THIẾT KẾ ĐƯỜNG ỐNG

STRUCTURALMECHANICS

OF

BURIED PIPES

Boca Raton London New York Washington, D.C.CRC Press

Reynold King WatkinsUtah State University

Logan, Utah

Loren Runar AndersonUtah State University

Logan, Utah

GENERAL NOTATION

GeometryA = 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.


CONTENTS

Chapter

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: Castigliano's Equation

Appendix B: Reconciliation of Formulas for Predicting Ring Deflection

Appendix C: Similitude Appendix D: Historical SketchAppendix E: Stress Analysis Appendix F: Strain Energy Analysis


PREFACE

Buried 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

ACKNOWLEDGMENT

Gratitude is expressed to Becky Hansen for her patient and expert preparation of manuscript.


Anderson, Loren Runar et al "ECONOMICS OF BURIED PIPES AND TANKS "Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Anderson, Loren Runar et al "INTRODUCTION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

CHAPTER 1 INTRODUCTION

Buried conduits existed in prehistory when caveswere protective habitat, and ganats (tunnels backunder mountains) were dug for water. The value ofpipes is found in life forms. As life evolved, themore complex the organism, the more vital andcomplex were the piping systems.

The earthworm lives in buried tunnels. His is ahigher order of life than the amoeba because he hasdeveloped a gut — a pipe — for food processingand waste disposal.

The Hominid, a higher order of life than theearthworm, is a magnificent piping plant. Thehuman piping system comprises vacuum pipes,pressure pipes, rigid pipes, flexible pipes — allgrown into place in such a way that flow is optimumand stresses are minimum in the pipes and betweenthe pipes and the materials in which they are buried.

Consider a community. A termite hill contains anintricate maze of pipes for transportation, ventilation,and habitation. But, despite its elegance, the termitepiping system can't compare with the piping systemsof a community of people. The average city dwellertakes for granted the services provided by city pipingsystems, and refuses to contemplate theconsequences if services were disrupted. Cities canbe made better only to the extent that piping systemsare made better. Improvement is slow becauseburied pipes are out-of-sight, and, therefore, out-of-mind to sources of funding for the infrastructure.

Engineering design requires knowledge of: 1.performance, and 2. limits of performance. Threegeneral sources of knowledge are:

SOURCES OF KNOWLEDGEExperience (Pragmatism)Experimentation (Empiricism)Principles (Rationalism)

In a phenomenon as complex as the soil-structureinteraction of buried pipes, all three sources must beutilized. There are too many variables; theinteraction is too complex (statically indeterminate tothe infinite degree); and the properties of soil are tooimprecise to rely on any one source of information.

Buried structures have been in use from antiquity.The ancients had only experience as a source ofknowledge. Nevertheless, many of their catacombs,ganats, sewers, etc., are still in existence. But theyare neither efficient nor economical, nor do we haveany idea as to how many failed before artisanslearned how to construct them.

The other two sources of knowledge are recent.Experimentation and principles required thedevelopment of soil mechanics in the twentiethcentury. Both experience and experimentation areneeded to verify principles, but principles are thebasic tools for design of buried pipes.

Complex soil-structure interactions are still analyzedby experimentation. But even experimentation ismost effective when based on principles — i.e.,principles of experimentation.

This text is a compendium of basic principles provento be useful in structural design of buried pipes.

Because the primary objective is design, the firstprinciple is the principle of design.

DESIGN OF BURIED PIPES

To design a buried pipe is to devise plans andspecifications for the pipe-soil system such thatperformance does not reach the limits ofperformance. Any performance requirement isequated to its limit divided by a safety factor, sf, i.e.:


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.


Performance LimitPerformance =

Safety Factor

Examples:

Stress = Strength/sf

Deformation = Deformation Limit/sf

Expenditures = Income/sf; etc.

If performance were exactly equal to the performance

limit, half of all installations would fail. A safety

factor, sf, is required. Designers must allow for

imperfections such as less-than-perfect construction,

overloads, flawed materials, etc. At present, safety

factors are experience factors. Future safety factors

must include probability of failure, and the cost of

failure — including risk and liability. Until then, a

safety factor of two is often used.

In order to find probability of failure, enough failures

are needed to calculate the standard deviation of

normal distribution of data.

NORMAL DISTRIBUTION

Normal distribution is a plot of many measurements

(observations) of a quantity with coordinates x and y,

where, see Figure 1-1,

x = abscissa = measurement of the quantity,

y = ordinate = number of measurements in any given

x-slot. A slot contains all measurements that are

closer to the given x than to the next higher x or the

next lower x. On the bar graph of data Figure 1-1, if

x = 680 kPa, the 680-slot contains all of x-values

from 675 to 685 kPa.

x) = the average of all measurements,

x = 3yx/ 3y,

n = total number of measurements = Ey,

w = deviation, w = x - x),

P = probability that measurement will fall

between ±w,

Pe = probability that a measurement will exceed

the failure level of xe (or fall below a

minimum level of xe ),

s = standard deviation = deviation within which

68.26 percent of all measurements fall (Ps =

68.26%).

P is the ratio of area within +w and the total area.

Knowing w/x, P can be found from Table 1.1. The

standard deviation s is important because: l. it is a

basis for comparing the precision of sets of

measurements, and 2. it can be calculated from

actual measurements; i.e.,

s = %3yw2/(n-1)

Standard deviation s is the horizontal radius of

gyration of area under the normal distribution curve

measured from the centroidal y axis. s is a deviation

of x with the same dimensions as x and w. An

important dimensionless variable (pi-term) is w/s.

Values are listed in Table 1-1. Because probability

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, the

probability that any measurement x will fall within

±w from the average, can be read from Table 1-1.

Likewise the probability of a failure, Pe , either

greater than an upper limit xe or less than a lower

limit, xe , can be read from the table. The deviation

of failure is needed; i.e., we = x e - x). Because pipe-

soil interaction is imprecise (large standard

deviation), it is prudent to design for a probability of

success of 90% (10% probability of failure) and to

include a safety factor. Probability analysis can be

accomplished conveniently by a tabular solution as

shown in the following example.

Example

The bursting pressure in a particular type of pipe has

been tested 24 times with data shown in Table 1-2.

What is the probability that an internal pressure of

0.8 MPa (120 psi or 0.8 MN/m2) will burst the pipe?

x = test pressure (MN/m2) at bursting

y = number of tests at each x

n = Gy = total number of tests


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 a

function of we /s that a value of x will fall outside of +we on either the +we or the -we .

we/s P Pe we/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 2.87

0.5 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 1.07

0.8 57.6 21.2 2.4 98.36 0.820.9 63.2 18.4 2.5 98.76 0.62

1.0 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 0.135

Table 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 yw2

(Mpa)* _ (MPa) (MPa) (MPa) (MPa)2

0.9 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 Σyw2

x = Σxy/n = 1.1 MPa

s = [Σyw2/(n-1)] = 0.125

*MPa is megapascal of pressure where a Pascal is N/m2; i.e., a megapascal is a million Newtons of forceper square meter of area. A Newton = 0.2248 lb. A square meter = 10.76 square ft.


From the data of Table 1-2,

x_

= Σxy/Σy = 26.4/24 = 1.1

s = \/Σyw2/(n-1) = \/0.36/23 = 0.125

w = x - x, so

we = (0.8 - 1.1) = -0.30 MN/m2 = deviation to failure pressure

we/s = 0.30/0.125 = 2.4.

From Table 1-1, interpolating, Pe= 0.82%.The probability that a pipe will fail by bursting

pressure less than 0.80 MN/m2 is Pe = 0.82 % orone out of every 122 pipe sections. Cost accountingof failures then follows.

The probability that the strength of any pipe sectionwill fall within a deviation of we = +0.3 MN/m2 is P

= 98.36%. It is noteworthy that P + 2Pe = 100%.

From probability data, the standard deviation can becalculated. From standard deviation, the zone of +wcan be found within which 90% of all measurementsfall. In this case w/s = w/0.125 for which P = 90%.From Table 1-1, interpolating for P = 90%, w/s =1.64%, and w = 0.206 MPa at 90% probability.

Errors (three classes)Mistake = blunder —

Remedies: double-check, repeat.Accuracy = nearness to truth —

Remedies: calibrate, repair, correct.Precision = degree of refinement —

Remedies: normal distribution, safety factor.

PERFORMANCE

Performance in soil-structure interaction isdeformation as a function of loads, geometry, andproperties of materials. Some deformations can bewritten in the form of equations from principles of

soil mechanics. The remainders involve suchcomplex soil-structure interactions that theinterrelationships must be found from experience orexperimentation. It is advantageous to write therelationships in terms of dimensionless pi-terms. SeeAppendix C. Pi-terms that have proven to be usefulare given names such as Reynold's number in fluidflow in conduits, Mach number in gas flow, influencenumbers, stability numbers, etc.

Pi-terms are independent, dimensionless groups offundamental variables that are used instead of theoriginal fundamental variables in analysis orexperimentation. The fundamental variables arecombined into pi-terms by a simple process in whichthree characteristic s of pi-terms must be satisfied.The starting point is a complete set of pertinentfundamental variables. This requires familiarity withthe phenomenon. The variables in the set must beinterdependent, but no subset of variables can beinterdependent. For example, force f, mass m, andacceleration a, could not be three of the fundamentalvariables in a phenomenon which includes othervariables because these three are not independent;i.e., f = ma. Only two of the three would beincluded as fundamental variables. Once theequation of performance is known, the deviation, w,can be found. Suppose r = f(x,y,z,...), then wr

2 =Mrx

2

wx2 + Mry

2w y2 + ... where w is a deviation at the

same given probability for all variables, such asstandard deviation with probability of 68%; mrx is thetangent to the r-x curve and wx is the deviation at agiven value of x. The other variables are treated inthe same way.

CHARACTERISTICS OF PI-TERMS

1. Number of pi-terms = (number of fundamentalvariables) minus (number of basic dimensions).

2. All pi-terms are dimensionless.

3. Each pi-term is independent. Independence isassured if each pi-term contains a fundamentalvariable not contained in any other pi-term.


_

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.


Pi-terms have two distinct advantages: fewervariables to relate, and the elimination of size effect.The required number of pi-terms is less than thenumber of fundamental variables by the number ofbasic dimensions. Because pi-terms aredimensionless, they have no feel for size (or anydimension) and can be investigated by model study.Once pi-terms have been determined, theirinterrelationships can be found either by theory(principles) or by experimentation. The results applygenerally because the pi-terms are dimensionless.Following is an example of a well-designedexperiment.

Example

Using experimental techniques, find the equation forinternal bursting pressure, P', for a thin-wall pipe.Start by writing the set of pertinent fundamentalvariables together with their basic dimensions, forceF and length L.

Basic Fundamental Variables Dimensions

P' = internal pressure FL-2

t = wall thickness LD = inside diameter of ring LS = yield strength of the

pipe wall material FL-2

These four fundamental variables can be reduced totwo pi-terms such as (P'/S) and (t/D). The pi-termswere written by inspection keeping in mind the threecharacteristics of pi-terms. The number of pi-termsis the number of fundamental variables, 4, minus thenumber of basic dimensions, 2, i.e., F and L. Thetwo pi-terms are dimensionless. Both areindependent because each contains a fundamentalvariable not contained in the other. Conditions forbursting can be investigated by relating only twovariables, the pi-terms, rather than interrelating theoriginal four fundamental variables. Moreover, theinvestigation can be performed on pipes of anyconvenient size because the pi-terms aredimensionless. Test results of a

small scale model study are plotted in Figure 1-2.The plot of data appears to be linear. Only the lastpoint to the right may deviate. Apparently the pipeis no longer thin-wall. So the thin-wall designationonly applies if t/D< 0.1. The equation of the plot isthe equation of a straight line, y = mx + b where y isthe ordinate, x is the abscissa, m is the slope, and bis the y-intercept at x = 0. For the case above,(P'/S) = 2(t/D), from which, solving for burstingpressure,

P = 2S/(D/t)

This important equation is derived by theoreticalprinciples under "Internal Pressure," Chapter 2.

PERFORMANCE LIMITS

Performance limit for a buried pipe is basically adeformation rather than a stress. In some cases it ispossible to relate a deformation limit to a stress(such as the stress at which a crack opens), butsuch a relationship only accommodates the designerfor whom the stress theory of failure is familiar. Inreality, performance limit is that deformation beyondwhich the pipe-soil system can no longer serve thepurpose for which it was intended. Theperformance limit could be a deformation in the soil,such as a dip or hump or crack in the soil surfaceover the pipe, if such a deformation is unacceptable.The dip or hump would depend on the relativesettlement of the soil directly over the pipe and thesoil on either side. See Figure 1-3.

But more often, the performance limit is excessivedeformation of the pipe which could cause leaks orcould restrict flow capacity. If the pipe collapsesdue to internal vacuum or external hydrostaticpressure, the restriction of flow is obvious. If, on theother hand, the deformation of the ring is slightly out-of-round, the restriction to flow is usually notsignificant. For example, if the pipe cross sectiondeflects into an ellipse such that the decrease of theminor diameter is 10% of the original circulardiameter, the decrease in cross-sectional area is only1%.


Figure 1-4 Typical performance limits of buried pipe rings due to external soil pressure.


The more common performance limit for the pipe isthat deformation beyond which the pipe cannot resistany increase in load. The obvious case is bursting ofthe pipe due to internal pressure. Less obvious andmore complicated is the deformation due to externalsoil pressure. Typical examples of performancelimits for the pipe are shown in Figure 1-4. Theseperformance limits do not imply collapse or failure.The soil generally picks up any increase in load byarching action over the pipe, thus protecting the pipefrom total collapse. The pipe may even continue toserve, but most engineers would prefer not todepend on soil alone to maintain the conduit crosssection. This condition is considered to be aperformance limit. The pipe is designed to withstandall external pressures. Any contribution of the soiltoward withstanding external pressure by archingaction is just that much greater margin of safety.The soil does contribute soil strength. On inspection,many buried pipes have been found in service eventhough the pipe itself has "failed." The soil holdsbroken clay pipes in shape for continued service.The inverts of steel culverts have been corroded oreroded away without failure. Cast iron bells havebeen found cracked. Cracked concrete pipes arestill in service, etc. The mitigating factor is theembedment soil which supports the conduit.

A reasonable sequence in the design of buried pipesis the following:

1. Plans for delivery of the product (distances,elevations, quantities, and pressures),

2. Hydraulic design of pipe sizes, materials,

3. Structural requirements and design of possiblealternatives,

4. Appurtenances for the alternatives,

5. Economic analysis, costs of alternatives,

6. Revision and iteration of steps 3 to 5,

7. Selection of optimum system.

With pipe sizes, pressures, elevations, etc., known

the structural design of the pipe can proceed in sixsteps as follows.

STEPS IN THE STRUCTURAL DESIGN OFBURIED PIPES

In order of importance:

1. Resistance to internal pressure, i.e., strength ofmaterials and minimum wall thickness;

2. Resistance to transportation and installation;

3. Resistance to external pressure and internalvacuum, i.e., ring stiffness and soil strength;

4. Ring deflection, i.e., ring stiffness and soilstiffness;

5. Longitudinal stresses and deflections;

6. Miscellaneous concerns such as flotation of thepipe, construction loads, appurtenances, ins tallationtechniques, soil availability, etc.

Environment, aesthetics, risks, and costs must beconsidered. Public relations and social impactcannot be ignored. However, this text deals onlywith structural design of the buried pipe.

PROBLEMS

1-1 Fluid pressure in a pipe is 14 inches of mercuryas measured by a manometer. Find pressure inpounds per square inch (psi) and in Pascals(Newtons per square meter)? Specific gravity ofmercury is 13.546.

(6.85 psi)(47.2 kPa)

1-2 A 100 cc laboratory sample of soil weighs 187.4grams mass. What is the unit weight of the soil inpounds per cubic ft? (117 pcf)

1-3 Verify the standard deviation of Figure 1-1. (s = 27.8 kPa)


1-4 From Figure 1-1, what is the probability thatany maximum daily pressure will exceed 784.5 kPa? (P = 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)

(b) What is the standard deviation? (8.38 Pa)

(c) What is the probable error? (+5.65 Pa)

1-6 Eleven 30 inch ID, non-reinforced concretepipes, Class 1, were tested in three-edge-bearing(TEB) test with results as follows: x = ultimate load in pounds per lineal ftx w w2

(lb/ft) (lb/ft) 35623125437534384188368837504188412536252938

(a) What is the average load, x, at failure?(x = 3727.5 lb/ft)

(b) What is the standard deviation?(s = 459.5 lb/ft)

(c) What is the probability that the load, x, at failureis less than the minimum specified strength of 3000lb/ft (pounds per linear ft)? (Pe = 5.68%)

Figure 1-5 Bar graphs of internal vacuum at collapseof thin-walled plastic pipes.

1-7 Fiberglass reinforced plastic (FRP) tanks weredesigned for a vacuum of 4 inches of mercury(4inHg). They were tested by internal vacuum forwhich the normal distribution of the results is shownas Series A in Figure 1-6. Two of 79 tanks failed atless than 4inHg. In Series B, the percent offiberglas was increased. The normal distributioncurve has the same shape as Series A, but is shifted1inHg to the right. What is the predicted probabilityof failure of Series B at or below 4 in Hg? (Pe = 0.17 % or one tank in every 590)

1-8 What is the probability that the vertical ringdeflection d = y/D of a buried culvert will exceed10% if the following measurements were made on23 culverts under identical conditions?Measured values of d (%)

6 9 6 6 5 68 5 4 6 7 73 6 7 5 4 56 7 8 7 5 (0.24 %)

1-9 The pipe stiffness is measured for many samplesof a particular plastic pipe. the average is 24 with astandard deviation of 3.

a) What is the probability that the pipe stiffness willbe less than 20? (Pe = 9.17 %)


e

b) What standard deviation is required if theprobability of a stiffness less than 20 is to bereduced to half its present value; i.e., less than4.585%? (s = 2.37)

1-10 A sidehill slope of cohesionless soil dips atangle 2. Write pi-terms for critical slope whensaturated.

1-11 Design a physical model for problem 1-10.

Figure 1-6 Normal distribution diagrams for fiberglass tanks designed for 4inHg vacuum.


Anderson, Loren Runar et al "PRELIMINARY RING DESIGN"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 2-1 Free-body-diagramof half of the pipe crosssection including internalpressure P’.

Equating rupturing force toresisting force, hoop stressin the ring is,

σ = P’(ID)2A

Figure 2-2 Common transportation/installation loads on pipes, called F-loads.


CHAPTER 2 PRELIMINARY RING DESIGN

The first three steps in the structural design of buriedpipes all deal with resistance to loads. Loads on aburied pipe can be complex, especially as the pipedeflects out-of-round. Analysis can be simplified ifthe cross section (ring) is assumed to be circular.For pipes that are rigid, ring deflection is negligible.For pipes that are flexible, ring deflection is usuallylimited by specification to some value not greaterthan five percent. Analysis of a circular ring isreasonable for the structural design of most buriedpipes. Analysis is prediction of structuralperformance. Following are basic principles foranalysis and design of the ring such that it cansupport the three most basic loads: internal pressure,transportation/installation, and external pressure.See Figures 2-1 and 2-2.

INTERNAL PRESSURE —(MINIMUM WALL AREA)

The first step in structural design of the ring is to findminimum wall area per unit length of pipe.

Plain pipe — If the pipe wall is homogeneous andhas smooth cylindrical surfaces it is plain (bare) andwall area per unit length is wall thickness. This isthe case in steel water pipes, ductile iron pipes, andmany plastic pipes.

Other pipes are corrugated or ribbed or compositepipes such as reinforced concrete pipes. For suchpipes, the wall area, A, per unit length of pipe is thepertinent quantity for design.

Consider a free-body-diagram of half of the pipewith fluid pressure inside. The maximum rupturingforce is P'(ID) where P' is the internal pressure andID is the inside diameter. See Figure 2-1. Thisrupturing force is resisted by tension, FA, in the wallwhere F is the circumferential tension stress in thepipe wall. Equating rupturing force to the resistingforce, F = P'(ID)/2A. Performance limit

is reached when stress, σ , equals yield strength, S.For design, the yield strength of the pipe wall isreduced by a safety factor,

σ = P'(ID)/2A = S/sf . . . . . (2.1)

where:σ = circumferential tensile, stress in the wall,P' = internal pressure,ID = inside diameter,OD = outside diameter,D = diameter to neutral surface,A = cross sectional area of the pipe wall per

unit length of pipe,S = yield strength of the pipe wall material,t = thickness of plain pipe walls, sf = safety factor.

This is the basic equation for design of the ring toresist internal pressure. It applies with adequateprecision to thin-wall pipes for which the ratio ofmean diameter to wall thickness, D/t, is greater thanten. Equation 2.1 can be solved for maximumpressure P' or minimum wall area A.

A = P'(ID)sf/2S = MINIMUM WALL AREA

For thick-wall pipes (D/t less than ten), thick-wallcylinder analysis may be required. See Chapter 6.Neglecting resistance of the soil, the performancelimit is the yield strength of the pipe. Once the ringstarts to expand by yielding, the diameter increases,the wall thickness decreases, and so the stress in thewall increases to failure by bursting.

Example

A steel pipe for a hydroelectric penstock is 51 inchID with a wall thickness of 0.219 inch. What is themaximum allowable head, h, (difference in elevationof the inlet and outlet) when the pipe is full of waterat no flow?


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 collapsing force to resisting force,ring compression stress is,

σ = P(OD)/2A

Figure 2-4 Free-body-diagram of half of the ring showing external radial pressure, P.


Given:E = 30(106)psi = modulus of elasticity,S = 36 ksi = yield strength,sf = 2 = safety factor,γ w = 62.4 lb/ft3 = unit weight of water,P' = hγ w = internal water pressure at outlet.

From Equation 2.1, σ = S/2 = P'(ID)/2A where A is0.219 square inches per inch of length of the pipe.Substituting in values, h = 357 ft.

TRANSPORTATION/INSTALLATION —MAXIMUM LINE LOAD ON PIPE

The second step in design is resistance to loadsimposed on the pipe during transportation andinstallation. The most common load is diametral F-load. See Figure 2-2. This load occurs when pipesare stacked or when soil is compacted on the sidesor on top of the pipe as shown.

If yield strength of the pipe material is exceeded dueto the F-load, either the pipe wall will crack or thecross section of the pipe will permanently deform.Either of these deformations (a crack is adeformation) may be unacceptable. So yieldstrength may possibly be a performance limit eventhough the ring does not collapse.

For some plastic materials, including mild steel,design for yield strength is overly conservative. Sowhat if yield strength is exceeded? A permanentdeformation (dent) in the ring is not necessarily pipefailure. In fact, the yield strength was probablyexceeded in the process of fabricating the pipe.

Some pipe manufacturers limit the F-load based ona maximum allowable ring deflection, d = ∆/D,where ∆ is the decrease in mean diameter D due toload F. Some plastics have a memory for excessivering deflection. In service, failure tends to occurwhere excess ive ring deflection occurred beforeinstallation. Increased ring stiffness decreases ringdeflection. It is not inconceivable that the ring canbe so flexible that it cannot even hold its circular

shape during placement of embedment. Oneremedy, albeit costly, is to hold the ring in shape bystulls or struts while placing embedment. It may beeconomical to provide enough ring stiffness to resistdeflection while placing the embedment. In anycase, ring deflection is a potential performance limitfor transportation/installation of pipes.

So two analyses are required for transportion andinstallation, with two corresponding performancelimits: yield strength, and ring deflection. See Figure2-3. In general, yield strength applies to rigid pipessuch as concrete pipes, and ring deflection applies toflexible pipes. See Figure 2-4.

Yield Strength Performance Limit

To analyze the yield strength performance limit,based on experience, pertinent fundamentalvariables may be written as follows:

fv's, Fundamental bd's, BasicVariables DimensionsF = transportation/installation FL-1

load (concentrated line load per unit length of pipe),

D = mean diameter of the pipe, LI = moment of inertia of the wall L3

cross section per unit length of pipe,

c = distance from the neutral axis Lof the wall cross section to the most remote wall surfacewhere the stress is at yield point.

S = yield strength of pipe wall FL-2 material

5 fv's - 2 bd's = 3 pi-terms.

The three pi-terms may be written by inspection. Atypical set is: (F/SD), (c/D), and (I/D3). This is onlyone of many possible sets of pi-terms. D is arepeating variable. Note that the pi-terms areindependent because each contains at least onefundamental variable that is not contained in any ofthe other pi-terms. All are dimensionless. Theinterrelationship of these three pi-terms can be


found either by experimentation or by analysis. Anexample of class ical analysis starts with circum-ferential stress σ = Mc/I where M is the maximumbending moment in the pipe ring due to load F. Butif stress is limited to yield strength, then S = Mc/Iwhere M = FD/2π based on ring analysis byCastigliano's theorem. See Appendix A, Table A-1.M is the maximum moment due to force F. Becauseit occurs at the location of F, there is no added ringcompression stress. Substituting in values andrearranging the fundamental variables into pi-term,

(F/SD) = 2π(D/c)(I/D3)

The three pi-terms are enclosed in parentheses. Disregarding pi-terms,

F = 2πSI/cD = F-load at yield strength, S.

For plain pipes, I = t3/12 and c = t/2 for which, I/c =t2/6 and, in pi-terms:

(F/SD) = π(t/D)2/3

Disregarding pi-terms,

F = πSt2/3D = F-load at yield strength S for plainpipes (smooth cylindrical surfaces). The modulus ofelasticity E has no effect on the F-load as long asthe ring remains circular. Only yield strength S is aperformance limit.

Ring Deflection Performance Limit

If the performance limit is ring deflection at theelastic limit, modulus of elasticity E is pertinent.Yield strength is not pertinent. For this case,pertinent fundamental variables and correspondingbasic dimensions are the following:

fv's, Fundamental bd's, Basic Variables Dimensionsd = ring deflection = /D -D = mean diameter of the L

pipeF = diametral line load FL-1

per unit length of pipeEI = wall stiffness FL

per unit length of pipe

where:∆ = decrease in diameter due to the F-load,E = modulus of elasticity,t = wall thickness for plain pipe,I = moment of inertia of wall cross section per

unit length of pipe = t3/12 for plain pipe.

4 fv's - 2 bd's = 2 pi-terms.

Two pi-terms, by inspection, are (d) and (FD2/EI).Again, the interrelationship of these pi-terms can befound either by experimentation or by analysis.Table 5-1 is a compilation of analyses of ringdeflections of pipes subjected to a few of thecommon loads. From Table A-1, ring deflection dueto F-loads is,

(d) = 0.0186 (FD2/EI) . . . . . (2.2)

This equation is already in pi-terms (parentheses).For plain pipes, for which I = t3/12 and c = t/2, thisequation for ring deflection is:

(d) = 0.2232 (F/ED) (D/t)3

The relationship between circumferential stress andring deflection is found by substituting from Table A-1, at yield stress, F = 2πSI/cD, where S is yieldstrength and c is the distance from the neutralsurface of the wall to the wall surface. Theresulting equation is:


(d) = 0.117 (σ /E) (D/c) . . . . . (2.3)

For plain pipes,

(d) = 0.234 (σ E) (D/t)

Note the introduction of a new pi-term, (σ /E). Thisrelationship could have been found byexperimentation using the three pi-terms inparentheses in Equation 2.3. Ring deflection at yieldstress, S, can be found from Equation 2.3 by settingσ = S. If ring deflection exceeds yield, the ring doesnot return to its original circular shape when the F-load is removed. Deformation is permanent. This isnot failure, but, for design, may be a performancelimit with a margin of safety.

The following equations summarize design of thepipe to resist transportation/installation loads.

For transportation/installation, the maximumallowable F-load and the corresponding ringdeflection, d, when circumferential stress is atyield strength, S, are found by the followingformulas.

Ring Strength

(F/SD) = 2π (D/c) (I/D3) . . . . . (2.4) For plain pipes, (F/SD) = π(t/D)2/3

Resolving, for plain pipes,

F = πSD(t/D)2/3.

Ring Deflection where d = (∆/D) due to F-load,is given by:

d = 0.0186 (FD2/EI), in terms of F-load . . . (2.5) d = 0.117 (σ /E) (D/c), in terms of stress, σ , or

d = 0.234(S/D)(D/t), for plain pipes with smoothcylindrical surfaces in terms of yield strength S.

Steel and aluminum pipe industries use an F-loadcriterion for transportation/installation. In Equation2.2 they specify a maximum flexibility factor FF =D2/EI. If the flexibility factor for a given pipe is lessthan the specified FF, then the probability oftransportation/installation damage is statistically lowenough to be tolerated.

For other pipes, the stress criterion is popular.When stress σ = yield strength S, the maximumallowable load is:

F = 2πSI/cD

For walls with smooth cylindrical surfaces,

F = πSt2/3D

In another form, for plain walls, the maximumallowable D/t is:

(D/t)2 = πSD/3F

For the maximum anticipated F-load, i.e. at yieldstrength, the minimum wall thickness term (t/D) canbe evaluated. Any safety factor could be small —approaching 1.0 — because, by plastic analysis,collapse does not occur just because thecircumferential stress in the outside surfacesreaches yield strength. To cause a plastic hinge(dent or cusp) the F-load would have to be increasedby three-halves.

Plastic pipe engineers favor the use of outsidediameter, OD, and a classification number called thedimension ratio, DR, which is simply DR = OD/t =(D+t)/t where D is mean diameter. Using thesedimensions, the F-load at yield is:

F = πSt/3(DR-1)

If the F-load is known, the required dimension ratioat yield strength is:

DR = (πSt/3F) + 1


Example

Unreinforced concrete pipes are to be stacked forstorage in vertical columns on a flat surface asindicated in Figure 2-2. The load on the bottom pipeis essentially an F-load. The following information isgiven:

ID = 30 inches = inside diameter,OD = 37.5 in = outside diameter,γ = 145 lb/ft3 = unit weight of concrete,F = 3727 lb/ft = F-load at fracture

+ s from tests where,s = + 460 lb/ft = standard deviation of the

ultimate F-load at fracture of the pipe.

a) How high can pipes be stacked if the F-load islimited to 3000 lb/ft? From the data, the weight ofthe pipe is 400 lb/ft. The number of pipes high in thestack is 3000/400 = 7.5. So the stack must belimited to seven pipes in height.

b) What is the probability that a pipe will break if thecolumn is seven pipes high? The seven pipe load atthe bottom of the stack is 7(400) = 2800 lb/ft. w =3727 - 2800 = 927 lb/ft which is the deviation of theseven-pipe load from the F-load. From Table 1-1,the probability of failure is 2.2% for the bottompipes. For all pipes in the stack, the probability isone-seventh as much or 0.315%, which is onebroken pipe for every 317 in the stack.

c) What is the circumferential stress in the pipe wallat an average F-load of 3727 lb/ft? From Equation2.4, F = πSD(t/D)2/3 where S = yield strength D/t =9, D = 51 inches. Solving, σ = 471 psi. This is goodconcrete considering that it fails in tension.

EXTERNAL PRESSURE —MINIMUM WALL AREA

Consider a free-body-diagram of half the pipe withexternal pressure on it. See Figure 2-4. Thevertical rupturing force is P(OD) where P is theexternal radial pressure assumed to be uniformly

distributed. OD is the outside diameter. Theresisting force is compression in the pipe wall, 2σ A,where σ is the circumferential stress in the pipe wall,called ring compression stress. Equating therupturing force to the resisting force, with stress atallowable, S/sf, the resulting equation is:

σ = P(OD)/2A = S/sf . . . . . (2.6)

This is the basis for design. Because of itsimportance, design by ring compression stress isconsidered further in Chapter 6.

The above analyses are based on the assumptionthat the ring is circular. If not, i.e., if deformationout-of-round is significant, then the shape of thedeformed ring must be taken into account. Butbasic deformation is an ellipse. See Chapter 3.

Example

A steel pipe for a hydroelectric penstock is 51inches in diameter (ID) with wall thickness of 0.219inch. It is to be buried in a good soil embedmentsuch that the cross section remains circular. Whatis the safety factor against yield strength, S = 36 ksi,if the external soil pressure on the pipe is 16 kips/ft2?For this pipe, OD = 51.44 inches, and A = t = 0.219inch. At 16 ksf, P = 111 psi. Substituting intoEquation 2.6, the safety factor is sf = 2.76. The soilpressure of 16 ksf is equivalent to about 150 feet ofsoil cover. See Chapter 3.

PROBLEMS

2-1 What is the allowable internal pressure in a 48-inch diameter 2-2/3 by 1/2 corrugated steel pipe, 16gage (0.064 inch thick)? (P' = 48.4 psi)

Given: D = 48 inches = inside diameter,t = 0.064 in = wall thickness,A = 0.775 in2/ft [AISI tables],S = 36 ksi = yield strength,


E = 30(106) psi,sf = 2 = safety factor.

2-2 What is the allowable internal pressure if areinforced conc rete pipe is 60 inch ID and has twocages comprising concentric hoops of half-inch steelreinforcing rods spaced at 3 inches in the wall whichis 6.0 inches thick? (P' = 78.5 psi)Given:S = 36 ksi = yield strength of steel,sf = 2 = safety factor,Ec = 3(106) psi = concrete modulus,Neglect tensile strength of concrete. 2-3 What must be the pretension force in the steelrods of Problem 2-2 if the pipe is not to leak atinternal pressure of 72 psi? Leakage through haircracks in the concrete appears as sweating.

(Fs = 2.9 kips) 2-4 How could the steel rods be pretensioned inProblem 2-3? Is it practical to pretension (or posttension) half-inch steel rods? How about smaller

diameter, high-strength wires? What about bond?How can ends of the rods (or wires) be fixed?

2-5 What is the allowable fresh water head (causinginternal pressure) in a steel pipe based on thefollowing data if sf = 2? (105 meters)ID = 3.0 meters,t = 12.5 mm = wall thickness,S = 248 MN/m2 = 36 ksi yield strength.

2-6 What maximum external pressure can beresisted by the RCP pipe of Problem 2-2 if the yieldstrength of the concrete in compression is 10 ksi,modulus of elasticity is E = 3000 ksi, and the internalpressure in the pipe is zero? See also Figure 2-5.

(P = 52 ksf, limited by the steel)

2-7 Prove that T = Pr for thin-walled circular pipe.See Figure 2-4.T = ring compression thrust,P = external radial pressure,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.


Anderson, Loren Runar et al "RING DEFORMATION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

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.


CHAPTER 3 RING DEFORMATION

Deformation of the pipe ring occurs under any load.For most buried pipe analyses, this deformation issmall enough that it can be neglected. For a fewanalyses, however, deformation of the ring must beconsidered. This is particularly true in the case ofinstability of the ring, as, for example, the hydrostaticcollapse of a pipe due to internal vacuum or externalpressure. Collapse may occur even though stresshas not reached yield strength. But collapse canoccur only if the ring deforms. Analysis of failuresrequires a knowledge of the shape of the deformedring.

For small ring deflection of a buried circular pipe, thebasic deflected cross section is an ellipse. Considerthe infinite medium with an imaginary circle shownin Figure 3-1 (top). If the medium is compressed(strained) uniformly in one direction, the circlebecomes an ellipse. This is easily demonstratedmathematic ally. Now suppose the imaginary circleis a flexible ring. When the medium is compressed,the ring deflects into an approximate ellipse withslight deviations. If the circumference of the ringremains constant, the ellipse must expand out intothe medium, increasing compressive stressesbetween ring and medium. See Figure 3-1 (bottom).The ring becomes a hard spot in the medium. Onthe other hand, if circumference of the ring isreduced, the ring becomes a soft spot and pressureis relieved between ring and medium. In either case,the basic deformation of a buried ring is an ellipse —slightly modified by the relative decreases in areaswithin the ring and without the ring. The shape isalso affected by non-uniformity of the medium. Forexample, if a concentrated reaction develops on thebottom of the ring, the ellipse is modified by a flatspot. Nevertheless, for small soil strains, the basicring deflection of a flexible buried pipe is an ellipse.Following are some pertinent approximategeometrical properties of the ellipse that aresufficiently accurate for most buried pipe analyses.Greater accuracy would require solutions of infiniteseries.

Geometry of the Ellipse

The equation of an ellipse in cartesian coordinates,x and y, is:

a2x2 + b2y2 = a2b2

where (See Figure 3-2):a = minor semi-diameter (altitude)b = major semi-diameter (base)r = radius of a circle of equal circumference

The circumference of an ellipse is π(a+b) whichreduces to 2πr for a circle of equal circumference.

In this text a and b are not used because the pipeindustry is more familiar with ring deflection, d.Ring deflection can be written in terms of semi-diameters a and b as follows:

d = ∆/D = RING DEFLECTION . . . . . (3.1)

where:∆ = decrease in vertical diameter of ellipse from

a circle of equal circumference,= 2r = mean diameter of the circle —

diameter to the centroid of wall cross-sectional areas,

a = r(1-d) for small ring deflections (<10%),b = r(1+d) for small ring deflections (<10%).

Assuming that circumferences are the same forcircle and ellipse, and that the vertical ring deflectionis equal to the horizontal ring deflection, area withinthe ellipse is Ae = Bab; and

Ae = πr2 (1 - d2)

The ratio of areas within ellipse and circle is:

Ar = A e / A o = ratio of areas.

See Figure 3-3.


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 A o within acircle of equal circumference) shown plotted as a function of ring deflection.


What of the assumption that the horizontal andvertical ring deflections are equal if thecircumferences are equal for circle and ellipse? Forthe circle, circumference is 2πr. For the ellipse,circumference is (b+a)(64-3R4)/(64-16R2), where Ris approximately R = (b-a)/(b+a). Only the firstterms of an infinite series are included in thisapproximate ellipse circumference. See texts onanalytical geometry. Equating circumferences of thecircle and the ellipse, and transforming the values ofa and b into vertical and horizontal values of ringdeflection, dy and dx, a few values of d y and thecorresponding dx are shown below for comparison.

Deviationdy (%) dx (%) (dy - dx)/dy

______ ______ __________ 0. 0. 0. 5.00 4.88 0.02410.00 9.522 0.04815.00 13.95 0.07020.00 18.116 0.094

For ring deflections of d = dy = 10%, thecorresponding dx is less than 10% by only4.8%(10%) = 0.48%. This is too small to besignificant in most calculations such as areas withinthe ellipse and ratios of radii.

Radii of curvature of the sides (spring lines) and thetop and bottom (crown and invert) of the ellipse are:

rx = r (1 - 3d + 4d2 - 4d3 + 4d4 - .....)

ry = r (1 + 3d + 4d2 + 4d3 + 4d4 + .....)

For ring deflection less than d = 10%, and neglectinghigher orders of d,

rx = r (1 - 3d), and ry = r (1 + 3d).

However, more precise, and almost as easy to use,are the approximate values:

rx = a2 / b = r (1 - d)2 / (1 + d)

ry = b2 / a = r (1 + d)2 / (1 - d)

An important property of the ellipse is the ratio ofradii rr = ry/rx, which is:

rr = (1 + d)3 / (1 - d)3 . . . . . (3.2)

where:rr = ratio of the maximum to minimum radii ofcurvature of the ellipse. See graph of Figure 3-4.

Measurement of Radius of Curvature

In practice it is often necessary to measure theradius of curvature of a deformed pipe. This can bedone from either inside or outside of the pipe. SeeFigure 3-5. Inside, a straightedge of known length Lis laid as a cord. The offset e is measured to thecurved wall at the center of the cord. Outside, e canbe found by laying a tangent of known length L andby measuring the offsets e to the pipe wall at eachend of the tangent. The average of these twooffsets is the value for e. Knowing the length of thecord, L, and the offset, e, the radius of curvature ofthe pipe wall can be calculated from the followingequation:

r = (4e2 + L2)/8e . . . . . (3.3)

It is assumed that radius of curvature is constantwithin cord length L. The calculated radius is to thesurface from which e measurements are made.

Example

An inspection reveals that a 72-inch corrugatedmetal pipe culvert appears to be flattened somewhaton top. From inside the pipe, a straightedge (cord)12 inches long is placed against the top, and the mid-ordinate offset is measured and found to be 11/32inch. What is the radius of curvature of the pipe ringat the top?

From Equation 3.3, r = (4e2 + L2)/8e. Substituting in

values and solving, ry = 52.5 inches which is theaverage radius within the 12-inch cord on the insideof the corrugated pipe. On the outside, the radius isgreater by the depth of the corrugations.


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 3.375

Figure 3-4 Ratio of radii,

rr = ry/rx = (1+d)3/(1-d)3,

(ry and rx are maximum and minimum radii, respectively,for ellipse) shown plotted as 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.


Ring Deflection Due to Internal Pressure

When subjected to uniform internal pressure, thepipe expands. The radius increases. Ring deflectionis equal to percent increase in radius;

d = ∆r/r = ∆D/D = 2πrε /2πr = ε

where:d = ring deflection (percent),∆r and ∆D are increases due to internal pressure,r = mean radius,D = mean diameter,ε = circumferential strain,E = modulus of elasticity = σ/ε.σ = circumferential stress = Eε = Ed.

But σ = P'(ID)/2A, from Equation 2.1,

where:P' = uniform internal pressure,ID = inside diameter,A = cross sectional area of wall per unit length.

Equating the two values for σ , and solving for d,

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.

Ring Deformation Due to External Loading

Computer software is available for evaluating thedeformation of a pipe ring due to any externalloading. Analysis is based on the energy method ofvirtual work according to Castigliano. Analysisprovides a component of deflection of some point Bon a structure with respect to a fixed point A. It isconvenient to select point A as the origin of fixedcoordinate axes — the axes are neither translatednor rotated. See Appendix A.

Example

Consider the quadrant of a circular cylinder shownin Figure 3-6. It is fixed along edge A-A-A, and isloaded with vertical line load Q along free edge B-B-B. What is the horizontal deflection of free edge Bwith respect to fixed edge A? This is a two-dimensional problem for which a slice of unit widthcan be isolated for analysis. Because A is fixed,the horizontal deflection of B with respect to A is xB

for which, according to Castigliano: xB = f (M/EI)(δM/δp)ds . . . . . (3.5)


where:xB = displacement of point B in the x-direction,EI = wall stiffness,E = modulus of elasticity,I = centroidal moment of inertia of the crosssection of the wall per unit length of cylinder,

M = moment of force about the neutral axis at C,

p = differential load (dummy load) applied atpoint B in the direction assumed for deflection,

ds = differential length along the slice, = rdθ

r = mean radius of the circular cylinder.

It is assumed that deflection is so small that radius rremains constant. It is also assumed that thedeflection 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 atB acting to the right assuming that deflection xB willbe in the x-direction. If the solution turns out to benegative, then the deflection is reversed. From thefree-body-diagram CB,

M = Qr(1-cosθ) + pr(sinθ)

M/ p = r(sinθ)

But because p approaches zero (differential),

M = Qr(1-cosθ)

ds = rdθ

Substituting into Equation 3.5,

xB = (Qr/EI) (1-cosθ) r(sinθ) rdθ

Integrating and substituting in limits of θ from 0 toπ/2,

xB = Qr3/2EI

This is one of a number of the most usefuldeflections of rings recorded in Table A-1.

PROBLEMS

3-1 A plain polyethylene pipe of 16-inch outsidediameter and DR = 15 is subjected to internalpressure of 50 psi. The surfaces are smooth andcylindrical (not ribbed or corrugated). DR(dimension ratio) = (OD)/t where t = wall thickness.Modulus of elasticity is 115 ksi. What is the ringdeflection? DR is dimension ratio = (OD)/t.

(d = 0.28%)

3-2 At ring deflection of 15%, and assuming thepipe cross section is an ellipse, what is the percenterror in finding the ratio of maximum to minimumradii of curvature by means of approximateEquation 3.2, rr = (1+d)3 / (1-d)3? (0.066%)

3-3 A 36 OD PVC buried pipeline is uncovered atone location. The top of the pipe appears to beflattened. A straight edge 200 mm long is laidhorizontally across the top and the vertical distancesdown to the pipe surface at each end of the straightedge are measured and found to be 9.2 and 9.4 mm.What is the radius of curvature of the outsidesurface of the pipe at the crown? Ry = 542 mm = 21.35 inches)

3-4 Assuming that the ring of problem 3-3 isdeflected into an ellipse, approximately what is thering deflection? Maximum ring deflection is usuallylimited to 5% according to specifications. (d = 5.74%)

3-5 What is the percent decrease in cross-sectionalarea inside the deflected pipe of problem 3-4 if thering deflection is d = 5.74%?

(0.33%)

3-6 What is the approximate ratio of maximum tominimum radii, rr, for an ellipse? (rr = 1.8)


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 / 2Et3)

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 = 3π 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)


Anderson, Loren Runar et al "SOIL MECHANICS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

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.


CHAPTER 4 SOIL MECHANICS

An elementary knowledge of basic principles of soilstresses is essential to understanding the structuralperformance of buried pipes. These principles areexplained in standard texts on soil mechanics. Afew are reviewed in the following paragraphsbecause of their special application to buried pipes.

VERTICAL SOIL PRESSURE P

For the analysis and design of buried pipes, externalsoil pressures on the pipes must be known. Verticalsoil pressure at the top of the pipe is caused by: 1.dead load, Pd , the weight of soil at the top of thepipe; and 2. live load, Pl , the effect of surface liveloads at the the top of the pipe. Figure 4-1 showsthese vertical soil pressures at the top of the pipe asfunctions of height of soil cover, H, for an HS=20truck axle load of 32 kips, and soil unit weight of 100pcf. Similar graphs are found in pipe handbookssuch as the AISI Handbook of Steel Drainage andHighway Construction Products. Soil unit weightcan be modified as necessary. Also other factorsmust be considered. What if a water table risesabove the top of the pipe, or the pipe deflects, or thesoil is not compacted, or is overcompacted? Forthese and other special cases, the followingfundamentals of soil mechanics may be useful.

If the embedment about a buried pipe is denselycompacted, vertical soil pressure at the top of thepipe is reduced by arching action of the soil over thepipe, like a masonry arch, that helps to support theload. To be conservative, arching action is usuallyignored. However, soil arching provides an addedmargin of safety. If the soil embedment is loose,vertical soil pressure at the top of the pipe may beincreased by pressure concentrations due to therelatively noncompressible area within the ring inloose, compressible soil. Pressure concentrationsdue to loose embedment cannot be ignored. Fordesign, either a pressure

concentration factor is needed, or minimum soildensity should be specified. Over a long period oftime, pressure concentrations on the pipe may bereduced by creep in the pipe wall (plastic pipes),earth vibrations, freeze-thaw cycles, wet-dry cycles,etc. The most rational soil load for design is verticalsoil pressure at the top of the pipe due to deadweight of soil plus the effect of live load with aspecification that the soil embedment be denserthan critical void ratio. Critical void ratio, roughly85% soil density (AASHTO T-99), is the void ratioat such density that the volume of the soil skeletondoes not decrease due to disturbance of soilparticles.

For design, the total vertic al soil pressure at the topof the pipe is:

P = Pd + Pl . . . . . (4.1)

where (see Figure 4-2)P = total vertical soil pressure at the level of

the top of the pipe Pd = dead load pressure due to weight of the

soil (and water content) Pl = vertical live load pressure at the level of

the top of the pipe due to surface loads.

This is a useful concept in the analysis of buriedpipes. Even rigid pipes are designed on this basis ifa load factor is included. See Chapter 12.

In fact, P is only one of the soil stresses. At a givenpoint in a soil mass, a precise stress analysis wouldconsider three (triaxial) direct stresses, threeshearing stresses, direct and shearing moduli in threedirections and three Poisson's ratios — with theadditional condition that soil may not be elastic. Theimprecisions of soil placement and soil compactionobfuscate the arguments for such rigor. Elasticanalysis may be adequate under some fewcircumstances. Superposition is usually adequatewithout concern for a combined stress analysisinvolving triaxial stresses and Poisson ratio. Basicsoil mechanics serves best.


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.


Dead Load Vertical Soil Pressure Pd

Dead load is vertical pressure due to the weight ofsoil at a given depth H. In the design of buriedpipes, H is the height of soil cover over a pipe. Total

pressure Pd is the weight of soil, including its watercontent, per unit area. See Figure 4-3.Intergranular (or effective) pressure Pd is thepressure felt by the soil skeleton when immersed inwater. The total and intergranular vertical stressesat the bottom of a submerged stratum can be relatedby the following stress equation:

σ = σ - u . . . . . (4.2)

whereσ = intergranular vertical soil stress (felt by the

soil when buoyed up by water),σ = total vertical soil stress = γ tH,u = pore water pressure = γ wH,γ t = total unit weight of soil and water,γ w = unit weight of water = 62.4 pcf.

Now consider more than one stratum of soil asshown in Figure 4-4. The total vertical dead load soilpressure Pd at the bottom of the strata is the sum ofthe loads imposed by all of the strata; i.e.,

Pd = Σγt H . . . . . (4.3)

whereγ t = total unit weight (wet weight) of soil

in a given stratum, andH = height of the same stratum.

Values of H for each soil stratum are provided bysoil borings. Values of γ t are simply the unitweights of representative soil samples including thewater content. If the soil samples are not available,from soil mechanics,

γ t = (G+Se)γw /(1+e) . . . . . (4.4)

whereG = specific gravity of soil grains, about 2.65,

S = degree of saturation = 1 when saturated,e = void ratio, from laboratory analysis,γ w = unit weight of water.

Table 4-1 is a summary of dead load soil stressesfrom which dead load pressure Pd can be found andcombined with live load pressure Pl. Live loadpressure is found from techniques described in theparagraphs to follow.

Intergranular vertical soil pressure P, at the bottomof multiple soil strata, is:

P = P - u = P - γ wh . . . . . (4.5)

whereP = vertical intergranular soil pressure,P = total dead plus live load pressures,h = height of water table above the pipe.

Total pressure is used to calculate ring compressionstress. Intergranular soil pressure is used to calcu-late ring deflection which is a function of soilcompression. As the soil is compressed, so is thepipe compressed — and in direct ratio. But soilcompression depends only on intergranular stresses.See Chapter 7.

Live Load Vertical Soil Pressure Pl

Live load soil pressure Pl is the vertical soil pressureat the top of the buried pipe due to surface loads.See Figure 4-5. For a single concentrated load W onthe surface, vertical soil pressure at point A at thetop of the pipe is:

σ = NW/H2 . . . . . (4.6)

whereW = concentrated surface load (dual-wheel)H = height of soil cover over the top of the pipeR = horizontal radius to stress, σ ,N = Boussinesq coefficient.

from the line of action of load W,N = Boussinesq coefficient = 3(H/R)5/2π.


_

__

_

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)


Figure 4-6 Chart for evaluating the vertical stress σ at a depth H below the corner A of a rectangular surfacearea loaded with a uniformly distributed pressure q. (After Newmark)


For a single wheel (or dual wheel) load, the maxi-mum stress σmax at A occurs when the wheel isdirectly over the pipe; i.e. R = 0, for which

σmax = 0.477 W/H2 . . . . . (4.7)

Load W can be assumed to be concentrated if depthH is greater than the maximum diameter or length ofthe surface loaded area.

For multiple wheel (or dual) loads, the maximumstress at point A, due to effects of all loads must beascertained. The trick is to position the wheel loadsso that the combined stress at A is maximum. Thiscan be done by trial.

The effect of a uniformly distributed surface loadcan be found by dividing the loaded surface area intoinfinitesimal areas and integrating to find the sum oftheir effects at some point at depth H. See Figure 4-6. Newmark performed such an integration andfound the vertical stress σ at a depth H belowcorner A of a rectangular area of greater length Land lesser breadth B, loaded with uniform pressureq. His neat solution is:

σ = Mq . . . . . (4.8)

where M is a coefficient which can be read on thechart of Figure 4-6 by entering with arguments L/Band B/H. If the stress due to pressure on an area isdesired below some point other than a corner, therectangular area can be expanded or subdividedsuch that point A is the common corner of a numberof areas. The maximum stress under a rectangulararea occurs below the center. See Figure 4-7. Therectangle is subdivided into four identical rectanglesof length L and breadth B as shown. The stress atpoint A is 4Mq, where M is found from theNewmark chart, Figure 4-6.

Alternatively, the Boussinesq equation can be usedwith less than five percent error if the concentratedload, Q = qBL, for each of the quadrants is assumedto act at the center of each quadrant. For this case,R = (L2+B2)/2. The resulting stress at A is 4NQ/H2

from Equation 4.6.

Example

What is the stress at point A below A' of Figure 4-8? The vertical stress σ at depth H below surfacepoint A is, by superposition:

σ = σ ' - Σ σ'" + σ "

whereσ ' = stress at corner A due to loaded area

L'xB' Σ σ'" = sum of stresses at corner A due to

loaded areas L'B" and L"B'σ " = stress at corner A due to loaded area

L"B'

Clearly, σ " due to area L"B" was subtracted twice,so must be added back once.

An occlusion in the soil mass, such as a pipe, violatesBoussinesq's assumptions of elasticity, continuity,compatibililty, and homogeneity. The pipe is a hardspot, a discontinuity. Soil is not elastic, nor homoge-neous, nor compatible when shearing planes form.Nevertheless, the Boussinesq assumptions areadequate for most present-day installation tech-niques. For most buried pipe design, it is sufficient,and conservative, to solve for Pl at the top of thepipe due to a single wheel load W at the surface byusing the Boussinesq equation with the radius R = 0.For additional wheel loads, simply add by superposi-tion the influence of other wheel loads at their radiiR.

Example

What is the maximum vertical soil stress at a depthof 30 inches due to the live load of a single axle HS-20 truck? Neglect surface paving. See Figures 4-9and 4-15. By trial, it can be shown that the point ofmaximum stress is point A under the center of onetire print. The rectangular tire prints are subdividedas shown for establishing a common corner A'. Theeffects of the left tire print and the right tire print areanalyzed separately, then combined. The length Land breadth B of each tire print are based on 104 psitire pressure. Use Newmark because H is less than3L.


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.


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.


Figure 4-10 Infinitesimal soil cube B and the corresponding Mohr circle which provides stresses on any planethrough B. Note the stresses σθ and τθ shown on the θ-plane. At soil slip, the circle is tangent to thestrength envelopes described below.

Figure 4-11 Shearing stress τ as a function of normal stress σ , showing a series of Mohr circles at soil slip,and the strength envelopes tangent to the Mohr circles.


Given:W = 32 k for HS-20 truck load (single axle),q = 104 psi,B = 7 inches,L = 22 inches,H = 30 inches.

For the left tire print:σL = 4Mq,L/B = 11/3.5 = 3.14,B/H = 3.5/30 = 0.12,From Figure 4.6, M = 0.018 and σL = 1.078 ksf

For the right tire print:σR = 2(M'-M")q, where,L'/B' = 83/3.5 = 23.7,B'/H' = 3.5/30 = 0.12,From Figure 4.6, M' = 0.02 (extrapolated)

L"/B" = 61/3.5 = 17.4,B"/H" = 3.5/30 = 0.12,From Figure 4.6, M" = 0.02 (extrapolated)

σR = r(M'- M")q = 0: σR = 0 ksf

At point A, σ = σ L + σ R; σ = 1.08 ksf

A rough check by Boussinesq is of interest becausethe results are conservatively higher and are moreeasily solved.

Given:W = 16 kips at the center of each tire print,H = 2.5 ft,RL = 0,RR = 6 ft,R/H = 2.4.From Figure 4.5, N = 0.004.

σL = 0.477 W/H2 = 1.22 ksfσR = NW/H2 = 0.01 ksf

At point A, σ = σL + σ R ; σ = 1.23 ksf

The Boussinesq solution is in error by 13.9%, but onthe high (conservative) side. Of interest is the small(negligible) effect of the right wheel load.

SOIL STRENGTH

Failure of a buried pipe is generally associated withfailure of the soil in which the pipe is buried. Theclassical, two-dimensional, shear-strength soil modelis useful for analysis. Analysis starts with an infini-tesimal soil cube on which stresses are known andthe orientation is given. The model comprises threeelements, the Mohr stress circle, orientation diagram,and strength envelopes.

Mohr Stress Circle

The Mohr stress circle is a plot of shearing stress, τ ,as a function of normal stress, σ , on all planes atangle θ through an infinitesimal soil cube B. SeeFigures 4-10 and 4-11. The sign convention iscompressive normal stress positive ( + ) andcounterclockwise shearing stress positive ( + ).The center of the circle is always on the σ -axis.Two additional points are needed to determine thecircle. They are (σx,τ xy) on a y-plane and (σ y,τyx)on an x-plane. These are known stresses on cubeB. An origin of planes always falls on the circle.τxy = -τyx from standard texts on solid mechanics.Any plane from the origin intersects the Mohrcircle at the stress coordinates acting on thatplane— which is correctly oriented if the followingprocedure is followed.

Orientation Diagram

Figure 4-10 shows infinitesimal cube B with the x-plane and y-plane identified and with the soil stressesacting on each plane. Cube B and its axes oforientation can be superimposed on the Mohr circlesuch that stress coordinates (where each plane inter-sects the Mohr circle) are the stresses on that plane.With cube B located on the Mohr circle as the originof axes, and with the axes correctly oriented, anyplane through B will intersect the Mohr circle at thepoint whose stress coordinates are the stressesacting on that plane, and all planes are correctlyoriented with respect to the original soil cube B.


Figure 4-12 Trigonometry for analysis of stresses at soil slip on shear planes, θf in cohesionless soil whichhas a soil friction angle of ϕ.


Another cube B' is shown with shearing stressesacting on it. For most pipe-soil interaction, onlyprincipal stresses are required as on cube B.

Strength Envelopes

Tangents to a series of Mohr circles plotted fromshear strength data are called strength envelopes.Shear strength circles are plotted from laboratorytests to failure (soil slip). See Figure 4-11. With thestrength envelopes known, the stresses at soil slipcan be evaluated. Suppose normal and shearingstresses are known on a specific plane at a specificpoint in a soil mass. These stresses, σ and τ, arethe coordinates of a point on the stress diagram. Ifthe point falls between the strength envelopes, soildoes not slip on that plane. But if the point fallsoutside of the strength envelopes, the soil slips onthat plane. For any cube B,If Mohr circle intersects the strength envelopes,soil slips on planes through the origin and pointsof intersection. Planes are correctly oriented.

The soil represented by Figure 4-11 has both cohe-sion (glued soil grains) and frictional resistance toshearing stress. Cohesion is the y-intercept, c.Even at zero normal stress the glue offers resistanceto shearing stress. But to this cohesion must beadded the frictional resistance which is normal stressσ times the coefficient of friction. The shearingstrength of the soil is the sum:

Strength = c + σ (tanϕ)

where ϕ is the soil friction angle and tan ϕ is theinternal coefficient of friction of the soil.

Cohesionless Soil Failure

For most buried pipes, the embedment around thepipe is specified to be cohesionless soil such as sandor gravel. For cohesionless soil, c = 0 and thestrength envelopes are idealized by straight lines asshown in Figure 4-12. At soil friction angle ϕ, asindicated, if any Mohr circle is tangent to thestrength envelopes, the soil slips at stress coordi-nates, τf and σ f , at the point of tangency. Assume

that an infinitesimal cube of soil is located at somepoint O in a pipe embedment with y-axis vertical andx-axis horizontal. The axes, shown dotted on Figure4-12, are planes through the cube. The shearingstresses on the vertical y-plane and the horizontal x-plane are zero. Therefore the normal stresses onthese planes are principal stresses, the minimum, σx,acting on the y-plane and the maximum, σy, on thex-plane. Any θ-plane through O (at angle θ) inter-sects Mohr circle at a point whose coordinates are

the normal stress σθ and shearing stress τθ on thatθ-plane.

The planes on which failure (soil slip) occurs arethose planes from O that intersect the Mohr circle atthe points of soil slip; i.e., where the Mohr circleintersects the strength envelopes. Angle θf on thecircumference intercepts the same arc as centralangle 2θf. But 2θ f = 90o + ϕ. Therefore,

θf = 45o + ϕ/2

θf is the angle of the failure plane, i.e., the plane ofsoil shear. The minimum principal stress is σ3 thefurthest point to the left on the Mohr circle. It actson a vertical y-plane shown dotted. The maximumprincipal stress is σ1 the furthest point to the right onthe Mohr circle and acts on a horizontal x-plane. Asexpected, the shearing stresses are zero on theplanes of principal stresses. σ2 is an intermediateprincipal stress at right angles to σ1 and σ 3. Criti-cality is related to σ1 and σ3, not σ2 and σ

1 or σ

2

and σ3 because the σ 1-σ 3 Mohr circle is the largestand approaches closer to the strength envelopes thaneither of the other Mohr circles. Because the planesthrough O are oriented, the failure planes in a triaxialtest specimen of cohesionless soil loaded as shown,would develop shearing planes called Lueders' linesat angles θ f. Note that the x and y axes are cor-rectly oriented and O is correctly located for thetriaxial test specimen. For most soil analyses, theprincipal stresses are horizontal or vertical. Foranalysis of failure of the embedment around aburied pipe, the relationship of the principal stresses,σ1 and σ 3 at soil slip becomes pertinent. See

Figure 4-12 from which, σ1 = X +


Figure 4-13 Analysis of stresses from the Mohr circle for cohesive soil (clay) with no frictional resistance(worst case) for which ϕ = 0.

Figure 4-14 Strength envelopes for cohesionless soil with the maximum principal stress acting horizontally ona y-plane, etc., and for which the infinitesimal soil cube is located at the right side of the Mohr circle.


Xsinϕ, and σ 3 = X - Xsinϕ. Eliminating X betweenthe two equations,

σ1 = σ 3(1+sinϕ)/(1-sinϕ). For convenience, letK = (1+sinϕ)/(1-sinϕ), then

K = σ1 /σ 3 . . . . . (4.9)

which is the ratio of maximum to minimum principalstresses at soil slip in cohesionless soil. The soil slips if σ1 > Kσ 3. σ 1 is called passiveresistance — the soil retreats from high pressure.The soil slips if σ3 < σ 1 /K. σ 3 is called activeresistance — soil advances against low pressure.

Cohesive Soil Failure

Under some circumstancess, pipes are buried incohesive soil such as clay. A saturated fat clay hasa negligibly small friction angle ϕ, but does havesignificant cohesion c. The strength envelopes areidealized by two horizontal straight lines spaced at 2cas shown in Figure 4-13. When the Mohr stresscircle is tangent to the strength envelopes, its diame-ter is 2c. The relationship between maximum andminimum principal stresses at soil slip, called deviatorstress, is,

σ1 - σ 3 = 2c . . . . . (4.10)

whereσ1 = maximum principal stressσ2 = intermediate principal stress, not criticalσ3 = minimum principal stressc = cohesion in shearing force per

unit area.

Because principal stresses are horizontal and verti-cal, cube O is oriented and superimposed on the leftside of the Mohr circle. The orientation is such thatvertical stress σ1 acts on a horizontal x-plane andthe horizontal stress σ3 acts on a vertical y-plane.The shear planes (slip), at points of tangency of theMohr circle to the strength envelopes, are at θf =45o. However, failure in clay is general shear; i.e.,viscous or plastic flow. Lueders' lines do not ap-

pear.

Soil Stress and Strength Analysis

Suppose that in cohesionless soil, at point O, themaximum principal stress σ1 is horizontal and theminimum principal stress σ3 is vertical. See Figure4-14. The vertical and horizontal planes on whichthese stresses act require that the infinitesimal cubebe superimposed on the Mohr circle with origin O atthe right side of the circle. At soil slip, σ1 /σ 3 =(1+sinϕ)/(1-sinϕ) and the shear planes are orientedas shown, i.e.,

θf = 45o - ϕ/2

In fact, the origin O may be located anywhere on theMohr circle depending on the orientation of theinfinitesimal soil cube on which the stresses act. Forthe purposes of this text, the principal stresses act onhorizontal and vertical planes.

It must be remembered that soil strength is based oneffective (intergranular) soil stresses . If the watertable is above the point at which soil strength is to becalculated, the water pressure u must be subtractedfrom total vertical stresses.

Example

Consider a point in cohesionless soil at the spring lineof a flexible pipe.

γ = 110 lb/ft3 = soil unit weight,ϕ = 30o = soil friction angle,c = 0 = cohesion,z = 10 ft = depth of soil at pipe spring lines.

a) What is the minimum stress σx of the pipe wallagainst the sidefill soil that would be required toprevent inward collapse of the pipe?

Because σy is the maximum principal stress and σ xis the minimum principal stress, at soil shear,

σ1 /σ 3 = K = (1+sinϕ)/(1-sinϕ) = 3


But σ1 = 10 ft(110 lb/ft3) = 1100 lb/ft2, so σ 3 = σ x =1100/3 = 36.7 lb/ft2 = horizontal stress against thesoil needed to prevent soil slip. It is called the activeresistance of soil as the soil advances into the pipe.If the pipe wall were to advance against the soil,maximum σx at soil slip would be σ x = 1100(3) =3300 lb/ft2. Because the soil retreats before theadvancing pipe wall, its resistance is passive resis-tance of soil.

b) Assume the soil backfill is saturated clay.γ = 139 lb/ft3,ϕ = 0,c = 420 lb/ft2,z = 10 ft = depth of soil at spring line of the pipe.

What is the minimum stress, σx of the pipe againstthe soil in order to prevent inward collapse of thepipe wall? Because the soil is clay, σ1 - σ 3 = 2c =840 lbft2. But σ1 = 130(10) = 1300 lb/ft2. At activesoil resistance,σ3 = σ x = 1300 - 840 = 460 lb/ft2

At passive soil resistance,σx = 1300 + 840 = 2140 lb/ft2

PROBLEMS

4-1 What is the vertical effective (intergranular) soilpressure at the level of the top of a pipe if the watertable is 17 ft above the top of the pipe and the soilsurface is 47 ft above the top of the pipe? The soilis dune sand for which:G = 2.7 = specific gravity of grains,e = 0.7 = void ratio,S = 0.1 = degree of saturation of soil above the

water table. (4.11 ksf)

4-2 What is the maximum principal stress σ1 actingvertically at soil slip in granular soil if horizontalstress is 600 gr/cm2 and if ϕ = 30o? (1800 gr/cm2)

4-3 What is the maximum principal stress actingvertically at failure in clay if ϕ = 0, σ3 = 4000 Paacting horizontally, and cohesion is c = 2800 Pa?

(σ1 = 9.6 kPa)

4-4 What is the total pressure at the top of a buriedpipe if height of soil cover is 3 ft, unit weight of soilis 140 pcf, and live load is an HS-20 truck (W = 16kips)? See Figure 4-15. (P = 1.3 ksf)

4-5 In Problem 4-4, an external water table rises tothe ground surface. What is the effective (intergra-nular) vertical soil pressure at the top of the buriedpipe? (P = 1.1 ksf)

4-6 Due to internal vacuum, the spring line of aburied flexible pipe bears against the sidefill sandwith stress of 3200 lb/ft2. What is minimum heightof soil above the spring line to prevent vertical pipecollapse? Soil friction angle is 35o and unit weight is115 lb/ft3. (H = 7.5 ft)

4-7 Suppose that a water table rises to the groundsurface in Problem 4-6. What is minimum soil coverto prevent vertical pipe collapse? Assume that thespecific gravity of the soil grains is 2.65. Soil frictionangle is 35o. (H = 16.5 ft)

4-8 If horizontal soil pressure is 3200 psf at passivesoil resistance, c = 0, and ϕ = 35o, what is the angleθf of the slip planes? (θ f = 27.5o)

4-9 Complete Table 4-2 for maximum verticaleffective stresses at depths of 5, 10, and 20 metersfor dead loads and for each of two surface liveloads, concentrated Q-load and uniformly distributedq-load, both of which are 100 kN loads. The surfacearea is a rectangle 4 x 10 meters.

4-10 In Problem 4-9, what are the vertical live loadstresses at 5, 10 and 20 m below the surface, at ahorizontal radius of 5 meters from the vertical line ofaction of the concentrated Q-load?

4-11 In Problem 4-9, what are the vertical live loadstresses at 5, 10, and 20 m below the q-load, but atpoints laterally 2 meters outboard from mid-length ofthe long side?

4-12 What is the error in Figure 4-1 for an H-20 liveload below the center of a 36x40-inch surface areaat a height of cover of 3 ft? (< 10%)


Anderson, Loren Runar et al "PIPE MECHANICS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

46 STRUCTURAL MECHANICS OF BURIED PIPES

Figure 5-1 Nomenclature used in the ring analysis of buried pipes.


CHAPTER 5 PIPE MECHANICS Theoretical mechanics is the analysis of forces andtheir effects on materials. In the case of buriedpipes, forces are statically indeterminate, and areoften indeterminable because the soil is not uniform.Internal pressures, if any, may also beindeterminable. Unknown soil loads are mitigated bythe ability of soil to arch over the pipe and relieve thepipe of some load. The effect of force on materialis deformation. Traditionally, force per unit area isstress, and deformation per unit length is strain.Design is the analysis of stresses or strains to makesure they do not exceed the maximum allowable. Maximum allowable occurs at performance limits.In the case of buried pipes, performance limit isusually excessive deformation; i.e., that deformationbeyond which performance is not acceptable.Excessive deformations include: buckling,collapsing, cracking, and tearing, as well asexcessive deformation of the pipe. Most useful,then, is the analysis of deformation. Somedeformations can be related to stresses such thatclassical stress theories can be used. Stress theoriesare more responsive to loads than are strain theories.But strain and strain energy theories are moreresponsive to deformation performance limits.Traditional stress theories are presented in this textwherever they contribute to understanding. Ingeneral, stresses are analyzed by theories ofelasticity. Clearly, performance of pipes is notlimited to the range of elasticity. The followingcomprises theoretical analyses of stresses, strains,and deformations.

Some basic simplifications are justified because ofinevitable imprecisions such as deviations of thegeometry, non-uniformities of the soil andindeterminable loads. Combined stress analysis isnot justified. Therefore, longitudinal analysis, andring analysis are each considered independently ofthe other. Concentrated loads are the worst caseloads, because loads are, in fact, distributed over afinite area. Ring instability is the worst case ofcollapse analysis because instability is reduced bythe interaction of ring stiffness and longitudinalstiffness.

LONGITUDINAL ANALYSIS

The two basic longitudinal analyses are axial andflexural. Axial analysis considers the longitudinaleffects of temperature changes, catenary tension,thrust at valves and elbows, and the Poisson effectof radial pressure. Flexural analysis considers thelongitudinal effect of beam bending.

Longitudinal beam analysis of buried pipes followsclassical procedures. Depending on the loads(weight of the pipe and its contents plus soil loads)and the reactions (high points or hard spots in thebedding), bending moment diagrams can be drawn,and deformations, strains, and stresses can beevaluated. Longitudinal analysis is discussed inChapter 14. For most buried pipes, either themanufacturer provides adequate longitudinalstrength, or the pipe is so flexible longitudinally thatit relieves itself of stress. Corrugated pipes, forexample, relieve themselves of longitudinal stressesby changing length and by beam bending thatconforms with uneven beddings. Lengths of pipesections are limited by manufacturers in order toprevent longitudinal failure.

RING ANALYSIS

Ring analysis considers stress, strain, deformation,and stability of the cross section (ring) cut by a planeperpendicular to the axis of the pipe. See Figure 5-1.

Stress

Stress theory provides an acceptable analysis forrigid rings. Deformation and strain theories providebetter analyses for flexible rings.

Circumferential stresses comprise: 1. hoop or ringcompression stress, and 2. moment stress or itsequivalent ring deformation stress. Circumferentialstress analysis is analogous to the stress analysis


Figure 5-2 Comparison of stress analyses of a short column and a pipe ring.


of an eccentrically-loaded short column, see Figure5-2, for which, within the elastic limit,

σ = F/A + Mc/I

whereσ = maximum stress in the most remote fibers,F = compressive load on the column,M = moment acting on the cut section,I/c = section modulus of wall.

For a pipe ring, by theory of elasticity,

σ = Pr/A + Mc/I . . . . . (5.1)

whereP = radial pressure,r = mean radius of the pipe,A = wall cross-sectional area per unit length,M = moment acting on the wall cross-section,I/c = section modulus of the wall per unit length .

For rigid rings, Equation 5.1 applies. Thrust, T, (=Pr) and moment, M, are functions of the soil loading.See Appendix A for values of T and M.

Example

Find stres, σ , at spring line of a ring loaded as shownin Figure 5-6a. From Appendix A, T = Pr and M =Pr2/4. Let m = r/t = ring flexibility. Substituting intoEquation 5.1, σ = Pm(1 + 3m/2).

For flexible rings, Equation 5.1 is more useful ifflexural stress Mc/I is written in terms of change inradius of the ring. From theory of elasticity,M/EI = dθ = 1/r - 1/r' where dθ is change in radiusof curvature. See Figure 5-3. Solving for M andsubstituting into Equation 5.1,

σ = Pr/A + Ecdθ . . . . . (5.2)

wheredθ = θ - θ ' = 1/r - 1/r',E = modulus of elasticity,c = distance from NS to the most remote fiber.

For a plain (bare) pipe, Equation 5.2 becomes,

σ = Pm + (E/m) (r'-r)/2r' . . . . . (5.3)

wherem = r/t = wall flexibility,r = mean radius,t = wall thickness.

Strain

Within the elastic limit, strain is ε = σ /E. Therefore,Equation 5.2 can be written as,

ε = Pr/AE + cdθ . . . . . (5.4)

where ε = circumferential strain in the surfaces of the

pipe wall,dθ = 1/r - 1/r'.

For a plain pipe with wall thickness, t,

ε = Pm/E + (r'-r)/2mr' . . . . . (5.5)

Deformation

For a flexible ring, deliberate control of ringdeformation is usually a better option than control ofsoil pressure. The best control is specification ofmaximum allowable ring deformation.

Where it is necessary to predict ring deformation,the basic ring deformation of a buried circular pipeis from circle to ellipse. See Figure 5-4.

Ring deflection from circle to ellipse decreasesradius of curvature at B by, dθ = 1/rx-1/r.

But from Figure 3-2,

rx = r(1-d)2/(1+d)

for small ring deflections — say less than 10%.


Figure 5-4 First mode ring deflection from a circle to an ellipse. Ring deflection is a function of the verticalsoil strain (compression) in the sidefill.


Substituting, and neglecting higher orders of d,for ellipse, by elastic analysis at spring lines

σ = Pr/A + (Ec/r)3d/(1-2d) . . . . . (5.6)comp deformation

term term

whered = ∆/D = ring deflection =∆y/D ~ ∆x/D.

For homogeneous plain pipe, wall thickness t, andmean radius r; m = r/t = wall flexibility. Stress is,

σ = Pm + 3Ed/2m(1-2d) . . . . . (5.7)

It is noteworthy from Equation 5.6 that thedeformation term is insignificant at small values of d(when maximum ring deflection is specified). If thepipe wall can yield without fracture (such as metalsand plastics), wall buckling or crushing does not occuruntil ring compression stress reaches yield strength.The only exception is instability caused by externalpressure when the ring is not constrained to nearlycircular shape. For flexible pipes, stability analysis isstiffness analysis — not stress analysis.

Stability

Ring stability is resistance to progressive (runaway)deformation due to persistent loads. The persistentloads may be caused by internal pressure, beamloading, or external pressure. Failure is usuallysudden and catastrophic. Failure due to internalpressure is runaway rupture because, at yield stress,the diameter of the ring increases and wall thicknessdecreases. Failure due to beam loading is fracture orbuckling of the pipe wherever the bending moment isexcessive. Failure due to external pressure iscollapse. The loading for progressive deformationmust be persistent; i.e., the load must bear against thepipe even as the pipe deforms away from the load.Persistent loads include constant or intermittentinternal pressure or vacuum, and gravity loads thatare not relieved by soil arching. The term, instability, most often implies collapse dueto external pressure, P. See Figure 5-5. Classical

analyses are available. For example, a non-constrained, circular, flexible ring will collapsecatastrophically under pressure if,

Pr3(1-ν2)/EI = 3, or PD3(1-ν2)/EI = 24

where ν = Poisson ratio. For most pipe design, third-dimensional effects enter in such that the effect ofν 2 is reduced and may be neglected. Conservatively,

Pr3/EI = 3 and PD3/EI = 24 . . . . . (5.8)

wherePr3/EI = ring stability number,P = critical uniform external pressure,r = mean radius = D/2,EI = wall stiffness per unit length of pipe,EI/r3 = ring stiffness,F/∆ = pipe stiffness,S = strength.

F/∆ = 53.77 EI/D3 = 6.72 EI/r3 . . . . . . . . . . . (5.9)

where F/∆, called pipe stiffness by the plastic pipeindustries, is the slope of the load-deflection diagramfrom a parallel plate test. See Figure 5-5. Thedeflected cross section is not an ellipse.

Ring stiffness, EI/r3, is that property of a circular ringwhich resists collapse caused by external pressure.EI/r3 is related to elasticity E — not to strength S.In that respect, it differs from section modulus andarc modulus, which are related to strength, SI/c.Ring stiffness can either be calculated or measuredfrom a parallel plate test in which a plot of F vs provides the slope F/∆, called pipe stiffness, fromwhich

EI/r3 = 0.149 F/∆. Classical unburied analysis is not responsive toburied pipe performance. If the pipe is buried(constrained), soil support has a major effect onstability. Pressure on the pipe is not uniform.Moreover, the buried pipe will be out-of-round. Itmay even have initial out-of-roundness, calledovality. For these reasons, stability is consideredfurther in Chapter 10.


~

Figure 5-5 Notation used in deriving the equation for external pressure, P, at collapse of a flexible, circular ring,based on pipe stiffness, F/∆ , from a parallel plate test (or three-edge bearing test).

Figure 5-6 Two soil loading assumptions for the analysis of rigid pipes.


Example

A steel pipe has a 51-inch mean diameter. Wallthickness is 0.187 inch, E = 30,000 ksi, and yieldstrength is 42 ksi. Neglecting Pm in Equation 5.3,what is the deformed radius of curvature r' at tensileyield stress on the inside surface? From Equation 5.3,σ = E(r'-r)/2mr'. Solving, r' = 41.25 inches.

What is r' at tensile yield on the outside surface?Equation 5.3 now becomes σ = E(r-r')/2mr'. Solving,r' = 18.45 inches.

Plastic Performance Limits

The limit of normal stress, σ , is strength S. Fordesign, σ = S/sf. Performance limit is yield stress for:internal pressure, ring compression, and longitudinalstress. However, for instability, the performance limitis ring collapse, which is a function of ring stiffness.

Ring stiffness, EI/r3, is derived from the theory ofelasticity. It is conservative. When mitigation orfailure analysis is needed, plastic theory may be moreappropriate. Plastic theory can be related to elastictheory by moment resistance as follows.

See Figure 5-7. In the center is a cross section(cross-hatched) of an element of pipe wall ofthickness t and of unit length along the pipe, locatedat the top of the pipe, point A. On the left is theelastic stress distribution due to ring deflection. Theresisting moment is Me = SI/c, where, I/c = sectionmodulus, and S = yield stress.

On the right is the plastic stress distribution. Theresisting moment is Mp = 3SI/2c. Elastic moment,Me, at surface yield stress, is not collapse. Once thesurface starts to yield, stresses within the wallthickness increase to the yield strength as shown atthe right of Figure 5-7. Performance limit is theidealized plastic moment,

Mp = 3Me /2 . . . . . (5.10) The ring is now buckling (plastic flow). Collapse is inprocess.

Corrugated Pipes:

Figure 5-8 is the wall of a corrugated pipe. Equation5.10 may be used as demonstrated in the followingexample.

Example

Figure 5-8 is a typical 6x2 or 3x1 corrugation.Values of section modulus are listed in industrymanuals — but are all based on elastic theory.What is the relationship of plastic theory to elastictheory for this corrugation?

For a single corrugation, section modulus, I/c, is notchanged if the corrugation is compressed horizontallyas shown to the right of the corrugation. But thesection modulus for the compressed corrugation isessentially the same as the rectangular equivalentsection for which Mp /Me = 1.5. From exactanalyses, the moment ratio can be as much as Mp

/Me = 1.7, but for design, it is conservative to hold toa ratio of 1.5.

Ribbed and Reinforced Pipes:

Plastic analysis of ribbed pipe walls follows the sameprocedure as the plain walls of Figure 5-7, butrequires location of the neutral surface, NS, andevaluation of moment of inertia, I, of the crosssection. See texts on mechanics of solids.

Plastic analysis of reinforced pipe walls can berelated to elastic analysis by transforming the pipewall cross section into its equivalent section in onematerial or the other. Procedure is then the same asfor ribbed pipe walls. The procedure fortransformation to equivalent section is described intexts on solid mechanics and on reinforced concretedesign. However, reinforced concrete pipescomprise steel which is somewhat plastic, andconcrete, which is not plastic. Therefore, plasticanalysis of reinforced concrete pipes is ofquestionable value. In general, rigid pipes should bedesigned by theories of elasticity, not theories ofplasticity.


Figure 5-7 Flexural stresses on a longitudinal section(cross-hatched) of pipe wall at A showing maximumelastic stress distribution to the left, and maximumplastic stress distribution to the right. The plasticresisting moment is 1.5 times the elastic resistingmoment.

Figure 5-8 Cross section of corrugated pipe wall,showing how it can be compressed horizontally to anequivalent rectangular section for evaluating sectionmodulus I/c.

Values of section modulus, I/c,per length of the pipe,are listed in industry manuals.


PROBLEMS

5-1 A thin-wall pipe is initially an ellipse with ringdeflection, d. What is the maximum moment in thering due to rerounding?

5-2 Find the moment at B on the rigid pipe of Figure5-6b if the vertical soil pressure is P and thehorizontal soil pressure is P/K; i.e., active horizontalsoil pressure (Appendix A).

5-3 If t = D/10 in Figure 5-6a, where and what isthe maximum tangential normal stress? Include ringcompression as well as flexural stresses.

(40 P at A, 43 P at B)

5-4 For a diametral line load F on a rigid pipe ofwall thickness t < D/10, what and where is themaximum tangential normal stress? Include ringcompression. (σ = 9.55F/t at location of load F)

5-5 From Equation 5.8 and the parallel plate load ofAppendix A, show that critical pressure on a flexiblecircular ring is P = 0.446F/∆, where F/∆ is the slopeof the F/-∆ diagram from a parallel plate test.

5-6 What is the maximum strain in a pipe ring if D/t= 20 and the ring is deflected from a circle into anellipse with ring deflection of d = 10%? Neglect thering compression strain. Consider only flexuralstrains. (ε=1.9%)

5-7 Find ring deflection at yield stress in a steel pipeif the ring deflects into an ellipse. Assume that ringcompression stress is negligible. (d = 9.9%) Given:D = 51 incht = 0.187 inchE = 30,000 ksiSy = 42 ksi

What can be said about ring deflection at plastichinging? Unstable? Indeterminable?

5-8 What is the external pressure on the pipe ofProblem 5-7 at collapse, if the pipe is not buried?

(2.9 psi) 5-9 What is external pressure at collapse of anunburied PVC pipe with ribs? The pipe is stiffenedby external ribs, Figure 5-9. (215 kPa)

Given:ID = 450 mm, smooth bore,t = 4 mm, wall thickness,OD = 500 mm, over the ribs,E = 3.5 GPa, modulus of elasticity.Ribs are 4 mm thick spaced at 50 mm.

5-10 What would be the external pressure atcollapse of the PVC pipe of Problem 5-9 if ID = 450mm and t = 4 mm, but without ribs? (4.8 kPa)

Figure 5-9. Wall cross section of an externally ribbed PVC pipe.


C−

Anderson, Loren Runar et al "RING STRESSES"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 6-1 Free-body-diagrams for analyzing hoop stresses in rigid and flexible rings with initial ovality.

Figure 6-2 Stress distribution across the wall of a thick-walled cylinder due to internal pressure (top) andexternal pressure (bottom). For this example, the outside diameter equals twice the inside diameter.


CHAPTER 6 RING STRESSES

For preliminary buried pipe design, stress analysisrequires only fundamental principles of pipemechanics. However, analysis in greater depth isoften essential. In all cases, performance limit isdeformation; i.e., rupture, wall crushing, wallbuckling, ring deflection, etc. In this chapter,performance limit is analyzed in terms of stress atthe point of excessive deformation.

Hoop Stress

Hoop stress due to internal pressure, P', in a thin-walled circular ring, from Equation 2.1, is:

σ = P'(ID)/2A . . . . . (6.1)

whereσ = hoop stress; i.e., circumferential stress in a

thin-walled pipe for which D/t > 10, ID = inside diameter,A = cross-sectional area of the pipe wall per unit

length of pipe = t for plain wall pipe,t = wall thickness for plain pipe,c = distance from neutral surface, of the wall to

the most remote surface,dθ = change in curvature = 1/r - 1/ro,r = deformed radius, ro = original radius,E = modulus of elasticity.

Now suppose that the pipe is not circular — it is out-of-round before installation — called ovality. SeeFigure 6-1. In the case of a rigid ring, the maximumhoop stress occurs at B on the maximum diameter(ID). This horizontal (ID) is called the span. If thelong axis is vertical, ID must be vertical.

In the case of a flexible pipe, Figure 6-1, themaximum hoop stress acts on the maximumdiameter. But hoop stress tends to round the pipe.If the ring is initially deformed, the circumferentialstress due to internal pressure, P', is, from Equation5-2, the sum of hoop stress and flexural stress; i.e.,σ = P'r/A + Ecdθ.

Because dθ is a function of loads on the ring, whichcould be complex, analyses can be complicated.However, for plastic pipes and elasto-plastic (metal)pipes, rupture does not occur until average hoopstress reaches yield. Therefore, the flexuralcomponent of stress is not an issue. Flexure adds tothe hoop stress at one surface of the wall, butsubtracts from it at the other. For flexible thin-walled pipes that are out-of-round, internal pressuretends to round the ring causing soil pressureconcentrations. However, most pipes are nearenough to circular when buried, that rerounding isnot an issue.

Circular Thick-walled Pipes

Analyses of circular thick-walled cylinders can befound in texts on solid mechanics. Thick-walledcylinders subjected to internal or external pressure,feel maximum tangential s tress , σ , on the inside ofthe wall. See Figure 6-2.

Internal Pressure, P':

σi = P'(a2 + b2)/(b2 - a2) . . . . . (6.2)tension on the inside surface,

σo = 2P'a2/(b2 - a2)tension on the outside surface,σav = P'a/t = average tangential stress, where subscripts i, o, and av, refer to inside, outside,and average; and:

σ = tangential stress (hoop stress, tension),b = outside radius,a = inside radius,P' = internal pressure,t = wall thickness = b-a.

External Pressure, P:

σi = 2Pb2/(b2 - a2) . . . . . (6.3)compression on the inside surface,


σo = P(b2 + a2)/(b2 - a2)

compression on the outside surface,

σav = Pb/t = average tangential stress.

Example 1

Figure 6-2 (top) shows the cross section of a thick-walled, high pressure pipe. What is the maximumtangential stress , σ , if OD = 2ID? From Equation6.2, σi = 5P'/3 = 5/3 rds σav.

If pressure is external, the maximum tangentialstress is still on the inside surface. See Figure 6-2(bottom). Because of the greater outside diameter,stress σi is greater due to external pressure than dueto an equal internal pressure. However,compressive yield strength is greater than tensileyield strength for many pipes.

Example 2

If OD = 2ID, Figure 6-2 (bottom), the ringcompression stress due to external pressure is, σi =8/5 the ring compression stress due to equal internalpressure.

Ring Compression Stress

Due to external pressure on a thin-walled pipe, ringcompression stress is,

σ = P(OD)/2A,

whereOD = maximum outside diameter,P = external pressure,A = wall area per unit length

= t for plain pipes.

Example 3

A PVC pipe, DR 41, is a storm sewer under 10 ft ofsoil. DR = OD/t = the standard dimension ratio.Ring deflection is less than 5% and can be

neglected. Dry unit weight of soil is 110 pcf.Saturated unit weight is 140 pcf. A water table canrise to 6 ft above the top of the pipe. What is thering compression stress, σ , in the pipe wall? σ = P(OD)2t = P(DR)/2,

whereP = vertical soil pressure on the pipeP = 4(110)psf + 6(140)psf = 1280 psf.The ring compression stress is σ = 182 psi.

Under some circumstances ring compression in thewall is not simply T = P(OD)/2. Consider a pipewith uniformly distributed pressure at the top and aline reaction (Class D) bedding, on the bottom. SeeFigure 6-3. Class D bedding is poor practice — buthappens. For this loading, ring compression thrustsT occur at A and B even though sidefill pressure iszero. Flexure occurs at A and B due to momentsM. Shear is zero because the load is symmetricalabout the vertical axis. Where thrust is known, thering compression stress is T/A, or T/t for plain pipes.Thrusts and moments are functions of loads, asdiscussed below.

Thrusts and Moments in the Ring

Thrusts T and moments M can be evaluated byenergy methods such as Castigliano's equation fordeflections due to loads:

δ = (M/EI)(ϑM/ϑp)rdθ

See Appendix A. δ is deflection in the direction ofa dummy load, p, (or dummy moment m forrotation). The dummy load or moment is applied atthe point where deflection, or rotation, is to be found.Assumptions are:

1. The ring is thin-walled, D/t > 10. Meandiameter D is used for analysis.

2. The pipe material is elastic.

3. Ring deflection, d, is small. Accuracy isadequate if d < 5%, or even 10% in some cases.


Figure 6-3 Buried pipe on a flat surface (left), showing the free-body-diagram for stress analysis (right). ThisClass D bedding is not recommended.

ExampleComplete a force analysis for Figure 6-3. Notation:D = 2r = mean diameter,T = ring compression thrust per unit

length,M = moment in the wall per unit length,P = vertical soil pressure,Q = PD = line reaction per unit length,t = wall thickness,I/c = section modulus per unit length,XB/A = horizontal shift of B with respect to

A,ΨB/A = change in tangent slopes of B with

respect to A when the ring is loaded. From Figure 6-3 (right), angles θ locate pointswhere thrusts T and moments M complete free-body-diagrams of segments of the ring and becomeunknowns for solution by equations of staticequilibrium plus equations of deflection from theCastigliano equation.

From Figure 6-3, with P known, five unknownsremain to be solved: TB, MB, TA, MA, and Q.Because three equations of static equilibrium areavailable, two additional equations are needed. Twoequations of deflection are:

ΨB/A = 0, and XB/A = 0.

As the ring deflects due to P, tangents at A and Bremain horizontal. Therefore ΨB/A = 0. Point Bshifts vertically, but not horizontally with respect toA. Therefore XB/A = 0. These two Castiglianoequations for deflection, together with threeequations of equilibrium, are solved simultaneouslyfor the unknowns:

TA = 0.1061 Pr compressionMA = 0.5872 Pr2

TB = 0.1061 Pr tensionMB = 0.2994 Pr2

Q = 2Pr


Figure 6-4 Values to which ring deflection, d, will be reduced after internal pressure P’ is applied to buriedsteel pipes (assuming initial ring deflection is greater than d).


This analysis is conservative. The theoretical linereaction Q is always worse than a soil bedding.Horizontal pressure of soil against the pipe providessome support. Measurements of soil stress revealdeviant stress patterns. In general, pressureconcentration shows up on the bottom due to a firmbedding. But this may be reversed if the bedding issoft and soil is compacted on top of the pipe. Ingeneral, pressure reduction shows up under thehaunches because of the difficulty of soil placement.But this may be reversed if concrete or low-slumpsoil cement is placed under the haunches or if thebedding is shaped by a V-cut. Compaction affectssoil pressure distribution. The more flexible the ring,the more uniform is soil pressure against the pipe.

Vertically compressible sidefill causes concentrationof pressures on the top and bottom of the pipe. Butan exception could occur if the pipe were located ina trench with firm sidewalls that support topfill byshear reactions. However, in time, shear breaksdown due to earth tremors, cycles of wetting anddrying, and changes in temperature. As a generalrule, the vertical dead load on top of a flexible pipeis (OD)γ H — called the soil prism load; i.e., theweight of a soil prism directly above the top of thepipe where γ is the unit weight of soil and H is theheight of soil cover above the top of the pipe. Thisgeneral rule may require a load factor for rigid pipedesign because the rigid ring may have to supportpart of the backfill within the trench if sidefill soil isnot compacted. If sidefill is compacted, the soilprism load may be adequate for rigid ring design.

For flexible ring design, the soil prism load isconservative. Normal pressures on the ring are nogreater than pressure, P, at the top because, 1. theflexible ring conforms with the soil, and 2. the soil isinvariably loose agains t the ring at the interface. Inplastic pipes, stress relaxation results in furtherreduction of normal pressure of the soil against thepipe.

The stiffer the ring, the greater are the pressureconcentrations on top and bottom when sidefills arecompressible. For a rigid pipe, well-compacted

sidefill is necessary if pressure concentrations are tobe avoided.

Combined Pressures

The question arises, what are the stresses in the wallof a pipe subjected to both internal and externalpressures? It would seem that external pressureshould be subtracted from internal pressure, or viseversa. For most installations, however, there willcome a time when either internal or externalpressure will not be acting. Therefore, the ring isusually designed for internal and external pressuresseparately. In the case of the flexible ring, becauseinternal pressure is usually not applied until after theexternal soil pressure is in place, ring deflection hasoccurred before the pipe is pressurized. If internalpressure is enough to partially re-round the ring,crescent gaps develop between the pipe and thesidefill. See Figure 6-4. Clearly the ring no longerneeds the support of the sidefill soil to retain itsshape. However, because of soil pressure on top,the pipe is not completely re-rounded. If thespecified allowable ring deflection is less than thering deflection with soil load on top, crescent gaps donot develop. Figure 6-4 shows test results for steelpipes. It is usually prudent to specify a minimumallowable ring deflection that is less than the value atwhich gaps would develop according to Figure 6-4.But even if crescent gaps develop, the ring does notcollapse for lack of side support when it isdepressurized. The ring may or may not deflect —and any ring deflection will be less than beforepressurization because soil particles tend to migrateinto the gaps.

Combined pressures include the effect of live loadpassing over a buried pipe as explained in Chapter 4,and Equation 4.1, P = Pd + Pl. If the water table isabove the pipe, the unit weight of soil is increased.Dead load pressure is found by soil mechanicsexplained in Chapter 4.

It may be concluded that internal pressure andexternal pressure are each analyzed separately.


Figure 6-5 Diagrams for force analysis of tanks buried to the top and subjected to internal test pressures.Internal and external pressures are analyzed separately and combined by perposition.


Re-rounding is seldom an issue for internal pressureanalysis, if pipes are held to nearly circular shapewhen installed. For design by ring compression, theprism load is the most reasonable load. The prismload is usually the total (not just the effective) load.

Combined stress analysis is rarely justified, but maybe required for rigid pipes — thick-walled and brittlebased on the familiar equation,

σ = T/A + Mc/I,

whereT/A = ring compression stress,Mc/I = flexural (bending) stress.

Example

In one city, acceptance for buried tanks is based onan internal pressure test when the tank is buried tothe top. See Figure 6-5a. What are the tangentialforce, TA, and moment, MA, at the top, point A?Shearing force, VA, is zero by symmetry. The soilis compacted sufficiently to prevent ring deflection.Therefore, the ring is fixed at B, Figure 6-5b. Theeffects of internal pressure and external soil loadcan be analyzed separately and then combined bysuperposition. Figure 6-5c is the free-body-diagramfor internal pressure analysis. Figure 6-5dintroduces the procedure for analyzing the effect ofsoil load. The moment at C due to the soil load isMs. It can be found by integrating the element ofsoil, shown cross-hatched, multiplied by its leverarm. The result is the moment at C (angle 2), due tothe soil load only. The equation is,

Ms = γ r3[(1/2)sin2θ - (1/2)sinθ - (1/4)sin2θsinθ - (1/3)cos3θ + (1/3)] . . . . . (6.4)

The five unknowns of Figure 6-5b require threeequations of equilibrium and two equations ofdeformation. From deformation, by symmetry, therelative rotation of A with respect to B is zero; i.e.,ψA/B = 0. The horizontal displacement of A withrespect to B is zero; i.e., χA/B = 0. From Castigliano,Appendix A, and Figure 6-6a showing a dummy

moment, m, at A in the assumed direction of rotationof A with respect to B,

ψA/B = (M/EI)( M/ m)rdθ = 0

M = MA + m - TAr(1-cosθ) + Ms . . . at point C. M/ m = 1, then m 0 in the M-equation.EI is wall stiffness and r is radius.

Substituting into Castigliano's equation, andintegrating within the limits for θ from 0 to π/2, thefirst equation of deformation becomes,

MA - 0.3634TAr + 0.0174γ r3 = 0 . . . . . (6.5)The second equation of deformation is zerohorizontal displacement of A with respect to B.Figure 6-6b shows a dummy force, p, in the assumeddirection of relative displacement of A with respectto B. From Castigliano,

χA/B = (M/EI)( M/ p)rdθ = 0

M = MA - (TA + p)r(1-cosθ) + Ms . . at point C. M/ p = -r(1-cosθ), then p 0 in M-equation.Substituting into Castigliano's equation andintegrating within limits for θ from 0 to π/2,the second equation of deformation becomes,

-MA + 0.6240TAr - 0.0320γ r3 = 0 . . . . . (6.6)

Equations 6.5 and 6.6 are solved simultaneously forthe two unknowns, MA and TA. Then by the threeequations of equilibrium, MB, VB, and TB areevaluated. The results are shown in Figure 6-6c.Soil load decreases hoop tension at A due to internalpressure by TA = 0.0560γ r2.

Ring Stress — Uses and Misuses

Stress is one basis of buried pipe design andanalysis. Most stress analyses are based on theoriesof elasticity for which yield stress is theperformance limit (failure). Elastic stress analysis isuseful in some cases such as hoop stress due tointernal pressure, ring compression stress due toexternal soil pressure, and the


Figure 6-6 Force analysis of a flexible circular cylinder buried to the top.


Boussinesq soil stress. Unfortunately, elasticanalysis is often misapplied. In buried pipe analysis,properties of materials are not elastic — either forthe pipe or the soil. Performance limits are notelastic.

Performance limit (failure) of the pipe invariablyoccurs beyond the elastic range. Performance limitis excessive deformation of pipe and soil. Excessivedeformation of the pipe may be fracture (leak), orcollapse, or it may be so much deformation thatcleaning tools cannot pass through the pipe, or thatappurtenances (fittings) are distressed. It is prudentto specify a maximum allowable ring deflection —but usually for reasons other than yield stress.

Performance limit of the soil is either excessivecompression or soil slip — a plane on whichshearing stress exceeds strength. From theories ofelasticity, the maximum shearing stresses in soilembedment around a flexible pipe occur on planesthrough the pipe axis at 45o with the horizontal.

Spangler's Iowa formula predicts ring deflectionbased on elastic pipe and elastic soil. Spangler, asoil engineer, included a horizontal elastic soilmodulus E' which he called "modulus of passiveresistance of soil." In fact, horizontal passiveresistance is Kσy where K = (1+sinφ)/(1-sinφ). Soilslip planes occur at (45o - φ/2) — not at 45o. E' isnot elastic, nor is it constant. E' varies with depth(degree of confinement), and with ring stiffness(movement of pipe wall into the sidefill).

Using the Mohr circle analysis, horizontal soilresistance is Kσy. Accordingly, soil slip planesshould occur at spring lines at angle (45o - φ/2) withthe horizontal. The analysis is conservative. Thesoil friction angle, φ, is not constant. It varies withdepth of cover and ring deflection. At spring lines,an infinitesimal cube of soil is subjected to two-dimensional compression — vertical and radial.Longitudinally it is confined. Under two-dimensional

compression, sidefill soil is densified such that itresists soil slip. (Three- dimensional compressioncan change carbon into diamonds.) In a controlledtest, planes of soil slip were observed in the sandembedment of a flexible ring. See Figure 6-7.Clearly, there are no soil slip planes at 45o or (45o -φ/2). There are no slip planes near the springlinesof flexible pipes because the soil is in two-dimensional compression.

PROBLEMS

6-1 Using Appendix A, what and where is themaximum compressive stress in a rigid pipe if D/t =5 and the load is uniform vertical pressure on topand bottom?

Figure 6-7 Soil slip planes in a sand embankment.Note lack of soil slip at the spring lines.


6-2 A thin-wall flexible circular pipe resting on a flatsurface is filled with water to the top. With noadditional internal pressure, what is the vertical ringdeflection? See Figure 6-8.

(Appen. A)

6-3 A 90o thin-wall arch is pinned (hinged) at thetwo ends and is loaded with uniform verticalpressure P. What are the reactions at the ends?

(B x = 0.832Pr, and By = 0.707Pr)

6-4 Plot the moment diagram for Problem 6-3.

6-5 What and where is the maximum moment inProblem 6-4?

6-6 If the arch of Problem 6-3 is thick-walled, forwhich t = r/5, what is the maximum circumferentialstress on the bottom surface of the arch at A?

6-7 What is the maximum circumferential stress atB in Problem 6-6?

6-8 What and where is the maximum stress in theplain pipe of Figure 6-3 if r/t = 10 and I/c = t2/6?

(At point A, σ =353P)

6-9 What height H of dense soil cover can be sup-ported by a 4-ft corrugated steel culvert? Assumethat ring deflection, d, is held by design specifi-

cation to a value small enough to be neglected.Data are, (101 ft or less?)Given:SoilType, SM, silty sandCompaction 80 percent (AASHTO T180)Ss = 1.0 = degree of saturation,G = 2.7 = specific gravity of soil grains,e = 0.4 = void ratio,sf = 1.0 at performance limit (wall crushing).PipeD = 48 inches = diameter,

t = 0.064 inch (16 gage)= specified wall thickness,

A = 0.775 in2/ft from AISI tables,I = 0.0227 in4/ft,S = 36 ksi at yield strength,E = 30(106) = modulus of elasticity,Corrugations, 2-2/3 x 1/2.

6-10 What height of dense soil cover can besupported by the pipe of Problem 6-9 if it issubjected to an internal vacuum of 12 psi?

(H = 88.5 ft)

6-11 What height of loose soil, e = 1.0, can besupported by the pipe of Problem 6-9 if the soilliquefies due to earth tremors?

(at r=24, N=14.9ft; at r= 24.25, H=15ft)

6-12 What wall thickness is required for a 2-2/3 x1/2 corrugated steel pipe, 36-inch diameter, under aforest road (HS-20 loading) with 2 ft of soil cover ifthe soil envelope is average granular soil compactedto at least 80 percent density (AASHTO T-180)?Assume that S = 36 ksi, and sf = 2 for the steel.Assume that the unit weight of soil is 120 pcf.

6-13 What wall thickness is required for a 1.0-meter bare (plain) steel pipe under 160 meters ofsaturated tailings with unit weight 140 pcf? A selectsoil envelope is specified. What should be theconditions for installation? Is steel pipe a goodoption? Under what conditions?


Anderson, Loren Runar et al "RING DEFLECTION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 7-1 Deflected ring showing notation for dimensions and for ring deflection analysis.

Figure 7-2 Segmented ring deflection showing the relationship between ring deflection, d, width, w, of thecrack, and wall thickness, t.


CHAPTER 7 RING DEFLECTION

Ring deflection is defined as the ratio of change invertical diameter to the original diameter, d = /D.See Figure 7-1. Diameter D is the diameter to theneutral surfaces of the cross section of the wall. Formost pipe analyses, it is sufficiently accurate to usethe mean diameter, (OD+ID)/2. The error of usingmean diameter increases for reinforced concretepipes, pipes with ribs or stiffener rings, etc. Ringdeflection is the result of: 1. inflation or deflation ofthe pipe, 2. flexing of the ring, 3. cracking of the ringinto segments, and 4. plastic hinging (or crushing) ofthe pipe walls. Ring deflections of rigid and flexiblepipes are two different phenomena. Each isanalyzed separately.

RIGID RING

Typical rigid pipes are concrete and vitrified claypipes. For rigid pipes, two basic modes of ringdeflection are elastic and segmented. Most rigidpipes are brittle. The limit of elastic deflection isreached when the pipe cracks into segments asshown in Figure 7-2. Because there is no such thingas a perfectly rigid pipe, the question arises, howmuch elastic ring deflection occurs in the rigid ring?For most rigid rings, elastic deflection is smallenough to be neglected. Hairline cracks are notcritical. Reduction in flow capacity is negligible.

Elastic ring deflection is calculated in the same wayfor both rigid and flexible pipes. Composite(reinforced) pipe walls require a transformed sectionfor analysis. This is true for reinforced concretepipes, but may also be true for materials withdifferent properties in compression and tension.Elastic ring deflections for various load conditionsare listed in Appendix A.

Example 1

A concrete pipe has ID = 36 and OD = 42 incheswith double cages of 1/4 inch steel reinforcing rodsspaced at 2 inches and located 0.6 inch from theinside and outside surfaces of the pipe. See Figure7-3

. What is the elastic ring deflection at yield stressof 1000 psi in the concrete for each of the threedifferent loading conditions shown? Cracks open iftensile stress is greater than 1000 psi. From thetransformed section, EI/r3 = 1651 psi; and fromdeflection equations in Appendix A, thecorresponding ring deflections are calculated andsummarized in Figure 7-3. None of these elastic ringdeflections is greater than 0.1%.

Segmented ring deflection is the result of cracksopening at spring lines, crown and invert. See Figure7-2. It is assumed that the segments are rigid. Ringdeflection can be calculated in terms of crack widthw. If the wall thickness is t and the neutral surfaceis at mid-thickness of the wall, the ring deflection isd = w/t. Allowing for some deflection of thesegments, and allowing for the possibility that neutralsurfaces are further from the pipe surfaces than t/2,the lower limit of ring deflection is greater than w/2t.Therefore,

td < w < 2td . . . . . (7.1)

which shows a range of widths of the crack as afunction of wall thickness and segmented ringdeflection. The relationship is not precise becausecracks are undependable. For example, two parallelcracks may open where only one is expected.

Example 2

Consider the same 36 inch ID reinforced concretepipe with three inch thick walls. If 0.01 inch widecracks open inside the pipe at the invert and crown,from Equation 7.1, ring deflection is between d =0.17% and d = 0.33%. Because of balancedplacement of steel, actual ring deflection may becloser to the upper limit, say, d = 0.3%.

Ring deflection is more the result of cracking than itis the result of elas tic deformation of the ring. Thesmall ring deflections justify design by rigid pipetheories. See Chapter 12.


Figure 7-3 Ring deflection at incipient cracking of a reinforced concrete pipe under three different loadingconditions. Note that the maximum deflection is d = 0.1%.


FLEXIBLE RING

As soil and surface loads are placed over a buriedflexible pipe, the ring tends to deflect — primarilyinto an ellipse with a decrease in vertical diameterand an almost equal (slightly less) increase inhorizontal diameter. Any deviation from ellipticalcross section is a secondary deformation which maybe the result of non-uniform soil pressure. Theincrease in horizontal diameter develops lateral soilsupport which increases the load-carrying capac ityof the ring. The decrease in vertical diameterpartially relieves the ring of load. The soil above thepipe takes more of the load in arching action overthe pipe — like a masonry arch. Both the increasein strength of the ring and the soil arching actioncontribute to structural integrity. Although some ringdeflection is beneficial, it cannot exceed a practicalperformance limit. Therefore the prediction of ringdeflection of buried flexible pipes is essential. Ringdeflection is elastic up to the formation of cracks orpermanent ring deformations. Clearly, the ring canperform with permanent deformations — and evenwith small crac ks, under some circumstances.Performance can surpass yield stress to thedetermination at which the ring becomes unstable.Instability is explained in Chapter 10.

The following analyses of ring deflection are basedon elastic theory for which the pertinent pi-termsare:d = ring deflection,ε = average sidefill soil settlement,d/ε = ring deflection term,Rs = stiffness ratio = E'D3/EI,

= ratio of soil stiffness E' to ring stiffness,EI/D3; or to pipe stiffness, F/∆ , where

F/∆ = 53.77 EI/D3.Notation:d = ∆ /D = ring deflection, = vertical decrease in ring diameter,D = original diameter of the flexible ring (more

precisely, the diameter to the neutral surfaces ofthe wall cross section),

∆ = vertical soil strain due to the anticipatedvertical soil pressure at the pipe springlines,

E' = soil modulus = slope of a secant on the stress-strain diagram from the point of initial verticaleffective soil pressure to the point of maximumvertical effective soil pressure,

E = modulus of elasticity of the pipe wall,I = centroidal moment of inertia of the pipe wall

cross section per unit length of the pipe.

Figure 7-4 is a graph of the ring deflection term as afunction of stiffness ratio. From the graph, ringdeflection can be found as follows. Enter Figure 7-4with a stiffness ratio, either Rs or Rs' and read outthe ring deflection term, d/ε . If the vertical soilstrain ε is known, ring deflection follows directlyfrom d/ε . Figure 7-4 represents tests and field datafor buried flexible pipes.

Vertical soil strain ε is predicted from laboratorycompression tests data such as the stress-straingraphs of Figure 7-5 for cohesionless siltly sand.

Soil stiffness E' is the slope of a secant to theanticipated soil pressure P on the stress-straindiagram for a specific soil density. Graphs can beprovided by soil test laboratories for the specificembedment to be used, and at the density to bespecified.

Ring stiffness contributes significant resistance toring deflection if Rs is less than about 300 (or Rs' isless than 6); i.e. low soil stiffness and high ringstiffness. For flexible pipes buried in good soil,stiffness ratio is usually greater than 300. Therefore, For design, ring deflection of flexible pipesburied in good soil is equal to (no greater than)the vertical strain (compression) of the sidefillsoil.

Circumstances arise under which the above rule isnot accurate. Equations for ring deflection are listedin Appendix A for a few loadings on rings of uniformwall thickness that are initially circular. If not, oneset of approximate adjustment factors is:

D to be multiplied by Dmax /Dmin


Figure 7-4 Ring deflection term as a function of stiffness ratio. The graph is a summary of 140 tests plottedat 90 percent level of confidence; i.e., 90 percent of test data fall below the graph.

Figure 7-5 Stress-strain relationship for typical cohesionless soil (silty sand). Ninety percent of all strains fallto the left of the graphs. Vertical pressure is effective (intergranular) soil pressure.Soil stiffness, E', is the slope of the secant from initial to ultimate effective soil pressures.


Rs = (SOIL STIFFNESS)/(RING STIFFNESS)

t to be multiplied by t min /tmax

Equations have been proposed for predicting ringdeflection of flexible pipes. One of these is the IowaFormula derived by M. G. Spangler. The IowaFormula is elegant and correctly derived, butdepends upon a number of factors which may bedifficult to evaluate. Such questionable factorsinclude a deflection lag factor, bedding factor, thehorizontal soil modulus, and the assumptions onwhich the Marston load is based. See Appendix Band Spangler (1973).

The horizontal soil modulus, E', is particularlytroublesome. It is based on theory of elasticitywhich is questionable. E' is not constant. In fact, E'is a function of the depth of burial and the horizontalcompression of the sidefill soil as the pipe expandsinto it.

The best procedure for design of buried flexiblepipes is to specify the allowable ring deflection, andthen make sure that vertical compression of thesidefill soil does not exceed allowable ring deflection.

The Iowa Formula, and other deflec tion equations,are approximate, but conservative. Some arecompared in Appendix B. However, within theprecision justified in most buried pipe analyses, theprocedures described in the following examples aremore relevant and understandable.

Example 1A corrugated plastic drain pipe (flexible) is to beburied in clean dry sand backfill that falls into placeat 80 percent density (AASHTO T-180). What ringdeflection is anticipated due to 10 ft of soil coverweighing 120 lb/ft3? Live load is neglected at thisdepth of cover. The stiffness ratio is larger than R's= 6, therefore, from the ring deflection graph ofFigure 7-5, ring deflection is not more than about,d = ε = 1%.

Example 2

What is the predicted ring deflection? A PVC pipeof

DR = 14 is to be placed in embedment compacted to80 percent density (AASHTO T-180) under 24 ft ofcohesionless silty sand at dry unit weight of 105lb/ft3, but with a water table at 9 ft below thesurface. Saturated unit weight is 132 lb/ft3. Fromthe Unibell (1882) Handbook of PVC Pipe, page159, values of PVC pipe stiffness for DR = 14 pipesvary from 815 to 1019 psi. Using, conservatively,the lower value, pipe stiffness is F/ = 815 psi. Thesoil stiffness is found from Figure 7-5. Because ringdeflection increases from zero at no soil pressure tomaximum at ultimate pressure, soil stiffness is theslope of the secant from the origin to the point ofultimate effective soil pressure on the stress-straindiagram. Vertical soil strain is a function ofeffective soil pressure (intergranular) — not totalpressure. Effective soil pressure is,

P = 15ft(132pcf) + 9ft(105pcf)- 15ft(62.4pcf) = 1.99ksf = 13.8psi

At 80% density, from Figure 7-5, the soil strain at1.99 ksf is ε = 1.85%. The soil stiffness is the slopeof the secant from 0 to 2 ksf on the 80% graph; i.e.,E' = 13.8psi/0.0185 = 747 psi. The resulting stiffnessratio is R's = E'/(F/ ) = 747/815 = 0.92. Enteringthe graph of Figure 7-4 with R's = 0.92, thecorresponding ring deflection term is d/ε = 0.48.Ring deflection is 48% of the vertical soil strain ε .Because soil strain is 1.85%, the predicted ringdeflection is, d = 1.85%(0.48) 1.0%.

If a straight pipe of elastic material and circularcross section is bent into a circular curve, the crosssection deforms into an ellipse. Ring deflection ofthe cross section is,

d = 2Z/3 + 71Z2/135 . . . . . (7.2)

where Z = 1.5(1-ν2)D4/16t2R2

d = ring deflection = ∆ /D,∆ = decrease in pipe diameter,ν = Poisson ratio,D = diameter of circular pipe,t = wall thickness,R = radius of the bend.


C-

C

CC

*Decrease in diameter is in the direction of theradius of the bend. See Chapter 14 for example.

REFERENCES

Spangler, M.G. (1973) and Handy, R.L. SoilEngineering, IEP, New York

Unibell (1982), Handbook of PVC Pipe

Watkins, R.K. (1974), Szpak, E., and Allman, W.B.,Structural design of polyethylene pipes subjected toexternal loads, Eng'rg Expr. Sta., USU.

PROBLEMS

7-1 What is the ultimate ring deflection of a steelwater pipe, ID = 36 inches and t = 1 inch, buriedunder saturated tailings which will rise ultimately to250 ft? For tailings, G = 2.7. Unconsolidated, e =0.7. When consolidated under H = 250 ft of tailings,e = 0.5. Assume a straight line variation of e withrespect to height above the pipe. Water table is atthe ground surface. (d = 11.8%)

7-2 A corrugated steel storm drain never flows full.Therefore the granular backfill soil is essentially dry.What is the ring deflection? Include HS-20 live load.Given; (0.8)Soil (granular)H = 4 ft = height of soil cover,G = 2.7 = specific gravity,e = 0.7 = void ratio,80% density (AASHTO T-180).

Steel pipe (corrugations 2 2/3 x 1/2)D = 48 inches = diameter,I = 0.0180 in4/ft (t = 0.052),E = 30(106) psi = modulus of elasticity.

7-3 What is the change in ring deflection of Prob-lem 7-2 if the soil cover is increased from 4 ft to 26ft using the same soil, same density? (d = 1.6%)

7-4 What is the probable ring deflection of anunreinforced concrete pipe, ID = 30 inch and wallthickness = 3.5 inches if a video from inside the pipereveals a 0.1-inch-wide crack in the crown?

7-5 What is the ring deflection of a steel pipe, OD= 26 inches and ID = 24 inches, if E = 30(106) ps iand the soil cover is H = 40 ft.? The soil unit weightis 100 lb/ft3 at 80% density (AASHTO T-180).

(Rs = 5.425; d = 0.064%)

7-6 Predict ring deflection of a plain steel pipe if:D = 10 ft,t = 0.5 inch,E = 30(106) psi.Soil is granular, 90% dense (AASHTO T-180)γ = 120 lb/ft3,H = 30 ft.

7-7 If the neutral surface is at the geometricalcenter of the wall of Figure 7-2, prove that the widthof the crack is approximately w = td; where w

= width of crackt = wall thicknessd = segmented ring deflection = decrease in vertical diameterD = diameter to neutral surface (NS)It is assumed that the wall crushes in compressionon one side of the neutral surface just as much as itstretches in tension on the other side before thecracks open. This is not true for all materials.

7-8 If the ring of Figure 7-2 is vitrified clay orunreinforced concrete, both of which are many timesstronger in compression than in tension, thecompression crushing zones in the wall are verysmall. As a worst case, assume no wall crushingand find the segmented ring deflection d if thecracks open 0.01 inch.

7-9 Assume that the ring of Figure 7-2 is reinforcedconcrete with a single steel wire cage in the centerof the wall. The wire is 1/4-inch diameter spaced at2 inches. What is the vertical diameter to the neutralsurfaces? (34.7 in)t = 3 inches Es = 30(106) psiID = 30 inches Ec = 3(106) psi


7-10 What is the horizontal diameter to the neutralsurfaces of Problem 7-9? What would be thedifference if the cracks were caused by uniform soilload on top and bottom (no sidefill)?

7-11 A high-density polyethylene (HDPE) pipe hasa minimum outside diameter (OD) of 6.60 inch anda maximum OD of 6.66 inch. The wall thicknessvaries from 0.83 maximum to 0.80 minimum. Whatis predicted ring deflection under the loading shownin Figure 7-3c if P = 5.0 kips per square ft? Assumethat long-term virtual modulus of elasticity for HDPEis 85 ksi.

7-12 What is the approximate ring deflection of thereinforced concrete pipe of Figure 7-6 when thewidth of crack at the crown is w = 0.06 inch?

7-13 If the load in Figure 7-6 is an F-load (parallelplate load), where is the neutral surface at the springlines?

7-14 Are the cracks in Problem 7-13 exactly equal

on the inside at crown and invert, and on the outsideat spring lines? Explain.

7-15 What is the maximum limit of ring deflectiondue to an F-load on a plastic pipe if permanent straindamage occurs at 1.75 percent strain on the outsidesurface? See Appendix A for deflection due to F-loading. (d = 12.78 13%)D = 0.5 meter,t = 16 mm,E = 300 ksi (2.07 GN/m2).

7-16 How high can plastic pipes be stacked ifmaximum allowable ring deflection is 10 percent?OD = 4.24 inches = outside diameter,ID = 3.92 inches = inside diameter,w = 1.08 lb/ft = pipe weight,E = 200 ksi = short term modulus of elasticity.

7-17 A flexible plastic pipe, DR = 31, for which F/∆= 90 psi, is buried in cohesionless soil with unitweight of 110 pcf at 80 percent density (AASHTOT-180). If the height of cover is 25 ft, what is thepredicted ring deflection?

Figure 7-6 Cross section of the wall of a reinforced concrete pipe, showing the transformed section inconcrete, and showing the procedure for finding the neutral surface (NS). It is assumed in this case thatconcrete can take no tension.


Anderson, Loren Runar et al "RING STIFFNESS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 8-1 Typical stress-strain diagram from laboratory tests showing:E = modulus of elasticity = slope within the elastic zoneE" = stiffness = slope at any point on the graph, or the average slope of the cord in some stress range such as A to B

Figure 8-2 Comparison of a leaf spring and a cylindrical spring showing the stiffness F/∆ for each.For the ring, F/∆ is called pipe stiffness and EI/D3 is called ring stiffness.


CHAPTER 8 RING STIFFNESS

It is noteworthy that analyses of ring deformationand ring deflection, Chapters 3 and 7, require aproperty of material called stiffness. Stiffness isdefined as resistance to deflection. Figure 8-1shows a typical stress-strain diagram. The abscissais strain which is deflection per unit length. Theordinate is stress which is load per unit area andwhich is resistance of the material to strain. Suchdiagrams come from laboratory tests. Stiffness ofthe material is the slope, E, of the stress-straindiagram at any particular point — at any particularstress. Stress-strain diagrams can be provided forshearing stress-strain as well as normal stress-strain.The initial linear portion of the diagram is the elasticzone within which stress causes no permanentdeformation. The material rebounds elastically. Theslope, which is constant, is called the modulus ofelasticity.

Notation:E = modulus of elasticity or stiffness,E = slope of the stress-strain diagram,σ = normal stress (or shearing stress τ ),ε = normal strain (or shearing strain γ ),I = centroidal moment of inertia of the cross-

sectional area (of the wall per unit length inthe case of a pipe),

L = length (as of a leaf spring),D = mean diameter of the ring,t = wall thickness of the ring (or cross-

sectional area A per unit length of pipe).F = concentrated load (diametral line load

per unit length in the case of a pipe)

The concept of stiffness is easily extended to aspring. See Figure 8-2. The leaf spring on the leftis deflected a vertical distance by load F. Thespring stiffness is the slope of the F/ diagram andis equal to F/ = 48EI/L3. See texts on mechanicsof materials. An interesting comparison can bedrawn between the leaf spring and a circular spring,or ring, on the right. The leaf spring is analogous tothe circular spring which, from Appendix A, has aspring stiffness of F/ = 53.77EI/D3. (Sp)ringstiffness is similar in form to the leaf spring stiffness.

It contains the term, EI/D3, which shows up in everyanalysis of ring deformation and deflection. EI/D3 isan important form of spring stiffness. For a pipewith a rectangular wall cross section, I = t3/12, thering stiffness becomes F/∆ = 4.48E/(D/t)3. D/t isanother form of ring stiffness. Stiffness can beexpressed in a variety of ways: EQUIVALENTQUANTITIES NOMENCLATUREF/∆ Pipe Stiffness= 53.77 (EI/D3) Ring Stiffness, EI/D3

= 4.48 E(t/D)3 where D/t = m-TERM= 4.48 E/(DR-1)3 where DR = (OD)/t

= Dimension Ratio

Units of stiffness can be reconciled by noting thatboth I and F are per unit length of pipe. In terms ofbasic dimensions, stiffnesses are:STIFFNESSRING = (EI/D3) FL-2(L4/L)L-3 = FL-2

PIPE = (F/∆) (F/L)L-1 = FL-2

FL-2 is the correct dimension for stiffness ofmaterial. Many properties of materials have thedimension FL-2. Check out, for example, strength,and bulk modulus. The m-term (D/t) and thedimension ratio (DR) are not properties of materialuntil multiplied by E, which has the dimension FL-2.In pipe analysis, EI is sometimes referred to as wallstiffness. In fact, EI is not a true stiffness becauseits dimension is not FL-2.

Beyond the zone of elasticity, stiffness E is still theslope of the F/∆ diagram. However, it is no longera constant. From the stress-strain diagram of Figure8-1, if the material is stressed to its ultimate wherethe slope is zero, it loses all stiffness and simplyflows. In the case of pipe stiffness, F/∆ oftenextends beyond the zone of elasticity. This is true,in particular, of plastic pipes — including metalswhich act as plastics after reaching yield stress. Forsome materials F/∆ is affected by temperatureand/or time. Pipe stiffness, F/∆, is preferred byplastic pipe industries because it can


Figure 8-3 Two methods of testing for pipe stiffness F/) .

Figure 8-4 Plots of data from two parallel plate tests on a 4-inch PVC sewer pipe.


be measured by a parallel plate test. To perform thetest, a length of pipe, usually longer than onediameter, on a flat surface is F-loaded as shown inFigure 8-3. As load F is applied in increments,corresponding deflections, ∆, are measured. Theplot of F vs ∆ provides F/∆ values (pipe stiffness)within any load limits based on temperature and time(rate of loading) of the test. A similar test is thethree-edge-bearing (TEB) test. See Figure 8-3.Double supports on the bottom position the pipe. Forpurposes of analysis, the TEB test is equivalent toa parallel plate test. It is the basis of design of rigidpipes. Plastic pipes are designed by pipe stiffnessdefined as slope of the secant from the origin to thepoint of five percent ring deflection on the F vs ∆ plot.

Figure 8-4 is a plot of data from two parallel platetests on a 4"PVC sewer pipe. The pipe stiffness(slope F/∆) is not constant. Ring deflection isusually limited by specification to a maximum of 5%,which justifies the slope of a secant from the originto five percent strain; i.e., F/∆ = 85 psi.

Based on values for F/∆, plastic pipe industries canevaluate the DR-term, the stiffness ratio, Rs, etc.For example, if E must be modified to serve in adifferent temperature than the parallel plate test, anadjusted value can be found for stiffness ratio, Rs =E'D3/EI, in predicting ring deflection.

Other pipe industries have their reasons for usingF/∆. The seams in riveted pipes or lock seams inspiral pipes allow enough slippage to affect EI/D3.The stiffness of mortar lined and/or coated pipes isaffected by hairline cracking of the mortar.Reinforced concrete pipes defy analysis of Rs, etc .Plastic pipe industries favor the use of dimensionratio, DR, which is defined as the ratio of averageoutside diameter to minimum wall thickness. Theoutside diameter is held constant in the extrusionmachine. Wall thickness is varied for the class ofpipe (wall strength) to be produced. Steel pipeindustries favor the use of the m-term which isdefined as the ratio of the wall thickness to themean diameter. Steel industries often use theinverse D/t-term, which is called ring flexibility.

Ring stiffness (EI/D3) is preferred by the steelindustries. E is a constant. Values for I can becalculated. Because D and t describe pipes, D/tfinds its way into much ring deflection analysis.

For corrugated pipes, tables of values are publishedfor moments of inertia, I. With I, E, and D known,the ring stiffness, EI/D3, can be calculated. Parallelplate tests, or TEB tests, on corrugated pipes are acheck against calculated ring stiffness, and oftenreveal differences. If ring stiffness is critical, testscan verify or modify the calculated values. Lengthsof test pipes should be greater than one diameter.Short test sections tend to twist, especially spirallycorrugated pipes.

Example 1

Figure 8-5 shows the results of parallel plate tests onspiral corrugated pipe. What is the differencebetween calculated ring stiffness and measured ringstiffness? From the F-∆ plot, F/∆ = 100 psi. Fromthe relationship, F/∆ = 53.77(EI/D3).

Measured EI/D3 = 1.9 psi. Calculated EI/D3 = 3.7 psi from the following:E = 30(106) psiD = 24.57 inches from Figure 8-5I = 1.892(10-3) in4/in from the AISI Handbook

of Steel Drainage & HighwayConstruction Products.

The measured value of ring stiffness is only abouthalf the calculated theoretical value for this pipe.

Example 2

Suppose the ring of Example 1 is deflected so muchthat the graph of Figure 8-5 is no longer linear. Ringstiffness is the slope of the tangent to the F-∆ plot atthe anticipated ring deflection. If the ring deflectsover a range, the ring stiffness is the slope of thecord between the ends of the range of ringdeflection. It must be pointed out, Ring deflectionbased only on elastic ring stiffness is not apertinent performance limit.


Figure 8-5 Parallel plate test of a 24 D, 2-2/3x1/2 corrugated steel, spiral lock-seam pipe.

Figure 8-6 Three-edge-bearing test of a cement-mortar-lined/cement-mortar-coated thin-wall steel pipe(CML/CMC). The lining is spun centrifugally on the inside. The coating is shot-crete with wire reinforcing.


PROBLEMS

8-1 Figure 8-4 is an F-∆ diagram for a 4-inch PVCsewer pipe DR30. Compare the measured pipes tiffness F/∆ and pipe stiffness calculated frompublished values of E = 400 ksi to 500 ksi.

8-2 Figure 8-6 is an F- diagram for CML/CMCpipe which was handled with care before testing.At each drop of the load-deflection diagram,cracking was audible and hair cracks appeared.What should be the pipe stiffness for designpurposes? What conditions should be specified?

(F/∆ = 450 psi)

8-3 Figure 8-7 is the result of a parallel plate teston spiral rib steel pipe. What is pipe stiffness?

(F/∆ = 72 psi)

8-4 Using average D, calculate ring stiffness EI/D3

for a reinforced concrete pipe if:ID =60 inches = inside diameter,OD = 72 inches = outside diameter,Es = 30(106) psi = steel modulus,Ec = 2(106) psi = concrete modulus.Reinforcing steel comprises 3/8-inch rebar hoopsrods in the center of the wall spaced at 3 inches.Assume that concrete takes no tension. (EI/D3 = 16 psi) 8-5 Calculate ring stiffness of the pipe for Problem8-4 if two cages of 3/8-inch rods are spaced at 3inches, but located 1.0 inch from the inside andoutside surfaces. Assume concrete takes notension. What about average D? (55psi)THIS PROBLEM SHOWS WHY RCP PIPE ISDESIGNED B Y TEB TESTS AND F/∆ — NOTBY RING STIFFNESS, EI/D 3.

plot will pass through the origin.Figure 8-7 Load deflection diagram for a spiral-rib pipe showing an initial zero correction such that the linear


Anderson, Loren Runar et al "NON-CIRCULAR CROSS SECTIONS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 9-1 Examples of non-circular cross sections of pipes used commonly in the corrugated steel pipeindustry, as described by AISI Handbook of Steel Drainage & Highway Construction Products.


CHAPTER 9 NON-CIRCULAR CROSS SECTIONS

If the pipe cross section is not circular, "ring"analysis must be modified. For most buried pipes, acircular cross section is the most effic ient shape.But even flexible circular rings deflect out-of-roundduring installation. Morever, a demand exists fornon-circular cross sections. Some typical examplesare shown in Figure 9-1. A standing demand existsin highway departments for culverts with reducedheight of cross section. Each inch of height of theculvert requires an enormous amount of soil to raisethe highway by that amount. The pipe arch and lowprofile arch are examples of efforts to servehighway demands for reduced heights of culverts.Multiple culverts serve to reduce heights, but alsoincrease costs, spread stream beds, and trap trash.

A pertinent variable for ring analysis is radius ofcurvature, r. The basic deflection of a flexible ringis from circle to ellipse, for which radii of curvatureare shown in Chapter 3. But non-ellipticaldeformation could make it necessary to measureradii of curvature. Techniques for measuring radiiare explained in Chapter 3.

Figure 9-2 shows the free-body-diagram of aninfinitesimal segment of pipe wall loaded by externalradial pressure P. The effects of bending moment(ring deformation) can be combined by superpositionas discussed in this chapter. Reactions are thrust Tin the pipe wall. From static equilibrium in thevertical direction, and noting that for the smalldifferential angle, sin(dθ/2) = (dθ/2), the equation ofvertical forces is Prdθ = 2Tdθ/2; and,

T = Pr . . . . . (9.1) whereT = tangential (circumferential) thrust in the

wall,P = external radial pressure (plus internal

vacuum),r = mean radius of curvature (assuming that

the pipe is thin-walled and cylindrical).

Mean radius is sufficiently accurate for thin-wallpipe analyses. Outside radius is more accurate —especially for thick-wall pipes.

RING COMPRESSION STRESS

For non-circular cross section, ring compressionstress is simply, σ = T/A; or, for plain pipes (smoothcylindrical surfaces, no ribs or corrugations, etc.), σ= T/t. "Ring compression" is a misnomer in non-circular pipes. Nevertheless, the expression ringcompression stress is understood to meancircumferential stress in the pipe wall.

RADIAL SOIL PRESSURE

Another pertinent variable is radial soil pressure P onthe pipe. From Equation 9.1, if thrust T is constant,P varies inversely as radius r. If the ring is flexible,the soil must be able to provide enough pressure Pfor equilibrium. It is conservative to neglect shearingstresses between soil and pipe. Shearing stressesreduce radial stresses. Moreover, any shearingstresses that develop during installation are easilybroken down by earth tremors, variations intemperature, rise and fall of the water table, wettingand drying of the soil, etc. Without shearing stress,thrust T is constant around the entire perimeter ofthe pipe. This is evident from Figure 9-3 where, forstatic equilibrium, T1 = T2 = T = constant thrustaround the entire perimeter. From Equation 9.1,

P1r1 = P2r2 = Pr = T . . . . . (9.2)

Wherever the radius r is small, the external pressureP is large. This introduces the very importantconcept that for a flexible non-circular cross section,the external soil-bearing capacity must be increasedwherever radii are decreased. If the corner plateson a pipe arch, Figure 9-4, have a radius equal toone-third the top radius, then the external normalpressure (radial soil support) must


Figure 9-2 Free-body-diagram of an infinitesimal segment of pipe from which ring compression thrust in thepipe wall is T = Pr.

Figure 9-3 Free-body-diagram of sections of a pipe wall of varying radii of curvature from which the ringcompression thrust is constant, T = P1r1 = P2r2 = Pr. Shearing stress between the soil and the pipe wall isneglected.


Figure 9-4 Typical cross section of a corrugated steel structural plate pipe arch showing radii of the top plate,corner plates, and bottom plate.

Figure 9-5 Elliptical cross section of a flexible ring showing the distribution of external pressure required forequilibrium.



rx

be three times as great as the pressure on the top ofthe pipe arch. The soil against the short-radiuscorner plates must have adequate bearing strength.It is noteworthy that only a little spreading of thecorner plates will allow reversal of curvature of thebottom plate if hydrostatic pressure should act on thebottom. Soil support at the corner plates isimperative.

If a circular cross section is deflected into an ellipse,then Pxrx = Pyry. See Figure 9-5. From Chapter 3,the ratio of radii is ry /rx = (b/a)3. But (b/a)3 =(1+d)3/(1-d)3, approximately. Therefore,

Px = Py(b/a)3 = Py(1+d)3/(1-d)3 . . . . . . . . . . . . (9.3)

wherea = r(1-d) = minimum semi-diameterb = r(1+d) = maximum semi-diameterd = /D = ring deflection

An accurate solution, from Chapter 3, is,

1 + 3d + 4d2 + 4d3 + ....) Px = Py . . . (9.4)

1 - 3d + 4d2 - 4d3 + ....)

The accuracy of Equation 9.4 is seldom justified.The following example illustrates the point.

Example 1

Assume that ring deflection of the elliptical crosssection is d = 10%. What is Px in terms of Py?

From Equation 9.3, Px = 1.826 Py.

From Equation 9.4, Px = 1.826 Py.

This many significant figures of accuracy is notjustified — either in practice or theory. It must beremembered that: the elliptical cross section is onlya theoretical assumption; shearing stresses areignored; the perimeter is assumed to be constant; thehorizontal and vertical ring deflections are assumedto be equal, etc. For an elliptical cross section,vertical ring deflection is slightly greater thanhorizontal ring deflection, but the difference is

negligible if ring deflection is less than about tenpercent. Equation 9.3 is accurate enough for mostring deflection analyses.

Example 2

A flexible pipe is deflected into an approximateellipse shown in Figure 9-5. Initial ring deflection isdo = 15.9%. If pressure on top is P = 1.0 ksf, whatis the required horizontal bearing capacity of thesidefill soil at the spring lines? The horizontalpressure Px at spring lines, from Equation 9.3, is,

Px = Py(ry/rx) = Py(1+d)3/(1-d)3 = 2.617 Py.

Horizontal bearing capacity of the soil at the springlines must be greater than 2.62 ksf. With a safetyfactor, specify soil-bearing capacity of 5 ksf. Sidefillmust be well compacted, otherwise the soil will be atincipient slip, and the ring at incipient collapse. Ringcompression stress is σ = Pyry/A, where A is wallcross-sectional area per unit length.

MEASURED CHANGE IN RADIUS

If the ring deflects into an ellipse, all that is neededto evaluate maximum and minimum radii ofcurvature is measurement of ring deflection. Theequations are shown in Figure 9-6.

For deformations other than ellipse, the change inradius can be evaluated from changes ∆e in themiddle ordinate e of a cord of length L. Figure 9-7shows the analysis, from which the approximateradius of curvature is r = L2/8e for small ratios of eto L. The change in radius from r to r' is found fromchange in the middle ordinate, ∆ e = e - e':

1/r - 1/r' = ∆ e/er . . . . . . . . . . . . . . . . . . . . (9.5)

Procedures for measuring e (either inside or outsidethe pipe) are described in Chapter 3. Minimumradius is pertinent to soil strength analysis; maximumradius is pertinent to ring stability, Chapter 10. Bothare pertinent to circumferential stress analysis.


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CIRCUMFERENTIAL STRESS

Assume that within cord length L, a pipe is initiallycircular and ring compression stress in the pipe wallis Pr/A. Now if the ring is deformed, the change inradius of curvature causes a change in Pr/A; andalso introduces a flexural stress, (E/m)(r'-r)/2r'. SeeEquation 5.3. The ring compression stress isessentially constant around the ring. However,flexural stress is maximum where change in radiusis greatest; i.e., where the change in middle ordinate, e, is greatest. Knowing the change in middleordinate, e, the circumferential stress within thecord length is,

σ = Pr/A + ( e/e)(Ec/r) . .. . . . . . . . . . . . . . . (9.6)

where (See Figure 9-7)e = middle ordinate for the original circle, e = L2/8r, or can be measured before the pipe

is deformed, ∆ e = change in middle ordinate due to ring

deformation,L = length of cordt = wall thicknessm = r/t = ring flexibilityr = initial radius of curvature at some location

before the pipe is deformed,r' = radius of curvature at the same location

after the pipe is deformed,A = cross-sectional area of the pipe wall per

unit length of pipe,P = radial pressure on the pipe,E = modulus of elasticity of the pipe,I/c = section modulus per unit length, e = change in middle ordinate due to ring

deformation,c = distance from the neutral surface of the

pipe wall to the most remote surface.

For a plain pipe, c = t/2, and (Ec/r) = (E/2m) to beused in Equation 9.6.From Equation 9.6, for any allowable stress, σ , orstrain, ε , and external pressure, P; the allowablechange in middle ordinate, ∆e/e, can be found. Ringdeformation must then be controlled so that themeasured ∆e/e does not exceed the allowable.

Internal pressure P' (no external constraint) causeshoop stress σ = P'r/A. A non-circular ring tries tore-round and change its radii. Equation 9.6 stillapplies. Changes in radii would have to bemeasured and then used to calculate flexuralstresses. Or knowing allowable stress, maximumallowable changes in radii, or changes in middleordinate from a cord can be used by inspectors forcontrol of the pipe shape during installation.

Internal pressure in a pipe with external constraint issometimes analyzed by neglecting any changes inradii of curvature. The presumption is that the soilis rigid. But, if the ring is constrained bycompressible soil, or by concentrated point loads andreactions, further analysis may be necessary. Afinite element analysis may be a good option.

For flexible pipes — even non-circular rings —flexural stresses due to change in radius, are notperformance limits in general. Brittle linings maypose an exception. Many common pipe materialscan yield without fracture. Consequently, theflexural stress term can be neglected. The ringsimply sustains permanent deformation withoutfracture or inversion.

In summary, the circumferential stress analysis ofnon-circular pipes is based on Equation 9.6 whichalways includes ring compression stress and,possibly, flexural stress.

Pr/A = ring compression stress (or hoop tension),

(∆ e/e)(Ec/r) = (∆ e/e)(E/2m) = (Mc/I) = flexural stress.

where (E/2m) = (Ec/r) = arc modulus — used if eitherchange in radius of curvature or ∆ e is known.

(I/c) = section modulus — used if the moment M isknown. The moment M can be evaluated fromcircumferential strains measured by electricalresistance strain gages positioned both inside andoutside the pipe at locations where critical momentis anticipated. Below yield, both thrust and momentcan be found from these strains.


Circumferential stress due to ring deformation isdependent upon either the section modulus or the arcmodulus.

For brittle pipes — the ring compression stress andring deformation stress must be combined foranalysis. Steel pipes with brittle linings or coatingsare not brittle pipes. Small cracks are not serious.

For flexible pipes — deformation is caused by thesoil. Changes in wall thickness make littledifference. If soil is placed such that ring deflectionis constant, the performance limit is wall crushing atyield stress due to ring compression stress, Pr/A.This is an important basis for design. See Chapter6. On the other hand, if ring deformation stressexceeds yield, the deformation is permanent. Butdeformation is not a performance limit until itbecomes excessive. For analysis of flexible buriedpipes, the stress due to ring deformation is not anappropriate basis for design. Design by ringcompression stress and design by ring deflection arethe two basic design procedures.

For corrugated and profile wall pipes, the com-bination of ring compression stress and ringdeformation stress may result in dimpling ofcorrugations, or plastic hinging. But for buried pipes,dimpling and incipient hinging are not collapse.

PROBLEMS

9-1 Derive Equation 9.6. Remember that flexuralstress is Mc/I and that M/EI = 1/r - 1/r' for a circlew here r is the original radius and r' is the deformedradius.

9-2 In order to check the assumption that a flexiblepipe, ID = 42, with ring deflection of d = 10%, is anellipse, what should be the middle ordinate inside thepipe to the spring line from a vertical cord (straightedge) that is 10 inches long? (e = 0.83)

9-3 Figure 9-8 shows the cross section of acorrugated steel culvert comprising circular panelswith radii as follows:

ry = 82.50 inches for top and bottom panels,rx = 26.25 inches = radius of side panels.

This culvert is to be installed under a highway withthe major diameter horizontal. Soil cover is 2 ftincluding an asphalt concrete pavement which isassumed to be flexible enough that the load is notspread by the pavement. The road is designed forHS-20 truck loads. Unit weight of soil is 135 pcf.Because the pipe is a drainage culvert, the watertable is never more than a few inches above theinvert of the pipe. What is the minimum sidefill soil-bearing capacity required at the spring lines if safetyfactor is to be 2.0?

(Px = 12.2 ksf)

9-4 In Problem 9-3, if the sidefill soil is cohesionlesswith a soil friction angle of φ = 35o, what is thesafety factor against soil shear failure?

(sf = 0.4 — incipient collapse)

9-5 The elliptical culvert of Problem 9-3 is rotated90o to serve as a livestock underpass. See Figure 9-9. At what soil friction angle of sidefill at the springlines would the culvert collapse by reversal ofcurvature of the 82.5-inch-radius side panels?Neglect H-20 surface live load. Why?

9-6 In cohesionless soil what soil friction angle isrequired to assure that shearing failure does notoccur in the embedment of a pipe arch for which:rc = 18 inches at corner plates,rt = 60 inches at top plate,rb = 180 inches at bottom plate,Corrugations are 2 2/3 x ½. Assume high fill..Neglect shearing stresses between the pipe and theembedment soil. (φ = 33o)

9-7 In Problem 9-6, what is the minimum rc in termsof rt if the soil friction angle is 30o?

(rc = rt /3)


Figure 9-8 Horizontal "ellipse" of corrugated steel plate to be used as a culvert and for which rx = 26.25inches and ry = 82.5 inches and showing the radial soil pressure acting on it.

Figure 9-9. Corrugated steel plate underpass forlivestock with H=2 feet of soil cover.

Figure 9-10. Cross section of a flexible circular ringthat has been deformed into an ellipse duringinstallation such that d=∆ /D=16% and for which therequired horizontal soil pressure is greater than thevertical soil pressure at the crown.


9-8 In Problem 9-7 discuss the performance limitsbased on soil-bearing capacity and soil compression.What about deformation of the ring due to soilcompression?

9-9 A corrugated steel culvert has the followingproperties:Corrugations 2 2/3 x 1/2D = 6 ftt = 0.1046 inchA = 1.356 in2/ftS = 36 ksi = yield strengthsf = 2d = ∆/DAssume elliptical cross section

What is the horizontal soil-bearing capacity requiredat the spring lines for this culvert if it is deflected byd = 16% into an ellipse under a fill height of 20 ft atsoil unit weight of 125 pcf? See Figure 9-10.

9-10 What is the approximate ring deflection ofProblem 9-9 if the embedment is silty sand

compacted to 80% density AASHTO T-180? (d = 2.3%)

9-11 What is the maximum ring compression stressin Problem 9-9? (σ = 6.4 ksi)

9-12 In Problem 9-9, what is the middle ordinatefrom a cord 12.65 inches long at yield point stress of36 ksi at the crown of the deflected pipe?

9-13 If ring deflection, d = 16%, in Problem 9-9 isincreased by an additional ∆d = 2%, what is therequired horizontal soil-bearing capacity?

9-14 See Figure 9-11. What vacuum in the soil isrequired at invert level of a small-scale model buriedpipe if the internal vacuum at collapse is to be thesame for the model and a large prototype pipe? Thewater table is at ground surface. Saturated unitweight of soil is 125 lb/ft3

t = 0.281 tm = 0.0359ID = 84 IDm =H = 48 Hm =

Figure 9-11. Hydrostatic pressures (and vacuum) in soil required such that P’m = P’.


Anderson, Loren Runar et al "RING STABILITY"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 10-1 Top — Elliptical ring with uniform radial pressure P acting on it.

Bottom — Free-body-diagram of half of the elliptical ring showing approximate ring compression thrust inthe walls due to pressure P.


CHAPTER 10 RING STABILITY

The performance limit of ring stability is instability.Ring instability is a spontaneous deformation thatprogresses toward inversion (reversal of curvature).At worst, instability is ring collapse. Buried pipescan invert only if the ring deflects and the soil slipsat the same time. Instability of buried pipes isanalyzed as a soil-structure interaction. Thestiffness of the ring resists inversion. Soil supportsthe ring by holding it in a stable (near circular)shape. Soil resists inversion of the ring.

Two basic modes of ring instability are: 1. ringcompression, i.e., wall crushing or buckling at yieldstress; and 2. ring deformation. See Figure 10-1.Each is analyzed separately. Instabilities of buriedand unburied rings are also analyzed separately.

UNBURIED RING COLLAPSE

From Chapters 2 and 5, external pressure at collapseof an unburied thin-walled, circular, elastic ring isfound from the equations:

Pr/A = σf RING COMPRESSION COLLAPSE

Pr3/EI = 3 R I N G D E F O R M A T I O NCOLLAPSEwhereP = external pressure at collapse,r = mean circular radius of ring,A = wall area per unit length,t = wall thickness of plain pipe,m = r/t = ring flexibility,do = initial ring deflection (ellipse),E = modulus of elasticity,I = moment of inertia of the cross-sectional

area of the wall per unit length,σf = yield strength,Pcr = critical pressure on the circular ring.

The ring compression collapse equation is a functionof ring flexibility and yield strength.

Performance limit is wall crushing. The ringdeformation collapse equation is a function of ringstiffness, EI/r3. Performance limit is inversion. Ringstiffness is related to pipe stiffness; i.e., F/∆ =53.77EI/D3. Pipe stiffness can be measured by aparallel plate test or three-edge-bearing test. Thering deformation collapse equation is based onassumptions that the ring is elastic, and that the pipeis restrained longitudinally. Longitudinal restraintresults in a plane stress analysis. The Poisson ratiois not included.

In a plane strain analysis, longitudinal stress is zero,the Poisson ratio is included, and the pressure atcollapse is,

Pcr = 3EI/r3(1-ν2),

wherePcr = critical pressure, i.e., P at collapse,ν = Poisson ratio = 0.27 for steel,EI/r3 = ring stiffness.

The difference in fluid pressures between top andbottom of the pipe is usually ignored, but may besignificant. For plane stress analysis of criticalpressure at collapse of circular, unburied pipes,

Pcr = 3EI/r3 . . . . . (10.1)

For plain pipes (not coated, lined, rib stiffened, orcorrugated), critical pressure at collapse is,

Pcr = E/4m3 . . . . . (10.2)

Moment of Inertia, I

In order to evaluate ring stiffness, EI/r3, the momentof inertia, I, must be known. For plain pipes, I =t3/12. For corrugated pipes, tables of values for Iare found in the manuals provided by manufacturers.For steel pipes with lining and coating, consider


Figure 10-2 Transformed section of a unit slice of mortar-lined and coated steel pipe wall, transformed intoits equivalent section in mortar, for evaluating moment of inertia. Shown on the right is the elastic stressdistribution.

Figure 10-3 Effective T-section , comprising stiffener ring and an effective width of pipe wall — oftenassumed to be 50t in steel pipes.


a unit slice of the wall. See Figure 10-2.Discounting conservatively the bond between mortarand steel, moment of inertia, I, is the sum of theseparate moments of inertia of steel, lining, andcoating. Because the mortar is the critical material,steel is transformed into its equivalent width, n, inmortar. n = Es /Em. For the layers,

Ic = tc3/12

Is = nts3/12

Il = tl3/12,

and I = Ic + Is + Il

For pipes with stiffener rings welded to the pipe, themoment of inertia is found from the effective T-section. See Figure 10-3. The procedure isdescribed in texts on mechanics of solids. For steelpipes, the T-section comprises the stiffener ring andan effective width of pipe wall — in steel usuallyassumed to be 50t. The pipe wall between effectiveT-sections is ignored in calculating I.

Elliptical Ring Instability

Instability of non-circular, unburied rings is difficultto analyze. However, analysis is available fromtexts on mechanics of solids for one important case— a ring that is initially elliptical with ring deflectiondo. Vacuum increases ring deflection. Stress isc ritical at the spring lines, B, where the ring issubjected to both maximum ring compression stressand maximum flexural stress. See Figure 10-1. Vacuum, P, at collapse is found from:

P2 - [σf /m + (1+6mdo)Pcr] P + σ f Pcr /m = 0. . . . . (10.3a)

Equation 10.3a is applied by plotting values of P asa function of m for given values of do and for aconstant σf. For design, a simplification is proposedby Murphy and Langner (1985),

P = Pcr /(1+40d) . . . . . (10.3b)

Example 1

A plain 18-inch high-density polyethylene pipe isout-of-round (elliptical) by five percent. It isunburied. What is the internal vacuum at collapse?

ID = 16.217 and t = 0.857,DR = 21 = OD/t = 2m+1,m = 10 = r/t,σf = 3.2 ksi = yield strength',E = 110 ksi = modulus of elasticity,Pcr = 27.5 psi = E/4m3 from Equation 10.2,do = 0.05 = initial ring deflection.

Substituting into Equation 10.3a, internal vacuum atcollapse is, P = 21.5 psi. From Equation 10.3b, P =9.2 psi with ample safety factor included.

Example 2

Calculate the vacuum at collapse of a mortar-linedand coated steel pipe, for which:

r = 25.5 inch (D = 51 inch for steel),tc = 0.75 inch, Ic = 0.03516 (cr)ts = 0.175 inch, nIs = 0.00335tl = 0.5 inch, Il = 0.01042Em = 4(106) psi,Es = 30(106) psi,n = 7.5 = Es /Em,σf = 10 ksi for mortar (critical),I = 0.04893 in3,d = 0 = ring deflection (negligible).

From Equation 10.1, Pcr = 3Σ(EmI/r3). Assuming themean radius of the steel is, rs = 25.5, then rc = 25.9625 and rl = 25.1625. Substituting values, Pcr = 32 psi. Example 3

What does Equation 10.3a reduce to if Pcr = P? It iseasily shown that all of the terms cancel except6mdoP, which must be zero. The only way this termcan be zero is if initial ring deflection is do = 0. Asexpected, for a circular ring, P = Pcr.


BURIED RING COLLAPSE

Stress Analysis of Elliptical Ring

Figure 10-1 is a half ring free-body-diagram. At Bthe ring compression stress is,

σc = P(OD)(1+d)/2A . . . . . (10.4)

and the ring deformation stress (flexural stress) is,

σd = Ec(1/r'x - 1/rx) . . . . . (10.5)

Notation:P = vertical external pressure,OD = outside diameter of the circular ring,do = initial ring deflection,d = ring deflection after P is applied,r'x = radius of curvature at B due to initial ring

deflection do,rx = radius of curvature at B after vacuum P is

applied,E = modulus of elasticity,c = distance from the neutral surface of the

wall to the most remote fiber (t/2 for aplain pipe),

σd = flexural stress caused by ring deformation.

Substituting in values of radii of curvature for theellipse from Chapter 3, the equation for flexural (ringdeflection) stress becomes:

σd = (Ec/r) 3(d-do) / (1-2d-2do) . . . . . (10.6)

The maximum stress is the sum of Equations 10.4and 10.6; i.e., σ = σ c + σ f. The maximum stress at Bin a buried plain pipe is,

σ = Pm(1+d) + (E/2m)3(d-do) / (1-2d-2do) . . . . . (10.7)

where m = r/t = ring flexibility of a plain pipe.Equation 10.7 can be solved to find vacuum P atyield stress for brittle (rigid) pipes.

But yield stress is not failure for plastics or elasto-

plastics (metals), for which wall crushing can occuronly after ring compression stress (not flexuralstress) reaches yield strength. See Chapter 5.Therefore, Equation 10-7 is limited. Initial ringdeflection, do, depends upon compression of sidefillwhich requires analysis of pipe-soil interaction.

Equations 10.4 to 10.7 are based on elastic theory.Under some circumstances, plastic theory isjustified. For plain pipes and corrugated pipes, theplastic moment (at plastic hinging) is 3/2 times themoment at yield stress by elastic theory.

Ring Deformation Collapse of Buried Pipes

For the following analyses, vacuum is negativepressure, p, inside the pipe plus positive externalhydrostatic pressure, u. Both affect ring collapse.

Rigid Pipes

Because ring deflection of rigid pipes is negligible,rigid pipes are analyzed by ring compression exceptthat vertical pressure on the pipe is P+p.Ring compression stress is,σ = (P+p)(OD)/2A . . . . . (10.8)

whereσ = ring compression stress in the pipe wall,P = total soil pressure at the top of the pipe,

including water pressure, u,p = internal vacuum,OD = outside diameter of the pipe,A = wall area per unit length of pipe.

Area A is a transformed section if the wall iscomposite such as concrete reinforced with steelbars. For design, the ring compression stress, σ ,from Equation 10.8 is equated to the strength of thepipe wall, σf, reduced by a safety factor.

In the case of a very large diameter pipe, it may benecessary to consider the change in pressure ofliquids (both inside and outside) throughout the depth


of the pipe. For example, if the pipe is empty, butthe water table is above the top of the pipe, it maybe prudent to apply Equation 10.8 to the bottom ofthe pipe where total pressure P, acting up on thebottom, is greater than prismatic soil pressure on topby the increase in hydrostatic pressure between thetop and bottom. Of course, water inside the pipe willnegate any increase in external hydrostatic pressure.It is noteworthy that internal vacuum and externalhydrostatic pressure have little effect on the openingof cracks in rigid pipes. The 0.01-inch crack is nota suitable performance limit.

Flexible Pipes

Collapse of buried flexible pipes is either: 1. wallcrushing (ring compression) or 2. inversion (ringdeformation). Collapse due to longitudinal bendingis not included in this analysis. Bending deforms thepipe cross section into an ellipse with the shortdiameter in the plane of the bend. Bending strengthis decreased. Bending failure is collapse. Followingare procedures for evaluating the vacuum at whicha buried flexible ring collapses.

If the ring could be held circular, analysis would besimple ring compression — the same as for a rigidpipe. But flexible ring analysis anticipates ringdeflection, do, before the vacuum is applied. Ringdeflection depends upon ring stiffness and stiffnessof the embedment soil. It is assumed that pipes areinitially circular and empty, and that coefficient offriction between the pipe and the backfill is zerobecause of the inevitable breakdown of shearingstresses due to earth tremors and changes intemperature, moisture, and pressures. Theembedment is assumed to be granular. The flexiblepipe is often assumed to be thin-walled; i.e., OD =ID = D = mean diameter of the pipe.

Performance limit is collapse which occurs if thering either crushes due to ring compression, orinverts due to sidefill soil slip at B. See Figure 10-4.

σf = (P+p)(1+d)OD/2A . . . . . (10.9)COLLAPSE BY WALL CRUSHINGwhereσf = ring compression stress at yield stress in

the pipe wall at B,A = wall area per unit length of pipe,t = wall thickness = A for plain pipes,OD = outside diameter,P = external pressure at top of pipe,p = internal vacuum,d = ∆ D = ring deflection.

Area, A, is used for transformed composite sections,or ribbed, or ring-stiffened or corrugated.

At ring deformation collapse the soil must slip inorder for the ring to deflect. See Figure 10-4. In theleft sketch, the vertical pressure includes soilpressure and vacuum; i.e. PA = P+p. Before it isburied, the ring is circular, but as backfill is placed,the ring deflects into an ellipse.

If the ring is flexible, and if shearing stressesbetween pipe and soil are neligible, vertical andhorizontal soil pressures are related as follows:

PArA = PBrB = Pr = constant

where Pr is the product of pressure and radius ofcurvature at any point on the circumference of thering. For an ellipse,

rA /rB = (1+d)3/(1-d)3.

Therefore,

PB = PA(1+d)3/(1-d)3 = PArr . . . . . (10.10)

whererA = mean radius of curvature at the top A,rB = mean radius of curvature at the side B,PA = pressure on the pipe at A,PB = pressure on the pipe at B,d = /D = initial ring deflection,rr = (1+d)3/(1-d)3 = ratio of radii.


Figure 10-4 UNSATURATED SOIL — (left) Free-body-diagram of an infinitesimal cube at spring line, B,showing the stresses at incipient soil slip. (right) Vertical soil pressure, Po, supported by the pipe due to ringstiffness.

Figure 10-5 Free-body-diagrams for finding the vertical deflection of point B by means of the Castiglianotheorem.


p

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But for any pipe stiffness, F/∆ (or equivalent ringstiffness, 53.77 EI/D3), the ring itself is able tosupport part of the vertical pressure as it deflects.That part of the vertical pressure supported by ringstiffness is Po shown in the sketch on the right ofFigure 10-4. For a given ring deflection, Po can becalculated by the Castigliano theorem from the free-body-diagram in Figure 10-5. From the sketch onthe left, the moment M at point B can be evaluatedby noting that the slope at B does not change duringdeflection. θBA = 0. Knowing M, Castigliano can beapplied again using the sketch on the right fromwhich the vertical deflection, YB, is evaluated for thevirtual load p due to the forces on the ring quadrant.Knowing YB, the ring deflection, d, can be found asa function of EI and Po, and from this relationship, Po

can be found as a function of d. The result is:

Po = 96(EI/D3)d/(1-2d)

For small ring deflections, it is conservative todisregard 2d in the denominator; whereupon,

Po = Ed/m3 = 96(EI/D3)d

wherePo = vertical pressure on top of the pipe

that can be supported by ringstiffness,

Ed/m3 = Po for a plain pipe,EI/D3 = ring stiffness = 0.0186 F/∆ ,F/∆ = pipe stiffness,I = moment of inertia of the pipe wall per

unit length of pipe = t3/12 for plainpipe,

t = wall thickness for plain pipe,D = mean diameter,r = mean radius of the circular pipe,m = r/t = ring flexibility,d = ring deflection.

Pressure Against Soil at Spring Lines The horizontal pressure of the pipe against the soil atB is reduced by Po; i.e.,

PB = (PA+p-Po)rr - p

See Figure 10-6 for free-body-diagram andassumptions. For a plain pipe, substituting in Po,

PB = (PA+uA+p-Ed/m3)rr - p . . . . . (10.11)

wherePB = horizontal pressure of pipe on soil,PA = vertical external soil pressure at A,p = internal vacuum,EI/D3 = ring stiffness,F/ = 53.77(EI/D3) = pipe stiffness,D = mean diameter of the circular ring,r = mean radius of the circular ring = D/2,t = thickness of the plain pipe wall,m = r/t = ring flexibility,d = ∆ /D = initial ring deflection,rr = (1+d)3/(1-d)3 = ratio of vertical and

horizontal radii (maximum and minimumradii of the ellipse).

If soil at B does not have adequate strength, the soilslips, and the ring inverts.

Strength of Soil at Spring Lines

Because most embedment is granular, the followingis analysis of strength for granular (cohesionless)sidefill. See Figure 10-6. The horizontal strength ofsoil at point B, at soil slip, is soil passive resistance, σx = Kσ y

where σx = horizontal effective soil stress at B,

y = vertical effective soil stress at B,K = ratio of horizontal to vertical effective

stresses at soil slip (ring collapse),K = (1+sinφ)/(1-sinφ),φ = friction angle of the embedment, for which

values can be obtained from tests.

σy can be evaluated at the spring lines by methodsof Chapter 4.


µA = rwh for floods at level h.

σ__

_

Figure 10-6 SATURATED SOIL — Free-body-diagram of an infinitesimal soil cube at B, showing thestresses acting on it at incipient soil slip, and showing the shear planes at soil slip.

Figure 10-7 Pressure diagram for analyzing critical hydrostatic pressure on the bottom of the pipe at inversionof the ring from the bottom. rr = ry /rx.


x x

x +u+p

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Vacuum at Collapse of Buried Pipes

The total horizontal soil pressure on the pipe atspring lines B at soil slip is,

PB = Kσ y + uB . . . . . (10.12)

wherePB = total horizontal pressure on soil at B,Kσy = horizontal effective soil slip stress at B,uB = hydrostatic pressure in the soil at B.

When the horizontal pressure PB from Equation10.11 is equal to PB from Equation 10.12, the soil ison the verge of slipping — instability. For plainpipes, including all of the pertinent variables, theequation of equilibrium of sidefill at soil slip is,

p'(r r-1) = Kσ y + uB - (PA - Ed/m3)rr . . . . . (10.13)COLLAPSE BY RING INVERSION

wherep' = vacuum at collapse,rr = (1+d)3/(1-d)3 = ratio of vertical to

horizontal radii of elliptical pipe,m = r/t,r = mean radius of the circular pipe,t = wall thickness for plain pipe,d = /D = initial ring deflection — usually due

to backfilling,K = (1+sinφ)/(1-sinφ) at passive resistance,φ = friction angle of the embedment,σy = vertical effective soil stress at B,uB = hydrostatic pressure (pore water pressure)

at B if a water table is above the pipe,PA = soil and water pressure at A,E = modulus of elasticity of pipe material,I = moment of inertia of the wall cross section

per unit length of pipe,E/m3 = 96EI/D3 where,EI/D3 = ring stiffness,F/∆ = pipe stiffness = 53.77EI/D3.

From Equation 10.13, the vacuum at ring collapsecan be calculated. E/m3 can be replaced by96EI/D3, or 1.7856F/∆ for other-than-plain pipes.

If the embedment is not cohesionless, as assumedin the above analysis, the same procedure may beused except that the relationship between horizontaland vertical stresses at soil slip must be evaluatedfor each particular embedment. In the case of idealcohesive soil, σ y - σ x = 2C, where C is the cohesionof the soil. See Chapter 4.

Below a groundwater table, the hydrostatic pressureon the bottom of the pipe is greater than on top.Figure 10-7 shows buoyant pressure on the bottom,γ w(h+H+D). An empty pipe tends to float, but inthis analysis, is assumed to be restrained by theeffective soil wedge on top. Collapse occurs fromthe bottom for large, empty, flexible pipes with awater table above the pipe.

Example 1A thin-wall, welded steel penstock is 51 inches indiameter with a wall thickness of 0.219 inch. It isburied in embedment of dry, uncompacted sand to aheight of two ft above the top of the pipe. Unitweight of the sand is 102 pcf. The soil friction angleis 25o. Ring deflection was not controlled duringbackfilling, so the average initial ring deflection is8%. What is the internal vacuum at collapse? Thepertinent data are:D = 2r = 51 inches,t = 0.219 inch = average wall thickness,E = 30(106) psi = modulus of elasticity,I = 875(10-6) in3 = moment of inertia of

the wall cross section per inch oflength of pipe.

EI/D3 = 0.198 psi = ring stiffness,d = 0.08 = ∆/D = initial ring deflection

before vacuum is applied,


_

_

_

_ _

H = 2 ft = soil cover,γ = 102 pcf = unit weight of soil,h = 0 — no water table,uB = 0 — no hydrostatic pressure at B,ϕ = 25o = soil friction angle at the spring lines,K = 2.464 = ratio of horizontal to vertical

effective soil stresses at soil slip.

Substituting values into Equation 10.13, p = 11.9 psi= vacuum at collapse. Increasing ϕ by compactingthe soil greatly increases the vacuum at collapse.

Example 2

Specifications for the penstock of Example 1 limitring deflection to 5%. Had ring deflection been 5%,what would be the internal vacuum at collapse?From Equation 10.13, p = 18.7 psi. Because themaximum vacuum (atmospheric pressure) is only14.7 psi, the pipe would not collapse. This illustratesthe importance of limiting ring deflection by soilcompaction if the pipe is to be subjected to vacuum.

Example 3

Solve Example 1 if ring stiffness is neglected; i.e. thepipe is so flexible that ring stiffness cannot bedepended upon to support any of the soil load. With8% ring deflection, what is the vacuum at collapse?From Equation 10.13, neglecting the ring stiffnessterm, Ed/m3, vacuum at collapse is, p = 8.0 psi. Ringstiffness does provide resistance to inversion in loosesoil.

STABILITY DESIGN AND ANALYSIS

Flexible Pipes Under High Fills With No InternalVacuum and No Water Table

One example of flexible pipes under high fills is drainpipes under sanitary landfills. Some sanitary landfillsreach heights of hundreds of feet. They requiredrainage by a system of pipes. The leachate may beso corrosive that plastic pipes are specified. Forplastic pipes, the embedment must be select, and

carefully placed and compacted so that ringdeflection will not be excessive.

Ring Compression:

Ring compression stress must be less than short-term yield. Long-term yield strength does not applybecause, under constant deflection in selectembedment, the plastic relaxes faster than the yieldstrength regresses. See Chapter 20.

Ring Inversion:

In evaluating critical pressure, P, it is assumed that:

1. The basic deformation is from circle to ellipse. 2. Friction between soil and pipe is negligible.3. There is no vacuum in the pipe and no watertable. Equation 10.13 applies. p = 0; uB = 0; and, for highfills , σy ≅ PA ≅ P. The critical pressure P is thefollowing in three different forms:

P = 8Ed/(DR-1)3(1-K/rr) P = Ed/m3(1-K/rr)

P = 1.7854 (F/ )d/(1-K/rr) . . . . . (10.14)

where K = (1+sinϕ)/(1-sinϕ), and r r = (1+d)3/(1-d)3.From Equations 10.14, it is clear that if K > rr, thereis no soil slip regardless of soil pressure P. In fact,P becomes negative.

Equation 10.14 is plotted in Figure 10-8. The verticalscale is dimensionless critical soil pressure term,P/(EI/r3). The soil friction angle of the sidefill is ϕ.The horizontal scale is ring deflection d. If thecritical soil pressure term, P/(EI/r3), and the ringdeflection term, d, locate a point below or to the leftof a soil friction curve (ϕ-curve), the buried pipe isstable. If the point falls above and to the right of aϕ-curve, collapse is incipient — not imminent butpossible. Collapse may progress over a period oftime, due to soil dynamics such as earth tremors,wetting and drying, etc.


Compaction of the embedment has a significanteffect on stability. No safety factor is included, butsoil arching action assures a margin of safety. Ifring deflection is more than 20%, Equation 10.14(Figure 10-8) loses accuracy because of non-elliptical ring deformation and ring stresses beyondelastic limit. Collapse is not a problem if ringdeflection is less than about 10% even for very loosegranular soil (ϕ = 15o). In fact, pipes with more than10% ring deflection are usually rejected for reasonsother than structural instability. Including a safetyfactor, if ring deflection is more than 10%, minimumsoil friction angle should be increased — say to ϕ =30°.

The conditions for stability are assured if the sidefillsare good granular soil, carefully compacted, and ifthe ring deflection is less than 10%. Underconditions where mitigation is sought, live loads andheight of soil cover can be limited.

Example

PVC piping is proposed for drainage under asanitary landfill that is to be 600 ft high. Unit weightof the landfill is 75 pounds per cubic feet. Fifteenyears are anticipated to complete the landfill. Thepiping is to serve for 100 years.

1. What dimension ratio (DR) is required? DR isthe ratio of outside pipe diameter to wall thickness.Assume a 15-year yield strength of 5000 psi forPVC. Safety factor is to be 1.5. From Equation10.9, ring compression stress is, σ = 0.5γ H(1+d)DR.For a long-term sanitary landfill, it is prudent, forcleaning the pipes, to hold ring deflection to nearlyzero by compacted select sidefill. Solving Equation10.14 with d = 0, and with a safety factor of 1.5; DR= 21.3. Specify PVC pipe SDR 21(200) ASTM D-2241. SDR is "standard dimension ratio." It isdefined the same as DR; i.e., SDR = OD/t. PVCpipes are resistant to corrosive leachate.

2. What is the maximum allowable ring deflection ifembedment is loose with soil friction angle ϕ = 15°?From the Uni-Bell Handbook , for SDR = 21, the

pipe stiffness is F/ = 234 psi for E = 400,000 psi.From Equation 10.14 if P/(F/ ) = 600(75)/234(144)= 1.34, and ϕ = 15o, ring deflection at incipientcollapse is d = 11.4%.

Ring deflection can be controlled by the quality anddensity of the sidefill. From laboratory tests, selectcrushed stone compacted to 95% density AASHTOT99 (70% relative density) will hold ring deflectionto less than d = 5% under 600 feet of cover at unitweight of 75 pcf.

With Soil Support — No Water Table or Vacuum

Performance limit is soil slip of the sidefill. At soilslip, the pressure of the pipe against the soil at springline is equal to passive soil resistance.

Prr = Kσ y . . . . . (10.15)

P is soil pressure at the top of the pipe. rr =(1+d)3/(1-d)3. From Equation 10.15, ring deflection,d, can be found at soil slip.

Example

What is the ring deflection of a flexible steel pipe atsidefill soil slip if D = 72 inches and H = 4 ft?Embedment is poor, granular, loose soil. Unit weightis γ = 100 pcf. The soil friction angle is assumed tobe ϕ = 15o from which, K = (1+sinϕ)/(1-sinϕ) = 1.7.P = γ H = 400 psf. σy = γ Z = (100pcf)(4ft+3ft) =700 psf. For the first trial, let Z = 7 ft. Substitutingthese values into Equation 10.15,Prr = (400lb/ft2)(1+d)3/(1-d)3 = 1189 lb/ft2.Solving, d = 18%. For a second trial, let Z =4ft+2.5ft = 6.5 ft to account for the 18% reductionin vertical diameter of the deflected ring. Thissolution yields d = 17%.

The above analysis is conservative because ringstiffness is ignored. Ring stiffness is included inFigure 10-8 which comprises graphs at soil slip ofsoil pressure as a function of ring deflection andsidefill soil friction angle. It is assumed that soil


Figure 10-8 UNSATURATED SOIL — Soil pressure term at soil slip as a function of ring deflection and soilfriction angle. Empty steel pipes — no internal vacuum. When used for design, a safety factor should beconsidered.

Figure 10-9 Conditions for collapse of a flexible ringin liquefied embedment.

Liquefied Soil — Soil can liquefy if it is saturated,and shaken, and if density is less than about 80%,AASHTO T-180. The concept of liquefaction is asfollows. Pour loose dry sand into a quart jar to thetop. Carefully fill to the top with water. Replacethe lid. Shake the jar, remove the lid and turn the jarupside down. Liquefied soil gushes out because thesand is "shaken down." Repeat the experiment, but,this time, densify (tamp) the sand in layers as it isplaced in the quart jar. Then fill to the top withwater, replace the lid, shake, remove the lid, and turnupside down. The wet sand "hangs up" in the jar. Ithas not liquefied. In fact, the soil strength hasincreased because the sand is "shaken up."


cover is high enough that H is essentially equal to Z.Figure 10-9 shows interrelationships of the pertinentvariables. Two important conclusions are:

1. Compaction of the embedment has a significanteffect on pressure, P, at soil slip.

2. Soil does not slip if ring deflection is less thanabout 10%. Therefore, maximum allowable ringdeflection is often limited by specification, to 5%,including safety factor — or less if otherperformance limits prevail.

Example

If height of cover is H = 12 ft, what is the ringdeflection of a steel pipe at soil slip? D = 72 inchesand t = 0.245 in. The embedment is poor, loose soilfor which γ = 100 pcf, and friction angle is φ = 15o.EI/r3 = 0.78 psi. P = 1200 psf = 8.33 psi. Thepressure term is P/(EI/r3) = 10.7. From Figure 10-9,d = 11%. No problem is anticipated if ringdeflection is less than 5% — even in this poor soil.

Flexible Pipes in Liquefied Soil Embedment

If the embedment liquefies when a circular pipe isempty, the ring may be subjected to the hydrostaticpressures shown in Figure 10-8. If flotation isprevented, catastrophic collapse occurs from thebottom according to the classical equation,

Pr3/EI = 3; or h = (E/4γ )(t/r)3 for plain pipe.

Example

What is the height, h, of water table above thebottom of a steel pipe in embedment so loose that itcan liquefy and cause catastrophic ring collapse?Pipe:D = 51 inches,t = 0.219,r/t = 117.Soil:γ = 125 pcf, saturated,

h = height of water table above invert,P = hγ

Solving, h = 5.4 ft. This illustrates the importance ofdensifying embedment soil — including soil underthe haunches — if a water table could rise in theembedment.

With Soil Support and Internal Vacuum— No Water Table

The performance limit for internal vacuum and/orexternal soil pressure is ring inversion. Embedmentusually prevents total collapse. Critical vacuum, p,is sensitive to radius of curvature. Ring deflectionreduces critical vacuum. Because vertical radius ofcurvature, ry, is greater than r; ring stiffness, EI/ry

3,is less than EI/r3, and the vacuum at collapse is lessfor a deflected ring than for a circular ring.

The stability analysis can include internal vacuum, p,and the resistance of ring stiffness which, for a plainpipe, is Ed/m3. The horizontal stresses on theinfinitesimal cube, B, of Figure 10-6 can be equatedto passive soil resistance (soil slip). Solving forvacuum, p, at soil slip,

p(rr-1) = Kσ y - (PA-Ed/m3)rr

UNSATURATED SOIL . . . . . (10.16)

For notation, see the more general form, Equation10.17. Figure 10-10 shows graphs of Equation 10.16for a plain steel pipe with D/t = 288, and ringdeflection d = 10%, in granular embedment with twofeet of cover. It is noteworthy that critical vacuumis increased significantly by compacting theembedment (increased soil friction angle, φ). Theeffect of soil unit weight on critical vacuum is small. Example

A plain steel pipe is 51 inches in diameter with a0.187 inch thick wall. D/t = 274.


_

D/t = 288

Figure 10-10 UNSATURATED SOIL — Example of graphs constructed for the design and analysis of thefollowing buried steel pipe:

D = 48 inches t = 0.167 inches

H = 2 ftγ = 100 pcf = soil unit weight


The height of soil cover is 2 ft. The soil is silty sand(SM) with soil friction angle φ = 25o (lightcompaction), and unit weight of about 100 pcf. Ifthe buried ring deflection is discovered to be d =10%, what is the internal vacuum at soil slip?Because D/t is close to 288, Figure 10-10 may beused. p = 8 psi. Had the soil been compacted suchthat φ = 35o, all else unchanged, the pipe could havewithstood a vacuum of 12 psi. And had the ringdeflection been only 5% in compacted soil, the pipecould have withstood a vacuum of 26 psi, which isabove atmospheric pressure. By ring compressionanalysis, the critical vacuum would be increasedtenfold. For the design of pipes to withstand internal vacuum,a safety factor of 1.5 is recommended. It is prudentto require that embedment soil be denser thancritical. Critical density can be evaluated in the soilslaboratory. Even without a water table, percolatingwater and earth tremors tend to shake loose soildown such that ring deflection could increase andreduce internal vacuum at collapse.

With Soil Support —With Water Table Above the Pipe:

If the water table is above the top of the pipe, thesoil is in no danger of liquefaction if density of theembedment is 90% Standard Proctor (ASTM D698or AASHTO T-99). The height of water table, h,above ground surface, adds to the internal vacuum.The worst case is an empty pipe with the watertable above ground surface (flood level). See Figure10-11. Critical vacuum includes water table abovethe pipe and effective soil pressure. Using thestability analysis of Figure 10-6, but including ringstiffness and vacuum and water table, the equationof stability is,

p(rr-1) = Kσ y + uB - (PA + πrγ w/2 - Ed/m3)rr SATURATED SOIL . . . . . (10.17)where:p = vacuum and/or pressure due to flood level h

above the pipe,

σy = effective vertical soil stress at B,PA = total vertical pressure at A,K = (1+sinφ)/(1-sinφ),φ = soil friction angle,uB = water pressure at B = (h+H+r)γ w,h = height of water table above ground surface,γ w = unit weight of water = 62.4 pcf,E = modulus of elasticity of steel = 30(10-6)psi,d = ring deflection (ellipse) = ∆ /D,D = circular diameter of the pipe,m = r/t = ring flexibility,r = D/2 = radius of the circular pipe,t = wall thickness,rr = ry /rx.

The term, (πrγ w /2), is uplift pressure equivalent tobuoyancy of the empty pipe. If the pipe is full ofwater, this term is dropped from Equation 10.17.

Noteworthy from Figure 10-11:

1. A water table reduces the critical vacuum.

2. The effect of D/t on p is minor for values of D/tgreater than 240. Soil becomes the primaryresistance to vacuum. The pipe is a lining.

3. The significant variables are ring deflection andsoil density.

Example

A steel pipe of diameter D = 51 inches and wallthickness t = 0.187 inch is buried under a soil coverof H = 4 ft. The embedment is loose granular soilwith saturated unit weight of 125 pcf and φ = 15o.The water table is at ground surface. Ringdeflection happens to be 10%. What is the internalvacuum at ring collapse? Substituting values intoEquation 10.17, the critical vacuum is p' = 0.4 psi,which leaves little margin of safety against collapsein a flood. If ring deflection had been held to 5%,even in this poor soil, the vacuum at collapse wouldhave been 3.8 psi which is equivalent to a flood 4.8ft above ground surface. In the example above, ifembedment had been compacted such


_

_


D/ t = 240

JMiller

JMiller

that saturated unit weight was 130 pcf and φ = 35o,vacuum at collapse would be P' = 6.9 psi. If ringdeflection were 5% in this compacted embedment,"vacuum" at collapse would exceed 20 psi.

REFERENCES

Murphy, C.E. and Langner, C.G., (1985) UltimatePipe Strength Under Bending," Proceedings ofASME 4th International Offshore and ArcticEngineering Symposium, New York, N.Y.

PROBLEMS

10-1 What does Equation 10.3 reduce to if do = 0,and ring compression stress term, σy /m, isnegligible? (P'r3/EI = 3)

10-2 Derive Equation 10.6 for flexural stress:σf = (Et/D)(3do-3d)/(1-2do-2d).

10-3 A 10-inch PVC pipe Schedule 80 is to serve asa buried conduit for telephone cables under a river.Suppose that for some reason the backfill is washedoff the pipe at one location. What is the head ofwater at which the pipe will collapse? From theUni-Bell Handbook, DR = 18.13, and F/∆ = 370 psi. (380 ft)

10-4 For a steel pipe penstock full of water,D = 96 inches in diameter ,t = 0.375 inch,m = 128 = r/t,E = 30(106) psi,The pipe is buried in an embedment of silt and finesand. The height of soil cover is 16 ft. At certaintimes during the year, the water table rises to 2 ftbelow the ground surface. The pipe stiffness is F/∆= 5.5 psi. The dry unit weight of soil is 100 pcf. Soilfriction angle is 15o. The maximum vacuum that canoccur is 11.4 psi. What is the elliptical

ring deflection at which collapse occurs? Specificgravity of soil grains is 2.65. (6.1%)

10-5 In Problem 10-4 what is the internal vacuum atcollapse if the ring deflection. is 3%? (39 psi)

10-6 In Problem 10-4, what pipe stiffness, F/∆,would be needed to prevent collapse with a safetyfactor of 2 w.r.t. vacuum if ring deflection is 5% andvacuum is 11.4 psi? (24 psi)

10-7 What is the allowable external hydrostaticpressure plus internal vacuum on 150 class PVCpipe of nominal 10 inch diameter? (818 psi)E = 400 ksi Assumptions:σf = 7 ksi Temperature, 40οFID = 9.82 in Safety factor = 2OD = 11.12 in No soil restraint

(Soil liquefies)

10-8 If the pipe of Problem 10-7 is 10% out-of-round (ellipse), what is the allowable externalhydrostatic pressure? (88 psi)

10-9 What is the external water head at collapse ofa 4D PVC pipe, SDR 26, if a sudden internalvacuum of 12 psi occurs at the same time as theexternal water head? D/t = 25. E = 400 ksi.

(90.5 ft)

10-10 What is the allowable external water head ona 20D polyethylene pipe, DR 32.5, if the pipe isunburied and has an initial ovality (ring deflection) of5%? From manufacturer's engineering data, shortterm E = 115 ksi.

10-11 What horizontal soil-bearing capacity isrequired for a 36D PVC pipe, SDR 41? Pipestiffness is F/∆ = 28 psi? Initial ring deflection dueto careless installation is 15.9% and the vertical soilload on top of the pipe is P = 1 ksf. The soil is dry.Safety factor is 1.0 at the point of collapse.


Anderson, Loren Runar et al "ENCASED FLEXIBLE PIPES"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 11-1 Inserted encased pipe (liner) showing how pressure develops between the liner and theencasement.

Figure 11-2 Basic deformation of the encased, flexible liner showing how the ring is blown against theencasement such that a gap opens and a blister forms in the liner.


CHAPTER 11 ENCASED FLEXIBLE PIPES

Encased pipes include the following:a) Flexible pipe with concrete cast about it,b) Pipe inserted into another pipe or a tunnel,c) Liner for deteriorated pipe,d) Liner in an encasement in which grout fills theannular space between pipe and encasement.

All have a common performance limit — ringinversion due to external pressure. A special caseis a pipe that floats in fluid (water or grout) in theencasement. For equations of analysis, seeAppendix A. Two basic analyses are: liner anddouble wall.

LINER ANALYSIS

The inside pipe is the liner. It is so flexiblecompared to the encasement that it can be analyzedas a flexible ring in a rigid encasement.Performance limit is inversion of the liner. Collapsemay be time dependent based on "creeping" ringdeformation under persistent pressure. Collapse issudden inversion. Resistance to inversion is ringstability — a function of yield strength, σf, ringstiffness, EI/r3, and ring deflection, d. It is assumedthat pressure persists against the ring.

Internal Pressure Failure of Encased Rings

Internal persistent pressure at fracture is a specialcase of instability. Once the ring starts to yield, thediameter increases and the rupturing force in thewall increases. If the liner is designed to take theinternal pressure, there is no failure. If the rigidencasement is designed to take the internal pressure,there is no failure. The liner is an innertube. If bothliner and encasement expand, each shares inresisting the internal pressure. For analysis, therelative resistances of each to expansion by internalpressure must be known. It is conservative todesign both encasement and liner so that each cantake the full internal pressure.

External Pressure at Inversion of Encased Rings

The performance limit is inversion of the liner due toexternal pressure. Because encasements usuallyleak, if the water table is above the encasement,water pressure builds up between encasement andliner. See Figure 11-1. Even if the liner is bonded tothe encasement, water pressure peels the liner awayfrom the encasement and deforms the liner asshown in Figure 11-2. A gap forms where the lineris unencased — over an arc no greater than 180o.The unencased section is a blister — betweencircular and approximately elliptical. Pressure P atinversion is greater than it would be if the liner werecompletely unencased. But part of the liner isunencased.

For a completely unencased liner, the classicalanalysis, from Chapter 10, is,

Pc rr3/EI = 3 . . . . . (11.1)

UNENCASED CIRCULAR RING COLLAPSE

Pc r applies to a circular ring. For a deformed ring, elliptical analysis, from Chapter10, is:

P2 - [σf /m+(1+6mdo)Pc r]P + σ f P c r /m = 0 . . . . . (10.3)

Notation is listed in the paragraph, "Arc Angle 2α."Figure 11-3 shows plots of Equation 10.3 for a plainsteel pipe and a plain PVC plastic pipe. Because thePVC dimension ratio is DR = OD/t, it follows that2m = DR-1 for ring flexibility of the PVC pipe.

From Figure 11-3 it is evident that the effect of ringdeflection on P is significant for an unencased ring.However, for an unencased blister in a liner, thedifference between circle and ellipse is notsignificant. One exception is a slip liner smaller indiameter than the encasement. In the following, theunencased section of liner is circular.


Steel Pipeσf = 42 ksi E = 30(103) ksi

PVC Pipe

σf = 4 ksiE = 400 ksi

Figure 11-3 Examples of external pressure at collapse of unencased pipes with elliptical cross sections.


Figure 11-4 is a summary of inversion mechanismsfor encased rings. If the gap is small (blister doesnot form) ring compression applies. If a blisterforms over the gap, it acts either as an arch o r abeam. Analysis of beam failure is classical. Anunconfined arch could potentially invert. If the gapis small, arch instability is analyzed for a circular arc.If the gap is large, the arch is elliptical. A simple,but conservative analysis is circular analysis usingthe maximum radius of curvature of the ellipse.

Arc Angle 2α

An unknown is the arc angle, 2α. Figure 11-5shows the collapse of a flexible, circular, hingedarch. From Timoshenko (1956),

Pr3/EI = (π/α) 2 - 1 . . . . . (11.2)

Notation: P = pressure on the blister at inversion,D = mean circular diameter of the liner,r = mean radius of the circular liner,t = wall thickness,m = r/t = ring flexibility,E = modulus of elasticity,σf = yield strength of the liner,∆ = decrease in minimum diameter,d = ∆/D = ring deflection,T = circumferential thrust in liner wall,M = moment due to ring deformation,α = half arc angle (Figure 11-5),β = blister angle (gap angle),σ = maximum circumferential stress,h = height of water table above the liner,H = height of soil cover over the pipe,I = moment of inertia of wall cross section.

Equation 11.2 can be applied to an encased flexiblering by selecting a portion of the ring that isequivalent to the circular hinged arch of Figure 11-5.Figure 11-6 shows the typical inversion of a flexibleliner. The blister can be seen developing in theblister arc, β. At the ends of the arc, the moment ismaximum, and plastic hinges can be seendeveloping. These become the gunwales of an

inverted "boat hull" that rises up into the pipe. Athird plastic hinge forms at the keel. Points of zeromoment are circles. Points of maximum momentare triangles.

Points B (circles) are points of tangency to theencasement — and so moment is zero. Points C (circles) are points of counterflexure —and so are hinge points — zero moment.Points ∆ (triangles) are plastic hinges — points ofmaximum moment that isolate the collapsemechanism in arc, β.

Points B, C, and ∆ are about equally spaced.Therefore, β = 2α, which is equivalent to the arcangle of Figure 11-5, and can be analyzed byEquation 11.2, from which critical P is,

P = E[(π/α) 2-1]/12m3 . . . . . (11.3)

Angle α is unknown. From tests, the arc angle forplastic liners is roughly, α = 30o to 45o. Equation11.3 neglects decrease in circumference due to ringcompression. In order to find α, worst caseassumptions are as follows:

Assumptions:1. There is no bond, interlocking, or frictionalresistance between liner and the encasement.2. There is no pressure inside the liner.3. The liner is subjected to external pressure P.Leaks in the encasement allow pressure on the linerdue to groundwater table. External pressureincludes any vacuum that may occur inside the liner.

4. The liner is flexible. Initially, it fits snuglyagainst the encasement. But it may shrink, leavingan annular space between liner and encasement.The liner is snug (but not press-fit) in theencasement.5. The liner may be plastic which can creep underpersistent pressure over a period of time.6. The cross section of the blister is circular.Third-dimensional (longitudinal) resistance to theformation of a blister is neglected.


Figure 11-4 Inversion mechanisms for a blister in a liner.


Figure 11-5 Collapse of a hinged circular arch subjected to uniform radial pressure, P.

Figure 11-6 Typical inversion of a flexible liner at critical pressure P. β = blister angle.


When subjected to external pressure, the linershrinks. Bond breaks down between liner andencasement. External pressure, P, distributes itselfaround the entire surface of the liner. At onelocation a gap opens between liner and encasement,and a "blister" forms in the liner. The blisterdevelops wherever the radius is maximum, or at thebottom of the liner where external hydrostaticpressure is greatest. Failure is inversion of theblister. From inside the pipe the inversion looks likea boat hull with the keel longitudinal.

Ring compression thrust, Pr, is constant all aroundthe ring. Assuming P is constant, ring compressionthrust is Pry, where ry is the greatest radius ofcurvature.

Analysis is prediction of minimum pressure P atinversion. See Figure 11-4. The rationale is tocalculate the decrease in perimeter of the liner, andto find ry as a function of the arc angle. It is thenpossible to find P, both when ring compression stressis at yield, and when the blister inverts. The lesserof these two P's is critical.

Notation:r = constrained mean radius of the liner ring,ry = maximum radius of the blister,OD = outside diameter of the unpressurized liner,t = thickness of the liner wall,A = area of the liner per unit length of pipe,A = t for a plain liner,δ = decrease in circumference of the liner ring,DR = dimension ratio of the liner = OD/t,α = half arc angle based on casing radius r,β = half arc angle based on blister radius, ry,E = modulus of elasticity of the liner,σ = ring compression stress in the liner,σf = yield stress, (at 50 years of persistent

pressure?).

Inversion is by one of three mechanisms shown inFigure 11-4:a) Wall crushing if ring compression exceeds yield,b) Arch inversion,

c) Beam failure if the blister is flat.

The ends of the beam are the plastic hinges at theedges of the blister. The moment capacity of plastichinges is, Mp = 3Me /2, where Me is the elasticmoment at yield stress; i.e., Me = σf I/c.

Wall Crushing Analysis

The liner buckles when ring compression stressequals yield; i.e., σ = σ f, where σ = Pr/A. For plainliners,

P = σf /(ry /t) . . . . . (11.4)

where ry is the maximum radius (which occurs atthe blister) where the maximum stress, is Pry /t. SeeFigure 11-7 (top). But Pr/t is constant all around thering. Therefore, the decrease in perimeter of theliner due to ring compression strain is,

δ = 2πrPr y /Et . . . . . (11.5)

The geometrical decrease in perimeter of the liner is:

δ = 2rα - 2ryβ . . . . . (11.6)

The blister width is AB = 2rsinα = 2r ysinβ , fromwhich ry = rsinα/sinβ. Equating the decreases inperimeter from Equations 11.5 and 11.6,

β/sinβ = α/sinα - πσ f /Esinα . . . . . (11.7) which can be solved by iteration for β in terms of α.Of interest is angle α at flat blister (straight beam atβ = 0), for which β/sinβ = 1.

From Figure 11-7, blister width AB = 2rsinα =2rysinβ. Substituting the resulting ry = rsinα /sinβinto Equation 11.4, critical pressure is,

P = σfsinβ/(r/t)sinα . . . . . (11.8)


Figure 11-7 (bottom) shows pressure P = f(α ). Theinversion sequence starts with ring compressionyielding (wall crushing), P = σf /(r/t), at α = β. Ablister forms at yield stress, σf. With its greaterradius, ry, the blister rises and the blister angledecreases. But yielding is not inversion. Yieldingprogresses down the wall crushing curve from rightto left — shown by arrows. Based only on wallcrushing, the blister inverts at the least value of α.In the example of Figure 11-7, α = 33o. But beforethe blister inverts by ring compression, it might invertby arch inversion or by beam failure. In Figure 11-7,the arch inverts at α = 34o.

Arch Inversion Analysis

From Equation 11.3, for a hinged circular arch,critical P is a function of arch angle α. From Figure11-6, α is a third of the gap angle, B-B, whichcannot exceed 180o. The plastic pipe example ofFigure 11-7 shows arch inversion as a heavy solidline. P increases as α decreases. But α decreasesonly by ring compression strain (down the wallcrushing curve). So α is critical where wallcrushing intersects the arch inversion curve.Inversion occurs at α = 34o, and P = 57 psi. Forvery conservative analysis, set α = 60o (one-third ofthe 180o upper limit of gap angle).

Beam Failure Analysis

For low strength materials, inversion may be beamfailure. At inversion, the blister cross section is afixed-ended beam. With the length known fromradius r and angle α, pressure P can be found fromthe equation of stress,

σ = Me c/I

where M = PL2/12 for a fixed-end beam . . . (11.9)

But plastic hinges form at Mp = 3Me /2, Substituting,

9σ f = 2Pr2sin2α /(I/c) . . . . . (11.10)

For plain liners (no ribs), I/c = t2/6, and (r/t) = m.Substituting into Equation 11.10,3σf = 4P(r/t)2sin2α, from which, at inversion

P = 3σf /4m2sin2α . . . . . (11.11)

Example

A hypothetical plastic liner has the followingproperties:DR = 51,m = r/t = 25,E = 400 ksi = virtual modulus of elasticity over

50 years of persistent pressure,σf = 4 ksi = yield strength at 50 years of

pressure.

Find: Persistent pressure P that would causeinversion at 50 years of service.

From Equation 11.7, due to ring compression strain,β/sinβ = α /sinα - πσ f /Esinα. For trial values of α,corresponding values of β are found by iteration. Ofcourse, ring compression stress approaches infinitywhen the blister is flat; i.e., when β/sinβ = 1.Equation 11.7 becomes, α /sinα - πσf /Esinα = 1.Solving, at inversion, α = 33o. But inversion occursby arch inversion or beam failure before α shrinks to33o. From Equation 11.8,

P = σf sinβ/msinα . . . . . (11.8)

Figure 11-7 is a plot of Equation 11.8, showing long-term critical P as a function of α. Quick wallc rushing would occur at P = 160 psi as shown.Over the long term, the plastic ring creeps, α isreduced, and, therefore, inversion pressure, P, isreduced. The amount of creep is found from long-term tests. For most plastics, failure is archinversion. From Figure 11-7 inversion occurs at α =34o. Critical pressure is, P = 57 psi.


Figure 11-7 Example of graphs of critical pressure, P, as a function of the half blister angle, α, for threecollapse mechanisms in a typical plastic pipe. Yielding (wall crushing) may be time dependent because ofplastic creep.


Only in rare cases, such as low strength orundersized liners, will inversion be beam failure. Asan exercise only, if a blister in the liner could flatteninto a beam, at α = 33o from Figure 11-7,from Equation 11.11, P = 16.2 psi at 50 years ofpersistent pressure.

The above procedure provides an approximateinversion analysis for liners. It is a limit analysis.Critical pressure at inversion is greater than thec alculated pressure because of longitudinalresistance, bond, etc. Compared to tests, analysis isconservative.

The analysis allows for modifications andinnovations. For example, when shrinkage of theliner is due to conditions other than externalpressure, such differences can be included in theanalysis. One modification is the virtual modulus ofelasticity E (not actual modulus — but, mistakenly,referred to as long-term modulus) that allows forcreep of the liner over a long period of time. Thiscreep causes a decrease in perimeter of the linerunder constant pressure.

If the liner is initially out-of-round, a gap will formwhere the radius is maximum. Using maximumradius instead of the circular radius, analysis of theliner can proceed. Radius of curvature can becalculated from a measured offset from themidlength of a cord placed across the curved sectionof liner.

The procedure can be programmed for computer.It can also be presented as tables and graphs which,for most overworked engineers, may be the bestpresentation.

Plastic pipes are often used as liners forrehabilitating damaged pipelines, usually in a fullcontact fit in the damaged encasement.

ExampleA folded PVC pipe is inserted, heated and inflated tobecome a liner in an 8 ID encasement. Find criticalpressure P.

P = external pressure at inversion,E = 420 ksi = modulus of elasticity,σf = 3 ksi = yield strength,DR = 35 = standard dimension ratio,m = 17 = r/t = (DR-1)/2,ν = 0.38 = Poisson ratio

In the short term, critical pressure is P = σf /m = 176psi. From eight tests, the critical pressure was 172psi with a standard deviation of 38 psi.

If unencased, the maximum external collapsepressure is only

P = E/4 m3(1-ν 2) = 25 psi

Clearly the observed collapse pressure is muchcloser to ring compression theory than to thehydrostatic collapse theory. It is noteworthy that thetest liners failed by bulging inward throughout ablister angle less than about β = 90o in Figure 11-6.For a blister angle of 60o, α = 30o. From Equation11.3, arch inversion occurs at P = 107 psi. But thisis based on E = 420 ksi. Over the long term, ifvirtual E were only two-thirds as great, criticalpressure would be only 71 psi.

COMPARISON OF ANALYTICAL METHODS

The following is a rough comparison of somemethods proposed for analyzing the constrainedflexible pipes subjected to external pressure. Soilcontributes significant constraint to the buriedflexible ring subjected to uniform external pressure.Two equations used in service for design of ringsthat are circular or nearly circular (encased), are asfollows.

a) One form of the AWWA C950 formula inAWWA-M11 (1989) is,

P2 = 0.593RwEsEI/0.149r3

whereRw = buoyancy factor = 1-0.33(h/H),Es = soil stiffness = secant modulus.


b) The other formula, proposed by the LargeDiameter Pipe Division of PPI, is similar to Equation41 in the Uni-Bell Handbook of PVC Pipe (1986),

P2 = 1.3225EPc r

wherePc r = E(r/ry)

3/4(1-ν2)m3,Pc r = collapse pressure of unconstrained pipe,E = modulus of e las t ic i ty a t operat ing

temperature of the pipe,t = mean wall thickness,D = mean diameter = 2r,r = radius assuming no ovalization (out-of-

roundness or initial ellipticity),ry = maximum radius of curvature of the

ovalized pipe,ν = Poisson ratio,Et = soil stiffness (tangent modulus) of the

embedment.

These two formulas are almost identical. For thefollowing comparison, it is assumed that:m = r/t,Rw = 0.67 = worst case with water table at the

ground surface,I = t3/12 for plain pipe,ry = r (circular cross section),ν = 0.38 = Poisson ratio for PVC.

Substituting these values,

P2 = 0.3748EsE/m3 AWWA

P2 = 0.3864EtE/m3 UNIBELL

If the secant soil modulus Es is about the same asthe tangent soil modulus Et, the two equations areessentially identical. Certainly the concepts are thesame. Weaknesses can be found in both, but mosttroublesome is reconciliation of basic concepts andlimits. For example, if either of the soil moduli Es orEt or the pipe modulus E approaches zero, thenpressure P approaches zero. Not so. However,there may be a range in which the formulas areaccurate. For purposes of comparison, these twoformulas are combined into one called the service

formula. It is as follows:

P2 = 0.38KEtE/m3 SERVICE FORMULA

whereP = pressure at collapse of the circular

ring,KEt = horizontal tangential soil modulus,Et = vertical tangential soil modulus,m = r/t = ratio of mean radius and wall

thickness = ring flexibility term.

Four Additional Equations For Comparison

1. If the ring is encased in rigid or relatively rigid soil,then the equation for critical uniform external soilpressure is

P = σf /m ENCASED

whereσf = yield strength of the pipe wall. This

is ring compression failure.

2. If the ring is unconstrained;

P = E/4(1-ν2)m3 UNCONSTRAINED

This is buckling failure by classical analysis.

3. If the ring is supported by soil in which theeffective cross section stiffness is back-calculatedfrom ring deflection tests,

P = 0.14E/m3 + 0.27KEt EMPIRICAL

4. If ring deflection can be predicted by the Iowaformula, then an effective cross section stiffness isavailable and

P = 0.22E/m3 + 0.16KEt IOWA FORMULA

If ring deflection is known in terms of load or can becalculated by a formula such as the Iowa formula, interms of load, the effective pipe stiffness F/∆ can


be calculated. From F/∆ , ring stiffness EI/D3 is0.0186F/∆. The load F is a parallel plate load, butF/∆ can be related to other loads such as the loadingassumed in the Iowa formula or the empirical(measured) soil pressures in ring deflection tests.From these relationships, the effective ring stiffnessEI/D3 can be calculated for the effective crosssection stiffness.

A comparison of the service formula with the otherfour is shown on Figure 11-8. This particularcomparison is for a circular steel pipe buried in soilfor which KEt = 700 psi. The service formula fallsin the middle of the pack. The encased equation isa limiting condition represented by the curve in theupper right-hand corner for yield strength of 42 ksi.The other limit is unconstrained. The empirical andIowa formulas traverse from an asymptoticunconstrained on the left to an asymptotic encasedon the right. This traverse is reasonable andindicates the increasing support of the soil as the ringflexibility term D/t increases. The service formulais less responsive. Limited testing of vacuum tocollapse of buried pipes seems to confirm theempirical equation.

Nevertheless, the service equation is preferred bydesigners who are wary of soil placement and wouldlike an increased safety factor whenever theembedment must be depended upon for support ofthe pipe. More precise methods are available forevaluating P. See Chapter 10.

DOUBLE-WALL PIPES AND TANKS

Double-wall pipes comprise a pipe within a pipe.The objective is usually to eliminate leakage fromthe outside pipe, but may also be to reduce frictionalresistance to fluid flow. Deteriorated pipes aresometimes rehabilitated by inserting pipes of smallerdiameter called slip liners. In the followingdiscussion, the slip liner is the "pipe." See Figure 11-9.The host pipe is the "casing." The samenomenclature applies to double wall

tanks, often called dual containment tanks. They aremanufactured with double walls to provide doubleprotection against leaks, and to provide a sensitivemeans to detect leakage from the inside tank beforethe soil is contaminated. A "sniffer" monitors thespace between tanks. Dual containment tanks areused extensively for underground storage ofpetroleum products (service stations) and hazardousproducts.

Due to fluid pressure between the pipe and thecasing, the pipe is forced to one side of the casingleaving a gap shown in Figure 11-10. Usually, butnot always, the gap is on the bottom where externalliquid pressure is the greatest, and where the pipemay have an increased radius of curvature due to itsown weight. The following assumptions areconservative.

Assumptions

1. The pipe is flexible. As the pipe is forced to oneside of the casing, if external pressure, P, increases,the contact angle increases until the pipe is incontact with 180o of casing. At 180o of contact,inversion of the flexible pipe is incipient. Above180o, the contact angle increases toward inversion ofthe pipe with little or no increase in pressure.

2. In fact, no pipe is perfectly flexible. Beam actionand shearing resistance can help to resist theexternal pressure. Performance limit is incipientinversion of the pipe. Resistance is ring stiffness —a function of moment of inertia of the wall crosssection and modulus of elasticity of the material.

3. External pressure is distributed all around thepipe. This implies that any bond between the pipeand casing is broken down — a conservativeassumption.

4. The casing is circular. The casing may increasein radius due to pressure, P, between casing andpipe. This occurs in some dual containment tanks.


Figure 11-8 Comparisons of the SERVICE FORMULA and four other equations for evaluating the criticaluniform external pressure at incipient inversion of a circular, cylindrical, steel pipe. The buried formulas arebased on a horizontal soil modulus of KE t = 700 psi.

Figure 11.9. Pipe used as a liner in a circular casing.


Figure 11-10. Pipe subjected to external pressure, forced into 180° of contact with the casing, on the vergeof incipient inversion. A gap forms over an approximate ellipse in which maximum and zero moments areequally spaced as shown. Arch A-A can be analyzed as a circular hinged arch.

5. That portion of the pipe in the gap (not in contactwith the casing) is assumed to be compressed into asemi-ellipse. See Figure 11-10. Conservativeanalysis is based on a 180o gap.

6. Points of maximum and zero moment are equallyspaced around the semi-ellipse. As a consequence,arc A-A is a 60o hinged arch that is assumed to becircular in order that it can be analyzed by classicalmethods.

NotationP = critical external pressure on pipe,r c = inside radius of the circular casing,rp = radius of the originally circular pipe,ry = maximum radius of the ellipse,E = modulus of elasticity,σf = yield stress,DR = dimension ratio,t = wall thickness,

A = wall cross-sectional area per unit length,I = moment of inertia of wall cross section,α = half arc angle of the critical hinged arch,a = minor semi-diameter of the ellipse,rc = major semi-diameter of the ellipse.

Equations

The perimeter of the ellipse is π(a+rc)

From the ellipse, Figure 11-11, ry = rc2/a.

From Timoshenko (1956) and Figure 11-10,Pry

3/EI = (π/α) 2 - 1.

For arc A-A, critical α = 30o; Pry3/EI = 35.

From geometry, the decrease in perimeter of thepipe is,

2πr p - πrc - (π/2)(a+rc)


From ring compression, the decrease in perimeter ofthe pipe is,

2π rp (Pry/EA)

Equating decreases in perimeter, and solving,

a = 4rp (1-Pry /EA) - 3rc

But from geometry of an ellipse, a = rc2/ry. Equating

values of a and rewriting,

ry2 - (EA/P)(1-3rc /4rp)ry = - (EA/4P)(rc

2/rp) . . . . . (11.12)

Equation 11.12 is a quadratic equation from which ry

can be evaluated for any assumed value for P.

Analysis

The critical analysis is the evaluation of pressure, P,at inversion of the pipe. If ry is known, P can befound. At α = 30o,

Pry3/EI = 35 . . . . . (11.13)

Maximum radius of curvature, ry, is found fromEquation 11.12. Unfortunately, ry is a function of P.The easiest analysis is by iteration. Assume a value,P'. The first assumption for P' may be found fromEquation 11.13 assuming that ry is equal to rc. FromEquation 11.12, the correct value for ry can then befound for the assumed P'. Substituting this ry intoEquation 11.13, it is possible to solve for P at radiusr y. The solved P is undoubtedly different from theoriginal assumed P'. They must be equal.Therefore, the next iteration is a repetition of theprocess assuming that the next P' is the previouslysolved P. Iterations continue until assumed P'equals solved P.

The procedure is as follows: 1. The "knowns" are values of rc, rp, E, and t. 2. Assume a value for P. Call it P'. Try P' fromEquation 11.13, if ry = rc.

3. Solve for ry in quadratic Equation 11.12. 4. Substituting ry into Equation 11.13, solve for P. 5. The solved value of P becomes the assumed P'for the next iteration.6. The process is repeated until the solved value ofP is equal to the assumed value, P', which is thecritical pressure at incipient inversion of the liner.

If pressure is persistent over the long term, plasticscreep. The quantity EA/P in Equation 11-12 mustthen be based on the "virtual" modulus of elasticity— not the real modulus. In the literature, virtualmodulus is defined, erroneous ly, as "long-term"modulus of elasticity.

It is possible to change the radius of the casing, rc,and to find corresponding value of P as describedabove. I f rp remains constant, P vs. rc can beplotted. If the casing remains circular, but isexpandable, the expansion of the casing can beplotted on the same coordinate axes of the graph ofP vs. rc. The intersection of the two plots is criticalpressure, P. For some tanks, such as fiberglass-wrapped dual-containment tanks, the casing is soflexible that it does not remain circular. It expandsunder internal pressure, but is deformed by the pipe(liner) which is forced into a semi-ellipse. The linerexerts non-uniform pressure on the casing such thatPxrc = Pry.

For dual-containment pipes and tanks with ribbedliners that are made of yield-sensitive materials suchas fiberglas, Timoshenko (1956) suggests theSouthwell solution, from which critical P is,

Pr/h = σf/[1+4(σ f /E)(r/h)2] . . . . . . . . . . . (11.14)

where h is the height of the ribbed section frominside diameter to outside diameter. Of course, theribs could be corrugations, etc. Example 1

What is the external pressure, P, at incipientinversion of a PVC pipe encased in a circular casingwith no gap before pressure is applied? For the pipe,


rp = 15 inches outside,E = 400,000 psi,σf = 4,000 psi,DR = 41,t = 0.732 inch,A = 0.732 in2/inch,I = 0.032685 in3 = t3/12,α = 30o (assumed critical),EI = 13,074 lb in,EA = 292,800 lb/in.

With no gap, rc = rp = 15 inches. Assume a valuefor P'. If ry = rc, then from Equation 11.13, P' =457,594/ry

2 = 135.58 psi. As the first assumption,try P' = 130 psi.

Assume P' = 130 psi. From Equation 11.12, ry = 15.42 inches. FromEquation 11.13, P = 124.71 psi.

Assume P' = 124.71 psi.From Equation 11.12, ry = 15.41 inches. FromEquation 11.13, P = 125.08 psi.

Assume P' = 125.08 psi.From Equation 11.14, ry = 15.41 inches. FromEquation 11.13, P = 125.05 psi. Only two iterationsare necessary. P = 125 psi at rc = rp.

It is noteworthy that, based on ring compression atyield stress (σf = 4000 psi), critical pressure is P =195.2 psi. For ring compression failure, the pipemust be restrained in a circular cross section. Withno restraint, the pipe will collapse if Pr3/EI = 3.Solving, P = 11.62 psi.

If, hypothetically, the PVC were yield sensitive(which it is not), critical pressure from Equation11.14 would be P = 11 psi. If the pipe wall wereribbed such that h = 1.5 inch, then P = 80 psi.

Example 2

Find P if rc is greater than rp. The pipe is smallerthan the casing to allow for insertion. Figure 11-12provides a "feel" for the difference between the

radius of casing, rc, and radii of the pipe, rp.Following procedure of Example 1, if rp = 15.0 andrc = 17, P = 13.4 psi. Figure 11-13 is a plot ofvalues.

The assumptions that: the pipe is elliptical in the gap,that arc angle α is 30o, and that inversion is onlytwo-dimensional, all reduce the precision of thevalues for P. Therefore, a safety factor is needed. Until test data are available, a safety factor of twois suggested for design.

REFERENCES

AWWA (1989), Steel Pipe — AWWA ManualM11, 3ed, American Water Works Association.

Timoshenko (1956), Strength of Materials, Part II,3 ed, D.Van Nostrand, p 189.

Uni-Bell (1986), Handbook of PVC Pipe, Uni-BellPVC Pipe Association.

PROBLEMS

11-1 What is the pressure at collapse of a PVC pipeDR 26 encased in concrete? Include the decreasein circumference of the pipe. Assume the pipe iscircular before the pressure is applied, and α = 30o.For PVC, E = 500 ksi, and σf = 5 ksi.

11-2 Derive Equation 11.3.

11-3 ID of an encasement is 8.5 inches. Assumingstrain due to creep is 5% percent over 50 years,what must be the DR for an 8D inch OD PVD linerif external pressure on it is 25 psi?E = 400 ksi virtual 50-year modulusσf = 4 ksi = 50-year strength

11-4 Plot a graph of external collapse pressure as afunction of collapse angle for a circular HDPE pipeDR 32.5 encased in concrete.


Figure 11-11. Quadrant of an ellipse,showing notation for analysis.

Figure 11-12. Sketch to scale of a pipe of 15- inchradius in casings of 15- 16- and 17-inch radii.

Figure 11-13. Examples of critical external pressure on a PVC pipe in a circular casing with larger insidediameter than the outside diameter of the pipe. The pipe is at incipient inversion when subjected to externalpressure, P.


11-5 Figure 11-14 is a standard egg-shaped sewerwith a plastic liner. The sewer leaks and thewater table rises 26 ft above the sewer. What isthe width, e, of the gap at C ? At 26 feet of head,

P = 11.27 psi,ro = 36 inches,

t = 0.945 inch,E' = 250,000 psi = virtual modulus at 50

years of persistent pressure,e = mid-ordinate width of gap,Circumference = 100 inches,Arc length BCD = 22.8 inches. Assume circulardeflection of arc BCD. (e = 0.46 inch)

Figure 11-14 Cross section of standard egg-shaped sewer showing (right) the external pressure on a liner.


Anderson, Loren Runar et al "RIGID PIPES"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 12-1 Procedure for conducting the three-edge-bearing (TEB) test on rigid pipes. The load at failureper unit length of pipe is called the D-LOAD.

Figure 12-2 Rigid pipe showing (left) how the ring is forced to support the weight of backfill, W, if the sidefillsoil is not compacted; and (right) the basis for the Marston load analysis.


CHAPTER 12 RIGID PIPES

Rigid pipes do not deflect enough for deflection toaffect soil pressure against the pipes. The soil is aload on the pipes. Rigid pipes include Portlandcement concrete pipes (both reinforced andunreinforced) and vitrified clay pipes. Other pipesmay be rigid under certain conditions. Cementmortar lined and cement mortar coated (CML/CMC)pipes perform as rigid pipes when buried in loose soilbecause pipe stiffness is relatively greater than soilstiffness. In densely compacted soil, CML/CMCpipes may be flexible, or, according to somedesigners, may be semi-rigid. Design of concrete pipes is described in theConcrete Pipe Design Manual, published by theAmerican Concrete Pipe Association [ACPA(1993)]. Pipe strengths are specified by standards.The ACPA has wisely left to the manufacturer theresponsibility of making pipe that meets thestandards. In general, performance limit islongitudinal cracking of the pipe wall due to internalor external pressures.

INTERNAL PRESSURE DESIGN

Designers often assume that concrete and clay cantake no tension. In fact, both can take tension.Nevertheless, unreinforced concrete or clay pipes areusually not designed to take internal pressure becausehoop strength is lost if longitudinal cracks form duringcuring or handling. With tension reinforcement, rigidpipes serve well as pressure pipes. Considerreinforced concrete pipes. If performance limit isleakage, the reinforcing steel must be pre-tensioned(or post-tensioned) such that the concrete is incompression before internal pressure is applied.When internal pressure is applied, the tensioned steeltakes additional tension and stretches. This relievescompression in the concrete. The concrete will notleak until it begins to take tension. Therefore, toavoid leakage, the steel must take the entire internalpressure. Pre-tension force in the steel is,

Ts = Pr/(1 + EsAs /EcAc) . . . . . (12.1)

whereTs = tension in the steel per unit length of pipe,P = internal pressure,r = inside radius,Ac = area of concrete per unit length,As = area of steel per unit length,σ = stress,E = modulus of elasticity.

A safety factor should be included. High strengthsteel is cost effective. Small diameter steel rodsincrease bond between steel and concrete.Procedure specifications for the manufacture ofprestressed pipe are usually left to the manufacturer.The pipeline engineer writes performancespecifications.

For typical reinforced concrete pipes, if Ac /As = 100and Es/Ec = 5, the pre-tension force is Ts = 0.95Pr; orsay, conservatively, Ts = Pr. With pressure, P, in thepipe, tensile force in the steel is doubled. It is prudentto check maximum stresses in the concrete and steel:σc = Pr/Ac

σs = 2Pr/As

EXTERNAL PRESSURE DESIGN

For rigid pipes, external pressure design is based onloads on the pipe — not stress or strain. See Figures12-1 and 12-2. The following analysis is historicaland simplistic. It is no longer proposed by ACPA, butis presented here as the basic rationale for analysis.For design,

APPLIED LOAD = ALLOWABLE LOAD. . . . . (12.2)

APPLIED LOAD = (Wl + Wd) . . . . . (12.3)


Variation of Lf from class C to class B is based on the width of the bedding.For Class B bedding, width of bedding is 0.6(OD).For Class C bedding, width of bedding is 0.5(OD)

Variation of Lf in class A is based on the percent of area of reinforcing steel:

Reinforced As = 1.0%, Lf = 4.8Reinforced As = 0.4%, Lf = 3.4Plain As = 0, Lf = 2.8

Theoretical values of load factors are based on moments at A, which are:For the TEB load, MA = 0.318 Wf rFor Class A bedding, MA = 0.125 Wf rFor Class D bedding, MA = 0.293 Wf r

Figure 12-3 Loadings on rigid pipes showing the three-edge-bearing test load (TEB) and the three soil loadingsidentified by the American Concrete Pipe Association, showing the original load factors, Lf, for each. Wf =load at failure. Failure is a longitudinal crack at point A.


whereWl = live load on the pipe, per unit length of

pipe,Wd = dead load,sf = safety factor,OD = outside diameter,ID = inside diameter,Lf = load factor,D-load is the load at failure in a three-edge-bearingtest = F-load per unit length. Lf = load factor, whichis determined by the bedding and by steel reinforcing.

The three-edge-bearing (TEB) test is conducted asshown in Figure 12-1. The TEB load at failure iscalled the D-load. In general, failure is the ultimate(maximum) load on the TEB test pipe. However, inreinforced concrete pipes, failure is often defined asthe TEB load at which longitudinal cracks open to awidth of 0.01 inch. The 0.01-inch crack came aboutin the 1930s when graduate student, Bill Schlick, wasinspecting reinforced concrete culverts in order toevaluate their performance. This task was assignedto him by his dean, Anson Marston, of the College ofEngineering, Iowa State College, Ames, Iowa. Theonly indication of inadequacy that Schlick couldidentify was cracking. So he put a half-inch-widestrip of 0.01 steel shim stock in his pocket, andproceeded to classify adequacy on the basis of crackwidths into which he could insert the 0.01-inch steel.This became the standard. It has proven to be betterthan happenstance. Cracks less than 0.01 inch tendto close by autogenous healing; i.e., by continuedhydration of the silica gel in the Portland Cement.Cracks greater than 0.01 inch can possibly allowoxygen to reach and corrode the reinforcing steel. In Equation 12.3, the live load Wl is the effect of liveload on the top of the pipe due to surface live loads.The wheel load is multiplied by an impact factor of1.5 for highway loadings. The dead load Wd is thevertical soil pressure on the pipe. It is usually takenas the weight of the prism of soil over the pipe.However, Figure 12-2 shows how the entire backfillload in a trench could be imposed on

the pipe if sidefill soil is not adequately compacted. Itis difficult to predict how much of the backfill load isimposed on the pipe. Anson Marston pioneered loadanalysis. The theoretical Marston load does notaccount for soil anomalies such as compaction of soildirectly against the top of the pipe. Arching action ofthe soil is ignored. A soil arch is formed if the sidefillis compacted. The soil arch supports much of thebackfill in the trench. At most, the pipe only has tosupport the prism of soil, γ H(OD), above it. In fact,a compacted soil arch relieves the pipe of essentiallyall of the vertical pressure except for loose soil in thefirst lift above the pipe. Soil arching can be assuredby compacting sidefill up to one soil lift above the topof the pipe; but avoiding compacting the first liftdirectly over the pipe. This result is backpacking,which protects the pipe from soil pressureconcentration, and develops a soil arch.

ALLOWABLE LOAD = FAILURE LOAD. . . . . (12.4)

FAILURE LOAD is based on the three-edge-bearingtest. The three-edge-bearing load at failure is the D-load. For unreinforced rigid pipes the D-load is themaximum load in lbs per ft of length of pipe. Forreinforced concrete pipes, the D-load is the load inpounds per ft of length of pipe per ft of ID. Whenthe pipe is buried, the soil load is less severe than theD-load. Therefore, a load factor, Lf, increases theallowable soil load above the D-load pipe strength.Figure 12-3 shows the four historical loads on rigidpipes. At left is a parallel plate load which, foranalysis, is tantamount to the TEB load. The otherthree are assumed to be soil loads in service.Horizontal soil support is neglected because the rigidring does not deflect and develop passive soil support.For each of the three bedding classes, theoreticalfailure load, Wf, is found from Appendix A. In allcases, Wf is greater than D-load; therefore,

Failure load, Wf for each bedding class is the D-load times its load factor, Lf.


Figure 12-4 Effect of diameter on load capacity on an equivalent beam that cracks at point A. OD is an overlyconservative beam length.

Figure 12-5 Backpacking, showing decreased vertical soil pressure on the ring, and limits of soil strength at thespring lines; i.e., active on the left and passive on the right.


Example

Sidefill soil support is to be included in the load factorfor Class A bedding. What is the revised theoreticalload factor? From Figure 12-3, load factor Lf forClass A bedding is 2.546. The critical moment is MA

= Wf r/8. Including sidefill support, soil pressure isthe third case of Appendix A for which MA = Wf r(1-K)/8. Sidefill support is at least active pressure forwhich K = (1-sinϕ)/(1+sinϕ). If ϕ = 30o, K = 1/3,and MA = Wf r/12, including sidefill support, therevised Lf = 3.820. The sidefill increases L f from2.546 to 3.820 — a significant 50% increase.

EVALUATION OF THE REQUIRED D-LOAD

Taking load factors into account, the rationale fordesign of rigid pipes is the equating of applied load toallowable load; i.e.,

(Wl + Wd) = (D-load)Lf for non-reinforced pipes, and

(Wl + Wd) = (D-load)Lf (ID)for reinforced concrete pipes.

Resolving these equations, the required D-loads forburied rigid pipes are:

D-load = (Wl + Wd)/Lf . . . . . (12.5)UNREINFORCED RIGID PIPES

D-load = (Wl + Wd)/Lf (ID) = P(OD)/(ID)Lf . . . . . (12.5)

REINFORCED CONCRETE PIPES

P is the vertical pressure on the pipe. Loads, W, arebased on complex pipe-soil interactions such as pipesettlement vs. soil settlement (positive or negativeprojecting pipe), compaction techniques, water table,bedding, etc. W is further complicated by boundaryconditions (trench vs. embankment), imperfect trenchconditions (compressible topfill), properties of thetrench wall soil, tunnels (pipes jacked-into-place), etc.

Recognizing the complexity as well as the importance

of the soil loading, before 1993, the AmericanConcrete Pipe Association (ACPA) published valuesfor load factor, Lf , based on the ACPA classificationof trench beddings shown at the bottom of Figure 12-3. Note how the load factors, Lf , are about the sameas theoretical values. The empirical ACPA loadfactors are based on the assumption that soil pressureon the top of the pipe is approximately uniform. It isthe bedding that causes pressure concentrations.Most engineers assume that Class D bedding isimpermissible, a term first proposed by Marston. Itis noteworthy that, in general, no safety factor isneeded in Equations 12.5. Margins of safety arealready in place — soil arching, horizontal support ofthe pipe by the sidefill, etc. An effective way for thedesigner to capitalize on these margins of safety is tospecify a select compacted embedment; and then toenforce it by inspection. Experienced installerscomply.

An additional margin of safety is provided by theACPA definition of load, W, based on outsidediameter. In fact, the mean diameter or insidediameter is more nearly correct. See Figure 12-4.Failure is a crack at A, due to a moment. The clearspan that causes the moment is ID, or possibly meandiameter — not OD. For reinforced concrete pipes,the D-load is conservatively multiplied by ID — notOD or mean diameter.

Example

What height of soil embankment is allowable over a24-inch vitrified clay pipe of standard strength if thesoil weight is 125 pcf and the bedding is Class B?The nominal pipe size is 24 inches. From Figure 12-3,the Class B load factor is 1.9. From Table 12-1, thestandard strength is 2600 lb/ft. Neglecting live load,and substituting values into Equation 12-5, theallowable height of soil is H = 19.76 ft; or say H = 20ft. Clearly, the effect of live load is negligible. Nosafety factor is needed.

Backpacking

The allowable load on a buried rigid pipe can be


Table 12-1 D-LOADS—ASTM Standards for Rigid Pipe Manufacturers.

Table 12-1 is still used for some conservative pipeline design and analysis. However, in its 1993 Concrete PipeTechnology Handbook, the American Concrete Pipe Association (ACPA) has abandoned the load factorconcept in favor of an ASCE design procedure. The ASCE procedure is based on tests and on finite elementanalysis that include sidefill soil support and boundary conditions — both pipe and trench — and on soil typeand compaction, etc.


doubled by backpacking. Backpacking iscompressible material against the pipe. Styrofoamhas been used. Uncompacted soil has been used.Bales of straw and leaves have been tried withquestionable success. The concept, called imperfecttrench method, is that backpacking is similar topacking used to protect fragile products for shipment.Organic material may be suspect, but uncompactedsoil is effective. Assuming that the embedment iscohesionless, soil failure (soil slip) is incipient if theratio of maximum to minimum principal stresses isgreater than K = (1+sinϕ)/(1-sinϕ), where ϕ = soilfriction angle.

The performance limit of a rigid pipe depends uponfailure of the sidefill soil. See the unit cube of soil Bat the spring lines in Figure 12-5. Soil failure can beeither active or passive. If the horizontal pressure. Px

of the pipe against the soil at B is less than σy /K, thesoil slips at active resistance, and the pipe wallcollapses inward. This is shown on the left side ofFigure 12-5. If the horizontal pressure, Px of the pipeagainst the soil at B is greater than σyK, the soil slipsat passive resistance, and the pipe wall collapsesoutward. This is shown on the right side of Figure12-5. If the height of backpacking is equal to the OD,and the stiffness is half as great as the embedment,the pressure on the pipe is half of what it would bewithout the backpacking. In Figure 12-5, half of thebackpacking is shown above, and half below the pipe.Accordingly, some designers conservatively specifyhalf a diameter of backpacking above and below.The compressibility should be no more than half thecompressibility of the embedment. In order toprevent passive soil slip at B, the embedment mustnot be excessively compressible. The backpackingmust retain soil arches over and under the pipe. Thisrationale is conservative.

An alternate evaluation of height of backpacking isthe classical equation for stresses around a hole withradial stress, σy, and tangential stress, σ x,

σx/σ y = (ρ2+r2)/(ρ2-r2) . . . . . (12.5)

See cube, C, at the top of an imaginary soil vault inFigure 12-5. What is radius ρ at which σx /σ y = 3,assuming that soil friction angle is 30o? The rationaleis that a soil vault forms over the backpacking. It isstable at such radius, ρ, that σx < Kσ y. Butbackpacking is needed to prevent soil particles fromfalling from the vault. From Equation 12.5, ρ =1.414r. With a safety factor of two, a good rule ofthumb for pipe protection is,

Height of backpacking should be at least half thepipe diameter. Backpacking under the pipe is notnecessary.

Another rule of thumb is, Backpacking permits twice the pressure P at topof the pipe. For deep burial, maximum height of soilover the pipe can be doubled.

ExampleUsing typical values, let backpacking pressure on topof the pipe be σy /2. For embedment at the springlines, K = 3. What are the limiting ratios of horizontalto vertical soil pressure on the pipe under high cover?See Figure 12-5.

For active soil pressure,RATIO = 2σx /σ y = 4/3; which is improbable becausethe rigid ring is usually stiff enough resist the 4/3 ratioof horizontal to vertical pressures.

For passive soil resistance,RATIO = 2σx/σ y = 6; which is impossible. σ y /2 isless than σx. The ring could not fail outward if σ y/2on top is less than σx on the side. Backpackingallows a broad range of tolerance.

MARSTON LOAD

In the analysis of soil loads on buried rigid pipes, theMarston load is still used by some pipeliners. Con-sider a rigid pipe in a trench as shown in Figure 12-2.The load, W, is the weight of backfill in the trenchminus the frictional resistance of the trench walls.


Figure 12-6 Comparison of soil pressures against rigid and flexible pipes.

Figure 12-7 Details of reinforced concrete pipes.


Based only on soil friction angles, the evaluation of Wis logical. Less logical are assumptions, such as:rigid trench walls, no sidefill soil support, no pipesettlement, no soil arch over the pipe, etc. Withoutsoil arching, the load on the pipe continues to increasedirectly as the height of soil cover in-creases. Thisrelationship has limitations. Marston providedadjustment factors, which are based on additionalassumptions. The imprecisions of soil properties,density, and placement, are usually greater than theprecision achieved by applying adjustment factors.For explanation of the Marston load, see Spangler(1973), Chapters 25 and 26.

ACPA DESIGN PROCEDUREThe above rationale for rigid pipe performance hasbeen adjusted and refined by ACPA to respond toneglected factors, and to refine simplistic assump-tions. Sidefill support is now included, soil types aretaken into account, etc.

COMPARISON OF SOIL PRESSURES ON RIGIDAND FLEXIBLE PIPESShown in Figure 12-6 are cross sections of a rigid anda flexible pipe. The ring deflects 6% duringbackfilling. Assuming elliptical deflection, hori-zontalpressure on the flexible pipe is Px = 1.5P.

Horizontal pressure on the rigid pipe is activepressure, P/3. The vertical pressure on top is greaterthan P, but not more than 2P. One exception ob-tainsif the first lift of soil over the pipe is heavilycompacted against the pipe, leaving pressure con-centration on top that could exceed 3P. The pres-sure on the bottom is often more critical because ofdifficulty in placing soil under the haunches. Theseproblems are anticipated by designers. UpdatedACPA (1993) software are available.

REFERENCESACPA (1993), Concrete Pipe TechnologyHandbook, American Concrete Pipe Association.

Spangler, M.G. and Handy, R.L., (1973), SoilEngineering, 3ed, IEP.

PROBLEMS12-1 What is the Marston load on the rigid pipe ofFigures 12-2 and 12-3 if the backfill soil is sandloosely dumped into place? Data are as follows:Pipe Trench OD = 36 inches, H = 8 ft,ID = 30 inches. B = 5 ft.Soilφ = 20o = friction angle,γ = 100 pcf.What is the required pipe strength (D-load)? Assumethat as soil is dumped into the trench, it slides inagainst the pipe and trench walls at active horizontalsoil pressure; i.e., the horizontal soil stress is σy /Kwhere K = (1+sinφ)/(1-sinφ) and φ = soil frictionangle.

12-2 A four-ft (ID) reinforced concrete pipe is toserve as a culvert under 2.5 ft of granular soil cover.If sf is 2 and bedding is Class B, what class pipe isrequired? t = 4.8 inches. Performance limit is 0.01inch crack. Assume HS-20 dual-wheel load and soilunit weight of 120 pcf. (Class IV)

12-3 Using a programmable computer for iterations,find the maximum allowable height of prismatic soilcover H, assuming γ = 125 pcf, for minimallyreinforced concrete culvert as shown in Figure 12-7,with only one cage of circumferential #3 rebarsspaced at 3 inches in the center of the 6- inch-thickwall, and with 26 longitudinal #3 rebars equallyspaced on the outside of the cage. Properties ofmaterials are listed on the Figure.

12-4 Using a PC find the area of cage rebars per ftof pipe length required for balanced design of the pipewall, i.e., σs/σ c = Ss /Sc in Problem 12-3.

12-5 If soil is carefully placed and compacted underthe haunches, what is the allowable height of soilcover in Problem 12-3 if full soil support of the ring isassured such that for the sidefill, the horizontal soilpressure is Px = Py /3?



Anderson, Loren Runar et al "Frontmatter"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 13-1 Soil stress models for minimum soil cover — free-body-diagrams of truncated pyramid and coneshowing shear planes at "punch-through" that direct the live load onto the pipe.

Figure 13-2 Truncated pyramid showing how a surface load, W, punches out a pyramid, which spreads the loadover a base area of (B+H)(L+H) at depth H.


CHAPTER 13 MINIMUM SOIL COVER

As the soil cover H decreases, the live load effecton the buried pipe increases. There exist minimumheights of soil cover H, less than which the surfacelive load may damage the pipe. Less evident is aminimum height of soil cover for dead load, weightof soil only, on buried pipes. Each of these cases isdiscussed for rigid and flexible rings in this chapter.Only cohesionless soil is considered. Vehicles aregenerally unable to maneuver on poor soil such aswet cohesive soil. They get stuck in the mud.

In many of the following analyses, the effect of thesurface live load on the pipe is based on thepyramid/cone model. The Boussinesq andNewmark procedures for calculating live loadpressure on the pipe are based on the assumptionthat the soil is elastic. The assumption is notadequate for failures of buried pipes. Failure ofpipes due to surface loads on less-than-minimum soilcover is punch-through. The pipe is not subjected tofailure load until the soil cover fails in shear aswheels punch through the soil cover and thenfracture or distort the buried pipe. Based on such amodel, the soil stress on a buried pipe is referred toas the pyramid/cone soil stress.

PYRAMID/CONE SOIL STRESS

Figure 13-1 shows the truncated pyramid/cone soilstress models. A surface live load can damage theburied pipe only after it punches through the soilcover. If the loaded surface area is circular, atruncated cone of soil is punched out. If the loadedsurface area is a rectangle, a truncated pyramid ofsoil is punched out. The truncated pyramid isimperfect because its sharp edges do not form.Nevertheless, by slight adjustment of the pyramidangle, the analysis can be made applicable. The tireprint of dual wheels is more nearly rectangular thancircular. Therefore the following analyses are basedon the pyramid.

Figure 13-2 represents a dual-wheel tire print over a

rectangular area of width B and length L. If thesurface load is great enough to punch throughgranular soil and damage the pipe, then shear planesmust form in the soil isolating a truncated pyramidwhich, like a pedestal, supports the load.

The total load on the pipe is the surface load W plusthe weight of the pyramid of soil. The weight of thepyramid is ignored because it is small compared toany surface load great enough to punch through.The vertical soil pressure on the pipe is load Wdivided by the base area of the pyramid. The angleθ which the shear planes make with the vertical isthe pyramid angle θ = 45o - ϕ/2 where ϕ is the soilfriction angle. At depth H, the base area over whichthe load is spread is (B + 2Htanθ)(L + 2Htanθ).From tests on cohesionless soil, the pyramid angle isroughly 35o for which the base area is approximately(B+H)(L+H). The precision is as good as can bejustified for typical installations. The results areconservative. Refinements may be forthcoming. Itfollows that at soil failure, the pressure on the pipe isthe pressure on the base of the pyramid; i.e.,

P = W/(B+H)(L+H) . . . . . (13.1)

For HS-20 dual-wheel load on a firm surface, B = 7and L = 22 inches with tire pressure of 105 psi.

NOTATION

P = vertical soil pressure at the level of the topof the pipe due to a surface load uniformlydistributed over a rectangular area,

W = weight of the surface load,γ = unit weight of soil,σy = stress at yield point,σ = ring compression stress,D = mean diameter of the pipe,r = mean radius of the pipe,c = distance from neutral surface of the pipe

wall cross section to the most remote fiberon the surface.


Figure 13-3 Sketch of a surface live wheel load W passing over a pipe buried in loose granular soil.

Figure 13-4 Flexible ring in the process of collapse under minimum dead load soil cover showing the loadwedges advancing against the ring and the lighter restraint wedges being lifted.


A = cross sectional area of the pipe wall perunit length of the pipe,

I = centroidal moment of inertia of the pipewall cross section per unit length of pipe,

M = moment in the wall of the ring due to ringdeformation,

T = circumferential thrust in the ring,S = compressive strength of the pipe wall,E = modulus of elasticity of the pipe material,H = installed height of cover of the soil,H' = rutted height of soil cover,H" = depth of the rut (See Figure 13-3),ρ = soil density in percent Standard Proctor

(AASHTO T-99) for the granular soilcover and the embedment.

For a dual truck wheel on a compacted soil surface,the tire print area is about 7 inches by 22-inc hesbased on typical tire pressures as follows.

Dual Load (kips) 5.5 7 9 16Tire Pres. (psi) 36 45 58 104

The loaded surface area can be adjusted fordifferent loads and different tire pressures. For atruck on pavement, the 7 x 22-inch contact area is inreasonable agreement with observations.

MINIMUM HEIGHT OF SOIL COVER

Minimum height of soil cover can be found bysolving Equation 13.1 for H if the surface load W isknown and if the allowable pressure P on the pipecan be evaluated for any given pipe and for anygiven performance limit, such as inversion or ringcompression at yield. The next problem is evaluationof the allowable pressure P. This must include ringcompressive strength, ring stiffness, and the criticallocation of the load. The remainder of the chapteris devoted to this problem.

Example

Consider a perfectly flexible ring. Suppose that load W on a highway truck dual is 10

kips. Vertical pressure P on the pipe at inversion is0.8 ksf. If B and L for the dual tire print are 7 and22 inches, respectively, from Equation 13.1, H = 28.6inches, which is the minimum granular soil cover forprotection of the pipe against collapse. This does notinclude a safety factor, but if it is based on theassumption that the pipe ring is perfectly flexible —like a chainlink watch band — a margin of safety isbuilt in depending on the ring stiffness. A perfectlyflexible ring is not practical. Moreover, longitudinalbeam strength of the pipe is ignored.

HEIGHT OF SOIL COVER

An unsuspected problem in the minimum coveranalysis is the definition of height of soil cover. Fora surfaced highway, the height of soil cover remainsconstant during passes of live loads. But duringconstruction, a heavy load crossing a buried pipeleaves ruts. See Figure 13-3. In fact, successivepasses of the load may increase the depths of theruts. If the depths of ruts approach a limit as thenumber of passes increases, the pipe-soil system isstable. But if the depths of the ruts continue toincrease with each pass of the surface load, it isobvious that the pipe feels increasingly adverse loadsand may be in the process of inversion by ratcheting;i.e., an additional increment of ring deformation witheach pass of the load. Whatever the ultimatedamage may be, a performance limit has beenexceeded. So minimum height of soil cover isdefined as that soil cover H, less than which thepipe-soil system becomes unstable upon multiplepasses of surface load W. The height of cover to beused in Equation 13.1 for soil stress on the pipe is H',the height of soil cover after the ruts have reachedtheir maximum depth, H". From Figure 13-3, H = H'+ H". From tests in moist, granular, well-graded siltysand (SM classification), following rain (fieldmoisture equivalent), the rut depths for dual wheelsare generally not greater than,

H" = 0.315(logW - 0.34)(103.9 - ρ) . . . . . (13.2)


Figure 13-5 Truncated pyramid punched through the minimum soil cover, H, by an approaching surface wheelload W. Shear planes form at a 1:2 slope. The inversion arc is 2α. Angle α is typically less than 45o.

Figure 13-6 Flexible ring in theprocess of collapse under minimumsoil cover due to semi-infinitesoil pressure P, showing the load"wedge" advancing against the ringand restraint "wedge" being liftedby the ring.


where H" is in inches, and W is in kips. Soil densityρ (in percent) is based on AASHTO T-99. Thefollowing table of values is from field tests fromwhich roughly 90% of the rut depths were less thanthe values listed. Equation 13.2 is not dimensionallyhomogeneous. It was achieved by regression fromplots of the following field data with tire pressures,p, as indicated.

H" = RUT DEPTH (inches)

p (psi) 36 45 58 104

ρ = Soil W = W = W = W =Density 5.5 7.0 9.0 16.0(%) kips kips kips kips

80 3.0 3.8 4.6 6.585 2.4 3.0 3.7 5.190 1.8 2.2 2.7 3.895 1.1 1.4 1.7 2.4

DEAD LOAD

Minimum cover of cohesionless soil over a buriedpipe exists if the pipe is unable to support thevariation in soil pressures around its perimeter. Theconcept is shown in Figure 13-4, which shows a toppressure of γ H, but a shoulder pressure greater thanγ H. If the pipe cannot support the difference inpressures, the shoulder wedges will slide in againstthe pipe, deforming the ring which, in turn, lifts upthe top wedges as shown. Collapse of the pipe iscatastrophic. If the pipe is rigid (brittle), collapse isfragmentation. If the pipe is flexible, equations ofstatic equilibrium of the soil wedges provide valuesof minimum soil cover H. For average granularbackfill, H turns out to be about D/10. Experimentsconfirm the above analysis of dead load collapse atH = D/10 for very flexible pipes under dry granularbackfill. A suggested allowable value of D/6 forminimum cover allows for a margin of conservatism.But this analysis is for a perfectly flexible ring. Infact, pipes have ring stiffness and so provideresistance to dead load collapse.

LIVE LOAD

The minimum cover of cohesionless soil is not basedon a location of live load directly over the crown ofthe pipe as in Figure 13-1. Rather, the criticallocation is an approaching load as shown in Figure13-5. The leading edge of the base area of thetruncated pyramid is at the crown. The ring tends todeform as indicated. Tests to failure of long spancorrugated steel arches prove that static surfaceloads symmetrically located over the crown can bemany times greater than the load on one side. Staticload failure is soil punch-through and ring collapse.

For fragile rigid pipes, failure is fracture of the pipeand possible collapse.

For flexible pipes, failure is inversion due todeformation as shown in Figure 13-5. Themechanism is downward deformation of the leftshoulder of the ring under the load and consequentialupward deflection of the right shoulder. For granularbackfill, the inversion angle is observed to be aboutα = 40o. For convenience, and to be conservative,the collapse angle is assumed to be α = 45o for thetruncated pyramid pressure shown in Figure 13-5.For the semi-infinite surface pressure shown inFigure 13-6, the load wedge contacts a full quadrantof the ring.

Analysis is evaluation of the maximum momentcaused by the live load. Dead loads are neglected.The weights of the wedges and the shear resistancebetween them are small compared to the live load.Moreover, in Figure 13-6, the weight of the loadwedge tends to balance the restraint wedge. Shearbetween pipe and soil is neglected. The ring is fixedat both ends of the collapse arch. Vertical soilpressure, P, becomes radial P, on a flexible ring.See Figure 13-7. Castigliano's equation is used tofind the reactions, the maximum moment, M, andthrust, T. See Appendix A. Maximum M is locatedby equating its derivative to zero. If wall crushing iscritical, thrust T is pertinent. If circumferentialstress is of interest,


σ = T/A + Mc/I . . . . . (13.3)

ELASTIC LIMIT

The thrust term is usually so small compared to themoment term, that it can be neglected.

It is much more likely that the critical performancelimit is inversion. Inversion is the result of plastichinging that ultimately generates a three linkmechanism. See Figure 13-7. The thrust term isrelatively small enough to be neglected. Plastichinging is a function of the Mc/I term, except that Mis at plastic limit — not elastic limit.

For plain pipes and corrugated pipes, the moment atplastic hinging is approximately 3/2 times the elasticmoment at yield stress. Therefore,

σ = 2Mc/3I . . . . . (13.4)PLASTIC HINGING

Truncated Pyramid Load (Surface Wheel Load)

For truncated pyramid pressure, the free-body-diagram is a fixed-ended 90o arch. See Figure 13-7.Dead load soil pressure is neglected. Live load soilpressure is constant radial pressure, P, over 45o leftof the crown, point A. From Castigliano's equation,the maximum moment occurs at the point ofminimum radius of curvature, about 12o to the rightof the crown, A, and is:

M = 0.022 Pr2 . . . . . (13.5)WHEEL LOAD

The circumferential thrust is T = γ Hr due to thedead weight of soil cover on the right side of thecrown. For design of the pipe based on yield stress,σf, the minimum required section modulus, I/c, is;

Figure 13-7 Free-body-diagram of the inversion arch for finding the maximum moment, M in terms ofpressure P due to a surface dual-wheel load W approaching a pipe under minimum soil cover H. Thelocations of potential plastic hinges are shown as circles starting at the location of maximum moment.


I/c = (0.022Pr2)sf/σf . . . . . (13.6)ELASTIC LIMIT

I/c = (0.015Pr2)sf/σf . . . . . (13.7)PLASTIC HINGING

where sf is safety factor. Tests show that I/c fromthese equations is conservative. A safety factor of1.5 is usually adequate, and does not need to begreater than 2 for highway culverts.

With M and T known, Equation 13.3 can be solvedfor the maximum stress whenever stress (or strain)is of interest, as in the case of bonded linings.

Semi-infinite Surface Pressure

For the semi-infinite uniform surface pressure ofFigure 13-6, the collapse arc under the load wedgeis 90o. Even though the ring is flexible, theassumption of constant radial pressure isconservative. From Castigliano's equation, themaximum moment is,

M = 0.08 Pr2 . . . . . (13.8)SEMI-INFINITE SURFACE LOAD

and is located at about 12o to the right of the crown.

T = γ Hr for use in Equation 13.3.

But because ring compression stress T/A is usuallysmall compared to the Mc/I stress, it can beneglected. Setting Mc/I = yield strength/safetyfactor, and solving for section modulus I/c,

I/c = (0.08Pr2)sf/σf . . . . . (13.9)ELASTIC LIMIT

I/c = (0.05Pr2)sf/σf . . . . . (13.10)PLASTIC HINGING

I/c is the required section modulus which can befound from tables of values for corrugated metalpipes and can be calculated for other pipes. Theequations for I/c are conservative, but are

responsive to the inversion of very flexible pipes withminimum soil cover when subjected to heavy surfaceloads. It is noteworthy that H does not appear inEquations 13.7 and 13.8. It is presumed that theheight of cover is already minimum as calculated byEquation 13.1 for punch-through of a truncatedpyramid. It is presumed that for a semi-infinitesurface pressure, at the level of the top of the pipe,P is equal to surface pressure.

Example 1

Assume that a pipe buried under minimum cover issubjected to a semi-infinite surface pressure suc hthat the edge of the base area of the load is at thecrown of a buried pipe. See Figure 13-6. Whatsection modulus I/c is required for the pipe wall?The pipe is to be a 6-ft diameter corrugated steelpipe with granular soil cover of 2 ft and semi-infinitesurface pressure of P = 900 psf. The yield strengthof the pipe is 36 ksi. Because the steel can yield,assume performance limit to be the formation ofplastic hinges starting at the point of maximummoment. Equation 13.10 applies. Assume a safetyfactor of 2. Substituting values, the required I/c =0.27 in3/ft. From the AISI Handbook of SteelDrainage and Highway Construction Products,a 3x1 corrugated steel pipe of 0.109-inch-thick steelis adequate. For this 3x1, the I/c is listed as 0.3358in3/ft.

Example 2

Consider Example 1 again, but for plain steel waterpipe. The section modulus I/c can be transformedinto required wall thickness from the relationships I= t3/12 and c = t/2. But from example 1, I/c = 0.27in3/ft. The required wall thickness is t = 0.367 inch.Specify wall thickness of 0.375 inch.

In the case of heavy surface wheel loads, it is oftenmore economical to increase the height of soil coverin order to reduce the possibility of inverting the pipe.Under some circumstances, stiffener rings can beattached to the ring to increase the section modulus.See Chapter 21.


Figure 13-8 AASHTO shandard H trucks.

Figure 13-9 Load-deflection diagrams from tests on 18D, HDPE corrugated pipe under 7 inches of granularsoil cover at 85% density (AASHTO T-99), showing permanent ring deflection d’ and rebound ring deflectiond”, and showing the zone of instability.


Example 3

Find the minimum cover of granular soil overcorrugated polyethylene pipe 18 inches in diameter(ID). The polyethylene is designated as HDPE(high density polyethylene). The soil cover iscompacted to 85% density (AASHTO T-99). Theyield strength of HDPE at sudden inversion is 3 ksi.The surface load is a highway truck dual wheel forwhich the area of the tire print is 7 inches by 22inches. The scheme is to substitute values of P fromEquation 13.1 into Equation 13.5, M = 0.022Pr2.Including values of r and I/c for 18 HDPE pipe, theequation becomes a quadratic,

(H+14.5in)2 = 56.25in2 + 25Win2/kip.

Solutions are:

W (kips) 5.5 7 9 16H' (inch) -0.6 0.7 2.3 6.9

W = dual wheel load (16 kips = HS-20 load).H' = rutted soil cover

A safety factor of two is usually applied to H'.Some specifications require a minimum of one ft ofcompacted granular backfill. For design of minimumcover, the safety factor is important because loadsare often dynamic—not just static.

The negative 0.6 at W = 5.5 kips, indicates that soilcover is not needed for such a light load. The pipecan carry a 5.5 kip dual-wheel load even though theruts expose the pipe. Of course, enough soil coverH should be provided to allow for rutting H", toprevent surface rocks from indenting the pipe, and toprevent crushing of the corrugations. A similaranalysis for 24 HDPE is almost identical to the tableabove for 18 HDPE. Apparently manufacturers arecareful to provide equivalent properties for theirpipes in all sizes. Installation techniques areprobably about the same for 18- and 24- inchdiameter corrugated HDPE pipes.

RING DEFLECTION

Example 3 is the result of field installations. Ringdeflections were observed under minimum cover.Dual-wheel loads were varied up to the standard H-20 load. Figure 13-8 is a sketch of standard truckweights and dimensions.

From the tests, it was found that ring deflectioncomprises two components: permanent ringdeflection d', and rebound ring deflection, d". Therebound ring deflection is elastic and rebounds fullyafter each pass of the dual-wheel load. Figure 13-9summarizes the ring deflections of 18 HDPE pipesburied under minimum cover of granular soil at 85%density (AASHTO T-99) and with 7 inc hes of soilcover before passes of the wheel loads. Thefollowing observations are noteworthy:

1. Ring deflection is less than 2% for the first passof the dual-wheel load.

2. For multiple passes, the rebound ring deflectionstabilizes if dual-wheel loads are less than roughly12.5 kips.

3. For dual-wheel loads greater than roughly 12.5kips, rebound ring deflections may increaseprogressively. This is defined as instability.

4. A probable zone of instability is suggested withoutenough test data to establish the boundariesprecisely.

It was found, in additional testing, that with 12 inchesof cover at 85% density before passes of the load,there was no indication of a zone of instability withdual-wheel loads up to 16 kips.

Example

Corrugated polyethylene tubing is used extensivelyfor drainage. The slotted pipes are embedded ingravel. Inside diameters are from 3 inches to 18inches. What is the ring deflection due to dual-


Figure 13-10 Results of tests showing ring deflection as a function of depth of cover term for buriedcorrugated polyethylene drain tubing subjected to AASHTO H-20 standard truck load. Recommendedminimum cover is indicated for each installation case.


wheel loads as a function of depth of soil cover?Tests were performed on 12-inch pipes and reportedto the American Society of Agricultural Engineers inJune 1980. Results and recom-mendations areshown in Figure 13-10.

Corrugated polyethylene pipes in larger diametersare available. These are attractive for use asculverts under roads. Additional concerns, such asdynamic loads, usually justify minimum cover greaterthan one foot.

FLOTATION

When pipes are buried in soil under water, theminimum height of cover to prevent flotation ofan empty pipe is about H = D/2. But the soil shouldbe denser than critical in order to preventliquefaction. See Chapter 21.

MINIMUM COVER FOR RIGID PIPES

There are two basic performance limits for buriedrigid pipes subjected to heavy surface loads withminimum soil cover. They are longitudinal fracturesand broken bells. Circumferential fractures canoccur, but less frequently. Half a bell can besheared circumferentially. Circumferential crackscan occur at midlength of the pipe acting as a beamunder a heavy wheel load if goodbedding/embedment support is provided under theends, but nowhere else.

Longitudinal Fractures

Longitudinal fractures occur if vertical pressure Pexceeds ring strength. The worst case location ofthe critical live surface load is directly above the pipe— not on approach as in the case of flexible pipes.See Figure 13-1. Minimum soil cover H is based onthe punch-through pyramid/cone. Longitudinalfractures occur at 12:00 and 6:00 o'clock and 9:00and 3:00 o'clock. This is not collapse of the pipe.Many gravity flow pipes serve even when cracked.

The soil envelope holds the ring in nearly circularshape. See Figure 7-2. But for some rigid pipes,such as pressure pipes, longitudinal cracks areunacceptable. Occasionally one longitudinal hairlinecrack occurs at 12:00 o'clock, or possibly at 6:00o'clock if the pipe is on a hard bedding. If theembedment is compacted select soil, a crack at12:00 o'clock might be caused by a surface wheelload or a conscientious constructor who compactsthe first layer above the pipe directly on top of thepipe. It is prudent to compact sidefills, but to leaveuncompacted the first layer directly over the pipe.If the pipe is a culvert or storm drain, a singlehairline crack is not performance limit. Goodembedment holds the pipe in shape such that thepipe is in ring compression, not flexure. It performslike the ancient brick sewers of Paris and Londonwhich function well, but may not be leak-proof.

Analysis is the same for minimum cover and formaximum external pressure on rigid pipes. SeeChapter 12. The vertical pressure is P = Pl + Pd

where the live load pressure, Pl, is found by thepyramid/cone theory. For minimum cover analysis,the dead load pressure, Pd, is negligible. The liveload pressure, Pl, is a function of height of cover, H.The minimum cover can be solved from the equationPc r = Pl where critical pressure Pcr is a function ofclass of bedding and class of pipe. See Chapter 12.

Example

A nonreinforced concrete pipe, 15 ID, bell andspigot, C-14, Class 3 is to be used as a storm drain.Outside diameter is OD = 19 inches. From tests, thethree-edge-bearing (TEB) strength is D-load = 3k/ft. What must be the minimum cover, H, if a CAT633 scraper is to pass over? The embedment iscrushed (angular) granular material passing 1/2 inchsieve. It contains less than 5% fines. The scraperwheel load is W = 50 kips on a circular tire printarea of 800 square inches. The diameter of thecircular tire print area is 32 inches. The live loadpressure at the top of the pipe is Pl =4(50k)/π (32+H)2. From Chapter 12, the critical


Figure 13-11 Rigid pipe cross section showing how voids are left under the haunches if soil is not shovedunder.

Figure 13-12 Bell end of a section of pipe that is subjected to live load pressure, but is simply supported byreactions, Q, on the ends of the contiguous pipe sections.


live load pressure is Pc r = 3kLf/ft(OD), where theload factor, L f , depends upon the class of bedding.Equating Pl = Pcr and solving for the minimum soilcover, H,

Bedding Lf HClass ___ (inches)A 2.5 12.0B 1.9 18.5C 1.5 24.8D 1.1 34.3D-load 1.0 37.6 (TEB test)

For a typical Class C bedding, the minimum cover isabout 25 inches. This is ring analysis. Therefore, itis assumed that the bedding is non-compressible.Horizontal ring support is neglected. The pipesection under live load is not pressed downwardbetween the two contiguous pipe sections such thatit acts as a beam.

Broken Bells

If a pipe section acts as a beam, the performancelimit is usually a broken bell. Under heavy live loadsand minimum soil cover, rigid pipes require soilsupport under the haunches. If soil is not actuallyshoved under the haunches, a void is left under thepipe. See Figure 13-11. For example, if the angle ofrepose of the embedment is ϕ' = 40o, the void iswider than half the outside diameter [0.643(OD)].Live load on the pipe could cause the top of the pipeto move downward either by cracking the pipe, or bypressing the pipe into the bedding. Under thehaunches, loose soil at its angle of repose, offerslittle resistance. As a pipe section is presseddownward, it becomes a simply supported beam withreactions, Q, at the ends of the pipe section. SeeFigure 13-12. It is this reaction, Q, that fractures thebell. Clay pipes and nonreinforced concrete pipesare vulnerable because of low tensile strength. Themaximum tensile stress is in the bell near the springline. Once cracked, a shard forms approximatelyone diameter in length as shown in Figure 13-12. Anapproximate analysis is the equating of this Q thatcan be withstood by the bell, to the Q reaction

caused by the live surface load on the pipe sectionacting as a beam.

ExampleWhat is the minimum cover, H, for the precedingexample if fracture is a broken bell? The length ofnonreinforced pipes is L = 8 ft. Assume (estimate)that the cross-sectional area of the thin part of thebell is A = 5 square inches. Tensile stress at thespring lines is σ = Q/2A, from which maximum Q =2Aσf , where σ f = tensile strength of the concrete.Tensile strength of the concrete is σf = 1.0 ksi. It isfound from the three-edge-bearing test from whichD-load = 3 k/ft. See Figure 13-13. Therefore,reaction Q at fracture is Q = 2Aσf = 10 kips. But Qis the reaction to pressure Pl on the pipe which iscaused by surface live load, W. From the punch-through cone analysis, Pl = 4W/π(32+H)2. Thepressurized area is going to be greater than half thepipe length. Therefore, with little error, it is assumedthat W is located at midspan, and that reaction Q =0.5P(OD)(32+H). Substituting for P and equatingthe two Q's,

W(OD)/π(32+H) = Aσf . . . . . . (13.11)

From Equation 13.11, the minimum soil cover is H= 28.5 inches.

In this analysis, the pipe section is a simple beamwith no support from the bedding or embedment.Minimum cover will be less than 28.5 dependingupon the support. This analysis is approximate.Therefore, it is prudent to specify good bedding andembedment, and to require a minimum cover ofthree feet for the impact loads of constructiontraffic. To place and compact embedment under thehaunches, a windrow of soil along the pipe can beshoved into place by laborers with J-bars workingfrom on top of the pipe, or by flushing the windrowunder the haunches with a high-pressure water jet,or by hand operated mechanical compactors. Someconstructors are now placing soil cement slurry withabout ten inch slump under the haunches. SeeChapter 16. Rigid pipes are usually strong enoughas beams to resist


TEB TEST

MA = (D-LOAD)r/π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . See Appendix A

If r = 8.5 inches and D-LOAD = 3,000 lb/ft

MA = 676.4 lb.

σf = 6M/t2 = 1,014.6 psi. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Tensile strength is σf = 1.0 ksi

Figure 13-13 Evaluation of tensile strength of a typical 15 inch nonreinforced Class 3 concrete pipe basedon three-edge-bearing (TEB) tests from which D-LOAD = 3,000 lb/ft.


circumferential fracture at midspan. Manufacturerslimit the length of pipes to prevent beam failures.The following example is an interesting comparisonof beam analyses.

Example

What is the minimum cover, H, for the pipe in theabove examples based on maximum longitudinaltensile stress, σ = Mc/I, in the bottom of a simplysupported beam? With a uniform load, w, atmidspan, M = wL2/8, where w is the load per unitlength of beam; i.e., w = P(OD). P =4W/π(32+H)2. I/c = (π/32)[(OD)4-(ID)4]/(OD). Iftensile strength is σf = 1 ksi, substituting values intothe equation, σf = M/(I/c), H = 26.2 inches. Failureby a broken bell is more critical but not by much.

PROBLEMS

13-1 One wheel of a scraper with rubber tirescarries a load of 25 kips. What is the diameter ofthe tire print on the ground surface if tire pressure is30 psi? (D = 32.6in)

13-2 What is the minimum height of cover ofgranular backfill compacted enough that rutting isnegligible, if it is to protect a 6x2 corrugateds tructural steel plate culvert of 0.1345-inch-thicks teel? The radius is r = 48 inches. The maximumdual-wheel load is 16 kips on an asphalt surface

which does not spread the wheel load. What aboutsafety factor? I = 0.938 in4/ft, and I/c = 0.879 in3/ft.

13-3 How do the solutions for Problem 13-2compare with solutions from the AISI Handbook ofSteel Drainage & Highway ConstructionProducts? Does the safety factor need to be two?

13-4 What is semi-infinite surface live load at failureof a flexible pipe buried in granular backfill usingplastic hinging analysis?

Pipe steel Soil3x1 corrugations Pit-run gravelt = 0.109 H = 15 inchesD = 96 inch diam. γ = 125pcf.

I/c = 0.3358 in3/ftσf = 36 ksi.

13-5 What is the minimum height of soil cover H fora nonreinforced concrete pipe Class 3 if an HS-20truck dual-wheel load passes over? W = 16 kips.

PipeOD = 19 inchesID = 15 inchesσf = 1.0 ksi

SoilCompacted granular soil.φ = 40o


(11.8 m)

(1.26 ksf)

(12, AISI, Handbook of steel drainage and highwayconstruction products, 1993, p. 258)

Anderson, Loren Runar et al "LONGITUDINAL MECHANICS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 14-1 Longitudinal stress due to decrease in temperature of a restrained pipe.

Figure 14-2 Longitudinal stress due to internal pressure (Poisson effect) in a restrained pipe.


CHAPTER 14 LONGITUDINAL MECHANICS

Longitudinal mechanics of buried pipes is theanalysis of longitudinal deformations compared toperformance limits of deformation. One excessivedeformation is fracture for which correspondingstrain limits can be identified. If the pipe changeslength enough to shear off appurtenances or to allowleakage of couplings, deformation is excessive andcorresponding strain limits can be identified. If thestrains can be evaluated, then corresponding stressescan be used as alternative bases for design. Theprincipal causes of longitudinal stress (strain) are:

1. Changes in temperature and pressure, whichcause relative lengthening or shortening of the pipewith respect to soil and thrust restraints,

2. Axial thrust is the result of internal pressure orvacuum at "thrustors" (valves, caps, reducers, wye's,tee's elbows, etc.),

3. Beam bending, which causes flexural stresses.Typical causes of beam bending are:a) Placement of pipe sections on timbers or moundsor piers for vertical alignment,b) Non-uniform settlement of the bedding,c) Side-hill soil creep or landslide, and massive soilmovement or settlement.

Each of the three causes of longitudinal stress isanalyzed separately. The results are combined foranalysis. It is convenient to separate analyses intotwo categories, gasketed pipe sections, andcontinuously welded pipes, such as welded steelwater pipes and welded polyethylene pipes.

Following are discussions of the three major causesof longitudinal stresses (and strains) in each of thetwo categories.

Longitudinally Restrained Pipes Temperature stresses occur when the pipe sectionis restrained at the ends such that it cannot lengthenor shorten. See Figure 14-1. For ordinary buried

pipes, temperature change would have to beunusually large to cause critical longitudinal stress.Temperature stress in an end-restrained pipe is:

σT = Eα (∆T) . . . . . (14.1)

whereσT = longitudinal stress in the restrained pipe

due to change in temperatureE = modulus of elasticityα = coefficient of thermal expansionT = change in temperature

In the case of end-restrained steel pipes (i.e., weldedjoints), in order to cause 36 ksi yield stress in thepipe wall, the change in temperature would have tobe roughly 185oF. Gasketed bell and spigot jointsavoid the problem if the pipe sections are not toolong.

The end-restrained pipe is also subjected tolongitudinal tension due to internal pressure. SeeFigure 14-2. When a rubber band is stretched, thethickness decreases. So also does the length of apipe section when inflated by internal pressure. Thisis called the Poisson effect. Longitudinal tensionstress is:

σP = νP'r/t . . . . . (14.2)

whereσP = longitudinal stress in the pipe due to

internal pressure,ν = Poisson ratio (in the range of 0.25 to 0.4

for most pipe materials),P' = internal pressure,r = inside radius of the pipe,t = wall thickness for plain pipe.

Values of σP are not usually critical. If steel pipesare capped or plugged and pressurized tocircumferential yield stress, P', the longitudinal stressis only half of yield stress.


If the buried pipe is welded and very long, it iseffectively restrained (by soil friction, if not endrestraints) and feels thrust (tension) due to decreasein temperature and internal pressure.

Example

A section of PVC pipe, 12D, DR 26 is installed on awarm day. When put in service as a water supplypipeline, temperature decreases 50oF and internalpressure increases to 160 psi. What is thelongitudinal stress if the pipe is restrainedlongitudinally?DR = 26 = OD/t,r/t = 12.5 = (DR-1)/2,E = 400 ksi = modulus of elasticity,α = 3(10-5)/oF = thermal coefficient,ν = 0.38 = Poisson ratio,P' = 160 psi = internal pressure,∆T = 50oF.

Combining Equations 14.1 and 14.2,σ = Eα∆ T + νP'(r/t)

Susbstituting values,σ = 600 psi + 760 psi = 1.36 ksi

At sudden rupture, yield stress is σf = 6 ksi. At 50years of persistent pressure, σf = 4 ksi. Ifrestrained, stresses relax over time to less than 1.36ksi. The temperature and Poisson effects areavoided by gasketed joints if the pipe sections arenot too long. If the sections are long, changes inlength due to the temperature and Poisson effectscause significant longitudinal stresses. As a pipesection changes length, it is partially restrained byfrictional resistance of the soil. As with a rope in atug-of-war, where tension applied by friction of thecontestants' hands cumulates to maximum atmidlength, so does thrust in a buried pipe sectioncumulate to maximum at midlength due to soilfriction. That thrust causes a longitudinal stress of,

σ = LHγ µ'/2t . . . . . (14.3)

whereσ = longitudinal stress in the pipe due to

frictional resistance of the soil,L = length of the pipe section,H = height of soil above the pipe,γ = unit weight of the soil cover,µ' = coefficient of friction, soil on pipe,t = wall thickness of plain pipe.

Some pipeline engineers assume values for thecoefficient of friction between granular soil and steelpipes as follows:

For tape-coated steel pipe, µ' = 0.2For mortar-coated steel pipe, µ' = 0.4

Example

What is the longitudinal stress at midlength of a pipethat shortens after it is buried? The pipe is a largediameter, tape-coated, gasketed steel pipecomprising 120-ft-long sections with 0.50-inch wallthickness. µ' = 0.2. H = 6 ft of granular soil coverat 125 pcf. Substituting into Equation 14.3, soilfriction can cause longitudinal stress up to 1.5 ksi.This is not impressively large, but it may combinewith other longitudinal stresses, such as beambending.

Beam Bending

See Figure 14-3. If a pipe that is initially straight isbent into a circular arc of radius R, longitudinalstrains develop in the outside fibers. If R is tooshort, the pipe buckles; i.e., crumples at a plastichinge. Strain is a better basis for analysis thanstress. In the following, both strain and stress areanalyzed because yield stress is performance limit insome cases. Due to beam bending, longitudinalstrain and stress (elastic theory) are:

ε = r/R, and σ = Er/R . . . . . (14.4)


whereε = maximum longitudinal strain — tension

outside of bend and compression inside,σ = maximum longitudinal stress,r = outside radius of the pipe cross section,R = longitudinal radius of the pipe axis on the

bend,E = modulus of elasticity,R' = initial radius of the bend if the pipe is

manufactured as a curved section.

Example

What is the minimum radius of the spool on whichpolyethylene gas pipe can be wound for shipping (orthe minimum longitudinal radius for loweringcontinuous pipe into a trench)? If allowable strain is2%, and pipe OD is 2 inches, from Equation 14.4, R = 50 inches.

If the pipe is manufactured initially to some meanradius R', then Equation 14.4 must take into accounta change in curvature as follows:

ε = r(1/R-l/R'), and σ = Er(1/R-1/R')

The mean radius of the bend can be measured byholding a cord (straightedge) of known length, s,along the inside of the bend as shown in Figure 14-4and by measuring the offset e at the middle of thecord. The mean radius R of the bend is,

R = s2/8e + e/2 + r . . . . . (14.5)

Knowing R, longitudinal stress (and strain) can beevaluated from Equation 14.4. The analysis aboveis useful for checking the installed radius of the bendagainst allowable. R can be calculated bymeasurements inside the pipe.

Longitudinal bending is caused by: 1. soil movement,and 2. non-uniform bedding.

Soil movement is caused by heavy surface loads,differential subgrade soil settlement, landslides, etc.Soil settlement can often be predicted.

Non-uniform bedding is inevitable despite specifi-cations calling for uniform bedding. Under soil loadsplus weight of the pipe and contents, the pipedeflects and causes longitudinal stress. Forreinforced pipes, manufacturers provide longitudinalreinforcement and limit the lengths of pipe sections.Longitudinal stress can be reduced by corrugatingthe pipe so that it flexes and conforms with thebedding rather than bridging over soft spots.

A bend in the pipe causes a moment, M, for whichlongitudinal stress is,

σ = Mr/I . . . . . (14.6)

where σ = longitudinal stress,r = outside radius of the pipe cross section,I = πtr3 = centroidal moment of inertia of plain

pipe cross section,t = wall thickness of plain pipe,M = moment at some point along the pipe

which acts as a beam.

Moment M can be analyzed from the loads andsupports on a pipe. The supports are intermittenthard spots along the pipe. If bedding were trulyuniform, the pipe could not deflect as a beam.However, bedding is never uniform, and so there isalways some deflection and some M. The maximumM from beam analysis is substituted into Equation14.6. For design, the maximum combinedlongitudinal stress must be less than the strengthreduced by a safety factor.

With few exceptions, pipes are remarkably stiffbeams that bridge over voids and soft spots in thesoil bedding. If reactions are only at the ends of pipesections or at midlength, as in Figure 14-5; themoment is maximum at midlength and may be foundfrom the equation, M/wL2 = 1/8. The reactions ofFigure 14-5 are extreme worst cases and areunlikely — one support per pipe section.

At the other extreme is a pipe on many closelyspaced reactions. This is unlikely. For example,


Figure 14-3 Longitudinal stress (and strain) in tension and compression due to bending of the pipe into alongitudinal mean radius R.

Figure 14-4 Technique for measuring longitudinal inside radius of curvature Ri in a bent pipe. (The sametechnique can be used inside the pipe to find the outside radius of curvature Ro.)

Figure 14-5 Two moment diagrams for the worst locations of one reaction per pipe section and the limits ofcase I for two reactions per pipe section. (These are the maximum possible moments.)


consider a large pipe on rollers on a level concretefloor, Figure 14-6. As the pipe is shoved on therollers, only two rollers carry the load and roll. Allothers are loose and remain at rest. The two highestof the rollers become the reactions. It is easilydemonstrated that the two reactions will have to beon different sides of the center of gravity of the load.See Figure 14-7. Two supports per pipe section isthe most probable number of reactions.

The above rationale reduces the number of reactionsfrom possibly one reaction, to more probably tworeactions per pipe section. Reactions would not belocated at the ends of pipe sections if bell holes areexcavated. Vertical alignment could not becontrolled by placing two reactions on the same sideof midlength.

If the maximum moment due to probable locations ofreactions can be found, then maximum stress can befound from Equation 14.6, and requirements can beestablished for allowable length of pipe sections andtheir longitudinal strength.

Figures 14-8 and 14-9 are dimensionless influencediagrams for likely locations of reactions. From thecritical influence number, M/wL2, maximumlongitudinal stress can be calculated.

For one reaction per section, the maximum momentis M = 0.125wL2, and is always at midlength. Theinfluence diagram is not shown.

The most probable number of reactions per sectionis two. These are hard spots assumed to beconcentrated reactions. There are an infinitenumber of relative locations of two reactions oneach pipe section. However, critical locations areincluded in the following three cases.

Case I. Reactions are at or near the ends of thebeam. This case is similar to one support per sectiondiscussed above. See Figure 14-5.

Case II. Reactions are at two points B equallyspaced from the ends of the pipe section by a

distance kL. See Figure 14-8.

Case III. Reactions are spaced at L/2 but the leftreaction is located a distance X from the left end ofthe beam. See Figure 14-9. It is noteworthy thatthe maximum moment occurs at X = 0.2L.

Pipe on Piles

An example of Case I is a buried pipe on piles (orbents) in order to maintain vertical alignment. Theneed for piles implies that the soil settles. If the soilsettles with respect to the pipe, the pipe lifts a wedgeof soil on top at a wedge angle of 45o + ϕ/2. Forcohesionless soil, the wedge angle is approximately1h:2v.

The load per unit length of pipe is the weight of soilplus the pipe and its contents.

Example

A 120-inch steel pipe with 0.75 wall thickness isburied on piles under 6 ft of soil at 120 pcf in a zonewhere the soil can settle. When the pipe is full ofwater, what is load, w, per unit length of pipe?


Figure 14-6 Large pipe section riding on a number of rollers of which two is the most probable number ofrollers actually supporting the pipe at one time.

Figure 14-7 Two reactions needed per pipe section for stability.


Figure 14-8 Influence diagram of moments for case II reactions which are spaced at equal distances fromthe pipe ends.

Figure 14-9 Influence diagram of moments for case III reactions which are separated by half the pipe lengthbut are at variable distances X from one pipe end.


Figure 14-10 Worst-case, simply supported, two-reaction beam (top), and the more probable sine-curvereactions (bottom), for which stresses and deflections are just four-tenths of the corresponding values for thesimply supported beam.

Figure 14-11 Beam deflection, y, of a pipe section that is supported at its ends.


ws = γ s(131.23 ft2) = 15.75 k/ft,wp = = 0.96 k/ft,ww = γ w(πr2) = 4.90 k/ft,

w = 21.6 k/ftw.

The above analyses are for concentrated reactions— idealized worst cases. In fact, reactions aredistributed over finite areas. Even if the pipe werenot on bedding between reactions, partial soil supportwould occur under the haunches where embedmentfalls in against the pipe. A rational distribution ofreactions is the sine-curve reaction shown in Figure14-10. It turns out that,The moments, strains, stresses, and deflectionsfor sine-curve reactions are four-tenths of thecorresponding values for concentrated reactions.

Exceptions may occur for adverse installations suchas buried pipes on piers. Even though specificationsrequire uniform bedding and compacted embedment,it is prudent for pipe manufacturers to limit thelengths of pipe sections or to provide adequatelongitudinal strength. Pipe manufacturer and pipelinedesigner should know about longitudinal stresses andthe strength of the pipe required to withstand them.

Longitudinal stresses are cumulative — beam stress,plus axial thrust due to special sections and soilmovement, plus soil friction due to temperature andinternal pressure.

Example

A 120-inch water pipe section of 0.75-inch-thicksteel is 120 ft long with gaskets (or slip couplings) ateach end. It is buried under 6 ft of soil, but the pipesection crosses a gulch where soil subsidence isanticipated. What is the stress due to bending? Thespan is simply supported. σ = Mc/I.whereM = wL2/8,w = 21.6 kips/ft from the above example,L = 120 ft,I = πr3t,t = 0.75 inch,

c = r = 60 inches.

Substituting values, σ = 55 ksi . Yield stress isexceeded. However, soil subsidence would have tobe 6.6 inches or more. With less subsidence, stressin the beam would be less than 55 ksi. Moreover,yield stress is not necessarily failure. Allowing foryield, the plastic moment at beam failure is increasedby 50%.

Longitudinal Deflection

Longitudinal stiffness of a pipe is EI = dM/dθ; whereM = resisting moment of the beam,E = modulus of elasticity,I = πr3t = moment of inertia of the pipe cross

section about its centroidal axis,θ = angle of circular bend of the pipe as a

beam.

Most designers relate stiffness to beam deflection ofa pipe in terms of either, 1. load on the pipe as abeam, or 2. maximum longitudinal stress at yield.These analyses provide a feel for how successfullya pipe bridges over soft spots in the bedding. SeeFigure 14-11.

Example

Consider a 120-inch diameter buried steel waterpipeline of 60-ft-long gasketed sections under 6 ft ofdry soil cover. If the soil subsides, what is thedeflection at midlength of a simply supported pipesection? Assume that,y = vertical deflection at midlength,D = 120 inches = 2r,t = 0.75 inch,L = 60 ft,H = 6 ft,γ = 120 pcf = 0.12kcf,w = 21.6 k/ft,I = πtr3,E = 30(106) psi,σy = 36 ksi = yield strength.


For the wedge load, pipe and contents, w = 21.6 k/ftfrom examples above. The deflection at midlengthfor a simply supported span is, y = 5wL4/384EI.Substituting values, y = 0.41 inch. Formulas forbeam deflection are available from texts onmechanics. The corresponding maximumlongitudinal stress is 13.5 ksi. This stress occurs ifthe ends are supported on hard spots, but inbetween, the soil subsides enough to open a gap of0.41 inch under the pipe. This is a longitudinally stiffpipe.

Deflection is proportional to the fourth power oflength. If the pipe section were 80 ft long instead of60 ft, the deflection would be increased by the fourthpower of 80/60 — a factor of 3.16. The 0.41-inchdeflection would become 1.3 inch, and the stress,which increases by the same factor, would be 43 ksi— greater than yield stress of 36 ksi. Allowing foryield, failure is plastic at two-thirds of 43, for whichthe safety factor is only 1.26 against plastic collapse.The fourth-power-of-L effect requires care thatspans over soil subsidence are not excessive.

GASKETED PIPE SECTIONS

Longitudinal stresses are inevitable in continuouslywelded pipes, but are often small enough to beneglected in the design of gasketed pipelines. It isassumed that gasketed joints transfer no moment.They do transfer shear. Adequate longitudinalstrength is taken for granted so long as specificationsinclude uniform bedding and compacted embedment.Pipe manufacturers are expected to provideadequate longitudinal pipe strength for ordinaryburied pipe conditions including handling, shipping,and installing. Ordinarily, the pipeline designer doesnot expect to investigate longitudinal stresses exceptfor adverse conditions such as beam action of aburied pipeline supported on piles or piers. Thepipeline designer accepts the manufacturer'srecommendation for lifting, stacking, stabbing ofjoints, etc., so longitudinal stresses will not beexcessive.Longitudinal thrust is generated at special sections(elbows, wyes, tees, valves, caps, reducers, etc.) by

internal pressure (or vacuum) and by change indirection of fluid flow. In continuously weldedpipelines, thrust is resisted by longitudinal stress inthe pipe. In gasketed pipelines, the longitudinalstress is usually negligible, but not always. Externalthrust restraints are required to resist longitudinalthrust. See Chapter 15.

If there is relative longitudinal movement of the pipeand embedment, soil friction against the pipe causeslongitudinal stress due to: 1. massive soil movement,2. pipe shortening or lengthing due to temperaturechange and internal pressure. Thrust restraintsreduce the relative movement.

Example

A gasketed pipeline crosses a soil embankment thatis settling. If the pipeline was straight at the time ofinstallation, it deflects downward into an arc(catenary) during soil settlement. Because the arclength is greater than the cord length, increase inlength must be accommodated by slight pull-out ofthe spigots. But in the process, each pipe sectionfeels tension due to soil friction. Highway engineerscompensate for pipe deflection under highwayembankments by installing the pipe with camber —reverse (upward) deflection so that the pipe isstraight after soil settlement.

An earth slide on a sidehill can induce a catenary ina pipeline with elongation and pull-out of spigots. Ifthe pipeline comprises gasketed sections, a seriousconsequence could be separation of spigots frombells at gasketed joints.

CONTINUOUS (WELDED) PIPES

Longitudinal stresses are inevitable in welded buriedpipelines. These stresses are caused by:

1. Special sections: valves, caps, reducers, tees,wyes, and bends (which include elbows) etc.,

2. Fixed terminals or longitudinal soil friction of thesoil against the pipe, and


3. Beam bending (longitudinal pipe deflection).

Each cause of stress is analyzed separately asfollows, and then combined if necessary.

1. Special Sections

Changes in direction of flow and internal pressure(or vacuum) cause longitudinal stresses. Forexample, at a cap or closed valve, internal pressureis resisted by longitudinal tension in the pipe. Fromequations of static equilibrium, the rupturing forceπr2P equals the resisting force 2πrtσ , or,

σ = PD/4t . . . . . (14.7)

whereσ = longitudinal stress,D = inside diameter = 2r,P = internal pressure or vacuum,t = wall thickness of plain pipe.

For a 10-ft pipe with 0.75-inch wall subjected tointernal pressure of 100 psi, the longitudinal stressdue to a cap or closed valve is σ = 4 ksi.Circumferential stress is twice as great. Thislongitudinal thrust is not critical except as itcontributes to other longitudinal stresses. Except forvery high velocity flow, the impulse effect, due tochange in direction of flow, is negligible. SeeChapter 15 on thrust restraints.

2. Fixed Terminals and Soil Friction Anything that causes the pipeline to lengthen orshorten results in longitudinal stress either becauseof fixed terminals or because of soil friction.Change in length is caused by temperature changeand internal pressure (or vacuum) in the pipe, and bymassive soil movements that change pipe alignmentor length.

Any change in temperature of a continuously weldedpipeline generates longitudinal strain and stress:

ε = α (∆T)σ = Eα (∆T) . . . . . (14.8)

See also Equation 14.1 whereε = longitudinal strain,α = coefficient of thermal expansion,T = change in temperature,E = modulus of elasticity,σ = longitudinal stress caused by strain.

Temperature stresses, by themselves, are not usuallycritical. However, notable exceptions can beidentified. A 9-ft welded steel water pipe waswelded up during a warm afternoon. It was coveredwith soil except for a short span of roughly 200 ft ata stream crossing. A drop in temperature that nightcaused sufficient tension stress to fracture the pipecircumferentially on the transition radius from pipe tobell.

Another example is a nuclear plant where a buried10-ft welded steel water pipe connected the reactorto the cooling tower in a straight run. When hotwater was introduced, the pipeline elongated andbroke out the cooling tower foundation to which itwas rigidly attached.

ExampleHow much longitudinal force, Q, can be developedby a D = 10-ft pipe with a t = 0.75 inch wall, end-restrained, if the temperature increases T = 50oF?From Equation 14.8 longitudinal temperature stressin the pipe wall is σ = Eα (∆T). The total force Q isstress times area of steel, so,

Q = Eα (∆T)πDt . . . . . (14.9)

Substituting in for E = 30(106) psi and α = 6.5 microunits per degree Fahrenheit for steel, Q = 2756 kips.This enormous force can break out a reinforcedconcrete foundation.

A successful remedy for eliminating thrust against afixed terminal is an extensible coupling; either a slipcoupling such as a gasketed coupling, or an insert ofbellows or corrugated pipe. The exten-


Figure 14-12 Free-body-diagram of a bend in the pipe caused by massive soil movement — depicted, in thiscase, as the very typical heavy wheel load over the pipe, and/or settlement of the backfill soil in an excavationunder the pipe.


sible insert reduces longitudinal stress in the pipe atthe terminal. But at some distance L' from thecoupling, stress is regenerated in the pipe due to soilfriction.

Any change in alignment of a straight continuouslywelded pipeline generates longitudinal stress byincreasing length. The change in alignment is usuallycaused by massive soil movement. If a pipe oflength L is initially straight, but then is deflected intoa circular arc of radius R by massive soil movement,the change in length ∆L is calculated fromtrigonometry,

∆L = R [2sin-1(L/2R) - (L/R)] . . . . . (14.10)

Strain is ε = ∆L/L; and stress is Eε . This stress canbe combined with other longitudinal stresses.

3. Beam Bending

When a continuously welded buried pipeline isdeflected, longitudinal stress is the sum of: 1. flexurestress, σF, due to longitudinal bending; and 2. tensionstress, σT, due to stretching the pipe around thebend. Longitudinal bending is often due to: 1. heavywheel loads passing over buried pipes oncompressible beddings, and 2. excavations underburied pipes not carefully backfilled to support thepipe as backfill settles. Worse is a combination ofboth — a heavy wheel load crossing over a poorlybackfilled excavation. See Figure 14-12.

If the radius of the bend, R, and the soil load, w, onthe pipe are known, longitudinal stresses are,

σF = Er/RσT = wR/A . . . . . (14.11)

where σF and σ T are longitudinal stresses due toflexure and tension, respectively,E = modulus of elasticity,r = outside pipe radius,w = soil load on pipe per unit length,R = radius of curvature of the bend,A = pipe cross-sectional area,

L = length of the bend in the pipe. Figure 14-12 shows restraints at the ends of thesection of length L. These may be approximatelycorrect for a straight taut pipe with high soil friction.Load, w, is the soil load. The restraints are points ofcounterflexure where moments are zero andreactions are rotated hinges. For this case, thelongitudinal stress is simply σF + σ T. But therestraints may slip. Analysis then becomescomplicated.

Example

A buried 4D PVC, DR-51 electrical conduit failedunder surface loads at the location where a sewerpipe crossed under in a 4-ft-wide trench which hadbeen excavated to greater depth than the conduitafter the conduit was in place. Fracture was acircumferential crack in the top of the pipe at thetrench wall. See Figure 14-12. What caused thefracture?

1. Fracture was the result of catenary tension andflexure. However, the stress due to the catenarywas found to be negligible.

2. The loads that caused fracture were persistent,repeated passes of vehicles.

3. The scenario for fracture was a pipe acting as abeam across a trench in which uncompacted backfillsettled and loaded the beam.

A decrease in temperature might contribute to thefracture. If the pipe temperature decreased 17oFafter installation, the tension stress would be 2 ksi.This is not a major stress compared to strength of 5ksi; but could combine with other tension stresses.

After the conduit had been installed at a depth of 1.5ft of soil cover, a sewer line was installed at a depthof 6.5 ft. The 4-ft-wide sewer trench was backfilledwithout compaction. Indeed, bedding for the pipebecame loose and void as backfill settled. Afterthat, concrete mix trucks passed over


Figure 14-13 Soil friction reaction generated by the contraction (or elongation) of the pipe through a distanceof L' from an extensible coupling to the point of effective restraint.

Figure 14-14 Reissner effect (elastic theory) showing the ring deflection, d, caused by bending a straight plainpipe into a longitudinal radius R.


as shown in Figure 14-12. Reactions at the trenchwall are unknown. Therefore, a worst-case limitanalysis assumes that the ends of the beam slip, butdo not rotate.

For analysis, consider the effect of dead load on afixed-ended beam, span of 4 ft, 1.5 ft of soil cover at100 pcf. For the pipe, OD = 4.13, t = 0.081, r =2.0245, and I/c = πr2t = 1.043 in3. Soil subsidencelifts a soil wedge above the spring lines at 1h:2vshear planes for which, at 100 pcf, weight is w = 194lb/ft of pipe length. Moment at the ends is M =wL2/12, for which stress is σ = 3 ksi. This is lessthan failure stress of 5 ksi. But a surface live loadof 16 kips over a 7x22-inch rectangle causes a loadon the pipe of w = 192 lb/ft, which is essentially thesame as dead load. Therefore, the combined stressis 6 ksi. This is upper limit. For actual stress lessthan quick yield, failure could be delayed over aperiod of time.

LONG PIPES

In the following, a pipe is long, straight, andunrestrained at its terminals by extensible couplings.It is so long that enough soil friction accumulatesfrom the ends to restrain a central portion of the pipeagainst change of length. See Figure 14-13. Soilfriction develops only as the pipe lengthens orshortens throughout some distance L' from thecouplings. From equations of static equilibrium, forany longitudinal stress,

L' = σ t/Pµ' . . . . . (14.12)

where

L' = distance along the pipe from the extensiblecoupling to the point of effective restraint,

σ = longitudinal stress in central restrainedportion due to temperature change andinternal pressure,

t = wall thickness for plain pipe,P' = internal pressure in the pipe,

P = mean effective soil pressure against thepipe,

γ = effective unit weight of soil,H = height of soil cover,E = modulus of elasticity,α = coefficient of thermal expansion,T = decrease in temperature,µ' = coefficient of friction of soil on pipe,ν = Poisson ratio.

Example

A 120-inch steel pipeline is to transport cold water toa municipality. An extensible Dresser coupling islocated at the treatment plant. If temperaturedecrease is 60oF and internal pressure is 125 psi,what is the distance L' from the Dresser coupling tothe point of effective restraint? The pipe is buriedunder 6 ft of soil cover.

P' = 125 psi,D = 120 inches,H = 6 ft,γ = 120 pcf,E = 30(103) ksi,α = 6.5 micro-units per degree F,t = 0.75 inch,T = 60oF,µ' = 0.2 for tape-wrapped pipe,ν = 0.3.

L' is found from Equation 14.12. For this flexiblesteel pipe, the effective radial soil pressure P is 0.72ksf. The longitudinal stress, due to temperaturedecrease and internal pressure (both tension) is:

σ = Eα (∆T) + νPD/2t = 14.7 ksi.

Substituting into Equation 14.12, L' = 919 ft. Beyond919 ft from the extensible coupling, longitudinalstress is the full 14.7 ksi.

If the allowable longitudinal stress is σ = 36/2 = 18ksi, from Equation 14.12, L' can be increased to1225 ft. It follows that, if σ exceeds 18 ksi for a


welded pipe, the maximum allowable spacingbetween extensible couplings should be 2250 ft. Inthis example, the coefficient of friction isquestionable. If µ' is greater than 0.2, allowablespacing decreases by a direct inverse ratio.

REISSNER EFFECT

When a pipe is bent, the cross section deforms intoan approximate ellipse. See Figure 14-14.Neglecting internal pressure, the ring deflection, d, isa function of radius of the bend, R, as follows:

d = 2Z/3 + 71Z2/135 . . . . . (14.13)

where Z = 1.5(1-ν2)D4/16t2R2

Example

What is the radius of the bend of a steel pipe if ringdeflection is found to be d = 5%?D = 109 inch = mean diameter,t = 0.25 inch,r = 54.5 = D/2,d = ∆ /D = ring deflection,∆ = change in diameter,R = radius of the bend,σf = yield stress,E = 30(106) psi = modulus of elasticity,ν = 0.25 = Poisson ratio.

Substituting, Z = 200(106)in2/R2.

From Equation 14.13, R = 4420 ft.

The maximum longitudinal stress is σ = Er/R = 30.8ksi.

PROBLEMS

Given: 12,000 ft of steel water pipe.D = 42 inches = diameter,t = 0.25-inch wall thickness,L = 60-ft-long pipe sections,

Bell and spigot joints,Coal tar enamel coating, felt wrapped,E = 30(106) psi,σf = 45(103) psi,ν = 0.25 = Poisson ratio,α = 6.5(10-6)/oF = coef. of therm. expan.Soil: granular, c = 0γ = 120 pcf,ϕ' = 30o = soil friction angle,µ' = tan 30o = soil on pipe friction,

= 0.4 for cement mortar,= 0.2 for tape.

Water table below pipe invert,H = 5 ft = soil cover,B = 6 ft = trench width,For vertical alignment, the pipe is positioned onmounds at the ends of each 60-ft section such thatthe pipe does not rest on bedding, but is partiallysupported by soil under the haunches.

14-1 What is the maximum longitudinal stress in thepipe due to beam bending under soil load only? Given: the same conditions except that, as analternative, all joints are welded up before thepipeline is backfilled.

14-2 What is the maximum longitudinal stress due toa decrease in temperature of 40oF after installationand due to internal pressure of 175 psi? (11.5 ksi)

14-3 What would be the maximum allowablesettlement of two contiguous mounds due tosubgrade soil subsidence? (y = 8.2 in)

14-4 If the soil cover is 5 ft, how far from a slipcoupling is full longitudinal load restraint establishedwith cold water in the line? (507 ft)

14-5 How many 60-ft lengths can be welded intoone section before the section is backfilled?Assume a significant temperature drop. Aftertemperature drop, contiguous sections are welded.

14-6 Design the weldments at the joints for theworst seismic conditions.


Given: A steel water main ruptured on ChristmasEve due to brittle fracture which started as acircumferential crack on top of the pipe, propagateddown to the spring lines, then along both spring lineslongitudinally. It occurred near midlength of a 100-ftsection of pipe under a railroad used to deliverchemicals to a water treatment plant on theMississippi River flood plain. The pipe had beenbored into place and grouted under the rails. Pitswere excavated on each end of the pipe section inorder to jack the pipe under the railroad. The pitswere backfilled with limerock with dry unit weight of135 to 140 pcf, and void ratio about 0.4 atcompacted density of about 90% AASHTO-T99.The steel pipe was 72-inch diameter and 0.375 wall.The saturated flood plain silt weighs 103 pcf. 14-7 At what settlement of the limerock backfill willthe pipe wall stress reach yield?

(4 inches)

14-8 What is the Poisson effect of internal pressureof 125 psi?

14-9 The 72-inch steel pipe is inside a 7-ft tunnelliner casing with grout between pipe and casing. Butthe casing and grout were opened at midspan toaccommodate a transverse 3-ft pipe at the top of the7-ft casing. What is the effect on the pipe of theopening in the top of the grout encasement atmidspan?

14-10 A pipeline is buried in tidewater soil. What isthe maximum stress in a 7-ft-diameter gasketed steelwater pipe of 20-ft-long sections ring stiffened, withwall thickness of 0.5 inch and positioned on pilebents spaced near each joint then buried under 6 ftof sand at 120 pcf?

14-11 What is the leakage area in the accumulatedgap in a stab joint in 10-ft-diameter flexible pipe ifthe tolerance is 3/32 inch in circumference?

(5.6 in2)

14-12 In Problem 14-11 what is the width of thegap? What gasket is needed? (0.06 inch)

14-13 What is the minimum longitudinal radius ofcurvature for laying a continuous polyethylene pipeinto an underwater trench from a barge?R = (D/2)(E/σy)

14-14 What is the ring deflection of the polyethylenepipe of Problem 14-13 at the minimum radius ofcurvature?

14-15 What is the minimum possible horizontalradius of curvature of PVC pipe without use ofelbows? OD = 24 in, DR = 32.5.

14-16 Find the maximum length of a gasketed pipesection at critical longitudinal temperature stress. T= 40oF, D = 10 ft, t = 0.5 in, c = 0, ϕ = ϕ' = 30o, H= 6 ft, α = 6.5(10-6) (225 ft)



14-19 Derive the expression for the shear load Q atthe bell and spigot for gasketed pipe sections oflength L under uniform soil load w per unit length ifthe reactions are spaced at L/2 and are a distance Xfrom the joint.

14-20 Prove that the sine-curve reactions causemaximum moments that are 0.4 times thecorresponding values for concentrated reactions.


Anderson, Loren Runar et al "THRUST RESTRAINTS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 15-2 Impulse thrust, Qi, at a bend in a pipe due to change in direction of flow, showing a free-vector-diagram of the components Fx and Fy.


CHAPTER 15 THRUST RESTRAINTS

A straight pipe section with end closures (valves orcaps) feels a longitudinal force when internalpressure is applied. In the case of static pressure,the longitudinal force F is simply the internalpressure times the area; i.e.,

F = PπD2/4

where F = longitudinal thrust in the pipe,P = internal pressure,D = inside diameter = 2r.

But suppose the pipe has gasketed joints betweenthe ends. See Figure 15-1. Now it cannot resist theforce F. Consequently, thrust restraints (thrustblocks) must be supplied at the ends of the pipe.

EVALUATION OF THRUST Q

Each thrust restraint must resist F without movingenough to allow a joint to leak. In addition to thepressure, P, suppose that the fluid is moving in thepipe due to pressure gradient, ∆ P. Force F isincreased by ∆F = ∆PπD2/4. For most buriedpipeline analyses, fluid friction, ∆F, is negligiblebecause the length between gaskets is short and iseasily resisted by soil friction on the pipe.

But now, suppose that the gasketed pipe is notstraight. A change in direction is introduced by anelbow (or bend). The sidewise thrust Q at the elbowis due to both pressure and the impulse of change indirection of flow. It is the vector sum of impulse andpressure forces, Qi and Qp, on the fluid at the bend.Each is found separately.

Impulse Thrust Qi

See free-vector-diagram, Figure 15-2.F = impulse force (vector),Qi = thrust due to impulse,

Qi = vector sum of impulse forces Fx and Fy,θ = offset angle of the bend,v = average velocity of fluid flow in the pipe.Fx = axial force on the fluid at the elbow,D = inside diameter,ρ = mass density of the fluid,(v-vcosθ) = change in the x-components of velocityas the column of fluid flows around the bend.

Fx and Fy can be found by the principle "impulseequals change in momentum." Both impulse andmomentum are vector quantities. Figure 15-2 showsa free-body-diagram of a column of fluid (cross-hatched). The area is πD2/4, and the length is (vdt).Impulse is force times time dt, and change inmomentum is mass times change in velocity. In thex-direction;

Impulse = Fxdt

Change in Momentum = (πD2/4)vdtρ(v-vcosθ).

Equating impulse to the change in momentum in thex-direction,

Fx = π(Dv)2 ρ(1-cosθ)/4

But Fx is only the x-component. In a similar mannerby equating the y-component of impulse to change inmomentum;

Fy = π(Dv)2 ρ(sinθ)/4

From the free-vector-diagram of Figure 15-2, theresultant of Fx and Fy is,

2Qi = π(Dv)2 ρsin(θ/2) . . . . . (15.1)

The angle between Qi and Fy is tan-1(Fx /Fy) = θ/2.Noting that the pipe is symmetrical about the Qi

vector, Equation 15.1 could have been writtendirectly, because the change in velocity in the Qi

direction is simply 2vsin(θ/2).


Figure 15-3 Pressure thrust-Qp at a bend (elbow) in a pipe due to internal pressure, P, showing the free-vector-diagram for calculating Qp.

Figure 15-4 Passive soil resistance on an elbow and on contiguous gasketed pipe sections showing how thesoil envelope can provide thrust restraint.


Pressure Thrust Qp

See Figure 15-3; whereD = inside diameter = 2r,P = internal fluid pressure,Qp = thrust due to internal pressure,θ = offset angle of the bend (elbow).

A free-body-diagram of the elbow with pressurizedfluid contents is shown cross-hatched. Neglectingthe small friction loss of flow around the bend, fromthe free-vector-diagram,

2Qp = πD2Psin(θ/2) . . . . . (15.2)

Qp is at an angle of θ/2 with the y-axis.Consequently, thrust, Q, is the sum, Qi+ Qp; i.e.,

2Q = πD2(P + v2ρ)sin(θ/2) . . . . . (15.3)

wherev = average velocity of fluid flow,ρ = mass density of the fluid,θ = offset angle of the bend.

Example

Find thrust-Q at a 90o elbow in a water pipe forwhich,θ = 90o,D = 30 inches,P = 200 psi = internal pressure,v = 15 ft/second = flow velocity,ρ = γ w /g = mass density of water,γ = 62.4 lb/ft3 = unit wt. of water,g = 32.2 ft/second2 = gravity.

Substituting into Equation 15.1, Qi = 3 kips.Substituting into Equation 15.2, Qp = 200 kips.Combined, Q = 203 kips. Impulse thrust, Qi isusually neglected.

If a large diameter pipe with high internal pressurehas an elbow with a large offset angle, θ, thrust-Q isenormous.

SPECIAL SECTIONS

Special sections redirect or alter flow. Examplesinclude elbows, wyes, tees, valves, reducers, caps,plugs, etc. The following analyses for elbows can beapplied to any special section. In every case, thrust,Q is the sum of impulse thrust, Qi, and pressurethrust, Qp.

COMMON THRUST RESTRAINTS

1. Welded or Bolted Joints at Special Sections

In a pressurized pipe, at a gasketed elbow, Q mustbe resisted by the soil or by a thrust restraint (thrustblock). For a welded elbow, Q is resisted by thepipe. Two analyses of a welded elbow follow.

a) If the contiguous pipes are unrestrained anduncapped (like a garden hose), normal force, F, andshearing force, S, act on the elbow. Analysis isconservative because soil resistance reduces F andS. σ = F/2πrt = average normal stress,τ = S/2πrt = average shearing stress.

From the equations of static equilibrium,

σ /P(r/t) = (1-cosθ) . . . . . (15.4)NORMAL STRESS TERM

τ /P(r/t) = sinθ . . . . . (15.5)SHEARING STRESS TERM

These stress terms are upper limits — twice theforce-per-unit-area — to account for eccentricity ofthe F-force and redistribution of stresses. Theoutside of a bend can stretch more than the inside.Therefore, stresses are greater on the inside. SeeProblem 15-12. Wall thickness is sometimesincreased for elbows. In general, greater wallthickness is not justified.

b) If the contiguous pipes are restrained andcapped, from the equations of equililbrium,longitudinal stress is,


σ = Pr/2t . . . . . (15.6)

This is only half as great as circumferential stress,and is independent of offset angle, θ. A moreprecise analysis would show that stress, σ , on theinside of the bend is increased slightly as the offsetangle, θ, is increased. Most pipes are ductile enoughthat the material "plastic-flows" at yield, and does notfail. Moreover, soil friction resists thrust. Inpractice, contiguous pipes are seldom capped.Longitudinal stress is not critical for isotropic plainsteel and plastic pipes. Of course, joints must beadequate.

For non-isotropic pipes (corrugated, ribbed, orwrapped with fiberglas or wire), longitudinal strengthmust be assured. Neglecting impulse force and soilresistance, for uncapped, unrestrained contiguouspipes:

At elbows, for longitudinal design,Pπr2(1-cosθ) = Aσf /sf . . . . . (15.7)

whereA = area of longitudinal fibers,σf = strength of the fibers.

At valves or caps (not at bends) for design,Pπr2 = Aσ f /sf . . . . . (15.8)

2. Embedment As Thrust Restraint

If thrust-Q is not large, the embedment is able todevelop adequate passive resistance. It may not benecessary to provide additional thrust restraint.Consider in Figure 15-4 the free-body-diagram of anelbow and one section of pipe on each side. Thejoints are gasketed so the pipe can take nolongitudinal force. Thrust-Q can be restrained onlyby the soil bearing against the pipe. The maximumsoil pressure bearing horizontally against the elbowis passive resistance Px at the average depth of soil,H + OD/2,

Px = (2H + OD)γ /2K

whereK = P/Px = (1-sinϕ)/(1+sinϕ),ϕ = soil friction angle,γ = unit weight of soil,OD = outside diameter,H = height of soil cover,L = length of pipe section.

The restraint capacity of soil against elbow is,Qelb = (area) times Px

where(area) = (OD)Lelb,Lelb = cord length (approximate) of elbow

from coupling to coupling as shown.

Multiplying (area) times Px,

Qelb = (2H + OD)γ LelbOD/2K

Added to this is the restraint capacity of the firstsection of pipe on each side of the elbow. Fullpassive resistance of the soil would be developed atthe elbow end of each section. At the opposite end,each pipe section could rotate, because of thegasket. But there would be no lateral movement.Passive soil resistance would not be developed. Acrude, but reasonable and conservative assumption,is that passive resistance varies linearly from Px atthe elbow end to zero at the opposite end. Due tosoil supporting the two pipe sections, the componentof restraint in the direction of Q is,

Qsecs = (OD)LPxcos(θ/2)

or, substituting for Px,

Qsecs = (OD)L(2H + OD)γ cos(θ/2)/2K

Combining the thrust restraints provided by theelbow and the two pipe sections,

Restraint-Q = OD(2H + OD)γ [Lelb + Lcos(θ/2)]/2K

. . . . . (15.9)


Rewriting Equation 15.3,

Thrust-Q = π(ID)2(P + v2ρ)sin(θ/2)/2. . . . . (15.10)

Equation 15.10 for thrust-Q was derived for ahorizontal bend. For a vertical bend (in a verticalplane), thrust-Q has a vertical component. If soilcover alone is to resist the upward component ofthrust-Q, then soil cover H must be great enoughthat soil weight can hold the pipe down. Aconservative restraint-Q for this vertical bend is,

Restraint-Q = OD(2H + OD)γ [Lelb + Lcos(θ/2)]/2

. . . . . (15.11)

This is the same as Equation 15.9 except that K iseliminated. For design, restraint-Q must be greaterthan thrust-Q. A safety factor should be included.

3. Thrust Block as Thrust Restraint

Thrust blocks are the most common restraints in usefor pressurized gasketed pipes. See Figure 15-5.Thrust blocks are usually concrete. A reasonableanalysis for design starts with the free-body-diagram. Assuming a cubical block, B = lengths of sides of the cube,γ = unit weight of soil,γ c = unit weight of the thrust block,jB = distance down to thrust-Q from the top of

the block,K = (1-sinϕ)/(1+sinϕ),ϕ = soil friction angle.

Other data are shown on the sketch. Friction on thesides of the block is undependable and isconservatively neglected.

Two modes of failure are considered: overturnabout point O, and slip. The conditions under whicheach mode controls are described by an example ofa cubical thrust block. Example — Assumptions

h = H/B = ratio of soil cover H to side B,j = ratio of distance between top of block and

thrust-Q, to side B,ϕ = 30° = soil friction angle,K = 1/3 = (1-sinϕ)/(1+sinϕ),γ = 120 pcf = unit weight of soil,γ c = 144 pcf = unit weight of concrete.

Taking the sum of the moments of force aboutoverturn fulcrum O,

Q/γ B3 = (2h + 1.10)/(1-j) OVERTURN . . . . (15.12)

Taking the sum of the horizontal forces,

Q/γ B3 = (3.577h + 2.193) SLIP. . . . . (15.13)

The dimensionless quantity Q/γ B3 is the thrust blockrestraint number. A table of values is shown asTable 15-1 for typical design based on theassumptions indicated.

Overturn

In order to design a cubical thrust block with thetypical soil properties assumed in the analysis above,it is only necessary to guess a trial value for B fromwhich values of h and j can be calculated. EnteringTable 15-1 with h and j, the restraint number, Q/γ B3

can be found in the overturn columns.

For a soil unit weight of γ = 120 pcf, Q/B3 = (120pcf)(restraint number)/sf. Solve for B. If not thesame as the assumed B, using the new B recalculatevalues for h and j. Enter Table 15-1 for a secondtrial solution of the restraint number from which anew value of B is calculated. If this new B isunchanged, then the answer has been found. If not,recycle the analysis with the new B.

Slip

The left of the two SLIP columns of Table 15-1


Figure 15-5 Free-body-diagram of a cubical thrust block.

Table 15-1 Values of cubical thrust block restraint number, Q/γ B3, for concrete at 144 pcf and soil at 120pcf and ϕ = 30o. No safety factor is included.


provides values for the restraint number fromEquation 15.13. The right of the two slip columns isthe minimum value of j at which slip is critical.

No safety factor is included. The analysis is soconservative, that safety factors need not be large.Nevertheless, the risk of failure may warrant asafety factor.

One novel concept is the thrust pin designed toconserve space. See Figure 15-6. It can be locatedinside the bend if necessary, tied with tendons.

4. TendonsInstead of thrust blocks or thrust pins, which restrainthe gasketed elbow by compression from outside thebend, restraint is by tension tendons inside the bendfastened to dead-men such as buried concreteblocks, boulders, beams, pins, etc. The tendonscould be rods, cables, wires, etc.

Or, instead of tying tendons to dead-men, they couldbe tied across the bend to corresponding pipe jointson either side of the elbow. See Figure 15-13.These "harp-strings" cannot resist the thrust withoutother restraints such as longitudinal friction betweensoil and pipes and soil bearing.

PIPES ON STEEP SLOPES

The analysis of thrust restraints, for pipes on slopes,is the same as above; but in addition, must includelongitudinal forces caused by gravity. For mostpipes, the length tends to shorten when the pipe is inservice because internal pressure increases andtemperature decreases. Therefore, it is goodpractice to design restraints such that the pipeshortens downhill. Frictional resistance toshortening is uphill and partially offsets the downhillcomponent of weight of the full pipe.

The thrust restraint (anchor) is clamped to the pipeuphill from it, and is slip-coupled to the pipe downhillfrom it. The downhill side is free to slip toward thenext anchor downhill. Expansion joints allow slip. For short pipe sections, com-mon sleeve-type

couplings allow adequate slip.

For most couplings, the pipe must be supported onboth sides of the coupling to assure alignment. Goodbackfill soil may provide alignment. In the case ofpoor backfill, or pipe on piers, two yokes on eachpier assure alignment as shown in Figure 15-7.Most couplings are not designed to resist longitudinalmoment or transverse shear. The allowable degreeof misalignment of a coupling is limited. Forexample, the allowable misalignment (offset angle)is about 3o for steel pipes 30 to 54 inch diameter.Manufacturers of couplings should be consulted forrestrictions and specific applications. If the pipe ison piers, couplings should not be located at midspansbetween piers.

Particular care is required for large pipes on steepslopes because of the difficulties of installation aswell as the additional loads on the thrust restraints.On slopes steeper than about 45°, the pipe is oftenplaced on piers above ground. The slope is toosteep to excavate a trench, too steep to hold the pipein position for welding and backfilling, and too steepto compact backfill. Moreover, a pipe on a steepslope may feel the downhill drag from creep ofsurface soil (downhill freezing and thawing) and thusoverload the anchor at the bottom of the slope.Slopes steeper than 45° are usually rock outcropsthat cannot be excavated without ripping or blasting.On steep slopes, or in inaccessible areas,construction might require lowering personnel,platforms, and equipment down the pipeline bycables. Rock pins are drilled and grouted into place.The pipe is laid downhill as the platform is lowered.Economics often favor service from an overheadcable on towers, or from helicopters. Helicoptersare expensive (one to two thousand dollars perhour), but can place pipes and piers quickly if groundcrews avoid delays. Under some conditions,tunneling may be an option.

Example

The following example identifies some of the manyproblems associated with pipelines on steep slopes.


Figure 15-6 Alternate concept of thrust restraint at an elbow in a pipeline provided by a thrust pin thatconserves space and quantity of concrete. The hole is bored, reinforcing steel is positionced, and concreteis cast into the bored hole.


A 30D steel penstock is to be installed in a remotearea in a cold climate. Much of the 4800-ft pipelineis on slopes less than 10°, in soil that can beexcavated for burial of the pipe to protect it fromavalanches, falling trees, and temperature extremes.However, part of the pipeline will be placed aboveground on piers and anchors in rock outcrops on asteep 48o decline. See Figure 15-7. The area isheavily timbered and inaccessible. Elbows in thepipeline will be anchored at the top and bottom of the48° slope. What are the requirements for piers andanchors?

It is cost effective to bring in pipes, equipment, andconcrete by helicopter. The required pipe wallthickness is 0.375 inch. If the maximum length of apipe section is 30 feet, the weight is about 3.6 kips,which is the load capacity of the helicopter. Boltedsleeve-type couplings are selected because theyeliminate field welding, which is slow and tortuous onthe steep slope. Couplings can be assembled quicklyand without heavy equipment. For 30-ft lengths ofpipe, ordinary sleeve couplings can accommodatethe longitudinal expansion and contraction due tolarge temperature changes above ground. A changein temperature of 100o Fahrenheit causes a quarterof an inch change in length in each 30-ft section ofpipe. If the pipe were to be installed in long sections,special expansion joints would be needed.

Before the pipes are flown in, piers or chairs mustbe in place and fixed. Unless deep foundations canbe provided, rock pins are required. Holes aredrilled into the rock at the uphill side of each pier.Deformed reinforcing rods (rebars) are grouted intothe drilled holes as shown in Figure 15-8. In order toassure bond, the depth of the holes should be at least50 diameters of the rod. If #8 rebars are used, thedepth of the holes must be 50 inches. In coldclimates, frost penetrates more than 50 inches, so10-ft depth is prudent. If piers are to be used,concrete placed from a helicopter requires atremie and good luck as well as skill. Forms for thepiers can be light-weight and reusable. Each pier

will be a monolithic casting. The upper ends of therebars may be threaded such that the bands thatclamp the pipe will also secure the pipe to the pier.

The amount of concrete may be reduced by usingprefabricated chairs. In this example, chairs forsupporting the pipe are to be flown in and fastened,either to pins drilled and grouted into the rock, or tosmall concrete footings anchored to the pins. Twopins must support the downhill components of weightof the chair and footing plus the pipe section full ofwater uphill from the pins, reduced by the frictionalresistance to the normal component of weight ofchair, footing, and one full pipe section. The chairweighs about 0.5 kips. If the pier requires two cubicyards of concrete it will weigh about 7.5 kips. If theweight of the full pipe is 430 lb/ft, it will weigh 13kips. The total weight is about 21 kips. The basiccoefficient of friction is the tangent of the soilfriction angle, say, 30°. The resulting downhillshearing load is 11 kips. See Figure 15-8. Theshearing stress on two #8 bars of 0.7854 in2 areaeach is 11/1.57 = 7 ksi which is not excessive.However, bearing of steel on grout should bechecked. The upper ends of the #8 bars arethreaded such that the chair can be bolted to thepins. If necessary, the downhill chair foot can bebolted to pins — one pin in the center of the foot, ortwo pins at the ends of the foot.

The bands that secure the pipe to the chair alsoclamp the pipe and prevent slipping of the band dueto the 11-kip shearing load. See Figure 15-9. Toprevent slip, each band must be tensioned to fourkips (assuming coefficient of friction between bandand pipe is one-third). It may be prudent to cleanand roughen the surface or to apply epoxy and "salt"the surface with carborundum dust. To tension thebands, with a margin of safety, use 3/4 bolts withtensile strength of 6.19 kips each and tensioned to 5kips. The band is a 1/4-inch steel strap, 4 incheswide. A large square washer is required todistribute the load and eliminate bending at thehole where band width is only about 3 inches.


Figure 15-7 Penstock on a slope showing examples of a concrete pier with two yokes for alignment of thecoupling, and anchors (thrust restraints) at the elbows.


P

CRC Employee

Figure 15-8 Chair on a 48o slope on a rock outcrop, showing rock pins used to secure the chair to the outcrop,and showing a bolt, at the downhill end of the diagonal brace, that serves as a pivot for slight rotation of thedownhill legs of the chair, eliminating the need for a slip band.


Figure 15-9 Partial details of a chair that can be flown onto location by helicopter and bolted to rock pins.


Adjustment of the fit of band-to-chair and chair-to-pins (or footing) is accomplished as follows. Boltholes are slotted to allow for horizontal adjustment,and shims are inserted to allow for verticaladjustment and rotation. Adjustment of longitudinaldisplacement of the downhill pipe section withrespect to the uphill pipe section is accomplished bypivots at the lower ends of the downhill chair legs.See Figure 15-9. A 7/8 bolt is adequate in doubleshear. The vertical leg must be cut to allow forslight rotation.

Thrust restraints (anchors) are required at theelbows shown in Figure 15-7. Anchor, A, at theinverted elbow is critical because the pins are intension. The depth of the six bored holes must bedetermined. If pressure in the pipe is 433 psi (1000ft static head), the uplift thrust at the elbow is Qp =158.56 kips. If mean velocity of flow is 15 fps, Qi =1.12 kips which can be neglected. If the weight ofelbow and anchor is assumed to be 50 kips, Q = 160- 50 = 110 kips. To hold the elbow down, each ofsix deformed steel bars must resist a pullout load of20 kips. Because the loads may not be equallydistributed, assume 30 kips. Select #11 rebars withcross-sectional areas of 1.485 in2 for which tensilestress is 20 ksi. If the shearing bond between thebar and the grout is 100 psi (equivalent to 50diameters of overlap), each bar must be grouted toa depth of 6.2 ft. Specify 12 ft of depth to assure amargin of safety. The hydrodynamic design of theelbow is within acceptable limits. See Chapter 18.The mean radius of the bend is greater than 2.5 pipediameters (about three diameters), the inside lengthof each mitred section is greater than half the piperadius, and the angle offsets of continuous mitredsections are only 7.5°.

The height of each chair must be custom fit to therock outcrop. A laser beam is useful for measuringup each chair. If the height is too great, towers mustbe designed. Sections of prefabricated towers canbe flown in. If concrete is a realistic alternative, thetower could be a vertical pipe filled with reinforcedconcrete.

EXPANSION INSERTS

An expansion/contraction insert allows for slightlongitudinal expansion or contraction withoutoverstressing the pipe or joints. Slip couplings andgaskets essentially eliminate all longitudinal thrust inthe pipe. However, under some circumstances, acorrugated pipe insert or bellows section can reducethrust to acceptable levels. Two questions arise: 1. What is the elongation of the insert per unit lengthof pipe at yield or endurance limit?2. What is the longitudinal force per unit length ofcircumference of the insert at yield stress or at theendurance limit?

Analysis is based on the free-body-diagrams ofFigure 15-10 for a corrugation, and Figure 15-11 fora bellows.

NotationT = thrust per unit circumference,x = elongation of insert due to T,b = pitch of corrugation (or bellows),c = depth of corrugation (or bellows),t = thickness of insert material,σ = maximum stress in the insert,E = modulus of elasticity,σf = yield stress.

From symmetry and static equilibrium, analysis isperformed on one-fourth of the pitch, b/4 as shownin Figure 15-10. Maximum stress is,

σ = T/t + 3Tc/t2 . . . . . (15.14)

For corrugations with ratios of b/c = 3, the axialterm, T/t, is less than 6% of the flexural term, 3Tc/t2,and may be neglected. This is mitigated by usingelastic theory for elasto-plastic pipes. The ratio ofaxial to flexural stress, is t/3c. If c = 2 and t =0.1345 inch, the ratio of axial to flexural stress is0.02. Axial stress is negligible.

Elongation, ∆ x, of a fourth of a corrugation due to


Figure 15-10 Procedure for analysis of a corrugation to find the relationship of thrust to elongation.

Figure 15-11 Procedure for analysis of a bellows to find the relationship of thrust to elongation.


T is found by the Castigliano equation, from which,for a full corrugation, x = 5.273(T/E)(c/t)3. For allpractical purposes,

x = 5(T/E)(c/t)3 . . . . . (15.15)

where x is elongation per corrugation length, b, dueto thrust T per unit of circumference. E is themodulus of elasticity and b/c = 3.

For bellows, from Castigliano's equation, theelongation of a repeating section is ∆ x =(3π/2)(T/E)(c/t)3. But a repeating section of bellowsis only 2/3rds as long as corrugation, b, of equaldepth, c. Therefore, compared to a 3x1 corrugation,the elongation of bellows per unit length ofcorrugation, b, is,

x = 7(T/E)(c/t)3 . . . . . (15.16)

The ratio of elongation per unit length of expansioninsert of bellows to corrugation is 7/5. Performancelimit of x is yield stress (or endurance limit).

For a corrugation,x = 5σf c2/3Et per corrugation length, b.

. . . . . (15.17)

For a bellows,x = 7σf c

2/3Et per corrugation length, b, . . . . . (15.18)

where depth, c, and thickness, t, are the same forbellows and corrugation.

Thrust, T, at yield stress or endurance limit, can befound from Equation 15.14. Neglecting axial stress,σf = 3Tc/t2.

Example

What is thrust, T, at yield stress for a 6x2corrugation if t = 0.1345 inch and σf = 36 ksi? T =σf t

2/3c = 108.5 lb/inch. The corresponding increasein length of the corrugation, from Equation 15.17, isx = 5σf (c

2/3Et) = 0.06 inch. Eight corrugations are

required to provide extension of a half inch.

For design by endurance limit, of concern is thematter of full reversal of stress or cyclic stresssuperimposed on a standing stress. Details arefound in texts on mechanics of solids. The analysesabove are conservative because a corrugated pipe orbellows is a three-dimensional problem — not just atwo-dimensional cross section of the corrugation orbellows. In fact, ring restraint is significant.

PROBLEMS

15-1 Prove that the angle of the impulse thrust, Qi,is = θ/2.

15-2 What forces act at an elbow in a water pipe?D = 6 ft = inside diameter,P' = 120 psi = internal pressure,θ = 60° = offset angle,v = 10 ft/second = average flow velocity,ϕ = 30° = soil friction angle,C = 0 = soil cohesion.

15-3 Design a thrust restraint for the 60° elbow inthe pipe of Problem 15-2. Assume the thrustrestraint is to be a solid cube (block) of reinforcedconcrete with its top at ground surface. Assumethrust-Q = 500 kips. γ = 120 pcf for soil and γc =144 pcf for the reinforced concrete.

15-4 What is the diameter of a thrust pin forProblem 15-2 assuming the maximum depth reachedby the boring auger is 25 ft?

15-5 On the 48° slope of Figure 15-8, theperpendicular distance to the bottom of the pipe fromrock pins must be 12 ft. Design a tower that can beflown in and onto which a chair can be attached.What about loads on the pins? Should the tower bevertical?

15-6 On the 48° slope what would be the problembetween the two anchors of Figure 15-7 if the pipe


were welded rather than coupled in 30-ft lengths.Assume that maximum change in temperaturecould be + or - 100°F from mean installationtemperature? Assume a coefficient of thermalexpansion of 6.5(106)/°F for the steel pipe.

15-7 If expansion joints are used at 90-ft spacing inthe welded pipe of problem 15-6, how muchexpansion (contraction) must be accommodated atthe expansion joints?

15-8 What is the longitudinal spring constant, T/x,for a 6x2 corrugation with steel thickness of 0.1345 inch? What is the elongation in percent atyield stress?

Given:Mortar-coated, gasketed steel pipe with a 90o elbowshown in Figure 15-12.ID = 60 inches,t = 0.375 inch,L = 30 ft,P' = 100 psi, internal pressure,v = 12 ft per second flow rate,µ' = 0.4 = coef. of frict. soil on pipe,H = 4 ft, height of soil cover,γ = 120 pcf = soil unit weight.

15-9 How many contiguous sections must bewelded on each side of the elbow to resist the thrust-Q?

15-10 What is the effect on the welded sections ofProblem 15-8 if the pipe shortens due to temperaturedecrease and pressure increase?

15-11 An alternative to the welded sections ofProblems 15-8 and 15-9 is a series of cross tiesshown in Figure 15-13. Design and show details foran adequate number of cross ties and clamps(bands) for attaching the ties to the pipes at thejoints. Neglect soil friction.

15-12 Stress-terms as functions of offset angle, θ,are shown below for worst-case (triangular) stressdistribution for which the maximum stress is twicethe average.NORMAL STRESS TERM IS σ /P(r/t),SHEAR STRESS TERM IS τ /P(r/t).

Ignore shear stress term because the pipe isrestrained by soil, and because maximum shearingstresses are near the neutral surface of the elbowwhere normal stresses are minimum. Compoundstress analysis is not justified. What is the maximumnormal (longitudinal) stress on this 90o elbow? [Pr/t]


Anderson, Loren Runar et al "EMBEDMENT"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 16-1 Embedment showing a compacted soil arch that supports load and protects the pipe.

Figure 16-2 Densely compacted "pedestal" of soil on top of a pipe showing the tendency to concentrate theMarston load on the pipe.


CHAPTER 16 EMBEDMENT

The soil in which a pipe is buried is not just load onthe pipe. Soil is a major component of the pipe-soilstructure.

Following are a few basic concepts that are usefulin evaluating the contribution of soil to the struc-turalperformance of buried pipe-soil structures.

1. Most undisturbed native soils are stable — evenlow-strength soils. They have settled in place; and,except for earthquakes and landslides, provide astable medium in which to bury pipes. Difference inweights of pipe and soil is usually not great. Insaturated soil, most pipes tend to float rather thansink.

2. The best buried pipe installations are those whichdisturb the native soil the least. A bored tunnel ofexact pipe OD into which the pipe is inserted, wouldcause the least disturbance. Microtunneling showspromise, with bore slightly greater than inserted pipe.A common installation is a narrow trench with onlyenough side clearance to align the pipe and to permitplacement of em-bedment. Regardless of trenchwidth or shape, the embedment is a transfer mediumthat fits the pipe to the trench and stabilizes pipe-soilinteraction.

3. Arching action of the soil helps to support theload. See Figure 16-1. The soil acts like a masonryarch. No cement is needed because the soil isconfined in compression. Soil protects the pipe.

4. In order to create a soil arch, the bedding mustbe compacted. Bedding provides abutments for thesoil arch. The sidefill is the soil arch. It must becompacted up and over the pipe.

5. If mechanical compactors are used, the soil archshould be compacted in lifts of less than one ft. onalternate sides of the pipe so that the compactionsurfaces are at the same elevation — balanced lifts.Soil should not be "pounded" directly on top of the

pipe. See Figure 16-2. To do so is to create apedestal that concentrates a Marston load on thepipe. The Marston worst-case load can be avoidedby compacting a soil arch.

6. Full contact of embedment against the pipeshould be achieved in order to:a) eliminate voids which could become channels ofgroundwater flow along the pipe (under thehaunches), and,b) reduce concentrations of soil pressure against thepipe.

As in all structural design, the buried structure hasthe basic objective of adequate performance atminimum cost. Minimum cost is a trade-off be-tween the cost of the structure and the cost ofinstallation. Installation costs include: a select soilenvelope if required; soil compaction, excavation,alignment, thrust restraints, cross-sectional shapecontrol, etc. Of course, the project cost also in-cludes liability, risk, service life, maintenance,repairs, replacement, overhead, insurance, bonds,etc.

At one extreme, an all-welded, non-corrosive, non-collapsible pipe could be designed which wouldrequire no installation costs beyond excavation andbackfilling by shoving-it-in. The cost of such a pipeis usually enormous. At the other extreme, a verylow-cost pipe could be designed to just resist internalpressure. But the pipe might be so flimsy that theembedment would have to be laid up particle byparticle like a masonry arch. The masonry archwould carry the loads and protect the flimsy pipe.The pipe might have to be retained by mandrels orstruts, in order to provide a form for laying up thesoil arch. The pipe is a liner for a masonry conduit.The cost of such installation would be enormous.

Somewhere between these two extremes is aminimum cost point. Pipe costs are available frommanufacturers. Cost figures for the soil embed-


ment and its placement require analysis by thedesign engineer. The basic soil property forembedment is density. For select embedment suchas pea gravel, compaction can be achieved bymerely moving the gravel into place in contact withthe pipe. For poor embedment, mechanicalcompaction may be required, and often a slowperiod for drying the soil to optimum moisturecontent for compaction. Following are suggestionsfor design of the embedment.

COMPACTION TECHNIQUES

The importance of soil compaction cannot beoveremphasized. In soil cell tests, the ring deflectionof flexible pipes 3 ft in diameter, in an embedment ofloose silty sand, was reduced to approximately halfby merely stomping soil under the haunches. Ringdeflection is direc tly related to vertical soilcompression. Under a given vertical soil pressure,loose soil compresses more than five times as muchas compacted soil. Below the water table, soildensity can be critical. At less than critical density,relative shifting of soil particles due to soilmovements (vibrations, tremors), tends to "shakedown" the soil grains into a smaller volume. If thisloose soil is saturated, the volume decrease of thesoil skeleton leaves only the non-compressible waterto carry the loads. The soil mass becomes liquefiedand the pipe may collapse.

On the other hand, at greater than critical density,any shifting of soil particles only tends to "shake-up"soil grains such that the soil volume tries to increase.But the confined, saturated embedment cannotincrease in volume. Consequently, intergranularstresses increase, and the shearing strengthincreases. Depending on the soil type, criticaldensity is no more than 85% (AASHTO T-99 orASTM D698). Below the water table, it is usuallyprudent to compact the soil to a density abovecritical. Ninety percent density is often specified toadd a margin of safety. The sidefill should be placedin balanced lifts to retain the cross section andalignment of the pipe.

A caveat is suggested in the use of water tocompact the soil. The soil must be free-draining andmust be dewatered such that seepage stresses helpto compact the soil. Flotation of the pipe must beavoided.

Another caveat applies to all compaction techniques.As the size of buried structures increases,contractors are prone to extrapolate thoseinstallation techniques they have learned byexperience with smaller structures. Installationtechniques cannot be scaled-up so simply. An antcan carry many times its own weight. An elephantcannot carry a load equal to its weight. A whalecannot carry itself, but must be buoyed up by water.From the similitude of scale-up, the unit weight ofsoil varies with length scale ratio. See Appendix C.For example, suppose a contractor has experience inbackfilling 6-ft-diameter flexible pipes with nodifficulty and less than 2% ring deflection. Now heis to backfill a 12-ft-diameter pipe. To scale-up to12-ft from his 6-ft pipe experience, he must imaginethat the soil he placed around the 6-ft pipe weighedtwice as much — like iron filings or ball bearings.Clearly, he would have to be more careful wheninstalling the 12-ft pipe to control ring deflection.

Following are techniques for compacting soil.

1. Select EmbedmentCarefully graded select soil falls into place atdensities greater than critical density. The onlyrequirement is to actually move the soil in against thepipe — including that hard-to-reach zone under thehaunches — in order to achieve intimate contactbetween embedment and pipe. 2. JettingSoil density greater than critical can be achieved byjetting. This technique is particularly attractive forsoil compaction about large buried structures. Soil isplaced in high lifts, such as 3 to 5 ft, or to the springline (mid-height) of large diameter pipes. A "stinger"pipe (1 inch? diameter, and 5 or 6 ft long, attached toa water hose) is injected vertically down to nearbottom of the soil lift. A high-pressure water jet


moves the soil into place at a density greater thancritical if the soil is free-draining and immediatelydewatered. Jet injections are made on a grid everyfew feet. Five-ft grids have been used successfullyfor 5- or 6-ft lifts of cohesionless soil. Gang jets canbe mounted on a tractor. See Figure 16-3. Theycan be injected into a lift of sidefill up to the springline. In order to fill holes left when jets arewithdrawn, the stingers are vibrated. A second liftup to the top is jetted in a similar manner. Thetechnique works well in sand.

3. FlushingSoil densities greater than critical can be achieved ifsoil is moved by a high-pressure water jet (fire hose)used to flush soil down a slope into place against thepipe. This method is shown schematically on theright side of Figure 16-3 where a windrow of soil isplaced adjacent to the pipe. A laborer with a high-pressure water jet plays the stream onto the insideslope of the windrow until a soil slide develops. Thissoil slide can be directed by the jet into place againstthe pipe with enough energy to fill in the voids. Ofcourse, the water must drain out rapidly for bestcompaction. Windrows are added on both sidessimultaneously in order to keep the soil in balance.This method is very effective in mountain soils thatwere deposited by the flushing action of tumblingstream flow.

4. Ponding (Flooding)The least effective method (yet often adequate) forcompaction is ponding or flooding. A lift of free-draining soil is placed up to the spring line of thepipe, then the soil is irrigated. Enough water mustbe applied that the lift of soil is saturated. The soilshould be free-draining and must be dewatered tosettle the soil. The pipe must not float out ofalignment. A second lift to the top of the structure,and ponded, is often specified. The compactionmechanism is downward seepage stress whichcompacts the soil. Soil is washed into voids andunder the haunches of the pipe.

5. High-Velocity ImpactSoil compaction as well as controlled placement can

be achieved by blowing, slinging, or dropping the soilinto place. With the proper gradation of soil particlesize, and with the proper amount of water, the soilhas the consistency of concrete. If this "concrete"is dropped from an adequate height, it will flow byimpact under the haunches of the pipe. It sets uplike low-grade concrete. Air-dry cohesionless soilwill ricochet and tumble under the haunches, atuniform density, if dropped from sufficient height.See Figure 16-4. Better control is achieved if theembedment is "shot-creted" into place or if dry soilis blown or slung into place.

6. VibrationLoose soil can be compacted by vibrating it withvibroplates and vibrating rollers on each soil lift.Some compaction of the embedment can beachieved by vibrating the pipe itself.

Concrete vibrators are designed for placement ofconcrete. The purpose is to "flow" the concrete intovoids, and to remove air pockets. Concretevibrators are effective in placement of embedmentaround pipes if enough water is mixed with the soilto form a viscous mix like concrete. The contractormay saturate a lift of sidefill and then settle it withconcrete vibrators. This technique places, but doesnot compact, the soil. Saturated soil is non-compressible; therefore, "non-compactable".

A method called saturated-internally-vibrated (SIV)is the vibration of the saturated embedment byconcrete vibrators. The method is expensive. If thesoil is not free-draining, particles flow into place, butsettle only under buoyant weight. The result is thesame as ponding. The soil gradation must becontrolled just as concrete aggregate is controlled.Flotation must be avoided.

7. Soil Cement (Flowable Fill) and SlurryUnder some circumstances, the best way to assuresupport under the haunches is by flowable fill (soilcement or slurry). The pipe is aligned on mounds.Flowable fill is poured into the haunch area on oneside of the pipe. Full contact is assured when theflowable fill rises on the other side of the pipe.


Figure 16-3 Compaction of backfill by jetting (left) and by flushing (right).

Figure 16-4 Soil compaction by high-velocity impact (drop).


Recommended slump is about 10 inches or aflowability of 12-inch diameter. Flowable fill can bemixed on site, or grout can be delivered in ready-mixtrucks. The minimum height of flowable fill is about60o of bottom arc (say D/10) above which the angleof repose of embedment fills in under the haunches.If flowable fill is required to a depth greater thanflotation depth, it can be poured in lifts. Someagencies specify compressive strength of 200 psi.Less strength (40 psi?) may be desirable to reducestress concentration, and to facilitate subsequentexcavations. Flowable fill should not shrinkexcessively. However, cracks do not cause distressof the pipe.

8. Mechanical CompactionMechanical compaction of the soil in lifts (layers) isan effective method for compaction. Mechanicalcompactors densify the soil by rolling, kneading,pressing, impacting, vibrating—or any combination.Sales instructions are available on mechanicalcompactors and on procedures such as optimumheights of soil lifts, moisture content, etc.Efficiencies of various compactors in various soilshave been studied. In order to retain shape andalignment of the structure, heavy equipment(compactors, loaders, scrapers, etc.) must notoperate close to the structure — especially flexiblestructures.

LIGHT AND HEAVY EQUIPMENT ZONES

If the buried structure is so flexible that heavycompactors can deform it, then only lightcompactors can be used close to it. Especiallyvulnerable are flexible structures with a large sideradii such as an acorn-shaped railway underpass andegg-shaped sewer. Heavy compactors must remainoutside of planes tangent to the structure andinclined at an angle less than 45o + ϕ/2 fromhorizontal. See Figure 16-5. Soil cover, H, greaterthan minimum is required above the structure. Theheavy equipment zone is often specified as shownon Figure 16-5. Operators should be reminded thata large structure gives a false illusion of strength. It

achieves its strength and stability only after theembedment has been placed about it. Because thestructure cannot resist high sidefill pressures duringsoil placement, operators should think, "If it were notthere, how far back from the edge of the sidefillwould I keep this equipment in order not to cause asoil slope failure?" The answer is found fromexperience and from the tangent plane concept. Amargin of safety is usually applied to the 45o+ϕ/2plane by specifying a 45o tangent plane. Theminimum cover, Hmin, for various types and weightsof equipment can be determined by the methodssuggested in Chapter 13. As a rule of thumb, theminimum soil cover should not be less than 3 ft forH-20 truck loads, D8 tractors, etc. For scrapers andsuper-compactors, 5 ft of soil may be a morecomfortable minimum.

TRENCH WIDTH

The trench only needs to be wide enough to align thepipe and to place embedment between pipe andtrench wall. If ring deflection is excessive, or if thepipe has less than minimum soil cover when surfaceloads pass over, the soil at the sides can slip. Ringinversion is incipient. If there is any possibility ofsoil liquefaction, the embedment should be denserthan critic al density. With a margin of safety, 90%standard density (AASHTO T-99 or ASTM D698)is often specified. In loose saturated soil,liquefaction can be caused by earth tremors. Soilcompaction may or may not be required dependingupon the quality of the embedment. For example,gravel falls into place at densities greater than 90%.Loss of embedment (piping) should be prevented.Piping is the wash-out of soil particles bygroundwater flow.

The Marston load on a pipe is the weight of backfillin the trench, reduced by frictional resistance of thetrench walls. The narrower the trench, the lighter isthe load on the pipe. The pipe has to be strongenough to support the load. Marston neglected thestrength contribution of the sidefill — both hori-zontal support of the pipe, and vertical support ofbackfill. Trenches are kept narrow for rigid pipes.


Figure 16-5 Light and heavy equipment zones during the placement of backfill soil showing, of particularconcern, the light equipment zone into which no heavy equipment is allowed.

Figure 16-6 Infinitesimal soil cube, B, at spring line, showing conditions for soil slip when Px = Kσ y.


When flexible pipes came on the market, Spanglerobserved that a flexible ring depends upon supportfrom sidefill soil. His observation led to theinference that, if the trench is excavated in poor soil,the trench walls cannot provide adequate horizontalsupport. The remedy appeared to be widerembedment (a wider trench) especially in poornative soil. In fact, a wide trench is seldom justified— either by experience or by principles of stability.

From Chapter 9, Pxrx = Pyry. If the deflected ringis elliptical, ry /rx = (1+d)3/(1-d)3. When ringstiffness is taken into account, Pxrx is less than Pyry.The contribution to the support of load P by ringstiffness can be included in the analysis if ringdeflection is significant. But ring stiffness is ignoredin conservative flexible pipe design.

As long as the ring is circular, theoretically, theembedment needs little horizontal strength.Practically, good sidefill adds a margin of safety.See Figure 16-6 where the infinitesimal soil cube Bis in equilibrium as long as pipe pressure Px does notexceed sidefill soil strength, σx. For stability,

Px < σ x = Kσ y . . . . . (16.1)

where K = (1+sinφ)/(1-sinφ), and φ is the frictionangle at soil slip. If sidefill soil is granular anddenser than critical, its friction angle is no less than30°, for which K = 3. From Equation 16.1, thesafety factor against soil slip is no less than threebecause Px = P.

Example

Suppose the trench walls are poor soil; with blowcount less than four. What should be the trenchwidth for a flexible pipe? See Figure 16-7. If thefriction angle of sidefill is φ = 35°, then the soil shearplane is at angle (45°-φ/2) = 27.5° and Px istransferred to the trench wall by a soil wedge with1:2 slopes as shown. If ring deflection is less than5% and the width of sidefill is half the pipe diameter,the pressure on the trench wall is about P/2 as

shown, and can be supported by trench walls with ablow count less than four. The trench width in poorsoil does not need to be greater than twice thediameter of the flexible pipe. The margin of safetyis increased by: stiffness of the ring, shearingresistance of soil on pipe, and arching action of thesoil. Both ring stiffness and ring deflection can beincluded in the analysis of Px, if greater accuracy isrequired.

In fact, the pressure, P/2 on the trench wall ofFigure 16-7 is only approximate. According to bothBoussinesq and Newmark, pressure on the trenchwall is not uniform, and the maximum pressure is0.7P. But these elastic analyses do not representeither particulate mechanics or passive resistance atpunch-through. Moreover, from experiments, thesoil wedge of Figure 16-7 does not remain intactduring punch-through. It is sheared into threewedges as discussed in Chapter 17.

As long as the pipe is nearly circular, in poor nativesoils, the trench does not need to be wider than halfa diameter on each side for both rigid and flexiblepipes. If ring deflection of a flexible pipe is no morethan 5%, the effect of ring deflection can beneglected. On a rigid pipe Pd is the Marston load.(Marston 1930) On a flexible pipe, Pd is more nearlythe prism load, γ H. (Spangler 1941 and 1973)Dead load is roughly three-fourths as great on aflexible pipe as on a rigid pipe.

The height of soil cover, H, is not a pertinent variablein the analysis of trench width. As soil load isincreased, the pressure on the pipe increases; but thestrength of the sidefill soil increases in directproportion. See Equation 16.1. A good rule of thumb for width of sidefill is: In poor soil, specify a minimum width of sidefillof half a diameter, D/2, from the pipe to the wallsof the trench, or from the pipe to the windrowslopes of the embedment in an embankment. SeeFigure 16-8.

In good soil, width of sidefill can be less, providedthat the embedment is placed at adequate density.


Figure 16-7 Approximate punch-through soil wedge showing how pressure P is transferred to the trench wallwhere it is distributed and reduced by roughly half in a trench of width 2D.

Figure 16-8 Cross-sectional sketch showing how the recommended width of embedment cover is D/2 for bothtrench and embankment if the installation is in poor soil.


Following are five concerns for embedments: 1. Wheel loads over a pipe with less than minimumsoil cover, 2. Water table above the pipe, and/or vacuum in thepipe,3. Migration of soil particles out of the embedment,4. Voids left by a trench shield, or sheet piling,5. Embedment on a sidehill.1. Wheel LoadsFigure 16-9 shows a wheel load on a pipe withminimum soil cover. The angle of punch-through incohesionless soil is about 1h:2v. The punch-throughis approximately a truncated cone for a single tire, ora truncated pyramid for a dual wheel. Verticalpressure on the pipe is P = Pd+Pl where Pd = γ H, isdead load, and Pl is live load. Pl = Q/(B+H)(L+H)for load, Q, on a rectangular tire print area, LxB(about 22x7 inches for HS-20 dual wheel). Sidefillsoil strength must support the pipe under this liveload. However, minimum cover of compactedgranular soil is about H = 1 ft for HS-20 dual wheel,and H = 3 ft for the single wheel of a scraper.Manufacturers of large steel pipes with mortarlinings recommend that a margin of safety of 1.5 ftbe added to the minimum cover. Recommendedminimum cover is 2.5 ft for HS-20 loads and 4.5 ftfor scraper wheel loads . With soil cover greaterthan minimum, wheel load pressure is less than Pl =W/2H2. In fact, a soil arch will support wheel loads.Punch-through does not occur. Trench width couldbe critical — but only if the sidefill embedment wereso poor that it could not support wheel loadsanyway.

2. Water TableSee Figure 16-10. When the water table is abovethe pipe, sidefill soil strength is effective (buoyant)strength, σx = K σ y. The effective vertical soilstress is σy = σ y - u, where u is the pore waterpressure; i.e., u = γ wh, where γ w is the unit weightof water and h is the height of the water table(head) above the spring line of the pipe. If the pipetends to float, for analysis, P is the hydrostaticbuoyant pressure on the bottom of the pipe, P =γ w(h+r), rather than soil pressure on top.

Soil particle migration is generally a function ofeither: a) groundwater flow that washes trench wallfines into the voids in a coarser embedment; or, b)wheel loads and earth tremors that shove or shakecoarser particles from the embedment into the finersoil of the trench wall. If fines migrate from trenchwall into embedment, the trench wall may settle, butthe pipe is unaffected. If embedment particlesmigrate into the trench wall, the shift in sidefillsupport may allow slight ring deflection. This couldoccur only if the trench wall soil is loose enough, orplastic enough, that the embedment particles canmigrate into it. Soil particle migration is unusual, butmust be considered. Remedies include: a)embedment with enough fines to filter out migratingparticles in groundwater flow; and, b) trench liners.Geotextile liners may be required under somecircumstances.

4. Trench Box (Voids in the Embedment)Soil should be in contact with the pipe in order toavoid piping (channels of groundwater flow) underthe haunches. Voids are avoided if the embedmentis flowable fill — a good idea when trench widthsare too narrow for placement of soil under thehaunches.

Flexible pipes tend to squat slightly into the bedding.The increased radius of the bottom, ry, must neitherbe great enough to cause spalling of the mortarlining, nor great enough that the relationship Pyry =Pxrx causes Px to exceed sidefill support capability.Constructors have clever ways to "chuck" soil downunder the haunches: J-bar it, or vibrate it, flush it,etc. It is sometimes proposed that the bottom of thetrench be shaped to fit the pipe. But the effort doesnot justify the cost and does not assure uniformsupport.

Voids left by the withdrawal of sheet piling or trenchshield do not affect the pipe if the tips of the piles orshield are above the spring lines of the pipe. SeeFigure 16-11.

5. Sidehill EmbedmentFigure 16-12 shows a pipe on a sidehill. What is

3. Soil Particle Migration


Figure 16-9 Dual-wheel load passing over a pipe buried under minimum soil cover showing a sidefill wedgeat incipient soil slip.

Figure 16-10 Free-body-diagram of an empty flexible pipe buried below the water table.


Figure 16-11 Comparison of soil slip planes whena trench box is withdrawn from trench bottom (leftside) and springlines (right side) where soil slipplanes are above the pipe and have no effect onthe pipe. Voids in the sidefill can occur and causering deflection if the trench box is at the bottom ofthe trench.

Figure 16-12 Pipe buried on an infinite slope of cohesionless soil, showing the infinitesimal cube for conjugatestresses and the resulting diagram of approximate stresses on the ring.


the minimum soil cover for stability? Cementstabilized soil may be one remedy. Assume that thepipe is flexible. With no cover H, the soil slopewould slide into the flexible pipe. From Rankine'sanalysis of conjugate axes,

cosi - (cos2i - cos2φ)1/2

F/P = K =cosi + (cos2i - cos2φ)1/2

. . . . . (16.2)

Soil pressures on the ring are shown at the left ofFigure 16-12. Py is based on prism load. Morereasonable is a wedge load lifted by the ring.Average Px = γ K(H+r). It acts slightly below thespring line. Using prism load, for ring stabilityγ Hcos2i = γ K(H+r), from which,

H(cos2i - K) = Kr . . . . . (16.3)

Example 1

For a worst case, assume that angle i is the angle ofrepose of the slope which is nearly equal to the soilfriction angle, φ. Let φ = 30o = i. K = 1, and, H =-r/sin2φ. H goes negative. The flexible ring isunstable. More reasonable is an uplifted wedgeload. But the height, H, of the wedge is unknown.A value for the load is assumed — say Py =γ Hcos2φ. Substituting into Equation 16.3, H = 2r =D. From this H, the wedge load can be calculatedto see if it is twice the prism load as assumed. Notmany iterations are justified because theassumptions are loose. If angle i is less than angleof repose, φ = 30o, then typical results for theconservative prism load would be:For i = 20o, H = 5.7r. For i = 10o, H = 1.9r. These are improbable upper limits, for this worst-case scenario. Iteration for wedge loads wouldresult in lower, more reasonable, values of H. Anyring stiffness would further reduce the values of H.

Example 2

How much live load, Pl, (in addition to dead load)can a flexible pipe resist if, for the embedment atslip, K = (1+sinφ)/(1-sinφ) = 2? At the spring line,

the vertical soil pressure is no less than the weight ofsoil, i.e., σz = γ (H + D/2). If the flexible pipe ringremains circular, the horizontal pressure against thesoil at soil slip is 2γ (H + D/2). If the circular ring isperfectly flexible, vertical soil pressure on the top ofthe pipe is equal to the horizontal pressure and isalso 2γ (H + D/2). Dead load pressure on the topof the pipe is γ H. So the pipe is capable of resistinga live load of Pl = 2γ (H + D/t) - γ H = γ (H + D).

REFERENCES

Marston, Anson (1930). The Theory of ExternalLoads on Closed Conduits in the Light of theLatest Experiments. Bul. 96, Engr. Exp. Sta., IowaState College.

Spangler, M.G. (1941). The Structural Design ofFlexible Pipe Culverts. Bul. 153, Engr. Exp. Sta.,Iowa State College.

Spangler, M.G. and R.L. Handy (1973). SoilEngineering, 3ed. Intex Educational Publishers.

Watkins, R.K. and M.G. Spangler (1958). Somecharacteristics of the modulus of passive resistanceof soil: A study in similitude. Proceedings,Highway Research Board 37, 576.

PROBLEMS

16-1 What will be the increase in ring deflection ofa very flexible pipe when the trench shield is pulledahead after the pipe has been embedded andbackfilled with dry sand to a height above the pipe ofabout H = 1+ meter? The pipe is 2 meters indiameter, in a trench 2.6 meters wide, with verticaltrench walls shored up by a shield with walls 100mm thick. The shield walls reach the level of thebottom of the pipe. The angle of repose (soil frictionangle) is 30o. Soil unit weight is 120 pcf.

16-2 For the flexible pipe of Problem 16-1, what isthe horizontal pressure against the trench walls atthe level of the spring lines?


16-3 At one location, a flexible storm drain happensto be embedded in soft clay at the sides (sidefill).See the sketch below, Figure 16-13. By analyzingthe free-body-diagram of an infinitesimal cube ofclay at the spring lines, what is the safety factoragainst collapse of the pipe due to sand backfillabove the pipe? Surface loads are not anticipated.The drain keeps water out of the sand backfill.

Assume that:Trench: Depth = 8 ft. (H = 6 ft)

Pipe: Diameter = 24 in., Pipe stiffness = 0.Clay: Unit wt = 120 pcf, Cohesion = 100 psf, ϕ = 0.Backfill: Specific gravity = 2.7, Void ratio = 0.4, ϕ = 30°.

16.4 What is the diameter of pipe in Problem 16-3at incipient collapse of the ring vertically? Assumethat the height of cover of the backfill sand is still 2meters.

Figure 16-13 Cross section of a very flexible pipe embedded in clay (left), and the Mohr circle for the clay(right).


Anderson, Loren Runar et al "PARALLEL PIPES AND TRENCHES"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 17-1 Pipe-clad soil column between parallel pipes showing minimum section AA, which must supportpart of the surface live load W plus the dead load, shown cross-hatched.


CHAPTER 17 PARALLEL PIPES AND TRENCHES

When buried pipes are installed in parallel, principlesof analysis for single pipes still apply. Soil covermust be greater than minimum. However, thedesign of parallel buried pipes requires an additionalanalysis for heavy surface loads. Consider a free-body-diagram of the pipe-clad soil column betweentwo parallel pipes. See Figure 17-1. Section AA isthe minimum cross section. This column mustsupport the full weight of the soil mass, showncross-hatched, plus part of the surface load Wshown as a live load pressure diagram. The soilcolumn is critical at its minimum section AA at thespring lines. For design, the strength of the columnat section AA must be greater than the vertical load.

STRENGTHS

Performance limits of the column are either: 1. ringcompression strength of the pipe wall; or 2.compressive strength (vertical passive resistance) ofthe soil at Section AA.

1. Ring compression strength of the pipe wall is yieldstrength σf. For steel pipes, σ f is usually 36 or 42ksi. For rigid pipes, σf is crushing strength of thewall. For plastic pipes, σf depends on temp-erature,s tress, and service life. Manufacturers publishvalues.

2. Strength of the soil is found as follows. Assumethat the embedment is granular and compacted. Soilstrength is vertical stress, σy, at slip. Horizontal soilstress is provided by the pipe walls. Approximatesoil strength may be found from triaxial soil tests inwhich interchamber pressure is equal to thehorizontal pressure, Px, of the pipe against the soil.For circular, flexible pipes at soil slip, Px = Pd = γ H.Live load pressure, P1, has no effect because thelive load is not directly above the pipe. If thereshould be a water table above Section AA,compressive soil strength at failure would be theeffective vertical σy confined by

horizontal σx = Kσ y. Px = P = σ x + u where u is thewater pressure at Section AA.

STRESSES

If bond between the soil and the pipe wall could beassured, the column would be analyzed as areinforced concrete column based on an equivalent(transformed) section. But bond between soil andpipe cannot be assured because of fluctuations intemperature, moisture, and loads, all of which tend tobreak down bond. It is assumed that bond is zero.Therefore, stresses in the pipe and soil are eachcalculated independently.

DESIGN OF PIPE

Before the soil column is analyzed, the pipe must beadequate. See Chapter 6. Design starts with thering compression equation,

P(OD)/2A = σf /sf,

whereOD = outside diameter of the pipe,A = pipe wall area per unit length of pipe,σf = ring compression strength of the wall,sf = safety factor,P = maximum vertical soil pressure on top of

the pipe.

For worst-case ring compression, live load W isdirectly above the pipe where P = P1 + Pd. The liveload effect, P1, can be found by Boussinesq orNewmark. If W is assumed to be a point load,according to Boussinesq, P1 = 0.477W/H2. SeeChapter 4. If live load W is assumed to be adistributed surface pressure, the Newmarkintegration can be used. Soil cover must be greaterthan minimum by the the pyramid/cone analysis ofChapter 13. In the following it is assumed that thepipe is adequate.


DESIGN OF SOIL COLUMN

The following analysis is for flexible pipes. Rigidpipes require modification of the procedure. SeeFigure 17-1. The vertical load supported by the twoflexible pipe walls at section AA is no less than2PD/2 = PD. So, in the design of the soil column, itis assumed, conservatively, that the pipe wallcladding takes a vertical load of PD. But this is onlypart of the total load. The remainder must besupported by the soil. The greatest load occurswhen the heavy live load W is centered abovesection AA — not over the top of the pipe. At thislocation, not only is the live load pressure maximum,but the portion supported by the pipe wall cladding isminimum. Pipe walls carry dead load, PdD = γ HD.Live load, Pl, on the pipes is small enough to beneglected. It is already supported by the ringstiffness required for minimum cover. What cannotbe neglected is the Boussinesq live load on sectionAA. Soil stress, σy, must be less than strength S'.Vertical stress is soil load divided by the cross-sectional area.

σy = Q'/X = S'/sf . . . . . (17.1)

whereσy = vertical soil stress on section AA,S' = vertical soil compression strength,X = width of section AA between pipes,sf = safety factor,Q' = Q - γ HD = load supported by the soil at

section AA = total load reduced by theload that is supported by the pipe walls,

γ = unit weight of soil,H = height of soil cover,D = diameter of the pipe = 2r,Q = vertical load on section AA = wd + w1.

Per unit length of pipe, Q is the sum of the deadweight of the cross-hatched soil mass wd, and thatportion w1 of the surface live load W that reachessection AA. See Figure 17-1. The dead load wd

per unit length (1) of pipe is soil unit weight times thecross-hatched area; i.e.,

wd = (1)[(X+2r)(H+r) - πr2/2]γ . . . . . (17.2)

The live load w 1 is the volume under the live loadpressure diagram of Figure 17-1 at section AA. Itis calculated by means of Boussinesq or Newmarkas described in Chapter 4. The pyramid/conepunch-through stress analysis does not applybecause the cover is not less than minimum. IfBoussinesq is justified, the live load w1 per unitlength is

w1 = 0.477WX/(H+r)2 . . . . . (17.3)

Example 1

What is the vertical soil stress at section AA ofFigure 17-1? The pipes are corrugated steel, 72-inch diameter, 2-2/3 by 1/2 corrugations, t = 0.0598,separated by X = 1.0 ft of soil, with 1.5 ft of soilcover at unit weight of 120 pcf. A surface wheelload of W = 20 kips is anticipated. From a ringcompression analysis, the 20-kip load can pass overeach pipe without exceeding yield stress of the pipe.Soil cover is greater than minimum. In order toevaluate soil stress at section AA, from Equation17.2, dead load on section AA is wd = 2.08 kips.

The live load, wl, can be evaluated by BoussinesqEquation 17.3, or by Newmark Figure 4-6. If thedual-wheel print is 1 ft by 2 ft, based on Newmark,w1 = 4MWX(1 ft)/2ft2, where,X = width of section AA = 1 ft.w1 = total live load on section AA,W = wheel load on the surface,M = f[(L/B), (B/H)] = coefficient from the

Newmark chart Figure 4-6, for eachquarter area of surface load W, where,

B = 0.5 ft,L = 1.0 ft,H = 4.5 ft = the Newmark H, which is thedepth to section AA = 1.5ft + 3 ft.

The Newmark denominator, 2 ft2, is the area ofsurface load W. Substituting values, L/B = 2, andB/H = 0.111, the Newmark M = 0.012, and w1 =0.48 kips. The total load on section AA is,

Q = 2.08 + 0.48 = 2.56 kips.


The load supported by the soil alone is,

Q' = Q - γ HD = 2.56 - 1.08 = 1.48 kips.

γ HD is the load supported by the pipe walls.Vertical soil stress on section AA is,

σy = 1.48/(1ft)(1ft) = 1480 psf.

Could Boussinesq have been used without significanterror? From Chapter 4, if H/B is greater than 3,Boussinesq is adequate. In this case, H/B is 4.5/0.5= 9. Let's see if it's adequate. From Equation 17.4,w1 = 0.477(20 kips)(1ft)(1ft)/(4.5ft)2 = 0.47 kipscompared to 0.48 kips by Newmark. There is noquestion that Boussinesq is adequate.

Example 2

What is the vertical soil strength at section AA forthe parallel pipes of the example above? The soilfriction angle is φ = 30o. The horizontal pressure ofthe pipe wall against the soil at section AA is σx =γ H = 180 psf. The vertical strength of the soil atslip is σxK where K = (1+sinφ)/(1-sinφ) = 3.Vertical soil strength is,

S' = 180(3) = 540 psf.

The vertical soil stress from Example 1 is 1480 psf— much greater than the soil strength, 540 psf. Itwould be necessary to: triple the space between theparallel pipes, or place concrete between the pipes,or specify stiffer pipes.

Tank spacing:For multiple parallel tanks, the following areminimum spacings which must be increased asneeded to accommodate deadmen or anchor slabs.Refer to Chapter 21 on tank anchors. If sufficientclearance must be allowed for deadmen to be setoutside of the tank shadow; Spacing betweenparallel tanks should be no less than one-fourthtank diameter. The live load is assumed to be HS-20 dual-wheel load with minimum soil cover of 2.25ft.

Rigid Pipes:Unlike flexible pipes, rigid pipes do not exertpressure, Px = P, against the soil. Total load, Q, issupported by the pipe walls in ring compression andthe soil in vertical passive resistance. It is possibleto analyze the equivalent section by column design.See texts on reinforced concrete. There is a greatdifference between the modulus of elasticity of thepipe wall and the modulus of elasticity(compressibility) of the soil.

Safety Factors:Analyses of the soil column with pipe wall cladding,are conservative. Longitudinal resistance of thepipes and soil cover is neglected. Also the archingaction of the soil cover is neglected. Safety factorscan be small.

PARALLEL TRENCH

Buried flexible pipes depend on the embedment forstability. Compacted soil at the sides supports andstiffens the top arch. What happens to a buriedflexible pipe when a trench is excavated parallel toit? What is the stability of the trench? At whatminimum separation between the pipe and theparallel trench will the pipe collapse? What are thevariables that influence collapse? Answers to thesequestions were the objectives of an experiment atUSU in 1968. In order to reduce the number ofvariables, ring stiffness was assumed to be zero.Results were conservative because no pipe has zeroring stiffness. For the most flexible plain steel pipes,D/t is less than 300. For the test pipes, D/t was 600in an attempt to approach zero stiffness. It wasnecessary to hold the pipes in shape on mandrelsduring placement of the backfill.

Vertical Trench Walls

Figure 17-2 is the cross section of a buried, flexiblepipe with an open cut vertical trench wall parallel toit. If trench wall AB is cut back closer and closer tothe buried pipe, side cover X decreases to the pointwhere the sidefill soil is no longer able to


Figure 17-2 Vertical trench wall parallel to a buried flexible pipe showing the soil wedge and shear planesthat form as the pipe collapses.

Figure 17-3 Formation of a soil prism on the pipe during ring deflection as the soil wedge is thrust into thetrench.


provide the lateral support required to retain theflexible ring. The ring deflects, thrusting out a soilwedge as indicated in Figure 17-3. As the ringdeflects, a soil prism breaks loose directly over thering. The soil prism collapses the flexible ring. Inorder to write pi-terms to investigate thisphenomenon, the pertinent fundamental variablesmust be identified. Ring stiffness is ignoredbecause the ring is flexible. In fact, at zero ringdeflection, the ring stiffness has no effect anyway.The remaining fundamental variables are:

Basic Fundamental Variables Dimensions

X = minimum side cover (minimum Lhorizontal separation betweenpipe and trench at collapse),

D = pipe diameter, LH = height of soil cover over L

the top of the pipe,Z = critical depth of trench in L

vertical cut (vertical sidewalls).

Critical depth, Z, is a convenient measure of soilstrength. It is defined as the maximum depth of atrench at which the walls stand in vertical cut. Atgreater depths the trench walls slip or cave in.Critical depth Z may be determined by excavationin the field, or it may be calculated from thedimensionless stability number, Zγ /C. See Figure17-6.

2C/γ Z = tan(45o - ϕ /2) . . . . . (17.4)

whereZ = critical depth of trench in vertical cut,γ = unit weight of soil (pcf),C = soil cohesion (psf),ϕ = soil friction angle of the trench wall.

Z can be found from Equation 17.4 if soilproperties, γ , C, and ϕ , are provided by laboratorytests.

To investigate the four fundamental variables,

three pi-terms are required. One possible set is(X/D), (H/D), and (H/Z). Tests show that (H/D)is not pertinent. Only (X/D) and (H/Z) remain aspertinent pi-terms. Results of the tests are asfollows.

For a vertical trench wall excavated parallel to aflexible pipe,

1. Failure is sudden and complete collapse.

2. The ring collapses under a free-standing prismof soil that breaks loose on top of the pipe.

3. If the ring has some stiffness, and if soil coverH is not great enough to collapse the ring, soil mayslough off the pipe into the trench. This is notconsidered failure because the soil can bereplaced during backfilling.

Test data are plotted in Figure 17-4, which shows(X/D) as a function of (H/Z). The best fit straightline equation is, X/D = 1.4(H/Z). The probableerror in X/D is plus or minus 0.1, so probable errorin side soil cover, X, is roughly plus or minus D/10.Because field conditions may be less reliable thanlaboratory conditions, the safety factor should betwo. Therefore, the minimum side cover, X, mightbe specified as,

X/D = 3H/Z . . . . . (17.5)

Of interest in Figure 17-4 are the data pointsindicated by squares. These do not representcollapse. The ring stiffness for the test pipes wasgreat enough that part of the shallow soil covermerely sloughed off the pipes after the soil wedgefell into the trenc h. If ring stiffness were to beincluded as a fundamental variable, ring deflectionwould have to be included. Then the coefficientof friction between pipe and soil should also beincluded as a fundamental variable.

If the pipe has significant ring stiffness, the heightof soil cover that it can support without collapsecan be found for uniform vertical pressure with


Figure 17-4 Cover term X/D as a function of soil strength term H/Z for a vertical trench wall excavatedparallel to a very flexible buried pipe.

Figure 17-5 Trench wall sloped at angle of repose for which the soil is stable, but the flexible ring requireseither some ring stiffness or a minimum cover.


no side support. See Appendix A, from whichmoment = Pr2/4 = σ I/c. For plain steel pipesbased on elastic theory, at yield stress,

P = 16σf I/cD2 . . . . . (17.6)

whereP = vertical soil pressure at collapse,I = moment of inertia of the wall cross

section,D = pipe diameter = 2r,c = half the distance to wall surface from

the neutral surface = t/2 for plain pipes,t = wall thickness of plain pipes,σf = yield strength of the pipe,m = D/t = ring flexibility for plain pipes.

P = (8σf /3m2) for plain pipes . . . (17.7)

Based on plastic theory (plastic hinging),

P = 24σf I/cD2 . . . . . (17.8)

P = 4σf /m2 for plain pipes . . . (17.9)

Vertical ring deflection at plastic hinging is,d = 0.01PD3/EI

whered = ∆/D = ring deflection,∆ = decrease in vertical diameter,P = vertical pressure on the ring,D = circular pipe diameter,EI = wall stiffness per unit length of pipe.

Sloped Trench Walls

Figure 17-5 shows a flexible pipe in cohesionlesssoil for which the slope is angle of repose ~ ϕ.Pressure distribution on the ring is triangular asshown. Maximum moment at A can be found byCastigliano's equation. However, it is sufficientlyaccurate to find equivalent moment MA = Pr2/4for average uniform pressure, Px = rγ . See

Figure 17-6 Rationale for finding critical depth, Z, of a vertical open cut in a trench wall in brittle soil withcohesion, C, and soil friction angle, ϕ ; 2C/γ Z = tan(45o - ϕ/2).


~

JMiller

JMiller

Appendix A. The required section modulus is,I/c = MA(sf)/σ f.where σf is yield stress.

EXCAVATION

Depth of the excavation must include"overexcavation" required to remove unstablesubbase material. It should be replaced byapproved bedding material. Some tankmanufacturers consider soil to be unstable if thecohesion is less than C = 750 psf based onunconfined compression test, or if the bearingcapacity is less than 3500 psf. In the field, bearingcapacity is adequate if an employee can walk onthe excavation floor without leaving footprints. Amuddy excavation floor can be choked with graveluntil it is stable. These are conservative criteriafor soil stability.

Of greater concern are OSHA safetyrequirements for retaining or sloping the walls ofthe trench. Excavations for tanks are usuallyshort enough that OSHA trench requirementsleave a significant margin of safety. Longi-tudinal,horizontal soil arching action is significant.

These criteria for bearing capacity and cohesionare equivalent to a vertical trench wall over 20 ftdeep. Bearing capacity of 3500 psf can supportmore than 29 ft of vertical trench depth at soil unitweight of 120 pcf. Cohesion of 750 psf cansupport a vertical open cut trench wall that is morethan 20 ft deep.

Critical Depth of Vertical Trench Wall

Granular soil with no cohesion cannot stand invertical cut. Much of the native soil in whichpipes and tanks are buried have cohesion.Therefore, the wall of the excavation can stand invertical open cut to some critical depth, Z. SeeFigure 17-6 (left). Greater depth will result in a"cave-in" starting at the bottom corner, O, where

the slope of the failure plane is (45o+ϕ /2). For atwo-dimensional trench analysis, the infinitesimalsoil cube, O, is subjected to vertical stress, γ Z,where,γ = soil unit weight,Z = critical depth of vertical trench wall,ϕ = soil friction angle,C = soil cohesion.

The Mohr circle is shown in Figure 17-6 (right).The orientation diagram (x-z) of planes on whichstresses act, is superimposed, showing the locationof the origin, O. The strength envelope slopes atsoil friction angle ϕ from the cohesive strength, C.At soil slip, the Mohr stress circle is tangent to thestrength envelope. From trigonometry,tan(45o - ϕ/2) = 2C/γ Z.This is the critical depth Equation 17.4.

From tests, Equation 17.4 provides a reasonableanalysis for brittle soil. If the soil is plastic, soil slipdoes not occur until shearing stresses reachshearing strength C. Consequently, in plastic soil,the critical depth equation is 2c/γ Z = 1. Belowthe water table, critical depth is essentiallydoubled. Example

What is the critical depth, Z, of a vertical, open-cut, trench wall if,C = 750 lbs/ft2,γ = 120 lbs/ft3,ϕ = 30o?

Substituting into Equation 17-4, Z = 22 ft. This isa lower limit if the soil has some plasticity (is notbrittle). Excavations for tanks are almost nevergreater than 20 feet.

Example

Suppose that a sloped trench wall exposes a pipeas shown in Figure 17-5. Pressure, Px, must beresisted by ring stiffness. What is the requiredwall thickness for a 72-inch plain steel pipe?


Assume the soil is granular with unit weight of 120pcf.

From Appendix A, the maximum moment in thering is M = Pxr

2/4. Ring resistance by elastictheory must be I/c = t2/6 = M/σf. Yield stress isσf = 42 ksi. Allowable stress is reduced from 42ksito 21 ksi. From Chapter 16, Px = γ r.Substituting values, t = 0.48 inch, for which D/t =150.

A little ring stiffness makes a big difference in thestability of a flexible ring on a sloped trench wall(sidehill). In the case of steel pipes, mortar liningsignificantly increases the ring stiffness.

The above rationale is based on an infinite slope.For a sloped trench wall, pressures on the ring areconsiderably less. For most pipes, the ringstiffness required for installation is adquate if itcomplies with the familiar rule of thumb accordingto which minimum cover is D/2. For a pipeparallel to a sloped trench wall in cohesionless soil,minimum cover is half a diameter to the slopedsurface of the trench wall.

PROBLEMS

17-1 What is the minimum allowable separation(side cover X) between a buried flexible plain steelpipe and a parallel trench if the trench walls arevertical and if:ϕ = 30 degrees = soil friction angleC = 400 psf = soil cohesionD = 6 ft = pipe diametert = 3/8 inch = pipe wall thicknessH = 4 ft = soil coverγ = 125 pcf = soil unit weight

(X = 6.5 ft)

17-2 Reconsider Problem 17-1 if the soil iscohesionless; i.e., c = 0.

17-3 A two meter plain steel pipe with wallthickness t = 10 mm, is buried in compacted finesand with a soil cover of 1 meter and with watertable at the surface on occasions. If a trench canbe exc avated with vertical sidewalls to amaximum depth of 4 meters in the soil at highwater table, what is the minimum separation X ofa trench excavated parallel to the pipe?

(X = 1.5 m)

17-4 Find vertical pressure at collapse of the pipeof Problem 17-3? Elastic yield σf = 290 MPa.

(19.3 kPa)

17-5 What would be the maximum soil cover ifthe pipe of Problem 17-3 is not to collapse, X isminimum 1.5 m, and P = 19.3 kPa? (H = 2.9 ft)

17-6 Prove that for a plain pipe, the vertical soilpressure at plastic hinging is 3/2 times the verticalsoil pressure at elastic limit. Assume yieldstrengths are the same for both elastic and plasticanalyses. Assume that the elastic limit is yieldstrength.

17-7 Two parallel 14-ft diameter flexible steelpipes are buried under H = 4 ft of dune sand, γ =120 pcf and ϕ = 25o, separated by X = 3 ft. Whatis the maximum allowable wheel load? DEADLOAD SOIL SLIP. So compact soil, ϕ = 0 38o,and increase X = 6 ft. (W = 70 kips)


Anderson, Loren Runar et al "SPECIAL SECTIONS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 18-1 Wye showing transition flow from a mainline (inflow) pipe to two branch (outflow) pipes. (In fact, flow could be in either direction.) Unit length of cone is x(cos Θ ).


CHAPTER 18 SPECIAL SECTIONS

Special sections in pipes are valves, tees, wyes,elbows, caps or plugs, transitions such as cones forchanges in diameter (i.e. change in flow velocity),and transitions between conduits of differing shapesor sizes such as transitions from a rectangularconduit to a circular pipe. Experience and expertiseare available from manufacturers of commonstandard specials. However pipeline engineers oftenneed an uncommon section. Following are somebasic rules and procedures for preliminary design ofspecials. As an example, consider a wye.

A wye (Y) is a bifurcation of the pipeline from alarger mainline pipe to two smaller branch pipes.See Figure 18-1. A wye may require a trifurcation,or branch pipes of different diameters, or at differentoffset angles, etc. Wyes can be either molded(warped surface) or mitered (circular cylinders orcones). The following example is a miteredbifurcation with equal offset angles. The basiccomponents are two truncated cones shown dottedwith large ends Di and small ends Do to match upwith the diameters of the inflow (mainline pipe) andthe outflow (branch pipes). The cones are cut andwelded together at the crotch to form the wye; andare then welded to the mainline and the branch pipesas shown. It is noteworthy that the crotch, theintersection of the two cut cones, is an ellipse in aplane. It is like a crotch seam in jeans. An ellipse iseasy to analyze and to fabricate. Because the cut isin a plane, the welded intersection lends itself toreinforcement by internal vane or external stiffenerring or crotch plate. For high pressure, fabricatorsfavor welding the crotch cuts to a heavy crotchplate, and welding stiffener rings to the outside ofthe welded miters.

Notation and Nomenclature

D = ID = inside diameter (nominal for steel)Di = inside diameter of inflow pipe

(mainline)

Do = inside diameter of outflow pipes (branch)

t = wall thicknessR = radius of bend in the pipe or coneδ = offset angle of contiguous mitered

sectionsL = length of section of pipeLT = length of truncated cone θ = angle between axes of each branch pipe

and the mainline pipe.

Consider the horizontal cross section 0-0 at thecrotch of Figure 18-1. A free-body-diagram of halfof the cross section shows a rupturing force ofpressure times the span of the cut. This force ismore than twice the rupturing force in each ring ofthe branch pipes and will cause ballooning of thecross section unless the rings are held together atthe center — either by a vane on the inside, or by acrotch plate on the outside. A crotch plate is a C-clamp with an elliptical inside cut as shown in Figure18-2. It is located at the plane of intersection(crotch cut) of the two branch pipes.

Force On Crotch Plate Due To Internal Pressure

See Figure 18-3. From ring analysis (Chapter 2) theforce to be resisted by the crotch plate from each ofthe cone walls at the vertex section A-A-A, is PrA

per unit length of the pipe. Per unit length of crotchplate (or vane), the force on the crotch platebecomes:

w = 2PrA(cosθ) . . . . . (18.1)

wherew = vertical force on crotch plate (vane)

per unit length of crotch plateP = internal pressurer = radius of the circular cone

(or circular cylinder in some cases) θ = offset angle of axes of branch cones

(or cylinders) from the axis of the mainline pipe


217

Figure 18-2 Development of the crotch cut and the crotch plate which, together with the stiffener rings, supports the hoop tension at the cuts (all cutsare elliptical).


David Alan Rech

Figure 18-3 Free-body-diagrams of cone corss sections showing where the crotch is cut, the hoop forces PrA

at section A-A-A and PrB at section B-B.


Figure 18-4 Free-body-diagram of one limb of the crotch plate showing an approximate procedure foranalyzing the forces on the limb assuming it to be a cantilever beam (bottom sketch) loaded at the free endof the statically indeterminate restraint Q of the stiffener rings.


α = angle on the cone cross section from the vertical axis to the crotch cut

Li = length of longitudinal element on the inside of the mitered cone section

But this force w oc curs only on section A-A-A atthe vertex of the crotch cut. At section B-B ofFigure 18-3, the hoop forces are not vertical. If thecrotch plate (or vane) resists vertical componentsonly, per unit length the vertical force on the crotchplate is: w' = 2PrB(cosθ)sin α ; where α is theangle, in the plane of section B-B, from vertical tothe intersection of the two branch pipes. Thehorizontal components of the hoop forces, PrB, arebalanced because of symmetry — i.e. because thepressures, diameters, and offset angles are equal inthe two branch pipes.

Figure 18-4 is a plot of w' throughout the length ofthe crotch cut of Figure 18-3. Clearly, the plot doesnot deviate significantly from a straight line.Therefore, if a straight line is assumed, angle αserves no purpose, and Equation 18.1 provides avalue for w for analyzing forces on the crotch plate.From the force analysis, the crotch plate can bedesigned.

Hydrodynamic Guidelines

In pressure lines of high velocity water flow, such aspenstocks for hydroelectric power plants, it isprudent to avoid sudden changes in velocity orsudden changes in direction of flow because ofturbulence and loss of energy. Guidelines used byfluid dynamicists for minimizing energy loss are asfollows. See Figures 18-5 to 7.

1. Keep the cross-sectional areas of the mainlinepipe nearly equal to the areas of the branch pipes.For a wye (bifurcation), Do

2 = Di2/2.

2. On mitered bends, keep the inside-of-bend offsetangles minimum. Inside offset angles are the criticalcause of turbulence. Inside offset angle shouldnever be greater than δ = 15o. It is preferable tokeep δ < 10o. Most engineers try to

keep δ < 7.5o or even < 6o for very high velocityflows.

3. Keep the radius of the bend greater than 2.5times the pipe diameter (or mean diameter of anymitered cone section); i.e. R > 2.5D = 5r. It ispreferable to keep R > 3D or even > 4D for veryhigh velocity flows. See Figure 18-7.

4. Keep the length, Li, on the inside of the bend ofeach mitered section, greater than half the meanradius of the section (pipe or cone).

5. Keep the cone taper angle minimum. The greaterthe taper angle, the shorter are the length s Li ofcontiguous cone (or pipe) sections. This means asharper bend (shorter radius R of bend). On theother hand, the smaller the taper angle, the longerthe crotch plate must be. Consequently, muchgreater loads must be supported by the cantileverlimbs of the crotch plate. The crotch plate, a criticalstructural element of the wye, presents a dilemma— the need for a large taper angle to keep thecrotch plate short, and the need for a small taperangle to keep the radius of the bend and the insidelengths, Li, within limits of hydrodynamic guidelines.From this point on, design is by trial. Therelationship of the hydrodynamic guidelines to thestructural integrity of the wye are best described byan example.

Example

Consider a penstock for a hydroelectric power plant.Suppose that the mainline pipe is 96-inc h steel pipe,bifurcated into two branch pipes to supply waterunder high pressure and high velocity to two equalsized turbines. For steel pipes, diameters are inside.Yield strength is 45 ksi. For preliminary design,including 100% surge, pressure is P = 225 psi.

First Trial

Start with a trial wye — say Figure 18-1. Try ataper angle of 7.5o, for which the truncated length ofthe cones is LT = 113.94 inches. To facilitatefabrication and welding, select the same steel


Figure 18-5 Wye with inside offset angles limited to 7.5o. Note that branches do not clear each other at thelower end.

Figure 18-6 Same wye, but mitered; i.e., cut near midlength of the cone, rotated 180o, and welded. Note thatthe branches now clear each other.


thickness for all of the pipes and cones at the wye.The hoop stress is maximum in the mainline pipewhere hoop stress is σ = PDi /2t. Therefore:

t = PDi(sf)/2S . . . . . (18.2)wheret = wall thickness (to be found)σ = hoop stress in the pipe wallP = internal pressure = 225 psiDi = inside diameter of the mainline

pipe = 96 inchesS = yield strength of steel = 45 ksisf = safety factor — say 1.5

Solving, t = 0.360 inch. This is not a standard, so trystandard t = 0.375 inch for analysis.

From guideline 1, Do2 = Di

2/2. So Do = 67.88 inchesfor the branch pipes. Specify diameters of thebranch (outflow) pipes to be a standard Do = 66inches. The ratio of areas, inflow to outflow, is1.058 — not bad. Moreover, the slight reduction inoutflow areas from perfect gives a slight increase inflow velocities into the turbines. This is desirablefrom the standpoint of turbine efficiency.

From guideline 2, the inside offset angle, δ, shouldbe less than about 7.5o. For the trial wye of Figure18-1, δ is greater than 7.5o — actually 15o. Ifreduced to 7.5o, the branch offset becomes θ = 15o

as shown in Figure 18-5. Obviously, the outflowpipes do not clear each other at the lower end. Inorder for the branches to clear each other, tworemedies are considered.

1. The taper angle could be reduced such that lengthof the cone is increased. But then, the length of thecrotch plate would have to be increased. That'sbad.

2. An alternative remedy might be to miter the conesas shown in Figure 18-6. The length of the crotchplate is increased only slightly. That's not so bad.The total offset angle from the axis of the mainlineto branches is θ low = 22.5o.

It is noteworthy that the inside offset angle from

tapered cone to pipe is less critical than the insideoffset angle for bends in pipes because the taper-to-cylinder transition is a symmetrical squeeze-down offlow. Bends are not symmetrical. If one or theother has to be mitigated, the taper (rather than theinside offset angle in a bend) is allowed to exceedthe recommended maximum.

Figure 18-7 shows how mitered bends are formed.Because a planar cut across any circular cone (orpipe) is a perfect ellipse, mitered bends can beachieved by cutting any cone (or pipe) at an angle ofδ/2 with the diameter, rotating one section 180o, andthen welding the cut. The ellipses match. Theresulting offset angle is δ.

It is not always necessary to miter the mainline pipe.See the mainline-to-cone cut in Figure 18-7. Whenthis particular cone tilts to angle θ = 15o, itshorizontal radius is approximately the same as theradius of the mainline pipe. Of course, the cut of thecone is an ellipse, but the ellipse is so nearly circular,that the cone and pipe can be pulled together forwelding. If the ring cut were mitered, the stiffenerrings would come in at some angle such as the ringcut angle of 6o shown on Figure 18-1. For thetransition, upper stiffener ring A is a circle — aneasy cut. See Figure 18-8. The lower stiffenerrings B (at the miter cuts in the cones) intersect atthe angles shown.

Second TrialFigure 18-8 is the second trial wye for analysis anddesign. Of primary concern is the crotch plate. Inthis case, the length of the crotch plate limbs is 109.5inches. This compares not too badly with a crotchlimb length of 94.8 inches for the first trial shown inFigure 18-1. A free-body-diagram of the force w onthe 109.5-inch limb can be calculated by means ofEquation 18.1. However, there are two values of θ,upper section and lower section. For the uppersection, cos θ = cos 15o = 0.966. For the lowersection, cos θ = cos 22.5o = 0.924. Not justified isany attempt to interrelate the two by applyingEquation 18.1 over the upper and lower sectionsseparately. Conservatively, we use the larger value,cos θ = 0.966, and we analyze the full


Figure 18-8 Second trial configuration of mitered wye showing the crotch plate and stiffener rings; and a fullcircle stiffener ring at A.


Figure 18-9 Mitered wye showing cross-hatched areas which, when multiplied by pressure P, are the loadsat the mitered cuts where stiffener rings and crotch plate are required.


109.5-inc h length of the limb as a single free-body-diagram with a straight line distribution of the w-force on the cantilever. The radius of the cones atthe vertex of the crotch cut is about 33.5 inches.Consequently w = 2Pr(cos 15o) = 14.6 kips per inch.With this information, forces on the crotch plate canbe found.

The above simplifications are justified by noting thatEquation 18.1, for finding w, applies not only to thecrotch plate, but to the stiffener rings as well. Infact, each stiffener ring is simply two crotch plateswith the ends of the limbs welded together. Allmitered cuts result in a w-force in the plane ofintersection of the two contiguous sections. Anypart of the mitered section that is not part of a fullring (tension hoop), when pressure P is applied, mustbe supported by a crotch plate or stiffener ring. Thisis shown in Figure 18-9. Areas shown cross-hatched, when multiplied by pressure P, representthe w-force distribution diagrams on each of the cutswhere crotch plate or stiffener rings are located.The areas are shown in the plane of the page, butrepresent the vertical w-force. The proof is evidentin the column of values at the right margin, all ofwhich, when multiplied by constant pressure P, aresimply Equation 18.1 for w. Clearly, rings at A andB must resist w-forces from the mitered joints aswell as interaction from the crotch plate. However,the areas at the A-cut and B-cut are smallcompared with the areas at the crotch cut and areusually ignored. Moreover, almost any reasonablestiffener ring at the A-cut can resist the w-forceacting on it. The w-force at B is insignificant. TheB ring only needs to help support the crotch plate.Moreover, the reduced radius rB at the B-cut, wherethe wall thickness is still 0.375 inch, results in amuch stronger cone at the B-cut than at the A-cut.

Crotch Plate Design

Figure 18-10 is a free-body-diagram of a cantileverrepresenting the crotch plate with the w-force andreactions at A, B, and O as shown. The reactions atA and B are the restraints by stiffener rings whichcan deform under Q-loads. Therefore the analysis

is statically indeterminate, depending upon the springconstants of the stiffener rings at A and B. For atrue ring, such as A, the spring constant is,

Q/∆ = 6.72 EI/r3 . . . . . (18.3)

whereQ = diametral load on the ring∆ = deflection of the diametral loadE = modulus of elasticity of the steel ringI = moment of inertia of the cross-sectional

area of the ring wallr = radius to the neutral surface of the ring

cross section.

The spring constants of ring B and the crotch plateare more difficult to evaluate because of theirshapes. A reasonable simplification of the crotchplate for preliminary design is to assume that ring Awill, at least, prevent rotation of the crotch platecantilever limb at section B. See Figure 18-11. Tothe left of section B, the crotch plate isapproximately a half ring, wherein section B doesnot rotate under load. Consequently, the springconstant for the crotch plate can be analyzed by anequivalent circular ring. To analyze the equivalentring, it is only necessary to neglect QA and to doublethe load QB on the equivalent ring. The springconstant can be calculated from Equation 18.3. It isnoteworthy that simulating the crotch plate by a ringwith twice the Q-load on it, we are assuming thatthe limb of the crotch plate does not rotate at B. Amore accurate analysis would prove that rotation atB is small.

Figure 18-12 is proposed as a reasonable trial crosssection of the crotch plate at section C-C. Try a1.5-inch by 30-inch plate for a web with a 1 x 12-inch flange on the outside. On the inside of the webare the steel cone walls double welded to the crotchplate and splitter plates securely welded to providethe equivalent of a flange and to provide abrasionresistance to head-on flow of water containingsediment. The crotch cross section shown is not astandard I-beam section. Nevertheless, assumingthat the walls of the cones and the splitter platescombine to provide an area


Figure 18-10 Forces on a cantilever beam which is used as an approximation of the upper limb of the crotchplate and showing (below) the general shape of the stiffener rings at A and B.


Figure 18-11 Upper limb of crotch plate showing how the spring constant for the load Q can be analyzedapproximately by means of an equivalent ring with a load of 2Q.


Figure 18-12 Trial cross section of crotch plate at Section C-C.


of steel equivalent to the 1x12 flange, anapproximate moment of inertia, I, can be easilycalculated.The spring constant for stiffener B may require acomputer analysis. However, at this point, in asequence of many simplifications, to simulate ring Bby an equivalent circular ring may be sufficient forpreliminary design. The basic function of ring B isto help support the crotch plate.A layout of the crotch cut can be accomplished byany of a number of methods for scribing an ellipse.Shown on Figure 18-13 is one method based onevaluation of the major and minor semi-diameters.A cutting plane is passed through the cone on theright at θ = 18.26o. θ can be found graphically, orby the law of sines for the small cross-hatchedtriangle shown. The major diameter is the length ofthe cutting plane. The minor diameter is the lengthof a line segment piercing the cone perpendicular tothe page at the geometrical center G of the ellipse.This can be found graphically as shown by theprojection below the cone. Or it can be found bytrigonometry. With the semi-diameters known, theelliptical crotch cut can be developed as shown onthe left. This procedure is based on two circlescentered at G with radii equal to the major and minorsemi-diameters. Any radial line from G intersectsthe two circles at the latitude and departure of apoint on the ellipse as shown. A few numericalvalues of latitudes and departures are shown at thelower left. This is the crotch cut.

No allowance is made yet for the thickness of thecone wall or the thickness of the crotch plate. If thecrotch cut is laid out on the inside of the cone, thecone wall thickness is not an issue. Depending onthe method used to lay out the cut on the cone, halfthe thickness of the crotch plate must be allowed inthe layout. Alternatively, a new crotch cut can beanalyzed by graphical techniques allowing for halfthe thickness of the crotch plate in the elliptical cutsin the cones. With the crotch cut drawn to scale, a trial crotchplate can be laid out. See Figure 18-14. Note thatthe crotch plate is cut well inside the crotch cut line.This allows ample surface for welding the cones and

the splitter plates to the crotch plate. These weldsmust be of excellent quality, for they must resist theentire hoop tension in the walls of the cones.

Stress analysis of the crotch plate and stiffener ringsstarts with assumed dimensions. See Figures 18-12,18-14, and 18-15. A free-body-diagram of the upperlimb of the crotch plate is represented as acantilever in Figure 18-14. With four unknowns, Mo,Qo, QB, and QA, it is statically indeterminate.Moreover, QA and QB are reactions from ringswhich deflect under the Q-loads. A reasonablesimplification is to assume that section BA is acantilever as shown at the bottom of Figure 18-14,and that the vertical deflections of points A and Bare equal. Solving, QA = 59.3 kips. The remainderof the load is QB = 800 - 59 = 741 kips to be dividedbetween crotch plate and "ring" B. This is aconservative simplification because the 800-kip totalforce is actually distributed — not concentrated —and to assume it to be concentrated at A and Bresults in higher stresses in the crotch plate and ringB than would the distributed w-force. The divisionof QB between crotch plate and ring B isproportional to the spring constants of the crotchplate and ring B. The data for the trial crosssections of crotch plate and ring B are shown inFigures 18-12 and 18-15. To assume that ring B iscircular is an approximation. The ring is not circular— neither inside cut nor outside cut. The twohalves of the ring are not even in the same plane.Nevertheless, using an average radius, r, anapproximate spring constant can be evaluated.From Equation 18.3, the ratio of spring constants forcrotch plate and ring B is,

(Q/∆) r = [9140/2(2293)](59/57.8)3 = 2.12.

From this ratio, the crotch plate must take 68% ofthe 741 kip load (506 kips), and ring B must take32% (235 kips). Stresses in the trial sections ofcrotch plate and ring B can be estimated becausethe moment in a ring due to a concentrated Q-loadis:

M = 0.318Qr . . . . . (18.4)


Figure 18-13 Graphical construction of the crotch cut based on the fact that it is an ellipse (constructed hereby one of various methods).


Figure 18-14 Dimensions of the proposed crotch plate and loads on the upper limb shown (above) for theentire limb, and (below) for the section BA between stiffener rings.


Figure 18-15 Trial cross sections for stiffener rings A and B.


where M = moments at locations of Q and at 90o to

Q-locations on the ringQ = concentrated diametral loadr = radius to neutral surface of the ring.

Crotch Plate

I/c = 571 in3

Ao = 75 in2

Qo = 506 kips r = 57.8 mean M = 20,333 kip inches σ = 2Q o /2Ao + M/(I/c)σ = 6.75 + 35.61 = 42.36 ksi

Ring B

I/c = 229 in3

AB = 38 in2

QB = 235 kipsr = 59 meanrmax = 70 inchesM = 5231 kip inches

σ = Q B /2AB + M/(I/c)σ = 3.09 + 22.84 = 25.94 ksi

Stress in the crotch plate is high; i.e., yield stress is45 ksi. However, considering that flexural yieldstress is not failure for pipes (failure is a plasticphenomenon, not an elastic phenomenon), andconsidering that surge pressures of 100% can bereduced by slow-closing valves and controlled-response turbines, this preliminary crotch platedesign seems acceptable. A final analysis may beadvisable.

The stress in ring B is greater than that typicalallowable, 45/2 = 22.5 ksi, with a safety factor of 2.Nevertheless, the same mitigating arguments apply.The preliminary design of ring B is acceptable.

The stress in ring A is

σ = QA /2A + M/(I/c) = 1.24 + 10.13 = 11.37 ksi

whereI/c = 104.24 in3

A = 24 in2

QA = 59 kipsr = 56 meanM = 0.318QA /(I/c) = 1056 kip inches.

Stress is low enough to justify elimination of theflange. Without the flange, neglecting contributionof the pipe wall, I/c > (16)2/6 = 42.67 in3, A > 16 in2,and σ < 1.85 + 24.75 = 26.60 ksi. Based on thesame arguments used for the crotch plate and ringB, the flange is not needed. However, there may besome question about web buckling in compression at9:00 and 3:00 o'clock. On the other hand, bucklingcould occur only when there is pressure in thepipeline, and pressure would reduce greatly thecompressive buckling stress in ring A. Forpreliminary design, the flange is to be eliminated.

Example

Figure 18-16 is a preliminary layout of a trifurcationin a water supply pipeline (penstock) for ahydroelectric power plant. Flow in the penstock isto be divided equally into three turbines. Rather thana fork with outflow pipe axes all in the same plane,a cluster of outflow pipes is proposed. From thestandpoint of hydrodynamics, it is an efficienttransition. The crotch plates are three heavy vanesat 120o all welded to a keel as shown. 120° sectionsof the transition cones are then welded to the crotchplates. The inflow is a 126-inch diameter steel pipe.The three outflows are all 72-inch diameter. Staticpressure is 106 psi, but with surge, it is designed for212 psi. For preliminary design, the proposed wallthickness is 0.375 inch. Yield strength of the pipewall is at least 42 ksi. It may not be necessary tomiter short cone sections if the trifurcation isencased in reinforced concrete. Even though thecones are long, crotch plates will not be required ifthe cones are supported from outside by thereinforced concrete casing. See Figure 18-17. Allhydrodynamic guidelines are met for high-pressure,high-velocity flow.


Figure 18-16 Layout of a cluster trifurcation from a 126-inch diameter intake to three 72-inch diameter outlets showing the top crotch plate on the leftand the bottom transition cone on the right.


Figure 18-17 Cross section of a hexagonal reinforced concrete encasement for the trifurcation of 126-D tothree 72-D pipes.


Figure 18-18 Branch in a steel penstock, showing a pair of crotch plates in pipes of equal diameter (left) anda stiffener ring that is a curved plate (right). For high pressure, it is possible to add a flat plate as showndotted. It may be advisable to contruct a small scale model — to check dimensions, to note any fabricationor assembly problems, and to test the special section to rupture.


BRANCH SECTIONS

Branch sections also require reinforcing by crotchplates or stiffener rings. Figure 18-18 is a sketch oftwo common branches which are explained in"Buried Steel Penstocks; Steel Plate EngineeringData — Volume 4," published by the American Ironand Steel Institute. Design procedures areexplained, and graphs are presented for detailing.

OTHER SPECIAL SECTIONS

Wyes and branches are not the only special sections.However, most special sections are variations of thewye. For example, a tee (T) is a junction of threepipes just as a bifurcation (wye) is a junction of threepipes. The junction can be reinforced with ringstiffeners and crotch plates in the same way as thewye.

The development of special sections has been mostlyempirical. Because specials are costly, the attitudeof manufacturers has been, "Make it stout. Ourproduct must not fail." For example, instead ofstiffening mitered sections with rings, the thicknessof the steel is increased. Competition is forcingreconsideration of this policy.

For costly specials, or specials for which failure iscostly, physical tests are performed. Small scalemodel tests are often adequate. The conditions forsimilitude discussed in Appendix C are not sufficientfor hydrodynamic model studies. However,hydrodynamic similitude is described in most texts onhydrodynamics. Hydrodynamicists have led theworld in physical model studies.

SAFETY FACTORS

Safety factors are based on monetary equivalents.Analysis of the cost of risk is outside the scope of

this text. However, once costs are known, thesafety factors (zones of safety) can be identified. Inlegal action, comparative damage is based on thepercent encroachment of each of the adversariesinto the zone of safety. For example, a buriedpressure pipe joint ruptures. To what percent didthe manufacturer, designer, welder, and installereach encroach into the zone of safety?

PROBLEMS

18-1 What are the major and minor diameters of thecrotch cut of Figure 18-1?

18-2 What should be the diameter of the three steeloutflow pipes of a trifurcation from a 3 meter steelinflow pipe if all outflow rates are equal and the flowvelocity is to remain constant? (1.732 m)

Given: mitered wye in a welded steel water pipelinefor which,Di = 51 inches = inside diameter, inflow pipeDo = 36 inches = inside diameter, outflow pipesP = 196 psi = design pressure, includes surgeTaper = 9o = taper angle of the transition cones

18-3 What should the wall thickness be if theallowable hoop stress in the steel is 20 ksi?

(t = 0.250 inches)

18-4 What is the length of the two transition cones?(LT = 47.35 inches)

18-5 What is the w-force on the crotch plate? (w = 6.6 k/in)

18-6 What should be the miter cut angle in thetransition cones if the outflow ends are to beseparated by 5 inches to allow for welding?


Anderson, Loren Runar et al "STRESS ANALYSIS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 19-1 F-load (parallel plate load) on a circular ring showing maximum principal stresses σx and σ y atpoints A and B. The parallel plate load is the basic test for failure due to external forces.


CHAPTER 19 STRESS ANALYSIS

Pipes are essential components of life. If quality oflife is to improve, it will be partly because of betterpiping systems that are fail-safe over longer servic elife. Pipe failure can be catastrophic. Failureanalysis is essential — not only for repair,assessment of damage, and allocation ofresponsibility; but for improvements in pipes and pipesystems of the future. Some (not all) failures can bestress analyzed. See Figure 19-1.

Stress failures of flexible pipes by bursting areanalyzed in Chapter 2. Wall crushing is analyzed inChapter 6. For these cases, reconstruction of failureis simple. Not so simple is the reconstruction of arigid pipe failure. Under some conditions, flexiblepipes fail as rigid pipes. For example, bursting of aflexible pipe sometimes occurs so suddenly as tocause brittle fracture instead of plastic yield. Waterhammer is often the cause. Bursting due to internalpressure may be affected by longitudinal stressand/or external loading. Failure usually occurs at ajoint or an appurtenance where stresses mayconcentrate or where strength of material may bedeficient. For purposes of design, analyses arebased on simple mechanics of stress anddeformation. But for failures, combined orc ompound stress analyses, with correspondingperformance limits, may be required. In fact, stresstheory is not the only theory for failure analysis.Nevertheless, questions about failure always seemto arise in terms of stresses and strains andperformance limits for tri-axially stressed elasticmaterials. Better theories are becoming availablefor plastic and stress-regressive materials.

Following are a few of the most common combinedand compound elastic stress analyses.

COMBINED STRESS ANALYSIS

Combined analysis is the addition of all stressesacting in the same direction at a point. Thesestresses may be either normal or shearing stresses.

The failure (critical) stresses are of greatest interest.Figure 19-1 shows an F-load (parallel plate load).Consider point A on the inside surface under the F-load. The maximum principal stress is caused byflexure,

σx = Mc/I

whereσx = maximum principal stressM = moment at section AI/c = section modulus of the pipe wall per unit

length of pipeA = cross-sectional area of the pipe wall per

unit length of pipeP' = internal pressure (or vacuum)P = vertical external soil pressureD = mean diameter of the pipe = 2rt = wall thickness for plain pipe wall

(smooth cylindrical surfaces)c = distance from neutral surface of the pipe wall to the most remote fiber = t/2 for plain piped = ∆/D = ring deflection∆ = decrease in vertical diameterS = normal yield strengthS' = shearing yield strengthτ = shearing stressE = modulus of elasticityC = cohesive strength of the materials.

But now suppose the pipe is subjected to internalpressure P' in addition to the F-load. Clearly σx isincreased by the additional stress P'r/A. At the crown,

σx = P'r/A + Mc/I POINT A

This is combined stress. The internal pressure couldbe negative; i.e., vacuum or external pressure. Butis stress σx at A the critical stress? At point B ringcompression stress due to the F-load is felt. With allstresses included, the critical combination ofstresses might occur at point B,


Figure 19-2 Procedure for superimposing the orientation diagram on the stress diagram in order to evaluatestresses on various planes through infinitesinal cube 0.


where ring compression stress F/2A due to the F-load, external fluid pressure P'r/A, and flexure stressMc/I all combine in a maximum negative(compression) principal stress. At springline,

σy = Pr/A + P'r/A + Mc/I POINT B

This stress at B could be more critical than themaximum stress at A, depending on tensile andcompressive strengths of the pipe wall. However,loads do not always occur simultaneously. Forexample, when an internal vacuum occurs, the ringcross section shrinks slightly and reduces ringcompression stress. As long as the ring is held inshape by the soil, the flexure stress is zero. Foranalyzing stress, a more useful formula than σ =Mc/I, is the approximate formula for σ based on ringdeflection. At springline,

σy= 8Ecd/D

The coefficient 8 is conservative. In fact, it variesas a function of ring deflection. If the ring remainselliptical, the coefficient varies:

from 6.7 at d = 5%to 7.5 at d = 10%to 8.0 at d = 12.5%.

For the concentrated F-load, the coefficient is 9because ring deflection is not quite elliptical. Butwhen the pipe is buried, the loads are distributed, andring deflection is essentially elliptical.

For combined stress analysis by elastic theory,failure is the equation of maximum stress (normal orshearing) to the corresponding yield strength ofmaterial. The assumption is that elastic limit isfailure — not necessarily true.

On what plane does fracture occur? The answer isuseful in reconstructing the cause of pipe failures.For example, on what plane does a compressionfracture occur in brittle material at point B in Figure19-1? An approximate answer can be found from aMohr circle analysis.

MOHR CIRCLE ANALYSIS

Appendix E describes techniques for construction,application and analysis of the Mohr stress circle,orientation diagram, and strength envelope. Theinterrelationship of the three diagrams isaccomplished by superposition onto a single diagram.

Construction

Stress Diagram:

1. Draw the free-body-diagram of stresses on aninfinitesimal unit cube, O. Figure 19-2a is anexample of cube O with numerical values ofstresses shown.

2. Draw the σ and τ axes, Figure 19-2b; and plot theMohr stress circle. Three points are required:

a) Center is on the σ -axis.b) Circle passes through point (σx, τ xy).c) Circle passes through point (σy, τ yx).

Note that τyx = -τ xy; and that the sign convention ispositive for normal stresses in compression, and forc ounterclockwise shearing couples. This is thecorrect sign convention. Tension (negative) is areduction of compressive intermolecular bond.

Orientation Diagram:

1. Orientation of coordinate axes x and y is the sameas for the unit cube. When superimposed on thestress diagram (Figure 19-2c), the y-axis intersectsthe stress circle at point (σy,-τ xy). The axes are stillparallel to the unit cube axes.

2. The origin of axes, O, is the location of the unitcube, which is shown superimposed and correctlyoriented on the stress axes of Figure 19-2c. Theorigin (cube O) always falls on the stress circle.


Figure 19-3 Strength envelopes.


Strength Envelopes:

For most soils and many construction materials,failure is slip on a shear plane. It is shown in Figure19-3a as slip of a block due to shearing force F. Atslip, F is equal to the sum of cohesion C (glue on theslip surface) and friction Ntanϕ, where ϕ is thefriction angle and N = σ A.

Dividing through by A and noting that F/A = S', theshearing stress at failure, called Coulomb strength,is,

S' = C + σ tanϕ = SHEARING STRENGTH

A plot of the Coulomb shearing strength on the σ -τ axes is the strength envelope. See Figure 19-3b.Any stress point outside of the strength envelope isfailure. The material slips (shears).

Application

1. Any plane through O perpendic ular to the page(seen as a line) is oriented the same as theinfinitesimal cube.

2. Any plane drawn through O intersects the stresscircle at a point whose stress coordinates are thestresses on that plane.

3. When a stress circle is tangent to the strengthenvelopes, shear slip is incipient on those planes thatintersect the stress circle at the points of tangency.These are the failure planes, at angle β. See Figure19-3c.

Analysis

1. The horizontal distance to the center of the stresscircle is the average of σx and σ y.

DISTANCE TO CENTER = (σx + σ y)/2

2. From Figure 19-2b, The radius of the Mohr circleis, by the Pythagorean theorem,

_________________ RADIUS OF CIRCLE = \/(σx - σ y)2/4 + (τ xy)2

3. Any central angle is twice the correspondingcircumferential angle where both angles interceptthe same arc.

The values of principal stresses, maximum shearingstresses, and the planes on which they act can becalculated by trigonometry from the superimposeddiagrams. As an example, for the unit cube ofFigure 19-2, the pertinent stresses are:

σ1 = 2500 psi = maximum principal stressσ3 = 500 psi = minimum principal stressτmax = 1000 psi = maximum shearing stress

Principal planes and planes of maximum shearingstress are dotted on Figure 19-2c.

Example

The vertical stresses at point B in the unreinforcedconcrete pipe of Figure 19-1 are at incipient failure.What would be the expected failure plane angle, β ,at B if failure occurred. β is the angle of bevel ofthe fracture. From the laboratory, tension strengthis σT = 2 ksi, and compression strength is σ C = 12ksi.

First, from the infinitesimal cube, O, draw the stressaxes, σ and τ , as shown in Figure 19-4, and plot thestress circles for both tension and compressionfailures. The tangents to the failure circles arestrength envelopes.

Second, superimpose cube B and its orientationdiagram onto the stress diagram. Keep orientationthe same — x-axis horizontal and y-axis vertical.Cube B must be located at a point where its axes(which are the principal planes through cube B)intersect the Mohr circle at the stresses on thoseprincipal planes. Cube B always falls on the Mohrcircle. Because both the tension load and the com-pression load act on horizontal planes, x-planes aredrawn through the failure stress points σt and σ c.


Figure 19-4 Mohr analysis at failure (cracking) at point B on the inside of an unreinforced concrete pipe atthe springline.


Horizontal stresses (on vertical planes) are both zeroso the y-axis is at zero stress.

The results are the angles β of failure planes throughO. From strength tests, the friction angle of thisconcrete is ϕ = 45.58°. From the Mohr diagram, theangle of the failure plane is β = 45° + ϕ /2 = 68°.

Example 2

If cohesionless soil is loaded vertically with stressσz, what is the minimum horizontal stress, at activesoil resistance, required to prevent shear failureplanes from developing in the soil? At what anglesdo shear failure planes develop if the horizontalresistance is not adequate? The soil friction angle is30°.

From Figure 19-5, a stress circle is drawn tangent tothe strength envelopes which, for cohesionless soil,are straight lines from zero stress at angles ϕ = 30°.The maximum principal stress is σz at the right sideof the stress circle. Because it acts on a horizontalplane, the x-plane is drawn through this point. Theminimum principal stress is σx at the left side of thestress circle as shown. Because it acts on a verticalplane, the z-plane is drawn through this point. Theintersection of the z and y axes is the origin wherethe infinitesimal cube O is located. Let the distanceto the center of the circle be X. The radius of thec ircle is Xsinϕ ; σ1 = X + Xsinϕ ; and σ 3 = X -Xsinϕ. The ratio is σ1 /σ 3 = K = (1+sinϕ)/(1-sinϕ)= 3. The horizontal stress required to prevent shearslip is σx = σ z /K = σ z /3.

Shear failure planes are at angle β from the origin,O, to the points of tangency of the failure stresscircle to the strength envelopes. These angles are β= 45° + ϕ/2. For ϕ = 30°, the failure planes are at +60o.

Seven different combined failures are shown inFigures 19-6 to 19-8. All of these failures wereobserved in asbestos cement pipes, 12-inch insidediameter. Such failures are typical of failures in

unreinforced brittle pipes in general. The sketchesof failure planes (bevel angles) are to scale.

The analyses above are only examples, but the sameprocedure can be followed to reconstruct combinedstress failures.

COMPOUND STRESS ANALYSIS

Occasionally the state of stresses at a point is socomplex that the mere combining of stresses in thesame direction is inadequate. Compound analysisinvestigates the critical stress resulting from multi-directional stresses, both normal and shearing. Forplanar stresses (biaxial) no new concepts areneeded. For triaxial stresses, it is usually preciseenough to consider separately each of the threeviews of the infinitesimal unit cube as a biaxial casein order to ascertain which view results in the largeststress circle; i.e., nearest to tangency with thestrength envelope. This becomes the critical casefor analysis and design. In general, compound stressanalysis is required for specific cases such as thefollowing.

Stress Risers

Stress risers are discontinuities that cause stressconcentrations. At discontinuities, concentratedstress sometimes exceeds the maximum allowable.Remedies include saddles, stiffener rings, and crotchplates. However, it may be sufficient to simplythicken the wall by means of a boss or plate. Failureanalysis due to stress risers starts with basic stressanalysis. See Appendix E.

The free-body-diagram for compound stress analysisis an infinitesimal cube on which six pairs of forcesact. Figure 19-9 is the cross section of wall of apressurized container. The maximum stress occurson the inside surface on cube Op. The notations forthree pairs of normal stresses and three pairs ofshearing stress couples are as follows.


Figure 19-5 Mohr analysis for evaluating minimum (active) resistance of soil at shearing failure of the soil.(Failure planes form at angles β.)


Figure 19-6 Sketches and descriptions of failure of asbestos cement (AC) pipes from field notes. Thesefractures are typical of brittle pipes.


Figure 19-7 Sketches and descriptions of failure of asbestos cement (AC) pipes from field notes. Thesefractures are typical of brittle pipes.


Figure 19-8 Sketches and description of failure of asbestos cement (AC) pipes from field notes. Thesefractures are typical of brittle pipes.


Figure 19-9 Cross-section of a segment of the wall of a pressurized container (top) with an infinitesimal cubeon the inside surface identified for stress analysis, and shown enlarged (bottom) with stresses acting on it.The front view (bottom right) is a typical free-body-diagram to be used for the Mohr stress analysis.


σr = radial stress acting on an r-plane(perpendicular to the radial stress),

σt = tangential stress acting on a t-plane(perpendicular to the tangential stress),

σz = longitudinal stress acting on a z-plane(perpendicular to the longitudinal stress),

τrt = two equal and opposite shearing couplesacting about the z-axis,

τ tz = two equal and opposite shearing couplesacting about the r-axis,

τrz = two equal and opposite shearing couplesacting about the t-axis.

The double subscripts for shearing stresses indicatethe direction of each stress and the plane in which itacts. In the discussions to follow, the shearingstresses are assumed to be zero. This assumption istrue only if, as is usually the case for pressurizedcontainers, the normal stresses on Op are principalstresses. The shearing stresses identified above arenot zero for such loads as torque on the container, orfor concentrated forces. For example, if torque T isapplied about the longitudinal axis of a thin-wall pipe,the resulting shearing stress is approximately τ tr =T/2πr2t. See texts on mechanics of materials.Shearing stresses can be taken into account bycompound stress analysis based on the Mohr circle.Figures 19-10 and 19-11 show points, Op, on theinside surfaces of pressurized containers, with theircorresponding Mohr circles. The Mohr stresscoordinates are normal stresses (abscissae) andshearing stress couples (ordinates). If the Op (originof planes) is superimposed on the Mohr circle withits axes oriented, Any plane drawn through the Op,intersects Mohr circle at the stress coordinates,normal, σ , and shearing, τ , acting on that plane.Strength envelopes can be shown on the same stressaxes. If stresses increase in the material such thatthe Mohr circle comes tangent to the strengthenvelope, failure is incipient. Failure planes arethose planes passing through the Op and intersecting

the circle at points of tangency of the circle to thestrength envelopes.

The ideal pressure container is a thin-wall spherewith a wall that performs as a membrane. A toyballoon is such a container. The ratio of volume tosurface area is greater for a sphere than for anyother container. The tangential (circumferential)stresses in the wall, σt, are all tension, and are equalin all directions. See Figure 19-10. This is a mostfavorable stress relationship. The only compressionstress in the wall of the container is radial pressure,P' = σ r, acting normal to the wall on the inside of thecontainer. The maximum shearing stress, τ , is halfthe difference between radial stress and tangentialstress; i.e., 2τ = σt - σ r , with due regard for signs.But the radial compression stress, σr, is so muchsmaller than the tangential tension stress, σt, that itis negligible. Therefore, σt = P'r/2t, and maximumshearing stress, τ , is half the tensile stress, σt. This,too, is a most favorable stress relationship of spherescompared to other vessels.

The fracture plane for thin-wall pressure containersis beveled at failure angle θf for which twopossibilities are shown on the Mohr stress circle ofFigure 19-10 for a hypothetical material. For steeland some plastics, the strength envelopes are nearlyhorizontal; i.e., ϕ = 0o.

Consequently, the failure planes are at θf = 45o. Forunreinforced concrete, the strength envelopes are atangles of approximately ϕ = 45o, for which θ f = 45o

+ ϕ /2 = 67.5o.

Structurally, cylindrical containers are less than halfas efficient as spheres. Longitudinal and tangentialstresses are not equal. See Figure 19-11.Tangential stresses are twice as large in a cylinderas in a sphere of the same diameter and wallthickness. And, of course, the ratio of volume tosurface area is less for a cylinder. When cylindersand spheres are used together, discontinuities occur,as, for example, between cylindrical tanks andhemispherical end closures. Similar stress


Figure 19-10 Mohr stress circle at failure of an infinitesimal cube located on the inside of a thin-wall spherewith internal pressure P'. The coordinate axes for the Mohr circle are positive for normal stresses (abscissae)in compression and are positive for shearing stresses (ordinates) in counterclockwise couples.


Figure 19-11 Mohr stress circle at failure of an infinitesimal cube, Op, on the inside surface of a pressurized,closed-ended cylinder, showing three orthogonal views of Op with the corresponding Mohr circles. Thelargest circle is critical; i.e., it will be the first to come tangent to the strength envelope.


σr

σ r

AT RADIUS, r,

TANGENTIAL STRESS σf = P(a/r)2(r2 + b2)/(b2 - a2)

LONGIDUTINAL STRESS σz = Pa2/(b2 - a2)

Figure 19-12 (top) Cross section of a thick-wall steel cylinder of diameter OD = 2(ID), and showing thevariation of tangential stress, σt, throughout the cylinder wall when subjected to internal pressure, P'. (bottom) Mohr circle with strength envelopes for steel (nearly horizontal), showing failure planes at 45o.


risers occur in wyes, tees, valves, reducers, etc.Pipes are sometimes capped when they are to beextended at a later date. Taps cause stressconcentrations on the edge of the tapped hole.Thick-wall containers feel higher tangential stress onthe inside than thin-wall hoop stress, σt = P'(ID)/2t.These stress risers are discussed in the following.

Thick-wall Cylinders

Thick-wall cylinders subjected to internal pressure,feel maximum tangential stress (hoop stress) on theinside of the wall:

σt (OD)2 + (ID)2

P' (OD)2 - (ID)2 . . . . . (19.1)

whereσt = naximum tangential stress (hoop stress)OD = outside diameterID = inside diameterP' = internal pressure

Example 1

Figure 19-12 is the cross section of a thick-wall,high-pressure pipe. What is the maximum tangentialstress, σt, and the maximum shearing stress, τ ? OD= 2(ID). From Equation 19.1, σt = 5P'/3. Theaverage hoop stress is σ = P'(ID)/2t = P'. Clearly,the maximum stress is 5/3 rds as great as theaverage. From the Mohr circle, maximum shearingstress is 4 P'/3. If the cylinder is steel, the strengthenvelopes are nearly horizontal as shown and thefailure planes are at 45o.

Longitudinal stress is calculated the same way forboth thick-wall and thin-wall cylinders. SeeChapters 14 and 15.

Radial stress is maximum on the inside of the wall,and is simply internal pressure, P', in compression.Tangential stress is maximum on the inside when thepipe is subjected to either internal pressure or

external pressure. For analysis of externalpressures on thick-wall pipes and tanks, refer totexts on mechanics of materials.

Taps

The most common attachments to pipes are tappedinto the pipes. Stress concentrations occur at holesin the walls of pressure containers. Taps arerequired for attaching corporation stops, smallerpipes, air-relief valves (ARVs), pressure gages, etc.Stresses concentrate in the wall around the tappedhole, and are critical if internal pressures are highand the pipe material is non-plastic. Plastics includemost pipe grade steel which is elasto-plastic. Plasticyields before it fractures. This is not true, however,for pipes subjected to impact loads or very lowtemperatures or repeated loadings. Under suchloads, even plastics can fail by brittle fracture. Oneremedy is to thicken the wall around the hole. Forexample, a bead or boss could be formed around theedge of the hole. See Figure 19-13. Engineers usea rule of thumb — the bead or boss must contain avolume equal to the volume of material cut away forthe hole.

From elastic theory (and tests) tangential stressesvary as shown in Figure 19-14 for a very large platewith a hole in the middle of it. The elastic equationsfor tangential stress σt, radial stress σ r, and shearingstress τ , on an infinitesimal cube in the plate, are:

2σt = σ o(1 + ρ2/r2) - σo(1 + 3ρ4/r4)cos 2θ . . . . . (19.2)

2σr = σo(1-ρ2/r2) + σo(1-4ρ2/r2+3ρ4/r4)cos 2θ . . . . . (19.3)

2τ = σo(1 + 2ρ2/r2 - 3ρ4/r4) sin 2θ) . . . . . (19.4)

where (See Figure 19-14)σr = radial stress with respect to the hole,σt = tangential stress, (perpendicular to radial

stress at any point),


Figure 19-14 Stresses in a plate with a hole in it, showing how stress channels crowd around the hole liketraffic around an excavation in the road, creating stress concentrations, Kσo, at the edge of the hole. If thehole is very small, K = 3 and σt = 3σ o.


τ = shearing stress,b = width of the plate,ρ = radius of the hole,r = radius to the stress point in the plate,θ = angle of the radius from the direction of

average stress, σo,σo = average uniform stress in the plate if no

hole were in it.

Of primary concern are stresses tangent to the edgeof the hole where r = ρ. If the plate is infinitelywide, ρ/b 0, and σo is unaffected by the hole.From Equations 19.2 to 19.4,

σt = -σ o, compression at θ = 0o . . . . . (19.5)

σt = 3 σ o, tension at θ = 90o . . . . . (19.6)

Shearing stresses, τ , are zero at the edge of the holeand increase to a maximum of τ = σ o /2 at greatdistances from the hole on planes at 45o with thelongitudinal axis.

Design of taps in pipes follows the above rationalefor uniaxial loading in the case of gasketed pressurepipes with no longitudinal stress. See Figure 19-15,top sketch. If the diameter of the hole is muchsmaller than the diameter of the pipe, Equations 19.5and 19.6 can be used. From Equation 19.6, themaximum stress at the edge of the hole tapped ingasketed pipes is tangential stress,

σt = 3P'(ID)/2A, tension at B . . . . . (19.7)GASKETED PRESSURE PIPES

whereσt = tangential stress at edge of hole,P' = pressure in the pipe,ID = inside diameter,A = cross-sectional area of the wall per unit

length of pipe,A = t for plain wall pipes.

For the case of a closed-end pipe or tank, see Figure19-15. In the bottom sketch, the tangential stress isanalyzed by combining Equations 19.5 and

19.6. For example, from Equation 19.5, thetangential stress at B is -σo /2 in compression due tolongitudinal stress, σo /2 in tension. At the samepoint B, from Equation 19.6, the tangential stress is3σo in tension due to hoop stress, σ o. Combining thetwo tangential stresses at B, for closed-end pipes,

σt = 5P'(ID)/4A, tension at B . . . . . (19.8)CLOSED-END PRESSURE PIPES

Example 1

What is the maximum tangential stress on the edgeof a small 1-inch hole (tap) in a 6-inch ID steel pipeif the wall thickness is 0.125 inch and internalpressure is 400 psi? Slip couplings, such as Dresseror Baker, eliminate longitudinal stress in the pipe.From Equation 19.7,

σ t = 3(400psi)(6 in)/2(0.125 in) = 28.8 ksi

If yield stress is 42 ksi, the safety factor is 42/28.8 =1.46. If P' is a repeating pressure, fatigue strengthof the steel should be considered. Fatigue strengthis usually lower than yield strength, especially if thepressure cycles are reversed, and the hole isthreaded. Threads are stress risers.

If the pipe is not gasketed, longitudinal stress, σz isusually tension, in which case stress risers at a tapare less than for a gasketed pipe.

This tap analysis is based on static pressure and asmall diameter tap in a large diameter pipe.Refinements are forthcoming. In the 1980s RolandW. Jeppson, at Utah State University, testedthreaded taps with corporation stops for serviceconnections in AWWA C-900 PVC pipes, 6 inchesin nominal diameter, subjected to cyclic internalwater pressure surges from 100 to 200 psi at a rateof 30 surges per minute. Loading was continualwithout down time during which plastic mightpartially recover. Dye in the water made theslightest crack visible. None of the five 0.75-inchtaps or the five 1-inch taps leaked after 1.5 million


Figure 19-15 Principal stresses and stress concentrations in the pipe wall and around a hole tapped in the wallof a pressure pipe: (top) gasketed pipe, and (bottom) closed-end pipe. Obviously, stress concentrations aremaximum around the hole in the gasketed pipe.


cycles of pressure. Similar surge tests wereperformed on larger 24-inch PVC pipes. Someleakage was noted. Results were only qualitative,but it was concluded that service life of a tappedlarge-diameter PVC pipe subjected to pressuresurges is only moderately less than an untappedpipe. However, both exceed normal expectedservice life. A different problem showed upserendipitously. The gasketed end closures for thetest sections leaked more than did the corporationstops.

Example 2

What is the maximum tangential stress at the edgeof the taps in Dr. Jeppson's pressure tests ofAWWA D-900 PVC pipe 6D? It is assumed(questionably) that the taps are small compared withthe pipe diameter.

OD = 6.900 inches average,t = 0.276 inch minimum,DR = 25 = dimension ratio,P' = 200 psi maximum.

Hoop stress in the wall of the pipe is,

σo = P'(ID)/2t = 200 psi(6.900-0.552)/2(0.276)σo = 2300 psi

Because the test pipe was gasketed, from Equation19.7, the maximum tangential stress is 3σo = 6.9 ksi.This is higher than the yield strength of PVC, andwould indicate that to assume that pipe diameter isinfinitely greater than hole diameter may be tooconservative. Moreover, Equation 19.7 is based onelastic limits. PVC is plastic.

If a tap, or any stress riser, is in any section otherthan a cylinder, the hoop stress, σo, must beevaluated for that specific section. For example, tofind the maximum "hoop" stress in the wall of avalve with a bonnet, σo = P'(ID)2t where ID is thegreatest width of a plane that can be passed throughthe bonnet and pipe.

Stress risers are not necessarily critical in pliantmaterials such as plastics — including steel.Because of a demand for corrosion-resistantpolyethylene pipes under sanitary landfills, andbecause those pipes must be perforated or slotted todrain leachate and collect methane gas, tests havebeen performed to ascertain the effects of smallperforations or circumferential slots (width of acircular saw blade) on the structural integrity of thepipe. Clearly, the strength of the ring is reduced bythe area of surface cut away. The ring stiffness isreduced also — but relatively less. When the ring isloaded and deformed into an ellipse, visible warpingat the edges of the perforations and the ends of thesawed slots is evidence of stress concentration.However, the plastic yields without fracture, andrelaxes. The integrity of the pipe is notcompromised.

End Closures

In this section, end closures are analyzed for stressrisers. However, any transition from a pipe toanother section (wye, tee, reducer, valve, etc.)results in similar stress risers, usually to a lesserdegree, and can be analyzed in a similar manner.

Consider the hemispherical cap on a thin-wallcylinder with internal pressure P' as shown onFigure 19-16. Notation is as follows.

P' = internal pressure,r = mean radius of thin-wall container,t = wall thickness,σr = radial stress on the inside of the wall,σt = tangential stress in the wall,σz = longitudinal stress in the wall,E = modulus of elasticity,ν = Poisson ratio,εr = radial strain = percent increase in radius,ε,t = tangential strain,εz = longitudinal strain,


Figure 19-16 Pressurized thin-wall container with hemispherical end-closures, showing:(bottom) how internal pressure, P', increases the radius by a percentage equal to the tangential strain; and (top) how shearing discontinuity at section A-A can be eliminated by reducing the wall thickness of thehemisphere, but leaving, instead, a reentrant corner which itself is a stress riser.


The tangential stresses are:

Cylinder, σt = P'/(r/t)Hemisphere, σt = 0.5P'(r/t)

If wall thicknesses are the same for cylinder andhemisphere, as shown at the left side of Figure 19-16, tangential stress in the hemisphere is only half asgreat as in the cylinder. Therefore, increase inradius is less in the sphere than in the cylinder; anda radial shearing stress occurs at section A-A. Thediscontinuity is a stress riser.

Tangential stresses would be equal if thehemispherical wall were only half as thick as thecylindrical wall. This is shown at the right side ofFigure 19-16. The shearing discontinuity is reducedbut is not eliminated. The problem is still thedifference between increases in radii. Neglectingthe relatively minor effect of radial pressure on thew all in thin-wall containers, tangential strain, εt, isthe percent increase in circumference, which is thepercent increase in radius — which is radial strain.Therefore, radial strain is εr = ε t, and stress-strainrelationships are Eεt = σ t - νσ z, as follows:

Cylinder, Eεt = σ t - νσ t /2

Eεt = P'(r/tc)(1 - 0.5ν) . . . . . . . . . . . . . . . . (19.9)

where tc is the thickness of the cylinder.

Hemisphere, Eεt = σ t

Eεt = P'(r/ts)(1 - ν) . . . . (19.10) where ts is the thickness of the hemisphere.

If the radial strains are to be equal, Equations 19.9and 19.10 must be equal; and the ratio of wallthicknesses would have to be,

ts /tc = (1 - ν)/2(1 - 0.5 ν)

For example, if Poisson ratio is ν = 0.25, the ratio ofwall thicknesses of hemisphere to cylinder is

0.4286, which is even less than the original ratio of0.5 required for equal tangential stresses. Theshearing discontinuity is essentially resolved.However, the connection now mates two differentwall thicknesses with a reentrant corner which isitself a stress riser. For brittle materials, thereentrant corner could be critical. But if the materialis pliant and a safety factor is included in design, thereentrant corner might be mitigated by a good weld,and/or by ductile pipe material for which the stressriser is not critical because the material yieldswithout fracture. Or it might be remedied byshaping the end closure to something other than ahemisphere. The end closures of rocket motors areellipsoidal to reduce stress risers.

Strength of Welded Joints in Steel Pipes

If the weld is a full-penetration butt weld,longitudinal strength is no less than the strength ofthe steel pipe. If there is any question about thewelding procedure, some designers assume strengthto be 90% of steel strength.

If the weld is a lap weld, and the gap is no greaterthan 0.125 inch, longitudinal strength of a singlewelded lap joint is about 75% of pipe strength. Thestrength of a double-welded lap joint is no less than80% of pipe strength. The forces on fillet welds oflap joints are shearing forces not couples. Thecurved surfaces of both bell and spigot preventmoments (couples) on the weld.

STRESSES IN STEEL PIPES

Steel pipes draw a disproportionate amount of stressanalysis because of advances in the analysis of steelstructures for which performance limit is yield stressby the elastic theory. For steel pipes, performancelimit is deformation (strain) — not elastic limit(yield). In fact, some steel pipes are strained wellabove elastic limit during the process of fabrication.Nevertheless, stress theories persist. Some commonstress analyses follow.


Figure 19-17 Standard tension test of steel, showing the stress-strain diagram with the elastic limit andultimate strength.

Figure 19-18 Standard tension test on steel, showing the Mohr circle at elastic limit in tension and incompression (dotted) and the strength envelopes and failure planes in shear (slip).


Performance Limit:

Performance limit is strength, σf, at elastic limit. It isoften referred to loosely as yield stress. It is foundfrom standard uniaxial stress-strain tests. SeeFigure 19-17. For pipe-quality steel, typicalproperties are as follows.

Properties of Pipe Steel:

σf = 42 ksi = stress (failure) at elastic limit,E = 30(103) ksi = modulus of elasticity,ε = F/E = strain below the elastic limit,εu = 21% = approximate elongation at fracture,ν = 0.27 to 0.30 = Poisson ratio,U = 15 lb.ft a 0oF = Charpy toughness.

For stress design, STRESS must be less thanSTRENGTH reduced by a safety factor.

Strength

From Figure 19-17, ultimate steel strength is greaterthan yield strength. For some analyses, thisdifference provides a margin of safety in addition tothe safety factor. For energy analyses, the marginof safety is much greater. Energy input, Ue, up toelastic limit per unit volume of the test specimen ofFigure 19-17, is the average force, P/2, times thedistance, ∆ , divided by volume, AL. If σf = 42 ksiand E = 30,000 ksi,

Ue = P∆ /2AL = (1/2)(P/A)(∆ /L) = 0.5σεUe = σf

2/2E. Ue = 29.4 ksi = RESILIENCE = Area under the stress-strain diagram up to

the elastic limit.

Energy up to ultimate strength, Uu, is the area underthe entire stress-strain diagram to elongation (strain)at failure — which is roughly 21%;

Uu = 8,800 ksi = TOUGHNESS = Area under the entire stress-strain diagram.

Ultimate energy, Uu, in steel is three hundred timesgreater than elastic energy, Ue. Design by elasticenergy (resilience) is extremely conservative. Theabove applies only to uniaxial stress. The energy ofcompound stresses is investigated in the following.

Stress:

Compound stress analysis is based on elastic theory(elastic limit). Compound analyses are seldomjustified for buried steel pipes. In the first place, oneprincipal stress is usually so much greater than eitherof the others that uniaxial stress analysis isadequate. In the second place, properties of theembedment are imprecise, loads are unpredictable,and pipe-soil interaction is statically indeterminate tothe infinite degree. In the third place, steel is elasto-plastic — not limited to the elastic limit. Yield stressis determined by tensile tests (not triaxial tests) thatprovide only normal failure stress, σf. Performancelimit is shearing failure — not tension failure.The shearing failure plane is beveled at 45o.

Mohr Stress Circle:

Figure 19-18 shows the Mohr stress circle for astandard tension test to failure. Ordinates areshearing stress, τ . Abscissae are normal stress, σ .For steel, it is assumed that tension is positive andclockwise shear is positive.

The infinitesimal cube is superimposed on the stresscircle by orienting the planes on which the principalstresses act. Any plane through the cubeintersects the Mohr circle at its own stresscoordinates. The shearing and normal stresses atthe point of intersection act on that plane. All planesare correctly oriented. For tests in compression, theshearing stress at yield is nearly the same as intension. See the dotted Mohr circle. Tangents tothe two Mohr failure circles are strengthenvelopes. Any shearing stress outside of


Figure 19-19 Compound stress analyses for steel pipes based on shearing strength. Mohr circles are shownfor principal stresses acting on an infinitesimal cube on the inside surface of the pipe wall.


the strength envelopes is failure. Failure planes areat 45o with the principal stresses.

Elastic Stress Analysis — Combined Stresses:

Stresses are added in combined stress analysis. Forexample, in Figure 19-19, σy = P'r/t - Pr/t, where P'is internal pressure and P is external soil pressure.However, for conservative design, P and P' areanalyzed separately. Longitudinal stress, σz, is thesum of longitudinal stresses due to temperaturedecrease, Poisson effect of internal pressure, andthrusts due to valves, elbows, etc. However,external loads such as longitudinal beam bending andthrust blocks may or may not contribute to combinedlongitudinal stresses. They must be considered on acase-by-case basis.

Elastic Stress Analysis — Compound Stresses:

Performance limit is shearing failure. Maximumshearing stress equals shearing strength; i.e., τ = τf.

In order to relate shearing stress at failure to thestandard tension test at failure, from the Mohr circlein Figure 19-18, τf = σ f /2.

Figure 19-19 shows an infinitesimal cube in theinside of a pipe wall with principal stresses acting onit. The horizontal stress is internal pressure P, whichis small compared to the other stresses. If the cubewere on the outside of the pipe wall, P would bezero. For analysis, σx = 0. Mohr circles are shownfor the three views of the infinitesimal cube. If thepipe is not subjected to torque or point loads,shearing stresses do not act on the cube, and thethree stresses are principal stresses. The front viewis critical. For analysis, shearing stress at failure ishalf the normal stress at yield, σf /2. Therefore, Pr/t= σf = 42 ksi. A safety factor (often two) is usedfor design; i.e., Pr/t = 21 ksi.

The side view is not critical. The critical circleshown is an improbable hypothetical condition

wherein longitudinal and circumferential stresses areof opposite sign. Vacuum in the pipe, a highexternal fill, and a high water table might combine tocause suc h stresses. But failure would probably becollapse not stress failure. Steel pipe collapse isusually a function of ring stiffness not yield stress.

The top view is critical only if longitudinal stress isexcessive or if internal pressure is zero or negative(vacuum). This usually requires collapse analysisnot stress analysis. See Chapter 10 on ring stability.Stresses are not added in compound stress analysis.

Figure 19-20 shows strength envelops of σy as afunction of σz at yield stress according to the theoryof elasticity. The shear stress analyses are dotted.A more accurate analysis is maximum strain energy,U = Σf (σ /2)ε . See Appendix F for the analysis ofstrain energy at failure. The strain energy results donot quite model test results for steel.

Huber, Hencky, von Mises Equation:

More accurate for steel is the Huber-Hencky-vonMises equation, which discounts the strain energythat only causes volume change. See Appendix Ffor the equation and Figure 19-20 for the plot of theelliptical strength envelope for two-dimensionalcompound stress. Most buried pipes do not requireHuber-Hencky-von Mises analysis. Because σx =P is relatively small, the two-dimensional analysis isadequate. As an example, if σz = σ f /2, according tothe Huber-Hencky-von Mises equation, σy =1.155σf. Clearly this allows a slightly greater stressthan yield, but the increase is small. It isconservative to design by uniaxial stress analysis;i.e., critical stress is σy = σ f at P = 0. In general, forburied pipes, compound stress analysis is not needed,and may be misleading because it is based on elasticlimits.

Collapse Analysis

For unburied steel pipes, collapse is Pr3/EI = 3,


Figure 19-20 Strength envelopes at elastic limit, σf.

where I is the moment of inertia of the wall crosssection per unit length of pipe. Collapse is a functionof external pressure, P, and ring stiffness, EI/r3 —not yield stress. Ring deflection and soil strength arepertinent but are usually controlled by specifications.

STRESSES IN CONCRETE PIPES

Concrete pipes are so complex that stress analysisis impractical. Performance limits are usually foundfrom tests.

FAILURE SURFACES

1. A crystalline surface with sharp edges indicatessudden fracture due to an instantaneous load such aswater hammer. High-strength steel (such as bolts)under high tension may fail by hydrogenembrittlement. Some plastics can fail by "brittlefracture" — especially under shock load and lowtemperatures.

2. A smooth, satin surface indicates quick fracture.

3. A long fracture surface (tear) indicates crackpropagation due to stress energy stored in the pipe.

4. A rippled surface — especially if oxydized nearthe pipe surface — indicates fatigue.

5. A hole in a pressure pipe abraded from outside-incould be caused by a leak out of which a high-pressure jet causes turbulence in the embedmentthat "sand-blasts" the pipe. The cutting action fromoutside is remarkably rapid.

6. Chevrons along a torn surface point toward thelocation where fracture was initiated.

7. Pock-marks of oxidized material indicatecorrosion. A stream bed appearance with loss ofmaterial inside on the invert indicates erosion.


PROBLEMS

19-1 An unreinforced concrete pipe has a wallthickness of t = 3 inches, ID = 36 inches, and lengthof L = 11 ft. A 12-kip dual-wheel load passes overmidspan. Soil cover is negligible. What is the waterpressure inside the pipe at failure if the soil has beenwashed out from around the pipe leaving it simplysupported on the ends? The joints are gasketed.Unit weight of concrete is 144 pcf, unit weight ofwater is 62.4 pcf, and strengths of the concrete are2 ksi in tension and 12 ksi in compression.

19-2 What is the maximum tangential stress in athick-wall, closed-end cylinder that has a small tapin it, if OD = 1.4(ID)?

19-3 A polyethylene pipe is to be installed under ariver. A tunnel, slightly larger than the pipe, isdirectionally drilled under the river and driller's mudis left in the tunnel to prevent collapse of the tunneluntil the pipe can be pulled through. As the pipe isbeing pulled into place, the driller's mud is replacedby grout. What is the maximum stress in the wall ofa polyethylene pipe subjected to the externalhydrostatic pressure of grout with unit weight 95 pcf,and depth of 80 ft? The pipe is empty such thatpressure inside is atmospheric.

DR = 9.5 = OD/t for this thick-wall pipe. Neglectlongitudinal force during the pull.

19-4 Resolve problem 19.3 while the polyethylenepipe is being pulled into place with a longitudinalstress of 500 psi. The pulling force is monitored tomake sure the pipe doesn't bind and break while it isbeing pulled into the tunnel.

19-5 Resolve problem 19.4 if the pipe is filled withdriller's mud as it is pulled into place. The unitweight of driller's mud is 75 pcf.

19-6 What is the effect of tapping, or perforating, orcircumferentially slotting the polyethylene pipe ofproblems 19.3 to 19.5? Small perforations or narrowsawed slots are common practice in fabricatingpolyethylene pipes to be used under sanitary landfillsto collect leachate and methane gas.

19-7 What should be the ratio of thicknesses of anend closure and thin-wall pressurized cylinder if theend closure is a hemisphere, fitted inside thecylinder, and reversed in direction (like the bottomsof spray paint cans)? What is required to withstandthe shearing discontinuity? What should be the sizeand type of weld? What might be the size anddesign of a restraining ring? Etc.


(P = 300 psi)

(σ t = 8 P')

(σ t = 280 psi)

(τ = 375 psi)

(σ t = 100 psi)

Anderson, Loren Runar et al "PLASTIC PIPES"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 20-1 Nomenclature for a plastic pipe buried in a select embedment within a trench and showing ringdeflection, d = ∆/D due to the vertical pressure of a landfill.

Figure 20-2 Typical tensile strength regression line for PVC at 70°F showing how strength decreases linearlywith time on a log-log plot.


CHAPTER 20 PLASTIC PIPES

The use of buried plastic pipes is widespread andincreasing. Plastic pipes are attractive for use inaggressive environments, and for collection andtransmission of liquids that are abrasive and/orcorrosive. Most small-diameter gas dstribution pipesare plastic. Corrugated plastic pipes are usedextensively for land drainage and culverts.

Plastics are relatively resistant to corrosion byaggressive chemicals and to abrasion by sediment influid flow. Plastic pipes are flexible, and conformwith soil in a favorable pipe-soil interaction thatreduces stress concentrations. Plastics can toleratedeformation. Plastics have high tolerance forfatigue stress (cyclic loading). Plastics relax underconstant deformation.

Plastic pipes are light weight. They have low flowresistance. They have long service life. They areeasily fabricated and easily joined in the field. It ispossible to manufacture plastic pipes in variousshapes, and with corrugations, ribs, fillets, etc.

On the other hand, compared to traditional pipematerials, plastics have low strength, and lowmodulus of elasticity. They may be sensitive totemperature — both too hot and too cold. Plasticshave high Poisson ratios and high coefficients ofthermal expansion. Questions arise as to the abilityof plastic pipes to perform under high pressure, bothinternal and external, and under heavy loads. Thesequestions are poignant because of the urgent needfor corrosion/abrasion-resistant piping for drainageunder high landfills (such as sanitary landfills) and inmines, where leachate is aggressive and sediment isabrasive.Following is a discussion of the conditions underwhich plastic pipes can resist internal pressures andexternal loads, and the effect of time on pipeperformance. Most of the performance data areempirical; i.e., provided by tests and by pipes inservice. As always, much useful data comes fromfailures. See Figure 20-1 for nomenclature.

UNIQUE PROPERTIES OF PLASTICS

A. Under Constant Stress:

1. Plastics Creep.Creep is time-dependent strain under persistentstress. Plastic creep progresses gradually toequilibrium or fracture. Persistent stress is constantor recurring stress.

2. Strength Regresses.Strength is stress at fracture. The time to fractureis roughly exponential, as shown in the log-log plot ofFigure 20-2.

B. Under Constant Deformation (Strain):

1. Stresses Relax.Over time stresses relax, at a decreasing rate , to astate of stress equilibrium in which stresses remainessentially constant, but less than the initial stresses.Strains are constant because deformation is heldconstant.

2. Strength Is Not Reduced.The ability to resist pressures or loads is not reducedover time under constant deformation (strain). Onlyunder persistent stress and the resulting creep doesstrength regress.

C. Stress-strain Relationship

1. Properties of Plastics Remain Pristine.For pipe design, plastics are non-degradable.Therefore, properties of material remain pristine(non-regressive) over time. The most pertinentproperties are strength, modulus of elasticity, andPoisson ratio — all of which remain pristine evenafter long-term persistent pressure and/or constantdeformation. The only exceptions occur underextreme temperatures, degradation of the plastics,and strength regression under long-term persistentstress.


Figure 20-3 Typical strength regression of plastics, showing short-term (pristine) strength that prevails up tothe point of long-term failure.

Figure 20-4 Typical stress-strain diagram for plastics, showing the effect due to creep over the long term, andthe resulting virtual modulus after the long term.


2. Rate of Stress Relaxation Exceeds Rate ofStrength Regression.Under constant strain, stresses relax at a faster ratethan the strength regresses. Therefore, failure doesnot occur. If a stressed plastic pipe is subjected toconstant deformation, such as a pressed fit in anencasement; and if the pipe does not fail at the timeit is squeezed into the encasement; then stresses inthe pipe will relax and the pipe will not fail over thelong term. Gross deformations, such as clamped-offends of plastic gas pipes, are evidence of the stress"relaxability" of plastic pipes.

Example

Figure 20-3 shows typical strength regression of aplastic pipe. If stress in the pipe is σ = S1, failurewill occur in one year. If σ = S50, failure will occurin 50 years. In this example, a pipe, under constantinternal pressure of σ = S50, is doomed to fail in 50years. However, suppose that in 30 years, it issuddenly subjected to increased internal pressure.What strength is available to resist this stress? FromFigure 20-3, the strength is S0 — pristine initialstrength.

3. Creep Depends Upon a Virtual Modulus.Some analysts use a virtual long-term modulus ofelasticity. This is not the true modulus. The truemodulus remains pristine. A serviceable definitionof modulus of elasticity in a plastic material isstiffness — the slope of a secant on the stress-strain diagram. An example is shown in Figure 20-4. Virtual modulus, E', can be used to predict long-term strain (creep) under persistent stress. It doesnot apply to collapse. Collapse is sudden — afunction of pristine stiffness of the plastic.

Example

A section of plastic pipe is a beam supporting aconstant load. What is the deflection in the longterm? Under constant load (persistent stress),plastic creeps, and deflection increases gradually.Deflection of a beam is inversely proportional to themodulus of elasticity, which is the slope of the

secant on the stress-strain diagram. Figure 20-4shows the true modulus, E, in the short-term. In thelong term, the virtual modulus is less. Virtualmodulus can be used to predict long-term deflectionof the beam under constant load.

4. Plastic Has a Memory for Distress.Analysts refer to a plastic's "memory for a load."Once plastic is distressed by a load, subsequent re-loading will tend to re-distress the plastic in the samepattern as the original distress. This is not a changein the pristine properties of the plastic. It is theconsequence of prestress that has become set bycreep over a period of time. The prestress isreversed when the load is removed and the plastictries to return to its original shape.

5. Time of Recovery Equals Time of Creep.Deformed plastics tend to return to their originalshape after the deformation is released. Someanalysts use a rule of thumb that the time requiredfor plastic to return to its original shape afterdeformation is equal to the time of deformationwhen stress relaxation was in progress. This impliesthat the distress memory fades in a period of timemore or less equal to the time of distress. Theconcept is faulty in that the plastic does not recovercompletely.

6. Performance Limits Are Different for Tensionand Compression.Under constant tension, plastics creep (elongate)and decrease in cross-sectional area. They fail atyield stress. Under constant compression, plasticscreep and increase in cross-sectional area such thatresistance to failure is increased. They do not fail attensile yield stress. Therefore, allowablecompression strength is ultra-conservative whenbased on tensile yield strength.

PLASTIC PIPE NOMENCLATURE

Some distinctive notation and nomenclature havearisen out of the development of plastic pipes.DR = dimension ratio = OD/t,SDR = standard dimension ratio,


t = wall thickness or the equivalent area per unit length of pipe,

OD = outside diameter,ID = inside diameter,D = mean diameter (to neutral surface)

= (OD - t) for plain pipe,r = mean radius = D/2,d = ring deflection = ∆/D ,∆ = vertical decrease in diameter,S = allowable stress in the plastic,P' = internal pressure,P = external soil pressure,p = internal vacuum plus external

hydrostatic pressure.

Some of the first commercially available plasticpipes were extruded with a constant OD. The wallthickness was adjusted for various pipe "series"(pressure ratings) by adjusting the ID. Blowmolding,spiral welding, and other manufacturing techniqueshave been developed, but the fixed OD is still thebasis for the ratings. DR = OD/t. The dimensionratio was more precisely defined as the mean ODdivided by the minimum t. For analysis, correctionfactors were introduced: (Odmin /ODmax)

3, and (tmin

/tmax)3. These correction factors are not accurate,

but introduce a margin of safety, and alert pipelinersto the need for precise pipe dimensions. WilliamAllman of DuPont found that the exponent, 3, is low.His tests indicated that 4.6 is more accurate.Improvement in the control of plastic pipedimensions has virtually eliminated the need forcorrection factors.

Based on the above notation, design equations forplastic pipes are rewritten in the following forms.

The Barlow formula for internal pressure is changedfrom σ = P'(ID)/2t, to,

P' = 2S/(DR-1) . . . . . (20.1)ALLOWABLE INTERNAL PRESSURE

A safety factor is included in allowable stress, S.The ring compression formula for external soilpressure is changed from σ = P(OD)/2t, to,

P = 2S/(DR) . . . . . (20.2)ALLOWABLE EXTERNAL SOIL PRESSURE

A safety factor, sf, is included in allowable stress, S,but is probably unnecessary. Compressive yieldstress is greater than tensile yield stress which isfound from tensile tests.

The formula for vacuum at collapse is changed frompD3/EI = 24, to,

p = 2E/(DR-1)3sf . . . . . (20.3)ALLOWABLE INTERNAL VACUUMincluding external fluid pressure.

PLASTIC PIPE PERFORMANCE

In the design and analysis of buried plastic pipes, thefollowing are concepts of performance.

1. In general, native undisturbed soil is stable. It hasbeen in place for a long time, and, except forearthquakes, landslides, etc., is a stable medium forpipe-soil interaction. Exceptions include soils thatare fluid such that the pipe sinks, or floats, orcollapses due to external fluid pressure.

2. The best buried pipe installations are those whichleast disturb the native soil. A bored tunnel ofprecise diameter into which the pipe is squeezed andinserted, would cause the least soil disturbance. Thepipe then serves as a tunnel liner. This concept isideal, but impractical. A more practical installationwith little disturbance of the native soil is a narrowtrench with only enough side clearance to align thepipe and to permit placement of select soil backfill.The backfill develops continuity between the pipeand native soil. Pipe-soil interaction is assured. Theflexible pipe is stabilized.

3. Once a plastic pipe is stabilized, its shape remainsnearly constant. Stresses relax, and the pipe doesnot fail over the long-term. An exception occursunder extreme ring deformation of profile or ribbedwalls. The highest stresses relax fastest, and shift


the neutral surface such that stresses increase onthe other side of the neutral surface. Theconsequence could be a long-term crack.

4. To design a plastic pipe for long-termperformance under persistent load, the regressed,long-term strength is used. An example is thedesign of pipe to withstand constant internalpressure. The plastic creeps over the long term.Therefore, diameter increases and wall thicknessdecreases. The virtual strength decreases. Thereverse occurs for constant external pressure.Compression causes plastic creep over the longterm. But the wall thickness increases. The virtualstrength increases. Long-term tensile strengthregression is not a correct basis (albeit conservative)for ring compression design. Long-term (virtual)modulus applies only to long-term deformation(creep) due to persistent load.

5. Under quick loading, the strengths and moduli ofelasticity are the initial, short-term values. Thisphenomenon is remarkable in that it is independentof previous stress history. For example, a pipesubjected to internal pressure over a long period oftime still maintains its initial quick-load strength andmodulus of elasticity. Consequently, water hammeranalysis is based on initial strength. Vacuumcollapse is analyzed by using initial modulus ofelasticity. See Chapter 10.

6. Because plastic pipes are flexible, they conformwith the soil. Pressure concentrations are relievedby distribution. See Figure 20-5. Not only areconcentrated stresses in the pipe relieved, but allstresses relax over time. The potential for soil slipmust be checked. See Chapter 10.

7. Arching action of the soil supports part of theload. See Figure 20-6. The soil acts as a masonryarch. No cement is needed because the pipeconfines the soil and retains the soil arch. The soilarch protects the pipe. But additional loads may becaused by surface vehicles, by differences betweenunit weights of pipe and soil, and by deflections of

the pipe — both longitudinal deflections and ringdeflections. The basic load on plastic pipes is thesoil prism load.

8. The flexible pipe is held in shape by the soil. Thestabilized ring itself becomes an arch that supportsvertical load. Without soil support, the ring wouldcollapse under relatively light load.

9. For plastic pipe under constant deformation (goodembedment), stresses in the pipe wall relax.Because the soil retains the ring in fixed shape, theplastic relaxes over time and relieves itself of someof the stresses in the pipe wall. Stresses relax at afaster rate than strengths regress under constantdeformation.

10. If the pipe is installed in a trench, embedment isa transfer medium completing the intimate fit of thepipe to the native soil. The value of the intimate fitis retention of the original pipe shape, and preventionof groundwater flow channels along the outside ofthe pipe.

11. In good embedment, ring deflection isapproximately equal to, or less than, the verticalstrain (compression) of the sidefill soil.

DESIGN OF BURIED PLASTIC PIPES

Successful performance of buried plastic pipes isbased on four performance limits:1. Circumferential strength (yield stress) — bothtensile (hoop) strength and ring compressionstrength,2. Deflection — both ring deflection and longitudinal(beam) deflection,3. Stability — as limited by incipient ring collapse, 4. Pipe-soil continuity — intimate contact betweenthe pipe and soil.

Circumferential Stress

Internal Pressure — For design, use Equation 20.1.If a pipe is subjected to sudden internal pressure,


Figure 20-5 Comparison of typical soil loads on a rigid ring and on a flexible ring, showing how the flexiblering deflects just enough to equalize the horizontal and vertical forces.


such as water hammer, the initial (short-term) tensilestrength may be used. For persistent internalpressure, the long-term tensile strength should beused. A safety factor of two is often used fordesign.

External Pressure — For design use Equation 20.2.If a plastic pipe is in good embedment, the initial(short-term) compressive strength may be used. Ifthe pipe does not fail during installation, it is not likelyto fail in the long-term. During installation, theperformance limit is wall crushing at 9:00 and 3:00o'clock when the ring compression stress reachesyield strength. See Figure 20-7. Compression yieldstrength is higher than tension yield. A safety factorof two based on tension yield is excessive, but oftenused. Tests are advisable for compression yieldstrength. Under high landfills, the effect of surfacelive loads on buried plastic pipes is negligible.

Ring deflection causes circumferential flexurals tresses which are also maximum at 9:00 and 3:00o'clock. Flexural stresses are compression on theinside of the pipe and tension on the outside. Wallcrushing can occur only when the pipe wall is atcompression yield stress throughout the entire wallthickness. Therefore, flexural stress does not affectwall crushing, but may affect stability and plastichinging. See Stability in the paragraphs to follow.

Example

A polyethylene pipe is buried under a landfill forwhich the vertical soil pressure on the pipe is 280psi. For this pipe, the dimension ratio is DR = 9.2.Ring deflection is not greater than 10% according toa "bullet" drawn through the pipe. What is thesafety factor for ring compression stress if the short-term yield strength of the polyethylene is 2300 psi.From Equation 20.2, the ring compression stress is1417 psi. The safety factor is sf = 2300/1417 = 1.6,which is adequate for short-term. The yield strengthis based on tension tests. Moreover, aftercompletion of the landfill, stresses relax if theembedment is good granular soil that holds constantthe cross-sectional shape of the pipe. Therefore the

safety factor, 1.6, is more than adequate for the longterm.

Deflection

Longitudinal (beam) deflection is usually of onlyminor concern. With careful placement of thebedding, the pipe does not deflect excessively as abeam. Pipe manufacturers specify a minimumlongitudinal radius of curvature, R. Excessivelongitudinal bending could cause a plastic hinge inthe beam. Under constant deformation over thelong term, longitudinal stresses relax.

Ring deflection, d = ∆/D , is of greater concern. SeeFigure 20-8. Excessive ring deflection can causeleaks at appurtenances and joints. It can contributeto instability of the ring. It can obstruct the passageof cleaning equipment and video cameras. Ringdeflection is usually limited by specification. Ringdeflection can be predicted by methods described inChapter 7.

Stability

Instability is incipient collapse of the ring. Incipientmeans that collapse is not inevitable, but that the ringis at its limit, and further resistance to collapsedepends upon soil arching action. Practically,enough stiffness is built into the flexible ring to holdit in shape during installation and loading. This pipestiffness contributes to the resistance of the pipe-soilsystem to collapse. See Chapters 10 and 11. Theappropriate pipe stiffness is the initial (short-term)pipe stiffness because collapse is a suddenphenomenon. Collapse of plastic pipes usuallyfollows the formation of plastic hinges at 9:00 and3:00 o'clock. Not only are plastic hinges a failuremechanism in the pipe, but the very short radius ofcurvature of the hinge results in such high horizontalsoil stresses at the spring lines that the soil may slip. Clearly the conditions for stability are assured ifgood granular soil is carefully compacted in the


Figure 20-7 Ring compression of a circular pipe loaded by soil pressure, P, showing the ring compressionstress distribution and the flexural (ring deflection) stress distribution across the wall.

Figure 20-8 Ring deflection, d = ∆D , for a typical plastic pipe cross section deflected into an ellipse due tothe typical horizontal and vertical soil pressures shown.


sidefills, and if the ring deflection is limited —usually to less than 10%. Under conditions wheremitigation is sought, the height of soil cover can berestricted.

Example

PVC piping is proposed for drainage under asanitary landfill to be 600 ft high. Unit weight of thelandfill is 75 pounds per cubic ft. Fifteen years areanticipated to complete the landfill. Service life forthe piping is to be at least 100 years.

a) What dimension ratio (DR) is required? DR isan inverse measure of the pipe stiffness. Assumethat the 15-year compressive yield strength of thePVC is 5000 psi. Actually it is greater than the 5000psi which is based on tensile yield. In good granularembedment, ring deflection is constant, and long-term stresses relax. Use a safety factor of sf = 1.5.From Equation 20.2, P = 312.5 psi, and ringcompression stress is, σ = 0.5 P(1+d)(DR)sf. For along-term sanitary landfill, it is prudent to hold ringdeflection to near zero by compaction of sidefill. Ifd 0, the equation above yields DR = 21.3. A goodselection is PVC pipe SDR 21(200) ASTM D 2241.SDR = OD/t = standard dimension ratio.

b) What is maximum allowable ring deflection if theembedment is loose (φ = 25°)? From the Uni-BellHandbook of PVC Pipe, published by the Uni-BellPVC Pipe Association, for SDR = 21, pipe stiffnessis F/∆ = 234 psi for E = 400,000 psi, and F/∆ = 292psi for E = 500,000 psi. Using 234 (to beconservative), EI/r3 = (F/∆) /6.72, and P/(EI/r3) = 9.Entering the graph, Figure 10-9, with φ = 25o, andP/(EI/r3) = 9, ring deflection at incipient collapse isabout d = 20%. Clearly, if ring deflection is held toless than d = 10%, the safety factor is greater thantwo.

Ring deflection can be controlled by the quality anddensity of the sidefill. From laboratory tests, selectcrushed stone compacted to 95% density AASHTOT-99 (70% relative density) will hold ring

deflection to less than d = 5% under 600 ft of soilcover at unit weight of 75 pcf.

Drain Pipe Caveats

1. The embedment must not dissolve, decompose, orplug up flow for 100 years. Limestone, for example,is dissolved by leachate in sanitary landfills, and canbe redeposited inside the pipe. Another example isloss of sidefill by migration of fine soil particles intothe drain pipe through the slots or perforations in thedrain pipe. Either the soil must be designed as afilter with well-graded soil particle size, orgeotextiles must be used to prevent soil particlemigration.

2. In the installation of corrugated high-densitypolyethylene pipes (HDPE), if the pipe is stretchedlongitudinally, pipe stiffness is reduced and couldresult in instability. Manufacturers caution thatincrease in length during installation of more than10% is to be avoided. Corrugated polyethylenepipes have the capability of short longitudinal radiusof curvature, R. The corrugated pipe can betransported on a spool. It can be installed downthrough a chute directly behind the excavatorbuckets of a trenching machine. The corrugatedpipe can bend around the short radius of a shoe, andinto the trench where it is immediately backfilled.But the pipe must not bind in the shoe and stretch.

Pipe-soil Continuity

Intimate contact between pipe and embedment helpsto assure position and alignment of the pipe. Ifgroundwater flow erodes channels along the pipe,intimate pipe-soil contact and soil density could belost. Soil contact retains the circular cross section ofthe pipe. By fixing the shape of the ring, stressesrelax over the long term.

Intimate soil contact is assured by using selectgranular embedment placed under the haunches of


Figure 20-9 Four shapes selected at random from many variations of plastic pipe shapes.


the pipe by shovel-slicing, J-barring, flushing, etc. Itis noteworthy that for structural integrity of pipeswith negligible ring deflection and significant ringstiffness, intimate soil contact may not be absolutelyessential. For example, in the case of plastic pipesof low DR in good embedment, the sidefill cansupport the soil load by arching action with orwithout contribution from the pipe.

The phenomenon is tantamount to boring a tunnelunder the landfill and inserting the pipe. The pipeonly has to support the talus that would slip againstit if the pipe could deflect slightly. Consequently, thestructural importance of intimate soil contact can bemitigated under some conditions. However, goodsidefill is essential to soil arching action.

Temperature

The properties of plastics are affected bytemperature. Pipe manufacturers can provide thenecessary design data. For example, if PVC pipe isto be used at a temperature of 120oF, it may beprudent to assume a modulus of elasticity of E = 400ksi rather than E = 500 ksi, and yield strength of 4ksi rather than 5 ksi. Decomposition of the biomassin sanitary landfills generates heat. Temperatures of120oF have been reported, but are not common.Similar adjustments apply to other plastics wheretemperature is of concern.

Example

Two samples of high-density polyethylene pipes(HDPE-SDR11), were heated to temperatures over140oF when buried under 12.7 ksf of vertical soilpressure. Under this pressure, ring deflection was7% and 9% for the two tests. The embedment wascompacted fine sand. There was no significantchange in ring deflection due to the increase intemperature. The ring deflection of one pipeincreased a fraction of 1% and the other decreaseda similar amount.

PLASTIC PIPE SPECIALTIES

A wide variety of special plastic pipes are becomingavailable for underground use. A random sample offour of them is shown in Figure 20-9. For design,each special pipe may require modifications of thedesign principles.

Example

Figure 20-9(a) shows an edge drain. Edge drainsare narrow, cylindrical, pipes that are placedvertically along the edges of paved surface slabs(such as highways and airport runways) to interceptand remove water that would otherwise saturate thesubbase soil and reduce the strength of the soil thatsupports the slab. In this case, the edge drain is 12inches high with 2 inches of soil cover. It is locatedalong the edge of a highway slab on which theheaviest load is an HS-20 dual-wheel load. What isthe test force for which this edge drain must bedesigned?

Figure 20-10(a) shows the worst probable loading.For worst-case analysis, it is assumed that the soil isloose enough that the dual-wheel load punches out atruncated pyramid as shown in Figure 20-10(b).Vertical soil pressure as a function of depth is,

Py = 2W/[2L+y)(B+y)] . . . . . (20.4)

where y = depth from soil surface,Py = vertical soil pressure at depth y,W = 16 kip load on rectangular tire print,L = width of the dual tire print,B = breath of dual tire print.

Equation 20.4 is based on the assumption that thesoil is granular and that the slopes of the pyramid areabout 2v:1h. At any depth, y, the horizontal (active)soil pressure on the edge drain is,


Figure 20-10 Worst-case load on an edge drain installed along a highway slab.

Figure 20-11 Pressure distribution and the resultant forces acting on an edge drain.


Px = Py /K . . . . . (20.5)

whereK = (1+sinφ)/(1-sinφ) = 3 if,φ = soil friction angle = 30o (assumed for this

soil),Px = horizontal pressure on the edge drain,Q = resultant horizontal force on the edge

drain assuming that the pressuredistribution is trapezoidal.

The Px diagram is shown dotted on Figure 20-11(left). If the coefficient of friction of soil on edgedrain is assumed to be φ' = 30o, the vertical shearingforce on the edge drain is V = Qtanφ'. Theresultant of Q and V is F as shown in Figure 20-11(right). A reasonable test for strength of the edgedrain would be the application of load F on a sampleof edge drain "sandwiched" between two parallelblocks with sandpaper surfaces. F is inclined atangle φ' and is located a distance, e, from the centerof the edge drain, such that e locates the centroid ofthe trapezoidal pressure diagram. The maximumtest force F on the edge drain must be greater thanthe HS-20 force F for which the edge drain isdesigned.

PROBLEMS

20-1 The measured pipe stiffness of an 8Dcorrugated polyethylene pipe is F/∆ = 70 psi. Thepipe is to be used under a sanitary landfill of unitweight γ = 75 lb per cubic ft. It is to be installedand backfilled below the water table. Therefore, foran instant, it may be subjected to hydrostaticexternal pressure of saturated sand at 120 lb percubic ft. What is the depth of liquefied sand

backfill at which the pipe will collapse? Assume forworst case that the pipe is empty. (37.5 ft)

20-2 A 6D HDPE pipe SDR9 is to be installedunder a river by directional boring. The pipe will be80 feet below river water surface. The river banksare 10 feet above the river water surface. Thebored hole is retained up to bank level by drillersmud (liquid) with unit weight of 80 pcf. The pipe isfilled with water and pulled through the bored hole inthe drillers mud. Then the drillers mud is displacedby grout that has a unit weight of 95 lb per cubic ft.After the grout sets, the water in the pipe is pumpedout leaving hydrostatic pressure of the river on theencased pipe through cracks in the grout. If the pipestiffness is F/∆ = 600 psi, what is the safety factoragainst collapse?

a) during installation?

b) during grouting?

c) due to external pressure of river water?

20-3 What is the ratio of longitudinal radius ofcurvature, R, to pipe radius, r, for corrugated HDPEpipe if maximum allowable elongation andcontraction are each limited to 10%? (R/r = 10)

20-4 From parallel block tests on the 12-inch edgedrain of the example above, test strength is F = 395lb/inch. What is the safety factor against failure ifthe edge drain is to be placed under 2 inches of soilcover with HS-20 load, W = 16 kips on a 22 x 7 inchtire print, as shown in Figure 20-10(a)? Frictionangles for the soil, φ, and for soil on the edge drain,φ', are both 30o. Assume trapezoidal pressurediagram as shown in Figure 20-11(a).

(F = 240 lb/inch; sf = 1.65)


Anderson, Loren Runar et al "EXTERNAL HYDROSTATICS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 21-1 Free-body-diagram of a pipe immersed in water, showing uplift (buoyant) force, Wb = πr2γ w,which causes the pipe to float if it is greater than the weight, W, of pipe and contents.


CHAPTER 21 EXTERNAL HYDROSTATICS

The analysis and design of buried pipes iscomplicated by external hydrostatic pressures.Chapter 10 describes the conditions for instabilityand collapse of pipes subjected to uniform, externalpressure. In fact, hydrostatic pressure is notuniform. Moreover, instability is only one of theproblems that arise from hydrostatic pressures(which include internal vacuum).

Due to external hydrostatic pressure, the pipe maycollapse, or crush, or float, or deform. Each of theseperformance limits requires its own analysis. Ringc ompression stress is based on total externalpressure including external hydrostatic pressure.Ring deflection is based on effective soil stress,which is total external stress minus the externalhydrostatic pressure. In Chapter 9 the effect ofradius of curvature of non-circular pipes isdiscussed. Clearly, external hydrostatic pressure onthe bottom plates of a pipe arch, with its long radiusof curvature, can invert the bottom plates. Externalhydrostatic pressure can cause a pipeline to float outof alignment.

Analysis and remedies for the problems of externalhydrostatic pressure are described in the following.

FLOTATION OF PIPELINES

What causes a pipe to float up out of verticalalignment, and what can be done to prevent it?

The cause of flotation is an uplift force (buoyancy),Wb, which is the weight, per unit length of pipe, ofthe displaced liquid. Figure 21-1 shows a pipeimmersed in water. If the buoyant force exceedsthe weight of the pipe and its contents, W, the pipewill float upward. For worst-case, the weight of thecontents is neglected if at any time the pipe isempty. Per unit length of pipe,

W = Wp + Wc . . . . (21.1)

NotationWb = buoyant (uplift) force on the pipe,W = weight of pipe and contents,Wp = weight of the pipe per unit length,Wc = weight of contents per unit length,Ws = buoyant weight of soil wedges above a

buried pipe per unit length of pipe,OD = outside pipe diameter,ID = inside pipe diameter,D = mean diameter,γ b = buoyant unit weight of the soil,γ w = unit weight of water, or of any other liquid

in which the pipe is immersed,sf = safety factor.

Per unit length of pipe, the buoyant force (upliftforce), Wb, is the weight of liquid displaced. Theliquid could be mud or grout.

Wb = π(OD)2γ w /4 . . . . (21.2)

For static equilibrium, the buoyant force must notexceed pipe weight reduced by a safety factor,

Wb = W/sf . . . . (21.3)

Design of a pipe immersed in water is simply amatter of substituting values into Equation 21.3.Performance limit is flotation of the pipe. Typicalremedies include weighting the pipe down withheavy collars, or thick walls, and holding the pipedown with straps and anchors.

Typically the immersed pipe is buried in soil beneatha body of water. If the density of the soil exceedscritical density, failure is the formation of soilwedges uplifted as shown in Figure 21-2. Criticaldensity occurs at that void ratio, greater than whichthe soil is so loose that it compresses whendisturbed, and less than which the soil is so densethat it expands when disturbed. Critical void ratio isroughly 85% density AASHTO T-99 (ASTM 698).The specification for minimum


allowable density is usually set at 90% density.Granular soil looser than critical does not break outwedges, but flows around the pipe as liquid or plasticthrough which the buoyant pipe rises. Thisphenomenon can be investigated by approximatetheory, but there are too many variables for aprecise model. A simplified theoretical analysisfollows. It is confirmed approximately byexperimentation.

Figure 21-2 is a free-body-diagram of the soilwedges. The soil wedges form a trapezoid of widthD at the bottom, height Z = H + D/2, and with soilslip planes sloped at (45o + ϕ /2) 2v:1h; butreduced by a semi-circle of diameter D. Thebuoyant weight of the soil wedges is,

Ws = γ b[Z(D + 0.5Z) - πD2/8]/sf . . . . . . . . (21.4)

In this equation, the slip slope, (45o + ϕ /2), is set attwo to one for approximate analysis of granularsoils. See Chapter 13.

In order to investigate the worst case, let Wc = 0 foran empty pipe, and, assuming the pipe to be verylight compared to the weight of the soil wedges, letWp = 0. If the weight of the pipe were significant,such as concrete pipe, its weight could be included.From the equations of static equilibrium:

Wb = Ws /sf . . . . (21.5)

The following is a simplified, worst-case example.

Example

How much cover H of compacted granular backfillis required over a buried pipe under water to preventflotation? Assume an empty pipe of negligibleweight; i.e., ID = D = OD; and with soil shear planeslopes at 2:1. Substituting values of Wb and Ws intoEquation 21.5,

γ wπD2/4 = γ b[Z(D + 0.5Z) - πD2/8]/sf,

where γ b can be found from soil mechanics, Chapter

4. For this example, assume γ b = γ w. Because Z =H = D/2, the relationship between H and sf can befound with results as follow:

sf = 1 2 3H = 0.33D 0.72D 1.05D

If control of the backfill is questionable, a veryconservative rule of thumb might be to specify soilcover equal to the pipe diameter. Considering theweight of the pipe and other mitigating features, theheight of cover should be at least half the pipediameter. But the soil must be compacted to densitygreater than critical — especially if the specifiedheight of cover is only H = D/2. Physical testsconfirm the critical D/2.

It was observed serendipitously, that in loose sand,a shock wave liquefies the sand (quick condition),and the pipe starts to float upward. This occurseven in compacted sand if, adjacent to the pipe, thesand is less dense than critical. Earth tremors(earthquakes) or shock waves liquefy the sand. Thepipe starts to float up. But as soon as the shockwave passes, soil particles in suspension settle backin place and stop the rise of the pipe. With repeatedshock waves, the pipe rises little by little until itbreaks free and floats to the surface. A very severeshock can conceivably shake up dense saturatedsand into a looser (quick) state by major soilupheaval — but only if enough time is allowed forwater to seep into the soil during the upheaval. Remedies include the use of gravel embedmentwhich has less tendency to liquefy when shockwaves pass through it. Gravel is attractive underwater where it is difficult to compact soil. Gravelfalls into place at density greater than critical. Fromexperience, even moderate shock waves cannotdislodge the pipe and cause it to float if the minimumsoil cover is equal to one pipe diameter and if the soilis well compacted. But how is compaction achievedunder water? If the soil is granular, it can bevibrated or jetted. But the surest technique is toplace select gravel or sand that is coarse enoughthat it achieves adequate density just by moving itinto place.


Flotation due to liquefaction is further reduced by theangle of approach of the shock wave. The wavefront rarely hits the entire pipeline at the sameinstant; i.e., at an angle of approach of 90o. At anyother angle, liquefaction occurs only at theintersection of the pipe and wave front. The lengthof the liquefied zone is usually short enough thatlongitudinal beam action prevents flotation.

STIFFENER RINGS

Stiffener rings on flexible pipes increase greatly theresistance to collapse by external hydrostaticpressure (and internal vacuum). Theoreticalanalysis is complex. However, empirical andexperimental data are becoming available. Limitedresearch has investigated the spacing of stiffenerrings on thin-wall steel pipes with very heavystiffener rings (infinitely stiff), immersed in water,with no soil support. The most pertinentfundamental variables are:

P' = vacuum at collapse,E = modulus of elasticity,D = diameter of the pipe,t = wall thickness,L = spacing of the rings.

From these five fundamental variables, all in termsof two basic dimensions (length and force), threedimensionless pi-terms can be written. Theinterrelationship of these pi-terms can beinvestigated either by analysis or by experimentation.The equation of one set of three pi-terms (inparentheses) can be written in the form,

(E/P') = f [(D/t), (L/D)]

Because the pi-terms are dimensionless, they haveno feel for size; so testing can be performed onsmall-scale physical models.

Example 1

Based on small-scale model study, what are theequations for collapse of thin-wall steel pipes withinfinitely stiff rings, when immersed in water (no soilsupport). See Figure 21-3

Beverage cans provided cylinders of uniform wallthickness, with correct similitude between thealuminum model and a steel prototype for which D/t= 492. The ends of the cans were cut out, and steelplugs, machined and gasketed, were inserted in theends to provide a seal for the vacuum, and tosimulate infinitely stiff rings.

The results of the experiment are shown in Figure21-4. The best fit curves through data points arelinear with their equations shown for correspondingranges on the graphs for D/t = 492. It would beprudent to include a safety factor that would mitigatethe fact that stiffener rings are not infinitely stiff, andthe fact that the pipes are not always as nearlycircular and cylindrical as the beverage cans.Performance limit in every case was collapse of themodel pipe by crumpling between the stiffener rings.

The general equation for collapse is:

106(P'/E) = (D/L)/m . . . . (21.6)

where m is the slope of the best-fit straight line.From the plots of data in Figure 21-4, collapseequations for stiffened steel pipes with D/t = 492,and E = 30(106) psi, are:

L/D Vacuum Equation0 to 5/3 P'= 14.3 D/L psi (21.7)5/3 to 10/3 P'= 8.6 D/L psi (21.8)over 10/3 P' = 0.504 psi (21.9)

These equations are conservative, especially towardthe left side of each equation range.


Figure 21-3 Flexible thin-wall pipe of diameter D with stiffener rings spaced at lengths L.

Figure 21-4 Plots of test data for critical vacuum term as a function of stiffener ring spacing term, showingbest-fit straight lines and their equations in three ranges. For these tests on steel pipes, D/t is a constant, 492.


In the range 0 < L/D < 5/3, Equation 21.7 lower lineis appropriate. In the range 5/3 < L/D < 10/3,Equation 21.8 upper line is more appropriate, but stillconservative. Equation 21.9 is the theoreticalsolution for an infinitely long pipe without stiffenerrings, i.e., P'D3/EI = 24. See Chapter 10.

Until more data become available, for steel pipes, itis reasonable to adjust the critical vacuums ofEquations 20.7, 20.8, and 20.9 (based on D/t = 492)to other values of D/t by a factor which is the cubeof the D/t ratio; i.e.,

P'/ P'492 = [492/(D/t)]3 . . . . (21.1)

For ring-stiffened steel pipes with D/t less than 240,this correction factor becomes so large thataccuracy is lost. For D/t = 240, the correctionfactor becomes P'/P'492 = 8.6. More accurate is thetank analysis, Equation 22.3, assuming that the endclosures of tanks are equivalent to stiffeners.Vacuum at collapse of tanks is,P' = 0.403E(1-ν2)-3/4(D/L))(t/r)5/2.

Example 2Stiffener rings are spaced at 20 ft on a 144-inch-diameter thin-wall steel pipe, of 0.375-inch-thicksteel, for which L/D = 20/12 and D/t = 384. Whatis the critical vacuum? Assuming excellent controlover imperfections, use Equation 21.7 from whichthe critical vacuum for D/t = 492 is,P' = 14.3(D/L) = 8.6 psi.

With the adjustment factor of the ratio of (D/t)3

from Equation 21.10, and with D/t = 384 for the 144-inch steel pipe,P' = 8.6(492/384)3 = 18 psi

which is higher than atmospheric, 14.7 psi.However, if external water pressure plus internalvacuum combine to a pressure greater thanatmospheric, and/or if control is less than excellent,then using the more conservative Equation 21.8, forD/L = 12/20 and with the adjustment factor ofEquation 21-10,P' = 8.3(D/L)(492/384)3 = 10.5 psi,

which is less than atmospheric pressure, but may betoo conservative. Based on the more accurate tankEquation 22.3,P' = 0.403E(1-ν2)-3/4(D/L)(t/r)5/2 = 15 psi.

Experimentation can start with two pi-terms, (P/E)= f(Lr/t2). Then flexibility of stiffeners (Pr3/EI) canbe included, then soil support (ϕ) etc. The exampleabove is unburied. When buried, the resistance toexternal hydrostatic pressure (plus vacuum)increases greatly depending upon compaction of thesoil. See Chapter 10.

Hydrostatic pressure is not limited to waterpressure. If concrete with unit weight of 145 pcf is"poured" about the 144-inch pipe up to the top, theexternal hydrostatic pressure on the bottom is 12 psi.The 20-ft spacing still appears adequate, but agreater margin of safety can be achieved if theconcrete is placed in two lifts of 6 ft each ratherthan one lift of 12 ft.

It is noteworthy that theoretical collapse vacuum fora long uns tiffened steel pipe of D/t = 492 (D = 144inches, t = 0.293 inch) from Equation 21.9, is P' =0.49 psi. With stiffener rings spaced at 20 ft on a144-inch pipe for which D/t = 492, from Equation21.8,P' = 8.3(D/L) = 5.0 psi.

Stiffeners provide more than ten times the resistanceto vacuum collapse than the plain pipe.

PROBLEMS

21-1 A two meter thin-wall plain steel pipe with awall thickness of t = 10 mm, is buried in com-pactedfine sand with a soil cover of 1 meter and watertable at the soil surface on occasions. The pipecould be empty. Is there any possibility of flotation?Under what conditions? What is the possibility ofsoil liquefaction? Does the probability of flotationincrease if the external water level is at flood stageabove ground surface? G = 2.65 = specific gravity,e = 0.4 = void ratio,


c = 0 = soil cohesion,ϕ = 30o = friction angle. (sf = 1.78)

21-2 What spacing of stiffener rings is advisable ifthe pipe of Problem 21-1 is to take a test vacuum of14.7 psi? (L = 0.7D)

21-3 Find the wall section modulus for Problem 21-1if stiffener rings are welded at 3 meter

spacing. Rings are 120 mm by 25 mm rectangularcross section with long axis perpendicular to the pipewall.E = 207 GPa = modulus of elasticity,S = 248 MPa = yield. (I/c = 3850 mm3)

21-4 What external hydrostatic pressure plus 14.7psi vacuum can the pipe of Problem 21-3 take? (P = 660 kPa)


Anderson, Loren Runar et al "BURIED TANKS AND SILOS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

CHAPTER 22 BURIED TANKS AND SILOS

A tank is generally either a circular cylinder withend closures, or a sphere. See Figure 22-1. Tanksare buried with the axis horizontal. Silos (and basinsand caissons) are buried with the axis vertical. Silosmay or may not have closures at the top. They areused as access ways, mine shafts, and bins orhoppers for storage of ore, coal, aggregate, etc.Bins have an outlet at the bottom for feeding aconveyor in a pipe or tunnel below. Tanks are usedto store fluids, but are also used as bunkers forhazardous materials.

ANALYSIS OF BURIED TANKS

Tanks are designed either for internal pressure orfor external loads. For worst-case internal pressure,external soil support is ignored. For worst-caseexternal pressure, it is usually assumed that the tankis empty, buried in poor soil with groundwater tableabove the tank and possibly a vacuum in the tank.The performance limit is collapse. An accuratemodel would be complicated. Even if a model weredevised, it could not be generalized, and it mightimply greater precision than can be justified. Soilproperties are major elements of the interaction, butcannot be quantified precisely under normalinstallation procedures. Boundary conditions areseldom precise. Therefore, basic principles ofengineering mechanics provide adequate analysis.Safety factors are required in any case. A free-body-diagram and principles of static equilibriumhelp in visualizing tank-soil interaction and indispelling misconceptions, some of which arise fromforensic arguments on assessment of responsibilityfor leaks in buried tanks. Leaks are analyzed inChapter 24. See nomenclature, Figure 22-2.

Maximum Internal Pressure

For worst-case analysis of internal pressure, it iscustomary to neglect support of the embedment.Analysis is based on principal stresses which are

critical on the inside surface of the tank. See theinfinitesimal cube in Figure 22-3. Because P' issmall compared with circumferential and longitudinalstresses, it is usually neglected. The maximumprincipal stress is the circumferential stress. For athick-wall tank (D/t < 10), stress analyses are foundin Chapter 19. For a thin-wall tank (D/t > 10), themaximum principal stress is the circumferentialstress. The following applies to thin-wall tanks.

Spheres

From mechanics of solids, hoop stress due to internalpressure is only half as great in a sphere as in acylinder of the same diameter and wall thickness.Therefore, for a sphere,

σ = P'(ID)/4A . . . . . (22.1)

whereσ = hoop stress,P' = internal pressure,ID = inside diameter,t = wall thickness,A = wall cross sectional area per unit length,A = t for plain cylinder wall.

Cylinders (Shells)

Cylindrical tanks are short pipes, for which thecritical circumferential stress is,

σ = P'(ID)/2A . . . . . (22.2)

Longitudinal stress is only half as great ascircumferential stress. End closures are stiffenerswhich help to resist internal pressure. But wheninflated, end closures cause stress concentrations inthe shell-to-head seam.

Equation 22.2 is reasonably accurate for design, butsafety factors are essential. End closures (heads)are analyzed separately.


Figure 22-1 Some typical buried tanks and silos.


Figure 22-2 Nomenclature for steel tanks (above) and buried tanks (below).


Maximum External Pressure

The following are worst-case analyses whichinclude burial in saturated soil. A major concern isflotation. In the following, it is assumed that the soilcover is sufficient to prevent flotation. FromChapter 21, soil cover should be at least half thetank diameter. The interaction of tank, soil, andgroundwater is complex. Strength of soil isdecreased under water and loads must includeexternal water pressures. The tank itself is complexbecause of the interaction of shell, heads, risers, andwelds. See Figure 22-2. Codes and standards areavailable from ASME (boiler code), UnderwritersLaboratory, and ASTM (tank standards).Recommendations are available from industries(waterworks, gas, and petroleum). Codes weredeveloped originally for tanks subjected to uniforminternal pressure. For tanks subjected to externalpressure, these codes only cover part of therequirements for performance.

Notation

Geometry:D = diameter of the circular shell,H = height of soil cover,h = elevation of water table,L = length of the tank (or height of silo),r = D/2 = radius of the circular cylinder,t = wall thickness,A = cross sectional wall area per unit length,R = radii of curvature of the head in a plane

that contains the axis of revolution.

Properties of Materials:E = modulus of elasticity of tank material,υ = Poisson ratio,ϕ = soil friction angle,S = yield stress.

Forces, Pressures, and Stresses:P = external pressure or internal vacuum,γ = unit weight of soil,σ = stress in the tank (Figure 22-3),Subscripts refer to directions.

Spheres

Collapse of spheres due to uniform external pressureis analyzed by marine engineers for bathyspheres.When soil support is included, analysis iscomplicated. Not only are circumferential stressesin a sphere half as great as in a cylinder, but thevertical soil pressure is less because two-way soilarching action (soil dome) is twice as effective as acylindrical soil arch. In the design of buried pipes,the benefit of soil arching action is usually neglected,but is a significant plus for conservative analysis. Inthe design of buried spheres, it may be prudent totest or evaluate the arching action of the soil dome.Little information is available on buried spheres.

Cylinders (Shells)

A theoretical analysis is available for uniformexternal pressure at collapse of cylindrical tankswith no soil support. In fact, soil provides support.Moreover, external water pressure is not uniform,but increases from top to bottom of the tank. R.Allan Reese (1993) investigated hydrostaticpressures on the bottoms of horizontal steel tanks atcollapse as the tanks were lowered in water. Heconcluded that it is sufficiently accurate to designtanks by assuming uniform external pressureaccording to an equation from Young (1989),

P = 0.807E(1-υ 2)-3/4(r/L)(t/r)5/2 . . . . . . . . . . (22.3)

where, in examples that follow:P = water pressure on the bottom of the tank at

collapse (sudden inversion),E = mod/elast = 30(106) psi for steel,t = wall thickness of plain wall,L = length of the tank,D = diameter of the tank,r = radius of of the shell,υ = Poisson ratio = 1/4 for steel.

It is better to use the pi-term (r/t) than the common(D/t) because radius includes out-of-roundness. At


Figure 22-3 Principal stresses on the inside of a tank wall sujected to internal pressure P'. The same stressanalysis applies for negative internal pressure (vacuum), with reversed signs.

Figure 22-4 Uniform external pressure on steel tanks at collapse — graphs of Equation 22.4.


some location on the shell, the radius may be greaterthan D/2. Designers use Equation 22.3 in thefollowing form for steel tanks buried in saturatedsoil:

P = 72(106)psi(D/L)(t/D)5/2 . . . . . (22.4)

Poisson ratio for steel is usually about ν = 0.27.Some designers use ν = 0.3 in Equation 22.4. Thedifferences are not significant; i.e.,

If ν = 0.25, the coefficient is 71.87,If ν = 0.27, the coefficient is 72.52,If ν = 0.30, the coefficient is 73.54.

If Poisson ratio is increased from 0.25 to 0.30, Pincreases by only 2.3%. Poisson ratio is oftenneglected. Without heads to support the shell, frompipe theory, at D/t = 575, P = 2E/(1-ν2)(D/t)3.

Example 1

What is the external pressure on the bottom of anempty 12,000-gallon steel tank at collapse if the tankis lowered in water until it collapses? Diameter D =96 inches, wall thickness t = 0.167, length L = 32 ft,D/t = 575, and L/D = 4.

Substituting into Equation 22.4, collapse pressure isP = 2.27 psi, which is equal to a depth of water of5.25 ft above the bottom of the tank. This is lessthan the diameter of the tank. Equations 22.3 and22.4 are not applicable. If wall thickness isincreased to 0.2391 inch, P = 5.56 psi. At collapse,the water surface is 4.83 ft above the tank.

Figure 22-4 shows graphs of Equation 22.4.Noteworthy are the effects of wall thickness andlength of tank on collapse pressure. For comparison,the bottom graph is pressure at collapse of pipes (noheads or ring stiffeners). Even though Equations22.3 and 22.4 are conservative, safety factors arerecommended because tanks are never perfectlycircular.

Heads (End Closures)

Heads are usually analyzed separately — notcompound head-shell analysis. The load is externalhydrostatic pressure plus any internal vacuum in thetank. If the water table is above the tank, analysisis sufficiently accurate if the pressure is assumed tobe the average pressure distributed uniformly overthe head. The shapes of heads vary fromhemisphere to dish (concave in or concave out) toflat heads (with or without stiffeners).

Hemispherical Heads

Hemispherical heads are easily analyzed by classicalmethods (Timoshenko, 1956). Circum-ferentialstress is half as great in a hemisphere as in acylinder of equal radius and wall thickness. Whenpressurized, the change in radius is not the same forhead and shell. For a cylinder, change in radius is(P/E)(r/t)(1-υ /2). For a sphere, change in radius is(P/2E)(r/t)(1-υ ). For equal values of P, E, υ , and r/t,the increase in radius is greater for the shell than forthe hemispherical head. If the head fits inside theshell, internal pressure tends to open a gap betweenshell and head. To avoid this, for many buried tanks,the head is a cap that fits on the outside of the shell.But, then, the possibility that the head might sheardown past the shell should be investigated.

Consider the case of uniform external pressure on asphere and cylinder of equal radii. Poisson ratio is1/4. If the hemisphere has the same thickness asthe cylinder, the tendency to decrease in radius is3/7ths as great as the cylinder. Stress between shelland head is less if the head is half as thick as theshell. See Chapter 24. This is not a priority for mosttank fabricators. Discontinuities in stress and strainat the head-to-shell joint are accommodated by agood fit and a good weld.

Some of the incompatibilities of shell andhemispherical head can be reduced by ellipsoidalheads or composite surfaces of revolution (dishes).


P = 0.34 psi.

For analysis, an infinitesimal element is isolated bytwo meridian cuts and two parallel cuts as shown inFigure 22-5. This element is a free-body-diagram onwhich stresses are related as follows:

σt /rt + σ m /rm = P/t . . . . . (22.5)

where:σt = tangential stress in direction of the parallel,σm = stress in direction of the meridian,rt = radius of curvature of the parallel,rm = radius of curvature of the meridian,P = external pressure on the head,t = thickness of the head,S = yield stress.

If the head is a hemisphere, from Equation 22.5,rt = rm, and P = 2St/r.

Example 2

Figure 22-6 shows a steel tank with the headattached to the shell by a reduced meridional radiusof curvature, rm. All of the steel is the samethickness. When ring compression stress in the shellreaches yield, what are the meridional and tangentialstresses at point A in the head?Given:t = 0.1875 inch,rt = 36 inches = D/2 = radius of the shell,rm = 12 inches,S = 36 ksi = yield stress.

At yield stress in the shell, external pressure is P =St/rt = 187.5 psi. Substituting values into Equation22.5, P/t = 1 ksi/inch, and

σt /36 + σ m /12 = 1000lb/in3 . . . . (22.6) Equation 22.6 contains two unknowns. A secondequation comes from a free-body-diagram of thehead isolated by a cutting plane of a parallel throughpoint A, Figure 22-6. Equating horizontal forces onthe head,

Pπr2 = σm 2πrt . . . . . (22.7)

from which σm = Pr/2t = 18 ksi compression.Substituting into Equation 22.6, the tangential stress

in the head is σt = -18 ksi tension. Stress at A inthe head is of concern because of the opposite sign.Shearing stress becomes critical. See Chapter 19.

Flat Heads

If the performance limit of a pressurized head isyield stress, analysis can be based either on platetheory or membrane theory, whichever gives thehigher stress. However, if the performance limit isdeformation (rupture), the membrane theory is moreresponsive. At tensile yield stress in a membranethe entire thickess is at yield stress. Rupture isincipient. On the other hand, at flexural yield stressin a plate, only one surface reaches yield stress —not the entire thickness. Rupture is not incipient.Surface yielding allows the disk to deform, tendingtoward a membrane with more uniform stressthroughout its thickness. Very thick disks (D/t<10),like thick-walled cylinders, justify plate analysis.

Membrane Analysis of Flat Heads

According to membrane theory, uniform pressure onone side deflects the disk into a segment of a spherewith radius R. See Figure 22-7.

Let:E = modulus of elasticity,r = radius of the disk,R = radius of the segment of a sphere formed by

pressurizing the disk,t = thickness of the disk,δ = deflection of the center of the disk,ε = tangential strain in the spherical membrane,σ = tensile stress in the membrane,θ = half angle of a meridional arc of a sphere,υ = Poisson ratio.

From geometry, the strain due to increase in lengthof r from a cord to an arc is:


Figure 22-5 Infinitesimal element of a surface of revolution subjected to external pressure, showing themeridional and tangential stresses and the corresponding radii of curvature.


Figure 22-6 Steel tank head with two different radii of curvature.

Figure 22-7 Geometry for analysis of stress and deflection in a thin-wall flat head (or bottom of a silo)deflected into a segment of a sphere by external pressure P. Stress on the welded knuckle is exacerbatedat the bottom where the deflection of head and shell both add to the flexure.


ε = Rθ/r - 1 . . . . . (22.8)

From elastic stress-strain relationship,ε = σ (1-υ )/E. For a sphere, σ = PR/2t. Substituting,

ε = PR(1-υ )/2Et . . . . . (22.9)

Equating the strains of Equations 22.8 and 22.9, andsolving for R,

R[2Etθ - Pr(1-υ )] = 2Etr . . . . . (22.10)

Because θ = sin-1(r/R), there are only two unknownsin Equation 22.10, R and P. The closed formsolution is messy, so solve for R by iteration.

From the elastic theory of thin-walled pressurevessels, at yield stress, σ = S, and,

P = 2tS/R . . . . . (22.11)

which can be solved if R is known. But fromEquation 22.9, R is a function of P. Therefore, asecond iteration is required to solve Equation 22.10for P.

Example 3

What is the stress in the flat heads of the steel tankof Example 1? Flat steel heads are usually thinenough to be analyzed as membranes, which deforminto segments of spheres when subjected to uniformpressure. By membrane analysis, the tensile stressis σ = PR/2t, where R is the radius of the sphericalsegment. In Example 1, for an assumed averagepressure on the head (conservatively high), of P =2.27 psi, R = 48/sin θ , where θ can be found byequating strains in the sphere; i.e.,

R(θ-sinθ)/r = PR(1-υ )/2Et,which reduces to

(θ-sinθ) = 3.59(10-6)P/psi.

Solving, θ = 2.0958o, R = 1312.5 inches, and tensilestress in the head is σ = 8.92 ksi. If yield stress is 36ksi, stress in the head is not critical. A typical rangeof length to diameter for buried 12,000-gallon steel

gasoline tanks is 1.76 < (L/D) < 5. Within thisrange, collapse due to uniform external hydrostaticpressure (plus internal vacuum) starts in the shell.Because heads are membranes in tension, they donot initiate collapse.

The head-to-shell weld must be of good qualitybecause the weld tends to flex as the head deforms.This is the worst at the bottom of the tank where theshell is sucked up and adds to the angle of flexure inthe weld and knuckle. See Figure 22-7. Considerthe tank of Example 1. Under pressure, the radiusr decreases by 1.2 thousandths for the head and 0.8for the shell. The relative difference of 0.4thousandths of an inch is negligible. The differencebetween uniform pressure and trapezoidal pressureon the head is also negligible. It is comparable totwo identical beams, one with uniform loading andthe other with triangular loading. The difference inmaximum stress is only 2.6%. Maximum deflectionsdiffer by only 0.15%. When water surface is abovethe tank, the pressure on the head can be analyzedas the average pressure uniformly distributed overthe head.

Plate Analysis of Flat Heads

Thick-wall heads (D/t<10) may justify analysis byplate theory. The rim of the disk is either fixed orsimply supported; i.e., with a gasket between theplate and the end of the shell. The fixed-edge analysis is most representative oftypical tanks with good connections between headand shell. Assume uniform average pressure is Pa v

= P. After Timoshenko (1956), the maximum stressby theory of elasticity occurs at the edge of the diskand is:

σ = (3/4)(r/t)2P . . . . . (22.12)

where equivalent thickness, t, is based on atransformed section. The maximum deflection is inthe center of the disk and is:


δ = (3/16)(P/E)(r/t)3(1-υ )r . . . . . (22.13)

The simply supported case is usually of noimportance. However, for comparison, in the simplysupported disk with uniform pressure, P, themaximum stress occurs in the center and is:

σ = (3/8)P(r/t)2(3+υ ) . . . . . (22.14)

If Poisson ratio is 1/4, the maximum stress in thesimply supported disk is 1.625 times as great as themaximum stress in the fixed-edge disk.

The maximum deflection is approximately four timesas great for the simply supported disk as for thefixed-edge disk.

Ribbed Flat Heads

Flat heads that are stiffened by ribs can be analyzedas plate disks with an equivalent wall thickness. Forthe equivalent disk, moment of inertia is t3/12 perunit width of the head. This can be equated to themoment of inertia of the cross section of the ribbedhead, and resolved for the equivalent thickness, t.Maximum stress occurs at the furthest surface fromthe neutral surface.

TANKS BURIED IN SATURATED SOIL

Burial of a tank subjects it to a broad range of soiland water loads and soil reactions. Indeed, if loadsinclude vehicles, or internal vacuum, or externalwater pressure, the soil becomes a support(reaction) rather than a load, and may be a moreimportant structural element than the tank itself.Decades of experience with millions of buried tanksprove the success of tank-soil interactions.Performance of tanks in good dry backfill ispredictable from both field experience and basicprinciples of mechanics. But if conditions areadverse, what are the performance limits? Whatare the conditions for collapse, and what can bedone to modify, or sometimes mitigate, poor

installation or poor saturated soil? Rigid Tanks

Rigid tanks buried below the water table can beanalyzed, conservatively, by neglecting support bythe embedment. The tank does not deflect withoutcracking. Therefore it does not rely upon the soil forsupport. For worst case, the analyses for uniformexternal pressure described above are adequate.For design of the heads, the plate theory applies withone caveat. If the head is reinforced concrete, themoment of inertia must be calculated for atransformed section as explained in texts onreinforced concrete design.

Flexible Tanks

Flexible buried tanks interact with the soil. Soilinsulates the tank from surface live loads. Soil is amajor structural component. The tank is a leak-proof liner that retains a shaped volume in the soil.Soil holds the shell in its circular shape and stabilizesit in its strongest structural configuration. The soil isnot so much a load to be resisted as it is a reactionto loads: traffic, vacuum, water table, etc. Evenpoor soil contributes some support. A commonmisconception is that, because the tank looks like asteel plate boiler which can withstand highpressures, then surely it can withstand a water tableabove the tank and careless installation (just kick inthe dirt). This concept is similar to the impressionthat a can of carbonated beverage which blows agusher when opened, must surely beable to support the weight of the consumer whosteps on the can after the contents are transferredfrom can to consumer. In fact, the beverage can isan approximate model of steel tanks. The samevacuum that collapses the empty can when theconsumer sucks on it can collapse a 12,000 gallonsteel tank if the sucker with a sealed straw ispersistent.

Spheres

Common examples of buried spheres are plasticseptic tanks. Typical 300-gallon septic tanks are


Figure 22-8 Membrane analysis of the flat bottom of a spherical PE septic tank to find tensile stress, σ , anddeflection at the center, δ, as functions of pressure, P.


made of polyethylene (PE) or fiberglas reinforcedplastic (FRP). They are 4.5 to 5 ft OD. See sketch,Figure 22-8. Before tanks are buried, somemanufacturers test load them up to maybe 8 kipsand check for leaks by holding, for a specifiednumber of minutes, an internal vacuum of up tomaybe 5 inches of mercury (5.7 ft of water headabove the bottom of the tank). It is known fromexperience that once a tank is buried, soil supportcan raise the allowable external water head to 7 ft;i.e., water table above the ground surface which is2 ft above the top of the tank. The bottom of thetank is critical. Not only is external water pressuremaximum, but the bottom is often made vulnerableby flattening it to provide a base for the tank asshown in Figure 22-8.

Example 4

A spherical polyethylene tank is about 51 inches indiameter with meridional ribs — like a pumpkin.The tank can resist a test vacuum of 5 inches ofmercury. The bottom is a disk, 21.25 inches indiameter. See Figure 22-8. If groundwater is atground surface, 6.33 ft above the bottom, and thetank is empty, external pressure on the disk is 2.74psi. Data for the disk follow.Given: bottom diskD = 21.25 = diameter,r = 10.625 = radius of disk,h = 6.33 ft water head,P = 2.74 psi = buoyant pressure on the disk,t = 0.35 = wall thickness,Q = 973 lb = total force on the disk,V = 14.58 lb/in = rim shear = Pr/2,E = 97 ksi = true modulus,E = 30 ksi = virtual long-term modulus,S = 2600 psi = short-term yield strength,S = 700 psi = long-term yield when indicated,υ = 0.4 = Poisson ratio,σ = stress,ε = strain.

From Equation 22.12, stress at the fixed rim of thedisk is 3.2 ksi, which exceeds yield. A plastic hingeforms at the rim, and the disk performs as a

membrane. Membrane analysis is based on strain,ε . From elastic theory, ε = σ1/E - υ (σ 2 + σ 3)/E.For most pressure vessels the strain effect ofpressure σ3 can be neglected. In a sphere, σ 2 = σ 1

= uniform membrane tensile stress, σ . The strainbecomes, ε = σ (1-υ )/E. From geometry, Figure 22-8, ε = (Rθ-r)/r = Rθ/r - 1

Equating the two strains (elastic and geometric) andsolving for R,R[2Etθ - Pr(1-υ )] = 2Etr . . . . . (22.15)

where θ = sin-1(r/R). Solving by iteration, radius R= 62 inches based on the long-term virtual E = 30ksi — not true E. Knowing the long-term R, stressis σ = PR/2t = 243 psi, which is less than the 700 psilong-term strength. Long-term deflection of thecenter of the membrane, from Figure 22-8, isapproximately δ = r2/2R = 0.9 inch.

Noteworthy is the short-term stress — before theplastic creeps. The radius, from Equation 22.15, isgreater before the plastic creeps when R = 92inches. The resulting stress is σ = 360 psi, which isgreater than long-term stress, 243psi, but is still lessthan the tensile strength of 700 psi. Cylinders

The heads of buried flexible cylindrical tanks can beanalyzed by assuming average uniform externalhydrostatic pressure. Soil support is lost because thehead deflects away from the soil. If the tank isplastic, soil particles fall in against the deflected headand ratcheting can occur due to cycles of "pumpingout" the tank. However, if design is based on long-term values for S and virtual E, stresses willultimately relax below the creep threshold, and theratcheting of deflection will cease. See Chapter 20on creep.

Unlike its heads, the shell feels both water pressureand soil pressure. It feels the weight of soil cover.Moreover, buoyant water pressure on the bottom ofthe shell causes it to expand against the sidefill


Figure 22-9 Water model -- conditions for similitude of external water pressures and internal vacuums inmodel and prototype, such that Pm = P.

Figure 22-10 Collapse modes for two 12,000-gallon steel tanks with primary flat spots on the bottoms, andsecondary flat spots on the shoulders; and showing crimped rings near mid-length where flat spots and plastichinges develop.


soil. The phenomenon is complex enough thatphysical tests are of value to the designer.

Example 5

What is the internal vacuum at collapse of 12,000-gallon thin-wall steel tanks buried in uncompactedsilty sand (Classification SM with 30% silt), 4 ft ofsoil cover, and water table at ground surface? Forreduced scale physical models based ondimensionless parameters that have no feel for size,design conditions are:

1. True scale model with proportional lengths,2. Same materials in model and prototype,3. Same pressures at corresponding locations in

model and prototype.

See Figure 22-9. External pressure on the bottom ofthe model tank must be equal to pressure on thebottom of the prototype tank. This is accomplishedby either a water model or a soil model. For awater model, Figure 22-9, water pressure is madeequal in model and prototype by pressurizing thewater at the bottom of the model test cell. Due toupward seepage stresses, the tank tries to float, butis restrained by soil on top held down by a porous lidwhich allows water to seep out of the cell but notsoil. Buoyant water pressures are equal on thebottoms of the tanks in model and prototype. For asoil model, a vacuum is applied at the bottom of themodel test cell. The soil is sucked down. It feelsheavier. The vacuum is predetermined such that soilpressures are equal at corresponding locations inmodel and prototype.

If buried steel tanks have good welds between theshell and flat heads, collapse does not begin in thehead. However, at some vacuum less than collapsevacuum, heads may be heard to snap through (oilcan) and then perform as membranes in tension.Collapse occurs in the shell. It is sudden and audiblewith a decrease in vacuum.

The soil of the test series in this example was looseand saturated. With 30% silt, the embedment was

mud. The performance limit was collapse des-cribed as follows.

Collapse usually starts as a flat spot on the bottom.See Figure 22-10. Secondary flat spots develop atthe shoulders such that the cross section of the tankis a crude delta with plastic hinges at the vertices.If L/D is less than about 4, the stiff heads effectivelyconstrain collapse to a single buckled ring nearmidlength. If L/D is greater than 5, the effect of thestiff ends is diminished at midlength and two buckledrings may develop. The results of the test aresummarized in the following. It is noted that theseresults are collapse — not leaks in welds, which areanalyzed in Chapter 24.

Vacuum, P', at collapse is shown on Figure 22-11for 12,000-gallon steel tanks, buried in loose sandysilt with 4 ft of soil cover and with water table at theground surface. See Watkins (1992). P' is not adimensionless pi-term. A correct pi-term would beP'/E where E is modulus of elasticity. But E isconstant for steel tanks, so E is dropped forconvenience in Figure 22-11.

1. Even in poor saturated soil these 12,000-gallons teel tanks withstand external water pressure plusinternal vacuum. Soil support is significant.

2. From the tests, L/D is not a major parameter forpredicting critical vacuum in tanks of equal volume.One graph serves all values of L/D. The increasedwater pressure on the bottom of the shell, due toincrease in diameter, is offset by the greater supportby the heads, which are closer together for tanks ofequal volume.

3. Collapse is sudden with an audible "whomp."Collapse may occur after years of service. Thetank may be deformed slowly by soil that settlesgradually, or by ratcheting whereby each cycle ofload deforms the tank and allows soil particles to slipin against the tank so that the tank cannot rebound.If the tank is plastic, stresses in the tank relax withtime after each cycle of load. Then


Figure 22-11 Conditions for collapse of 12,000-gallon steel tanks buried in silty sand with water table atground surface 4-feet above the tank; and showing the vacuum, P', at collapse and the equivalent height ofwater table, h, to scale. For design, a safety factor is needed. Standard deviation of h is plus or minus 1-foot.


particles slip in against the tank. Energy is storeduntil, at some future time, the tank may eitherrupture or invert.

4. In service, collapse occurs less often than leaks atwelds. Leaks are analyzed in Chapter 24.

Dual-wall Tanks

Dual containment vessels are attractive for under-ground storage of hazardous fluids. A tank within atank is not only double protection against leaks, butalso provides a convenient means for detecting leaksin the inside tank by a "sniffer" that monitors theannular space between tanks. See Chapter 11.

ANALYSIS OF BURIED SILOS

Ring Analysis

Ring analysis of buried silos is based on horizontalsoil pressure and external water pressure.Horizontal soil pressure is statically indeterminate.Assume that embedment is of good quality,compacted, and in contact with the silo. For simpleanalysis, assume that external soil pressure is active.This is very conservative because of horizontalarching action of the soil around the silo.Theoretically, soil pressure could be greater thanactive if the silo ring could expand due to thePoisson effect of longitudinal compression, and dueto internal pressure of the contents. Realistically,ring expansion is negligible. Active soil pressure isan upper limit. In fact, arching action of select, well-compacted, granular embedment significantlyreduces soil pressure against the silo. The circular,vertical hole in the soil (like a bored well) isremarkably stable.

Active horizontal soil pressure is vertical soilpressure times K, where K = (1-sinϕ)/(1+sinϕ).Below the water table, active soil pressure isvertical, effective (buoyed) soil pressure times K.Water pressure must be added for ring compression

analysis if the silo is empty. Ring buckling is not anissue because there is no horizontal ring deflection.If the silo is held in circular shape, performance limitis ring compression yield strength.

Longitudinal (Vertical) Analysis

Longitudinal stress analysis is based on the verticalfriction on the outside of the silo plus the verticalfriction load due to the column of material as it isdrawn down inside the silo. If the silo is very high,the weight of the silo itself may be added.

Figure 22-12 is a worst-case scenario. It is assumedthat the silo rests on a heavy foundation that doesnot settle. It is assumed that the soil is placedwithout enough compaction to prevent compressiondue to its own weight plus surface live loads. Thesurface live loads may tend to deflect the top of thesilo ring and break down horizontal arching action ofthe soil. So it may be prudent to cast a reinforcedconcrete collar about the top of the silo if heavy liveloads (such as ore trucks) are anticipated. Thecollar should not be attached to the silo. If heavylive loads are anticipated, it would be prudent tocompact the soil. But if the soil is not compacted,surface wheel loads cause radial soil pressure Kγ zagainst the silo with a resulting shearing stress ofKγ sµsz; where

σ = normal soil stress against the wall,σAV = average normal soil stress at depth L/2,τAV = average shearing stress of soil on

the silo,σz = vertical stress in the silo wall at the base,K = (1-sinϕ)/(1+sinϕ) for active soil resistance,ϕ = soil friction angle,γ s = unit weight of the soil,γ c = unit weight of the silo, contents,µs = coefficient of friction, soil on silo,L = length (height) of the silo,D = mean diameter of the silo,t = wall thickness of the silo for plain wall,Vs = total shearing load due to compressible soil,Vc = total shearing load of the contents of the

silo.


Figure 22-12 Worst-case scenario for buried silos showing how compressible external soil and the drawingdown of the contents both cause shearing loads on silos. Loads can be reduced by slip couplings in the silo,corrugations, and soil compaction.


Shearing stresses can be integrated over the entiresurface of the silo to obtain the total vertical load,Vs, at the base of the silo. If the soil is cohesionless,the average horizontal soil stress against the silo is,σAV = Lγ sK/2, and the entire shearing load due tothe soil against the silo wall is,

Vs = πDLσAV µs = πDL2γ sKµs /2 . . . . . . . (22.16)

This is a limiting value. In fact, shearing load on thesilo forces the silo downward slightly and soreverses the direction of shearing stress near thebottom of the silo. This case can be analyzed byconventional tec hniques. One example is a solutionby C.J. Costantino (Ref. 5) and Longinow.

However, with a good foundation, the silo does notmove downward very much and so for a worst case,the upward shearing stresses are neglected and thehorizontal plane of zero shear is at the bottom of thesilo as shown in Figure 22-12.

For a worst case assume that the contents are beingdrawn down through the hopper and gate. Draw-down loosens the material (contents) in the hopper.If the material has enough strength (cohesion and/orfriction angle) to "dome" over the hopper at thebottom of the silo during draw-down, the entirecontents of the silo are supported vertically byshearing stresses between contents and silo wall;i.e., the contents are hung up on the silo wall; and,

Vc = γ cL πD2/4 . . . . . (22.17)

Now if the silo has plain walls, the vertical normalstress at the base of the silo wall is σz = V/πDtwhere V = Vs + Vc, and

σz = L(2Lγ sKµs + Dγ c)/4t . . . . . (22.18)

Example 6

Figure 22-12 is a steel silo with plain walls. For thissilo, the weight of the silo itself is negligible.

Silo:D = 12 ft = diameter,L = 30 ft length (height),t = 0.375 inch = wall thickness.

Soil:γ s = 120 pcf = soil unit weight,c = 0, cohesionless soil,K = 1/3 = (1-sinϕ)/(1+sinϕ),ϕ = 30° = soil friction angle,µs = 0.5 coefficient of friction soil on pipe.

Silo contents, coal:γ c = 60 pcf.

What is the vertical stress in the steel wall at thebottom of the silo? Substituting the above data intoEquation 22.18, σz = 3.2 ksi. If yield strength is 36ksi, the safety factor is 11.

REFERENCES

1. Reese, R. Allan (1993), Full Scale BucklingStudy of Steel Underground Storage Tanks, SteelTank Institute, Lake Zurich, IL.

2. Young, Warren C. (1984), Roark's Formulas forStress and Strain, 6th ed., McGraw-Hill, 1989, table35.

3. Timoshenko, Stephen P. (1956), Strength ofMaterials, Part II, 3rd ed., D. Van Nostrand.

4. Watkins, Reynold K. (1992) , StructuralPerformance of Buried Steel Tanks, Sponsored bySteel Tank Institute.

5. Costantino, C.J. and Longinow, A., The Theoryof Limiting Equilibrium for AxisymmetricProblems; A Comparison with Experiment on SiloSkin Friction, IIT Research Institute, Chicago,Illinois.


Figure 22-13 Silo buried in sand showing the steel cutting edge through which saturated sand is sucked duringlowering of the silo. The same procedure is used for sinking caissons.


PROBLEMS

22-1 From principles of engineering mechanics,based on theory of elasticity, derive an equation forfinding the difference in the decrease in radius of asphere and a cylinder of equal radius and equal wallthickness under the same external pressure.

22-2 A cylindrical steel tank with flat heads is testedby internal vacuum. What is the vacuum atcollapse? (5 psi)

D = 84 inches = diameter,t = 0.167 inch = wall thickness,L = 18 ft = length of the tank.

22-3. The tank of Problem 22-2 is buried horizontallyin poor soil with height of cover H = 4 ft. If thewater table is at the ground surface, what is anapproximate estimate of the internal vacuum atcollapse? (0.2 psi)

22-4. Find the volume of the tank of 22-3.

22-5 Before installation, a steel cylindrical tank withflat heads is tested for both leaks and resistance toexternal hydrostatic pressure by means of a singletest. An internal vacuum is applied and held for aspecified number of minutes. The internal vacuummust not be great enough to collapse the tank. Whatis the vacuum, P', at collapse? Given: L = 42 ft, D= 9 ft, t = 3/16 inch, volume is about 20,000 gallons. (2 psi)

22-6 Design the most economical buried, plain steelsilo for loading coal slack onto a conveyor belt. H-20 trucks back in to dump the coal. There is apotential for a truck dual to roll onto the rim of thes ilo. A grizzly protects the top opening againsttrucks falling in. What should be done about thetruck problem? Safety factor is 2. Find t.

Given:Soil Coal Siloc = 0 c = 0 D = 12 ftγ s = 120 pcf γ c = 80 pcf L = 36 ftϕ = 30° ϕ = 40o σ f = 36 ksisf = 2 yield

22-7 Reconsider Problem 22-6. Plot the verticalshearing load due to surface live load on the outsideof the silo wall as a function of depth.

22-8 Silos are to be sunk into sand by excavatingsaturated sand from inside the cylinders by means ofa suction tube. Each silo comprises concrete ringsthat are 2 meters long with tongue-and-groove jointssealed by mastic. See Figure 22-13. The concreteis reinforced by wire mesh which is structurallynegligible. After the silo is in place, the bottom isplugged and sealed by a cast-in place floor. The silois empty. Water table is at ground surface.Allowable compressive stress in the concrete(reduced by a safety factor) is fc = 5 ksi.

a) What wall thickness is required for the silo? (7.5 mm)

b) What is the maximum vertical stress in theconcrete?

c) What is the potential for flotation of the silo?Available wall thickness is t = 150 mm.

Given:Silo SoilOD = 2 meters e = 0.56L = 10 m G = 2.65

ϕ = 34° µ = 0.5

Concreteγ = 22 kN/m3 Sea Waterfc = 34.5 MPa G = 1.0256

22-9 Plot a graph of Equation 22.3 for P/E atcollapse if L/D = 1 for a circular cylindrical tankwith rigid heads. Poisson ratio is 0.28.


Anderson, Loren Runar et al "FLOTATION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 23-1 Buried tank, 92D, tied down to a concrete anchor by straps, showing the buoyant uplift, Ww, ofthe empty tank (weight of tank neglected), and the effective resistance, Ws, of the soil wedge with 2-ft of soilcover. The anchor and straps must prevent flotation.

Figure 23-2 12-kilogallon, 92D tank tied down to a slab, showing the net uplift, ∆W , the first-try location ofhold down straps, and the moment diagram.


CHAPTER 23 FLOTATION

Buried tanks are used extensively at servicestations. Unfortunately, a groundwater table is nota valid reason to relocate service stations.Therefore, when tanks are below the water table,holddowns, weights, etc., may be required toprevent flotation. High soil cover can preventflotation, but may not be economical. Reinforcedconcrete pavement over a tank helps to resistflotation. Holddowns require anchors — a concreteslab or deadmen. When water table is a problem,soil at the bottom of the excavation is so wet that aconcrete slab is used as a platform on which towork. In some cases, two longitudinal footings(deadmen) may be adequate anchors. The tank istied to the anchors by straps. See Figure 23-1.

Holddowns

Straps are generally used instead of cables and steelrods because straps minimize stress in the tank orpipe, and damage to the coating. Questions arise,how many straps are needed, what size, and whatspacing? Soil cover provides weight. Anchorsprovide weight and soil resistance. What is the soilresistance? Two mechanisms are: soil wedge andsoil bearing capacity.

Soil Wedge If the embedment is granular and compacted, afloating tank must lift a soil wedge. See Figure 23-1.If buoyant force of the tank exceeds the effectiveweight of the soil wedge, the anchors must restrainthe difference. See Chapter 4 for effective unitweight of soil.

Example 1

12,000-gallon tanks are common at service stations.Suppose that a 12-kilogallon steel tank is buriedunder 2 ft of soil cover. How much anchorage isrequired if the tank is empty when the water tablerises to or above the ground surface?

TankD = 92 inchesL = 35 ftSoil γ d = 100 pcf = dry unit weight of soil,e = 0.5 = void ratio*,γ b = 58.4 pcf = γ d - γ w/(1+e),ϕ = 23o = soil friction angle from lab tests,θf = 56.5o = soil slip angle = 45o + ϕ/2,H = 2.0 ft = soil cover.

* G = (1+e)100/62.4 = 2.4 = specific gravity of soilgrains. The specific gravity of most soil is in therange of 2.65 to 2.7. Therefore, this soil probablycontains organic matter.

What force must be resisted by anchors?

∆W = Ww - Ws . . . . (23.1)

whereWw = 2.881 kips/ft = buoyant uplift force per unit

length of tank,Ws = 2.579 kips/ft = effective soil wedge

(ballast) on top per unit length at θf = 45o

+ ϕ/2,∆W = 0.302 kips/ft to be restrained by straps.For this 35-ft-long tank,

The buoyant uplift force is 101 kips.The resisting (ballast) force is 90 kips.

Neglecting the weight of the tank and the soilwedges at the ends of the tank, the force to beresisted by anchors is 11 kips.

Example 2

For most analyses of granular embedments,engineers use a generalized soil wedge with slipplane slope of 2v:1h. In Example 1, assuming thegeneralized soil wedge at effective (buoyant) unitweight,


Ww = 2.881 kips/ft = buoyant uplift force per unitlength of tank,

Ws = 2.257 kips/ft = effective generalized soilwedge (ballast) on top,

∆W = 0.624 kips/ft to be resisted by straps.The buoyant uplift force is 101 kips.The resisting soil wedge is 79 kips.

Neglecting soil wedges at the ends of the tank, thetotal uplift force to be restrained by the straps is 22kips. The conservatism (22 compared to 11) isjustified because of conservative assumptions — inparticular, the generalized soil wedge. Figure 23-1shows soil slip planes from the tank spring lines tothe ground surface. Tests show that planes are wellestablished near the tank, but are not wellestablished at the ground surface. In fact, the"plane" may be more nearly a spiral cylinder thatbreaks out on the ground surface at a width lessthan the 15 ft shown. For the generalized soilwedge (2v:1h), the break-out width at the surface is13.5 ft. Shearing stress of soil on pipe is neglected,weight of the tank is ignored, and soil liquefaction isnot considered. If there is any possibility of soilliquefaction, flotation is of concern.

Soil Bearing

If a laborer sinks into the mud while walking on it,soil bearing capacity may be more critical than thesoil wedge for resisting flotation. Pressure on alaborer's shoe print is roughly 2 ksf. Just as alaborer's foot sinks into mud, so does a buoyant tankrise through mud. It is assumed that the soil bearingcapacity is the same — up or down.

Example 3

In the examples above, suppose that the soil is sopoor that bearing capacity is only 250 lb/ft2. Soilresistance is Ws = (92D tank diameter)(250 lb/ft2) =1.917 kips/ft. Neglecting weight of the tank,Ww = 2.881 kips/ft,Ws = 1.917 kips/ft,∆W = 0.964 kips/ft to be resisted by straps.The buoyant uplift force is 101 kips.

The resisting soil force is 67 kips.The total force on the anchors is 35(0.964) = 34kips. The force in each of four straps is 8.5 kips.

Design of Straps

Continuing the example above, as a first try, use twoslings (four straps) made of 0.25 x 1.25 hot rolledsteel for which yield strength is 42 ksi and allowableis 21 ksi. See Figure 23-2. If the force in each ofthe four straps is 8.5 kips, the resulting stress in eachstrap is σ = 27 ksi — too high. It would be prudentto increase the size or number of straps. Try fourstraps, 0.25 x 1.75. The stress is σ = 19.4 ksi . Thisis acceptable unless fasteners are critical. How arethe straps attached to the slab? How are the strapstightened? How much prestress is required?Prestress reduces soil slip when the tank tries tofloat. But prestress could cause flat spots on thebottom of the tank. Prestress causes shearingstresses on the tank due to a "tight cinch". Fordouble shear (both sides of the strap), shearingstress in the tank wall is,

τ = T/tD . . . . (23.2)

whereT = 8.5 kips tension in the strap, D = 92 inches,t = 0.187 = tank cylinder wall thickness (thin

wall assumed).

Substituting values, τ = 494 psi. Even if prestressshould double the shearing stress, failure isimprobable.

The next concern is longitudinal stress, especially atjoints, due to bending of the tank when held down bystraps. Suppose straps are located tentatively at 7 ftfrom the ends of the 35-ft-long tank. See Figure 23-2. If, from the soil bearing case above, the resultinguplift force is ∆W = 964 lbs/ft, the moment diagramis maximum at B where M = 23.62 kip ft. Themaximum longitudinal stress, σ , occurs at the topand bottom of the tank at B, where,


σ = M/(I/c) . . . . (23.3)

Substituting values including I/c = πr2t; longitudinalstress is σ = 230 psi. Unless circumferential weldsnear B are bad, longitudinal stress is of no concern.

Safety factors are essential because soil-structureinteraction is complex and variable. In the aboveexample, soil wedges at the ends of the tank areneglected. Should they be included? Probably notif the soil wedges of Figure 23-1 are not truewedges but curve upward on a short concavesection of a spiral. A spiral is more probable ifgranular soil is loose. If the soil is plastic or hasmore than about 10% fines, bearing capacity may becritical. Bearing capacity must resist the verticalforce on the slab (or deadman) that "pulls" it upthrough the soil. For granular soils with less than10% fines, the bearing capacity exceeds two kip/ft2,depending on degree of compaction. For plastic soilor soil with a high fraction of fines or organic matter,the bearing capacity can be less.

Design of Anchors

Anchors can be rock bolts, concrete slabs,reinforced concrete beams (deadmen), etc. Slabsand deadmen are discussed in the following. Atypical slab is reinforced concrete approximately thesame width and length as the tank, and a minimumof 8-inches thick. See Figure 23-3. Deadmen canbe two or more transverse beams on which the tankis positioned. More generally, deadmen are twolongitudinal beams located as shown in Figure 23-4.

Holddown capability is the effective weight of thedeadman (or slab) plus resistance of soil — eitherthe effective weights of soil wedges (in this case,soil prisms) or the soil bearing capacity. Followingis the procedure for analysis.

Soil Prism

With shearing planes at 2v:1h, the soil lifted by theslab is the volume of prisms (partly dotted in Figure23-3) times the effective unit weight of soil. The soilis assumed to be granular and compacted.

Example

Wedge Analysis

Consider the 12 kilogallon tank, D = 7.667 ftdiameter and L = 35 ft long, with soil cover H = 2 ft.The effective unit weight of the embedment, fromthe example above, is 58.4 lb/ft3. What is the safetyfactor against flotation?

1. Try a slab. See Figure 23-3.L = 35 ft = length,2X = 8 ft = width,t = 8 inches = thickness.

The 8 x 35-ft slab should be reinforced so that it canresist the concentrated forces of the straps. Find thereaction to tension in the four straps. The volume ofeach of the two soil prisms is 704 ft3. The effectiveweight of the two prisms is 2(704)(58.4) = 82 kips.When added to the 79-kip effective weight of thegeneralized soil wedge on top of the tank, theresistance to flotation is 161 kips. Because thebuoyant uplift force is 101 kips, the safety factoragainst flotation is 165/101 = 1.6.

2. Try longitudinal deadmen. See Figure 23-4.Assume that the deadmen are as long as the tank,i.e., 35 ft, and that,a = 12 inches,b = 12 inches,r = 46 inches,h = 24 inches,j = 35 inches = (b+r/2),γ = 58.4 pcf = effective (buoyant) unit weight

of soil from the above examples.


Figure 23-3 Holddown straps to a slab, showing the soil prisms (partly dotted) that help to resist the upliftforce of the straps. Total strap resistance is the effective (buoyant) weights of the slab and soil prisms.

Figure 23-4 Holddown straps to longitudinal deadmen, showing (dotted) the prism shear surface AB and thespiral shear surface AC for less-than-select-soil.


The area of the soil prism lifted by each deadman isroughly j(2r+h) = 28.2 ft2. Each prism is 35 ft long.Therefore, the volume of each prism is 987 ft3. Theeffective weight of each prism is 115 kips. For twoprisms, this should be doubled. However, in less-than-excellent embedment, the soil shear surface isnot plane AB, but is more nearly spiral AC, forwhich the volume to be lifted is roughly half of theprism. Therefore the total effective weight of soillifted by deadmen is 115 kips. When added to the79-kip effective weight of the soil wedge on top ofthe tank, resistance to flotation is 194 kips. Thebuoyant uplift force of the tank is 101 kips. Thesafety factor against flotation is 194/101 = 1.9.

Soil Bearing Analysis

If the saturated soil is so poor that a laborer sinksinto the mud, the tank and deadmen (or slab) couldfloat upward through the mud. Suppose the bearingcapacity of the mud is 250 lb/ft2.

1. Try a slab.If the slab is 35 ft long by 8 ft wide, half the area iseffectively resisted by the soil bearing capacity.Adding the buoyant weight of an 8-inch-thickconcrete slab at 81 pcf, resistance to tension in thestraps is (4 ft)(35 ft)(0.25 k/ft2) + 8(35)2/3)(0.081)= 35 + 15 = 50 kips. Resistance of the soil cover is(8 ft)(35 ft)(250 lb/ft2) = 70 kips. The sum of thetwo is 120 kips. Buoyant uplift force is 101 kips.The safety factor against flotation is 120/101. sf =1.2. If there is any possibility of soil liquefaction, itwould be prudent to specify select embedment.

2. Try two longitudinal deadmen. If the deadmen are 35 ft long and the cross sectionsare 1 ft x 1 ft, the soil bearing resistance plusbuoyant weight of the deadmen is 2(35)(0.250) +2(35)(0.081) = 23 kips. Added to the 70-kipresistance of the soil cover, the total resistance is 93kips. Buoyant uplift force is 101 kips. When empty,the tank will float in the mud.

One possible analysis of the resistance of deadmenmight start with inversion of the classical soil bearingrationale. As a deadman rises, a soil wedge formson top and shoves soil outward in general shear.This is explained in texts on soil mechanics. Thedeadman “plows” its way up through the soil.

PROBLEMS

23-1 A 20-kilogallon steel tank, 9-ft diameter, by 42-ft length, is buried under H = 2 ft of granular soil.Design straps and a reinforced concrete slab 9 ft x42 ft x 9 inches thick, to prevent flotation with asafety factor of 1.2 It is possible for a water tableto reach the surface at a time when the tank isempty. Assume effective unit weight of thesubmerged soil is 60 lb/ft3. a) What must be thenumber and size of steel straps if allowable tensilestrength is 20 ksi? What must be the thickness ofthe slab to prevent flotation if unit weight ofconcrete is 142.2 lb/ft3? (What is the submergedunit weight of concrete?)


Anderson, Loren Runar et al "LEAKS IN BURIED PIPES AND TANKS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 24-1 Response (flattening) of a flexible ring due to a hard spot in the embedment.

Figure 24-2 Joggle joint weld showing longitudinal leverage due to an external force that causes a hinge toform and rotate.


CHAPTER 24 LEAKS IN BURIED PIPES AND TANKS

Performance limits for buried structures areexcessive deformations. One such deformation is aleak. Because of leaks, not only is product lost, butthe soil (or the product) may become contaminated.Craters and landslides have been caused by majorleaks. Gas and oil spills and fires could be theconsequences of leaks in fuel lines and fuel tanks.

Leaks may occur at gasketed joints due to pinchedgaskets or rolled gaskets, or due to grit undergaskets. Leaks may occur through cracks. A leakin a high-pressure pipeline is a fluid jet that canbackwash sand against the pipe, sand-blast the pipe,wear through the pipe wall from the outside, andcause a blow-out. Because the cylinder is in contactwith embedment, deformation of the cylinder isconcomitant with movement of the soil — soil slip orsoil compression. The soil provides support forvertical loads and for the cylinder. But poor soilcarelessly placed, like a misfit shoe, stresses thecylinder. Installers of buried pipes and tanks usetechniques that minimize deformation of the cylinder.The track record of the installer is important. Someburied fuel tank agencies now require certification ofinstallers.

WELDS IN TANKS

If a pressure-tested cylinder leaks after it is buried,the leak is due possibly to deformation of a weldsuch as a flat spot or a leverage hinge. See Figures24-1 and 24-2. Deformations that cause weldfractures are usually local and usually occur in thecylinder. Analysis is similar for both pipes andtanks, but because tanks are more complicated, thefollowing examples involve tanks.

If collapse is the performance limit, heads stiffen theshell enough that walls are typically thinner in tanksthan in pipes. Consequently, under identical burialconditions, tank shells are more sensitive than pipesto local deformations caused by non-uniform loads.

A comparison of the sensitivities of tank shells andpipes requires similitude of the model and prototype.It is then possible to compare the pressures atequivalent deformations. These pressures areinverse measures of sensitivity. Let the tank be theprototype and the pipe be the model. Relativesensitivity indicates the level of care required forhandling and installing a prototype tank based on amodel pipe experience.

Similitude

Similitude is achieved by writing the equation ofdeformation in terms of dimensionless pi-terms; andthen by equating corresponding pi-terms for shelland pipe. See Appendix C. The pressure thatcauses critical deformation is a function of thefollowing pertinent fundamental variables.

Fundamental variables:P = external pressure on the tank,Q = external force on the tank,δ = any and all deflections or movements,S = yield strength of the wall,E = modulus of elasticity,E = 30(106) psi for steel,D = mean diameter,r = radius,w = width of joggle joint overlap,x = transition length from maximum to

minimum radius of curvature,I = moment of inertia of the wall cross

section,I = t3/12 for the plain wall tank cylinder,L = length of the tank,t = thickness of the shell,V = head shear force,σ = normal stress,τ = shearing stress,ν = Poisson ratio = 0.25 for steel.Subscript, r, refers to ratio of model to prototype.

Pertinent dimensionless pi-terms are:


(P/E) = external pressure term,(δ/r) = deformation term,(r/t) = ring flexibility term,(L/r) = length term for the tank,(rr) = ratio of maximum radius to minimum

radius at flat spots or out-of-roundness,(1-ν2) = 15/16 = Poisson term, whereν = Poisson ratio (assume 1/4),(d) = ring deflection = ratio of decrease in

vertical diameter to original diameter.

Following are the sensitivities of a prototype steeltank and a model pipe for two cases: without soilsupport, and with soil support.

I. Without soil support

Deformation is collapse under uniform externalpressure. The classical equations are:

TANK:(P/E)(L/r)(r/t)5/2 = 0.8/(1-ν2)3/4 = 0.84 . . . . (24.1)

PIPE:(P/E)(r/t)3 = 1/4(1-ν2) = 0.234 . . . . (24.2)

CORRUGATED PIPE: (P/E)(r3/I) = 3/(1-ν2) = 3.33 . . . . (24.3)

Example 1

For a particular steel tank, L/r = 42ft/4.5ft, and r/t =54in/0.1875in. Its collapse pressure is to becompared with a corrugated steel pipe for which r =78 inches and I = 0.938 in4/ft. What is the relativesensitivity; i.e., ratio of pressures, Pr, at collapse.From the quotients of Equations 24.3 and 24.1, Pr =8.24. The pressure at collapse of the pipe is eighttimes as great as the pressure at collapse of thetank. Without soil support, the prototype tank iseight times as sensitive to collapse as the pipe. Theheads of the tank provide significant stiffness, but agood soil embedment is essential.

II. With Soil Support

Sensitivity is deformation of the ring due to a hardspot in the soil. See Figure 23-1. From Appendix A,localized ring deformation is of the form, δ =kQr2/EI, where Q is the concentrated force, k is aconstan t , and r2/EI is the flexibility factor — ahandling factor with upper limits recommended byAISI (1994). In dimensionless parameters, (δ/r) =k(Qr/EI). In Example 1 above, at equal values of(δ/r) for model and prototype, the ratio of loads, forprototype and model is, Qr = Ir/rr = 68. The tank is68 times as sensitive as the pipe to hard spots in theembedment. Racks must not bear against the tank.Bedding must be uniform.

Deformations at welds cause most of the structuralleaks in buried tanks. For uniform internal orexternal pressure, circumferential welds must resistlongitudinal stress — but longitudinal stress is onlyhalf as great as circumferential stress. Butt weldscan be made nearly as resilient as the parent metal,and they can tolerate deformation. But they areexpensive. Consequently, longitudinal joints areoften joggle joints — usually welded on the outsideonly. See Figure 24-2. If the seam is not seriouslydeformed, such welds are adequate. When the weldis deformed, three conditions develop which couldfracture the weld: leverage, shear, and gap.

Leverage — A hard spot force on a joggle jointcauses a leverage hinge as shown in Figure 24-2.From tests, a joggle joint in 1/4-inch steel, stickwelded with E-6024 rod, loaded as a simplysupported beam of 3-inch span; fractured when a lineforce at mid-span reached 230 lb per inch of weld.Stresses were concentrated at the corner of theweld which became vulnerable to leverage — atypical cause of leaks in fuel tanks.

Shear — When the cylinder is deformed,c ircumferential shearing stress, τ , develops on theneutral surface of the wall. If wall thickness isdoubled, and the deformation remains the same, the


shearing stress is doubled. In joggle joints, shear inthe weld is increased even more. For the typicaljoggle joint of Figure 24-3,

τ = Ewt2(1/ro - 1/r)/2bx . . . . (24.4)

where x is the length of transition from minimum tomaximum radius of curvature. The transition isusually visible as a localized crimp for which length,x, is very short — an inch or two — at the ends ofa flat spot.

Example 2

A flat spot occurs in the joggle joint of a steel tank.What is the shearing stress in the weld? Shearingyield stress is usually about 20 ksi.

Given:E = 30(106) psi,w = 1 inch = minimum recommended pene-

tration of the joggle joint spigot into the bell,t = 0.1875 inch (3/16 inch),ro = original radius of curvature,

= 54 inches (minimum),r = infinity at the flat spot (maximum),x = length of transition from minimum to

maximum radius.

Substituting into Equation 24.4, τ = 52 kips/x(inch).If the transition length is x = 2.5 inches, shearingstress in the weld is 20 ksi (shearing yield stress).For 3/16 steel, it is more likely that x is less than 2.5inches. In such a case, the weld yields and couldcrack if its ductility limit is exceeded.

If the "flat spot" were not flat, but had a radius ofcurvature twice the original tank radius, the shearingstress would be τ = 26 kips/x(inch) which is half ofthe stress at a flat spot. If the radius of curvaturewere inverted, shearing stress in the weld would begreater At the ends of a flat spot, the radius ofcurvature is less than the original radius. Therefore,actual shearing stress is greater than the valuesabove.

If the ring were deflected into an ellipse, see Figure

24-4,

τ = (Ext/πr)(1/rx - 1/ry) . . . . . (24.5)

where the minimum and maximum radii of curva-ture are:

rx = r(1-d)2/(1+d),ry = r(1+d)2/(1-d) . . . . . (24.6)

Substituting the values from Example 2, ringdeflection at yield is 76.8%. Clearly, some smallelliptical ring deflection is not a cause of shearingcracks in welds. It is noteworthy that ring deflec-tion of the shell causes other complications such ashead shear (guillotine) and increased potential forinversion, both of which could contribute to crackedwelds. These are analyzed separately under "HeadShear" and "Inversion Analysis."

Gap — In forming a joggle joint, one end of the canis deformed into a spigot of smaller diameter suchthat it can be inserted into the mating can. SeeFigure 24-5. This usually leaves a gap as shown.When a hard spot in the soil deforms the ring, thegap tends to narrow under the hard spot, and widenat other locations. Widening of the gap may crackthe weld or, at least, compound the effect ofshearing stress caused by change of radius. Worsethan the elliptical ring show n is the accumulation ofgap at the crimped ends of a flat spot — againwhere shearing stress is maximum. Once a weld iscracked, the crack propagates and opens. Also, if ahard spot bears against the joggled can, but not themating can, the crack tends to open. In either case,the crack widens and leakage increases.

In order to minimize the effect of a gap, typicalstandards require that outside circumference of thejoggle spigot be 1/32 to 3/32 inch smaller than theinside circumference of the mating bell. Even 3/32-inch difference can cause a gap to accumulate if thejoint is welded continuously without first "tackwelding" or "wedging" the gap with shims (or screwdrivers). Large gaps are sometimes “slugged;” i.e.,the gap is partially filled with a bar


Figure 24-5 Joggle joint in a tank showing how the gap becomes narrower under hard spots and wider at softspots in the embedment. Wide gaps also tend to accumulate at the ends of continuous welds.


of reinforcing steel before it is welded. "Slugged"welds are to be avoided.

Installers of tanks, especially long tanks, try to avoidhard spots in the embedment which might cause flatspots in the tank. Longitudinal deflections of tanksmust be restricted by leveling the bedding and bycarefully placing embedment under the haunches ofthe tank. A flat hard bedding, rocks in theembedment, frozen soil, and loose soil under thehaunches — all can cause flat spots in tank shells.

Inversion Analysis

A major problem with flat spots is the potential forinversion. For conservative analysis, the flat spot isa beam with fixed ends. See Figure 24-6. Themaximum moment is at the ends where Mmax =PL2/12. But at the formation of a plastic hinge, Mp

= 3SI/2t = St2/4. Equating,

(L/t)2 = 3S/P . . . . . (24.7)

From the geometry of Figure 24-6,

(L/t) = (D/t)sin(θ/2) . . . . . (24.8)

where θ is the arc angle of the flat spot betweenplastic hinges. Eliminating L/t between the twoequations, and substituting a known value for S,pressure, P, can be found as a function of angle, θ,and ring flexibility, (D/t).

Figure 24-7 shows graphs of solutions for steelcylinders for which S = 36 ksi. The analysis isconservative because soil support is neglected.Arching action of the top of the cylinder is alsoneglected even though the "flat spot" may not becompletely flat. Noteworthy is the increase inpressure P at inversion when D/t is decreased. Acommon upper limit for plain pipe is D/t = 288. Formortar-lined pipes, upper limits are usually not morethan D/t = 240. Noteworthy also is the increase inangle θ as D/t is decreased. Soil support increasesas θ increases, but the beam analysis losesaccuracy. The beam inversion model is reasonably

accurate up to roughly θ = 45o. Beyond that, astability analysis is more relevant because ofincreased soil support and arching of the top of thepipe.

Head Shear

Heads stiffen the shell, but they also cause headshear when, under soil load, a head shears downpast the flexible shell like a guillotine. The shelldeflects easily under backfill load. But heads remaincircular. This causes distress in the head-to-shellwelds at the bottom of the tank where head sheartends to curl the flange (knuckle) and crack the weldas shown in Figure 24-8. If the seam is a jogglejoint, a leverage hinge may form. For analysis, theeffective shearing load on each head is soil pressure,P, acting over an area of a tank diameter timesroughly one diameter longitudinally. P = Pd+Pl

where Pd is dead load pressure and Pl is live loadpressure. Shearing load is:

V = PD2 . . . . . (24.9)

Reaction is developed under the shell. If sidefill soilcompresses vertically, the shell deflects, and theheads shear down past it. Analysis must considerweld strength and resistance of the flange tobending. Distress in the weld is exacerbated bydeflections of the head and shell both of which bendthe flange. Unfortunately, the 90o flange angle isbent (reduced) the most on the invert where headshear is greatest.

Example 3

Consider the steel tank, D = 9 ft, L = 42 ft, t = 3/16inch. What is the shearing force, V, between thehead and the shell due to a soil cover of 3 ft withunit weight of 120 pcf? From Equation 24.9, V =29.16 kips — enough to distort a 3/16-thick knuckleand weld.

From field experience, the shear load, V, is also feltat the first joint from the head. Many of the leaks insteel fuel tanks are cracks in the bottom at this joint.The joint between the first and second cans


Figure 24-6 Model for beam analysis of a flat spot (exaggerated) in a flexible cylinder, showing the momentdiagram and the geometry. The flat spot is shown here at the top of the cylinder, but may occur anywhere.When a flat spot does occur, it is usually at the invert.

Figure 24-7 Pressure at beam inversion of flat spots on steel cylinders as a function of ring flexibility, D/t,and angle of the flat spot, θ, (beam length).


Figure 24-8 Failure of a circumferential weld due to a stiff head that shears down past a flexible shell. Steeltank standards call for a minimum of 1.5-inch flange and 0.5-inch penetration into the shell.


does not have the benefit of a head to give itstrength. Yet it may be subjected to much of theshear force, V.

Precautions

In order to avoid cracks in welds, attention should bedirected to the following:

1. Careful handling and installation in order to avoidflat spots,

2. Well-compacted embedment that does not com-press or slip under anticipated loads,

3. Sound welds with enough toughness that they canyield without cracking under slight deformation.

4. Control of internal vacuum and high externalwater table.

Additional safeguards to prevent or mitigate leaksinclude the following:

1. Double containment tanks or coated tanks,

2. Sniffer systems such as a vent between theproduct tank and the double containment tank.

3. Diligent monitoring of contents to discern any loss,

4. Sensor devices in the path of any possible leakageplume,

5. Control of surface loads and high water table.

Longitudinal Beam Action

Stress in the welds can be exacerbated bylongitudinal beam action. See Figure 24-9. The

Figure 24-9 Typical conditions for longitudinal stresses in a tank caused by concentrated supports — on theends of the tank (top) and at the tie-downs (bottom) due to a high water table or flood.


most common distress is tension in the bottom of atank that is simply supported at the ends, but is notbedded. Cracked welds may occur on the bottomnear mid-length. Analysis is the standard procedurefor finding maximum stress and deflection atmidlength of beams. Figure 24-9 also shows a caseof concentrated reactions at tie-downs that result inlongitudinal stresses at the top and bottom of thetank during a flood. Long tanks with single weldjoints and with concentrated reactions are in greatestdanger of cracked welds.

Tests for Leaks

It is common practice to test tanks for leaks beforethey are buried — usually before they leave thefabrication plant. One such test is internal pressure— typically 5 or 6 psi. The pressure valve is thenclosed, and pressure in the tank is monitored for aspecified length of time in minutes. Drop in pressureis a measure of the amount of leakage, and may bethe basis for specifying allowable leakage. Soapywater is often poured over the tank so that leaks canbe located from soap bubbles. However, internalpressure tests do not indicate resistance of the tankto vacuum. A vacuum test is more responsive toresistance of the tank to vacuum (externalpressure). A vacuum test can measure the amountof leakage.

Safety Factors

Often overlooked in risk analysis is the time toleakage. Compression and consolidation ofembedment may increase with time. For example,if traff ic compresses the sidefill in increments, thetank feels incremental ring deflection (and possiblybeam deflection). Stresses increase until, at somefuture date, a weld fails — often with an audible"thud." The safety factor should also include thepotential for assessing encroachments into themargin of safety (safety zone). In the event of adispute, the safety zone is like a demilitarized zone.

It separates antagonists and accommodates minorencroachments. Legal exposure is measured by theextent of encroachment into the safety zone.Adjudication of contributory negligence is based onthe percent encroachment by each of the parties:owner, engineer, manufacturer, installer, etc.

Specifications (procedure or performance) canestablish the relative responsibilities of all parties andthe exposure of each in the safety zone. Theresponsibilities are intensified by public concern forcontainment — especially of hazardous materials.For example, the time to leakage must often begreater than the traditional 50 years service life.Control of leakage plumes and procedures fordecontamination become important.

LEAKS IN PIPES

The more common leaks in pipes are the follow ing.

Rigid Pipes

Rigid pipes are sensitive to: 1. water hammer; 2.leaks at gaskets; and 3. movement of pipe sections.

1. Water Hammer If the pipes are brittle, water hammer can causelongitudinal cracks. If the water hammer isrepetitive, fatigue strength may control analysis.Fatigue analysis is found in texts on strength ofmaterials.

2. Gaskets A tiny leak past a gasket in a high-pressure pipe canultimately result in a major break. A rolled gasketcan be blown out of the joint ("fish-mouth") and leak.Sand under the gasket can cause a leak. A crackedbell can allow leakage past the gasket. The leak issmall at first, but in time saturates the soil.Turbulence of the high-pressure jet swirling againstthe pipe "sand-blasts" and wears away the pipe,from the outside surface in, until a large leak opens.From tests, the sand-blasted leak increases


in size at an exponential rate. Such leaks should beidentified as soon as possible by timely monitoringof the pipeline.

Even if the embedment is select gravel, a leak on topof the pipe can wash particles down from thebackfill into the embedment where they are caughtup in the jet turbulence and proceed to sand-blast thepipe.

3. Movement of Pipe Sections Sections of rigid pipe are connected by gaskets.Lengths are short. If pipe sections shift out ofalignment, the sidefill soil must serve as a thrustrestraint. A problem arises when the sidefill soil iswashed away by a leak, or loses its supportingstrength when saturated.

Example

An asbestos cement (AC) pipeline follows a circularcurve. It is positioned on select bedding, but sidefilland backfill are wind-blown silty sand. What is thesafety factor against separation of a joint? Safetyfactor is the sidefill soil strength divided by thesidewise thrust of two contiguous pipe sections onthe circular curve. See Figure 24-10.Pipe:L = 13 ft lay length of sections,ID = 24 inches = inside diameter,t = 3 inches = wall thickness,R = 457 ft = radius of curve,P' = 350 psi = internal pressure,z = 1.25 ft = outside radius,Q = thrust.Soil:P = soil pressure on pipe,H = 1.75 ft = soil cover,γ = 100 pcf = soil unit weight,σf = 400 psf = average horizontal soil strength,

saturated and loaded by 3 ft of soil.

The safety factor is sf = σf /P.From trigonometry, angle θ = 1.63o.

Figure 24-10 Contiguous sections in gasketedpressure pipeline that are out of alignment by offsetangle, θ; showing thrust, Q, and the approximate soilresistance diagrams, P(OD).


From Chapter 15, thrust is Q = 4 kips, which mustbe resisted by the two P(OD) diagrams shown.Taking moment about point A, P = 3Q/2L(OD) =416 psf. Because soil stength is 400 psf, the safetyfactor is just under 1.0. Failure is incipient.

In this example, there are two possible causes of abreak: sand-blasting and movement of contiguouspipe sections. It is noteworthy that according toplans, angle, θ = 1.63o. However, some of theinstalled angle offsets in the pipeline might begreater.

Flexible Pipes

Leaks in flexible pipes can occur at welded jointsw hen the pipe is deflected either radially, orlongitudinally. Consider steel. Steel pipes areflexible and high strength. Steel pipes are oftenspecified for high-pressure transmission. Stressrisers at welds can cause cracks. One typical stressriser is the angle offset as discussed above for rigidpipe sections. An example of offset angles inwelded steel pipes is a mitered joint. Figure 24-11 isan exaggerated mitered joint. Due to pressure, P',inside the pipe, Q = 2P'πr2sinθ. The impulse, Qi,due to change in direction of flow, is neglectedbecause it is relatively small. The averagelongitudinal stress in the pipe is Pr/2t(sin2θ).Hydraulic design limits the offset angle to 15o inorder to minimize turbulence. Q is resisted by thepipe wall on which components of force in shear, V,and thrust, T, are:V = P'πr2 sinθcosθ = 0.4066 Pr2 at θ = 15o,T = P'πr2 sin2θ = 0.0535 Pr2 at θ = 15o.

If the pipes were rigid, forces, T, Q, and T areconcurrent at point C. But steel pipes are not rigid.Longitudinal strain will tend to rotate BB' slightly.Rotation is restrained by the pipe on either side ofthe elbow. The resulting moment is Te; i.e., thrust,T, in the pipe wall times its offset, e, from the neutralaxis. Maximum e occurs when cut BB' is moved upto the weld such that B' falls on A'. Now thelongitudinal stress diagram is a triangle withmaximum stress at A' where e = r/3 and

longitudinal stress is twice the average stress.Because the elbow can deform slightly, and becausestresses diverge (Saint Venant principle), longitudinalstress at A' may be less than twice the averagestress, but is greater than average stress. Suchrationale is moot because yield stress at A' does notcause fracture as long as strains do not approach21% (elongation) for steel with a good weld.

Of greater conc ern is the loss of hoop strength dueto the skew cuts (miters) at the joint. Hoops are cutwithin the triangles shown dotted in Figure 24-11.The lost hoop strength must be supplied by the seam.It the pipe were a membrane, the resulting forces onthe seam would be stress triangles shown in Figure24-12. Hoop stresses greater than yield wouldcause similar stress distributions. Figure 24-12 is a30o skew for which ovality of the elliptical seam is d= 7.2%. The maximum force per unit length ofseam is Max w = Prcosθ. This is worst-case designbecause the pipe is not a membrane.

When stresses are near yield, a stiffener ring at theseam can resist the w-forces. The stiffener ringperforms like a crotch plate in a wye. For hoopstresses less than yield, the seam itself provides anadequate stiffener ring. The mating pipe wallsintersect in a V which has significant ring stiffness.Because the pipe is not a membrane, wall shearstrength resists much of the force on the seam. Formost mitered joints, if angle offsets are no greaterthan 15o, pipe designers increase the wall thicknessof the mitered pipe sections, and specify fullpenetration butt welds. Mitered joints are commonin steel pipes.

Strengths of lap welds are less than the steel pipe.Roger Brockenbrough (1990) of USX tested lapjoints and reported longitudinal strengths at about:

Single weld, 75% of pipe strength,Double weld, 83% of pipe strength.

The percentages are referred to as weldefficiencies.


Figure 24-11 Free-body-diagram of a mitered joint (elbow) in a pipe, showing the free-vector-diagram offorces on the elbow and the resulting upper limit of stress at A’.

Figure 24-12 Forces, w, on the mitered seam due to loss of strength of the fully circular hoops that have beencut. This is a conservative analysis based on membrane theory. A stiffener ring on the seam can resist theseforces.


REFERENCES

AISI (1994), Handbook of Steel Drainage &Highway Construction Products, 4ed, AmericanIron and Steel Institute.

Brockenbrough, Roger L. (1990), Journal ofStructural Engineering, Vol. 116, No. 7 July 1990ASCE.

PROBLEMS

24-1. What is the theoretical bending moment(resistance) to flexure of a weld in a joggle joint in3/16-inch steel plate? See the leverage hinge ofFigure 24-2. Weld strength is 20 ksi. (117 lb)

24-2. Given, a 42-ft-long steel tank, 9 ft diameter,3/16 wall thickness, that comprises seven 6-ft-longcans connected by joggle joints. It is lowered into anexcavation and positioned on pre-leveled timbers atthe ends of the tank. Soil is end-dumped (looseembedment). Considering the inadequate soilsupport, what is the height of soil cover at weldfailure? Tensile yield stress of welded seams is 20ksi. Unit weight of soil is 100 pcf. Unit weight ofthe contents (gasoline) is 42.4 pcf. Unit weight ofsteel is 490 pcf. (11.2 ft)

24-3. What is the maximum stress in the weld ofproblem 24-2 if the tank is tied down by two straps12 ft from each end, and the tank is submerged inwater (flood) while empty and before it isbackfilled? (780 psi)


Anderson, Loren Runar et al "LONG SPAN STRUCTURES"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 25-1 Two of many typical long-span structures — horizontal ellipse (top), and an inverted pearshape (bottom) to serve as a grade separation for a road over a railroad.


CHAPTER 25 LONG SPAN STRUCTURES

The term "Long Span" refers to corrugated steelpipes in large diameters. The "pipes" are more oftennon-circular than circular. The most commonshapes are pipe arch, horizontal ellipse, low profilearch, high profile arch, underpass, and the invertedpear. They serve as small bridges and gradeseparations. The structure comprises corrugatedsteel plates (structural plates) that are boltedtogether on site. The distinguishing feature is thelow ring flexibility because of the large radii.Moreover, because the plates are bolted together,ring flexibility is less than the theoretical values listedin the AISI Handbook of Steel Drainage &Highway Construction Products. From one fieldtest the actual ring stiffness was roughly 80% oftheoretical.

For standard corrugated circular steel pipes, amaximum flexibility factor, FF, is recommended forhandling and installation; i.e.,

FF = D2/EI . . . . . (25.1)

whereE = modulus of elasticity = 30(106) psi,D = diameter; i.e., horizontal span (inches),I = moment of inertia of the wall (in4/inch).

From the AISI Manual, recommended maximumvalues of FF for ordinary installations are:

FF = 0.0433 in/lb for factory-made pipe with riveted,welded, or helical seams.

FF = 0.0200 in/lb for field-assembled pipe withbolted seams.

Despite the maximum value of FF = 0.0200 for field-assembled pipes long-span pipes can exceed 0.02in/lb if embedment is carefully placed andc ompacted. For example, for a 30-ft span of 6x2structural plate with t = 0.218 inch; the theoreticalI = 0.127 in3; and the flexibility factor is FF = 0.035

in/lb. At 80% of theoretical I, FF = 0.043. If thespan is increased to 50 ft for this example, FF =0.118 in/lb. Fifty-foot spans are in service. Thesuccess is based on careful installation. For shapesother than circular, AISI publishes modificationfactors.

From principles of similitude, FF is not a correctproperty for buried pipes with distributed soilpressure agains t the ring. FF is correct for aconcentrated force on the ring which is moretypical of handling loads than of buried soilpressures.

Some design procedures base ring flexibility on theIowa formula. See Appendix B. They propose thatthe EI term (structure) in the denominator mustexceed some recommended percentage of the0.061E'r3 term (soil). This is intended to beminimum ring stiffness for soil-structure interaction.In fact, the Iowa formula only predicts ringdeflection — not a ring stiffness required forinstallation. Of course, designers specify amaximum allowable ring deflection — but for otherreasons than ease of installation. Without anelaborate scaffold of stulls, struts, and ties to hold theshape of the structure during installation, a minimumring stiffness is desirable. So required ring stiffnessis an economical trade-off with cost of installation.

The Iowa formula applies to circular rings. Mostlong-span structures are not circular. Nevertheless,as in the case of circular rings, the greater thestiffness, the less will be the care required to holdthe structure in shape during installation.

Performance Limits

The basic performance limits after installation arediscussed in Chapters 9, 10, and 13. These includering compression, soil support, minimum cover, etc.But for very flexible long-span structures, additionalprecautions are required during installation. Failures


have occurred during construction of the structureand placement of the embedment. Installationperformance limits include: 1. shape of thestructure, 2. unstable soil-structure interaction, and3. minimum soil cover.

1. Shape The structure must be held in shape duringplacement of the embedment soil. Shape can bemonitored by hanging plumb bobs inside atappropriate locations such as the crown, and atchanges in radius of the plates. Because plumbbobs are affected by wind, a laser beam is a betterdatum for monitoring the shape. Measurements bysteel tape at any angle from the laser beam to pre-identified points on the structure can be monitored.The cross section should be monitored at a numberof stations throughout the length of the structurebecause corrugated structures can deflectlongitudinally.

In the case of horizontal ellipses, and profile arches,the top radius is usually a circular arc of about 80o.However, the arc angle can vary. For pipe arches,the top arc angle is greater than 80o.

Installers resort to various techniques for holdingthe structure in shape during placement of the soil.Once the structure is assembled on a plane surfacebedding, it is essential that support be continuous.

a) Preshaping the bedding to fit the structure hasbeen tried. The procedure is tedious and imperfect.Continuous support is not assured. In some cases,the structure has been dragged longitudinally a fewfeet fore and aft to seat (preshape) the bedding.

b) Well-graded soil can be flushed under thestructure from windrows of soil by means of high-pressure jets (usually water — sometimes air).Laborers do not respond enthusiastically to therequirement of ramming soil under the bottom plateswith 2x4 studs.

c) Flowable soil cement is an increasinglypopular option. See Chapter 16.

An alternate option is elimination of the bottom

plates which are replaced by footings. See profilearches in Figure 9-1. The footings must be able toresist the thrust Pry due to pressure P on top of thestructure where the radius is ry. Because thrust Pry

on the footings is at an angle, subbase support of thefootings must take into account shear on, andoverturning of, the footings, as well as verticalbearing capacity of the subbase.

Once the structure is bedded, soil is placed andcompacted in lifts as described in Chapter 16,keeping soil lifts balanced on the sides of thestructure in order to prevent sidewise movement.Heavy compactors must be kept outside of the 45o

tangent plane as described in Chapter 16. In spite ofcare in placing embedment, soil pressure on thesides of the structure causes the top of the structureto hump up (uplift). The uplift must be limited andcontrolled. Manufacturers recommend allowablepercentages of uplift during installation. Part of theuplift can be reversed when soil is placed over thetop of the structure — but don't depend on it.Sidefill holds the structure close to its "uplifted"shape.

In some cases, the structure is stulled, strutted,and/or tied to hold it in shape during placement ofembedment soil. Care is required to prevent damageto the structure such as dents or perforations bystulls and struts when backfill is placed on top of thestructure. Tie wires (diagonal and horizontal) can beof such a gage that monitoring includes tension (oreven the first break) in horizontal wires. Oneproject monitored tension in the tie wires by the pitchsound of the wire when plucked.

If the structure approaches the limits of its uplift, awindrow of soil can be placed (by crane) on top. Oradditional tie wires can be placed from the crown tothe bottom plates or diagonally from the crown to thecorner plates or footings.

2. Stability Embedment soil must be of good quality in order


to assure performance during floods (high watertable), earth tremors, excessive surface loads, etc.Where groundwater is a problem, soil should haveadequate strength (bearing capacity) both whensaturated and when dry. It must be dense enoughthat it will not liquefy by earth tremors. Theseprecautions are imperative for the small radii at thesides of horizontal ellipses and low profile arches.The precautions are reversed for high profile arches,underpasses, and inverted pears. Because of thelarge radius of curvature of the sides, a very smallhorizontal soil pressure can deform the structure.During placement of sidefill, there is no topfill toresist uplift. In such cases, horizontal struts can beplaced in the structure to prevent horizontaldeflection. In some cases the side plates can be tiedback to deadmen when deflection becomessignificant. If uplift begins to approach themaximum allowable during placement of sidefill, awindrow of soil placed on top of the structure canarrest or reduce uplift.

Any scaffolding (stulls, struts, ties) inside thestructure should be removed after stability isachieved, but before high soil cover (topfill) is placedover the structure, and before heavy surface loadspass over. The basic criterion for stability isminimum cover. Of course, compaction of topfilldirectly over the structure must be done carefully.See Chapter 16.

3. Minimum Cover Stabilization of the structure is minimum soil cover toprotect the structure from surface wheel loads.Minimum cover is analyzed in Chapter 13 forcircular flexible pipes. The same analyses apply tolong spans where the critical radius is the radius ofthe top plates. For analysis by ring compression, thediameter, D, is the span. Two additional concernsmust be investigated: multiple axle loads anddistribution of wheel loads on the structure.

a) Multiple axle loads can affect soil pressuredistribution on top of the structure. It may beprudent to analyze the effect of wheel spacing onaxles — especially if the load could be moving

longitudinally along the pipe. Wheel spacing mightjustify a finite element analysis or a Castiglianoanalysis with a more complex soil pressure diagramthan was used for circular pipes in Chapter 13.

b) In Chapter 13, critical pressure on the pipedue to surface live loads was based on the rationalethat a truncated pyramid is punched through the soilcover. The live load effect on the pipe was uniformpressure over half of the ring. In the case of long-spans, the top arch may be so large that thepunched-through pressure area is smaller than thetop arch. More complicated analysis may berequired.

ExampleAn inverted pear was designed as a railroad gradeseparation. Nomenclature and dimensions areshown in Figure 25-2. Data are:Structure: 6x2 corrugated sectional plateD = 28 ft = span,ry = 25 ft = top radius,σf = 36 ksi = steel strength,E = 30(106) psi elastic modulus,t = 0.218 = nominal steel thickness,A = 3.2 in2/ft = wall cross-sectional area, I = 1.523 in4/ft = theoretical moment of inertia,

from the AISI Handbook of SteelDrainage & Highway ConstructionProducts,

S = 1.376 in3/ft = I/c.Soil:γ = 100 pcf = soil unit weight,ϕ = 30o,H = 3 ft of soil cover,Load:

a) Find dual-wheel load, W, if it is assumed tobe concentrated at midspan.

σ = PD/2A where P = Pd + Pl

Pd = 300 psf dead load of soil cover,Pl = 0.12 W/ft2 = 0.477W/H2 by Boussinesq,

Substituting values and solving, W = 66 kips. If thewheel load is HS-20 load (16 kips), the safety


Figure 25-2 Nomenclature and dimensions for a long-span, pear-shaped structure to serve as a road gradeseparation. The dimensions are used in the example and problems in this chapter. This structure meetsclearance requirements of the American Railroad Engineering Association (AREA) a) If the specified soilcover is 3 ft, what is the allowable surface load, W, based on ring compression?


3'

JMiller

factor in ring compression is 4. What about multipleaxles?

b) During installation, with compacted select soilcover of only 2 ft, there arises a need for a dual-wheel truck to pass over the structure. What is theallowable dual-wheel load, W, based on slip of soilwedges? If the structure is not deformed, stiffnessis negligible. From the dimensions of Figure 25-2,soil pressures against the structure are as shown inFigure 25-3 (top). Critical soil wedges are shown in Figure 25-3 (bottom). Ignoring shear, the soilresistance at point C on the corner plates (shoulder)is 4.745 ksf = 33 psi. Assuming that the dual tireprint is 7x22 inches, and the truncated pyramidslopes at 1h:2v, the live load area at A is(7+24)(22+24) = 1426 in2. At punch-through,

P = Pd + Pl = 1.39 psi + W/1426in2

Because Pr is constant all around the perimeter of aflexible structure, at the corner , C, Pc = 4.17P.Substituting values and solving, W = 9.3 kips. Allowan axle load of 18 kips to cross — with care andrestraint of corners (shoulders).

Safety Factor

Fortunately, the concerns for long-span structuresare worst-case. Soil placement and compaction isusually done with care. Analysis neglectslongitudinal soil arching. Neglected also, is thelongitudinal beam strength of the corrugatedstructure. Despite the accordion configuration ofcorrugations, the structure has some longitudinalstrength. Therefore, safety factors can be small. Inminimum cover tests, the punch-through load isgreater by a factor of two than is predicted by soilslip analysis. A surface load test was performed ona long-span structure, 6x2 corrugation, 0.218structural plate, 12-ft top radius, with 20 inches ofwell-compacted select soil cover. A single-axledual-wheel load punched through when the axleweight was finally raised to 168 kips. Failure wascatastrophic.

PROBLEMS

25-1 What is the maximum dual-wheel load at soilslip for the above example of a pear-shapedstructure if soil cover is assumed to be the specifiedminimum H = 3 ft? The tire print is 7x22 inches.a) Check ring compression at A,b) Ring compression at maximum span,c) What is load, W, at minimum cover?

(W = 7.6 kips)

25-2 Problem 25-1 considers only the top arch.Now consider the corner (shoulder) plates of 6 ftradius. What is the maximum dual-wheel load, W,if the shoulder plates are supported by soil of 100pcf and friction angle of 30o? Analyze for bothhorizontal equilibrium and vertical equilibrium.Include the hold-down force of the top arch on theshoulder plates. (W = 1.6 kips)

25-3 A long-span ellipse is a bridge over a stream.During a flood, the bridge is partially plugged,causing the water table to overflow the roadsurface. What must be the friction angle of thegranular soil at the spring lines when an HS-20 dual-wheel load of 16 kips passes over? See sketch.Inside the ellipse is essentially empty.

(ϕ = 32.6o)


Figure 25-3 Pear-shaped long-span showing soil pressures (top) and the pressure diagrams for the soil wedgeslip analysis (bottom). The soil cover of H = 2 ft is less than minimum in this example.


Anderson, Loren Runar et al "NON-CIRCULAR LININGS AND COATINGS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 26-1 Two important deformations of buried flexible pipes.

Figure 26-2 Crack at the invert of the lining caused by increase in radius of curvature.


CHAPTER 26 NON-CIRCULAR LININGS AND COATINGS

Linings and coatings are not always circular in crosssection. Non-circular pipes are discussed in Chapter9. Circular linings (encased in host pipes) arediscussed in Chapter 11. Following is a discussionof the effects of non-circularity on linings andcoatings.

Coatings prevent external corrosion of the host pipeand stiffen the ring. The performance limit isexcessive cracking. Linings prevent leaks andstiffen the ring. Two basic performance limits oflinings are cracking and collapse (inversion).Following are two analyses of non-circular liningsand coatings, rigid and flexible.

RIGID LININGS AND COATINGS

Rigid linings and coatings are applied to thin-walls teel pipes that are to be buried. The linings andcoatings are of high-strength Portland cementmortar. Purposes of the linings and coatings are:

1. to provide adequate stiffness of the pipe in orderto handle and transport the pipe, and to install andbackfill; and,

2. to protect the steel against corrosion and againstabrasion caused by sediment in the fluid flow andpossibly by cavitation.

For design, the first two performance limits are hoopstrength against internal pressure and ringcompression strength against external pressure. Athird performance limit is ring stiffness. The pipemust be stiff enough to be handled and installed. Afourth performance limit is excessive cracking of themortar. Pipe engineers consider performance limitto be crack widths greater than 1/16 inch. This is arule of thumb that is intended to avoid the break-outof shards in the lining and to prevent circulation ofcorrosive fluid through the crack to the steel.Shards are usually caused by a dent or inversion inthe pipe wall. Cracks in the coating that are widerthan 1/16 inch might allow electrically conductive

groundwater and transient electric al currents to getto the steel.

Some pipe engineers still hold to the veryconservative 0.01-inch crack as performance limit.Compared with the 1/16 crack (0.0625) the safetyfactor is more than 6. Clearly, a broad safety zoneis available for mitigation.

A paradox arises between ring stiffness and mortarcracking. Because the mortar is thicker than thesteel, ring stiffness is affected primarily by thicknessof the mortar. However, the thicker the mortar, thewider are the cracks at any given ring deflection.The optimum coating thickness between desirablestiffness and undesirable crack width comes fromexperience. Both the minimum stiffness and themaximum crack width are related to ringdeformation. Of course, ring deformation alsodepends upon loads and embedment soil.

Ring deformations in the following analyses areeither: a) ellipse, or b) D-bedding. See Figure 26-1.The ellipse is the basic deformation under verticalsoil pressure. The D-bedding (so-called by rigid pipeindustries) is a worst-case deformation (usuallylabeled "impermissible"). The pressure, P, can begreater than the soil prism load when surface liveloads pass over, or when a deflected pipe isrerounded by internal pressure.

Notationr = radius to the neutral surface of circular pipe,r' = maximum (or minimum) radius,d = ring deflection = ∆/D where D = 2r,∆ = vertical decrease in diameter of deflected

ring,t = thickness of each lamina in the wall,w = width of a crack in the mortar (See Figure

26-2), c = distance from the neutral surface of the wall

to the exposed mortar surface. For worst-case analysis, this is the thickness of mortar,either lining or coating.


It is sufficiently accurate to assume that the steel isthe neutral surface. Whether a radius to the neutralsurface is to the inside or the outside surface of thesteel makes little difference.

Crack Width From the geometry of Figure 26-3, assumingconservatively that the crack penetrates to the steel,c is the coating mortar thickness, and the crackwidth is w = c∆θ where ∆θ = (1/r'-1/r)inch.Substituting with r in inches,

w = c(1/r - 1/r')inch2 . . . . . (26.1)

Radius of CurvatureThe maximum radius of curvature, r', is at 6:00o'clock as follows:

Ellipse, r'/r = (1+d)2/(1-d) . . . . . (26.2)

D-bedding, r/r' = 1 - 5d . . . . . (26.3)

For a derivation of Equation 26.3, see Figure 26-1a.From Appendix A, the maximum moment is at theinvert, B, and is, M = 0.5872Pr2

From mechanics of solids,M/EI = 1/r - 1/r'whereI = mom. of inertia,E = mod. of elasticity,r = circular radius,r' = radius at B.

Substituting for M, and solving,r/r' = 1 - 0.5872(Pr3/EI)

From Appendix A,d = 0.116(Pr3/EI)whered = ring deflection.

In terms of d, eliminating (Pr3/EI),

r/r' = 1 - 5d

The minimum radius of curvature for the ellipse is atspring lines (9:00 and 3:00 o'clock) where, fromFigure 26-1a, r' = rx and,

r'/r = (1-d)2/(1+d) . . . . . (26.4)

The minimum radius, rx, for the D-bedding does notdiffer significantly from an ellipse.

CRACKS IN MORTAR LINING

Structural cracks in the lining can be performancelimits if shards break out. Potential shards can bedetected by inspection and by rapping on the lininginside of the pipe. The lining is usually high-qualitymortar. Density is achieved by centrifugallyspinning the sand-cement slurry into place. Water issqueezed out. The inside of the pipe is a satin-smooth, cement-rich surface. Hair cracks developif the lining dries out. For storing and transportingpipe sections, the ends are often sealed with sheetsof plastic to reduce drying.

Hair cracks may develop during backfilling becauseof ring deflection. Pressure in the pipe rerounds thepipe and closes deflection cracks. Cracks narrowerthan 1/16 inch are closed by autogenous healingwhen the mortar is wet. Autogenous healing is theformation of silica gel by hydration. Small cracks inthe lining usually don't penetrate to the steel.

Width of Cracks in Mortar Lining Cracks open when the ring is deflected. Typicallythe widest cracks are inside the pipe at 6:00 and12:00 o'clock. See Figure 26-2. It is reasonable toassume that the inside surface of the steel is theneutral surface. In the following analyses, twodeformations are compared: elliptical, and D-bedding. From Equations 26.1, 26.2, and 26.3, therelationship between width of crack, w, and ringdeflection, d, can be found. If the ring is deformed,the relationship between w and r’ may be moredependable. From inside the pipe, r’ can be foundby laying a horizontal cord (straightedge) at B,


Figure 26-3 Free-body-diagram of a section of pipe wall at spring line. The notation shown is used in analysisof crack width, w, in the mortar coating. w = c∆θ.

Figure 26-4 Exposed cracks in a coating that is deflected into an ellipse, shown exaggerated.


and measuring the vertical middle ordinate from thecord to the lining surface. Then r' = L2/8e where Lis cord length, and e is the middle ordinate from cordto pipe surface. It is assumed that radius, r', iscircular.

Caveat Once the lining is cracked, a plastic hinge may formin the steel with a radius smaller than the meanradius subtended by the cord. The plastic hinge canbe analyzed from yield stress of the steel and theradius of curvature. A plastic hinge is recognizedvisually as the crimped rim around a dent.

Example:What are the widths of cracks in the lining of a pipeat the invert? The pipe is nominal 42-inch (inside)diameter.r = 21.5 inches to the neutral surface (assumed to beat the inside surface of the steel),c = 0.5 inch = thickness of the lining.

From Equations 26.1, 26.2, and 26.3, in terms of ringdeflection, d, the maximum radii of curvature, r', andcorresponding widths of crack, w, are listed in Table26-I.

CRACKS IN MORTAR COATING

The coating on some steel pipes is tape.Improvements in the protective qualities have madetape attractive. However, if ring stiffness is ofconcern, or if increased weight of pipe is desired,then mortar coating is used.

Excessive cracking is performance limit. Excessivecracking allows corrosion of the steel fromaggressive groundwater and electrical currents inthe soil. As with linings, cracks are caused either byring deflection or by flat spots (dents).

Ring Deflection of Ellipse and D-Bedding:As a result of ring deflection, the widest cracks inthe coating occur near spring lines. Figure 26-4 isan exaggerated sketch. For worst-case analysis, itis assumed that the neutral surface is near enough to

the outside surface of the steel that the coating isentirely in tension. It is noteworthy that thedeflected radius at spring line is roughly the same forthe ellipse and the D-bedding. The ellipse isanalyzed in the following.

Example:What is the width of a crack in the 1.5-inch mortarcoating on the following pipe.ID = 72 inches = nominal diameter,D = 74 inches = outside diameter of steel

(assumed neutral surface, NS)w = width of crack,r = 37 = initial circular radius of the NS,r' = deflected radius of elliptical NS at spring

lines,c = 1.5 from NS to the outside surface of

coating (coating thickness),d = ring deflection (percent) of elliptical cross

section.

From Figure 26-3, w = c∆θ where∆θ = θ '-θ = (1/r'-1/r)inch.

w = c(1/r'-1/r)inch . . . . . (26.5)

where r' = rx in Figure 26-1a. This r' is approxi-mately the same for both the ellipse and the D-bedding of Figure 26-1. For an ellipse, the deflectedradius, r', is,

r'/r = (1-d)2/(1+d) . . . . . (26.4)

Substituting values into Equation 26.1 and neglectingthe small value of d2,

w = (r/c)[3d/(1-2d)] = K(r/c) . . . . . (26.6)

where K is a ring deflection factor in brackets, [ ].See Table 26-II for values.

In Table 26-II, e is the middle ordinate from atwelve inch cord on the inside of the pipe to theinside surface of the pipe. See Figure 26-5. Theneutral surface (NS) is approximately at the outsidesurface of the steel where r = 37 inches. If the


Table 26-1. LINING CRACK WIDTHS AS FUNCTIONS OF RING DEFLECTIONat the invert of 42D pipe with 0.5-inch mortar lining.

Elliptical deformation D-bedding deformationd (%) r' (inch) w (in) r'(inch) w (in)1 22.154 0.0007 22.632 0.00122 22.825 0.0014 23.889 0.00233 23.515 0.0020 25.294 0.00354 24.223 0.0026 26.875 0.00475 24.951 0.0032 28.667 0.00586 25.700 0.0038 30.714 0.00707 26.468 0.0044 33.077 0.00818 27.258 0.0049 35.833 0.00939 28.070 0.0054 39.091 0.0105

Cracks in the D-bedding deformation are wider by roughly two to one. A 0.01-inch crack in D-bedding wouldoccur at ring deflection of d = 8.6%. A more dependable equivalent is the long radius of roughly 37 foundby measuring middle ordinate from a cord (straightedge) at the invert.

Table 26-2. COATING CRACK WIDTHS — ELLIPSE DEFLECTIONat the spring line, of 72D pipe, 1.5-inch mortar coating.

d K w r' e(%) (inch) (inch) (inch)0 0.0000 37.00 0.5001 0.0306 0.0012 35.90 0.5162 0.0625 0.0025 34.84 0.5323 0.0957 0.0039 33.80 0.5494 0.1304 0.0053 32.79 0.5665 0.1667 0.0068 31.80 0.5846 0.2045 0.0083 30.84 0.6037* 0.2442 0.0099 29.91 0.62325** 1.5000 0.0608 16.65 1.150

* w = 0.01-inch crack.** w = 1/16-inch crack. More precisely, d = 25.34%, but such precision is not justified. The inaccuraciesof these approximate ellipses increase as ring deflections increase. Moreover, the assumption of ellipse isquestionable — especially after the mortar becomes cracked at large ring deflections.K = 3d/(1-2d) = ring deflection factor for ellipse.r' = radius at springline.e = middle ordinate at the invert from a 12-inch cord to the lining.


thickness of steel plus lining is 1.0 inch, the insideradius is r'-1 inch for this 72D pipe with 1.5-inch-thick coating. In this example, from geometry, interms of the middle ordinate, e, from a 12-inch cordto the lining,

r'-1 = 18inch2/e . . . . . (26.7)

Equation 26.7 provides a means from inside the pipeto estimate the width of cracks in the coating. Forthis example, if inside the pipe, e = 0.625, fromEquation 26.7, r' = 29.8 in, which is then substitutedinto Equation 26.5 to find that w = 0.01 inch. A listof values appears in Table II for this particularexample. The procedure applies to other mortar-coated pipes.

Excessive Crack Widths

What is excessive crack width? For linings,experienced pipeline engineers say "any crack widerthan 1/16 inch." A more conservative specificationis the 0.01-inch crack. Water in the pipe causes thelining to expand and close cracks. Autogenoushealing (hydration of silicates in the mortar) sealssmall cracks. For coatings, also, the 0.01-inch crackis conservative. For 1/16 inch coating cracks,conditions for corrosion are affected by: electricalground current, the water table, and water quality.

Cracks in coatings do not penetrate to the steel if theonly loads are soil pressure. It might be argued thatwhen the pipe is pressurized internally, the steelstretches, and cracks in the coating widen andpenetrate to the steel. However, pressure in thepipe rerounds the pipe. So cracks in the coating arenarrowed by the pressure.

It is noteworthy that cracks are widest when thepipe is emptied after hydrotests. In service, withpressure in the line, cracks are narrow. Cycles ofdewatering cause pipe-soil interaction to stabilize atcrack widths that are less than they wereimmediately after hydrotests. Over time, the soilmigrates into place against the pressure-stiffened

pipe. Thereafter the soil holds the pipe in its nearly-circular shape. Consequently, service life is notshortened by cycles of dewatering.

Flat Spots

Dents in the pipe are often called "flat spots" eventhough they are not flat. Inside the pipe it is possibleto measure the radius of curvature of the "flat spot."Then from Equation 26.1 the width of cracks in themortar can be estimated. An alternate rationale isFigure 26-6 which is a grossly exaggerated sketch ofa flat spot in a coating. The crack at B in the middleof the flat spot is not a problem because it opens tosteel — not to groundwater. The two cracks at theends of the flat spot each open at roughly half themiddle crack width. Therefore, if the lining is half asthick as the coating, the crack widths, w, in thecoating are roughly equal to the crack widthmeasured in the lining at the middle of the flat spot.If the flat spot is circular rather than long, cracks inthe lining appear as a starburst. Of course, crackwidths cannot be predicted precisely. The aboverationale is conservative.

Flexible Linings and Coatings

Development of flexible linings and coatings, such asepoxies and tapes, is significant. Not only iscorrosion resistance achieved, but resistance toimpacts and pipe deformations is impressive. Abackhoe tooth can dent a mortar-lined steel pipeleaving a starburst in the mortar lining, but withoutdamage to the tape coating.

Inversion Analysis

One cause of inversion is a concentrated reaction, Q, (hard spot). See Figure 26-7. From Appendix A,the moment at B is, MB = 0.3Qr. For worst-caseanalysis, mortar-to-steel bond is discounted. Forinversion analysis, see Figure 26-8. In the examplethat follows, the data are:


Figure 26-5 Procedure for finding ring deflection of an elliptical pipe by laying a 12-inch cord, at the springline and measuring e.

Figure 26-6 Exaggerated sketch of flat spot in a mortar coating.


Figure 26-8 Equivalent wall section in mortar for stiffness analysis of the mortar coating at invert, B.


Q = P(OD) + W per unit length,P = soil pressure on the pipe,W = wt (pipe + contents) per unit length,σf = 1200 psi = yield strength of mortar,n = 7.5 = Es /Em,Es = 30(106) psi = mod/elast, steel,Em = 4(106) psi = mod/elast, mortar,tl = 0.75 = thickness of the lining,ts = 0.32 = thickness of the steel,tc = 1.50 = thickness of the coating,rl = 36.38 = radius to center of lining,rs = 36.91 = radius to center of steel,rc = 37.82 = radius to center of coating.

Example:What is the line reaction, Q, at inversion? First, findthe moment resisting ring stiffness of the criticalmortar coating as a fraction of the total ring stiffnesswhich must resist moment, MB = Qr/4. See Figure26-7 and Appendix A. Total ring stiffness =ΣEmb(t/r)3/12.

b t r Emb(t/r)3

Lining 1 0.75 36.38 35Steel 7.5 0.32 36.91 20Coating 1 1.50 37.82 250

TOTAL 305

The last column, Emb(t/r)3, is ring stiffness, 12EI/r3,where moment of inertia is I = bt3/12. For ratios, the12 is factored out.

Moment in the Coating: (Coating is critical)Mc = (250/305)MB = 0.82MB = Qr/4. Coating yields(i.e., cracks) at Mc = σ f (tc)

2/6 = 450 lbin/in. Since r= 37.82 for the coating, Q = 570 lb/ft. Q is thecritical line load that cracks the coating. The weightof pipe and contents is W = 2370 lb/ft. Clearly, thereaction must not be the concentrated Q-reaction ofFigure 26-7. The pipe must be supported by soilunder the haunches. As soon as the coating cracksat the invert, the lining and steel let go, and the Q-reaction is distributed over an area.

Remedies

Coating Ideally, wherever cracks are excessive in the mortarcoating, the pipe could be uncovered and re-roundedby internal pressure. Then the embedment could becarefully re-compacted while the pipe is circular. Ifit is not practical to pressurize the pipe, it may bepossible to reround the pipe by carefully monitoringring deflection during re-compaction of sidefill soil.

Lining Linings tend to expand in a moist environmentthereby reducing crack widths. One basic concernis cracks so wide (>1/16 inch) that water couldcirculate through cracks to the steel. Another basicconcern is shards of lining that might break out.Break-out would require suction such as theBernoulli effect. Loose shards can be detected bythe flat, hollow sound when the lining is rapped witha hard object at locations of multiple cracks, parallelor starburst. Wide cracks may be associated withinversions (Q-reactions) or flat spots (dents). If thepipe is not uniformly bedded, but is propped up onhigh spots, and if ring deflection is excessive duringinstallation, hydrotests tend to reround the pipe byforcing it down against the high spots. The resultcould be inversion at the high spots, and the potentialfor shards to form.

General Uniform bedding is essential. Ring deflection duringinstallation should be limited by specification. Goodembedment should be placed with care. The reasonfor an embedment is remembered as P5 — packingfor placing, positioning and protecting the pipe. Infact, the embedment is part of the conduit — not justpressure on a pipe.

Example 1A tape-coated, mortar-lined steel pipe is installedand hydrotested. Then it is inspected and found to


have excessive ring deflection and flat spots atvarious locations under the haunches. Maximumallowable ring deflection was conservativelyspecified as 3%. The embedment had been placedby covering the pipe with sand, then jetting down tothe level of the invert with high-pressure water jetsthat flush soil under the haunches. What causedexcessive ring deflection and flat spots? What canbe done to remedy or mitigate it?

Class pipe: 150 to 200D = 43 inches to neutral surface of the wall,t = 0.175 = thickness of the steel,tl = 0.500 = thickness of the lining,E = 30(106) psi = mod/elast, steel,El = 3(106) psi = mod/elast, mortar lining,ν = 0.3 = Poisson ratio,rs = 21.59 = mean radius of steel,rm = 21.25 = mean radius of mortar lining,r = 21.5 = approximate radius to NS,t = 0.2645 = wall thickness of equivalent

unlined steel pipe.

Design of the Pipe — OK

1. Hoop tension stress under internal pressure is notexcessive.

2. Ring compression stress due to external soilpressure is OK.

3. Composite ring stiffness is approximately EI/r3

= 4.59 psi. For the steel only, EI/r3 = 1.33 psi. Forthe mortar lining only, EI/r3 = 3.26 psi. Thecomposite ring stiffness is the sum, 4.59 psi. It isassumed that there is no bond between mortar liningand steel. Because there may be bond, and becauseshrinkage cracks are disregarded, values for ringstiffness are not precise. The composite ringstiffness of 4.59 is equivalent to D/t = 163 forunlined steel pipes. Steel pipe engineers recommenda maximum D/t = 288, or, with care in installation, upto D/t = 325. Ring stiffness is OK.

4. Ring deflection is limited by specification to amaximum of 3% to be sure that the lining will not

crack excessively. Small cracks (less than 1/16-inch-wide) close in time by autogenous healing(hydration of the silicates in the cement.)

5. Cracks should be no wider than 1/16 inch. Figure26-2 shows a crack in the lining at the invert. Figure26-9 shows crack width, w, as a function of ringdeflection, d, when the pipe is deflected into anellipse. The ordinate at right is the ratio of radii,maximum ry to mean circular, r. The width of crack,w, can be found either as a function of the ratio ofradii, ry /r; or as a function of ring deflection, d. Themaximum radius, ry, of a flat spot or inversion in thepipe can be found by measuring the ordinate to thepipe wall from the middle of a cord of known length.In this pipe (r = 21.5, t = 0.59 lining to center of thesteel), for a 1/16- inch crack to open, the pipe wallmust invert to a radius of ry = -17 in. See Figure26-10. For cracks to occur in the steel, radius rymust be less than 2.5t. Manufacturers recommenda lower limit of ry = 7t.

6. Longitudinal stresses are caused by temperaturechange, internal pressure change, and longitudinalbending moment. Beam action is not anticipated ifthe pipe is supported by soil bedding and soil underthe haunches. The longitudinal design is OK.

Installation and Hydrotesting

1. The buried pipe-soil interaction has stabilized.The sidefill is compacted to 90% standard density.Soil arching action has been achieved. No increasesin deformations or stresses are anticipated. Themaximum ring deflections and flat spots occurredduring installation or hydrotesting.

2. Flat spots were discovered in the invert. SeeFigure 26-11. The result could be disbonding andspalling of the lining due the Bernoulli lift and tovibrations (turbulence) in water flow. Long flatspots should be rerounded and lining replaced ifwidth of the flat spot is greater than roughly b = 5inches. Circular flat spots should be rerounded andlining replaced if diameter of the flat spot is greater


Figure 26-9 Width of crack, w, in the lining at the invert (bottom) of the pipe due to ring deflection fromcircular radius to invert radius (or maximum measured), and from circle to ellipse. For this pipe, r = 21.5, t= 0.59 to center of the steel.


Figure 26-10 Inversion of the invert of a pipe showing (top) the circular ring radius, r, and the inverted(negative) radius, ry, of the invert at 1/16-inch crack in the half-inch-thick mortar lining. r = 21.5, ry = 17);and showing (bottom) the critical radius of steel at cracking if critical strain is ε = 20%.


Figure 26-11. Flat spot at the invert of a pipe.


than roughly b = 7 inches. For different data, espe-cially for pressures greater than 100 psi, values forb must be recalculated.

3. Voids were left under the pipe during installation.Figure 26-12 shows a void between reactions underthe pipe such that the pipe performs as a beam. Iflength of the beam is greater than about L = 58 ft,longitudinal stress in the steel exceeds yield. Loosesoil under the haunches reduces beam stress.Compacted soil under haunches eliminates beamaction completely.

4. Voids under the pipe contribute to inversion of thepipe invert. If loads and reactions are as shown inFigure 26-12, the line reaction is F = wL/LB. If soilcover is 6 ft at 110 pcf, and the pipe is full of water,w = 3076 lb/ft. With some loose soil support underthe haunches at angle of repose, the net load isreduced to 40% (or less), and w = 1.2 k/ft.Assuming L/LB = 2, the reactions are 2.4 k/ft. Ifreactions are line reactions as shown under the pipecross section at the right of Figure 26-12, inversionoccurs at F = 1.4 k/ft. Clearly, 2.4 k/ft would invertthe pipe. If the reaction is distributed over a beddingangle of 90o, inversion occurs at roughly F = 3 k/ft.With the more reasonably distributed reaction,inversion would be unlikely — until, that is, the pipeis hydrotested.

5. Inversion is possible after the pipe is hydrotested(125% of its designed internal pressure) if ringdeflection is significant. If ring deflection is 3%,internal pressure rerounds the pipe and increasesvertical diameter by 1.3 inch. Load on this pipe isincreased by an additional 1 k/ft as the pipe tries tolift a soil wedge or as it compresses soil as shown onthe top and bottom of the cross section of Figure 26-13. Assuming L/LB = 2, the reactions are F = 3.2k/ft. Inversion of the invert occurs at roughly F = 3k/ft. Now inversion is possible.

\Structural Performance and Performance LimitsDuring Operation

1. Operation causes pipe-soil interaction to come

into equilibrium. Water pressure rerounds the pipeand leaves gaps on the sides. See Figure 26-13.Turbulent flow of water vibrates the pipe and shakessoil down unto the gaps. Cycles of dewateringreduce ring deflection to an equilibrium somewherebetween zero and the ring deflection at hydrotest. Iffailures do not occur during installation orhydrotesting or as a result of breakout of shardswithin a short period of operating time, failure willprobably not occur.

2. The specified ring deflection, d = 3%, is intendedto protect the mortar lining of the pipe. Ringdeflection, per se, is not so much a performance limitas it is a condition that contributes to flat spots andcracking of the lining.

Remedies for Flat Spots

1. Flat spots should be identified and located. In thispipe the critical flat spots are those long flat spotsgreater than 6 inches wide, and those circular flatspots greater than 7 inches in diameter. If voidsunder the pipe on either side of a flat spot areextensive, they should be filled with low-strength,flowable, soil cement. If voids are filled beforewater (weight and pressure) is in the pipe, thereaction force, F, will be reduced significantly. SeeFigure 26-12. The cause of flat spots will beeliminated.

2. Flat spots should be inspected to ascertain if theradius of curvature is so small that the steel couldcrack and weaken the hoop tensile strength. Thecritical radius for pipe steel is ry = 2.5t. See Figure26-10 (bottom). Steel engineers recommend thatradius of curvature be greater than seven times wallthickness.

3. Wide cracks (>1/16 inch) or concentrations ofcracks that might indicate disbonding and potentialbreaking out of shards, should be repaired.

4. Legal remedies, such as warranties over anextended period of time, could be investigated.


Figure 26-12 Pipe as a beam on reactions, showing the F-force at the reactions, and showing the momentdiagram for a fixed-ended beam.

Figure 26-13 Soil pressure on pipe due to rerounding by internal pressure.


Example 2What are the critical radius of curvature and the ringdeflection in terms of critical crack width? Ringdeflection is not exclusively critical, but is ac onvenient basis for relating other conditions whichare critical such as critical crack width. Criticalradius of curvature can cause critical crack width.

Given:r = radius of circular pipe,ry = maximum radius of curvature,d = ring deflection = ∆/D ,w = width of crack in lining,t = thickness of lining.

Find:ry/r = ratio of radii.

From geometry of an ellipse,

ry /r = (1+d)2/(1-d) . . . . . (26.8)

Find:w = width of crack.

From the mechanics of pipering analysis,

w = t(1/r - 1/ry) . . . . . (26.9)

The two equations, 26.8 and 26.9 interrelate thethree variables, w, ry, and d. Because d applies toelliptical ring deflection, it is limited as a criterion forstructural integrity. As long as soil support under thepipe is uniform, ring deflection is a relevant criterionfor structural integrity. Ring deflection should belimited by specifications . Pipeline engineers specifyand always anticipate that the pipe will be supporteduniformly.

However, in case of flat spots or inversions, the ratioof radii, ry /r, is a more relevant criterion forstructural integrity of the pipe than is ring deflection.This presumes that support is not uniform under thepipe. The reactions are hard spots. The ratio ofradii, ry /r, is a means of

estimating deficiency of the pipeline in terms ofcracks in the lining and reduced service life. It isalso a basis for estimating the extent ofencroachment into the margin of safety (safetyfactor) by the installer of the pipeline in cases wheresupport under the pipe is not reasonably uniform.

Example 3What is the rerounding of a 42D cement mortarlined pipe due to internal pressure?Data:ID = 42 inches,t = 0.175 = steel thickness,t' = 0.5 = thickness of mortar lining,B = breadth of the flat spot (or diameter),σf = 42 ksi = yield stress of steel,P = 100 psi = internal pressure that rerounds the

ring,The moment in the steel at plastic hinging is 3/2times the moment at yield stress.

For some flat spots the longitudinal length is greaterthan the circumferential breadth. If the length isapproximately equal to the breadth, the flat spot isroughly circular. For the examples that follow, eachis analyzed separately.

Long Flat SpotAt plastic hinging, (B/d)2 = 3σf /P. If P = 100 psiand σf = 42 ksi, B = 6.21 inches.

If the long flat spot is wider than about 6.21 inches,internal pressure of 100 psi will reround the steel.The lining is too brittle and cracked to reroundwithout spalling. Therefore, long flat spots widerthan about 5 or 6 inches should be rerounded and thelining should be chipped out and reapplied. Forinternal pressure greater than 100 psi, the breadth ofthe critical flat spot decreases.

Circular Flat Spot

For circular flat spots, the above equation is roughly(B/d)2 = 6σf /P. Solving, the critical diameter is B =8.8 inches. Therefore, circular flat


spots greater in diameter than about 7 inches shouldbe rerounded and the lining should be chipped outand reapplied.

Rerounding by jacking requires care. A line force of80 lb/in can deflect the ring by about 3% if the pipeis not buried. The 100 psi internal pressure requiredto reround the steel becomes 621 lb/in over the widthof the critical 6.21-inch flat spot. It might be prudentto apply the rerounding force over a smaller area, orby a ball peen tip, and then peen the flat spot into acircular cross section. The reaction end of the jackshould be supported over a large area — oncompressible material that conforms with, anddoesn't concentrate pressure on, the liner. This mayrequire longitudinal timber lagging or strong-backs,and possibly, with compressible packing betweentimber and pipe lining. It may be well to rehearsethe procedure using 6x6 or 8x8 timbers.

EGG-SHAPED LINING

In order to achieve improved hydraulics in variableflow pipes such as sewer pipes, the cross section ofthe pipes is egg-shaped. See Figure 26-14.Rehabilitation of such pipes results in linings withvariable radii. Following is a discussion of suchlinings. Tests are recommended because of themany variables in egg-shaped linings. It isnoteworthy that the following analysis has oneserendipitous feature. From similitude, thefundamental variables can be combined intodimensionless pi-terms which have no feel for size.Therefore, the observations (measurements) on themodel apply to any size of "standard egg" if the samematerials are used in a test model as in theprototype. One size of model can predictperformance of many sizes of prototype. Theconditions for similitude are:

1. Corresponding materials are the same inmodel and prototype.

2. Corresponding lengths are to scale, anglesequal. The model is true-to-scale.

3. Corresponding pressures are equal.

Deflections on the model are true-to-scaledeflections of the prototype.

Rationale

When external pressure is applied to an egg-shapedlining, the lining feels circumferential compression.The circumference shortens, and a gap formsbetween the lining and the host pipe at the locationof maximum radius of curvature — in this case, atone of the long-radius sides. Pressure in the gap"blows" the rest of the lining into a tight fit againstthe host pipe. The radius of lining in the gapincreases, ring compression increases, and pressureof lining against host pipe increases. Failure isreversal of curvature of lining in the gap. If externalpressure persists, the lining crumples.

Physical Tests

1. Physical tests are important especially if thelining is plastic and creeps. Beam deflection isapproximate and usually does not occur untilcollapse is in process. This should be watched for intest programs. Compressive yield stress in the long-radius arc is a pertinent condition for collapse.Stress is a function of deflection of the arc which isa function of shortening of the circumference of thelining which, in turn, is a function of externalpressure and modulus of elasticity. Theory ofelasticity is only approximate for plastics andrequires physical tests.

2. Performance limits can be identified by tests. Itis not incontrovertible that deflection of the longradius of the egg lining is the only performance limitfor design. In test linings, snap-through occurs atyield stress after a period of time. Snap-through isthe onset of collapse.

3. The primary performance limit is excessivedeformation. The most pertinent fundamentalvariable is deformation. Failure is excessivedeformation — reversal of curvature and collapse.


Figure 26-14 Cross section of the standard egg host pipe in a lining test (taken from laboratory notes).

Figure 26-15 Critical, long-radius are BCD, showing the basic geometry of the center plane of the lining.Lengths are in inches.


Figure 26-16 Critical are BCD, showing the gap between dotted and solid lines caused by external pressure.Lengths apply to the center plane of the lining.


4. Excessive deformations include leaks in the lining,and wall crushing when circumferential compressionstress exceeds yield (based on yield strength), arcsnap-through (based on the modulus of elasticity),and, maybe, beam failure (based on both yieldstrength and modulus of elasticity). Other pertinentfundamental variables are from geometry andpressure. The three categories of pertinentvariables are: geometry, loads, and properties ofmaterials (modulus, strength, and virtual long-termmodulus). Poisson ratio is only of minor concern inplastics. Time is a function of loads (pressure) andcreep (geometry).

5. Time is a pertinent fundamental variable becauseplastics creep and change shape under persistentpressure. The effect of time is reduction of thevirtual modulus of elasticity. Data are available onthe relationship of time and virtual modulus.Therefore, the effect of time can be included inanalytical models by using a "virtual modulus." Thevirtual modulus is of value only in calculating theshortening of the circumference due to creep overthe long term. Persistent pressure includesintermittent pressure. Of course, the rate ofreduction of virtual modulus is slowed if pressuresare intermittent. The plastic tends to reboundpartially between applications of the intermittentpressures.

6. The properties of plastics remain pristine. Thetrue modulus for snap-through, is short-term modulus— not virtual modulus.

7. The precision of models must be taken intoconsideration. Classical equations are usually basedon elasticity. Linings are plastic — not elastic.

Analysis of the Standard Egg Lining

Following is the rationale for performance of aclose-fit plastic lining encased in a rigid standard egghost pipe when pressure is applied between liningand host pipe. Worst-case assumptions are:

1. The lining is cylindrical — no elbows or otherspecial sections.

2. There is no circumferential shearing stress on thelining. This is a worst-case assumption that isconservative, but is as accurate as can be justified.

3. External pressure on the egg-shaped lining isperpendicular to the surface. In service there maybe spots where the lining bonds to the host pipe, butbond tends to break down in time.

4. The circumference of the lining shortens underexternal pressure. A gap forms between lining andpipe at the location of greatest radius of curvature.The lining conforms with the host pipe except at thegap. See Figures 26-15 and 26-16.

5. The circumferential thrust, T, in the wall of thelining is constant all around the circumference of thelining, and is,

T = PR = Piri = constant,

where, at any point on the lining,T = circumferential thrust per unit length,P = external pressure,r = radius of the center plane of the lining

where the gap will form (greatest radius),R = radius of lining at the gap after gap has

formed,t = wall thickness,σ = PR/t = ring compression stress.

6. Decrease in circumference is a function ofcircumferential strain (stress divided by virtualmodulus of elasticity). It is time dependent.

7. Because the material is plastic — not elastic —virtual modulus is the slope of the secant on thestress-strain diagram over time (allowing for creep)from zero to the compression stress, σ = PR/t. Thisis approximate and conservative because stress-strain diagrams are in tension. The lining is incompression. The plastic lining "creeps" over aperiod of time when subjected to persistent


pressure. But creep reduces the circumferencewhich increases the radius of curvature and thewidth of the gap. It is a progressive sequence thatconverges as shown in the following example.

Example 4. 12 mm LINING UNDER 7 FT OFHEAD OF GROUNDWATER

What is the increase in radius of curvature from theoriginal r to deflected R due to external pressure onthe lining? What is the gap width, ∆e , of thefollowing standard egg-shaped lining?

Notation:H = 36 inches = inside height of the

standard egg host pipe,B = 24 inches = inside width of the

standard egg host pipe,P = 3 psi = pressure in the gap = 7 ft of

head at 62.4 pcf,t = 0.4724 = 12 mm = wall thickness of

the lining,C = 93 inches = mean circumference of

the lining (92.91),ro = 36 = outside radius of the long-radius

sides of the lining,r = 35.7638 = ro - t/2 = mean radius (to

center plane) of long-radius sides,R = deflected mean radius of the long

radius side at the gap,σf = 5000 psi = yield stress,E = 500 ksi = mod/elast (short-term),E' = 250 ksi = virtual mod/elast (long-

term),ψ = 36.1388o = See Figure 26-14,σ = 229 psi = Pr o /t = circumferential

stress in the wall,Arc BC = 11.2789 = r(ψ )/2 = mean arc length

before deflection,Cord BC = 11.0928 = rsin(18.0694o) = cord

length, See Figure 26-15.e = mid ordinate from the cord to the arc,∆e = width of the gap = er - eR. See

Figure 26-16.Arc BCD is assumed to be circular.

Find: R for deflected ArcBCD —SHORT TERM E = 500 ksi

First trial r = 35.7638∆C = ProC/Et = 0.0425 inchArcBC = 11.2789 - 0.0425/2 = 11.2576sin θ = 11.0928/R (1)θ = ArcBC/R = 11.2576/R (2)

Solving by iteration (values of θ must be equal.),

Trial R sin θ (2) sin θ (1)50 0.2233 0.221940 0.2777 0.277338 0.2919 0.2919R = 38

Second trial r = 38; ro = 38.2362∆C = ProC/Et = 0.0452ArcBC = 11.2789 - 0.0452/2 = 11.2562sin θ = 11.0928/R (1)θ = ArcBC/R = 11.256 (2)

Trial R sin θ (2) sin θ(1)40 0.2777 0.277339 0.2846 0.284438.5 0.2882 0.2881R = 38.7, MAX SHORT TERM

Find: GAP, ∆e . See Figure 26.16.∆e = er - eR (3)∆e = 61.5251[(1/r) - (1/R)] = 0.13 ∆e = 1/8 in.

where:(cordBC) = 11.0928er = (cordBC)2/2r = 61.5251/rr = 35.7638eR = (cordBC)2/2R = 61.5251/RR = 38.7

Find: σ = P(R+t/2)/t = 3(38.7+0.2362)/0.4724.σ = 247 psi . The safety factor is 20 based on yieldstrength of 5 ksi.


Find: P at arc snap-through (reversal of curvature).From texts on mechanics of materials,P = (k2 - 1)(EI/R3)where (See Appendix A),k = 15 at θ = 17o, 17.2 15o,

8.62 30o,I = t3/12, and θ = sin-1(cordBC/R)θ = sin-1(11.0928/38.7) = 16.66o = 17o

Substituting values,P = 16.98 psi, but only approximately.P = 17 psi. The safety factor is 5.7 based onpressure P = 3 psi.

Find: R for deflected ArcBCD —LONG TERM E = 250 ksi.

First trial r = 35.7638.∆C = ProC/Et = 0.0850 inchArcBC = 11.2789 - 0.0850/2 = 11.2364sin θ = 11.0928/R (1)θ = 11.2364/R (2)

Continuing the solution by iteration (Values of θmust be equal.),

R = 41.5, MAX LONG TERM

∆e = 1/4 inch

σ = 265 psi . Because yield strength is 5ksi, thesafety factor is 19 for circumferential (ringcompression) stress in the wall.

Find: P at arc snap-through (reversal of curvature)

P = (k2 - 1)(EI/R3)where k = 15, roughly, at θ = 15.5o.θ = sin-1(11.0928/41.5) = 8.5o

I = t3/12,t = 0.4724,R = 41.5,R/t = 87.85,E = 250 ksi,Substituting values, P = 13.77 psi.

P = 14 psiThis is equivalent to a head of 32 ft. The long-termsafety factor is 4.6 based on a head of 7 ft.Closure

The rationale described above is approximate, butconservative. Overlooked are such contributions tostructural integrity of the lining as the following.

1. Longitudinal action of the cylindrical lining helpsto reduce deflections. The above rationale considersonly two dimensions — the cross section.

2. Ring stiffness of the lining helps to reducedeflections. Ring stiffness begins to affect thedeflection as soon as deflection begins. Beforedeflection, stiffness has no effect on performance.

3. The moduli of elasticity are based on tensiontests. Compression moduli are greater than tensionmoduli.

4. The yield strength is based on tension tests.Compression yield strength is greater.

PROBLEMS

26-1. Find width of crack in the coating of theexample in CRACKS IN MORTAR LINING.

26-2. Find width of crack in the lining of the examplein CRACKS IN MORTAR COATING.

26-3. What is the radius of curvature of a mortar-lined steel pipe at the invert where a single 0.01-inch crack opens in the mortar? The mortar is 1.0inch thick and the ID of the thin-wall steel pipe is 36 inches. The pipe is subjected to externalhydrostatic pressure, but not to internal pressure.

26-4. The mortar-lined pipe of Problem 26-1 leaks.It is to be lined with a close-fit high-densitypolyethylene (HDPE) lining that is 0.9 inch thick.The long-radius arc is 60o. What is the persistentexternal pressure on the HDPE lining at snap-through in 50 years?


Anderson, Loren Runar et al "RISERS "Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure 27-1 Notation for risers, showing radial pressure on the left side and shearing stress (drag-down) onthe right side that is caused by soil compression.


CHAPTER 27 RISERS

Risers are basically pipes that are buried vertically,or nearly so. In fact, some risers are inclined. Theyare called risers because they usually "rise" from aburied tank or pipe. Risers serve many purposes:access (such as manholes and mine shafts),cleanout, ventilation, collection of gas (methane fromsanitary landfills), standpipes (for water pressurecontrol), bins (for feeding underground conveyors),accumulators (to collect entrapped air in waterpipes), etc. Most risers are cylinders (usually pipes).See Figure 27-1. Basic concerns are ringcompression and longitudinal (vertical) thrust. Thecritical location of both is usually at, or near, thebottom of the riser.

Ring Compression

From Chapter 6, ring compression stress is,

fc = rσ x /t

wherefc = circumferential stress in thin-wall riser,σx = external radial pressure against riser,r = outside radius of curvature of the riser,t = wall thickness of the riser.

For design, fc must be less than the yield stress ofthe riser. The safety factor is needed becausepressure, σx, is sensitive to soil properties and to soilplacement, which never assures uniform pressure.Because of soil arching action, σx is neither activesoil pressure, nor radial elastic stress. These arelimits only. At the lower limit, if a vertical hole werebored into the ground, and the riser carefully slippeddown into it, σx would be zero down to some depthbelow which the free-standing hole collapses (cave-in) under the soil weight. Above this collapse depth,the only pressure on the riser is hydrostatic pressureif a water table is above the collapse depth. In sucha case, stability analysis applies as discussed inChapter 10. The "vacuum" in the pipe is externalhydrostatic pressure. The question, of course, is

critical depth of the free-standing bored hole. Thesurest procedure is a test hole.At upper limit, if the soil is cohesionless, the riserfeels radial active pressure,

σ = Kσ z

whereσx = radial pressure on the riser at depth z, K = (1+sinϕ)/(1-sinϕ) from Mohr circle,ϕ = soil friction angle,σz = equivalent vertical stress caused by

compaction of the soil.

Active pressure is assumed if the soil is loose andslides into place against the riser. If the soil iscompacted, σz is roughly equivalent to theprecompression stress at reversal of curvature ofthe stress-strain diagram for the compacted soil.See Figure 27-2. This analysis is upper limitbecause arching action (σx) of the soil around thepipe is ignored. In fact, arching action is significant.The designer can get a feel for the effect of archingaction by an elastic analysis.

Elastic theory provides a conservative stressanalysis. Radial pressure against the riser at depth,z, is σx. Principal stresses on an infinitesimal cubeof soil are shown in Figure 27-3. From elastictheory, strains are:

Eεz = σ z - ν(σ x + σ y)Eεy = σ y - ν(σ x + σ z)Eεx = σ x - ν(σ y + σ z)

where:E = modulus of elasticity,ν = Poisson ratio,ε = strains in the directions indicated,σ = principal stresses on the infinitesimal soil

cube in the directions indicated.

It is reasonable to assume that horizontal strains, εx

and εy, are zero because the soil is confined


Figure 27-2 Sketch of stress-strain diagrams for soil, showing precompression stresses located wherecurvature reverses. Precompression stresses are approximately equivalent to the effect of compaction (soildensity). With no compaction (70% density?) curvature does not reverse.

Figure 27-3 Infinitesimal soil cube at the risersurface, showing the principal stresses. σz is thevertical soil stress. σx is the radial soil pressure onthe riser. σy is the circumferential stress whichdevelops soil arching action.

Figure 27-4 Horizontal stress, σx, is equal to thebearing capacity of the soil that resists horizontalmovement of the riser into the soil. This is themaximum stress that develops when the riserdeflects laterally into the soil.


horizontally. Therefore, σy = σ x. Solving forpressure against the riser,

σx = νσ z /(1-ν) . . . . . (27.1)

According to elastic theory, pressure on the riser issensitive to Poisson ratio, <, as follows:

ν σx Applied to:0.00 0 cork, trash.0.1 0.11σz

0.2 0.25σz

0.3 0.43σz

0.33 0.50σz Assume 0.5σz for waste.0.4 0.67σz for some plastics.0.5 1.00σz constant volume elastics.

The radial stress, σx, varies from zero to σz. Somedesigners use the elastic model with a Poisson ratioof 0.33, for which radial pressure on the riser is σx

= σz/2. But this rationale applies only to elasticmaterial. Soil is not elastic. Based on active soilpressure against the riser,

σx = σ z(1-sinϕ)/(1+sinϕ)

If soil friction angle is ϕ = 30o, σx = σz/3. Althoughconservative, the coefficient, 1/3, is more reasonablethan the 1/2 used by some designers.

If the riser deflects laterally, movement of the risercauses passive soil resistance. From Chapter 4, atpassive soil resistance,

σx = σ z (1+sinϕ)/(1-sinϕ)

If soil friction angle is ϕ = 30o, σx = 3σz.

This is analyzed as pressure on one side of the riser.See Figure 27-4. The only reaction to this force iscantilever action which may be more critical thanring compression if the riser can deflect laterally.Design is classical analysis for a vertical cantilever.It may be necessary to locate points ofcounterflexure in the deflected riser.

Thrust

Thrust is the vertical force on the riser. It is causedby the frictional "drag-down" as the soil compresses.Thrust depends upon: pressure against the riser, thecoefficient of friction of the soil on riser, and therelative movement (compression) of the soil withrespect to the riser. A safety factor is required.Analysis of the drag-down force is similar to theanalysis of silos in Chapter 22. The following designprocedure is conservative because soil archingaround the riser is neglected in calculating drag-down. In the following it is assumed that soil iscohesionless. Two soil conditions are analyzed,without compaction and with compaction. It isassumed that soil is confined horizontally such that(radial) strains are zero. Therefore, σy = σ x. This ismodeled by a confined compression test. See Figure27-3. For design, vertical stress, Q/A, must be lessthan the yield stress, fc, reduced by a safety factor.

Q/A = fc /(sf) . . . . . (27.2)

Notation:Q = total drag-down load on the riser,A = cross-sectional area of the riser = 2πrt,fc = longitudinal yield stress in the riser,fz = longitudinal (vertical) stress in the riser,r = mean radius of the riser,t = wall thickness of the riser,σx = radial (active) soil pressure on the riser,σz = vertical soil pressure at depth, z,z = depth to soil cube (Figure 3),K = ratio of σx /σ z at soil slip,

= (1- sinϕ)/(1+sinϕ) for granular soil,ϕ = soil friction angle,µ = coefficient of friction, soil on riser,sf = safety factor.

If the soil is cohesive, K must be modified. SeeChapter 4. The vertical stress, fz, in the riser atdepth, z, is an upper limit because horizontal archingof the soil, σy, is ignored. Q is the total


drag-down force of the soil on the riser as soilcompresses. It is assumed that the riser is fixed inlength, and is supported on a base that does notsettle. See Figure 27-2. At depth, z, vertical stressin the riser wall caused by drag-down is,

fz = Q/A = Kzµσ z /2t . . . . . (27.3)UNCOMPACTED SOIL

whereτ = µσx = vertical shearing stress on riser,σx = Kσz = radial soil pressure on the riser,Q = πrzτ = πrzµK σz = drag-down force.

During backfilling, soil "slides" into place against theriser such that radial pressure in Equation 27.3 isroughly active soil pressure; i.e., σx = Kσ z

K = (1-sinϕ)/(1+sinϕ) . . . . . (27.4)

Uncompacted soil The active radial pressure is σx = Kσ z where σ z isthe vertical soil stress at depth z.

Example 1

Figure 27-1 shows a riser in uncompacted soil.Therefore, the σx-diagram is a triangle to somedepth, z. What is the drag-down force, Qs due toshearing stresses? The volume under the σx-diagram is simply the area under the triangle, zσx/2,times the circumference of the riser, 2πr.Therefore, Qs = µπ rzσ x.

Compacted soil If cohesionless embedment is uniformly compacted(in lifts), it is reasonable to assume that σz is theequivalent precompression stress of compaction. Ifsoil is compacted throughout the entire height of theriser, radial soil pressure on the riser is constant —not zero to maximum from top-to-bottom.Therefore, the 2 in the denominator of Equation 27-3cancels, and,

fz = Kzµσ z /t . . . . . (27.5)COMPACTED SOIL

σz is the precompression stress for samples of thecompacted soil. It can be found from stress-straindiagrams of laboratory compression tests oncompacted soil.

CAVEAT — Usually, compacted soil does notcompress with respect to the riser. Therefore,shearing stresses, τ , do not develop. The verypurpose of compaction is to prevent relativesettlement of soil with respect to the riser. Theabove rationale applies to "unusual" cases of relativemovement.

Structural Analysis of the Pipe

Figure 27-1 (right) shows loads on the pipe ring dueto soil and riser. The ring can be analyzed by closedform integration. See Appendix A. Such analysesare conservative, however, because arching of thesoil is ignored, both longitudinal and circumferential.Moment, thrust, and shear in the ring make possiblethe stress analysis of the ring. Plastic analysis ismore relevant than elastic analysis. Stressconcentrations at the intersection of riser and pipeare critical.

Surface Live Loads

If a live surface load can pass over the riser, theriser must support the entire load. If a concentrateddual-wheel load is located at a point on the rim ofthe riser, the riser can be analyzed as a short columnwith both an axial load and a moment due to theeccentricity of load from the neutral axis of the riser.The maximum stress is usually at or near the top. Itmust be less than yield stress reduced by a safetyfactor. Localized buckling could reduce the criticalload, P. The maximum stress, from classicalmechanics, is the familiar,

fc = P/A + Mc/I = 3P/2πrt, approximate . . . . . (27.6)

whereP = dual-wheel load,A = 2πrt = cross sectional area of the riser,


M = Pr = moment of force on the riser,I/c = πtr3/r = πtr2.

If the live surface load is not on the riser, but islocated on soil adjacent to the riser, the problems arevertical force on the riser, and non-uniform radialpressures.Ring Compression:The ring compression pressure, σx, under the wheelload may greater than the passive resistance of thesoil around the rim. The riser rim could invert.Approximate analysis is possible, but is usually notjustified.

ThrustIf live loads are anticipated, either the riser must beable to support the loads, or loads must be kept offthe riser. Manholes in roads support wheel loads.

It is common practice to place a collar around therim of the riser. See Figure 27-5. The collar is notattached to the riser. Therefore, it bears on soil.The collar usually rises above the surface like a curbso wheel loads do not roll onto the riser. See Figure27-5. The collar keeps the wheel load far enoughaway from the riser that radial soil pressure is notexcessive. Vertical soil stress, σz, next to the risercan be calculated using the Boussinesq procedure.See Chapter 4. Radial stress is σx = Kσz. Radialstress is reduced by increasing distance, R, from theriser. See Figure 27-6.

Example 2Assume that an HS-20 dual- wheel load is P = 16kips with tire pressure of 105 psi is located atdistance, R, from the riser. The depth to theinfinitesimal soil cube is Z. Assume that the soil iscohesionless with a soil friction angle of 30o forwhich K = 1/3. Values from Boussinesq are plottedin Figure 27-6. Clearly, σx is reduced dramaticallyas R increases. Shown for comparison is a plot forR = 0; i.e., the wheel is on soil at the edge of theriser. Theoretically, σx approaches infinity on thesoil surface where Z = 0 because Boussinesqassumes a point load. In fact, the dual-wheel load isspread over an area. The maximum pressure is

simply tire pressure, 105 psi, for which σx = 35 psi =K(105 psi).

The effect of non-uniform radial stresses on the ringcan be analyzed, if necessary, by the Castiglianoequation. See Appendix A.

Drag-down shearing stress caused by the wheelload is, τ = µσx.

See Figure 27-1. Again, the problem is evaluation ofradial stresses, σx; and, again, for cohesionless soil,the major distinction is between uncompacted andcompacted soil; i.e., between active soil pressure,Equation 27.3, and equivalent precompressed soilpressure, Equation 27.5. The drag-down shearingforce is µ times the volume under the σx-diagram;i.e.,

Qs = µ σ xdA . . . . . . . . (27.7)

See Figure 27-5. For a concentrated wheel load, P,radial soil stress, σx, at depth Z varies with radii, R,from the wheel load (P-axis) to the riser. Moreover,σx is the radial component of the Boussinesq stressat R and Z. The boundaries of the pressurized areato be integrated depend upon ring stiffness. It isreasonable to assume an area equal to Z times one-eighth of the circumference (45o arc). At eachdepth, Z, radial stress, σx , is K times the verticalstress, σz, and is assumed to be constant around the45o arc at depth Z.

Example 3

A riser is buried in uncompacted soil. Roughly whatis the drag-down force, Qs, due to the wheel load P= 16 kips at R = 10 inches from the edge of theriser? Radius of the riser is r = 16 inches.

From Figure 27-6 by counting squares, the areaunder the (R=10) curve is roughly 100 lb/in. A 45o

arc of the riser is πr/4 = 12.5 inches. The stressvolume is approximately (12.5 in)(100 lb/in) = 1.25kips. The shearing drag-down force is 1.25(µ) kips.If the coefficient of friction is µ =


Figure 27-5 Collar for preventing excessive soil pressures on the riser due to surface wheel loads, showingprincipal stresses on an infinitesimal soil cube at the riser surface and at depth, Z.

Figure 27-6 Radial pressures, σx, acting on the riser, as a function of depth, Z, and distance, R, from a 16-kip wheel load to the riser.


0.5, the drag-down force is Qs = 625 lbs. If R isdecreased to 5 inches, the area under the σx-curveis roughly 2.5 times as large. The down-drag forceis (2.5)(625 lbs) = 1.56 kips, a quarter of the wheelload. Keep wheel loads away from the edge of theriser.

Risers in Sanitary Landfills

Sanitary landfills pose unique problems because ofcompressibility of waste material. Properties of thewaste vary. Unit weight of waste is typically γ = 75pcf. Some engineers design for 80 pcf to includevariable water content and non-homogeneity.Vertical compression can be as much as 30%.Temperature can vary from below freezing to 120o

F. Landfills spread horizontally as the height rises.The rise is gradual (over many years). Design ofrisers must take into account lateral deflection due tospreading of the sanitary landfill, and drag-down(friction as the waste compresses).

Drag-down For lack of accurate values, the drag-down"coefficient" is sometimes set at 0.5; i.e., half thevertical pressure of the waste at any given depth.The result is a triangular drag-down shear diagram(τ -diagram) as shown on the right side of the riser inFigure 27-1 where τ is 0.5(σ z), and where σ z = γ z.Because sanitary landfills are usually high, z can bea large value. To relieve the riser of large drag-down thrust, and to protect the riser, a "chimney" ofcompacted, select backfill is sometimes specifiedwith the riser serving as the "flue." An alternativecould be a telescoping riser. See Figure 27-7. Itmay be prudent to provide flanges on each stick ofpipe. Analysis shows that friction should hold thesticks in place. However, variations in temperaturecause the sticks to lengthen and shorten, both ofwhich cause incremental creep downward. Flangesresist downward creep. Flanges can be placedanywhere on the larger diameter telescoping sticks.Some designers use pipes with the bell on one end toserve as a flange for the larger diameter sticks.Flanges on the smaller diameter sticks must be far

enough from the ends to allow telescoping byinsertion of the smaller diameter pipe into the largerdiameter pipe.

The required bearing area of the flanges can beanalyzed as follows. Area times bearing capacity ofthe soil (trash) must exceed the frictional drag-downplus the weight of each stick. This design procedureis conservative because downward creep andtelescoping both reduce or eliminate drag-down.Safety factors are not needed.

Resistance to Live Loads When a live load passes over a riser supported by aburied pipe or tank, the question arises as to howmuch of the load is felt. If the riser is non-compressible (no slip joints or corrugated sections),the entire live load could be supported by the pipe ortank. If the riser is compressible, it shortens underthe live load. The load, or part of the load, issupported by frictional resistance of the waste. Thatfrictional resistance is analyzed the same way asdrag-down, except that it is reversed in direction.

Lateral Deflection If insertion clearance is not adequate between thelarger and smaller sticks, lateral deflection of theriser may bind contiguous sticks. At worst, thebending moment could fracture the ends. Lateraldeflection can be accommodated by shortening thelengths of the sticks and the insertions, and byincreasing the annular space between the smallerand larger diameter sticks. Gaskets at the insertionsmay or may not be required to prevent influx ofleachate or soil. The length of the sticks may bedetermined by the rate of rise of the sanitary landfill.Telescoping sticks facilitate rise, but the stick addedfor each lift must be limited in length depending onthe method of support. A mound of select granularsoil around the base of the added stick mayadequately support short sticks. Otherwise, tiewires are needed.

Foundation for the Riser In some cases, the weight of the riser and the drag-


Figure 27-7 Diagrammatic sketch of a telescopingriser to accommodate compression of the waste insanitary landfills, shown here with flanges to resistincremental creep downward. The flanges do notnecessarily have to be at midlength of the sticks.


David Alan Rech

down force are supported by a foundation thatstraddles the pipe or tank. This is typical of heavy-duty risers such as shafts through which ore oraggregate is dropped onto a conveyor belt or skip.In addition to drag-down, both inside and outside theriser, vibrator motors are attached to the riser tokeep the product flowing.

In many cases, the riser is supported directly on thepipe or tank. See Figure 27-8, which shows a riserthrough which crushed stone is dropped onto aconveyor belt in a 120-inch-diameter corrugatedsteel pipe. Views of the opening are shown frominside the pipe. Pipe rings are cut by the opening.Therefore, the ring compression which supportsexternal loads is lost. Thrust, Q, from the riser mustbe supported by rings adjacent to those that are cut.In the example of Figure 27-7, a collar of plates iswelded to the inside of the pipe. The collar isequivalent to a boss around the hole where stressesare concentrated. The adjacent rings mustwithstand their own ring compression plus the ringcompression of the cut rings. In some cases, a steelframe outside the pipe replaces the rings that arecut. If the riser thrust is supported on the pipe, itmay be prudent to analyze the ring by Castigliano'sequation or by finite element analysis.

Example 4

A telescoping riser,PVC SDR 26, standsvertically in waste.D = 18.000,D' = 18.462,∆L = 14 inches,wt = 24.5 lb/ft,γ = 80 pcf waste,τ = 0.5σ x ,L = 11 ft,Df = flange diam.

Find:Maximum angle offset,allowed if the sanitarylandfill should spread.

tanθ = (D'-D)/∆Lθ = 1.89o

Offset = 4.4 inchesin 11 ft pipe length.

Risers on Slopes Following is a rational procedure for the structuraldesign of a riser (or any pipe) when buried on anincline.

Notation and Assumed Data — Assumptions areconservative. For the following examples, data are:Soil:Landfill trash —γ = 100 pcf = unit weight of landfill,(usually

closer to 75 pcf),µ = 1/4 = coefficient of friction of landfill on

pipe,H = 100 ft = height of landfill cover over the

pipe,P = γ H = 10 ksf = vertical pressure on the pipe.

Select granular embedment soil —γ = 125 pcf = unit weight compacted,ϕ = 30o = embedment soil friction angle,ε = vertical soil compression at pressure P,µ = 1/3 = coefficient of friction of embedment

against pipe.

Pipes:σ = stress,σf = yield stress (assumed to be failure),F/∆ = pipe stiffness by parallel plate test,F/∆ = 53.77EI/D3 by analysis,D = mean diameter of the pipe, (also used for

nominal diameter),ID = 12 inches = inside diameter,OD = outside diameter = ID + 2t,E = modulus of "elasticity",t = wall thickness,I = t3/12 = moment of inertia of the longitudinal

cross section of the pipe wall about itsneutral axis.

Assumed Data —12D PVC Schedule 80, DR = 18.56


Figure 27-9 Buried pipe on a 1v:3h slope, showing the soil pressure against the elliptical pipe cross section cutby a vertical plane. Because Px < P, the potential for soil slip is less than for a circular pipe.

Figure 27-10 Equivalent horizontal soil supports for embankment and trench conditions. In both cases, theembedment is confined.


σf = 4000 psi (long-term tension), = 7500 psi pristine,E = 370,000 psi,OD = 12.75,ID = 11.48,D = 12.11,t = 0.687,d = 7.5% allowable,F/∆ = 300 psi (by analysis).12D HDPE, SDR 11σf = 1600 psi (50-yr tension), = 3200 psi pristine,E = 135,000 psi,OD = 12.75,ID = 10.432,D = 11.59,t = 1.159,F/∆ = 600 psi (by analysis).

1. Ring compression stress, σ = P(OD)cosθ/2t, mustbe less than σf. See Figure 27-9. P is the verticalsoil pressure at the depth of the pipe. Because cosθ= 0.95 on the 1v:3h slope, it is close enough to unityto ignore in the examples that follow. On a slope,the vertical section is an ellipse for which the ratio ofhorizontal to vertical radii is greater than 1.Therefore, for stability, less soil support is requiredon the sides of the ellipse than on the sides of thecircular pipe. A horizontal pipe is the worst case foranalysis.

Example 5

What are the safety factors for PVC and HDPE inring compression? Neglecting cosθ = 0.95 (forworst case cosθ = 1), assume horizontal pipes. Thelimit of ring compression stress is usually assumedto be the allowable, σf, which is yield stress intension — always less than compression.PVC, σ = 1300 psi. Safety factor is nearly 6.HDPE, σ = 760 psi. Safety factor is over 4.

What is the height of cover at yield stress in the pipewall?PVC, H = 580 ft.HDPE, H = 420 ft.

2. Ring deflection, d = ∆/D , must be less than theallowable d set by a specification (based on crackedjoints, accessibility of tools in the pipe, etc.) to lessthan 7.5% in this example. Upper bounds for ringdeflection.

a) Ring deflection is no greater than verticalcompression, ε , of the sidefill embedment. Valuesfor ε are found by confined compression tests in asoils laboratory. For typical select embedmentcompacted to 90%, standard Proctor density is ε <3%. Therefore, in select, compacted embedment,there is little concern for ring deflection. See Figure27-10. Ring deflection that is assumed equal to soilstrain is conservatively high because the stiffness ofthe ring is neglected.

b) Worst-case ring deflection due to assumedvertical pressure, P, shown below, can be calculated,by elastic analysis d = 0.56P/(F/∆) . This theoreticalring deflection is an upper bound because horizontalsoil support is neglected. Accordingly, upper boundsfor ring deflection are:PVC, d = 13%HDPE, d = 6.5%

sketch here

Suppose the pipes were heavier, say PVC Schedule120 and HDPE, SDR 9,PVC-120 d = 4%HDPE-9 d = 2%SOIL SUPPORT WOULD NOT BE NEEDED.With horizontal soil support, heavier pipes are notjustified. Horizontal soil support reduces ringdeflections. These theoretical ring deflections are


even more conservative because soil stiffness andarching action of the soil are neglected.

3. Longitudinal slippage of the pipe down the slopemust be prevented. Slippage is caused by thedownhill component of the weight of the pipe and bycompression of the landfill over time. Shearingforces of landfill on the pipe tend to drag the pipedownhill. See Figure 27-9. Also longitudinalexpansion and contraction of the pipe, due tochanges in temperature and pressure, causeincremental “creep” down the slope.

a) Downhill slippage is prevented by thrust restraintssuch as thrust blocks, collars, and flanges,that are secured to the slope. Until the slope isretained by landfill, the slope must be less than theangle of repose of the soil. For a granular soil slope,the angle of repose is usually greater than 30o. Theslope at 1v:3h is 18.4o — less than 30o.

b) Downhill slippage is prevented by a good beddingwith a coefficient of friction of soil on pipe greaterthan the coefficient of friction of landfill on pipe.See Figure 27-9. Assuming the worst-case loadingof Example 5, the longitudinal friction force isroughly, F = P(OD)µ cosθ. The only variable forfriction force of the landfill and friction force of thebedding is µ , which should be less for the landfillthan for the bedding. A good bedding in contactwith the pipe is required.

5. Conditions for conservatism and mitigation includeneglect of horizontal support of the pipe by the soil.Such neglect may be justified for a rigid ring. For aflexible ring, horizontal soil support reduces bothring compression and ring deflection. Soil arching protects pipes. See Figure 27-10.

Plastic Pipes:a) Short-term yield strength is a pertinent, butconservative, property of plastic pipes. Long-termyield strength is not pertinent if the embedment holdsthe ring in shape. The ring conforms with the soiland stresses relax faster than the long-term strengthregresses. At constant deformation, if

failure does not occur when full load is applied,failure won't occur. But if stress suddenly increasesbefore the time to long-term strength regression, thestrength of plastic is still pristine.

b) If stresses are persistent, and the ring is not heldin shape by soil, the strength of the plastic regressesover time. An example is a pipe buried in fluid atconstant pressure. A good soil embedment assuresconstant deformation such that stress relaxationprevails over strength regression.

c) The rate of loading in a sanitary landfill isincremental and slow (years) — not sudden asassumed in these examples. Slow loading allowstime for stress relaxation.

d) Yield stresses are usually reported for hoopstrength in TENSION, not COMPRESSION. Ringcompression strength is greater than the tensionyield stresses assumed in these analyses.

e) Ring analysis on a slope is less severe thananalysis on a horizontal plane.

Soil:a) Vertical soil compression is pertinent in checkingthe predicted upper bound of ring deflection.

b) Active and passive resistance of soil becomespertinent if the pipe ring is so flexible that ringbuckling becomes an issue. Such is not the case inthese examples. The pipes are not flexible enough.

c) Active and passive resistance of particulate soilare functions of the soil friction angle, ϕ ; i.e., K =(1+sinϕ)/(1-sinϕ), where K is the ratio of orthogonalprincipal normal stresses in the soil at soil slip.Shearing planes form. Soil slip is not an issue forthese plastic pipes because ring deflection is small;i.e., ring stiffness is substantial.

d) On a slope, the frictional drag-down force oflandfill on the pipe is usually less than the fric tionalresistance of the bedding because the coefficient offriction on top is less.


PROBLEMS

27-1. Find the minimum Df in Example 4 required tosupport a small pipe from creeping downward into alarge pipe. Bearing capacity of the waste is 800 psf.Frictional resistance to creep is negligible.

27-2. What live load, W, can be supported on top ofthe riser if the top 11-ft pipe is small with Df =

20, and if live load forces the pipe downwardreversing friction?

27-3. Suppose the first (bottom) riser pipe ofExample 4 is a small pipe 20 ft long, no flange,embedded to the top in 20 ft of loose waste.Estimate the bearing force on the base of this pipe.Is the pipe weight significant?

(Q = πrLγ = 38 kips)

27-4. Derive Equations 27.1 and 27.3.


(Df = 19.64)

Anderson, Loren Runar et al "ANALYSIS OF BURIED STRUCTURES BY THE FINITE ELEMENT METHOD"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

CHAPTER 28 ANALYSIS OF BURIED STRUCTURES BY THE FINITE ELEMENTMETHOD

Introduction

The finite element method was introduced as a toolfor engineering applications by Turner et al. (1956)for the solution of stress analysis problems relatedprimarily to the aircraft industry. Since that time thefinite element method has become a useful andaccepted tool in many areas of civil engineering.Applications in geotechnical engineering includestatic and dynamic stress analysis of various soil andsoil-structure systems, seepage analysis includinggroundwater modeling, and consolidation analysisincluding both magnitude and rate of settlement.Stress analysis applications in geotechnicalengineering for static and dynamic loading wasintroduced in the late 1960s and early 1970s andincluded such applications as: static analysis ofstresses and movements in embankments [Kulhawy,et al. (1969); Duncan, (1972); Kulhawy andDuncan, (1972)]; earthquake stress analysis ofembankments [Clough and Chopra, (1966)];earthquake response analysis [Idriss, et al. (1974)],and soil-structure interaction [Clough, (1972)].

Katona, et al. (1976) pioneered the application of thefinite element method for the solution of buried pipeproblems. Their FHWA-sponsored projectproduced the well-known public domain computerprogram CANDE (Culvert ANalysis and DEsign).CANDE has been upgraded several times and isnow available for use on a PC. Others also madeearly contributions in the use of the finite elementmethod for buried structures problems, Katona,(1982); Leonards, et al. (1982); Sharp, et al. (1984);and Sharp, et al. (1985); TRB Record 1008. The basic idea behind the finite element method forstress analysis is that a continuum is represented bya number of elements connected only at the

element nodal points (joints), as shown by the two-dimensional representation of a buried pipe in Figure28-1. A structural analysis of the finite elementassemblage can be made in a manner similar to thestructural analysis of a building. The processinvolves solving for the nodal displacements andthen, based on the nodal displacements, the stressesand strains within each element of the assemblagecan be determined. The elements shown in Figure28-1 are the basic structural units of the soil-pipecontinuum, just as beams and columns are the basicstructural units of building frames. Each element iscontinuous and stresses and strains can be evaluatedat any point within the element. The majordifference between the analysis of a continuum anda framed structure is that even though the finiteelement representation of a continuum is onlyconnected to adjacent elements at its nodal points, itis necessary to maintain displacement compatibilitybetween adjacent elements. Special shape functionsare used to relate displacements along the elementboundaries to the nodal displacements and to specifythe displacement compatibility between adjacentelements. Once the continuum has been idealized asshown on Figure 28-1, an exact structural analysis ofthe system is performed using the stiffness methodof analysis [Zienkiewicz, (1977); Gere and Weaver,(1980); Dunn, Anderson and Kiefer, (1980)].

Note in Figure 28-1 that only half of the soil pipesystem is represented. The analysis results for theother half of the pipe can be obtained by symmetryas long as the geometry, properties, and loadingconditions are symmetrical. The boundaryconditions along the line of symmetry must beproperly established to model the full systembehavior. Taking advantage of symmetrysignificantly reduces the size of the problem thatmust be solved as discussed in the next section.


Figure 28-1 Mesh representing symmetric pipe-soil system.


Most geotechnical engineering applications can besolved using a two-dimensional idealization as shownin Figure 28-1. However, there are some problemsthat must recognize the three-dimensional nature ofthe problem. Figure 28-2 shows a finite elementrepresentation of a buried cylindrical tank. Themodel was used to investigate the development ofleaks in the tank at the junction between thecylindrical walls and the end plates on the tank.Again, symmetry was used to minimize the size ofthe problem.

Basic Principles of the Finite Element Analysis

Equation 28.1 represents the equilibrium equations,in matrix form, for each node of a finite elementassemblage such as the one shown on Figure 28-1.After applying boundary conditions (identifyingnodes with fixed or restricted movement) the systemof equations given by Equation 28.1 can be solvedfor the unknown nodal displacements represented bythe vector {d}. These displacements can in turn beused to evaluate element stresses and strains.

[K] {d} = {f} . . . . . (28.1)

where[K] = the global stiffness matrix {d} = the nodal displacement vector and {f} = the nodal load vector.

The stiffness matrix [K] relates the nodaldisplacements to nodal forces. The elements of themartix are functions of the structural geometry, theelement dimensions, the elastic properties of theelements, and the element shape functions. The sizeof the stiffness matrix depends on the number ofdegrees of freedom at each node and the number ofnodes. Thus, the more nodes that are used torepresent a contiuum, the larger the system ofsimultaneous equations that must be solved. Takingadvantage of symmetry, as discussed in the previoussection, can significantly reduce the number ofequations that must be solved. A completederivation

of the finite element method for soil-structureinteraction problems is presented by Nyby (1981).

A finite element analysis of a soil-structureinteraction system, such as a buried pipe, is different from a finite element analysis of a simplelinearly elastic continuum in several ways.

1. The soil has a nonlinear stress-strain relationship.2. Different element types must be used torepresent the pipe and the soil.3. It may be necessary to allow movement betweenthe soil and the walls of the pipe, requiring the use ofan interface element.4. Very flexible pipes may involve large displace-ments for which the solution may be geometricallynonlinear.

Nonlinear soil properties

The stress strain behavior of soil is nonlinear.Therefore, large load increments can lead tosignificant errors in evaluating stress and strainwithin a soil mass. The stress-strain relationshipshould be determined from the results of laboratorytests on representative soil samples. Duncan et al.(1980) suggested a method for describing the stress-strain characteristics of soil using hyperbolicparameters. They also presented typical values forsoil that can be used if the results of laboratory testsare not available. Care should always be exercisedwhen using “typical” values.

The Duncan soil model is often used forgeotechnical engineering applications. (Duncan, etal. 1980) The original development of hyperbolicstress-strain theory that is used by the Duncan soilmodel was presented by Kondner and Zelasko(1963). The soil model assumes that the stress-strain properties of soil can be modeled using ahyperbolic relationship. A thorough discussion of theDuncan soil model is presented in Duncan et al.(1980).

Figure 28-3 shows a typical nonlinear stress-strain


Figure 28-2 Mesh for a buried tank which requires three-dimensional iterations.


Figure 28-3 Hyperbolic representation of a stress-strain curve [after Duncan et al. (1980)].


c urve and the corresponding hyperbolictransformation that is often used as a convenientw ay to represent the stress-strain properties.(Duncan et al. 1980). In Figure 28-3, the value ofthe initial tangent modulus Et is a function of theconfining pressure. Figure 28-3 shows the changein the tangent modulus that occurs as strainincreases. For a given constant value of confiningpressure, the value of the elastic modulus is afunction of the percent of mobilized strength of thesoil, or the stress level. As the stress levelapproaches unity (100% of the available strength ismobilized) the value of the modulus of elasticityapproaches zero. The Mohr-Coulomb strengththeory of soil indicates that the strength of the soil isalso dependent on confining pressure, as shown inFigure 28-4. Figure 28-5 shows the logarithmicrelationship between the initial tangent modulusversus confining pressure. The soil model combinesthe relationship of variation of initial tangent moduluswith confining pressure and the variation of elasticitywith stress level to evaluate the tangent modulus ofelasticity at any given stress condition. The equationthat is used to evaluate the modulus of elasticity asa function of confining

pressure strength is:

whereEt = tangent elastic modulus

Pa = atmospheric pressureK = an elastic modulus constantn = elastic modulus exponentσ1 = major principal stressσ3 = minor principal stress (confining pressure),

andRf = failure ratio

The soil model as presented in Duncan et al. (1980)also uses a hyperbolic model for the bulk modulus.The hyperbolic relationship for the bulk modulus issimilar to the initial elastic modulus relationship,where the bulk modulus is exponentially related tothe confining pressure. This particular soil modeldoes not allow for dilatency of the soil duringstraining. The equation that is used which relatesthe bulk modulus to confining pressure is:

where B = the bulk modulusKb = bulk modulus constant, andm = bulk modulus exponent.

The two equations given above are used to evaluatethe strain-dependent elasticity parameters that arerequired in the stiffness matrix. Poisson's ratio andthe shear modulus are both computed based onequations developed in classical theory of elasticity.

Shear failure is tested by evaluating the stress levelbefore the modulus of elasticity is computed. If thestress level is computed to be more than 95% of thestrength, the modulus of elasticity is computed basedon a stress level of 0.95. This result is a low modulusof elasticity. The bulk modulus is unaffected, thusmodeling a high resistance to volumetriccompression in shear. A test must also beperformed to evaluate if tension failure has occurredwhen computing the elastic parameters. If theconfining pressure is negative, then the soil elementis in tension failure, and the elastic parameters needto be set to very small values, thus simulating atension condition.Figure 28-6 shows a stress-strain curve for a soilsample in a triaxial shear test. The loading sequencefor the sample was to increase the vertical stressuntil the sample had undergone initial strain, then tounload the sample, and


Figure 28-4 Variation of strength with confining pressure [after Duncan et al. (1980)]

Figure 28-5 Variation of initial tangent modulus with confining pressure [after Duncan et al. (1980)]


382

Figure 28-6 Deviator stress versus strain for a triaxial soil sample showing primary loading, unloading, andreloading.

finally to reload the sample until failure. In Figure26-6 it can be seen that the sample has a nonlinearstress-strain response on primary loading. Theunloading and reloading characteristics below theprevious maximum past pressure, however, do notfollow the initial primary curve; they show aninelastic response. After reloading beyond themaximum past pressure, the stress-strain curveagain follows the initial nonlinear primary loadingcurve.

Duncan et al. (1980) discuss the behavior of soil onunloading and reloading in comparison with that onprimary loading. The soil stiffness is reported to be1.3 to 3.0 times greater when in theoverconsolidated range. Volumetric strain isreported to be unaffected by stress history. Triaxialtesting for unloading and reloading generally showsthat the magnitudes of the hyperbolic constant andexponent depend on whether the soil is in primaryloading or unloading and reloading. It is necessaryto model stress history for each soil element in afinite element analysis in order to model initialdeformation of the pipe due to compaction. Some ofthe soil elements should respond in the reboundrange because of compaction until the surcharge

pressures exceed compactive loading pressures.For other applications, such as pipe rerounding, soilelements must respond appropriately as the pipererounds when internal pressure is added to theloading sequence. Because the soil is much stifferin the rebound range, the pipe deformation isdependent on the stress history of the soil. Not all ofthe soil elements in the finite-element mesh willrespond to the rebound range at any given time asthe pipe rerounds or as the compaction loads aremodeled. Thus, it is necessary to monitor the stresshistory of each soil element during the analysis andto use appropriate stiffness parameters depending onthe current stresses of each element.

The stress history of the soil elements can bemonitored by evaluating the position of the center ofMohr’s circle for each element (the average stress).The average stress at any load increment iscompared with the maximum average stress fromprevious increments. If the average current stressis less than the maximum previous stress, the soilelastic modulus is computed by using the unloadingparameters. Likewise, the soil elements aremonitored in the rebound range and will convert tothe primary loading curve when the average current


stresses exceed the maximum past average stress.This method allows simulation of soil elementresponse for any soil element on either rebound orprimary loading, depending on the loading conditionsand soil response. Two additional soil parametersare required as inputs for the Duncan soil model thataccount for the behavior of the soil in the unloadingand reloading range, and the maximum pasteffective stress (preconsolidation stress orcompaction stress) must also be specified.

When using the finite element method to solvegeotechnical engineering problems the nonlinearstress-strain conditions are generally accommo-dated by adding loads in increments and adjustingthe soil properties according to the magnitude of thestrain. If an embankment is being constructed, it isnecessary to follow the construction sequence byadding the soil layers in increments and thenadjusting the soil properties after each layer isadded. Each new layer is first represented as a loadto determine the increase in stress within theembankment. After determining the stress increasefrom the new layer, additional elements are added tothe finite element mesh to represent the newmaterial. This allows the FEA program to follow thenonlinear stress-strain properties of the soil. Thestiffness matrix [K] in Equation 28-1 is a function ofthe material properties, the geometry of the elementand the shape functions that are used to describe thestress-strain behavior at the edges of the elements.The stiffness matrix is initially formed using thebeginning soil properties and geometry of theelement. As loads are added to the soil structure thesoil deforms and the soil properties change, and thus,the stiffness matrix must be adjusted to reflect thenew soil properties.

Interface Elements

In the finite element analysis of buried pipes, thepipe is generally modeled using beam elements inwhich shear, moment, and thrust can be representedat the ends of each element. The nodes of the pipeelements are connected to the adjacent soil elements

at their common nodal points. In some cases,however, it may be necessary to allow slip to occurbetween the pipe and the soil. This can beaccommodated in the finite element analysis byplacing "interface" elements between the pipe nodesand the soil element nodes. These interfaceelements have essentially no size, but kinematicallyallow movement between nodes when a specifiedfriction force is exceeded.

Geometric Nonlinearity

As described above in the section on nonlinear soilproperties, the stiffness matrix [K] in Equation 28.1is a function of the material properties of eachelement, the geometry of the element and theelement shape function. As the soil deforms underadded loads the geometry of the finite element meshchanges. If these changes are small (smalldisplacement theory) the stiffness matrix does nothave to be reformulated after each load inc rement.However, if the pipe is very flexible thedeformations can be large and it is necessary toreformulate the geometry of the finite element meshafter each loading increment. This is referred to asgeometric nonlinearity.

Construction of the Stiffness Matrix

The stiffness matrix is composed of several parts.One component is a constitutive matrix relatingstress to strain through the elasticity parameters.Another component relates element strains to nodaldisplacements through the strain-displacementmatrix. This matrix is computed based on elementtypes, shape functions, and nodal coordinates. It isnot within the scope of this report to derive theabove mentioned relationships.

Beam, bar, and soil elements each have their ownparticular stiffness matrices. This comes about dueto basic engineering mechanics principles. A beamelement is a three-force element and a bar is a two-force element. Both beam and bar elements arecalled one-dimensional elements, their strain-


displacement matrix is derived based on theappropriate shape functions and their cross-sectionalarea, length, and angle of inclination. A soil elementis a two-dimensional element. It does not transmitmoment stresses. The strain-displacement matrix isderived based on the coordinates of each nodewhich comprise the element, and the shape functionsthat are used to describe the deformationcharacteristics of the soil elements.

A finite element analysis of a continuum involves thesolution of the system of equations shown asEquation 28.1. If the continuum is a linear elasticsystem and the displacements are small then thestiffness matrix, [K], in Equation 28.1 can begenerated once and then used over and over againfor solving the problem for different loadingconditions. The load vector {f} changes fordifferent loading conditions but the stiffness matrixremains the same. Furthermore, the load of interestcan be applied in one increment since the materialproperties are linearly elastic. However, if thematerial properties are non-linear (strain dependent)then the stiffness matrix changes as the loadincreases and it is not possible to apply the load tothe system in one large increment. The load must,therefore, be added in increments and the stiffnessmatrix modified with each loading increment. If theproblem involves large displacement, which may bethe case for very flexible pipes, then the geometrycomponent to the stiffness matrix must change asthe system deforms. The global stiffness matrix forthe entire system is made up from contributions fromall of the elements in the assemblage. Beam andbar elements tend to be linear elastic and the soilelements are non-linear (strain dependent). Theindividual element stiffness matrices are inserted intothe overall global stiffness matrix by consideringequilibrium at each node in the finite elementassemblage.

Required Input

Execution of a finite element program requires theuser to prepare data that includes the mesh

geometry, material properties and loading conditions.Some programs have an automatic mesh generationoption that signif icantly reduces the effort requiredto input the data. The data that are required includenodal coordinates, element data, material properties, interface properties, nodal link properties,construction sequence information, preexistingelement stresses, strains, and displacements, andexternal loading information. Boundary conditionsare input. The boundary conditions indicate whethera node is free to displace in the x or y direction or inrotation.

Material information.. Material information foreach specified material type must be input. Soil,structure (bars and beams), and interface materialswill each have separate input requirements. Thematerial properties that are needed for bar elementsinclude the cross-sectional area, modulus ofelasticity, and weight/unit length. Beam materialsalso require moment of inertia, shear area , Poisson'sratio, and distances from extreme fibers to theneutral axis. Soil material properties that are neededinclude all the parameters that are used in theDuncan soil model described in a previous section.

Additionally, Mohr-Coulomb strength information,unit weight, and the lateral earth pressure coefficientare generally necessary for each soil type. Thenodal link elements behave like springs. Normal andshear spring coefficients and orientation angles arerequired for the nodal link elements. Interfaceelements are similar to nodal links except that springcoefficients are nonlinear. Normal and shear springcoefficients, a modulus exponent, wall friction angle,and an adhesion value are necessary on eachinterface element material type.

Construction information. The constructionsequence is modeled by adding layers of soilelements or placing the structure (beam elements).Soil layer construction information is indicated byspecifying the sequence at which the layers are tobe added relative to sequences of adding externalloads. Structural placement information is indicated


by specifying the soil layer that immediately followsthe structure placement and the largest soil layernumber that is adjacent to the structure. Soil layerdata that are needed for each layer are the largestand smallest soil element numbers within the soillayer, largest and smallest newly placed nodenumbers within the soil layer, and a series of nodeswhich define the top surface of the new layer.

Preexisting stresses. An important feature whensolving non-linear problems, such as soil-structureinteraction problems, is the ability to specifypreexisting stresses. These stresses may be in thesoil, structure, or interface elements. Preexistingstresses are stresses which are already in placebefore any construction layers or external loadingforces are added to the system. For some problemsit may also be important to specify preexistingstrains or displacements. The "preexisting stress"concept is very convenient when performing a seriesof analyses. The use of preexisting stresses, strains,and displacements essentially defines the stresscondition for the preexisting elements. Constructionsequences, therefore, need only be modeled once fora given mesh and soil configuration. Afterwards,the preexisting stresses resulting from thatconstruction simulation can be input for the entiremesh, and the subsequent analyses can beperformed by adding only combinations of externalloads to the mesh.

External loads. External loads are generally input aseither concentrated loads or uniform loads. Theirinput is rather simple. Each loading sequence musthave the number of concentrated loads and numberof uniform loads that are to be used. Concentratedloads are specified by denoting the node numberwhich will receive the load, and the x and ycomponents of the point load. Uniform loads arespecified for each element that will receive theuniform load.

Output

Typical output from a finite element analysis

includes a summary of the input data and the resultsof the analysis in terms of stresses, strains, anddisplacements. The input summary includes elementand node information, material properties,construction and load sequencing, preexistingelement information, and initial stresses used forestimating the initial elastic parameters.

Nodal displacements include the total displacementsfor the x, y and rotation components, and theincremental displacements and rotations for eachparticular loading increment..

The structural responses that are listed include themoment, shear, and thrust for each node of eachstructural member. The listing contains theincremental structural forces and the total structuralforces from the accumulated incremental forces.

The soil element strain information includes the soilelement strains in x and y direction and the shearstrain. The principal strains can also be listed foreach element.

Soil element stresses that can be output include thehorizontal and vertical stresses, shear stresses, andprincipal stresses. The angle of orientation of theorigin of planes with respect to the principal plane,the ratio of major to minor principle stress, andstress levels can also be printed out for eachelement. The stress levels that are output indicatethe stress condition of each element. If thecomputed stress is greater than the strength of thematerial, then the element has undergone a localshear failure, and the elastic parameters would haverequired adjustment. If tension stresses werecomputed in the soil, the the element would haveundergone a tension failure. The element elasticityparameters would again need to be adjusted to allowthe displacements that would occur for a soilelement in tension.

Summary

The finite element method is a powerful tool for


stress analysis of complex systems. It has beenparticularly useful in geotechnical engineering for thesolution of a wide variety of problems including soil-structure interaction problems such as the analysisof buried structures. Its application in solving soil-structure interaction problems requires anunderstanding of basic engineering mechanics andan understanding of soil behavior. Judgement isrequired in conducting and reviewing the results offinite element analysis of geotechnical problems.Comparing the results of a finite element analysissolution with measurements made on a physicalsystem is important whenever possible. The powerof the method lies in the ability to solve complexsystems and in being able to look at many differentloading conditions and system configurations.However, never accept the results at face valuewithout a thorough critical review.

REFERENCES

J.M. Duncan, P. Byrne, K.S. Wong, and P. Mabry.Strength, stress-strain and bulk-modulus parametersfor finite element analyses of stresses andmovements in soil masses. GeotechnicalEngineering Report UCB/GT/80-01. University ofCalifornia, Berkeley, 1980.

F.H.Kulhawy, J.M.Duncan, and H.B.Seed (1969),Finite element analysis of stresses and movementsin embankments during construction. GeotechnicalEngineering Report TE-69-4. University ofCalifornia, 1969

M . G . K a t o n a , J . B . F o r r e s t , R . J . O d e l l o ,andJ.R.Allgood (1976), CANDE—a modern

approach for the structural design and analysis ofburied culverts, Report FHWA-RD-77-5. FHWA,U.S. Department of Transportation, 1976.

G.A.Leonards, T.H.Wu, and C.H.Juang (1982),Predicting performance of buried conduits. ReportFHWA/IN/JHRP-81\3. FHWA, U.S. Department ofTransportation, 1982

K.D.Sharp, F.W. Kiefer, L.R.Anderson, andE.Jones (1984), Soils Testing Report forapplications of finite element analysis of FRP pipeperformance: Soils Testing Report. BuriedStructures Laboratory, Utah State University,Logan, UT, 1984

I.S.Dunn, L.R.Anderson, and F.W.Kiefer, (1980)Fundamentals of Geotechnical Analysis, Wiley, 1980

B.W.Nyby and L.R.Anderson (1981), Finite elementanalysis of soil-structure interaction.

Proceedings of the International Conference onFinite Element Methods (H.Guangqian andY.K.Cheung, eds.) . Science Press, Beijing China,1982.

K.D.Sharp, L.R.Anderson, A.P.Moser, andM.J.Warner (1984), Applications of finite elementanalysis of FRP pipe performance. BuriedStructures Laboratory, Utah State University,Logan, UT, 1984


Anderson, Loren Runar et al "APPLICATION OF FINITE ELEMENT ANALYSIS TO A BURIED PIPE"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

CHAPTER 29 APPLICATION OF FINITE ELEMENT ANALYSIS TO A BURIED PIPE

Introduction

As discussed in Chapter 28, the finite-elementanalysis of buried flexible pipes requires capabilitiesgenerally not included in most finite-element analysisapplications. Many types of buried pipes (such asfiberglas-reinforced plastic pipe) are very flexible,thus, requiring a finite-element analysis of thesystem to accommodate large deflections.

The sensitivity of the pipe and the soil properties tocompaction loading may be an importantconsideration. The stress history of each soilelement should be monitored at each loadingincrement to determine whether the element is in astate of primary loading, unloading or reloading, andthen appropriate parameters need to be used foreach increment of the analysis.

Properly accounting for all elements of the soilstructure system makes the finite-element method auseful tool for the analysis of buried flexible pipeswhen subjected to various installation conditions,backfill material types, surcharge loadings, andinternal pressurization.

This chapter illustrates the application of the finite-element method to the solution of a soil structureinteraction problem. The response as computed bya finite-element analysis (FEA) of a buried flexiblefiberglas-reinforced plastic (FRP) pipe, whensubjected to various installation and static loadingconditions, was compared with measured strains anddeflections taken from physical tests in a soil box atthe Buried Structures Laboratory at Utah StateUniversity, Sharp et al. (1985). The FEA modeledthe actual soil box installation conditions.

The approach that was taken in the study was tosimulate the backfill and loading conditions that hadbeen used in the soil-box tests in order to comparethe predicted response of the pipe from FEA resultswith the measured response. This required that the

four soils that were used in the soil box be tested forengineering properties, including triaxial testing atseveral densities for evaluation of the hyperbolicparameters for the Duncan soil model. Results fromthe tests are described by Sharp et al. (1984). In thefollowing discussion, only the results of theapplications of the finite-element program to theinstallation conditions for silty sand are presented.The results of the remaining applications areincluded in the report by Sharp et al. (1984).

Determination of Duncan Soil Parameters

The silty sand that was used in the soil-box tests ischaracterized as a nonplastic material with about40% passing the 0.075-mm sieve and about 10%clay-size particles. Maximum dry density is 124.7lb/ft3 and the optimum water content is 9.5 % basedon AASHTO T-99 compaction.

Triaxial shear tests on the silty sand were performedby using samples compacted at water contentssimilar to those used in soil-box tests. Figure 29-1 isthe sketch of a "soil box." Elastic modulus and bulkmodulus parameters are required in the FEA foreach density. Testing of the clean granularmaterials (washed sand and gravel) was performedby using saturated samples, and the volume changewas monitored by measuring the volume of waterextruded or imbibed in the samples during drainedshear.

Because the stress-strain and strength properties ofthe silty sand and clay are dependent on drainage,density, compaction water content, and watercontent at shear, it was necessary for the soilparameters to represent field conditions as much aspossible. Therefore, the silty sand and the clay weretested by using unsaturated undrained conditions.The triaxial device was not equipped to


Figure 29-1 Sketch of the large USU soil cell (soil box) for investigating pipe-soil interaction of buried pipes.The cell is an elliptical steel cylinder with horizontal radius of curvature at the spring lines equal to three timesthe vertical radius at the invert. The ellipse reduces boundary effects by simulating a one-third ratio ofhorizontal to vertical soil stresses as load is applied. Boundary effects are negligible for pipes up to about 60inches (1.5 m) in diameter. The cell is 15 ft (4.6 m) wide, 18 ft (5.5 m) high to the hydraulic cylinders, and22 ft (6.7 m) long. Vertical pressure is applied by 50 hydraulic cylinders on ten beams. The beams arepinned at one end such that they can be tilted up out of the way for installing pipes. This is the larger of twosoil cells used, primarily, for investigating performance of pipes under high soil pressures, and for comparisonwith finite element analyses.


measure volumetric strain of undrained samples.Therefore, bulk modulus parameters were notmeasured for the fine-grained soils (silty sand andclay).

Triaxial testing was performed on the silty sand atthree different densities (95%, 80%, and 77%relative compaction based on Standard Proctor).Stress-strain curves were obtained for threeconfining pressures within the range used in the soil-box tests for each density. The triaxial testingprocedure involved preparing the sample withcompaction techniques similar to those in the fieldand by applying the deviator by initial loading,unloading, and reloading to failure. This resulted indata that were interpreted with procedures outlinedby Duncan et al. (1980) for determination of shearstrength and the hyperbolic parameters for theelastic moduli.

The unloading and reloading data were alsoevaluated to obtain the rebound parameters for eachdensity.

Because data were not obtained for the hyperbolicbulk modulus parameters, an FEA sensitivity studywas performed using soil-box test results to calibratethe silty-sand data using 90 and 80% relativecompaction. The elastic modulii for the 90% relativec ompaction were obtained by interpolating themeasured values from the 95%, 80%, and 77%percent relative compaction data. Sensitivity studieswere performed by using a 90% homogeneousrelative compaction and an 80% homogeneousrelative compaction in the finite-element mesh todetermine the bulk modulus parameters. Thesesensitivity studies show that the shape of the load-deflection curve can be adjusted by modifying thebulk modulus exponent. Pipe strain plots can also beadjusted because of complex interrelationshipsbetween hyperbolic elastic and bulk modulusparameters in conjunction with the shear strength ofthe soil. Table 29-1 shows the final values for thesoil parameters that were used for the silty sand at90% and 80% relative compaction. A more

complete description of sensitivity is contained in thereport by Sharp et al. (1984).

Finite-Element Modeling

The modeling of the fiberglass pipe performance insilty sand consisted of several installation conditionsand 10 and 100 psi pipe stiffnesses (ASTM D-2412). Installation conditions included homogeneouscompaction at 90 and 80% relative compaction, poorhaunches, and soft crown. In general the poor-haunch and soft-crown conditions were obtained bynot compacting the soil in those areas. Figure 29-2shows the finite-element mesh and soil materials ortypes that were used in this study. The poor-haunchcondition used 90% relative compaction for all soiltypes except that in the haunches (soil type 6). Thesoft-crown condition used 90% relative compactionfor all soil types except that in the area from theshoulders to the crown of the pipe, shown in Figure29-2 as soil type 7. These installation conditionswere used because of the way in which materialswere placed around the pipe during installation.When the pipe has low stiffness, it is difficult tocompact the fill material over the crown untilsufficient cover has been placed. Also, extra effortis required to compact the soil in the haunches, so itwas desired to model installation conditions in whichthere was loose material in the haunches. The othersoil types that are shown were included toinvestigate effects due to split installation, differentfoundation materials, and other types of installations.

The finite-element modeling scheme consisted oftwo phases. The first modeled constructionincrements without compaction simulation. Fourinstallation conditions were modeled for pipes withstiffnesses of 10 and 100 psi. The secondincorporated a compaction simulation on eachconstruction increment before the next constructionincrement was added. Three installation conditionswere modeled by using the compaction simulation.Table 29-2 shows the installation conditions thatwere used for the silty sand and indicates those thatincluded compaction simulation.


Table 29-1. Soil Parameters for Silty Sand

RCStand

densitylb/in3

ϕdeg

∆deg

cpsi

K n Rf Kb m Ko Kur nur

90% 0.065 30 0. 8.3 480 .44 .75 80 .38 .48 720 .44

80% 0.058 30 0. 3.5 350 .28 .89 15 .40 .37 525 .28

Note: New report in Duncan et al., 1998) for definition of parameter.

Figure 29-2 Finite-element mesh for buried pipe installation, including pipe coordinate system.


TABLE 29-2. Installation Conditions for Silty Sand

Condition Pipe Stiffness(psi)

CompactionSimulation

1. 90% homogeneous 10. 100 yes

2. 90% backfill with soft top @ 80% RC 10. 100 yes

3. 90% backfill with soft top @ 99% RC 10 no

4. 90% backfill with haunches @ 80% RC 10. 100 yes

5. 90% backfill with haunches @ 79% RC 10 no

6. 90% backfill with haunches @ 77% RC 10 no

7. 90% backfill with top and haunches @ 85% RC

10 no

8. 80% homogeneous 10. 100 no

Soil elements in the foundation and up to the springline (soil materials 1 through 3) were treated aspreexisting elements having stresses and strainspredefined at the time of program execution. In theconstruction sequence used in the first phase,placement of the remainder of soil 3 and all of soils4 and 7 was simulated as the first constructionincrement. The second construction incrementcompleted the mesh by placing soil material S.

The second phase or the modeling incorporatedcompaction simulation after each constructionsequence. The compaction simulation involvedaddition and removal of compaction loads at the endof the first and second construction increments. Thefirst compaction load was added to the first layer ofsoil material 4. The second compaction loading wasplaced on the completed mesh over soil material 5.I t was found that the first loading sequence wascritical in inducing initial ovalization of the pipe. Itwas not possible to load directly over the pipe (soilmaterial 7) without causing structural failure andlarge unrealistic deformation of the pipe and soilbecause of an unstable condition. This result

appears to be in accordance with the actualinstallation conditions, in which it was found thateffective soil compaction could not be performedover the pipe until sufficient cover had been placed.Also, it was not possible to add compaction loads onthe soil before placement of the first increment.When loads were added adjacent to the spring lineon soil material 3, pressures caused excessivedeformation of the pipe.

For compaction simulation, a uniform static load of10 psi was used corresponding to the type ofcompaction equipment used in the soil-box test. Amore rigorous compaction sequence would havebeen to load each soil element individually with alarger pressure, which would result in a bettersimulation of the compaction process. The load of10 psi was an equivalent surface pressure over alarge area. The compaction load was added in twoincrements of 5 psi each. After the compaction loadhad been placed, a sequence of unloading wasfollowed. A series of unloading steps in smallpressure increments was followed until the elements


in the top row of the mesh approached a tensioncondition with negative confining pressure.

Small increments were used in the loading andunloading sequence in order to assure that the soilelements were being evaluated correctly for eitherthe primary loading or rebound parameters.Magnitudes that were too large for compactionloading might have caused poor convergence andincorrect evaluation of the appropriate soil responsemodel. It was not possible with the silty sand toremove the same load magnitude that was placedwithout causing tension failures in all elements in themesh.

This was not the case with the clay analysis,however, because it was numerically possible toremove the same quantity of compaction load thatwas placed. Although it may seem invalid to modelcompaction loading without removing the same loadthat was placed, it must be pointed out that thesolution is an incremental loading procedure. Thetotal load vector is not evaluated at each iteration.Only the total stresses and strains in each elementare evaluated. Thus although success at unloadingthe elements with the same compaction loadmagnitude was not achieved, it was possible toinduce stress history in the soil due to compactionloads, which resulted in allowable soil stresses,strains, and deformations.

The inability to unload the soil element completelywithout complete tension failure might be attributedto numerical approximations with the finite-elementtechnique. Problems arise in the soil model inevaluating Poisson's ratio when the soil is in theunloading and reloading range. Poisson's ratio iscomputed by using the theory of elasticityrelationships between bulk modulus and elasticmodulus. When the elastic modulus increases as inunloading and reloading, Poisson's ratio is computedat its minimum of 0.0 if the bulk modulus does notincrease proportionately. Behavior of the bulk modulus is difficult to determineon unloading and reloading relative to primaryloading. Sharp et al. (1984) tested bulk modulus

behavior in triaxial shear of saturated drainedgranular material for primary loading as well asrebound loading. It was not possible to concludehow volumetric deformation behaves as a functionof stress history for the coarse-grained material. Inaddition, the granular material exhibited dilation at asmall strain (a response that cannot beaccommodated within the theory of elasticity).

RESULTS OF FEA MODELING

The results of the applications of FEA werecompared with the measured response from the soil-box tests. The comparisons that follow are for apipe with 10 psi pipe stiffness. Soil box compactionconditions that were used for comparisons were90% relative compaction with homogeneousconditions, 90% relative compaction with poorhaunches, and 80% relative compaction withhomogeneous conditions. In the soilbox testing,every attempt was made to achieve homogeneousconditions. However, as noted, the flexible nature of the pipedoes not always allow for thorough compaction inthe haunches and around the shoulders and crown ofthe pipe. Therefore, for the homogeneous conditionsthat were attempted in the soil box there wasactually some variation in density. When the pipewas installed with poor haunches in the soil box, noattempt was made to compact the soil. Finite-element modeling of homogeneous and poor-haunchconditions is better defined because, numerically, allsoil elements in a homogeneous condition haveidentical stress-strain properties.Comparisons of the FEA results with those of thesoil-box tests are made by using pipe-strain andload-deflection results. The pipe-strain plots indicatethe bending strain in the outside fibers (tension ispositive) and thrust strain around the circumferenceof the pipe for a given surcharge pressure. Theload-deflection plots indicate the vertical andhorizontal ring deflections (the ratios of change invertical and horizontal diameters to initial diameter)versus surcharge pressure. In the soil-box tests, theload-deflection plots are referenced to the deformedstate of the pipe after


Figure 29-3 Vertical soil pressure versus pipe deflection for (curve A) soil-box data. 90 percent relativecompaction, silty sand and (curve B) FEA, no compaction simulation.

Figure 29-4 Pipe strains as functions of circumferential position, conditions as in Figure 29-3.


compaction. In the FEA plots, the reference of ringdeflection is based on the initial undeformedcondition. Thus in the comparison of figures thatfollow, the zero point of deflection should beconsidered when direct comparisons of loaddeflections between the FEA and the soil-box testsare performed. Pipe-strain plots for both soil-boxand FEA results are referenced from the sameunstrained condition. The pipe-strain plots showbending and thrust strain versus position on the pipe.Zero degrees on the pipe is at the invert, 90 degreesis at the spring line, and 180 degrees is at the crownas shown in Figures 29-2 and 29-3. The values forpipe strain from 180 to 360 degrees are symmetricwith 0 to 180 degrees for the FEA because the FEAmesh presented here used an axis of symmetry forthe analysis of symmetric bedding.

Homogeneous Installation at 90% RelativeCompaction

Figures 29-3 and 29-4 show the soil-box test resultsfor a 10-psi pipe installed with homogeneouscompaction at 90% of Standard Proctor maximumdry density. Physical pipe date are:

Curve Parameter A BStiffness (psi) 10 10Thickness (in.) 0.285 0.300Surface pressure (psi) 40.9 50.0Vertical deflection (t) 5.53 4.82Horizontal deflection (1) 3.74 2.52

Figure 29-3 shows the load-deflection curve andFigure 29-4 shows pipe strain versus location on thepipe for a surcharge pressure of 48.9 psi.Noteworthy features of these results are the shapeof the load-deflection curve, relative magnitudes ofhorizontal and vertical ring deflections, and theshapes and magnitudes of bending and thrust strain.This condition was modeled with FEA in severalways. Figures 29-3 and 29-4 also show the resultsfrom the FEA using homogeneous 90% relativecompaction and no compaction simulations. These

figures show a similarity in the general shape of theload-deflection curve. The pipe-strain plots in Figure29-4 indicates that the magnitudes of pipe strain forthis case at a surface pressure of 50.0 psi arecomparable, but several maxima and points ofinflection are missing. The magnitude of ringdeflection at the 50 psi surface pressure is alsocomparable within one-half of 1% of the measureddeflection.

Figures 29-5 and 29-6 show the results of the FEAfor the homogeneous dense soil condition withcompaction simulation during construction. Thephysical pipe data are:

Curve Parameter A BStiffness (psi) 10 10Thickness (in.) 0.285 0.300Surface pressure (psi) 48.9 50.0Vertical deflection (t) 5.53 5.42Horizon. deflection (1) 3.74 3.14

The load-deflection plot in Figure 29-5 has lost someof the initial steepness, but the difference betweenvertical and horizontal deflection is maintained.Compared to Figure 29-3, the magnitudes ofdeflection are similar. Figure 29-6 shows the pipe-strain plots from the compaction simulation at asurface pressure of 50.0 psi. Comparison of Figure29-6 with the measured values from the soil-boxresults in Figure 29-4 shows that compactionsimulation improved correlation. In fact, the generalshape, maxima, and magnitudes all compare well.This result is the best comparison between any soil-box test and FEA result.

Additional comparisons included soft elements in theshoulders of the pipe. Because techniques did notallow compaction above the pipe, a theoreticallyhomogeneous installation would still have soil of alesser density at the crown. FEA results for thiscondition included various degrees of compactionsimulation. From additional load-deflection and pipe-strain plots it is evident that with the soft-crownanalyses the pipe strain at the


Figure 29-5 Vertical soil pressure versus pipe deflection for (curve A) soil-box data, 90% relativecompaction, silty sand and (curve B) FEA with compaction simulation.

Figure 29-6 Pipe strains as functions of circumferential location, conditions as in Figure 29-5.


135-degree location increases. See Figure 29-3.This is due to decreased stiffness of the soil in theshoulders which allows for more bendingdeformation in the pipe. Compaction simulation forthe soft-crown condition decreases the bendingstrains and ring deformations because the soilresponds initially in the rebound range and inhibitsdeformation at the low pressures. Becausecompaction simulation does not include adding loadsdirectly over the pipe at the first constructionincrement, a soft-crown condition is actually createdin the homogeneous case. The soil at the crown isuncompacted and does not respond to the stifferrebound modulus at the lower pressure ranges asdoes the surrounding soil elements that receivecompaction loads directly.

Poor-Haunch Installation at 90% RelativeCompaction

Figures 29-7 and 29-8 show the results for the poor-haunch installation with the silty sand in the soil-boxtests. The physical pipe data are:

Curve Parameter A BStiffness (psi) 10 10Thickness (in.) 0.285 0.300Surface pressure (psi) 35.5 30.0Vertical deflection (%) 3.14 2.21Horizontal deflection (%) 1.30 1.09

Figure 29-7 shows the load-deflection response andFigure 29-8 shows the pipe strain around the pipe fora surface pressure of 35.5 psi. Again the initialsteepness of the load-deflection curve, the relativemagnitudes between the vertical and horizontaldeflections, and the shapes and magnitude of thestrain plots are noteworthy. The bending strains arehigher at the 30 to 45 degree locations on the pipebecause of the lack of support in the haunch area.Also, a comparison of the homogeneous installationand the poor-haunch installation in Figures 29-6 and29-8, respectively, shows noticeable differences in

the pipe-strain plots from soil-box tests. Figures 29-7 and 29-8 also show the FEA results for the poor-haunch condition without compaction simulation.The load-deflection plot shows similar behavior , yetthe deformations are larger in the FEA results. Thepipe-strain plot shows a peak of large strain at 45degrees and low strains from spring line to crownthat are similar to the soil-box results. Figures 29-9and 29-10 show the FEA results for poor hauncheswith compaction simulation. The load-deflectionplots show larger deflections and the pipe-strainplots show larger strains from spring line to crownthan in Figure 28-8. However, strain at the invert ofthe pipe with compaction simulation was closer tothe measured results than to the FEA results that didnot include compaction simulation. The physicalpipe data for Figures 29-9 and 29-10 are:


Homogeneous Installation with 80% RelativeRelative Compaction

Figures 29-11 and 29-12 show the soil-box resultsfor 80% relative compaction homogeneous installa-tion. The physical pipe data are:


The vertical and horizontal deflections are similarthroughout the test, which includes ellipticaldeformation as shown in Figure 29-11. Figures 29-11 and 29-12 also show the results from the FEA


Figure 29-7 Vertical soil pressure versus pipe deflection for (curve A) soil-box data, 90% relativecompaction, silty sand, and poor haunch support; and (curve B) FEA, no compaction simulation, and poorhaunch support.



Figure 29-9 Vertical soil pressure versus pipe deflection for (curve A) soil-box data, 90% relative compaction,silty sand, and poor haunch support; and (curve B) FEA with compaction simulation and poor haunch support.



Figure 29-11 Vertical soil pressure versus pipe deflection for (curve A) soil-box data, 80% relativecompaction and (curve B) FEA, no compaction simulation.



for the 80% relative compaction homogeneouscondition. Although the load-deflection curve showsmore deformation with the loose material than withthe dense material, the actual comparison of soil-boxtests with FEA tests shows that the FEA does notcompare as well for the looser soil condition. Thepipe-strain plots in Figure 29-12 also confirm thepoorer comparison. In terms of magnitude of themaximum strain, there is correlation, but the overallshape of the pipe-strain plots do not match measuredvalues as well as the analyses with 90% density.

DISCUSSION OF RESULTS

Soil compaction simulation for the FEA response ofthe FRP pipe improves the comparison with soil-boxtests for homogeneous soil. For nonhomogeneoussoil installation conditions, compaction simulation didnot improve the comparison of FEA with soil-boxtests enough to justify the additional computationaleffort required.

The FEA results were generally better for dense soilinstallation conditions than for loose soil conditions.This is due to a combination of numerical difficultieswith the finite-element method and lack of similitudein the soil-box model that arises with loose soilconditions. Entries in the stiffness matrix becomesensitive to the magnitudes of the elastic and bulkmoduli at low soil stiffness. In order to achievelarger deflections, lower values of the bulk modulusare required. This, however, can result in singularmatrix warnings, which indicates that entries in thestiffness matrix will not produce reliable results.

The geometric nonlinear analysis (wherein theformulation of the stiffness matrix accounts for thenodal deflections at each loading increment) doesnot significantly change the results for installationcondition modeling. The inclusion of the geometricnonlinear analysis would generally predict higherdeflections. For example, an analysis that did notinclude geometric nonlinearities might predict avertical ring deflection of 7%. The same conditionsincluding geometric nonlinearities would predict ring

deflections of around 8%. However, for the othertypes of loading conditions (such as rerounding), theformulation of the stiffness matrix must reflect theshape of the pipe.

SUMMARY AND CONCLUSIONS

Good correlation of finite-element modeling offlexible pipes with test data requires modelingc apabilities not readily available in most computerprograms. Such capabilities include analysis of thestress history of the soil elements to determinewhether each element is in primary loading or inunloading and reloading, modification of the iterationscheme to better model the soil response when thereis a change from one stress condition to another, andlarge-deflection theory by modifying nodalcoordinates after each load increment. In addition,postprocessing plotting routines are needed tographically analyze the pipe response to each loadingcondition. The development of these featuresmakes possible the analysis of flexible pipe undercompaction simulation, surcharge pressures,rerounding caused by internal pressure, and variousinstallation conditions. The results of analyses forvarious installation conditions show the effects ofshoulder and haunch support on the pipe and suggestthat these conditions can be considered in FEAanalysis of pipe and installation conditions.

The results of the USU study of four soil types andvarious loading conditions, show a good correlationbetween FEA results and the measured responsesfrom physical model tests in a soil-box. The finite-element method can be used for analyzingperformance of buried flexible pipes with variousinstallation conditions, soil types and densities,loading conditions, and pipe sizes and stiffnesses.The cost of FEA is less than physical testing.However, calibration of the FEA requires resultsfrom physical tests. The two techniques usedtogether are applicable and cost effective foranalyzing buried flexible pipe performance.


REFERENCES

Duncan, J.M. (1979). Behavior and design of long-span metal culverts. Journal of GeotechnicalEngineering, ASCE, Vol. 105, No. GT3, March1979.

Duncan, J.M. (1980), P.Byrne, K.S.Wong, andP.Mabry. Strength, stress-strain and bulk modulusparameters for finite element analyses of stressesand movements in soil masses. GeotechnicalEngineering Report UCB/GT/80-81. University ofCalifornia, Berkeley, 1980.

Katona, M.G. (1976), J.B.Forrest, R.J.Odello, andJ.R.Allgood. CANDE—A modern approach for thestructural design and analysis of buried culverts.Report FHWA-RD-77-5, FHWA, U.S.Departmentof Transportation, 1976.

Katona, M.G. (1982). Effects of frictional slippageof soil-structure interfaces of buried culverts. InTransportation Research Record 878, TRB, NationalResearch Council, Washington, D.C., 1982, pp 8-10.

Knight, G.K. (1983), and A.P.Moser. The structuralresponse of fiberglass reinforced plastic pipe underearth loadings. Buried Structures Laboratory, UtahState University, Logan, 1983.

Kulhawy, F.N. (1969), J.M.Duncan, and H.B.Seed.Finite element analysis of stresses and movementsin embankments during construction. GeotechnicalEngineering Report TE-69-4. University ofCalifornia, Berkeley, 1969.

Leonards, G.A. (1982), T.H.Wu, and C.H.Juang.Predic ting performance of buried conduits. ReportFHWA/IN/JHRP-81/3. FHWA, U.S. Departmentof Transportation, 1982.

Medrano, (1984), A.P.Moser, and O.K.Shupe.Performance of fiberglass reinforced plastic pipe tovarious soil loads and conditions. Buried StructuresLaboratory, Utah State University, Logan, 1094.

Nyby, D.W. (1981). Finite element analysis of soil-structure interaction. Ph.D. dissertation, Utah StateUniversity, Logan, 1981.

Ozawa, Y. (1973),and J.M.Duncan. ISBILD: Acomputer program for analysis of static stresses andmovements in embankments. GeotechnicalEngineering Report, University of California,Berkeley, 1973.

Sharp, K.D. (1984), F.W.Kiefer, L.R.Anderson, andE.Jones. Soils testing report for applications of finiteelement analysis of FRP pipe performance. SoilsTesting Report, Buried Structures Laboratory, UtahState University, Logan, 1984.

Sharp, K.D. (1984), L.R.Anderson, A.P.Moser, andM.J.Warner. Applications of finite element analysisof FRP pipe performance. Buried StructuresLaboratory, Utah State University, Logan, 1984.

Sharp, K.B. (1985), L.R.Anderson, A.P.Moser, andR.R.Bishop. Finite element analysis applied to theresponse of buried FRP pipe under variousinstallation conditions. In Transportation Research

Record 1008, Transportation Research Board,National Research Council, pp 63-72, 1985.

Wilson, E.L. (1963). Finite element analysis of two-dimensional structures. Ph.D. dissertation,University of California, Berkeley 1963.


Anderson, Loren Runar et al "ECONOMICS OF BURIED PIPES AND TANKS "Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

CHAPTER 30 ECONOMICS OF BURIED PIPES AND TANKS

The engineer is responsible for cost effectiveness.Cost effectiveness of any project is the extent towhich the returns on investment outweigh the costs.Costs are money expenditures; but costs also includetime, effort, overhead, insurance, warranty, publicrelations, etc. Returns include money, but alsoreputation, etc. All costs and returns must bereduced to monetary equivalents. And for analysis,all monetary equivalents must be reduced to thesame basis — a present worth, P; or a periodicpayment (annually), R; or a future lump sum, S. Forplanning and design, it is customary to compare themonetary equivalents of costs for all possiblealternatives after reducing them to the same basis ofmonetary equivalence — usually equal periodicpayment, R, (or sometimes present worth, P). Forsale of a project, or settlement of claims, presentworth, P, is usually the best monetary equivalent.The present worth of a project that is to continueforever by repeated replacements is calledcapitalized cost, P . The interest on capitalized costprovides the periodic payments. Monetaryequivalence depends upon time and interest ratesbecause the value of money is not constant.

VALUE OF MONEY

Nomenclature

P = present worth, an amount of money now,S = sum, the value after n periods of time at

interest rate i,R = equal periodic payments over n periods at

interest rate i, that accumulates sum S orrepays present worth P,

P = capitalized cost (interest = R),i = interest rate, compounded at the end of

each pay period,n = number of pay periods of time (usually in

years).

Derivation of Equations

The interrelationships of money values, P, S, and R,for n periods at interest rate i compounded at theend of each period, are as follows.

S = P + Pi + i(P+Pi) + i[P+Pi+i(P+Pi)] + ....

S = P(1+i) + Pi(1+i) + Pi(1+i)2 + Pi(1+i)3 + ...+ Pi(1+i)n-1 . . . . . (30.1)

Rewriting,S = P[(1+i)2 + i(1+i)2 + i(1+i)3 + i(1+i)4 + ...+ i(1+i)n-1] . . . . . (30.2)

Multiplying both sides of Equation 30.1 by (1+i)

S(1+i) = P[(1+i)2 + i(1+i)2 + i(1+i)3 + i(1+i)4 + ...+ i(1+i)n] . . . . . (30.3)

Subtracting Equation 30.2 from Equation 30.3,

S = P(1+i)n . . . . . (30.4)

Consider each R of a series of periodic payments tobe a separate P. For this analysis, let R be due atthe beginning of each pay period. Then, evaluatingand adding up the sums of all of the R's in the series,while working backward from the last R-payment tothe first from one year past the last R,

S = R(1+i) + R(1+i)2 + R(1+i)3 + ...+ R(1+i)n . . . . . (30.5)

Multiplying both sides by (1+i),

S(1+i) = R(1+i)2 + R(1+i)3 + ...+ R(1+i)n+1

. . . . . (30.6)

Subtracting Equation 30.5 from Equation 30.6 yieldsEquation 30.7. Eliminating S between Equations30.4 and 30.7 yields Equation 30.8.


If R is at the beginning of each pay period,

Si = R[(1+i)n+1 - (1+i)] . . . . . (30.7)

Pi(1+i)n = R[(1+i)n+1 - (1+i)] . . . . . (30.8)

S = P(1+i)n . . . . . (30.9)

P = R(1+1/i) . . . . . (30.10)

The three variables P, R, and S, can be interrelatedby the three Equations, 30.7, 30.8, and 30.9, for anygiven interest rate, i, and any number of pay periods,n. Equation 30.10 is capitalized cost, P , which isthe present worth of all series payments, R, if theproject is to continue on forever by replacements;i.e., for n = oo .

If R is at the end of each pay period,

Si = R[(1+i)n - 1] . . . . . (30.11)

Pi(1+i)n = R[(1+i)n - 1] . . . . . (30.12)

S = P(1+i)n . . . . . (30.13)

P = R/i . . . . . (30.14)

The three variables P, R, and S, can be interrelatedby the three Equations, 30.11, 30.12, and 30.13 forany given interest rate, i, and any number of payperiods, n. Equation 30.14 is capitalized cost, P ,which is the present worth of all periodic payments,R, if the projec t is to continue on forever; i.e. forinfinite time, n = oo .

In general, R is used as a basis for comparingalternatives for least cost. P is used as a basis forsale or for settlement of a claim.

Example 1

The purchase cost of tanks in a buried tank farm is$1,000,000. Installation costs are $100,000 per yearat year end for 4 years. What is the annual end-of-period payment R to pay off the project in fouryears? R is annual installation cost plus the end-of-

year repayment of purchase cost, P = $1,000,000from Equation 30.12 in n = 4 years at i = 7%interest.

R = $100,000+ $1,000,000(0.07)(1.07)4/[(1.07)4 - 1]

R = $100,000 + $295,228 = $395,228

Find the present worth, P, if R = $395,228. FromEquation 30.12,

P = R[(1+i)n - 1]/i(1+i)n

= $395,228[(1.07)4 - 1]/(0.07)(1.07)4

P = $1,338,721

If the life of the tanks is 50 years, and payments aremade over the 50-year life rather than 4 years, whatis the annual end-of-year payment? From Equation30.12, if P = $1,338,721, i = 7%, and n = 50 payperiods,

R = $1,338,721(0.07)(1.07)50/[(1.07)50 - 1]

R = $97,003.50

If the tank farm must be replaced every 50 years,find the capitalized cost. From Equation 30.14, P =R/i, where R = $97,003.50 and i = 7%. P =$97,003.50/0.07 = $1,385,764.32.

P = $1,385,764

MONETARY EQUIVALENCE

Monetary equivalence includes overhead and alldirect costs such as purchases and installation. Butmonetary equivalence also includes maintenance andthe cost of risk. Risk includes public relations,insurance, and legal counsel, in addition to the costof replacement. Example 2

Purchase price of a buried tank is $10,000.Installation cost is $40,000. Maintenance is $500 peryear. The cost of a leak (replacement, damage, and


liability) is estimated to be $100,000. For thisanalysis, assume that insurance covering leaks canbe obtained for $2000 per year. The tank isdesigned for 50 years of service. If the interest rateis 8%, what is the equivalent annual R at beginning-of-pay periods for the project? This R can becompared with the R's for alternatives such as moreexpensive (dual-containment) tanks with lessprobability of a leak. For beginning-of-pay periods,utilizing Equation 30.8,

R = $500 + $2,000+ $50,000(0.08)(1.08)50/[(1.08)51 - 1.08]

R = $500 + $4,000 + $3784 = $6284.

R = $6284

What is the present worth, P, of the project in theevent of sale or legal action? Already known fromExample 2 is R = $6284. From Equation 30.8,

P(0.08)(1.08)50 = $6284[(1.08)51 - 1.08]

P(3.7521) = $6284[49.5737] = $311,521.39

P = $83,025

COST OF FAILURE

The cost of failure could be staggering, depending ondamage and liabilities, but the probability of failuremay be remote. The monetary equivalence offailure in any pay period is the approximate cost offailure times the probability of failure. Twotechniques for finding the monetary equivalence offailure follow.

Probability of a 100-Year Event:A 100-year event (or any time-indexed event) isdefined as the average period of time betweenoccurrences of an event of magnitude greater thansome given magnitude. If there is no periodicity, anevent can occur at any time. For example, theperiodicity of storms, floods, and earthquakes hasnot yet been predicted with confidence, despitepromising new techniques. Available only are

records of previous events with magnitudes andtimes noted. The best monetary equivalent of thetime-indexed magnitude of an event is an insurancepremium — a series payment, Rc r, which is equal tothe cost of the event times the probability that theevent will occur in any one pay period. Areasonable profit for the insurance carrier is added.For an insurance carrier, with many portfolios, theprobability that a 100-year event will occur in anyone year is 1/100. The unpredictable timing of eachis balanced out in the totality of portfolios. Themonetary equivalent of the magnitude of an event isPc r. The probability is 1/nc r. The periodic paymentto cover the event i s Rcr = Pc r/nc r , where nc r is thetime index (100 years, etc.).

Example 3The tank of Example 2 will float out of itsembedment if it is empty when a 50 year floodoccurs. The cost of flotation is tank replacement at$50,000 plus public liability estimated to be $100,000.What insurance premium would be justified ifreasonable profit for the carrier is 30%? Assumingflotation of an empty tank, from Example 2, the costof the disaster is the purchase price plus installation,plus liability; i.e., Pcr = $10,000 + $40,000 + $100,000= $150,000. But the probability of a flood that couldcause flotation in any one time period is only one in50 years; i.e., 1/ncr = 0.02. R = ($150,000)(0.02) =$3000. Including a 30% profit for the insurancecarrier, the insurance premium is $3900. This isbased on the assumption that the tank is alwaysempty. But what is the probability that the tank isempty if it is used for gasoline storage at a servicestation?

Example 4Find the probability, Pe, that a tank will float if itis not always empty. The subscript, e,distinguishes probability from present worth. If thetank is used to store gasoline at a service station, itwill be refilled as soon as it is empty. Assume thatgasoline is pumped out at a constant rate betweenrefillings. The first problem is that the tank mayfloat with some gasoline in it. What is the leastamount of gasoline in the tank that acts as ballastand


Figure 30-1 Buried tank (top) showing the level of the contents (cross-hatched) lower than which the tankwill float; and (bottom) the volume of contents as a function of time between refills, showing (cross-hatched)the time, between refills, during which flatation can occur.

Figure 30-2 Systems diagram ofa buried tank under worst-caseconditions of external loads.


keeps the tank from floating? See the cross-hatchedareas of Figure 30-1. Assume saturated soil withflood level at or above ground surface. The tankvolume is 12,000 gallons.D = 7 ft,L = 42 ft,WT = 8 kips = weight of the tank,wT = 190 lb/ft = tank wt/unit length,H = 2 ft = height of soil cover,γ s = 57.6 pcf = unit weight of soil,ws = 1980 lb/ft = buoyant soil wt/unit length,w = 2170 lb/ft = total wt/ft acting down,w = 2400 lb/ft = buoyant force acting up,∆w = 230 lb/ft = wt/ft of gasoline at a level lower

than which the tank floats,w = 1616 lb/ft = wt/ft of gasoline in the tank

when the tank is full (unit weight of gasoline= 42 pcf),

T = time between refillings,∆T = time during which the tank can float (with

less than 230 lb/ft of gasoline ballast).

What is the probability, Pe, that the tank will floatconsidering gasoline is in the tank betweenrefillings? Pe = ∆T/T , where ∆T = (230/1616)T =0.1423T. Pe = 0.1423.

Probability of Failure by Fault Tree

A justifiable R, either for insurance coverage orelimination of the probability of failure, is the cost offailure times the probability that it will occur in anypay period. A model for estimating probability offailure is the fault tree — a logic diagram that startswith an undesired event (failure) and traces backthrough all of the causes (faults) assessing andrelating the probability of each fault to the probabilityof failure. An example illustrates the fault tree.

Example 5

A buried fuel tank costs $50,000 installed. A leakcould result in liabilities up to $100,000. What is theprobability of a leak?

System DiagramThe first step is a diagram of the system. SeeFigure 30-2. The soil cover, H, is minimum for anHS-20 dual-wheel load. It is understood that thetank meets specifications for quality.

Fault Tree Construction A fault tree is a trunk and root system. The trunk isthe event (failure), and the roots are the faults(causes of failure). Figure 30-3 is a table ofsymbols. In this example, the failure is a fuelLEAK. It is shown at the top of the fault tree ofFigure 30-4 in a rectangle indicating that it is anevent. The two causes (faults) considered in thisexample, are shown in a horizontal row below theevent. They are connected to LEAK through anand gate (bullet) indicating that both POOREMBEDMENT and EXCESSIVE LOAD arenecessary to cause the leak. The probabilities, Pe,of the faults must be multiplied together to getthrough the and gate. POOR EMBEDMENT is anevent (rectangle) which could be the result of aHARD SPOT or LOOSE SOIL in the embedment.Because either loose soil or a hard spot is sufficientto cause a leak, they are connected to POOREMBEDMENT through an or gate (arrowhead).The probabilities, Pe, of the faults must be addedtogether to get through the or gate.

HARD SPOT is a diamond, a basic faultundeveloped. If the constructor had a history ofproblems with hard spots, a probability might bedeveloped. Without such a history, probability canonly be assumed and varied to determine the effectof hard spots on the probability of a leak. This couldserve to evaluate encroachment of the constructor’shard spots into the safety zone (safety factor zone).Or from the history (track record) of theconstructor, suppose that repair of a hard spotshows up in the bedding of one out of every 20tanks. Probability is 0.05.

LOOSE SOIL is a circle because it is a basic fault,but developed. From geotechnical tests, the soil wasfound to be loose in one out of every 40 tanks,(0.025).


Figure 30-4 Example of a Fault Tree for finding the probability of a leak in a tank that is buried in poorembedment, and is subjected to excessive loads.


With loose soil in one out of every 40 tanks (0.025),and hard spots in one out of every 20 tanks (0.050),the Pe's are shown in Figure 30-4 for HARD SPOTand LOOSE SOIL. They are added togetherthrough the or gate yielding a value of Pe = 0.075 =probability of a POOR EMBEDMENT.

EXCESSIVE LOAD is an event (rectangle) thatmight be caused by e ither a 50-YEAR FLOOD orLIVE LOAD. If either one or the other issufficient, the two are connected to EXCESSIVELOAD through an or gate.

The 50-YEAR FLOOD is a triangle because it isdeveloped elsewhere. See Example 3. Theprobability is shown on 50-YEAR FLOOD; Pe =0.02 in any payment period (year). The LIVELOAD is a diamond (undeveloped). Why notassume that a ready-mix concrete truck could passover the tank inadvertently — say once every 40years? Then the probability in any payment period(year) is Pe = 0.025 as shown on LIVE LOAD.

The probability of EXCESSIVE LOAD is the sumof the Pe's for the 50-YEAR FLOOD and LIVELOAD. For EXCESSIVE LOAD, Pe = 0.02 +0.025 = 0.045.

The probability of a LEAK in any pay period is theproduct of the Pe's for POOR EMBEDMENT andEXCESSIVE LOAD. Pe = 0.075(0.045). Theprobability of a leak is,Pe = 0.003375 = 1/296 or approximately one in every300 tanks per year.

This 1-in-300 probability of a leak per year mightjustify either a periodic payment to reduce thepossibility of a leak, or insurance premiums thatcover the leak.

It is noteworthy that the above simple example doesnot include such effects as contents of the tank, timerelated consolidation of sidefill soil, vacuum in thetank, and liquefaction of the embedment. A morecomplex fault tree might be justified.

Example 6

What beginning-of-period payment, R, can bejustified to cover the cost of a leak in the tank ofExample 5 if the cost of the leak is $150,000. Theprobability of a leak is 1/296 = 0.003375 in any oneyear. R = $150,000(3.375)10-3.R = $506.25 per year.

SAFETY FACTORS

Simply defined, a safety factor is the ratio ofperformance limit to performance. Not so simple isa number for that safety factor. For above-groundstructures, classical performance is measured bystress in the material. Performance limit is strength.For buried structures, performance is, usually,deformation — including leaks and excessivemovement of the soil or structure. Failure is usuallyreduced to a monetary equivalent. What is themonetary equivalent of failure? Are there anymitigations for failure? What are the encroachmentsinto the safety zone (demilitarized zone)? And, inthe event of legal action, who is responsible for theencroachments? What are the monetary equivalentsof encroachments? Whose are the responsibilities?Legal counsel usually becomes involved in thesequestions. However, information upon whichcounsel directs legal proceedings must come fromthe engineer. Reduction to monetary equivalenceoften comes from the engineer. Comparative orcontributory encroachment into the safety factorzone requires engineering knowledge.

Example 7

A 12,000-gallon tank for storing gasoline, leaked justafter installation. The leak was a circumferentialcrack on the bottom at midspan. The tank wasburied in dry soil (no water table). What are therelative responsibilities for the leak (in percent) ofthe engineer, manufacturer, and installer? Neglectcontributory negligence of other parties in thishypothetical case. It is known that an HS-20 axleload of 32 kips passed over the tank at midspan.See


Figure 30-2. The dual wheels were separated by 6ft.D = 84 inches,L = 42 ft,t = 0.6 inch,H = 2 ft,γ = 120 pcf = unit weight of dry soil,γ G = 42 pcf = unit weight of gasoline,σf = 5 ksi = yield strength of the tank wall,wT = 200 lb/ft = weight per unit length.

The safety factor is two based on yield strength of5 ksi. Design stress is 2.5 ksi. Therefore the safetyzone is from 2.5 to 5 ksi.

Weight per unit length of tank is the tank full ofgasoline, plus prismatic soil load on top. If the tankis a simply supported beam, including live load, thelongitudinal stress in the bottom at midspan is 3.83ksi.

ENGINEER specified compacted soil with noorganics or large rocks. The tank was designed for40% of simply supported beam stresses — assumingthat the bedding would provide some support. Butthe beam was simply supported. Therefore, designencroachment is (3.83-2.50)/(5-2.5) = 0.53.

MANUFACTURER supplied a tank with wallstrength of 4 ksi according to post-leak tests — not5 ksi as advertised. The encroachment is (5-4)/(5-2.5) = 0.4.

INSTALLER leveled the tank on timbers at theends thereby forcing the tank to act as a simplysupported beam with no bedding. There was nocompaction under the haunches. Had a beddingreduced stresses to 40% of the simplysupported beam stresses, the maximum stress in thetank would be (0.4)(3.83) = 1.53 ksi. Theencroachment is (3.83-1.53)/(5-1.53) = 0.66.

Percent Encroachment into the Safety Zone(Safety zone = 0.53+0.40+0.66=1.59):

ENGINEER = 33%MANUFACTURER = 25%INSTALLER = 42%

PROBLEMS

30-1 What is the probability of a leak in a farm of10 buried tanks if the probability of a leak in any onetank is 1/300 during any pay period?

(1/30)

30-2 Two tanks are under consideration for aproject. Tank A costs $10,000 plus installation costof $5,000. Service life is 30 years if an insurancepremium of $400 is paid at the beginning of eachyear. Tank B costs $15,000 plus installation cost of$7,000 and service life of 50 years. What are theequal series payments, R, at year end for the twoalternatives? Interest rate is 8%.

(A, $1764; B, $1798)

30-3 If the tanks of Problem 30-2 are to bereplaced by identical tanks at the end of eachservice life, what are the capitalized costs, P , oftanks A and B?

(A, $22,055; B, $22,479)

30-4 The head of a tank leaks near the bottom. H= 2 ft, D = 84 inches, t = 0.187 inch. Findcontributory negligence of:Manufacturer — insertion is only 0.3 inch.Constructor — a timber is left under the end of thetank for vertical alignment.Owner — a 16-kip wheel load passed over the tankthe evening before the leak was detected.


Anderson, Loren Runar et al "CASTIGLIANO’S EQUATION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

Figure A-1 Virtual load q on a bridge -- useful for calculating the deflection of the bridge at the locationof q, in the direction of q, due to the truck load Q.

Figure A-2 Free-body-diagrams for writing moment M for the Q-load (left) and moment m for the virtualq-load (right) for a ring.

APPENDIX A CASTIGLIANO'S EQUATION

A powerful method for calculating deflection is themethod of virtual work. Consider the beam(bridge) of Figure A-1. We would like to knowthe deflection of the beam at midspan due to atruck load Q located as shown. I locate myself atmidspan before the truck reaches the bridge. I amthe virtual load, q. Comes now the truck. As itreaches the location shown, I feel myself beinglowered by the deflection of the bridge at midspandue to the truck load Q. I do virtual work becauseof deflection, ∆ , of the bridge by the truck. Myvirtual work is q∆ . But this virtual work acts onthe bridge which stores my virtual work as(potential) virtual energy. The stored virtualenergy is mdθ, where m is the moment at anypoint on the bridge due to the virtual load, and dθis the change in curvature of the bridge at thatpoint due to the truck load. But dθ = Mdx/EI. Sothe virtual stored energy is mMdx/EI. Equatingvirtual work to virtual stored energy,

q∆ =f oL mMdx/EI

whereq = virtual load, (Let q = 1.)m = moment in the beam due to load q,M = moment in the beam due to load Q,E = modulus of elasticity of material,I = moment of inertia of the cross section of

the beam about its horizontal neutralsurface,

L = length of the beam.

Because the virtual load was set equal to unity, theresulting equation for deflection of the bridge atmidspan due to the truck is, ∆ =f o

L mMdx/EI

integrated over the length of the beam. M is theequation of the moment as a function of x to beintegrated over the length of the beam. x isdistance from an assumed origin of axes. Thedummy moment, m, is the equation of moment asa function of q throughout the length of the beamin terms of x measured from the same origin of

axes. Clearly, the two moment equations requireanalyses of two separate free-body-diagrams ofthe beam.

Deflection by virtual work is not limited to astraight beam. It can be a curved beam or a ringfor which equations for moments M and m arewritten in terms of radius and angle, θ.Deflections can also be found for shearing loadsand thrust as well as moments. In fact, becauseenergy is scalar, the deflection due to moments,thrust and shear can be found by adding theenergies of all of the load elements.

Castigliano observed that it is usually easier toapply the virtual work equation by use of theLeibnitz rule which allows differentiation under theintegral sign as follows:

∆ =f oL (M/EI)( M/ q)ds

CASTIGLIANO EQUATIONwhere∆ = deflection of the beam at the location of

the virtual load q in the direction of q. q isa differential that approaches zero at thelocation where the deflection is to becalculated (q can also be a differentialmoment for calculating the angle ofrotation of the beam at the location of q.),

M = equation of the moment at any point dueto both q and Q,

EI = stiffness of the beam,ds = differential distance along the beam (dx

for the beam). The advantage of Castigliano over virtual work isa reduction of analysis. Only one free-body-diagram is required with both the virtual load andthe applied load in place. For small deflections,Castigliano's equation is well adapted to ringsbecause M can be written for the applied loadsand pressures plus the virtual load in terms ofstiffness EI, angle θ, and ds. EI is constant, andds = rdθ where r is the radius of the ring. SeeFigure A-2.


Figure A-3 F-load on a ring showing a quadrant of the ring as a free-body-diagram for evaluation of themoment at A by Castigliano's equation.

Figure A-4 Free-body-diagram of a quadrant of a ring subjected to an F-load, showing the notation forevaluation of the vertical deflection of A with respect to B.


Because of symmetry, it is convenient to considerhalf of the ring.

Example 1

What is the moment at A due to an F-load on aring? The free-body-diagram is a quadrant shownin Figure A-3. Let Q = F/2 for convenience.Moment MA is an unknown in addition to the threereactions at B. The quadrant is staticallyindeterminate to the first degree. Therefore, anequation of deflection is required in addition to thethree equations of static equilibrium. In this case,it is possible to find MA by means of a singleequation of deflection. Note that the relativerotation of A with respect to B is zero, i.e., ψA/B =0. Applying m at A in the direction of rotation ofψA/B, from Castigliano,

ψA/B = (M/EI)( M/ m)rdθ = 0

M = Qsinθ - MA - m, where m 0 M/ m = -1

Substituting into Castigliano's equation,

0 = (Qrsinθ-MA)dθ = [Qrcosθ+MAθ],from the limits 0 < θ < π/2.

Substituting the limits, MA = 2Q/π.

Example 2

Knowing MA, what is ring deflection, d, ofExample 1 due to the F-load? For a free-body-diagram, use the quadrant redrawn in Figure A-4.d = ∆/D = (yA/B)/r; where yA/B is verticaldisplacement of A with respect to B due to thehalf load, Q = F/2. From Castigliano, yA/B = (M/EI)( M/ q)rdθ

M = (F/2 + q)rsinθ - 2Q/π; where q 0. M/ q = r sinθ.

Substituting into Castigliano's equation,

yA/B = (Qr3/EI) (sin2θ - (2sinθ)/π)dθ.

yA/B = Qr3/EI[θ/2 - (sin2θ)/4+2cosθ)/π],from the limits 0 < θ < π/2.

Substituting limits, yA/B = (Qr3/EI)(π/4 - 2/π)

But d = (yA/B)/r, so d = 0.0186Fr2/EI

This is one of the solutions in Table A-1, asummary of moments, thrusts, and ring deflectionsfor rings subjected to typical loads.

Example 3

A ring from a profile rib HDPE pipe is cut andloaded as shown in the sketch below. Theobjective is to find any possibility for long termwall buckling at B due to constant load, F. Whatis the gap, 2x, as a function of load, F? The righthalf of ring AB is analyzed by Castigliano'sequation,

x = (M/EI)( M/ p)rdθ

M = (F+p)r(1-cosθ); where p 0. M/ p) = r(1-cosθ)

x = (Fr3/EI) (1-2cosθ+cos2θ)dθ

Integrating,

x = (Fr3/EI)[θ-2sinθ+θ/2+(sin2θ)/4]

Substituting limits, 0 < θ < π,

x = (Fr3/EI)(3π/2)


Table A-1 Mechanical Analyses of Thin-wall Rings With Symmetrical Loads


Table A-2 Circular Arcs (Partial Circular Rings)


Anderson, Loren Runar et al "RECONCILIATION OF FORMULAS FOR RING DEFLECTION"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

APPENDIX B RECONCILIATION OF FORMULAS FOR RING DEFLECTION

M. G. Spangler was the first to predic t the ringdeflection of buried flexible circular pipes. His IowaFormula is:

∆x = Df KsWcr3/(EI+0.061E'r3)

where (See Figure B-1.)∆x = horizontal increase in diameter due to

vertical soil pressure P,P = vertical soil pressure on top of pipe,∆ = vert ical decrease in diameter which

Spangler assumed to be equal to ∆ x,d = ring deflection = ∆/D ,D = mean circular diameter of the pipe,r = D/2 = mean radius of the pipe,Df = deflection lag factor which can be

incorporated into a time-dependent soilmodulus and can be ignored (Spanglerassumed Df = 1.5.),

Ks = bedding factor. Spangler found it can varyfrom 0.083 to 0.110 depending upon thebedding angle, α . (A reasonable assumptionis Ks = 0.1.)

Figure B-1 Assumed loads on a flexible ring inSpangler's Iowa Formula.

Wc = Marston load per unit length,(Wc PD for most flexible pipes.)

ε = vertical soil strain, (compression) in sidefillsoil due to pressure P,

EI = stiffness of the pipe wall per unit length(See Figure B-2.)

E' = soil modulus. (Spangler defines this ashorizontal modulus of soil reaction, E", atthe spring lines. It is more relevant to relatering deflection to vertical soil modulus fromconfined compression tests.)

Rs = E'/(EI/D3) = stiffness ratio = ratio of soilstiffness E' to ring stiffness EI/D3.Rs = 53.77R's.

R's = E'/(F/∆) = stiffness ratio based on pipestiffness, F/∆ .

The Iowa Formula can be written in the followingform:

d/ε = Rs/(80+0.061Rs)IOWA FORMULA

This is the relationship between two dimensionlessvariables: d/ε = ring deflection term, and Rs =stiffness ratio. This relationship is shown on FigureB-3. The graph approaches a horizontal asymptoteat d/ε = 1.64. But empirical ring deflections do notexceed vertical compression of the sidefill soil.Therefore, d/ε does not exceed unity. Morereasonable is the empirical graph of Figure B-3which is expressed by the formula:

d/ε = Rs/(30+Rs)EMPIRICAL FORMULA

The empirical formula is an upper 90 percentile ofring deflections from tests at USU in the 1960s.


Figure B-2 Pertinent notation for calculating the ring deflection of buried flexible pipes.

Figure B-3 Comparison of the Iowa Formula and an Upper Limit Empirical graph for predicting ringdeflection of buried flexible pipes.


Iowa Formula

The Iowa Formula was derived to predict horizontalring deflection of buried flexible pipes. Figure B-1is the basis for the derivation. Bedding angle, α ,could vary from 0 to 180o. Horizontal passive soilresistance acted over a parabola subtended by anarc of 100o. If an arc of 90o seemed reasonable,100o must be conservative. The passive resistancewas better represented by a sine curve throughoutthe height of the ring, but a sine curve appeared tobe more difficult to integrate. Integrations wereperformed using virtual work to solve the twodeformation equations required for analysis. Threeequations were available from static equilibrium.With load, Wc, and soil resistance, h, known, fiveunknowns can be solved. The mathematicalanalysis by Spangler was elegant and correct aftera modification of definition of h. A critical variablewas the soil resistance, h, assumed to be a functionof horizontal soil modulus of elasticity, E-prime (E').E' was assumed to be constant for any soil anddensity (compaction). For details see text bySpangler, M.G. and Handy, R.L., Soil Engineering,3rd Ed., IEP, New York 1975.

E-prime (E')

E-prime (E') is the horizontal soil modulus in theIowa Formula. In the derivation of the formula itwas assumed that all materials are elastic and thatE' is constant for any given soil embedment. In fact,E' is not constant. Pipe-soil interaction is not elastic.Flexible pipes are best analyzed by mechanics ofplastics. Soil can vary from viscous through plasticto granular (best analyzed by the mechanics ofparticulates). Values of E' were investigated atUSU in 1996 and 1997. E' was found to be afunction of soil depth (confinement), ring stiffness,and soil type and density. Figure B-4 is a graph ofE' in silty sand (Unified Classification SM). FigureB-5 is a proposal for conservative values of E' forsilty sand. Other soils require similar tests toevaluate E'.

EXTERNAL PRESSURE DESIGN BASED ONRING STIFFNESS (BUCKLING)

Like the Iowa Formula, some ring design equationsare based on elastic stiffnesses of the ring and thesoil. These equations are subject to the samecautions as any equation based on elasticity. Twosuch design equations are considered in thefollowing.One equation is simplified AWWA C950:

P = \/R wEs(EI)/(0.149r3) . . . . (B.1)whereP = uniform collapse pressure, Rw

= buoyancy factor = 1- 0.33(H'/H), where H'<H,

H = height of fill over pipe,H' = height of water table over pipe,Es = soil stiffness (secant modulus),E = long-term (virtual) pipe mod,r = mean radius of pipe = D/2,I = mom/iner of wall cross section.

The other equation is from "Industrial Brochure" bythe Large Diameter Pipe Division of PPI. It issimilar to Equation 41 in the Uni-Bell Handbook,

P = 1.15 \/E tp . . . . . (B.2)where

p(1-ν2)(D/t)3(r'/r)3 = 2Ep = collapse pressure on circ pipe,E = modulus at operating temperature,t = mean wall thickness,D = mean diameter = 2r,r = mean circular radius,r' = maximum radius of curvature,ν = Poisson ratio,Et = soil stiffness (tangent modulus).

In order to compare Equations B.1 and B.2, it isassumed that:

m = r/t = D/2t = ring flexibility,Rw = 0.67 (Water table at surface),I = t3/12 (plain pipe),r' = r (circular pipe cross section),ν = 0.38 = Poisson ratio for PVC.


JMiller

JMiller

Figure B-4 E' for DRY SILTY SAND from USU model studies of 1996. If the soil is saturated, values ofE' are reduced to one-half.

Figure B-5 Proposed conservative values of Er for silty sand (Unified Classification SM).


Substituting values,

AWWA P = 0.61\/E sE/m3

Uni-Bell P = 0.62 \/EtE/m3

If the secant soil modulus Es is about the same asthe tangent soil modulus Et, Equations B.1 and B.2are comparable. The Uni-Bell equation assumes awater table at ground surface. AWWA disregardsovality, but includes height of soil cover and watertable.

The range of applicability of these two equations islimited. If either of the soil moduli approaches zero,P approaches zero. Not so. Clearly poor soil is outof range.

For further comparisons, Equations B.1 and B.2 arerewritten as service equation,

SERVICE, P = 0.62 \/KEtE/m3

whereP = pressure on circular pipe,KEt = horizontal tangential soil modulus,K = (1-sinϕ)/(1+sinϕ) = 1/3 if,ϕ = soil friction angle

= 30o for cohesionless soil,m = r/t = ratio of mean radius and wall thickness

for plain pipe.

If the ring is encased in relatively rigid soil, the soilacts as a casing at yield, S,

ENCASED, P = S/m

This is ring compression failure.

If the ring is nonconstrained; i.e., not buried,

NONCONSTRAINED, P = E/4(1-ν 2)m3

This is buckling failure.

If the ring is supported by soil in which the cross-section stiffness is calculated from ring deflectiontests, then, from an equation for the empirical curveof Figure B-3,

EMPIRICAL, P =0.14E/m3 + 0.27KEt

If ring deflection can be predicted by the IowaFormula, then effective wall stiffness can becalculated and,

IOWA, P = 0.22E/m3 + 0.16KEt

Figure B-6 is a comparison of values of critical P bythe above equations of elasticity.

Figure B-6 Comparison of external pressure design procedures for ring compression (ENCASED), top twographs, and for elastic collapse, bottom graphs.


JMiller

JMiller

JMiller

Anderson, Loren Runar et al "SIMILITUDE"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

APPENDIX C SIMILITUDE

Engineering is basically design and analysis; withdue regard to cost, risk, safety, etc. In the following,that which is designed is a buried pipe. The analysisis a model that predicts performance. Performancemust not exceed some performance limits. Themajor concern, usually, is the model. Mathematicalmodels are convenient. Physical, small-scale modelsare used occasionally. The most dependable modelsare full-scale prototypes in service. Mathematicalmodels are written to describe prototypeperformance because it is impractical to perform afull-scale prototype study for every buried pipe to beinstalled. The set of principles upon which a modelcan be related to the prototype for predictingprototype performance, is called similitude.Similitude applies to all models — mathematical,small-scale, and prototype. There are three basicsteps in achieving similitude.1. Fundamental variables (fv's) — a list of all of thevariables that affect the phenomenon. All of the fv'smust be uniquely interdependent. However, nosubset of the fv's can be uniquely interdependent.For example, force, mass, and acceleration cannotall be used as fundamental variables in a morecomplex phenomenon, because force equals masstimes acceleration. Therefore the subset is uniquelyinterdependent. Only two of the three fundamentalvariables could be used in the phenomenon to beinvestigated.

2. Basic dimensions (bd's) — the dimensions inwhich the fv's can be written. The basic dimensionsfor buried pipes are usually force (F), distance (L);and sometimes time (T) and temperature.

3. Pi-terms (pi's) — combinations of the fv's thatmeet the following three requirements;a) The number of pi's must be at least the number offv's minus the number of bd's.b) The pi-terms must all be dimensionless.c) No subset of pi's can be interdependent. This isassured if each pi-term contains a fundamentalvariable not contained in any other pi-term.

The pi-terms can be written by inspection.

Example

Write a set of pi-terms for investigating themaximum wheel load W that can pass over a buriedflexible pipe without denting the top of the pipe. SeeFigure C-1 for a graphical model. Following thethree pi-term requirements;

fv's bd's

W = wheel load FEI = wall stiffness FLH = height of soil cover LP = all pressures FL-2

D = pipe diameter LE' = soil modulus FL-2

γ = soil unit weight FL-3

7 fv's - 2 bd's = 5 pi's required.

pi's (W/E'D2) π 1

(EI/D3) π 2

(H/D) π 3

(P/E') π 4

((D/E') π 5

This set of five pi's, by inspection, is not the onlypossible set. If this set is not convenient forinvestigating the phenomenon, a different set can bewritten. In this case, the maximum wheel load isgiven by a mathematical model:

π 1 = f(π2, π3, π4, π5) . . . . . (C.1)if we can somehow find the relationship of pi-terms.Principles of physics provide one possibility.Prototype studies provide the writing of empiricalequations of best-fit lines through plots of data. Ifsmall scale model studies are to be used, EquationC.1 must describe the performance of both modeland prototype. Therefore, the model must bedesigned such that corresponding pi-terms on theright side of


Figure C-1 Sketch of a physical model for evaluating that wheel load passing over a buried flexible pipe thatdents the top of the pipe.


Equation C.1 are set equal for model and prototype.This can be accomplished, even for small-scalemodels, because the pi-terms are dimensionless, and,therefore have no feel for size — or any otherdimension, for that matter. The subscript mdesignates model. In order to design the model,design conditions (dc's) are:

1. (EI/D3)m = (EI/D3)2. (H/D)m = (H/D)3. (P/E')m = (P/E')4. (γ D/E')m = (γ D/E')

Using subscript r to represent the ratio of prototypeto model, each of the design conditions can be metaccording to the following:

1. (EI)r = (Dr)3 where Dr is scale ratio

2. (Hr) = (Dr) geometrical similarity

3. (Pr) = (E'r)

4. (γ r) = (E'r)/(Dr)

Because soil is a complex material, it would beconvenient if the same soil could be placed andcompacted in the same way in both model andprototype. The result is that E'r = 1, and γ r = 1. Butnow design conditions 3 and 4 are not met. Fromdesign condition 3, Pr = 1. Therefore, all pressuresP must be the same in the model as at correspondingpoints in the prototype. For example, tire pressuresmust be the same in model and prototype. The soilpressure must be the same at corresponding depthsin the model and prototype. But this is impossiblefor a small scale model if the soil has the same unitweight. One solution is to test the model in along-arm centrifuge such that centrifugal force plus

gravity increases the effective unit weight of the soilin the model. Another approximate solution is todraw seepage stresses down through the model (air,or water if the soil is to be saturated) in order toincrease the effective unit weight of the model soil.For most minimum soil cover studies, it turns out thatthe effect of the unit weight of the soil is negligibleso dc 4 is ignored. Once the model is designed andbuilt, from tests, the weight W can be observedwhen the buried pipe is dented.

The prediction equation (pe) is the equation of thepi-terms on the left sides of Equation C-1 for modeland prototype, i.e.,

(W/E'D2) = (W/E'D2)m

If E'r = 1, then the prediction equation is:

W = Wm(Dr)2

where Dr is the scale ratio of prototype to model. Ifthe scale ratio is 5 (i.e., 1:5 model to prototype), thenthe load W on the prototype that will dent the buriedpipe is 25 times the load Wm on the model that dentsthe model buried pipe.

In order to write the mathematical model (equation)for the phenomenon, enough tests must be made toprovide data plots for π 1 = f(π2) with π3 heldconstant and for π 1 = f'(π3) with π2 held constant.From the best-fit lines plotted through the data, anequation of combination can be written for π 1 =f[(π 2), (π3)]. This becomes the mathematicalmodel.

In fact, design condition 3 may not be significant.Just keep tire pressures the same in model andprototype. Then the mathematical model is simplythe equation of the best-fit plot of π 1 = f(π2). Ofcourse it can be written in terms of the originalfundamental variables.


Anderson, Loren Runar et al "HISTORICAL SKETCH"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

429

APPENDIX D HISTORICAL SKETCH

Pipeline engineering dates from prehistory. Theganats of ancient Persia were underground tunnelsbored back under the mountains to collect freshwater for the cities on the plains. The catacombs ofEgypt were remarkable underground conduits.Medieval Paris and London had brick-lined sewers.The subway tunnels of Saint Louis, long sinceabandoned, are rediscovered as engineers study lightrail systems. The technology of buried pipes of thepast arose from experience — including failures.

The modern approach to buried pipeline engineeringbegan in the early 1920s by Anson Marston, Deanof Engineering at Iowa State College. Each springhe saw the plight of Iowa farmers as they boggeddown in quagmires of mud on the rural roads. Hisconcern became a rallying cry, "Let's get Iowa outof the mud." Because of this effort, Marston wasnamed the first Chairman of the Highway ResearchBoard. As such, he reasoned, correctly, that thefirst step toward adequate roads was drainage.That meant buried drain pipes, and a procedure fordesigning buried drain pipes. He proposed a theoryfor predicting soil loads on buried rigid pipes. Thestrengths of the pipes were determined by crushingsamples of the pipe between parallel plates. Fordesign, the soil loads on the buried pipes had to beless than the parallel plate loads that caused failure.But how much less? Tests were needed. Marstonassigned the testing to a student, M.G. Spangler,who was instructed to bury samples of rigid pipe andmeasure the soil loads on them. The objective wasto relate parallel plate loads to soil loads at pipefailure, and thus provide a design procedure forhighway pipes and culverts.

During the time Spangler was testing rigid pipes,flexible corrugated steel pipes appeared on themarket. Spangler realized that for flexible pipes, aparallel plate test was not representative of fieldconditions. In the field, soil at the sides of the buriedpipe supports the pipe and resists deflection. SoSpangler derived the Iowa Formula for predicting

the ring deflection of buried flexible pipes. Theformula was based on: 1. the Marston soil load onthe pipe; 2. ring stiffness and 3. soil stiffness whichSpangler called the modulus of passive resistance ofsoil. The Iowa Formula required a number ofadjustments such as deflection time lag factor,bedding angle, and load factors. The load was soonchanged from the Marston load to prismatic soil loadplus the influence of live load. Both soil and pipewere assumed to be elastic. The boundariesincluded a plane of equal settlement which wasaffected by trench or embankment condition, andpositive or negative projection. The Iowa Formulawas published in 1941 in the Iowa EngineeringExperiment Station Bulletin 153.

Spangler was convinced that buried corrugated steelculverts invert at the top when ring deflection isabout 20%. So he applied a safety factor of fourand proposed that buried flexible pipes be limited to5% ring deflection. Kelly of Armco Corporationattempted to apply the Iowa Formula to corrugatedsteel pipes. But the formula broke down. With 5%ring deflection and all else constant in the formula,Kelly plotted height of soil cover as a function ofpipe diameter. The result was that in diameters over5 ft, the allowable height of soil cover increased asthe diameter increased. This seemed irrational. TheIowa Formula was abandoned.

In 1957, Spangler's student R.K. Watkins,discovered that Spangler's modulus of passiveresistance of soil had to be redefined in order to bea correct property of material. A modified IowaFormula overcame the irrationality demonstrated byKelly. It was published in 1958 in the Proceedingsof the Highway Research Board.

Pipeline agencies commenced to publish values forthe corrected soil modulus, now called the modulusof soil reaction. Published values were excessivelyconservative. They became a catch-all for themany assumptions in the Iowa formula, and for


STRUCTURAL MECHANICS OF BURIED PIPES430

incautions in installation. Published values of the soilmodulus, at best, yielded only a rough, conservativeestimate of ring deflection.

From field and laboratory testing, Watkins found themodulus of soil reaction to be elusive andundependable — a sidewise modulus based onvertical soil loading. It was not constant. E-prime isa function of depth of soil cover (confinement) andring stiffness. Similar findings were reported byDuncan, Molin and others.

The many factors and assumptions required to solvethe Iowa Formula made the prediction of ringdeflection less precise than direct prediction of ringdeflection based on vertical compression of thesidefill soil. Ring deflection is related to vertical soilcompression and to the ratio of soil stiffness to ringstiffness. Soil stiffness is found by standardlaboratory compression tests. Ring stiffness is aform of the spring constant for a diametral line loadon the ring. For flexible pipes buried in goodgranular soil, the stiffness ratio is so large that theinfluence of ring stiffness is negligible and the soilalone determines ring deflection;

Research at USU showed that: 1. Ring deflectionof buried flexible pipes is equal to (or less than)vertical compression of the embedment (the sidefill).2. For high soil cover, pipe "failure" is not necessarilyring deflection (Spangler's 20%), nor is it necessarilyMarston's parallel plate load. The pipe wall canbuckle or crush by ring compression at deflectionless than 20%. In fact, the wall can buckle or crushwhen deflection is zero. A pipe with high ratio ofwall strength to stiffness, such as a thin-wall steelpipe, may buckle at less than 20% ring deflection. Apipe with low ratio of wall strength to stiffness, suchas plastic pipe, may crush at less than 20% ringdeflection.

These observations have since been confirmed byfinite element analyses, and by tests — especially onlarge diameter buried flexible steel pipes.

For flexible pipes in select soil envelopes, engineers

can predict ring deflection as a function of verticalsoil strain. Whoever uses the Iowa Formula mustreduce the number of variables by substitutingaverage or assumed values for those variables thathave the least effect on the result. The IowaFormula is not a procedure for design. It is anapproximate procedure for predicting ringdeflection.

Ring deflection has proven to be an important limit tobe specified. Other important performance limitsinclude soil slip, ring collapse, and ring compressionstress as reported by White and Layer. Ringcompression is described in Chapter 6.

Other models for analysis have been proposed byHoeg, Luscher, Meyerhof, and others. Most arebased on elastic theory.

An elegant analysis of elastic soil embedment waspresented by Burns, J.Q., and Richards, R.M.,Attenuation of stresses for buried cylinders ,Proceedings of the Symposium on Soil StructureInteraction, University of Arizona, Sept. 1964.Both pipe and soil were assumed to be elastic. Oneanalysis was for full bond between soil and pipe, andthe other was for zero friction between soil and pipe.They provide a "feel" for pipe-soil interaction.

Performance limits require special analysis for avariety of embedment conditions, such asbackpacking and encasements, and for new pipematerials and configurations — especially plastics.Stresses in plastic pipes relax under constant strain,and creep (even to a long-term regressed strength)under constant stress. Clearly, pipe-soil interactionbecomes complex. Basic principles of engineeringmechanics of materials are proving to be the mostdependable tools for analysis. Worst-case condi-tions are assumed. Greater precision is not justifiedbecause of imprecisions in soils and in installations.

Because of their versatility and corrosion resistance,plastic pipes have increased and dominated someburied pipe markets since World War II. Bombingof German cities destroyed not only the industries


APPENDIX D HISTORICAL SKETCH 431

that provided steel for guns (and pipes), but also thewater supply pipelines that served the cities. Indesperation, one quick remedy seemed to be PVCpipes. The Germans had led in processing andfabricating PVC (polyvinyl-chloride). PVC pipeswere successful. Other plastic pipes soon came onthe market.

With computers available for complex pipe analyses,with new pipe configurations and materials on themarket, and with an urgent and sustained need forburied pipes, present-day technology is only a primerfor future design of buried pipes.


APPENDIX E STRESS ANALYSIS

Figure E-1 shows a stressed medium with aninfinitesimal cube, Op. Op is the free-body-diagram for stress analysis. A maximum of sixpairs of stresses can act on Op, three normal stresses(direct stresses) and three couples (shearing stresspairs). The stresses all occur in pairs inequilibrium. Notation is as follows:

σx = normal stress in the x-direction acting on ayz-plane,

σy = normal stress in the y-direction acting on ayz-plane,

σz = normal stress in the z-direction acting onan xy plane,

τx = each of two equal and opposite shearingcouples acting about an x-axis in yz-planes,

τy = each of two equal and opposite shearingcouples acting about a y-axis in xz-planes

τz = each of two equal and opposite shearingcouples acting about a z-axis in xy planes.

The resultant normal and shearing stresses on anyplane that passes through the cube can be found bya stress circle proposed by Otto Mohr and found intexts on mechanics of solids. The Mohr circle isapplied separately to each of the three orthogonalviews of the cube. The Mohr circle, Figure E-1,applies to a view of the xy-plane (front view). Thetwo couples are shown as shearing stress pairs onsurfaces of the cube that are at right angles to eachother. The shearing couple on the top and bottomis equal and opposite to the shearing couple on thesides in order to satisfy equilibrium conditions.Because the couples are equal, the subscript isdropped for shearing stress τ . Analysis comprisesthree steps.

A. The Mohr stress circle is plotted by first drawingthe stress axes, σ and τ , as shown. Three pointsestablish the stress circle on the axes. From thestresses on the infinitesimal cube, points on thestress axes are plotted as follows:

1. Plot the normal and shearing stress point,(σ , τ ) acting on the y-plane,

2. Plot the normal and shearing stress point,(σy, τ ) acting on the x-plane,

3. Locate the center of the circle on the σ -axis.

By simply connecting the two plotted points, thecenter of the circle is located on the σ -axis.Contrary to popular sign convention, compressionis positive for normal stress; and counterclockwiseis positive for couples (shearing stresses).Molecular bonding forces hold the materialtogether in compression. Tension is simply areduction in compressive bonds in the material.

B. The orientation diagram is x and y, representingx and y planes, shown dotted on the infinitesimalcube. The orientation diagram is superimposed onthe stress diagram. The x-plane is drawn throughpoint (σy,τ ) representing the stresses on that x-plane; and the y-plane is drawn through point (σx,τ ) representing the stresses on that y-plane. Theorigin, Op, (called the origin of planes) always fallson the Mohr stress circle. The origin of planes isactually the infinitesimal cube superimposed on thestress diagram. Any plane through the Op is aplane through the infinitesimal cube. Analysis isthe following:

ANY PLANE DRAWN FROM THE ORIGIN OFPLANES, Op, AT SOME ANGLE θ ,INTERSECTS THE MOHR CIRCLE AT THENORMAL AND SHEARING STRESSESACTING ON THAT 2-PLANE.

The θ-plane is correctly oriented with respect to theoriginal x-y coordinate axes of the cube. Clearly,maximum and minimum normal stresses (calledprincipal stresses) and maximum shearing stressescan be found on the circle. It is common practice toidentify the maximum principal stress as thefurthest point to the right on the circle, and theminimum principal stress as the furthest point to


Figure E-1 Stress analysis for the front view of infinitesimal cube Op in a stressed medium, showing thethree superimposed diagrams: Mohr stress circle, orientation diagram x-y, and strength envelopes.


the left. Maximum shearing stresses are the highestand lowest point on the circle. The highest is plus(counterclockwise couple) and the lowest isnegative (clockwise couple).

Of passing interest are some important stresstheorems which can be demonstrated by the Mohrcircle. Shearing stresses are zero on the planes ofprincipal stresses (called principal planes).Principal stresses (maximum and minimum) act onplanes that are perpendicular to each other.Maximum shearing stresses are equal to the averageof the principal stresses, and act on perpendicularplanes at 45o with the principal planes. Planes ofmaximum shearing stresses are at right angles toeach other, but one shearing stress is positive andthe other is negative. The normal stresses are equalon both planes of maximum shearing stress. Theplanes on which stresses act are the dotted linesfrom Op to the stress points on the Mohr circle.The numerical values of stresses can be foundeither by drawing to scale the Mohr circle andmeasuring values, or by trigonometry. Followingare some useful trigonometric hints.

1. The center of the circle is located on the σ -axis at (σx + σ y)/2.

2. The square of the radius of the circle is[(σx - σ y)2 + τ 2] by the Pythagorean theorem.

3. If a central angle drawn from the center ofthe circle intercepts the same arc segment as acircumferential angle drawn from Op, the centralangle is twice the corresponding circumferential

angle. For example, on Figure E-1, the failureangle, β, is half the corresponding central angle of2β = 90o + ϕ − α Because ϕ and α are known, thefailure angle β can be evaluated.

C. The strength envelopes are the limits of stressesin the material. If stresses increase to the pointwhere the stress circle touches the strengthenvelope, the material fails on failure plane, β, fromOp to the point of tangency. Strength envelopescan be determined in the laboratory by loading thematerial to failure under tri-axial loads. See FigureE-2. A Mohr circle is drawn for each of thefailures. Tangents to the circles become thestrength envelopes. Of interest is the theoreticalorigin at the intersection of the strength envelopes.Ideally it represents the sum of all the bondingforces in the material -- and represents the absolutemaximum tensile strength of perfect strands ofmolecules. In fact, perfect strands are unrealistic.Nevertheless, in the case of steel, the ideal origin ofstrength envelopes is so far to the left, that thestrength envelopes are essentially horizontal in therange of strengths of steel in typical usage. FigureE-3 depicts strength envelopes for steel. Becausethe strength envelopes are essentially parallel, yieldstress (failure) in tension is almost the same as incompression. Shown on the figure, in solid line, isa Mohr circle for tension failure. With the Oplocated as shown, failure planes are at 45o. It ispossible to see these slip planes in some specimensof failed steel. They are called Leuder's lines. Intypical failures of thin-wall pipes or tankssubjected to excessive internal pressure, the fracturesurface is beveled at roughly 45o.


Figure E-2 Development of strength envelopes by drawing tangents to Mohr circles plotted from triaxialtests to failure.

Figure E-3 Sketch depicting strength evelopes for steel, showing how strength envelopes are parallelbecause yield stress, σy, is approximately equal in both tension and compression; and showing the 45o

failure planes.


Anderson, Loren Runar et al "STRAIN ENERGY ANALYSIS"Structural Mechanics of Buried PipesBoca Raton: CRC Press LLC,2000

437

APPENDIX F STRAIN ENERGY ANALYSIS

An infinitesimal cube is subjected to triaxial principalstresses. The general case would include shearingstresses, but the cube can be oriented such that allstresses are principal stresses. Moreover, for pipes,principal stresses can usually be identified directly.Shearing stresses are zero on plains of principalstresses.

U = strain energy (scalar),σ = normal stress,σf = yield strength,ε = normal strain,E = mudulus of elasticity,ν = Poisson ratio (0.27 for steel).Subscripts, 1, 2, and 3 refer to maximum, inter-mediate, and minimum values.

Strain energy is U = Σf σεdε . As stress increasesfrom 0 to σ within the elastic range, the strainenergy is average stress times strain, i.e.:

U = Σσε/2 . . . . . (F.1)

From mechanics of materials,ε1 = σ 1/E-ν(σ 2+σ 3)/E. Strain energy for each stresscan be written as follows.

Strain Strain Energy TermEε1 = σ1-ν(ε 2+ε 3) 2EU1 = σ1

2-νσ1(σ2+σ 3)Eε2 = σ2-ν(ε 1+ε 3) 2EU2 = σ2

2- νσ2(σ1+σ 3)Eε3 = σ3-ν(ε 1+ε 2) 2EU3 = σ3

2- νσ3(σ1+σ 2)

Because energy is scalar, the three strain energy

terms can be added. The sum is,2EU = σ1

2+σ22+σ3

2-2ν (σ1σ 2+σ 1σ 3+σ 2σ 3)

If yield stress is the performance limit, the total EUmust be equated to the strain energy term at tensileyield stress: Uf = σ f εf /2 = σ f2 /2E. Rewriting, 2EUf

= σf2 = 2EU.

σ12 + σ2

2 + σ32 - 2ν(σ1σ2 + σ 2σ 3 + σ 3σ 1) = σ f2

. . . . . (F.2)

For most pipe materials, the strain energy of volumechange should not be included in the total strainenergy at failure (yield stress). Subtracted out ofEquation A.2, the result is the Huber-Hencky-vonMises equation:

σ12+σ2

2+σ32 - (σ1σ 2+σ 2σ 3+σ 3σ 1) = σ f2 . . . . . (F.3)

For pipes, the square of the smallest stress, σ3, isoften ignored with results:

σ12 + σ2

2 - σ1σ 2 = σ f2 . . . . . (F.4)

On coordinate axes a plot of σ1 vs. σ 2, at elasticlimit, is the von Mises ellipse as shown below.

Figure F-1 Infinitesimal cube showing principalstresses acting on it. ©2000 CRC Press LLC

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