Structural mechanics of buried pipes 2395

Basic Fundamental Variables Dimensions P' = internal pressure FL-2 t = wall thickness L D = inside diameter of ring L S = yield strength of the pipe wall material FL-2 These four fundame

STRUCTURAL MECHANICS

OF BURIED PIPES

Boca Raton London New York Washington, D.C.

CRC Press

Reynold King Watkins

Utah State University Logan, Utah

Loren Runar Anderson

Utah State University Logan, Utah

GENERAL NOTATION

Geometry

A = cross sectional wall area per unit length of pipe,

r = radius of curvature of the pipe (tank) cylinder,

Forces, Pressures, and Stresses

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 Sketch Appendix E: Stress Analysis Appendix F: Strain Energy Analysis

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 systems require no fuel for pumping Ships transport 250 ton-miles per gallon Rails transport 125 ton-miles per gallon Trucks transport 10 ton-miles per gallon Aircraft transport less than 10 ton-miles per gallon of fuel.

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 and damage to the products that are transported.

The structural mechanics of buried pipes can be complicated an interaction of soil and pipe each with vastly different properties Imprecisions in properties of the soil embedment are usually so great that complicated 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 mechanics and basic scientific principles prevail.

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 Pipes

Boca Raton: CRC Press LLC,2000

Anderson, Loren Runar et al "INTRODUCTION"

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000

CHAPTER 1 INTRODUCTION

Buried conduits existed in prehistory when caves

were protective habitat, and ganats (tunnels back

under mountains) were dug for water The value of

pipes is found in life forms As life evolved, the

more complex the organism, the more vital and

complex were the piping systems

The earthworm lives in buried tunnels His is a

higher order of life than the amoeba because he has

developed a gut — a pipe — for food processing

and waste disposal

The Hominid, a higher order of life than the

earthworm, is a magnificent piping plant The

human piping system comprises vacuum pipes,

pressure pipes, rigid pipes, flexible pipes — all

grown into place in such a way that flow is optimum

and stresses are minimum in the pipes and between

the pipes and the materials in which they are buried

Consider a community A termite hill contains an

intricate maze of pipes for transportation, ventilation,

and habitation But, despite its elegance, the termite

piping system can't compare with the piping systems

of a community of people The average city dweller

takes for granted the services provided by city piping

systems, and refuses to contemplate the

consequences if services were disrupted Cities can

be made better only to the extent that piping systems

are made better Improvement is slow because

buried pipes are 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 Three

general sources of knowledge are:

Buried structures have been in use from antiquity

The ancients had only experience as a source of

knowledge 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 the

development 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 analyzed

by experimentation But even experimentation ismost effective when based on principles — i.e.,principles of experimentation

This text is a compendium of basic principles proven

to 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 FactorExamples:

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

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 Thestandard deviation s is important because: l it is abasis for comparing the precision of sets ofmeasurements, and 2 it can be calculated fromactual measurements; i.e.,

s = %3yw2/(n-1) Standard deviation s is the horizontal radius ofgyration of area under the normal distribution curvemeasured from the centroidal y axis s is a deviation

of x with the same dimensions as x and w Animportant 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 standarddeviation can be calculated from test data, theprobability 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 , eithergreater than an upper limit xe or less than a lowerlimit, xe, can be read from the table The deviation

of failure is needed; i.e., we = x e- x) Because soil interaction is imprecise (large standarddeviation), it is prudent to design for a probability ofsuccess of 90% (10% probability of failure) and toinclude a safety factor Probability analysis can beaccomplished conveniently by a tabular solution asshown in the following example

pipe-Example

The bursting pressure in a particular type of pipe hasbeen tested 24 times with data shown in Table 1-2.What is the probability that an internal pressure of0.8 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

©2000 CRC Press LLC

Table 1-1 Probability P as a function of w/s that a value of x will fall within +w, and probability Pe a s afunction of we /s that a value of x will fall outside of +weon either the +we or the -we

From the data of Table 1-2,

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 % or

one out of every 122 pipe sections Cost accounting

of failures then follows

The probability that the strength of any pipe section

will 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 be

calculated From standard deviation, the zone of +w

can be found within which 90% of all measurements

fall 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 is

deformation as a function of loads, geometry, and

properties of materials Some deformations can be

written in the form of equations from principles of

soil mechanics The remainders involve suchcomplex soil-structure interactions that theinterr elationships 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 ar e 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 wr2 =

Mrx2

wx2 + Mry2wy2 + where w is a deviation at thesame 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: fewer

variables to relate, and the elimination of size effect

The required number of pi-terms is less than the

number of fundamental variables by the number of

basic dimensions Because pi-terms are

dimensionless, they have no feel for size (or any

dimension) and can be investigated by model study

Once pi-terms have been determined, their

interrelationships can be found either by theory

(principles) or by experimentation The results apply

generally because the pi-terms are dimensionless

Following is an example of a well-designed

experiment

Example

Using experimental techniques, find the equation for

internal bursting pressure, P', for a thin-wall pipe

Start by writing the set of pertinent fundamental

variables together with their basic dimensions, force

F and length L

Basic Fundamental Variables Dimensions

P' = internal pressure FL-2

t = wall thickness L

D = inside diameter of ring L

S = yield strength of the

pipe wall material FL-2

These four fundamental variables can be reduced to

two pi-terms such as (P'/S) and (t/D) The pi-terms

were written by inspection keeping in mind the three

characteristics of pi-terms The number of pi-terms

is the number of fundamental variables, 4, minus the

number of basic dimensions, 2, i.e., F and L The

two pi-terms are dimensionless Both are

independent because each contains a fundamental

variable not contained in the other Conditions for

bursting can be investigated by relating only two

variables, the pi-terms, rather than interrelating the

original four fundamental variables Moreover, the

investigation can be performed on pipes of any

convenient size because the pi-terms are

dimensionless 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 pipe

is 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 b

is 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 of ten, the performance limit is excessivedeformation of the pipe whic h 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 is

that deformation beyond which the pipe cannot resist

any increase in load The obvious case is bursting of

the pipe due to internal pressure Less obvious and

more complicated is the deformation due to external

soil pressure Typical examples of performance

limits for the pipe are shown in Figure 1-4 These

performance limits do not imply collapse or failure

The soil generally picks up any increase in load by

arching action over the pipe, thus protecting the pipe

from total collapse The pipe may even continue to

serve, but most engineers would prefer not to

depend on soil alone to maintain the conduit cross

section This condition is considered to be a

performance limit The pipe is designed to withstand

all external pressures Any contribution of the soil

toward withstanding external pressure by arching

action is just that much greater margin of safety

The soil does contribute soil strength On inspection,

many buried pipes have been found in service even

though the pipe itself has "failed." The soil holds

broken clay pipes in shape for continued service

The inverts of steel culverts have been corroded or

eroded away without failure Cast iron bells have

been found cracked Cracked concrete pipes are

still in service, etc The mitigating factor is the

embedment soil which supports the conduit

A reasonable sequence in the design of buried pipes

is 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 possible

alternatives,

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

PROBLEMS1-1 Fluid pressure in a pipe is 14 inches of mercury

as 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 that

any maximum daily pressure will exceed 784.5

1-5 Figure 1-5 shows bar graph for internal vacuum

at collapse of a sample of 58 thin-walled plastic

pipes

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 concrete

pipes, Class 1, were tested in three-edge-bearing

(TEB) test with results as follows:

x = ultimate load in pounds per lineal ft

(s = 459.5 lb/ft)(c) What is the probability that the load, x, at failure

is less than the minimum specified strength of 3000

lb/ft (pounds per linear ft)? (Pe = 5.68%)

Figure 1-5 Bar graphs of internal vacuum at collapse

of 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 shown

as 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 probability

of 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 on

23 culverts under identical conditions?

of a particular plastic pipe the average is 24 with astandard deviation of 3

a) What is the probability that the pipe stiffness will

be less than 20? (Pe = 9.17 %)

e

b) What standard deviation is required if the

probability of a stiffness less than 20 is to be

reduced to half its present value; i.e., less than

4.585%? (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 Pipes

Boca Raton: CRC Press LLC,2000

Figure 2-1 Free-body-diagram

of half of the pipe cross

section including internal

pressure P’

Equating rupturing force to

resisting force, hoop stress

in the ring is,

s = 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 buried

pipes all deal with resistance to loads Loads on a

buried pipe can be complex, especially as the pipe

deflects out-of-round Analysis can be simplified if

the 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 usually

limited by specification to some value not greater

than five percent Analysis of a circular ring is

reasonable for the structural design of most buried

pipes Analysis is prediction of structural

performance Following are basic principles for

analysis and design of the ring such that it can

support 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 find

minimum wall area per unit length of pipe

Plain pipe — If the pipe wall is homogeneous and

has smooth cylindrical surfaces it is plain (bare) and

wall area per unit length is wall thickness This is

the case in steel water pipes, ductile iron pipes, and

many plastic pipes

Other pipes are corrugated or ribbed or composite

pipes such as reinforced concrete pipes For such

pipes, the wall area, A, per unit length of pipe is the

pertinent quantity for design

Consider a free-body-diagram of half of the pipe

with fluid pressure inside The maximum rupturing

force is P'(ID) where P' is the internal pressure and

ID is the inside diameter See Figure 2-1 This

rupturing force is resisted by tension, FA, in the wall

where F is the circumferential tension stress in the

pipe wall Equating rupturing force to the resisting

force, F = P'(ID)/2A Performance limit

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

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

T his 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 increas es,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 inch

ID with a wall thickness of 0.219 inch What is themaximum allowable head, h, (difference in elevation

of the inlet and outlet) when the pipe is full of water

at 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,

s = P(OD)/2A

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

E = 30(106)psi = modulus of elasticity,

S = 36 ksi = yield strength,

sf = 2 = safety factor,

gw = 62.4 lb/ft3 = unit weight of water,

P' = hgw = internal water pressure at outlet

From Equation 2.1, s = S/2 = P'(ID)/2A where A is

0.219 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 loads

imposed on the pipe during transportation and

installation The most common load is diametral

F-load See Figure 2-2 This load occurs when pipes

are stacked or when soil is compacted on the sides

or on top of the pipe as shown

If yield strength of the pipe material is exceeded due

to the F-load, either the pipe wall will crack or the

cross section of the pipe will permanently deform

Either of these deformations (a crack is a

deformation) may be unacceptable So yield

strength may possibly be a performance limit even

though the ring does not collapse

For some plastic materials, including mild steel,

design for yield strength is overly conservative So

what if yield strength is exceeded? A permanent

deformation (dent) in the ring is not necessarily pipe

failure In fact, the yield strength was probably

exceeded in the process of fabricating the pipe

Some pipe manufacturers limit the F-load based on

a maximum allowable ring deflection, d = D/D,

where D is the decrease in mean diameter D due to

load F Some plastics have a memory for excessive

ring deflection In service, failure tends to occur

where excess ive ring deflection occurred before

installation Increased ring stiffness decreases ring

deflection It is not inconceivable that the ring can

be 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 Dimensions

F = transportation/installation FL-1

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

D = mean diameter of the pipe, L

I = moment of inertia of the wall L3

cross section per unit length

of pipe,

c = distance from the neutral axis L

of 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 An

example of class ical analysis starts with

circum-ferential stress s = Mc/I where M is the maximum

bending moment in the pipe ring due to load F But

if stress is limited to yield strength, then S = Mc/I

where M = FD/2p based on ring analysis by

Castigliano's theorem See Appendix A, Table A-1

M is the maximum moment due to force F Because

it occurs at the location of F, there is no added ring

compression stress Substituting in values and

rearranging the fundamental variables into pi-term,

(F/SD) = 2p(D/c)(I/D3)

The three pi-terms are enclosed in parentheses

Disregarding pi-terms,

F = 2pSI/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) = p(t/D)2/3

Disregarding pi-terms,

F = pSt2/3D = F-load at yield strength S for plain

pipes (smooth cylindrical surfaces) The modulus of

elasticity E has no effect on the F-load as long as

the ring remains circular Only yield strength S is a

performance limit

Ring Deflection Performance Limit

If the performance limit is ring deflection at the

elastic limit, modulus of elasticity E is pertinent

Yield strength is not pertinent For this case,

pertinent fundamental variables and corresponding

basic dimensions are the following:

fv's, Fundamental bd's, Basic Variables Dimensions

d = ring deflection = /D

-D = mean diameter of the L

pipe

F = diametral line load FL-1

per unit length of pipe

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 due

to 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)3The relationship between circumferential stress andring deflection is found by substituting from Table A-

1, at yield stress, F = 2pSI/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 (s /E) (D/c) (2.3)

For plain pipes,

(d) = 0.234 (s E) (D/t)

Note the introduction of a new pi-term, (s /E) This

relationship could have been found by

experimentation using the three pi-terms in

parentheses in Equation 2.3 Ring deflection at yield

stress, S, can be found from Equation 2.3 by setting

s = S If ring deflection exceeds yield, the ring does

not return to its original circular shape when the

F-load is removed Deformation is permanent This is

not failure, but, for design, may be a performance

limit with a margin of safety

The following equations summarize design of the

pipe to resist transportation/installation loads

For transportation/installation, the maximum

allowable F-load and the corresponding ring

deflection, d, when circumferential stress is at

yield strength, S, are found by the following

formulas

Ring Strength

(F/SD) = 2p (D/c) (I/D3) (2.4)

For plain pipes, (F/SD) = p(t/D)2/3

Resolving, for plain pipes,

F = pSD(t/D)2/3

Ring Deflection where d = (D/D) due to F-load,

is given by:

d = 0.0186 (FD2/EI), in terms of F-load (2.5)

d = 0.117 (s /E) (D/c), in terms of stress, s , or

d = 0.234(S/D)(D/t), for plain pipes with smooth

cylindrical 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 stre s s s = yield strength S, the maximumallowable load is:

F = 2pSI/cDFor walls with smooth cylindrical surfaces,

F = pSt2/3D

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

(D/t)2 = pSD/3FFor the maximum anticipated F-load, i.e at yieldstrength, the minimum wall thickness term (t/D) can

be 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 increased

by 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 = pSt/3(DR-1)

If the F-load is known, the required dimension ratio

at yield strength is:

DR = (pSt/3F) + 1

Unreinforced concrete pipes are to be stacked for

storage in vertical columns on a flat surface as

indicated in Figure 2-2 The load on the bottom pipe

is essentially an F-load The following information is

+ 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 is

limited to 3000 lb/ft? From the data, the weight of

the pipe is 400 lb/ft The number of pipes high in the

stack is 3000/400 = 7.5 So the stack must be

limited to seven pipes in height

b) What is the probability that a pipe will break if the

column is seven pipes high? The seven pipe load at

the bottom of the stack is 7(400) = 2800 lb/ft w =

3727 - 2800 = 927 lb/ft which is the deviation of the

seven-pipe load from the F-load From Table 1-1,

the probability of failure is 2.2% for the bottom

pipes For all pipes in the stack, the probability is

one-seventh as much or 0.315%, which is one

broken pipe for every 317 in the stack

c) What is the circumferential stress in the pipe wall

at an average F-load of 3727 lb/ft? From Equation

2.4, F = pSD(t/D)2/3 where S = yield strength D/t =

9, D = 51 inches Solving, s = 471 psi This is good

concrete considering that it fails in tension

EXTERNAL PRESSURE —

MINIMUM WALL AREA

Consider a free-body-diagram of half the pipe with

external pressure on it See Figure 2-4 The

vertical rupturing force is P(OD) where P is the

external radial pressure assumed to be unifor mly

distributed OD is the outside diameter Theresisting force is compression in the pipe wall, 2s A,where s 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:

s = 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 What

is 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 into

Equation 2.6, the safety factor is sf = 2.76 The soil

pressure of 16 ksf is equivalent to about 150 feet ofsoil cover See Chapter 3

PROBLEMS2-1 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(10) psi,

sf = 2 = safety factor

2-2 What is the allowable internal pressure if a

reinforced conc rete pipe is 60 inch ID and has two

cages comprising concentric hoops of half-inch steel

reinforcing rods spaced at 3 inches in the wall which

is 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 steel

rods of Problem 2-2 if the pipe is not to leak at

internal pressure of 72 psi? Leakage through hair

cracks in the concrete appears as sweating

(Fs = 2.9 kips) 2-4 How could the steel rods be pretensioned in

Problem 2-3? Is it practical to pretension (or post

tension) 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 Pipes

Boca 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 ring

is 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 is

small enough that it can be neglected For a few

analyses, however, deformation of the ring must be

considered This is particularly true in the case of

instability of the ring, as, for example, the hydrostatic

collapse of a pipe due to internal vacuum or external

pressure Collapse may occur even though stress

has not reached yield strength But collapse can

occur only if the ring deforms Analysis of failures

requires a knowledge of the shape of the deformed

ring

For small ring deflection of a buried circular pipe, the

basic deflected cross section is an ellipse Consider

the infinite medium with an imaginary circle shown

in Figure 3-1 (top) If the medium is compressed

(strained) uniformly in one direction, the circle

becomes an ellipse This is easily demonstrated

mathematic ally Now suppose the imaginary circle

is a flexible ring When the medium is compressed,

the ring deflects into an approximate ellipse with

slight deviations If the circumference of the ring

remains constant, the ellipse must expand out into

the medium, increasing compressive stresses

between ring and medium See Figure 3-1 (bottom)

The ring becomes a hard spot in the medium On

the other hand, if circumference of the ring is

reduced, the ring becomes a soft spot and pressure

is 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 areas

within the ring and without the ring The shape is

also affected by non-uniformity of the medium For

example, if a concentrated reaction develops on the

bottom of the ring, the ellipse is modified by a flat

spot Nevertheless, for small soil strains, the basic

ring deflection of a flexible buried pipe is an ellipse

Following are some pertinent approximate

geometrical properties of the ellipse that are

sufficiently accurate for most buried pipe analyses

Greater accuracy would require solutions of infinite

series

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 p(a+b) whi c hreduces to 2pr 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/D = RING DEFLECTION (3.1)

where:

D = 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 deflection

is equal to the horizontal ring deflection, area withinthe ellipse is Ae = Bab; and

Ae= pr2 (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 Ao within a

circle of equal circumference) shown

plotted as a function of ring deflection

What of the assumption that the horizontal and

vertical ring deflections are equal if the

circumferences are equal for circle and ellipse? For

the circle, circumference is 2pr For the ellipse,

circumference is (b+a)(64-3R4)/(64-16R2), where R

is approximately R = (b-a)/(b+a) Only the first

terms of an infinite series are included in this

approximate ellipse circumference See texts on

analytical geometry Equating circumferences of the

circle and the ellipse, and transforming the values of

a and b into vertical and horizontal values of ring

deflection, dy and dx, a few values of dy and the

correspondin g dx are shown below for comparison

For ring deflections of d = dy = 10%, the

corresponding dx is less than 10% by only

4.8%(10%) = 0.48% This is too small to be

significant in most calculations such as areas within

the ellipse and ratios of radii

Radii of curvature of the sides (spring lines) and the

top and bottom (crown and invert) of the ellipse are:

However, more precise, and almost as easy to use,

are the approximate values:

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 L

is laid as a cord The offset e is measured to thecurved wall at the center of the cord Outside, e can

be found by laying a tangent of known length L and

by 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 somewhat

on top From inside the pipe, a straightedge (cord)

12 inches long is placed against the top, and the ordinate offset is measured and found to be 11/32inch What is the radius of curvature of the pipe ring

mid-at the top?

From Equation 3.3, r = (4e2 + L2)/8e Substituting invalues and solving, ry = 52.5 inches which is theaverage radius within the 12-inch cord on the inside

of the corrugated pipe On the outside, the radius isgreater by the depth of the corrugations

(ry and rxare 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 length

L and the middle ordinate e

Ring Deflection Due to Internal Pressure

When subjected to uniform internal pressure, the

pipe expands The radius increases Ring deflection

is equal to percent increase in radius;

d = Dr/r = DD/D = 2pre/2pr = e

where:

d = ring deflection (percent),

Dr and DD are increases due to internal pressure,

r = mean radius,

D = mean diameter,

e = circumferential strain,

E = modulus of elasticity = s/e

s = circumferential stress = Ee = Ed

But s = 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 s , and solving for d,

d = P'(ID)/2AE (3.4)

Figure 3-6 Quadrant of a circular cylinder fixed at

the 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 B

on 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

ExampleConsider the quadrant of a circular cylinder shown

in 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 xBfor which, according to Castigliano:

xB = f (M/EI)(dM/dp)ds (3.5)

xB = displacement of point B in the x-direction,

EI = wall stiffness,

E = modulus of elasticity,

I = centroidal moment of inertia of the cross

section of the wall per unit length of cylinder,

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

p = differential load (dummy load) applied at

point B in the direction assumed for deflection,

ds = differential length along the slice, = rdq

r = mean radius of the circular cylinder

It is assumed that deflection is so small that radius r

remains constant It is also assumed that the

deflection is due to moment M, flexure — not to

shear or axial loads In Figure 3-6, consider arc CB

as a free-body-diagram Apply the dummy load p at

B acting to the right assuming that deflection xB will

be in the x-direction If the solution turns out to be

negative, then the deflection is reversed From the

Substituting into Equation 3.5,

xB = (Qr/EI) (1-cosq) r(sinq) rdq

Integrating and substituting in limits of q from 0 to

p/2,

xB = Qr3/2EI

This is one of a number of the most useful

deflections 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 minimum

radii of curvature by means of approximate

Equation 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

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 is

E Elastic limit is not exceeded Use the Castiglianoequation (y = 3PL4 / 2Et3)

3-8 A half of a circular ring is loaded at the crown

by 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 load For flexible rings, the F-load test is called aparallel plate test What is the ring deflection ifelastic limit is not exceeded?

(See Table A-1)

Anderson, Loren Runar et al "SOIL MECHANICS"

Structural Mechanics of Buried Pipes

Boca 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 on

a pipe buried at depth of soil cover, H, in soil of 100 pcf unit weight Pressure is minimum at 5 or 6 ft ofcover