Numerical Methods in Soil Mechanics 15.PDF

Numerical Methods in Soil Mechanics 15.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 "THRUST RESTRAINTS" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 15-2 Impulse thrust, Qi, at a bend in a pipe due to change in direction of flow, showing a free-vectordiagram of the components Fx and Fy ©2000 CRC Press LLC CHAPTER 15 THRUST RESTRAINTS A straight pipe section with end closures (valves or caps) feels a longitudinal force when internal pressure is applied In the case of static pressure, the longitudinal force F is simply the internal pressure times the area; i.e., F = PπD2/4 where F = P = D = longitudinal thrust in the pipe, internal pressure, inside diameter = 2r But suppose the pipe has gasketed joints between the ends See Figure 15-1 Now it cannot resist the force F Consequently, thrust restraints (thrust blocks) must be supplied at the ends of the pipe Qi = vector sum of impulse forces Fx and Fy, θ = offset angle of the bend, v = average velocity of fluid flow in the pipe Fx = axial force on the fluid at the elbow, D = inside diameter, ρ = mass density of the fluid, (v-vcosθ) = change in the x-components of velocity as the column of fluid flows around the bend Fx and Fy can be found by the principle "impulse equals change in momentum." Both impulse and momentum are vector quantities Figure 15-2 shows a free-body-diagram of a column of fluid (crosshatched) The area is πD2/4, and the length is (vdt) Impulse is force times time dt, and change in momentum is mass times change in velocity In the x-direction; Impulse = Fxdt EVALUATION OF THRUST Q Change in Momentum = (πD2/4)vdtρ(v-vcosθ) Each thrust restraint must resist F without moving enough to allow a joint to leak In addition to the pressure, P, suppose that the fluid is moving in the pipe due to pressure gradient, ∆ P Force F is increased by ∆ F = ∆ PπD2/4 For most buried pipeline analyses, fluid friction, ∆F, is negligible because the length between gaskets is short and is easily resisted by soil friction on the pipe But now, suppose that the gasketed pipe is not straight A change in direction is introduced by an elbow (or bend) The sidewise thrust Q at the elbow is due to both pressure and the impulse of change in direction of flow It is the vector sum of impulse and pressure forces, Qi and Qp, on the fluid at the bend Each is found separately Impulse Thrust Qi See free-vector-diagram, Figure 15-2 F = impulse force (vector), Qi = thrust due to impulse, ©2000 CRC Press LLC Equating impulse to the change in momentum in the x-direction, Fx = π(Dv)2 ρ(1-cosθ)/4 But Fx is only the x-component In a similar manner by equating the y-component of impulse to change in momentum; Fy = π(Dv)2 ρ(sinθ)/4 From the free-vector-diagram of Figure 15-2, the resultant of Fx and Fy is, 2Qi = π(Dv)2 ρsin(θ/2) (15.1) The angle between Qi an d Fy is tan-1(Fx /Fy) = θ/2 Noting that the pipe is symmetrical about the Qi vector, Equation 15.1 could have been written directly, because the change in velocity in the Qi direction is simply 2vsin(θ/2) Figure 15-3 Pressure thrust-Q p at a bend (elbow) in a pipe due to internal pressure, P, showing the freevector-diagram for calculating Qp Figure 15-4 Passive soil resistance on an elbow and on contiguous gasketed pipe sections showing how the soil envelope can provide thrust restraint ©2000 CRC Press LLC Pressure Thrust Qp SPECIAL SECTIONS See Figure 15-3; where D = inside diameter = 2r, P = internal fluid pressure, Qp = thrust due to internal pressure, θ = offset angle of the bend (elbow) Special sections redirect or alter flow Examples include elbows, wyes, tees, valves, reducers, caps, plugs, etc The following analyses for elbows can be applied to any special section In every case, thrust, Q is the sum of impulse thrust, Qi, and pressure thrust, Qp A free-body-diagram of the elbow with pressurized fluid contents is shown cross-hatched Neglecting the small friction loss of flow around the bend, from the free-vector-diagram, 2Qp = πD2Psin(θ/2) (15.2) Qp is at an angle of θ/2 with the y-axis Consequently, thrust, Q, is the sum, Qi+ Qp; i.e., 2Q = πD2(P + v2ρ)sin(θ/2) where v = ρ = θ = (15.3) average velocity of fluid flow, mass density of the fluid, offset angle of the bend COMMON THRUST RESTRAINTS Welded or Bolted Joints at Special Sections In a pressurized pipe, at a gasketed elbow, Q must be resisted by the soil or by a thrust restraint (thrust block) For a welded elbow, Q is resisted by the pipe Two analyses of a welded elbow follow a) If the contiguous pipes are unrestrained and uncapped (like a garden hose), normal force, F, and shearing force, S, act on the elbow Analysis is conservative because soil resistance reduces F and S σ = F/2πrt = average normal stress, τ = S/2πrt = average shearing stress Example From the equations of static equilibrium, Find thrust-Q at a 90o elbow in a water pipe for which, θ = 90o, D = 30 inches, P = 200 psi = internal pressure, v = 15 ft/second = flow velocity, ρ = γ w /g = mass density of water, γ = 62.4 lb/ft3 = unit wt of water, g = 32.2 ft/second2 = gravity Substituting into Equation 15.1, Qi = kips Substituting into Equation 15.2, Qp = 200 kips Combined, Q = 203 kips Impulse thrust, Qi is usually neglected If a large diameter pipe with high internal pressure has an elbow with a large offset angle, θ, thrust-Q is enormous ©2000 CRC Press LLC σ /P(r/t) = (1-cosθ) NORMAL STRESS TERM (15.4) τ /P(r/t) = sinθ SHEARING STRESS TERM (15.5) These stress terms are upper limits — twice the force-per-unit-area — to account for eccentricity of the F-force and redistribution of stresses The outside of a bend can stretch more than the inside Therefore, stresses are greater on the inside See Problem 15-12 Wall thickness is sometimes increased for elbows In general, greater wall thickness is not justified b) If the contiguous pipes are restrained and capped, from the equations of equililbrium, longitudinal stress is, σ = Pr/2t (15.6) This is only half as great as circumferential stress, and is independent of offset angle, θ A more precise analysis would show that stress, σ , on the inside of the bend is increased slightly as the offset angle, θ, is increased Most pipes are ductile enough that the material "plastic-flows" at yield, and does not fail Moreover, soil friction resists thrust In practice, contiguous pipes are seldom capped Longitudinal stress is not critical for isotropic plain steel and plastic pipes Of course, joints must be adequate For non-isotropic pipes (corrugated, ribbed, or wrapped with fiberglas or wire), longitudinal strength must be assured Neglecting impulse force and soil resistance, for uncapped, unrestrained contiguous pipes: At elbows, for longitudinal design, Pπr2(1-cosθ) = Aσ f /sf (15.7) where A = area of longitudinal fibers, σf = strength of the fibers At valves or caps (not at bends) for design, Pπr2 = Aσ f /sf (15.8) Embedment As Thrust Restraint If thrust-Q is not large, the embedment is able to develop adequate passive resistance It may not be necessary to provide additional thrust restraint Consider in Figure 15-4 the free-body-diagram of an elbow and one section of pipe on each side The joints are gasketed so the pipe can take no longitudinal force Thrust-Q can be restrained only by the soil bearing against the pipe The maximum soil pressure bearing horizontally against the elbow is passive resistance Px at the average depth of soil, H + OD/2, Px = (2H + OD)γ /2K ©2000 CRC Press LLC where K = ϕ = γ = OD = H = L = P/Px = (1-sinϕ)/(1+sinϕ), soil friction angle, unit weight of soil, outside diameter, height of soil cover, length of pipe section The restraint capacity of soil against elbow is, Qelb = (area) times Px where (area) Lelb = = (OD)Lelb, cord length (approximate) of elbow from coupling to coupling as shown Multiplying (area) times Px, Qelb = (2H + OD)γ LelbOD/2K Added to this is the restraint capacity of the first section of pipe on each side of the elbow Full passive resistance of the soil would be developed at the elbow end of each section At the opposite end, each pipe section could rotate, because of the gasket But there would be no lateral movement Passive soil resistance would not be developed A crude, but reasonable and conservative assumption, is that passive resistance varies linearly from Px at the elbow end to zero at the opposite end Due to soil supporting the two pipe sections, the component of restraint in the direction of Q is, Qsecs = (OD)LP xcos(θ/2) or, substituting for Px, Qsecs = (OD)L(2H + OD)γ cos(θ/2)/2K Combining the thrust restraints provided by the elbow and the two pipe sections, Restraint-Q = OD(2H + OD)γ [Lelb + Lcos(θ/2)]/2K (15.9) h j Rewriting Equation 15.3, Thrust-Q = π(ID)2(P + v2ρ)sin(θ/2)/2 (15.10) Equation 15.10 for thrust-Q was derived for a horizontal bend For a vertical bend (in a vertical plane), thrust-Q has a vertical component If soil cover alone is to resist the upward component of thrust-Q, then soil cover H must be great enough that soil weight can hold the pipe down A conservative restraint-Q for this vertical bend is, Restraint-Q = OD(2H + OD)γ [Lelb + Lcos(θ/2)]/2 = = ϕ = K = γ = γc = H/B = ratio of soil cover H to side B, ratio of distance between top of block and thrust-Q, to side B, 30° = soil friction angle, 1/3 = (1-sinϕ)/(1+sinϕ), 120 pcf = unit weight of soil, 144 pcf = unit weight of concrete Taking the sum of the moments of force about overturn fulcrum O, Q/γ B3 = (2h + 1.10)/(1-j) OVERTURN (15.12) Taking the sum of the horizontal forces, (15.11) Q/γ B3 = (3.577h + 2.193) This is the same as Equation 15.9 except that K is eliminated For design, restraint-Q must be greater than thrust-Q A safety factor should be included Thrust Block as Thrust Restraint Thrust blocks are the most common restraints in use for pressurized gasketed pipes See Figure 15-5 Thrust blocks are usually concrete A reasonable analysis for design starts with the free-bodydiagram Assuming a cubical block, B = lengths of sides of the cube, γ = unit weight of soil, γ c = unit weight of the thrust block, jB = distance down to thrust-Q from the top of the block, K = (1-sinϕ)/(1+sinϕ), ϕ = soil friction angle Other data are shown on the sketch Friction on the sides of the block is undependable and is conservatively neglected T wo modes of failure are considered: overturn about point O, and slip The conditions under which each mode controls are described by an example of a cubical thrust block Example — Assumptions ©2000 CRC Press LLC SLIP (15.13) The dimensionless quantity Q/γ B3 is the thrust block restraint number A table of values is shown as Table 15-1 for typic al design based on the assumptions indicated Overturn In order to design a cubical thrust block with the typical soil properties assumed in the analysis above, it is only necessary to guess a trial value for B from which values of h and j can be calculated Entering Table 15-1 with h and j, the restraint number, Q/γ B3 can be found in the overturn columns For a soil unit weight of γ = 120 pcf, Q/B3 = (120 pcf)(restraint number)/sf Solve for B If not the same as the assumed B, using the new B recalculate values for h and j Enter Table 15-1 for a second trial solution of the restraint number from which a new value of B is calculated If this new B is unchanged, then the answer has been found If not, recycle the analysis with the new B Slip The left of the two SLIP columns of Table 15-1 Figure 15-5 Free-body-diagram of a cubical thrust block Table 15-1 Values of cubical thrust block restraint number, Q/γ B3, for concrete at 144 pcf and soil at 120 pcf and ϕ = 30o No safety factor is included ©2000 CRC Press LLC provides values for the restraint number from Equation 15.13 The right of the two slip columns is the minimum value of j at which slip is critical No safety factor is included The analysis is so conservative, that safety factors need not be large Nevertheless, the risk of failure may warrant a safety factor One novel concept is the thrust pin designed to conserve space See Figure 15-6 It can be located inside the bend if necessary, tied with tendons Tendons Instead of thrust blocks or thrust pins, which restrain the gasketed elbow by compression from outside the bend, restraint is by tension tendons inside the bend fastened to dead-men such as buried concrete blocks, boulders, beams, pins, etc The tendons could be rods, cables, wires, etc Or, instead of tying tendons to dead-men, they could be tied across the bend to corresponding pipe joints on either side of the elbow See Figure 15-13 These "harp-strings" cannot resist the thrust without other restraints such as longitudinal friction between soil and pipes and soil bearing PIPES ON STEEP SLOPES The analysis of thrust restraints, for pipes on slopes, is the same as above; but in addition, must include longitudinal forces caused by gravity For most pipes, the length tends to shorten when the pipe is in service because internal pressure increases and temperature decreases Therefore, it is good practice to design restraints such that the pipe shortens downhill Frictional resistance to shortening is uphill and partially offsets the downhill component of weight of the full pipe The thrust restraint (anchor) is clamped to the pipe uphill from it, and is slip-coupled to the pipe downhill from it The downhill side is free to slip toward the next anchor downhill Expansion joints allow slip For short pipe sections, com-mon sleeve-type ©2000 CRC Press LLC couplings allow adequate slip For most couplings, the pipe must be supported on both sides of the coupling to assure alignment Good backfill soil may provide alignment In the case of poor backfill, or pipe on piers, two yokes on each pier assure alignment as shown in Figure 15-7 Most couplings are not designed to resist longitudinal moment or transverse shear The allowable degree of misalignment of a coupling is limited For example, the allowable misalignment (offset angle) is about 3o for steel pipes 30 to 54 inch diameter Manufacturers of couplings should be consulted for restrictions and specific applications If the pipe is on piers, couplings should not be located at midspans between piers Particular care is required for large pipes on steep slopes because of the difficulties of installation as well as the additional loads on the thrust restraints On slopes steeper than about 45°, the pipe is often placed on piers above ground The slope is too steep to excavate a trench, too steep to hold the pipe in position for welding and backfilling, and too steep to compact backfill Moreover, a pipe on a steep slope may feel the downhill drag from creep of surface soil (downhill freezing and thawing) and thus overload the anchor at the bottom of the slope Slopes steeper than 45° are usually rock outcrops that cannot be excavated without ripping or blasting On steep slopes, or in inaccessible areas, construction might require lowering personnel, platforms, and equipment down the pipeline by cables Rock pins are drilled and grouted into place The pipe is laid downhill as the platform is lowered Economics often favor service from an overhead cable on towers, or from helicopters Helicopters are expensive (one to two thousand dollars per hour), but can place pipes and piers quickly if ground crews avoid delays Under some conditions, tunneling may be an option Example The following example identifies some of the many problems associated with pipelines on steep slopes Figure 15-6 Alternate concept of thrust restraint at an elbow in a pipeline provided by a thrust pin that conserves space and quantity of concrete The hole is bored, reinforcing steel is positionced, and concrete is cast into the bored hole ©2000 CRC Press LLC A 30D steel penstock is to be installed in a remote area in a cold climate Much of the 4800-ft pipeline is on slopes less than 10°, in soil that can be excavated for burial of the pipe to protect it from avalanches, falling trees, and temperature extremes However, part of the pipeline will be placed above ground on piers and anchors in rock outcrops on a steep 48o decline See Figure 15-7 The area is heavily timbered and inaccessible Elbows in the pipeline will be anchored at the top and bottom of the 48° slope What are the requirements for piers and anchors? will be a monolithic casting The upper ends of the rebars may be threaded such that the bands that clamp the pipe will also secure the pipe to the pier It is cost effective to bring in pipes, equipment, and concrete by helicopter The required pipe wall thickness is 0.375 inch If the maximum length of a pipe section is 30 feet, the weight is about 3.6 kips, which is the load capacity of the helicopter Bolted sleeve-type couplings are selected because they eliminate field welding, which is slow and tortuous on the steep slope Couplings can be assembled quickly and without heavy equipment For 30-ft lengths of pipe, ordinary sleeve couplings can accommodate the longitudinal expansion and contraction due to large temperature changes above ground A change in temperature of 100o Fahrenheit causes a quarter of an inch change in length in each 30-ft section of pipe If the pipe were to be installed in long sections, special expansion joints would be needed The amount of concrete may be reduced by using prefabricated chairs In this example, chairs for supporting the pipe are to be flown in and fastened, either to pins drilled and grouted into the rock, or to small concrete footings anchored to the pins Two pins must support the downhill components of weight of the chair and footing plus the pipe section full of water uphill from the pins, reduced by the frictional resistance to the normal component of weight of chair, footing, and one full pipe section The chair weighs about 0.5 kips If the pier requires two cubic yards of concrete it will weigh about 7.5 kips If the weight of the full pipe is 430 lb/ft, it will weigh 13 kips The total weight is about 21 kips The basic coefficient of friction is the tangent of the soil friction angle, say, 30° The resulting downhill shearing load is 11 kips See Figure 15-8 The shearing stress on two #8 bars of 0.7854 in area each is 11/1.57 = ksi which is not excessive However, bearing of steel on grout should be checked The upper ends of the #8 bars are threaded such that the chair can be bolted to the pins If necessary, the downhill chair foot can be bolted to pins — one pin in the center of the foot, or two pins at the ends of the foot Before the pipes are flown in, piers or chairs must be in place and fixed Unless deep foundations can be provided, rock pins are required Holes are drilled into the rock at the uphill side of each pier Deformed reinforcing rods (rebars) are grouted into the drilled holes as shown in Figure 15-8 In order to assure bond, the depth of the holes should be at least 50 diameters of the rod If #8 rebars are used, the depth of the holes must be 50 inches In cold climates, frost penetrates more than 50 inches, so 10-ft depth is prudent If piers are to be used, concrete placed from a helicopter requires a tremie and good luck as well as skill Forms for the piers can be light-weight and reusable Each pier The bands that secure the pipe to the chair also clamp the pipe and prevent slipping of the band due to the 11-kip shearing load See Figure 15-9 To prevent slip, each band must be tensioned to four kips (assuming coefficient of friction between band and pipe is one-third) It may be prudent to clean and roughen the surface or to apply epoxy and "salt" the surface with carborundum dust To tension the bands, with a margin of safety, use 3/4 bolts with tensile strength of 6.19 kips each and tensioned to kips The band is a 1/4-inch steel strap, inches wide A large square washer is required to distribute the load and eliminate bending at the hole where band width is only about inches ©2000 CRC Press LLC P Figure 15-7 Penstock on a slope showing examples of a concrete pier with two yokes for alignment of the coupling, and anchors (thrust restraints) at the elbows ©2000 CRC Press LLC Figure 15-8 Chair on a 48o slope on a rock outcrop, showing rock pins used to secure the chair to the outcrop, and showing a bolt, at the downhill end of the diagonal brace, that serves as a pivot for slight rotation of the downhill legs of the chair, eliminating the need for a slip band ©2000 CRC Press LLC Figure 15-9 Partial details of a chair that can be flown onto location by helicopter and bolted to rock pins ©2000 CRC Press LLC Adjustment of the fit of band-to-chair and chair-topins (or footing) is accomplished as follows Bolt holes are slotted to allow for horizontal adjustment, and shims are inserted to allow for vertical adjustment and rotation Adjustment of longitudinal displacement of the downhill pipe section with respect to the uphill pipe section is accomplished by pivots at the lower ends of the downhill chair legs See Figure 15-9 A 7/8 bolt is adequate in double shear The vertical leg must be cut to allow for slight rotation Thrust restraints (anchors) are required at the elbows shown in Figure 15-7 Anchor, A, at the inverted elbow is critical because the pins are in tension The depth of the six bored holes must be determined If pressure in the pipe is 433 psi (1000 ft static head), the uplift thrust at the elbow is Qp = 158.56 kips If mean velocity of flow is 15 fps, Qi = 1.12 kips which can be neglected If the weight of elbow and anchor is assumed to be 50 kips, Q = 160 - 50 = 110 kips To hold the elbow down, each of six deformed steel bars must resist a pullout load of 20 kips Because the loads may not be equally distributed, assume 30 kips Select #11 rebars with cross-sectional areas of 1.485 in for which tensile stress is 20 ksi If the shearing bond between the bar and the grout is 100 psi (equivalent to 50 diameters of overlap), each bar must be grouted to a depth of 6.2 ft Specify 12 ft of depth to assure a margin of safety The hydrodynamic design of the elbow is within acceptable limits See Chapter 18 The mean radius of the bend is greater than 2.5 pipe diameters (about three diameters), the inside length of each mitred section is greater than half the pipe radius, and the angle offsets of continuous mitred sections are only 7.5° The height of each chair must be custom fit to the rock outcrop A laser beam is useful for measuring up each chair If the height is too great, towers must be designed Sections of prefabricated towers can be flown in If concrete is a realistic alternative, the tower could be a vertical pipe filled with reinforced concrete ©2000 CRC Press LLC EXPANSION INSERTS An expansion/contraction insert allows for slight longitudinal expansion or contraction without overstressing the pipe or joints Slip couplings and gaskets essentially eliminate all longitudinal thrust in the pipe However, under some circumstances, a corrugated pipe insert or bellows section can reduce thrust to acceptable levels Two questions arise: What is the elongation of the insert per unit length of pipe at yield or endurance limit? What is the longitudinal force per unit length of circumference of the insert at yield stress or at the endurance limit? Analysis is based on the free-body-diagrams of Figure 15-10 for a corrugation, and Figure 15-11 for a bellows Notation T = x = b = c = t = σ = E = σf = thrust per unit circumference, elongation of insert due to T, pitch of corrugation (or bellows), depth of corrugation (or bellows), thickness of insert material, maximum stress in the insert, modulus of elasticity, yield stress From symmetry and static equilibrium, analysis is performed on one-fourth of the pitch, b/4 as shown in Figure 15-10 Maximum stress is, σ = T/t + 3Tc/t2 (15.14) For corrugations with ratios of b/c = 3, the axial term, T/t, is less than 6% of the flexural term, 3Tc/t2, and may be neglected This is mitigated by using elastic theory for elasto-plastic pipes The ratio of axial to flexural stress, is t/3c If c = and t = 0.1345 inch, the ratio of axial to flexural stress is 0.02 Axial stress is negligible Elongation, ∆ x, of a fourth of a corrugation due to Figure 15-10 Procedure for analysis of a corrugation to find the relationship of thrust to elongation Figure 15-11 Procedure for analysis of a bellows to find the relationship of thrust to elongation ©2000 CRC Press LLC T is found by the Castigliano equation, from which, for a full corrugation, x = 5.273(T/E)(c/t)3 For all practical purposes, x = 5(T/E)(c/t)3 (15.15) where x is elongation per corrugation length, b, due to thrust T per unit of circumference E is the modulus of elasticity and b/c = For bellows, from Castigliano's equation, the elongation of a repeating section is ∆ x = (3π/2)(T/E)(c/t)3 But a repeating section of bellows is only 2/3rds as long as corrugation, b, of equal depth, c Therefore, compared to a 3x1 corrugation, the elongation of bellows per unit length of corrugation, b, is, x = 7(T/E)(c/t) (15.16) The ratio of elongation per unit length of expansion insert of bellows to corrugation is 7/5 Performance limit of x is yield stress (or endurance limit) For a corrugation, x = 5σf c 2/3Et per corrugation length, b (15.17) For a bellows, x = 7σf c 2/3Et per corrugation length, b, (15.18) where depth, c, and thickness, t, are the same for bellows and corrugation Thrust, T, at yield stress or endurance limit, can be found from Equation 15.14 Neglecting axial stress, σf = 3Tc/t2 Example What is thrust, T, at yield stress for a 6x2 corrugation if t = 0.1345 inch and σf = 36 ksi? T = σf t2/3c = 108.5 lb/inch The corresponding increase in length of the corrugation, from Equation 15.17, is x = 5σf (c 2/3Et) = 0.06 inch Eight corrugations are ©2000 CRC Press LLC required to provide extension of a half inch For design by endurance limit, of concern is the matter of full reversal of stress or cyclic stress superimposed on a standing stress Details are found in texts on mechanics of solids The analyses above are conservative because a corrugated pipe or bellows is a three-dimensional problem — not just a two-dimensional cross section of the corrugation or bellows In fact, ring restraint is significant PROBLEMS 15-1 Prove that the angle of the impulse thrust, Qi, is = θ/2 15-2 What forces act at an elbow in a water pipe? D = ft = inside diameter, P' = 120 psi = internal pressure, θ = 60° = offset angle, v = 10 ft/second = average flow velocity, ϕ = 30° = soil friction angle, C = = soil cohesion 15-3 Design a thrust restraint for the 60° elbow in the pipe of Problem 15-2 Assume the thrust restraint is to be a solid cube (block) of reinforced concrete with its top at ground surface Assume thrust-Q = 500 kips γ = 120 pcf for soil and γ c = 144 pcf for the reinforced concrete 15-4 What is the diameter of a thrust pin for Problem 15-2 assuming the maximum depth reached by the boring auger is 25 ft? 15-5 On the 48° slope of Figure 15-8, the perpendicular distance to the bottom of the pipe from rock pins must be 12 ft Design a tower that can be flown in and onto which a chair can be attached What about loads on the pins? Should the tower be vertical? 15-6 On the 48° slope what would be the problem between the two anchors of Figure 15-7 if the pipe ©2000 CRC Press LLC were welded rather than coupled in 30-ft lengths Assume that maximum change in temperature could be + or - 100°F from mean installation temperature? Assume a coefficient of thermal expansion of 6.5(106)/°F for the steel pipe 15-12 Stress-terms as functions of offset angle, θ, are shown below for worst-case (triangular) stress distribution for which the maximum stress is twice the average NORMAL STRESS TERM IS σ /P(r/t), SHEAR STRESS TERM IS τ /P(r/t) 15-7 If expansion joints are used at 90-ft spacing in the welded pipe of problem 15-6, how much expansion (contraction) must be accommodated at the expansion joints? 15-8 What is the longitudinal spring constant, T/x, for a 6x2 corrugation with steel thickness of 0.1345 inch? What is the elongation in percent at yield stress? Given: Mortar-coated, gasketed steel pipe with a 90o elbow shown in Figure 15-12 ID = 60 inches, t = 0.375 inch, L = 30 ft, P' = 100 psi, internal pressure, v = 12 ft per second flow rate, µ' = 0.4 = coef of frict soil on pipe, H = ft, height of soil cover, γ = 120 pcf = soil unit weight 15-9 How many contiguous sections must be welded on each side of the elbow to resist the thrustQ? 15-10 What is the effect on the welded sections of Problem 15-8 if the pipe shortens due to temperature decrease and pressure increase? 15-11 An alternative to the welded sections of Problems 15-8 and 15-9 is a series of cross ties shown in Figure 15-13 Design and show details for an adequate number of cross ties and clamps (bands) for attaching the ties to the pipes at the joints Neglect soil friction ©2000 CRC Press LLC Ignore shear stress term because the pipe is restrained by soil, and because maximum shearing stresses are near the neutral surface of the elbow where normal stresses are minimum Compound stress analysis is not justified What is the maximum normal (longitudinal) stress on this 90o elbow? [Pr/t] ... H = L = P/Px = (1-sinϕ)/(1+sinϕ), soil friction angle, unit weight of soil, outside diameter, height of soil cover, length of pipe section The restraint capacity of soil against elbow is, Qelb... restrained only by the soil bearing against the pipe The maximum soil pressure bearing horizontally against the elbow is passive resistance Px at the average depth of soil, H + OD/2, Px = (2H... change in velocity in the Qi direction is simply 2vsin(θ/2) Figure 15-3 Pressure thrust-Q p at a bend (elbow) in a pipe due to internal pressure, P, showing the freevector-diagram for calculating