Numerical Methods in Soil Mechanics 06.PDF

Numerical Methods in Soil Mechanics 06.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 "RING STRESSES"

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000

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

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

CHAPTER 6 RING STRESSES

For preliminary buried pipe design, stress analysis

requires only fundamental principles of pipe

mechanics However, analysis in greater depth is

often essential In all cases, performance limit is

deformation; i.e., rupture, wall crushing, wall

buckling, ring deflection, etc In this chapter,

performance limit is analyzed in terms of stress at

the point of excessive deformation

Hoop Stress

Hoop stress due to internal pressure, P', in a

thin-walled circular ring, from Equation 2.1, is:

where

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

thin-walled pipe for which D/t > 10,

ID = inside diameter,

A = cross-sectional area of the pipe wall per unit

length of pipe = t for plain wall pipe,

t = wall thickness for plain pipe,

c = distance from neutral surface, of the wall to

the most remote surface,

dq = change in curvature = 1/r - 1/ro,

r = deformed radius, ro = original radius,

E = modulus of elasticity

Now suppose that the pipe is not circular — it is

out-of-round before installation — called ovality See

Figure 6-1 In the case of a rigid ring, the maximum

hoop stress occurs at B on the maximum diameter

(ID) This horizontal (ID) is called the span If the

long axis is vertical, ID must be vertical

In the case of a flexible pipe, Figure 6-1, the

maximum hoop stress acts on the maximum

diameter But hoop stress tends to round the pipe

If the ring is initially deformed, the circumferential

stress due to internal pressure, P', is, from Equation

5-2, the sum of hoop stress and flexural stress; i.e.,

s = P'r/A + Ecdq

Because dq is a function of loads on the ring, which could be complex, analyses can be complicated However, for plastic pipes and elasto-plastic (metal) pipes, rupture does not occur until average hoop stress reaches yield Therefore, the flexural component of stress is not an issue Flexure adds to the hoop stress at one surface of the wall, but subtracts from it at the other For flexible thin-walled pipes that are out-of-round, internal pressure tends to round the ring causing soil pressure concentrations However, most pipes are near enough to circular when buried, that rerounding is not an issue

Circular Thick-walled Pipes

Analyses of circular thick-walled cylinders can be found in texts on solid mechanics Thick-walled cylinders subjected to internal or external pressure, feel maximum tangential s t r e s s , s , on the inside of the wall See Figure 6-2

Internal Pressure, P':

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

so = 2P'a2/(b2 - a2) tension on the outside surface,

sav = P'a/t = average tangential stress, where subscripts i, o, and av, refer to inside, outside, and average; and:

s = tangential stress (hoop stress, tension),

b = outside radius,

a = inside radius, P' = internal pressure,

t = wall thickness = b-a

External Pressure, P:

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

so = P(b2 + a2)/(b2 - a2)

compression on the outside surface,

sav = Pb/t = average tangential stress

Example 1

Figure 6-2 (top) shows the cross section of a

thick-walled, high pressure pipe What is the maximum

tangential stre s s , s , if OD = 2ID? From Equation

6.2, si = 5P'/3 = 5/3 rds sav

If pressure is external, the maximum tangential

stress is still on the inside surface See Figure 6-2

(bottom) Because of the greater outside diameter,

stress si is greater due to external pressure than due

to an equal internal pressure However,

compressive yield strength is greater than tensile

yield strength for many pipes

Example 2

If OD = 2ID, Figure 6-2 (bottom), the ring

compression stress due to external pressure is, si =

8/5 the ring compression stress due to equal internal

pressure

Ring Compression Stress

Due to external pressure on a thin-walled pipe, ring

compression stress is,

s = P(OD)/2A,

where

OD = maximum outside diameter,

P = external pressure,

A = wall area per unit length

= t for plain pipes

Example 3

A PVC pipe, DR 41, is a storm sewer under 10 ft of

soil DR = OD/t = the standard dimension ratio

Ring deflection is less than 5% and can be

neglected Dry unit weight of soil is 110 pcf Saturated unit weight is 140 pcf A water table can rise to 6 ft above the top of the pipe What is the ring compression stress, s , in the pipe wall?

s = P(OD)2t = P(DR)/2,

where

P = vertical soil pressure on the pipe

P = 4(110)psf + 6(140)psf = 1280 psf

The ring compression stress is s = 182 psi

Under some circumstances ring compression in the wall is not simply T = P(OD)/2 Consider a pipe with uniformly distributed pressure at the top an d a line reaction (Class D) bedding, on the bottom See

Figure 6-3 Class D bedding is poor practice — but happens For this loading, ring compression thrusts

T occur at A and B even though sidefill pressure is zero Flexure occurs at A and B due to moments

M Shear is zero because the load is symmetrical about the vertical axis Where thrust is known, the ring compression stress is T/A, or T/t for plain pipes Thrusts and moments are functions of loads, as discussed below

Thrusts and Moments in the Ring Thrusts T and moments M can be evaluated by energy methods such as Castigliano's equation for deflections due to loads:

d = (M/EI)(JM/Jp)rdq

See Appendix A d is deflection in the direction of

a dummy load, p, (or dummy moment m for rotation) The dummy load or moment is applied at the point where deflection, or rotation, is to be found Assumptions are:

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

2 The pipe material is elastic

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

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

Example

Complete a force analysis for Figure 6-3

Notation:

D = 2r = mean diameter,

T = ring compression thrust per unit

length,

M = moment in the wall per unit length,

P = vertical soil pressure,

Q = PD = line reaction per unit length,

t = wall thickness,

I/c = section modulus per unit length,

XB/A = horizontal shift of B with respect to

A,

YB/A = change in tangent slopes of B with

respect to A when the ring is loaded

From Figure 6-3 (right), angles q locate points

where thrusts T and moments M complete

free-body-diagrams of segments of the ring and become

unknowns for solution by equations of static

equilibrium plus equations of deflection from the

Castigliano equation

From Figure 6-3, with P known, five unknowns remain to be solved: TB, MB, TA, MA, and Q Because three equations of static equilibrium are available, two additional equations are needed Two equations of deflection are:

YB/A = 0, and XB/A = 0

As the ring deflects due to P, tangents at A and B remain horizontal Therefore YB/A = 0 Point B shifts vertically, but not horizontally with respect to

A Therefore XB/A = 0 These two Castigliano equations for deflection, together with three equations of equilibrium, are solved simultaneously for the unknowns:

TA = 0.1061 Pr compression

MA = 0.5872 Pr2

TB = 0.1061 Pr tension

MB = 0.2994 Pr2

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

This analysis is conservative The theoretical line

reaction Q is always worse than a soil bedding

Horizontal pressure of soil against the pipe provides

some support Measurements of soil stress reveal

deviant stress patterns In general, pressure

concentration shows up on the bottom due to a firm

bedding But this may be reversed if the bedding is

soft and soil is compacted on top of the pipe In

general, pressure reduction shows up under the

haunches because of the difficulty of soil placement

But this may be reversed if concrete or low-slump

soil cement is placed under the haunches or if the

bedding is shaped by a V-cut Compaction affects

soil pressure distribution The more flexible the ring,

the more uniform is soil pressure against the pipe

Vertically compressible sidefill causes concentration

of pressures on the top and bottom of the pipe But

an exception could occur if the pipe were located in

a trench with firm sidewalls that support topfill by

shear reactions However, in time, shear breaks

down due to earth tremors, cycles of wetting and

drying, and changes in temperature As a general

rule, the vertical dead load on top of a flexible pipe

is (OD)gH — called the soil prism load; i.e., the

weight of a soil prism directly above the top of the

pipe where g is the unit weight of soil and H is the

height of soil cover above the top of the pipe This

general rule may require a load factor for rigid pipe

design because the rigid ring may have to support

part of the backfill within the trench if sidefill soil is

not compacted If sidefill is compacted, the soil

prism load may be adequate for rigid ring design

For flexible ring design, the soil prism load is

conservative Normal pressures on the ring are no

greater than pressure, P, at the top because, 1 the

flexible ring conforms with the soil, and 2 the soil is

invariably loose agains t the ring at the interface In

plastic pipes, stress relaxation results in further

reduction of normal pressure of the soil against the

pipe

The stiffer the ring, the greater are the pressure

concentrations on top and bottom when sidefills are

compressible For a rigid pipe, well-compacted

sidefill is necessary if pressure concentrations are to

be avoided

Combined Pressures

The question arises, what are the stresses in the wall

of a pipe subjected to both internal and external pressures? It would seem that external pressure should be subtracted from internal pressure, or vise versa For most installations, however, there will come a time when either internal or external pressure will not be acting Therefore, the ring is usually designed for internal and external pressures separately In the case of the flexible ring, because internal pressure is usually not applied until after the external soil pressure is in place, ring deflection has occurred before the pipe is pressurized If internal pressure is enough to partially re-round the ring, crescent gaps develop between the pipe and the sidefill See Figure 6-4 Clearly the ring no longer needs the support of the sidefill soil to retain its shape However, because of soil pressure on top, the pipe is not completely re-rounded If the specified allowable ring deflection is less than the ring deflection with soil load on top, crescent gaps do not develop Figure 6-4 shows test results for steel pipes It is usually prudent to specify a minimum allowable ring deflection that is less than the value at which gaps would develop according to Figure 6-4 But even if crescent gaps develop, the ring does not collapse for lack of side support when it is depressurized The ring may or may not deflect — and any ring deflection will be less than before pressurization because soil particles tend to migrate into the gaps

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

It may be concluded that internal pressure and external pressure are each analyzed separately

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

Re-rounding is seldom an issue for internal pressure

analysis, if pipes are held to nearly circular shape

when installed For design by ring compression, the

prism load is the most reasonable load The prism

load is usually the total (not just the effective) load

Combined stress analysis is rarely justified, but may

be required for rigid pipes — thick-walled and brittle

based on the familiar equation,

s = T/A + Mc/I,

where

T/A = ring compression stress,

Mc/I = flexural (bending) stress

Example

In one city, acceptance for buried tanks is based on

an internal pressure test when the tank is buried to

the top See Figure 6-5a What are the tangential

force, TA, and moment, MA, at the top, point A?

Shearing force, VA, is zero by symmetry The soil

is compacted sufficiently to prevent ring deflection

Therefore, the ring is fixed at B, Figure 6-5b The

effects of internal pressure and external soil load

can be analyzed separately and then combined by

superposition Figure 6-5c is the free-body-diagram

for internal pressure analysis Figure 6-5d

introduces the procedure for analyzing the effect of

s oil load The moment at C due to the soil load is

Ms It can be found by integrating the element of

soil, shown cross-hatched, multiplied by its lever

arm The result is the moment at C (angle 2), due to

the soil load only The equation is,

Ms = gr3[(1/2)sin2q - (1/2)sinq - (1/4)sin2qsinq

- (1/3)cos3q + (1/3)] (6.4)

The five unknowns of Figure 6-5b require three

equations of equilibrium and two equations of

deformation From deformation, by symmetry, the

relative rotation of A with respect to B is zero; i.e.,

yA/B = 0 The horizontal displacement of A with

respect to B is zero; i.e., cA/B = 0 From Castigliano,

Appendix A, and Figure 6-6a showing a dummy

moment, m, at A in the assumed direction of rotation

of A with respect to B,

yA/B = (M/EI)( M/ m)rdq = 0

M = MA + m - TAr(1-cosq) + Ms at point C M/ m = 1, then m 0 in the M-equation

EI is wall stiffness and r is radius

Substituting into Castigliano's equation, and integrating within the limits for q from 0 to p/2, the first equation of deformation becomes,

MA - 0.3634TAr + 0.0174gr3 = 0 (6.5) The second equation of deformation is zero horizontal displacement of A with respect to B

Figure 6-6b shows a dummy force, p, in the assumed direction of relative displacement of A with respect

to B From Castigliano,

cA/B = (M/EI)( M/ p)rdq = 0

M = MA - (TA + p)r(1-cosq) + Ms at point C M/ p = -r(1-cosq), then p 0 in M-equation

Substituting into Castigliano's equation and integrating within limits for q from 0 to p/2,

the second equation of deformation becomes, -MA+ 0.6240TAr - 0.0320gr3 = 0 (6.6) Equations 6.5 and 6.6 are solved simultaneously for the two unknowns, MA and TA Then by the three equations of equilibrium, MB, VB, and TB a r e evaluated The results are shown in Figure 6-6c Soil load decreases hoop tension at A due to internal pressure by TA = 0.0560gr2

Ring Stress — Uses and Misuses

Stress is one basis of buried pipe design and analysis Most stress analyses are based on theories

of elasticity for which yield stress is the performance limit (failure) Elastic stress analysis is useful in some cases such as hoop stress due to internal pres sure, ring compression stress due to external soil pressure, and the

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