Numerical Methods in Soil Mechanics 02.PDF

Numerical Methods in Soil Mechanics 02.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 "PRELIMINARY RING DESIGN"

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000

Figure 2-1 Free-body-diagram

of half of the pipe cross

section including internal

pressure P’

Equating rupturing force to

resisting force, hoop stress

in the ring is,

s = P’(ID)2A

Figure 2-2 Common transportation/installation loads on pipes, called F-loads

CHAPTER 2 PRELIMINARY RING DESIGN

The first three steps in the structural design of buried

pipes all deal with resistance to loads Loads on a

buried pipe can be complex, especially as the pipe

deflects out-of-round Analysis can be simplified if

the cross section (ring) is assumed to be circular

For pipes that are rigid, ring deflection is negligible

For pipes that are flexible, ring deflection is usually

limited by specification to some value not greater

than five percent Analysis of a circular ring is

reasonable for the structural design of most buried

pipes Analysis is prediction of structural

performance Following are basic principles for

analysis and design of the ring such that it can

support the three most basic loads: internal pressure,

transportation/installation, and external pressure

See Figures 2-1 and 2-2

INTERNAL PRESSURE —

(MINIMUM WALL AREA)

The first step in structural design of the ring is to find

minimum wall area per unit length of pipe

Plain pipe — If the pipe wall is homogeneous and

has smooth cylindrical surfaces it is plain (bare) and

wall area per unit length is wall thickness This is

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

many plastic pipes

Other pipes are corrugated or ribbed or composite

pipes such as reinforced concrete pipes For such

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

pertinent quantity for design

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

with fluid pressure inside The maximum rupturing

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

ID is the inside diameter See Figure 2-1 This

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

where F is the circumferential tension stress in the

pipe wall Equating rupturing force to the resisting

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

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

s = P'(ID)/2A = S/sf (2.1)

where:

s = circumferential tensile, stress in the wall, P' = internal pressure,

ID = inside diameter,

OD = outside diameter,

D = diameter to neutral surface,

A = cross sectional area of the pipe wall per

unit length of pipe,

S = yield strength of the pipe wall material,

t = thickness of plain pipe walls,

sf = safety factor

T his is the basic equation for design of the ring to resist internal pressure It applies with adequate precision to thin-wall pipes for which the ratio of mean diameter to wall thickness, D/t, is greater than ten Equation 2.1 can be solved for maximum pressure P' or minimum wall area A

A = P'(ID)sf/2S = MINIMUM WALL AREA

For thick-wall pipes (D/t less than ten), thick-wall cylinder analysis may be required See Chapter 6 Neglecting resistance of the soil, the performance limit is the yield strength of the pipe Once the ring starts to expand by yielding, the diameter increas es, the wall thickness decreases, and so the stress in the wall increases to failure by bursting

Example

A steel pipe for a hydroelectric penstock is 51 inch

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

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

at no flow?

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

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

S = 36 ksi = yield strength,

sf = 2 = safety factor,

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

P' = hgw = internal water pressure at outlet

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

0.219 square inches per inch of length of the pipe

Substituting in values, h = 357 ft

TRANSPORTATION/INSTALLATION —

MAXIMUM LINE LOAD ON PIPE

The second step in design is resistance to loads

imposed on the pipe during transportation and

installation The most common load is diametral

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

are stacked or when soil is compacted on the sides

or on top of the pipe as shown

If yield strength of the pipe material is exceeded due

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

cross section of the pipe will permanently deform

Either of these deformations (a crack is a

deformation) may be unacceptable So yield

strength may possibly be a performance limit even

though the ring does not collapse

For some plastic materials, including mild steel,

design for yield strength is overly conservative So

what if yield strength is exceeded? A permanent

deformation (dent) in the ring is not necessarily pipe

failure In fact, the yield strength was probably

exceeded in the process of fabricating the pipe

Some pipe manufacturers limit the F-load based on

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

where D is the decrease in mean diameter D due to

load F Some plastics have a memory for excessive

ring deflection In service, failure tends to occur

where excess ive ring deflection occurred before

installation Increased ring stiffness decreases ring

deflection It is not inconceivable that the ring can

be so flexible that it cannot even hold its circular

shape during placement of embedment One remedy, albeit costly, is to hold the ring in shape by stulls or struts while placing embedment It may be economical to provide enough ring stiffness to resist deflection while placing the embedment In any case, ring deflection is a potential performance limit for transportation/installation of pipes

So two analyses are required for transportion and installation, with two corresponding performance limits: yield strength, and ring deflection See Figure 2-3 In general, yield strength applies to rigid pipes such as concrete pipes, and ring deflection applies to flexible pipes See Figure 2-4

Yield Strength Performance Limit

To analyze the yield strength performance limit, based on experience, pertinent fundamental variables may be written as follows:

F = transportation/installation FL-1

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

D = mean diameter of the pipe, L

I = moment of inertia of the wall L3

cross section per unit length

of pipe,

c = distance from the neutral axis L

of the wall cross section to the most remote wall surface where the stress is at yield point

S = yield strength of pipe wall FL-2

material

5 fv's - 2 bd's = 3 pi-terms

The three pi-terms may be written by inspection A typical set is: (F/SD), (c/D), and (I/D3) This is only one of many possible sets of pi-terms D is a repeating variable Note that the pi-terms are independent because each contains at least one fundamental variable that is not contained in any of the other pi-terms All are dimensionless The interrelationship of these three pi-terms can be

found either by experimentation or by analysis An

example of class ical analysis starts with

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

bending moment in the pipe ring due to load F But

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

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

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

M is the maximum moment due to force F Because

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

compression stress Substituting in values and

rearranging the fundamental variables into pi-term,

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

The three pi-terms are enclosed in parentheses

Disregarding pi-terms,

F = 2pSI/cD = F-load at yield strength, S

For plain pipes, I = t3/12 and c = t/2 for which, I/c =

t2/6 and, in pi-terms:

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

Disregarding pi-terms,

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

pipes (smooth cylindrical surfaces) The modulus of

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

the ring remains circular Only yield strength S is a

performance limit

Ring Deflection Performance Limit

If the performance limit is ring deflection at the

elastic limit, modulus of elasticity E is pertinent

Yield strength is not pertinent For this case,

pertinent fundamental variables and corresponding

basic dimensions are the following:

pipe

F = diametral line load FL-1

per unit length of pipe

per unit length of pipe

where:

D = decrease in diameter due to the F-load,

E = modulus of elasticity,

t = wall thickness for plain pipe,

I = moment of inertia of wall cross section per

unit length of pipe = t3/12 for plain pipe

4 fv's - 2 bd's = 2 pi-terms

Two pi-terms, by inspection, are (d) and (FD2/EI) Again, the interrelationship of these pi-terms can be found either by experimentation or by analysis Table 5-1 is a compilation of analyses of ring deflections of pipes subjected to a few of the common loads From Table A-1, ring deflection due

to F-loads is, (d) = 0.0186 (FD2/EI) (2.2)

This equation is already in pi-terms (parentheses) For plain pipes, for which I = t3/12 and c = t/2, this equation for ring deflection is:

(d) = 0.2232 (F/ED) (D/t)3

The relationship between circumferential stress and ring deflection is found by substituting from Table

A-1, at yield stress, F = 2pSI/cD, where S is yield strength and c is the distance from the neutral surface of the wall to the wall surface The resulting equation is:

(d) = 0.117 (s /E) (D/c) (2.3)

For plain pipes,

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

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

relationship could have been found by

experimentation using the three pi-terms in

parentheses in Equation 2.3 Ring deflection at yield

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

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

not return to its original circular shape when the

F-load is removed Deformation is permanent This is

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

limit with a margin of safety

The following equations summarize design of the

pipe to resist transportation/installation loads

For transportation/installation, the maximum

allowable F-load and the corresponding ring

deflection, d, when circumferential stress is at

yield strength, S, are found by the following

formulas

Ring Strength

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

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

Resolving, for plain pipes,

F = pSD(t/D)2/3

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

is given by:

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

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

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

cylindrical surfaces in terms of yield strength S

Steel and aluminum pipe industries use an F-load criterion for transportation/installation In Equation 2.2 they specify a maximum flexibility factor FF =

D2/EI If the flexibility factor for a given pipe is less than the specified FF, then the probability of transportation/installation damage is statistically low enough to be tolerated

For other pipes, the stress criterion is popular When stre s s s = yield strength S, the maximum allowable load is:

F = 2pSI/cD For walls with smooth cylindrical surfaces,

F = pSt2/3D

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

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

be evaluated Any safety factor could be small — approaching 1.0 — because, by plastic analysis, collapse does not occur just because the circumferential stress in the outside surfaces reaches yield strength To cause a plastic hinge (dent or cusp) the F-load would have to be increased

by three-halves

Plastic pipe engineers favor the use of outside 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 dimensions, the F-load at yield is:

F = pSt/3(DR-1)

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

at yield strength is:

DR = (pSt/3F) + 1

Unreinforced concrete pipes are to be stacked for

storage in vertical columns on a flat surface as

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

is essentially an F-load The following information is

given:

ID = 30 inches = inside diameter,

OD = 37.5 in = outside diameter,

g = 145 lb/ft3 = unit weight of concrete,

F = 3727 lb/ft = F-load at fracture

+ s from tests where,

s = + 460 lb/ft = standard deviation of the

ultimate F-load at fracture of the pipe

a) How high can pipes be stacked if the F-load is

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

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

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

limited to seven pipes in height

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

column is seven pipes high? The seven pipe load at

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

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

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

the probability of failure is 2.2% for the bottom

pipes For all pipes in the stack, the probability is

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

broken pipe for every 317 in the stack

c) What is the circumferential stress in the pipe wall

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

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

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

concrete considering that it fails in tension

EXTERNAL PRESSURE —

MINIMUM WALL AREA

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

external pressure on it See Figure 2-4 The

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

external radial pressure assumed to be unifor mly

distributed OD is the outside diameter The resisting force is compression in the pipe wall, 2s A, where s is the circumferential stress in the pipe wall, called ring compression stress Equating the rupturing force to the resisting force, with stress at allowable, S/sf, the resulting equation is:

s = P(OD)/2A = S/sf (2.6)

This is the basis for design Because of its importance, design by ring compression stress is considered further in Chapter 6

The above analyses are based on the assumption that the ring is circular If not, i.e., if deformation out-of-round is significant, then the shape of the deformed ring must be taken into account But basic deformation is an ellipse See Chapter 3 Example

A steel pipe for a hydroelectric penstock is 51 inches in diameter (ID) with wall thickness of 0.219 inch It is to be buried in a good soil embedment such that the cross section remains circular What

is the safety factor against yield strength, S = 36 ksi,

if the external soil pressure on the pipe is 16 kips/ft2? For this pipe, OD = 51.44 inches, and A = t = 0.219 inch At 16 ksf, P = 111 psi Substituting into

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

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

PROBLEMS 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) Given:

D = 48 inches = inside diameter,

t = 0.064 in = wall thickness,

A = 0.775 in2/ft [AISI tables],

S = 36 ksi = yield strength,

E = 30(10) psi,

sf = 2 = safety factor

2-2 What is the allowable internal pressure if a

reinforced conc rete pipe is 60 inch ID and has two

cages comprising concentric hoops of half-inch steel

reinforcing rods spaced at 3 inches in the wall which

is 6.0 inches thick? (P' = 78.5 psi)

Given:

S = 36 ksi = yield strength of steel,

sf = 2 = safety factor,

Ec = 3(106) psi = concrete modulus,

Neglect tensile strength of concrete

2-3 What must be the pretension force in the steel

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

internal pressure of 72 psi? Leakage through hair

cracks in the concrete appears as sweating

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

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

tension) half-inch steel rods? How about smaller

diameter, high-strength wires? What about bond? How can ends of the rods (or wires) be fixed? 2-5 What is the allowable fresh water head (causing internal pressure) in a steel pipe based on the following data if sf = 2? (105 meters)

ID = 3.0 meters,

t = 12.5 mm = wall thickness,

S = 248 MN/m2 = 36 ksi yield strength

2-6 What maximum external pressure can be resisted by the RCP pipe of Problem 2-2 if the yield strength of the concrete in compression is 10 ksi, modulus of elasticity is E = 3000 ksi, and the internal pressure in the pipe is zero? See also Figure 2-5

(P = 52 ksf, limited by the steel)

2-7 Prove that T = Pr for thin-walled circular pipe See Figure 2-4

T = ring compression thrust,

P = external radial pressure,

r = radius (more precisely, outside radius)

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