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  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, 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 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
  25. 25. Given: shape during placement of embedment. OneE = 30(106)psi = modulus of elasticity, remedy, albeit costly, is to hold the ring in shape byS = 36 ksi = yield strength, stulls or struts while placing embedment. It may besf = 2 = safety factor, economical to provide enough ring stiffness to resistgw = 62.4 lb/ft3 = unit weight of water, deflection while placing the embedment. In anyP = hg w = internal water pressure at outlet. case, ring deflection is a potential performance limit for transportation/installation of pipes.From Equation 2.1, s = S/2 = P(ID)/2A where A is0.219 square inches per inch of length of the pipe. So two analyses are required for transportion andSubstituting in values, h = 357 ft. installation, with two corresponding performance limits: yield strength, and ring deflection. See Figure 2-3. In general, yield strength applies to rigid pipesTRANSPORTATION/INSTALLATION — such as concrete pipes, and ring deflection applies toMAXIMUM LINE LOAD ON PIPE flexible pipes. See Figure 2-4.The second step in design is resistance to loads Yield Strength Performance Limitimposed on the pipe during transportation andinstallation. The most common load is diametral F- To analyze the yield strength performance limit,load. See Figure 2-2. This load occurs when pipes based on experience, pertinent fundamentalare stacked or when soil is compacted on the sides variables may be written as follows:or on top of the pipe as shown. fvs, Fundamental bds, BasicIf yield strength of the pipe material is exceeded due Variables Dimensionsto the F-load, either the pipe wall will crack or the F = transportation/installation FL-1cross section of the pipe will permanently deform. load (concentrated line loadEither of these deformations (a crack is a per unit length of pipe),deformation) may be unacceptable. So yield D = mean diameter of the pipe, Lstrength may possibly be a performance limit even I = moment of inertia of the wall L3though the ring does not collapse. cross section per unit length of pipe,For some plastic materials, including mild steel, c = distance from the neutral axis Ldesign for yield strength is overly conservative. So of the wall cross section towhat if yield strength is exceeded? A permanent the most remote wall surfacedeformation (dent) in the ring is not necessarily pipe where the stress is at yield point.failure. In fact, the yield strength was probably S = yield strength of pipe wall FL-2exceeded in the process of fabricating the pipe. material 5 fvs - 2 bds = 3 pi-terms.Some pipe manufacturers limit the F-load based ona maximum allowable ring deflection, d = D/D, The three pi-terms may be written by inspection. Awhere D is the decrease in mean diameter D due to typical set is: (F/SD), (c/D), and (I/D3). This is onlyload F. Some plastics have a memory for excessive one of many possible sets of pi-terms. D is aring deflection. In service, failure tends to occur repeating variable. Note that the pi-terms arewhere excess ive ring deflection occurred before independent because each contains at least oneinstallation. Increased ring stiffness decreases ring fundamental variable that is not contained in any ofdeflection. It is not inconceivable that the ring can the other pi-terms. All are dimensionless. Thebe so flexible that it cannot even hold its circular interrelationship of these three pi-terms can be©2000 CRC Press LLC
  26. 26. found either by experimentation or by analysis. An fvs, Fundamental bds, Basicexample of class ical analysis starts with circum- Variables Dimensionsferential stress s = Mc/I where M is the maximum d = ring deflection = /D -bending moment in the pipe ring due to load F. But D = mean diameter of the Lif stress is limited to yield strength, then S = Mc/I pipewhere M = FD/2p based on ring analysis by F = diametral line load FL-1Castiglianos theorem. See Appendix A, Table A-1. per unit length of pipeM is the maximum moment due to force F. Because EI = wall stiffness FLit occurs at the location of F, there is no added ring per unit length of pipecompression stress. Substituting in values andrearranging the fundamental variables into pi-term, where: D = decrease in diameter due to the F-load,(F/SD) = 2p(D/c)(I/D3) E = modulus of elasticity, t = wall thickness for plain pipe,The three pi-terms are enclosed in parentheses. I = moment of inertia of wall cross section perDisregarding pi-terms, unit length of pipe = t3/12 for plain pipe.F = 2pSI/cD = F-load at yield strength, S. 4 fvs - 2 bds = 2 pi-terms.For plain pipes, I = t3/12 and c = t/2 for which, I/c = Two pi-terms, by inspection, are (d) and (FD2/EI).t2/6 and, in pi-terms: Again, the interrelationship of these pi-terms can be found either by experimentation or by analysis.(F/SD) = p(t/D)2/3 Table 5-1 is a compilation of analyses of ring deflections of pipes subjected to a few of theDisregarding pi-terms, common loads. From Table A-1, ring deflection due to F-loads is,F = pSt2/3D = F-load at yield strength S for plainpipes (smooth cylindrical surfaces). The modulus of (d) = 0.0186 (FD2/EI) . . . . . (2.2)elasticity E has no effect on the F-load as long asthe ring remains circular. Only yield strength S is a This equation is already in pi-terms (parentheses).performance limit. For plain pipes, for which I = t3/12 and c = t/2, this equation for ring deflection is:Ring Deflection Performance Limit (d) = 0.2232 (F/ED) (D/t)3If the performance limit is ring deflection at the The relationship between circumferential stress andelastic limit, modulus of elasticity E is pertinent. ring deflection is found by substituting from Table A-Yield strength is not pertinent. For this case, 1, at yield stress, F = 2pSI/cD, where S is yieldpertinent fundamental variables and corresponding strength and c is the distance from the neutralbasic dimensions are the following: surface of the wall to the wall surface. The resulting equation is:©2000 CRC Press LLC
  27. 27. (d) = 0.117 (s /E) (D/c) . . . . . (2.3) Steel and aluminum pipe industries use an F-load criterion for transportation/installation. In EquationFor plain pipes, 2.2 they specify a maximum flexibility factor FF = D2/EI. If the flexibility factor for a given pipe is less(d) = 0.234 (s E) (D/t) than the specified FF, then the probability of transportation/installation damage is statistically lowNote the introduction of a new pi-term, (s /E). This enough to be tolerated.relationship could have been found byexperimentation using the three pi-terms in For other pipes, the stress criterion is popular.parentheses in Equation 2.3. Ring deflection at yield When stress s = yield strength S, the maximumstress, S, can be found from Equation 2.3 by setting allowable load is:s = S. If ring deflection exceeds yield, the ring doesnot return to its original circular shape when the F- F = 2pSI/cDload is removed. Deformation is permanent. This isnot failure, but, for design, may be a performance For walls with smooth cylindrical surfaces,limit with a margin of safety. F = pSt2/3DThe following equations summarize design of thepipe to resist transportation/installation loads. In another form, for plain walls, the maximum allowable D/t is:For transportation/installation, the maximumallowable F-load and the corresponding ring (D/t)2 = pSD/3Fdeflection, d, when circumferential stress is atyield strength, S, are found by the following For the maximum anticipated F-load, i.e. at yieldformulas. strength, the minimum wall thickness term (t/D) can be evaluated. Any safety factor could be small — approaching 1.0 — because, by plastic analysis,Ring Strength collapse does not occur just because the circumferential stress in the outside surfaces(F/SD) = 2p (D/c) (I/D3) . . . . . (2.4) reaches yield strength. To cause a plastic hingeFor plain pipes, (F/SD) = p(t/D)2/3 (dent or cusp) the F-load would have to be increased by three-halves.Resolving, for plain pipes, Plastic pipe engineers favor the use of outsideF = pSD(t/D) /3. 2 diameter, OD, and a classification number called the dimension ratio, DR, which is simply DR = OD/t = (D+t)/t where D is mean diameter. Using theseRing Deflection where d = (D/D) due to F-load, dimensions, the F-load at yield is:is given by: F = pSt/3(DR-1) 2d = 0.0186 (FD /EI), in terms of F-load . . . (2.5) If the F-load is known, the required dimension ratiod = 0.117 (s /E) (D/c), in terms of stress, s , or at yield strength is:d = 0.234(S/D)(D/t), for plain pipes with smooth DR = (pSt/3F) + 1cylindrical surfaces in terms of yield strength S.©2000 CRC Press LLC
  28. 28. Example distributed. OD is the outside diameter. The resisting force is compression in the pipe wall, 2s A,Unreinforced concrete pipes are to be stacked for where s is the circumferential stress in the pipe wall,storage in vertical columns on a flat surface as called ring compression stress. Equating theindicated in Figure 2-2. The load on the bottom pipe rupturing force to the resisting force, with stress atis essentially an F-load. The following information is allowable, S/sf, the resulting equation is:given: s = P(OD)/2A = S/sf . . . . . (2.6)ID = 30 inches = inside diameter,OD = 37.5 in = outside diameter, This is the basis for design. Because of itsg = 145 lb/ft3 = unit weight of concrete, importance, design by ring compression stress isF = 3727 lb/ft = F-load at fracture considered further in Chapter 6. + s from tests where,s = + 460 lb/ft = standard deviation of the The above analyses are based on the assumption ultimate F-load at fracture of the pipe. that the ring is circular. If not, i.e., if deformation out-of-round is significant, then the shape of thea) How high can pipes be stacked if the F-load is deformed ring must be taken into account. Butlimited to 3000 lb/ft? From the data, the weight of basic deformation is an ellipse. See Chapter 3.the pipe is 400 lb/ft. The number of pipes high in thestack is 3000/400 = 7.5. So the stack must be Examplelimited to seven pipes in height. A steel pipe for a hydroelectric penstock is 51b) What is the probability that a pipe will break if the inches in diameter (ID) with wall thickness of 0.219column is seven pipes high? The seven pipe load at inch. It is to be buried in a good soil embedmentthe bottom of the stack is 7(400) = 2800 lb/ft. w = such that the cross section remains circular. What3727 - 2800 = 927 lb/ft which is the deviation of the is the safety factor against yield strength, S = 36 ksi,seven-pipe load from the F-load. From Table 1-1, if the external soil pressure on the pipe is 16 kips/ft2?the probability of failure is 2.2% for the bottom For this pipe, OD = 51.44 inches, and A = t = 0.219pipes. For all pipes in the stack, the probability is inch. At 16 ksf, P = 111 psi. Substituting intoone-seventh as much or 0.315%, which is one Equation 2.6, the safety factor is sf = 2.76. The soilbroken pipe for every 317 in the stack. pressure of 16 ksf is equivalent to about 150 feet of soil cover. See Chapter 3.c) What is the circumferential stress in the pipe wallat an average F-load of 3727 lb/ft? From Equation2.4, F = pSD(t/D)2/3 where S = yield strength D/t = PROBLEMS9, D = 51 inches. Solving, s = 471 psi. This is goodconcrete considering that it fails in tension. 2-1 What is the allowable internal pressure in a 48- inch diameter 2-2/3 by 1/2 corrugated steel pipe, 16 gage (0.064 inch thick)? (P = 48.4 psi)EXTERNAL PRESSURE —MINIMUM WALL AREA Given: D = 48 inches = inside diameter,Consider a free-body-diagram of half the pipe with t = 0.064 in = wall thickness,external pressure on it. See Figure 2-4. The A = 0.775 in 2/ft [AISI tables],vertical rupturing force is P(OD) where P is the S = 36 ksi = yield strength,external radial pressure assumed to be uniformly©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
  34. 34. What of the assumption that the horizontal and An important property of the ellipse is the ratio ofvertical ring deflections are equal if the radii rr = ry/rx, which is:circumferences are equal for circle and ellipse? Forthe circle, circumference is 2pr. For the ellipse, rr = (1 + d)3 / (1 - d)3 . . . . . (3.2)circumference is (b+a)(64-3R4)/(64-16R2), where Ris approximately R = (b-a)/(b+a). Only the first where:terms of an infinite series are included in this rr = ratio of the maximum to minimum radii ofapproximate ellipse circumference. See texts on curvature of the ellipse. See graph of Figure 3-4.analytical geometry. Equating circumferences of thecircle and the ellipse, and transforming the values ofa and b into vertical and horizontal values of ring Measurement of Radius of Curvaturedeflection, dy and dx, a few values of d y and thecorresponding dx are shown below for comparison. In practice it is often necessary to measure the radius of curvature of a deformed pipe. This can be Deviation done from either inside or outside of the pipe. Seedy (%) dx (%) (dy - dx)/dy Figure 3-5. Inside, a straightedge of known length L______ ______ __________ is laid as a cord. The offset e is measured to the 0. 0. 0. curved wall at the center of the cord. Outside, e can 5.00 4.88 0.024 be found by laying a tangent of known length L and10.00 9.522 0.048 by measuring the offsets e to the pipe wall at each15.00 13.95 0.070 end of the tangent. The average of these two20.00 18.116 0.094 offsets is the value for e. Knowing the length of the cord, L, and the offset, e, the radius of curvature ofFor ring deflections of d = dy = 10%, the the pipe wall can be calculated from the followingcorresponding dx is less than 10% by only equation:4.8%(10%) = 0.48%. This is too small to besignificant in most calculations such as areas within r = (4e2 + L2)/8e . . . . . (3.3)the ellipse and ratios of radii. It is assumed that radius of curvature is constantRadii of curvature of the sides (spring lines) and the within cord length L. The calculated radius is to thetop and bottom (crown and invert) of the ellipse are: surface from which e measurements are made.rx = r (1 - 3d + 4d2 - 4d3 + 4d4 - .....) Examplery = r (1 + 3d + 4d2 + 4d3 + 4d4 + .....) An inspection reveals that a 72-inch corrugated metal pipe culvert appears to be flattened somewhatFor ring deflection less than d = 10%, and neglecting on top. From inside the pipe, a straightedge (cord)higher orders of d, 12 inches long is placed against the top, and the mid- ordinate offset is measured and found to be 11/32rx = r (1 - 3d), and ry = r (1 + 3d). inch. What is the radius of curvature of the pipe ring at the top?However, more precise, and almost as easy to use,are the approximate values: From Equation 3.3, r = (4e2 + L2)/8e. Substituting in values and solving, ry = 52.5 inches which is therx = a2 / b = r (1 - d)2 / (1 + d) average radius within the 12-inch cord on the inside of the corrugated pipe. On the outside, the radius isry = b2 / a = r (1 + d)2 / (1 - d) greater by the depth of the corrugations.©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
  36. 36. Ring Deflection Due to Internal Pressure Ring Deformation Due to External LoadingWhen subjected to uniform internal pressure, the Computer software is available for evaluating thepipe expands. The radius increases. Ring deflection deformation of a pipe ring due to any externalis equal to percent increase in radius; loading. Analysis is based on the energy method of virtual work according to Castigliano. Analysisd = Dr/r = DD/D = 2pre /2pr = e provides a component of deflection of some point B on a structure with respect to a fixed point A. It iswhere: convenient to select point A as the origin of fixedd = ring deflection (percent), coordinate axes — the axes are neither translatedDr and DD are increases due to internal pressure, nor rotated. See Appendix A.r = mean radius,D = mean diameter,e = circumferential strain, ExampleE = modulus of elasticity = s/e.s = circumferential stress = Ee = Ed. Consider the quadrant of a circular cylinder shown in Figure 3-6. It is fixed along edge A-A-A, and isBut s = P(ID)/2A, from Equation 2.1, loaded with vertical line load Q along free edge B-B- B. What is the horizontal deflection of free edge Bwhere: with respect to fixed edge A? This is a two-P = uniform internal pressure, dimensional problem for which a slice of unit widthID = inside diameter, can be isolated for analysis. Because A is fixed,A = cross sectional area of wall per unit length. the horizontal deflection of B with respect to A is xB for which, according to Castigliano:Equating the two values for s , and solving for d, xB = f (M/EI)(dM/dp)ds . . . . . (3.5)d = P(ID)/2AE . . . . . (3.4)Figure 3-6 Quadrant of a circular cylinder fixed atthe crown A-A-A with Q-load at the spring line, B-B-B, showing a slice isolated for analysis.©2000 CRC Press LLC
  37. 37. where: PROBLEMSxB = displacement of point B in the x-direction,EI = wall stiffness, 3-1 A plain polyethylene pipe of 16-inch outsideE = modulus of elasticity, diameter and DR = 15 is subjected to internalI = centroidal moment of inertia of the cross pressure of 50 psi. The surfaces are smooth andsection of the wall per unit length of cylinder, cylindrical (not ribbed or corrugated). DR (dimension ratio) = (OD)/t where t = wall thickness.M = moment of force about the neutral axis at C, Modulus of elasticity is 115 ksi. What is the ring deflection? DR is dimension ratio = (OD)/t.p = differential load (dummy load) applied at (d = 0.28%)point B in the direction assumed for deflection,ds = differential length along the slice, = rdq 3-2 At ring deflection of 15%, and assuming the pipe cross section is an ellipse, what is the percentr = mean radius of the circular cylinder. error in finding the ratio of maximum to minimum radii of curvature by means of approximateIt is assumed that deflection is so small that radius r Equation 3.2,remains constant. It is also assumed that the rr = (1+d)3 / (1-d)3? (0.066%)deflection is due to moment M, flexure — not toshear or axial loads. In Figure 3-6, consider arc CBas a free-body-diagram. Apply the dummy load p at 3-3 A 36 OD PVC buried pipeline is uncovered atB acting to the right assuming that deflection xB will one location. The top of the pipe appears to bebe in the x-direction. If the solution turns out to be flattened. A straight edge 200 mm long is laidnegative, then the deflection is reversed. From the horizontally across the top and the vertical distancesfree-body-diagram CB, down to the pipe surface at each end of the straight edge are measured and found to be 9.2 and 9.4 mm.M = Qr(1-cosq) + pr(sinq) What is the radius of curvature of the outside surface of the pipe at the crown?M/ p = r(sinq) Ry = 542 mm = 21.35 inches)But because p approaches zero (differential), 3-4 Assuming that the ring of problem 3-3 isM = Qr(1-cosq) deflected into an ellipse, approximately what is the ring deflection? Maximum ring deflection is usuallyds = rdq limited to 5% according to specifications. (d = 5.74%)Substituting into Equation 3.5,xB = (Qr/EI) (1-cosq) r(sinq) rdq 3-5 What is the percent decrease in cross-sectional area inside the deflected pipe of problem 3-4 if theIntegrating and substituting in limits of q from 0 to ring deflection is d = 5.74%?p/2, (0.33%)xB = Qr3/2EI 3-6 What is the approximate ratio of maximum toThis is one of a number of the most useful minimum radii, rr, for an ellipse? (rr = 1.8)deflections of rings recorded in Table A-1.©2000 CRC Press LLC
  38. 38. 3-7 A horizontal, rectangular plate is a cantileverbeam loaded by a uniform vertical pressure, P, andsupported (fixed) along one edge. What is thevertical deflection of the opposite edge? Thethickness of the plate is t, the length measured fromthe fixed edge is L, and the modulus of elasticity isE. Elastic limit is not exceeded. Use the Castiglianoequation. (y = 3PL4 / 2Et 3)3-8 A half of a circular ring is loaded at the crownby an F-load (load per unit length of the cylinder).The reactions are rollers at the spring lines B, asshown. If the wall stiffness is EI, what is thevertical deflection of point A? (yA = 0.1781 Fr3/EI)3-9 What is the vertical ring deflection of the hingedarch of problem 3-8 if it is loaded with a uniformvertical pressure P instead of the F-load?3-10 The top and bottom halves of the circularcylinder of problem 3-8 are symmetrical. If thespring lines of the two halves are hinged together,what is the ring deflection due to the F-load and anequal and opposite reaction at the bottom? (d = 0.1781 Fr2/EI)3-11 Sections of pipe are tested by applying an F-load. For flexible rings, the F-load test is called aparallel plate test. What is the ring deflection ifelastic limit is not exceeded? [d = 0.0186F/(EI/D3)D]3-12 Find EI = f(Q/x) at point B for the ring cut at Aand loaded by force, Q. (EI = 3p Qr/x)3.13 A pipe in a casing floats when liquid grout isintroduced between pipe and casing. Find themoment, thrust and shear at crown and invert. (See Table A-1)©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
  41. 41. CHAPTER 4 SOIL MECHANICSAn elementary knowledge of basic principles of soil concentration factor is needed, or minimum soilstresses is essential to understanding the structural density should be specified. Over a long period ofperformance of buried pipes. These principles are time, pressure concentrations on the pipe may beexplained in standard texts on soil mechanics. A reduced by creep in the pipe wall (plastic pipes),few are reviewed in the following paragraphs earth vibrations, freeze-thaw cycles, wet-dry cycles,because of their special application to buried pipes. etc. The most rational soil load for design is vertical soil pressure at the top of the pipe due to dead weight of soil plus the effect of live load with aVERTICAL SOIL PRESSURE P specification that the soil embedment be denser than critical void ratio. Critical void ratio, roughlyFor the analysis and design of buried pipes, external 85% soil density (AASHTO T-99), is the void ratiosoil pressures on the pipes must be known. Vertical at such density that the volume of the soil skeletonsoil pressure at the top of the pipe is caused by: 1. does not decrease due to disturbance of soildead load, Pd , the weight of soil at the top of the particles.pipe; and 2. live load, Pl , the effect of surface liveloads at the the top of the pipe. Figure 4-1 shows For design, the total vertic al soil pressure at the topthese vertical soil pressures at the top of the pipe as of the pipe is:functions of height of soil cover, H, for an HS=20truck axle load of 32 kips, and soil unit weight of 100 P = Pd + Pl . . . . . (4.1)pcf. Similar graphs are found in pipe handbookssuch as the A ISI Handbook of Steel Drainage and where (see Figure 4-2)Highway Construction Products. Soil unit weight P = total vertical soil pressure at the level ofcan be modified as necessary. Also other factors the top of the pipemust be considered. What if a water table rises Pd = dead load pressure due to weight of theabove the top of the pipe, or the pipe deflects, or the soil (and water content)soil is not compacted, or is overcompacted? For Pl = vertical live load pressure at the level ofthese and other special cases, the following the top of the pipe due to surface loads.fundamentals of soil mechanics may be useful. This is a useful concept in the analysis of buriedIf the embedment about a buried pipe is densely pipes. Even rigid pipes are designed on this basis ifcompacted, vertical soil pressure at the top of the a load factor is included. See Chapter 12.pipe is reduced by arching action of the soil over thepipe, like a masonry arch, that helps to support the In fact, P is only one of the soil stresses. At a givenload. To be conservative, arching action is usually point in a soil mass, a precise stress analysis wouldignored. However, soil arching provides an added consider three (triaxial) direct stresses, threemargin of safety. If the soil embedment is loose, shearing stresses, direct and shearing moduli in threevertical soil pressure at the top of the pipe may be directions and three Poissons ratios — with theincreased by pressure concentrations due to the additional condition that soil may not be elastic. Therelatively noncompressible area within the ring in imprecisions of soil placement and soil compactionloose, compressible soil. Pressure concentrations obfuscate the arguments for such rigor. Elasticdue to loose embedment cannot be ignored. For analysis may be adequate under some fewdesign, either a pressure circumstances. Superposition is usually adequate without concern for a combined stress analysis involving triaxial stresses and Poisson ratio. Basic soil mechanics serves best.©2000 CRC Press LLC
  42. 42. Figure 4-2 Vertical soil pressure P at the level of the top of a buried pipe where P = Pl + Pd , showing liveload pressure Pl and dead load pressure Pd superimposed.Figure 4-3 A single stratum of saturated soil with water table at the top (buoyant case) showing vertical stressat depth H.©2000 CRC Press LLC
  43. 43. Dead Load Vertical Soil Pressure Pd S = degree of saturation = 1 when saturated, e = void ratio, from laboratory analysis,Dead load is vertical pressure due to the weight of gw = unit weight of water.soil at a given depth H. In the design of buriedpipes, H is the height of soil cover over a pipe. Total Table 4-1 is a summary of dead load soil stresses from which dead load pressure Pd can be found andpressure Pd is the weight of soil, including its water combined with live load pressure Pl. Live loadcontent, per unit area. See Figure 4-3. pressure is found from techniques described in theIntergranular (or effective) pressure Pd is the paragraphs to follow.pressure felt by the soil skeleton when immersed inwater. The total and intergranular vertical stresses Intergranular vertical soil pressure P, at the bottomat the bottom of a submerged stratum can be related of multiple soil strata, is:by the following stress equation: __ P = P - u = P - g wh . . . . . (4.5)s=s-u . . . . . (4.2)where _ where_ P = vertical intergranular soil pressure,s = intergranular vertical soil stress (felt by the P = total dead plus live load pressures, soil when buoyed up by water), h = height of water table above the pipe.s = total vertical soil stress = g tH,u = pore water pressure = g wH, Total pressure is used to calculate ring compressiongt = total unit weight of soil and water, stress. Intergranular soil pressure is used to calcu-gw = unit weight of water = 62.4 pcf. late ring deflection which is a function of soil compression. As the soil is compressed, so is theNow consider more than one stratum of soil as pipe compressed — and in direct ratio. But soilshown in Figure 4-4. The total vertical dead load soil compression depends only on intergranular stresses.pressure Pd at the bottom of the strata is the sum of See Chapter 7.the loads imposed by all of the strata; i.e.,Pd = Sgt H . . . . . (4.3) Live Load Vertical Soil Pressure Plwhere Live load soil pressure Pl is the vertical soil pressuregt = total unit weight (wet weight) of soil at the top of the buried pipe due to surface loads. in a given stratum, and See Figure 4-5. For a single concentrated load W onH = height of the same stratum. the surface, vertical soil pres sure at point A at the top of the pipe is:Values of H for each soil stratum are provided bysoil borings. Values of g t are simply the unit s = NW/H2 . . . . . (4.6)weights of representative soil samples including thewater content. If the soil samples are not available, wherefrom soil mechanics, W = concentrated surface load (dual-wheel) H = height of soil cover over the top of the pipeg t = (G+Se) g w /(1+e) . . . . . (4.4) R = horizontal radius to stress, s , N = Boussinesq coefficient.where from the line of action of load W,G = specific gravity of soil grains, about 2.65, N = Boussinesq coefficient = 3(H/R) 5/2p.©2000 CRC Press LLC
  44. 44. Figure 4-4 Multiple strata(three strata with the claystratum divided into two atthe water table) showing thetotal vertical dead load soilpressure Pd at the bottom(level of the top of the pipe).Figure 4-5 Vertical soil pressure at depth H (at the level of the top of a pipe) and at radius R from the lineof action of a concentrated surface load W. (After Boussinesq)©2000 CRC Press LLC
  45. 45. ©2000 CRC Press LLC
  46. 46. Figure 4-6 Chart for evaluating the vertical stress s at a depth H below the corner A of a rectangular surfacearea loaded with a uniformly distributed pressure q. (After Newmark)©2000 CRC Press LLC
  47. 47. For a single wheel (or dual wheel) load, the maxi- Examplemum stress smax at A occurs when the wheel isdirectly over the pipe; i.e. R = 0, for which What is the stress at point A below A of Figure 4- 8? The vertical stress s at depth H below surf a c esmax = 0.477 W/H2 . . . . . (4.7) point A is, by superposition:Load W can be assumed to be concentrated if depth s = s - S s" + s "H is greater than the maximum diameter or length of wherethe surface loaded area. s = stress at corner A due to loaded area LxBFor multiple wheel (or dual) loads, the maximum S s" = sum of stresses at corner A due tostress at point A, due to effects of all loads must be loaded areas LB" and L"Bascertained. The trick is to position the wheel loads s" = stress at corner A due to loaded areaso that the combined stress at A is maximum. This L"Bcan be done by trial. Clearly, s " due to area L"B" was subtracted twice,The effect of a uniformly distributed surface load so must be added back once.can be found by dividing the loaded surface area intoinfinitesimal areas and integrating to find the sum of An occlusion in the soil mass, such as a pipe, violatestheir effects at some point at depth H. See Figure 4- Boussinesqs assumptions of elasticity, continuity,6. Newmark performed such an integration and compatibililty, and homogeneity. The pipe is a hardfound the vertical stress s at a depth H below spot, a discontinuity. Soil is not elastic, nor homoge-corner A of a rectangular area of greater length L neous, nor compatible when shearing planes form.and lesser breadth B, loaded with uniform pressure Nevertheless, the Boussinesq assumptions areq. His neat solution is: adequate for most present-day installation tech- niques. For most buried pipe design, it is sufficient,s = Mq . . . . . (4.8) and conservative, to solve for Pl at the top of the pipe due to a single wheel load W at the surface bywhere M is a coefficient which can be read on the using the Boussinesq equation with the radius R = 0.chart of Figure 4-6 by entering with arguments L/B For additional wheel loads, simply add by superposi-and B/H. If the stress due to pressure on an area is tion the influence of other wheel loads at their radiidesired below some point other than a corner, the R.rectangular area can be expanded or subdividedsuch that point A is the common corner of a number Exampleof areas. The maximum stress under a rectangulararea occurs below the center. See Figure 4-7. The What is the maximum vertical soil stress at a depthrectangle is subdivided into four identical rectangles of 30 inches due to the live load of a single axle HS-of length L and breadth B as shown. The stress at 20 truck? Neglect surface paving. See Figures 4-9point A is 4Mq, where M is found from the and 4-15. By trial, it can be shown that the point ofNewmark chart, Figure 4-6. maximum stress is point A under the center of one tire print. The rectangular tire prints are subdividedAlternatively, the Boussinesq equation can be used as shown for establishing a common corner A. Thewith less than five percent error if the concentrated effects of the left tire print and the right tire print areload, Q = qBL, for each of the quadrants is assumed analyzed separately, then combined. The length Lto act at the center of each quadrant. For this case, and breadth B of each tire print are based on 104 psiR = (L2+B2)/2. The resulting stress at A is 4NQ/H2 tire pressure. Use Newmark because H is less thanfrom Equation 4.6. 3L.©2000 CRC Press LLC
  48. 48. Figure 4-7 Procedure for subdividing a rectangular surface area such that the stress below the center at depthH is the sum of the stresses below the common corners A of the four quadrants.Figure 4-8 Subdivision of the loaded surface area, LxB, for evaluation of vertical stress, P, at depth, H, underpoint A.©2000 CRC Press LLC
  49. 49. Figure 4-9 Single axle HS-20 truck load showing typical tire prints for tire pressure of 104 psi, and showingthe Newmark subdivision for evaluating vertical soil stress under the center of one tire print.©2000 CRC Press LLC
  50. 50. Figure 4-10 Infinitesimal soil cube B and the corresponding Mohr circle which provides stresses on any planethrough B. Note the stresses sq and t q shown on the q -plane. At soil slip, the circle is tangent to thestrength envelopes described below.Figure 4-11 Shearing stress t as a function of normal stress s , showing a series of Mohr circles at soil slip,and the strength envelopes tangent to the Mohr circles.©2000 CRC Press LLC
  51. 51. Given: SOIL STRENGTHW = 32 k for HS-20 truck load (single axle),q = 104 psi, Failure of a buried pipe is generally associated withB = 7 inches, failure of the soil in which the pipe is buried. TheL = 22 inches, classical, two-dimensional, shear-strength soil modelH = 30 inches. is useful for analysis. Analysis starts with an infini- tesimal soil cube on which stresses are known andFor the left tire print: the orientation is given. The model comprises threesL = 4Mq, elements, the Mohr stress circle, orientation diagram,L/B = 11/3.5 = 3.14, and strength envelopes.B/H = 3.5/30 = 0.12,From Figure 4.6, M = 0.018 and sL = 1.078 ksf Mohr Stress CircleFor the right tire print:sR = 2(M-M")q, where, The Mohr stress circle is a plot of shearing stress, t ,L/B = 83/3.5 = 23.7, as a function of normal stress, s , on all planes atB/H = 3.5/30 = 0.12, angle q through an infinitesimal soil cube B. SeeFrom Figure 4.6, M = 0.02 (extrapolated) Figures 4-10 and 4-11. The sign convention is compressive normal stress positive ( + ) andL"/B" = 61/3.5 = 17.4, counterclockwise shearing stress positive ( + ).B"/H" = 3.5/30 = 0.12, The center of the circle is always on the s -axis.From Figure 4.6, M" = 0.02 (extrapolated) Two additional points are needed to determine the circle. They are (sx, t xy ) on a y-plane and (s y , ty x)sR = r(M- M")q = 0: sR = 0 ksf on an x-plane. These are known stresses on cube B. An origin of planes always falls on the circle.At point A, s = s L + s R; s = 1.08 ksf t xy = -ty x from standard texts on solid mechanics. Any plane from the origin intersects the MohrA rough check by Boussinesq is of interest because circle at the stress coordinates acting on thatthe results are conservatively higher and are more plane— which is correctly oriented if the followingeasily solved. procedure is followed.Given:W = 16 kips at the center of each tire print, Orientation DiagramH = 2.5 ft,RL = 0, Figure 4-10 shows infinitesimal cube B with the x-RR = 6 ft, plane and y-plane identified and with the soil stressesR/H = 2.4. acting on each plane. Cube B and its axes ofFrom Figure 4.5, N = 0.004. orientation can be superimposed on the Mohr circle such that stress coordinates (where each plane inter-sL = 0.477 W/H2 = 1.22 ksf sects the Mohr circle) are the stresses on that plane.sR = NW/H2 = 0.01 ksf With cube B located on the Mohr circle as the origin of axes, and with the axes correctly oriented, anyAt point A, s = sL + s R ; s = 1.23 ksf plane through B will intersect the Mohr circle at the point whose stress coordinates are the stressesThe Boussinesq solution is in error by 13.9%, but on acting on that plane, and all planes are correctlythe high (conservative) side. Of interest is the small oriented with respect to the original soil cube B.(negligible) effect of the right wheel load. ©2000 CRC Press LLC
  52. 52. Figure 4-12 Trigonometry for analysis of stresses at soil slip on shear planes, q f in cohesionless soil whichhas a soil friction angle of j.©2000 CRC Press LLC