THIẾT KẾ ĐƯỜNG ỐNG

STRUCTURAL
MECHANICS
OF
BURIED PIPES
Reynold King Watkins
Utah State University
Logan, Utah

Loren Runar Anderson
Utah State University
Logan, Utah

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

GENERAL NOTATION

Geometry
A = cross sectional wall area per unit length of pipe,
B = breadth of the trench
D = 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 Stresses
P = 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 Materials
c = cohesion of soil,
E = modulus of elasticity of pipe (tank) material,
S = allowable stress (strength) of material,
γ = unit weight of material,
ν = Poisson ratio,
ϕ = soil friction angle.

©2000 CRC Press LLC

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 Sketch
Appendix 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 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.

"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 Pipes
Boca Raton: CRC Press LLC,2000

Anderson, Loren Runar et al "INTRODUCTION"

CHAPTER 1 INTRODUCTION

Buried conduits existed in prehistory when caves In a phenomenon as complex as the soil-structure
were protective habitat, and ganats (tunnels back interaction of buried pipes, all three sources must be
under mountains) were dug for water. The value of utilized. There are too many variables; the
pipes is found in life forms. As life evolved, the interaction is too complex (statically indeterminate to
more complex the organism, the more vital and the infinite degree); and the properties of soil are too
complex were the piping systems. imprecise to rely on any one source of information.

The earthworm lives in buried tunnels. His is a Buried structures have been in use from antiquity.
higher order of life than the amoeba because he has The ancients had only experience as a source of
developed a gut — a pipe — for food processing knowledge. Nevertheless, many of their catacombs,
and waste disposal. ganats, sewers, etc., are still in existence. But they
are neither efficient nor economical, nor do we have
The Hominid, a higher order of life than the any idea as to how many failed before artisans
earthworm, is a magnificent piping plant. The learned how to construct them.
human piping system comprises vacuum pipes,
pressure pipes, rigid pipes, flexible pipes — all The other two sources of knowledge are recent.
grown into place in such a way that flow is optimum Experimentation and principles required the
and stresses are minimum in the pipes and between development of soil mechanics in the twentieth
the pipes and the materials in which they are buried. century. Both experience and experimentation are
needed to verify principles, but principles are the
Consider a community. A termite hill contains an basic tools for design of buried pipes.
intricate maze of pipes for transportation, ventilation,
and habitation. But, despite its elegance, the termite Complex soil-structure interactions are still analyzed
piping system can't compare with the piping systems by experimentation. But even experimentation is
of a community of people. The average city dweller most effective when based on principles — i.e.,
takes for granted the services provided by city piping principles of experimentation.
systems, and refuses to contemplate the
consequences if services were disrupted. Cities can This text is a compendium of basic principles proven
be made better only to the extent that piping systems to be useful in structural design of buried pipes.
are made better. Improvement is slow because
buried pipes are out-of-sight, and, therefore, out-of- Because the primary objective is design, the first
mind to sources of funding for the infrastructure. principle is the principle of design.

Engineering design requires knowledge of: 1.
performance, and 2. limits of performance. Three DESIGN OF BURIED PIPES
general sources of knowledge are:
To design a buried pipe is to devise plans and
SOURCES OF KNOWLEDGE specifications for the pipe-soil system such that
Experience (Pragmatism) performance does not reach the limits of
Experimentation (Empiricism) performance. Any performance requirement is
Principles (Rationalism) equated to its limit divided by a safety factor, sf, i.e.:


Figure 1-1 Bar graph of maximum peak daily pressures in a water supply pipeline over a period of 1002 days
with its corresponding normal distribution curve shown directly below the bar graph.


Performance Limit s = standard deviation = deviation within which
Performance =
Safety Factor 68.26 percent of all measurements fall (Ps =
Examples: 68.26%).
Stress = Strength/sf
Deformation = Deformation Limit/sf P is the ratio of area within +w and the total area.
Expenditures = Income/sf; etc. Knowing w/x, P can be found from Table 1.1. The
standard deviation s is important because: l. it is a
If performance were exactly equal to the performance basis for comparing the precision of sets of
limit, half of all installations would fail. A safety measurements, and 2. it can be calculated from
factor, sf, is required. Designers must allow for actual measurements; i.e.,
imperfections such as less-than-perfect construction,
overloads, flawed materials, etc. At present, safety s = %3yw2/(n-1)
factors are experience factors. Future safety factors
must include probability of failure, and the cost of Standard deviation s is the horizontal radius of
failure — including risk and liability. Until then, a gyration of area under the normal distribution curve
safety factor of two is often used. measured from the centroidal y axis. s is a deviation
of x with the same dimensions as x and w. An
In order to find probability of failure, enough failures important dimensionless variable (pi-term) is w/s.
are needed to calculate the standard deviation of Values are listed in Table 1-1. Because probability
normal distribution of data. P is the ratio of area within ±w and the total area, it
is also a dimensionless pi-term. If the standard
deviation can be calculated from test data, the
NORMAL DISTRIBUTION probability that any measurement x will fall within
±w from the average, can be read from Table 1-1.
Normal distribution is a plot of many measurements Likewise the probability of a failure, Pe , either
(observations) of a quantity with coordinates x and y, greater than an upper limit xe or less than a lower
where, see Figure 1-1, limit, xe , can be read from the table. The deviation
)
of failure is needed; i.e., we = x e - x. Because pipe-
x = abscissa = measurement of the quantity, soil interaction is imprecise (large standard
deviation), it is prudent to design for a probability of
y = ordinate = number of measurements in any given success of 90% (10% probability of failure) and to
x-slot. A slot contains all measurements that are include a safety factor. Probability analysis can be
closer to the given x than to the next higher x or the accomplished conveniently by a tabular solution as
next lower x. On the bar graph of data Figure 1-1, if shown in the following example.
x = 680 kPa, the 680-slot contains all of x-values
from 675 to 685 kPa. Example
)
x = the average of all measurements, The bursting pressure in a particular type of pipe has
x = 3yx/3y, been tested 24 times with data shown in Table 1-2.
n = total number of measurements = Ey, What is the probability that an internal pressure of
)
w = deviation, w = x - x, 0.8 MPa (120 psi or 0.8 MN/m2) will burst the pipe?
P = probability that measurement will fall
between ±w, x = test pressure (MN/m2) at bursting
Pe = probability that a measurement will exceed y = number of tests at each x
the failure level of xe (or fall below a n = Gy = total number of tests
minimum level of xe ),


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 +w e on either the +w e or the -w e .

w e /s P Pe w e /s P Pe
____ (%) (%) ____ (%) (%)
0.0 0.0 50.0 1.5 86.64 6.68
0.1 8.0 46.0 1.6 89.04 5.48
0.2 15.9 42.1 1.7 91.08 4.46
0.3 23.6 38.2 1.8 92.82 3.59
0.4 31.1 34.5 1.9 94.26 2.87

0.5 38.3 30.9 2.0 95.44 2.28
0.6 45.1 27.4 2.1 96.42 1.79
0.6745 50.0 25.0 2.2 97.22 1.39
0.7 51.6 24.2 2.3 97.86 1.07

0.8 57.6 21.2 2.4 98.36 0.82
0.9 63.2 18.4 2.5 98.76 0.62

1.0 68.26 15.9 2.6 99.06 0.47
1.1 72.9 13.6 2.7 99.30 0.35
1.2 78.0 11.5 2.8 99.48 0.26
1.3 80.6 9.7 2.9 99.62 0.19
1.4 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 tabular
solution of the average bursting pressure and its standard deviation.

x y xy w yw yw 2
(Mpa)* _ (MPa) (MPa) (MPa) (MPa) 2
0.9 2 1.8 -0.2 -0.4 0.08
1.0 7 7.0 -0.1 -0.7 0.07
1.1 8 8.8 0.0 0.0 0.00
1.2 4 4.8 +0.1 +0.4 0.04
1.3 2 2.6 +0.2 +0.4 0.08
1.4 1 1.4 +0.3 +0.3 0.09
Sums 24 26.4 0.36
n Σxy Σyw 2

x = Sxy/n = 1.1 MPa

s = [ Syw 2/(n-1)] = 0.125

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


From the data of Table 1-2, soil mechanics. The remainders involve such
complex soil-structure interactions that the
_
x = Σxy/Sy = 26.4/24 = 1.1 interrelationships must be found from experience or
experimentation. It is advantageous to write the
s = /Syw /(n-1) = /0.36/23
2
= 0.125 relationships in terms of dimensionless pi-terms. See
Appendix C. Pi-terms that have proven to be useful
_
w = x - x, so are given names such as Reynold's number in fluid
flow in conduits, Mach number in gas flow, influence
w e = (0.8 - 1.1) = -0.30 MN/m2 numbers, stability numbers, etc.
= deviation to failure pressure
Pi-terms are independent, dimensionless groups of
w e/s = 0.30/0.125 = 2.4. fundamental variables that are used instead of the
original fundamental variables in analysis or
From Table 1-1, interpolating, Pe= 0.82%. experimentation. The fundamental variables are
The probability that a pipe will fail by bursting combined into pi-terms by a simple process in which
three characteristic s of pi-terms must be satisfied.
pressure less than 0.80 MN/m2 is Pe = 0.82 % or
The starting point is a complete set of pertinent
one out of every 122 pipe sections. Cost accounting
fundamental variables. This requires familiarity with
of failures then follows.
the phenomenon. The variables in the set must be
interdependent, but no subset of variables can be
The probability that the strength of any pipe section
interdependent. For example, force f, mass m, and
will fall within a deviation of w e = +0.3 MN/m2 is P
acceleration a, could not be three of the fundamental
= 98.36%. It is noteworthy that P + 2Pe = 100%. variables in a phenomenon which includes other
variables because these three are not independent;
From probability data, the standard deviation can be i.e., f = ma. Only two of the three would be
calculated. From standard deviation, the zone of +w included as fundamental variables. Once the
can be found within which 90% of all measurements equation of performance is known, the deviation, w,
fall. In this case w/s = w/0.125 for which P = 90%. can be found. Suppose r = f(x,y,z,...), then w r2 =
From Table 1-1, interpolating for P = 90%, w/s = Mrx2
1.64%, and w = 0.206 MPa at 90% probability. w x2 + Mry 2w y 2 + ... where w is a deviation at the
same given probability for all variables, such as
Errors (three classes) standard deviation with probability of 68%; mrx is the
Mistake = blunder — tangent to the r-x curve and wx is the deviation at a
Remedies: double-check, repeat. given value of x. The other variables are treated in
Accuracy = nearness to truth — the same way.
Remedies: calibrate, repair, correct.
Precision = degree of refinement —
Remedies: normal distribution, safety factor. CHARACTERISTICS OF PI-TERMS

1. Number of pi-terms = (number of fundamental
PERFORMANCE variables) minus (number of basic dimensions).

Performance in soil-structure interaction is 2. All pi-terms are dimensionless.
deformation as a function of loads, geometry, and
properties of materials. Some deformations can be 3. Each pi-term is independent. Independence is
written in the form of equations from principles of assured if each pi-term contains a fundamental
variable 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 equation
for bursting pressure P' in plain pipe. Plain (or bare) pipe has smooth cylindrical surfaces with constant wall
thickness — 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 the
surface 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 small scale model study are plotted in Figure 1-2.
variables to relate, and the elimination of size effect. The plot of data appears to be linear. Only the last
The required number of pi-terms is less than the point to the right may deviate. Apparently the pipe
number of fundamental variables by the number of is no longer thin-wall. So the thin-wall designation
basic dimensions. Because pi-terms are only applies if t/D< 0.1. The equation of the plot is
dimensionless, they have no feel for size (or any the equation of a straight line, y = mx + b where y is
dimension) and can be investigated by model study. the ordinate, x is the abscissa, m is the slope, and b
Once pi-terms have been determined, their is the y-intercept at x = 0. For the case above,
interrelationships can be found either by theory (P'/S) = 2(t/D), from which, solving for bursting
(principles) or by experimentation. The results apply pressure,
generally because the pi-terms are dimensionless.
Following is an example of a well-designed P = 2S/(D/t)
experiment.
This important equation is derived by theoretical
principles under "Internal Pressure," Chapter 2.
Example

Using experimental techniques, find the equation for PERFORMANCE LIMITS
internal bursting pressure, P', for a thin-wall pipe.
Start by writing the set of pertinent fundamental Performance limit for a buried pipe is basically a
variables together with their basic dimensions, force deformation rather than a stress. In some cases it is
F and length L. possible to relate a deformation limit to a stress
(such as the stress at which a crack opens), but
Basic such a relationship only accommodates the designer
Fundamental Variables Dimensions for whom the stress theory of failure is familiar. In
reality, performance limit is that deformation beyond
P' = internal pressure FL-2 which the pipe-soil system can no longer serve the
t = wall thickness L purpose for which it was intended. The
D = inside diameter of ring L performance limit could be a deformation in the soil,
S = yield strength of the such as a dip or hump or crack in the soil surface
pipe wall material FL-2 over the pipe, if such a deformation is unacceptable.
The dip or hump would depend on the relative
settlement of the soil directly over the pipe and the
These four fundamental variables can be reduced to soil on either side. See Figure 1-3.
two pi-terms such as (P'/S) and (t/D). The pi-terms
were written by inspection keeping in mind the three But more often, the performance limit is excessive
characteristics of pi-terms. The number of pi-terms deformation of the pipe whic h could cause leaks or
is the number of fundamental variables, 4, minus the could restrict flow capacity. If the pipe collapses
number of basic dimensions, 2, i.e., F and L. The due to internal vacuum or external hydrostatic
two pi-terms are dimensionless. Both are pressure, the restriction of flow is obvious. If, on the
independent because each contains a fundamental other hand, the deformation of the ring is slightly out-
variable not contained in the other. Conditions for of-round, the restriction to flow is usually not
bursting can be investigated by relating only two significant. For example, if the pipe cross section
variables, the pi-terms, rather than interrelating the deflects into an ellipse such that the decrease of the
original four fundamental variables. Moreover, the minor diameter is 10% of the original circular
investigation can be performed on pipes of any diameter, the decrease in cross-sectional area is only
convenient size because the pi-terms are 1%.
dimensionless. Test results of a


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


The more common performance limit for the pipe is the structural design of the pipe can proceed in six
that deformation beyond which the pipe cannot resist steps as follows.
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 STEPS IN THE STRUCTURAL DESIGN OF
soil pressure. Typical examples of performance BURIED PIPES
limits for the pipe are shown in Figure 1-4. These
performance limits do not imply collapse or failure. In order of importance:
The soil generally picks up any increase in load by
arching action over the pipe, thus protecting the pipe 1. Resistance to internal pressure, i.e., strength of
from total collapse. The pipe may even continue to materials and minimum wall thickness;
serve, but most engineers would prefer not to
depend on soil alone to maintain the conduit cross 2. Resistance to transportation and installation;
section. This condition is considered to be a
performance limit. The pipe is designed to withstand 3. Resistance to external pressure and internal
all external pressures. Any contribution of the soil vacuum, i.e., ring stiffness and soil strength;
toward withstanding external pressure by arching
action is just that much greater margin of safety. 4. Ring deflection, i.e., ring stiffness and soil
The soil does contribute soil strength. On inspection, stiffness;
many buried pipes have been found in service even
though the pipe itself has "failed." The soil holds 5. Longitudinal stresses and deflections;
broken clay pipes in shape for continued service.
The inverts of steel culverts have been corroded or 6. Miscellaneous concerns such as flotation of the
eroded away without failure. Cast iron bells have pipe, construction loads, appurtenances, ins tallation
been found cracked. Cracked concrete pipes are techniques, soil availability, etc.
still in service, etc. The mitigating factor is the
embedment soil which supports the conduit. Environment, aesthetics, risks, and costs must be
considered. Public relations and social impact
A reasonable sequence in the design of buried pipes cannot be ignored. However, this text deals only
is the following: with structural design of the buried pipe.

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

2. Hydraulic design of pipe sizes, materials, 1-1 Fluid pressure in a pipe is 14 inches of mercury
as measured by a manometer. Find pressure in
3. Structural requirements and design of possible pounds per square inch (psi) and in Pascals
alternatives, (Newtons per square meter)? Specific gravity of
mercury is 13.546.
4. Appurtenances for the alternatives, (6.85 psi)(47.2 kPa)

5. Economic analysis, costs of alternatives, 1-2 A 100 cc laboratory sample of soil weighs 187.4
grams mass. What is the unit weight of the soil in
6. Revision and iteration of steps 3 to 5, pounds per cubic ft? (117 pcf)

7. Selection of optimum system. 1-3 Verify the standard deviation of Figure 1-1.
(s = 27.8 kPa)
With pipe sizes, pressures, elevations, etc., known


1-4 From Figure 1-1, what is the probability that
any maximum daily pressure will exceed 784.5
kPa? (Pe = 0.62%)
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) Figure 1-5 Bar graphs of internal vacuum at collapse
of thin-walled plastic pipes.
(b) What is the standard deviation? (8.38 Pa)

(c) What is the probable error? (+5.65 Pa) 1-7 Fiberglass reinforced plastic (FRP) tanks were
designed for a vacuum of 4 inches of mercury
1-6 Eleven 30 inch ID, non-reinforced concrete (4inHg). They were tested by internal vacuum for
pipes, Class 1, were tested in three-edge-bearing which the normal distribution of the results is shown
(TEB) test with results as follows: as Series A in Figure 1-6. Two of 79 tanks failed at
x = ultimate load in pounds per lineal ft less than 4inHg. In Series B, the percent of
x w w2 fiberglas was increased. The normal distribution
(lb/ft) (lb/ft) curve has the same shape as Series A, but is shifted
3562 1inHg to the right. What is the predicted probability
3125 of failure of Series B at or below 4 in Hg?
4375 (Pe = 0.17 % or one tank in every 590)
3438
4188 1-8 What is the probability that the vertical ring
3688 deflection d = y/D of a buried culvert will exceed
3750 10% if the following measurements were made on
4188 23 culverts under identical conditions?
4125 Measured values of d (%)
3625 6 9 6 6 5 6
2938 8 5 4 6 7 7
3 6 7 5 4 5
(a) What is the average load, x, at failure? 6 7 8 7 5 (0.24 %)
(x = 3727.5 lb/ft)
(b) What is the standard deviation? 1-9 The pipe stiffness is measured for many samples
(s = 459.5 lb/ft) of a particular plastic pipe. the average is 24 with a
(c) What is the probability that the load, x, at failure standard deviation of 3.
is less than the minimum specified strength of 3000
lb/ft (pounds per linear ft)? (Pe = 5.68%) a) What is the probability that the pipe stiffness will
be less than 20? (Pe = 9.17 %)


b) What standard deviation is required if the 1-10 A sidehill slope of cohesionless soil dips at
probability of a stiffness less than 20 is to be angle 2. Write pi-terms for critical slope when
reduced to half its present value; i.e., less than saturated.
4.585%? (s = 2.37)
1-11 Design a physical model for problem 1-10.

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


Anderson, Loren Runar et al "PRELIMINARY RING DESIGN"

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 is reached when stress, s , equals yield strength, S.
pipes all deal with resistance to loads. Loads on a For design, the yield strength of the pipe wall is
buried pipe can be complex, especially as the pipe reduced by a safety factor,
deflects out-of-round. Analysis can be simplified if
the cross section (ring) is assumed to be circular. s = P'(ID)/2A = S/sf . . . . . (2.1)
For pipes that are rigid, ring deflection is negligible.
For pipes that are flexible, ring deflection is usually where:
limited by specification to some value not greater s = circumferential tensile, stress in the wall,
than five percent. Analysis of a circular ring is P' = internal pressure,
reasonable for the structural design of most buried ID = inside diameter,
pipes. Analysis is prediction of structural OD = outside diameter,
performance. Following are basic principles for D = diameter to neutral surface,
analysis and design of the ring such that it can A = cross sectional area of the pipe wall per
support the three most basic loads: internal pressure, unit length of pipe,
transportation/installation, and external pressure. S = yield strength of the pipe wall material,
See Figures 2-1 and 2-2. t = thickness of plain pipe walls,
sf = safety factor.

INTERNAL PRESSURE — T his is the basic equation for design of the ring to
(MINIMUM WALL AREA) resist internal pressure. It applies with adequate
precision to thin-wall pipes for which the ratio of
The first step in structural design of the ring is to find mean diameter to wall thickness, D/t, is greater than
minimum wall area per unit length of pipe. ten. Equation 2.1 can be solved for maximum
pressure P' or minimum wall area A.
Plain pipe — If the pipe wall is homogeneous and
has smooth cylindrical surfaces it is plain (bare) and A = P'(ID)sf/2S = MINIMUM WALL AREA
wall area per unit length is wall thickness. This is
the case in steel water pipes, ductile iron pipes, and For thick-wall pipes (D/t less than ten), thick-wall
many plastic pipes. cylinder analysis may be required. See Chapter 6.
Neglecting resistance of the soil, the performance
Other pipes are corrugated or ribbed or composite limit is the yield strength of the pipe. Once the ring
pipes such as reinforced concrete pipes. For such starts to expand by yielding, the diameter increas es,
pipes, the wall area, A, per unit length of pipe is the the wall thickness decreases, and so the stress in the
pertinent quantity for design. wall increases to failure by bursting.

Consider a free-body-diagram of half of the pipe
with fluid pressure inside. The maximum rupturing Example
force is P'(ID) where P' is the internal pressure and
ID is the inside diameter. See Figure 2-1. This A steel pipe for a hydroelectric penstock is 51 inch
rupturing force is resisted by tension, FA, in the wall ID with a wall thickness of 0.219 inch. What is the
where F is the circumferential tension stress in the maximum allowable head, h, (difference in elevation
pipe wall. Equating rupturing force to the resisting of the inlet and outlet) when the pipe is full of water
force, F = P'(ID)/2A. Performance limit at no flow?


Figure 2-3 Free-body-diagrams of the ring subjected to the concentrated F-load, and showing pertinent
variables 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.


Given: shape during placement of embedment. One
E = 30(106)psi = modulus of elasticity, remedy, albeit costly, is to hold the ring in shape by
S = 36 ksi = yield strength, stulls or struts while placing embedment. It may be
sf = 2 = safety factor, economical to provide enough ring stiffness to resist
gw = 62.4 lb/ft3 = unit weight of water, deflection while placing the embedment. In any
P' = hg w = internal water pressure at outlet. case, ring deflection is a potential performance limit
for transportation/installation of pipes.
From Equation 2.1, s = S/2 = P'(ID)/2A where A is
0.219 square inches per inch of length of the pipe. So two analyses are required for transportion and
Substituting in values, h = 357 ft. installation, with two corresponding performance
limits: yield strength, and ring deflection. See Figure
2-3. In general, yield strength applies to rigid pipes
TRANSPORTATION/INSTALLATION — such as concrete pipes, and ring deflection applies to
MAXIMUM LINE LOAD ON PIPE flexible pipes. See Figure 2-4.

The second step in design is resistance to loads Yield Strength Performance Limit
imposed on the pipe during transportation and
installation. The most common load is diametral F- To analyze the yield strength performance limit,
load. See Figure 2-2. This load occurs when pipes based on experience, pertinent fundamental
are stacked or when soil is compacted on the sides variables may be written as follows:
or on top of the pipe as shown.
fv's, Fundamental bd's, Basic
If yield strength of the pipe material is exceeded due Variables Dimensions
to the F-load, either the pipe wall will crack or the F = transportation/installation FL-1
cross section of the pipe will permanently deform. load (concentrated line load
Either of these deformations (a crack is a per unit length of pipe),
deformation) may be unacceptable. So yield D = mean diameter of the pipe, L
strength may possibly be a performance limit even I = moment of inertia of the wall L3
though the ring does not collapse. cross section per unit length
of pipe,
For some plastic materials, including mild steel, c = distance from the neutral axis L
design for yield strength is overly conservative. So of the wall cross section to
what if yield strength is exceeded? A permanent the most remote wall surface
deformation (dent) in the ring is not necessarily pipe where the stress is at yield point.
failure. In fact, the yield strength was probably S = yield strength of pipe wall FL-2
exceeded in the process of fabricating the pipe. material
5 fv's - 2 bd's = 3 pi-terms.
Some pipe manufacturers limit the F-load based on
a maximum allowable ring deflection, d = D/D, The three pi-terms may be written by inspection. A
where D is the decrease in mean diameter D due to typical set is: (F/SD), (c/D), and (I/D3). This is only
load F. Some plastics have a memory for excessive one of many possible sets of pi-terms. D is a
ring deflection. In service, failure tends to occur repeating variable. Note that the pi-terms are
where excess ive ring deflection occurred before independent because each contains at least one
installation. Increased ring stiffness decreases ring fundamental variable that is not contained in any of
deflection. It is not inconceivable that the ring can the other pi-terms. All are dimensionless. The
be so flexible that it cannot even hold its circular interrelationship of these three pi-terms can be


found either by experimentation or by analysis. An fv's, Fundamental bd's, Basic
example of class ical analysis starts with circum- Variables Dimensions
ferential stress s = Mc/I where M is the maximum d = ring deflection = /D -
bending moment in the pipe ring due to load F. But D = mean diameter of the L
if stress is limited to yield strength, then S = Mc/I pipe
where M = FD/2p based on ring analysis by F = diametral line load FL-1
Castigliano's theorem. See Appendix A, Table A-1. per unit length of pipe
M is the maximum moment due to force F. Because EI = wall stiffness FL
it occurs at the location of F, there is no added ring per unit length of pipe
compression stress. Substituting in values and
rearranging the fundamental variables into pi-term, where:
D = decrease in diameter due to the F-load,
(F/SD) = 2p(D/c)(I/D3) E = modulus of elasticity,
t = wall thickness for plain pipe,
The three pi-terms are enclosed in parentheses. I = moment of inertia of wall cross section per
Disregarding pi-terms, unit length of pipe = t3/12 for plain pipe.

F = 2pSI/cD = F-load at yield strength, S. 4 fv's - 2 bd's = 2 pi-terms.

For plain pipes, I = t3/12 and c = t/2 for which, I/c = Two pi-terms, by inspection, are (d) and (FD2/EI).
t2/6 and, in pi-terms: Again, the interrelationship of these pi-terms can be
found either by experimentation or by analysis.
(F/SD) = p(t/D)2/3 Table 5-1 is a compilation of analyses of ring
deflections of pipes subjected to a few of the
Disregarding pi-terms, common loads. From Table A-1, ring deflection due
to F-loads is,
F = pSt2/3D = F-load at yield strength S for plain
pipes (smooth cylindrical surfaces). The modulus of (d) = 0.0186 (FD2/EI) . . . . . (2.2)
elasticity E has no effect on the F-load as long as
the ring remains circular. Only yield strength S is a This equation is already in pi-terms (parentheses).
performance limit. For plain pipes, for which I = t3/12 and c = t/2, this
equation for ring deflection is:

Ring Deflection Performance Limit (d) = 0.2232 (F/ED) (D/t)3

If the performance limit is ring deflection at the The relationship between circumferential stress and
elastic limit, modulus of elasticity E is pertinent. ring deflection is found by substituting from Table A-
Yield strength is not pertinent. For this case, 1, at yield stress, F = 2pSI/cD, where S is yield
pertinent fundamental variables and corresponding strength and c is the distance from the neutral
basic dimensions are the following: surface of the wall to the wall surface. The
resulting equation is:


(d) = 0.117 (s /E) (D/c) . . . . . (2.3) Steel and aluminum pipe industries use an F-load
criterion for transportation/installation. In Equation
For plain pipes, 2.2 they specify a maximum flexibility factor FF =
D2/EI. If the flexibility factor for a given pipe is less
(d) = 0.234 (s E) (D/t) than the specified FF, then the probability of
transportation/installation damage is statistically low
Note the introduction of a new pi-term, (s /E). This enough to be tolerated.
relationship could have been found by
experimentation using the three pi-terms in For other pipes, the stress criterion is popular.
parentheses in Equation 2.3. Ring deflection at yield When stress s = yield strength S, the maximum
stress, S, can be found from Equation 2.3 by setting allowable load is:
s = S. If ring deflection exceeds yield, the ring does
not return to its original circular shape when the F- F = 2pSI/cD
load is removed. Deformation is permanent. This is
not failure, but, for design, may be a performance For walls with smooth cylindrical surfaces,
limit with a margin of safety.
F = pSt2/3D
The following equations summarize design of the
pipe to resist transportation/installation loads. In another form, for plain walls, the maximum
allowable D/t is:
For transportation/installation, the maximum
allowable F-load and the corresponding ring (D/t)2 = pSD/3F
deflection, d, when circumferential stress is at
yield strength, S, are found by the following For the maximum anticipated F-load, i.e. at yield
formulas. strength, the minimum wall thickness term (t/D) can
be evaluated. Any safety factor could be small —
approaching 1.0 — because, by plastic analysis,
Ring Strength collapse does not occur just because the
circumferential stress in the outside surfaces
(F/SD) = 2p (D/c) (I/D3) . . . . . (2.4) reaches yield strength. To cause a plastic hinge
For plain pipes, (F/SD) = p(t/D)2/3 (dent or cusp) the F-load would have to be increased
by three-halves.
Resolving, for plain pipes,
Plastic pipe engineers favor the use of outside
F = pSD(t/D) /3.
2
diameter, OD, and a classification number called the
dimension ratio, DR, which is simply DR = OD/t =
(D+t)/t where D is mean diameter. Using these
Ring Deflection where d = (D/D) due to F-load, dimensions, the F-load at yield is:
is given by:
F = pSt/3(DR-1)
2
d = 0.0186 (FD /EI), in terms of F-load . . . (2.5)
If the F-load is known, the required dimension ratio
d = 0.117 (s /E) (D/c), in terms of stress, s , or at yield strength is:

d = 0.234(S/D)(D/t), for plain pipes with smooth DR = (pSt/3F) + 1
cylindrical surfaces in terms of yield strength S.


Example distributed. OD is the outside diameter. The
resisting force is compression in the pipe wall, 2s A,
Unreinforced concrete pipes are to be stacked for where s is the circumferential stress in the pipe wall,
storage in vertical columns on a flat surface as called ring compression stress. Equating the
indicated in Figure 2-2. The load on the bottom pipe rupturing force to the resisting force, with stress at
is essentially an F-load. The following information is allowable, S/sf, the resulting equation is:
given:
s = P(OD)/2A = S/sf . . . . . (2.6)
ID = 30 inches = inside diameter,
OD = 37.5 in = outside diameter, This is the basis for design. Because of its
g = 145 lb/ft3 = unit weight of concrete, importance, design by ring compression stress is
F = 3727 lb/ft = F-load at fracture considered further in Chapter 6.
+ s from tests where,
s = + 460 lb/ft = standard deviation of the The above analyses are based on the assumption
ultimate F-load at fracture of the pipe. that the ring is circular. If not, i.e., if deformation
out-of-round is significant, then the shape of the
a) How high can pipes be stacked if the F-load is deformed ring must be taken into account. But
limited to 3000 lb/ft? From the data, the weight of basic deformation is an ellipse. See Chapter 3.
the pipe is 400 lb/ft. The number of pipes high in the
stack is 3000/400 = 7.5. So the stack must be Example
limited to seven pipes in height.
A steel pipe for a hydroelectric penstock is 51
b) What is the probability that a pipe will break if the inches in diameter (ID) with wall thickness of 0.219
column is seven pipes high? The seven pipe load at inch. It is to be buried in a good soil embedment
the bottom of the stack is 7(400) = 2800 lb/ft. w = such that the cross section remains circular. What
3727 - 2800 = 927 lb/ft which is the deviation of the is the safety factor against yield strength, S = 36 ksi,
seven-pipe load from the F-load. From Table 1-1, if the external soil pressure on the pipe is 16 kips/ft2?
the probability of failure is 2.2% for the bottom For this pipe, OD = 51.44 inches, and A = t = 0.219
pipes. For all pipes in the stack, the probability is inch. At 16 ksf, P = 111 psi. Substituting into
one-seventh as much or 0.315%, which is one Equation 2.6, the safety factor is sf = 2.76. The soil
broken pipe for every 317 in the stack. pressure of 16 ksf is equivalent to about 150 feet of
soil cover. See Chapter 3.
c) What is the circumferential stress in the pipe 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 = PROBLEMS
9, D = 51 inches. Solving, s = 471 psi. This is good
concrete considering that it fails in tension. 2-1 What is the allowable internal pressure in a 48-
inch diameter 2-2/3 by 1/2 corrugated steel pipe, 16
gage (0.064 inch thick)? (P' = 48.4 psi)
EXTERNAL PRESSURE —
MINIMUM WALL AREA Given:
D = 48 inches = inside diameter,
Consider a free-body-diagram of half the pipe with t = 0.064 in = wall thickness,
external pressure on it. See Figure 2-4. The A = 0.775 in 2/ft [AISI tables],
vertical rupturing force is P(OD) where P is the S = 36 ksi = yield strength,
external radial pressure assumed to be uniformly


E = 30(106) psi, diameter, high-strength wires? What about bond?
sf = 2 = safety factor. How can ends of the rods (or wires) be fixed?

2-2 What is the allowable internal pressure if a 2-5 What is the allowable fresh water head (causing
reinforced conc rete pipe is 60 inch ID and has two internal pressure) in a steel pipe based on the
cages comprising concentric hoops of half-inch steel following data if sf = 2? (105 meters)
reinforcing rods spaced at 3 inches in the wall which ID = 3.0 meters,
is 6.0 inches thick? (P' = 78.5 psi) t = 12.5 mm = wall thickness,
Given: S = 248 MN/m2 = 36 ksi yield strength.
S = 36 ksi = yield strength of steel,
sf = 2 = safety factor, 2-6 What maximum external pressure can be
Ec = 3(106) psi = concrete modulus, resisted by the RCP pipe of Problem 2-2 if the yield
Neglect tensile strength of concrete. strength of the concrete in compression is 10 ksi,
modulus of elasticity is E = 3000 ksi, and the internal
2-3 What must be the pretension force in the steel pressure in the pipe is zero? See also Figure 2-5.
rods of Problem 2-2 if the pipe is not to leak at (P = 52 ksf, limited by the steel)
internal pressure of 72 psi? Leakage through hair
cracks in the concrete appears as sweating. 2-7 Prove that T = Pr for thin-walled circular pipe.
(Fs = 2.9 kips) See Figure 2-4.
2-4 How could the steel rods be pretensioned in T = ring compression thrust,
Problem 2-3? Is it practical to pretension (or post P = external radial pressure,
tension) half-inch steel rods? How about smaller r = radius (more precisely, outside radius).

Figure 2-5 Equivalent diagrams for uniform external soil pressure on a pipe, showing (on the right) the more
convenient form for analysis.


Anderson, Loren Runar et al "RING DEFORMATION"

Figure 3-1 (top) Vertical compression (strain) in a medium transforms an imaginary circle into an ellipse with
decreases in circumference and area.

(bottom) Now if a flexible ring is inserted in place of the imaginary ellipse and then is allowed to expand such
that 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. Geometry of the Ellipse
For most buried pipe analyses, this deformation is
small enough that it can be neglected. For a few The equation of an ellipse in cartesian coordinates,
analyses, however, deformation of the ring must be x and y, is:
considered. This is particularly true in the case of
instability of the ring, as, for example, the hydrostatic a2x2 + b2y2 = a2b2
collapse of a pipe due to internal vacuum or external
pressure. Collapse may occur even though stress where (See Figure 3-2):
has not reached yield strength. But collapse can a = minor semi-diameter (altitude)
occur only if the ring deforms. Analysis of failures b = major semi-diameter (base)
requires a knowledge of the shape of the deformed r = radius of a circle of equal circumference
ring.
The circumference of an ellipse is p(a+b) which
For small ring deflection of a buried circular pipe, the reduces to 2pr for a circle of equal circumference.
basic deflected cross section is an ellipse. Consider
the infinite medium with an imaginary circle shown In this text a and b are not used because the pipe
in Figure 3-1 (top). If the medium is compressed industry is more familiar with ring deflection, d.
(strained) uniformly in one direction, the circle Ring deflection can be written in terms of semi-
becomes an ellipse. This is easily demonstrated diameters a and b as follows:
mathematic ally. Now suppose the imaginary circle
is a flexible ring. When the medium is compressed, d = D/D = RING DEFLECTION . . . . . (3.1)
the ring deflects into an approximate ellipse with
slight deviations. If the circumference of the ring where:
remains constant, the ellipse must expand out into D = decrease in vertical diameter of ellipse from
the medium, increasing compressive stresses a circle of equal circumference,
between ring and medium. See Figure 3-1 (bottom). = 2r = mean diameter of the circle —
The ring becomes a hard spot in the medium. On diameter to the centroid of wall cross-
the other hand, if circumference of the ring is sectional areas,
reduced, the ring becomes a soft spot and pressure a = r(1-d) for small ring deflections (<10%),
is relieved between ring and medium. In either case, b = r(1+d) for small ring deflections (<10%).
the basic deformation of a buried ring is an ellipse —
slightly modified by the relative decreases in areas Assuming that circumferences are the same for
within the ring and without the ring. The shape is circle and ellipse, and that the vertical ring deflection
also affected by non-uniformity of the medium. For is equal to the horizontal ring deflection, area within
example, if a concentrated reaction develops on the the ellipse is Ae = Bab; and
bottom of the ring, the ellipse is modified by a flat
spot. Nevertheless, for small soil strains, the basic Ae = pr2 (1 - d2)
ring deflection of a flexible buried pipe is an ellipse.
Following are some pertinent approximate The ratio of areas within ellipse and circle is:
geometrical properties of the ellipse that are
sufficiently accurate for most buried pipe analyses.
A r = A e / A o = ratio of areas.
Greater accuracy would require solutions of infinite
series.
See Figure 3-3.


Figure 3-2 Some approximate properties of an ellipse that are pertinent to ring analyses of pipes where d is
the 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 An important property of the ellipse is the ratio of
vertical ring deflections are equal if the radii rr = ry/rx, which is:
circumferences are equal for circle and ellipse? For
the circle, circumference is 2pr. For the ellipse, rr = (1 + d)3 / (1 - d)3 . . . . . (3.2)
circumference is (b+a)(64-3R4)/(64-16R2), where R
is approximately R = (b-a)/(b+a). Only the first where:
terms of an infinite series are included in this rr = ratio of the maximum to minimum radii of
approximate ellipse circumference. See texts on curvature of the ellipse. See graph of Figure 3-4.
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 Measurement of Radius of Curvature
deflection, dy and dx, a few values of d y and the
corresponding dx are shown below for comparison. In practice it is often necessary to measure the
radius of curvature of a deformed pipe. This can be
Deviation done from either inside or outside of the pipe. See
dy (%) dx (%) (dy - dx)/dy Figure 3-5. Inside, a straightedge of known length L
______ ______ __________ is laid as a cord. The offset e is measured to the
0. 0. 0. curved wall at the center of the cord. Outside, e can
5.00 4.88 0.024 be found by laying a tangent of known length L and
10.00 9.522 0.048 by measuring the offsets e to the pipe wall at each
15.00 13.95 0.070 end of the tangent. The average of these two
20.00 18.116 0.094 offsets is the value for e. Knowing the length of the
cord, L, and the offset, e, the radius of curvature of
For ring deflections of d = dy = 10%, the the pipe wall can be calculated from the following
corresponding dx is less than 10% by only equation:
4.8%(10%) = 0.48%. This is too small to be
significant in most calculations such as areas within r = (4e2 + L2)/8e . . . . . (3.3)
the ellipse and ratios of radii.
It is assumed that radius of curvature is constant
Radii of curvature of the sides (spring lines) and the within cord length L. The calculated radius is to the
top and bottom (crown and invert) of the ellipse are: surface from which e measurements are made.

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

ry = r (1 + 3d + 4d2 + 4d3 + 4d4 + .....) An inspection reveals that a 72-inch corrugated
metal pipe culvert appears to be flattened somewhat
For ring deflection less than d = 10%, and neglecting on top. From inside the pipe, a straightedge (cord)
higher orders of d, 12 inches long is placed against the top, and the mid-
ordinate offset is measured and found to be 11/32
rx = r (1 - 3d), and ry = r (1 + 3d). inch. What is the radius of curvature of the pipe ring
at the top?
However, more precise, and almost as easy to use,
are the approximate values: From Equation 3.3, r = (4e2 + L2)/8e. Substituting in
values and solving, ry = 52.5 inches which is the
rx = a2 / b = r (1 - d)2 / (1 + d)
average radius within the 12-inch cord on the inside
of the corrugated pipe. On the outside, the radius is
ry = b2 / a = r (1 + d)2 / (1 - d)
greater 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 1.618
10 1.826
12 2.062
15 2.476
20 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 length
L and the middle ordinate e.


Ring Deflection Due to Internal Pressure Ring Deformation Due to External Loading

When subjected to uniform internal pressure, the Computer software is available for evaluating the
pipe expands. The radius increases. Ring deflection deformation of a pipe ring due to any external
is equal to percent increase in radius; loading. Analysis is based on the energy method of
virtual work according to Castigliano. Analysis
d = Dr/r = DD/D = 2pre /2pr = e provides a component of deflection of some point B
on a structure with respect to a fixed point A. It is
where: convenient to select point A as the origin of fixed
d = ring deflection (percent), coordinate axes — the axes are neither translated
Dr and DD are increases due to internal pressure, nor rotated. See Appendix A.
r = mean radius,
D = mean diameter,
e = circumferential strain, Example
E = modulus of elasticity = s/e.
s = circumferential stress = Ee = Ed. Consider the quadrant of a circular cylinder shown
in Figure 3-6. It is fixed along edge A-A-A, and is
But s = P'(ID)/2A, from Equation 2.1, loaded with vertical line load Q along free edge B-B-
B. What is the horizontal deflection of free edge B
where: with respect to fixed edge A? This is a two-
P' = uniform internal pressure, dimensional problem for which a slice of unit width
ID = inside diameter, can be isolated for analysis. Because A is fixed,
A = cross sectional area of wall per unit length. the horizontal deflection of B with respect to A is xB
for which, according to Castigliano:
Equating the two values for s , and solving for d,
xB = f (M/EI)(dM/dp)ds . . . . . (3.5)

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

Figure 3-6 Quadrant of a circular cylinder fixed at
the crown A-A-A with Q-load at the spring line, B-
B-B, showing a slice isolated for analysis.


where: PROBLEMS
xB = displacement of point B in the x-direction,
EI = wall stiffness, 3-1 A plain polyethylene pipe of 16-inch outside
E = modulus of elasticity, diameter and DR = 15 is subjected to internal
I = centroidal moment of inertia of the cross pressure of 50 psi. The surfaces are smooth and
section of the wall per unit length of cylinder, cylindrical (not ribbed or corrugated). DR
(dimension ratio) = (OD)/t where t = wall thickness.
M = moment of force about the neutral axis at C, Modulus of elasticity is 115 ksi. What is the ring
deflection? DR is dimension ratio = (OD)/t.
p = differential load (dummy load) applied at (d = 0.28%)
point B in the direction assumed for deflection,

ds = differential length along the slice, = rdq 3-2 At ring deflection of 15%, and assuming the
pipe cross section is an ellipse, what is the percent
r = mean radius of the circular cylinder. error in finding the ratio of maximum to minimum
radii of curvature by means of approximate
It is assumed that deflection is so small that radius r Equation 3.2,
remains constant. It is also assumed that the rr = (1+d)3 / (1-d)3? (0.066%)
deflection is due to moment M, flexure — not to
shear or axial loads. In Figure 3-6, consider arc CB
as a free-body-diagram. Apply the dummy load p at 3-3 A 36 OD PVC buried pipeline is uncovered at
B acting to the right assuming that deflection xB will one location. The top of the pipe appears to be
be in the x-direction. If the solution turns out to be flattened. A straight edge 200 mm long is laid
negative, then the deflection is reversed. From the horizontally across the top and the vertical distances
free-body-diagram CB, down to the pipe surface at each end of the straight
edge are measured and found to be 9.2 and 9.4 mm.
M = Qr(1-cosq) + pr(sinq) What is the radius of curvature of the outside
surface of the pipe at the crown?
M/ p = r(sinq) Ry = 542 mm = 21.35 inches)

But because p approaches zero (differential),
3-4 Assuming that the ring of problem 3-3 is
M = Qr(1-cosq) deflected into an ellipse, approximately what is the
ring deflection? Maximum ring deflection is usually
ds = rdq limited to 5% according to specifications.
(d = 5.74%)
Substituting into Equation 3.5,

xB = (Qr/EI) (1-cosq) r(sinq) rdq 3-5 What is the percent decrease in cross-sectional
area inside the deflected pipe of problem 3-4 if the
Integrating and substituting in limits of q from 0 to ring deflection is d = 5.74%?
p/2, (0.33%)

xB = Qr3/2EI
3-6 What is the approximate ratio of maximum to
This is one of a number of the most useful minimum radii, rr, for an ellipse? (rr = 1.8)
deflections of rings recorded in Table A-1.


3-7 A horizontal, rectangular plate is a cantilever
beam loaded by a uniform vertical pressure, P, and
supported (fixed) along one edge. What is the
vertical deflection of the opposite edge? The
thickness of the plate is t, the length measured from
the fixed edge is L, and the modulus of elasticity is
E. Elastic limit is not exceeded. Use the Castigliano
equation. (y = 3PL4 / 2Et 3)

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, as
shown. If the wall stiffness is EI, what is the
vertical deflection of point A?
(yA = 0.1781 Fr3/EI)

3-9 What is the vertical ring deflection of the hinged
arch of problem 3-8 if it is loaded with a uniform
vertical pressure P instead of the F-load?

3-10 The top and bottom halves of the circular
cylinder of problem 3-8 are symmetrical. If the
spring lines of the two halves are hinged together,
what is the ring deflection due to the F-load and an
equal 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 a
parallel plate test. What is the ring deflection if
elastic 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 A
and loaded by force, Q. (EI = 3p Qr/x)

3.13 A pipe in a casing floats when liquid grout is
introduced between pipe and casing. Find the
moment, thrust and shear at crown and invert.
(See Table A-1)


Anderson, Loren Runar et al "SOIL MECHANICS"

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 of
cover.


CHAPTER 4 SOIL MECHANICS

An elementary knowledge of basic principles of soil concentration factor is needed, or minimum soil
stresses is essential to understanding the structural density should be specified. Over a long period of
performance of buried pipes. These principles are time, pressure concentrations on the pipe may be
explained in standard texts on soil mechanics. A reduced by creep in the pipe wall (plastic pipes),
few are reviewed in the following paragraphs earth vibrations, freeze-thaw cycles, wet-dry cycles,
because of their special application to buried pipes. etc. The most rational soil load for design is vertical
soil pressure at the top of the pipe due to dead
weight of soil plus the effect of live load with a
VERTICAL SOIL PRESSURE P specification that the soil embedment be denser
than critical void ratio. Critical void ratio, roughly
For the analysis and design of buried pipes, external 85% soil density (AASHTO T-99), is the void ratio
soil pressures on the pipes must be known. Vertical at such density that the volume of the soil skeleton
soil pressure at the top of the pipe is caused by: 1. does not decrease due to disturbance of soil
dead load, Pd , the weight of soil at the top of the particles.
pipe; and 2. live load, Pl , the effect of surface live
loads at the the top of the pipe. Figure 4-1 shows For design, the total vertic al soil pressure at the top
these vertical soil pressures at the top of the pipe as of the pipe is:
functions of height of soil cover, H, for an HS=20
truck axle load of 32 kips, and soil unit weight of 100 P = Pd + Pl . . . . . (4.1)
pcf. Similar graphs are found in pipe handbooks
such as the A ISI Handbook of Steel Drainage and where (see Figure 4-2)
Highway Construction Products. Soil unit weight P = total vertical soil pressure at the level of
can be modified as necessary. Also other factors the top of the pipe
must be considered. What if a water table rises Pd = dead load pressure due to weight of the
above the top of the pipe, or the pipe deflects, or the soil (and water content)
soil is not compacted, or is overcompacted? For Pl = vertical live load pressure at the level of
these and other special cases, the following the top of the pipe due to surface loads.
fundamentals of soil mechanics may be useful.
This is a useful concept in the analysis of buried
If the embedment about a buried pipe is densely pipes. Even rigid pipes are designed on this basis if
compacted, vertical soil pressure at the top of the a load factor is included. See Chapter 12.
pipe is reduced by arching action of the soil over the
pipe, like a masonry arch, that helps to support the In fact, P is only one of the soil stresses. At a given
load. To be conservative, arching action is usually point in a soil mass, a precise stress analysis would
ignored. However, soil arching provides an added consider three (triaxial) direct stresses, three
margin of safety. If the soil embedment is loose, shearing stresses, direct and shearing moduli in three
vertical soil pressure at the top of the pipe may be directions and three Poisson's ratios — with the
increased by pressure concentrations due to the additional condition that soil may not be elastic. The
relatively noncompressible area within the ring in imprecisions of soil placement and soil compaction
loose, compressible soil. Pressure concentrations obfuscate the arguments for such rigor. Elastic
due to loose embedment cannot be ignored. For analysis may be adequate under some few
design, either a pressure circumstances. Superposition is usually adequate
without concern for a combined stress analysis
involving triaxial stresses and Poisson ratio. Basic
soil mechanics serves best.


Figure 4-2 Vertical soil pressure P at the level of the top of a buried pipe where P = Pl + Pd , showing live
load 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 stress
at depth H.


Dead Load Vertical Soil Pressure Pd S = degree of saturation = 1 when saturated,
e = void ratio, from laboratory analysis,
Dead load is vertical pressure due to the weight of gw = unit weight of water.
soil at a given depth H. In the design of buried
pipes, H is the height of soil cover over a pipe. Total Table 4-1 is a summary of dead load soil stresses
from which dead load pressure Pd can be found and
pressure Pd is the weight of soil, including its water
combined with live load pressure Pl. Live load
content, per unit area. See Figure 4-3.
pressure is found from techniques described in the
Intergranular (or effective) pressure Pd is the
paragraphs to follow.
pressure felt by the soil skeleton when immersed in
water. The total and intergranular vertical stresses
Intergranular vertical soil pressure P, at the bottom
at the bottom of a submerged stratum can be related
of multiple soil strata, is:
by the following stress equation:
_
_ P = P - u = P - g wh . . . . . (4.5)
s=s-u . . . . . (4.2)

where _
where
_ P = vertical intergranular soil pressure,
s = intergranular vertical soil stress (felt by the
P = total dead plus live load pressures,
soil when buoyed up by water),
h = height of water table above the pipe.
s = total vertical soil stress = g tH,
u = pore water pressure = g wH,
Total pressure is used to calculate ring compression
gt = total unit weight of soil and water,
stress. Intergranular soil pressure is used to calcu-
gw = unit weight of water = 62.4 pcf.
late ring deflection which is a function of soil
compression. As the soil is compressed, so is the
Now consider more than one stratum of soil as
pipe compressed — and in direct ratio. But soil
shown in Figure 4-4. The total vertical dead load soil
compression depends only on intergranular stresses.
pressure Pd at the bottom of the strata is the sum of
See Chapter 7.
the loads imposed by all of the strata; i.e.,

Pd = Sgt H . . . . . (4.3) Live Load Vertical Soil Pressure Pl

where Live load soil pressure Pl is the vertical soil pressure
gt = total unit weight (wet weight) of soil at the top of the buried pipe due to surface loads.
in a given stratum, and See Figure 4-5. For a single concentrated load W on
H = height of the same stratum. the surface, vertical soil pres sure at point A at the
top of the pipe is:
Values of H for each soil stratum are provided by
soil borings. Values of g t are simply the unit s = NW/H2 . . . . . (4.6)
weights of representative soil samples including the
water content. If the soil samples are not available, where
from soil mechanics, W = concentrated surface load (dual-wheel)
H = height of soil cover over the top of the pipe
g t = (G+Se) g w /(1+e) . . . . . (4.4) R = horizontal radius to stress, s ,
N = Boussinesq coefficient.
where from the line of action of load W,
G = specific gravity of soil grains, about 2.65, N = Boussinesq coefficient = 3(H/R) 5/2p.


Figure 4-4 Multiple strata
(three strata with the clay
stratum divided into two at
the water table) showing the
total vertical dead load soil
pressure 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 line
of action of a concentrated surface load W. (After Boussinesq)


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


For a single wheel (or dual wheel) load, the maxi- Example
mum stress smax at A occurs when the wheel is
directly over the pipe; i.e. R = 0, for which What is the stress at point A below A' of Figure 4-
8? The vertical stress s at depth H below surf a c e
smax = 0.477 W/H2 . . . . . (4.7) point A is, by superposition:

Load W can be assumed to be concentrated if depth
s = s ' - S s'" + s "
H is greater than the maximum diameter or length of where
the surface loaded area. s' = stress at corner A due to loaded area
L'xB'
For multiple wheel (or dual) loads, the maximum S s'" = sum of stresses at corner A due to
stress at point A, due to effects of all loads must be loaded areas L'B" and L"B'
ascertained. The trick is to position the wheel loads s" = stress at corner A due to loaded area
so that the combined stress at A is maximum. This L"B'
can be done by trial.
Clearly, s " due to area L"B" was subtracted twice,
The effect of a uniformly distributed surface load so must be added back once.
can be found by dividing the loaded surface area into
infinitesimal areas and integrating to find the sum of An occlusion in the soil mass, such as a pipe, violates
their effects at some point at depth H. See Figure 4- Boussinesq's assumptions of elasticity, continuity,
6. Newmark performed such an integration and compatibililty, and homogeneity. The pipe is a hard
found the vertical stress s at a depth H below spot, a discontinuity. Soil is not elastic, nor homoge-
corner A of a rectangular area of greater length L neous, nor compatible when shearing planes form.
and lesser breadth B, loaded with uniform pressure Nevertheless, the Boussinesq assumptions are
q. His neat solution is: adequate for most present-day installation tech-
niques. For most buried pipe design, it is sufficient,
s = Mq . . . . . (4.8) and conservative, to solve for Pl at the top of the
pipe due to a single wheel load W at the surface by
where M is a coefficient which can be read on the using the Boussinesq equation with the radius R = 0.
chart of Figure 4-6 by entering with arguments L/B For additional wheel loads, simply add by superposi-
and B/H. If the stress due to pressure on an area is tion the influence of other wheel loads at their radii
desired below some point other than a corner, the R.
rectangular area can be expanded or subdivided
such that point A is the common corner of a number Example
of areas. The maximum stress under a rectangular
area occurs below the center. See Figure 4-7. The What is the maximum vertical soil stress at a depth
rectangle is subdivided into four identical rectangles of 30 inches due to the live load of a single axle HS-
of length L and breadth B as shown. The stress at 20 truck? Neglect surface paving. See Figures 4-9
point A is 4Mq, where M is found from the and 4-15. By trial, it can be shown that the point of
Newmark chart, Figure 4-6. maximum stress is point A under the center of one
tire print. The rectangular tire prints are subdivided
Alternatively, the Boussinesq equation can be used as shown for establishing a common corner A'. The
with less than five percent error if the concentrated effects of the left tire print and the right tire print are
load, Q = qBL, for each of the quadrants is assumed analyzed separately, then combined. The length L
to act at the center of each quadrant. For this case, and breadth B of each tire print are based on 104 psi
R = (L2+B2)/2. The resulting stress at A is 4NQ/H2 tire pressure. Use Newmark because H is less than
from Equation 4.6. 3L.


Figure 4-7 Procedure for subdividing a rectangular surface area such that the stress below the center at depth
H 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, under
point A.


Figure 4-9 Single axle HS-20 truck load showing typical tire prints for tire pressure of 104 psi, and showing
the 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 plane
through B. Note the stresses sq and t q shown on the q -plane. At soil slip, the circle is tangent to the
strength envelopes described below.

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


Given: SOIL STRENGTH
W = 32 k for HS-20 truck load (single axle),
q = 104 psi, Failure of a buried pipe is generally associated with
B = 7 inches, failure of the soil in which the pipe is buried. The
L = 22 inches, classical, two-dimensional, shear-strength soil model
H = 30 inches. is useful for analysis. Analysis starts with an infini-
tesimal soil cube on which stresses are known and
For the left tire print: the orientation is given. The model comprises three
sL = 4Mq, elements, the Mohr stress circle, orientation diagram,
L/B = 11/3.5 = 3.14, and strength envelopes.
B/H = 3.5/30 = 0.12,
From Figure 4.6, M = 0.018 and sL = 1.078 ksf
Mohr Stress Circle
For the right tire print:
sR = 2(M'-M")q, where, The Mohr stress circle is a plot of shearing stress, t ,
L'/B' = 83/3.5 = 23.7, as a function of normal stress, s , on all planes at
B'/H' = 3.5/30 = 0.12, angle q through an infinitesimal soil cube B. See
From Figure 4.6, M' = 0.02 (extrapolated) Figures 4-10 and 4-11. The sign convention is
compressive normal stress positive ( + ) and
L"/B" = 61/3.5 = 17.4, counterclockwise shearing stress positive ( + ).
B"/H" = 3.5/30 = 0.12, The center of the circle is always on the s -axis.
From Figure 4.6, M" = 0.02 (extrapolated) Two additional points are needed to determine the
circle. They are (sx, t xy ) on a y-plane and (s y , ty x)
sR = r(M'- M")q = 0: sR = 0 ksf on an x-plane. These are known stresses on cube
B. An origin of planes always falls on the circle.
At point A, s = s L + s R; s = 1.08 ksf t xy = -ty x from standard texts on solid mechanics.
Any plane from the origin intersects the Mohr
A rough check by Boussinesq is of interest because circle at the stress coordinates acting on that
the results are conservatively higher and are more plane— which is correctly oriented if the following
easily solved. procedure is followed.

Given:
W = 16 kips at the center of each tire print, Orientation Diagram
H = 2.5 ft,
RL = 0, Figure 4-10 shows infinitesimal cube B with the x-
RR = 6 ft, plane and y-plane identified and with the soil stresses
R/H = 2.4. acting on each plane. Cube B and its axes of
From Figure 4.5, N = 0.004. orientation can be superimposed on the Mohr circle
such that stress coordinates (where each plane inter-
sL = 0.477 W/H2 = 1.22 ksf sects the Mohr circle) are the stresses on that plane.
sR = NW/H2 = 0.01 ksf With cube B located on the Mohr circle as the origin
of axes, and with the axes correctly oriented, any
At point A, s = sL + s R ; s = 1.23 ksf plane through B will intersect the Mohr circle at the
point whose stress coordinates are the stresses
The Boussinesq solution is in error by 13.9%, but on acting on that plane, and all planes are correctly
the high (conservative) side. Of interest is the small oriented with respect to the original soil cube B.
(negligible) effect of the right wheel load.


Figure 4-12 Trigonometry for analysis of stresses at soil slip on shear planes, q f in cohesionless soil which
has a soil friction angle of j.


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