Numerical Methods in Soil Mechanics 21.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 "EXTERNAL HYDROSTATICS" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 21-1 Free-body-diagram of a pipe immersed in water, showing uplift (buoyant) force, Wb = πr2γ w, which causes the pipe to float if it is greater than the weight, W, of pipe and contents ©2000 CRC Press LLC CHAPTER 21 EXTERNAL HYDROSTATICS The analysis and design of buried pipes is complicated by external hydrostatic pressures Chapter 10 describes the conditions for instability and collapse of pipes subjected to uniform, external pressure In fact, hydrostatic pressure is not uniform Moreover, instability is only one of the problems that arise from hydrostatic pressures (which include internal vacuum) Due to external hydrostatic pressure, the pipe may collapse, or crush, or float, or deform Each of these performance limits requires its own analysis Ring c ompression stress is based on total external pressure including external hydrostatic pressure Ring deflection is based on effective soil stress, which is total external stress minus the external hydrostatic pressure In Chapter the effect of radius of curvature of non-circular pipes is discussed Clearly, external hydrostatic pressure on the bottom plates of a pipe arch, with its long radius of curvature, can invert the bottom plates External hydrostatic pressure can cause a pipeline to float out of alignment Analysis and remedies for the problems of external hydrostatic pressure are described in the following FLOTATION OF PIPELINES What causes a pipe to float up out of vertical alignment, and what can be done to prevent it? The cause of flotation is an uplift force (buoyancy), Wb, which is the weight, per unit length of pipe, of the displaced liquid Figure 21-1 shows a pipe immersed in water If the buoyant force exceeds the weight of the pipe and its contents, W, the pipe will float upward For worst-case, the weight of the contents is neglected if at any time the pipe is empty Per unit length of pipe, W = Wp + Wc ©2000 CRC Press LLC (21.1) Notation Wb = W = Wp = Wc = Ws = OD ID D γb γw = = = = = sf = buoyant (uplift) force on the pipe, weight of pipe and contents, weight of the pipe per unit length, weight of contents per unit length, buoyant weight of soil wedges above a buried pipe per unit length of pipe, outside pipe diameter, inside pipe diameter, mean diameter, buoyant unit weight of the soil, unit weight of water, or of any other liquid in which the pipe is immersed, safety factor Per unit length of pipe, the buoyant force (uplift force), Wb, is the weight of liquid displaced The liquid could be mud or grout Wb = π(OD)2γ w /4 (21.2) For static equilibrium, the buoyant force must not exceed pipe weight reduced by a safety factor, Wb = W/sf (21.3) Design of a pipe immersed in water is simply a matter of substituting values into Equation 21.3 Performance limit is flotation of the pipe Typical remedies include weighting the pipe down with heavy collars, or thick walls, and holding the pipe down with straps and anchors Typically the immersed pipe is buried in soil beneath a body of water If the density of the soil exceeds critical density, failure is the formation of soil wedges uplifted as shown in Figure 21-2 Critical density occurs at that void ratio, greater than which the soil is so loose that it compresses when disturbed, and less than which the soil is so dense that it expands when disturbed Critical void ratio is roughly 85% density AASHTO T-99 (ASTM 698) The specification for minimum allowable density is usually set at 90% density Granular soil looser than critical does not break out wedges, but flows around the pipe as liquid or plastic through which the buoyant pipe rises This phenomenon can be investigated by approximate theory, but there are too many variables for a precise model A simplified theoretical analysis follows It is confirmed approximately by experimentation Figure 21-2 is a free-body-diagram of the soil wedges The soil wedges form a trapezoid of width D at the bottom, height Z = H + D/2, and with soil slip planes sloped at (45o + ϕ /2) 2v:1h; but reduced by a semi-circle of diameter D The buoyant weight of the soil wedges is, Ws = γ b[Z(D + 0.5Z) - πD2/8]/sf (21.4) In this equation, the slip slope, (45o + ϕ /2), is set at two to one for approximate analysis of granular soils See Chapter 13 In order to investigate the worst case, let Wc = for an empty pipe, and, assuming the pipe to be very light compared to the weight of the soil wedges, let Wp = If the weight of the pipe were significant, such as concrete pipe, its weight could be included From the equations of static equilibrium: Wb = Ws /sf (21.5) The following is a simplified, worst-case example Example How much cover H of compacted granular backfill is required over a buried pipe under water to prevent flotation? Assume an empty pipe of negligible weight; i.e., ID = D = OD; and with soil shear plane slopes at 2:1 Substituting values of Wb and Ws into Equation 21.5, γ wπD2/4 = γ b[Z(D + 0.5Z) - πD2/8]/sf, where γ b can be found from soil mechanics, Chapter ©2000 CRC Press LLC For this example, assume γ b = γ w Because Z = H = D/2, the relationship between H and sf can be found with results as follow: sf = H = 0.33D 0.72D 1.05D If control of the backfill is questionable, a very conservative rule of thumb might be to specify soil cover equal to the pipe diameter Considering the weight of the pipe and other mitigating features, the height of cover should be at least half the pipe diameter But the soil must be compacted to density greater than critical — especially if the specified height of cover is only H = D/2 Physical tests confirm the critical D/2 It was observed serendipitously, that in loose sand, a shock wave liquefies the sand (quick condition), and the pipe starts to float upward This occurs even in compacted sand if, adjacent to the pipe, the sand is less dense than critical Earth tremors (earthquakes) or shock waves liquefy the sand The pipe starts to float up But as soon as the shock wave passes, soil particles in suspension settle back in place and stop the rise of the pipe With repeated shock waves, the pipe rises little by little until it breaks free and floats to the surface A very severe shock can conceivably shake up dense saturated sand into a looser (quick) state by major soil upheaval — but only if enough time is allowed for water to seep into the soil during the upheaval Remedies include the use of gravel embedment which has less tendency to liquefy when shock waves pass through it Gravel is attractive under water where it is difficult to compact soil Gravel falls into place at density greater than critical From experience, even moderate shock waves cannot dislodge the pipe and cause it to float if the minimum soil cover is equal to one pipe diameter and if the soil is well compacted But how is compaction achieved under water? If the soil is granular, it can be vibrated or jetted But the surest technique is to place select gravel or sand that is coarse enough that it achieves adequate density just by moving it into place Flotation due to liquefaction is further reduced by the angle of approach of the shock wave The wave front rarely hits the entire pipeline at the same instant; i.e., at an angle of approach of 90o At any other angle, liquefaction occurs only at the intersection of the pipe and wave front The length of the liquefied zone is usually short enough that longitudinal beam action prevents flotation STIFFENER RINGS Stiffener rings on flexible pipes increase greatly the resistance to collapse by external hydrostatic pressure (and internal vacuum) Theoretical analysis is complex However, empirical and experimental data are becoming available Limited research has investigated the spacing of stiffener rings on thin-wall steel pipes with very heavy stiffener rings (infinitely stiff), immersed in water, with no soil support The most pertinent fundamental variables are: P' E D t L = = = = = vacuum at collapse, modulus of elasticity, diameter of the pipe, wall thickness, spacing of the rings From these five fundamental variables, all in terms of two basic dimensions (length and force), three dimensionless pi-terms can be written The interrelationship of these pi-terms can be investigated either by analysis or by experimentation The equation of one set of three pi-terms (in parentheses) can be written in the form, (E/P') = f [(D/t), (L/D)] Because the pi-terms are dimensionless, they have no feel for size; so testing can be performed on small-scale physical models ©2000 CRC Press LLC Example Based on small-scale model study, what are the equations for collapse of thin-wall steel pipes with infinitely stiff rings, when immersed in water (no soil support) See Figure 21-3 Beverage cans provided cylinders of uniform wall thickness, with correct similitude between the aluminum model and a steel prototype for which D/t = 492 The ends of the cans were cut out, and steel plugs, machined and gasketed, were inserted in the ends to provide a seal for the vacuum, and to simulate infinitely stiff rings The results of the experiment are shown in Figure 21-4 The best fit curves through data points are linear with their equations shown for corresponding ranges on the graphs for D/t = 492 It would be prudent to include a safety factor that would mitigate the fact that stiffener rings are not infinitely stiff, and the fact that the pipes are not always as nearly circular and cylindrical as the beverage cans Performance limit in every case was collapse of the model pipe by crumpling between the stiffener rings The general equation for collapse is: 106(P'/E) = (D/L)/m (21.6) where m is the slope of the best-fit straight line From the plots of data in Figure 21-4, collapse equations for stiffened steel pipes with D/t = 492, and E = 30(106) psi, are: L/D Vacuum to 5/3 P'= 14.3 D/L psi 5/3 to 10/3 P'= 8.6 D/L psi over 10/3 P' = 0.504 psi Equation (21.7) (21.8) (21.9) These equations are conservative, especially toward the left side of each equation range Figure 21-3 Flexible thin-wall pipe of diameter D with stiffener rings spaced at lengths L Figure 21-4 Plots of test data for critical vacuum term as a function of stiffener ring spacing term, showing best-fit straight lines and their equations in three ranges For these tests on steel pipes, D/t is a constant, 492 ©2000 CRC Press LLC In the range < L/D < 5/3, Equation 21.7 lower line is appropriate In the range 5/3 < L/D < 10/3, Equation 21.8 upper line is more appropriate, but still conservative Equation 21.9 is the theoretical solution for an infinitely long pipe without stiffener rings, i.e., P'D3/EI = 24 See Chapter 10 Until more data become available, for steel pipes, it is reasonable to adjust the critical vacuums of Equations 20.7, 20.8, and 20.9 (based on D/t = 492) to other values of D/t by a factor which is the cube of the D/t ratio; i.e., P'/ P'492 = [492/(D/t)]3 (21.1) For ring-stiffened steel pipes with D/t less than 240, this correction factor becomes so large that accuracy is lost For D/t = 240, the correction factor becomes P'/P'492 = 8.6 More accurate is the tank analysis, Equation 22.3, assuming that the end closures of tanks are equivalent to stiffeners Vacuum at collapse of tanks is, P' = 0.403E(1-ν2)-3/4(D/L))(t/r)5/2 Example Stiffener rings are spaced at 20 ft on a 144-inchdiameter thin-wall steel pipe, of 0.375-inch-thick steel, for which L/D = 20/12 and D/t = 384 What is the critical vacuum? Assuming excellent control over imperfections, use Equation 21.7 from which the critical vacuum for D/t = 492 is, P' = 14.3(D/L) = 8.6 psi With the adjustment factor of the ratio of (D/t)3 from Equation 21.10, and with D/t = 384 for the 144inch steel pipe, P' = 8.6(492/384)3 = 18 psi which is higher than atmospheric, 14.7 psi However, if external water pressure plus internal vacuum combine to a pressure greater than atmospheric, and/or if control is less than excellent, then using the more conservative Equation 21.8, for D/L = 12/20 and with the adjustment factor of Equation 21-10, P' = 8.3(D/L)(492/384)3 = 10.5 psi, ©2000 CRC Press LLC which is less than atmospheric pressure, but may be too conservative Based on the more accurate tank Equation 22.3, P' = 0.403E(1-ν2)-3/4(D/L)(t/r)5/2 = 15 psi Experimentation can start with two pi-terms, (P/E) = f(Lr/t2) Then flexibility of stiffeners (Pr3/EI) can be included, then soil support (ϕ) etc The example above is unburied When buried, the resistance to external hydrostatic pressure (plus vacuum) increases greatly depending upon compaction of the soil See Chapter 10 Hydrostatic pressure is not limited to water pressure If concrete with unit weight of 145 pcf is "poured" about the 144-inch pipe up to the top, the external hydrostatic pressure on the bottom is 12 psi The 20-ft spacing still appears adequate, but a greater margin of safety can be achieved if the concrete is placed in two lifts of ft each rather than one lift of 12 ft It is noteworthy that theoretical collapse vacuum for a long uns tiffened steel pipe of D/t = 492 (D = 144 inches, t = 0.293 inch) from Equation 21.9, is P' = 0.49 psi With stiffener rings spaced at 20 ft on a 144-inch pipe for which D/t = 492, from Equation 21.8, P' = 8.3(D/L) = 5.0 psi Stiffeners provide more than ten times the resistance to vacuum collapse than the plain pipe PROBLEMS 21-1 A two meter thin-wall plain steel pipe with a wall thickness of t = 10 mm, is buried in com-pacted fine sand with a soil cover of meter and water table at the soil surface on occasions The pipe could be empty Is there any possibility of flotation? Under what conditions? What is the possibility of soil liquefaction? Does the probability of flotation increase if the external water level is at flood stage above ground surface? G = 2.65 = specific gravity, e = 0.4 = void ratio, c ϕ = = = soil cohesion, 30o = friction angle (sf = 1.78) 21-2 What spacing of stiffener rings is advisable if the pipe of Problem 21-1 is to take a test vacuum of 14.7 psi? (L = 0.7D) 21-3 Find the wall section modulus for Problem 21-1 if stiffener rings are welded at meter ©2000 CRC Press LLC spacing Rings are 120 mm by 25 mm rectangular cross section with long axis perpendicular to the pipe wall E = 207 GPa = modulus of elasticity, S = 248 MPa = yield (I/c = 3850 mm3) 21-4 What external hydrostatic pressure plus 14.7 psi vacuum can the pipe of Problem 21-3 take? (P = 660 kPa) ... research has investigated the spacing of stiffener rings on thin-wall steel pipes with very heavy stiffener rings (infinitely stiff), immersed in water, with no soil support The most pertinent fundamental... steel plugs, machined and gasketed, were inserted in the ends to provide a seal for the vacuum, and to simulate infinitely stiff rings The results of the experiment are shown in Figure 21-4 The... the plain pipe PROBLEMS 21-1 A two meter thin-wall plain steel pipe with a wall thickness of t = 10 mm, is buried in com-pacted fine sand with a soil cover of meter and water table at the soil