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 ©2000 CRC Press LLC CHAPTER 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 is reached when stress, s , equals yield strength, S For design, the yield strength of the pipe wall is reduced by a safety factor, INTERNAL PRESSURE — (MINIMUM WALL AREA) 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 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 ©2000 CRC Press LLC s = P'(ID)/2A = S/sf where: s = P' = ID = OD = D = A = S t sf = = = (2.1) circumferential tensile, stress in the wall, internal pressure, inside diameter, outside diameter, diameter to neutral surface, cross sectional area of the pipe wall per unit length of pipe, yield strength of the pipe wall material, thickness of plain pipe walls, safety factor 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 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 ©2000 CRC Press LLC Given: E = S = sf = gw = P' = 30(106)psi = modulus of elasticity, 36 ksi = yield strength, = safety factor, 62.4 lb/ft3 = unit weight of water, hg w = 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 Fload 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 ©2000 CRC Press LLC 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: fv's, Fundamental bd's, Basic Variables Dimensions 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 fv's - bd's = 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 circumferential 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, fv's, Fundamental Variables d = ring deflection = /D D = mean diameter of the pipe F = diametral line load per unit length of pipe EI = wall stiffness per unit length of pipe where: D = E = t = I = bd's, Basic Dimensions L FL-1 FL decrease in diameter due to the F-load, modulus of elasticity, wall thickness for plain pipe, moment of inertia of wall cross section per unit length of pipe = t3/12 for plain pipe F = 2pSI/cD = F-load at yield strength, S fv's - bd's = pi-terms For plain pipes, I = t3/12 and c = t/2 for which, I/c = t2/6 and, in 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, (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 (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: Ring Deflection Performance Limit (d) = 0.2232 (F/ED) (D/t)3 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: The relationship between circumferential stress and ring deflection is found by substituting from Table A1, 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: ©2000 CRC Press LLC (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 Fload is removed Deformation is permanent This is not failure, but, for design, may be a performance limit with a margin of safety 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 stress s = yield strength S, the maximum allowable load is: F = 2pSI/cD For walls with smooth cylindrical surfaces, F = pSt2/3D 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) For plain pipes, (F/SD) = p(t/D)2/3 (2.4) 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 Resolving, for plain pipes, F = pSD(t/D) /3 Ring Deflection where d = (D/D) due to F-load, is given by: 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) d = 0.0186 (FD /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 ©2000 CRC Press LLC If the F-load is known, the required dimension ratio at yield strength is: DR = (pSt/3F) + Example 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: 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 ID OD g F s = = = = 30 inches = inside diameter, 37.5 in = outside diameter, 145 lb/ft3 = unit weight of concrete, 3727 lb/ft = F-load at fracture + s from tests where, = + 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 (2.6) This is the basis for design Because of its importance, design by ring compression stress is considered further in Chapter 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 Example 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 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 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 PROBLEMS 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 uniformly ©2000 CRC Press LLC 2-1 What is the allowable internal pressure in a 48inch 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 in 2/ft [AISI tables], S = 36 ksi = yield strength, E sf = 30(106) psi, = = 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 inches in the wall which is 6.0 inches thick? (P' = 78.5 psi) Given: S = 36 ksi = yield strength of steel, sf = = 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 ©2000 CRC Press LLC ... 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... or minimum wall area A 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... 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