internal pressure acts to reround the pipe and, therefore, Eq. 3-10 overpredicts the actual long-term deflection for pressurized pipe. Safe allowable deflections for pressurized pipe are given in Table 3-11. Spangler and Handy (13) give equations for correcting deflection to account for rerounding.

Table 3-11 Safe Deflection Limits for Pressurized Pipe DR or SDR

Safe Deflection as % of Diameter

32.5

7.5

26

7.5

21

7.5

17

6.0

13.5

6.0

11

5.0

9

4.0

7.3

3.0

* Based on Long-Term Design Deflection of Buried Pressurized Pipe given in ASTM F1962.

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Chapter 6 219


Compressive Ring Thrust

Earth pressure exerts a radial-directed force around the circumference of a pipe that results in a compressive ring thrust in the pipe wall. (This thrust is exactly opposite to the tensile hoop thrust induced when a pipe is pressurized.) See Figure 3-1B. Excessive ring compressive thrust may lead to two different performance limits: crushing of the material or buckling (loss of stability) of the pipe wall. See Figure 3-1C. This section will discuss crushing, and the next section will discuss buckling. As is often the case, the radial soil pressure causing the stress is not uniform around the pipe’s circumference. However, for calculation purposes it is assumed uniform and equal to the vertical soil pressure at the pipe crown. Pressure pipes often have internal pressure higher than the radial pressure applied by the soil. As long as there is pressure in the pipe that exceeds the external pressure, the net thrust in the pipe wall is tensile rather than compressive, and wall crush or buckling checks are not necessary. Whether one needs to check this or not can be quickly determined by simply comparing the internal pressure with the vertical soil pressure.

67

Crushing occurs when the compressive stress in the wall exceeds the compressive yield stress of the pipe material. Equations 3-13 and 3-14 give the compressive stress 67 resulting from( earth DRload pressure for conventional extruded DR pipe and P E + and P L )live S =F894 profile wall PE Pipe: Eq. 2-13 for ASTM (3-13)

( P E + P L ) DO 288A ( P E + P L ) DO S= Where 288A due to earth load, psf PE = vertical soil pressure (3-14)

Where:

288 ( P E + P L ) DR S = 288

S=

Eq. 2-13

Eq. 2-14 Eq. 2-14

PL = vertical soil pressure due to live-load, psf

PSE ==pipe vertical soil pressure wall compressive stress, lb/in2 due to earth load, psf Where: PDR soilDpressure due to live-load, psf Dimension Ratio, L ==vertical O/t 2 S = pipe wall compressive stress, D = pipe outside diameter (for profile pipe DOto = Dearth 2HP), load, in I +lb/in PE =O vertical soil pressure due psf D=I == pipe inside diameter,Ratio, in /t Dimension D O PDR vertical soil pressure due to live-load, psf L wall height, in diameter (for profile2 pipe DO = DI + 2HP), in == profile pipe outside SD=HOPpipe wall compressive stress, lb/in 2 AI = = profile wall average diameter, cross-sectional area, D pipe inside DR =(Obtain Dimension Ratio, D /tin in /in the profile wall area fromOthe manufacturer of the profile pipe.) profile wall height, in (for profile pipe DO = DI + 2HP), in DHOP==pipe outside diameter (Note: These equations contain a factor of 144 in the denominator for correct units conversions.) A = profile wall average area, in2/in DI = pipe inside diameter,cross-sectional in HP = profile wall height, in A = profile walla average cross-sectional area, in2/in units conversions.) (Note: These equations contain factor of 144 in the denominator for correct

Note: These equations contain a factor of 144 in the denominator for correct units conversions.)

on 2-14 may overstate the wall stress in profile pipe. Ring deflection in profile pe induces arching. The "Deep Fill Installation" section of this chapter discusses gn and for calculating earth pressure resulting frominarching, 2-14gives may equations overstate the wall stress the in profile pipe. Ring deflection profile 1/16/09 154-264.indd 219

9:57:10 AM

220 Chapter 6


68 Equation 3-14PE may overstate the wall stress in profile 2406 800 pipe. Ring deflection in profile wall pipe induces arching. The “Deep Fill Installation” section of this chapter 68 discusses arching and gives equations for calculating the earth pressure resulting from arching, PRD. PRD is given by Equation 3-23 and may be substituted for PE to long-termdetermine compressive stress value should be PE 2406 800reduced the wall compressive stress when arching occurs. for elevated temperature

The pipeline operation. Temperature design factors used for hydrostatic pressure may be The compressive stress in the pipe wall can be compared to the pipe material used, i.e. 0.5 @ 140°F. Additional temperature design factors may be obtained by allowable compressive stress. If the calculated compressive stress exceeds the reference to Table 1-11 in Section 1 of this chapter. stress, then a lower DR (heavier wall thickness) or heavier profile wall is The long-termallowable compressive stress value should be reduced for elevated temperature required. pipeline operation. Temperature design factors used for hydrostatic pressure may be used, i.e. 0.5 @ 140°F. Ring Compression ExampleAdditional temperature design factors may be obtained by Allowable Compressive Stress reference to Table 1-11 in Section 1 of this chapter. Allowable long-term compressive stress values for the several PE material Find the pipe wall compressive stress in a DRChapter 32.5 HDPE pipe buried under 46 ft of designation codes can ring be found in Appendix, 3. cover. The ground water level is at the surface, the saturated weight of the insitu siltyRing Compression Example 3 long-term compressive stress value should be reduced for elevated temperature . clay soil is 120The lbs/ft pipeline operation. Temperature design factors used for hydrostatic pressure may be used. compressive See temperature ring re-rating or adjustment in the Appendix, Chapter 3. wall stress in a DR factors 32.5 HDPE pipe buried under

Find the pipe 46 ft of SOLUTION: Find the vertical earth pressure acting on the pipe. Use Equation 2-1. cover. The ground water level is at the surface, the saturated weight of the insitu silty3 Compression Example . clay soil is 120Ring lbs/ft Find the pipe wall compressive ringto stress a DR 32.5 PE4710 pipe underthe 46 water Although the net soil pressure is equal theinbuoyant weight ofburied the soil, ft ofacting cover. The waterTherefore level is at thethe surface, saturated weight the insitu pressure is also onground the pipe. totalthepressure (waterofand earth load) SOLUTION: Find thesoil vertical earth 3. pressure acting on the pipe. Use Equation 2-1. silty-clay is 120 lbs/ft can be found using the saturated unit weight of the soil. SOLUTION: Find the vertical earth pressure acting on the pipe. Use Equation 3-1.

Although the net soil pressure is equal to the buoyant weight of the soil, the water Although soilpipe. pressure is equal to the buoyant weight of the soil, the water pressure is also actingtheonnet the total pressure (water and earth load) pcf)(46 ft) the =Therefore 5520Therefore psf thethe P E = (120 pressure is also acting on pipe. total pressure (water and earth can be found using the saturated unit weight of the soil. load) can be found using the saturated unit weight of the soil. Next, solve for the compressive stress.

Next, solve for the compressive stress. P E = (120 pcf)(46 ft) = 5520 psf

(5520 lb / ft 2 )(32.5) S = = 623 lb / inch2 Next, solve for the compressive stress. 288 The compressive stress is well below the allowable limit of 1150 psi for the PE4710 2 2 (5520 lb / )(32.5) ft 2 material given in Appendix, Chapter 3./ compressive is the within the =1000 lb/in allowable stress for HDPE S = stress 623 lb

The Table 2-12.

288

inch

given in

Constrained (Buried) Pipe Wall Buckling

Excessive compressive stress (or thrust) may cause the pipe wall to become unstable 2 and buckle. Buckling from ring compressive initiates locally a large The compressive stress isWall within the 1000 lb/instress allowable stressas for HDPE given Constrained (Buried) Pipe Buckling Table 2-12. “dimple,” and then grows to reverse curvature followed by structural collapse. Resistance to buckling is proportional to the wall thickness divided by the diameter

in

Excessive compressive stress (or thrust) may cause the pipe wall to become unstable and buckle. Buckling from ring compressive stress initiates locally as a large "dimple," Constrained (Buried) Pipe Wall Buckling and then grows to reverse curvature followed by structural collapse. Resistance to buckling is proportional to the wall thickness divided by the diameter raised to a power. Excessive compressive stress (or thrust) may cause the pipe wall to become unstable Therefore the Buried pipe has an added 154-264.indd 220 lower the DR, the higher the resistance. 1/16/09 9:57:10 AM

Chapter 6 221


69 raised to a power. Therefore the lower the DR, the higher the resistance. Buried pipe

Non-pressurized pipes or gravity flow pipes are most likely to have a net compressive has an added resistance due to support (or constraint) from the surrounding soil. 69 stress in the pipe wall and, therefore, the allowable buckling pressure should be pipes flow pipes most likely to have a net calculated andNon-pressurized compared to theor gravity total (soil and are ground water) pressure. For most stress in fluid the pipe wall and,in therefore, the exceeds allowable buckling pressure pressure pipe compressive applications, the pressure the pipe the external pressure, Non-pressurized pipes or gravity flow pipes aretotal most likely to have a pressure. net compressive should be calculated and compared to the (soil and ground water) and the net stress in the pipe wall is tensile. Buckling needs only be considered for that stress in the For pipe wall and, the allowable buckling pressure should be mostunder pressure pipetherefore, applications, theduring fluid pressure in the pipe exceeds theconstruction time the pipe is not pressure, such as and immediately after calculated andexternal compared to the total (soil and ground water) pressure. pressure, and the net stress in the pipe wall is tensile. Buckling needs For most and during system shut-downs. pressure pipe only applications, the pressure pipe pressure, exceedssuch theasexternal pressure, be considered forfluid that time the pipeinisthe not under during and and the net stress in the pipe wall is tensile. Buckling needs only be considered for that immediately after construction and during system shut-downs and, in cases in This chapter gives two equations for calculating buckling. The modified Luscher time the pipe is notaunder pressure, as during and immediately after construction which surge pressure eventsuch can produce a temporary negative internal pressure. Equation for buried pipes that are beneath the ground water level, subject to vacuum and duringis system shut-downs. Under these circumstances the pipe will react much stiffer to buckling as its pressure, or under live load with a shallow cover. These forces act to increase even the modulus is higher under short term loading. When designing, select a modulus slightest eccentricity in the pipe wallthebynegative following deformation inward. While soil the duration of external pressure. For pipemodified that are This chapter appropriate gives twoforequations for calculating buckling. The Luscher pressure alone can create instability, due soilto is less likely to should followbedeformation inward, negative surge, given to to vacuum Equation is forsubjected buried to pipes thatpressure are beneath the consideration ground water level, subject particularly if selecting it is granular. So, dry ground buckling is only considered for deep that gives pipe sufficient collapse strength to resist pressure, or under livea DR load with a the shallow cover.unconstrained These forces act to increase even the applications and is given by the Moore-Selig Equation found in the section, “Buckling of the full applied negative without support deformation for the soil. This inward. is to insure While soil slightest eccentricity in the pipepressure wall by following Pipes in Deep,against Dry Fills.” affects thatsoil resultisinless the embedment developing itsinward, pressure alone can construction create instability, likely to material follow not deformation full design strength. particularly if it is granular. So, dry ground buckling is only considered for deep applications and givengives by the Equation found The in the section, “Buckling of Thisischapter two Moore-Selig equations for calculating buckling. modified Luscher Luscher Constrained Buckling Below Ground Water Level Pipes in Equation Deep,Equation Dryfor Fills.” is for buried pipes that are beneath the ground water level, subject to vacuum pressure, or under live load with a shallow cover. These forces act to

For pipes below the even ground watereccentricity level, operating full or deformation partial vacuum, or increase the slightest in the pipeunder wall by a following subject to live load, Luscher’s equation may be used to determine the allowable inward. soil pressure alone canBelow create instability, soil is less likely to follow Luscher Equation forWhile Constrained Buckling Ground Water Level constrained buckling pressure. Equation and So, 2-16 forbuckling DR and profile pipe deformation inward, particularly if it2-15 is granular. dry are ground is only respectively. considered for deep applications and is given by the Moore-Selig Equation found in For pipes below the ground water level, operating under a full or partial vacuum, or the section, “Buckling of Pipes in Deep, Dry Fills”. subject to live load, Luscher’s equation may be used to determine the allowable constrained buckling pressure. Equation 2-15 and 2-16 are for DR and profile pipe Luscher Equation for Constrained Buckling Below Ground Water Level respectively. For pipes below the ground water level, operating under a full or partial vacuum, 5.65 E or subject to live load, Luscher’s equation may be used to determine the allowable B ′E ′ Eq. 2-15 PWC =bucklingRpressure. constrained Equation 3-15 and 3-16 are for DR and profile pipe 3 N 12( DR − 1) respectively. (3-15)

(3-16)

PWC =

5.65 E RB ′E ′ N 12( DR − 1) 3

Eq. 2-15

PWC =

5.65 EI RB ′E ′ 3 N DM

Eq. 2-16

PWC =

5.65 EI RB ′E ′ 3 N DM

Eq. 2-16

Where: 154-264.indd 221

PWC = allowable constrained buckling pressure, lb/in2

1/16/09 9:57:10 AM

222 Chapter 6


70

Where H GWbuckling pressure, lb/in2 PWC = allowable constrained

R = 1 - 0.33

N = safety factor

H

H GW H R = buoyancy reduction factor HWhere GW = height of ground water above pipe, ft = buoyancy reduction factor Where:HR = depth of cover, ft HGW = height of ground water above pipe, ft RH == depth buoyancy of cover, ft reduction factor HGW = height of ground water above pipe, ft H(3-18) = depth of cover, ft 1 B′ = (-0.065H) 1+ 4 e (3-17)

Where:

R = 1 - 0.33

1 Where B′ = log base number, 2.71828 Where: e = natural 1+ 4 e(-0.065H)

70

Eq. 2-17 Eq. 2-17

Eq. 2-18 Eq. 2-18

E’ = soil reaction modulus, psi

eE== apparent natural log base number, 2.71828 modulus of elasticity, psi E'DR== soil reaction modulus, psi Dimension Ratio Where:EI ==pipe 3/12, if solid wallpsi apparent modulus of(telasticity, wall moment of inertia, in4/in construction) DR =Mean Dimension diameter 2z ornumber, DO – t), in eD=M =natural log(Dbase 2.71828 I +Ratio IE' = soil pipe wall modulus, moment psi of inertia, in4/in (t3/12, if solid wall reaction Although buckling occurs rapidly, long-term external pressure can gradually construction) E = apparent modulus of elasticity, psi deform the pipe to the point of instability. This behavior is considered viscoelastic DR Dimension Ratio DM == Mean diameter (DI + 2z or DO – t), in and can be accounted for in Equations 3-15 and 3-164by using the apparent modulus I = pipe wall moment of inertia, in /in (t3/12, if solid wall of elasticity value for the appropriate time and temperature of the loading. For construction) instance, a vacuum event is resisted by the short-term value of the modulus whereas gh buckling occurs rapidly, long-term external gradually deform the Dcontinuous diameter or Dpressure – resisted t), in can M = Mean I + 2zwould Obe ground water (D pressure by the 50 year value. For o the point of instability. This behavior is considered viscoelastic and can be modulus values see Appendix, Chapter 3. nted for in Equations 2-15 and 2-16 by using the apparent modulus of elasticity For pipes buried with less than 4 ftoforthe a full diametercan of cover, Equations and for appropriate time andlong-term temperature loading. For instance, a3-15 vacuum gh the buckling occurs rapidly, external pressure gradually deform the 3-16 may have limited applicability. In this case the designer may want to use is resisted by the short-term value of the modulus whereas continuous ground o the point of instability. This behavior is considered viscoelastic and can be Equations 3-39 and pressure be resisted by 3-40. the 50by year value. modulus valuesofsee Table nted for inwould Equations 2-15 and 2-16 using the For apparent modulus elasticity for the appropriate time and temperature of the loading. For instance, a vacuum The designer should apply a safety factor commensurate with the application. A is resisted by safety the short-term value the whereas continuous ground factor of 2.0 has beenof used formodulus thermoplastic pipe. pressure would be resisted by the 50 year value. For modulus values pes buried with less than 4 ft or a full diameter of cover, Equations 2-15see andTable 2-16 The allowable constrained buckling pressure should be compared to the total ave limited applicability. In this case the designer may want to use Equations 2-39 vertical stress acting on the pipe crown from the combined load of soil, and ground 40. water or floodwater. It is prudent to check buckling resistance against a ground pes buried withwater less level thanfor4aft100-year-flood. or a full diameter of cover,theEquations 2-16 In this calculation total vertical2-15 stressand is typically ave limited applicability. In this case the designer may want to use Equations 2-39 esigner should taken applyasathe safety commensurate withplus thethe application. prism factor load pressure for saturated soil, fluid pressureA of safety any 40.2.0 has beenfloodwater of used forabove thermoplastic pipe. the ground surface.

esigner apply abuckling safety factor commensurate with the application. safety llowableshould constrained pressure should be compared to the totalAvertical of 2.0 has been used for thermoplastic pipe. acting on the pipe crown from the combined load of soil, and ground water or water. It is prudent to check buckling resistance against a ground water level for a 154-264.indd 222

1/16/09 9:57:11 AM

pressure for saturated soil, plus the fluid pressure of any floodwater above the ombined pressureequation, from soil, ground 2-39, water, and can thenoperate to use with the nstrained Equation to and verifyvacuum, that the pipe d surface.buckling nstrained buckling equation, Equation 2-39, to verify that the pipe can operate with acuum independent of any soil support or soil load, in case construction does not R pipes operating under a vacuum, it is customary to use Equation 2-15 to check acuum of any soilground support orload soiland load, in case construction not6 Chapter op the independent full soil support. Where vacuum is short-term, such then as2-15 during water ombined pressure from soil, water, vacuum, and to does use the R pipes operating under a vacuum, it is customary to use Equation check Design of PEto Piping Systems op the full soil support. Where vacuum load is short-term, such as during water mer events two calculations with Equation 2-14 are necessary. First determine nstrained to and verifyvacuum, that the pipe ombined buckling pressureequation, from soil,Equation ground 2-39, water, and can thenoperate to use with theif mer events two calculations with Equation 2-14 are necessary. First determine pe is sufficient for equation, the water and a long-term modulus; acuum independent of ground any soil support orsoil soil load, that in using case doeswith notif nstrained buckling Equation 2-39, topressure verify the construction pipe can operate pe the is sufficient for theis ground water andcombined soil pressure using a long-term modulus; determine thesupport. pipe sufficient for the ground water, soil and op full ifsoil Where vacuum short-term, such as pressure during water acuum independent of any soil support orload soil is load, in case construction does not determine if the pipe is sufficient for the combined ground water, soil pressure and um using the short-term mer events two support. calculations withmodulus. Equationload 2-14is are necessary. determine op loading the full soil Where vacuum short-term, suchFirst as during waterif um usingforthe modulus. pe loading is sufficient theshort-term ground water and soil pressure usingtoause long-term modulus; mer events two For calculations with Equation 2-14 necessary. First determine if DR pipes operating under a vacuum, itare is customary Equation 3-15 to determine if the pipe is sufficient for the combined ground water, soil pressure and theground combinedwater pressure from soil, ground water, andavacuum, and then to pe is sufficient check for the and soil pressure using long-term modulus; um loadingifusing the theshort-term unconstrained buckling equation, Equation 3-39, to verify the pipe and determine theuse pipe is sufficientmodulus. for the combined ground water, soilthat pressure trained Buckling Example can operate with the modulus. vacuum independent of any soil support or soil load, in case m loading using the short-term trained Bucklingconstruction Example does not develop the full soil support. Where vacuum load is short-

223

such as during water hammer events two calculations with Equationbuckling 3-14 a 36" SDR 26term, HDPE pipe have satisfactory resistance to constrained necessary. First determine if the pipe soil is sufficient fortothe ground water buckling and soil a 36" Buckling SDR HDPE pipe have resistance constrained rained Example installed with26are 18 ft of cover in a satisfactory compacted embedment. Assume ground 2 pressure using a long-term modulus; then determine if the pipe is sufficient for the installed with and 18 ftanofE'cover in a compacted soil embedment. Assume ground to the surface of 1500 lb/in . rained Buckling Example 2 pressure and vacuum loading using the short-term combined ground water, soil . to 36" the surface an E'pipe of 1500 a SDR 26and HDPE havelb/in satisfactory resistance to constrained buckling modulus.

with2618HDPE ft of cover in a satisfactory compacted resistance soil embedment. Assumebuckling ground ainstalled 36" SDR pipe have to constrained 2 UTION: Solve Equation 2-15. Since this is a long-term loading condition, the . to the surface and an E' of 1500 lb/in installed with 18 ft of cover in Example a compacted soil embedment. Assume ground Constrained Buckling 2 tothis UTION: Solve Equation 2-15. Since is a long-term loading condition, the sto relaxation modulus can be assumed be 28,200 psi. Soil cover, H, and ground . have satisfactory resistance to constrained buckling the surface and E'SDR of 1500 lb/inpipe Does an a 36” 26 PE4710 s height, relaxation can assumed to be 28,200 Soil factor, cover, B', H, and ground HGWmodulus , are both 18befeet. soil psi. support is found as when installed with 18Therefore, ft of cover in athe compacted soil embedment? Assume ground , are both 18 feet. Therefore, the soil support factor, B', is found as height, H s; GW UTION: Solve water Equation 2-15. and Since this a 2long-term loading condition, the to the surface an E’ of 1500islb/in . s; s relaxation modulus can be assumed tothis be 28,200 psi. Soil cover, H, and ground UTION: Solve Equation Since3-15. is this a long-term condition, 12-15. SOLUTION: Solve Equation Since is a long-termloading loading condition, the the B = = 0.446 c , are both 18 feet. Therefore, the soil support factor, B', is found as height, H -(0.065)(18) GW 1 relaxation modulus can assumed to be psi. Soil cover, H,Appendix and ground 4 ebe 50 stress relaxation modulus for28,200 PE4710 material is given in the to B ′year = 1+ = 0.446 s; -(0.065)(18) both 18 feet.psi.Therefore, support factor, B', isboth found as height, HGW, are 1+34as , are 18 feet. Chapter Soil cover, H,the and soil ground water height, HGW e 29,000 s; Therefore, the 1 soil support factor, B’, is found as follows; B = = found 0.446 as follows: ′ he bouyancy reduction R, is 1-(0.065)(18) 1+ 4factor, e Bc = = found 0.446 as follows: he bouyancy reduction factor, R, is 18 1+ 4 e-(0.065)(18) R = 1- 0.33 18 = 0.67 18 = 0.67 R = 1- 0.33 he bouyancy reduction factor, R, factor, is found as asfollows: and the bouyancy reduction R, is found follows: 18 he bouyancy reduction factor, 18 R, is found as follows: = the 1- 0.33 = 0.67long-term constrained buckling pressure: Equation 2-15 Rfor allowable 18 Equation 2-15 Rfor allowable = the 1- 0.33 = 0.67long-term constrained buckling pressure: 18 Solve Equation 3-15 for the allowable long-term constrained buckling pressure:

5.65 0.67(0.446 )1500(28,200) Equation 2-15 P for the allowable long-term constrained buckling pressure: WC = 5.65 0.67(0.446)1500(28,200) ( 329,000) 2 12 ( 26 1 ) Equation 2-15 for = allowable long-term3 constrained buckling pressure: PWCthe 2 12(26 − 1) 72 5.65 0.67(0.446)1500(28,200) PWC = 3 23.5 23 = 3387 3340 psf PWC =5.65 0.67(0.446 )15−00(28,200) 2 .2 psi 12(26 1) P WC = = 23 . 2 psi = 3340 psf P WC 2 12(26 1) 3 earth pressure and ground water pressure applied to the pipe is found using The earth pressure and ground water pressure applied to the pipe is found using ion 2-1 (prism load) with a saturated soil weight. The saturated soil weight being (prism load) with 233-1 .2water. psi = 3340 psf a saturated soil weight. The saturated soil weight PWC et weight of bothEquation soil= and being the net weight of both soil and water. PWC = 23.2 psi = 3340 psf lb P E = (120)(18) = 2160 2 ft

are this with the constrained buckling pressure. Since PWC exceeds PE, DR 26 atisfactory resistance to constrained pipe buckling. 154-264.indd 223

1/16/09 9:57:12 AM

224

Chapter 6


74

Installation Category # 2: Shallow Cover Vehicular Loading Compare this with the constrained buckling pressure. Since P

exceeds PE, DR 26

The Standard has Installation methodology assumes that the pipeWCbehaves primarily as a satisfactory resistance to constrained pipe buckling. "membrane" structure, that is, the pipe is almost perfectly flexible with little ability to resist bending.Installation At shallow cover #2: depths, especially those Loading less than one pipe diameter, Category Shallow Cover Vehicular membrane action may not fully develop, andassumes surcharge orpipe livebehaves loadsprimarily place a bending The Standard Installation methodology that the load on the pipe crown. In this case the pipe’s flexural stiffness carries part of the load as a “membrane” structure, that is, the pipe is almost perfectly flexible with little and prevents ability the pipe crown from dimpling under the Equation to resist bending. At shallow cover inward depths, especially thoseload. less than one pipe 2-19, published by diameter, Watkinsmembrane [14] gives the soil pressure that can be supported at the pipe action may not fully develop, and surcharge or live loads place crown by the acombination ofthe the pipe’s stiffness resistance) bending load on pipe crown.flexural In this case the pipe’s(bending flexural stiffness carries and the soil’s internal part resistance heaving addition to under checking of the loadagainst and prevents the pipeupward. crown from In dimpling inward the load.Watkins' (14) formula, the designer should check deflection using Equations 2-10 or 2-11, pipe wall Equation 3-19, published by Watkins gives the soil pressure that can be supported compressive at stress Equations 2-13of or 2-14,flexural and stiffness pipe wall buckling using the pipeusing crown by the combination the pipe’s (bending Equations 2-15 or 2-16. resistance) and the soil’s internal resistance against heaving upward. In addition to checking Watkins’ formula, the designer should check deflection using Equations 3-10 oris3-11, pipe wall compressive stress using or 3-14, pipe wall Watkins' equation recommended only where the Equations depth of3-13 cover is and greater than buckling using Equations 3-15 or 3-16. half of the pipe diameter and the pipe is installed at least 18 inches below the

oneroad surface. In other words, it is recommended that the pipe regardless of diameter always Watkins’ equation is recommended only where the depth of cover is greater than be at least 18”one-half beneath the road surface where areatlive present; of the pipe diameter and the pipe isthere installed leastloads 18 inches below themore may be required depending the words, properties of the pipethat and installation. Indiameter some cases, road surface.on In other it is recommended the pipe regardless of lesser cover depths be18”sufficient therewhere is athere reinforced concrete always bemay at least beneath thewhere road surface are live loads present; cap or a reinforced concrete pavement over the pipe. Equation 2-19and may be used more may be requiredslab depending on the properties of the pipe installation. In for both DR pipe and profile pipe. Seecover definition of “A” below. where there is a reinforced concrete some cases, lesser depths may be sufficient cap or a reinforced concrete pavement slab over the pipe. Equation 3-19 may be used for both DR pipe and profile pipe. See definition of “A” below.

(3-19)

PWAT =

2

12 w(KH ) 7387(I)  w H S - DO  + 2  MAT N S DO 288A  N S DO c 

Eq. 2-19

Where

Where:PWAT = allowable live load pressure at pipe crown for pipes with one diameter or less of cover, psf w = unit weight of soil, lb/ft3

154-264.indd

= outside allowable load pressure at pipe crown for pipes with one PDWAT diameter, live in O = pipe diameter or less of cover, psf H = depth of cover, ft 3 3 /12 forlb/ft moment of inertia DR pipe), in4/in wI = pipe = wall unit weight of(tsoil, 2 /in, for profile pipe or wall thickness (in) for DR pipe wall average cross-sectional area, inin DA O= profile = pipe outside diameter, (obtain the profile from the manufacturer of the profile pipe.) H = depth of cover, ft to wall centroid, in Ic ==outer fiber pipe wall moment of inertia (t3/12 for DR pipe), in4/in c = HP – z for profile pipe and c = 0.5t for DR pipe, in A = profile wall average cross-sectional area, in2/in, for profile pipe or HP = profile wall height, in wall thickness (in) for DR pipe z = pipe wall centroid, in cS = = material outeryield fiber to wall centroid, in strength, lb/in2, Use 3000 PSI for PE3408 MAT c = HP – z for profile pipe and c = 0.5t for DR pipe, in HP = profile wall height, in z = pipe wall centroid, in SMAT = material yield strength, lb/in2, USE 3000 PSI FOR pe3408 NS = safety factor K = passive earth pressure coefficient 224 1/16/09 9:57:12 AM

K=

1 + SIN( I ) 1 - SIN( I )



Ɏ =

angle of internal friction, deg

NS = safety factor K = passive earth pressure coefficient

75

75

quation 2-19 is for a point load applied to the pipe crown. Wheel loads should be etermined using a point load method such as given by Equations 2-2 (Timeoshenko) (3-20) 2-4 (Boussinesq). K = 1 + SIN( I ) Eq. 2-20 1 - SIN( I ) 1 + SIN( φ )

K=

Eq. 2-20

hen a pipe is installed internalshallow friction, degcover below an unpaved surface, rutting can occur 1 - SIN ( φ =) angle ofwith hich will not only reduce cover depth, but also increase the impact factor. Ɏ =


Equation 3-19 is for a point load applied to the pipe crown. Wheel loads should be determined usingdeg a point load method such as given by Equations 3-2 (Timoshenko) angle of internal friction, hallow Cover Example or 3-4 (Boussinesq).

a pipe is installed with shallow cover below an unpaved surface, rutting can Equation 2-19When is for a point load applied to the pipe crown. Wheel loads should be etermine the safety factor against flexural failure pipe accompanied occur which not only reduce cover depth, of but the also increase the impact factor.by soil determined using a pointwill load method such as given by Equations 2-2 (Timeoshenko) eave, for a 36" RSC 100 F894 profile pipe 3.0 feet beneath an H20 wheel load. or 2-4 (Boussinesq). Shallow Coverwith Example ssume asphalt surface granular embedment. for a an point load applied to the pipe crown. Wheel loads should be safetyas factor againstby flexural failure of the pipe(Timeoshenko) accompanied by a point load Determine methodthesuch given Equations 2-2

When a pipe issoil installedforwith cover below an unpaved surface, rutting a 36”shallow RSC 100 F894 profile 3.0 feet an H20 wheel load.can occur OLUTION: The liveheave, load pressure acting at the pipe crown of beneath the pipe can be found using q). which will not only reduce cover depth, also embedment. increase the impact factor. Assume an asphalt surface withbut granular quation 2-4, the Boussinesq point load equation. At 3.0 feet of cover the highest live ad pressure occurs directly a single and SOLUTION: Theunder live load pressurewheel acting at the equals: crown of the pipe can be found

stalled with shallow cover an unpaved using Equation 3-4,below the Boussinesq point load surface, equation. Atrutting 3.0 feet ofcan coveroccur the Shallow Cover Example y reduce cover depth, butpressure also increase theunder impact factor. highest live load occurs directly a single wheel and equals: 3

(3)(2.0)(16000)(3.0 ) = 1697 psf P= Determine pthe safety factor against flexural failure of the pipe accompanied by soil WAT L 5 S 2 (3.0 ) heave, for a 36" RSC 100 F894 profile pipe 3.0 feet beneath an H20 wheel load. xample Assume an asphalt surface with granular embedment. Where If = 2.0

afety factorWhere: against flexural failure of atthe W = 16,000 SOLUTION: The live lbs load pressure acting thepipe crownaccompanied of the pipe canby be soil found using I = 2.0 H = 3.0 ft f RSC 100 F894 profile pipe 3.0 feet beneath an H20 wheel load. Equation 2-4, the Boussinesq point load equation. At 3.0 feet of cover the highest live w = 120 Wpcf=directly 16,000 lbs a single wheel and equals: pressure occurs under altload surface with granular embedment. H = 3.0 ft

120 pcf is to be compared with the value in Equation 3-19. To solve The w live=load pressure pressure acting the crown of are the pipe can be found using Equation 3-19, the at following parameters required: 3

live load (3)(2.0)(1 6000)(3.0 )At 3.0 feet of cover the highest live 4 /in point = inload Ip = 0.171 heBoussinesq live load pressure is to equation. be compared with the value in Equation 2-19. To solve WAT 5 2S (3.0 ) 2single A = 0.470 in /in curs directly under a wheel and equals: quation 2-19, the following parameters are required:

T

=

HP = 2.02 in (Profile Wall Height) DO = DI+2*h = 36.00+2*2.02 = 40.04 in Where: Z = 0.58 I =in 0.171 in4/in 3 C = h-z = 1.44 )in (3)(2.0)(16000)(3.0 If = 2.0 S = 3000 5 psi

2π (3.0 φ = )30 deg.W = 16,000 lbs H = 3.0 ft w = 120 pcf

ere:

The live load pressure is to be compared with the value in Equation 2-19. To solve 1/16/09 9:57:12 AM If =154-264.indd 2.02-19,225the following parameters are required: Equation

C == 3000 h-z = psi 1.44 in SO = DI+2*h = 36.00+2*2.02 = 40.04 in D S = 3000 psi Determine the earth Z pressure coefficient: ĭ 30 deg. ==0.58 in ĭ == h-z 30 deg. C = 1.44 in 226 Chapter 6 S = 3000 psi of PEpressure Piping Systems coefficient: Determine theDesign earth 1+ sin (30) 1+ 0.5 Determine theKearth pressure coefficient: = 30 deg. = ĭ = = 3.0 1-sin(30) 1- 0.5 1+ sin(30) coefficient: 1+ 0.5 Determine theKearth pressure = 1+ sin(30) = 1+ 0.5 = 3.0 K = 1-sin(30) = 1- 0.5 = 3.0 The live load pressure incipient failure Determine1thesin earth pressureto coefficient: (30) 10.5 equals: 1+sin(30) 1+ 0.5 K = = = 3.0 The live load pressure incipient to equals: 1-sin(30) 1-failure 0.5 2 The live load pressure incipient failure equals: (12)120(3. 0 *to3.0 120(40.04)3.0 * 0.171 ) 7387 + (3000 ) PWAT = 2 40.04 (1.44) 288 * 0.470 The live load pressure40.04 incipient to failure equals: 2 The live load pressure incipient0 *to3.0 failure equals: (12)120(3. 120(40.04)3.0 * 0.171 ) 7387 PWAT = (12)120(3.0 * 3.0 )2 +7387 *2 0.171 (3000 - 120(40.04)3.0 ) 40.04 + 40.04 2 (1.44)(3000 - 288 * 0.470 ) PWAT = 1584 = 4498 psf PWAT = 2904 +40.04 40.04 (1.44) 288 * 0.470 2 (12)120(3.0 * 3.0 ) 7387 * 0.171 120(40.04)3.0 + (3000 ) PWAT = 2 = 2904 + 1584 = 4498 psf PWAT 40.04 40.04 (1.44) 288 * 0.470 The resulting safety 2904 +equals: 1584 = 4498 psf PWAT = factor The resulting safety factor equals:

The resulting safety factor+ equals: = 2904 1584 = 4498 psf PWAT P 4498 The resulting safety factor N = WAT = equals: = 2.65 1697 pL 4498 PWAT The resulting safety factor equals: N = = = 2.65 4498 P WAT Installation Category #3: Deep Fill Installation N = p L = 1697 = 2.65 The performance limits for pipes in a deep fill are the same as for any buried pipe. 1697 pL They P include: 4498 Installation Category # 3: Deep Fill Installation N = WAT = = 2.65 1. Compressive ring thrust stress 1697 pL The performance limits a deep fill are the same as for any buried pipe. They Installation Category #for 3:pipes DeepinFill Installation 2. Ring deflection Installation Category # 3: Deep Fill Installation include: 3. Constrained pipe wall buckling

The performance limits for pipes in a deep fill are the same as for any buried pipe. They (1) ringinFill thrust stress, The compressive suggested calculation method forfill pipe deepsame fill applications involves the pipe. They The performance limits#for pipes a deep areinthe as for any buried Installation Category 3: Deep Installation include: (2) ring deflection, and for each performance limit that are different than introduction of design routines include: (3) constrained pipe wall buckling those previously given. (1) compressive ring stress, The performance limits for pipes in thrust a deep fill are the same as for any buried pipe. They (1) compressive ring thrust stress, (2) ring deflection, and Compressive ring thrust is calculated using soil arching. The arching calculation include: (2) ring deflection, and (3) constrained pipe wall buckling may calculation also be used formethod profile pipe designs standard applications. Profile The suggested for pipeinin deeptrench fill applications involves the (3) constrained pipe wall buckling (1) compressive ring thrust stress, pipes are relatively low stiffness pipes where significant arching maydifferent occur at than those introduction of design routines for each performance limit that are relatively shallow depths of cover. (2) ring deflection, and previously given. The suggested calculation pipe method for pipe in deep fill applications involves the (3) constrained wall buckling a calculation depth of around 50for feeteach or for so itperformance becomes to use Spangler’s equation The suggested method pipe inimpractical deep applications involves the introduction ofAtdesign routines limitfill that are different than those published in this chapter because it neglects the significant duethan to those introduction ofasdesign routines for each performance limit thatload arereduction different previously given. arching and the inherent stiffening of the embedment and consequential increase in previously given. The suggested calculation method for pipe in deep fill applications involves the E’ due to the increased lateral earth pressure applied to the embedment. This section introduction of design routines for each performance limit that are different than those gives an alternate deflection equation for use with PE pipes. It was first introduced previously given. (1) by Watkins et al. for metal pipes, but later Gaube extended its use to include PE pipes. (15)

154-264.indd 226

1/16/09 9:57:13 AM

Chapter 6 227


Where deep fill applications are in dry soil, Luscher’s equation (Eq. 3-15 or 3-16) may often be too conservative for design as it considers a radial driving force from ground water or vacuum. Moore and Selig(17) developed a constrained pipe wall buckling equation suitable for pipes in dry soils, which is given in a following section. Considerable care should be taken in the design of deeply buried pipes whose failure may cause slope failure in earthen structures, or refuse piles or whose failure may have severe environmental or economical impact. These cases normally justify the use of methods beyond those given in this Chapter, including finite element analysis and field testing, along with considerable professional design review. Compressive Ring Thrust and the Vertical Arching Factor

The combined horizontal and vertical earth load acting on a buried pipe creates a radially-directed compressive load acting around the pipe’s circumference. When a PE pipe is subjected to ring compression, thrust stress develops around the pipe hoop, and the pipe’s circumference will ever so slightly shorten. The shortening permits “thrust arching,” that is, the pipe hoop thrust stiffness is less than the soil hoop thrust stiffness and, as the pipe deforms, less load follows the pipe. This occurs much like the vertical arching described by Marston.(18) Viscoelasticity enhances this effect. McGrath(19) has shown thrust arching to be the predominant form of arching 78 with PE pipes. Burns and Richard(6) have published equations that give the resulting stress occurring in a pipe due to arching. As discussed above, the arching is usually 78 considered when calculating the ring compressive stress in profile pipes. For deeply − 1 S A buried pipes the Burns and Richard’s equations to derive Eq. 2-21 = 0.88 (19) − 0has .71simplified VAFMcGrath S Aby + Equation 2.5 a vertical arching factor as given 3-21.

S A 1 S A 2.5 VAF = Vertical Arching Factor Where VAF = Vertical Arching Factor Hoop Thrust Stiffness Ratio Where: SSA = = Hoop Thrust Stiffness Ratio (3-21)

Where:

VAF

Eq. 2-21

0.88 0.71

A

VAF = Vertical Arching Factor 1.43 MStiffness S rCENT SA (3-22) = Hoop=Thrust Ratio SA EA Where 1.43to M S rCENT rCENT = radius centroidal axis of pipe, in

S A=

Eq. 2-22 Eq. 2-22

M = one-dimensional EAmodulus of soil, psi Where: E =s apparent modulus of elasticity of pipe material, psi (See Appendix, Chapter 3) A= profile wall average in2 /in, wall thickness rCENT = radius tocross-sectional centroidalarea, axis oforpipe, in (in) for DR pipe

Where: Ms= one-dimensional modulus of soil, psi E = apparent modulus of elasticity of pipe material, psi rCENT = radius to centroidal axis of pipe, in A= profile wall average cross-sectional area, in2/in, or wall Ms= one-dimensional modulus of soil, psi thickness (in) for DR pipe E apparent modulus of elasticity of pipe material, psi 154-264.indd =227 1/16/09

9:57:14 AM

228 Chapter 6


79 One-dimensional modulus values for soil can be obtained from soil testing, geotechnical texts, or Table 3-12 which gives typical values. The typical values in Table 3-12 were obtained by converting values from McGrath (20). 80 100

6000

2900

1300

Table 3-12 Typical Values of Ms, One-Dimensional Modulus of Soil

6500

3200

1450

Gravelly Sand/Gravels Gravelly Sand/Gravels Gravelly Sand/Gravels and extended fromVertical values given1 (psi) by McGrath [20]. For depths not shown in McGrath [20], the Soil Stress 95% Std. Proctor (psi) 90% Std. Proctor (psi) Std. Proctor (psi) s were approximated using the hyperbolic soil model with appropriate values for 85% K and n where 10 3000 1600 550 d K=200, K=100, and K=45 for 95% Proctor, 90% Proctor, and 85% Proctor, respectively. 20 3500 1800 650 Soil Stress (psi) = [ soil depth (ft) x soil density (pcf)]/144 40

4200

2100

800

60

5000

2500

1000

80

6000

2900

1300

100

6500

3200

1450

* Adapted and extended from values given by McGrath (20). For depths not shown in McGrath(20), the MS values were approximated using the hyperbolic soil model with appropriate values for K and n where n=0.4 and K=200, K=100, and K=45 for 95% Proctor, 90% Proctor, and 85% Proctor, respectively.

Soil Stress (psi) = [ soil depth (ft) x soil density (pcf)]/144 dial directed earthVertical pressure can be found by multiplying the prism load (pressure) vertical arching factor as shown in Eq. 2-23.

Where:

1

The radial directed earth pressure can be found by multiplying the prism load (pressure) by the vertical arching factor as shown in Eq. 3-23. (3-23)

P RD = (VAF)wH

Eq. 2-23

Where PRD = radial directed earth pressure, lb/ft 2

w = unit weight of soil, pcf

PRD = radial directed earth pressure, lb/ft2 H = depth of cover, ft w = unit weight of soil, pcf H = depth of cover, ft

The ring compressive stress in the pipe wall can be found by substituting PRD from Equation 3-23 for PE in Equation 3-13 for DR pipe and Equation 3-14 for profile wall pipe.

g compressiveEarth stress in the pipe wall can be found by substituting PRD from Pressure Example n 2-23 for PE in Equation 2-13 for DR pipe and Equation 2-14 for profile wall

Determine the earth pressure acting on a 36” profile wall pipe buried 30 feet deep. The following properties are for one unique 36” profile pipe made from PE3608 material. Other 36” profile pipe may have different properties. The pipe’s crossEarth Pressure sectional Example area, A, equals 0.470 inches2/inch, its radius to the centroidal axis is 18.00 inches plus 0.58 inches, and its apparent modulus is 27,000 psi. Its wall height is 2.02 ine the radial earth acting on(2.02 a 36" RSC profilethe wall buriedin30 in andpressure its DO equals 36 in +2 in) or 40.04100 in. Assume pipepipe is installed a 2 granular soil compacted 90% Standard Proctor (Ms /inch, = 1875 psi), the insitutosoil its radius ep. The pipe'sclean cross-sectional area, A,toequals 0.470 inches is as stiff as the embedment, the backfill weighs 120 pcf. is (Where the excavation troidal axis is 18.00 inches plus 0.58 and inches, and its modulus 28,250 psi. Its

ght is 2.02 in and its DO equals 36 in +2 (2.02 in) or 40.04 in. Assume the pipe is d in a clean granular soil compacted to 90% Standard Proctor (Ms = 1875 psi), tu soil is as stiff as the embedment, and the backfill weighs 120 pcf. (Where the tion is in a stable trench, the stiffness of the insitu soil can generally be ignored alculation.) 154-264.indd 228

1/16/09 9:57:14 AM

Chapter 6 229


is in a stable trench, the stiffness of the insitu soil can generally be ignored in this calculation.) The following series of equations calculates the hoop compressive stress, S, in the pipe wall due to the earth pressure applied by the soil above the pipe. The earth pressure is reduced from the prism load by the vertical arching factor. (From Equation 3-22)

1.43( 1875

lbs 2

)(18.58 inch)

inch = 3.93 3.75 S A = 1.43( 1875 lbs )(18.58 inch) 2 lbs 2 inch (28250 inch )(0.470 ) 2 = 3.75 SA = lbs inch 2 inch lbs 22 )(18.58 1.43( 1875 inch) inch (28250 inch )(0.470 ) 2 S AA = = 3.75 3 . 75 1 inch 2 inch 2 (From Equation VAF = 0.883-21) - 0.71 =inch 0.57 lbs (28250 33.75 )(0.470 + 2.5 inch ) 22.75 − 1 inch VAF = 0.88 - 0.71 = 0. 57 0.56 lb 33.75 +2.5 . 75 1 57 ( 2052 2 = 0. 120 pcf)(30 ft) = P RD = 0.88 - 0.71 VAF = 0.57 ft lb 3.75 + 2.5 (From 3-23) (120 pcf)(30 = 0.57 ft) = 2052 2 P RD Equation ft lb 2016 P RD RD = 0.57(120 pcf)(30 ft) = 2052 PRD DO 2052 psf (40.04 in ) ft 22 S 607 psi d 1000 psi A 288 (0psf .470 in.204/ in )) P288 D 2052 ( 40 in RD O S = Equation = 607 psi ≤ 1000 psi (From 3-14)= 288 A 288 (0.470 in 2 / in) PRD 2052 psf (40.04 in) RD DO O 596 S 607 psi d 1000 psi 288 A 288 (0.470 in 22 / in) Ring Deflection of Pipes Using Watkins-Gaube Graph (Allowable compressive stress per Table C.1, Appendix to Chapter 3)

80 80 80 80

Ring Deflection of Pipes Using Watkins-Gaube Graph R. Watkins [1]Deflection developed an extremely straight-forward approach to calculating pipe Ring of Pipes Using Watkins-Gaube Graph Ring Deflection of Pipes Using Watkins-Gaube Graph deflection in a fill that does not rely on E’. It is based on the concept that the deflection (1) developed an extremely extremely straight-forward approach to calculating R. Watkins R. Watkins [1] developed an straight-forward approach to calculating pipe of a pipe embedded ininaa layer ofdoes soilnotisrely proportional to the compression or settlement of pipe deflection fill that on E’. It is based on the concept that the deflection in aand fill that does not relyof onproportionality E’. It is basedis on the concept that the deflection the soil layer that the constant a function of the relative stiffness deflection of ain pipe embedded in a is layer of soil is proportional to the compression or R. [1] developed an of extremely straight-forward approach to calculating pipe of aWatkins pipe embedded asoil. layer soil proportional to the compression or settlement of between the pipe and Watkins used laboratory testing establish graph settlement of the soil layer and that the constant of proportionality isto a function oftheand deflection in a fill that does not rely on E’. It is based on the concept that deflection the soil layer and that the constant of proportionality is aDfunction of the relativeranges stiffness proportionality constants, called Deformation ,laboratory for the stiffness of Fcompression the relative stiffness between the pipe and soil.Factors, Watkins testing of a pipe embedded in asoil. layer of soil is used proportional to used the ortosettlement of between the pipe and Watkins laboratory testing to establish and graph metal pipes. Gaube [15, 16] extended Watkins' work by testing to include PE pipes. In establish and graph proportionality constants, called Deformation Factors, DF relative , for the soil layer and that the constant of proportionality is a function of the stiffness proportionality constants, called Deformation Factors, D , for the stiffness ranges of F (15, 16) order to predict deflection, designer first determines the amount of compression in the stiffness ranges of the metal pipes. Gaube extended Watkins’ work by testing between the pipe and soil. Watkins used laboratory testing to establish and graph metal pipes. Gaube [15,the 16]pipe extended Watkins' work by testing geotechnical to include PEequations. pipes. In the layer of soil in which is installed using conventional to include PE pipes. Incalled order toDeformation predict deflection, the designer determines proportionality Factors, Dthe for the stiffness ranges of F, first F order to predictconstants, deflection, the designer first multiplied determines amount of compression in Then, deflection equals the soil compression by the D factor. This bypasses F the amount of compression in the layer of soil in which the pipe is installed using metal pipes. Gaube [15, 16] extended Watkins' work by testing to include PE pipes. In the layer of soil in which the pipe is installed using conventional geotechnical equations. some of the inherent problems associated with using E' values. The designer using the conventional geotechnical equations. Then, deflection equals the soil compression order to predict deflection, the designer first determines the amount of compression in Then, deflection equals the soil compression multiplied by the D factor. This bypasses F Watkins-Gaube Graph select conservative soil modulus values to accommodate multiplied by theshould Dthe This bypasses some of the inherent problems associated F factor. the layer of soil in which pipe is installed using conventional geotechnical equations. some of the inherent problemsTwo associated with using E' values. The designer using the variance due to installation. otherE’,factors to consider using this method is with using the soil reaction modulus, values. The designer using WatkinsF factor. Then, deflection equals the soilselect compression multiplied by thewhen Dthe This bypasses F Watkins-Gaube Graph should conservative soil modulus values to accommodate that it of assumes a constant Deformation Factor independent ofThe depth of cover andthe it Gaube Graph (Figure 3-6) should select conservative soil values todesigner some the inherent problems associated with using E'modulus values. using variance due to installation. Two other factors to consider when using this method is does not address the effect of the presence of ground water on the Deformation Factor. accommodate variance due to installation. Two other factors to consider when using Watkins-Gaube shouldDeformation select conservative soil modulusofvalues that it assumes Graph a constant Factor independent depthtoofaccommodate cover and it variance due to installation. Two other factors to consider when this method does not address the effect of the presence of ground water on the using Deformation Factor.is To the Watkins-Gaube the designer first determines the relative stiffness thatuse it assumes a constant Graph, Deformation Factor independent of depth of cover and it does not pipe address of is thegiven presence ground Factor, water on Deformation . Equation 2-24Factor. and 2between andthe soil,effect which by theofRigidity RFthe To are usefor the Graph, designer first determines the relative stiffness 25 DRWatkins-Gaube pipe and for profile pipethe respectively: andAM2between pipe and soil, which is given by the Rigidity Factor, RF. Equation 2-24 154-264.indd 229 1/16/09 9:57:15

to predict deflection, the designer first determines the amount of compression in yer of soil in which the pipe is installed using conventional geotechnical equations. deflection equals the soil compression multiplied by the DF factor. This bypasses 230 Chapter 6 of the inherent Design problems with using E' values. The designer using the of PE Pipingassociated Systems ns-Gaube Graph should select conservative soil modulus values to accommodate ce due to installation. Two other factors to consider when using this method is assumes a constant Deformation Factor independent of depth of cover and it not address the effect of the presence of ground water on the Deformation Factor.

this method is that it assumes a constant Deformation Factor independent of depth of cover and it does not address the effect of the presence of ground water on the Watkins-Gaube Graph, the designer first determines the relative stiffness Deformation Factor. 81

e the en pipe and soil, which is given by the Rigidity Factor, RF. Equation 2-24 and 2To use Watkins-Gaube Graph, the designer first determines the relative stiffness e for DR pipe and for the profile pipe respectively:

between pipe and soil, which 3 is given by the Rigidity Factor, R F. Equation 3-24 and E D S m 3-25 are for DRRpipe Eq. 2-25 = and for profile pipe respectively: F

(3-24)

Where:

EI 12 E S ( DR − 1) 3 RF = E

81 Eq. 2-24

3

E S D m Ratio DR =RDimension Eq. 2-25 F= ES = SecantEImodulus of the soil, psi E = Apparent modulus of elasticity of pipe material, psi IWhere = Pipe Ratio wall moment of inertia of pipe, in4/in DR = Dimension Where: D Mean diameter (DI + 2z or DO – t), in ESm= = Secant modulus of the soil, psi DR = Dimension Ratioof elasticity of pipe material, psi E = Apparent modulus ES = Secant modulus of the soil, I = Pipe wall moment of inertia of pipe, in4psi /in ecant modulus of the soil may be obtained or psi from a geotechnical E = Apparent modulus offrom pipe testing material, Dm = Mean diameter (DI + 2zof or elasticity DO – t), in 4 eer’s evaluation. In lieu of a precise determination, the soil modulus may be I = Pipe wall moment of inertia of pipe, in /in from Table 2-14 by the following equation d to theDone-dimensional modulus, MS, D m = Mean diameter (DI + 2z or O – t), in The secantratio. modulus of the soil may be obtained from testing or from a geotechnical e ȝ is the soil's Poisson (3-25)

engineer’s evaluation. In lieu of a precise determination, the soil modulus may related to the modulus, , from Table by the following thebesoil may beone-dimensional obtained from testingMSor from a 3-12 geotechnical equation where µ is the soil’s Poisson ratio. In lieu of a precise determination, the soil modulus may be

modulus of valuation. (3-26) modulus, M e one-dimensional S, from P )(1-Table (1+ 2 P ) 2-14 by the following equation = Eq. 2-26 E M S S e soil's Poisson ratio. (1- P ) Table 3-13 Typical range of Poisson’s Ratio for Soil (Bowles (21)) Soil Type (1+ P )(1- 2 P ) Table 2-15: Typical Ratio Ratio, for μSoil (Bowles [21]) = M S range of Poisson’s Poisson’s Eq. 2-26 E S Saturated Clay 0.4-0.5 (1- P ) Unsaturated Soil Type Clay Sandy Clay

Silt Saturated Clay Sand (Dense)

0.1-0.3 Poisson 0.2-0.3

0.3-0.35 0.2-0.4

Ratio, ȝ

0.4-0.5

Coarse Sand (Void Ratio 0.4-0.7) 0.15 Soil Unsaturated Clay 0.1-0.3 le 2-15: Typical range of Poisson’s Ratio for (Bowles [21]) Fine-grained Sand (Void Ratio 0.4-0.7)

Sandy Clay Soil Type Silt Saturated Clay Sand (Dense) Unsaturated Clay Coarse Sand (Void Ratio 0.4Sandy Clay0.7)

Silt Fine-grained Sand (Void Ratio 0.4-0.7) Sand (Dense) 154-264.indd 230

0.25

0.2-0.3 Poisson Ratio, ȝ 0.3-0.35 0.4-0.5 0.2-0.4 0.1-0.3 0.15 0.2-0.3 0.3-0.35 0.2-0.4

0.25 1/16/09 9:57:16 AM

Chapter 6 231


82

soil layer surrounding the pipe bears the entire load of the overburden above it out arching. Therefore, settlement (compression) of the soil layer is proportional to prism load and Next, not the thedesigner radial determines directed earth pressure.Factor, SoilDF strain, İS, may be the Deformation , by entering the Watkinsmined from geotechnical analysis or from the following equation: Gaube Graph with the Rigidity Factor. See Fig. 3-6. The Deformation Factor is the proportionality constant between vertical deflection (compression) of the soil layer

wH can be Eq. 2-27 ε S = containing the pipe and the deflection of the pipe. Thus, pipe deflection 144obtained ES by multiplying the proportionality constant DF times the soil settlement.

82

Where:If DF is less than 1.0 in Fig. 3-6, use 1.0.

The soil The layer surrounding thethepipe bearsthethe entire load of the overburden above it surrounding entire load of the overburden above it w = soil unitlayer weight of soil,settlement pcf pipe bears without arching. Therefore, (compression) of thelayer soilislayer is proportional to without arching. Therefore, settlement (compression) of the soil proportional H = depth of cover (height of fill above pipe crown), ft the prismto the load andload not radial directedearth earth pressure. Soil strain, εS, may be İS, may be prism andthe notof the radial pressure. Soil strain, Es =from secant modulus the soil,directed psi determined geotechnical analysis or from the following equation: determined from geotechnical analysis or from the following equation: 82

(3-27)

S

determined from geotechnical analysis or from the following equation:

Where: wH Where HS = S w = unit weight144 ofEsoil, pcf

Eq. 2-27

w = unit weight of soil, pcf Where: H = depth of cover (height of fill above pipe crown), ft = weight depth ofpcfcover (height of fill above pipe crown), ft wH = unit of soil, Es = secant modulus of theofsoil, psi(height of fill above pipe crown), ft H = depth cover E = secant modulus of the soil, psi E =ssecant modulus of the soil, psi

s The designer can find the pipe deflection as a percent of the diameter by multiplying the soil strain, in percent, byasthe deformation factor: The designer can find the pipe deflection a percent of the diameter by multiplying the

5

DeformationFactor, Factor, DDFF Deformation

soil strain, in percent, by the deformation factor: The designer can find the pipe deflection as a percent of the diameter by multiplying the soil strain, in percent, by the deformation factor:

Deformation Factor, DF

Deformation Factor, DF

wH the pipe as The the soil layer surrounding bears entire loadof of the aboveby it multiplying the designer can find deflection a the percent theoverburden diameter Eq. 2-27 Hpipe S = Therefore, settlement (compression) of the soil layer is proportional to without arching. train, in percent, by the deformation factor: 144 E the prism load and not the S radial directed earth pressure. Soil strain, İ , may be

10

50

5

10

10050

500 500 10001000

100

5000 10,000

5000 10,000

R(RigidityRigidity Factor) Factor, RF

Rigidity Factor, R R(Rigidity Factor) Figure 2-6: Watkins-Gaube Graph Rigidity Factor, RF Figure 3-6 Watkins-Gaube Graph 'X (100) = D H Figure 2-6: Watkins-Gaube Graph D F

F

M

(3-28)

∆X

Eq. 2-28

S

(100) = D ε

Eq. 2-28

Where: ǻ X/DM multiplied F S by 100 gives percent deflection.

DM Where

multiplied by 100 givesby percent Where:∆X/D ǻ MX/D 100deflection. gives percent deflection. M multiplied

5

10

50

100

500

1000

5000 10,000

R(Rigidity Factor) Rigidity Factor, RF Figure 2-6: Watkins-Gaube Graph 154-264.indd 231

'X

1/16/09 9:57:16 AM

the of 12% a 6" or SDR pipecompacted under 140 atft 90% of fillofwith granular embedment ns –deflection Gaube Calculation Technique containing less11fines, standard proctor. The fill weighs 75 ining 12% or less fines, compacted at 90% of standard proctor. The fill weighs 75 pcf. UTION: First, calculate the vertical soil pressure equation, Eq. 2-1.

232 Chapter 6 the of a 6" SDR 11 pipe under 140 ft of fill with granular embedment ns –deflection Gaube Calculation Technique Design of PE Piping Systems Eq.fines, 2-1: = wH the E ining 12% or less at vertical 90% of soil standard proctor. The fill SOLUTION: First,Pcompacted calculate pressure equation, Eq.weighs 2-1. 75 3 UTION: First, calculate the vertical soil pressure equation, Eq. 2-1. (75lb/ft ft) under 140 ft of fill with granular embedment PEa= 6" the deflection of SDR )(140 11 pipe Eq. 2-1: PE = wH 2 ining 12% or less fines, at 90% PE = 10,500 psi of standard proctor. The fill weighs 75 Eq. 2-1: Pcompacted wH or 72.9 E =lb/ft UTION: First, calculate thePvertical soil 3pressure )(140 ft) equation, Eq. 2-1. E = (75lb/ft 3 Application of the Watkins-Gaube Calculation Technique Example of the )(140 ft) P E = (75lb/ft 2 P E = 10,500 lb/ft or 72.9 psi = wH Eq. 2-1: P 2 E Find the deflection 6” SDRTable 11 pipe 2-14 from ft MS is obtained by interpolation from andPE4710 equals 2700.under The140secant P lb/ftof aor 72.9pressure psi made UTION: First, calculate the vertical soil equation, Eq.materials 2-1. E = 10,500 3 of fill with granular embedment containing 12% or less fines, compacted at 90% of lus can be found Poisson = (75lb/ft a )(140 ft) Ratio of 0.30 PEassuming standard proctor. The fill weighs 75 pcf. The MS Eq. is obtained from =by wH 2-1: PE psi 2700 (1 2interpolation 0or.30 )(1 psi 2(0.30 )) Table 2-14 and equals 2700. The secant lb/ft 72.9 E = 10,500 MS is modulus obtainedP by interpolation from Table 2-14 and of equals E S be found 2005 psi can Poisson 0.30 2700. 3 assuming SOLUTION: First, calculate theavertical soil Ratio pressure equation, Eq. 3-1. The secant ft) P E = (75lb/ft )(140 ( 1 0 . 30 ) lus can be found assuming a Poisson Ratio of 0.30 Eq. 3-1: PE = wH 2 2700 psi (1 0.30)(1 2(0.30)) 10,500 lb/ft or 72.9Table psi 2-14 and equals2005 Eft) psi The secant E = interpolation 3)(140 S(1 0from MS is obtainedP 2700. 2700 psi .30)(1 2((10 .30 PEby = (75lb/ft 0.))30) 2005 psi E lus can be found Poisson PES=assuming 10,500 lb/ft 2 or a 72.9 (1psi 0.30)Ratio of 0.30

Mgidity obtained by interpolation Table 2-14 2(0.30 2700 psi (Equation 1 0from .30)(12-24. )) and equals 2700. The secant S is factor is obtained from The MS is obtained by interpolation from Table 3-12 and equals 2700. E Sassuming 2005 psiThe secant modulus can be found lus can be found a Poisson Ratio of 0.30 assuming a Poisson’s Ratio (1 of 0.30. 03.30) 12 (2005) (11 1) The rigidity is psi obtained = 2700 = R F factor (1 + 0.30from )(18−52Equation 2(0.30)) 2-24. 28250 ES = gidity factor is obtained from Equation 2-24. 3 = 2005 psi (112 .30)5) (11 1) − (0200 = 1) 3 = 852 12 (2005R)F(11 28250 = = 8 52 RF gidity factor is obtained from Equation 2-24. The rigidity factor is obtained from Equation 3-24. 28250 Figure 2-6, the deformation factor is found to be 1.2. The soil strain is calculated 3 12 (2005) (11 1) uation 2-27. = 830 82-24. 52 RF = gidity Using factor is obtained Equation Figure 2-6, from the deformation factor is found to be 1.2. The soil strain is calculated 29,000 28250 75pcf * 140ft 3is found to be 1.2. The soil strain is calculated Figure the deformation factor by2-6, Equation 2-27. 12 (2005) (11 − 1)x 100 = 3.6% HRS ==Figure 52deformation factor is found to be 1.2. The soil strain is calculated by 3-6, the average uation 2-27. Using lbs value=of8the F 28250 144 * 200 5 75pcf Equation 3-27. 2 * 140ftto be 1.2. The soil strain is calculated inch is found Figure 2-6, the deformation factor = x 100 = 3.6% 75pcf *H S140ft lbs uation 2-27. H S = x 100 = 3.6% 144 * 200 5 2 lbs inch 144 * 2005factor Figure 2-6, the deformation 2 is found to be 1.2. The soil strain is calculated 75pcf * 140ft inch uation 2-27. H S = x 100 = 3.6% lbsthe soil strain by the deformation factor: eflection is found by144 multiplying 5 by multiplying The deflection*is200 found the soil strain by the deformation factor: 2 75pcf * 140ft inch = • 100 = 3.6% ε'SX is found by multiplying The deflection lbs3.6 = 4.4% the soil strain by the deformation factor: (100) = 1.2* * 2005 the eflection is found multiplying 2 soil strain by the deformation factor: D Mby144 'Xinch (100) = 1.2* 3.6 = 4.4% 'X Moore-Selig for Constrained Buckling in Dry Ground D M 3.6 eflection is found by multiplying the soil strain by the deformation factor: (100) =Equation 1.2* = 4.4% As discussed previously, a compressive thrust stress exists in buried pipe. When this DM thrust 'X stress approaches a critical value, the pipe can experience a local instability eflection is found by multiplying the collapse. soil4.4% strain the section deformation factor:Luscher’s (100) = 1.2* 3.6 = or large deformation and In anby earlier of this chapter, D M equation was given for constrained buckling under ground water. Moore and Selig(17) ∆X used an alternate approach called the continuum theory to develop design have (100) = 1.2* 3.6 = 4.4% equations for contrained buckling due to soil pressure (buckling of embedded pipes). DM The particular version of their equations given below is more appropriate for dry applications than Luscher’s equation. Where ground water is present, Luscher’s equation should be used.

154-264.indd 232

1/16/09 9:57:18 AM

quation.

Where ground water is present, Luscher’s equation should be Chapter 6 233


Selig Equation for critical buckling pressure follows: (Critical buckling the pressure at which buckling will occur. A safety factor should be

The Moore-Selig Equation for critical buckling pressure follows: (Critical buckling pressure is the pressure at which buckling will occur. A safety factor should be provided.) (3-29)

Where:

PCR =

1 2 2.4 M R H (EI )3 ( E*S )3 DM

Eq. 2-29

Where PCR = Critical constrained buckling pressure, psi

ϕ = Calibration Factor, 0.55 for granular soils

PCR = Critical constrained buckling pressure, psi R H = Geometry Factor ĳ = Calibration Factor, 0.55 granular E = Apparent modulus of elasticityfor of pipe material, psisoils 4 3 RH = Geometry Factor I = Pipe wall moment of Inertia, in /in (t /12, if solid wall construction) ES* = ES /(1-modulus μ) E = Apparent of elasticity of pipe material, psi 4 3 E = Secant modulus of the soil, psi of Inertia, in /in (t /12, if solid wall I = SPipe wall moment μs = Poisson’s Ratio of Soil (Consult a textbook on soil for values. Bowles (1982) gives typical values construction) for sand and rock ranging from 0.1 to 0.4.) ES* = EThe S/(1-ȝ) geometry factor is dependent on the depth of burial and the relative stiffness modulus of the soil, psi ES = Secant between the embedment soil and the insitu soil. Moore has shown that for deep ȝs = Poisson's Ratio of Soil burials in uniform fills, RH equals 1.0.

85 85 85

SOLUTION: Critical Buckling Example ry factor is dependent on the depth of burial and the relative stiffness Determine the critical buckling pressure and safety factor against buckling for the 2000 lbs embedment soil insitu soil. Moore has shown that for deep burials *and the SOLUTION: 6” = 11 pipe (5.987” = 2860 mean diameter) in the previous example. E SSDR 2 s,SOLUTION: RH equals 1.0. (1- 0.3) inch SOLUTION: 2000 lbs E **S = 2000 = 2860 lbs 2 0.3) = 2860 inch = (11 2 2 2.4* lbs ling Example E S (10.3)0.55* 1.0 (28250* inch 0.018 )3 (2860 )3 = 354 PCR = 2 5.987 inch 1 2 2.4* 0.55* he critical buckling pressure and1.0 safety factor against buckling forlbsthe 6" (29000 0.018 )31 (2860 )32 = 358 354 lbs2 2 P CR = 2.4* 0.55* 1.0 (28250* in the previous example. in 5.987 (28250* 0.018 )3 (2860 )3 = 354 inch CR =against buckling: Determine thePS.F. 2 5.987 inch Determine the Safety Factor against buckling:

Determine the S.F. against buckling: 358 144 354* PCR buckling: Determine theS.F. S.F. =against = = 4.9 140*75 PE 354* 144 PCR S.F. = = = 4.9 354* 144 Installation #4: Shallow Cover Flotation Effects P E Category S.F. = PCR = 140*75 = 4.9 140*75 Shallow cover some special considerations for flexible pipes. As already P E presents

discussed, full soil structure interaction (membrane effect) may not occur, and live loads are carried in part by the bending stiffness of the pipe. Even if the pipe has sufficient strength to carry live load, the cover depth may not be sufficient to prevent

Installation Category # 4: Shallow Cover Flotation Effects

Shallow cover presents special considerations for flexible pipes. As already Installation Category # 4:some Shallow Cover Flotation Effects Installation Category # 4: Shallow Cover Flotation Effects discussed, full soil structure interaction (membrane effect) may not occur, and live loads 154-264.indd 233

1/27/09 11:20:22 AM

234

Chapter 6


the pipe from floating upward or buckling if the ground becomes saturated with ground water. This section addresses: • Minimum soil cover requirements to prevent flotation • Hydrostatic buckling (unconstrained) Design Considerations for Ground Water Flotation High ground water can float buried pipe, causing upward movement off-grade as well as catastrophic upheaval. This is not an issue for plastic pipes alone. Flotation of 86 metal or concrete pipes may occur at shallow cover when the pipes are empty.

Flotation occurs when the ground water surrounding the pipe produces a buoyant upport around the and theof pipe to buckle from the external forcepipe greater thanallow the sum the downward forces provided by the soilhydrostatic weight, soil friction, the weight of the pipe, and the weight of its contents. In addition to ure. the disruption occurring due to off-grade movements, flotation may also cause reduction of soil support around the pipe and allow the pipe to buckle ion is generallysignificant not a design consideration for buried pipe where the pipeline runs from the external hydrostatic pressure.

nearly full of liquid or where ground water is always below the pipe invert. Where conditions are Flotation not met,is generally a quick not "rule of thumb" is thatforpipe buried in soil a design buried pipe where thehaving pipeline a consideration 3 ated unit weightruns of at 120fulllb/ft withorat least 1½ pipe diameters of the cover fullleast or nearly of liquid where ground water is always below pipe will Where conditions aresoils not met, quicklesser “rule of cover, thumb” ground is that pipe oat. However, ifinvert. burial is inthese lighter weight orawith water buried in soil having a saturated unit weight of at least 120 lb/ft3 with at least 1½ on should be checked. pipe diameters of cover will not float. However, if burial is in lighter weight soils or with lesser cover, ground water flotation should be checked.

ematically the relationship between the buoyant force and the downward forces is in Equation 2-30. Refer to Figure 2-7. For an empty pipe,force flotation will occur if Mathematically the relationship between the buoyant and the downward forces is given in Equation 3-30. Refer to Figure 3-7. For an empty pipe, flotation will occur if:

(3-30)

F B > W P + W S + W D + WL

Eq. 2-30

Where

Where:FB = buoyant force, lb/ft of pipe WP = pipe weight, lb/ft of pipe

FWBS ==weight buoyant lb/ft of saturatedforce, soil above pipe,of lb/ftpipe of pipe W pipe weight, lb/ftlb/ft of ofpipe WDP==weight of dry soil above pipe, pipe W weight saturated soil above pipe, lb/ft of pipe WLS== weight of liquid of contents, lb/ft of pipe WD = weight of dry soil above pipe, lb/ft of pipe WL = weight of liquid contents, lb/ft of pipe

154-264.indd 234

1/16/09 9:57:19 AM

FB = WP = WS = WD = WL =

buoyant force, lb/ft of pipe pipe weight, lb/ft of pipe weight of saturated soil above pipe, lb/ftDesign of ofpipe PE Piping Systems weight of dry soil above pipe, lb/ft of pipe weight of liquid contents, lb/ft of pipe

Chapter 6 235

87 87 87

gure 2-7: Schematic of Ground Water Flotation Forces Figure 3-7 Schematic of Ground Water Flotation Forces

==

SS

22

ZZGG and FFB B empty ddOO For a 1 ft length of pipe running 44 submerged, the upward buoyant force is: (3-31)

F B = ωG

Where: Where:

π

4

dO

2

Eq. Eq.2-31 2-31

Eq. 2-31

of pipe totally submerged, the upward buoyant force is: Where Where:

pipeoutside outsidediameter, diameter,ftft O== pipe dO = pipe outsideddO diameter, ft

ȦȦGG=of=ground specific specific weightofofground groundwater water ω G = specific weight water weight 33 outside diameter, ft dO = pipe (fresh (fresh water water = = 62.4 62.4 lb/ft lb/ft ) ) (fresh water = 62.4 lb/ft 3) ȦG = specific ground water 3) of (sea water water ==64.0 64.0 lb/ft lb/ft3)3) (sea water =weight 64.0(sea lb/ft (fresh water = 62.4 lb/ft3) (sea water = 64.0 lb/ft3) The pipe average pipe weight, W in lbs/ft maybe be obtained obtained from manufacturers’ The Theaverage average pipeweight, weight, W WPPin inPlbs/ft lbs/ft may may be obtained from from manufacturers’ manufacturers’literature literatureoror literature or from Equation 3-32 or from the Table of Weights in the Appendix to from fromEquation Equation 2-32. 2-32. Chapter. This calculation is based on the use of a pipe material density pipe weight, Wthis in lbs/ft may be obtained from manufacturers’ literature orof P 0.955 gm/cc. n 2-32. (3-32) (1.06 .06 DR .12) ) DR11.12 2 2(1 Eq. Eq.2-32 2-32 59 59.6.6 WWP P SSddOO 22 DR DR 2 (1.06 ⋅ DR − 1.12 ) Eq. 2-32 59.6 W P = πd O DR 2 Equation Equation2-33 2-33gives givesthe theweight weightofofsoil soilper perlineal linealfoot footofofpipe. pipe.

3 gives the weight of soil per lineal foot of pipe. 154-264.indd 235 == (H (H- - ) )

Eq. Eq.2-33 2-33

1/27/09 11:22:42 AM

Sd O 2

WP

236

(1.06 DR 1.12) 59.6 DR 2

Eq. 2-32

Chapter 6


88

ion 2-33 gives the weight of soil per lineal foot of pipe. Table 2-16: Saturated and Dry Soil Unit Weight Equation 3-33 gives the weight of soil per lineal foot of pipe. (3-33)

Soil Type

Unit Weight, lb/ft3

W D = Z d (H - H S ) d O

Eq. 2-33

Saturated

Where

Dry

88

118-150 weight of dry soil, pcf (See Table 3-14 for typical values.) 93-144 Where: ω d = unit Table 2-16: Saturated and Dry Soil Unit Weight

nds & Gravel

H = depth of cover, ft

3 ȦHdS ==levelunit weight of dry soil,ft pcfUnit (See Table 2-16 for typical 87-131 37-112 Weight, lb/ft of ground water saturation above pipe, values.) Soil Type Glacial Till 131-150 106-144 Dry H = depth of cover, ft Saturated Table 3-14 level of ground water118-150 saturation above pipe, ft 93-144 HSaturated S&=Gravel Sands and Dry Soil Unit Weight

lts & Clays

ushed Rock

119-137

Silts & Clays nic Silts & Clay Soil Type

Glacial Till

81-112

94-125

Unit Weight, lb/ft3

87-131

37-112

Dry, the weight of31-94 Saturated, unit weight of saturated soil above the ground water, pcf 131-150 pipe, lbs per foot of pipe

106-144

119-137

94-125

ω S

Crushed Rock Sands & Gravel

118-150

Silts & Clays2 O Organic Silts & Till Clay Glacial S S G Crushed Rock

87-131

· § d (4 - S ) +131-150 W = ( Z - Z )¨¨ d O H S ¸¸ 81-112 119-137 8 ¹ © Organic Silts & Clay

81-112

ω d

93-144 37-112 106-144 94-125

Eq.31-94 2-34

31-94

Where: ȦS = saturated unit weight of soil, pcf (3-34)   d O 2 (4 - π ) Eq. 2-34 + d O H S  W S = ( ω S - ω G ) 8   ea is submerged, the soil particles are buoyed by their immersion in the r. The effectiveWhere weight of submerged soil, (ȦS – ȦG), is the soil's saturated ω weight of soil, pcf Where: ȦS =water. saturated weight ofa soil, less the density S =ofsaturated the unit ground Forunit example, soil pcf of 120 pcf When an area is submerged, the soil buoyed by their immersion nit weight has an effective weight of 57.6 pcfparticles when are completely immersed in in the ground water. The effective weight of submerged soil, (W – W ), is the soil’s S G 62.4 = 57.6 pcf). unit weight less the density of the ground water. For example, a soil of When an areasaturated is submerged, the soil particles are buoyed by their immersion in the 120 pcf saturated unit weight has an effective weight of 57.6 pcf when completely ground water. The effective weight of liquid submerged (ȦS – ȦG), is the soil's saturated 35 gives the weight per lineal foot of the in a fullsoil, pipe. immersed in water (120 - 62.4 = 57.6 pcf). unit weight less the density of the ground water. For example, a soil of 120 pcf saturated unit Equation weight 3-35 hasgives an effective 57.6 pcf when immersed in the weight weight per linealof foot of the liquid in a completely full pipe. water (120 - 62.4 (3-35)= 57.6 pcf). S d '2 Eq. 2-35 W L = ZL 4 Equation 2-35 gives the weight per lineal foot of the liquid in a full pipe. Where

Where:WL = weight of the liquid in the pipe, lb/ft

ω L = unit weight of liquid in the pipe, pcf

π the d ' pipe, lb/ft WL = weight of the in Wliquid L = ωL ȦL = unit weight of liquid in4 the pipe, pcf

ll, the liquid weight is 154-264.indd 236

2

Eq. 2-35

Where: WL = weight of the liquid in the pipe, lb/ft Ȧ = unit weight of liquid in the pipe, pcf

1/27/09 11:26:22 AM

Chapter 6 237


89

W L = ZL

S d '2

89 Eq. 2-36

8

89

and if half-full, the liquid weight is (3-36)

S d' 2 W L = ZL π d ' Where 8 = 2

WL

ωL

Eq. 2-36

ȦL =8 unit weight of the liquid in the pipe, lb/ft3 d’ = pipe inside diameter, ft

Where

Eq. 2-36

Where unit weight thelb/ft liquid in the pipe, lb/ft3 ȦL =of the 3 L = unit weight liquid in theofpipe, Where

ω

d’ = pipe inside diameter, ft For liquid levels between empty ftand half-full (0% to 50%), or between half-full and full d’ = pipe inside diameter, 3 (50% to 100%), the Ȧ following formulas provide an liquid approximate liquid weight an weight of the in the pipe, lb/ftwith L = unit accuracy of about +10%. Please refer to Figure 2-8. For liquid levels between empty and half-full (0% to 50%), or between half-full and d’ = pipe inside diameter, ft For liquid levels between empty and half-full (0% to 50%), or between half-full and full to 100%), the following provide approximate (50% to 100%),full the(50% following formulas provide anformulas approximate liquidanweight with an liquid weight accuracy of about +10%. Please refer to Figure 2-8. with an accuracy of about ±10%. Please refer to Figure 3-8.

quid levels between empty and half-full (0% to 50%), or between half-full and full to 100%), the following formulas provide an approximate liquid weight with an acy of about +10%. Please refer to Figure 2-8.

Figure 2-8: Flotation and Internal Liquid Levels Figure 2-8: Flotation and Internal Liquid Levels

Figure 3-8 Flotation and Internal Liquid Levels

For a liquid level between empty and half-full, the weight of the liquid in the pipe is approximately For a liquid level between empty and half-full, the weight of the liquid in the pipe is For a liquid level between empty and half-full, the weight of the liquid in the pipe is approximately approximately (3-37)

3

4 hl d ' - hl + 0.392 W L = Z L3 4 hl 3d ' - hl hl + 0.392 W L = ZL 3 hl

Eq. 2-37 Eq. 2-37

Figure andinInternal Liquid Levels Where2-8: Where:Flotation hl = liquid level pipe, ft hl = liquid level in pipe, ft

Where: hl = liquid level in pipe, ft

90 For a liquid level between half-full and full, the weight of the liquid in the pipe is For a liquid level between half-full and full, the weight of the liquid in the pipe is Forapproximately a liquid level between half-full and full, the weight of the liquid in the pipe is approximately approximately liquid level between empty and half-full, the weight of the liquid in the pipe is

ximately

(3-38)

· § Sd ' 2 = - 1.573 he ¸¸ W L Z L ¨¨ ¹ © 4 3

Where:4 hhle = d '-h - hl l + 0.392 W L = ωL 3 hl

Eq. 2-38 Eq. 2-37

nstrained Pipe Wall Buckling (Hydrostatic Buckling)

Where: hl = liquid level in pipe, ft equation for buckling given in this section is here to provide assistance when ning shallow cover applications. However, it may be used to calculate the buckling 237 1/27/09 liquid level half-full and full, the weight of the liquid pipe is ance of154-264.indd abovebetween grade pipes subject to external air pressure due in to the an internal

11:32:44 AM

238

  πd ' 2 - 1.573 he  W L = ω L    4

Eq. 2-38

Chapter 6

90


Where: he = d '-hl

· § Sd ' 2 = Z L ¨¨Buckling - 1.573 nstrained Pipe (Hydrostatic Buckling) W L Wall he ¸¸ 4 ¹ © Where

Eq. 2-38

equation for buckling in this section is here to provide assistance when Where: he = d '-hgiven l ning shallow cover applications. However, it may be used to calculate the buckling ance of above grade pipes subject to external air pressure due to an internal Unconstrained Pipe Wall Bucklingand (Hydrostatic for Wall submerged pipes in lakes or ponds, for pipesBuckling) placed in casings without dm, Pipe Buckling (Hydrostatic Buckling) encasement. The equation for buckling given in this section is here to provide assistance when designing shallow cover applications. However, it may be used to calculate the

for buckling given in this section is here to provide assistance when buckling resistance of above grade pipes subject to external air pressure to an nstrained are pipes that are not constrained soil embedment ordue concrete low coverpipe applications. However, it may be used tobycalculate the buckling internal vacuum, for submerged pipes in lakes or ponds, and for pipes placed in ement. Abovepipes ground pipestoare unconstrained, as are pipes placed in a casing above grade subject external air pressure due to an internal casings without grout encasement. o grouting.pipes Buried pipe or may be considered essentially where the ubmerged in lakes ponds, and for pipes placed inunconstrained casings without unding soil does not significantly increase its constrained buckling byresistance beyond its Unconstrained pipe are pipes that are not soil embedment or concrete ment. strained strength. This can happen where the depth of cover is insufficient to encasement. Above ground pipes are unconstrained, as are pipes placed in a casing nt the pipe fromprior floating slightly upward andbebreaking with the embedment to grouting. Buried pipe may consideredcontact essentially unconstrained where d pipe are pipes that are not constrained by soil embedment or concrete its springline.theGround water, flooding, or vacuum cause buckling surrounding soil does not significantly increase itscan buckling resistance beyondof Above ground pipes are unconstrained, as are pipes placed in a casing strained pipe. its unconstrained strength. This can happen where the depth of cover is insufficient ng. Buried pipe may be considered essentially unconstrained where the to prevent the pipe from floating slightly upward and breaking contact with the oil does not significantly increase its buckling resistance beyond its embedment below its springline. Ground water, flooding, or vacuum can cause cial case of This unconstrained buckling to as buckling may strength. can happen wherereferred the depth of "upward" cover is insufficient to happen buckling of unconstrained pipe. allow buried pipe. Upward buckling occurs when lateral pressure due to ground pe from floating slightly upward and breaking contact with the embedment or vacuum pushes thecase sides of the or pipe inwardreferred while forcing the pipe crown A special of unconstrained buckling to as “upward” buckling ingline. Ground water, flooding, vacuum can cause buckling of mayand oilpipe. above it upward. lookspipe. likeUpward pipe deflection rotated degrees.) happen (Collapse for shallow buried buckling occurs when 90 lateral pressure A to ground water or vacuum the sides of theispipe inward while s susceptible todue upward buckling where pushes the cover depth insufficient to forcing restrain the pipe crown the soil above it upward. lookscover like pipe d crown movement. It hasand been suggested that a(Collapse minimum of deflection four feet is e of unconstrained buckling referred to as "upward" buckling may happen 90 degrees.) A pipeto is susceptible upward buckling; buckling where the coverlarger depth ed before soil rotated support contributes averting to upward however, ried pipe. Upward buckling occurs when lateral pressure due to ground is insufficient to restrain crown movement. has been suggested that a ter pipe may require as much as a upward diameter and a half toItdevelop full support. um pushes the minimum sides ofcover the pipe inward while forcing the pipe crown and of four feet is required before soil support contributes to averting it upward. (Collapse looks like pipe deflection rotated 90 degrees.) A upward buckling; however, larger diameter pipe may require as much as a diameter nservative design for shallow cover buckling is to assume notosoil support, and ptible to upward buckling where the cover depth is insufficient restrain and a half to develop full support. nmovement. the pipe using the unconstrained pipe wall buckling equation. In lieu It has been suggested that a minimum cover of four feet isof this, a Acontributes conservative for shallow cover buckling is tobe assume no soil support, ete cap,support sufficient to resist design upward deflection, may also placed over the and pipe e soil to averting upward buckling; however, larger to design the pipe using the unconstrained pipe wall buckling equation. In lieu of hen the pipe may be designed using Luscher's equation for constrained buckling. may require as much as a diameter and a half to develop full support.

this, a concrete cap, sufficient to resist upward deflection, may also be placed over the pipe and then the pipe may be designed using Luscher’s equation for constrained 2-39 for andshallow 2-40 give the buckling allowableisunconstrained buckling eions design cover to assume nopipe soilwall support, andpressure buckling.

Repipe and profile pipe, respectively. using the unconstrained pipe wall buckling equation. In lieu of this, a Equations 3-39 and 3-40 give the allowable unconstrained pipe wall sufficient to resist upward deflection, may also be placed over the buckling pipe pressure for DR pipe and profile pipe, respectively. ipe may be designed using Luscher's equation for constrained buckling. 3

(3-39)

f O 2E  1  Eq. 2-39   WU = 9 and 2-40 give thePallowable N S (1unconstrained - µ 2 )  DR - 1  pipe wall buckling pressure nd profile pipe, respectively.

PWU = 154-264.indd 238

f O 2E § 1 · ¨ ¸ N S (1 - P 2 ) © DR - 1 ¹

3

Eq. 2-39 1/16/09 9:57:22 AM

PWU =

fO 24EI N S (1 - P 2 ) D3M



91

Where:

PWU = allowable unconstrained pipe wall buckling pressure, psi (3-40)= Dimension fO 24EI DR Ratio Eq. 2-40 PWU = E = apparent N S (1modulus - µ 2 ) D3M of elasticity of pipe material, psi fO = Ovality Correction Factor, Figure 2-9 NWhere = safety factor S pipe wall pressure, psi IPWU ==allowable Pipe unconstrained wall moment ofbuckling inertia, in4/in Dimension Ratio ȝDR = = Poisson's ratio Where: E = apparent modulus of elasticity of pipe material, psi DM = Mean diameter, (DI + 2z or DO -t), in Correction Factor, Figureunconstrained 3-9 allowable pipe wall buckling pressure, psi WU = DfOI = Ovality = P pipe inside diameter, in NS = safety factor DR = Dimension Ratio z = wall-section centroidal distance from inner fiber of pipe, in I = Pipe wall moment of inertia, in4/in E = apparent modulus of elasticity of pipe material, psi μ = Poisson’s ratio fO = Ovality Correction Factor, Figure 2-9 DM = Mean diameter, (DI + 2z or DO -t), in NS = safety factor ugh buckling occurs long-term external pressure can gradually deform the DI = piperapidly, inside diameter, in 4 I = Pipe wall moment of inertia, in /in = wall-section centroidal from inner of pipe, in (obtain from pipe producer) o the point of zinstability. This distance behavior is fiber considered viscoelastic and can be ȝ = Poisson's ratio nted for in EquationsD2-39 and 2-40 by using the apparent modulus of elasticity M = Mean diameter, (DI + 2z or DO -t), in Although buckling occurs rapidly, long-term external pressure can gradually deform for the appropriate time and temperature of theinspecific application as given in inside diameter, DI = pipe the pipe to the point of instability. This behavior is considered viscoelastic and 2-6. For Poisson's zratio, typically usedistance a valuefrom of 0.45 long-term = designers wall-section centroidal innerforfiber of pipe, in can be accounted for in Equations 3-39 and 3-40 by using the apparent modulus of g on polyethylene pipe, and 0.35 for short-term loading.

elasticity value for the appropriate time and temperature of the specific application as given in the Appendix, Chapter 3. For Poisson’s ratio, use a value of 0.45 for all PE yAlthough or deflection of the pipe rapidly, diameterlong-term increasesexternal the local radius of the buckling occurs pressure cancurvature graduallyofdeform pipe materials.

the wall and thus reduces buckling resistance. Ovality is typically reported as the pipe to the point of orinstability. This behavior is considered viscoelastic and can be deflection ofor: the pipe diameter increases the local radius of curvature of ntage reductionOvality in pipe diameter accounted for the in pipe Equations and 2-40 by resistance. using theOvality apparent modulus ofaselasticity wall and 2-39 thus reduces buckling is typically reported value for the appropriate time andin temperature the percentage reduction pipe diameter or:of the specific application as given in Table 2-6. For Poisson's ratio, designers typically use a value of 0.45 for long-term (3-41) § DI - D MIN · loading on polyethylene pipe, andN0.35 ¨¨ short-term ¸¸ loading. Eq. 2-41 %DEFLECTIO = 100for D I © ¹ Ovality or deflection of the pipe diameter increases the local radius of curvature of the Where: pipe wall and Where thus reduces buckling resistance. Ovality is typically reported as the DI = pipe inside diameter, in = pipe inside or: diameter, in percentage reductionDin diameter I pipe DMIN = pipe minimum inside diameter, in DMIN = pipe minimum inside diameter, in  D - D MIN   %DEFLECTION = 100 I DI  

Eq. 2-41

Where: DI = pipe inside diameter, in DMIN = pipe minimum inside diameter, in

154-264.indd 239

1/16/09 9:57:22 AM

240

Chapter 6


92

Figur e 2-9: Ovalit y Com pens ation Facto r, ƒ0 The desig ner should compare the critical buckling pressure with the actual anticipated pressure, 3-9 Ovality Compensation Factor, ƒØ and apply Figure a safety factor commensurate with their assessment of the application. A safety factor of 2.5 is common, but specific circumstances may warrant a higher or lower safety For large-diameter pipe, the anticipated pressure may Thefactor. designer should compare thesubmerged critical buckling pressure with the actual be conservatively calculated by determining the height of water from the pipe invert anticipated pressure, and apply a safety factor commensurate with their assessment rather than from the pipe crown.

of the application. A safety factor of 2.5 is common, but specific circumstances may warrant a higher or lower safety factor. For large-diameter submerged pipe, the anticipated pressure may be conservatively calculated by determining the height of Ground Water Flotation Example water from the pipe invert rather than from the pipe crown.

Find the allowable flood Flotation water level above a 10" DR 26 HDPE pipe installed with only 2 Ground Water Example ft of cover. Assume the pipe has 3 percent ovality due to shipping, handling, and the allowable flood water level above a 10” DR 26 PE4710 pipe installed with installationFind loads. only 2 ft of cover. Assume the pipe has 3 percent ovality due to shipping, handling, and installation loads.

SOLUTION: Use Equation pipe wallbuckling bucklingpressure pressure depends thethe SOLUTION: Use Equation 2-39. 3-39. TheThe pipe wall dependsupon upon duration of the water level above the pipe. If the water level is long lasting, then duration of the water level above the pipe. If the water level is constant, then a alongterm valuelong-term of the stress beshould used, be butused, if the value relaxation of the stressmodulus relaxationshould modulus butwater if the level water rises only occasionally, a only shorter term elastic modulus bemodulus applied.may be applied. level rises occasionally, a shorter termmay elastic Case (a): For the long lasting water above the pipe, the stress relaxation modulus at 50 year, 73ºF is approximately 29,000 lb/in2 for a typical PE4710 material. Assuming Case (a): 3% Forovality the constant water above stress relaxationlong-term modulus at 50 (fO equals 0.76) and a 2.52the to 1pipe, safetythe factor, the allowable year, 73°ҢFpressure, is approximately 28,200 lb/in for a typical P3408 material. Assuming 3% PWU is given by:

93

ovality (fO equals 0.76) and a 2.5 to 1 safety factor, the allowable long-term pressure, 3 PWU is given by: (0.76) 2 (28,200) (29,000)  1 

PWU =

psig= (3.2 1.4 psi 3.2 ftft-hd) − Hd  =1.4  2.5 (1 - 0.452 )  26 - 1 

Case (b): Flooding conditions are occasional happenings, usually lasting a few days to a week or so. However, ground water elevations may remain high for several weeks 154-264.indd 240 1/16/09 9:57:22 following a flood. The 1000 hour (41.6 days) elastic modulus value has been used to AM

3

(0.76) 2 (28,200)  1   = 1.4 psi = 3.2 ft − Hd  PWU = 2.5 (1 - 0.452 )  26 - 1 

Chapter 6 241


Case (b): Flooding conditions are occasional happenings, usually lasting a few days to a week or so. However, ground water elevations may remain high for several weeks following a flood. The 1000 conditions hour (41.6 days) elastic modulus value been Case (b): Flooding are occasional happenings, usually lastinghas a few days used to approximate the duration. to a expected week or so. flood However, ground water elevations may remain high for several weeks following a flood. The 1000 hour elastic modulus value has been used to approximate the expected flood duration. 3

(0.76) 2(43,700) (46,000)  1  head) 2.1 psi = 45.2 .9 ft. ft −(ofHd   = 2.2 PWU = 2.5 (1 - 0.452 )  26 - 1 

SECTION 3: THERMAL DESIGN CONSIDERATIONS INTRODUCTION Like most materials, polyethylene is affected by changing temperature. Unrestrained,

154-264.indd 241

1/16/09 9:57:23 AM

242

Chapter 6


Section 4 Thermal Design Considerations Introduction Similar to all thermoplastics, the engineering behavior of PE can be significantly affected by temperature. An increase in temperature causes a decrease in strength and in apparent modulus. A decrease in temperature results in opposite effects. For effective pipeline design these effects must be adequately recognized. In the case of pressure pipe the highest operating temperature is limited by the practical consideration of retaining sufficient long-term strength or maintaining the pressure rating that is sufficient for the intended application. That maximum temperature is generally 140°F (60°C). De-rating factors for up to 140°F are presented in the Appendix to Chapter 3. If higher temperatures are being considered, the pipe supplier should be consulted for additional information. In the case of buried applications of non-pressure pipe, in which the embedment material provides a significant support against pipe deformation, the highest operating temperature can be higher –sometimes, as high 180°F (~82°C). The temperature re-rating factors for apparent modulus of elasticity, which are presented in the Appendix, Chapter 3, can be used for the re-rating of a pipe’s 73°F pipe stiffness for any other temperature between -20 to 140°F (-29 to 60°C). For temperatures above 140°F the effect is more material dependent and the pipe supplier should be consulted. A beneficial feature of PE pipe is that it retains much of its toughness even at low temperatures. It can be safely handled, installed and operated even in sub-freezing conditions. The formation of ice in the pipe will restrict or, stop flow but not cause pipe breakage. Although under sub-freezing conditions PE pipe is somewhat less tough it is still much tougher that most other pipe materials.

Strength and Stress/Strain Behavior As discussed earlier in this Handbook, the engineering properties of PE material are affected by the magnitude of a load, the duration of loading, the environment and the operating temperature. And, also as discussed earlier, the standard convention is to report the engineering properties of PE piping materials based on a standard environment – which is water – and, a standard temperature – which is 73°F (23°C). A design for a condition that departs from this convention requires that an appropriate accommodation be made. This Section addresses the issue of the effect of a different temperature than that of the base temperature.

154-264.indd 242

1/16/09 9:57:23 AM

Chapter 6 243


To properly consider the affect of temperature on strength and, on stress/strain properties this must be done based on actually observed long-term strength behavior. Tables which are presented in an Appendix to Chapter 3, list temperature adjustment factors that have been determined based on long-term evaluations. Thermal Expansion/Contraction Effects Fused PE pipe joints are fully restrained. The pipe and the fused joints can easily accommodate the stress induced by changes in temperature. In general thrust restraints and mechanical expansion joints are not required in a fully fused PE piping system. However, thrust restraint may be necessary where PE pipe is connection to other ‘bell and spigot’ end pipe. Design for this condition is addressed later in this chapter and in PPI’s TN-36. Because the coefficient of thermal expansion for PE is significantly larger than that of non-plastics, considerations relating to the potential effects of thermal expansion/ contraction may include: • Piping that is installed when it is warm may cool sufficiently after94 installation to generate significant tensile forces. Thus, the final connection should be made after the pipe has equilibrated to its operating temperature.

experience greater expansion and contraction than many other mechanical joint • Unrestrained pipe may shrink enough so that itHowever, pulls out from ncreasing or decreasing (respectively) temperatures. itsalow that does not provide sufficient pull-out resistance. Methods used to connect PE eases the challenge of arresting this movement, and very often end pipe should against pull-out that is either inherent to the joint employed to eliminate the provide effectsrestraint of temperature changes.

design or additional mechanical restraint. See Chapter 9. (Note –specially designed thrust blocks may be needed to restrain movement when mechanical joints are in installed lineand withoperated PE pipes.) in sub-freezing conditions. Ice in the

e can be or stop flow, but not cause pipe breakage. In sub-freezing conditions, • Unrestrained pipe that is exposed to significant temperature swings will in some ot as impact resistant as it is at room temperature. In all cases, one combination, expand and contract, deflect laterally, or apply compressive or tensile unloading guidelines in the handling and storage section of the PPI loads to constraints or supports. dbook chapter “Inspections, Tests, and Safety Considerations” that A mitigating factor is PE’s relatively low modulus of elasticity, which greatly reduces ing devices to safely unload polyethylene piping products. the thrust that is generated by a restrained expansion/contraction. This thrust imposes no problem on thermal fusion connections.

ermal Effects

See Chapter 8 for additional information on designing above grade pipelines for thermal effects.

hange in length for an unrestrained pipe placed on a frictionless etermined from Unrestrained Equation 3-1.Thermal Effects

The theoretical change in length for an unrestrained pipe placed on a frictionless surface can be determined from Equation 4-1. (4-1)

∆L = L α ∆T

Eq. 3-1

ere:

L

= =

pipeline length change, in pipe length, ft

154-264.indd 243

1/16/09 9:57:23 AM

244

Chapter 6


Where ∆ L = pipeline length change, ft

L = pipe length, ft

α = thermal expansion coefficient, in/in/ºF ∆ T = temperature change,ºF The coefficient of thermal expansion for PE pipe material is approximately 1 x 10-4 in/in/˚F. As a “rule of thumb,” temperature change for unrestrained PE pipe is about “1/10/100,” that is, 1 inch for each 10˚F temperature change for each 100 foot of pipe. A temperature rise results in a length increase while a temperature drop results in a length decrease. End Restrained Thermal Effects

A length of pipe that is restrained or anchored on both ends and one placed on a frictionless surface will exhibit a substantially different reaction 95 to temperature change than the unrestrained pipe discussed above. If the pipe is restrained in a straight line between two points and the temperature decreases, the pipe will stress is created alongtothe pipe.inThe magnitude this canorbe attempt decrease length. Because theof ends arestress restrained anchored, length quation 3-2. change cannot occur, so a longitudinal tensile stress is created along the pipe. The magnitude of this stress can be determined using Equation 4-2. (4-2)

V

E D 'T

Eq. 3-2

are as definedWhere above, terms and are as defined above, and = =

σ = longitudinal stress in pipe, psi longitudinal stress in pipe, psi E = apparent modulus elasticity of pipe material, psi apparent modulus elasticity of pipe material, psi

47

The value of the apparent modulus of elasticity of the pipe material has a large / ft 2 P L = 2890lb impact on the calculated stress. As with all thermoplastic materials, PE’s modulus, apparent modulus of elasticity of isthe pipe material hasand a the large and therefore its stiffness, dependent on temperature duration of the culated stress.applied As with all thermoplastic materials, polyethylene’s load. Therefore, the appropriate elastic modulus should be selected based on foreBoussinesq its stiffness, istwo dependent on temperature the duration of it is important Equation these variables. When determining theand appropriate time interval, herefore, the appropriate elastic modulus should be selected based to consider that heat transfer occurs at relatively slow rates through the wall of PE bles.The When determining thetemperature appropriate timedointerval, it is important Boussinesq Equation gives thechanges pressure at occur any point inBecause a soilthe mass under a pipe; therefore not rapidly. temperature 96 surface load. The Boussinesq Equation may be used to find the pressure eat concentrated transfer occurs at relatively slow rates through the wall of change does not happen rapidly, the average temperature is often chosen for the transmitted from a wheel load todo a point that is not alongBecause the line of the action of the load. therefore temperature changes not occur rapidly. modulus selection.

Pavement effectsrapidly, are neglected. e does not happen chosen (4-3) F V AP the average temperature is oftenEq. 3-3 ection. Where terms are as defined above, and

are as defined

3I f Ww H F = end thrust, lb PL = 5 above, and 2π /4)(DO AP = area of pipe cross section,( r 2 – Di2) in2 3

Eq. 2-4

1: Modulus Elasticity for HDPE Pipe Material = Apparent end thrust, lb at Various Temperatures = area of pipe cross section,(ʌ/4)(DO2 – Di2) in2 Where:

PE 3408 Apparent Elastic Modulus†, 1000 psi (MPa), at Temperature, °F (°C) 2 9) 0 (-18) 40PL(4) 60 (16) 73 (23)due 100 (38)load120 140 (60) = vertical soil pressure to live lb/ft(49) 260.0 170.0 130.0 110.0 100.0 65.0 50.0 load, lbthe Ww =towheel 3 can(1793) also be used determine compressive and thrust (1172) (896) (758) (690) stress (448) (345) 154-264.indd 244

1/27/09 11:36:22 AM

P

= =

end thrust, lb area of pipe cross section,(ʌ/4)(DO2 – Di2) in2 Chapter 6 245


nd 3-3 can also be used to determine the compressive stress and thrust m a temperature increase.

gth change of polyethylene pipe during temperature changes is greater Equations 4-2 and 4-3 can also be used to determine the compressive stress and r materials, thethrust amount of force required to restrain the movement is (respectively) from a temperature increase. ts lower modulus of elasticity.

Although the length change of PE pipe during temperature changes is greater than many other materials, the amount of force required to restrain the movement is less decreases from weather or operating conditions, a longitudinal because of its lower modulus of elasticity.

erature velops along the pipe that can be determined using Equation 3-2. The As operating pipeline temperature decreases from weather or operating conditions, e stress for pipe at its pressure rating is determined using a

longitudinal tensile stress develops along the pipe that can be determined using Equation 4-2. The allowable tensile stress for pipe operating at its pressure rating is determined by the HDS for that temperature. The HDS is that of the pipe material for the base temperature at 73˚F (23˚C) times the temperature adjustment factor listed Eq. 3-4 in Appendix, Chapter 3. (4-4)

allow =

DB = F =

HDS xxDF FT σ allow == HDB

Where HDS = Hydrostatic Design Stress, psi (Table 1-1)

Allowable tensile stress at 73°F, lb/in2 FT = Temperature factor (See Appendix, Chapter 3) Hydrostatic Design Basis, psi (Table 1-1)* Design Factor, from Table 1-2

Equation 4-3 is used to determine the thrust load applied to structural anchoring

manufacturer should devices.be consulted for HDB values for temperatures han 73°F.

sed

During temperature increase, the pipeline attempts to increase in length, but is restrained by mechanical guides that direct longitudinal compressive thrust to structural anchors that prevent length increase. This in turn creates a longitudinal to determine the thrust load applied to astructural compressive stress in the pipe and thrust load anchoring against the structural anchors. The compressive stress that develops in the pipe and is resisted by the structural anchors is determined using Equation 4-2. Compressive stress should not exceed the compressive stress per Appendix in 3. but is increase, allowable the pipeline attempts totheincrease inChapter length,

ture echanical guides that direct longitudinal compressive thrust to structural vent length increase. This in turn Systems creates a longitudinal compressive Above Ground Piping e and a thrustThe load against the structural anchors. The compressive design considerations for PE piping systems installed above ground are ops in the pipeextensive and is and, resisted by the structural anchors is determined therefore, are addressed separately in the Handbook chapter on above 3-2. Compressive stress should not exceed the allowable compressive ground applications for PE pipe. 2-12 in Section 2 of this chapter. Buried Piping Systems A buried pipe is generally well restrained by soil loads and will experience very little lateral movement. However, longitudinal end loads may result that need to be addressed. Transitions to other pipe materials that use the bell and spigot assembly technique will need to be calculated using the thrust load as delivered by the pressure

154-264.indd 245

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97 246

Chapter 6


e Ground Piping Systems

plus the for potential of the load due to temperature fixing the endare design considerations polyethylene piping systems changes. installedMerely above ground of the PE to the mating material may result stream joints pullingHandbook apart sive and, therefore, are addressed separately in in theupPPI Engineering unless those connections arePE restrained. er on “Above Ground Applications for Pipe.” The number of joints that need to be

ed Piping

restrained to prevent bell and spigot pull out may be calculated using techniques as recommended by the manufacturer of the alternate piping material. Equation 4-3 may be used to calculate the total thrust load due to temperature change. Systems

Low thrust capacity connections to manholes or other piping systems as will be presentwell in many no pressure systems may its be addressed via a with ried pipe is generally restrained bygravity soil flow friction along length, and longitudinalchange, thrust anchor as shown in Fig. 4-1. The size of the thrust block will rate or low temperature soil such friction alone is usually sufficient to prevent vary depending on soil conditions and the thrust load as calculated via Equation 4-3. mal expansion and contraction movement. Consequently, a buried polyethylene

Figure 4-1 Longitudinal Thrust Anchor

will usually experience a change in internal stress rather than dimensional change movement. A very significant temperature decrease may exceed soil friction aint, and apply Conclusion contraction thrust loads to pipeline appurtenances. Longitudinal anchoring may used to underground connections that have limited Thebe durability andprotect visco-elastic nature of modern PE piping materials makes these ance to longitudinal movement, arearray usually not requiredsuch unless great products ideally suited forbut a broad of piping applications as: potable erature change water is anticipated. mains and service lines, natural gas distribution, oil and gas gathering, force

main sewers, gravity flow lines, industrial and various mining piping. To this end, fundamental design considerations such as fluid flow, burial design and thermal response were presented within this chapter in an effort to provide guidance to the piping system designer on the use of these tough piping materials in the full array of potential piping applications.

3-1: Longitudinal Thrust Anchoramount of background ForFigure the benefit of the pipeline designer, a considerable information and/or theory has been provided within this chapter. However, the designer should also keep in mind that the majority of pipeline installations fall within the be criteria for the AWWA Design presented in Section 3 friction can employed to arrest theWindow effectsapproach of operating temperature

ally, the soil ges. In smaller diameter pipe, the pipe is usually “snaked” from side to side within tch to assist in the soil anchoring.

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Chapter 6 247


of this chapter. Pipeline installations that fall within the guidelines for the AWWA Window, may be greatly simplified in matters relating to the design and use of flexible PE piping systems. While every effort has been made to be as thorough as possible in this discussion, it also should be recognized that these guidelines should be considered in light of specific project, installation and/or service needs. For this reason, this chapter on pipeline design should be utilized in conjunction with the other chapters of this Handbook to provide a more thorough understanding of the design considerations that may be specific to a particular project or application using PE piping systems. The reader is also referred to the extensive list of references for this chapter as additional resources for project and or system analysis and design. References for Section 1 Kerr, Logan, “Water Hammer Problems in Engineering Design”, Consulting Engineering, May 1985 Howard; A., “The Reclamation E’ Table, 25 Years Later”, Plastic Pipe XIII, Washington, D. C., Oct. 2-5, 2006. Marshall, G. P., Brogden, S., “Evaluation of the Surge and Fatigue Resistances of PVC and PE Pipe Materials for use in the U.K. Water Industry”, Plastics Pipe X, September ’98, Gotenberg, Sweden. UK Water Industry IGN 4-37-02, “Design Against Surge And Fatigue Conditions For Thermoplastics Pipes”, March 1999.

References for Section 2 1. Jeppson, Roland W., Analysis of Flow in Pipe Networks, Ann Arbor Science, Ann Arbor, MI. 2. Distribution Network Analysis, AWWA Manual M32, American Water Works Association, Denver, CO. 3. ASTM D 2513, Standard Specification for Thermoplastic Gas Pipe, Tubing and Fittings, American Society for Testing and Materials, West Conshohocken, PA. 4. ASTM D 2737, Standard Specification for PE (PE) Tubing, American Society for Testing and Materials, West Conshohocken, PA. 5. ASTM D 2447, Standard Specification for PE (PE) Plastic Pipe, Schedules 40 and 80, Based on Outside Diameter, American Society for Testing and Materials, West Conshohocken, PA. 6. ASTM D 3035, Standard Specification for PE (PE) Plastic Pipe (DR-PR) Based on Controlled Outside Diameter, American Society for Testing and Materials, West Conshohocken, PA. 7. ASTM F 714, Standard Specification for PE (PE) Plastic Pipe (SDR-PR) Based on Controlled Outside Diameter, American Society for Testing and Materials, West Conshohocken, PA. 8. ANSI/AWWA C901, AWWA Standard for PE (PE) Pressure Pipe and Tubing, 1/2 In.(13 mm) Through 3 In. (76 mm) for Water Service, American Water Works Association, Denver, CO 9. ANSI/AWWA C906, AWWA Standard for PE (PE) Pressure Pipe and Fittings, 4 In. Through 63 In. for Water Distribution, American Water Works Association, Denver, CO. 10. API Specification 15LE, Specification for PE Line Pipe (PE), American Petroleum Institute, Washington DC. 11. ASTM D 2104, Standard Specification for PE (PE) Plastic Pipe, Schedule 40, American Society for Testing and Materials, West Conshohocken, PA. 12. ASTM D 2239, Standard Specification for PE (PE) Plastic Pipe (SIDR-PR) Based on Controlled Inside Diameter, American Society for Testing and Materials, West Conshohocken, PA. 13. ASTM F 894, Standard Specification for PE (PE) Large Diameter Profile Wall Sewer and Drain Pipe, American Society for Testing and Materials, West Conshohocken, PA. 14. PPI TR-22, PE Piping Distribution Systems for Components of Liquid Petroleum Gases, Plastics Pipe Institute, Irving, TX. 15. Nayyar, Mohinder L. (1992). Piping Handbook , 6th Edition, McGraw-Hill, New York, NY. 16. Iolelchick, I.E., Malyavskaya O.G., & Fried, E. (1986). Handbook of Hydraulic Resistance, Hemisphere Publishing Corporation. 17. Moody, L.F. (1944). Transactions, Volume 6, American Society of Mechanical Engineers (ASME), New York, NY. 18. Swierzawski, Tadeusz J. (2000). Flow of Fluids, Chapter B8, Piping Handbook, 7th edition, Mohinder L. Nayyar, McGraw-Hill, New York, NY. 19. Lamont, Peter A. (1981, May). Common Pipe Flow Formulas Compared with the Theory of Roughness, Journal of the American Water Works Association, Denver, CO. 20. Flow of Fluids through Valves, Fittings and Pipe. (1957). Crane Technical Paper No 410, the Crane Company, Chicago, IL. 21. Chen, W.F., & J.Y. Richard Liew. (2002). The Civil Engineering Handbook, 2 nd edition, CRC Press, Boca Raton, FL.

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248

Chapter 6


22. Bowman, J.A. (1990). The Fatigue Response of Polyvinyl Chloride and PE Pipe Systems, Buried Plastics Pipe Technology, ASTM STP 1093, American Society for Testing and Materials, Philadelphia. 23. Marshall, GP, S. Brogden, & M.A. Shepherd, Evaluation of the Surge and Fatigue Resistance of PVC and PE Pipeline Materials for use in the UK Water Industry, Proceedings of Plastics Pipes X, Goteborg, Sweden. 24. Fedossof, F.A., & Szpak, E. (1978, Sept 20-22). Cyclic Pressure Effects on High Density PE Pipe, Paper presented at the Western Canada Sewage Conference, Regian, Saskatoon, Canada. 25. Parmakian, John. (1963). Waterhammer Analysis, Dover Publications, New York, NY. 26. Thompson, T.L., & Aude, T.C. (1980). Slurry Pipelines, Journal of Pipelines, Elsevier Scientific Publishing Company, Amsterdam. 27. Handbook of Natural Gas Engineering. (1959). McGraw-Hill, New York, NY. 28. AGA Plastic Pipe Manual for Gas Service. (2001). American Gas Association, Washington DC. 29. ASCE Manuals and Reports on Engineering Practice No. 60. (1982). Gravity Sewer Design and Construction, American Society of Civil Engineers, New York, NY. 30. Hicks, Tyler G. (1999). Handbook of Civil Engineering Calculations, McGraw-Hill, New York, NY. 31. PPI TR-14, Water Flow Characteristics of Thermoplastic Pipe, Plastics Pipe Institute, Irving, TX. 32. Kerr, Logan, Water Hammer Problems in Engineering Design, Consulting Engineering, May 1985.

References for Section 3 1. Watkins, R.K., Szpak, E., & Allman, W.B. (1974). Structural Design of PE Pipes Subjected to External Loads, Engr. Experiment Station, Utah State Univ., Logan. 2. AWWA (2006), PE Pipe Design and Installation, M55, American Water Works Association, Denver, CO. 3. Howard, A.K. (1996). Pipeline Installation, Relativity Printing, Lakewood, Colorado,ISBN 0-9651002-0-0. 4. Spangler, M.G. (1941). The Structural Design of Flexible Pipe Culverts, Bulletin 153, Iowa Engineering Experiment Station, Ames, IA. 5. Watkins, R.K., & Spangler, M.G. (1958). Some Characteristics of the Modulus of Passive Resistance of Soil—A Study in Similitude, Highway Research Board Proceedings 37:576-583, Washington. 6. Burns, J.Q., & Richard, R.M. (1964). Attenuation of Stresses for Buried Cylinders, Proceedings of the Symposium on Soil Structure Interaction, pp.378-392, University of Arizona, Tucson. 7. Katona, J.G., Forrest, F.J., Odello, & Allgood, J.R. (1976). CANDE—A Modern Approach for the Structural Design and Analysis of Buried Culverts, Report FHWA-RD-77-5, FHWA, US Department of Transportation. 8. Howard, A.K. (1977, January). Modulus of Soil Reaction Values for Buried Flexible Pipe, Journal of the Geotechnical Engineering Division, ASCE, Vol. 103, No GT 1. 9. Petroff, L.J. (1995). Installation Technique and Field Performance of PE, Profile Pipe, Proceedings 2 nd Intl. Conference on the Advances in Underground Pipeline Engineering, ASCE, Seattle. 10. Duncan, J.M., & Hartley, J.D. (1982). Evaluation of the Modulus of Soil Reaction, E’, and Its Variation with Depth, Report N o. UCB/GT/82-02, University of California, Berkeley. 11. Howard, A.K. (1981). The USBR Equation for Predicting Flexible Pipe Deflection, Proceedings Intl. Conf. On Underground Plastic Pipe, ASCE, New Orleans, LA. 12. Janson, L.E. (1991). Long-Term Studies of PVC and PE Pipes Subjected to Forced Constant Deflection, Report No. 3, KP-Council, Stockholm, Sweden. 13. Spangler, M.G., & Handy, R.L. (1982). Soil Engineering, 4th ed., Harper & Row, New York. 14. Watkins, R.K. (1977). Minimum Soil Cover Required Over Buried Flexible Cylinders, Interim Report, Utah State University, Logan, UT. 15. Gaube, E. (1977, June). Stress Analysis Calculations on Rigid PE and PVC Sewage Pipes, Kunstoffe, Vol.67, pp. 353-356, Germany. 16. Gaube, E., & Muller, W. (1982, July). Measurement of the long-term deformation of PE pipes laid underground, Kunstoffe, Vol. 72, pp. 420-423, Germany. 17. Moore, I. D., & Selig, E. T. (1990). Use of Continuum Buckling Theory for Evaluation of Buried Plastic Pipe Stability, Buried Plastic Pipe Technology, ASTM STP 1093, Philadelphia. 18. Marston, A. (1930). Iowa Engineering Experiment Station, Bulletin No. 96. 19. McGrath, T. (1994). Analysis of Burns & Richard Solution for Thrust in Buried Pipe, Simpson Gumpertz & Heger, Inc, Cambridge, Mass. 20. McGrath, T.J. (1998). Replacing E’ with the Constrained Modulus in Flexible Pipe Design, proceedings Pipeline Div. Conf. Pipelines in the Constructed Environment, ASCE, San Diego, CA. 21. Bowles, J.E. (1982). Foundation Analysis and Design, 3 rd ed., McGraw-Hill Book Company, New York.

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Chapter 6 249


Appendix A.1 Pipe Weights And Dimensions (Dips) (Black)

OD Nominal in.

3

4

6

154-264.indd 249

Actual in.

3.960

4.800

6.900

DR*

Pipe inside diameter (d)

Minimum Wall Thickness (t)

Weight (w)

in.

in.

lb. per foot

7

2.76

0.566

2.621

9

3.03

0.440

2.119

11

3.20

0.360

1.776

13.5

3.34

0.293

1.476

15.5

3.42

0.255

1.299

17

3.47

0.233

1.192

21

3.56

0.189

0.978

26

3.64

0.152

0.798

32.5

3.70

0.122

0.644

7

3.35

0.686

3.851

9

3.67

0.533

3.114

11

3.87

0.436

2.609

13.5

4.05

0.356

2.168

15.5

4.14

0.310

1.909

17

4.20

0.282

1.752

21

4.32

0.229

1.436

26

4.41

0.185

1.172

32.5

4.49

0.148

0.946

7

4.81

0.986

7.957

9

5.27

0.767

6.434

11

5.57

0.627

5.392

13.5

5.82

0.511

4.480

15.5

5.96

0.445

3.945

17

6.04

0.406

3.620

21

6.20

0.329

2.968

26

6.34

0.265

2.422

32.5

6.45

0.212

1.954

1/16/09 9:57:25 AM

250 Chapter 6


OD Nominal in.

8

10

12

14

154-264.indd 250

Actual in.

9.050

11.100

13.200

15.300

DR*



Weight (w)

in.

in.

lb. per foot

7

6.31

1.293

13.689

9

6.92

1.006

11.069

11

7.31

0.823

9.276

13.5

7.63

0.670

7.708

15.5

7.81

0.584

6.787

17

7.92

0.532

6.228

21

8.14

0.431

5.106

26

8.31

0.348

4.166

32.5

8.46

0.278

3.361

7

7.74

1.586

20.593

9

8.49

1.233

16.652

11

8.96

1.009

13.955

13.5

9.36

0.822

11.595 10.210

15.5

9.58

0.716

17

9.72

0.653

9.369

21

9.98

0.529

7.681

26

10.19

0.427

6.267

32.5

10.38

0.342

5.056

7

9.20

1.886

29.121

9

10.09

1.467

23.548

11

10.66

1.200

19.734

13.5

11.13

0.978

16.397

15.5

11.39

0.852

14.439

17

11.55

0.776

13.250 10.862

21

11.87

0.629

26

12.12

0.508

8.863

32.5

12.34

0.406

7.151

7

10.67

2.186

39.124

9

11.70

1.700

31.637

11

12.35

1.391

26.513

13.5

12.90

1.133

22.030

15.5

13.21

0.987

19.398

17

13.39

0.900

17.801

21

13.76

0.729

14.593

26

14.05

0.588

11.907

32.5

14.30

0.471

9.607

1/16/09 9:57:25 AM

Chapter 6 251


OD Nominal in.

16

18

20

24

154-264.indd 251

Actual in.

17.400

19.500

21.600

25.800

DR*



Weight (w)

in.

in.

lb. per foot

7

12.13

2.486

50.601

9

13.30

1.933

40.917

11

14.05

1.582

34.290

13.5

14.67

1.289

28.492

15.5

15.02

1.123

25.089

17

15.23

1.024

23.023

21

15.64

0.829

18.874

26

15.98

0.669

15.400

32.5

16.26

0.535

12.425

7

13.59

2.786

63.553

9

14.91

2.167

51.390

11

15.74

1.773

43.067

13.5

16.44

1.444

35.785

15.5

16.83

1.258

31.510

17

17.07

1.147

28.916

21

17.53

0.929

23.704

26

17.91

0.750

19.342

32.5

18.23

0.600

15.605

7

15.06

3.086

77.978

9

16.51

2.400

63.055

11

17.44

1.964

52.842

13.5

18.21

1.600

43.907

15.5

18.65

1.394

38.662

17

18.91

1.271

35.479

21

19.42

1.029

29.085

26

19.84

0.831

23.732

32.5

20.19

0.665

19.147

11

20.83

2.345

75.390

13.5

21.75

1.911

62.642

15.5

22.27

1.665

55.159

17

22.58

1.518

50.618

21

23.20

1.229

41.495

26

23.70

0.992

33.858

32.5

24.12

0.794

27.317

1/16/09 9:57:25 AM

252

Chapter 6


OD Nominal in.

30

Actual in.

32.000

DR*



Weight (w)

in.

in.

lb. per foot

13.5

26.97

2.370

96.367

15.5

27.62

2.065

84.855

17

28.01

1.882

77.869

21

28.77

1.524

63.835

26

29.39

1.231

52.086

32.5

29.91

0.985

42.023

* These DRs (7.3, 9, 11, 13.5, 17, 21, 26, 32.5) are from the standard dimension ratio (SDR) series established by ASTM F 412.51

154-264.indd 252

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Chapter 6 253


Appendix A.2 PIPE WEIGHTS AND DIMENSIONS (IPS) (BLACK) Pipe inside diameter (d)


Weight (w)

DR

in.

in.

lb. per foot

OD Nominal in.

1/2

3/4

1

1 1/4

154-264.indd 253

Actual in.

0.840

1.050

1.315

1.660

7

0.59

0.120

0.118

7.3

0.60

0.115

0.114

9

0.64

0.093

0.095

9.3

0.65

0.090

0.093

11

0.68

0.076

0.080

11.5

0.69

0.073

0.077

7

0.73

0.150

0.184

7.3

0.75

0.144

0.178

9

0.80

0.117

0.149

9.3

0.81

0.113

0.145

11

0.85

0.095

0.125

11.5

0.86

0.091

0.120

7

0.92

0.188

0.289

7.3

0.93

0.180

0.279

9

1.01

0.146

0.234

9.3

1.02

0.141

0.227

11

1.06

0.120

0.196

11.5

1.07

0.114

0.188

7

1.16

0.237

0.461

7.3

1.18

0.227

0.445

9

1.27

0.184

0.372

9.3

1.28

0.178

0.362

11

1.34

0.151

0.312

11.5

1.35

0.144

0.300

13.5

1.40

0.123

0.259

1/16/09 9:57:25 AM

254

Chapter 6




Weight (w)

DR

in.

in.

lb. per foot

OD Nominal in.

1 1/2

2

3

154-264.indd 254

Actual in.

1.900

2.375

3.500

7

1.32

0.271

0.603

7.3

1.35

0.260

0.583

9

1.45

0.211

0.488

9.3

1.47

0.204

0.474

11

1.53

0.173

0.409

11.5

1.55

0.165

0.393

13.5

1.60

0.141

0.340

15.5

1.64

0.123

0.299

7

1.66

0.339

0.943

7.3

1.69

0.325

0.911

9

1.82

0.264

0.762

9.3

1.83

0.255

0.741

11

1.92

0.216

0.639

11.5

1.94

0.207

0.614

13.5

2.00

0.176

0.531

15.5

2.05

0.153

0.467

17

2.08

0.140

0.429

7

2.44

0.500

2.047

7.3

2.48

0.479

1.978

9

2.68

0.389

1.656

9.3

2.70

0.376

1.609

11

2.83

0.318

1.387

11.5

2.85

0.304

1.333

13.5

2.95

0.259

1.153

15.5

3.02

0.226

1.015

17

3.06

0.206

0.932

21

3.15

0.167

0.764

26

3.21

0.135

0.623

1/16/09 9:57:25 AM

Chapter 6 255




Weight (w)

DR

in.

in.

lb. per foot

OD Nominal in.

4

5

6

154-264.indd 255

Actual in.

4.500

5.563

6.625

7

3.14

0.643

3.384

7.3

3.19

0.616

3.269

9

3.44

0.500

2.737

9.3

3.47

0.484

2.660

11

3.63

0.409

2.294

11.5

3.67

0.391

2.204

13.5

3.79

0.333

1.906

15.5

3.88

0.290

1.678

17

3.94

0.265

1.540

21

4.05

0.214

1.262

26

4.13

0.173

1.030

32.5

4.21

0.138

0.831

7

3.88

0.795

5.172

7.3

3.95

0.762

4.996

9

4.25

0.618

4.182

9.3

4.29

0.598

4.065

11

4.49

0.506

3.505

11.5

4.54

0.484

3.368

13.5

4.69

0.412

2.912

15.5

4.80

0.359

2.564

17

4.87

0.327

2.353

21

5.00

0.265

1.929

26

5.11

0.214

1.574

32.5

5.20

0.171

1.270

7

4.62

0.946

7.336

7.3

4.70

0.908

7.086

9

5.06

0.736

5.932

9.3

5.11

0.712

5.765

11

5.35

0.602

4.971

11.5

5.40

0.576

4.777

13.5

5.58

0.491

4.130

15.5

5.72

0.427

3.637

17

5.80

0.390

3.338

21

5.96

0.315

2.736

26

6.08

0.255

2.233

32.5

6.19

0.204

1.801

1/16/09 9:57:26 AM

256

Chapter 6




Weight (w)

DR

in.

in.

lb. per foot

OD Nominal in.

8

10

12

154-264.indd 256

Actual in.

8.625

10.750

12.750

7

6.01

1.232

12.433

7.3

6.12

1.182

12.010

9

6.59

0.958

10.054

9.3

6.66

0.927

9.771

11

6.96

0.784

8.425

11.5

7.04

0.750

8.096

13.5

7.27

0.639

7.001

15.5

7.45

0.556

6.164

17

7.55

0.507

5.657

21

7.75

0.411

4.637

26

7.92

0.332

3.784

7

7.49

1.536

19.314

7.3

7.63

1.473

18.656

9

8.22

1.194

15.618

9.3

8.30

1.156

15.179

11

8.68

0.977

13.089

11.5

8.77

0.935

12.578

13.5

9.06

0.796

10.875

15.5

9.28

0.694

9.576

17

9.41

0.632

8.788

21

9.66

0.512

7.204

26

9.87

0.413

5.878

32.5

10.05

0.331

4.742

7

8.89

1.821

27.170

7.3

9.05

1.747

26.244

9

9.75

1.417

21.970

9.3

9.84

1.371

21.353

11

10.29

1.159

18.412

11.5

10.40

1.109

17.693

13.5

10.75

0.944

15.298

15.5

11.01

0.823

13.471

17

11.16

0.750

12.362

21

11.46

0.607

10.134

26

11.71

0.490

8.269

32.5

11.92

0.392

6.671

1/16/09 9:57:26 AM

Chapter 6 257




Weight (w)

DR

in.

in.

lb. per foot

OD Nominal in.

14

16

18

154-264.indd 257

Actual in.

14.000

16.000

18.000

7

9.76

2.000

32.758

7.3

9.93

1.918

31.642

9

10.70

1.556

26.489

9.3

10.81

1.505

25.745

11

11.30

1.273

22.199

11.5

11.42

1.217

21.332

13.5

11.80

1.037

18.445

15.5

12.09

0.903

16.242

17

12.25

0.824

14.905

21

12.59

0.667

12.218

26

12.86

0.538

9.970

32.5

13.09

0.431

8.044

7

11.15

2.286

42.786

7.3

11.35

2.192

41.329

9

12.23

1.778

34.598

9.3

12.35

1.720

33.626

11

12.92

1.455

28.994

11.5

13.05

1.391

27.862

13.5

13.49

1.185

24.092

15.5

13.81

1.032

21.214

17

14.00

0.941

19.467

21

14.38

0.762

15.959

26

14.70

0.615

13.022

7

12.55

2.571

54.151

7.3

12.77

2.466

52.307

9

13.76

2.000

43.788

9.3

13.90

1.935

42.558

11

14.53

1.636

36.696

11.5

14.68

1.565

35.263

13.5

15.17

1.333

30.491

15.5

15.54

1.161

26.849

17

15.76

1.059

24.638

21

16.18

0.857

20.198

26

16.53

0.692

16.480

32.5

16.83

0.554

13.296

1/16/09 9:57:26 AM

258

Chapter 6


OD Nominal in.

20

22

24

154-264.indd 258

Actual in.

20.000

22.000

24.000

DR



Weight (w)

in.

in.

lb. per foot

7

13.94

2.857

66.853

7.3

14.19

2.740

64.576

9

15.29

2.222

54.059

9.3

15.44

2.151

52.541

11

16.15

1.818

45.304

11.5

16.31

1.739

43.535

13.5

16.86

1.481

37.643

15.5

17.26

1.290

33.146

17

17.51

1.176

30.418

21

17.98

0.952

24.936

26

18.37

0.769

20.346

32.5

18.70

0.615

16.415

9

16.82

2.444

65.412

9.3

16.98

2.366

63.574

11

17.76

2.000

54.818

11.5

17.94

1.913

52.677

13.5

18.55

1.630

45.548

15.5

18.99

1.419

40.107

17

19.26

1.294

36.805

21

19.78

1.048

30.172

26

20.21

0.846

24.619

32.5

20.56

0.677

19.863

9

18.35

2.667

77.845

9.3

18.53

2.581

75.658

11

19.37

2.182

65.237

11.5

19.58

2.087

62.690

13.5

20.23

1.778

54.206

15.5

20.72

1.548

47.731

17

21.01

1.412

43.801

21

21.58

1.143

35.907

26

22.04

0.923

29.299

32.5

22.43

0.738

23.638

1/16/09 9:57:26 AM

Chapter 6 259


OD Nominal in.

28

30

32

36

42

154-264.indd 259

Actual in.

28.000

30.000

32.000

36.000

42.000

DR



Weight (w)

in.

in.

lb. per foot

11

22.60

2.545

88.795

11.5

22.84

2.435

85.329

13.5

23.60

2.074

73.781

15.5

24.17

1.806

64.967

17

24.51

1.647

59.618

21

25.17

1.333

48.874

26

25.72

1.077

39.879

32.5

26.17

0.862

32.174

11

24.22

2.727

101.934

11.5

24.47

2.609

97.954

13.5

25.29

2.222

84.697

15.5

25.90

1.935

74.580

17

26.26

1.765

68.439

21

26.97

1.429

56.105

26

27.55

1.154

45.779

32.5

28.04

0.923

36.934

13.5

26.97

2.370

96.367

15.5

27.62

2.065

84.855

17

28.01

1.882

77.869

21

28.77

1.524

63.835

26

29.39

1.231

52.086

32.5

29.91

0.985

42.023

15.5

31.08

2.323

107.395

17

31.51

2.118

98.553

21

32.37

1.714

80.791

26

33.06

1.385

65.922

32.5

33.65

1.108

53.186

15.5

36.26

2.710

146.176

17

36.76

2.471

134.141

21

37.76

2.000

109.966

26

38.58

1.615

89.727

32.5

39.26

1.292

72.392

1/16/09 9:57:26 AM

260 Chapter 6


OD


Weight (w)

in.

in.

lb. per foot

Nominal in.

Actual in.

DR 17

42.01

2.824

175.205

48

48.000

21

43.15

2.286

143.629

54

154-264.indd 260


54.000

26

44.09

1.846

117.194

32.5

44.87

1.477

94.552

21

48.55

2.571

181.781

26

49.60

2.077

148.324

32.5

50.48

1.662

119.668

1/16/09 9:57:26 AM

Chapter 6 261


Appendix A.3 List of Design Chapter Variables

154-264.indd 261

υ ρ μ

=

kinematic viscosity, ft2/sec

=

fluid density, lb/ft3

=

dynamic viscosity, lb-sec/ft2

Δv

=

Sudden velocity change, ft/sec

a

=

Wave velocity (celerity), ft/sec

AC

=

Cross-sectional area of pipe bore, ft2

ac

=

contact area, ft2

A

=

profile wall average cross-sectional area, in2/in, for profile pipe or wall thickness (in) for DR pipe

AS

=

Area of pipe cross-section or (∏/4) (DO2 – Di2), in2

AP

=

area of the outside wall of the pipe, 100 in2

C

=

Hazen-Williams Friction Factor, dimensionless ,see table 1-7.

c

=

outer fiber to wall centroid, in

CV

=

percent solids concentration by volume

CW

=

percent solids concentration by weight

DA

=

pipe average inside diameter, in

DF

=

Design Factor, from Table 1-2

d’

=

Pipe inside diameter, ft

DI

=

Pipe inside diameter, in

DM

=

Mean diameter (DI+2z or DO-t), in

DMIN

=

pipe minimum inside diameter, in

Do

=

pipe outside diameter, in

dO

=

pipe outside diameter, ft

DR

=

Dimension Ratio, DO/t

E

=

Apparent modulus of elasticity for pipe material, psi

e

=

natural log base number, 2.71828

E’

=

Modulus of soil reaction, psi

Ed

=

Dynamic instantaneous effective modulus of pipe material (typically 150,000 psi for PE pipe)

EN

=

Native soil modulus of soil reaction, psi

ES

=

Secant modulus of the soil, psi

ES*

=

ES/(1-μ)

f

=

friction factor (dimensionless, but dependent upon pipe surface roughness and Reynolds number)

F

=

end thrust, lb

FB

=

buoyant force, lb/ft

FL

=

velocity coefficient (Tables 1-14 and 1-15)

fO

=

Ovality Correction Factor, Figure 2-9

FS

=

Soil Support Factor

FT

=

Service Temperature Design Factor, from Table 1-11

g

=

Constant gravitational acceleration, 32.2 ft/sec2

HP

=

profile wall height, in

H

=

height of cover, ft

hl

=

liquid level in the pipe, ft

HGW

=

ground water height above pipe, ft

h1

=

pipeline elevation at point 1, ft

1/16/09 9:57:26 AM

262 Chapter 6


h1

=

inlet pressure, in H2O

hU

=

upstream pipe elevation, ft

h2

=

pipeline elevation at point 2, ft

h2

=

outlet pressure, in H2O

dD

=

downstream pipe elevation, ft

HDB

=

Hydrostatic Design Basis, psi

hE

=

Elevation head, ft of liquid

hf

=

friction (head) loss, ft. of liquid

HS

=

level of ground water saturation above pipe, ft

IV

=

Influence Value from Table 2-5

I

=

Pipe wall moment of inertia, in4/in

IDR

=

If

154-264.indd 262

ID -Controlled Pipe Dimension Ratio impact factor

k

=

kinematic viscosity, centistokes

KBULK

=

Bulk modulus of fluid at working temperature

KBED

=

Bedding factor, typically 0.1

K

=

passive earth pressure coefficient

K’

=

Fittings Factor, Table 1-5

KP

=

permeability constant (Table 1-13)

LEFF

=

Effective Pipeline length, ft.

L

=

Pipeline length, ft

LDL

=

Deflection lag factor

Δ L

=

pipeline length change, in

M

=

horizontal distance, normal to the pipe centerline, from the center of the load to the load edge, ft

Ms

=

one-dimensional modulus of soil, psi

n

=

roughness coefficient, dimensionless

N

=

horizontal distance, parallel to the pipe centerline, from the center of the load to the load edge, ft

NS

=

safety factor

P

=

Internal Pressure, psi

PW

=

perimeter wetted by flow, ft

p1

=

inlet pressure, lb/in2 absolute

p2

=

outlet pressure, lb/in2 absolute

PA

=

pipe internal pressure, atmospheres (1 atmosphere = 14.7 lb/in2 )

PC

=

Pressure Class

PCR

=

Critical constrained buckling pressure, psi

PE

=

vertical soil pressure due to earth load, psf

Pf

=

friction (head) loss, psi

PL

=

vertical soil pressure due to live load, psf

POS

=

Occasional Surge Pressure

PRD

=

radial directed earth pressure, lb/ft2

PRS

=

Recurring Surge Pressure

Ps

=

Transient surge pressure, psig

PWAT

=

Allowable live load pressure at pipe crowm for pipes with one diameter or less of cover, psf

PWC

=

allowable constrained buckling pressure, lb/in2

PWU

=

allowable unconstrained pipe wall buckling pressure, psi

1/16/09 9:57:27 AM

Chapter 6 263


154-264.indd 263

pi

=

Pressure due to sub-area i lb/ft2

Q

=

flow rate, gpm

QFPS

=

flow, ft3/sec

Qh

=

flow, standard ft3/hour

qP

=

volume of gas permeated, cm3 (gas at standard temperature and pressure)

rH

=

hydraulic radius, ft

r

=

distance from the point of load application to pipe crown, ft

R

=

buoyancy reduction factor

rCENT

=

radius to centroidal axis of pipe, in

Re

=

Reynolds number, dimensionless

RH

=

Geometry Factor

RSC

=

Ring Stiffness Constant, lb/ft

r T

=

equivalent radius, ft

RF

=

Rigidity factor, dimensions

s

=

liquid density, gm/cm3

SH

=

hydraulic slope, ft/ft

S

=

pipe wall compressive stress, lb/in2

SMAT

=

material yield strength, lb/in2

SA

=

Hoop Thrust Stiffness Ratio

Sg

=

gas specific gravity

SL

=

carrier liquid specific gravity

SM

=

slurry mixture specific gravity

SS

=

solids specific gravity

t

=

minimum wall thickness, in

t’

=

wall thickness, mils

TCR

=

Critical time, seconds

V

=

flow velocity, ft/sec

VAF

=

Vertical Arching Factor

VC

=

critical settlement velocity, ft/sec

ν

=

kinematic viscosity. ft2/sec

VMin

=

approximate minimum velocity, ft/sec

w

=

unit weight of soil, pcf

w

=

unit weight of soil, lb/ft3

WD

=

weight of dry soil above pipe, lb/ft of pipe

Ww

=

wheel load, lb

WL

=

weight of liquid contents, lb/ft of pipe

WL

=

weight of the liquid in contacts, lb/ft of pipe

WP

=

Working Pressure, psi

WP

=

pipe weight, lb/ft of pipe

WPR

=

Working Pressure Rating, psi

wS

=

distributed surcharge pressure acting over ground surface, lb/ft2

WS

=

weight of saturated soil above pipe, lb/ft of pipe

ζ

=

dynamic viscosity, centipoises

Z

=

Centroid of wall section, in

Z

=

Pipe wall centroid, in

Zi

=

wall-section centroidal distance from inner fiber of pipe, in

α

=

thermal expansion coefficient, in/in/ºF

1/16/09 9:57:27 AM

264 Chapter 6


Δ L

=

length change, in

Δ T

=

temperature change, ºF

Δ X

=

Horizontal deflection, in

ΔV

=

Sudden velocity change., ft/sec

=

absolute roughness, ft.

ε ε s

154-264.indd 264

=

Soil strain

Θ

=

elapsed time, days

μs μ σ σallow ϕ ω D ω G ω L ω S φ Γ

=

Poisson’s Ratio of Soil

=

Poisson’s ratio

=

longitudinal stress in pipe, psi

=

Allowable tensile stress at 73ºF, lb/in

=

Calibration Factor, 0.55 for granular soils change in psi

=

unit weight of dry soil,lb/ft3 (See Table 2-16 for typical values.)

=

unit weight of groundwater lb/ft3

=

unit weight of liquid in the pipe, lb/ft3

=

unit weight of saturated soil, pcf lb/ft3

=


=

Dynamic viscosity, lb-sec/ft2

1/16/09 9:57:27 AM