Numerical Methods in Soil Mechanics 06.PDF Numerical Methods in Geotechnical Engineering contains the proceedings of the 8th European Conference on Numerical Methods in Geotechnical Engineering (NUMGE 2014, Delft, The Netherlands, 18-20 June 2014). It is the eighth in a series of conferences organised by the European Regional Technical Committee ERTC7 under the auspices of the International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE). The first conference was held in 1986 in Stuttgart, Germany and the series has continued every four years (Santander, Spain 1990; Manchester, United Kingdom 1994; Udine, Italy 1998; Paris, France 2002; Graz, Austria 2006; Trondheim, Norway 2010). Numerical Methods in Geotechnical Engineering presents the latest developments relating to the use of numerical methods in geotechnical engineering, including scientific achievements, innovations and engineering applications related to, or employing, numerical methods. Topics include: constitutive modelling, parameter determination in field and laboratory tests, finite element related numerical methods, other numerical methods, probabilistic methods and neural networks, ground improvement and reinforcement, dams, embankments and slopes, shallow and deep foundations, excavations and retaining walls, tunnels, infrastructure, groundwater flow, thermal and coupled analysis, dynamic applications, offshore applications and cyclic loading models. The book is aimed at academics, researchers and practitioners in geotechnical engineering and geomechanics.
Anderson, Loren Runar et al "RING STRESSES" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 6-1 Free-body-diagrams for analyzing hoop stresses in rigid and flexible rings with initial ovality Figure 6-2 Stress distribution across the wall of a thick-walled cylinder due to internal pressure (top) and external pressure (bottom) For this example, the outside diameter equals twice the inside diameter ©2000 CRC Press LLC CHAPTER RING STRESSES For preliminary buried pipe design, stress analysis requires only fundamental principles of pipe mechanics However, analysis in greater depth is often essential In all cases, performance limit is deformation; i.e., rupture, wall crushing, wall buckling, ring deflection, etc In this chapter, performance limit is analyzed in terms of stress at the point of excessive deformation Hoop Stress Hoop stress due to internal pressure, P', in a thinwalled circular ring, from Equation 2.1, is: s = P'(ID)/2A (6.1) where s = hoop stress; i.e., circumferential stress in a thin-walled pipe for which D/t > 10, ID = inside diameter, A = cross-sectional area of the pipe wall per unit length of pipe = t for plain wall pipe, t = wall thickness for plain pipe, c = distance from neutral surface, of the wall to the most remote surface, dq = change in curvature = 1/r - 1/ro, r = deformed radius, ro = original radius, E = modulus of elasticity Now suppose that the pipe is not circular — it is outof-round before installation — called ovality See Figure 6-1 In the case of a rigid ring, the maximum hoop stress occurs at B on the maximum diameter (ID) This horizontal (ID) is called the span If the long axis is vertical, ID must be vertical In the case of a flexible pipe, Figure 6-1, the maximum hoop stress acts on the maximum diameter But hoop stress tends to round the pipe If the ring is initially deformed, the circumferential stress due to internal pressure, P', is, from Equation 5-2, the sum of hoop stress and flexural stress; i.e., s = P'r/A + Ecdq ©2000 CRC Press LLC Because dq is a function of loads on the ring, which could be complex, analyses can be complicated However, for plastic pipes and elasto-plastic (metal) pipes, rupture does not occur until average hoop stress reaches yield Therefore, the flexural component of stress is not an issue Flexure adds to the hoop stress at one surface of the wall, but subtracts from it at the other For flexible thinwalled pipes that are out-of-round, internal pressure tends to round the ring causing soil pressure concentrations However, most pipes are near enough to circular when buried, that rerounding is not an issue Circular Thick-walled Pipes Analyses of circular thick-walled cylinders can be found in texts on solid mechanics Thick-walled cylinders subjected to internal or external pressure, feel maximum tangential stress, s , on the inside of the wall See Figure 6-2 Internal Pressure, P': si = P'(a2 + b2)/(b2 - a2) tension on the inside surface, (6.2) so = 2P'a2/(b2 - a2) tension on the outside surface, sav = P'a/t = average tangential stress, where subscripts i, o, and av, refer to inside, outside, and average; and: s b a P' t = tangential stress (hoop stress, tension), = outside radius, = inside radius, = internal pressure, = wall thickness = b-a External Pressure, P: si = 2Pb2/(b2 - a2) compression on the inside surface, (6.3) so = P(b2 + a2)/(b2 - a2) compression on the outside surface, sav = Pb/t = average tangential stress Example Figure 6-2 (top) shows the cross section of a thickwalled, high pressure pipe What is the maximum tangential stress, s , if OD = 2ID? From Equation 6.2, si = 5P'/3 = 5/3 rds sav If pressure is external, the maximum tangential stress is still on the inside surface See Figure 6-2 (bottom) Because of the greater outside diameter, stress si is greater due to external pressure than due to an equal internal pressure However, compressive yield strength is greater than tensile yield strength for many pipes Example If OD = 2ID, Figure 6-2 (bottom), the ring compression stress due to external pressure is, si = 8/5 the ring compression stress due to equal internal pressure neglected Dry unit weight of soil is 110 pcf Saturated unit weight is 140 pcf A water table can rise to ft above the top of the pipe What is the ring compression stress, s , in the pipe wall? s = P(OD)2t = P(DR)/2, where P = vertical soil pressure on the pipe P = 4(110)psf + 6(140)psf = 1280 psf The ring compression stress is s = 182 psi Under some circumstances ring compression in the wall is not simply T = P(OD)/2 Consider a pipe with uniformly distributed pressure at the top and a line reaction (Class D) bedding, on the bottom See Figure 6-3 Class D bedding is poor practice — but happens For this loading, ring compression thrusts T occur at A and B even though sidefill pressure is zero Flexure occurs at A and B due to moments M Shear is zero because the load is symmetrical about the vertical axis Where thrust is known, the ring compression stress is T/A, or T/t for plain pipes Thrusts and moments are functions of loads, as discussed below Thrusts and Moments in the Ring Ring Compression Stress Thrusts T and moments M can be evaluated by energy methods such as Castigliano's equation for deflections due to loads: Due to external pressure on a thin-walled pipe, ring compression stress is, d = (M/EI)(JM/Jp)rdq s = P(OD)/2A, where OD = maximum outside diameter, P = external pressure, A = wall area per unit length = t for plain pipes See Appendix A d is deflection in the direction of a dummy load, p, (or dummy moment m for rotation) The dummy load or moment is applied at the point where deflection, or rotation, is to be found Assumptions are: The ring is thin-walled, D/t > 10 diameter D is used for analysis Mean Example A PVC pipe, DR 41, is a storm sewer under 10 ft of soil DR = OD/t = the standard dimension ratio Ring deflection is less than 5% and can be ©2000 CRC Press LLC The pipe material is elastic Ring deflection, d, is small Accuracy is adequate if d < 5%, or even 10% in some cases Figure 6-3 Buried pipe on a flat surface (left), showing the free-body-diagram for stress analysis (right) This Class D bedding is not recommended Example Complete a force analysis for Figure 6-3 Notation: D = 2r = mean diameter, T = ring compression thrust per unit length, M = moment in the wall per unit length, P = vertical soil pressure, Q = PD = line reaction per unit length, t = wall thickness, I/c = section modulus per unit length, XB/A = horizontal shift of B with respect to A, YB/A = change in tangent slopes of B with respect to A when the ring is loaded From Figure 6-3, with P known, five unknowns remain to be solved: T B, MB, T A, MA, and Q Because three equations of static equilibrium are available, two additional equations are needed Two equations of deflection are: From Figure 6-3 (right), angles q locate points where thrusts T and moments M complete freebody-diagrams of segments of the ring and become unknowns for solution by equations of static equilibrium plus equations of deflection from the Castigliano equation TA MA TB MB Q ©2000 CRC Press LLC YB/A = 0, and XB/A = As the ring deflects due to P, tangents at A and B remain horizontal Therefore YB/A = Point B shifts vertically, but not horizontally with respect to A Therefore XB/A = These two Castigliano equations for deflection, together with three equations of equilibrium, are solved simultaneously for the unknowns: = = = = = 0.1061 Pr compression 0.5872 Pr2 0.1061 Pr tension 0.2994 Pr2 2Pr Figure 6-4 Values to which ring deflection, d, will be reduced after internal pressure P’ is applied to buried steel pipes (assuming initial ring deflection is greater than d) ©2000 CRC Press LLC This analysis is conservative The theoretical line reaction Q is always worse than a soil bedding Horizontal pressure of soil against the pipe provides some support Measurements of soil stress reveal deviant stress patterns In general, pressure concentration shows up on the bottom due to a firm bedding But this may be reversed if the bedding is soft and soil is compacted on top of the pipe In general, pressure reduction shows up under the haunches because of the difficulty of soil placement But this may be reversed if concrete or low-slump soil cement is placed under the haunches or if the bedding is shaped by a V-cut Compaction affects soil pressure distribution The more flexible the ring, the more uniform is soil pressure against the pipe Vertically compressible sidefill causes concentration of pressures on the top and bottom of the pipe But an exception could occur if the pipe were located in a trench with firm sidewalls that support topfill by shear reactions However, in time, shear breaks down due to earth tremors, cycles of wetting and drying, and changes in temperature As a general rule, the vertical dead load on top of a flexible pipe is (OD)g H — called the soil prism load; i.e., the weight of a soil prism directly above the top of the pipe where g is the unit weight of soil and H is the height of soil cover above the top of the pipe This general rule may require a load factor for rigid pipe design because the rigid ring may have to support part of the backfill within the trench if sidefill soil is not compacted If sidefill is compacted, the soil prism load may be adequate for rigid ring design For flexible ring design, the soil prism load is conservative Normal pressures on the ring are no greater than pressure, P, at the top because, the flexible ring conforms with the soil, and the soil is invariably loose agains t the ring at the interface In plastic pipes, stress relaxation results in further reduction of normal pressure of the soil against the pipe The stiffer the ring, the greater are the pressure concentrations on top and bottom when sidefills are compressible For a rigid pipe, well-compacted ©2000 CRC Press LLC sidefill is necessary if pressure concentrations are to be avoided Combined Pressures The question arises, what are the stresses in the wall of a pipe subjected to both internal and external pressures? It would seem that external pressure should be subtracted from internal pressure, or vise versa For most installations, however, there will come a time when either internal or external pressure will not be acting Therefore, the ring is usually designed for internal and external pressures separately In the case of the flexible ring, because internal pressure is usually not applied until after the external soil pressure is in place, ring deflection has occurred before the pipe is pressurized If internal pressure is enough to partially re-round the ring, crescent gaps develop between the pipe and the sidefill See Figure 6-4 Clearly the ring no longer needs the support of the sidefill soil to retain its shape However, because of soil pressure on top, the pipe is not completely re-rounded If the specified allowable ring deflection is less than the ring deflection with soil load on top, crescent gaps not develop Figure 6-4 shows test results for steel pipes It is usually prudent to specify a minimum allowable ring deflection that is less than the value at which gaps would develop according to Figure 6-4 But even if crescent gaps develop, the ring does not collapse for lack of side support when it is depressurized The ring may or may not deflect — and any ring deflection will be less than before pressurization because soil particles tend to migrate into the gaps Combined pressures include the effect of live load passing over a buried pipe as explained in Chapter 4, and Equation 4.1, P = Pd + Pl If the water table is above the pipe, the unit weight of soil is increased Dead load pressure is found by soil mechanics explained in Chapter It may be concluded that internal pressure and external pressure are each analyzed separately Figure 6-5 Diagrams for force analysis of tanks buried to the top and subjected to internal test pressures Internal and external pressures are analyzed separately and combined by perposition ©2000 CRC Press LLC Re-rounding is seldom an issue for internal pressure analysis, if pipes are held to nearly circular shape when installed For design by ring compression, the prism load is the most reasonable load The prism load is usually the total (not just the effective) load Combined stress analysis is rarely justified, but may be required for rigid pipes — thick-walled and brittle based on the familiar equation, s = T/A + Mc/I, where T/A = ring compression stress, Mc/I = flexural (bending) stress Example In one city, acceptance for buried tanks is based on an internal pressure test when the tank is buried to the top See Figure 6-5a What are the tangential force, T A, and moment, MA, at the top, point A? Shearing force, VA, is zero by symmetry The soil is compacted sufficiently to prevent ring deflection Therefore, the ring is fixed at B, Figure 6-5b The effects of internal pressure and external soil load can be analyzed separately and then combined by superposition Figure 6-5c is the free-body-diagram for internal pressure analysis Figure 6-5d introduces the procedure for analyzing the effect of s oil load The moment at C due to the soil load is Ms It can be found by integrating the element of soil, shown cross-hatched, multiplied by its lever arm The result is the moment at C (angle 2), due to the soil load only The equation is, Ms = g r3[(1/2)sin 2q - (1/2)sinq - (1/4)sin2qsinq - (1/3)cos 3q + (1/3)] (6.4) The five unknowns of Figure 6-5b require three equations of equilibrium and two equations of deformation From deformation, by symmetry, the relative rotation of A with respect to B is zero; i.e., yA/B = The horizontal displacement of A with respect to B is zero; i.e., cA/B = From Castigliano, Appendix A, and Figure 6-6a showing a dummy ©2000 CRC Press LLC moment, m, at A in the assumed direction of rotation of A with respect to B, yA/B = (M/EI)( M/ m)rdq = M = MA + m - TAr(1-cosq) + Ms at point C M/ m = 1, then m in the M-equation EI is wall stiffness and r is radius Substituting into Castigliano's equation, and integrating within the limits for q from to p/2, the first equation of deformation becomes, MA - 0.3634TAr + 0.0174g r3 = (6.5) The second equation of deformation is zero horizontal displacement of A with respect to B Figure 6-6b shows a dummy force, p, in the assumed direction of relative displacement of A with respect to B From Castigliano, cA/B = (M/EI)( M/ p)rdq = M = MA - (TA + p)r(1-cosq) + Ms at point C M/ p = -r(1-cosq), then p in M-equation Substituting into Castigliano's equation and integrating within limits for q from to p/2, the second equation of deformation becomes, -MA + 0.6240TAr - 0.0320g r3 = (6.6) Equations 6.5 and 6.6 are solved simultaneously for the two unknowns, MA and T A Then by the three equations of equilibrium, MB, VB, and T B are evaluated The results are shown in Figure 6-6c Soil load decreases hoop tension at A due to internal pressure by TA = 0.0560g r2 Ring Stress — Uses and Misuses Stress is one basis of buried pipe design and analysis Most stress analyses are based on theories of elasticity for which yield stress is the performance limit (failure) Elastic stress analysis is useful in some cases such as hoop stress due to internal pres sure, ring compression stress due to external soil pressure, and the Figure 6-6 Force analysis of a flexible circular cylinder buried to the top ©2000 CRC Press LLC Boussinesq soil stress Unfortunately, elastic analysis is often misapplied In buried pipe analysis, properties of materials are not elastic — either for the pipe or the soil Performance limits are not elastic Performance limit (failure) of the pipe invariably occurs beyond the elastic range Performance limit is excessive deformation of pipe and soil Excessive deformation of the pipe may be fracture (leak), or collapse, or it may be so much deformation that cleaning tools cannot pass through the pipe, or that appurtenances (fittings) are distressed It is prudent to specify a maximum allowable ring deflection — but usually for reasons other than yield stress Performance limit of the soil is either excessive compression or soil slip — a plane on which s hearing stress exceeds strength From theories of elasticity, the maximum shearing stresses in soil embedment around a flexible pipe occur on planes through the pipe axis at 45o with the horizontal compression, sidefill soil is densified such that it resists soil slip (Three- dimensional compression can change carbon into diamonds.) In a controlled test, planes of soil slip were observed in the sand embedment of a flexible ring See Figure 6-7 Clearly, there are no soil slip planes at 45o or (45o f/2) There are no slip planes near the springlines of flexible pipes because the soil is in twodimensional compression PROBLEMS 6-1 Using Appendix A, what and where is the maximum compressive stress in a rigid pipe if D/t = and the load is uniform vertical pressure on top and bottom? Spangler's Iowa formula predicts ring deflection based on elastic pipe and elastic soil Spangler, a s oil engineer, included a horizontal elastic soil modulus E' which he called "modulus of passive resistance of soil." In fact, horizontal passive resistance is Ksy where K = (1+sinf)/(1-sinf) Soil slip planes occur at (45o - f/2) — not at 45o E' is not elastic, nor is it constant E' varies with depth (degree of confinement), and with ring stiffness (movement of pipe wall into the sidefill) Using the Mohr circle analysis, horizontal soil resistance is Ksy Accordingly, soil slip planes should occur at spring lines at angle (45o - f/2) with the horizontal The analysis is conservative The soil friction angle, f, is not constant It varies with depth of cover and ring deflection At spring lines, an infinitesimal cube of soil is subjected to twodimensional compression — vertical and radial Longitudinally it is confined Under two-dimensional ©2000 CRC Press LLC Figure 6-7 Soil slip planes in a sand embankment Note lack of soil slip at the spring lines ©2000 CRC Press LLC 6-2 A thin-wall flexible circular pipe resting on a flat surface is filled with water to the top With no additional internal pressure, what is the vertical ring deflection? See Figure 6-8 (Appen A) 6-3 A 90o thin-wall arch is pinned (hinged) at the two ends and is loaded with uniform vertical pressure P What are the reactions at the ends? (B x = 0.832Pr, and By = 0.707Pr) cation to a value small enough to be neglected Data are, (101 ft or less?) Given: Soil Type, SM, silty sand Compaction 80 percent (AASHTO T180) Ss = 1.0 = degree of saturation, G = 2.7 = specific gravity of soil grains, e = 0.4 = void ratio, sf = 1.0 at performance limit (wall crushing) Pipe D = 48 inches = diameter, t = 0.064 inch (16 gage) = specified wall thickness, A = 0.775 in 2/ft from AISI tables, I = 0.0227 in 4/ft, S = 36 ksi at yield strength, E = 30(106) = modulus of elasticity, Corrugations, 2-2/3 x 1/2 6-10 What height of dense soil cover can be supported by the pipe of Problem 6-9 if it is subjected to an internal vacuum of 12 psi? (H = 88.5 ft) 6-4 Plot the moment diagram for Problem 6-3 6-5 What and where is the maximum moment in Problem 6-4? 6-6 If the arch of Problem 6-3 is thick-walled, for which t = r/5, what is the maximum circumferential stress on the bottom surface of the arch at A? 6-7 What is the maximum circumferential stress at B in Problem 6-6? 6-8 What and where is the maximum stress in the plain pipe of Figure 6-3 if r/t = 10 and I/c = t2/6? (At point A, σ =353P) 6-9 What height H of dense soil cover can be supported by a 4-ft corrugated steel culvert? Assume that ring deflection, d, is held by design specifi- ©2000 CRC Press LLC 6-11 What height of loose soil, e = 1.0, can be supported by the pipe of Problem 6-9 if the soil liquefies due to earth tremors? (at r=24, N=14.9ft; at r= 24.25, H=15ft) 6-12 What wall thickness is required for a 2-2/3 x 1/2 corrugated steel pipe, 36-inch diameter, under a forest road (HS-20 loading) with ft of soil cover if the soil envelope is average granular soil compacted to at least 80 percent density (AASHTO T-180)? Assume that S = 36 ksi, and sf = for the steel Assume that the unit weight of soil is 120 pcf 6-13 What wall thickness is required for a 1.0meter bare (plain) steel pipe under 160 meters of saturated tailings with unit weight 140 pcf? A select soil envelope is specified What should be the conditions for installation? Is steel pipe a good option? Under what conditions? ... invariably loose agains t the ring at the interface In plastic pipes, stress relaxation results in further reduction of normal pressure of the soil against the pipe The stiffer the ring, the greater... Longitudinally it is confined Under two-dimensional ©2000 CRC Press LLC Figure 6-7 Soil slip planes in a sand embankment Note lack of soil slip at the spring lines ©2000 CRC Press LLC 6-2 A thin-wall... ring design, the soil prism load is conservative Normal pressures on the ring are no greater than pressure, P, at the top because, the flexible ring conforms with the soil, and the soil is invariably