Successfully reported this slideshow.
Upcoming SlideShare
×

THIẾT KẾ ĐƯỜNG ỐNG

5,649 views

Published on

Tài liệu dành cho các bạn học chuyên ngành đường ống

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Cảm ơn tinh thần chia sẻ của bạn ,quyển này rất căn bản ..

Are you sure you want to  Yes  No

THIẾT KẾ ĐƯỜNG ỐNG

1. 1. STRUCTURAL MECHANICS OFBURIED PIPES Reynold King Watkins Utah State University Logan, Utah Loren Runar Anderson Utah State University Logan, Utah CRC Press Boca Raton London New York Washington, D.C.
2. 2. ©2000 CRC Press LLC
3. 3. ©2000 CRC Press LLC
4. 4. GENERAL NOTATIONGeometryA = cross sectional wall area per unit length of pipe,B = breadth of the trenchD = pipe (tank) diameter,H = height of soil cover,h = height of water table,L = length of tank or pipe section,r = radius of curvature of the pipe (tank) cylinder,R = radius of a bend in the pipe,t = thickness of the wall,x = horizontal coordinate axis,y = vertical coordinate axis,z = longitudinal axis (with exceptions),β = angle of soil shear plane.Forces, Pressures, and StressesP = external pressure on the pipe or tank,P = internal pressure,p = vacuum in the pipe or tank,Q = concentrated force,W = surface wheel load,γ = unit weights,σ = direct (normal) stress,τ = shearing stress.Subscripts refer to directions of forces and stresses.Properties of Materialsc = cohesion of soil,E = modulus of elasticity of pipe (tank) material,S = allowable stress (strength) of material,γ = unit weight of material,ν = Poisson ratio,ϕ = soil friction angle.©2000 CRC Press LLC
5. 5. CONTENTSChapter 1 Introduction 2 Preliminary Ring Design 3 Ring Deformations 4 Soil Mechanics 5 Pipe Mechanics 6 Ring Stresses 7 Ring Deflection 8 Ring Stiffness 9 Non-circular Cross Sections 10 Ring Stability 11 Encased Flexible Pipes 12 Rigid Pipes 13 Minimum Soil Cover 14 Longitudinal Mechanics 15 Thrust Restraints 16 Embedment 17 Parallel Pipes and Trenches 18 Special Sections 19 Stress Analysis 20 Plastic Pipes 21 External Hydrostatics 22 Buried Tanks and Silos 23 Flotation 24 Leaks in Buried Pipes and Tanks 25 Long-Span Structures 26 Non-circular Linings and Coatings 27 Risers 28 Analysis of Buried Structures by the Finite Element Method 29 Application of Finite Element Analysis to a Buried Pipe 30 Economics of Buried Pipes and Tanks Appendix A: Castiglianos Equation Appendix B: Reconciliation of Formulas for Predicting Ring Deflection Appendix C: Similitude Appendix D: Historical Sketch Appendix E: Stress Analysis Appendix F: Strain Energy Analysis©2000 CRC Press LLC
6. 6. PREFACEBuried pipes are an important medium of transportation. Only open channels are less costly to construct.On the average, pipelines transport over 500 ton-miles of product per gallon of fuel. Gravity systemsrequire no fuel for pumping. Ships transport 250 ton-miles per gallon. Rails transport 125 ton-miles pergallon. Trucks transport 10 ton-miles per gallon. Aircraft transport less than 10 ton-miles per gallon offuel.Buried pipelines are less hazardous, and less offensive environmentally than other media of transportation.They produce less contamination, eliminate evaporation into the atmosphere, and generally reduce loss anddamage to the products that are transported.The structural mechanics of buried pipes can be complicated -- an interaction of soil and pipe each withvastly different properties. Imprecisions in properties of the soil embedment are usually so great thatcomplicated analyses are not justified. This text is a tutorial primer for designers of buried structures --most of which are pipes. Complicated theories are minimized. Fundamentals of engineering mechanicsand basic scientific principles prevail."Science is understanding gained by deliberate inquiry." -- Philip Handler ACKNOWLEDGMENTGratitude is expressed to Becky Hansen for her patient and expert preparation of manuscript.©2000 CRC Press LLC
7. 7. ©2000 CRC Press LLC
8. 8. Anderson, Loren Runar et al "ECONOMICS OF BURIED PIPES AND TANKS "Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000
9. 9. Anderson, Loren Runar et al "INTRODUCTION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000
10. 10. CHAPTER 1 INTRODUCTIONBuried conduits existed in prehistory when caves In a phenomenon as complex as the soil-structurewere protective habitat, and ganats (tunnels back interaction of buried pipes, all three sources must beunder mountains) were dug for water. The value of utilized. There are too many variables; thepipes is found in life forms. As life evolved, the interaction is too complex (statically indeterminate tomore complex the organism, the more vital and the infinite degree); and the properties of soil are toocomplex were the piping systems. imprecise to rely on any one source of information.The earthworm lives in buried tunnels. His is a Buried structures have been in use from antiquity.higher order of life than the amoeba because he has The ancients had only experience as a source ofdeveloped a gut — a pipe — for food processing knowledge. Nevertheless, many of their catacombs,and waste disposal. ganats, sewers, etc., are still in existence. But they are neither efficient nor economical, nor do we haveThe Hominid, a higher order of life than the any idea as to how many failed before artisansearthworm, is a magnificent piping plant. The learned how to construct them.human piping system comprises vacuum pipes,pressure pipes, rigid pipes, flexible pipes — all The other two sources of knowledge are recent.grown into place in such a way that flow is optimum Experimentation and principles required theand stresses are minimum in the pipes and between development of soil mechanics in the twentieththe pipes and the materials in which they are buried. century. Both experience and experimentation are needed to verify principles, but principles are theConsider a community. A termite hill contains an basic tools for design of buried pipes.intricate maze of pipes for transportation, ventilation,and habitation. But, despite its elegance, the termite Complex soil-structure interactions are still analyzedpiping system cant compare with the piping systems by experimentation. But even experimentation isof a community of people. The average city dweller most effective when based on principles — i.e.,takes for granted the services provided by city piping principles of experimentation.systems, and refuses to contemplate theconsequences if services were disrupted. Cities can This text is a compendium of basic principles provenbe made better only to the extent that piping systems to be useful in structural design of buried pipes.are made better. Improvement is slow becauseburied pipes are out-of-sight, and, therefore, out-of- Because the primary objective is design, the firstmind to sources of funding for the infrastructure. principle is the principle of design.Engineering design requires knowledge of: 1.performance, and 2. limits of performance. Three DESIGN OF BURIED PIPESgeneral sources of knowledge are: To design a buried pipe is to devise plans andSOURCES OF KNOWLEDGE specifications for the pipe-soil system such that Experience (Pragmatism) performance does not reach the limits of Experimentation (Empiricism) performance. Any performance requirement is Principles (Rationalism) equated to its limit divided by a safety factor, sf, i.e.:©2000 CRC Press LLC
11. 11. Figure 1-1 Bar graph of maximum peak daily pressures in a water supply pipeline over a period of 1002 dayswith its corresponding normal distribution curve shown directly below the bar graph.©2000 CRC Press LLC
12. 12. Performance Limit s = standard deviation = deviation within whichPerformance = Safety Factor 68.26 percent of all measurements fall (Ps =Examples: 68.26%). Stress = Strength/sf Deformation = Deformation Limit/sf P is the ratio of area within +w and the total area. Expenditures = Income/sf; etc. Knowing w/x, P can be found from Table 1.1. The standard deviation s is important because: l. it is aIf performance were exactly equal to the performance basis for comparing the precision of sets oflimit, half of all installations would fail. A safety measurements, and 2. it can be calculated fromfactor, sf, is required. Designers must allow for actual measurements; i.e.,imperfections such as less-than-perfect construction,overloads, flawed materials, etc. At present, safety s = %3yw2/(n-1)factors are experience factors. Future safety factorsmust include probability of failure, and the cost of Standard deviation s is the horizontal radius offailure — including risk and liability. Until then, a gyration of area under the normal distribution curvesafety factor of two is often used. measured from the centroidal y axis. s is a deviation of x with the same dimensions as x and w. AnIn order to find probability of failure, enough failures important dimensionless variable (pi-term) is w/s.are needed to calculate the standard deviation of Values are listed in Table 1-1. Because probabilitynormal distribution of data. P is the ratio of area within ±w and the total area, it is also a dimensionless pi-term. If the standard deviation can be calculated from test data, theNORMAL DISTRIBUTION probability that any measurement x will fall within ±w from the average, can be read from Table 1-1.Normal distribution is a plot of many measurements Likewise the probability of a failure, Pe , either(observations) of a quantity with coordinates x and y, greater than an upper limit xe or less than a lowerwhere, see Figure 1-1, limit, xe , can be read from the table. The deviation ) of failure is needed; i.e., we = x e - x. Because pipe-x = abscissa = measurement of the quantity, soil interaction is imprecise (large standard deviation), it is prudent to design for a probability ofy = ordinate = number of measurements in any given success of 90% (10% probability of failure) and tox-slot. A slot contains all measurements that are include a safety factor. Probability analysis can becloser to the given x than to the next higher x or the accomplished conveniently by a tabular solution asnext lower x. On the bar graph of data Figure 1-1, if shown in the following example.x = 680 kPa, the 680-slot contains all of x-valuesfrom 675 to 685 kPa. Example)x = the average of all measurements, The bursting pressure in a particular type of pipe hasx = 3yx/3y, been tested 24 times with data shown in Table 1-2.n = total number of measurements = Ey, What is the probability that an internal pressure of )w = deviation, w = x - x, 0.8 MPa (120 psi or 0.8 MN/m2) will burst the pipe?P = probability that measurement will fall between ±w, x = test pressure (MN/m2) at burstingPe = probability that a measurement will exceed y = number of tests at each x the failure level of xe (or fall below a n = Gy = total number of tests minimum level of xe ),©2000 CRC Press LLC
13. 13. Table 1-1 Probability P as a function of w/s that a value of x will fall within +w, and probability Pe a s afunction of we /s that a value of x will fall outside of +w e on either the +w e or the -w e .w e /s P Pe w e /s P Pe____ (%) (%) ____ (%) (%)0.0 0.0 50.0 1.5 86.64 6.680.1 8.0 46.0 1.6 89.04 5.480.2 15.9 42.1 1.7 91.08 4.460.3 23.6 38.2 1.8 92.82 3.590.4 31.1 34.5 1.9 94.26 2.870.5 38.3 30.9 2.0 95.44 2.280.6 45.1 27.4 2.1 96.42 1.790.6745 50.0 25.0 2.2 97.22 1.390.7 51.6 24.2 2.3 97.86 1.070.8 57.6 21.2 2.4 98.36 0.820.9 63.2 18.4 2.5 98.76 0.621.0 68.26 15.9 2.6 99.06 0.471.1 72.9 13.6 2.7 99.30 0.351.2 78.0 11.5 2.8 99.48 0.261.3 80.6 9.7 2.9 99.62 0.191.4 83.8 8.1 3.0 99.74 0.135Table 1-2 Pressure data from identical pipes tested to failure by internal bursting pressure, and a tabularsolution of the average bursting pressure and its standard deviation. x y xy w yw yw 2(Mpa)* _ (MPa) (MPa) (MPa) (MPa) 20.9 2 1.8 -0.2 -0.4 0.081.0 7 7.0 -0.1 -0.7 0.071.1 8 8.8 0.0 0.0 0.001.2 4 4.8 +0.1 +0.4 0.041.3 2 2.6 +0.2 +0.4 0.081.4 1 1.4 +0.3 +0.3 0.09Sums 24 26.4 0.36 n Σxy Σyw 2x = Sxy/n = 1.1 MPas = [ Syw 2/(n-1)] = 0.125*MPa is megapascal of pressure where a Pascal is N/m2; i.e., a megapascal is a million Newtons of forc eper square meter of area. A Newton = 0.2248 lb. A square meter = 10.76 square ft. ©2000 CRC Press LLC
14. 14. From the data of Table 1-2, soil mechanics. The remainders involve such complex soil-structure interactions that the_x = Σxy/Sy = 26.4/24 = 1.1 interrelationships must be found from experience or experimentation. It is advantageous to write thes = /Syw /(n-1) = /0.36/23 2 = 0.125 relationships in terms of dimensionless pi-terms. See Appendix C. Pi-terms that have proven to be useful _w = x - x, so are given names such as Reynolds number in fluid flow in conduits, Mach number in gas flow, influencew e = (0.8 - 1.1) = -0.30 MN/m2 numbers, stability numbers, etc. = deviation to failure pressure Pi-terms are independent, dimensionless groups ofw e/s = 0.30/0.125 = 2.4. fundamental variables that are used instead of the original fundamental variables in analysis orFrom Table 1-1, interpolating, Pe= 0.82%. experimentation. The fundamental variables areThe probability that a pipe will fail by bursting combined into pi-terms by a simple process in which three characteristic s of pi-terms must be satisfied.pressure less than 0.80 MN/m2 is Pe = 0.82 % or The starting point is a complete set of pertinentone out of every 122 pipe sections. Cost accounting fundamental variables. This requires familiarity withof failures then follows. the phenomenon. The variables in the set must be interdependent, but no subset of variables can beThe probability that the strength of any pipe section interdependent. For example, force f, mass m, andwill fall within a deviation of w e = +0.3 MN/m2 is P acceleration a, could not be three of the fundamental= 98.36%. It is noteworthy that P + 2Pe = 100%. variables in a phenomenon which includes other variables because these three are not independent;From probability data, the standard deviation can be i.e., f = ma. Only two of the three would becalculated. From standard deviation, the zone of +w included as fundamental variables. Once thecan be found within which 90% of all measurements equation of performance is known, the deviation, w,fall. In this case w/s = w/0.125 for which P = 90%. can be found. Suppose r = f(x,y,z,...), then w r2 =From Table 1-1, interpolating for P = 90%, w/s = Mrx21.64%, and w = 0.206 MPa at 90% probability. w x2 + Mry 2w y 2 + ... where w is a deviation at the same given probability for all variables, such asErrors (three classes) standard deviation with probability of 68%; mrx is theMistake = blunder — tangent to the r-x curve and wx is the deviation at a Remedies: double-check, repeat. given value of x. The other variables are treated inAccuracy = nearness to truth — the same way. Remedies: calibrate, repair, correct.Precision = degree of refinement — Remedies: normal distribution, safety factor. CHARACTERISTICS OF PI-TERMS 1. Number of pi-terms = (number of fundamentalPERFORMANCE variables) minus (number of basic dimensions).Performance in soil-structure interaction is 2. All pi-terms are dimensionless.deformation as a function of loads, geometry, andproperties of materials. Some deformations can be 3. Each pi-term is independent. Independence iswritten in the form of equations from principles of assured if each pi-term contains a fundamental variable not contained in any other pi-term.©2000 CRC Press LLC
15. 15. Figure 1-2 Plot of experimental data for the dimensionless pi-terms (P/S) and (t/D) used to find the equationfor bursting pressure P in plain pipe. Plain (or bare) pipe has smooth cylindrical surfaces with constant wallthickness — not corrugated or ribbed or reinforced.Figure 1-3 Performance limits of the soil showing how settlement of the soil backfill leaves a dip in thesurface over a flexible (deformed) pipe and a hump and crack in the surface over a rigid (undeformed) pipe.©2000 CRC Press LLC
16. 16. Pi-terms have two distinct advantages: fewer small scale model study are plotted in Figure 1-2.variables to relate, and the elimination of size effect. The plot of data appears to be linear. Only the lastThe required number of pi-terms is less than the point to the right may deviate. Apparently the pipenumber of fundamental variables by the number of is no longer thin-wall. So the thin-wall designationbasic dimensions. Because pi-terms are only applies if t/D< 0.1. The equation of the plot isdimensionless, they have no feel for size (or any the equation of a straight line, y = mx + b where y isdimension) and can be investigated by model study. the ordinate, x is the abscissa, m is the slope, and bOnce pi-terms have been determined, their is the y-intercept at x = 0. For the case above,interrelationships can be found either by theory (P/S) = 2(t/D), from which, solving for bursting(principles) or by experimentation. The results apply pressure,generally because the pi-terms are dimensionless.Following is an example of a well-designed P = 2S/(D/t)experiment. This important equation is derived by theoretical principles under "Internal Pressure," Chapter 2.ExampleUsing experimental techniques, find the equation for PERFORMANCE LIMITSinternal bursting pressure, P, for a thin-wall pipe.Start by writing the set of pertinent fundamental Performance limit for a buried pipe is basically avariables together with their basic dimensions, force deformation rather than a stress. In some cases it isF and length L. possible to relate a deformation limit to a stress (such as the stress at which a crack opens), but Basic such a relationship only accommodates the designerFundamental Variables Dimensions for whom the stress theory of failure is familiar. In reality, performance limit is that deformation beyondP = internal pressure FL-2 which the pipe-soil system can no longer serve thet = wall thickness L purpose for which it was intended. TheD = inside diameter of ring L performance limit could be a deformation in the soil,S = yield strength of the such as a dip or hump or crack in the soil surface pipe wall material FL-2 over the pipe, if such a deformation is unacceptable. The dip or hump would depend on the relative settlement of the soil directly over the pipe and theThese four fundamental variables can be reduced to soil on either side. See Figure 1-3.two pi-terms such as (P/S) and (t/D). The pi-termswere written by inspection keeping in mind the three But more often, the performance limit is excessivecharacteristics of pi-terms. The number of pi-terms deformation of the pipe whic h could cause leaks oris the number of fundamental variables, 4, minus the could restrict flow capacity. If the pipe collapsesnumber of basic dimensions, 2, i.e., F and L. The due to internal vacuum or external hydrostatictwo pi-terms are dimensionless. Both are pressure, the restriction of flow is obvious. If, on theindependent because each contains a fundamental other hand, the deformation of the ring is slightly out-variable not contained in the other. Conditions for of-round, the restriction to flow is usually notbursting can be investigated by relating only two significant. For example, if the pipe cross sectionvariables, the pi-terms, rather than interrelating the deflects into an ellipse such that the decrease of theoriginal four fundamental variables. Moreover, the minor diameter is 10% of the original circularinvestigation can be performed on pipes of any diameter, the decrease in cross-sectional area is onlyconvenient size because the pi-terms are 1%.dimensionless. Test results of a©2000 CRC Press LLC
17. 17. Figure 1-4 Typical performance limits of buried pipe rings due to external soil pressure.©2000 CRC Press LLC
18. 18. The more common performance limit for the pipe is the structural design of the pipe can proceed in sixthat deformation beyond which the pipe cannot resist steps as follows.any increase in load. The obvious case is bursting ofthe pipe due to internal pressure. Less obvious andmore complicated is the deformation due to external STEPS IN THE STRUCTURAL DESIGN OFsoil pressure. Typical examples of performance BURIED PIPESlimits for the pipe are shown in Figure 1-4. Theseperformance limits do not imply collapse or failure. In order of importance:The soil generally picks up any increase in load byarching action over the pipe, thus protecting the pipe 1. Resistance to internal pressure, i.e., strength offrom total collapse. The pipe may even continue to materials and minimum wall thickness;serve, but most engineers would prefer not todepend on soil alone to maintain the conduit cross 2. Resistance to transportation and installation;section. This condition is considered to be aperformance limit. The pipe is designed to withstand 3. Resistance to external pressure and internalall external pressures. Any contribution of the soil vacuum, i.e., ring stiffness and soil strength;toward withstanding external pressure by archingaction is just that much greater margin of safety. 4. Ring deflection, i.e., ring stiffness and soilThe soil does contribute soil strength. On inspection, stiffness;many buried pipes have been found in service eventhough the pipe itself has "failed." The soil holds 5. Longitudinal stresses and deflections;broken clay pipes in shape for continued service.The inverts of steel culverts have been corroded or 6. Miscellaneous concerns such as flotation of theeroded away without failure. Cast iron bells have pipe, construction loads, appurtenances, ins tallationbeen found cracked. Cracked concrete pipes are techniques, soil availability, etc.still in service, etc. The mitigating factor is theembedment soil which supports the conduit. Environment, aesthetics, risks, and costs must be considered. Public relations and social impactA reasonable sequence in the design of buried pipes cannot be ignored. However, this text deals onlyis the following: with structural design of the buried pipe.1. Plans for delivery of the product (distances,elevations, quantities, and pressures), PROBLEMS2. Hydraulic design of pipe sizes, materials, 1-1 Fluid pressure in a pipe is 14 inches of mercury as measured by a manometer. Find pressure in3. Structural requirements and design of possible pounds per square inch (psi) and in Pascalsalternatives, (Newtons per square meter)? Specific gravity of mercury is 13.546.4. Appurtenances for the alternatives, (6.85 psi)(47.2 kPa)5. Economic analysis, costs of alternatives, 1-2 A 100 cc laboratory sample of soil weighs 187.4 grams mass. What is the unit weight of the soil in6. Revision and iteration of steps 3 to 5, pounds per cubic ft? (117 pcf)7. Selection of optimum system. 1-3 Verify the standard deviation of Figure 1-1. (s = 27.8 kPa)With pipe sizes, pressures, elevations, etc., known©2000 CRC Press LLC
19. 19. 1-4 From Figure 1-1, what is the probability thatany maximum daily pressure will exceed 784.5kPa? (Pe = 0.62%)1-5 Figure 1-5 shows bar graph for internal vacuumat collapse of a sample of 58 thin-walled plasticpipes.x = collapse pressure in Pascals, Pa.(Least increment is 5 Pa.)y = number that collapsed at each value of x.(a) What is the average vacuum at collapse? (75.0 Pa) Figure 1-5 Bar graphs of internal vacuum at collapse of thin-walled plastic pipes.(b) What is the standard deviation? (8.38 Pa)(c) What is the probable error? (+5.65 Pa) 1-7 Fiberglass reinforced plastic (FRP) tanks were designed for a vacuum of 4 inches of mercury1-6 Eleven 30 inch ID, non-reinforced concrete (4inHg). They were tested by internal vacuum forpipes, Class 1, were tested in three-edge-bearing which the normal distribution of the results is shown(TEB) test with results as follows: as Series A in Figure 1-6. Two of 79 tanks failed atx = ultimate load in pounds per lineal ft less than 4inHg. In Series B, the percent ofx w w2 fiberglas was increased. The normal distribution(lb/ft) (lb/ft) curve has the same shape as Series A, but is shifted3562 1inHg to the right. What is the predicted probability3125 of failure of Series B at or below 4 in Hg?4375 (Pe = 0.17 % or one tank in every 590)34384188 1-8 What is the probability that the vertical ring3688 deflection d = y/D of a buried culvert will exceed3750 10% if the following measurements were made on4188 23 culverts under identical conditions?4125 Measured values of d (%)3625 6 9 6 6 5 62938 8 5 4 6 7 7 3 6 7 5 4 5(a) What is the average load, x, at failure? 6 7 8 7 5 (0.24 %) (x = 3727.5 lb/ft)(b) What is the standard deviation? 1-9 The pipe stiffness is measured for many samples (s = 459.5 lb/ft) of a particular plastic pipe. the average is 24 with a(c) What is the probability that the load, x, at failure standard deviation of 3.is less than the minimum specified strength of 3000lb/ft (pounds per linear ft)? (Pe = 5.68%) a) What is the probability that the pipe stiffness will be less than 20? (Pe = 9.17 %)©2000 CRC Press LLC
20. 20. b) What standard deviation is required if the 1-10 A sidehill slope of cohesionless soil dips atprobability of a stiffness less than 20 is to be angle 2. Write pi-terms for critical slope whenreduced to half its present value; i.e., less than saturated.4.585%? (s = 2.37) 1-11 Design a physical model for problem 1-10.Figure 1-6 Normal distribution diagrams for fiberglass tanks designed for 4inHg vacuum.©2000 CRC Press LLC
21. 21. Anderson, Loren Runar et al "PRELIMINARY RING DESIGN"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000
22. 22. Figure 2-1 Free-body-diagramof half of the pipe crosssection including internalpressure P’.Equating rupturing force toresisting force, hoop stressin the ring is,s = P’(ID)2AFigure 2-2 Common transportation/installation loads on pipes, called F-loads. ©2000 CRC Press LLC
23. 23. CHAPTER 2 PRELIMINARY RING DESIGNThe first three steps in the structural design of buried is reached when stress, s , equals yield strength, S.pipes all deal with resistance to loads. Loads on a For design, the yield strength of the pipe wall isburied pipe can be complex, especially as the pipe reduced by a safety factor,deflects out-of-round. Analysis can be simplified ifthe cross section (ring) is assumed to be circular. s = P(ID)/2A = S/sf . . . . . (2.1)For pipes that are rigid, ring deflection is negligible.For pipes that are flexible, ring deflection is usually where:limited by specification to some value not greater s = circumferential tensile, stress in the wall,than five percent. Analysis of a circular ring is P = internal pressure,reasonable for the structural design of most buried ID = inside diameter,pipes. Analysis is prediction of structural OD = outside diameter,performance. Following are basic principles for D = diameter to neutral surface,analysis and design of the ring such that it can A = cross sectional area of the pipe wall persupport the three most basic loads: internal pressure, unit length of pipe,transportation/installation, and external pressure. S = yield strength of the pipe wall material,See Figures 2-1 and 2-2. t = thickness of plain pipe walls, sf = safety factor.INTERNAL PRESSURE — T his is the basic equation for design of the ring to(MINIMUM WALL AREA) resist internal pressure. It applies with adequate precision to thin-wall pipes for which the ratio ofThe first step in structural design of the ring is to find mean diameter to wall thickness, D/t, is greater thanminimum wall area per unit length of pipe. ten. Equation 2.1 can be solved for maximum pressure P or minimum wall area A.Plain pipe — If the pipe wall is homogeneous andhas smooth cylindrical surfaces it is plain (bare) and A = P(ID)sf/2S = MINIMUM WALL AREAwall area per unit length is wall thickness. This isthe case in steel water pipes, ductile iron pipes, and For thick-wall pipes (D/t less than ten), thick-wallmany plastic pipes. cylinder analysis may be required. See Chapter 6. Neglecting resistance of the soil, the performanceOther pipes are corrugated or ribbed or composite limit is the yield strength of the pipe. Once the ringpipes such as reinforced concrete pipes. For such starts to expand by yielding, the diameter increas es,pipes, the wall area, A, per unit length of pipe is the the wall thickness decreases, and so the stress in thepertinent quantity for design. wall increases to failure by bursting.Consider a free-body-diagram of half of the pipewith fluid pressure inside. The maximum rupturing Exampleforce is P(ID) where P is the internal pressure andID is the inside diameter. See Figure 2-1. This A steel pipe for a hydroelectric penstock is 51 inchrupturing force is resisted by tension, FA, in the wall ID with a wall thickness of 0.219 inch. What is thewhere F is the circumferential tension stress in the maximum allowable head, h, (difference in elevationpipe wall. Equating rupturing force to the resisting of the inlet and outlet) when the pipe is full of waterforce, F = P(ID)/2A. Performance limit at no flow?©2000 CRC Press LLC
24. 24. Figure 2-3 Free-body-diagrams of the ring subjected to the concentrated F-load, and showing pertinentvariables for yield strength and ring deflection.Equating the collapsingforce to resisting force,ring compression stress is,s = P(OD)/2AFigure 2-4 Free-body-diagram of half of the ring showing external radial pressure, P.©2000 CRC Press LLC
29. 29. E = 30(106) psi, diameter, high-strength wires? What about bond?sf = 2 = safety factor. How can ends of the rods (or wires) be fixed?2-2 What is the allowable internal pressure if a 2-5 What is the allowable fresh water head (causingreinforced conc rete pipe is 60 inch ID and has two internal pressure) in a steel pipe based on thecages comprising concentric hoops of half-inch steel following data if sf = 2? (105 meters)reinforcing rods spaced at 3 inches in the wall which ID = 3.0 meters,is 6.0 inches thick? (P = 78.5 psi) t = 12.5 mm = wall thickness,Given: S = 248 MN/m2 = 36 ksi yield strength.S = 36 ksi = yield strength of steel,sf = 2 = safety factor, 2-6 What maximum external pressure can beEc = 3(106) psi = concrete modulus, resisted by the RCP pipe of Problem 2-2 if the yieldNeglect tensile strength of concrete. strength of the concrete in compression is 10 ksi, modulus of elasticity is E = 3000 ksi, and the internal2-3 What must be the pretension force in the steel pressure in the pipe is zero? See also Figure 2-5.rods of Problem 2-2 if the pipe is not to leak at (P = 52 ksf, limited by the steel)internal pressure of 72 psi? Leakage through haircracks in the concrete appears as sweating. 2-7 Prove that T = Pr for thin-walled circular pipe. (Fs = 2.9 kips) See Figure 2-4.2-4 How could the steel rods be pretensioned in T = ring compression thrust,Problem 2-3? Is it practical to pretension (or post P = external radial pressure,tension) half-inch steel rods? How about smaller r = radius (more precisely, outside radius).Figure 2-5 Equivalent diagrams for uniform external soil pressure on a pipe, showing (on the right) the moreconvenient form for analysis.©2000 CRC Press LLC
30. 30. Anderson, Loren Runar et al "RING DEFORMATION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000
31. 31. Figure 3-1 (top) Vertical compression (strain) in a medium transforms an imaginary circle into an ellipse withdecreases in circumference and area.(bottom) Now if a flexible ring is inserted in place of the imaginary ellipse and then is allowed to expand suchthat its circumference remains the same as the original imaginary circle, the medium in contact with the ringis compressed as shown by infinitesimal cubes at the spring lines, crown and invert.©2000 CRC Press LLC
32. 32. CHAPTER 3 RING DEFORMATIONDeformation of the pipe ring occurs under any load. Geometry of the EllipseFor most buried pipe analyses, this deformation issmall enough that it can be neglected. For a few The equation of an ellipse in cartesian coordinates,analyses, however, deformation of the ring must be x and y, is:considered. This is particularly true in the case ofinstability of the ring, as, for example, the hydrostatic a2x2 + b2y2 = a2b2collapse of a pipe due to internal vacuum or externalpressure. Collapse may occur even though stress where (See Figure 3-2):has not reached yield strength. But collapse can a = minor semi-diameter (altitude)occur only if the ring deforms. Analysis of failures b = major semi-diameter (base)requires a knowledge of the shape of the deformed r = radius of a circle of equal circumferencering. The circumference of an ellipse is p(a+b) whichFor small ring deflection of a buried circular pipe, the reduces to 2pr for a circle of equal circumference.basic deflected cross section is an ellipse. Considerthe infinite medium with an imaginary circle shown In this text a and b are not used because the pipein Figure 3-1 (top). If the medium is compressed industry is more familiar with ring deflection, d.(strained) uniformly in one direction, the circle Ring deflection can be written in terms of semi-becomes an ellipse. This is easily demonstrated diameters a and b as follows:mathematic ally. Now suppose the imaginary circleis a flexible ring. When the medium is compressed, d = D/D = RING DEFLECTION . . . . . (3.1)the ring deflects into an approximate ellipse withslight deviations. If the circumference of the ring where:remains constant, the ellipse must expand out into D = decrease in vertical diameter of ellipse fromthe medium, increasing compressive stresses a circle of equal circumference,between ring and medium. See Figure 3-1 (bottom). = 2r = mean diameter of the circle —The ring becomes a hard spot in the medium. On diameter to the centroid of wall cross-the other hand, if circumference of the ring is sectional areas,reduced, the ring becomes a soft spot and pressure a = r(1-d) for small ring deflections (<10%),is relieved between ring and medium. In either case, b = r(1+d) for small ring deflections (<10%).the basic deformation of a buried ring is an ellipse —slightly modified by the relative decreases in areas Assuming that circumferences are the same forwithin the ring and without the ring. The shape is circle and ellipse, and that the vertical ring deflectionalso affected by non-uniformity of the medium. For is equal to the horizontal ring deflection, area withinexample, if a concentrated reaction develops on the the ellipse is Ae = Bab; andbottom of the ring, the ellipse is modified by a flatspot. Nevertheless, for small soil strains, the basic Ae = pr2 (1 - d2)ring deflection of a flexible buried pipe is an ellipse.Following are some pertinent approximate The ratio of areas within ellipse and circle is:geometrical properties of the ellipse that aresufficiently accurate for most buried pipe analyses. A r = A e / A o = ratio of areas.Greater accuracy would require solutions of infiniteseries. See Figure 3-3.©2000 CRC Press LLC
33. 33. Figure 3-2 Some approximate properties of an ellipse that are pertinent to ring analyses of pipes where d isthe ring deflection and ry and rx are the maximum and minimum radii of curvature, respectively.Figure 3-3 Ratio of areas, Ar = Ae /Ao(Ae within an ellipse and Ao within acircle of equal circumference) shownplotted as a function of ring deflection.©2000 CRC Press LLC
35. 35. d (%) rr 0 1.000 1 1.062 2 1.128 3 1.197 4 1.271 5 1.350 6 1.434 8 1.61810 1.82612 2.06215 2.47620 3.375Figure 3-4 Ratio of radii,rr = ry/rx = (1+d)3/(1-d)3,(ry and rx are maximum andminimum radii, respectively,for ellipse) shown plottedas a function of ring deflection d.Figure 3-5 Procedure for calculating the radius of curvature of a ring from measurements of a cord of lengthL and the middle ordinate e.©2000 CRC Press LLC
39. 39. Anderson, Loren Runar et al "SOIL MECHANICS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000
40. 40. Figure 4-1 Vertical soil pressure under one pair of dual wheels of a single axle HS-20 truck load, acting ona pipe buried at depth of soil cover, H, in soil of 100 pcf unit weight. Pressure is minimum at 5 or 6 ft ofcover.©2000 CRC Press LLC