Numerical Methods in Soil Mechanics 19.PDF

Numerical Methods in Soil Mechanics 19.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 "STRESS ANALYSIS" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 19-1 F-load (parallel plate load) on a circular ring showing maximum principal stresses σx and σ y at points A and B The parallel plate load is the basic test for failure due to external forces ©2000 CRC Press LLC CHAPTER 19 STRESS ANALYSIS Pipes are essential components of life If quality of life is to improve, it will be partly because of better piping systems that are fail-safe over longer servic e life Pipe failure can be catastrophic Failure analysis is essential — not only for repair, assessment of damage, and allocation of responsibility; but for improvements in pipes and pipe systems of the future Some (not all) failures can be stress analyzed See Figure 19-1 Stress failures of flexible pipes by bursting are analyzed in Chapter Wall crushing is analyzed in Chapter For these cases, reconstruction of failure is simple Not so simple is the reconstruction of a rigid pipe failure Under some conditions, flexible pipes fail as rigid pipes For example, bursting of a flexible pipe sometimes occurs so suddenly as to cause brittle fracture instead of plastic yield Water hammer is often the cause Bursting due to internal pressure may be affected by longitudinal stress and/or external loading Failure usually occurs at a joint or an appurtenance where stresses may concentrate or where strength of material may be deficient For purposes of design, analyses are based on simple mechanics of stress and deformation But for failures, combined or c ompound stress analyses, with corresponding performance limits, may be required In fact, stress theory is not the only theory for failure analysis Nevertheless, questions about failure always seem to arise in terms of stresses and strains and performance limits for tri-axially stressed elastic materials Better theories are becoming available for plastic and stress-regressive materials Following are a few of the most common combined and compound elastic stress analyses COMBINED STRESS ANALYSIS Combined analysis is the addition of all stresses acting in the same direction at a point These stresses may be either normal or shearing stresses ©2000 CRC Press LLC The failure (critical) stresses are of greatest interest Figure 19-1 shows an F-load (parallel plate load) Consider point A on the inside surface under the Fload The maximum principal stress is caused by flexure, σx = Mc/I where σx = maximum principal stress M = moment at section A I/c = section modulus of the pipe wall per unit length of pipe A = cross-sectional area of the pipe wall per unit length of pipe P' = internal pressure (or vacuum) P = vertical external soil pressure D = mean diameter of the pipe = 2r t = wall thickness for plain pipe wall (smooth cylindrical surfaces) c = distance from neutral surface of the pipe wall to the most remote fiber = t/2 for plain pipe d = ∆/D = ring deflection ∆ = decrease in vertical diameter S = normal yield strength S' = shearing yield strength τ = shearing stress E = modulus of elasticity C = cohesive strength of the materials But now suppose the pipe is subjected to internal pressure P' in addition to the F-load Clearly σ x is increased by the additional stress P'r/A At the crown, σx = P'r/A + Mc/I POINT A This is combined stress The internal pressure could be negative; i.e., vacuum or external pressure But is stress σx at A the critical stress? At point B ring compression stress due to the F-load is felt With all stresses included, the critical combination of stresses might occur at point B, Figure 19-2 Procedure for superimposing the orientation diagram on the stress diagram in order to evaluate stresses on various planes through infinitesinal cube ©2000 CRC Press LLC where ring compression stress F/2A due to the Fload, external fluid pressure P'r/A, and flexure stress Mc/I all combine in a maximum negative (compression) principal stress At springline, σy = Pr/A + P'r/A + Mc/I POINT B This stress at B could be more critical than the maximum stress at A, depending on tensile and compressive strengths of the pipe wall However, loads not always occur simultaneously For example, when an internal vacuum occurs, the ring cross section shrinks slightly and reduces ring compression stress As long as the ring is held in shape by the soil, the flexure stress is zero For analyzing stress, a more useful formula than σ = Mc/I, is the approximate formula for σ based on ring deflection At springline, σy= 8Ecd/D The coefficient is conservative In fact, it varies as a function of ring deflection If the ring remains elliptical, the coefficient varies: from to to 6.7 at d = 5% 7.5 at d = 10% 8.0 at d = 12.5% For the concentrated F-load, the coefficient is because ring deflection is not quite elliptical But when the pipe is buried, the loads are distributed, and ring deflection is essentially elliptical For combined stress analysis by elastic theory, failure is the equation of maximum stress (normal or shearing) to the corresponding yield strength of material The assumption is that elastic limit is failure — not necessarily true On what plane does fracture occur? The answer is useful in reconstructing the cause of pipe failures For example, on what plane does a compression fracture occur in brittle material at point B in Figure 19-1? An approximate answer can be found from a Mohr circle analysis ©2000 CRC Press LLC MOHR CIRCLE ANALYSIS Appendix E describes techniques for construction, application and analysis of the Mohr stress circle, orientation diagram, and strength envelope The interrelationship of the three diagrams is accomplished by superposition onto a single diagram Construction Stress Diagram: Draw the free-body-diagram of stresses on an infinitesimal unit cube, O Figure 19-2a is an example of cube O with numerical values of stresses shown Draw the σ and τ axes, Figure 19-2b; and plot the Mohr stress circle Three points are required: a) Center is on the σ -axis b) Circle passes through point (σx, τ xy) c) Circle passes through point (σy, τ yx) Note that τ yx = -τ xy; and that the sign convention is positive for normal stresses in compression, and for c ounterclockwise shearing couples This is the correct sign convention Tension (negative) is a reduction of compressive intermolecular bond Orientation Diagram: Orientation of coordinate axes x and y is the same as for the unit cube When superimposed on the stress diagram (Figure 19-2c), the y-axis intersects the stress circle at point (σy,-τ xy) The axes are still parallel to the unit cube axes The origin of axes, O, is the location of the unit cube, which is shown superimposed and correctly oriented on the stress axes of Figure 19-2c The origin (cube O) always falls on the stress circle Figure 19-3 Strength envelopes ©2000 CRC Press LLC Strength Envelopes: For most soils and many construction materials, failure is slip on a shear plane It is shown in Figure 19-3a as slip of a block due to shearing force F At slip, F is equal to the sum of cohesion C (glue on the slip surface) and friction Ntanϕ, where ϕ is the friction angle and N = σ A Dividing through by A and noting that F/A = S', the shearing stress at failure, called Coulomb strength, is, S' = C + σ tanϕ = SHEARING STRENGTH A plot of the Coulomb shearing strength on the σ τ axes is the strength envelope See Figure 19-3b Any stress point outside of the strength envelope is failure The material slips (shears) _ RADIUS OF CIRCLE = \ (σx - σ y )2/4 + (τ xy )2 / Any central angle is twice the corresponding circumferential angle where both angles intercept the same arc The values of principal stresses, maximum shearing stresses, and the planes on which they act can be calculated by trigonometry from the superimposed diagrams As an example, for the unit cube of Figure 19-2, the pertinent stresses are: σ1 = 2500 psi = maximum principal stress σ3 = 500 psi = minimum principal stress τ max = 1000 psi = maximum shearing stress Principal planes and planes of maximum shearing stress are dotted on Figure 19-2c Example Application Any plane through O perpendic ular to the page (seen as a line) is oriented the same as the infinitesimal cube Any plane drawn through O intersects the stress circle at a point whose stress coordinates are the stresses on that plane When a stress circle is tangent to the strength envelopes, shear slip is incipient on those planes that intersect the stress circle at the points of tangency These are the failure planes, at angle β See Figure 19-3c Analysis The horizontal distance to the center of the stress circle is the average of σx and σ y DISTANCE TO CENTER = (σx + σ y)/2 From Figure 19-2b, The radius of the Mohr circle is, by the Pythagorean theorem, ©2000 CRC Press LLC The vertical stresses at point B in the unreinforced concrete pipe of Figure 19-1 are at incipient failure What would be the expected failure plane angle, β, at B if failure occurred β is the angle of bevel of the fracture From the laboratory, tension strength is σT = ksi, and compression strength is σ C = 12 ksi First, from the infinitesimal cube, O, draw the stress axes, σ and τ , as shown in Figure 19-4, and plot the stress circles for both tension and compression failures The tangents to the failure circles are strength envelopes Second, superimpose cube B and its orientation diagram onto the stress diagram Keep orientation the same — x-axis horizontal and y-axis vertical Cube B must be located at a point where its axes (which are the principal planes through cube B) intersect the Mohr circle at the stresses on those principal planes Cube B always falls on the Mohr circle Because both the tension load and the compression load act on horizontal planes, x-planes are drawn through the failure stress points σt and σ c Figure 19-4 Mohr analysis at failure (cracking) at point B on the inside of an unreinforced concrete pipe at the springline ©2000 CRC Press LLC Horizontal stresses (on vertical planes) are both zero so the y-axis is at zero stress unreinforced brittle pipes in general The sketches of failure planes (bevel angles) are to scale The results are the angles β of failure planes through O From strength tests, the friction angle of this concrete is ϕ = 45.58° From the Mohr diagram, the angle of the failure plane is β = 45° + ϕ /2 = 68° The analyses above are only examples, but the same procedure can be followed to reconstruct combined stress failures Example COMPOUND STRESS ANALYSIS If cohesionless soil is loaded vertically with stress σz, what is the minimum horizontal stress, at active soil resistance, required to prevent shear failure planes from developing in the soil? At what angles shear failure planes develop if the horizontal resistance is not adequate? The soil friction angle is 30° Occasionally the state of stresses at a point is so complex that the mere combining of stresses in the same direction is inadequate Compound analysis investigates the critical stress resulting from multidirectional stresses, both normal and shearing For planar stresses (biaxial) no new concepts are needed For triaxial stresses, it is usually precise enough to consider separately each of the three views of the infinitesimal unit cube as a biaxial case in order to ascertain which view results in the largest stress circle; i.e., nearest to tangency with the strength envelope This becomes the critical case for analysis and design In general, compound stress analysis is required for specific cases such as the following From Figure 19-5, a stress circle is drawn tangent to the strength envelopes which, for cohesionless soil, are straight lines from zero stress at angles ϕ = 30° The maximum principal stress is σz at the right side of the stress circle Because it acts on a horizontal plane, the x-plane is drawn through this point The minimum principal stress is σx at the left side of the stress circle as shown Because it acts on a vertical plane, the z-plane is drawn through this point The intersection of the z and y axes is the origin where the infinitesimal cube O is located Let the distance to the center of the circle be X The radius of the c ircle is Xsinϕ ; σ1 = X + Xsinϕ ; and σ = X Xsinϕ The ratio is σ /σ = K = (1+sinϕ)/(1-sinϕ) = The horizontal stress required to prevent shear slip is σx = σ z /K = σ z /3 Shear failure planes are at angle β from the origin, O, to the points of tangency of the failure stress circle to the strength envelopes These angles are β = 45° + ϕ/2 For ϕ = 30°, the failure planes are at + 60o Seven different combined failures are shown in Figures 19-6 to 19-8 All of these failures were observed in asbestos cement pipes, 12-inch inside diameter Such failures are typical of failures in ©2000 CRC Press LLC Stress Risers Stress risers are discontinuities that cause stress concentrations At discontinuities, concentrated stress sometimes exceeds the maximum allowable Remedies include saddles, stiffener rings, and crotch plates However, it may be sufficient to simply thicken the wall by means of a boss or plate Failure analysis due to stress risers starts with basic stress analysis See Appendix E The free-body-diagram for compound stress analysis is an infinitesimal cube on which six pairs of forces act Figure 19-9 is the cross section of wall of a pressurized container The maximum stress occurs on the inside surface on cube Op The notations for three pairs of normal stresses and three pairs of shearing stress couples are as follows Figure 19-5 Mohr analysis for evaluating minimum (active) resistance of soil at shearing failure of the soil (Failure planes form at angles β.) ©2000 CRC Press LLC σr σr Figure 19-11 Mohr stress circle at failure of an infinitesimal cube, Op, on the inside surface of a pressurized, closed-ended cylinder, showing three orthogonal views of Op with the corresponding Mohr circles The largest circle is critical; i.e., it will be the first to come tangent to the strength envelope ©2000 CRC Press LLC AT RADIUS, r, TANGENTIAL STRESS σf = P(a/r)2(r2 + b2)/(b2 - a2) LONGIDUTINAL STRESS σz = Pa2/(b2 - a2) Figure 19-12 (top) Cross section of a thick-wall steel cylinder of diameter OD = 2(ID), and showing the variation of tangential stress, σt, throughout the cylinder wall when subjected to internal pressure, P' (bottom) Mohr circle with strength envelopes for steel (nearly horizontal), showing failure planes at 45o ©2000 CRC Press LLC risers occur in wyes, tees, valves, reducers, etc Pipes are sometimes capped when they are to be extended at a later date Taps cause stress concentrations on the edge of the tapped hole Thick-wall containers feel higher tangential stress on the inside than thin-wall hoop stress, σt = P'(ID)/2t These stress risers are discussed in the following Thick-wall Cylinders Thick-wall cylinders subjected to internal pressure, feel maximum tangential stress (hoop stress) on the inside of the wall: σt P' (OD)2 + (ID)2 (OD)2 - (ID)2 where σt = OD = ID = P' = (19.1) naximum tangential stress (hoop stress) outside diameter inside diameter internal pressure Example Figure 19-12 is the cross section of a thick-wall, high-pressure pipe What is the maximum tangential stress, σt, and the maximum shearing stress, τ ? OD = 2(ID) From Equation 19.1, σt = 5P'/3 The average hoop stress is σ = P'(ID)/2t = P' Clearly, the maximum stress is 5/3 rds as great as the average From the Mohr circle, maximum shearing stress is P'/3 If the cylinder is steel, the strength envelopes are nearly horizontal as shown and the failure planes are at 45o Longitudinal stress is calculated the same way for both thick-wall and thin-wall cylinders See Chapters 14 and 15 Radial stress is maximum on the inside of the wall, and is simply internal pressure, P', in compression Tangential stress is maximum on the inside when the pipe is subjected to either internal pressure or ©2000 CRC Press LLC external pressure For analysis of external pressures on thick-wall pipes and tanks, refer to texts on mechanics of materials Taps The most common attachments to pipes are tapped into the pipes Stress concentrations occur at holes in the walls of pressure containers Taps are required for attaching corporation stops, smaller pipes, air-relief valves (ARVs), pressure gages, etc Stresses concentrate in the wall around the tapped hole, and are critical if internal pressures are high and the pipe material is non-plastic Plastics include most pipe grade steel which is elasto-plastic Plastic yields before it fractures This is not true, however, for pipes subjected to impact loads or very low temperatures or repeated loadings Under such loads, even plastics can fail by brittle fracture One remedy is to thicken the wall around the hole For example, a bead or boss could be formed around the edge of the hole See Figure 19-13 Engineers use a rule of thumb — the bead or boss must contain a volume equal to the volume of material cut away for the hole From elastic theory (and tests) tangential stresses vary as shown in Figure 19-14 for a very large plate with a hole in the middle of it The elastic equations for tangential stress σt, radial stress σ r, and shearing stress τ , on an infinitesimal cube in the plate, are: 2σt = σ o(1 + ρ2/r2) - σo(1 + 3ρ4/r4)cos 2θ (19.2) 2σr = σo(1-ρ2/r2) + σo(1-4ρ2/r2+3ρ4/r4)cos 2θ (19.3) 2τ = σo(1 + 2ρ2/r2 - 3ρ4/r4) sin 2θ) (19.4) where (See Figure 19-14) σr = radial stress with respect to the hole, σt = tangential stress, (perpendicular to radial stress at any point), Figure 19-14 Stresses in a plate with a hole in it, showing how stress channels crowd around the hole like traffic around an excavation in the road, creating stress concentrations, Kσo, at the edge of the hole If the hole is very small, K = and σt = 3σ o ©2000 CRC Press LLC τ b ρ r θ σo = = = = = shearing stress, width of the plate, radius of the hole, radius to the stress point in the plate, angle of the radius from the direction of average stress, σo, = average uniform stress in the plate if no hole were in it Of primary concern are stresses tangent to the edge of the hole where r = ρ If the plate is infinitely wide, ρ/b 0, and σ o is unaffected by the hole From Equations 19.2 to 19.4, σt = -σ o, compression at θ = 0o (19.5) σt = σ o, tension at θ = 90o (19.6) Shearing stresses, τ , are zero at the edge of the hole and increase to a maximum of τ = σ o /2 at great distances from the hole on planes at 45o with the longitudinal axis Design of taps in pipes follows the above rationale for uniaxial loading in the case of gasketed pressure pipes with no longitudinal stress See Figure 19-15, top sketch If the diameter of the hole is much smaller than the diameter of the pipe, Equations 19.5 and 19.6 can be used From Equation 19.6, the maximum stress at the edge of the hole tapped in gasketed pipes is tangential stress, σt = 3P'(ID)/2A, tension at B (19.7) GASKETED PRESSURE PIPES where σt = P' = ID = A = A tangential stress at edge of hole, pressure in the pipe, inside diameter, cross-sectional area of the wall per unit length of pipe, = t for plain wall pipes For the case of a closed-end pipe or tank, see Figure 19-15 In the bottom sketch, the tangential stress is analyzed by combining Equations 19.5 and ©2000 CRC Press LLC 19.6 For example, from Equation 19.5, the tangential stress at B is -σo /2 in compression due to longitudinal stress, σo /2 in tension At the same point B, from Equation 19.6, the tangential stress is 3σo in tension due to hoop stress, σ o Combining the two tangential stresses at B, for closed-end pipes, σt = 5P'(ID)/4A, tension at B (19.8) CLOSED-END PRESSURE PIPES Example What is the maximum tangential stress on the edge of a small 1-inch hole (tap) in a 6-inch ID steel pipe if the wall thickness is 0.125 inch and internal pressure is 400 psi? Slip couplings, such as Dresser or Baker, eliminate longitudinal stress in the pipe From Equation 19.7, σ t = 3(400psi)(6 in)/2(0.125 in) = 28.8 ksi If yield stress is 42 ksi, the safety factor is 42/28.8 = 1.46 If P' is a repeating pressure, fatigue strength of the steel should be considered Fatigue strength is usually lower than yield strength, especially if the pressure cycles are reversed, and the hole is threaded Threads are stress risers If the pipe is not gasketed, longitudinal stress, σz is usually tension, in which case stress risers at a tap are less than for a gasketed pipe This tap analysis is based on static pressure and a small diameter tap in a large diameter pipe Refinements are forthcoming In the 1980s Roland W Jeppson, at Utah State University, tested threaded taps with corporation stops for service connections in AWWA C-900 PVC pipes, inches in nominal diameter, subjected to cyclic internal water pressure surges from 100 to 200 psi at a rate of 30 surges per minute Loading was continual without down time during which plastic might partially recover Dye in the water made the slightest crack visible None of the five 0.75-inch taps or the five 1-inch taps leaked after 1.5 million Figure 19-15 Principal stresses and stress concentrations in the pipe wall and around a hole tapped in the wall of a pressure pipe: (top) gasketed pipe, and (bottom) closed-end pipe Obviously, stress concentrations are maximum around the hole in the gasketed pipe ©2000 CRC Press LLC cycles of pressure Similar surge tests were performed on larger 24-inch PVC pipes Some leakage was noted Results were only qualitative, but it was concluded that service life of a tapped large-diameter PVC pipe subjected to pressure surges is only moderately less than an untapped pipe However, both exceed normal expected service life A different problem showed up serendipitously The gasketed end closures for the test sections leaked more than did the corporation stops Example What is the maximum tangential stress at the edge of the taps in Dr Jeppson's pressure tests of AWWA D-900 PVC pipe 6D? It is assumed (questionably) that the taps are small compared with the pipe diameter OD t DR P' = = = = 6.900 inches average, 0.276 inch minimum, 25 = dimension ratio, 200 psi maximum Hoop stress in the wall of the pipe is, σo σo = P'(ID)/2t = 200 psi(6.900-0.552)/2(0.276) = 2300 psi Because the test pipe was gasketed, from Equation 19.7, the maximum tangential stress is 3σo = 6.9 ksi This is higher than the yield strength of PVC, and would indicate that to assume that pipe diameter is infinitely greater than hole diameter may be too conservative Moreover, Equation 19.7 is based on elastic limits PVC is plastic If a tap, or any stress riser, is in any section other than a cylinder, the hoop stress, σo, must be evaluated for that specific section For example, to find the maximum "hoop" stress in the wall of a valve with a bonnet, σo = P'(ID)2t where ID is the greatest width of a plane that can be passed through the bonnet and pipe ©2000 CRC Press LLC Stress risers are not necessarily critical in pliant materials such as plastics — including steel Because of a demand for corrosion-resistant polyethylene pipes under sanitary landfills, and because those pipes must be perforated or slotted to drain leachate and collect methane gas, tests have been performed to ascertain the effects of small perforations or circumferential slots (width of a circular saw blade) on the structural integrity of the pipe Clearly, the strength of the ring is reduced by the area of surface cut away The ring stiffness is reduced also — but relatively less When the ring is loaded and deformed into an ellipse, visible warping at the edges of the perforations and the ends of the sawed slots is evidence of stress concentration However, the plastic yields without fracture, and relaxes The integrity of the pipe is not compromised End Closures In this section, end closures are analyzed for stress risers However, any transition from a pipe to another section (wye, tee, reducer, valve, etc.) results in similar stress risers, usually to a lesser degree, and can be analyzed in a similar manner Consider the hemispherical cap on a thin-wall cylinder with internal pressure P' as shown on Figure 19-16 Notation is as follows P' r t σr σt σz E ν εr ε,t εz = = = = = = = = = = = internal pressure, mean radius of thin-wall container, wall thickness, radial stress on the inside of the wall, tangential stress in the wall, longitudinal stress in the wall, modulus of elasticity, Poisson ratio, radial strain = percent increase in radius, tangential strain, longitudinal strain, Figure 19-16 Pressurized thin-wall container with hemispherical end-closures, showing: (bottom) how internal pressure, P', increases the radius by a percentage equal to the tangential strain; and (top) how shearing discontinuity at section A-A can be eliminated by reducing the wall thickness of the hemisphere, but leaving, instead, a reentrant corner which itself is a stress riser ©2000 CRC Press LLC The tangential stresses are: Cylinder, Hemisphere, σt = P'/(r/t) σt = 0.5P'(r/t) If wall thicknesses are the same for cylinder and hemisphere, as shown at the left side of Figure 1916, tangential stress in the hemisphere is only half as great as in the cylinder Therefore, increase in radius is less in the sphere than in the cylinder; and a radial shearing stress occurs at section A-A The discontinuity is a stress riser Tangential stresses would be equal if the hemis pherical wall were only half as thick as the cylindrical wall This is shown at the right side of Figure 19-16 The shearing discontinuity is reduced but is not eliminated The problem is still the difference between increases in radii Neglecting the relatively minor effect of radial pressure on the w all in thin-wall containers, tangential strain, ε t, is the percent increase in circumference, which is the percent increase in radius — which is radial strain Therefore, radial strain is ε r = ε t, and stress-strain relationships are Eε t = σ t - νσz, as follows: Cylinder, Eε t = σ t - νσt /2 Eε t = P'(r/tc)(1 - 0.5ν) (19.9) where tc is the thickness of the cylinder Hemisphere, Eε t = σ t Eε t = P'(r/ts)(1 - ν) 0.4286, which is even less than the original ratio of 0.5 required for equal tangential stresses The shearing discontinuity is essentially resolved However, the connection now mates two different wall thicknesses with a reentrant corner which is itself a stress riser For brittle materials, the reentrant corner could be critical But if the material is pliant and a safety factor is included in design, the reentrant corner might be mitigated by a good weld, and/or by ductile pipe material for which the stress riser is not critical because the material yields without fracture Or it might be remedied by shaping the end closure to something other than a hemisphere The end closures of rocket motors are ellipsoidal to reduce stress risers Strength of Welded Joints in Steel Pipes If the weld is a full-penetration butt weld, longitudinal strength is no less than the strength of the steel pipe If there is any question about the welding procedure, some designers assume strength to be 90% of steel strength If the weld is a lap weld, and the gap is no greater than 0.125 inch, longitudinal strength of a single welded lap joint is about 75% of pipe strength The strength of a double-welded lap joint is no less than 80% of pipe strength The forces on fillet welds of lap joints are shearing forces not couples The curved surfaces of both bell and spigot prevent moments (couples) on the weld (19.10) STRESSES IN STEEL PIPES where ts is the thickness of the hemisphere If the radial strains are to be equal, Equations 19.9 and 19.10 must be equal; and the ratio of wall thicknesses would have to be, ts /tc = (1 - ν)/2(1 - 0.5 ν) For example, if Poisson ratio is ν = 0.25, the ratio of wall thicknesses of hemisphere to cylinder is ©2000 CRC Press LLC Steel pipes draw a disproportionate amount of stress analysis because of advances in the analysis of steel structures for which performance limit is yield stress by the elastic theory For steel pipes, performance limit is deformation (strain) — not elastic limit (yield) In fact, some steel pipes are strained well above elastic limit during the process of fabrication Nevertheless, stress theories persist Some common stress analyses follow Figure 19-17 Standard tension test of steel, showing the stress-strain diagram with the elastic limit and ultimate strength Figure 19-18 Standard tension test on steel, showing the Mohr circle at elastic limit in tension and in compression (dotted) and the strength envelopes and failure planes in shear (slip) ©2000 CRC Press LLC Performance Limit: Performance limit is strength, σf, at elastic limit It is often referred to loosely as yield stress It is found from standard uniaxial stress-strain tests See Figure 19-17 For pipe-quality steel, typical properties are as follows Properties of Pipe Steel: σf E ε εu ν U = = = = = = 42 ksi = stress (failure) at elastic limit, 30(103) ksi = modulus of elasticity, F/E = strain below the elastic limit, 21% = approximate elongation at fracture, 0.27 to 0.30 = Poisson ratio, 15 lb.ft a 0oF = Charpy toughness For stress design, STRESS must be less than STRENGTH reduced by a safety factor Ultimate energy, Uu, in steel is three hundred times greater than elastic energy, Ue Design by elastic energy (resilience) is extremely conservative The above applies only to uniaxial stress The energy of compound stress es is investigated in the following Stress: Compound stress analysis is based on elastic theory (elastic limit) Compound analyses are seldom justified for buried steel pipes In the first place, one principal stress is usually so much greater than either of the others that uniaxial stress analysis is adequate In the second place, properties of the embedment are imprecise, loads are unpredictable, and pipe-soil interaction is statically indeterminate to the infinite degree In the third place, steel is elastoplastic — not limited to the elastic limit Yield stress is determined by tensile tests (not triaxial tests) that provide only normal failure stress, σf Performance limit is shearing failure — not tension failure The shearing failure plane is beveled at 45o Strength From Figure 19-17, ultimate steel strength is greater than yield strength For some analyses, this difference provides a margin of safety in addition to the safety factor For energy analyses, the margin of safety is much greater Energy input, Ue, up to elastic limit per unit volume of the test specimen of Figure 19-17, is the average force, P/2, times the distance, ∆ , divided by volume, AL If σf = 42 ksi and E = 30,000 ksi, Ue Ue = P∆ /2AL = (1/2)(P/A)(∆ /L) = 0.5σε = σf2/2E Ue = 29.4 ksi = RESILIENCE = Area under the stress-strain diagram up to the elastic limit Energy up to ultimate strength, Uu, is the area under the entire stress-strain diagram to elongation (strain) at failure — which is roughly 21%; Uu = 8,800 ksi = TOUGHNESS = Area under the entire stress-strain diagram ©2000 CRC Press LLC Mohr Stress Circle: Figure 19-18 shows the Mohr stress circle for a standard tension test to failure Ordinates are shearing stress, τ Abscissae are normal stress, σ For steel, it is assumed that tension is positive and clockwise shear is positive The infinitesimal cube is superimposed on the stress circle by orienting the planes on which the principal stresses act Any plane through the cube intersects the Mohr circle at its own stress coordinates The shearing and normal stresses at the point of intersection act on that plane All planes are correctly oriented For tests in compression, the shearing stress at yield is nearly the same as in tension See the dotted Mohr circle Tangents to the two Mohr failure circles are strength envelopes Any shearing stress outside of Figure 19-19 Compound stress analyses for steel pipes based on shearing strength Mohr circles are shown for principal stresses acting on an infinitesimal cube on the inside surface of the pipe wall ©2000 CRC Press LLC the strength envelopes is failure Failure planes are at 45o with the principal stresses Elastic Stress Analysis — Combined Stresses: Stresses are added in combined stress analysis For example, in Figure 19-19, σy = P'r/t - Pr/t, where P' is internal pres sure and P is external soil pressure However, for conservative design, P and P' are analyzed separately Longitudinal stress, σz , is the sum of longitudinal stresses due to temperature decrease, Poisson effect of internal pressure, and thrusts due to valves, elbows, etc However, external loads such as longitudinal beam bending and thrust blocks may or may not contribute to combined longitudinal stresses They must be considered on a case-by-case basis wherein longitudinal and circumferential stresses are of opposite sign Vacuum in the pipe, a high external fill, and a high water table might combine to cause suc h stresses But failure would probably be collapse not stress failure Steel pipe collapse is usually a function of ring stiffness not yield stress The top view is critical only if longitudinal stress is excessive or if internal pressure is zero or negative (vacuum) This usually requires collapse analysis not stress analysis See Chapter 10 on ring stability Stresses are not added in compound stress analysis Figure 19-20 shows strength envelops of σy as a function of σz at yield stress according to the theory of elasticity The shear stress analyses are dotted A more accurate analysis is maximum strain energy, U = Σ f (σ /2)ε See Appendix F for the analysis of strain energy at failure The strain energy results not quite model test results for steel Elastic Stress Analysis — Compound Stresses: Performance limit is shearing failure Maximum shearing stress equals shearing strength; i.e., τ = τf In order to relate shearing stress at failure to the standard tension test at failure, from the Mohr circle in Figure 19-18, τ f = σ f /2 Figure 19-19 shows an infinitesimal cube in the inside of a pipe wall with principal stresses acting on it The horizontal stress is internal pressure P, which is small compared to the other stresses If the cube were on the outside of the pipe wall, P would be zero For analysis, σx = Mohr circles are shown for the three views of the infinitesimal cube If the pipe is not subjected to torque or point loads, shearing stresses not act on the cube, and the three stresses are principal stresses The front view is critical For analysis, shearing stress at failure is half the normal stress at yield, σf /2 Therefore, Pr/t = σf = 42 ksi A safety factor (often two) is used for design; i.e., Pr/t = 21 ksi The side view is not critical The critical circle shown is an improbable hypothetical condition ©2000 CRC Press LLC Huber, Hencky, von Mises Equation: More accurate for steel is the Huber-Hencky-von Mises equation, which discounts the strain energy that only causes volume change See Appendix F for the equation and Figure 19-20 for the plot of the elliptical strength envelope for two-dimensional compound stress Most buried pipes not require Huber-Hencky-von Mises analysis Because σx = P is relatively small, the two-dimensional analysis is adequate As an example, if σz = σ f /2, according to the Huber-Hencky-von Mises equation, σy = 1.155σf Clearly this allows a slightly greater stress than yield, but the increase is small It is conservative to design by uniaxial stress analysis; i.e., critical stress is σy = σ f at P = In general, for buried pipes, compound stress analysis is not needed, and may be misleading because it is based on elastic limits Collapse Analysis For unburied steel pipes, collapse is Pr3/EI = 3, Figure 19-20 Strength envelopes at elastic limit, σf where I is the moment of inertia of the wall cross section per unit length of pipe Collapse is a function of external pressure, P, and ring stiffness, EI/r — not yield stress Ring deflection and soil strength are pertinent but are usually controlled by specifications A smooth, satin surface indicates quick fracture A long fracture surface (tear) indicates crack propagation due to stress energy stored in the pipe A rippled surface — especially if oxydized near the pipe surface — indicates fatigue STRESSES IN CONCRETE PIPES Concrete pipes are so complex that stress analysis is impractical Performance limits are usually found from tests A hole in a pressure pipe abraded from outside-in could be caused by a leak out of which a highpressure jet causes turbulence in the embedment that "sand-blasts" the pipe The cutting action from outside is remarkably rapid FAILURE SURFACES A crystalline surface with sharp edges indicates sudden fracture due to an instantaneous load such as water hammer High-strength steel (such as bolts) under high tension may fail by hydrogen embrittlement Some plastics can fail by "brittle fracture" — especially under shock load and low temperatures ©2000 CRC Press LLC Chevrons along a torn surface point toward the location where fracture was initiated Pock-marks of oxidized material indicate corrosion A stream bed appearance with loss of material inside on the invert indicates erosion PROBLEMS 19-1 An unreinforced concrete pipe has a wall thickness of t = inches, ID = 36 inches, and length of L = 11 ft A 12-kip dual-wheel load passes over midspan Soil cover is negligible What is the water pressure inside the pipe at failure if the soil has been washed out from around the pipe leaving it simply supported on the ends? The joints are gasketed Unit weight of concrete is 144 pcf, unit weight of water is 62.4 pcf, and strengths of the concrete are ksi in tension and 12 ksi in compression (P = 300 psi) 19-2 What is the maximum tangential stress in a thick-wall, closed-end cylinder that has a small tap (σ t = P') in it, if OD = 1.4(ID)? 19-3 A polyethylene pipe is to be installed under a river A tunnel, slightly larger than the pipe, is directionally drilled under the river and driller's mud is left in the tunnel to prevent collapse of the tunnel until the pipe can be pulled through As the pipe is being pulled into place, the driller's mud is replaced by grout What is the maximum stress in the wall of a polyethylene pipe subjected to the external hydrostatic pressure of grout with unit weight 95 pcf, and depth of 80 ft? The pipe is empty such that pressure inside is atmospheric ©2000 CRC Press LLC DR = 9.5 = OD/t for this thick-wall pipe Neglect longitudinal force during the pull (σ t = 280 psi) 19-4 Resolve problem 19.3 while the polyethylene pipe is being pulled into place with a longitudinal stress of 500 psi The pulling force is monitored to make sure the pipe doesn't bind and break while it is (τ = 375 psi) being pulled into the tunnel 19-5 Resolve problem 19.4 if the pipe is filled with driller's mud as it is pulled into place The unit weight of driller's mud is 75 pcf (σ t = 100 psi) 19-6 What is the effect of tapping, or perforating, or circumferentially slotting the polyethylene pipe of problems 19.3 to 19.5? Small perforations or narrow sawed slots are common practice in fabricating polyethylene pipes to be used under sanitary landfills to collect leachate and methane gas 19-7 What should be the ratio of thicknesses of an end closure and thin-wall pressurized cylinder if the end closure is a hemisphere, fitted inside the cylinder, and reversed in direction (like the bottoms of spray paint cans)? What is required to withstand the shearing discontinuity? What should be the size and type of weld? What might be the size and design of a restraining ring? Etc ... radial strain = percent increase in radius, tangential strain, longitudinal strain, Figure 19-16 Pressurized thin-wall container with hemispherical end-closures, showing: (bottom) how internal... Neglecting the relatively minor effect of radial pressure on the w all in thin-wall containers, tangential strain, ε t, is the percent increase in circumference, which is the percent increase in. .. small 1-inch hole (tap) in a 6-inch ID steel pipe if the wall thickness is 0.125 inch and internal pressure is 400 psi? Slip couplings, such as Dresser or Baker, eliminate longitudinal stress in the