Numerical Methods in Soil Mechanics 11.PDF

Numerical Methods in Soil Mechanics 11.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 "ENCASED FLEXIBLE PIPES" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 11-1 Inserted encased pipe (liner) showing how pressure develops between the liner and the encasement Figure 11-2 Basic deformation of the encased, flexible liner showing how the ring is blown against the encasement such that a gap opens and a blister forms in the liner ©2000 CRC Press LLC CHAPTER 11 ENCASED FLEXIBLE PIPES Encased pipes include the following: a) Flexible pipe with concrete cast about it, b) Pipe inserted into another pipe or a tunnel, c) Liner for deteriorated pipe, d) Liner in an encasement in which grout fills the annular space between pipe and encasement All have a common performance limit — ring inversion due to external pressure A special case is a pipe that floats in fluid (water or grout) in the encasement For equations of analysis, see Appendix A Two basic analyses are: liner and double wall LINER ANALYSIS The inside pipe is the liner It is so flexible compared to the encasement that it can be analyzed as a flexible ring in a rigid encasement Performance limit is inversion of the liner Collapse may be time dependent based on "creeping" ring deformation under persistent pressure Collapse is sudden inversion Resistance to inversion is ring stability — a function of yield strength, σf, ring stiffness, EI/r3, and ring deflection, d It is assumed that pressure persists against the ring External Pressure at Inversion of Encased Rings The performance limit is inversion of the liner due to external pressure Because encasements usually leak, if the water table is above the encasement, water pressure builds up between encasement and liner See Figure 11-1 Even if the liner is bonded to the encasement, water pressure peels the liner away from the encasement and deforms the liner as shown in Figure 11-2 A gap forms where the liner is unencased — over an arc no greater than 180o The unencased section is a blister — between circular and approximately elliptical Pressure P at inversion is greater than it would be if the liner were completely unencased But part of the liner is unencased For a completely unencased liner, the classical analysis, from Chapter 10, is, Pc rr3/EI = (11.1) UNENCASED CIRCULAR RING COLLAPSE Pc r applies to a circular ring For a deformed ring, elliptical analysis, from Chapter 10, is: P2 - [σf /m+(1+6mdo)Pc r]P + σ f P Internal Pressure Failure of Encased Rings Internal persistent pressure at fracture is a special case of instability Once the ring starts to yield, the diameter increases and the rupturing force in the wall increases If the liner is designed to take the internal pressure, there is no failure If the rigid encasement is designed to take the internal pressure, there is no failure The liner is an innertube If both liner and encasement expand, each shares in resisting the internal pressure For analysis, the relative resistances of each to expansion by internal pressure must be known It is conservative to design both encasement and liner so that each can take the full internal pressure ©2000 CRC Press LLC cr /m = (10.3) Notation is listed in the paragraph, "Arc Angle 2α." Figure 11-3 shows plots of Equation 10.3 for a plain steel pipe and a plain PVC plastic pipe Because the PVC dimension ratio is DR = OD/t, it follows that 2m = DR-1 for ring flexibility of the PVC pipe From Figure 11-3 it is evident that the effect of ring deflection on P is significant for an unencased ring However, for an unencased blister in a liner, the difference between circle and ellipse is not significant One exception is a slip liner smaller in diameter than the encasement In the following, the unencased section of liner is circular Steel Pipe σf = 42 ksi E = 30(103) ksi PVC Pipe σf = ksi E = 400 ksi Figure 11-3 Examples of external pressure at collapse of unencased pipes with elliptical cross sections ©2000 CRC Press LLC Figure 11-4 is a summary of inversion mechanisms for encased rings If the gap is small (blister does not form) ring compression applies If a blister forms over the gap, it acts either as an arch o r a beam Analysis of beam failure is classical An unconfined arch could potentially invert If the gap is small, arch instability is analyzed for a circular arc If the gap is large, the arch is elliptical A simple, but conservative analysis is circular analysis using the maximum radius of curvature of the ellipse Arc Angle 2α An unknown is the arc angle, 2α Figure 11-5 shows the collapse of a flexible, circular, hinged arch From Timoshenko (1956), Pr3/EI = (π/α) - Points B (circles) are points of tangency to the encasement — and so moment is zero Points C (circles) are points of counterflexure — and so are hinge points — zero moment Points ∆ (triangles) are plastic hinges — points of maximum moment that isolate the collapse mechanism in arc, β Points B, C, and ∆ are about equally spaced Therefore, β = 2α, which is equivalent to the arc angle of Figure 11-5, and can be analyzed by Equation 11.2, from which critical P is, (11.2) Notation: P = pressure on the blister at inversion, D = mean circular diameter of the liner, r = mean radius of the circular liner, t = wall thickness, m = r/t = ring flexibility, E = modulus of elasticity, σf = yield strength of the liner, ∆ = decrease in minimum diameter, d = ∆/D = ring deflection, T = circumferential thrust in liner wall, M = moment due to ring deformation, α = half arc angle (Figure 11-5), β = blister angle (gap angle), σ = maximum circumferential stress, h = height of water table above the liner, H = height of soil cover over the pipe, I = moment of inertia of wall cross section Equation 11.2 can be applied to an encased flexible ring by selecting a portion of the ring that is equivalent to the circular hinged arch of Figure 11-5 Figure 11-6 shows the typical inversion of a flexible liner The blister can be seen developing in the blister arc, β At the ends of the arc, the moment is maximum, and plastic hinges can be seen developing These become the gunwales of an ©2000 CRC Press LLC inverted "boat hull" that rises up into the pipe A third plastic hinge forms at the keel Points of zero moment are circles Points of maximum moment are triangles P = E[(π/α) 2-1]/12m3 (11.3) Angle α is unknown From tests, the arc angle for plastic liners is roughly, α = 30o to 45o Equation 11.3 neglects decrease in circumference due to ring compression In order to find α, worst case assumptions are as follows: Assumptions: There is no bond, interlocking, or frictional resistance between liner and the encasement There is no pressure inside the liner The liner is subjected to external pressure P Leaks in the encasement allow pressure on the liner due to groundwater table External pressure includes any vacuum that may occur inside the liner The liner is flexible Initially, it fits snugly against the encasement But it may shrink, leaving an annular space between liner and encasement The liner is snug (but not press-fit) in the encasement The liner may be plastic which can creep under persistent pressure over a period of time The cross section of the blister is circular Third-dimensional (longitudinal) resistance to the formation of a blister is neglected Figure 11-4 Inversion mechanisms for a blister in a liner ©2000 CRC Press LLC Figure 11-5 Collapse of a hinged circular arch subjected to uniform radial pressure, P Figure 11-6 Typical inversion of a flexible liner at critical pressure P β = blister angle ©2000 CRC Press LLC When subjected to external pressure, the liner shrinks Bond breaks down between liner and encasement External pressure, P, distributes itself around the entire surface of the liner At one location a gap opens between liner and encasement, and a "blister" forms in the liner The blister develops wherever the radius is maximum, or at the bottom of the liner where external hydrostatic pressure is greatest Failure is inversion of the blister From inside the pipe the inversion looks like a boat hull with the keel longitudinal Ring compression thrust, Pr, is constant all around the ring Assuming P is constant, ring compression thrust is Pry , where ry is the greatest radius of curvature Analysis is prediction of minimum pressure P at inversion See Figure 11-4 The rationale is to calculate the decrease in perimeter of the liner, and to find r y as a function of the arc angle It is then possible to find P, both when ring compression stress is at yield, and when the blister inverts The lesser of these two P's is critical c) Beam failure if the blister is flat The ends of the beam are the plastic hinges at the edges of the blister The moment capacity of plastic hinges is, Mp = 3Me /2, where Me is the elastic moment at yield stress; i.e., Me = σf I/c Wall Crushing Analysis The liner buckles when ring compression stress equals yield; i.e., σ = σ f, where σ = Pr/A For plain liners, P = σf /(ry /t) (11.4) where ry is the maximum radius (which occurs at the blister) where the maximum stress, is Pry /t See Figure 11-7 (top) But Pr/t is constant all around the ring Therefore, the decrease in perimeter of the liner due to ring compression strain is, δ = 2πrPr y /Et (11.5) The geometrical decrease in perimeter of the liner is: Notation: r = constrained mean radius of the liner ring, ry = maximum radius of the blister, OD = outside diameter of the unpressurized liner, t = thickness of the liner wall, A = area of the liner per unit length of pipe, A = t for a plain liner, δ = decrease in circumference of the liner ring, DR = dimension ratio of the liner = OD/t, α = half arc angle based on casing radius r, β = half arc angle based on blister radius, r y , E = modulus of elasticity of the liner, σ = ring compression stress in the liner, σf = yield stress, (at 50 years of persistent pressure?) Inversion is by one of three mechanisms shown in Figure 11-4: a) Wall crushing if ring compression exceeds yield, b) Arch inversion, ©2000 CRC Press LLC δ = 2rα - 2ryβ (11.6) The blister width is AB = 2rsin α = 2r ysin β , from which ry = rsinα/sinβ Equating the decreases in perimeter from Equations 11.5 and 11.6, β/sin β = α/sinα - πσ f /Esin α (11.7) which can be solved by iteration for β in terms of α Of interest is angle α at flat blister (straight beam at β = 0), for which β/sin β = From Figure 11-7, blister width AB = 2rsin α = 2rysin β Substituting the resulting ry = rsin α /sin β into Equation 11.4, critical pressure is, P = σfsin β/(r/t)sin α (11.8) Figure 11-7 (bottom) shows pressure P = f(α ) The inversion sequence starts with ring compression yielding (wall crushing), P = σf /(r/t), at α = β A blister forms at yield stress, σf With its greater radius, ry, the blister rises and the blister angle decreases But yielding is not inversion Yielding progresses down the wall crushing curve from right to left — shown by arrows Based only on wall crushing, the blister inverts at the least value of α In the example of Figure 11-7, α = 33o But before the blister inverts by ring compression, it might invert by arch inversion or by beam failure In Figure 11-7, the arch inverts at α = 34o Arch Inversion Analysis From Equation 11.3, for a hinged circular arch, critical P is a function of arch angle α From Figure 11-6, α is a third of the gap angle, B-B, which cannot exceed 180o The plastic pipe example of Figure 11-7 shows arch inversion as a heavy solid line P increases as α decreases But α decreases only by ring compression strain (down the wall crushing curve) So α is critical where wall crushing intersects the arch inversion curve Inversion occurs at α = 34o, and P = 57 psi For very conservative analysis, set α = 60o (one-third of the 180o upper limit of gap angle) Beam Failure Analysis For low strength materials, inversion may be beam failure At inversion, the blister cross section is a fixed-ended beam With the length known from radius r and angle α, pressure P can be found from the equation of stress, σ = Me c/I where M = PL2/12 for a fixed-end beam (11.9) But plastic hinges form at Mp = 3Me /2, Substituting, ©2000 CRC Press LLC 9σ f = 2Pr2sin 2α /(I/c) (11.10) For plain liners (no ribs), I/c = t2/6, and (r/t) = m Substituting into Equation 11.10, 3σf = 4P(r/t)2sin 2α, from which, at inversion P = 3σf /4m2sin 2α (11.11) Example A hypothetical plastic liner has the following properties: DR = 51, m = r/t = 25, E = 400 ksi = virtual modulus of elasticity over 50 years of persistent pressure, σf = ksi = yield strength at 50 years of pressure Find: Persistent pressure P that would cause inversion at 50 years of service From Equation 11.7, due to ring compression strain, β/sin β = α /sin α - πσ f /Esin α For trial values of α, corresponding values of β are found by iteration Of course, ring compression stress approaches infinity when the blister is flat; i.e., when β/sin β = Equation 11.7 becomes, α /sin α - πσf /Esin α = Solving, at inversion, α = 33o But inversion occurs by arch inversion or beam failure before α shrinks to 33o From Equation 11.8, P = σf sin β/msin α (11.8) Figure 11-7 is a plot of Equation 11.8, showing longterm critical P as a function of α Quick wall c rushing would occur at P = 160 psi as show n Over the long term, the plastic ring creeps, α is reduced, and, therefore, inversion pressure, P, is reduced The amount of creep is found from longterm tests For most plastics, failure is arch inversion From Figure 11-7 inversion occurs at α = 34o Critical pressure is, P = 57 psi Figure 11-7 Example of graphs of critical pressure, P, as a function of the half blister angle, α, for three collapse mechanisms in a typical plastic pipe Yielding (wall crushing) may be time dependent because of plastic creep ©2000 CRC Press LLC Only in rare cases, such as low strength or undersized liners, will inversion be beam failure As an exercise only, if a blister in the liner could flatten into a beam, at α = 33o from Figure 11-7, from Equation 11.11, P = 16.2 psi at 50 years of persistent pressure P E σf DR m ν The above procedure provides an approximate inversion analysis for liners It is a limit analysis Critical pressure at inversion is greater than the c alculated pressure because of longitudinal resistance, bond, etc Compared to tests, analysis is conservative In the short term, critical pressure is P = σf /m = 176 psi From eight tests, the critical pressure was 172 psi with a standard deviation of 38 psi The analysis allows for modifications and innovations For example, when shrinkage of the liner is due to conditions other than external pressure, such differences can be included in the analysis One modification is the virtual modulus of elasticity E (not actual modulus — but, mistakenly, referred to as long-term modulus) that allows for creep of the liner over a long period of time This creep causes a decrease in perimeter of the liner under constant pressure P = E/4 m3(1-ν 2) = 25 psi If the liner is initially out-of-round, a gap will form where the radius is maximum Using maximum radius instead of the circular radius, analysis of the liner can proceed Radius of curvature can be calculated from a measured offset from the midlength of a cord placed across the curved section of liner The procedure can be programmed for computer It can also be presented as tables and graphs which, for most overworked engineers, may be the best presentation = external pressure at inversion, = 420 ksi = modulus of elasticity, = ksi = yield strength, = 35 = standard dimension ratio, = 17 = r/t = (DR-1)/2, = 0.38 = Poisson ratio If unencased, the maximum external collapse pressure is only Clearly the observed collapse pressure is much closer to ring compression theory than to the hydrostatic collapse theory It is noteworthy that the test liners failed by bulging inward throughout a blister angle less than about β = 90o in Figure 11-6 For a blister angle of 60 o , α = 30o From Equation 11.3, arch inversion occurs at P = 107 psi But this is based on E = 420 ksi Over the long term, if virtual E were only two-thirds as great, critical pressure would be only 71 psi COMPARISON OF ANALYTICAL METHODS The following is a rough comparison of some methods proposed for analyzing the constrained flexible pipes subjected to external pressure Soil contributes significant constraint to the buried flexible ring subjected to uniform external pressure Two equations used in service for design of rings that are circular or nearly circular (encased), are as follows Plastic pipes are often used as liners for rehabilitating damaged pipelines, usually in a full contact fit in the damaged encasement a) One form of the AWWA C950 formula in AWWA-M11 (1989) is, Example A folded PVC pipe is inserted, heated and inflated to become a liner in an ID encasement Find critical pressure P P2 = 0.593RwEsEI/0.149r3 where Rw = buoyancy factor = 1-0.33(h/H), Es = soil stiffness = secant modulus ©2000 CRC Press LLC b) The other formula, proposed by the Large Diameter Pipe Division of PPI, is similar to Equation 41 in the Uni-Bell Handbook of PVC Pipe (1986), formula It is as follows: P2 = 1.3225EP c r where Pc r = E(r/ry)3/4(1-ν2)m3, Pc r = collapse pressure of unconstrained pipe, E = modulus of e l a s t i c i t y a t o p e r a t i n g temperature of the pipe, t = mean wall thickness, D = mean diameter = 2r, r = radius assuming no ovalization (out-ofroundness or initial ellipticity), ry = maximum radius of curvature of the ovalized pipe, ν = Poisson ratio, Et = soil stiffness (tangent modulus) of the embedment where P = KEt Et m = = = P2 = 0.38KE tE/m3 pressure at collapse of the circular ring, horizontal tangential soil modulus, vertical tangential soil modulus, r/t = ratio of mean radius and wall thickness = ring flexibility term Four Additional Equations For Comparison If the ring is encased in rigid or relatively rigid soil, then the equation for critical uniform external soil pressure is P = σf /m These two formulas are almost identical For the following comparison, it is assumed that: m = r/t, Rw = 0.67 = worst case with water table at the ground surface, I = t3/12 for plain pipe, ry = r (circular cross section), ν = 0.38 = Poisson ratio for PVC Substituting these values, where σf = ENCASED yield strength of the pipe wall This is ring compression failure If the ring is unconstrained; P = E/4(1-ν2)m3 UNCONSTRAINED This is buckling failure by classical analysis P2 = 0.3748EsE/m3 AWWA P2 = 0.3864EtE/m3 UNIBELL If the secant soil modulus Es is about the same as the tangent soil modulus Et, the two equations are essentially identical Certainly the concepts are the same Weaknesses can be found in both, but most troublesome is reconciliation of basic concepts and limits For example, if either of the soil moduli Es or Et or the pipe modulus E approaches zero, then pressure P approaches zero Not so However, there may be a range in which the formulas are accurate For purposes of comparison, these two formulas are combined into one called the service ©2000 CRC Press LLC SERVICE FORMULA If the ring is supported by soil in which the effective cross section stiffness is back-calculated from ring deflection tests, P = 0.14E/m3 + 0.27KE t EMPIRICAL If ring deflection can be predicted by the Iowa formula, then an effective cross section stiffness is available and P = 0.22E/m3 + 0.16KE t IOWA FORMULA If ring deflection is known in terms of load or can be calculated by a formula such as the Iowa formula, in terms of load, the effective pipe stiffness F/∆ can be calculated From F/∆ , ring stiffness EI/D3 is 0.0186F/∆ The load F is a parallel plate load, but F/∆ can be related to other loads such as the loading assumed in the Iowa formula or the empirical (measured) soil pressures in ring deflection tests From these relationships, the effective ring stiffness EI/D3 can be calculated for the effective cross section stiffness A comparison of the service formula with the other four is shown on Figure 11-8 This particular comparis on is for a circular steel pipe buried in soil for which KEt = 700 psi The service formula falls in the middle of the pack The encased equation is a limiting condition represented by the curve in the upper right-hand corner for yield strength of 42 ksi The other limit is unconstrained The empirical and Iowa formulas traverse from an asymptotic unconstrained on the left to an asymptotic encased on the right This traverse is reasonable and indicates the increasing support of the soil as the ring flexibility term D/t increases The service formula is less responsive Limited testing of vacuum to collapse of buried pipes seems to confirm the empirical equation Nevertheless, the service equation is preferred by designers who are wary of soil placement and would like an increased safety factor whenever the embedment must be depended upon for support of the pipe More precise methods are available for evaluating P See Chapter 10 DOUBLE-WALL PIPES AND TANKS Double-wall pipes comprise a pipe within a pipe The objective is usually to eliminate leakage from the outside pipe, but may also be to reduce frictional resistance to fluid flow Deteriorated pipes are sometimes rehabilitated by inserting pipes of smaller diameter called slip liners In the following discussion, the slip liner is the "pipe." See Figure 11-9 The host pipe is the "casing." The same nomenclature applies to double wall ©2000 CRC Press LLC tanks, often called dual containment tanks They are manufactured with double walls to provide double protection against leaks, and to provide a sensitive means to detect leakage from the inside tank before the soil is contaminated A "sniffer" monitors the space between tanks Dual containment tanks are used extensively for underground storage of petroleum products (service stations) and hazardous products Due to fluid pressure between the pipe and the casing, the pipe is forced to one side of the casing leaving a gap shown in Figure 11-10 Usually, but not always, the gap is on the bottom where external liquid pressure is the greatest, and where the pipe may have an increased radius of curvature due to its own weight The following assumptions are conservative Assumptions The pipe is flexible As the pipe is forced to one side of the casing, if external pressure, P, increases, the contact angle increases until the pipe is in contact with 180o of casing At 180o of contact, inversion of the flexible pipe is incipient Above 180o, the contact angle increases toward inversion of the pipe with little or no increase in pressure In fact, no pipe is perfectly flexible Beam action and shearing resistance can help to resist the external pressure Performance limit is incipient inversion of the pipe Resistance is ring stiffness — a function of moment of inertia of the wall cross section and modulus of elasticity of the material External pressure is distributed all around the pipe This implies that any bond between the pipe and casing is broken down — a conservative assumption The casing is circular The casing may increase in radius due to pressure, P, between casing and pipe This occurs in some dual containment tanks Figure 11-8 Comparisons of the SERVICE FORMULA and four other equations for evaluating the critical uniform external pressure at incipient inversion of a circular, cylindrical, steel pipe The buried formulas are based on a horizontal soil modulus of KEt = 700 psi Figure 11.9 Pipe used as a liner in a circular casing ©2000 CRC Press LLC Figure 11-10 Pipe subjected to external pressure, forced into 180° of contact with the casing, on the verge of incipient inversion A gap forms over an approximate ellipse in which maximum and zero moments are equally spaced as shown Arch A-A can be analyzed as a circular hinged arch That portion of the pipe in the gap (not in contact with the casing) is assumed to be compressed into a semi-ellipse See Figure 11-10 Conservative analysis is based on a 180o gap Points of maximum and zero moment are equally spaced around the semi-ellipse As a consequence, arc A-A is a 60o hinged arch that is assumed to be circular in order that it can be analyzed by classical methods A I α a rc = = = = = wall cross-sectional area per unit length, moment of inertia of wall cross section, half arc angle of the critical hinged arch, minor semi-diameter of the ellipse, major semi-diameter of the ellipse Equations The perimeter of the ellipse is π(a+r c) From the ellipse, Figure 11-11, ry = rc2/a Notation P = critical external pressure on pipe, r c = inside radius of the circular casing, rp = radius of the originally circular pipe, ry = maximum radius of the ellipse, E = modulus of elasticity, σf = yield stress, DR = dimension ratio, t = wall thickness, ©2000 CRC Press LLC From Timoshenko (1956) and Figure 11-10, Pry3/EI = (π/α) - For arc A-A, critical α = 30o; Pry3/EI = 35 From geometry, the decrease in perimeter of the pipe is, 2πr p - πrc - (π/2)(a+rc) From ring compression, the decrease in perimeter of the pipe is, 2π rp (Pry/EA) Equating decreases in perimeter, and solving, a = 4rp (1-Pry /EA) - 3rc But from geometry of an ellipse, a = rc2/ry Equating values of a and rewriting, ry2 - (EA/P)(1-3rc /4rp)ry = - (EA/4P)(rc2/rp) (11.12) Equation 11.12 is a quadratic equation from which ry can be evaluated for any assumed value for P Analysis The critical analysis is the evaluation of pressure, P, at inversion of the pipe If ry is known, P can be found At α = 30o, Pry3/EI = 35 (11.13) Maximum radius of curvature, ry, is found from Equation 11.12 Unfortunately, ry is a function of P The easiest analysis is by iteration Assume a value, P' The first assumption for P' may be found from Equation 11.13 assuming that ry is equal to rc From Equation 11.12, the correct value for ry can then be found for the assumed P' Substituting this ry into Equation 11.13, it is possible to solve for P at radius r y The solved P is undoubtedly different from the original assumed P' They must be equal Therefore, the next iteration is a repetition of the process assuming that the next P' is the previously solved P Iterations continue until assumed P' equals solved P The procedure is as follows: The "knowns" are values of rc, rp, E, and t Assume a value for P Call it P' Try P' from Equation 11.13, if ry = rc ©2000 CRC Press LLC Solve for ry in quadratic Equation 11.12 Substituting ry into Equation 11.13, solve for P The solved value of P becomes the assumed P' for the next iteration The process is repeated until the solved value of P is equal to the assumed value, P', which is the critical pressure at incipient inversion of the liner If pres sure is persistent over the long term, plastics creep The quantity EA/P in Equation 11-12 must then be based on the "virtual" modulus of elasticity — not the real modulus In the literature, virtual modulus is defined, erroneous ly, as "long-term" modulus of elasticity It is possible to change the radius of the casing, rc, and to find corresponding value of P as described above I f rp remains constant, P vs rc can be plotted If the casing remains circular, but is expandable, the expansion of the casing can be plotted on the same coordinate axes of the graph of P vs rc The intersection of the two plots is critical pressure, P For some tanks, such as fiberglasswrapped dual-containment tanks, the casing is so flexible that it does not remain circular It expands under internal pressure, but is deformed by the pipe (liner) which is forced into a semi-ellipse The liner exerts non-uniform pressure on the casing such that Pxrc = Pry For dual-containment pipes and tanks with ribbed liners that are made of yield-sensitive materials such as fiberglas, Timoshenko (1956) suggests the Southwell solution, from which critical P is, Pr/h = σf/[1+4(σ f /E)(r/h)2] (11.14) where h is the height of the ribbed section from inside diameter to outside diameter Of course, the ribs could be corrugations, etc Example What is the external pressure, P, at incipient inversion of a PVC pipe encased in a circular casing with no gap before pressure is applied? For the pipe, rp E σf DR t A I α EI EA = = = = = = = = = = 15 inches outside, 400,000 psi, 4,000 psi, 41, 0.732 inch, 0.732 in 2/inch, 0.032685 in = t3/12, 30o (assumed critical), 13,074 lb in, 292,800 lb/in radius of casing, rc, and radii of the pipe, rp Following procedure of Example 1, if rp = 15.0 and r c = 17, P = 13.4 psi Figure 11-13 is a plot of values The assumptions that: the pipe is elliptical in the gap, that arc angle α is 30o, and that inversion is only two-dimensional, all reduce the precision of the values for P Therefore, a safety factor is needed Until test data are available, a safety factor of two is suggested for design With no gap, rc = rp = 15 inches Assume a value for P' If ry = rc, then from Equation 11.13, P' = 457,594/ry2 = 135.58 psi As the first assumption, try P' = 130 psi REFERENCES AWWA (1989), Steel Pipe — AWWA Manual M11, 3ed, American Water Works Association Assume P' = 130 psi From Equation 11.12, ry = 15.42 inches Equation 11.13, P = 124.71 psi From Assume P' = 124.71 psi From Equation 11.12, r y = 15.41 inches Equation 11.13, P = 125.08 psi From Timoshenko (1956), Strength of Materials, Part II, ed, D.Van Nostrand, p 189 Assume P' = 125.08 psi From Equation 11.14, ry = 15.41 inches From Equation 11.13, P = 125.05 psi Only two iterations are necessary P = 125 psi at rc = rp It is noteworthy that, based on ring compression at yield stress (σf = 4000 psi), critical pressure is P = 195.2 psi For ring compression failure, the pipe must be restrained in a circular cross section With no restraint, the pipe will collapse if Pr3/EI = Solving, P = 11.62 psi If, hypothetically, the PVC were yield sensitive (which it is not), critical pressure from Equation 11.14 would be P = 11 psi If the pipe wall were ribbed such that h = 1.5 inch, then P = 80 psi Uni-Bell (1986), Handbook of PVC Pipe, Uni-Bell PVC Pipe Association PROBLEMS 11-1 What is the pressure at collapse of a PVC pipe DR 26 encased in concrete? Include the decrease in circumference of the pipe Assume the pipe is circular before the pressure is applied, and α = 30o For PVC, E = 500 ksi, and σf = ksi 11-2 Derive Equation 11.3 11-3 ID of an encasement is 8.5 inches Assuming strain due to creep is 5% percent over 50 years, what must be the DR for an 8D inch OD PVD liner if external pressure on it is 25 psi? E = 400 ksi virtual 50-year modulus σf = ksi = 50-year strength Example Find P if rc is greater than rp The pipe is smaller than the casing to allow for insertion Figure 11-12 provides a "feel" for the difference between the ©2000 CRC Press LLC 11-4 Plot a graph of external collapse pressure as a function of collapse angle for a circular HDPE pipe DR 32.5 encased in concrete Figure 11-11 Quadrant of an ellipse, showing notation for analysis Figure 11-12 Sketch to scale of a pipe of 15- inch radius in casings of 15- 16- and 17-inch radii Figure 11-13 Examples of critical external pressure on a PVC pipe in a circular casing with larger inside diameter than the outside diameter of the pipe The pipe is at incipient inversion when subjected to external pressure, P ©2000 CRC Press LLC 11-5 Figure 11-14 is a standard egg-shaped sewer with a plastic liner The sewer leaks and the water table rises 26 ft above the sewer What is the width, e, of the gap at C ? At 26 feet of head, P ro = = 11.27 psi, 36 inches, t E' = = 0.945 inch, 250,000 psi = virtual modulus at 50 years of persistent pressure, e = mid-ordinate width of gap, Circumference = 100 inches, Arc length BCD = 22.8 inches Assume circular deflection of arc BCD (e = 0.46 inch) Figure 11-14 Cross section of standard egg-shaped sewer showing (right) the external pressure on a liner ©2000 CRC Press LLC ... 2rsin α = 2rysin β Substituting the resulting ry = rsin α /sin β into Equation 11.4, critical pressure is, P = σfsin β/(r/t)sin α (11.8) Figure 11-7 (bottom) shows pressure P = f(α ) The inversion... that may occur inside the liner The liner is flexible Initially, it fits snugly against the encasement But it may shrink, leaving an annular space between liner and encasement The liner is snug... ring stiffness, EI/r3, and ring deflection, d It is assumed that pressure persists against the ring External Pressure at Inversion of Encased Rings The performance limit is inversion of the liner