Numerical Methods in Soil Mechanics 24.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 "LEAKS IN BURIED PIPES AND TANKS" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 24-1 Response (flattening) of a flexible ring due to a hard spot in the embedment Figure 24-2 Joggle joint weld showing longitudinal leverage due to an external force that causes a hinge to form and rotate ©2000 CRC Press LLC CHAPTER 24 LEAKS IN BURIED PIPES AND TANKS Performance limits for buried structures are excessive deformations One such deformation is a leak Because of leaks, not only is product lost, but the soil (or the product) may become contaminated Craters and landslides have been caused by major leaks Gas and oil spills and fires could be the consequences of leaks in fuel lines and fuel tanks Leaks may occur at gasketed joints due to pinched gaskets or rolled gaskets, or due to grit under gaskets Leaks may occur through cracks A leak in a high-pressure pipeline is a fluid jet that can backwash sand against the pipe, sand-blast the pipe, wear through the pipe wall from the outside, and cause a blow-out Because the cylinder is in contact with embedment, deformation of the cylinder is concomitant with movement of the soil — soil slip or soil compression The soil provides support for vertical loads and for the cylinder But poor soil carelessly placed, like a misfit shoe, stresses the cylinder Installers of buried pipes and tanks use techniques that minimize deformation of the cylinder The track record of the installer is important Some buried fuel tank agencies now require certification of installers WELDS IN TANKS If a pressure-tested cylinder leaks after it is buried, the leak is due possibly to deformation of a weld such as a flat spot or a leverage hinge See Figures 24-1 and 24-2 Deformations that cause weld fractures are usually local and usually occur in the cylinder Analysis is similar for both pipes and tanks, but because tanks are more complicated, the following examples involve tanks If collapse is the performance limit, heads stiffen the shell enough that walls are typically thinner in tanks than in pipes Consequently, under identical burial conditions, tank shells are more sensitive than pipes to local deformations caused by non-uniform loads ©2000 CRC Press LLC A comparison of the sensitivities of tank shells and pipes requires similitude of the model and prototype It is then possible to compare the pressures at equivalent deformations These pressures are inverse measures of sensitivity Let the tank be the prototype and the pipe be the model Relative sensitivity indicates the level of care required for handling and installing a prototype tank based on a model pipe experience Similitude Similitude is achieved by writing the equation of deformation in terms of dimensionless pi-terms; and then by equating corresponding pi-terms for shell and pipe See Appendix C The pressure that causes critical deformation is a function of the following pertinent fundamental variables Fundamental variables: P = external pressure on the tank, Q = external force on the tank, d = any and all deflections or movements, S = yield strength of the wall, E = modulus of elasticity, E = 30(106) psi for steel, D = mean diameter, r = radius, w = width of joggle joint overlap, x = transition length from maximum to minimum radius of curvature, I = moment of inertia of the wall cross section, I = t3/12 for the plain wall tank cylinder, L = length of the tank, t = thickness of the shell, V = head shear force, s = normal stress, t = shearing stress, n = Poisson ratio = 0.25 for steel Subscript, r, refers to ratio of model to prototype Pertinent dimensionless pi-terms are: (P/E) (d/r) (r/t) (L/r) (rr) = = = = (1-n2) n (d) = = = external pressure term, = deformation term, ring flexibility term, length term for the tank, ratio of maximum radius to minimum radius at flat spots or out-of-roundness, 15/16 = Poisson term, where Poisson ratio (assume 1/4), ring deflection = ratio of decrease in vertical diameter to original diameter Following are the sensitivities of a prototype steel tank and a model pipe for two cases: without soil support, and with soil support II With Soil Support Sensitivity is deformation of the ring due to a hard spot in the soil See Figure 23-1 From Appendix A, localized ring deformation is of the form, d = kQr2/EI, where Q is the concentrated force, k is a constan t , a n d r2/EI is the flexibility factor — a handling factor with upper limits recommended by AISI (1994) In dimensionless parameters, (d/r) = k(Qr/EI) In Example above, at equal values of (d/r) for model and prototype, the ratio of loads, for prototype and model is, Qr = Ir/rr = 68 The tank is 68 times as sensitive as the pipe to hard spots in the embedment Racks must not bear against the tank Bedding must be uniform I Without soil support Deformation is collapse under uniform external pressure The classical equations are: TANK: (P/E)(L/r)(r/t)5/2 = 0.8/(1-n2)3/4 = 0.84 (24.1) PIPE: (P/E)(r/t)3 = 1/4(1-n2) = 0.234 (24.2) CORRUGATED PIPE: (P/E)(r3/I) = 3/(1-n2) = 3.33 (24.3) Deformations at welds cause most of the structural leaks in buried tanks For uniform internal or external pressure, circumferential welds must resist longitudinal stress — but longitudinal stress is only half as great as circumferential stress Butt welds can be made nearly as resilient as the parent metal, and they can tolerate deformation But they are expensive Consequently, longitudinal joints are often joggle joints — usually welded on the outside only See Figure 24-2 If the seam is not seriously deformed, such welds are adequate When the weld is deformed, three conditions develop which could fracture the weld: leverage, shear, and gap Example For a particular steel tank, L/r = 42ft/4.5ft, and r/t = 54in/0.1875in Its collapse pressure is to be compared with a corrugated steel pipe for which r = 78 inches and I = 0.938 in 4/ft What is the relative sensitivity; i.e., ratio of pressures, Pr, at collapse From the quotients of Equations 24.3 and 24.1, Pr = 8.24 The pressure at collapse of the pipe is eight times as great as the pressure at collapse of the tank Without soil support, the prototype tank is eight times as sensitive to collapse as the pipe The heads of the tank provide significant stiffness, but a good soil embedment is essential ©2000 CRC Press LLC Leverage — A hard spot force on a joggle joint causes a leverage hinge as shown in Figure 24-2 From tests, a joggle joint in 1/4-inch steel, stick welded with E-6024 rod, loaded as a simply supported beam of 3-inch span; fractured when a line force at mid-span reached 230 lb per inch of weld Stresses were concentrated at the corner of the weld which became vulnerable to leverage — a typical cause of leaks in fuel tanks Shear — When the cylinder is deformed, c ircumferential shearing stress, t , develops on the neutral surface of the wall If wall thickness is doubled, and the deformation remains the same, the shearing stress is doubled In joggle joints, shear in the weld is increased even more For the typical joggle joint of Figure 24-3, 24-4, t = Ewt2(1/ro - 1/r)/2bx where the minimum and maximum radii of curvature are: (24.4) where x is the length of transition from minimum to maximum radius of curvature The transition is usually visible as a localized crimp for which length, x, is very short — an inch or two — at the ends of a flat spot Example A flat spot occurs in the joggle joint of a steel tank What is the shearing stress in the weld? Shearing yield stress is usually about 20 ksi Given: E = 30(106) psi, w = inch = minimum recommended penetration of the joggle joint spigot into the bell, t = 0.1875 inch (3/16 inch), ro = original radius of curvature, = 54 inches (minimum), r = infinity at the flat spot (maximum), x = length of transition from minimum to maximum radius Substituting into Equation 24.4, t = 52 kips/x(inch) If the transition length is x = 2.5 inches, shearing stress in the weld is 20 ksi (shearing yield stress) For 3/16 steel, it is more likely that x is less than 2.5 inches In such a case, the weld yields and could crack if its ductility limit is exceeded If the "flat spot" were not flat, but had a radius of curvature twice the original tank radius, the shearing stress would be t = 26 kips/x(inch) which is half of the stress at a flat spot If the radius of curvature were inverted, shearing stress in the weld would be greater At the ends of a flat spot, the radius of curvature is less than the original radius Therefore, actual shearing stress is greater than the values above If the ring were deflected into an ellipse, see Figure ©2000 CRC Press LLC t = (Ext/pr)(1/r x - 1/ry ) rx = r(1-d)2/(1+d), ry = r(1+d)2/(1-d) (24.5) (24.6) Substituting the values from Example 2, ring deflection at yield is 76.8% Clearly, some small elliptical ring deflection is not a cause of shearing cracks in welds It is noteworthy that ring deflection of the shell causes other complic ations such as head shear (guillotine) and increased potential for inversion, both of which could contribute to cracked welds These are analyzed separately under "Head Shear" and "Inversion Analysis." Gap — In forming a joggle joint, one end of the can is deformed into a spigot of smaller diameter such that it can be inserted into the mating can See Figure 24-5 This usually leaves a gap as shown When a hard spot in the soil deforms the ring, the gap tends to narrow under the hard spot, and widen at other locations Widening of the gap may crack the weld or, at least, compound the effect of shearing stress caused by change of radius Worse than the elliptical ring show n is the accumulation of gap at the crimped ends of a flat spot — again where shearing stress is maximum Once a weld is cracked, the crack propagates and opens Also, if a hard spot bears against the joggled can, but not the mating can, the crack tends to open In either case, the crack widens and leakage increases In order to minimize the effect of a gap, typical standards require that outside circumference of the joggle spigot be 1/32 to 3/32 inch smaller than the inside circumference of the mating bell Even 3/32inch difference can cause a gap to accumulate if the joint is welded continuously without first "tack welding" or "wedging" the gap with shims (or screw drivers) Large gaps are sometimes “slugged;” i.e., the gap is partially filled with a bar Figure 24-5 Joggle joint in a tank showing how the gap becomes narrower under hard spots and wider at soft spots in the embedment Wide gaps also tend to accumulate at the ends of continuous welds ©2000 CRC Press LLC of reinforcing steel before it is welded "Slugged" welds are to be avoided Installers of tanks, especially long tanks, try to avoid hard spots in the embedment which might cause flat spots in the tank Longitudinal deflections of tanks must be restricted by leveling the bedding and by carefully placing embedment under the haunches of the tank A flat hard bedding, rocks in the embedment, frozen soil, and loose soil under the haunches — all can cause flat spots in tank shells accurate up to roughly q = 45o Beyond that, a stability analysis is more relevant because of increased soil support and arching of the top of the pipe Head Shear A major problem with flat spots is the potential for inversion For conservative analysis, the flat spot is a beam with fixed ends See Figure 24-6 The maximum moment is at the ends where Mmax = PL2/12 But at the formation of a plastic hinge, Mp = 3SI/2t = St2/4 Equating, Heads stiffen the shell, but they also cause head shear when, under soil load, a head shears down past the flexible shell like a guillotine The shell deflects easily under backfill load But heads remain circular This causes distress in the head-to-shell welds at the bottom of the tank where head shear tends to curl the flange (knuckle) and crack the weld as shown in Figure 24-8 If the seam is a joggle joint, a leverage hinge may form For analysis, the effective shearing load on each head is soil pressure, P, acting over an area of a tank diameter times roughly one diameter longitudinally P = Pd+P l where Pd is dead load pressure and Pl is live load pressure Shearing load is: (L/t)2 = 3S/P V = PD2 Inversion Analysis (24.7) From the geometry of Figure 24-6, (L/t) = (D/t)sin(q/2) (24.8) where q is the arc angle of the flat spot between plastic hinges Eliminating L/t between the two equations, and substituting a known value for S, pressure, P, can be found as a function of angle, q, and ring flexibility, (D/t) Figure 24-7 shows graphs of solutions for steel cylinders for which S = 36 ksi The analysis is conservative because soil support is neglected Arching action of the top of the cylinder is also neglected even though the "flat spot" may not be completely flat Noteworthy is the increase in pressure P at inversion when D/t is decreased A common upper limit for plain pipe is D/t = 288 For mortar-lined pipes, upper limits are usually not more than D/t = 240 Noteworthy also is the increase in angle q as D/t is decreased Soil support increases as q increases, but the beam analysis loses accuracy The beam inversion model is reasonably ©2000 CRC Press LLC (24.9) Reaction is developed under the shell If sidefill soil compresses vertically, the shell deflects, and the heads shear down past it Analysis must consider weld strength and resistance of the flange to bending Distress in the weld is exacerbated by deflections of the head and shell both of which bend the flange Unfortunately, the 90o flange angle is bent (reduced) the most on the invert where head shear is greatest Example Consider the steel tank, D = ft, L = 42 ft, t = 3/16 inch What is the shearing force, V, between the head and the shell due to a soil cover of ft with unit weight of 120 pcf? From Equation 24.9, V = 29.16 kips — enough to distort a 3/16-thick knuckle and weld From field experience, the shear load, V, is also felt at the first joint from the head Many of the leaks in steel fuel tanks are cracks in the bottom at this joint The joint between the first and second cans Figure 24-6 Model for beam analysis of a flat spot (exaggerated) in a flexible cylinder, showing the moment diagram and the geometry The flat spot is shown here at the top of the cylinder, but may occur anywhere When a flat spot does occur, it is usually at the invert Figure 24-7 Pressure at beam inversion of flat spots on steel cylinders as a function of ring flexibility, D/t, and angle of the flat spot, q, (beam length) ©2000 CRC Press LLC Figure 24-8 Failure of a circumferential weld due to a stiff head that shears down past a flexible shell Steel tank standards call for a minimum of 1.5-inch flange and 0.5-inch penetration into the shell ©2000 CRC Press LLC does not have the benefit of a head to give it strength Yet it may be subjected to much of the shear force, V Additional safeguards to prevent or mitigate leaks include the following: Double containment tanks or coated tanks, Precautions Sniffer systems such as a vent between the product tank and the double containment tank In order to avoid cracks in welds, attention should be directed to the following: Diligent monitoring of contents to discern any loss, Careful handling and installation in order to avoid flat spots, Sensor devices in the path of any possible leakage plume, Well-compacted embedment that does not compress or slip under anticipated loads, Control of surface loads and high water table Sound welds with enough toughness that they can yield without cracking under slight deformation Longitudinal Beam Action Control of internal vacuum and high external water table Stress in the welds can be exacerbated by longitudinal beam action See Figure 24-9 The Figure 24-9 Typical conditions for longitudinal stresses in a tank caused by concentrated supports — on the ends of the tank (top) and at the tie-downs (bottom) due to a high water table or flood ©2000 CRC Press LLC most common distress is tension in the bottom of a tank that is simply supported at the ends, but is not bedded Cracked welds may occur on the bottom near mid-length Analysis is the standard procedure for finding maximum stress and deflection at midlength of beams Figure 24-9 also shows a case of concentrated reactions at tie-downs that result in longitudinal stresses at the top and bottom of the tank during a flood Long tanks with single weld joints and with concentrated reactions are in greatest danger of cracked welds Tests for Leaks It is common practice to test tanks for leaks before they are buried — usually before they leave the fabrication plant One such test is internal pressure — typically or psi The pressure valve is then closed, and pressure in the tank is monitored for a specified length of time in minutes Drop in pressure is a measure of the amount of leakage, and may be the basis for specifying allowable leakage Soapy water is often poured over the tank so that leaks can be located from soap bubbles However, internal pres sure tests not indicate resistance of the tank to vacuum A vacuum test is more responsive to resistance of the tank to vacuum (external pressure) A vacuum test can measure the amount of leakage It separates antagonists and accommodates minor encroachments Legal exposure is measured by the extent of encroachment into the safety zone Adjudication of contributory negligence is based on the percent encroachment by each of the parties: owner, engineer, manufacturer, installer, etc Specifications (procedure or performance) can establish the relative responsibilities of all parties and the exposure of each in the safety zone The responsibilities are intensified by public concern for containment — especially of hazardous materials For example, the time to leakage must often be greater than the traditional 50 years service life Control of leakage plumes and procedures for decontamination become important LEAKS IN PIPES The more common leaks in pipes are the follow ing Rigid Pipes Rigid pipes are sensitive to: water hammer; leaks at gaskets; and movement of pipe sections Safety Factors Water Hammer If the pipes are brittle, water hammer can cause longitudinal cracks If the water hammer is repetitive, fatigue strength may control analysis Fatigue analysis is found in texts on strength of materials Often overlooked in risk analysis is the time to leakage Compression and consolidation of embedment may increase with time For example, if traffic compresses the sidefill in increments, the tank feels incremental ring deflection (and possibly beam deflection) Stresses increase until, at some future date, a weld fails — often with an audible "thud." The safety factor should also include the potential for assessing encroachments into the margin of safety (safety zone) In the event of a dispute, the safety zone is like a demilitarized zone Gaskets A tiny leak past a gasket in a high-pressure pipe can ultimately result in a major break A rolled gasket can be blown out of the joint ("fish-mouth") and leak Sand under the gasket can cause a leak A cracked bell can allow leakage past the gasket The leak is small at first, but in time saturates the soil Turbulence of the high-pressure jet swirling against the pipe "sand-blasts" and wears away the pipe, from the outside surface in, until a large leak opens From tests, the sand-blasted leak increases ©2000 CRC Press LLC in size at an exponential rate Such leaks should be identified as soon as possible by timely monitoring of the pipeline Even if the embedment is select gravel, a leak on top of the pipe can wash particles down from the backfill into the embedment where they are caught up in the jet turbulence and proceed to sand-blast the pipe Movement of Pipe Sections Sections of rigid pipe are connected by gaskets Lengths are short If pipe sections shift out of alignment, the sidefill soil must serve as a thrust restraint A problem arises when the sidefill soil is washed away by a leak, or loses its supporting strength when saturated Example An asbestos cement (AC) pipeline follows a circular curve It is positioned on select bedding, but sidefill and backfill are wind-blown silty sand What is the safety factor against separation of a joint? Safety factor is the sidefill soil strength divided by the sidewise thrust of two contiguous pipe sections on the circular curve See Figure 24-10 Pipe: L = 13 ft lay length of sections, ID = 24 inches = inside diameter, t = inches = wall thickness, R = 457 ft = radius of curve, P' = 350 psi = internal pressure, z = 1.25 ft = outside radius, Q = thrust Soil: P = soil pressure on pipe, H = 1.75 ft = soil cover, g = 100 pcf = soil unit weight, sf = 400 psf = average horizontal soil strength, saturated and loaded by ft of soil The safety factor is sf = sf /P From trigonometry, angle q = 1.63o ©2000 CRC Press LLC Figure 24-10 Contiguous sections in gasketed pressure pipeline that are out of alignment by offset angle, q; showing thrust, Q, and the approximate soil resistance diagrams, P(OD) From Chapter 15, thrust is Q = kips, which must be resisted by the two P(OD) diagrams shown Taking moment about point A, P = 3Q/2L(OD) = 416 psf Because soil stength is 400 psf, the safety factor is just under 1.0 Failure is incipient In this example, there are two possible causes of a break: sand-blasting and movement of contiguous pipe sections It is noteworthy that according to plans, angle, q = 1.63o However, some of the installed angle offsets in the pipeline might be greater Flexible Pipes Leaks in flexible pipes can occur at welded joints w hen the pipe is deflected either radially, or longitudinally Consider steel Steel pipes are flexible and high strength Steel pipes are often specified for high-pressure transmission Stress risers at welds can cause cracks One typical stress riser is the angle offset as discussed above for rigid pipe sections An example of offset angles in welded steel pipes is a mitered joint Figure 24-11 is an exaggerated mitered joint Due to pressure, P', inside the pipe, Q = 2P'pr2sinq The impulse, Qi, due to change in direction of flow, is neglected because it is relatively small The average longitudinal stress in the pipe is Pr/2t(sin q) Hydraulic design limits the offset angle to 15o in order to minimize turbulence Q is resisted by the pipe wall on which components of force in shear, V, and thrust, T, are: V = P'pr2 sinqcosq = 0.4066 Pr2 at q = 15o, 2 T = P'pr sin q = 0.0535 Pr2 at q = 15o If the pipes were rigid, forces, T, Q, and T are concurrent at point C But steel pipes are not rigid Longitudinal strain will tend to rotate BB' slightly Rotation is restrained by the pipe on either side of the elbow The resulting moment is Te; i.e., thrust, T, in the pipe wall times its offset, e, from the neutral axis Maximum e occurs when cut BB' is moved up to the weld such that B' falls on A' Now the longitudinal stress diagram is a triangle with maximum stress at A' where e = r/3 and ©2000 CRC Press LLC longitudinal stress is twice the average stress Because the elbow can deform slightly, and because stresses diverge (Saint Venant principle), longitudinal stress at A' may be less than twice the average stress, but is greater than average stress Such rationale is moot because yield stress at A' does not cause fracture as long as strains not approach 21% (elongation) for steel with a good weld Of greater conc ern is the loss of hoop strength due to the skew cuts (miters) at the joint Hoops are cut within the triangles shown dotted in Figure 24-11 The lost hoop strength must be supplied by the seam It the pipe were a membrane, the resulting forces on the seam would be stress triangles shown in Figure 24-12 Hoop stresses greater than yield would cause similar stress distributions Figure 24-12 is a 30o skew for which ovality of the elliptical seam is d = 7.2% The maximum force per unit length of seam is Max w = Prcosq This is worst-case design because the pipe is not a membrane When stresses are near yield, a stiffener ring at the s eam can resist the w-forces The stiffener ring performs like a crotch plate in a wye For hoop stresses less than yield, the seam itself provides an adequate stiffener ring The mating pipe walls intersect in a V which has significant ring stiffness Because the pipe is not a membrane, wall shear strength resists much of the force on the seam For most mitered joints, if angle offsets are no greater than 15o, pipe designers increase the wall thickness of the mitered pipe sections, and specify full penetration butt welds Mitered joints are common in steel pipes Strengths of lap welds are less than the steel pipe Roger Brockenbrough (1990) of USX tested lap joints and reported longitudinal strengths at about: Single weld, 75% of pipe strength, Double weld, 83% of pipe strength The percentages efficiencies are referred to as weld Figure 24-11 Free-body-diagram of a mitered joint (elbow) in a pipe, showing the free-vector-diagram of forces on the elbow and the resulting upper limit of stress at A’ Figure 24-12 Forces, w, on the mitered seam due to loss of strength of the fully circular hoops that have been cut This is a conservative analysis based on membrane theory A stiffener ring on the seam can resist these forces ©2000 CRC Press LLC REFERENCES AISI (1994), Handbook of Steel Drainage & Highway Construction Products, 4ed, American Iron and Steel Institute Brockenbrough, Roger L (1990), Journal of Structural Engineering, Vol 116, No July 1990 ASCE 24-2 Given, a 42-ft-long steel tank, ft diameter, 3/16 wall thickness, that comprises seven 6-ft-long cans connected by joggle joints It is lowered into an excavation and positioned on pre-leveled timbers at the ends of the tank Soil is end-dumped (loose embedment) Considering the inadequate soil support, what is the height of soil cover at weld failure? Tensile yield stress of welded seams is 20 ksi Unit weight of soil is 100 pcf Unit weight of the contents (gasoline) is 42.4 pcf Unit weight of steel is 490 pcf (11.2 ft) PROBLEMS 24-1 What is the theoretical bending moment (resistance) to flexure of a weld in a joggle joint in 3/16-inch steel plate? See the leverage hinge of Figure 24-2 Weld strength is 20 ksi (117 lb) ©2000 CRC Press LLC 24-3 What is the maximum stress in the weld of problem 24-2 if the tank is tied down by two straps 12 ft from each end, and the tank is submerged in water (flood) while empty and before it is backfilled? (780 psi) ... joggle joint spigot into the bell, t = 0.1875 inch (3/16 inch), ro = original radius of curvature, = 54 inches (minimum), r = infinity at the flat spot (maximum), x = length of transition from minimum... inch smaller than the inside circumference of the mating bell Even 3/32inch difference can cause a gap to accumulate if the joint is welded continuously without first "tack welding" or "wedging"... on a joggle joint causes a leverage hinge as shown in Figure 24-2 From tests, a joggle joint in 1/4-inch steel, stick welded with E-6024 rod, loaded as a simply supported beam of 3-inch span; fractured