Numerical Methods in Soil Mechanics 22.PDF

Numerical Methods in Soil Mechanics 22.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 "BURIED TANKS AND SILOS"

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000

CHAPTER 22 BURIED TANKS AND SILOS

A tank is generally either a circular cylinder with

end closures, or a sphere See Figure 22-1 Tanks

are buried with the axis horizontal Silos (and basins

and caissons) are buried with the axis vertical Silos

may or may not have closures at the top They are

used as access ways, mine shafts, and bins or

hoppers for storage of ore, coal, aggregate, etc

Bins have an outlet at the bottom for feeding a

conveyor in a pipe or tunnel below Tanks are used

to store fluids, but are also used as bunkers for

hazardous materials

ANALYSIS OF BURIED TANKS

Tanks are designed either for internal pressure or

for external loads For worst-case internal pressure,

external soil support is ignored For worst-case

external pressure, it is usually assumed that the tank

is empty, buried in poor soil with groundwater table

above the tank and possibly a vacuum in the tank

The performance limit is collapse An accurate

model would be complicated Even if a model were

devised, it could not be generalized, and it might

imply greater precision than can be justified Soil

properties are major elements of the interaction, but

cannot be quantified precisely under normal

installation procedures Boundary conditions are

seldom precise Therefore, basic principles of

engineering mechanics provide adequate analysis

Safety factors are required in any case A

free-body-diagram and principles of static equilibrium

help in visualizing tank-soil interaction and in

dispelling misconceptions, some of which arise from

forensic arguments on assessment of responsibility

for leaks in buried tanks Leaks are analyzed in

Chapter 24 See nomenclature, Figure 22-2

Maximum Internal Pressure

For worst-case analysis of internal pressure, it is

customary to neglect support of the embedment

Analysis is based on principal stresses which are

critical on the inside surface of the tank See theinfinitesimal cube in Figure 22-3 Because P' issmall compared with circumferential and longitudinalstresses, it is usually neglected The maximumprincipal stress is the circumferential stress For athick-wall tank (D/t < 10), stress analyses are found

in Chapter 19 For a thin-wall tank (D/t > 10), themaximum principal stress is the circumferentialstress The following applies to thin-wall tanks.Spheres

From mechanics of solids, hoop stress due to internalpressure is only half as great in a sphere as in acylinder of the same diameter and wall thickness.Therefore, for a sphere,

σ = P'(ID)/4A (22.1)

where

σ = hoop stress,P' = internal pressure,

ID = inside diameter,

t = wall thickness,

A = wall cross sectional area per unit length,

A = t for plain cylinder wall

Cylinders (Shells) Cylindrical tanks are short pipes, for which thecritical circumferential stress is,

σ = P'(ID)/2A (22.2)Longitudinal stress is only half as great ascircumferential stress End closures are stiffenerswhich help to resist internal pressure But wheninflated, end closures cause stress concentrations inthe shell-to-head seam

Equation 22.2 is reasonably accurate for design, butsafety factors are essential End closures (heads)are analyzed separately

Figure 22-1 Some typical buried tanks and silos.

Figure 22-2 Nomenclature for steel tanks (above) and buried tanks (below).

Maximum External Pressure

The following are worst-case analyses which

include burial in saturated soil A major concern is

flotation In the following, it is assumed that the soil

cover is sufficient to prevent flotation From

Chapter 21, soil cover should be at least half the

tank diameter The interaction of tank, soil, and

groundwater is complex Strength of soil is

decreased under water and loads must include

external water pressures The tank itself is complex

because of the interaction of shell, heads, risers, and

welds See Figure 22-2 Codes and standards are

available from ASME (boiler code), Underwriters

Laboratory, and ASTM (tank standards)

Recommendations are available from industries

(waterworks, gas, and petroleum) Codes were

developed originally for tanks subjected to uniform

internal pressure For tanks subjected to external

pressure, these codes only cover part of the

requirements for performance

Notation

Geometry:

D = diameter of the circular shell,

H = height of soil cover,

h = elevation of water table,

L = length of the tank (or height of silo),

r = D/2 = radius of the circular cylinder,

t = wall thickness,

A = cross sectional wall area per unit length,

R = radii of curvature of the head in a plane

that contains the axis of revolution

Forces, Pressures, and Stresses:

P = external pressure or internal vacuum,

γ = unit weight of soil,

σ = stress in the tank (Figure 22-3),

Subscripts refer to directions

Spheres

Collapse of spheres due to uniform external pressure

is analyzed by marine engineers for bathyspheres.When soil support is included, analysis iscomplicated Not only are circumferential stresses

in a sphere half as great as in a cylinder, but thevertical soil pressure is less because two-way soilarching action (soil dome) is twice as effective as acylindrical soil arch In the design of buried pipes,the benefit of soil arching action is usually neglected,but is a significant plus for conservative analysis Inthe design of buried spheres, it may be prudent totest or evaluate the arching action of the soil dome.Little information is available on buried spheres

Cylinders (Shells)

A theoretical analysis is available for uniformexternal pressure at collapse of cylindrical tankswith no soil support In fact, soil provides support.Moreover, external water pressure is not uniform,but increases from top to bottom of the tank R.Allan Reese (1993) investigated hydrostaticpressures on the bottoms of horizontal steel tanks atcollapse as the tanks were lowered in water Heconcluded that it is sufficiently accurate to designtanks by assuming uniform external pr essureaccording to an equation from Young (1989),

P = 0.807E(1-υ 2)-3/4(r/L)(t/r)5/2 (22.3)where, in examples that follow:

P = water pressure on the bottom of the tank atcollapse (sudden inversion),

E = mod/elast = 30(106) psi for steel,

t = wall thickness of plain wall,

L = length of the tank,

D = diameter of the tank,

r = radius of of the shell,

υ = Poisson ratio = 1/4 for steel

It is better to use the pi-term (r/t) than the common(D/t) because radius includes out-of-roundness At

Figure 22-3 Principal stresses on the inside of a tank wall sujected to internal pressure P' The same stressanalysis applies for negative internal pressure (vacuum), with reversed signs

Figure 22-4 Uniform external pressure on steel tanks at collapse — graphs of Equation 22.4

some location on the shell, the radius may be greater

than D/2 Designers use Equation 22.3 in the

following form for steel tanks buried in saturated

soil:

P = 72(106)psi(D/L)(t/D)5/2 (22.4)

Poisson ratio for steel is usually about ν = 0.27

Some designers use ν = 0.3 in Equation 22.4 The

differences are not significant; i.e.,

If ν = 0.25, the coefficient is 71.87,

If ν = 0.27, the coefficient is 72.52,

If ν = 0.30, the coefficient is 73.54

If Poisson ratio is increased from 0.25 to 0.30, P

increases by only 2.3% Poisson ratio is often

neglected Without heads to support the shell, from

pipe theory, at D/t = 575, P = 2E/(1-ν2)(D/t)3

Example 1

What is the external pressure on the bottom of an

empty 12,000-gallon steel tank at collapse if the tank

is lowered in water until it collapses? Diameter D =

96 inches, wall thickness t = 0.167, length L = 32 ft,

D/t = 575, and L/D = 4

Substituting into Equation 22.4, collapse pressure is

P = 2.27 psi, which is equal to a depth of water of

5.25 ft above the bottom of the tank This is less

than the diameter of the tank Equations 22.3 and

22.4 are not applicable If wall thickness is

increased to 0.2391 inch, P = 5.56 psi At collapse,

the water surface is 4.83 ft above the tank

Figure 22-4 shows graphs of Equation 22.4

Noteworthy ar e the effects of wall thickness and

length of tank on collapse pressure For comparison,

the bottom graph is pressure at collapse of pipes (no

heads or ring stiffeners) Even though Equations

22.3 and 22.4 are conservative, safety factors a r e

recommended because tanks are never perfectly

circular

Heads (End Closures)

Heads are usually analyzed separately — notcompound head-shell analysis The load is externalhydrostatic pressure plus any internal vacuum in thetank If the water table is above the tank, analysis

is sufficiently accurate if the pressure is assumed to

be the average pressure distributed uniformly overthe head The shapes of heads vary fromhemisphere to dish (concave in or concave out) toflat heads (with or without stiffeners)

Hemispherical Heads

Hemispherical heads are easily analyzed by classicalmethods (Timoshenko, 1956) Circum-ferentialstress is half as great in a hemisphere as in acylinder of equal radius and wall thickness Whenpressurized, the change in radius is not the same forhead and shell For a cylinder, change in radius is(P/E)(r/t)(1-υ /2) For a sphere, change in radius is(P/2E)(r/t)(1-υ ) For equal values of P, E, υ , and r/t,the increase in radius is greater for the shell than forthe hemispherical head If the head fits inside theshell, internal pressure tends to open a gap betweenshell and head To avoid this, for many buried tanks,the head is a cap that fits on the outside of the shell.But, then, the possibility that the head might sheardown past the shell should be investigated

Consider the case of uniform external pressure on asphere and cylinder of equal radii Poisson ratio is1/4 If the hemisphere has the same thickness asthe cylinder, the tendency to decrease in radius is3/7ths as great as the cylinder Stress between shelland head is less if the head is half as thick as theshell See Chapter 24 This is not a priority for mosttank fabricators Discontinuities in stress and strain

at the head-to-shell joint are accommodated by agood fit and a good weld

Some of the incompatibilities of shell andhemispherical head can be reduced by ellipsoidalheads or composite surfaces of revolution (dishes)

P = 0.34 psi

For analysis, an infinitesimal element is isolated by

two meridian cuts and two parallel cuts as shown in

Figure 22-5 This element is a free-body-diagram on

which stresses are related as follows:

σt /rt + σ m /rm = P/t (22.5)

where:

σt = tangential stress in direction of the parallel,

σm = stress in direction of the meridian,

rt = radius of curvature of the parallel,

rm = radius of curvature of the meridian,

P = external pressure on the head,

t = thickness of the head,

S = yield stress

If the head is a hemisphere, from Equation 22.5,

rt = rm, and P = 2St/r

Example 2

Figure 22-6 shows a steel tank with the head

attached to the shell by a reduced meridional radius

of curvature, rm All of the steel is the same

thickness When ring compression stress in the shell

reaches yield, what are the meridional and tangential

stresses at point A in the head?

Given:

t = 0.1875 inch,

rt = 36 inches = D/2 = radius of the shell,

rm = 12 inches,

S = 36 ksi = yield stress

At yield stress in the shell, external pressure is P =

St/rt = 187.5 psi Substituting values into Equation

22.5, P/t = 1 ksi/inch, and

σt /36 + σ m /12 = 1000lb/in3 (22.6)

Equation 22.6 contains two unknowns A second

equation comes from a free-body-diagram of the

head isolated by a cutting plane of a parallel through

point A, Figure 22-6 Equating horizontal forces on

the head,

Pπr2 = σm 2πrt (22.7)

from which σm = Pr/2t = 18 ksi compr ession.Substituting into Equation 22.6, the tangential stress

in the head is σt = -18 ksi tension Stress at A in

the head is of concern because of the opposite sign.Shearing stress becomes critical See Chapter 19.Flat Heads

If the performance limit of a pressurized head isyield stress, analysis can be based either on platetheory or membrane theory, whichever gives thehigher stress However, if the performance limit isdeformation (rupture), the membrane theory is moreresponsive At tensile yield stress in a membranethe entire thickess is at yield stress Rupture isincipient On the other hand, at flexural yield stress

in a plate, only one surface reaches yield stress —not the entire thickness Rupture is not incipient.Surface yielding allows the disk to deform, tendingtoward a membrane with more uniform stressthroughout its thickness Very thick disks (D/t<10),like thick-walled cylinders, justify plate analysis Membrane Analysis of Flat Heads

According to membrane theory, uniform pressure onone side deflects the disk into a segment of a spherewith radius R See Figure 22-7

Let:

E = modulus of elasticity,

r = radius of the disk,

R = radius of the segment of a sphere formed by

pressurizing the disk,

t = thickness of the disk,

δ = deflection of the center of the disk,

ε = tangential strain in the spherical membrane,

σ = tensile stress in the membrane,

θ = half angle of a meridional arc of a sphere,

υ = Poisson ratio

From geometry, the strain due to increase in length

of r from a cord to an arc is:

Figure 22-5 Infinitesimal element of a surface of revolution subjected to external pressure, showing themeridional and tangential stresses and the corresponding radii of curvature.

Figure 22-6 Steel tank head with two different radii of curvature.

Figure 22-7 Geometry for analysis of stress and deflection in a thin-wall flat head (or bottom of a silo)deflected into a segment of a sphere by external pressure P Stress on the welded knuckle is exacerbated

at the bottom where the deflection of head and shell both add to the flexure

ε = Rθ/r - 1 (22.8)

From elastic stress-strain relationship,

ε = σ (1-υ )/E For a sphere, σ = PR/2t Substituting,

ε = PR(1-υ )/2Et (22.9)

Equating the strains of Equations 22.8 and 22.9, and

solving for R,

R[2Etθ - Pr(1-υ )] = 2Etr (22.10)

Because θ = sin-1(r/R), there are only two unknowns

in Equation 22.10, R and P The closed form

solution is messy, so solve for R by iteration

From the elastic theory of thin-walled pressure

vessels, at yield stress, σ = S, and,

which can be solved if R is known But from

Equation 22.9, R is a function of P Therefore, a

second iteration is required to solve Equation 22.10

for P

Example 3

What is the stress in the flat heads of the steel tank

of Example 1? Flat steel heads are usually thin

enough to be analyzed as membranes, which deform

into segments of spheres when subjected to uniform

pressure By membrane analysis, the tensile stress

is σ = PR/2t, where R is the radius of the spherical

segment In Example 1, for an assumed average

pressure on the head (conservatively high), of P =

2.27 psi, R = 48/sinθ, where θ can be found by

equating strains in the sphere; i.e.,

R(θ-sinθ)/r = PR(1-υ )/2Et,

which reduces to

(θ-sinθ) = 3.59(10-6)P/psi

Solving, θ = 2.0958o, R = 1312.5 inches, and tensile

stress in the head is σ = 8.92 ksi If yield stress is 36

ksi, stress in the head is not critical A typical range

of length to diameter for buried 12,000-gallon steel

gasoline tanks is 1.76 < (L/D) < 5 Within thisrange, collapse due to uniform external hydrostaticpressure (plus internal vacuum) starts in the shell.Because heads are membranes in tension, they donot initiate collapse

The head-to-shell weld must be of good qualitybecause the weld tends to flex as the head deforms.This is the worst at the bottom of the tank where theshell is sucked up and adds to the angle of flexure inthe weld and knuckle See Figure 22-7 Considerthe tank of Example 1 Under pressure, the radius

r decreases by 1.2 thousandths for the head and 0.8for the shell The relative difference of 0.4thousandths of an inch is negligible The differencebetween uniform pressure and trapezoidal pressure

on the head is also negligible It is comparable totwo identical beams, one with uniform loading andthe other with triangular loading The difference inmaximum stress is only 2.6% Maximum deflectionsdiffer by only 0.15% When water surface is abovethe tank, the pressure on the head can be analyzed

as the average pressure uniformly distributed overthe head

Plate Analysis of Flat HeadsThick-wall heads (D/t<10) may justify analysis byplate theory The rim of the disk is either fixed orsimply supported; i.e., with a gasket between theplate and the end of the shell

The fixed-edge analysis is most representative oftypical tanks with good connections between headand shell Assume uniform average pressure is Pa v

= P After Timoshenko (1956), the maximum stress

by theory of elasticity occurs at the edge of the diskand is:

σ = (3/4)(r/t)2P (22.12)where equivalent thickness, t, is based on atransformed section The maximum deflection is inthe center of the disk and is: