Numerical Methods in Soil Mechanics 13.PDF

Numerical Methods in Soil Mechanics 13.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 "Frontmatter" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 13-1 Soil stress models for minimum soil cover — free-body-diagrams of truncated pyramid and cone showing shear planes at "punch-through" that direct the live load onto the pipe Figure 13-2 Truncated pyramid showing how a surface load, W, punches out a pyramid, which spreads the load over a base area of (B+H)(L+H) at depth H ©2000 CRC Press LLC CHAPTER 13 MINIMUM SOIL COVER As the soil cover H decreases, the live load effect on the buried pipe increases There exist minimum heights of soil cover H, less than which the surface live load may damage the pipe Less evident is a minimum height of soil cover for dead load, weight of soil only, on buried pipes Each of these cases is discussed for rigid and flexible rings in this chapter Only cohesionless soil is considered Vehicles are generally unable to maneuver on poor soil such as wet cohesive soil They get stuck in the mud In many of the following analyses, the effect of the surface live load on the pipe is based on the pyramid/cone model The Boussinesq and Newmark procedures for calculating live load pressure on the pipe are based on the assumption that the soil is elastic The assumption is not adequate for failures of buried pipes Failure of pipes due to surface loads on less-than-minimum soil cover is punch-through The pipe is not subjected to failure load until the soil cover fails in shear as wheels punch through the soil cover and then fracture or distort the buried pipe Based on such a model, the soil stress on a buried pipe is referred to as the pyramid/cone soil stress PYRAMID/CONE SOIL STRESS Figure 13-1 shows the truncated pyramid/cone soil stress models A surface live load can damage the buried pipe only after it punches through the soil cover If the loaded surface area is circular, a truncated cone of soil is punched out If the loaded surface area is a rectangle, a truncated pyramid of soil is punched out The truncated pyramid is imperfect because its sharp edges not form Nevertheless, by slight adjustment of the pyramid angle, the analysis can be made applicable The tire print of dual wheels is more nearly rectangular than circular Therefore the following analyses are based on the pyramid Figure 13-2 represents a dual-wheel tire print over a ©2000 CRC Press LLC rectangular area of width B and length L If the surface load is great enough to punch through granular soil and damage the pipe, then shear planes must form in the soil isolating a truncated pyramid which, like a pedestal, supports the load The total load on the pipe is the surface load W plus the weight of the pyramid of soil The weight of the pyramid is ignored because it is small compared to any surface load great enough to punch through The vertical soil pressure on the pipe is load W divided by the base area of the pyramid The angle θ which the shear planes make with the vertical is the pyramid angle θ = 45o - ϕ/2 where ϕ is the soil friction angle At depth H, the base area over which the load is spread is (B + 2Htanθ )(L + 2Htanθ ) From tests on cohesionless soil, the pyramid angle is roughly 35o for which the base area is approximately (B+H)(L+H) The precision is as good as can be justified for typical installations The results are conservative Refinements may be forthcoming It follows that at soil failure, the pressure on the pipe is the pressure on the base of the pyramid; i.e., P = W/(B+H)(L+H) (13.1) For HS-20 dual-wheel load on a firm surface, B = and L = 22 inches with tire pressure of 105 psi NOTATION P = vertical soil pressure at the level of the top of the pipe due to a surface load uniformly distributed over a rectangular area, W = weight of the surface load, γ = unit weight of soil, σy = stress at yield point, σ = ring compression stress, D = mean diameter of the pipe, r = mean radius of the pipe, c = distance from neutral surface of the pipe wall cross section to the most remote fiber on the surface Figure 13-3 Sketch of a surface live wheel load W passing over a pipe buried in loose granular soil Figure 13-4 Flexible ring in the process of collapse under minimum dead load soil cover showing the load wedges advancing against the ring and the lighter restraint wedges being lifted ©2000 CRC Press LLC A = I = M = T S E H H' H" ρ = = = = = = = cross sectional area of the pipe wall per unit length of the pipe, centroidal moment of inertia of the pipe wall cross section per unit length of pipe, moment in the wall of the ring due to ring deformation, circumferential thrust in the ring, compressive strength of the pipe wall, modulus of elasticity of the pipe material, installed height of cover of the soil, rutted height of soil cover, depth of the rut (See Figure 13-3), soil density in percent Standard Proctor (AASHTO T-99) for the granular soil cover and the embedment For a dual truck wheel on a compacted soil surface, the tire print area is about inches by 22-inc hes based on typical tire pressures as follows Dual Load (kips) Tire Pres (psi) 5.5 16 36 45 58 104 The loaded surface area can be adjusted for different loads and different tire pressures For a truck on pavement, the x 22-inch contact area is in reasonable agreement with observations MINIMUM HEIGHT OF SOIL COVER Minimum height of soil cover can be found by solving Equation 13.1 for H if the surface load W is known and if the allowable pressure P on the pipe can be evaluated for any given pipe and for any given performance limit, such as inversion or ring compression at yield The next problem is evaluation of the allowable pressure P This must include ring compressive strength, ring stiffness, and the critical location of the load The remainder of the chapter is devoted to this problem kips Vertical pressure P on the pipe at inversion is 0.8 ksf If B and L for the dual tire print are and 22 inches, respectively, from Equation 13.1, H = 28.6 inches, which is the minimum granular soil cover for protection of the pipe against collapse This does not include a safety factor, but if it is based on the assumption that the pipe ring is perfectly flexible — like a chainlink watch band — a margin of safety is built in depending on the ring stiffness A perfectly flexible ring is not practical Moreover, longitudinal beam strength of the pipe is ignored HEIGHT OF SOIL COVER An unsuspected problem in the minimum cover analysis is the definition of height of soil cover For a surfaced highway, the height of soil cover remains constant during passes of live loads But during construction, a heavy load crossing a buried pipe leaves ruts See Figure 13-3 In fact, successive passes of the load may increase the depths of the ruts If the depths of ruts approach a limit as the number of passes increases, the pipe-soil system is stable But if the depths of the ruts continue to increase with each pass of the surface load, it is obvious that the pipe feels increasingly adverse loads and may be in the process of inversion by ratcheting; i.e., an additional increment of ring deformation with each pass of the load Whatever the ultimate damage may be, a performance limit has been exceeded So minimum height of soil cover is defined as that soil cover H, less than which the pipe-soil system becomes unstable upon multiple passes of surface load W The height of cover to be used in Equation 13.1 for soil stress on the pipe is H', the height of soil cover after the ruts have reached their maximum depth, H" From Figure 13-3, H = H' + H" From tests in moist, granular, well-graded silty sand (SM classification), following rain (field moisture equivalent), the rut depths for dual wheels are generally not greater than, Example H" = 0.315(logW - 0.34)(103.9 - ρ) Consider a perfectly flexible ring Suppose that load W on a highway truck dual is 10 ©2000 CRC Press LLC (13.2) Figure 13-5 Truncated pyramid punched through the minimum soil cover, H, by an approaching surface wheel load W Shear planes form at a 1:2 slope The inversion arc is 2α Angle α is typically less than 45o Figure 13-6 Flexible ring in the process of collapse under minimum soil cover due to semi-infinite soil pressure P, showing the load "wedge" advancing against the ring and restraint "wedge" being lifted by the ring ©2000 CRC Press LLC where H" is in inches, and W is in kips Soil density ρ (in percent) is based on AASHTO T-99 The following table of values is from field tests from which roughly 90% of the rut depths were less than the values listed Equation 13.2 is not dimensionally homogeneous It was achieved by regression from plots of the following field data with tire pressures, p, as indicated H" = RUT DEPTH (inches) p (psi) 36 45 58 104 ρ = Soil Density (%) W= 5.5 kips W= 7.0 kips W= 9.0 kips W= 16.0 kips 80 85 90 95 3.0 2.4 1.8 1.1 3.8 3.0 2.2 1.4 4.6 3.7 2.7 1.7 6.5 5.1 3.8 2.4 DEAD LOAD Minimum cover of cohesionless soil over a buried pipe exists if the pipe is unable to support the variation in soil pressures around its perimeter The concept is shown in Figure 13-4, which shows a top pressure of γ H, but a shoulder pressure greater than γ H If the pipe cannot support the difference in pressures, the shoulder wedges will slide in against the pipe, deforming the ring which, in turn, lifts up the top wedges as shown Collapse of the pipe is catastrophic If the pipe is rigid (brittle), collapse is fragmentation If the pipe is flexible, equations of static equilibrium of the soil wedges provide values of minimum soil cover H For average granular backfill, H turns out to be about D/10 Experiments confirm the above analysis of dead load collapse at H = D/10 for very flexible pipes under dry granular backfill A suggested allowable value of D/6 for minimum cover allows for a margin of conservatism But this analysis is for a perfectly flexible ring In fact, pipes have ring stiffness and so provide resistance to dead load collapse ©2000 CRC Press LLC LIVE LOAD The minimum cover of cohesionless soil is not based on a location of live load directly over the crown of the pipe as in Figure 13-1 Rather, the critical location is an approaching load as shown in Figure 13-5 The leading edge of the base area of the truncated pyramid is at the crown The ring tends to deform as indicated Tests to failure of long span corrugated steel arches prove that static surface loads symmetrically located over the crown can be many times greater than the load on one side Static load failure is soil punch-through and ring collapse For fragile rigid pipes, failure is fracture of the pipe and possible collapse For flexible pipes, failure is inversion due to deformation as shown in Figure 13-5 The mechanism is downward deformation of the left shoulder of the ring under the load and consequential upward deflection of the right shoulder For granular backfill, the inversion angle is observed to be about α = 40o For convenience, and to be conservative, the collapse angle is assumed to be α = 45o for the truncated pyramid pressure shown in Figure 13-5 For the semi-infinite surface pressure shown in Figure 13-6, the load wedge contacts a full quadrant of the ring Analysis is evaluation of the maximum moment caused by the live load Dead loads are neglected The weights of the wedges and the shear resistance between them are small compared to the live load Moreover, in Figure 13-6, the weight of the load wedge tends to balance the restraint wedge Shear between pipe and soil is neglected The ring is fixed at both ends of the collapse arch Vertical soil pressure, P, becomes radial P, on a flexible ring See Figure 13-7 Castigliano's equation is used to find the reactions, the maximum moment, M, and thrust, T See Appendix A Maximum M is located by equating its derivative to zero If wall crushing is critical, thrust T is pertinent If circumferential stress is of interest, σ = T/A + Mc/I (13.3) ELASTIC LIMIT The thrust term is usually so small compared to the moment term, that it can be neglected It is much more likely that the critical performance limit is inversion Inversion is the result of plastic hinging that ultimately generates a three link mechanism See Figure 13-7 The thrust term is relatively small enough to be neglected Plastic hinging is a function of the Mc/I term, except that M is at plastic limit — not elastic limit For plain pipes and corrugated pipes, the moment at plastic hinging is approximately 3/2 times the elastic moment at yield stress Therefore, σ = 2Mc/3I PLASTIC HINGING Truncated Pyramid Load (Surface Wheel Load) For truncated pyramid pressure, the free-bodydiagram is a fixed-ended 90o arch See Figure 13-7 Dead load soil pressure is neglected Live load soil pressure is constant radial pressure, P, over 45o left of the crown, point A From Castigliano's equation, the maximum moment occurs at the point of minimum radius of curvature, about 12o to the right of the crown, A, and is: M = 0.022 Pr2 WHEEL LOAD (13.5) The circumferential thrust is T = γ Hr due to the dead weight of soil cover on the right side of the crown For design of the pipe based on yield stress, σf, the minimum required section modulus, I/c, is; (13.4) Figure 13-7 Free-body-diagram of the inversion arch for finding the maximum moment, M in terms of pressure P due to a surface dual-wheel load W approaching a pipe under minimum soil cover H The locations of potential plastic hinges are shown as circles starting at the location of maximum moment ©2000 CRC Press LLC I/c = (0.022Pr2)sf/σf ELASTIC LIMIT (13.6) I/c = (0.015Pr2)sf/σf PLASTIC HINGING (13.7) where sf is safety factor Tests show that I/c from these equations is conservative A safety factor of 1.5 is usually adequate, and does not need to be greater than for highway culverts responsive to the inversion of very flexible pipes with minimum soil cover when subjected to heavy surface loads It is noteworthy that H does not appear in Equations 13.7 and 13.8 It is presumed that the height of cover is already minimum as calculated by Equation 13.1 for punch-through of a truncated pyramid It is presumed that for a semi-infinite surface pressure, at the level of the top of the pipe, P is equal to surface pressure Example With M and T known, Equation 13.3 can be solved for the maximum stress whenever stress (or strain) is of interest, as in the case of bonded linings Semi-infinite Surface Pressure For the semi-infinite uniform surface pressure of Figure 13-6, the collapse arc under the load wedge is 90o Even though the ring is flexible, the assumption of constant radial pressure is conservative From Castigliano's equation, the maximum moment is, M = 0.08 Pr2 SEMI-INFINITE SURFACE LOAD (13.8) and is located at about 12o to the right of the crown T = γ Hr for use in Equation 13.3 But because ring compression stress T/A is usually small compared to the Mc/I stress, it can be neglected Setting Mc/I = yield strength/safety factor, and solving for section modulus I/c, I/c = (0.08Pr2)sf/σf ELASTIC LIMIT (13.9) I/c = (0.05Pr2)sf/σf PLASTIC HINGING (13.10) I/c is the required section modulus which can be found from tables of values for corrugated metal pipes and can be calculated for other pipes The equations for I/c are conservative, but are ©2000 CRC Press LLC Assume that a pipe buried under minimum cover is s ubjected to a semi-infinite surface pressure suc h that the edge of the base area of the load is at the crown of a buried pipe See Figure 13-6 What section modulus I/c is required for the pipe wall? The pipe is to be a 6-ft diameter corrugated steel pipe with granular soil cover of ft and semi-infinite surface pressure of P = 900 psf The yield strength of the pipe is 36 ksi Because the steel can yield, assume performance limit to be the formation of plastic hinges starting at the point of maximum moment Equation 13.10 applies Assume a safety factor of Substituting values, the required I/c = 0.27 in3/ft From the AISI Handbook of Steel Drainage and Highway Construction Products, a 3x1 corrugated steel pipe of 0.109-inch-thick steel is adequate For this 3x1, the I/c is listed as 0.3358 in3/ft Example Consider Example again, but for plain steel water pipe The section modulus I/c can be transformed into required wall thickness from the relationships I = t3/12 and c = t/2 But from example 1, I/c = 0.27 in3/ft The required wall thickness is t = 0.367 inch Specify wall thickness of 0.375 inch In the case of heavy surface wheel loads, it is often more economical to increase the height of soil cover in order to reduce the possibility of inverting the pipe Under some circumstances, stiffener rings can be attached to the ring to increase the section modulus See Chapter 21 Figure 13-8 AASHTO shandard H trucks Figure 13-9 Load-deflection diagrams from tests on 18D, HDPE corrugated pipe under inches of granular soil cover at 85% density (AASHTO T-99), showing permanent ring deflection d’ and rebound ring deflection d”, and showing the zone of instability ©2000 CRC Press LLC Example RING DEFLECTION Find the minimum cover of granular soil over corrugated polyethylene pipe 18 inches in diameter (ID) The polyethylene is designated as HDPE (high density polyethylene) The soil cover is compacted to 85% density (AASHTO T-99) The yield strength of HDPE at sudden inversion is ksi The surface load is a highway truck dual wheel for which the area of the tire print is inches by 22 inches The scheme is to substitute values of P from Equation 13.1 into Equation 13.5, M = 0.022Pr2 Including values of r and I/c for 18 HDPE pipe, the equation becomes a quadratic, Example is the result of field installations Ring deflections were observed under minimum cover Dual-wheel loads were varied up to the standard H20 load Figure 13-8 is a sketch of standard truck weights and dimensions (H+14.5in)2 = 56.25in + 25Win 2/kip Solutions are: W (kips) 5.5 H' (inch) -0.6 0.7 2.3 16 6.9 W = dual wheel load (16 kips = HS-20 load) H' = rutted soil cover A safety factor of two is usually applied to H' Some specifications require a minimum of one ft of compacted granular backfill For design of minimum cover, the safety factor is important because loads are often dynamic—not just static The negative 0.6 at W = 5.5 kips, indicates that soil cover is not needed for such a light load The pipe can carry a 5.5 kip dual-wheel load even though the ruts expose the pipe Of course, enough soil cover H should be provided to allow for rutting H", to prevent surface rocks from indenting the pipe, and to prevent crushing of the corrugations A similar analysis for 24 HDPE is almost identical to the table above for 18 HDPE Apparently manufacturers are careful to provide equivalent properties for their pipes in all sizes Installation techniques are probably about the same for 18- and 24- inch diameter corrugated HDPE pipes ©2000 CRC Press LLC From the tests, it was found that ring deflection comprises two components: permanent ring deflection d', and rebound ring deflection, d" The rebound ring deflection is elastic and rebounds fully after each pass of the dual-wheel load Figure 13-9 summarizes the ring deflections of 18 HDPE pipes buried under minimum cover of granular soil at 85% density (AASHTO T-99) and with inc hes of soil cover before passes of the wheel loads The following observations are noteworthy: Ring deflection is less than 2% for the first pass of the dual-wheel load For multiple passes, the rebound ring deflection stabilizes if dual-wheel loads are less than roughly 12.5 kips For dual-wheel loads greater than roughly 12.5 kips, rebound ring deflections may increase progressively This is defined as instability A probable zone of instability is suggested without enough test data to establish the boundaries precisely It was found, in additional testing, that with 12 inches of cover at 85% density before passes of the load, there was no indication of a zone of instability with dual-wheel loads up to 16 kips Example Corrugated polyethylene tubing is used extensively for drainage The slotted pipes are embedded in gravel Inside diameters are from inches to 18 inches What is the ring deflection due to dual- Figure 13-10 Results of tests showing ring deflection as a function of depth of cover term for buried corrugated polyethylene drain tubing subjected to AASHTO H-20 standard truck load Recommended minimum cover is indicated for each installation case ©2000 CRC Press LLC wheel loads as a function of depth of soil cover? Tests were performed on 12-inch pipes and reported to the American Society of Agricultural Engineers in June 1980 Results and recom-mendations are shown in Figure 13-10 Corrugated polyethylene pipes in larger diameters are available These are attractive for use as culverts under roads Additional concerns, such as dynamic loads, usually justify minimum cover greater than one foot FLOTATION When pipes are buried in soil under water, the minimum height of cover to prevent flotation of an empty pipe is about H = D/2 But the soil should be denser than critical in order to prevent liquefaction See Chapter 21 MINIMUM COVER FOR RIGID PIPES There are two basic performance limits for buried rigid pipes subjected to heavy surface loads with minimum soil cover They are longitudinal fractures and broken bells Circumferential fractures can occur, but less frequently Half a bell can be sheared circumferentially Circumferential cracks can occur at midlength of the pipe acting as a beam under a heavy wheel load if good bedding/embedment support is provided under the ends, but nowhere else Longitudinal Fractures Longitudinal fractures occur if vertical pressure P exceeds ring strength The worst case location of the critical live surface load is directly above the pipe — not on approach as in the case of flexible pipes See Figure 13-1 Minimum soil cover H is based on the punch-through pyramid/cone Longitudinal fractures occur at 12:00 and 6:00 o'clock and 9:00 and 3:00 o'clock This is not collapse of the pipe Many gravity flow pipes serve even when cracked ©2000 CRC Press LLC The soil envelope holds the ring in nearly circular shape See Figure 7-2 But for some rigid pipes, such as pressure pipes, longitudinal cracks are unacceptable Occasionally one longitudinal hairline crack occurs at 12:00 o'clock, or possibly at 6:00 o'clock if the pipe is on a hard bedding If the embedment is compacted select soil, a crack at 12:00 o'clock might be caused by a surface wheel load or a conscientious constructor who compacts the first layer above the pipe directly on top of the pipe It is prudent to compact sidefills, but to leave uncompacted the first layer directly over the pipe If the pipe is a culvert or storm drain, a single hairline crack is not performance limit Good embedment holds the pipe in shape such that the pipe is in ring compression, not flexure It performs like the ancient brick sewers of Paris and London which function well, but may not be leak-proof Analysis is the same for minimum cover and for maximum external pressure on rigid pipes See Chapter 12 The vertical pressure is P = Pl + Pd where the live load pressure, Pl, is found by the pyramid/cone theory For minimum cover analysis, the dead load pressure, Pd, is negligible The live load pressure, Pl, is a function of height of cover, H The minimum cover can be solved from the equation Pc r = Pl where critical pressure Pcr is a function of class of bedding and class of pipe See Chapter 12 Example A nonreinforced concrete pipe, 15 ID, bell and spigot, C-14, Class is to be used as a storm drain Outside diameter is OD = 19 inches From tests, the three-edge-bearing (TEB) strength is D-load = k/ft What must be the minimum cover, H, if a CAT 633 scraper is to pass over? The embedment is crushed (angular) granular material passing 1/2 inch sieve It contains less than 5% fines The scraper wheel load is W = 50 kips on a circular tire print area of 800 square inches The diameter of the circular tire print area is 32 inches The live load pressure at the top of the pipe is Pl = 4(50k)/π (32+H) From Chapter 12, the critical Figure 13-11 Rigid pipe cross section showing how voids are left under the haunches if soil is not shoved under Figure 13-12 Bell end of a section of pipe that is subjected to live load pressure, but is simply supported by reactions, Q, on the ends of the contiguous pipe sections ©2000 CRC Press LLC live load pressure is Pc r = 3kLf/ft(OD), where the load factor, Lf , depends upon the class of bedding Equating Pl = Pcr and solving for the minimum soil cover, H, Bedding Class A B C D D-load Lf _ 2.5 1.9 1.5 1.1 1.0 H (inches) 12.0 18.5 24.8 34.3 37.6 (TEB test) For a typical Class C bedding, the minimum cover is about 25 inches This is ring analysis Therefore, it is assumed that the bedding is non-compressible Horizontal ring support is neglected The pipe section under live load is not pressed downward between the two contiguous pipe sections such that it acts as a beam Broken Bells If a pipe section acts as a beam, the performance limit is usually a broken bell Under heavy live loads and minimum soil cover, rigid pipes require soil support under the haunches If soil is not actually shoved under the haunches, a void is left under the pipe See Figure 13-11 For example, if the angle of repose of the embedment is ϕ' = 40o, the void is wider than half the outside diameter [0.643(OD)] Live load on the pipe could cause the top of the pipe to move downward either by cracking the pipe, or by pressing the pipe into the bedding Under the haunches, loose soil at its angle of repose, offers little resistance As a pipe section is pressed downward, it becomes a simply supported beam with reactions, Q, at the ends of the pipe section See Figure 13-12 It is this reaction, Q, that fractures the bell Clay pipes and nonreinforced concrete pipes are vulnerable because of low tensile strength The maximum tensile stress is in the bell near the spring line Once cracked, a shard forms approximately one diameter in length as shown in Figure 13-12 An approximate analysis is the equating of this Q that can be withstood by the bell, to the Q reaction ©2000 CRC Press LLC caused by the live surface load on the pipe section acting as a beam Example What is the minimum cover, H, for the preceding example if fracture is a broken bell? The length of nonreinforced pipes is L = ft Assume (estimate) that the cross-sectional area of the thin part of the bell is A = square inches Tensile stress at the spring lines is σ = Q/2A, from which maximum Q = 2Aσf , where σ f = tensile strength of the concrete Tensile strength of the concrete is σf = 1.0 ksi It is found from the three-edge-bearing test from which D-load = k/ft See Figure 13-13 Therefore, reaction Q at fracture is Q = 2Aσf = 10 kips But Q is the reaction to pressure Pl on the pipe which is caused by surface live load, W From the punchthrough cone analysis, Pl = 4W/π(32+H) The pressurized area is going to be greater than half the pipe length Therefore, with little error, it is assumed that W is located at midspan, and that reaction Q = 0.5P(OD)(32+H) Substituting for P and equating the two Q's, W(OD)/π(32+H) = Aσf (13.11) From Equation 13.11, the minimum soil cover is H = 28.5 inches In this analysis, the pipe section is a simple beam with no support from the bedding or embedment Minimum cover will be less than 28.5 depending upon the support This analysis is approximate Therefore, it is prudent to specify good bedding and embedment, and to require a minimum cover of three feet for the impact loads of construction traffic To place and compact embedment under the haunches, a windrow of soil along the pipe can be shoved into place by laborers with J-bars working from on top of the pipe, or by flushing the windrow under the haunches with a high-pressure water jet, or by hand operated mechanical compactors Some constructors are now placing soil cement slurry with about ten inch slump under the haunches See Chapter 16 Rigid pipes are usually strong enough as beams to resist TEB TEST MA = (D-LOAD)r/π See Appendix A If r = 8.5 inches and D-LOAD = 3,000 lb/ft MA = 676.4 lb σf = 6M/t2 = 1,014.6 psi Tensile strength is σf = 1.0 ksi Figure 13-13 Evaluation of tensile strength of a typical 15 inch nonreinforced Class concrete pipe based on three-edge-bearing (TEB) tests from which D-LOAD = 3,000 lb/ft ©2000 CRC Press LLC circumferential fracture at midspan Manufacturers limit the length of pipes to prevent beam failures The following example is an interesting comparison of beam analyses Example What is the minimum cover, H, for the pipe in the above examples based on maximum longitudinal tensile stress, σ = Mc/I, in the bottom of a simply supported beam? With a uniform load, w, at midspan, M = wL2/8, where w is the load per unit length of beam; i.e., w = P(OD) P = 4W/π(32+H) I/c = (π/32)[(OD)4-(ID)4]/(OD) If tensile strength is σf = ksi, substituting values into the equation, σf = M/(I/c), H = 26.2 inches Failure by a broken bell is more critical but not by much PROBLEMS 13-1 One wheel of a scraper with rubber tires carries a load of 25 kips What is the diameter of the tire print on the ground surface if tire pressure is 30 psi? (D = 32.6in) 13-2 What is the minimum height of cover of granular backfill compacted enough that rutting is negligible, if it is to protect a 6x2 corrugated s tructural steel plate culvert of 0.1345-inch-thick s teel? The radius is r = 48 inches The maximum dual-wheel load is 16 kips on an asphalt surface ©2000 CRC Press LLC which does not spread the wheel load What about safety factor? I = 0.938 in 4/ft, and I/c = 0.879 in3/ft (11.8 m) 13-3 How the solutions for Problem 13-2 compare with solutions from the AISI Handbook of Steel Drainage & Highway Construction Products? Does the safety factor need to be two? (12, AISI, Handbook of steel drainage and highway construction products, 1993, p 258) 13-4 What is semi-infinite surface live load at failure of a flexible pipe buried in granular backfill using plastic hinging analysis? Pipe steel Soil 3x1 corrugations Pit-run gravel t = 0.109 H = 15 inches D = 96 inch diam γ = 125pcf I/c = 0.3358 in 3/ft σf = 36 ksi (1.26 ksf) 13-5 What is the minimum height of soil cover H for a nonreinforced concrete pipe Class if an HS-20 truck dual-wheel load passes over? W = 16 kips Pipe OD = 19 inches ID = 15 inches σf = 1.0 ksi Soil Compacted granular soil φ = 40o ... minimum soil cover due to semi-infinite soil pressure P, showing the load "wedge" advancing against the ring and restraint "wedge" being lifted by the ring ©2000 CRC Press LLC where H" is in inches,... W passing over a pipe buried in loose granular soil Figure 13-4 Flexible ring in the process of collapse under minimum dead load soil cover showing the load wedges advancing against the ring and... for the maximum stress whenever stress (or strain) is of interest, as in the case of bonded linings Semi-infinite Surface Pressure For the semi-infinite uniform surface pressure of Figure 13-6,