Numerical Methods in Soil Mechanics 14.PDF

Numerical Methods in Soil Mechanics 14.PDF Numerical Methods in Geotechnical Engineering contains the proceedings of the 8th European Conference on Numerical Methods in Geotechnical Engineering (NUMGE 2014, Delft, The Netherlands, 18-20 June 2014). It is the eighth in a series of conferences organised by the European Regional Technical Committee ERTC7 under the auspices of the International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE). The first conference was held in 1986 in Stuttgart, Germany and the series has continued every four years (Santander, Spain 1990; Manchester, United Kingdom 1994; Udine, Italy 1998; Paris, France 2002; Graz, Austria 2006; Trondheim, Norway 2010). Numerical Methods in Geotechnical Engineering presents the latest developments relating to the use of numerical methods in geotechnical engineering, including scientific achievements, innovations and engineering applications related to, or employing, numerical methods. Topics include: constitutive modelling, parameter determination in field and laboratory tests, finite element related numerical methods, other numerical methods, probabilistic methods and neural networks, ground improvement and reinforcement, dams, embankments and slopes, shallow and deep foundations, excavations and retaining walls, tunnels, infrastructure, groundwater flow, thermal and coupled analysis, dynamic applications, offshore applications and cyclic loading models. The book is aimed at academics, researchers and practitioners in geotechnical engineering and geomechanics.

Anderson, Loren Runar et al "LONGITUDINAL MECHANICS"

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000

Figure 14-1 Longitudinal stress due to decrease in temperature of a restrained pipe.

Figure 14-2 Longitudinal stress due to internal pressure (Poisson effect) in a restrained pipe

CHAPTER 14 LONGITUDINAL MECHANICS

Longitudinal mechanics of buried pipes is the

analysis of longitudinal deformations compared to

performance limits of def ormation One excessive

deformation is fracture for which corresponding

strain limits can be identified If the pipe changes

length enough to shear off appurtenances or to allow

leakage of couplings, deformation is excessive and

corresponding strain limits can be identified If the

strains can be evaluated, then corresponding stresses

can be used as alternative bases for design The

principal causes of longitudinal stress (strain) are:

1 Changes in temperature and pressure, which

cause relative lengthening or shortening of the pipe

with respect to soil and thrust restraints,

2 Axial thrust is the result of internal pressure or

vacuum at "thrustors" (valves, caps, reducers, wye's,

tee's elbows, etc.),

3 Beam bending, which causes flexural stresses

Typical causes of beam bending are:

a) Placement of pipe sections on timbers or mounds

or piers for vertical alignment,

b) Non-uniform settlement of the bedding,

c) Side-hill soil creep or landslide, and massive soil

movement or settlement

Each of the three causes of longitudinal stress is

analyzed separately The results are combined for

analysis It is convenient to separate analyses into

two categories, gasketed pipe sections, and

continuously welded pipes, such as welded steel

water pipes and welded polyethylene pipes

Following are discussions of the three major causes

of longitudinal stresses (and strains) in each of the

two categories

Longitudinally Restrained Pipes

Temperature stresses occur when the pipe section

is restrained at the ends such that it cannot lengthen

or shorten See Figure 14-1 For ordinary buried

pipes, temperature change would have to be unusually large to cause critical longitudinal stress Temperature stress in an end-restrained pipe is:

σT = Eα (∆T) (14.1)

where

σT = longitudinal stress in the restrained pipe

due to change in temperature

E = modulus of elasticity

α = coefficient of thermal expansion

T = change in temperature

In the case of end-restrained steel pipes (i.e., welded joints), in order to cause 36 ksi yield stress in the pipe wall, the change in temperature would have to

be roughly 185oF Gasketed bell and spigot joints avoid the problem if the pipe sections are not too long

T he end-restrained pipe is also subjected to longitudinal tension due to internal pressure See

Figure 14-2 When a rubber band is stretched, the thickness decreases So also does the length of a pipe section when inflated by internal pressure This

is called the Poisson effect Longitudinal tension stress is:

σP = νP'r/t (14.2)

where

σP = longitudinal stress in the pipe due to

internal pressure,

ν = Poisson ratio (in the range of 0.25 to 0.4

for most pipe materials), P' = internal pressure,

r = inside radius of the pipe,

t = wall thickness for plain pipe

Values of σP are not usually critical If steel pipes are capped or plugged and pressurized to circumferential yield stress, P', the longitudinal stress

is only half of yield stress

If the buried pipe is welded and very long, it is

effectively restrained (by soil friction, if not end

restraints) and feels thrust (tension) due to decrease

in temperature and internal pressure

Example

A section of PVC pipe, 12D, DR 26 is installed on a

warm day When put in service as a water supply

pipeline, temperature decreases 50oF and internal

pressure increases to 160 psi What is the

longitudinal stress if the pipe is restrained

longitudinally?

DR = 26 = OD/t,

r/t = 12.5 = (DR-1)/2,

E = 400 ksi = modulus of elasticity,

α = 3(10-5)/oF = thermal coefficient,

ν = 0.38 = Poisson ratio,

P' = 160 psi = internal pressure,

∆T = 50oF

Combining Equations 14.1 and 14.2,

σ = Eα∆T + νP'(r/t)

Susbstituting values,

σ = 600 psi + 760 psi = 1.36 ksi

At sudden rupture, yield stress is σf = 6 ksi At 50

years of persistent pressure, σf = 4 ksi If

restrained, stresses relax over time to less than 1.36

ksi The temperature and Poisson effects are

avoided by gasketed joints if the pipe sections are

not too long If the sections are long, changes in

length due to the temperature and Poisson effects

cause significant longitudinal stresses As a pipe

section changes length, it is partially restrained by

frictional resistance of the soil As with a rope in a

tug-of-war, where tension applied by friction of the

contestants' hands cumulates to maximum at

midlength, so does thrust in a buried pipe section

cumulate to maximum at midlength due to soil

friction That thrust causes a longitudinal stress of,

σ = LHγµ'/2t (14.3)

where

σ = longitudinal stress in the pipe due to

frictional resistance of the soil,

L = length of the pipe section,

H = height of soil above the pipe,

γ = unit weight of the soil cover,

µ' = coefficient of friction, soil on pipe,

t = wall thickness of plain pipe

Some pipeline engineers assume values for the coefficient of friction between granular soil and steel pipes as follows:

For tape-coated steel pipe, µ' = 0.2 For mortar-coated steel pipe, µ' = 0.4 Example

What is the longitudinal stress at midlength of a pipe that shortens after it is buried? The pipe is a large diameter, tape-coated, gasketed steel pipe comprising 120-ft-long sections with 0.50-inch wall thickness µ' = 0.2 H = 6 ft of granular soil cover

at 125 pcf Substituting into Equation 14.3, soil friction can cause longitudinal stress up to 1.5 ksi This is not impressively large, but it may combine with other longitudinal stresses, such as beam bending

Beam Bending See Figure 14-3 If a pipe that is initially straight is bent into a circular arc of radius R, longitudinal strains develop in the outside fibers If R is too short, the pipe buckles; i.e., crumples at a plastic hinge Strain is a better basis for analysis than stress In the following, both strain and stress are analyzed because yield stress is performance limit in some cases Due to beam bending, longitudinal strain and stress (elastic theory) are:

ε = r/R, and σ = Er/R (14.4)

ε = maximum longitudinal strain — tension

outside of bend and compression inside,

σ = maximum longitudinal stress,

r = outside radius of the pipe cross section,

R = longitudinal radius of the pipe axis on the

bend,

E = modulus of elasticity,

R' = initial radius of the bend if the pipe is

manufactured as a curved section

Example

What is the minimum radius of the spool on which

polyethylene gas pipe can be wound for shipping (or

the minimum longitudinal radius for lowering

continuous pipe into a trench)? If allowable strain is

2%, and pipe OD is 2 inches, from Equation 14.4,

R = 50 inches

If the pipe is manufactured initially to some mean

radius R', then Equation 14.4 must take into account

a change in curvature as follows:

ε = r(1/R-l/R'), and σ = Er(1/R-1/R')

The mean radius of the bend can be measured by

holding a cord (straightedge) of known length, s,

along the inside of the bend as shown in Figure 14-4

and by measuring the offset e at the middle of the

cord The mean radius R of the bend is,

R = s2/8e + e/2 + r (14.5)

Knowing R, longitudinal stress (and strain) can be

evaluated from Equation 14.4 The analysis above

is useful for checking the installed radius of the bend

against allowable R can be calculated by

measurements inside the pipe

Longitudinal bending is caused by: 1 soil movement,

and 2 non-uniform bedding

Soil movement is caused by heavy surface loads,

differential subgrade soil settlement, landslides, etc

Soil settlement can often be predicted

Non-uniform bedding is inevitable despite specifi-cations calling for uniform bedding Under soil loads plus weight of the pipe and contents, the pipe deflects and causes longitudinal stress For reinforced pipes, manufacturers provide longitudinal reinforcement and limit the lengths of pipe sections Longitudinal stress can be reduced by corrugating the pipe so that it flexes and conforms with the bedding rather than bridging over soft spots

A bend in the pipe causes a moment, M, for which longitudinal stress is,

where

σ = longitudinal stress,

r = outside radius of the pipe cross section,

I = πtr3 = centroidal moment of inertia of plain

pipe cross section,

t = wall thickness of plain pipe,

M = moment at some point along the pipe

which acts as a beam

Moment M can be analyzed from the loads and supports on a pipe The supports are intermittent hard spots along the pipe If bedding were truly uniform, the pipe could not deflect as a beam However, bedding is never uniform, and so there is always some deflection and some M The maximum

M from beam analysis is substituted into Equation 14.6 For design, the maximum combined longitudinal stress must be less than the strength reduced by a safety factor

With few exceptions, pipes are remarkably stiff beams that bridge over voids and soft spots in the soil bedding If reactions are only at the ends of pipe sections or at midlength, as in Figure 14-5; the moment is maximum at midlength and may be found from the equation, M/wL2 = 1/8 The reactions of

Figure 14-5 are extreme worst cases and are unlikely — one support per pipe section

At the other extreme is a pipe on many closely spaced reactions This is unlikely For example,

Figure 14-3 Longitudinal stress (and strain) in tension and compression due to bending of the pipe into a longitudinal mean radius R

Figure 14-4 Technique for measuring longitudinal inside radius of curvature Ri in a bent pipe (The same technique can be used inside the pipe to find the outside radius of curvature Ro.)

Figure 14-5 Two moment diagrams for the worst locations of one reaction per pipe section and the limits of case I for two reactions per pipe section (These are the maximum possible moments.)

consider a large pipe on rollers on a level concrete

floor, Figure 14-6 As the pipe is shoved on the

rollers, only two rollers carry the load and roll All

others are loose and remain at rest The two highest

of the rollers become the reactions It is easily

demonstrated that the two reactions will have to be

on different sides of the center of gravity of the load

See Figure 14-7 Two supports per pipe section is

the most probable number of reactions

The above rationale reduces the number of reactions

from possibly one reaction, to more probably two

reactions per pipe section Reactions would not be

located at the ends of pipe sections if bell holes are

excavated Vertical alignment could not be

controlled by placing two reactions on the same side

of midlength

If the maximum moment due to probable locations of

reactions can be found, then maximum stress can be

found from Equation 14.6, and requirements can be

established for allowable length of pipe sections and

their longitudinal strength

Figures 14-8 and 14-9 are dimensionless influence

diagrams for likely locations of reactions From the

critical influence number, M/wL2, maximum

longitudinal stress can be calculated

For one reaction per section, the maximum moment

is M = 0.125wL2, and is always at midlength The

influence diagram is not shown

The most probable number of reactions per section

is two These are hard spots assumed to be

concentrated reactions There are an infinite

number of relative locations of two reactions on

each pipe section However, critical locations are

included in the following three cases

Case I Reactions are at or near the ends of the

beam This case is similar to one support per section

discussed above See Figure 14-5

Case II Reactions are at two points B equally

spaced from the ends of the pipe section by a

distance kL See Figure 14-8

Case III Reactions are spaced at L/2 but the left reaction is located a distance X from the left end of the beam See Figure 14-9 It is noteworthy that the maximum moment occurs at X = 0.2L

Pipe on Piles

An example of Case I is a buried pipe on piles (or bents) in order to maintain vertical alignment The need for piles implies that the soil settles If the soil settles with respect to the pipe, the pipe lifts a wedge

of soil on top at a wedge angle of 45o + ϕ/2 For cohesionless soil, the wedge angle is approximately 1h:2v

The load per unit length of pipe is the weight of soil plus the pipe and its contents

Example

A 120-inch steel pipe with 0.75 wall thickness is buried on piles under 6 ft of soil at 120 pcf in a zone where the soil can settle When the pipe is full of water, what is load, w, per unit length of pipe?

Figure 14-6 Large pipe section riding on a number of rollers of which two is the most probable number of rollers actually supporting the pipe at one time

Figure 14-7 Two reactions needed per pipe section for stability

Figure 14-8 Influence diagram of moments for case II reactions which are spaced at equal distances from the pipe ends

Figure 14-9 Influence diagram of moments for case III reactions which are separated by half the pipe length but are at variable distances X from one pipe end

Figure 14-10 Worst-case, simply supported, two-reaction beam (top), and the more probable sine-curve reactions (bottom), for which stresses and deflections are just four-tenths of the corresponding values for the simply supported beam

Figure 14-11 Beam deflection, y, of a pipe section that is supported at its ends

ws = γs(131.23 ft2) = 15.75 k/ft,

ww = γw(πr2) = 4.90 k/ft,

w = 21.6 k/ftw

The above analyses are for concentrated reactions

— idealized wor st cases In fact, reactions are

distributed over finite areas Even if the pipe were

not on bedding between reactions, partial soil support

would occur under the haunches where embedment

falls in against the pipe A rational distribution of

reactions is the sine-curve reaction shown in Figure

14-10 It turns out that,

The moments, strains, stresses, and deflections

for sine-curve reactions are four-tenths of the

corresponding values for concentrated reactions.

Exceptions may occur for adverse installations such

as buried pipes on piers Even though specifications

require uniform bedding and compacted embedment,

it is prudent for pipe manufacturers to limit the

lengths of pipe sections or to provide adequate

longitudinal strength Pipe manufacturer and pipeline

designer should know about longitudinal stresses and

the strength of the pipe required to withstand them

Longitudinal stresses are cumulative — beam stress,

plus axial thrust due to special sections and soil

movement, plus soil friction due to temperature and

internal pressure

Example

A 120-inch water pipe section of 0.75-inch-thick

steel is 120 ft long with gaskets (or slip couplings) at

each end It is buried under 6 ft of soil, but the pipe

section crosses a gulch where soil subsidence is

anticipated What is the stress due to bending? The

span is simply supported σ = Mc/I

where

M = wL2/8,

w = 21.6 kips/ft from the above example,

L = 120 ft,

I = πr3t,

t = 0.75 inch,

c = r = 60 inches

Substituting values, σ = 55 ksi Yield stress is exceeded However, soil subsidence would have to

be 6.6 inches or more With less subsidence, stress

in the beam would be less than 55 ksi Moreover, yield stress is not necessarily failure Allowing for yield, the plastic moment at beam failure is increased

by 50%

Longitudinal Deflection

Longitudinal stiffness of a pipe is EI = dM/dθ; where

M = resisting moment of the beam,

E = modulus of elasticity,

I = πr3t = moment of inertia of the pipe cross

section about its centroidal axis,

θ = angle of circular bend of the pipe as a

beam

Most designers relate stiffness to beam deflection of

a pipe in terms of either, 1 load on the pipe as a beam, or 2 maximum longitudinal stress at yield These analyses provide a feel for how successfully

a pipe bridges over soft spots in the bedding See

Figure 14-11 Example Consider a 120-inch diameter buried steel water pipeline of 60-ft-long gasketed sections under 6 ft of dry soil cover If the soil subsides, what is the deflection at midlength of a simply supported pipe section? Assume that,

y = vertical deflection at midlength,

D = 120 inches = 2r,

t = 0.75 inch,

L = 60 ft,

H = 6 ft,

γ = 120 pcf = 0.12kcf,

w = 21.6 k/ft,

I = πtr3,

E = 30(106) psi,

σy = 36 ksi = yield strength