Numerical Methods in Soil Mechanics 14.PDF

Numerical Methods in Soil Mechanics 14.PDF Numerical Methods in Geotechnical Engineering contains the proceedings of the 8th European Conference on Numerical Methods in Geotechnical Engineering (NUMGE 2014, Delft, The Netherlands, 18-20 June 2014). It is the eighth in a series of conferences organised by the European Regional Technical Committee ERTC7 under the auspices of the International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE). The first conference was held in 1986 in Stuttgart, Germany and the series has continued every four years (Santander, Spain 1990; Manchester, United Kingdom 1994; Udine, Italy 1998; Paris, France 2002; Graz, Austria 2006; Trondheim, Norway 2010). Numerical Methods in Geotechnical Engineering presents the latest developments relating to the use of numerical methods in geotechnical engineering, including scientific achievements, innovations and engineering applications related to, or employing, numerical methods. Topics include: constitutive modelling, parameter determination in field and laboratory tests, finite element related numerical methods, other numerical methods, probabilistic methods and neural networks, ground improvement and reinforcement, dams, embankments and slopes, shallow and deep foundations, excavations and retaining walls, tunnels, infrastructure, groundwater flow, thermal and coupled analysis, dynamic applications, offshore applications and cyclic loading models. The book is aimed at academics, researchers and practitioners in geotechnical engineering and geomechanics.

Anderson, Loren Runar et al "LONGITUDINAL MECHANICS" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 14-1 Longitudinal stress due to decrease in temperature of a restrained pipe Figure 14-2 Longitudinal stress due to internal pressure (Poisson effect) in a restrained pipe ©2000 CRC Press LLC CHAPTER 14 LONGITUDINAL MECHANICS Longitudinal mechanics of buried pipes is the analysis of longitudinal deformations compared to performance limits of deformation One excessive deformation is fracture for which corresponding strain limits can be identified If the pipe changes length enough to shear off appurtenances or to allow leakage of couplings, deformation is excessive and corresponding strain limits can be identified If the strains can be evaluated, then corresponding stresses can be used as alternative bases for design The principal causes of longitudinal stress (strain) are: Changes in temperature and pressure, which cause relative lengthening or shortening of the pipe with respect to soil and thrust restraints, pipes, temperature change would have to be unusually large to cause critical longitudinal stress Temperature stress in an end-restrained pipe is: σT = Eα (∆T) where σT = E α T = = = (14.1) longitudinal stress in the restrained pipe due to change in temperature modulus of elasticity coefficient of thermal expansion change in temperature Axial thrust is the result of internal pressure or vacuum at "thrustors" (valves, caps, reducers, wye's, tee's elbows, etc.), In the case of end-restrained steel pipes (i.e., welded joints), in order to cause 36 ksi yield stress in the pipe wall, the change in temperature would have to be roughly 185oF Gasketed bell and spigot joints avoid the problem if the pipe sections are not too long Beam bending, which causes flexural stresses Typical causes of beam bending are: a) Placement of pipe sections on timbers or mounds or piers for vertical alignment, b) Non-uniform settlement of the bedding, c) Side-hill soil creep or landslide, and massive soil movement or settlement T he end-restrained pipe is also subjected to longitudinal tension due to internal pressure See Figure 14-2 When a rubber band is stretched, the thickness decreases So also does the length of a pipe section when inflated by internal pressure This is called the Poisson effect Longitudinal tension stress is: Each of the three causes of longitudinal stress is analyzed separately The results are combined for analysis It is convenient to separate analyses into two categories, gasketed pipe sections, and continuously welded pipes, such as welded steel water pipes and welded polyethylene pipes σP = νP'r/t ν = Following are discussions of the three major causes of longitudinal stresses (and strains) in each of the two categories P' r t = = = Longitudinally Restrained Pipes Temperature stresses occur when the pipe section is restrained at the ends such that it cannot lengthen or shorten See Figure 14-1 For ordinary buried Values of σP are not usually critical If steel pipes are capped or plugged and pressurized to circumferential yield stress, P', the longitudinal stress is only half of yield stress ©2000 CRC Press LLC where σP = (14.2) longitudinal stress in the pipe due to internal pressure, Poisson ratio (in the range of 0.25 to 0.4 for most pipe materials), internal pressure, inside radius of the pipe, wall thickness for plain pipe If the buried pipe is welded and very long, it is effectively restrained (by soil friction, if not end restraints) and feels thrust (tension) due to decrease in temperature and internal pressure Example A section of PVC pipe, 12D, DR 26 is installed on a warm day When put in service as a water supply pipeline, temperature decreases 50oF and internal pressure increases to 160 psi What is the longitudinal stress if the pipe is restrained longitudinally? DR = 26 = OD/t, r/t = 12.5 = (DR-1)/2, E = 400 ksi = modulus of elasticity, α = 3(10-5)/oF = thermal coefficient, ν = 0.38 = Poisson ratio, P' = 160 psi = internal pressure, ∆T = 50oF Combining Equations 14.1 and 14.2, σ = Eα∆ T + νP'(r/t) Susbstituting values, σ = 600 psi + 760 psi = 1.36 ksi At sudden rupture, yield stress is σf = ksi At 50 years of persistent pressure, σ f = ksi If restrained, stresses relax over time to less than 1.36 ksi The temperature and Poisson effects are avoided by gasketed joints if the pipe sections are not too long If the sections are long, changes in length due to the temperature and Poisson effects cause significant longitudinal stresses As a pipe section changes length, it is partially restrained by frictional resistance of the soil As with a rope in a tug-of-war, where tension applied by friction of the contestants' hands cumulates to maximum at midlength, so does thrust in a buried pipe section cumulate to maximum at midlength due to soil friction That thrust causes a longitudinal stress of, σ = LHγ µ'/2t ©2000 CRC Press LLC where σ = L H γ µ' t = = = = = longitudinal stress in the pipe due to frictional resistance of the soil, length of the pipe section, height of soil above the pipe, unit weight of the soil cover, coefficient of friction, soil on pipe, wall thickness of plain pipe Some pipeline engineers assume values for the coefficient of friction between granular soil and steel pipes as follows: For tape-coated steel pipe, For mortar-coated steel pipe, Example What is the longitudinal stress at midlength of a pipe that shortens after it is buried? The pipe is a large diameter, tape-coated, gasketed steel pipe comprising 120-ft-long sections with 0.50-inch wall thickness µ' = 0.2 H = ft of granular soil cover at 125 pcf Substituting into Equation 14.3, soil friction can cause longitudinal stress up to 1.5 ksi This is not impressively large, but it may combine with other longitudinal stresses, such as beam bending Beam Bending See Figure 14-3 If a pipe that is initially straight is bent into a circular arc of radius R, longitudinal strains develop in the outside fibers If R is too short, the pipe buckles; i.e., crumples at a plastic hinge Strain is a better basis for analysis than stress In the following, both strain and stress are analyzed because yield stress is performance limit in some cases Due to beam bending, longitudinal strain and stress (elastic theory) are: ε = r/R, and σ = Er/R (14.3) µ' = 0.2 µ' = 0.4 (14.4) where ε = σ r R = = = E R' = = maximum longitudinal strain — tension outside of bend and compression inside, maximum longitudinal stress, outside radius of the pipe cross section, longitudinal radius of the pipe axis on the bend, modulus of elasticity, initial radius of the bend if the pipe is manufactured as a curved section Non-uniform bedding is inevitable despite specifications calling for uniform bedding Under soil loads plus weight of the pipe and contents, the pipe deflects and causes longitudinal stress For reinforced pipes, manufacturers provide longitudinal reinforcement and limit the lengths of pipe sections Longitudinal stress can be reduced by corrugating the pipe so that it flexes and conforms with the bedding rather than bridging over soft spots A bend in the pipe causes a moment, M, for which longitudinal stress is, Example What is the minimum radius of the spool on which polyethylene gas pipe can be wound for shipping (or the minimum longitudinal radius for lowering continuous pipe into a trench)? If allowable strain is 2%, and pipe OD is inches, from Equation 14.4, R = 50 inches σ = Mr/I If the pipe is manufactured initially to some mean radius R', then Equation 14.4 must take into account a change in curvature as follows: t M ε = r(1/R-l/R'), and σ = Er(1/R-1/R') Moment M can be analyzed from the loads and supports on a pipe The supports are intermittent hard spots along the pipe If bedding were truly uniform, the pipe could not deflect as a beam However, bedding is never uniform, and so there is always some deflection and some M The maximum M from beam analysis is substituted into Equation 14.6 For design, the maximum combined longitudinal stress must be less than the strength reduced by a safety factor The mean radius of the bend can be measured by holding a cord (straightedge) of known length, s, along the inside of the bend as shown in Figure 14-4 and by measuring the offset e at the middle of the cord The mean radius R of the bend is, R = s2/8e + e/2 + r (14.5) Knowing R, longitudinal stress (and strain) can be evaluated from Equation 14.4 The analysis above is useful for checking the installed radius of the bend against allowable R can be calculated by measurements inside the pipe Longitudinal bending is caused by: soil movement, and non-uniform bedding Soil movement is caused by heavy surface loads, differential subgrade soil settlement, landslides, etc Soil settlement can often be predicted ©2000 CRC Press LLC where σ = r = I = = = (14.6) longitudinal stress, outside radius of the pipe cross section, πtr3 = centroidal moment of inertia of plain pipe cross section, wall thickness of plain pipe, moment at some point along the pipe which acts as a beam With few exceptions, pipes are remarkably stiff beams that bridge over voids and soft spots in the soil bedding If reactions are only at the ends of pipe sections or at midlength, as in Figure 14-5; the moment is maximum at midlength and may be found from the equation, M/wL2 = 1/8 The reactions of Figure 14-5 are extreme worst cases and are unlikely — one support per pipe section At the other extreme is a pipe on many closely spaced reactions This is unlikely For example, Figure 14-3 Longitudinal stress (and strain) in tension and compression due to bending of the pipe into a longitudinal mean radius R Figure 14-4 Technique for measuring longitudinal inside radius of curvature Ri in a bent pipe (The same technique can be used inside the pipe to find the outside radius of curvature Ro.) Figure 14-5 Two moment diagrams for the worst locations of one reaction per pipe section and the limits of case I for two reactions per pipe section (These are the maximum possible moments.) ©2000 CRC Press LLC consider a large pipe on rollers on a level concrete floor, Figure 14-6 As the pipe is shoved on the rollers, only two rollers carry the load and roll All others are loose and remain at rest The two highest of the rollers become the reactions It is easily demonstrated that the two reactions will have to be on different sides of the center of gravity of the load See Figure 14-7 Two supports per pipe section is the most probable number of reactions The above rationale reduces the number of reactions from possibly one reaction, to more probably two reactions per pipe section Reactions would not be located at the ends of pipe sections if bell holes are excavated Vertical alignment could not be controlled by placing two reactions on the same side of midlength distance kL See Figure 14-8 Case III Reactions are spaced at L/2 but the left reaction is located a distance X from the left end of the beam See Figure 14-9 It is noteworthy that the maximum moment occurs at X = 0.2L Pipe on Piles An example of Case I is a buried pipe on piles (or bents) in order to maintain vertical alignment The need for piles implies that the soil settles If the soil settles with respect to the pipe, the pipe lifts a wedge of soil on top at a wedge angle of 45o + ϕ/2 For cohesionless soil, the wedge angle is approximately 1h:2v If the maximum moment due to probable locations of reactions can be found, then maximum stress can be found from Equation 14.6, and requirements can be established for allowable length of pipe sections and their longitudinal strength Figures 14-8 and 14-9 are dimensionless influence diagrams for likely locations of reactions From the critical influence number, M/wL2, maximum longitudinal stress can be calculated For one reaction per section, the maximum moment is M = 0.125wL2, and is always at midlength The influence diagram is not shown The most probable number of reactions per section is two These are hard spots assumed to be concentrated reactions There are an infinite number of relative locations of two reactions on each pipe section However, critical locations are included in the following three cases Case I Reactions are at or near the ends of the beam This case is similar to one support per section discussed above See Figure 14-5 Case II Reactions are at two points B equally spaced from the ends of the pipe section by a ©2000 CRC Press LLC The load per unit length of pipe is the weight of soil plus the pipe and its contents Example A 120-inch steel pipe with 0.75 wall thickness is buried on piles under ft of soil at 120 pcf in a zone where the soil can settle When the pipe is full of water, what is load, w, per unit length of pipe? Figure 14-6 Large pipe section riding on a number of rollers of which two is the most probable number of rollers actually supporting the pipe at one time Figure 14-7 Two reactions needed per pipe section for stability ©2000 CRC Press LLC Figure 14-8 Influence diagram of moments for case II reactions which are spaced at equal distances from the pipe ends Figure 14-9 Influence diagram of moments for case III reactions which are separated by half the pipe length but are at variable distances X from one pipe end ©2000 CRC Press LLC Figure 14-10 Worst-case, simply supported, two-reaction beam (top), and the more probable sine-curve reactions (bottom), for which stresses and deflections are just four-tenths of the corresponding values for the simply supported beam Figure 14-11 Beam deflection, y, of a pipe section that is supported at its ends ©2000 CRC Press LLC w s = γ s(131.23 ft2) wp = w w = γ w(πr2) w = 15.75 k/ft, = 0.96 k/ft, = 4.90 k/ft, = 21.6 k/ftw The above analyses are for concentrated reactions — idealized worst cases In fact, reactions are distributed over finite areas Even if the pipe were not on bedding between reactions, partial soil support would occur under the haunches where embedment falls in against the pipe A rational distribution of reactions is the sine-curve reaction shown in Figure 14-10 It turns out that, The moments, strains, stresses, and deflections for sine-curve reactions are four-tenths of the corresponding values for concentrated reactions Exceptions may occur for adverse installations such as buried pipes on piers Even though specifications require uniform bedding and compacted embedment, it is prudent for pipe manufacturers to limit the lengths of pipe sections or to provide adequate longitudinal strength Pipe manufacturer and pipeline designer should know about longitudinal stresses and the strength of the pipe required to withstand them Longitudinal stresses are cumulative — beam stress, plus axial thrust due to special sections and soil movement, plus soil friction due to temperature and internal pressure Example A 120-inch water pipe section of 0.75-inch-thick steel is 120 ft long with gaskets (or slip couplings) at each end It is buried under ft of soil, but the pipe section crosses a gulch where soil subsidence is anticipated What is the stress due to bending? The span is simply supported σ = Mc/I where M = wL2/8, w = 21.6 kips/ft from the above example, L = 120 ft, I = πr3t, t = 0.75 inch, ©2000 CRC Press LLC c = r = 60 inches Substituting values, σ = 55 ksi Yield stress is exceeded However, soil subsidence would have to be 6.6 inches or more With less subsidence, stress in the beam would be less than 55 ksi Moreover, yield stress is not necessarily failure Allowing for yield, the plastic moment at beam failure is increased by 50% Longitudinal Deflection Longitudinal stiffness of a pipe is EI = dM/dθ; where M = resisting moment of the beam, E = modulus of elasticity, I = πr3t = moment of inertia of the pipe cross section about its centroidal axis, θ = angle of circular bend of the pipe as a beam Most designers relate stiffness to beam deflection of a pipe in terms of either, load on the pipe as a beam, or maximum longitudinal stress at yield These analyses provide a feel for how successfully a pipe bridges over soft spots in the bedding See Figure 14-11 Example Consider a 120-inch diameter buried steel water pipeline of 60-ft-long gasketed sections under ft of dry soil cover If the soil subsides, what is the deflection at midlength of a simply supported pipe section? Assume that, y = vertical deflection at midlength, D = 120 inches = 2r, t = 0.75 inch, L = 60 ft, H = ft, γ = 120 pcf = 0.12kcf, w = 21.6 k/ft, I = πtr3, E = 30(106) psi, σy = 36 ksi = yield strength For the wedge load, pipe and contents, w = 21.6 k/ft from examples above The deflection at midlength for a simply supported span is, y = 5wL4/384EI Substituting values, y = 0.41 inch Formulas for beam deflection are available from texts on mechanics The corresponding maximum longitudinal stress is 13.5 ksi This stress occurs if the ends are supported on hard spots, but in between, the soil subsides enough to open a gap of 0.41 inch under the pipe This is a longitudinally stiff pipe Deflection is proportional to the fourth power of length If the pipe section were 80 ft long instead of 60 ft, the deflection would be increased by the fourth power of 80/60 — a factor of 3.16 The 0.41-inch deflection would become 1.3 inch, and the stress, which increases by the same factor, would be 43 ksi — greater than yield stress of 36 ksi Allowing for yield, failure is plastic at two-thirds of 43, for which the safety factor is only 1.26 against plastic collapse The fourth-power-of-L effect requires care that spans over soil subsidence are not excessive GASKETED PIPE SECTIONS Longitudinal stresses are inevitable in continuously welded pipes, but are often small enough to be neglected in the design of gasketed pipelines It is assumed that gasketed joints transfer no moment They transfer shear Adequate longitudinal strength is taken for granted so long as specifications include uniform bedding and compacted embedment Pipe manufacturers are expected to provide adequate longitudinal pipe strength for ordinary buried pipe conditions including handling, shipping, and installing Ordinarily, the pipeline designer does not expect to investigate longitudinal stresses except for adverse conditions such as beam action of a buried pipeline supported on piles or piers The pipeline designer accepts the manufacturer's recommendation for lifting, stacking, stabbing of joints, etc., so longitudinal stresses will not be excessive Longitudinal thrust is generated at special sections (elbows, wyes, tees, valves, caps, reducers, etc.) by ©2000 CRC Press LLC internal pressure (or vacuum) and by change in direction of fluid flow In continuously welded pipelines, thrust is resisted by longitudinal stress in the pipe In gasketed pipelines, the longitudinal stress is usually negligible, but not always External thrust restraints are required to resist longitudinal thrust See Chapter 15 If there is relative longitudinal movement of the pipe and embedment, soil friction against the pipe causes longitudinal stress due to: massive soil movement, pipe shortening or lengthing due to temperature change and internal pressure Thrust restraints reduce the relative movement Example A gasketed pipeline crosses a soil embankment that is settling If the pipeline was straight at the time of installation, it deflects downward into an arc (catenary) during soil settlement Because the arc length is greater than the cord length, increase in length must be accommodated by slight pull-out of the spigots But in the process, each pipe section feels tension due to soil friction Highway engineers compensate for pipe deflection under highway embankments by installing the pipe with camber — reverse (upward) deflection so that the pipe is straight after soil settlement An earth slide on a sidehill can induce a catenary in a pipeline with elongation and pull-out of spigots If the pipeline comprises gasketed sections, a serious consequence could be separation of spigots from bells at gasketed joints CONTINUOUS (WELDED) PIPES Longitudinal stresses are inevitable in welded buried pipelines These stresses are caused by: Special sections: valves, caps, reducers, tees, wyes, and bends (which include elbows) etc., Fixed terminals or longitudinal soil friction of the soil against the pipe, and Beam bending (longitudinal pipe deflection) Each cause of stress is analyzed separately as follows, and then combined if necessary Special Sections Changes in direction of flow and internal pressure (or vacuum) cause longitudinal stresses For example, at a cap or closed valve, internal pressure is resisted by longitudinal tension in the pipe From equations of static equilibrium, the rupturing force πr2P equals the resisting force 2πrtσ , or, σ = PD/4t where σ = D = P = t = (14.7) longitudinal stress, inside diameter = 2r, internal pressure or vacuum, wall thickness of plain pipe For a 10-ft pipe with 0.75-inch wall subjected to internal pressure of 100 psi, the longitudinal stress due to a cap or closed valve is σ = ksi Circumferential stress is twice as great This longitudinal thrust is not critical except as it contributes to other longitudinal stresses Except for very high velocity flow, the impulse effect, due to change in direction of flow, is negligible See Chapter 15 on thrust restraints Fixed Terminals and Soil Friction ε = α (∆T) σ = Eα (∆T) (14.8) See also Equation 14.1 where ε = longitudinal strain, α = coefficient of thermal expansion, T = change in temperature, E = modulus of elasticity, σ = longitudinal stress caused by strain Temperature stresses, by themselves, are not usually critical However, notable exceptions can be identified A 9-ft welded steel water pipe was welded up during a warm afternoon It was covered with soil except for a short span of roughly 200 ft at a stream crossing A drop in temperature that night caused sufficient tension stress to fracture the pipe circumferentially on the transition radius from pipe to bell Another example is a nuclear plant where a buried 10-ft welded steel water pipe connected the reactor to the cooling tower in a straight run When hot water was introduced, the pipeline elongated and broke out the cooling tower foundation to which it was rigidly attached Example How much longitudinal force, Q, can be developed by a D = 10-ft pipe with a t = 0.75 inch wall, endrestrained, if the temperature increases T = 50oF? From Equation 14.8 longitudinal temperature stress in the pipe wall is σ = Eα (∆T) The total force Q is stress times area of steel, so, Q = Eα (∆T)πDt Anything that causes the pipeline to lengthen or shorten results in longitudinal stress either because of fixed terminals or because of soil friction Change in length is caused by temperature change and internal pressure (or vacuum) in the pipe, and by massive soil movements that change pipe alignment or length Substituting in for E = 30(106) psi and α = 6.5 micro units per degree Fahrenheit for steel, Q = 2756 kips This enormous force can break out a reinforced concrete foundation Any change in temperature of a continuously welded pipeline generates longitudinal strain and stress: A successful remedy for eliminating thrust against a fixed terminal is an extensible coupling; either a slip coupling such as a gasketed coupling, or an insert of bellows or corrugated pipe The exten- ©2000 CRC Press LLC (14.9) Figure 14-12 Free-body-diagram of a bend in the pipe caused by massive soil movement — depicted, in this case, as the very typical heavy wheel load over the pipe, and/or settlement of the backfill soil in an excavation under the pipe ©2000 CRC Press LLC sible insert reduces longitudinal stress in the pipe at the terminal But at some distance L' from the coupling, stress is regenerated in the pipe due to soil friction Any change in alignment of a straight continuously welded pipeline generates longitudinal stress by increasing length The change in alignment is usually caused by massive soil movement If a pipe of length L is initially straight, but then is deflected into a circular arc of radius R by massive soil movement, the change in length ∆L is calculated from trigonometry, ∆L = R [2sin -1(L/2R) - (L/R)] (14.10) Strain is ε = ∆L/L; and stress is Eε This stress can be combined with other longitudinal stresses Beam Bending When a continuously welded buried pipeline is deflected, longitudinal stress is the sum of: flexure stress, σF, due to longitudinal bending; and tension stress, σT, due to stretching the pipe around the bend Longitudinal bending is often due to: heavy wheel loads passing over buried pipes on compressible beddings, and excavations under buried pipes not carefully backfilled to support the pipe as backfill settles Worse is a combination of both — a heavy wheel load crossing over a poorly backfilled excavation See Figure 14-12 If the radius of the bend, R, and the soil load, w, on the pipe are known, longitudinal stresses are, σF = Er/R σT = wR/A (14.11) where σF and σ T are longitudinal stresses due to flexure and tension, respectively, E = modulus of elasticity, r = outside pipe radius, w = soil load on pipe per unit length, R = radius of curvature of the bend, A = pipe cross-sectional area, ©2000 CRC Press LLC L = length of the bend in the pipe Figure 14-12 shows restraints at the ends of the s ection of length L These may be approximately correct for a straight taut pipe with high soil friction Load, w, is the soil load The restraints are points of counterflexure where moments are zero and reactions are rotated hinges For this case, the longitudinal stress is simply σF + σ T But the restraints may slip Analysis then becomes complicated Example A buried 4D PVC, DR-51 electrical conduit failed under surface loads at the location where a sewer pipe crossed under in a 4-ft-wide trench which had been excavated to greater depth than the conduit after the conduit was in place Fracture was a circumferential crack in the top of the pipe at the trench wall See Figure 14-12 What caused the fracture? Fracture was the result of catenary tension and flexure However, the stress due to the catenary was found to be negligible The loads that caused fracture were persistent, repeated passes of vehicles The scenario for fracture was a pipe acting as a beam across a trench in which uncompacted backfill settled and loaded the beam A decrease in temperature might contribute to the fracture If the pipe temperature decreased 17oF after installation, the tension stress would be ksi This is not a major stress compared to strength of ksi; but could combine with other tension stresses After the conduit had been installed at a depth of 1.5 ft of soil cover, a sewer line was installed at a depth of 6.5 ft The 4-ft-wide sewer trench was backfilled without compaction Indeed, bedding for the pipe became loose and void as backfill settled After that, concrete mix trucks passed over Figure 14-13 Soil friction reaction generated by the contraction (or elongation) of the pipe through a distance of L' from an extensible coupling to the point of effective restraint Figure 14-14 Reissner effect (elastic theory) showing the ring deflection, d, caused by bending a straight plain pipe into a longitudinal radius R ©2000 CRC Press LLC as shown in Figure 14-12 Reactions at the trench wall are unknown Therefore, a worst-case limit analysis assumes that the ends of the beam slip, but not rotate For analysis, consider the effect of dead load on a fixed-ended beam, span of ft, 1.5 ft of soil cover at 100 pcf For the pipe, OD = 4.13, t = 0.081, r = 2.0245, and I/c = πr2t = 1.043 in3 Soil subsidence lifts a soil wedge above the spring lines at 1h:2v shear planes for which, at 100 pcf, weight is w = 194 lb/ft of pipe length Moment at the ends is M = wL2/12, for which stress is σ = ksi This is less than failure stress of ksi But a surface live load of 16 kips over a 7x22-inch rectangle causes a load on the pipe of w = 192 lb/ft, which is essentially the same as dead load Therefore, the combined stress is ksi This is upper limit For actual stress less than quick yield, failure could be delayed over a period of time LONG PIPES In the following, a pipe is long, straight, and unrestrained at its terminals by extensible couplings It is so long that enough soil friction accumulates from the ends to restrain a central portion of the pipe against change of length See Figure 14-13 Soil friction develops only as the pipe lengthens or shortens throughout some distance L' from the couplings From equations of static equilibrium, for any longitudinal stress, L' = σ t/Pµ' (14.12) = σ = t P' = = = γ H E α T µ' ν = = = = = = = mean effective soil pressure against the pipe, effective unit weight of soil, height of soil cover, modulus of elasticity, coefficient of thermal expansion, decrease in temperature, coefficient of friction of soil on pipe, Poisson ratio Example A 120-inch steel pipeline is to transport cold water to a municipality An extensible Dresser coupling is located at the treatment plant If temperature decrease is 60oF and internal pressure is 125 psi, what is the distance L' from the Dresser coupling to the point of effective restraint? The pipe is buried under ft of soil cover P' D H γ E α t T µ' ν = = = = = = = = = = 125 psi, 120 inches, ft, 120 pcf, 30(103) ksi, 6.5 micro-units per degree F, 0.75 inch, 60oF, 0.2 for tape-wrapped pipe, 0.3 L' is found from Equation 14.12 For this flexible steel pipe, the effective radial soil pressure P is 0.72 ksf The longitudinal stress, due to temperature decrease and internal pressure (both tension) is: σ = Eα (∆T) + νPD/2t = 14.7 ksi where L' P distance along the pipe from the extensible coupling to the point of effective restraint, longitudinal stress in central restrained portion due to temperature change and internal pressure, wall thickness for plain pipe, internal pressure in the pipe, ©2000 CRC Press LLC Substituting into Equation 14.12, L' = 919 ft Beyond 919 ft from the extensible coupling, longitudinal stress is the full 14.7 ksi If the allowable longitudinal stress is σ = 36/2 = 18 ksi, from Equation 14.12, L' can be increased to 1225 ft It follows that, if σ exceeds 18 ksi for a welded pipe, the maximum allowable spacing between extensible couplings should be 2250 ft In this example, the coefficient of friction is questionable If µ' is greater than 0.2, allowable spacing decreases by a direct inverse ratio REISSNER EFFECT When a pipe is bent, the cross section deforms into an approximate ellipse See Figure 14-14 Neglecting internal pressure, the ring deflection, d, is a function of radius of the bend, R, as follows: d = 2Z/3 + 71Z2/135 (14.13) where Z = 1.5(1-ν2)D4/16t2R2 Example What is the radius of the bend of a steel pipe if ring deflection is found to be d = 5%? D = 109 inch = mean diameter, t = 0.25 inch, r = 54.5 = D/2, d = ∆ /D = ring deflection, ∆ = change in diameter, R = radius of the bend, σf = yield stress, E = 30(106) psi = modulus of elasticity, ν = 0.25 = Poisson ratio Substituting, Z = 200(106)in 2/R2 From Equation 14.13, R = 4420 ft The maximum longitudinal stress is σ = Er/R = 30.8 ksi PROBLEMS Given: 12,000 ft of steel water pipe D = 42 inches = diameter, t = 0.25-inch wall thickness, L = 60-ft-long pipe sections, ©2000 CRC Press LLC Bell and spigot joints, Coal tar enamel coating, felt wrapped, E = 30(106) psi, σf = 45(103) psi, ν = 0.25 = Poisson ratio, α = 6.5(10-6)/oF = coef of therm expan Soil: granular, c = γ = 120 pcf, ϕ' = 30o = soil friction angle, µ' = tan 30o = soil on pipe friction, = 0.4 for cement mortar, = 0.2 for tape Water table below pipe invert, H = ft = soil cover, B = ft = trench width, For vertical alignment, the pipe is positioned on mounds at the ends of each 60-ft section such that the pipe does not rest on bedding, but is partially supported by soil under the haunches 14-1 What is the maximum longitudinal stress in the pipe due to beam bending under soil load only? Given: the same conditions except that, as an alternative, all joints are welded up before the pipeline is backfilled 14-2 What is the maximum longitudinal stress due to a decrease in temperature of 40oF after installation and due to internal pressure of 175 psi? (11.5 ksi) 14-3 What would be the maximum allowable settlement of two contiguous mounds due to subgrade soil subsidence? (y = 8.2 in) 14-4 If the soil cover is ft, how far from a slip coupling is full longitudinal load restraint established with cold water in the line? (507 ft) 14-5 How many 60-ft lengths can be welded into one section before the section is backfilled? Assume a significant temperature drop After temperature drop, contiguous sections are welded 14-6 Design the weldments at the joints for the worst seismic conditions Given: A steel water main ruptured on Christmas Eve due to brittle fracture which started as a circumferential crack on top of the pipe, propagated down to the spring lines, then along both spring lines longitudinally It occurred near midlength of a 100-ft section of pipe under a railroad used to deliver chemicals to a water treatment plant on the Mississippi River flood plain The pipe had been bored into place and grouted under the rails Pits were excavated on each end of the pipe section in order to jack the pipe under the railroad The pits were backfilled with limerock with dry unit weight of 135 to 140 pcf, and void ratio about 0.4 at compacted density of about 90% AASHTO-T99 The steel pipe was 72-inch diameter and 0.375 wall The saturated flood plain silt weighs 103 pcf 14-11 What is the leakage area in the accumulated gap in a stab joint in 10-ft-diameter flexible pipe if the tolerance is 3/32 inch in circumference? (5.6 in 2) 14-7 At what settlement of the limerock backfill will the pipe wall stress reach yield? (4 inches) 14-15 What is the minimum possible horizontal radius of curvature of PVC pipe without use of elbows? OD = 24 in, DR = 32.5 14-8 What is the Poisson effect of internal pressure of 125 psi? 14-16 Find the maximum length of a gasketed pipe section at critical longitudinal temperature stress T = 40oF, D = 10 ft, t = 0.5 in, c = 0, ϕ = ϕ' = 30o, H = ft, α = 6.5(10-6) (225 ft) 14-9 The 72-inch steel pipe is inside a 7-ft tunnel liner casing with grout between pipe and casing But the casing and grout were opened at midspan to accommodate a transverse 3-ft pipe at the top of the 7-ft casing What is the effect on the pipe of the opening in the top of the grout encasement at midspan? 14-10 A pipeline is buried in tidewater soil What is the maximum stress in a 7-ft-diameter gasketed steel water pipe of 20-ft-long sections ring stiffened, with wall thickness of 0.5 inch and positioned on pile bents spaced near each joint then buried under ft of sand at 120 pcf? ©2000 CRC Press LLC 14-12 In Problem 14-11 what is the width of the gap? What gasket is needed? (0.06 inch) 14-13 What is the minimum longitudinal radius of curvature for laying a continuous polyethylene pipe into an underwater trench from a barge? R = (D/2)(E/σy ) 14-14 What is the ring deflection of the polyethylene pipe of Problem 14-13 at the minimum radius of curvature? 14-17 Derive Equation 14.3 14-18 Derive Equation 14.5 14-19 Derive the expression for the shear load Q at the bell and spigot for gasketed pipe sections of length L under uniform soil load w per unit length if the reactions are spaced at L/2 and are a distance X from the joint 14-20 Prove that the sine-curve reactions cause maximum moments that are 0.4 times the corresponding values for concentrated reactions  ... longitudinal pipe strength for ordinary buried pipe conditions including handling, shipping, and installing Ordinarily, the pipeline designer does not expect to investigate longitudinal stresses except for... coupling to the point of effective restraint, longitudinal stress in central restrained portion due to temperature change and internal pressure, wall thickness for plain pipe, internal pressure in. .. wound for shipping (or the minimum longitudinal radius for lowering continuous pipe into a trench)? If allowable strain is 2%, and pipe OD is inches, from Equation 14.4, R = 50 inches σ = Mr/I