Horizontal Directional Drilling - Energy Piping Systems

Chapter 12 421

Horizontal Directional Drilling

Chapter 12

Horizontal Directional Drilling Introduction The Horizontal Directional Drilling (HDD) Industry has experienced so much growth in the past two decades that HDD has become commonplace as a method of installation. One source reported that the number of units in use increased by more than a hundredfold in the decade following 1984. This growth has been driven by the benefits offered to utility owners (such as the elimination of traffic disruption and minimal surface damage) and by the ingenuity of contractors in developing this technology. To date, HDD pipe engineering has focused on installation techniques, and rightfully so. In many cases, the pipe experiences its maximum lifetime loads during the pullback operation. The purpose of this chapter is to acquaint the reader with some of the important considerations in selecting the proper PE pipe. Proper selection of pipe involves consideration not only of installation design factors such as pullback force limits and collapse resistance, but also of the long-term performance of the pipe once installed in the bore-hole. The information herein is not all-inclusive; there may be parameters not discussed that will have significant bearing on the proper engineering of an application and the pipe selection. For specific projects, the reader is advised to consult with a qualified engineer to evaluate the project and prepare a specification including recommendations for design and installation and for pipe selection. The reader may find additional design and installation information in ASTM F1962, “Standard Guide for Use of MaxiHorizontal Directional Drilling for Placement of PE Pipe or Conduit Under Obstacles, Including River Crossings,” and in the ASCE Manual of Practice 108, “Pipeline Design for Installation by Directional Drilling.”

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Background Some of the earliest uses of large diameter PE pipe in directional drilling were for river crossings. These are major engineering projects requiring thoughtful design, installation, and construction, while offering the owner the security of deep river bed cover with minimum environmental damage or exposure, and no disruption of river traffic. PE pipe is suited for these installations because of its scratch tolerance and the fused joining system which gives a zero-leak-rate joint with design tensile capacity equal to that of the pipe. To date, directional drillers have installed PE pipe for gas, water, and sewer mains; communication conduits; electrical conduits; and a variety of chemical lines. These projects involved not only river crossings but also highway crossings and right-of-ways through developed areas so as not to disturb streets, driveways, and business entrances.

PE Pipe for Horizontal Directional Drilling This chapter gives information on the pipe selection and design process. It is not intended to be a primer on directional drilling. The reader seeking such information can refer to the references of this chapter. Suggested documents are the “MiniHorizontal Directional Drilling Manual” (1) and the “Horizontal Directional Drilling Good Practices Guidelines” (2) published by the North American Society for Trenchless Technology (NASTT). Horizontal Directional Drilling Process Knowledge of the directional drilling process by the reader is assumed, but some review may be of value in establishing common terminology. Briefly, the HDD process begins with boring a small, horizontal hole (pilot hole) under the crossing obstacle (e.g. a highway) with a continuous string of steel drill rod. When the bore head and rod emerge on the opposite side of the crossing, a special cutter, called a back reamer, is attached and pulled back through the pilot hole. The reamer bores out the pilot hole so that the pipe can be pulled through. The pipe is usually pulled through from the side of the crossing opposite the drill rig. Pilot Hole Pilot hole reaming is the key to a successful directional drilling project. It is as important to an HDD pipeline as backfill placement is to an open-cut pipeline. Properly trained crews can make the difference between a successful and an unsuccessful drilling program for a utility. Several institutions provide operatortraining programs, one of which is University of Texas at Arlington Center for Underground Infrastructure Research and Education (CUIRE). Drilling the pilot hole

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establishes the path of the drill rod (“drill-path”) and subsequently the location of the PE pipe. Typically, the bore-head is tracked electronically so as to guide the hole to a pre-designed configuration. One of the key considerations in the design of the drill-path is creating as large a radius of curvature as possible within the limits of the right-of-way, thus minimizing curvature. Curvature induces bending stresses and increases the pullback load due to the capstan effect. The capstan effect is the increase in frictional drag when pulling the pipe around a curve due to a component of the pulling force acting normal to the curvature. Higher tensile stresses reduce the pipe’s collapse resistance. The drill-path normally has curvature along its vertical profile. Curvature requirements are dependent on site geometry (crossing length, required depth to provide safe cover, staging site location, etc.) But, the degree of curvature is limited by the bending radius of the drill rod and the pipe. More often, the permitted bending radius of the drill rod controls the curvature and thus significant bending stresses do not occur in the pipe. The designer should minimize the number of curves and maximize their radii of curvature in the right-of-way by carefully choosing the entry and exit points. The driller should also attempt to minimize extraneous curvature due to undulations (dog-legs) from frequent overcorrecting alignment or from differences in the soil strata or cobbles. Pilot Hole Reaming The REAMING operation consists of using an appropriate tool to open the pilot hole to a slightly larger diameter than the carrier pipeline. The percentage oversize depends on many variables including soil types, soil stability, depth, drilling mud, borehole hydrostatic pressure, etc. Normal over-sizing may be from 1.2 to 1.5 times the diameter of the carrier pipe. While the over-sizing is necessary for insertion, it means that the inserted pipe will have to sustain vertical earth pressures without significant side support from the surrounding soil. Prior to pullback, a final reaming pass is normally made using the same sized reamer as will be used when the pipe is pulled back (swab pass). The swab pass cleans the borehole, removes remaining fine gravels or clay clumps and can compact the borehole walls. Drilling Mud Usually a “drilling mud” such as fluid bentonite clay is injected into the bore during cutting and reaming to stabilize the hole and remove soil cuttings. Drilling mud can be made from clay or polymers. The primary clay for drilling mud is sodium montmorillonite (bentonite). Properly ground and refined bentonite is added to fresh water to produce a “mud.” The mud reduces drilling torque, and gives stability and support to the bored hole. The fluid must have sufficient gel strength to keep cuttings suspended for transport, to form a filter cake on the borehole wall that

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contains the water within the drilling fluid, and to provide lubrication between the pipe and the borehole on pullback. Drilling fluids are designed to match the soil and cutter. They are monitored throughout the process to make sure the bore stays open, pumps are not overworked, and fluid circulation throughout the borehole is maintained. Loss of circulation could cause a locking up and possibly overstressing of the pipe during pullback. Drilling muds are thixotropic and thus thicken when left undisturbed after pullback. However, unless cementitious agents are added, the thickened mud is no stiffer than very soft clay. Drilling mud provides little to no soil side-support for the pipe. Pullback The pullback operation involves pulling the entire pipeline length in one segment (usually) back through the drilling mud along the reamed-hole pathway. Proper pipe handling, cradling, bending minimization, surface inspection, and fusion welding procedures need to be followed. Axial tension force readings, constant insertion velocity, mud flow circulation/exit rates, and footage length installed should be recorded. The pullback speed ranges usually between 1 to 2 feet per minute. Mini-Horizontal Directional Drilling The Industry distinguishes between mini-HDD and conventional HDD, which is sometimes referred to as maxi-HDD. Mini-HDD rigs can typically handle pipes up to 10” or 12” diameter and are used primarily for utility construction in urban areas, whereas HDD rigs are typically capable of handling pipes as large as 48”diamter. These machines have significantly larger pullback forces ranging up to several hundred thousand pounds. General Guidelines The designer will achieve the most efficient design for an application by consulting with an experienced contractor and a qualified engineer. Here are some general considerations that may help particularly in regard to site location for PE pipes: 1. Select the crossing route to keep it to the shortest reasonable distance. 2. Find routes and sites where the pipeline can be constructed in one continuous length; or at least in long multiple segments fused together during insertion. 3. Although compound curves have been done, try to use as straight a drill path as possible. 4. Avoid entry and exit elevation differences in excess of 50 feet; both points should be as close as possible to the same elevation.

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5. Locate all buried structures and utilities within 10 feet of the drill-path for miniHDD applications and within 25 feet of the drill-path for maxi-HDD applications. Crossing lines are typically exposed for exact location. 6. Observe and avoid above-ground structures, such as power lines, which might limit the height available for construction equipment. 7. The HDD process takes very little working space versus other methods. However, actual site space varies somewhat depending upon the crossing distance, pipe diameter, and soil type. 8. Long crossings with large diameter pipe need bigger, more powerful equipment and drill rig. 9. As pipe diameter increases, large volumes of drilling fluids must be pumped, requiring more/larger pumps and mud-cleaning and storage equipment. 10. Space requirements for maxi-HDD rigs can range from a 100 feet wide by 150 feet long entry plot for a 1000 ft crossing up to 200 feet wide by 300 feet long area for a crossing of 3000 or more feet. 11. On the pipe side of the crossing, sufficient temporary space should be rented to allow fusing and joining the PE carrier pipe in a continuous string beginning about 75 feet beyond the exit point with a width of 35 to 50 feet, depending on the pipe diameter. Space requirements for coiled pipe are considerably less. Larger pipe sizes require larger and heavier construction equipment which needs more maneuvering room (though use of PE minimizes this). The initial pipe side “exit” location should be about 50’ W x 100’ L for most crossings, up to 100’ W x 150’ L for equipment needed in large diameter crossings. 12. Obtain “as-built” drawings based on the final course followed by the reamer and the installed pipeline. The gravity forces may have caused the reamer to go slightly deeper than the pilot hole, and the buoyant pipe may be resting on the crown of the reamed hole. The as-built drawings are essential to know the exact pipeline location and to avoid future third party damage. Safety Safety is a primary consideration for every directionally drilled project. While this chapter does not cover safety, there are several manuals that discuss safety including the manufacturer’s Operator’s Manual for the drilling rig and the Equipment Manufacturer’s Institute (EMI) Safety Manual: Directional Drilling Tracking Equipment. (3)

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Geotechnical Investigation Before any serious thought is given to the pipe design or installation, the designer will normally conduct a comprehensive geotechnical study to identify soil formations at the potential bore sites. The purpose of the investigation is not only to determine if directional drilling is feasible, but to establish the most efficient way to accomplish it. With this information the best crossing route can be determined, drilling tools and procedures selected, and the pipe designed. The extent of the geotechnical investigation often depends on the pipe diameter, bore length and the nature of the crossing. Refer to ASTM F1962, Guide for Use of Maxi-Horizontal Directional Drilling for Placement of Polyethylene Pipe or Conduit Under Obstacles, Including River Crossings (4) and ASCE MOP 108, Pipeline Design for Installation by Horizontal Directional Drilling (5) for additional information. During the survey, the geotechnical consultant will identify a number of relevant items including the following: a. Soil identification to locate rock, rock inclusions, gravelly soils, loose deposits, discontinuities and hardpan. b. Soil strength and stability characteristics c. Groundwater (Supplemental geotechnical data may be obtained from existing records, e.g. recent nearby bridge constructions, other pipeline/cable crossings in the area.) For long crossings, borings are typically taken at 700 ft intervals. For short crossings (1000 ft or less), as few as three borings may suffice. The borings should be near the drill-path to give accurate soil data, but sufficiently far from the borehole to avoid pressurized mud from following natural ground fissures and rupturing to the ground surface through the soil-test bore hole. A rule-of -thumb is to take borings at least 30 ft to either side of bore path. Although these are good general rules, the number, depth and location of boreholes is best determined by the geotechnical engineer. Geotechnical Data For River Crossings River crossings require additional information such as a study to identify river bed, river bed depth, stability (lateral as well as scour), and river width. Typically, pipes are installed to a depth of at least 20 ft below the expected future river bottom, considering scour. Soil borings for geotechnical investigation are generally conducted to 40 ft below river bottom.

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Summary The best conducted projects are handled by a team approach with the design engineer, bidding contractors and geotechnical engineer participating prior to the preparation of contract documents. The geotechnical investigation is usually the first step in the boring project. Once the geotechnical investigation is completed, a determination can be made whether HDD can be used. At that time, design of both the PE pipe and the installation can begin. The preceding paragraphs represent general guidance and considerations for planning and designing an HDD PE pipeline project. These overall topics can be very detailed in nature. Individual HDD contractors and consultant engineering firms should be contacted and utilized in the planning and design stage. Common sense along with a rational in-depth analysis of all pertinent considerations should prevail. Care should be given in evaluating and selecting an HDD contractor based upon successful projects, qualifications, experience and diligence. A team effort, strategic partnership and risk-sharing may be indicated. Product Design: PE Pipe DR Selection After completion of the geotechnical investigation and determination that HDD is feasible, the designer turns attention to selecting the proper pipe. The proper pipe must satisfy all hydraulic requirements of the line including flow capacity, working pressure rating, and surge or vacuum capacity. These considerations have to be met regardless of the method of installation. Design of the pipe for hydraulic considerations can be found in Chapter 6. For HDD applications, in addition to the hydraulic requirements, the pipe must be able to withstand (1) pullback loads which include tensile pull forces, external hydrostatic pressure, and tensile bending stresses, and (2) external service loads (post-installation soil, groundwater, and surcharge loads occurring over the life of the pipeline). Often the load the pipe sees during installation such as the combined pulling force and external pressure will be the largest load experienced by the pipe during its life. The remainder of this document will discuss the DR (Dimension Ratio) selection based on pullback and external service loads. (PE pipe is classified by DR. The DR is the “dimension ratio” and equals the pipe’s outer diameter divided by the minimum wall thickness.) A more detailed explanation of the DR concept is provided in Chapter 5. While this chapter gives guidelines to assist the designer, the designer assumes all responsibility for determining the appropriateness and applicability of the equations and parameters given in this chapter for any specific application. Directional drilling is an evolving technology, and industry-wide design protocols are still developing. Proper design requires considerable professional judgment beyond the scope of this chapter. The designer is advised to consult ASTM F 1962, Guide for Use of Maxi-Horizontal Directional Drilling for Placement of Polyethylene Pipe or Conduit

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Under Obstacles, Including River Crossings (4) when preparing an HDD design. This methodology is applied to designing municipal water pipe crossings as shown in Petroff (6). Normally, the designer starts the DR selection process by determining the DR requirement for the internal pressure. The designer will then determine if this DR is sufficient to withstand earth, live, and groundwater service loads. If so, then the installation (pullback) forces are considered. Ultimately, the designer chooses a DR that will satisfy all three requirements: the pressure, the service loads, and the pullback load. Although there can be some pipe wall stresses generated by the combination of internal pressurization and wall bending or localized bearing, generally internal pressure and external service load stresses are treated as independent. This is permissible primarily since PE is a ductile material and failure is usually driven by the average stress rather than local maximums. There is a high safety factor applied to the internal pressure, and internal pressurization significantly reduces stresses due to external loads by re-rounding. (One exception to this is internal vacuum, which must be combined with the external pressure.)

Figure 1 Borehole Deformation

Design Considerations for Net External Loads This and the following sections will discuss external buried loads that occur on directionally drilled pipes. One important factor in determining what load reaches the pipe is the condition of the borehole, i.e. whether it stays round and open or collapses. This will depend in great part on the type of ground, the boring techniques, and the presence of slurry (drilling mud and cutting mixture). If the borehole does not deform (stays round) after drilling, earth loads are arched around the borehole and little soil pressure is transmitted to the pipe. The pressure acting on the pipe is the hydrostatic pressure due to the slurry or any groundwater present. The slurry itself may act to keep the borehole open. If the borehole collapses or deforms substantially, earth

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nes, consideration should be given to the time the line sits fter construction. This may be several months. Most directionally contain fluid will have a static head, which will remain in the lineChapter 12 s head may be subtracted from the external pressureHorizontal dueDirectional to Drilling er load. The designer should keep in mind that the external load ith time, for example, flooding.

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pressure will be applied to the pipe. The resulting pressure could exceed the slurry ROUNDWATER PRESSURE

pressure unless considerable tunnel arching occurs above the borehole. Where no tunnel arching occurs, the applied external pressure is equal to the combined earth, reach whenandthe deforms andinor, contacts can be the usedpipe to establish theborehole net external pressure as it isthe groundwater, live-load pressure. For river crossings, unconsolidated river bed nt soil loadsoils, transmitted toanticipated the pipe will depend on the extent of little arching is . The applied pressure likely equals theofdifferential pressure between the inside and outside of the thegeostatic (sometimesbetween called the prism load). In consolidated soils,soil. archingEarth above the d the relativestress stiffness the pipe and the borehole andcomplexity, the applied pressure will likely less than the geostatic ot be uniform. Duemaytooccur, this there is be not a simple stress, even after total collapse of the borehole crown onto the pipe. If the soil deposit ting earth load to height ofnet cover. Groundwater loading will occur borehole condition, pressure is defined bywith little or is a stiff clay,the cemented, orexternal partially lithified, the borehole may stay open uations can be used to establish the net external pressure or, as it is e deforms ornonot; the only question is borehole): whether or not slurry med/collapsed borehole) orthis Eq. 2the (open deformation. In case, applied pressure is likely to be the just the slurry head or alled, themay differential between the what insideloads and outside of the and thus in fact pressure control design. Thus, reach the groundwater head.

the stability borehole. Pon addition PofI theto(1) the overt external pressures such as slurry head and groundwater, E PGW PSUR In following equations can be used to establish the net external pressure or, as it is internal vacuum in the pipe results in an increase in external pressure due to the sometimes called, differential between the inside on the borehole condition, the net pressure external pressure is defined by outside of the rmining the removal loads reaching the pipeinside depends detailed ofthe atmospheric pressure from the pipe. On on the other hand, aand positive pipe. internal pressure in the pipe external pressure.engineer The following borehole) or may Eq.mediate 2 (open borehole): edeformed/collapsed soil, the designer may wish to consult athe geotechnical can be used to establish the net external pressure or, as it is sometimes determining equations earth and groundwater loads. called, the differential pressure between the inside andexternal outside of the pipe. Depending on the borehole pressure is defined by PN PE PPNGW PMUD PSUR PIPI (1) condition, the net(2) on the borehole the net external is defined by either Eq. 1Depending (deformed/collapsed borehole) or Eq.pressure 2 (open borehole): e either - Groundwater Pressure Onlycondition, Eq. 1 (deformed/collapsed borehole) or Eq. 2 (open borehole):

=lled Netstable external pressure, PN PE round Ppsi PSUR deforms PI (1) if it(1)remains little after drilling. For GW and External pressure due to earth pressure, psi in competent rock will typically result in a stable borehole. Stable = Groundwater height of riverpressure water), Ppressure PMUD (including PI slurry the (2) occur in some(2) soils the exerts sufficient to N where d and open hole. Since the deformations around the hole are Where = Surcharge and live psi are negligible. The external load sures transmitted to loads, the pipe P N = Net external pressure, psi = Net external pressure, psithe P PNin PMUD PI of due (2) nternal pressure, psi (negative event vacuum) peN consists only of the hydrostatic pressure to the slurry or PE = External pressure due to earth pressure, psi P = External pressure due to earth pressure, psi pressure of(including drilling slurry or pressure pressure height of river water), psigroundwater r,=ifE Hydrostatic present. PEquation gives thethehydrostatic due to GW = Groundwater4 = Groundwater pressure (including the height of river water), P GWif slurry can PSUR =carry Surchargeshear and live loads, psi ure, stress, psi grout. Standing water should psi be added to the =surface Netpsiexternal Where: PI =P Internal pressure, (negative in thepressure, event of vacuum) N psi PMUD pressure ofpressure drilling slurry or groundwater ifpressure, slurry can carry shear stress, psi PE= Hydrostatic = External due to pressure, earth PSUR Surcharge and live loads, er, and= surcharge pressures usedpsi in Eq. 1 are discussed in a psi (Earth, ground water, and surcharge pressures used in Eq. 1 are discussed in a following section of this (including the height of river water), PGW = Groundwater Internal chapter.) pressure, psi (negativepressure in the event of vacuum) PI = chapter.) this g H W W psi of drilling slurry or groundwater PMUD = Hydrostatic (3) (4) PGW g=pressure MUD H B 2 P in = Surcharge and live loads, psi PMUD SUR pressure, if slurry can carry stress, psi 2 shear 144 in 2 (3)in the event of vacuum) psi (negative PI = Internal 144 2pressure, ft ft Where = Hydrostatic of drilling slurryinor Pfluid MUD pressure Hydrostatic due used topressure ground surface water, nd water, andg surcharge pressures in Eq.and 1 are discussed a groundwater W = MUD = Unit weight of slurry (drilling mud and cuttings), pcf if slurry can carry shear stress, psi i tion of this chapter.) H B pressure, = Elevation difference between lowest point in borehole and entry or exit pit, ft (144 is included for units conversion.) g MUD H B P = (Earth, Unit weight of slurry (drilling mud2 and pressures cuttings), pcf MUD ground water, and surcharge used in Eq. 1 are discussed in a in (3) = following Elevationsection difference between lowest point in borehole and 144 of this 12chapter.) 2 ft entry or exit pit, ft (144 is included for units conversion.) g MUD HB PMUD in 2 (3) 144will2 give careful the net external pressure, the designer 421-461.indd 429 1/16/09 10:15:04 AM ft = Unit weight of slurry (drilling mud and cuttings), pcf g

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When calculating the net external pressure, the designer will give careful consideration to enumerating all applied loads and their duration. In fact, most pipelines go through operational cycles that include (1) unpressurized or being drained, (2) operating at working pressure, (3) flooding, (4) shutdowns, and (5) vacuum and peak pressure events. As each of these cases could result in a different net external pressure, the designer will consider all phases of the line’s life to establish the design cases. In addition to determining the load, careful consideration must be given to the duration of each load. PE pipe is viscoelastic, that is, its effective properties depend on duration of loading. For instance, an HDD conduit resists constant groundwater and soil pressure with its long-term apparant modulus stiffness. On the other hand, an HDD force-main may be subjected to a sudden vacuum resulting from water hammer. When a vacuum occurs, the net external pressure equals the sum of the external pressure plus the vacuum. Since surge is instantaneous, it is resisted by the pipe’s short-term apparant modulus,, which can be four times higher than the longterm apparent modulus. For pressure lines, consideration should be given to the time the line sits unpressurized after construction. This may be several months. Most directionally drilled lines that contain fluid will have a static head, which will remain in the line once filled. This head may be subtracted from the external pressure due to earth/ groundwater load. The designer should keep in mind that the external load also may vary with time, for example, flooding. Earth and Groundwater Pressure Earth loads can reach the pipe when the borehole deforms and contacts the pipe. The amount of soil load transmitted to the pipe will depend on the extent of deformation and the relative stiffness between the pipe and the soil. Earth loading may not be uniform. Due to this complexity, there is not a simple equation for relating earth load to height of cover. Groundwater loading will occur whether the hole deforms or not; the only question is whether or not the slurry head is higher and thus may in fact control design. Thus, what loads reach the pipe will depend on the stability of the borehole. The designer may wish to consult a geotechnical engineer for assistance in determining earth and groundwater loads, as the loads reaching the pipe depend on the nature of the soil. Stable Borehole - Groundwater Pressure Only A borehole is called stable if it remains round and deforms little after drilling. For instance, drilling in competent rock (rock that can be drilled without fracturing and

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- Groundwater Pressure Only Groundwater Pressure Only

Chapter 12 led stable if it remains round and deforms little after drilling. For Horizontal Directional Drilling dinstable if it remains and result deforms after drilling. For competent rock willround typically in alittle stable borehole. Stable competent rock will typically result in a stable borehole. Stable ccur in some soils where the slurry exerts sufficient pressure to in some slurry exerts sufficient to are durand opensoils hole.where Sincethe the deformations aroundpressure the hole and hole. Since deformations around hole are ures open transmitted to thethe pipe are negligible. Thethe external load collapsing) will typically result in a stable borehole. Stable boreholes may occur in es transmitted to the pipe are negligible. The external load pe consists only of the hydrostatic pressure due to the slurry some soils where the slurry exerts sufficient pressure to maintain a roundor and open consists only of the hydrostatic pressure due to the slurry , if present. hole. Equation gives the hydrostatic pressure dueor to Since the 4 deformations around the hole are small, soil pressures transmitted f grout. present.Standing Equation 4 negligible. gives water the hydrostatic pressure to the pipesurface are The external load applied toadded the pipe due consists only of should be to to the grout. Standing surface water should added iftopresent. the Equation the hydrostatic pressure due to the slurry or be the groundwater,

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4 gives the hydrostatic pressure due to groundwater or drilling slurry. Standing surface water should be added to the groundwater.

gWHW (4) PGW =g W H W 2 (4) PGW = in 144 in 2 144 2ft 2 ft Where fluid pressure due to ground and surface water, W = Hydrostatic P = Hydrostatic fluid and surface water, psi = Hydrostatic GW fluid pressurepressure dueduetoto ground ground and surface water, (4)

gw = Unit weight of water, pcf

HW = Height to free water surface above pipe, ft (144 is included for correct units conversion.) Borehole Deforms/Collapse With Arching Mobilized

12

When the crown 12 of the hole deforms sufficiently to place soil above the hole in the plastic state, arching is mobilized. In this state, hole deformation is limited. If no soil touches the pipe, there is no earth load on the pipe. However, when deformation is sufficient to transmit load to the pipe, it becomes the designer’s chore to determine how much earth load is applied to the pipe. At the time of this writing, there have been no published reports giving calculation methods for finding earth load on directionally drilled pipes. Based on the successful performance of directionally drilled PE pipes, it is reasonable to assume that some amount of arching occurs in many applications. The designer of HDD pipes may gain some knowledge from the approaches developed for determining earth pressure on auger bored pipes and on jacked pipes. It is suggested that the designer become familiar with all of the assumptions used with these methods. For additional information on post installation design of directionally drilled pipelines see Petroff (9). O’Rourke et. al. (7) published an equation for determining the earth pressure on auger bored pipes assuming a borehole approximately 10% larger than the pipe. In this model, arching occurs above the pipe similar to that in a tunnel where zones of loosened soil fall onto the pipe. The volume of the cavity is eventually filled with soil that is slightly less dense than the insitu soil, but still capable of transmitting soil load. This method of load calculation gives a minimal loading. The method published here is more conservative. It is based on trench type arching as opposed to tunnel arching and is used by Stein (8) to calculate loads on jacked pipe. In Stein’s model, the maximum earth load (effective stress) is found using the modified form of Terzaghi’s equation given by Eq. 6., Petroff (9). External groundwater pressure must be added to the effective earth pressure. Stein and O’Rourke’s methods

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dod published published here here isismore more conservative. conservative. It Itisisbased onontrench trench posed to tunnel arching and is used by Stein tobased calculate loads osed posed to to tunnel tunnel arching arching and and is is used used by by Stein Stein to to calculate calculate loads loads Stein’s model the maximum earth load (effective stress) is Stein’s Stein’s model model the themaximum maximum earth earthload load (effective (effective isis odified form Terzhaghi’s equation given by Eq. stress) 6.stress) External 432 of Chapter 12 Horizontal Directional Drilling dified odified form form ofof Terzhaghi’s Terzhaghi’s equationgiven givenby byEq. Eq.6.6.External External ure must be added to the equation effective earth pressure. Stein and re ure must must be be added added to to the the effective effective earth earth pressure. pressure. Stein Stein and and ds should only be considered where the depth of cover is sds should should only only bebeconsidered considered where where the depth depth cover coverisis lop arching (typically exceeding fivethe (5) pipeofof diameters), lop p arching arching (typically exceeding exceeding five five (5) (5) pipe pipehas diameters), diameters), ch as traffic(typically loads are insignificant, the soil sufficient h ch as as traffic traffic loads loads are are insignificant, insignificant, the the soil soil has has sufficient sufficient should only be considered where the depth of cover is sufficient o transmit arching, and [?conditions are] confirmed tobydevelop a arching (typically and exceeding five (5) pipe diameters), loads such as traffic transmit transmit arching, and [?conditions [?conditions are] are] confirmed confirmed byby a a loads eer. Usingarching, the equations given in Stein, thedynamic external pressure are insignificant, the soil has sufficient internal friction to transmit arching, er. eer.Using Usingthe theequations equationsgiven givenininStein, Stein,the theexternal externalpressure pressure and conditions are confirmed by a geotechnical engineer. § KH C § d ·· 1 exp ¨¨ -in2Stein, thetan ¸¸ Using the equations given external is given below: ¨ ¸ pressure Kg H B 2 © ¹¹ SE C © (5) P (5) HH k Kg EV Kg SESE C 2C (5) (5) PEV PEV in § KH CKH C§ d · § d · · 2 2 144 1inin exp ¨ tan¸ ¨ ¸ ¸¸ 2 2¨ ¨ -B2 tan 144 144 ft B © 2 ¹© 2 ¹ ¹ © 2 2 k ft ft (6) (6) KH KHC §dδd· 1 − exp  - 22 C tan tan¨ ¸ BB ©22¹  k = earth pressure, psi PEV = external (6) C KH  dδ  gSE = effective soil2weight, tanpcf   B 2 HC = depth of cover, ft PEV = external earth pressure, psi (6) k = arching factor Where gSE = effective soil weight, pcf PE =width, external earth B = “silo” ft pressure, psi of cover, ft C = depth g = effective soil weight, pcf degrees PEV =H external earth pressure, psi SE of wall friction, d = angle (For HDD, d = f) k = arching factor H = depth of cover, ft 13 C soil weight, gSE =f effective pcf = angle of internal 13 13ft friction, degrees k = arching factor = “silo” width, depth of cover, ft coefficient given by: HC =B K = earth pressure B = “silo” width, ftfriction, d = angle of wall degrees (For HDD, d = f) k = arching δfactor = angle of wall friction, degrees (For HDD, δ = f) f = angle of internal friction, degrees B = “silo” width, f = angleftof internal friction, f ·degrees 2§ K = earth pressure coefficient given by: d = f) K tan 45 ¨ ¸ given by: (For K = earth pressure coefficient d = angle of wall friction, degrees HDD, 2 © ¹ f = angle of internal friction, degrees f· § K = earth pressure K tan 2coefficient ¨ 45 ¸ given by: 2¹ © o” width should be estimated based on the application. It varies between f 2 e diameter Aonconservative approach is to K =and tan 45 width −borehole The “silo” estimated based the application. It varies between the the should bediameter. 2   pipe diameterthe and the borehole diameter. A conservative approach is to assume the is the silo width equals borehole diameter. (The effective soil weight o” width should be estimated based on the application. It isvaries between silo width equals the borehole diameter. (The effective soil weight the dry unit unit weight of the soil for soil above the groundwater level, it is the e diameter and the diameter. A conservative approach is to of theborehole soil for soil above the groundwater level, it is the saturated unit weight ed unit weightweight less the weight of water for soil below the groundwater lessequals the weightthe of water for soil below the groundwater theshould silo width borehole diameter. (The level.) effective weight is dth be estimated based on the application. It varies soil between unit weight of the soil for soil above the groundwater level, isit to is the ameter and the borehole A conservative approach Borehole Collapsediameter. with Prism Load ed unit weight less weight diameter. of water for soil below soil the weight groundwater siloCollapse width equals thethe borehole is le with Prism Loadin the soil above (The In the event that arching the pipeeffective breaks down, considerable earth loading occur on the pipe. In the that arching doeslevel, not occur, weight of the soilmayfor soil above theevent groundwater it the is upper the limit on the load is the weight of the soil prism (P = g H ) above the pipe. The E SE C nit weight less theinweight of above water the for soil groundwater event that arching the soil pipe below breaksthe down, considerable le Collapse with Prism prism load is mostLoad likely to develop in shallow applications subjected to live loads, ading may occur on the pipe. In the event that arching does not occur, the mit on the load is the weight of the soil prism (PE = ySEHC) [do you mean event that arching in the soil above the pipe breaks down, considerable ollapse ove the with pipe.Prism The Load prism load is most likely to develop in shallow ading may occur on the pipe. In the event that arching does not occur, the ions subjected to live loads, boreholes in unconsolidated sediments such = ySEHconsiderable mit onarching the loadinisthe thesoil weight of the soil prism (PEdown, C) [do you mean that above the pipe breaks crossings, and holes subjected to dynamic loads. The “prism” load is 421-461.indd 432 1/27/09 12:41:21 PM ove the pipe. The prism load is most likely to develop in shallow

that arching in the soil above the pipe breaks down, considerable hat arching in the pipe breaks down, g may occur on the soil pipe.above In the the event that arching doesconsiderable not occur, the may occur on the pipe. In the event that arching does not the you mean n the load is the weight of the soil prism (PE = ySEHC) [dooccur, Chapter 12 433 Horizontal Directional Drilling = y H ) [do you mean the load is the weight of the soil prism (P E SE C the pipe. The prism load is most likely to develop in shallow he pipe. The prism load is most to develop in shallow on of Earth Groundwater Pressure subjected toand live loads, boreholes in likely unconsolidated sediments such ubjected to live loads, boreholes in unconsolidated sediments such ssings, and holes subjected to dynamic loads. The “prism” load is sings, and isholes subjected dynamic loads.its The “prism” must load isbe undwater present in theto soil formation, pressure 7. boreholes in unconsolidated suchinasa in river some river crossings,one and holes 7. or in the external load term. For sediments instance, crossing can subjected to dynamic loads. The “prism” load is given by Eq. 7. h reasonable confidence g SEthat H C the directionally drilled pipe is subjected (7) P g SE HGroundwater hmbination pressure of from the sediments abovePressure it(7) combined with the water Earth E and C 2 P (7) in bination of Earth Pressure E and Groundwater 2 144 in 2 144 ft ere groundwater is present in the soil formation, its pressure must be 2 ft WHERE e groundwater is present in the soil formation,in its pressure must ounted for inPthe external load term. For instance, a river crossing onebe can E = earth pressure on pipe, psi unted for in the external load term. For instance, in a river crossing one can P = earth pressure on pipe, psi E on of with Earth and Groundwater ume reasonable confidence that the directionally drilled pipe is subjected gSE = effective weight of soil, pcf Pressure PEgSE =with earth pressure on pipe, psi me reasonable confidence that the directionally drilled pipewith is subjected = effective weight of soil, pcf HC = soil height above pipe crown, ft the earth pressure from the sediments above it combined the water geSE = effective weight of soil, pcf earth pressure from the sediments above it combined with the H = soil height above pipe crown, ft (Note: 144 is included for units conversion.) Clevel atisor present undwater in thesurface soil formation, its pressure must be water ater below ground ssure. H = soil height above pipe crown, ft C sure. is included for units for(Note: in the144 external load of term. For instance, in a river crossing one can Combination Earth andconversion.) Groundwater Pressure (Note: 144 is included for units conversion.) h reasonableWhere confidence the pipe is subjected groundwater theWsoil H C directionally formation, g B H W that isg Dpresent in H g W Hdrilled W its pressure must be accounted P P (8) E GW for in the external load term. For instance, river crossing one can the assume with 2 h pressure from the sediments above itin acombined with water in reasonable confidence that the directionally drilled pipe is subjected to the earth 14 144above ft 2 it combined with the water pressure. pressure from the sediments 14 ground se(1): Water level at or below surface (1): Water level at or below ground surface Case 1 Water level at or below ground surface

g H + g (H − H ) + g W H W PE + PGW =g B HB W W+ g D (DH C −C H W2 )W+ g W H (8) W PE + PGW = (8) in ater level at or below ground surface 144 in 2 2 ater level at or above ground surface 144 (i.e. pipe in river bottom) 2ft ft Case 2 Water level at or above ground surface (i.e. pipe in river bottom) g B H W g D H C H W g W H W PE (9)PGW (8) gBHC g W H 2 W PE PGW (9) in 144 in 2 2 144 ft ft 2 WHERE se (2): Water Hlevel at or above ground surface (i.e. pipe in river bottom) waterground above pipe springline, ft W = Height (2): Water level at orof Ground above surface (i.e. pipe in river bottom) (8)

HC = height of cover, ft

gB = buoyant weight of soil, pcf g H + gWHW H W = Height Ground water above ft Ppcf (9) g B HB C +Cpipe g W H2springline, gW =of weight of water, E + PGW = W P + P = (9) in Height Cover, HC =level gDof = dry unit weightEft of soil, GW pcf Water at or above ground surface 144 (i.e. in river bottom) in 2pipe gB = buoyant weight of soil, pcf 144 2ft 2 Live Loads ft gW = weight of water, pcf g Hor other g Wvehicles H W are significant for pipe at shallow depths loads from trucks gD = dry unitWheel weight soil,BpcfC PE they Pof (9) GW 2 cut trenching or directional drilling. The wheel whether are installed by open in Height ofpipe Ground water above pipe ft size, ere: H W =applied to the depends on the vehicle weight, thespringline, tire pressure and 144water 2 = Height of Ground above pipe springline, ft e: HHW load = Height Cover, ftft pavement and distance from the pipe to the speed, of surface smoothness, Cvehicle Height of Cover, ft HCg =point of loading.weight In order to proper soil structure interaction, pipe subject to ofdevelop soil, pcf B = buoyant g = buoyant weight of soil, pcf vehicular loading should be installed at least 18” or one diameter B s from trucks otherof vehicles forpipepipe at (whichever shallowis gW = or weight water, pcfare significant gare weight ofby water, pcf Wg == of unit Ground water above ft drillingThe herHthey open trenching or directional dry weight ofcut soil, pcfpipe springline, W = Height D installed g = dry unit weight of soil, pcf D Cover, ft HC = Height applied to the of pipe depends on the vehicle weight, the tire pressure gB =speed, buoyantsurface weight smoothness, of soil, pcf pavement and distance from the ehicle e Loads Loads gW = of water, pcfto develop proper soil structure interaction, point ofweight loading. In order 421-461.indd 433 10:15:05 AM gDloads = dry unit soil, pcf teel to vehicular loading bevehicles installedare at least 18" orforone pipe fromweight trucksofshould or other significant pipe at 1/16/09 shallow

434 Chapter 12


larger) under the road surface. Generally, HDD pipes are always installed at a deeper depth so as to prevent inadvertent returns from occurring during the boring. The soil pressure due to live load such as an H20 wheel load can be found in Tables 3-3 and 3-4 in Chapter 6 or can be calculated using one of the methods in Chapter 6. To find the total pressure applied to the pipe, add the soil pressure due to live load, PL, to the earth pressure, PE. See Example 1 in Appendix A. Performance Limits Hydrostatic Buckling or Collapse

Ring Deformation

Figure 2 Performance Limits of HDD Pipe Subjected to Service Loads

Performance Limits of HDD Installed Pipe The design process normally consists of calculating the loads applied to the pipe, selecting a trial pipe DR, then calculating the safety factor for the trial DR. If the safety factor is adequate, the design is sufficient. If not, the designer selects a lower DR and repeats the process. The safety factor is established for each performance limit of the pipe by taking the ratio of the pipe’s ultimate strength or resistance to the applied load. External pressure from earth load, groundwater, vacuum and live load applied to the HDD pipe produces (1) a compressive ring thrust in the pipe wall and (2) ring bending deflection. The performance limit of unsupported PE pipe subjected to compressive thrust is ring buckling (collapse). The performance limit of a PE pipe subjected to ring bending (a result of non-uniform external load, i.e. earth load) is ring deflection. See Figure 2. Viscoelastic Behavior Both performance limits are proportional to the apparent modulus of elasticity of the PE material. For viscoelastic materials like PE, the modulus of elasticity is a timedependent property, that is, its value changes with time under load. A newly applied load increment will cause a decrease in apparent stiffness over time. Unloading will

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Chapter 12 435


result in rebounding or an apparent gain in stiffness. The result is a higher resistance to short term loading than to long-term loading. Careful consideration must be given to the duration and frequency of each load, so that the performance limit associated with that load can be calculated using PE material properties representative of that time period. The same effects occur with the pipe’s tensile strength. For instance, during pullback, the pipe’s tensile yield strength decreases with pulling time, so the safe (allowable) pulling stress is a function of time under load, and temperature. Typical safe pull tensile stress values for MDPE and HDPE are given in Table 1. Consult the manufacturer for specific applications. The values are given as a function of the duration of continuous loading. For pipe temperatures (not outside air temperatures) other than 73ºF, multiply the value in Table 1 by the temperature compensating multipliers found in Table B.1.2 of the Appendix to Chapter 3. The Safe Pull Load at 12 hours is given for a variety of pipe sizes and DR’s in Tables 3 and 4 (3xxx material) and Tables 5 and 6 (4xxx material) in a following section, “Tensile Stress During Pullback”.

Table 1 Safe Pull Tensile Stress @ 73º F Duration (Hours)

Typical Safe Pull Stress (psi) @ 73ºF PE2xxx (PE2406)

PE3xxx (PE3408)

PE4xxx (PE4710)

0.5

1100

1400

1500

1

1050

1350

1400

12

850

1100

1150

24

800

1050

1100

The safe pull stress is the stress at 3% strain. For strains less than 3% the pipe will essentially have complete strain recovery after pullback. The stress values in Table 1 were determined by multiplying 3% times the apparent tensile modulus from the Appendix to Chapter 3 adjusted by a 0.60 factor to account for the high stress level during pullback.

Ring Deflection (Ovalization) Non-uniform pressure acting on the pipe’s circumference such as earth load causes bending deflection of the pipe ring. Normally, the deflected shape is an oval. Ovalization may exist in non-rerounded coiled pipe and to a lesser degree in straight lengths that have been stacked, but the primary sources of bending deflection of directionally drilled pipes is earth load. Slight ovalization may also occur during pullback if the pipe is pulled around a curved path in the borehole. Ovalization reduces the pipe’s hydrostatic collapse resistance and creates tensile

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436

Chapter 12


bending stresses in the pipe wall. It is normal and expected for buried PE pipes to undergo ovalization. Proper design and installation will limit ovalization (or as it is often called “ring deflection”) to prescribed values so that it has no adverse effect on the pipe. Ring Deflection Due to Earth Load As discussed previously, insitu soil characteristics and borehole stability determine to great extent the earth load applied to directionally drilled pipes. Methods for calculating estimated earth loads, when they occur, are given in the previous section on “Earth and Groundwater Pressure.” Since earth load is non-uniform around a pipe’s circumference, the pipe will undergo ring deflection, i.e. a decrease in vertical diameter and an increase in horizontal diameter. The designer can check to see if the selected pipe is stiff enough to limit deflection and provide an adequate safety factor against buckling. (Buckling is discussed in a later section of this chapter.) The soil surrounding the pipe may contribute to resisting the pipe’s deflection. Formulas used for entrenched pipe, such as Spangler’s Iowa Formula, are likely not applicable as the HDD installation is different from installing pipe in a trench where the embedment can be controlled. In an HDD installation, the annular space surrounding the pipe contains a mixture of drilling mud and cuttings. The mixture’s consistency or stiffness determines how much resistance it contributes. Consistency (or stiffness) depends on several factors including soil density, grain size and the presence of groundwater. Researchers have excavated pipe installed by HDD and observed some tendency of the annular space soil to return to the condition of the undisturbed native soil. See Knight (11) and Ariaratnam (12). It is important to note that the researched installations were located above groundwater, where excess water in the mud-cuttings slurry can drain. While there may be consolidation and strengthening of the annular space soil particularly above the groundwater level, it may be weeks or even months before significant resistance to pipe deflection develops. Until further research establishes the soil’s contribution to resisting deflection, one option is to ignore any soil resistance and to use Equation 10 which is derived from ring deflection equations published by Watkins and Anderson (13). (Coincidentally, Equation 10 gives the same deflection as the Iowa Formula with an E’ of zero.) Spangler’s Iowa formula is discussed in Chapter 6. The design deflection limits for directionally drilled pipe are given in Table 2. Design deflection limits are for use in selecting a design DR. Field deflection measurements of directionally drilled pressure pipe are normally not made. Design deflection must be limited to control buckling resistance.

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above the groundwater level, it may be weeks or even months before resistance to pipe deflection develops. Until further research the soil’s contribution to resisting deflection, one option is to ignore Chapter 12 Horizontal Directional Drilling sistance and to use Equation 10 which is derived from ring deflection published by Watkins and Anderson (1995). (Coincidentally, Equation e same deflection as the Iowa Formula with an E’ of zero.) (10)

0.0125PE y Δy = D E

437

(10)

12 (DR - 1) 3 WHERE

imit (%y/D) Δy = vertical ring7.5 6.0 6.0 6.0 5.0 4.0 3.0 deflection, in re: y = ring ring deformation, in pplications D = pipe diameter, in D = pipe Pdiameter, inpsi E = Earth pressure, DRpressure, = Pipe Dimension psi Ratio = Earth PEpressure mits for applications are equal to 1.5 times the short-term deflection limits E = apparent modulus of elasticity, psi (Refer to Appendix, Chapter 3, Engineering Properties, for the DR = Pipe Dimension Ratio X2.1 of ASTM F-714. appropriate value for the Material Designation Code of the PE pipe being used and the applicable service E = modulus ofconditions) elasticity, psi * To obtain ring deflection in percent, multiply Δy/D by 100. ections are for use in selecting DR and for field quality control. (Field ng deflection in percent, y/D by 100. eflections exceeding the multiply design deflection do not necessarily indicate Table 2 over-strainedDesign pipe. In this case, an engineering analysis of such pipe Deflection Limits of Buried Polyehtylene Pipe, Long Term, %* ction Limits (Ovality Limits) erformed before acceptance.) DR or SDR

21

17

15.5

13.5

11

9

7.3

Deflection Limit (% Δy/D) Non-Pressure Applications

7.5 7.5 7.5 7.5 7.5 nned or ring deflection (in percent)7.5is limited by the 7.5 pipe wall strain,, the Buckling Deflection Limit (%Δy/D) aulic capacity, and the pipe’s7.5geometric stability. Jansen observed 6.0 6.0 6.0 5.0 4.0 3.0 Pressure Applications ,ernal pressure-rated pipe, to Guide soil only, “no upper limit * Design deflectionsubjected limits ASTM pipe F1962, forpressure Use offrom Maxi-Horizontal Directional pressure applied toperthe either earth andDrilling livefor load, Placement of PE Pipe or Conduit Under Obstacles, Including River Crossings. ctical design point of view seems to exist for the bending strain.” On r, or the drilling slurry creates a ring compressive hoop stress in the and, pipes is are subject to soil induced If the pressurized external pressure increased to strains a point from whereboth the hoop stress Unconstrained Buckling and internal pressure. The combined strain may produce a critical value, there is a sudden and large inward deformation ofhigh, the Uniform external pressure applied to the pipe either from earth and live load, outer-fiber tensile stress. However, as the internal pressure is called buckling. Constraining the pipe by embedding it in soil or or the drilling slurrybending creates a ring compressive hoop stress Due thegrout pipe tends to re-round the strain is reduced. to us will groundwater, increase the and pipe’s buckling strength and allowin the it to pipe’s wall. If the external pressure is increased to a point where the hoop stress al for external combinedpressure strain (bending and hoop tensile), it is conservative gher than if unconstrained. However, as noted in to a reaches a critical value, there is a sudden and large inward deformation of the pipe tion of pressure pipes to less than non-pressure pipes. In lieu of an ction it is notwall, likely pipes installed below the groundwater level calledthat buckling. Constraining the pipe by embedding it in soil or cementitious lation for allowable deflection limits, the strength limits in 3 can beand used. grout willfrom increase thesurrounding pipe’s buckling andTable allow it to withstand higher significant support the mud-cuttings mixture for external pressure than if unconstrained. However, as noted in a previous section e groundwater support may take considerable time to develop. is not likely that Table installed groundwater level will acquire 3 itbelow until further itresearch is pipes available is the conservative to assume no significant support from the surrounding mud-cuttings mixture and for pipe above Deflection of Buried Polyehtylene Longequation, Term, %*may om the soil.Limits The following equation, known Pipe, as Levy’s groundwater support may take considerable time to develop. Therefore, until 17 it is conservative 15.5 21 external 13.5 9 7.3 o determine further the allowable pressure (or11nonegative internal research is available to assume constraint from the soil. or unconstrained pipe. . equation, known as Levy’s equation, may be used to determine the The following Limit (%y/D) allowable external 7.5pressure 7.5(or negative 7.5 internal 7.5 pressure) 7.5 for unconstrained 7.5 7.5pipe. ure Applications (11) 2E 1 3 fO PUC ( ) = (11 UA 2 (1 - m μ ) DR - 1 N

Pua = Allowable unconstrained pressure, psi 19 E = Modulus of elasticity (apparent), psi m = 421-461.indd Poisson’s Ratio 437

1/16/09 10:15:06 AM

I

ν

m

f

D

n

= = = = = =

= =

30,000 psi for HDPE at 73.4 F (23 C), 50 year loading Pipe wall moment of inertia, in4/in for solid wall polyethylene t3/12 438 Chapter 12 Horizontal Directional Drilling Poisson’s ratio, 0.45 for polyethylene Mean diameter, inches (inside diameter plus one wall thickness) Ovality compensation factor, dimensionless (see Figure 10-1) D −D min Where Deflection = × 100 where P% uc = Allowable unconstrained buckling pressure D E = Apparent modulus of elasticity, psi (Refer to Appendix, Chapter 3, Engineering Properties, Pipe average diameter, for the appropriate value forin the Material Designation Code of the PE pipe being used and the applicable service conditions.) Pipe minimum diameter, in μ = Poisson’s Ratio = 0.45 for all PE pipe materials

DR = Dimension Ratio (Do/t), where Do =Outside Pipe Diameter and t = Minimum Wall Thickness

10-1 fo = Ovality compensationFigure factor (see figure 3)

N = Safety factor, generally 2.0 or higherCorrection Factor, f % Deflection vs. Ovality

Figure 3 Ovality Compensation Factor=fo

he buckling pressure of a dimension ratio (DR) series polyethylene pipe (i.e., a discussion of buckling see the section in Chapter 6 titled olid wall pipesForofa detailed different diameters but with the same ratio of specified outside “Unconstrained Pipe Wall Buckling (Hydrostatic Buckling). Note that the(22) apparent minimum wall thickness), the following variation of Love's equation , Eq. 10-2, is modulus of elasticity is a function of the duration of the anticipated load. When selecting a modulus to use in Equation 11 consideration should be given to internal pressurization of the line. When the pressure in the pipe exceeds the external pressure due to earth and live load, groundwater and/or slurry, the stress in the pipe wall reverses from compressive to tensile stress and collapse will not occur. For determining the pipe’s resistance to buckling during pullback, an additional reduction for tensile stresses is required, which is discussed in a later section of this chapter. Wall Compressive Stress The compressive stress in the wall of a directionally drilled PE pipe rarely controls design and it is normally not checked. However, it is included here because in some

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ormally not checked. However, it is included here because in es such as directional drilling at very deep depths such as in Chapter 12 trol design. Horizontal Directional Drilling

439

e applied to a buried pipe creates a compressive thrust stress When the pipe is pressurized, the stress is reduced due to the creating tensile thrust Thedrilling net stress can be such positive special cases stresses. such as directional at very deep depths as in landfills it nding on themay depth cover. Buried pressure lines may be controlof design. ompressive stress when applied shut todown or creates whena compressive experiencing The earth pressure a buried pipe thrust stress e usually short-term not typically in the pipeconditions wall. When theand pipe isare pressurized, the stress considered is reduced due to the pressure creating tensile thrust stresses.of Thepolyolefins net stress can be positive or sign, since internal the short-term design stress is negative depending on stress. the depth of cover. Buried lines may be er than the long-term design Pipes with pressure large depths ofsubject to net compressive stress when shut down or when experiencing vacuum. These are ng at low pressures may have net compressive stresses in the usually short-term conditions and are not typically considered significant for design, owing equation can be used to determine the net compressive since the short-term design stress of polyolefins is considerably higher than the longterm design stress. Pipes with large depths of cover and operating at low pressures may have net compressive stresses in the pipe wall. The following equation can be used to determine the net compressive stress: (12)

Sc =

PS D O PD 2t 288t

(12)

Where SC = Compressive wall stress, psi

= Earthstress, load pressures, psf CompressivePSwall psi D = Pipe outside diameter, O Earth load pressures, psf in t = Wall thickness, in Pipe outside Pdiameter, in = (Positive) internal pressure, psi Wall thickness,Din = Mean diameter, DO-t, in Positive) internal pressure, psi The D compressive Mean diameter, O-t, in wall stress should be kept less than the allowable compressive

stress of the material. For PE4710 PE pipe grade resins, 1150 psi is a safe allowable stress. For other materials see the Appendix of Chapter 3.

wall stress should be kept less than the allowable compressive

EXAMPLE CALCULATIONS An example calculation for selecting the DR for an HDD pipe is given in Appendix A.

Installation Design Considerations After determining 22 the DR required for long-term service, the designer must determine if this DR is sufficient for installation. Since installation forces are so significant, a lower DR (stronger pipe) may be required. During pullback the pipe is subjected to axial tensile forces caused by the frictional drag between the pipe and the borehole or slurry, the frictional drag on the ground surface, the capstan effect around drill-path bends, and hydrokinetic drag. In addition, the pipe may be subjected to external hoop pressures due to net external fluid head and bending stresses. The pipe’s collapse resistance to external pressure given in Equation 2 is reduced by the axial pulling force. Furthermore,

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440

Chapter 12


the drill path curvature may be limited by the pipe’s bending radius. (Torsional forces occur but are usually negligible when back-reamer swivels are properly designed.) Considerable judgment is required to predict the pullback force because of the complex interaction between pipe and soil. Sources for information include experienced drillers and engineers, programs such as DRILLPATH (14) and publications such as ASTM F1962 and ASCE MOP 108, “Pipeline Design for Installation by Horizontal Directional Drilling”. Typically, pullback force calculations are approximations that depend on considerable experience and judgment. The pullback formulas given herein and in DRILLPATH and ASTM F1962 are based on essentially an “ideal” borehole. The ideal borehole behaves like a rigid tunnel with gradual curvature, smooth alignment (no dog-legs), no borehole collapses, nearly complete cuttings removal, and good slurry circulation. The ideal borehole may be approached with proper drilling techniques that achieve a clean bore fully reamed to its final size before pullback. The closer the bore is to ideal; the more likely the calculated pullback force will match the actual. Because of the large number of variables involved and the sensitivity of pullback forces to installation techniques, the formulas presented in this document are for guidelines only and are given only to familiarize the designer with the interaction that occurs during pullback. Pullback values obtained should be considered only as qualitative values and used only for preliminary estimates. The designer is advised to consult with an experienced driller or with an engineer familiar with calculating these forces. The following discussion assumes that the entry and exit pits of the bore are on the same, or close to the same, elevation. For an overview, see Svetlik (15). Pullback Force Large HDD rigs can exert between 100,000 lbs. to 500,000 lbs. pull force. The majority of this power is applied to the cutting face of the reamer device/tool, which precedes the pipeline segment into the borehole. It is difficult to predict what portion of the total pullback force is actually transmitted to the pipeline being inserted. The pulling force which overcomes the combined frictional drag, capstan effect, and hydrokinetic drag, is applied to the pull-head and first joint of PE pipe. The axial tensile stress grows in intensity over the length of the pull. The duration of the pullload is longest at the pull-nose. The tail end of the pipe segment has zero applied tensile stress for zero time. The incremental time duration of stress intensity along the length of the pipeline from nose to tail causes a varying degree of recoverable elastic strain and viscoelastic stretch per foot of length along the pipe. The DR must be selected so that the tensile stress in the pipe wall due to the pullback force, does not exceed the permitted tensile stress for the pipe material. Increasing the pipe wall thickness will allow for a greater total pull-force. Even though the

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elected so that the tensile stress due to the pullback force does ermitted tensile stress for the pipe. Increasing the pipe wall Chapter 12 Horizontal Directional Drilling ow for a greater total pull-force, but the thicker wall also ght per foot of the pipe in direct proportion. Hence, thicker wall essarily reduce stress, only increase the absolute value of the ge. The designer should carefully check all proposed DR’s. thicker wall increases the weight per foot of the pipe, the pullback force within the

esistance

441

bore itself is not significantly affected by the increased weight. Hence, thicker wall pipe generally reduces stress. The designer should carefully check all proposed DR’s.

Drag Resistance o pullback in Frictional the borehole depends primarily on the frictional Pipe resistance to pullback in the depends primarily the frictional force ween the pipe and the borehole orborehole the pipe and the on ground created between the pipe and the borehole or the pipe and the ground surface in the y area, the frictional drag between pipe and drilling slurry, the entry area, the frictional drag between pipe and drilling slurry, the capstan effect bends, and atthe weight of the Equation gives resistance the bends, and the weight of thepipe. pipe. Equation 13 gives13 the frictional or e or required required pullingpulling forceforce forforpipe pulled in straight, level bores pipe pulled in straight, level bores or across level ground. Equation 13, gives the frictional resistance or required pulling force for pipe pulled und. in straight, level bores or across level ground. (See Kirby et al. (16)). (13)

FP = mWB L

(13)

pulling force,Where Flbs P = pulling force, lbs coefficient of mfriction between pipe slurry or (typically 0.40) = coefficient of friction between pipe and and slurry (typically(typically 0.25) or between 0.25) pipe and ground w = net downward (or upward) force on pipe, lb/ft B between pipe and ground (typically 0.40) L = length, ft net downward (or upward) force on pipe, lb/ft ength, ft When a slurry is present, WB equals the buoyant force on the pipe minus the weight

of the pipe and its contents, if any. Filling the pipe with fluid significantly reduces force andbuoyant thus the pulling force. pipe has a density is buoyancy the upward force ofPEthe pipe and near its that of present, wB the water. If the pipe is installed “dry” (empty) using a closed nose-pull head, the pipe e pipe with fluid significantly reduces the buoyancy force and will want to “float” on the crown of the borehole leading to the sidewall loading and frictional drag through the buoyancy-per-foot force and the wetted soil to pipe coefficient of friction. Most major pullbacks are done “wet”. That is, the pipeline is filled with water as it starts to descend into the bore (past the breakover point). 24 Water is added through a hose or small pipe inserted into the pullback pipe. (See the calculation examples.)

Note: The buoyant force pushing the empty pipe to the borehole crown will cause the PE pipe to “rub” the borehole crown. During pullback, the moving drill mud lubricates the contact zone. If the drilling stops, the pipe stops, or the mud flow stops, the pipe - slightly ring deflected by the buoyant force - can push up and squeeze out the lubricating mud. The resultant “start-up” friction is measurably increased. The pulling load to loosen the PE pipe from being “stuck” in the now decanted (moist) mud can be very high. This situation is best avoided by using thicker (lower DR) pipes, doing “wet” pulls, and stopping the pull only when removing drill rods.

421-461.indd 441

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ing stiffness pipes, inserting full pipe not empty pipe, and progress.] 442

Chapter 12


orehole, the force can be factored into horizontal and vertical y et al.(3) shows an additional frictional force that occurs in Force the pressureCapstan required by the borehole to keep the steel pipe For curves in the borehole, the force can be factored into horizontal and vertical with a radius of curvature similar to that used for steel pipe, components. Huey et al. shows an additional frictional force that occurs in steel nsignificant forpipe PE pipe. For very tight bends, it may be prudent due to the pressure required by the borehole to keep the steel pipe curved. For The frictionalbores resistance a similar pull isto that compounded bythese theforces are with a radiusduring of curvature used for steel pipe, the pipe is pulled around curve ortight bend creating an angle q, them. In insignificant for PE a pipe. For very bends, it may be prudent to consider addition to this force, the capstan effect increases frictional resistance when nding of the forces due to the direction of the pulling vectors. pulling along a curved path. As the is pulled around a curve or bend creating FC, due to the capstan effect is pipe given in Eq. 14. Equations 13 an angle q, there is a compounding of the forces due to the direction of the pulling vectors. d recursivelyThe topulling the pipe for each section along the pullback force, FC, due to the capstan effect is given in Eq. 14. Equations 13 and n in Figure 4.14This method is credited to each Larry Slavin, Outside are applied recursively to the pipe for section along the pullback distance as ervices, Inc. Rockaway, N.J. shown in Figure 4. This method is credited to Larry Slavin, Outside Plant Consulting (17)

Services, Inc. Rockaway, N.J. (14)

Fc = e mq (mWB L)

(14)

Where

Natural logarithm base (e=2.71828) e = Natural logarithm base (e=2.71828) m = coefficient of friction coefficient of friction angle of bend in pipe, radians angle of bendq =in pipe, radians wB = weight of pipe or buoyant force on pipe, lbs/ft weight of pipe or buoyant force on pipe, lbs/ft L = Length of pull, ft Length of pull, ft

25

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a m g Wp L1 L 2 L 3 L 4

a F1 m b Wb L 2 Wb H m g Wp L 2 expm g a

Chapter 12 443


m g a Wb L 3 expm b[What a m gisW"W p Lp3" exp above?]

(mg aexp )(m g m expH Wp (aL 1m + LW L 3 + L 4 )) 2 +L b F3 m b Wb L 4F1=W b b g p 4 expm g a F = exp(m a )(F + m W L + W H − m W L exp(m a ))

b b 2 b g p 2 g H = Depth of 2bore (ft)b 1 Fi = Pull Force on pipe at Point i (lb) F3 = F2 + m b Wb L 3 − exp(m b a )(m g Wp L 3 exp(m g a )) Li = Horizontal distance of Pull from point to point (ft) m = Coeff. of friction (ground (g) and borehole (b)) F4 = exp(m b b )(F3 + m b Wb L 4 − Wb H − exp(m b a )(m g Wp L 4 exp(m g a ))) W = Pipe weight (p) and Buoyant pipe weight (b) (lb/ft) a, b = Entry Where and Exit angles (radians)

Where:

H = Depth of bore (ft)

H = Depth of bore (ft) Fi = Pull Force on pipe at Point i (lb) Fi = Pull Force on pipe at Point i (lb) Li = Horizontal distance of Pull from point Li = Horizontal distance of Pull from point to point (ft)

to point (ft) m = Coeff. of friction (ground (g) and borehole (b)) m = Coeff. of friction (ground (g) and borehole (b)) W = Pipe 4 weight (p) and Buoyant pipe weight (b) (lb/ft) Wp = Weight of pipeFigure (lb/ft) a, b = Entry angles (radians) Wb = Buoyant force on pipe minus weightand ofCalculation pipeExit and contents (lb/ft) Estimated Pullback Force a, b = Entry and Exit angles (radians)

ic Force

Figure 4 Estimated Pullback Force Calculation

Figure 4

ng, pipe movement is resisted Estimated by the drag forceForce of theCalculation drilling fluid. Pullback Hydrokinetic Force kinetic force is difficult to estimate and depends on the drilling slurry, During pulling, pipe movement is resisted by the drag force of the drilling fluid. Hydrokinetic Force rate pipe pullback rate, and borehole and pipe sizes. Typically, the This hydrokinetic force is difficult to estimate and depends on the drilling slurry, c pressure is estimated bepullback in therate, 30and toborehole 60 kPa tosizes. 8 psi) range. slurry flow rateto pipe and(4 pipe Typically, the

During pulling, pipe movement is resisted by the drag force of the drilling fluid. This hydrokinetic force is difficult to estimate and depends on the drilling slurry, (15) πp (D pipe 2 2 slurry pullback rate, and borehole and pipe FHK flow p rate (15)sizes. Typically, the H - OD ) hydrokinetic 8 pressure is estimated to be in the 30 to 60 kPa (4 to 8 psi) range. hydrokinetic pressure is estimated to be in the 30 to 60 kPa (4 to 8 psi) range.

p Where 2 FHK = hydrokinetic force, FHK = p (D H - OD 2 ) FHK = hydrokinetic force, lbs lbs 8 p = hydrokinetic pressure, p = hydrokinetic pressure, psipsi DH = borehole diameter, inin diameter, DH = borehole Where: FHKin= hydrokinetic force, lbs OD = pipe outside diameter, OD = pipe outside diameter, in p = hydrokinetic pressure, psi ull back force, FT, then isDthe combined pullback diameter, in force, FP, plus the H = borehole ASCE MOP 108 suggests a different method for calculating the hydrokinetic OD = pipe outside diameter, c force, FHK. For the example shown in Figure 4, Fin F4. drag P equals

TRESS

(15)

force. suggests the external surface area of the pipe bypullback a fluid dragforce, FP, plus TheIttotal pullmultiplying back force, FT, then is the combined (18) coefficient of 0.025force, lb/in2 after Puckett . The total pullshown back force, FT, then 4, is the hydrokinetic FHK. For the example in Figure FP equals F4. DURING PULLBACK combined pullback force, FP, plus the hydrokinetic force, FHK. For the example shown in Figure 4, FP equals F4. TENSILE STRESS DURING PULLBACK

the

um outer fiber tensile stress should not exceed the safe pull stress. um outer fiberThe tensile stress obtained by taking the of the thesafe pull stress. maximum outerisfiber tensile stress should notsum exceed

The maximum outer fiber tensile stress is obtained by taking the sum of the

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444

Chapter 12


ss in the pipeTensile due toStress the pullback force, the hydrokinetic pulling force, During Pullback nsile bendingThestress due to pipe curvature. During pullback it is maximum outer fiber tensile stress should not exceed the safe pull stress. The o monitor the maximum pulling outer force and tostress useisaobtained “weakbylink” astheatensile pipestress of fiber tensile taking (such the sum of mechanical break-away or other failsafepulling method to prevent in the pipe due toconnector the pullback force, the hydrokinetic force, and the tensile ng the pipe. bending stress due to pipe curvature. During pullback it is advisable to monitor the pulling force and to use a “weak link” (such as a pipe of higher DR) mechanical breakaway connector or other failsafe method to prevent over-stressing the pipe.

stress occurring in the pipe wall during pullback is given by Eq. 16.

The tensile stress occurring in the pipe wall during pullback is given by Eq. 16. (16)

st =

E D FT + T OD 2R πt (D OD - t)

(16)

Where

sT = Axial tensile stress, psi sT = Axial tensile stress, psi FT = Total pulling force, lbs FT = Total pulling force, lbs t = Minimum wall thickness, in in t = Minimum wall thickness, DOD = Outer diameter of pipe, in diameter of pipe, in DOD = Outer ET = Time-dependent apparent modulus, psi (Refer to Appendix, Chapter 3, Engineering Properties, ET = Time-dependent tensile modulus, psi Code of the PE pipe being used and the for the appropriate value for the Material Designation applicable service conditions R = Minimum radius of curvature in bore path, in R = Minimum radius of curvature in bore path, in

ensile stress due to the pulling forces should not exceed the safe pull The axial tensile stress due to the pulling force should not exceed the pipe’s safe pull s in Table 5 load. canAsbe used,in aorprevious the designer can strength calculate a safe pull discussed section, the tensile of PE pipe is load-rate Time under load is an important considerationsection, in selectingthe the appropriate on a differentsensitive. pull time. As discussed in a previous tensile strengthsensitive, to use in calculating safe pull load. During pullback, the pulling PE pipe is tensile load-rate and the therefore values of “safe” pull force is not continually applied to the pipe, as the driller must stop pulling after h might be satisfactory for sliplining or insert renewal where the pull extracting each drill rod in order to remove the rod from the drill string. The net osed for a maximum of 30 min. to 60 min may not be satisfactory for result is that the pipe moves the length of the drill rod and then stops until the drilling. With extracted directional drilling, the istime duration of stress intensity rod is removed. Pullback an incremental (discrete) process rather than nger--between 4 hours to 24 hours. The “safe” pullload is time a continuous process. The pipe is not subjected to a constant tensile force and thus maymin. relax some between pulls. Apull one-hour modulus value might in be safe Hence, the 60 or less “safe” loadapparent (to limit elongation the for design, however, a 12-hour value will normally minimize “stretching” of the tion of the pipeline where the pull force is largest), is inappropriate for pipeline. Tables 3 through 6 give safe pull loads for PE pipes based on a 12-hour tion pulls. Table 2 gives safe tensile stress values for time intervals. value. The safe pull force also referred to as the allowable tensile load in the Tables ur value will3 through normally keep the pull-nose and 6 is based on the minimum pipe wall“stretch” thickness andlow may be foundavoid using erniation of the HDPE pipeline. safe values gas Equation 17. (The safe pull Allowable load may also be foundpullback using the average wallfor thickness. Check with the manufacture for the average wall values.) Allowable safe pullback ven in ASTM F-1807, “Practice for Determining Allowable Tensile valuesGas for gasPipe pipe are given inPull-In ASTM F-1807, “Practice for Determining olyethylene (PE) during Installation.” Tables 5Allowable and 6 Tensile Load for Polyehtylene (PE) Gas Pipe during Pull-In Installation”. pull loads for HDPE pipes.

ck, pipe may take several hours (typically equal to the duration of the ver from the axial strain. When pulled from the reamed borehole, the hould be pulled out about 3% longer than the total length of the pull. 421-461.indd strain will 444recover immediately and the viscoelastic stretch will1/16/09

10:15:07 AM

Chapter 12 445


(17)

WHERE FS = Safe Pull Force (lbs)

!

TALLOW = Safe Pull Stress (psi) DOD = Outside Diameter (in) DR = Dimension Ratio

After pullback, pipe may take several hours (typically equal to the duration of the pull) to recover from the axial strain. When pulled from the reamed borehole, the pull-nose should be pulled out about 3% longer than the total length of the pull. The elastic strain will recover immediately and the viscoelastic stretch will “remember” its original length and recover overnight. One does not want to come back in the morning to discover the pull-nose sucked back below the borehole exit level due to stretch recovery and thermal-contraction to an equilibrium temperature. In the worst case, the driller may want to pull out about 4% extra length (40 feet per 1000 feet) to insure the pull-nose remains extended beyond the borehole exit.

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446

Chapter 12


TABLE 3 PE 3xxx 12 hour Pull IPS Size Safe Pull Force, lbs Size

Nom. OD

9

11

13.5

17

1.25

1.660

940

787

653

527

1.5

1.900

1232

1030

855

690

2

2.375

1924

1610

1336

1079

3

3.500

4179

3497

2902

2343

4

4.500

6908

5780

4797

3872

6

6.625

14973

12529

10398

8393

8

8.625

25377

21235

17623

14225

10

10.750

39423

32988

27377

22098

12

12.750

55456

46404

38511

31086

14

14.000

66863

55949

46432

37480

16

16.000

87331

73076

60646

48954

18

18.000

110528

92487

76756

61957

20

20.000

136454

114182

94760

76490

22

22.000

165110

138160

114660

92553

24

24.000

196494

164422

136454

110146

26

26.000

230608

192967

160144

129268

28

28.000

267450

223796

185729

149920

30

30.000

307022

256909

213210

172102

32

32.000

N.A.

292305

242585

195814

34

34.000

N.A.

329985

273856

221056

36

36.000

N.A.

369949

307022

247827

42

42.000

N.A.

N.A.

417891

337321

48

48.000

N.A.

N.A.

N.A.

440582

54

54.000

N.A.

N.A.

N.A.

N.A.

*Tables are based on the Minimum Wall Thickness of Pipe

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Chapter 12 447


TABLE 4 PE 3xxx 12 hour Pull DIPS Size Safe Pull Force, lbs Size

Nom. OD

9

11

13.5

17

4

4.800

7860

6577

5458

4406

6

6.900

16241

13590

11279

9104

8

9.050

27940

23379

19403

15662

10

11.100

42031

35171

29188

23561

12

13.200

59440

49738

41277

33319

14

15.300

79856

66822

55456

44764

16

17.400

103282

86424

71724

57895

18

19.500

129717

108544

90081

72713

20

21.600

159160

133182

110528

89218

24

25.800

227074

190010

157690

127287

30

32.000

349323

292305

242585

195814

36

38.300

N.A.

418730

347506

280506

42

44.500

N.A.

N.A.

469121

378673

48

50.800

N.A.

N.A.

N.A.

493483


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448 Chapter 12


TABLE 5 PE 4xxx 12 hour Pull IPS Size Safe Pull Force, lbs Size

Nom. OD

9

11

13.5

17

1.25

1.660

983

822

682

551

1.5

1.900

1287

1077

894

722

2

2.375

2012

1683

1397

1128

3

3.500

4369

3656

3034

2449

4

4.500

7222

6043

5015

4048

6

6.625

15653

13098

10870

8774

8

8.625

26531

22200

18424

14872

10

10.750

41214

34487

28621

23103

12

12.750

57977

48513

40262

32499

14

14.000

69902

58492

48543

39184

16

16.000

91300

76398

63403

51179

18

18.000

115552

96691

80244

64773

20

20.000

142657

119372

99067

79967

22

22.000

172615

144440

119871

96760

24

24.000

205426

171896

142657

115152

26

26.000

241090

201739

167424

135144

28

28.000

279607

233969

194172

156735

30

30.000

320978

268587

222901

179925

32

32.000

N.A.

305592

253612

204715

34

34.000

N.A.

344985

286304

231104

36

36.000

N.A.

386765

320978

259092

42

42.000

N.A.

N.A.

436886

352654

48

48.000

N.A.

N.A.

N.A.

460609

54

54.000

N.A.

N.A.

N.A.

N.A.


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Chapter 12 449


TABLE 6 PE 4xxx 12 hour Pull DIPS Size Safe Pull Force, lbs Size

Nom. OD

9

11

13.5

17

4

4.800

8217

6876

5706

4606

6

6.900

16980

14208

11791

9518

8

9.050

29210

24442

20285

16374

10

11.100

43942

36770

30515

24632

12

13.200

62141

51998

43154

34834

14

15.300

83486

69859

57977

46799

16

17.400

107977

90353

74984

60527

18

19.500

135613

113478

94176

76018

20

21.600

166395

139235

115552

93273

24

25.800

237395

198647

164858

133073

30

32.000

365201

305592

253612

204715

36

38.300

N.A.

437764

363302

293256

42

44.500

N.A.

N.A.

490445

395886

48

50.800

N.A.

N.A.

N.A.

515914


External Pressure During Installation During pullback it is reasonable to assume that the borehole remains stable and open and that the borehole is full of drilling slurry. The net external pressure due to fluid in the borehole, then, is the slurry head, PMUD. This head can be offset by pulling the pipe with an open nose or filling the pipe with water for the pullback. However, this may not always be possible, for instance when installing electrical conduit. In addition to the fluid head in the borehole, there are also dynamic sources of external pressure: 1. If the pulling end of the pipe is capped, a plunger action occurs during pulling which creates a mild surge pressure. The pressure is difficult to calculate. The pipe will resist such an instantaneous pressure with its relatively high shortterm modulus. If care is taken to pull the pipe smoothly at a constant speed, this calculation can be ignored. If the pipe nose is left open, this surge is eliminated. 2. External pressure will also be produced by the frictional resistance of the drilling mud flow. Some pressure is needed to pump drilling mud from the reamer tool into the borehole, then into the pipe annulus, and along the pipe length while conveying reamed soil debris to the mud recovery pit. An estimate of this short term hydrokinetic pressure may be calculated using annular flow pressure loss formulas borrowed from the oil well drilling industry. This external pressure is dependent upon specific drilling mud properties, flow rates, annular opening, and hole configuration. This is a short-term installation condition. Thus, PE pipe’s short-term external differential pressure capabilities are compared to the actual

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onsideration of the dynamic or hydrokinetic pressure, PHK, the designer will nsideration of the dynamic or hydrokinetic pressure, PHK, the designer will additional external pressure to the slurry head: additional external pressure to the slurry head: 450 Chapter 12


+ PPHK -- PP1 MUD PPNN = PPMUD HK 1

(17) (17)

re the terms have been defined previously. re the terms have been defined previously.

short-term total external pressure during this installation condition.

stance to External Collapse Pressure During Under Collapse normal conditions, the annular-flow back pressure component is less stance to External Pressure During back Installation than 4-8 psi. back Installation

In consideration of the dynamic or hydrokinetic pressure, PHK, the designer will add

allowable external pressure equation, Eq.11, with the appropriate additionalbuckling external pressure to the slurry head: allowable external buckling pressure equation, Eq.11, with the appropriate dependent modulus value can be used to calculate the pipe’s resistance to (18) PN = PMUD + Pcan HK - Pbe I dependent modulus value used to calculate the pipe’s resistance to external pressure, PN, given by Eq.17 during pullback. The following external pressure, PN, given by Eq.17 during pullback. The following Where the terms have been defined previously. ctions in strength should be taken: ctions in strength should be taken: Resistance to External Collapse Pressure During Pullback Installation

The allowable external buckling pressure equation, Eq.11, with the appropriate apparent modulus (see chapter 3- Appendix) value can be used to calculate the during pullback. The can be pipe’s resistance external pressure, PN, given by Eq.18 he tensile pulling forceto the reduces the buckling resistance. This he tensile pulling force reduces the buckling resistance. This can be reductions in strength should be taken: unted for by following an additional reduction factor, fR. The pulling load in the pipe unted for by an additional reduction factor, fR. The pulling load in the pipe The tensile force reduces the buckling resistance. This can be accounted by es a hoop strain as pulling described by Poisson’s ratio. The hoop strainforreduces es a hoop strain as described byFRPoisson’s ratio. The hoopa hoop strain reduces . The pulling load in the pipe creates strain an additional reduction factor, buckling resistance. Multiply Eq.11 by the reduction factor, fR to obtain the buckling resistance. Multiply Eq.11 byhoop thestrain reduction fR to obtain the as described by Poisson’s ratio. The reduces the factor, buckling resistance. wable externalMultiply buckling pressure during pullback. Eq.11pressure by the reduction factor,pullback. FR to obtain the allowable external buckling able external buckling during pressure during pullback.

re re

(19)

(20)

F = FRR

(5.57 - (r + 1.09) 2 ) - 1.09 (5.57 - (r 1.09) 2 ) - 1.09

s r = s TT r 2S 2S

(18) (18) (19) (19)

s = Where calculated tensile stress during pullback (psi) sTT = scalculated tensile stress during pullback (psi) = calculated tensile stress during pullback (psi) T s = ssafe pull stress (psi) = safe pull stress (psi) s = safe pull stress (psi)

r = tensile stress ratio

e the pullback time is typically several hours, modulus value consistent Since the pullback time is typically several hours, a a value consistent e the pullback time is typically several hours, amodulus modulus value with consistent the pullback time can be selected from Appendix, Chapter 3. the pullback time can be selected from Table 2. he pullback time can be selected from Table 2. Bending Stress HDD river crossings incorporate radii-of-curvature, which allow the PE pipe to cold bend within its elastic limit. These bends are so long in radius as to be well within the flexural bending capability of SDR 11 PE pipe which can be cold bent to 25 times 29SDR 11 PE pipe, the radius of curvature could its nominal OD (example: for a 12” 29 be from infinity down to the minimum of 25 feet, i.e., a 50-foot diameter circle). Because the drill stem and reaming rod are less flexible, normally PE can bend easily

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Chapter 12 451


to whatever radius the borehole steel drilling and reaming shafts can bend because these radii are many times the pipe OD. However, in order to minimize the effect of ovaling some manufacturers limit the radius of curvature to a minimum of 40 to 50 times the pipe diameter. As in a previous section, the tensile stress due to bending is included in the calculations. Thermal Stresses and Strains HDD pipeline crossings generally become fully restrained in the axial direction as progressive sedimentation and soil consolidation occur within the borehole. The rate at which restraint occurs depends on the soil and drilling techniques and can take from a few hours to months. This assumption is valid for the vast majority of soil conditions, although it may not be completely true for each and every project. During pipe installation, the moving pipeline is not axially restrained by the oversize borehole. However, the native soil tends to sediment and embed the pipeline when installation velocity and mud flow are stopped, thus allowing the soil to grip the pipeline and prevent forward progress or removal. Under such unfortunate stoppage conditions, many pipelines may become stuck within minutes to only a few hours. The degree to which the pipeline will be restrained after completed installation is in large part a function of the sub-surface soil conditions and behavior, and the soil pressure at the depth of installation. Although the longitudinal displacement due to thermal expansion or contraction is minimal, the possibility of its displacement should be recognized. The PE pipe should be cut to length only after it is in thermal equilibrium with the surrounding soil (usually overnight). In this way the “installed” versus “operating” temperature difference is dropped to nearly zero, and the pipe will have assumed its natural length at the existing soil/water temperature. Additionally, the thermal inertia of the pipe and soil will oppose any brief temperature changes from the flow stream. Seasonal temperature changes happen so slowly that actual thermally induced stresses are usually insignificant within PE for design purposes. Torsion Stress A typical value for torsional shear stress is 50% of the tensile strength. Divide the transmitted torque by the wall area to get the torsional shear stress intensity. During the pullback and reaming procedure, a swivel is typically used to separate the rotating cutting head assembly from the pipeline pull segment. Swivels are not 100% efficient and some minor percent of torsion will be transmitted to the pipeline. For thick wall PE pipes of SDR 17, 15.5, 11, 9 and 7, this torsion is not significant and usually does not merit a detailed engineering analysis.

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452

Chapter 12


EXAMPLE CALCULATIONS Example Calculations are given in Appendix A and B.

References 1. Sener, E.M. & Stein, R. (1995). Mini-Horizontal Directional Drilling Manual, North American Society for Trenchless Technology (NASTT), Reston, Virginia. 2. Horizontal Directional Drilling Good Practices Guidelines, North American Society for Trenchless Technology (NASTT), Reston, Virginia. 3. Safety Manual: Directional Drilling Tracking Equipment, Equipment Manufacturer’s Institute (EMI). 4. ASTM F1962, Guide for Use of Maxi-Horizontal Directional Drilling for Placement of Polyethylene Pipe or Conduit Under Obstacles, Including River Crossings, ASTM, West Conshohocken, PA. 5. ASCE Manuals and Reports on Engineering Practice No. 108, “Pipeline Design for Installation by Horizontal Directional Drilling”, ASCE, Reston, Virginia (2005). 6. Petroff, L.J. (2006). Designing Polyethylene Water Pipe for Directional Drilling Applications Using ASTM F1962, NASTT No-Dig Conference, Nashville, TN. 7. O’Rourke, T.D., El-Gharbawy, S.L. & Stewart, H.E. (1991). Soil Loads at Pipeline Crossings, ASCE Specialty Conference on Pipeline Crossings, Denver, CO.. 8. Stein, D., Mollers, K. & Bielecki, R (1989) Microtunnelling, Ernest & Sohn, Berlin.. 9. Petroff, L.J. (1999). Guidelines for Design of Directionally-Drilled Polyethylene Pipe, International Plastic Pipe Fuel Gas Symposium, PPI, AGA, & GTI, New Orleans, LA. 10. Spangler, M. G. & Handy, R. L. (1973). Soil Engineering, Intext, Harper and Row, New York, NY. 11. Knight, M.A., Duyvestyn, G., & Gelinas, M. (2001, Sept). Excavation of surface installed pipeline, J. Infrastructural Systems, Vol. 7, no 3, ASCE. 12. Ariaratnam, S.T. (2001). Evaluation of the Annular Space Region in Horizontal Directional Drilling Installations, Arizona State University. 13. Watkins, R.K. & Anderson, L.R. (1995). Structural Mechanics of Buried Pipes, Dept. of Civil and Environmental Engineering, Utah State University, Logan, UT. 14. DRILLPATH (1996). Infrasoft L.L.C., Maurer Technology. 15. Svetlik, H. (1995, March). Design Considerations for PE Pipe Used in Directional Drilling, No-Dig Engineering, Vol.2, No.3. 16. Kirby, M.J., Kramer, S.R., Pittard, G.T., & Mamoun, M. (1996). Design Guidelines and Procedures for Guided Horizontal Drilling, Proceedings of the International No-Dig ‘96 Conf., New Orleans, LA. 17. Huey, D.P., Hair, J.D., & McLeod, K.B. (1996). Installation Loading and Stress Analysis Involved with Pipelines Installed by Horizontal Directional Drilling, No-Dig ‘96 Conf., New Orleans, LA. 18. Puckett, J.S. (2003). Analysis of Theoretical versus Actual HDD Pulling Loads, ASCE International Conference on Pipeline Engineering and Construction, Baltimore, MD.

Appendix A Design Calculation Example for Service Loads (Post-Installation) Example 1

A 6” IPS DR 11 PE4710 pipe is being pulled under a railroad track. The minimum depth under the track is 10 ft. Determine the safety factor against buckling.

Given Parameters OD = 6.625 in Nominal Pipe OD

DR = 11 Pipe Dimension Ratio

H = 10 ft. Max. Borehole Depth

gs = 120 lbf/ft 3

Unit Weight of Soil

PLive = 1,100 Ibf/ft 2 E-80 Live Load

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Chapter 12 453


Wheel loading from train will be applied for several minutes without relaxation. crossing may minutes accumulate. A relaxation. conservative choice for the l loading from Repetitive train will betrains applied for several without apparentmay modulus is the 1000-hour modulus. choice for the itive trains crossing accumulate. A conservative ent modulus is the 1000-hour modulus. PE Material Parameters Wheel loading E from willpsi be applied forLong-Term several minutes without relaxation. 43,700 ȝ= 0.45 Poisson’s Ration mid =train Wheel loading from train will be appliedAfor conservative several minutes without relaxation. Repetitive trains crossing may accumulate. choice for the 43,700 psi ȝ= 0.45 Long-Term Poisson’s Ration Repetitive trains crossing may accumulate. A conservative choice for the pparent modulus the Live 1000-hour modulus. on Pipe (Assuming that the apparent Soilisand Load Pressure earth load equals the modulus is the 1000-hour modulus. See Appendix of Chapter 3 Table B.1.1. loadon is Pipe perhaps too conservative exceptload for aequals calculation nd Live Load prism Pressure (Assuming that the earth the involving dynamic E = 46,000 psi Eload 43,700 psi ȝ= 0.45 Long-Term Poisson’s Ration mid mid = is surface loading.) perhaps too conservative except for a calculation involving dynamic e loading.) 2all PE pipe 2 Ratio = 0.45 for Soil and Live Load the earth 1 ftPipe /144(Assuming inmaterials Pthat = 15.97 psi load equals the Pμ == Poisson’s (gPressure SH+ PLive)on 2 Soil and2too rism load is perhaps conservative except for a calculation involving dynamic Live Load Pressure on Pipe (Assuming that the earth load equals the prism P = 15.97 psi SH+ PLive) 1 ft /144 in urface loading.) load isDeflection perhaps too conservative except soil for a calculation dynamic surface Ring resulting from and live involving load pressures assuming no side loading.) support is given 10.pressures assuming no side Deflection resulting from soil by andequation live load 2 2 ) P1 =ft(g /144 in P = 15.97 psi Prt=is(ggiven SH+ P Liveequation H + P ) 1 /144 by 10. s Live

Ring Deflection Presulting from soil and live load pressures assuming no side = 16.0 psi upport is given by equation 10. y 0.0125P O O Deflection resulting and live load pressures assuming no side support y Ring 0.0125 P D fromEsoil mid O = is given by equation 10. O 3 E mid D 12DR 1 )3 P 12 Δy y (DR − 0.10125 O

O

D

E mid

%y/D = 5.482 Percent deflection from soil loads 3 12DR 1 %y/D = 5.482 Percent deflection from soil loads Determine critical unconstrained buckling pressure based on deflection from % Δ y / D = 5.1 Percent deflection from soil loads loading and safety factorpressure using Eq.based 11 mine critical unconstrained buckling on deflection from Percent deflection from soil loads %y/D = 5.482 g and safety factor usingcritical Eq. 11 Determine unconstrained buckling pressure based on deflection from foloading = 0.56and Ovality compensation safety factor using Eq. 11 factor for 5.5% ovality from Figure 3 Determine critical unconstrained buckling pressure Figure based3 on deflection from 56 Ovality compensation for 5.5% fo = 0.58 Ovalityfactor compensation factor forovality 5.1% ovalityfrom from Figure 3 oading and safety factor using Eq. 11 2E mid 1 PUC ( )3 f O 2 2E mid 1 DR 1 ( 1 m ) 3 ( factor )for f 5.5% ovality from Figure 3 compensation o = 0.56 Ovality P UC = μ22 ) DR − 1 O (1- m

PUC = 68.4 psi 2E mid 1 P61.37 (Critical ) 3 safety f O factor) PCritical psibuckling buckling pressure (no safety factor) UC UC =unconstrained 2 pressure (nounconstrained ( 1 - m ) DR 1 61.37 psi Critical unconstrained buckling pressure (no safety factor) PUC SFcr SFcr = 3.84 Safety factor against buckling PUC P PUC = 61.37 psiSF =Critical unconstrained buckling pressure (no safety factor) factor against buckling P SFcrcr= 4.33.84 Safety Safety factor against buckling Example 2: A 6” IPS DR 13.5 HDPE pipe is being pulled under a small river for PUC use as aniselectrical duct. under At its lowest the pipe will be 18 feet SFcr 2: A 6” IPS DR 13.5 HDPE ple pipe being pulled a smallpoint, river for SFcr = 3.84 Safety factor against buckling P below the river surface. Assume the slurry weight is equal to 75 use as an electrical duct. At its lowest point, the pipe will be 18 feet Ib/cu.ft. The the ductslurry is empty theto pull. below the river surface. Assume weightduring is equal 75 Calculate a) the ExampleIb/cu.ft. 2: A 6”The IPS DR is being under a small riveragainst for maximum pulling force and the safety buckling for duct13.5 is HDPE empty pipe during the pulled pull. b)Calculate a)factor the use as an electrical duct. At its lowest point, the pipe will be 18 feet the pipe. Assume that the pipe’s ovality is 3% and that the pulling maximum pulling force and b) the safety factor against buckling for below the river surface. Assume the slurry weight is equal to 75 time willpipe’s not exceed the pipe. Assume that the ovality 10 is hours. 3% and that the pulling Ib/cu.ft. The duct is empty during the pull. Calculate a) the time will not exceed 10 hours. maximum pulling force and b) the safety factor against buckling for 421-461.indd 453

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454

Chapter 12


Example 2

A 6” IPS DR 13.5 PE4710 pipe is being pulled under a small river for use as an Solution:

electrical duct. At its lowest point, the pipe will be 18 feet below the river surface. Assume the slurry is equal to 75 Ib/cu.ft. The duct is empty during the pull. Solution: Calculate the safe pullweight strength or allowable tensile load. Calculate a) the maximum pulling force and b) the safety factor against buckling for the pipe.the Assume the pipe’s ovality is 3% and that the pulling safe pull strength or allowable tensile load.time will not OD =Calculate 6.625in. Pipethat outside diameter Solution: exceed 10 hours. DR = 13.5 Pipe dimension ratio

6.625in. diameter Tallow =OD 1150=psi TypicalPipe safeoutside pull stress for HDPE for 12-hour pull duration Calculate the safe pull strength or allowable tensile load. Solution DR = 13.5 Pipe dimension ratio the safe pull strength or allowable tensile load. TCalculate 1 psi 1 Typical safe pull stress for HDPE for 12-hour pull duration allow = 1150 2 OD = 6.625in. Pipe ) outside diameter Fs STOD OD (- Pipe outside diameter allow= 6.625in. 2 DR DR DR = -13.5 Pipe dimension ratio DR = 13.5 Pipe dimension ratio 1 1 2 = 1150 psi Typical stress HDPE for1. 12-hour pull duration TTFallow S - stress2for)safe T OD ( = 1150 psi - Typical safe pull PE4710pull for 12-hour pull for duration. See Table allow s allow 4 DRpull DR strength for 6î IPS DR 13.5 HDPE pipe assuming Fs = 1.088 x 10 lbf Safe 10-hour maximum pull duration 1 1 2 πTallow xOD ) strength for 6î IPS DR 13.5 HDPE pipe assuming 104( lbf Safe pull FFss == 1.088 2 DR DR 10-hour duration Step 1: Determine the criticalmaximum buckling pull pressure during Installation for the 4 Fs = 1.088 x 10 lbf pipe (include tensile reduction factor assuming the frictional drag 4 1.088 x6”10 lbf13.5Safe strength for 6î pull IPS DR 13.5 HDPE pipe assuming FSafe for IPS DR PE pipepull assuming 12-hour maximum duration. s =pull Step 1:force Determine the critical pressure during Installation for the during pull in 1000 psibuckling longitudinal pipe stress) Also see Table 5 for safe pull results force. 10-hour maximum pull duration pipe (include tensile reduction factor assuming the frictional drag Step 1psi during pullmodulus results inof1000 psi longitudinal pipe E = 57,500 Apparent elasticity (for 10 hours at stress) 73 degrees F) Step 1: the critical Determine the critical buckling pressure during Installation for the Determine buckling pressure during Installation for the pipe (include ȝ = 0.45 Poisson’s ratio (long term value) pipe (include the tensile reduction factor assuming the frictional drag reduction factor assuming frictional drag during pull results 1000 psi Etensile = 57,500 psiOvality Apparent modulus of elasticity (for 10 in hours at 73 degrees F) fo = 0.76 compensation factor (for 3% ovality) during pull results in 1000 psi longitudinal pipe stress) longitudinal pipe stress) ȝ = 0.45 Poisson’s ratio (long term value) foE = 0.76 Ovality factor 3% ovality) = 63,000 psi - Apparent of elasticity (for 12 hours1000 at 73 degrees F) (for R = 0.435 Tensile ratio modulus (based on compensation assumed psi pull stress calculation) Eμ = 57,500 psi Apparent modulus of elasticity (for 10 hours at 73 degrees F) = Poisson’s Ratio = 0.45 for all PE materials ȝf == 0.76 0.45 Poisson’s ratio (long term value) - Ovality compensation factor (for 3% ovality) Ro = 0.435 Tensile ratio (based on assumed 1000 psi pull stress calculation) fo = 0.76 Ovality compensation factor (for 3% ovality)

(based 5.57Tensile r 1.09 1.09 onf assumed 0.71 1000 Tensile Reduction Factor Rf R= 0.435 ratio psi pull stress calculation) R 2

R = 0.435 - Tensile ratio (based on assumed2 1000 psi pull stress calculation) f f R 0.71 1 5.573 r 1.09 1.09 fR2E = 0.71 R

Critical unconstrained Tensile Reduction Factor buckling pressure for DR 13.5 ) f f P 39.90 ( Pcr O R CR (1 - P 2 ) DR - 1 Criticalsafety unconstrained pipe without factor 2E= 5.57 −1 (r + 31.09)2 − 1.09 f f = 0 . 71 Tensile Reduction Factor buckling pressure for DR 13.5 R ) fO fR PCR R39.90 ( Pcr 2 DR 1 P (1 ) without safety Step 2: Determine expected loads on pipe (assume onlypipe static drilling fluidfactor Critical unconstrained 1 on 3pipe, and borehole intact with no buckling 2E Factor head acting soil loading) Tensile Reduction pressure for DR 13.5 =43.71 2 ( ) ⋅ fO ⋅ fR PCR = 39.90 PPcr CR = 2: Step Determine expected loads on pipe (assume only static drilling fluid (1 -3 µ ) DR - 1 pipe without safety factor Critical unconstrained buckling pressure for DR 13.5 pipe without safety factor headfluid acting on pipe, H and borehole intact with soil loading) gslurry = 75 lbf/ft , drilling weight = 18 ft, Maximum boreno depth Step 2: Determine expected loads on pipe (assume only static drilling fluid 3 gslurry = 75 lbf/ft , drilling fluid weight H = 18 ft, Maximum depth Total staticbore drilling 1ft 2 head acting on pipe, and borehole intact with no soil loading) ) P 9.37 psi Pslurry Hg slurry ( fluid head pressure if slurry 144in 2 static drilling 2 drilled Total from surface gPslurry = 75 lbf/ft( 3,1ft drilling fluid weight H = 18 ft, Maximum bore depth ) P 9.37 psi Hg fluid head pressure if slurry slurry slurry 2 144in drilled fromduring surface Step 3: Determine the resulting safety factor against critical buckling Total static drilling 1ft 2 Installation ) Pslurry = 9.37 psi Pslurry = Hg slurry ( fluid head 1/16/09 pressure ifAM 421-461.indd 454 10:15:09 2 Step 3: Determine the resulting safety factor against critical buckling during

1 .209 1.09 f R f 0R .710.71 Tensile Tensile Reduction Factor r 1r.09 f R f R 5.575.57 1.09 Reduction Factor 2

Chapter 12 455


2E 2E 1 13 3 39.90 ) f)O f RO f R PCR PCR 39.90 ( ( Pcr Pcr -1 -1 (1 - P(12 )- P 2 )DR DR Step StepStep 2: 22:

Critical unconstrained Critical unconstrained buckling pressure for DR buckling pressure for DR 13.513.5 without safety factor pipepipe without safety factor

Determine expected loads on pipe (assume static drilling Determine expected loads on pipe (assume onlyonly static drilling fluidfluid

Determine expected loads on onlyborehole static drilling fluid head acting onloading) head acting on (assume pipe, intact no soil head acting onpipe pipe, andand borehole intact withwith no soil loading) pipe, and borehole intact with no soil loading) 3

33 ggslurry 75 lbf/ft , drilling fluid weight H =H18= ft, 18Maximum ft, Maximum depth 75 lbf/ft drilling fluid weight gslurry = 75==lbf/ft , ,drilling fluid weight borebore depth slurry

H = 18 ft, Maximum bore depth

1ft 2 1ft 2 ) Pslurry9.379.37 Hg(slurry ( 2 ) P2 slurry psi psi PslurryPslurry Hg slurry 144in 144in

Total static drilling Total static drilling pressure fluidfluid headhead pressure if if drilled from surface drilled from surface

Pslurry = 9.36 psi Total static drilling fluid head pressure if drilled from surface

StepStep 3: 3: Step 3

Determine resulting safety factor against critical buckling during Determine the the resulting safety factor against critical buckling during Installation Installation

Determine the resulting safety factor against critical buckling duringSafety installation Safety factor against factor against

SFCR

P P PCR SFCRSF=CR =CR CR SFCR PslurryPslurry Pslurry

= 4.25 = 4.25 SFCR SFCR 4.25

critical buckling during critical buckling during pull.pull.

34 34 SFCR = 4.67 3: 3: Determine Determine the the safety safety factor factor for for long long termterm performance performance the the Example Example 3: Example Determine the safety factor for long term performance for the for for Safety factor against critical buckling during pull communication communication ductduct in example in Assume example 2. Assume 2. Assume are are 10 10 feetfeet of of communication duct in example 2. there arethere 10there feet of Example 3 riverbed riverbed deposits deposits above above the the borehole borehole having having a saturated a saturated unit unit riverbed deposits above 3the 3 borehole having a saturated unit 3 Determine safety long-term performance for the communication duct weight weight offactor of for 110 Ib/ft Ib/ft .deep, (18. (18 feet feet deep, deep, 3% 3% initial initial ovality) ovality) weight ofthe 110 Ib/ft .110 (18 feet 3% initial ovality) Solution:

in Example 2. Assume there are 10 feet of riverbed deposits above the borehole having a saturated unit weight of 110 lb/ft3. (18 feet deep, 3% initial ovality) Solution: Solution:

Solution 1: 1: Determine Determine pipe pipe soil soil loadload (Warning (Warning requires requires Input of ovality of ovality Step 1: StepStep Determine the pipe the soilthe load (Warning requires Input of Input ovality compensation compensation in step in step 4.) 4.) Step 1 compensation in step 4.)

E

Determine the pipe soil load (Warning: Requires input of ovality compensation E long E long =psi 28,200 = 28,200 psi psi termLong Long termterm apparent apparent modulus modulus 4. long in = step 28,200 Long apparent modulus

3 3 3 = 62.4 glbf/ft. 62.4 lbf/ft. lbf/ft. Unit weight weight of water of water w =gpsi w gw =E long 62.4 Unit weight ofUnit water = 29,000 - Long-term apparent modulus H =3 H = ftweight 18Max. ftofMax. borehole borehole depth depth = 62.4 -18 Unit water H g=w 18 ftlbf/ft. Max. borehole depth 3 3 3 g = g 110 = 110 lbf/ft. lbf/ft. Saturated Saturated unit weight weight of sediments of sediments s - Borehole s 18 ft Max. depth Saturated unit weight ofunit gs H== 110 lbf/ft. sediments GW 18= ft18unitftweight Groundwater Groundwater height 110 lbf/ft. -=Saturated of sediments s = 18 GWg= ft 3GW Groundwater height height C = C 10ft. = 10ft. Height Height of soil of soil cover cover GW = 18 ft Groundwater height C = 10ft. Height of soil cover C = 10ft. -OD Height of6.625 soil cover OD = = 6.625 in in Nominal Nominal OD OD OD = 6.625 in Nominal pipe OD pipepipe OD = 6.625 in -DR Nominal pipe OD DR = 13.5 = 13.5 Pipe Pipe dimension dimension ratioratio DR = 13.5 Pipe dimension ratio DR = 13.5 - µ Pipe dimension ratio = 0.45 Long Term Long Long Term Term Poisson’s Poisson’s ratioratio µ = 0.45 = µ0.45 Poisson’s ratio

μ = Poisson’s Ratio = 0.45 for all PE materials

Psoil

2 1ft 2 1ft 2 1ft load load on pipe on pipe fromfrom 10’ 10’ of of = = = 3.30 ) P2 psi ) P=soil3.30 psiPrism psi loadPrism P P ( g ( g g ) g C ) ( C on Prism pipe from 10’ of soil (g S - gsoilW ) soil C (S SW 2 W) Psoil ( 23.30 144in 144in saturated saturated cover cover (including (including 144in saturated cover (including buoyant buoyant force force on submerged on Psoil = 3.30 psi buoyant force on submerged submerged soil) soil) soil) Prism load on pipe from 10’ of saturated cover (including soil) buoyant force on submerged

2: 2: Calculate the the ringring deflection deflection resulting from soil soil loads loads assuming no no Step 2: StepStep Calculate theCalculate ring deflection resulting fromresulting soil from loads assuming noassuming sideside support. support. side support. 421-461.indd 455

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Psoil

(g S - g W ) C (

1ft ) Psoil 144in 2

456 Chapter 12


3.30 psi

Prism load on pipe from 10’ of saturated cover (including buoyant force on submerged soil)

Step 2: Calculate the ring deflection resulting from soil loads assuming n Pside P SFCR 4.254.25 SFCRSFCR CR CRsupport. SFCR PslurryPslurry Step 2

Calculate the ring deflection resulting from soil loads assuming no side support. Example Example 3: 3: Determine Determine the the safety safety factor factor for for longlong termterm performance performance for for the the communication communication ductduct in example in example 2. Assume 2. Assume there there are are 10 10 feetfeet of of 0 . 0125 P 100 u u riverbed deposits deposits above the the borehole borehole having having a saturated a saturated unitunit soil above Percent deflection from soil % (Δy/D) =riverbed 3 3 weight weight of 110 of 110 Ib/ft Ib/ft . (18 . (18 feet feet deep, deep, 3% 3% initial initial ovality) ovality) loads %y/D = 3.43 %(y/D) = E long [ ] 12 (DR - 1) 3 Solution: Solution: % (Δy/D) = 3.33 Percent deflection from soil loads

t = OD/DR t =0.491 inpipe StepStep 1:t = OD/DR 1: Determine Determine the the pipe soil soil loadload (Warning (Warning requires requires Input Input of ovality of ovality t =0.491 in compensation compensation in step in step 4.) 4.) Step 3: Determine the long-term hydrostatic loads on the pipe Step 3 Determine hydrostatic loads on term the pipe E long E long = the 28,200 =long-term 28,200 psi psi Long Long term apparent apparent modulus modulus 3 3 gw =gw62.4 = 62.4 lbf/ft. lbf/ft. UnitUnit weight weight of water of waterExternal pressure due to GW = 11.09 ) +Max. Psoil PWborehole PW = H( =H18= ft18Max. ft borehole depth depth groundwater head 2.31 ft/psi 3 3 GW External pressure due to g g = 110 lbf/ft. lbf/ft. Saturated Saturated unit unit weight weight of sediments of sediments s = s110 ) Psoil PW 11.09 ( PP = 11.09 WW groundwater headdue to GW pressure GWGW 18= ft18 3ft Groundwater Groundwater height heightExternal 2=.31 ft/psi due to groundwater ) Psoil head PW Unit 11.09weight of drilling fluid PW =pressure (75 gExternal lb/cu.ft. groundwater head slurry C = C 10ft. = 10ft. Height Height of soil of soil cover cover GW External pressure due to .31 ft/psi ) P 35 PWNominal 11.09 PW OD ( 2OD soil GW External pressure = 6.625 = 6.625 in in Nominal pipe pipe OD OD groundwater headdue to 3 .31lb/cu.ft. ft/psi ) + Psoil PW = 11.09 = ( 275 PgWslurry Unit weight of drillinggroundwater fluid head DR2.DR =3113.5 = 13.5 PipePipe dimension dimension ratioratio ft/psi 3 1ft 2 Unit weight of drilling fluid gUnit 75 lb/cu.ft. External pressure due to slurry = = ) P 9.37 psi Pslurry g H ( weight ofslurry drilling fluid µ= µ 0.45 = 0.45 Long Long Term Term Poisson’s Poisson’s ratioratio slurry slurry 2 3 Unit weight of drilling fluid head g slurry 75 lb/cu.ft.144in 2 3 1ft External pressure due to slurry weight g slurry lb/cu.ft. 9.37 psi of drilling fluid Pslurry= 75g slurry H (2 2 2 ) Pslurry Unit 1ft 144in 1ft head External pressure Prism Prism load load on pipe on pipe fromfrom 10’due 10’ of to of slurry ) PP) Pfor 3.30 psi Psoil Psoil(Pgslurry CW, therefor )( H C ( 2 use 9.37psi psi g-)gslurry >ggPSWslurry buckling load S -( soil3.30 W 1ft 22 2 soilW slurry 144in 144in head saturated saturated cover cover (including (including External pressure due to slurry Pslurry g slurry H ( 1ft 2 ) Pslurry 9.37 psi External pressure due to slurry ! P for buckling load P P , therefor use 144in buoyant buoyant force force on submerged on submerged ) WPslurry = 9.37 psi head P gslurry Pslurry psi H ( W ==9.37 slurry slurry 2 Step 4: Determine critical unconstrained buckling pressure based on 144in head PW ! Ppressure use External due to slurry headPW for buckling load soil)soil) slurry , therefor deflection from loading PW ! P4:slurry , therefor use PW for buckling load Step Determine critical unconstrained buckling pressure based on PW for buckling load , therefor usethe StepStep 2:PW2:> Pslurry Calculate Calculate the ring ring deflection deflection resulting resulting from from soil soil loads loads assuming assuming no no deflection loading Step Determine critical unconstrained pressure on 5% Ovalityfrom Compensation based buckling on 3% initial ovalitybased and 2% fo = 0.4:64 side side support. support. Therefor use PW for buckling load deflection loading Step 4: Determine critical unconstrained buckling pressure based on deflection from Step 4: 64 Determine critical unconstrained pressure based fo = 0. 5% Ovality Compensation based buckling on 3% initial ovality and on 2% deflection from loading Step 4 deflection from loading deflection based on 3% initial ovality and 2% fo = 0. 642E long 5% Ovality 1 3 Compensation Determine critical (unconstrained buckling pressure based deflection from loading = ) f P 23.17 psi on on deflection Critical unconstrained O UC 5% Ovality Compensation based 3% initial ovality and 2% foP=UC0.= 64 2 DR -1 −percent (12E m 5% ) Ovality Five Compensation based on 3.3% deflection with an additional factor for fofo== 0.0.6464 Ovality Compensation based on 3% initial ovality 2% buckling pressure (noand safety 1 3 longdeflection conservatism. fO PUC 23.17 psi PUC0 . 0125 02E . 0125 Psoil u 2deflection u( Pusoil100 u) 100 Critical unconstrained Percent Percent deflection deflection fromfrom soil soil factor 1 -1 3 (1 long m ) ( DR ) f P 23.17 psi P buckling pressure (no safety Critical unconstrained UC O UC 2 loads loads %y/D %y/D = 3.43 = 3.43 %(y/D) %(y/D) = = 2E DR1 - 1) 3 f Elong mlongE) long ( P 23.17 psi PUC[ ([12E factor ] ] buckling pressure (no Critical unconstrained safety O UC 22 31 3 3 long μ(DR DR 1 m ) 1) 1) = ( ) f P 23.17 psi PUC =12(1(DR P12 Critical unconstrained O UC 2 factor buckling Safety factor against pressure of buckling pressure (no safety SF CR =(1 −UCm ) DR - 1 SF CR = 2.08 buckling pressure (no safety PUC = 23.83 P psiW highest load (slurry) factor P UCt =0.491 t = OD/DR t = Critical OD/DR t =0.491 in in Safety factor against factor buckling pressure of buckling pressure (no safety factor) SF CR SF CRunconstrained 2.08 PP highestfactor load against (slurry) buckling pressure of Safety SF SF CR SFCRCR= 2.14 UCW 2.08 PPUC StepStep 3: Determine 3: Determine the the long-term long-term hydrostatic hydrostatic loads loads on the on the pipe pipe buckling pressure of highest load (slurry) W Safety factor against SF CR SF CR 2.08 PPUC Safety factor against highest load (slurry) buckling pressure of SF CR = 2.08 SF CR = W PW highest load (slurry) 421-461.indd 456

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fo = 0. 64

PUC =

5% Ovality Compensation based on 3% initial ovality and 2% deflection

2E long

1 3 ) fO (1 − m ) DR - 1

SF CR =

2

(

P UC PW

PUC = 23.17 psi

SF CR = 2.14 2.08

Chapter 12 457


Critical unconstrained buckling pressure (no safety factor

Safety factor against buckling pressure of highest load (slurry)

Safety Factor against buckling pressure of highest load groundwater head (11.09 psi)

APPENDIX B Design Calculations Example for Pullback Force Example 1

APPENDIX B: DESIGN CALCULATIONS EXAMPLE FOR PULLBACK FORCE

Find the estimated force required to pull back pipe for the above theoretical river crossing using Slavin’s Method. Determine the safety factor against collapse. Assume the PE pipe is 35 ft deep and approximately 870 ft long with a 10 deg. entry angle and a 15 deg. exit angle. Actual pullback force will vary depending on backreamer size, selection, and use; bore hole staying open; soil conditions; lubrication with bentonite; driller expertise; and other application circumstances.

36 Pipe Properties Outside Diameter OD = 24 in - Long-term Modulus - Elong = 29,000 psi, PE4710 Material

Standard Dimension Ratio DR = 12 - 12 hr Modulus - E 24hr =63,000 psi

Minimum wall thickness t = 2.182 in - Poisson’s ratio (long term) - µ = 0.45 - Safe Pull Stress (12 hr) - s pb = 1,150 psi

Path Profile H = 35 ft Depth of bore

gin = 10 deg Pipe entry angle gex = 15 deg Pipe exit angle L1 = 100 ft Pipe drag on surface (This value starts at total length of pull, approximately 870 ft. then decreases with time. Assume 100 ft remaining at end of pull)

Lcross = 870 ft

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pull, approximately 870 ft. then decreases with time. Assume 100 ft remaining at end of pull) L

458 Chapter 12

= 870 ft

cross Drilling Horizontal Directional

Path length (Determine L2 and L4): Average Radius of Curvature for Path at Pipe Entry radians Path length (Determine L2 and L4)

gin

is

given

in

Average Radius of Curvature for Path at Pipe Entry gin is given in radians

R avg in = 2H/g in avgin

2

R avgin = 2.298 × 10 3 ft

Ravg in = 2.298 x 103 ft

Average Radius offor Curvature for Path at Pipe Exit Average Radius of Curvature Path at Pipe Exit 2 R agex 2H/g ex R agex 1.021 u 10 3 ft 2 Ravg R agex = 1.021 × 10 3 ft agexex = 2H/g ex Horizontal Required to 37 Achieve Depth or Rise to the Surface at Ravg ex = 1.021 Distance x 10 3 ft Pipe Entry Horizontal Distance Required to Achieve Depth orto RiseAchieve to the SurfaceDepth at Pipe Entry Horizontal Distance Required or Rise to the Surface at Pipe Entry L 2 2H/g in 2H/g LL2 2= = 401.07 ft in

L 2 401.07ft L 2 = 401.07ft

Horizontal Distance Required or Rise to the Surface at Horizontal Distance Required to Achieve Depth orto RiseAchieve to the SurfaceDepth at Pipe Exit Pipe Exit Distance Required to Achieve Depth or Rise to the Surface at Horizontal Where L2 & LExit Pipe 4 = horizontal transition distance at bore exit & entry respectively.

L 4 2H/gex L 4 = 2H/gAxial ex Determine Bending Stress

L 4 267.38ft L 4 = 267.38ft

R = Ravg ex - Min. & for L4Drill=pathhorizontal transition distance at bore exit & entry Where: L2 Radius R = 1.021 x 10 3 ft respectively. Where: L2 & L4 = horizontal transition distance at bore exit & entry OD = 24 in respectively. Radius of curvature should exceed 40 times the pipe outside diameter to prevent ring collapse. Determine Axial Bending Stress: r = 40 OD Determine r = 80Axial ft Okay Bending Stress: Min. Radius for Drill path R > r R = Ravgex 3 R= = 1.021 Ravgex x 10 ft Min. Radius for Drill path R Bending strain

3 R = =1.021 OD 24 inx 10 ft ea = 9.79 x 10 -4 in/in = 24 in should exceed 40 times the pipe outside diameter to RadiusOD of curvature Where of prevent ring collapse.should exceed 40 times the pipe outside diameter to Radius curvature ea = bending strain, in/in prevent ring collapse. OD = outside diameter of pipe, in r = 40 OD R = minimum radius of curvature, ft = 80 40 ft OD rr = Okay. R>r Bending stress r =strain 80 ft Okay. R>r Bending ea = OD/2R

Sa = E12hrea

Bending Sa = 61.68 psistrain ea = 9.79 x 10-4 in/in ea = OD/2R wheree = OD/2R ea = 9.79 x 10-4 in/in a Sa = bending stress, psi Where: ea = bending strain, in/in Where: ea ==bending in/inof pipe, in OD outside strain, diameter Bending stress Bending stress 421-461.indd 458 sa = E24hrea

OD outside diameter of pipe, in ft R == minimum radius of curvature, R = minimum radius of curvature, ft sa = 55.32 psi

1/16/09 10:15:11 AM

Chapter 12 459


Find Pulling Force Weight of Empty Pipe Pw =3.61x10 -2 lbf/in3 ga = 0.95 gb = 1.5 wa = πOD 2 (DR-1/DR 2)r w ga 12 in/ft wa = 61.54 lbf/ft

Net Upward Buoyant Force on Empty Pipe Surrounded by Mud Slurry Wb = π(OD 2 /4) r w gb 12 in/ft - wa wb = 232.41 lbf/ft

Where rw = density of water, lb/in3 ga = specific gravity of the pipe material

gb = specific gravity of the mud slurry wa = weight of empty pipe, Ibf/ft wb = π(OD2 /4)rwgb 12in/ft - wa Determine pullback force acting on pipe See figure:

L1 = 100 ft - va = 0.4 L2 = 401.07ft - vb = 0.25 L3 = 200ft - σ = gin - σ = 10 deg = 0.175 radians L 4 = 267.38 - β = gex - β = 15 deg = 0.262 radians L3 = Lcross- L 2 - L 4 - L 3 = 201.549ft TA = exp (va σ) [va wa (L1 + L 2 + L 3 + L 4)] TA = 2.561 x 104 Ibf TB = exp (vb σ) (TA + vb [wb] L2 + wb H - va wa L 2 exp (vb σ)) TB = 4.853 x 104 Ibf TC =TB + vb [wb] L 3 - exp(vb σ) (va wa L 3.exp (va σ)) TC = 5.468 x 104 Ibf TD = exp(vb σ) [TC + vb [wb] L 4 - wb H - exp(vb σ) (va wa L 4 exp (vb σ))] TD = 5.841 x 104 Ibf Where TA = pull force on pipe at point A, Ibf

TB = pull force on pipe at point B. Ibf TC = pull force on pipe at point C, Ibf TD = pull force on pipe at point D, Ibf L1 = pipe on surface, ft L2 = horizontal distance to achieve desired depth, ft L3 = additional distance traversed at desired depth, ft

421-461.indd 459

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460

Chapter 12


L 4 = horizontal distance to traversed rise to surface, ftat desired depth, ft L3 = additional distance va = coefficient of friction applicable surface before L4 = horizontal distance to riseattothesurface, ftthe pipe enters bore hole v = coefficient of friction applicable within the lubricated bore hole orbefore after the (wet) exits b va = coefficient of friction applicable at the surface thepipe pipe σ = bore hole angle at pipe entry, radians enters bore hole β = bore holeof angle at pipe exit, radians vb = coefficient friction applicable within the lubricated bore hole (refer to figure at start of this appendix) or after the (wet) pipe exits ı = bore hole angle at pipe entry, radians Hydrokinetic ȕ = bore hole angle Pressure at pipe exit, radians

okinetic

∆P = 10 psi Dh = 1.5 0D (refer to figure at Dh = 36in ∆T = ∆P (π/8) (Dh2 - OD2) Pressure ∆T = 2.82 x 103Ibf

start of this appendix)

¨P = 10 psi Dh = 36in Dh = 1.5 0D Where: 2 increment, Ibf ∆T = 2pulling force ¨T = ¨P (ʌ/8) (Dh - OD ) ¨T = 2.82 x 103Ibf ∆P = hydrokinetic pressure, psi e: ¨T = pulling force increment, Ibf Dh = back reamed hole diameter, in ¨P = hydrokinetic pressure, psi Dh = back reamed hole diameter, in

Compare Axial Tensile Stress with Allowable Tensile Stress During Pullback of 1,150 psi: (Assume the pull takes several hours and use 12 hours safe pull stress.)

pare Axial Tensile Stress with Allowable Tensile Stress Average Axial Stress Acting on Pipe Cross-section at Points A, B. C, DDuring Pullback of s1 = 190.13 psi <1,150 psi OK psi: Average

s2 = 343.40 psi <1,150 psi OK = 384.55 psi <1,150 psi Axials3Stress Acting onOKPipe s4 = 409.48 psi <1,150 psi OK

s1 = (Ti + ∆T)

(

Cross-section at Points A, B. C, D

1 DR 2 ) ( ) πOD2 DR - 1

Where

, TB, TC, TD (Ibf) sTi1==TA190.13 psi <1,100 psi OK si = corresponding stress, psi s2 = 343.40 psi <1,100 psi OK sBreakaway psi <1,100 psi OK 3 = 384.55 links should be set so that pullback force applied to pipe does not exceed 1,150 psi stress. sID4==OD409.48 psi <1,100 psi OK - 2t Fb = s pb (π/4)(OD 2 - ID 2)

x 10,5 Ibf Where: Ti = TFb A, =T1.64 B, T C TD (Ibf)

si = corresponding stress, psi factor against ring collapse during pullback Determine safety External Hydraulic Load External static head Breakaway links should bepressure set so that pullback force applied to pipe does Pha = (1.5) (62.4 lbf/ft3) (H) not exceed 1,100 psi stress. Pha = 22.75 psi

ID = OD - 2t Fb = spb (ʌ/4)(OD2 - ID2)

Fb = 1.64 x 105 Ibf

mine safety factor against ring collapse during pullback 421-461.indd 460

1/27/09 12:50:17 PM

Chapter 12 461


Combine static head with hydrokinetic pressure Peffa = Pha + ∆P Peffa = 32.75 psi

Critical collapse pressure Resistance to external hydraulic load during pullback fo = 0.76 Ovality compensation factor (for 3% ovality) r = S 4/2SPb r = 0.178 Tensile ratio (based on 1,150 psi pull stress calculation) Tensile reduction factor PCR = 108 psi

Safety factor against collapse SF = Pcr/Pha F = 4.75

Where Pha = applied effIective pressure due to head of water of drilling Pcr = calculated critical buckling pressure found by solving Equation 11 multiplied by Equation 19 for 24” DR11, psi SF = Safety Factor

421-461.indd 461

1/16/09 10:15:12 AM

Horizontal Directional Drilling - Energy Piping Systems

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